In_a_Nutshell_Princeton_A._Zee_Quantum

Qluantum FieldTheory inaNutshell ‘Second Edition p< ee A ee] Praise for the first edition “Quantum field theory is an extraordinarily beautiful subject, but it can be an intimidating one. The profound and deeply physical concepts it embodies can get lost, to the beginner,amidst its technicalities. In this book, Zee imparts the wisdom of an experienced andremarkably creative practitioner in a user-friendly style. I wish something like it had beenavailable when I was a student.” —Frank Wilczek, Massachusetts Institute of Technology “Finally! Zee has written a ground-breaking quantum field theory text based on the course I made him teach when I chaired the Princeton physics department. With utmost clarityhe gives the eager student a light-hearted and easy-going introduction to the multifacetedwonders of quantum field theory. I wish I had this book when I taught the subject.” —Marvin L. Goldberger, President, Emeritus, California Institute of Technology “This book is filled with charming explanations that students will find beneficial.” —Ed Witten, Institute for Advanced Study “This book is perhaps the most user-friendly introductory text to the essentials of quantum field theory and its many modern applications. With his physically intuitive approach,Professor Zee makes a serious topic more reachable for beginners, reducing the conceptualbarrier while preserving enough mathematical details necessary for a firm grasp of thesubject.” —Bei Lok Hu, University of Maryland “Like the famous Feynman Lectures on Physics, this book has the flavor of a good blackboard lecture. Zee presents technical details, but only insofar as they serve the largerpurpose of giving insight into quantum field theory and bringing out its beauty.” —Stephen M. Barr, University of Delaware “This is a fantastic book—exciting, amusing, unique, and very valuable.” —Clifford V . Johnson, University of Durham “Tony Zee explains quantum field theory with a clear and engaging style. For budding or seasoned condensed matter physicists alike, he shows us that field theory is a nourishing nut to be cracked and savored.” —Matthew P. A. Fisher, Kavli Institute for Theoretical Physics “I was so engrossed that I spent all of Saturday and Sunday one weekend absorbing half the book, to my wife’s dismay. Zee has a talent for explaining the most abstruse and arcane concepts in an elegant way, using the minimum number of equations (the jokes and anecdotes help)....I wish this were available when I was a graduate student. Buy the book, keep it by your bed, and relish the insights delivered with such flair and grace.” —N. P. Ong, Princeton University What readers are saying “Funny, chatty, physical: QFT education transformed!! This text stands apart from others in so many ways that it’s difficult to list them all ....T h e exposition is breezy and chatty. The text is never boring to read, and is at times very, very funny. Puns and jokes abound,as do anecdotes....A book which is much easier, and more fun, to read than any of the others. Zee’s skills as a popular physics writer have been used to excellent effect in writingthis textbook....Wholeheartedly recommended.” —M. Haque “A readable, and rereadable instant classic on QFT ....A ta n introductory level, this type of book—with its pedagogical (and often very funny) narrative—is priceless. [It] is fullof fantastic insights akin to reading the Feynman lectures. I have since used QFT in a Nutshell as a review for [my] year-long course covering all of Peskin and Schroder, and have been pleasantly surprised at how Zee is able to preemptively answer many of theopen questions that eluded me during my course ....I value QFT in a Nutshell the same way I do the Feynman lectures....I t ’ sa text to teach an understanding of physics.” —Flip Tanedo “One of those books a person interested in theoretical physics simply must own! A real scientific masterpiece. I bought it at the time I was a physics sophomore and that was thebest choice I could have made. It was this book that triggered my interest in quantum fieldtheory and crystallized my dreams of becoming a theoretical physicist....T h e main goal of the book is to make the reader gain real intuition in the field. Amazin g...amusing... real fun. What also distinguishes this book from others dealing with a similar subjectis that it is written like a tale....I feel enormously fortunate to have come across this book at the beginning of my adventure with theoretical physics ....D e f initely the best quantum field theory book I have ever read.” —Anonymous “I have used Quantum Field Theory in a Nutshell as the primary text....Ia m immensely pleased with the book, and recommend it highly ....D o n ’ tl e tt h e‘ damn the torpedoes, full steam ahead’ approach scare you off. Once you get used to seeing the physics quickly,I think you will find the experience very satisfying intellectually.” —Jim Napolitano “This is undoubtedly the best book I have ever read about the subject. Zee does a fantastic job of explaining quantum field theory, in a way I have never seen before, and I have read most of the other books on this topic. If you are looking for quantum field theoryexplanations that are clear, precise, concise, intuitive, and fun to read—this is the book for you.” —Anonymous “One of the most artistic and deepest books ever written on quantum field theory. Amazing...extremely pleasant...al o to f very deep and illuminating remarks....I recommend the book by Zee to everybody who wants to get a clear idea what good physicsis about.” —Slava Mukhanov “Perfect for learning field theory on your own—by far the clearest and easiest to follow book I’ve found on the subject.” —Ian Z. Lovejoy “A beautifully written introduction to the modern view of field s...breezy and enchanting, leading to exceptional clarity without sacrificing depth, breadth, or rigorof content....[It] passes my test of true greatness: I wish it had been the first book onthis topic that I had found.” —Jeffrey D. Scargle “A breeze of fresh air...a real literary gem which will be useful for students who make their first steps in this difficult subject and an enjoyable treat for experts, who will findnew and deep insights. Indeed, the Nutshell is like a bright light source shining among tall and heavy trees—the many more formal books that exist—and helps seeing the forestas a whole!...I have been practicing QFT during the past two decades and with all my experience I was thrilled with enjoyment when I read some of the sections.” —Joshua Feinberg “This text not only teaches up-to-date quantum field theory, but also tells readers how research is actually done and shows them how to think about physics. [It teaches thingsthat] people usually say ‘cannot be learned from books.’ [It is] in the same style as Fearful Symmetry and Einstein ’s Universe . All three books...a r e classics.” —Yu Shi “I belong to the [group of ] enthusiastic laymen having enough curiosity and insistence... but lacking the mastery of advanced math and physics....I really could not see the forest for the trees. But at long last I got this book!” —Makay Attila “More fun than any other QFT book I have read. The comparisons to Feynman’s writings made by several of the reviewers seem quite apt....H i s enthusiasm is quite infectious....I doubt that any other book will spark your interest like this one does.” —Stephen Wandzura “I’m having a blast reading this book. It’s both deep and entertaining; this is a rare breed, indeed. I usually prefer the more formal style (big Landau fan), but I have to say that whenZee has the talent to present things his way, it’s a definite plus.” —Pierre Jouvelot “Required reading for QFT: [it] heralds the introduction of a book on quantum field theory that you can sit down and read. My professor’s lectures made much more sense as Ifollowed along in this book, because concepts were actually EXPLAINED, not just workedout.” —Alexander Scott “Not your father’s quantum field theory text: I particularly appreciate that things are motivated physically before their mathematical articulation....M o s t especially though, the author’s ‘heuristic’ descriptions are the best I have read anywhere. From them alonethe essential ideas become crystal clear.” —Dan Dill Q uantum Field Theory in a Nutshell This page intentionally left blank Quantum Field Theory in a Nutshell SECOND EDITION A. Zee PRINCETON UNIVERSITY PRESS.PRINCETON AND OXFORD Copyright © 2010 by Princeton University Press Published by Princeton University Press, 41 William Street, Princeton, New Jersey 08540In the United Kingdom: Princeton University Press,6 Oxford Street, Woodstock, Oxfordshire OX20 1TW All Rights ReservedLibrary of Congress Cataloging-in-Publication Data Zee, A. Quantum field theory in a nutshell / A. Zee.—2nd ed. p. cm. Includes bibliographical references and index.ISBN 978-0-691-14034-6 (hardcover : alk. paper) 1. Quantum field theory. I. Title.QC174.45.Z44 2010530.14 /prime3—dc22 2009015469 British Library Cataloging-in-Publication Data is availableThis book has been composed in Scala LF with ZzT EX by Princeton Editorial Associates, Inc., Scottsdale, Arizona Printed on acid-free paper.press.princeton.eduPrinted in the United States of America1 0987654321 T o my parents, who valued education above all else This page intentionally left blank Contents Preface to the First Edition xv Preface to the Second Edition xix Convention, Notation, and Units xxv IPart I: Motivation and Foundation I.1 Who Needs It? 3 I.2 Path Integral Formulation of Quantum Physics 7 I.3 From Mattress to Field 17 I.4 From Field to Particle to Force 26 I.5 Coulomb and Newton: Repulsion and Attraction 32 I.6 Inverse Square Law and the Floating 3-Brane 40 I.7 Feynman Diagrams 43 I.8 Quantizing Canonically 61 I.9 Disturbing the Vacuum 70 I.10 Symmetry 76 I.11 Field Theory in Curved Spacetime 81 I.12 Field Theory Redux 88 IIPart II: Dirac and the Spinor II.1 The Dirac Equation 93 II.2 Quantizing the Dirac Field 107 II.3 Lorentz Group and Weyl Spinors 114 II.4 Spin-Statistics Connection 120 xii | Contents II.5 Vacuum Energy, Grassmann Integrals, and Feynman Diagrams for Fermions 123 II.6 Electron Scattering and Gauge Invariance 132 II.7 Diagrammatic Proof of Gauge Invariance 144 II.8 Photon-Electron Scattering and Crossing 152 III Part III: Renormalization and Gauge Invariance III.1 Cutting Off Our Ignorance 161 III.2 Renormalizable versus Nonrenormalizable 169 III.3 Counterterms and Physical Perturbation Theory 173 III.4 Gauge Invariance: A Photon Can Find No Rest 182 III.5 Field Theory without Relativity 190 III.6 The Magnetic Moment of the Electron 194 III.7 Polarizing the Vacuum and Renormalizing the Charge 200 III.8 Becoming Imaginary and Conserving Probability 207 IV Part IV: Symmetry and Symmetry Breaking IV .1 Symmetry Breaking 223 IV .2 The Pion as a Nambu-Goldstone Boson 231 IV .3 Effective Potential 237 IV .4 Magnetic Monopole 245 IV .5 Nonabelian Gauge Theory 253 IV .6 The Anderson-Higgs Mechanism 263 IV .7 Chiral Anomaly 270 VPart V: Field Theory and Collective Phenomena V .1 Superfluids 283 V .2 Euclid, Boltzmann, Hawking, and Field Theory at Finite Temperature 287 V .3 Landau-Ginzburg Theory of Critical Phenomena 292 V .4 Superconductivity 295 V .5 Peierls Instability 298 V .6 Solitons 302 V .7 Vortices, Monopoles, and Instantons 306 VI Part VI: Field Theory and Condensed Matter VI.1 Fractional Statistics, Chern-Simons Term, and Topological Field Theory 315 VI.2 Quantum Hall Fluids 322 Contents | xiii VI.3 Duality 331 VI.4 The σModels as Effective Field Theories 340 VI.5 Ferromagnets and Antiferromagnets 344 VI.6 Surface Growth and Field Theory 347 VI.7 Disorder: Replicas and Grassmannian Symmetry 350 VI.8 Renormalization Group Flow as a Natural Concept in High Energy and Condensed Matter Physics 356 VII Part VII: Grand Unification VII.1 Quantizing Yang-Mills Theory and Lattice Gauge Theory 371 VII.2 Electroweak Unification 379 VII.3 Quantum Chromodynamics 385 VII.4 Large NExpansion 394 VII.5 Grand Unification 407 VII.6 Protons Are Not Forever 413 VII.7 SO(10) Unification 421 VIII Part VIII: Gravity and Beyond VIII.1 Gravity as a Field Theory and the Kaluza-Klein Picture 433 VIII.2 The Cosmological Constant Problem and the Cosmic Coincidence Problems 448 VIII.3 Effective Field Theory Approach to Understanding Nature 452 VIII.4 Supersymmetry: A Very Brief Introduction 461 VIII.5 A Glimpse of String Theory as a 2-Dimensional Field Theory 469 Closing Words 473 NPart N N.1 Gravitational Waves and Effective Field Theory 479 N.2 Gluon Scattering in Pure Yang-Mills Theory 483 N.3 Subterranean Connections in Gauge Theories 497 N.4 Is Einstein Gravity Secretly the Square of Yang-Mills Theory? 513 More Closing Words 521 Appendix A: Gaussian Integration and the Central Identity of Quantum Field Theory 523 Appendix B: A Brief Review of Group Theory 525 xiv | Contents Appendix C: Feynman Rules 534 Appendix D: Various Identities and Feynman Integrals 538 Appendix E: Dotted and Undotted Indices and the Majorana Spinor 541 Solutions to Selected Exercises 545 Further Reading 559 Index 563 Preface to the First Edition As a student, I was rearing at the bit, after a course on quantum mechanics, to learn quantum field theory, but the books on the subject all seemed so formidable. Fortunately,I came across a little book by Mandl on field theory, which gave me a taste of the subjectenabling me to go on and tackle the more substantive texts. I have since learned that otherphysicists of my generation had similar good experiences with Mandl. In the last three decades or so, quantum field theory has veritably exploded and Mandl would be hopelessly out of date to recommend to a student now. Thus I thought of writinga book on the essentials of modern quantum field theory addressed to the bright and eagerstudent who has just completed a course on quantum mechanics and who is impatient tostart tackling quantum field theory. I envisaged a relatively thin book, thin at least in comparison with the many weighty tomes on the subject. I envisaged the style to be breezy and colloquial, and the choiceof topics to be idiosyncratic, certainly not encyclopedic. I envisaged having many shortchapters, keeping each chapter “bite-sized.” The challenge in writing this book is to keep it thin and accessible while at the same time introducing as many modern topics as possible. A tough balancing act! In the end,I had to be unrepentantly idiosyncratic in what I chose to cover. Note to the prospectivebook reviewer: You can always criticize the book for leaving out your favorite topics. I donot apologize in any way, shape, or form. My motto in this regard (and in life as well),taken from the Ricky Nelson song “Garden Party,” is “You can’t please everyone so yougotta please yourself.” This book differs from other quantum field theory books that have come out in recent years in several respects. I want to get across the important point that the usefulness of quantum field theory is far from limited to high energy physics, a misleading impression my generation of theoreticalphysicists were inculcated with and which amazingly enough some recent textbooks on xvi | Preface to the First Edition quantum field theory (all written by high energy physicists) continue to foster. For instance, the study of driven surface growth provides a particularly clear, transparent, and physicalexample of the importance of the renormalization group in quantum field theory. Insteadof being entangled in all sorts of conceptual irrelevancies such as divergences, we havethe obviously physical notion of changing the ruler used to measure the fluctuatingsurface. Other examples include random matrix theory and Chern-Simons gauge theoryin quantum Hall fluids. I hope that condensed matter theory students will find this bookhelpful in getting a first taste of quantum field theory. The book is divided into eight parts, 1 with two devoted more or less exclusively to condensed matter physics. I try to give the reader at least a brief glimpse into contemporary developments, for example, just enough of a taste of string theory to whet the appetite. This book is perhapsalso exceptional in incorporating gravity from the beginning. Some topics are treated quitedifferently than in traditional texts. I introduce the Faddeev-Popov method to quantizeelectromagnetism and the language of differential forms to develop Yang-Mills theory, forexample. The emphasis is resoundingly on the conceptual rather than the computational. The only calculation I carry out in all its gory details is that of the magnetic moment of theelectron. Throughout, specific examples rather than heavy abstract formalism will befavored. Instead of dealing with the most general case, I always opt for the simplest. I had to struggle constantly between clarity and wordiness. In trying to anticipate and to minimize what would confuse the reader, I often find that I have to belabor certain pointsmore than what I would like. I tried to avoid the dreaded phrase “It can be shown tha t...”a s much as possible. Otherwise, I could have written a much thinner book than this! There are indeed thinnerbooks on quantum field theory: I looked at a couple and discovered that they hardly explainanything. I must confess that I have an almost insatiable desire to explain. As the manuscript grew, the list of topics that I reluctantly had to drop also kept growing. So many beautiful results, but so little space! It almost makes me ill to think about all thestuff (bosonization, instanton, conformal field theory, etc., etc.) I had to leave out. As onecolleague remarked, the nutshell is turning into a coconut shell! Shelley Glashow once described the genesis of physical theories: “Tapestries are made by many artisans working together. The contributions of separate workers cannot bediscerned in the completed work, and the loose and false threads have been covered over.” Iregret that other than giving a few tidbits here and there I could not go into the fascinatinghistory of quantum field theory, with all its defeats and triumphs. On those occasionswhen I refer to original papers I suffer from that disconcerting quirk of human psychologyof tending to favor my own more than decorum might have allowed. I certainly did notattempt a true bibliography. 1Murray Gell-Mann used to talk about the eightfold way to wisdom and salvation in Buddhism (M. Gell-Mann and Y . Ne’eman, The Eightfold Way ). Readers familiar with contemporary Chinese literature would know that the celestial dragon has eight parts. Preface to the First Edition | xvii The genesis of this book goes back to the quantum field theory course I taught as a beginning assistant professor at Princeton University. I had the enormous good fortuneof having Ed Witten as my teaching assistant and grader. Ed produced lucidly writtensolutions to the homework problems I assigned, to the extent that the next year I wentto the chairman to ask “What is wrong with the TA I have this year? He is not half asgood as the guy last year!” Some colleagues asked me to write up my notes for a muchneeded text (those were the exciting times when gauge theories, asymptotic freedom,and scores of topics not to be found in any texts all had to be learned somehow) but awiser senior colleague convinced me that it might spell disaster for my research career.Decades later, the time has come. I particularly thank Murph Goldberger for urging meto turn what expository talents I have from writing popular books to writing textbooks. Itis also a pleasure to say a word in memory of the late Sam Treiman, teacher, colleague,and collaborator, who as a member of the editorial board of Princeton University Presspersuaded me to commit to this project. I regret that my slow pace in finishing the bookdeprived him of seeing the finished product. Over the years I have refined my knowledge of quantum field theory in discussions with numerous colleagues and collaborators. As a student, I attended courses on quan-tum field theory offered by Arthur Wightman, Julian Schwinger, and Sidney Coleman. Iwas fortunate that these three eminent physicists each has his own distinctive style andapproach. The book has been tested “in the field” in courses I taught. I used it in my field theory course at the University of California at Santa Barbara, and I am grateful to some ofthe students, in particular Ted Erler, Andrew Frey, Sean Roy, and Dean Townsley, forcomments. I benefitted from the comments of various distinguished physicists who readall or parts of the manuscript, including Steve Barr, Doug Eardley, Matt Fisher, MurphGoldberger, Victor Gurarie, Steve Hsu, Bei-lok Hu, Clifford Johnson, Mehran Kardar, IanLow, Joe Polchinski, Arkady Vainshtein, Frank Wilczek, Ed Witten, and especially JoshuaFeinberg. Joshua also did many of the exercises. Talking about exercises: You didn’t get this far in physics without realizing the absolute importance of doing exercises in learning a subject. It is especially important that you domost of the exercises in this book, because to compensate for its relative slimness I haveto develop in the exercises a number of important points some of which I need for laterchapters. Solutions to some selected problems are given. I will maintain a web page http://theory.kitp.ucsb.edu/~zee/nuts.html listing all the errors, typographical and otherwise, and points of confusion that will undoubtedly cometo my attention. I thank my editors, Trevor Lipscombe, Sarah Green, and the staff of Princeton Editorial Associates (particularly Cyd Westmoreland and Evelyn Grossberg) for their advice and forseeing this project through. Finally, I thank Peter Zee for suggesting the cover painting. This page intentionally left blank Preface to the Second Edition What one fool could understand, another can. —R. P. Feynman1 Appreciating the appreciators It has been nearly six years since this book was published on March 10, 2003. Since authorsoften think of books as their children, I may liken the flood of appreciation from readers,students, and physicists to the glorious report cards a bright child brings home fromschool. Knowing that there are people who appreciate the care and clarity crafted into thepedagogy is a most gratifying feeling. In working on this new edition, merely looking atthe titles of the customer reviews on Amazon.com would lighten my task and quicken mypace: “Funny, chatty, physical. QFT education transformed!,” “A readable, and re-readableinstant classic on QFT ,” “A must read book if you want to understand essentials in QFT ,”“One of the most artistic and deepest books ever written on quantum field theory,” “Perfectfor learning field theory on your own,” “Both deep and entertaining,” “One of those booksa person interested in theoretical physics simply must own,” and so on. In a Physics T oday review, Zvi Bern, a preeminent younger field theorist, wrote: Perhaps foremost in his mind was how to make Quantum Field Theory in a Nutshell as much fun as possible....I have not had this much fun with a physics book since reading The Feynman Lectures on Physics ....[This is a book] that no student of quantum field theory should be without. Quantum Field Theory in a Nutshell is the ideal book for a graduate student to curl up with after having completed a course on quantum mechanics. But, mainly, it is for anyone whowishes to experience the sheer beauty and elegance of quantum field theory. A classical Chinese scholar famously lamented “He who knows me are so few!” but here Zvi read my mind. Einstein proclaimed, “Physics should be made as simple as possible, but not any simpler.” My response would be “Physics should be made as fun as possible, but not 1R. P. Feynman, QED: The Strange Theory of Light and Matter, p. xx. xx | Preface to the Second Edition any funnier.” I overcame the editor’s reluctance and included jokes and stories. And yes, I have also written a popular book Fearful Symmetry about the “sheer beauty and elegance” of modern physics, which at least in that book largely meant quantum field theory. I wantto share that sense of fun and beauty as much as possible. I’ve heard some people say that“Beauty is truth” but “Beauty is fun” is more like it. I had written books before, but this was my first textbook. The challenges and rewards in writing different types of book are certainly different, but to me, a university professordevoted to the ideals of teaching, the feeling of passing on what I have learned andunderstood is simply incomparable. (And the nice part is that I don’t have to hand outfinal grades.) It may sound corny, but I owe it, to those who taught me and to thoseauthors whose field theory texts I studied, to give something back to the theoretical physicscommunity. It is a wonderful feeling for me to meet young hotshot researchers who hadstudied this text and now know more about field theory than I do. How I made the book better: The first text that covers the twenty-first century When my editor Ingrid Gnerlich asked me for a second edition I thought long and hardabout how to make this edition better than the first. I have clarified and elaborated hereand there, added explanations and exercises, and done more “practical” Feynman diagramcalculations to appease those readers of the first edition who felt that I didn’t calculateenough. There are now three more chapters in the main text. I have also made the “mostaccessible” text on quantum field theory even more accessible by explaining stuff thatI thought readers who already studied quantum mechanics should know. For example,I added a concise review of the Dirac delta function to chapter I.2. But to the guy onAmazon.com who wanted complex analysis explained, sorry, I won’t do it. There is a limit.Already, I gave a basically self-contained coverage of group theory. More excitingly, and to make my life more difficult, I added, to the existing eight parts (of the celestial dragon), a new part consisting of four chapters, covering field theoretichappenings of the last decade or so. Thus I can say that this is the first text since the birthof quantum field theory in the late 1920s that covers the twenty-first century. Quantum field theory is a mature but certainly not a finished subject, as some stu- dents mistakenly believe. As one of the deepest constructs in theoretical physics and allencompassing in its reach, it is bound to have yet unplumbed depths, secret subterraneanconnections, and delightful surprises. While many theoretical physicists have moved pastquantum field theory to string theory and even string field theory, they often take the limitin which the string description reduces to a field description, thus on occasion revealingpreviously unsuspected properties of quantum field theories. We will see an example inchapter N.4. My friends admonished me to maintain, above all else, the “delightful tone” of the first edition. I hope that I have succeeded, even though the material contained in part N is “hotoff the stove” stuff, unlike the long-understood material covered in the main text. I alsoadded a few jokes and stories, such as the one about Fermi declining to trace. Preface to the Second Edition | xxi As with the first edition, I will maintain a web site http://theory.kitp.ucsb.edu/~zee/ nuts2.html listing the errors, typographical or otherwise, that will undoubtedly come tomy attention. Encouraging words In the quote that started this preface, Feynman was referring to himself, and to you! Ofcourse, Feynman didn’t simply understand the quantum field theory of electromagnetism,he also invented a large chunk of it. To paraphrase Feynman, I wrote this book for foolslike you and me. If a fool like me could write a book on quantum field theory, then surelyyou can understand it. As I said in the preface to the first edition, I wrote this book for those who, having learned quantum mechanics, are eager to tackle quantum field theory. During a sabbaticalyear (2006–07) I spent at Harvard, I was able to experimentally verify my hypothesis thata person who has mastered quantum mechanics could understand my book on his or herown without much difficulty. I was sent a freshman who had taught himself quantummechanics in high school. I gave him my book to read and every couple of weeks or sohe came by to ask a question or two. Even without these brief sessions, he would haveunderstood much of the book. In fact, at least half of his questions stem from the holesin his knowledge of quantum mechanics. I have incorporated my answers to his fieldtheoretic questions into this edition. As I also said in the original preface, I had tested some of the material in the book “in the field” in courses I taught at Princeton University and later at the University of California atSanta Barbara. Since 2003, I have been gratified to know that it has been used successfullyin courses at many institutions. I understand that, of the different groups of readers, those who are trying to learn quantum field theory on their own could easily get discouraged. Let me offer you somecheering words. First of all, that is very admirable of you! Of all the established subjectsin theoretical physics, quantum field theory is by far the most subtle and profound. Byconsensus it is much much harder to learn than Einstein’s theory of gravity, which in factshould properly be regarded as part of field theory, as will be made clear in this book. Sodon’t expect easy cruising, particularly if you don’t have someone to clarify things for youonce in a while. Try an online physics forum. Do at least some of the exercises. Remember:“No one expects a guitarist to learn to play by going to concerts in Central Park or byspending hours reading transcriptions of Jimi Hendrix solos. Guitarists practice. Guitaristsplay the guitar until their fingertips are calloused. Similarly, physicists solve problems.” 2 Of course, if you don’t have the prerequisites, you won’t be able to understand this or anyother field theory text. But if you have mastered quantum mechanics, keep on truckingand you will get there. 2N. Newbury et al., Princeton Problems in Physics with Solutions, Princeton University Press, Princeton, 1991. xxii | Preface to the Second Edition The view will be worth it, I promise. My thesis advisor Sidney Coleman used to start his field theory course saying, “Not only God knows, I know, and by the end of the semester,you will know.” By the end of this book, you too will know how God weaves the universeout of a web of interlocking fields. I would like to change Dirac’s statement “God is amathematician” to “God is a quantum field theorist.” Some of you steady truckers might want to ask what to do when you get to the end. Dur- ing my junior year in college, after my encounter with Mandl, I asked Arthur Wightmanwhat to read next. He told me to read the textbook by S. S. Schweber, which at close to athousand pages was referred to by students as “the monster” and which could be extremelyopaque at places. After I slugged my way to the end, Wightman told me, “Read it again.”Fortunately for me, volume I of Bjorken and Drell had already come out. But there is wis-dom in reading a book again; things that you miss the first time may later leap out at you.So my advice is “Read it again.” Of course, every physics student also knows that differentexplanations offered by different books may click with different folks. So read other fieldtheory books. Quantum field theory is so profound that most people won’t get it in onepass. On the subject of other field theory texts: James Bjorken kindly wrote in my much-used copy of Bjorken and Drell that the book was obsolete. Hey BJ, it isn’t. Certainly, volume Iwill never be pass ´e. On another occasion, Steve Weinberg told me, referring to his field theory book, that “I wrote the book that I would have liked to learn from.” I could equallywell say that “I wrote the book that Iwould have liked to learn from.” Without the least bit of hubris, I can say that I prefer my book to Schweber’s. The moral here is that if youdon’t like this book you should write your own. I try not to do clunky I explained my philosophy in the preface to the first edition, but allow me a few morewords here. I will teach you how to calculate, but I also have what I regard as a higher aim,to convey to you an enjoyment of quantum field theory in all its splendors (and by “all” Imean not merely quantum field theory as defined by some myopic physicists as applicableonly to particle physics). I try to erect an elegant and logically tight framework and put alight touch on a heavy subject. In spite of the image conjured up by Zvi Bern of some future field theorist curled up in bed reading this book, I expect you to grab pen and paper and work. You could doit in bed if you want, but work you must. I intentionally did not fill in all the steps; itwould hardly be a light touch if I do every bit of algebra for you. Nevertheless, I have donealgebra when I think that it would help you. Actually, I love doing algebra, particularlywhen things work out so elegantly as in quantum field theory. But I don’t do clunky. Ido not like clunky-looking equations. I avoid spelling everything out and so expect you to have a certain amount of “sense.” As a small example, near the end of chapter I.10 Isuppressed the spacetime dependence of the fields ϕ aandδϕa. If you didn’t realize, after Preface to the Second Edition | xxiii some 70 pages, that fields are functions of where you are in spacetime, you are quite lost, my friend. My plan is to “keep you on your toes” and I purposely want you to feel puzzledoccasionally. I have faith that the sort of person who would be reading this book can alwaysfigure it out after a bit of thought. I realize that there are at least three distinct groups ofreaders, but let me say to the students, “How do you expect to do research if you have tobe spoon-fed from line to line in a textbook?” Nuts who do not appreciate the Nutshell In the original preface, I quoted Ricky Nelson on the impossibility of pleasing everyone and so I was not at all surprised to find on Amazon.com a few people whom one of my friendscalls “nuts who do not appreciate the Nutshell .” My friends advise me to leave these people alone but I am sufficiently peeved to want to say a few words in my defense, no matter hownutty the charge. First, I suppose that those who say the book is too mathematical cancelout those who say the book is not mathematical enough. The people in the first group arenot informed, while those in the second group are misinformed. Quantum field theory does not have to be mathematical. I know of at least three Field Medalists who enjoyed the book. A review for the American Mathematical Society offeredthis deep statement in praise of the book: “It is often deeper to know why something istrue rather than to have a proof that it is true.” (Indeed, a Fields Medalist once told me thattop mathematicians secretly think like physicists and after they work out the broad outlineof a proof they then dress it up with epsilons and deltas. I have no idea if this is true onlyfor one, for many, or for all Fields Medalists. I suspect that it is true for many.) Then there is the person who denounces the book for its lack of rigor. Well, I happen to know, or at least used to know, a thing or two about mathematical rigor, since I wrote mysenior thesis with Wightman on what I would call “fairly rigorous” quantum field theory.As we like to say in the theoretical physics community, too much rigor soon leads to rigormortis. Be warned. Indeed, as Feynman would tell students, if this ain’t rigorous enoughfor you the math department is just one building over. So read a more rigorous book. It isa free country. More serious is the impression that several posters on Amazon.com have that the book is too elementary. I humbly beg to differ. The book gives the impression of being elementarybut in fact covers more material than many other texts. If you master everything in theNutshell, you would know more than most professors of field theory and could start doing research. I am not merely making an idle claim but could give an actual proof. All theingredients that went into the spinor helicity formalism that led to a deep field theoreticdiscovery described in part N could be found in the first edition of this book. Of course,reading a textbook is not enough; you have to come up with the good ideas. As for he who says that the book does not look complicated enough and hence can’t be a serious treatment, I would ask him to compare a modern text on electromagnetism withMaxwell’s treatises. xxiv | Preface to the Second Edition Thanks In the original preface and closing words, I mentioned that I learned a great deal of quan- tum field theory from Sidney Coleman. His clarity of thought and lucid exposition havealways inspired me. Unhappily, he passed away in 2007. After this book was published, Ivisited Sidney on different occasions, but sadly, he was already in a mental fog. In preparing this second edition, I am grateful to Nima Arkani-Hamed, Yoni Ben-Tov, Nathan Berkovits, Marty Einhorn, Joshua Feinberg, Howard Georgi, Tim Hsieh, BrendanKeller, Joe Polchinski, Yong-shi Wu, and Jean-Bernard Zuber for their helpful comments.Some of them read parts or all of the added chapters. I thank especially Zvi Bern andRafael Porto for going over the chapters in part N with great care and for many usefulsuggestions. I also thank Craig Kunimoto, Richard Neher, Matt Pillsbury, and RafaelPorto for teaching me the black art of composing equations on the computer. My editorat Princeton University Press, Ingrid Gnerlich, has always been a pleasure to talk toand work with. I also thank Kathleen Cioffi and Cyd Westmoreland for their meticulouswork in producing this book. Last but not least, I am grateful to my wife Janice for herencouragement and loving support. Convention, Notation, and Units For the same reason that we no longer use a certain king’s feet to measure distance, we use natural units in which the speed of light cand the Dirac symbol /planckover2piare both set equal to 1. Planck made the profound observation that in natural units all physical quantities can beexpressed in terms of the Planck mass M Planck≡1//radicalbig GNewton /similarequal1019Gev. The quantities cand/planckover2piare not so much fundamental constants as conversion factors. In this light, I am genuinely puzzled by condensed matter physicists carrying around Boltzmann’s constantk, which is no different from the conversion factor between feet and meters. Spacetime coordinates x μare labeled by Greek indices (μ =0, 1, 2, 3 ) with the time coordinate x0sometimes denoted by t. Space coordinates xiare labeled by Latin indices (i=1, 2, 3 ) and ∂μ≡∂/∂xμ. We use a Minkowski metric ημνwith signature ( +,−,−,−) so that η00=+ 1. We write ημν∂μϕ∂νϕ=∂μϕ∂μϕ=(∂ϕ)2=(∂ϕ/∂t)2−/summationtext i(∂ϕ/∂xi)2. The metric in curved spacetime is always denoted by gμν, but often I will also use gμνfor the Minkowski metric when the context indicates clearly that we are in flat spacetime. Since I will be talking mostly about relativistic quantum field theory in this book I will without further clarification use a relativistic language. Thus, when I speak of momentum,unless otherwise specified, I mean energy and momentum. Also since /planckover2pi=1, I will not distinguish between wave vector kand momentum, and between frequency ωand energy. In local field theory I deal primarily with the Lagrangian density Land not the La- grangian L=/integraltext d 3xL. As is common practice in the literature and in oral discussion, I will often abuse terminology and simply refer to Las the Lagrangian. I will commit other minor abuses such as writing 1 instead of Ifor the unit matrix. I use the same symbol ϕfor the Fourier transform ϕ(k) of a function ϕ(x) whenever there is no risk of confu- sion, as is almost always the case. I prefer an abused terminology to cluttered notation and unbearable pedantry. The symbol ∗denotes complex conjugation, and † hermitean conjugation: The former applies to a number and the latter to an operator. I also use the notation c.c. and h.c. Often xxvi | Convention, Notation, and Units when there is no risk of confusion I abuse the notation, using † when I should use ∗.F o r instance, in a path integral, bosonic fields are just number-valued fields, but neverthelessI write ϕ †rather than ϕ∗. For a matrix M, then of course M†andM∗should be carefully distinguished from each other. I made an effort to get factors of 2 and πright, but some errors will be inevitable. Q uantum Field Theory in a Nutshell This page intentionally left blank Part I Motivation and Foundation This page intentionally left blank I.1 Who Needs It? Who needs quantum field theory? Quantum field theory arose out of our need to describe the ephemeral nature of life. No, seriously, quantum field theory is needed when we confront simultaneously the two great physics innovations of the last century of the previous millennium: special relativityand quantum mechanics. Consider a fast moving rocket ship close to light speed. You needspecial relativity but not quantum mechanics to study its motion. On the other hand, tostudy a slow moving electron scattering on a proton, you must invoke quantum mechanics,but you don’t have to know a thing about special relativity. It is in the peculiar confluence of special relativity and quantum mechanics that a new set of phenomena arises: Particles can be born and particles can die. It is this matter ofbirth, life, and death that requires the development of a new subject in physics, that ofquantum field theory. Let me give a heuristic discussion. In quantum mechanics the uncertainty principle tells us that the energy can fluctuate wildly over a small interval of time. According to specialrelativity, energy can be converted into mass and vice versa. With quantum mechanics andspecial relativity, the wildly fluctuating energy can metamorphose into mass, that is, intonew particles not previously present. Write down the Schr ¨odinger equation for an electron scattering off a proton. The equation describes the wave function of one electron, and no matter how you shakeand bake the mathematics of the partial differential equation, the electron you followwill remain one electron. But special relativity tells us that energy can be converted tomatter: If the electron is energetic enough, an electron and a positron (“the antielectron”)can be produced. The Schr ¨odinger equation is simply incapable of describing such a phenomenon. Nonrelativistic quantum mechanics must break down. You saw the need for quantum field theory at another point in your education. T oward the end of a good course on nonrelativistic quantum mechanics the interaction betweenradiation and atoms is often discussed. You would recall that the electromagnetic field is 4 | I. Motivation and Foundation Figure I.1.1 treated as a field; well, it is a field. Its Fourier components are quantized as a collection of harmonic oscillators, leading to creation and annihilation operators for photons. Sothere, the electromagnetic field is a quantum field. Meanwhile, the electron is treated as apoor cousin, with a wave function /Psi1(x) governed by the good old Schr ¨odinger equation. Photons can be created or annihilated, but not electrons. Quite aside from the experimentalfact that electrons and positrons could be created in pairs, it would be intellectually moresatisfying to treat electrons and photons, as they are both elementary particles, on the samefooting. So, I was more or less right: Quantum field theory is a response to the ephemeral nature of life. All of this is rather vague, and one of the purposes of this book is to make these remarks more precise. For the moment, to make these thoughts somewhat more concrete, let usask where in classical physics we might have encountered something vaguely resemblingthe birth and death of particles. Think of a mattress, which we idealize as a 2-dimensionallattice of point masses connected to each other by springs (fig. I.1.1). For simplicity, letus focus on the vertical displacement [which we denote by q a(t)] of the point masses and neglect the small horizontal movement. The index asimply tells us which mass we are talking about. The Lagrangian is then L=1 2(/summationdisplay am˙q2 a−/summationdisplay a,bkabqaqb−/summationdisplay a,b,cgabcqaqbqc−...) (1) Keeping only the terms quadratic in q(the “harmonic approximation”) we have the equa- tions of motion m¨qa=−/summationtext bkabqb. T aking the q’s as oscillating with frequency ω,w e have/summationtext bkabqb=mω2qa. The eigenfrequencies and eigenmodes are determined, respec- tively, by the eigenvalues and eigenvectors of the matrix k. As usual, we can form wave packets by superposing eigenmodes. When we quantize the theory, these wave packets be-have like particles, in the same way that electromagnetic wave packets when quantizedbehave like particles called photons. I.1. Who Needs It? | 5 Since the theory is linear, two wave packets pass right through each other. But once we include the nonlinear terms, namely the terms cubic, quartic, and so forth in the q’s in (1), the theory becomes anharmonic. Eigenmodes now couple to each other. A wave packetmight decay into two wave packets. When two wave packets come near each other, theyscatter and perhaps produce more wave packets. This naturally suggests that the physicsof particles can be described in these terms. Quantum field theory grew out of essentially these sorts of physical ideas.It struck me as limiting that even after some 75 years, the whole subject of quantum field theory remains rooted in this harmonic paradigm, to use a dreadfully pretentiousword. We have not been able to get away from the basic notions of oscillations and wavepackets. Indeed, string theory, the heir to quantum field theory, is still firmly founded onthis harmonic paradigm. Surely, a brilliant young physicist, perhaps a reader of this book,will take us beyond. Condensed matter physics In this book I will focus mainly on relativistic field theory, but let me mention here thatone of the great advances in theoretical physics in the last 30 years or so is the increasinglysophisticated use of quantum field theory in condensed matter physics. At first sight thisseems rather surprising. After all, a piece of “condensed matter” consists of an enormousswarm of electrons moving nonrelativistically, knocking about among various atomic ionsand interacting via the electromagnetic force. Why can’t we simply write down a giganticwave function /Psi1(x 1,x2,... ,xN), where xjdenotes the position of the jth electron and N is a large but finite number? Okay, /Psi1is a function of many variables but it is still governed by a nonrelativistic Schr ¨odinger equation. The answer is yes, we can, and indeed that was how solid state physics was first studied in its heroic early days (and still is in many of its subbranches). Why then does a condensed matter theorist need quantum field theory? Again, let us first go for a heuristic discussion, giving an overall impression rather than all the details. Ina typical solid, the ions vibrate around their equilibrium lattice positions. This vibrationaldynamics is best described by so-called phonons, which correspond more or less to thewave packets in the mattress model described above. This much you can read about in any standard text on solid state physics. Furthermore, if you have had a course on solid state physics, you would recall that the energy levelsavailable to electrons form bands. When an electron is kicked (by a phonon field say) froma filled band to an empty band, a hole is left behind in the previously filled band. Thishole can move about with its own identity as a particle, enjoying a perfectly comfortableexistence until another electron comes into the band and annihilates it. Indeed, it was with a picture of this kind that Dirac first conceived of a hole in the “electron sea” as theantiparticle of the electron, the positron. We will flesh out this heuristic discussion in subsequent chapters in parts V and VI. 6 | I. Motivation and Foundation Marriages T o summarize, quantum field theory was born of the necessity of dealing with the marriage of special relativity and quantum mechanics, just as the new science of string theory isbeing born of the necessity of dealing with the marriage of general relativity and quantummechanics. I.2 Path Integral Formulation of Quantum Physics The professor’s nightmare: a wise guy in the class As I noted in the preface, I know perfectly well that you are eager to dive into quantum field theory, but first we have to review the path integral formalism of quantum mechanics. Thisformalism is not universally taught in introductory courses on quantum mechanics, buteven if you have been exposed to it, this chapter will serve as a useful review. The reason Istart with the path integral formalism is that it offers a particularly convenient way of goingfrom quantum mechanics to quantum field theory. I will first give a heuristic discussion,to be followed by a more formal mathematical treatment. Perhaps the best way to introduce the path integral formalism is by telling a story, certainly apocryphal as many physics stories are. Long ago, in a quantum mechanics class,the professor droned on and on about the double-slit experiment, giving the standardtreatment. A particle emitted from a source S(fig. I.2.1) at time t=0 passes through one or the other of two holes, A 1andA2, drilled in a screen and is detected at time t=Tby a detector located at O. The amplitude for detection is given by a fundamental postulate of quantum mechanics, the superposition principle, as the sum of the amplitude for theparticle to propagate from the source Sthrough the hole A 1and then onward to the point Oand the amplitude for the particle to propagate from the source Sthrough the hole A2 and then onward to the point O. Suddenly, a very bright student, let us call him Feynman, asked, “Professor, what if we drill a third hole in the screen?” The professor replied, “Clearly, the amplitude forthe particle to be detected at the point Ois now given by the sum of three amplitudes, the amplitude for the particle to propagate from the source Sthrough the hole A 1and then onward to the point O, the amplitude for the particle to propagate from the source S through the hole A2and then onward to the point O, and the amplitude for the particle to propagate from the source Sthrough the hole A3and then onward to the point O.” The professor was just about ready to continue when Feynman interjected again, “What if I drill a fourth and a fifth hole in the screen?” Now the professor is visibly losing his 8 | I. Motivation and Foundation SOA1 A2 Figure I.2.1 patience: “All right, wise guy, I think it is obvious to the whole class that we just sum over all the holes.” T o make what the professor said precise, denote the amplitude for the particle to propagate from the source Sthrough the hole Aiand then onward to the point Oas A(S→Ai→O). Then the amplitude for the particle to be detected at the point Ois A(detected at O)=/summationdisplay iA(S→Ai→O) (1) But Feynman persisted, “What if we now add another screen (fig. I.2.2) with some holes drilled in it?” The professor was really losing his patience: “Look, can’t you see that youjust take the amplitude to go from the source Sto the hole A iin the first screen, then to the hole Bjin the second screen, then to the detector at O, and then sum over all iandj?” Feynman continued to pester, “What if I put in a third screen, a fourth screen, eh? What if I put in a screen and drill an infinite number of holes in it so that the screen is no longerthere?” The professor sighed, “Let’s move on; there is a lot of material to cover in thiscourse.” SOA1 A2 A3B1 B2 B3 B4 Figure I.2.2 I.2. Path Integral Formulation | 9 SO Figure I.2.3 But dear reader, surely you see what that wise guy Feynman was driving at. I especially enjoy his observation that if you put in a screen and drill an infinite number of holes in it,then that screen is not really there. Very Zen! What Feynman showed is that even if therewere just empty space between the source and the detector, the amplitude for the particleto propagate from the source to the detector is the sum of the amplitudes for the particle togo through each one of the holes in each one of the (nonexistent) screens. In other words,we have to sum over the amplitude for the particle to propagate from the source to thedetector following all possible paths between the source and the detector (fig. I.2.3). A(particle to go from StoOin time T)= /summationdisplay (paths )A/parenleftbig particle to go from StoOin time Tfollowing a particular path/parenrightbig (2) Now the mathematically rigorous will surely get anxious over how/summationtext (paths )is to be defined. Feynman followed Newton and Leibniz: T ake a path (fig. I.2.4), approximate itby straight line segments, and let the segments go to zero. You can see that this is just likefilling up a space with screens spaced infinitesimally close to each other, with an infinitenumber of holes drilled in each screen. Fine, but how to construct the amplitude A(particle to go from StoOin time Tfollowing a particular path)? Well, we can use the unitarity of quantum mechanics: If we know theamplitude for each infinitesimal segment, then we just multiply them together to get theamplitude of the whole path. S O Figure I.2.4 10 | I. Motivation and Foundation In quantum mechanics, the amplitude to propagate from a point qIto a point qFin timeTis governed by the unitary operator e−iHT, where His the Hamiltonian. More precisely, denoting by |q/angbracketrightthe state in which the particle is at q, the amplitude in question is just /angbracketleftqF|e−iHT|qI/angbracketright. Here we are using the Dirac bra and ket notation. Of course, philosophically, you can argue that to say the amplitude is /angbracketleftqF|e−iHT|qI/angbracketrightamounts to a postulate and a definition of H. It is then up to experimentalists to discover that His hermitean, has the form of the classical Hamiltonian, et cetera. Indeed, the whole path integral formalism could be written down mathematically start- ing with the quantity /angbracketleftqF|e−iHT|qI/angbracketright, without any of Feynman’s jive about screens with an infinite number of holes. Many physicists would prefer a mathematical treatment withoutthe talk. As a matter of fact, the path integral formalism was invented by Dirac preciselyin this way, long before Feynman. 1 A necessary word about notation even though it interrupts the narrative flow: We denote the coordinates transverse to the axis connecting the source to the detector by q, rather thanx, for a reason which will emerge in a later chapter. For notational simplicity, we will think of qas 1-dimensional and suppress the coordinate along the axis connecting the source to the detector. Dirac’s formulation Let us divide the time TintoNsegments each lasting δt=T/N . Then we write /angbracketleftqF|e−iHT|qI/angbracketright=/angbracketleftqF|e−iHδte−iHδt ...e−iHδt|qI/angbracketright Our states are normalized by /angbracketleftq/prime|q/angbracketright=δ(q/prime−q)withδthe Dirac delta function. (Recall thatδis defined by δ(q)=/integraltext∞ −∞(dp/2π)eipqand/integraltext dqδ(q) =1. See appendix 1.) Now use the fact that |q/angbracketrightforms a complete set of states so that/integraltext dq|q/angbracketright/angbracketleftq|= 1. T o see that the normalization is correct, multiply on the left by /angbracketleftq/prime/prime|and on the right by |q/prime/angbracketright, thus obtaining/integraltext dqδ(q/prime/prime−q)δ(q −q/prime)=δ(q/prime/prime−q/prime). Insert 1 between all these factors of e−iHδtand write /angbracketleftqF|e−iHT|qI/angbracketright =(N−1/productdisplay j=1/integraldisplay dqj)/angbracketleftqF|e−iHδt|qN−1/angbracketright/angbracketleftqN−1|e−iHδt|qN−2/angbracketright.../angbracketleftq2|e−iHδt|q1/angbracketright/angbracketleftq 1|e−iHδt|qI/angbracketright (3) Focus on an individual factor /angbracketleftqj+1|e−iHδt|qj/angbracketright. Let us take the baby step of first eval- uating it just for the free-particle case in which H=ˆp2/2m. The hat on ˆpreminds us that it is an operator. Denote by |p/angbracketrightthe eigenstate of ˆp, namely ˆp|p/angbracketright=p|p/angbracketright. Do you re- member from your course in quantum mechanics that /angbracketleftq|p/angbracketright=eipq? Sure you do. This 1For the true history of the path integral, see p. xv of my introduction to R. P. Feynman, QED: The Strange Theory of Light and Matter. I.2. Path Integral Formulation | 11 just says that the momentum eigenstate is a plane wave in the coordinate representa- tion. (The normalization is such that/integraltext (dp/2π)|p/angbracketright/angbracketleftp|= 1. Again, to see that the nor- malization is correct, multiply on the left by /angbracketleftq/prime|and on the right by |q/angbracketright, thus obtaining/integraltext (dp/2π)eip(q/prime−q)=δ(q/prime−q).) So again inserting a complete set of states, we write /angbracketleftqj+1|e−iδt( ˆp2/2m)|qj/angbracketright=/integraldisplaydp 2π/angbracketleftqj+1|e−iδt( ˆp2/2m)|p/angbracketright/angbracketleftp|qj/angbracketright =/integraldisplaydp 2πe−iδt(p2/2m)/angbracketleftqj+1|p/angbracketright/angbracketleftp|qj/angbracketright =/integraldisplaydp 2πe−iδt(p2/2m)eip(qj+1−qj) Note that we removed the hat from the momentum operator in the exponential: Since the momentum operator is acting on an eigenstate, it can be replaced by its eigenvalue. Also,we are evidently working in the Heisenberg picture. The integral over pis known as a Gaussian integral, with which you may already be familiar. If not, turn to appendix 2 to this chapter. Doing the integral over p, we get (using (21)) /angbracketleftqj+1|e−iδt( ˆp2/2m)|qj/angbracketright=/parenleftbigg−im 2πδt/parenrightbigg1 2 e[im(qj+1−qj)2]/2δt =/parenleftbigg−im 2πδt/parenrightbigg1 2 eiδt(m/2 )[(qj+1−qj)/δt ]2 Putting this into (3) yields /angbracketleftqF|e−iHT|qI/angbracketright=/parenleftbigg−im 2πδt/parenrightbiggN 2/parenleftBiggN−1/productdisplay k=1/integraldisplay dqk/parenrightBigg eiδt(m/2 )/Sigma1N−1 j=0[(qj+1−qj)/δt ]2 withq0≡qIandqN≡qF. We can now go to the continuum limit δt→0. Newton and Leibniz taught us to replace [(qj+1−qj)/δt ]2by˙q2, andδt/summationtextN−1 j=0by/integraltextT 0dt. Finally, we define the integral over paths as /integraldisplay Dq(t) =lim N→∞/parenleftbigg−im 2πδt/parenrightbiggN 2/parenleftBiggN−1/productdisplay k=1/integraldisplay dqk/parenrightBigg We thus obtain the path integral representation /angbracketleftqF|e−iHT|qI/angbracketright=/integraldisplay Dq(t) ei/integraltextT 0dt1 2m˙q2 (4) This fundamental result tells us that to obtain /angbracketleftqF|e−iHT|qI/angbracketrightwe simply integrate over all possible paths q(t) such that q(0)=qIandq(T)=qF. As an exercise you should convince yourself that had we started with the Hamiltonian for a particle in a potential H=ˆp2/2m+V(ˆq)(again the hat on ˆqindicates an operator) the final result would have been /angbracketleftqF|e−iHT|qI/angbracketright=/integraldisplay Dq(t) ei/integraltextT 0dt[1 2m˙q2−V( q) ](5) 12 | I. Motivation and Foundation We recognize the quantity1 2m˙q2−V( q) as just the Lagrangian L(˙q,q). The Lagrangian has emerged naturally from the Hamiltonian! In general, we have /angbracketleftqF|e−iHT|qI/angbracketright=/integraldisplay Dq(t) ei/integraltextT 0dtL(˙q,q)(6) T o avoid potential confusion, let me be clear that tappears as an integration variable in the exponential on the right-hand side. The appearance of tin the path integral measure Dq(t) is simply to remind us that qis a function of t(as if we need reminding). Indeed, this measure will often be abbreviated to Dq. You might recall that/integraltextT 0dtL(˙q,q)is called the action S(q) in classical mechanics. The action Sis a functional of the function q(t) . Often, instead of specifying that the particle starts at an initial position qIand ends at a final position qF, we prefer to specify that the particle starts in some initial state Iand ends in some final state F. Then we are interested in calculating /angbracketleftF|e−iHT|I/angbracketright, which upon inserting complete sets of states can be written as /integraldisplay dqF/integraldisplay dqI/angbracketleftF|qF/angbracketright/angbracketleftqF|e−iHT|qI/angbracketright/angbracketleftqI|I/angbracketright, which mixing Schr ¨odinger and Dirac notation we can write as /integraldisplay dqF/integraldisplay dqI/Psi1F(qF)∗/angbracketleftqF|e−iHT|qI/angbracketright/Psi1I(qI). In most cases we are interested in taking |I/angbracketrightand|F/angbracketrightas the ground state, which we will denote by |0/angbracketright. It is conventional to give the amplitude /angbracketleft0|e−iHT|0/angbracketrightthe name Z. At the level of mathematical rigor we are working with, we count on the path integral /integraltext Dq(t) ei/integraltextT 0dt[1 2m˙q2−V( q) ]to converge because the oscillatory phase factors from different paths tend to cancel out. It is somewhat more rigorous to perform a so-called Wick rotationto Euclidean time. This amounts to substituting t→−itand rotating the integration contour in the complex tplane so that the integral becomes Z=/integraldisplay Dq(t) e−/integraltextT 0dt[1 2m˙q2+V( q) ], (7) known as the Euclidean path integral. As is done in appendix 2 to this chapter with ordinary integrals we will always assume that we can make this type of substitution with impunity. The classical world emerges One particularly nice feature of the path integral formalism is that the classical limit of quantum mechanics can be recovered easily. We simply restore Planck’s constant /planckover2piin (6): /angbracketleftqF|e−(i//planckover2pi )HT|qI/angbracketright=/integraldisplay Dq(t) e(i//planckover2pi)/integraltextT 0dtL(˙q,q) and take the /planckover2pi→0 limit. Applying the stationary phase or steepest descent method (if you don’t know it see appendix 3 to this chapter) we obtain e(i//planckover2pi)/integraltextT 0dtL(˙qc,qc), where qc(t)is the “classical path” determined by solving the Euler-Lagrange equation (d/dt)(δL/δ ˙q)− (δL/δq) =0 with appropriate boundary conditions. I.2. Path Integral Formulation | 13 Appendix 1 For your convenience, I include a concise review of the Dirac delta function here. Let us define a function dK(x)by dK(x)≡/integraldisplayK 2 −K 2dk 2πeikx=1 πxsinKx 2(8) for arbitrary real values of x. We see that for large Kthe even function dK(x) is sharply peaked at the origin x=0, reaching a value of K/2πat the origin, crossing zero at x=2π/K , and then oscillating with ever decreasing amplitude. Furthermore, /integraldisplay∞ −∞dx dK(x)=2 π/integraldisplay∞ 0dx xsinKx 2=2 π/integraldisplay∞ 0dy ysiny=1 (9) The Dirac delta function is defined by δ(x)=limK→∞dK(x). Heuristically, it could be thought of as an infinitely sharp spike located at x=0 such that the area under the spike is equal to 1. Thus for a function s(x) well-behaved around x=awe have /integraldisplay∞ −∞dx δ(x −a)s(x) =s(a) (10) (By the way, for what it is worth, mathematicians call the delta function a “distribution,” not a function.) Our derivation also yields an integral representation for the delta function that we will use repeatedly in this text: δ(x)=/integraldisplay∞ −∞dk 2πeikx(11) We will often use the identity /integraldisplay∞ −∞dx δ(f (x))s(x) =/summationdisplay is(xi) |f/prime(xi)|(12) where xidenotes the zeroes of f( x) (in other words, f( xi)=0 and f/prime(xi)=df (xi)/dx .) T o prove this, first show that/integraltext∞ −∞dx δ(bx)s(x) =/integraltext∞ −∞dxδ(x) |b|s(x)=s(0)/|b|. The factor of 1 /bfollows from dimensional analysis. (T o see the need for the absolute value, simply note that δ(bx) is a positive function. Alternatively, change integration variable to y=bx: forbnegative we have to flip the integration limits.) T o obtain (12), expand around each of the zeroes of f( x) . Another useful identity (understood in the limit in which the positive infinitesimal εtends to zero) is 1 x+iε=P1 x−iπδ(x) (13) T o see this, simply write 1 /(x+iε)=x/(x2+ε2)−iε/(x2+ε2), and then note that ε/(x2+ε2)as a function ofxis sharply spiked around x=0 and that its integral from −∞ to∞is equal to π. Thus we have another representation of the Dirac delta function: δ(x)=1 πε x2+ε2(14) Meanwhile, the principal value integral is defined by /integraldisplay dxP1 xf( x)=lim ε→0/integraldisplay dxx x2+ε2f( x) (15) 14 | I. Motivation and Foundation Appendix 2 I will now show you how to do the integral G≡/integraltext+∞ −∞dxe−1 2x2. The trick is to square the integral, call the dummy integration variable in one of the integrals y, and then pass to polar coordinates: G2=/integraldisplay+∞ −∞dx e−1 2x2/integraldisplay+∞ −∞dy e−1 2y2 =2π/integraldisplay+∞ 0dr re−1 2r2 =2π/integraldisplay+∞ 0dw e−w=2π Thus, we obtain /integraldisplay+∞ −∞dx e−1 2x2=√ 2π (16) Believe it or not, a significant fraction of the theoretical physics literature consists of varying and elaborating this basic Gaussian integral. The simplest extension is almost immediate: /integraldisplay+∞ −∞dx e−1 2ax2=/parenleftbigg2π a/parenrightbigg1 2 (17) as can be seen by scaling x→x/√a. Acting on this repeatedly with −2(d/da) we obtain /angbracketleftx2n/angbracketright≡/integraltext+∞ −∞dx e−1 2ax2x2n /integraltext+∞ −∞dx e−1 2ax2=1 an(2n−1)(2n−3)...5.3.1 (18) The factor 1 /anfollows from dimensional analysis. T o remember the factor (2n−1)!!≡(2n−1)(2n−3)...5. 3.1 imagine 2 npoints and connect them in pairs. The first point can be connected to one of (2n−1)points, the second point can now be connected to one of the remaining (2n−3)points, and so on. This clever observation, due to Gian Carlo Wick, is known as Wick’s theorem in the field theory literature. Incidentally, field theorists usethe following graphical mnemonic in calculating, for example, /angbracketleftx 6/angbracketright: Write /angbracketleftx6/angbracketrightas/angbracketleftxxxxxx/angbracketright and connect the x’s, for example 〈〉xxxxxx The pattern of connection is known as a Wick contraction. In this simple example, since the six x’s are identical, any one of the distinct Wick contractions gives the same value a−3and the final result for /angbracketleftx6/angbracketrightis just a−3times the number of distinct Wick contractions, namely 5 .3.1=15. We will soon come to a less trivial example, with distinct x’s, in which case distinct Wick contraction gives distinct values. An important variant is the integral /integraldisplay+∞ −∞dx e−1 2ax2+Jx=/parenleftbigg2π a/parenrightbigg1 2 eJ2/2a(19) T o see this, take the expression in the exponent and “complete the square”: −ax2/2+Jx=−(a/2)(x2− 2Jx/a) =−(a/2)(x−J/a)2+J2/2a. The xintegral can now be done by shifting x→x+J/a , giving the factor of (2π/a)1 2. Check that we can also obtain (18) by differentiating with respect to Jrepeatedly and then setting J=0. Another important variant is obtained by replacing JbyiJ: /integraldisplay+∞ −∞dx e−1 2ax2+iJx=/parenleftbigg2π a/parenrightbigg1 2 e−J2/2a(20) I.2. Path Integral Formulation | 15 T o get yet another variant, replace aby−ia : /integraldisplay+∞ −∞dx e1 2iax2+iJx=/parenleftbigg2πi a/parenrightbigg1 2 e−iJ2/2a(21) Let us promote ato a real symmetric NbyNmatrix Aijandxto a vector xi(i,j=1,... ,N). Then (19) generalizes to /integraldisplay+∞ −∞/integraldisplay+∞ −∞.../integraldisplay+∞ −∞dx1dx2...dxNe−1 2x.A.x+J.x=/parenleftbigg(2π)N det[A]/parenrightbigg1 2 e1 2J.A−1.J(22) where x.A.x=xiAijxjandJ.x=Jixi(with repeated indices summed.) T o derive this important relation, diagonalize Aby an orthogonal transformation Oso that A=O−1.D.O, where Dis a diagonal matrix. Call yi=Oijxj. In other words, we rotate the coordinates in the N-dimensional Euclidean space we are integrating over. The expression in the exponential in the integrand then becomes −1 2y.D.y+(OJ) .y. Using/integraltext+∞ −∞.../integraltext+∞ −∞dx1...dxN=/integraltext+∞ −∞.../integraltext+∞ −∞dy1...dyN, we factorize the left-hand side of (22) into a product of Nintegrals, each of the form/integraltext+∞ −∞dyie−1 2Diiy2 i+(OJ) iyi. Plugging into (19) we obtain the right hand side of (22), since (OJ) .D−1.(OJ)=J.O−1D−1O.J=J.A−1.J(where we use the orthogonality of O). (T o make sure you got it, try this explicitly for N=2.) Putting in some i’s (A→−iA,J→iJ), we find the generalization of (22) /integraldisplay+∞ −∞/integraldisplay+∞ −∞.../integraldisplay+∞ −∞dx1dx2...dxNe(i/2)x.A.x+iJ.x =/parenleftbigg(2πi)N det[A]/parenrightbigg1 2 e−(i/2)J.A−1.J(23) The generalization of (18) is also easy to obtain. Differentiate (22) ptimes with respect to Ji,Jj,...Jk, and Jl, and then set J=0. For example, for p=1 the integrand in (22) becomes e−1 2x.A.xxiand since the integrand is now odd in xithe integral vanishes. For p=2 the integrand becomes e−1 2x.A.x(xixj), while on the right hand side we bring down A−1 ij. Rearranging and eliminating det[ A] (by setting J=0 in (22)), we obtain /angbracketleftxixj/angbracketright=/integraltext+∞ −∞/integraltext+∞ −∞.../integraltext+∞ −∞dx1dx2...dxNe−1 2x.A.xxixj /integraltext+∞ −∞/integraltext+∞ −∞.../integraltext+∞ −∞dx1dx2...dxNe−1 2x.A.x=A−1 ij Just do it. Doing it is easier than explaining how to do it. Then do it for p=3 and 4. You will see immediately how your result generalizes. When the set of indices i,j,... ,k,lcontains an odd number of elements, /angbracketleftxixj...xkxl/angbracketright vanishes trivially. When the set of indices i,j,... ,k,lcontains an even number of elements, we have /angbracketleftxixj...xkxl/angbracketright=/summationdisplay Wick(A−1)ab...(A−1)cd (24) where we have defined /angbracketleftxixj...xkxl/angbracketright =/integraltext+∞ −∞/integraltext+∞ −∞.../integraltext+∞ −∞dx1dx2...dxNe−1 2x.A.xxixj...xkxl /integraltext+∞ −∞/integraltext+∞ −∞.../integraltext+∞ −∞dx1dx2...dxNe−1 2x.A.x(25) and where the set of indices {a,b,... ,c,d}represent a permutation of {i,j,... ,k,l}. The sum in (24) is over all such permutations or Wick contractions. For example, /angbracketleftxixjxkxl/angbracketright=(A−1)ij(A−1)kl+(A−1)il(A−1)jk+(A−1)ik(A−1)jl (26) (Recall that A, and thus A−1, is symmetric.) As in the simple case when xdoes not carry any index, we could connect the x’s in/angbracketleftxixjxkxl/angbracketrightin pairs (Wick contraction) and write a factor (A−1)abif we connect xatoxb. Notice that since /angbracketleftxixj/angbracketright=(A−1)ijthe right hand side of (24) can also be written in terms of objects like /angbracketleftxixj/angbracketright. Thus, /angbracketleftxixjxkxl/angbracketright=/angbracketleftxixj/angbracketright/angbracketleftxkxl/angbracketright+/angbracketleftxixl/angbracketright/angbracketleftxjxk/angbracketright+/angbracketleftxixk/angbracketright/angbracketleftxjxl/angbracketright. 16 | I. Motivation and Foundation Please work out /angbracketleftxixjxkxlxmxn/angbracketright; you will become an expert on Wick contractions. Of course, (24) reduces to (18) for N=1. Perhaps you are like me and do not like to memorize anything, but some of these formulas might be worth memorizing as they appear again and again in theoretical physics (and in this book). Appendix 3 T o do an exponential integral of the form I=/integraltext+∞ −∞dqe−(1//planckover2pi)f (q)we often have to resort to the steepest-descent approximation, which I will now review for your convenience. In the limit of /planckover2pismall, the integral is dominated by the minimum of f( q) . Expanding f( q)=f( a)+1 2f/prime/prime(a)(q−a)2+O[(q−a)3] and applying (17) we obtain I=e−(1//planckover2pi)f (a)/parenleftbigg2π/planckover2pi f/prime/prime(a)/parenrightbigg1 2 e−O(/planckover2pi1 2)(27) Forf( q) a function of many variables q1,..., qNand with a minimum at qj=aj, we generalize immediately to I=e−(1//planckover2pi)f (a)/parenleftbigg(2π/planckover2pi)N detf/prime/prime(a)/parenrightbigg1 2 e−O(/planckover2pi1 2)(28) Heref/prime/prime(a) denotes the NbyNmatrix with entries [f/prime/prime(a)]ij≡(∂2f/∂qi∂qj)|q=a. In many situations, we do not even need the factor involving the determinant in (28). If you can derive (28) you are well on your way tobecoming a quantum field theorist! Exercises I.2.1 Verify (5). I.2.2 Derive (24). I.3 From Mattress to Field The mattress in the continuum limit The path integral representation Z≡/angbracketleft 0|e−iHT|0/angbracketright=/integraldisplay Dq(t) ei/integraltextT 0dt[1 2m˙q2−V( q) ](1) (we suppress the factor /angbracketleft0|qf/angbracketright/angbracketleftqI|0/angbracketright; we will come back to this issue later in this chapter) which we derived for the quantum mechanics of a single particle, can be generalized almostimmediately to the case of Nparticles with the Hamiltonian H=/summationdisplay a1 2maˆp2 a+V(ˆq1,ˆq2,... ,ˆqN). (2) We simply keep track mentally of the position of the particles qawitha=1, 2, ... ,N. Going through the same steps as before, we obtain Z≡/angbracketleft 0|e−iHT|0/angbracketright=/integraldisplay Dq(t) eiS(q)(3) with the action S(q)=/integraldisplayT 0dt/parenleftBig/summationdisplay a1 2ma˙q2 a−V[q1,q2,... ,qN]/parenrightBig . The potential energy V( q1,q2,...,qN)now includes interaction energy between particles, namely terms of the form v(qa−qb), as well as the energy due to an external potential, namely terms of the form w(qa). In particular, let us now write the path integral description of the quantum dynamics of the mattress described in chapter I.1, with the potential V( q 1,q2,..., qN)=/summationdisplay ab1 2kab(qa−qb)2+... We are now just a short hop and skip away from a quantum field theory! Suppose we are only interested in phenomena on length scales much greater than the lattice spacingl(see fig. I.1.1). Mathematically, we take the continuum limit l→0. In this limit, we can 18 | I. Motivation and Foundation replace the label aon the particles by a two-dimensional position vector /vectorx, and so we write q(t,/vectorx)instead of qa(t). It is traditional to replace the Latin letter qby the Greek letter ϕ. The function ϕ(t,/vectorx)is called a field. The kinetic energy/summationtext a1 2ma˙q2 anow becomes/integraltext d2x1 2σ(∂ϕ/∂t)2. We replace/summationtext aby/integraltext d2x/l2and denote the mass per unit area ma/l2byσ. We take all the ma’s to be equal; otherwise σwould be a function of /vectorx, the system would be inhomogeneous, and we would have a hard time writing down a Lorentz-invariant action (see later). We next focus on the first term in V. Assume for simplicity that kabconnect only nearest neighbors on the lattice. For nearest-neighbor pairs (qa−qb)2/similarequall2(∂ϕ/∂x)2+... in the continuum limit; the derivative is obviously taken in the direction that joins the lattice sitesaandb. Putting it together then, we have S(q)→S(ϕ)≡/integraldisplayT 0dt/integraldisplay d2xL(ϕ) =/integraldisplayT 0dt/integraldisplay d2x1 2/braceleftBigg σ/parenleftbigg∂ϕ ∂t/parenrightbigg2 −ρ/bracketleftBigg/parenleftbigg∂ϕ ∂x/parenrightbigg2 +/parenleftbigg∂ϕ ∂y/parenrightbigg2/bracketrightBigg −τϕ2−ςϕ4+.../bracerightBigg (4) where the parameter ρis determined by kabandl. The precise relations do not concern us. Henceforth in this book, we will take the T→∞ limit so that we can integrate over all of spacetime in (4). We can clean up a bit by writing ρ=σc2and scaling ϕ→ϕ/√σ, so that the combination (∂ϕ/∂t)2−c2[(∂ϕ/∂x)2+(∂ϕ/∂y)2] appears in the Lagrangian. The parameter cevidently has the dimension of a velocity and defines the phase velocity of the waves on our mattress. We started with a mattress for pedagogical reasons. Of course nobody believes that the fields observed in Nature, such as the meson field or the photon field, are actuallyconstructed of point masses tied together with springs. The modern view, which I will callLandau-Ginzburg, is that we start with the desired symmetry, say Lorentz invariance if wewant to do particle physics, decide on the fields we want by specifying how they transformunder the symmetry (in this case we decided on a scalar field ϕ), and then write down the action involving no more than two time derivatives (because we don’t know how toquantize actions with more than two time derivatives). We end up with a Lorentz-invariant action (setting c=1) S=/integraldisplay ddx/bracketleftbigg1 2(∂ϕ)2−1 2m2ϕ2−g 3!ϕ3−λ 4!ϕ4+.../bracketrightbigg (5) where various numerical factors are put in for later convenience. The relativistic nota- tion(∂ϕ)2≡∂μϕ∂μϕ=(∂ϕ/∂t)2−(∂ϕ/∂x)2−(∂ϕ/∂y)2was explained in the note on convention. The dimension of spacetime, d, clearly can be any integer, even though in our mattress model it was actually 3. We often write d=D+1 and speak of a ( D+1)- dimensional spacetime. We see here the power of imposing a symmetry. Lorentz invariance together with the insistence that the Lagrangian involve only at most two powers of ∂/∂t immediately tells us I.3. From Mattress to Field | 19 that the Lagrangian can only have the form1L=1 2(∂ϕ)2−V( ϕ) withVsome function of ϕ. For simplicity, we now restrict Vto be a polynomial in ϕ, although much of the present discussion will not depend on this restriction. We will have a great deal more to say aboutsymmetry later. Here we note that, for example, we could insist that physics is symmetricunder ϕ→−ϕ, in which case V( ϕ) would have to be an even polynomial. Now that you know what a quantum field theory is, you realize why I used the letter q to label the position of the particle in the previous chapter and not the more common /vectorx. In quantum field theory, /vectorxis a label, not a dynamical variable. The /vectorxappearing in ϕ(t,/vectorx) corresponds to the label ainq a(t)in quantum mechanics. The dynamical variable in field theory is not position, but the field ϕ. The variable /vectorxsimply specifies which field variable we are talking about. I belabor this point because upon first exposure to quantum field theorysome students, used to thinking of /vectorxas a dynamical operator in quantum mechanics, are confused by its role here. In summary, we have the table q→ϕ a→/vectorx (6) qa(t)→ϕ(t,/vectorx)=ϕ(x) /summationtext a→/integraltext dDx Thus we finally have the path integral defining a scalar field theory in d=(D+1)dimen- sional spacetime: Z=/integraldisplay Dϕei/integraltext ddx(1 2(∂ϕ)2−V( ϕ) )(7) Note that a (0 +1)-dimensional quantum field theory is just quantum mechanics. The classical limit As I have already remarked, the path integral formalism is particularly convenient for taking the classical limit. Remembering that Planck’s constant /planckover2pihas the dimension of energy multiplied by time, we see that it appears in the unitary evolution operator e(−i//planckover2pi)HT. T racing through the derivation of the path integral, we see that we simply divide the overallfactor iby/planckover2pito get Z=/integraldisplay Dϕe(i//planckover2pi)/integraltext d4xL(ϕ)(8) 1Strictly speaking, a term of the form U(ϕ)(∂ϕ)2is also possible. In quantum mechanics, a term such as U(q)(dq/dt)2in the Lagrangian would describe a particle whose mass depends on position. We will not consider such “nasty” terms until much later. 20 | I. Motivation and Foundation In the limit /planckover2pimuch smaller than the relevant action we are considering, we can evaluate the path integral using the stationary phase (or steepest descent) approximation, as I explainedin the previous chapter in the context of quantum mechanics. We simply determine theextremum of/integraltext d 4xL(ϕ). According to the usual Euler-Lagrange variational procedure, this leads to the equation ∂μδL δ(∂μϕ)−δL δϕ=0 (9) We thus recover the classical field equation, exactly as we should, which in our scalar field theory reads (∂2+m2)ϕ(x) +g 2ϕ(x)2+λ 6ϕ(x)3+...=0 (10) The vacuum In the point particle quantum mechanics discussed in chapter I.2 we wrote the path integral for /angbracketleftF|e−iHT|I/angbracketright, with some initial and final state, which we can choose at our pleasure. A convenient and particularly natural choice would be to take |I/angbracketright=|F/angbracketrightto be the ground state. In quantum field theory what should we choose for the initial and finalstates? A standard choice for the initial and final states is the ground state or the vacuumstate of the system, denoted by |0/angbracketright, in which, speaking colloquially, nothing is happening. In other words, we would calculate the quantum transition amplitude from the vacuum tothe vacuum, which would enable us to determine the energy of the ground state. But thisis not a particularly interesting quantity, because in quantum field theory we would like tomeasure all energies relative to the vacuum and so, by convention, would set the energyof the vacuum to zero (possibly by having to subtract a constant from the Lagrangian).Incidentally, the vacuum in quantum field theory is a stormy sea of quantum fluctuations,but for this initial pass at quantum field theory, we will not examine it in any detail. Wewill certainly come back to the vacuum in later chapters. Disturbing the vacuum We might enjoy doing something more exciting than watching a boiling sea of quantum fluctuations. We might want to disturb the vacuum. Somewhere in space, at some instant in time, we would like to create a particle, watch it propagate for a while, and then annihilateit somewhere else in space, at some later instant in time. In other words, we want to setup a source and a sink (sometimes referred to collectively as sources) at which particlescan be created and annihilated. T o see how to do this, let us go back to the mattress. Bounce up and down on it to create some excitations. Obviously, pushing on the mass labeled by ain the mattress corresponds to adding a term such as J a(t)qato the potential V( q1,q2,... ,qN). More generally, I.3. From Mattress to Field | 21 JJJ ? t x Figure I.3.1 we can add/summationtext aJa(t)qa. When we go to field theory this added term gets promoted to/integraltext dDxJ(x)ϕ(x) in the field theory Lagrangian, according to the promotion table (6). This so-called source function J(t,/vectorx)describes how the mattress is being disturbed. We can choose whatever function we like, corresponding to our freedom to push on themattress wherever and whenever we like. In particular, J(x) can vanish everywhere in spacetime except in some localized regions. By bouncing up and down on the mattress we can get wave packets going off here and there (fig. I.3.1). This corresponds precisely to sources (and sinks) for particles. Thus, wereally want the path integral Z=/integraldisplay Dϕei/integraltext d4x[1 2(∂ϕ)2−V( ϕ) +J(x)ϕ(x) ](11) Free field theory The functional integral in (11) is impossible to do except when L(ϕ)=1 2[(∂ϕ)2−m2ϕ2] (12) The corresponding theory is called the free or Gaussian theory. The equation of motion (9) works out to be (∂2+m2)ϕ=0, known as the Klein-Gordon equation.2Being linear, it can be solved immediately to give ϕ(/vectorx,t)=ei(ωt−/vectork./vectorx)with ω2=/vectork2+m2(13) 2The Klein-Gordon equation was actually discovered by Schr ¨odinger before he found the equation that now bears his name. Later, in 1926, it was written down independently by Klein, Gordon, Fock, Kudar, de Donder, and Van Dungen. 22 | I. Motivation and Foundation In the natural units we are using, /planckover2pi=1 and so frequency ωis the same as energy /planckover2piω and wave vector /vectorkis the same as momentum /planckover2pi/vectork. Thus, we recognize (13) as the energy- momentum relation for a particle of mass m, namely the sophisticate’s version of the layperson’s E=mc2. We expect this field theory to describe a relativistic particle of mass m. Let us now evaluate (11) in this special case: Z=/integraldisplay Dϕei/integraltext d4x{1 2[(∂ϕ)2−m2ϕ2]+Jϕ}(14) Integrating by parts under the/integraltext d4xand not worrying about the possible contribution of boundary terms at infinity (we implicitly assume that the fields we are integrating over falloff sufficiently rapidly), we write Z=/integraldisplay Dϕei/integraltext d4x[−1 2ϕ(∂2+m2)ϕ+Jϕ ](15) You will encounter functional integrals like this again and again in your study of field theory. The trick is to imagine discretizing spacetime. You don’t actually have to do it:Just imagine doing it. Let me sketch how this goes. Replace the function ϕ(x) by the vector ϕ i=ϕ(ia) withian integer and athe lattice spacing. (For simplicity, I am writing things as if we were in 1-dimensional spacetime. More generally, just let the index i enumerate the lattice points in some way.) Then differential operators become matrices.For example, ∂ϕ(ia) →(1/a) (ϕ i+1−ϕi)≡/summationtext jMijϕj, with some appropriate matrix M. Integrals become sums. For example,/integraltext d4xJ(x)ϕ(x) →a4/summationtext iJiϕi. Now, lo and behold, the integral (15) is just the integral we did in (I.2.23) /integraldisplay+∞ −∞/integraldisplay+∞ −∞.../integraldisplay+∞ −∞dq1dq2...dqNe(i/2)q.A.q+iJ.q =/parenleftbigg(2πi)N det[A]/parenrightbigg1 2 e−(i/2)J.A−1.J(16) The role of Ain (16) is played in (15) by the differential operator −(∂2+m2). The defining equation for the inverse, A.A−1=IorAijA−1 jk=δik, becomes in the continuum limit −(∂2+m2)D(x−y)=δ(4)(x−y) (17) We denote the continuum limit of A−1 jkbyD(x−y)(which we know must be a function ofx−y, and not of xandyseparately, since no point in spacetime is special). Note that in going from the lattice to the continuum Kronecker is replaced by Dirac. It is very usefulto be able to go back and forth mentally between the lattice and the continuum. Our final result is Z(J)=Ce−(i/2)/integraltext/integraltext d4xd4yJ(x)D(x −y)J(y)≡CeiW(J)(18) withD(x) determined by solving (17). The overall factor C, which corresponds to the overall factor with the determinant in (16), does not depend on Jand, as will become clear in the discussion to follow, is often of no interest to us. I will often omit writing Caltogether. Clearly, C=Z(J=0)so that W(J) is defined by Z(J)≡Z(J=0)eiW(J)(19) I.3. From Mattress to Field | 23 Observe that W(J) =−1 2/integraldisplay/integraldisplay d4xd4yJ(x)D(x −y)J(y) (20) is a simple quadratic functional of J. In contrast, Z(J) depends on arbitrarily high powers ofJ. This fact will be of importance in chapter I.7. Free propagator The function D(x) , known as the propagator, plays an essential role in quantum field theory. As the inverse of a differential operator it is clearly closely related to the Green’sfunction you encountered in a course on electromagnetism. Physicists are sloppy about mathematical rigor, but even so, they have to be careful once in a while to make sure that what they are doing actually makes sense. For the integralin (15) to converge for large ϕwe replace m 2→m2−iεso that the integrand contains a factor e−ε/integraltext d4xϕ2, where εis a positive infinitesimal we will let tend to zero.3 We can solve (17) easily by going to momentum space and multiplying together four copies of the representation (I.2.11) of the Dirac delta function δ(4)(x−y)=/integraldisplayd4k (2π)4eik(x−y)(21) The solution is D(x−y)=/integraldisplayd4k (2π)4eik(x−y) k2−m2+iε(22) which you can check by plugging into (17): −(∂2+m2)D(x−y)=/integraldisplayd4k (2π)4k2−m2 k2−m2+iεeik(x−y)=/integraldisplayd4k (2π)4eik(x−y)=δ(4)(x−y)asε→0. Note that the so-called iεprescription we just mentioned is essential; otherwise the integral giving D(x) would hit a pole. The magnitude of εis not important as long as it is infinitesimal, but the positive sign of εis crucial as we will see presently. (More on this in chapter III.8.) Also, note that the sign of kin the exponential does not matter here by the symmetry k→−k. T o evaluate D(x) we first integrate over k0by the method of contours. Define ωk≡ +/radicalbig /vectork2+m2with a plus sign. The integrand has two poles in the complex k0plane, at ±/radicalBig ω2 k−iε, which in the ε→0 limit are equal to +ωk−iεand−ωk+iε. Thus for ε positive, one pole is in the lower half-plane and the other in the upper half plane, and so as we go along the real k0axis from −∞ to+∞ we do not run into the poles. The issue is how to close the integration contour. Forx0positive, the factor eik0x0is exponentially damped for k0in the upper half-plane. Hence we should extend the integration contour extending from −∞ to+∞ on the real 3As is customary, εis treated as generic, so that εmultiplied by any positive number is still ε. 24 | I. Motivation and Foundation axis to include the infinite semicircle in the upper half-plane, thus enclosing the pole at −ωk+iεand giving −i/integraltextd3k (2π)32ωke−i(ω kt−/vectork./vectorx). Again, note that we are free to flip the sign of/vectork. Also, as is conventional, we use x0andtinterchangeably. (In view of some reader confusion here in the first edition, I might add that I generally use x0withk0andtwith ωk;k0is a variable that can take on either sign but ωkis a positive function of /vectork.) Forx0negative, we do the opposite and close the contour in the lower half-plane, thus picking up the pole at +ωk−iε. We now obtain −i/integraltext (d3k/(2π)32ωk)e+i(ω kt−/vectork./vectorx). Recall that the Heaviside (we will meet this great and aptly named physicist in chapter IV .4) step function θ(t) is defined to be equal to 0 for t< 0 and equal to 1 for t> 0. As for whatθ(0)should be, the answer is that since we are proud physicists and not nitpicking mathematicians we will just wing it when the need arises. The step function allows us topackage our two integration results together as D(x)=−i/integraldisplayd3k (2π)32ωk[e−i(ω kt−/vectork./vectorx)θ(x0)+ei(ωkt−/vectork./vectorx)θ(−x0)] (23) Physically, D(x) describes the amplitude for a disturbance in the field to propagate from the origin to x. Lorentz invariance tells us that it is a function of x2and the sign of x0(since these are the quantities that do not change under a Lorentz transformation). We thus expectdrastically different behavior depending on whether xis inside or outside the lightcone defined by x 2=(x0)2−/vectorx2=0. Without evaluating the d3kintegral we can see roughly how things go. Let us look at some cases. In the future cone, x=(t,0)witht> 0,D(x)=−i/integraltext (d3k/(2π)32ωk)e−iωkta superpo- sition of plane waves and thus D(x) oscillates. In the past cone, x=(t,0)witht< 0, D(x)=−i/integraltext (d3k/(2π)32ωk)e+iωktoscillates with the opposite phase. In contrast, for xspacelike rather than timelike, x0=0, we have, upon interpret- ingθ(0)=1 2(the obvious choice; imagine smoothing out the step function), D(x)= −i/integraltext (d3k/(2π)32/radicalbig /vectork2+m2)e−i/vectork./vectorx. The square root cut starting at ±im tells us that the characteristic value of |/vectork|in the integral is of order m, leading to an exponential decay ∼e−m|/vectorx|, as we would expect. Classically, a particle cannot get outside the lightcone, but a quantum field can “leak” out over a distance of order m−1by the Heisenberg uncertainty principle. Exercises I.3.1 Verify that D(x) decays exponentially for spacelike separation. I.3.2 Work out the propagator D(x) for a free field theory in (1+1)-dimensional spacetime and study the large x1behavior for x0=0. I.3.3 Show that the advanced propagator defined by Dadv(x−y)=/integraldisplayd4k (2π)4eik(x−y) k2−m2−isgn(k0)ε I.3. From Mattress to Field | 25 (where the sign function is defined by sgn (k0)=+ 1i fk0>0 and sgn (k0)=− 1i fk0<0) is nonzero only ifx0>y0. In other words, it only propagates into the future. [Hint: both poles of the integrand are now in the upper half of the k0-plane.] Incidentally, some authors prefer to write (k0−ie)2−/vectork2−m2instead ofk2−m2−isgn(k0)εin the integrand. Similarly, show that the retarded propagator Dret(x−y)=/integraldisplayd4k (2π)4eik(x−y) k2−m2+isgn(k0)ε propagates into the past. I.4 From Field to Particle to Force From field to particle In the previous chapter we obtained for the free theory W(J) =−1 2/integraldisplay/integraldisplay d4xd4yJ(x)D(x −y)J(y) (1) which we now write in terms of the Fourier transform J(k)≡/integraltext d4xe−ikxJ(x) : W(J) =−1 2/integraldisplayd4k (2π)4J(k)∗ 1 k2−m2+iεJ(k) (2) [Note that J(k)∗=J(−k) forJ(x) real.] We can jump up and down on the mattress any way we like. In other words, we can choose any J(x) we want, and by exploiting this freedom of choice, we can extract a remarkable amount of physics. Consider J(x)=J1(x)+J2(x), where J1(x) andJ2(x) are concentrated in two local regions 1 and 2 in spacetime (fig. I.4.1). Then W(J) contains four terms, of the form J∗ 1J1, J∗ 2J2,J∗ 1J2, andJ∗ 2J1. Let us focus on the last two of these terms, one of which reads W(J) =−1 2/integraldisplayd4k (2π)4J2(k)∗ 1 k2−m2+iεJ1(k) (3) We see that W(J) is large only if J1(x) andJ2(x) overlap significantly in their Fourier transform and if in the region of overlap in momentum space k2−m2almost vanishes. There is a “resonance type” spike at k2=m2, that is, if the energy-momentum relation of a particle of mass mis satisfied. (We will use the language of the relativistic physicist, writing “momentum space” for energy-momentum space, and lapse into nonrelativisticlanguage only when the context demands it, such as in “energy-momentum relation.”) We thus interpret the physics contained in our simple field theory as follows: In region 1 in spacetime there exists a source that sends out a “disturbance in the field,” whichis later absorbed by a sink in region 2 in spacetime. Experimentalists choose to call this I.4. From Field to Particle to Force | 27 J1tJ2 x Figure I.4.1 disturbance in the field a particle of mass m. Our expectation based on the equation of motion that the theory contains a particle of mass mis fulfilled. A bit of jargon: When k2=m2,kis said to be on mass shell. Note, however, that in (3) we integrate over all k, including values of kfar from the mass shell. For arbitrary k,i ti s a linguistic convenience to say that a “virtual particle” of momentum kpropagates from the source to the sink. From particle to force We can now go on to consider other possibilities for J(x) (which we will refer to generically as sources), for example, J(x)=J1(x)+J2(x), where Ja(x)=δ(3)(/vectorx−/vectorxa). In other words, J(x) is a sum of sources that are time-independent infinitely sharp spikes located at /vectorx1and /vectorx2in space. (If you like more mathematical rigor than is offered here, you are welcome to replace the delta function by lumpy functions peaking at /vectorxa. You would simply clutter up the formulas without gaining much.) More picturesquely, we are describing two massive lumps sitting at /vectorx1and/vectorx2on the mattress and not moving at all [no time dependence in J(x) ]. What do the quantum fluctuations in the field ϕ, that is, the vibrations in the mattress, do to the two lumps sitting on the mattress? If you expect an attraction between the twolumps, you are quite right. As before, W(J) contains four terms. We neglect the “self-interaction” term J 1J1since this contribution would be present in Wregardless of whether J2is present or not. We want to study the interaction between the two “massive lumps” represented by J1andJ2. Similarly we neglect J2J2. 28 | I. Motivation and Foundation Plugging into (1) and doing the integral over d3xandd3ywe immediately obtain W(J) =−/integraldisplay/integraldisplay dx0dy0/integraldisplaydk0 2πeik0(x−y)0/integraldisplayd3k (2π)3ei/vectork.(/vectorx1−/vectorx2) k2−m2+iε(4) (The factor 2 comes from the two terms J2J1andJ1J2. ) Integrating over y0we get a delta function setting k0to zero (so that kis certainly not on mass shell, to throw the jargon around a bit). Thus we are left with W(J) =/parenleftbigg/integraldisplay dx0/parenrightbigg/integraldisplayd3k (2π)3ei/vectork.(/vectorx1−/vectorx2) /vectork2+m2(5) Note that the infinitesimal iεcan be dropped since the denominator /vectork2+m2is always positive. The factor (/integraltext dx0)should have filled us with fear and trepidation: an integral over time, it seems to be infinite. Fear not! Recall that in the path integral formalism Z=CeiW(J) represents /angbracketleft0|e−iHT|0/angbracketright=e−iET, where Eis the energy due to the presence of the two sources acting on each other. The factor (/integraltext dx0)produces precisely the time interval T. All is well. Setting iW=−iET we obtain from (5) E=−/integraldisplayd3k (2π)3ei/vectork.(/vectorx1−/vectorx2) /vectork2+m2(6) The integral is evaluated in an appendix. This energy is negative! The presence of two delta function sources, at /vectorx1and/vectorx2, has lowered the energy. (Notice that for the two sources infinitely far apart, we have, as we might expect, E=0: the infinitely rapidly oscillating exponential kills the integral.) In other words, two like objects attract each other by virtueof their coupling to the field ϕ. We have derived our first physical result in quantum field theory! We identify Eas the potential energy between two static sources. Even without doing the integral, we see by dimensional analysis that the characteristic distance beyond whichthe integral goes to zero is given by the inverse of the characteristic value of k, which is m. Thus, we expect the attraction between the two sources to decrease rapidly to zero over the distance 1 /m. The range of the attractive force generated by the field ϕis determined inversely by the massmof the particle described by the field. Got that? The integral is done in the appendix to this chapter and gives E=−1 4πre−mr(7) The result is as we expected: The potential drops off exponentially over the distance scale 1/m. Obviously, dE/dr > 0: The two massive lumps sitting on the mattress can lower the energy by getting closer to each other. What we have derived was one of the most celebrated results in twentieth-century physics. Yukawa proposed that the attraction between nucleons in the atomic nucleus isdue to their coupling to a field like the ϕfield described here. The known range of the nuclear force enabled him to predict not only the existence of the particle associated with I.4. From Field to Particle to Force | 29 this field, now called the πmeson1or the pion, but its mass as well. As you probably know, the pion was eventually discovered with essentially the properties predicted by Yukawa. Origin of force That the exchange of a particle can produce a force was one of the most profound concep-tual advances in physics. We now associate a particle with each of the known forces: forexample, the photon with the electromagnetic force and the graviton with the gravitationalforce; the former is experimentally well established and while the latter has not yet beendetected experimentally hardly anyone doubts its existence. We will discuss the photon andthe graviton in the next chapter, but we can already answer a question smart high schoolstudents often ask: Why do Newton’s gravitational force and Coulomb’s electric force bothobey the 1 /r 2law? We see from (7) that if the mass mof the mediating particle vanishes, the force produced will obey the 1 /r2law. If you trace back over our derivation, you will see that this comes from the fact that the Lagrangian density for the simplest field theory involves two powersof the spacetime derivative ∂(since any term involving one derivative such as ϕ∂ ϕ is not Lorentz invariant). Indeed, the power dependence of the potential follows simply from dimensional analysis:/integraltext d 3k(ei/vectork./vectorx/k2)∼1/r. Connected versus disconnected We end with a couple of formal remarks of importance to us only in chapter I.7. First, note that we might want to draw a small picture fig. (I.4.2) to represent the integrandJ(x)D(x −y)J(y) inW(J) : A disturbance propagates from ytox(or vice versa). In fact, this is the beginning of Feynman diagrams! Second, recall that Z(J)=Z(J=0)∞/summationdisplay n=0[iW(J)]n n! For instance, the n=2 term in Z(J)/Z(J =0)is given by 1 2!/parenleftbigg −i 2/parenrightbigg2/integraldisplay/integraldisplay/integraldisplay/integraldisplay d4x1d4x2d4x3d4x4D(x 1−x2) D(x 3−x4)J(x1)J(x2)J(x3)J(x4) The integrand is graphically described in figure I.4.3. The process is said to be discon- nected: The propagation from x1tox2and the propagation from x3tox4proceed inde- pendently. We will come back to the difference between connected and disconnected inchapter I.7. 1The etymology behind this word is quite interesting (A. Zee, Fearful Symmetry : see pp. 169 and 335 to learn, among other things, the French objection and the connection between meson and illusion). 30 | I. Motivation and Foundation yx Figure I.4.2 x1x2 x4 x3 Figure I.4.3 I.4. From Field to Particle to Force | 31 Appendix Writing /vectorx≡(/vectorx1−/vectorx2)andu≡cosθwithθthe angle between /vectorkand/vectorx, we evaluate the integral in (6) in spherical coordinates (with k=|/vectork|andr=|/vectorx|): I≡1 (2π)2/integraldisplay∞ 0dk k2/integraldisplay+1 −1dueikru k2+m2=2i (2π)2ir/integraldisplay∞ 0dk ksinkr k2+m2(8) Since the integrand is even, we can extend the integral and write it as 1 2/integraldisplay∞ −∞dk ksinkr k2+m2=1 2i/integraldisplay∞ −∞dk k1 k2+m2eikr. Since ris positive, we can close the contour in the upper half-plane and pick up the pole at +im , obtaining (1/2i)(2πi)(im/2 im)e−mr=(π/2)e−mr. Thus, I=(1/4πr)e−mr. Exercise I.4.1 Calculate the analog of the inverse square law in a (2 +1)-dimensional universe, and more generally in a(D+1)-dimensional universe. I.5 Coulomb and Newton: Repulsion and Attraction Why like charges repel We suggested that quantum field theory can explain both Newton’s gravitational force and Coulomb’s electric force naturally. Between like objects Newton’s force is attractive whileCoulomb’s force is repulsive. Is quantum field theory “smart enough” to produce thisobservational fact, one of the most basic in our understanding of the physical universe?You bet! We will first treat the quantum field theory of the electromagnetic field, known as quantum electrodynamics or QED for short. In order to avoid complications at this stageassociated with gauge invariance (about which much more later) I will consider insteadthe field theory of a massive spin 1 meson, or vector meson. After all, experimentally all weknow is an upper bound on the photon mass, which although tiny is not mathematicallyzero. We can adopt a pragmatic attitude: Calculate with a photon mass mand set m=0a t the end, and if the result does not blow up in our faces, we will presume that it is OK. 1 Recall Maxwell’s Lagrangian for electromagnetism L=−1 4FμνFμν, where Fμν≡∂μAν −∂νAμwithAμ(x) the vector potential. You can see the reason for the important overall minus sign in the Lagrangian by looking at the coefficient of (∂0Ai)2, which has to be positive, just like the coefficient of (∂0ϕ)2in the Lagrangian for the scalar field. This says simply that time variation should cost a positive amount of action. I will now give the photon a small mass by changing the Lagrangian to L=−1 4FμνFμν +1 2m2AμAμ+AμJμ. (The mass term is written in analogy to the mass term m2ϕ2in the scalar field Lagrangian; we will see shortly that the sign is correct and that this term indeed leads to a photon mass.) I have also added a source Jμ(x) ,which in this context is more familiarly known as a current. We will assume that the current is conserved so that∂ μJμ=0. 1When I took a field theory course as a student with Sidney Coleman this was how he treated QED in order to avoid discussing gauge invariance. I.5. Coulomb and Newton | 33 Well, you know that the field theory of our vector meson is defined by the path integral Z=/integraltext DA eiS(A)≡eiW(J)with the action S(A)=/integraldisplay d4xL=/integraldisplay d4x{1 2Aμ[(∂2+m2)gμν−∂μ∂ν]Aν+AμJμ} (1) The second equality follows upon integrating by parts [compare (I.3.15)]. By now you have learned that we simply apply (I.3.16). We merely have to find the inverse of the differential operator in the square bracket; in other words, we have to solve [(∂2+m2)gμν−∂μ∂ν]Dνλ(x)=δμ λδ(4)(x) (2) As before [compare (I.3.17)] we go to momentum space by defining Dνλ(x)=/integraldisplayd4k (2π)4Dνλ(k)eikx Plugging in, we find that [ −(k2−m2)gμν+kμkν]Dνλ(k)=δμ λ, giving Dνλ(k)=−gνλ+kνkλ/m2 k2−m2(3) This is the photon, or more accurately the massive vector meson, propagator. Thus W(J) =−1 2/integraldisplayd4k (2π)4Jμ(k)∗−gμν+kμkν/m2 k2−m2+iεJν(k) (4) Since current conservation ∂μJμ(x)=0 gets translated into momentum space as kμJμ(k)=0, we can throw away the kμkνterm in the photon propagator. The effective action simplifies to W(J) =1 2/integraldisplayd4k (2π)4Jμ(k)∗ 1 k2−m2+iεJμ(k) (5) No further computation is needed to obtain a profound result. Just compare this result to (I.4.2). The field theory has produced an extra sign. The potential energy between twolumps of charge density J 0(x) is positive. The electromagnetic force between like charges is repulsive! We can now safely let the photon mass mgo to zero thanks to current conservation. [Note that we could not have done that in (3).] Indeed, referring to (I.4.7) we see that thepotential energy between like charges is E=1 4πre−mr→1 4πr(6) T o accommodate positive and negative charges we can simply write Jμ=Jμ p−Jμ n.W e see that a lump with charge density J0 pis attracted to a lump with charge density J0 n. Bypassing Maxwell Having done electromagnetism in two minutes flat let me now do gravity. Let us move on to the massive spin 2 meson field. In my treatment of the massive spin 1 meson field I 34 | I. Motivation and Foundation took a short cut. Assuming that you are familiar with the Maxwell Lagrangian, I simply added a mass term to it and took off. But I do not feel comfortable assuming that you areequally familiar with the corresponding Lagrangian for the massless spin 2 field (the so-called linearized Einstein Lagrangian, which I will discuss in a later chapter). So here I willfollow another strategy. I invite you to think physically, and together we will arrive at the propagator for a massive spin 2 field. First, we will warm up with the massive spin 1 case. In fact, start with something even easier: the propagator D(k)=1/(k 2−m2)for a massive spin 0 field. It tells us that the amplitude for the propagation of a spin 0 disturbanceblows up when the disturbance is almost a real particle. The residue of the pole is a propertyof the particle. The propagator for a spin 1 field D νλcarries a pair of Lorentz indices and in fact we know what it is from (3): Dνλ(k)=−Gνλ k2−m2(7) where for later convenience we have defined Gνλ(k)≡gνλ−kνkλ m2(8) Let us now understand the physics behind Gνλ. I expect you to remember the concept of polarization from your course on electromagnetism. A massive spin 1 particle has threedegrees of polarization for the obvious reason that in its rest frame its spin vector can pointin three different directions. The three polarization vectors ε (a) μare simply the three unit vectors pointing along the x,y, and zaxes, respectively (a =1, 2, 3 ):ε(1) μ=(0, 1, 0, 0 ), ε(2) μ=(0, 0, 1, 0 ),ε(3) μ=(0, 0, 0, 1 ). In the rest frame kμ=(m,0 ,0 ,0 )and so kμε(a) μ=0 (9) Since this is a Lorentz invariant equation, it holds for a moving spin 1 particle as well. Indeed, with a suitable normalization condition this fixes the three polarization vectorsε (a) μ(k) for a particle with momentum k. The amplitude for a particle with momentum kand polarization ato be created at the source is proportional to ε(a) λ(k), and the amplitude for it to be absorbed at the sink is proportional to ε(a) ν(k). We multiply the amplitudes together to get the amplitude for propagation from source to sink, and then sum over the three possible polarizations.Now we understand the residue of the pole in the spin 1 propagator D νλ(k): It represents/summationtext aε(a) ν(k) ε(a) λ(k). T o calculate this quantity, note that by Lorentz invariance it can only be a linear combination of gνλandkνkλ. The condition kμε(a) μ=0 fixes it to be proportional togνλ−kνkλ/m2. We evaluate the left-hand side for kat rest with ν=λ=1, for instance, and fix the overall and all-crucial sign to be −1. Thus /summationdisplay aε(a) ν(k)ε(a) λ(k)=−Gνλ(k)≡−/parenleftbigg gνλ−kνkλ m2/parenrightbigg (10) We have thus constructed the propagator Dνλ(k)for a massive spin 1 particle, bypassing Maxwell (see appendix 1). Onward to spin 2! We want to similarly bypass Einstein. I.5. Coulomb and Newton | 35 Bypassing Einstein A massive spin 2 particle has 5 (2 .2+1=5, remember?) degrees of polarization, char- acterized by the five polarization tensors ε(a) μν(a=1, 2, ... ,5)symmetric in the indices μ andνsatisfying kμε(a) μν=0 (11) and the tracelessness condition gμνε(a) μν=0 (12) Let’s count as a check. A symmetric Lorentz tensor has 4 .5/2=10 components. The four conditions in (11) and the single condition in (12) cut the number of components down to10−4−1=5, precisely the right number. (Just to throw some jargon around, remember how to construct irreducible group representations? If not, read appendix B.) We fix thenormalization of ε μνby setting the positive quantity/summationtext aε(a) 12(k)ε(a) 12(k)=1. So, in analogy with the spin 1 case we now determine/summationtext aε(a) μν(k)ε(a) λσ(k). We have to construct this object out of gμνandkμ, or equivalently Gμνandkμ. This quantity must be a linear combination of terms such as GμνGλσ,Gμνkλkσ, and so forth. Using (11) and (12) repeatedly (exercise I.5.1) you will easily find that /summationdisplay aε(a) μν(k)ε(a) λσ(k)=(GμλGνσ+GμσGνλ)−2 3GμνGλσ (13) The overall sign and proportionality constant are determined by evaluating both sides for kat rest (for μ=λ=1 and ν=σ=2, for instance). Thus, we have determined the propagator for a massive spin 2 particle Dμν,λσ(k)=(GμλGνσ+GμσGνλ)−2 3GμνGλσ k2−m2(14) Why we fall We are now ready to understand one of the fundamental mysteries of the universe: Why masses attract. Recall from your courses on electromagnetism and special relativity that the energy or mass density out of which mass is composed is part of a stress-energy tensor Tμν. For our purposes, in fact, all you need to remember is that Tμνis a symmetric tensor and that the component T00is the energy density. If you don’t remember, I will give you a physical explanation in appendix 2. T o couple to the stress-energy tensor, we need a tensor field ϕμνsymmetric in its two indices. In other words, the Lagrangian of the world should contain a term like ϕμνTμν. This is in fact how we know that the graviton, the particle responsible for gravity, has spin 2,just as we know that the photon, the particle responsible for electromagnetism and hence 36 | I. Motivation and Foundation coupled to the current Jμ, has spin 1. In Einstein’s theory, which we will discuss in a later chapter, ϕμνis of course part of the metric tensor. Just as we pretended that the photon has a small mass to avoid having to discuss gauge invariance, we will pretend that the graviton has a small mass to avoid having to discussgeneral coordinate invariance. 2Aha, we just found the propagator for a massive spin 2 particle. So let’s put it to work. In precise analogy to (4) W(J) =−1 2/integraldisplayd4k (2π)4Jμ(k)∗−gμν+kμkν/m2 k2−m2+iεJν(k) (15) describing the interaction between two electromagnetic currents, the interaction between two lumps of stress energy is described by W(T) = −1 2/integraldisplayd4k (2π)4Tμν(k)∗(GμλGνσ+GμσGνλ)−2 3GμνGλσ k2−m2+iεTλσ(k)(16) From the conservation of energy and momentum ∂μTμν(x)=0 and hence kμTμν(k)=0, we can replace Gμνin (16) by gμν. (Here as is clear from the context gμνstill denotes the flat spacetime metric of Minkowski, rather than the curved metric of Einstein.) Now comes the punchline. Look at the interaction between two lumps of energy density T00. We have from (16) that W(T) =−1 2/integraldisplayd4k (2π)4T00(k)∗1+1−2 3 k2−m2+iεT00(k) (17) Comparing with (5) and using the well-known fact that (1+1−2 3)>0, we see that while like charges repel, masses attract. T rumpets, please! The universe It is difficult to overstate the importance (not to speak of the beauty) of what we havelearned: The exchange of a spin 0 particle produces an attractive force, of a spin 1 particlea repulsive force, and of a spin 2 particle an attractive force, realized in the hadronic stronginteraction, the electromagnetic interaction, and the gravitational interaction, respectively.The universal attraction of gravity produces an instability that drives the formation ofstructure in the early universe. 3Denser regions become denser yet. The attractive nuclear force mediated by the spin 0 particle eventually ignites the stars. Furthermore, the attractiveforce between protons and neutrons mediated by the spin 0 particle is able to overcomethe repulsive electric force between protons mediated by the spin 1 particle to form a 2For the moment, I ask you to ignore all subtleties and simply assume that in order to understand gravity it is kosher to let m→0. I will give a precise discussion of Einstein’s theory of gravity in chapter VIII.1. 3A good place to read about gravitational instability and the formation of structure in the universe along the line sketched here is in A. Zee, Einstein ’s Universe (formerly known as An Old Man ’s T oy ). I.5. Coulomb and Newton | 37 variety of nuclei without which the world would certainly be rather boring. The repulsion between likes and hence attraction between opposites generated by the spin 1 particle allowelectrically neutral atoms to form. The world results from a subtle interplay among spin 0, 1, and 2.In this lightning tour of the universe, we did not mention the weak interaction. In fact, the weak interaction plays a crucial role in keeping stars such as our sun burning at asteady rate. Time differs from space by a sign This weaving together of fields, particles, and forces to produce a universe rich withpossibilities is so beautiful that it is well worth pausing to examine the underlying physicssome more. The expression in (I.4.1) describes the effect of our disturbing the vacuum(or the mattress!) with the source J, calculated to second order. Thus some readers may have recognized that the negative sign in (I.4.6) comes from the elementary quantummechanical result that in second order perturbation theory the lowest energy state alwayshas its energy pushed downward: for the ground state all the energy denominators have the same sign.In essence, this “theorem” follows from the property of 2 by 2 matrices. Let us set the ground state energy to 0 and crudely represent the entirety of the other states by a singlestate with energy w> 0. Then the Hamiltonian including the perturbation veffective to second order is given by H=/parenleftBiggwv v 0/parenrightBigg Since the determinant of H(and hence the product of the two eigenvalues) is manifestly negative, the ground state energy is pushed below 0. [More explicitly, we calculate theeigenvalue εwith the characteristic equation 0 =ε(ε−w)−v 2≈−(wε+v2), and hence ε≈−v2 w.] In different fields of physics, this phenomenon is variously known as level repulsion or the seesaw mechanism (see chapter VII.7). Disturbing the vacuum with a source lowers its energy. Thus it is easy to understand that generically the exchange of a particles leads to an attractive force. But then why does the exchange of a spin 1 particle produces a repulsion between like objects? The secret lies in the profundity that space differs from time by a sign, namely,thatg 00=+ 1 while gii=− 1 fori=1, 2, 3. In (10), the left-hand side is manifestly positive forν=λ=i. T aking kto be at rest we understand the minus sign in (10) and hence in (4). Roughly speaking, for spin 2 exchange the sign occurs twice in (16). Degrees of freedom Now for a bit of cold water: Logically and mathematically the physics of a particle with mass m/negationslash=0 could well be different from the physics with m=0. Indeed, we know from classical 38 | I. Motivation and Foundation electromagnetism that an electromagnetic wave has 2 polarizations, that is, 2 degrees of freedom. For a massive spin 1 particle we can go to its rest frame, where the rotation grouptells us that there are 2 .1+1=3 degrees of freedom. The crucial piece of physics is that we can never bring the massless photon to its rest frame. Mathematically, the rotation groupSO( 3)degenerates into SO( 2), the group of 2-dimensional rotations around the direction of the photon’s momentum. We will see in chapter II.7 that the longitudinal degree of freedom of a massive spin 1 meson decouples as we take the mass to zero. The treatment given here for the interactionbetween charges (6) is correct. However, in the case of gravity, the 2 3in (17) is replaced by 1 in Einstein’s theory, as we will see chapter VIII.1. Fortunately, the sign of the interactiongiven in (17) does not change. Mute the trumpets a bit. Appendix 1 Pretend that we never heard of the Maxwell Lagrangian. We want to construct a relativistic Lagrangian for a massive spin 1 meson field. T ogether we will discover Maxwell. Spin 1 means that the field transforms as a vectorunder the 3-dimensional rotation group. The simplest Lorentz object that contains the 3-dimensional vector isobviously the 4-dimensional vector. Thus, we start with a vector field A μ(x). That the vector field carries mass mmeans that it satisfies the field equation (∂2+m2)Aμ=0 (18) A spin 1 particle has 3 degrees of freedom [remember, in fancy language, the representation jof the rotation group has dimension (2j+1); here j=1.] On the other hand, the field Aμ(x) contains 4 components. Thus, we must impose a constraint to cut down the number of degrees of freedom from 4 to 3. The only Lorentz covariantpossibility (linear in A μ)is ∂μAμ=0 (19) It may also be helpful to look at (18) and (19) in momentum space, where they read (k2−m2)Aμ(k)=0 and kμAμ(k)=0. The first equation tells us that k2=m2and the second that if we go to the rest frame kμ=(m,/vector0) thenA0vanishes, leaving us with 3 nonzero components Aiwithi=1, 2, 3. The remarkable observation is that we can combine (18) and (19) into a single equation, namely (gμν∂2−∂μ∂ν)Aν+m2Aμ=0 (20) Verify that (20) contains both (18) and (19). Act with ∂μon (20). We obtain m2∂μAμ=0, which implies that ∂μAμ=0 . (At this step it is crucial that m/negationslash=0 and that we are not talking about the strictly massless photon.) We have thus obtained (19 ); using (19) in (20) we recover (18). We can now construct a Lagrangian by multiplying the left-hand side of (20) by +1 2Aμ(the1 2is conventional but the plus sign is fixed by physics, namely the requirement of positive kinetic energy); thus L=1 2Aμ[(∂2+m2)gμν−∂μ∂ν]Aν (21) Integrating by parts, we recognize this as the massive version of the Maxwell Lagrangian. In the limit m→0w e recover Maxwell. A word about terminology: Some people insist on calling only Fμνa field and Aμa potential. Conforming to common usage, we will not make this fine distinction. For us, any dynamical function of spacetime is a field. I.5. Coulomb and Newton | 39 Appendix 2: Why does the graviton have spin 2? First we have to understand why the photon has spin 1. Think physically. Consider a bunch of electrons at rest inside a small box. An observer moving by sees the box Lorentz-Fitzgerald contracted and thus a highercharge density than the observer at rest relative to the box. Thus charge density J 0(x) transforms like the time component of a 4-vector density Jμ(x). In other words, J/prime0=J0/√ 1−v2. The photon couples to Jμ(x) and has to be described by a 4-vector field Aμ(x) for the Lorentz indices to match. What about energy density? The observer at rest relative to the box sees each electron contributing mto the energy enclosed in the box. The moving observer, on the other hand, sees the electrons moving and thus each having an energy m/√ 1−v2. With the contracted volume and the enhanced energy, the energy density gets enhanced by two factors of 1 /√ 1−v2, that is, it transforms like the T00component of a 2-indexed tensor Tμν. The graviton couples to Tμν(x) and has to be described by a 2-indexed tensor field ϕμν(x) for the Lorentz indices to match. Exercise I.5.1 Write down the most general form for/summationtext aε(a) μν(k)ε(a) λσ(k)using symmetry repeatedly. For example, it must be invariant under the exchange {μν↔λσ}. You might end up with something like AGμνGλσ+B(GμλGνσ+GμσGνλ)+C(Gμνkλkσ+kμkνGλσ) +D(kμkλGνσ+kμkσGνλ+kνkσGμλ+kνkλGμσ)+Ekμkνkλkσ (22) with various unknown A,... ,E. Apply kμ/summationtext aε(a) μν(k)ε(a) λσ(k)=0 and find out what that implies for the constants. Proceeding in this way, derive (13). I.6 Inverse Square Law and the Floating 3-Brane Why inverse square? In your first encounter with physics, didn’t you wonder why an inverse square force law and not, say, an inverse cube law? In chapter I.4 you learned the deep answer. When amassless particle is exchanged between two particles, the potential energy between thetwo particles goes as V( r)∝/integraldisplay d3kei/vectork./vectorx1 /vectork2∝1 r(1) The spin of the exchanged particle controls the overall sign, but the 1 /rfollows just from dimensional analysis, as I remarked earlier. Basically, V( r) is the Fourier transform of the propagator. The /vectork2in the propagator comes from the (∂iϕ.∂iϕ)term in the action, where ϕdenotes generically the field associated with the massless particle being exchanged, and the(∂iϕ.∂iϕ)form is required by rotational invariance. It couldn’t be /vectorkor/vectork3in (1); /vectork2is the simplest possibility. So you can say that in some sense ultimately the inverse squarelaw comes from rotational invariance! Physically, the inverse square law goes back to Faraday’s flux picture. Consider a sphere of radius rsurrounding a charge. The electric flux per unit area going through the sphere varies as 1 /4πr 2. This geometric fact is reflected in the factor d3kin (1). Brane world Remarkably, with the tiny bit of quantum field theory I have exposed you to, I can already take you to the frontier of current research, current as of the writing of this book. In stringtheory, our (3+1)-dimensional world could well be embedded in a larger universe, the way a(2+1)-dimensional sheet of paper is embedded in our everyday (3+1)-dimensional world. We are said to be living on a 3 brane. So suppose there are nextra dimensions, with coordinates x 4,x5,... ,xn+3. Let the characteristic scales associated with these extra coordinates be R. I can’t go into the I.6. Inverse Square Law | 41 different detailed scenarios describing what Ris precisely. For some reason I can’t go into either, we are stuck on the 3 brane. In contrast, the graviton is associated intrinsically withthe structure of spacetime and so roams throughout the (n+3+1)-dimensional universe. All right, what is the gravitational force law between two particles? It is surely not your grandfather’s gravitational force law: We Fourier transform V( r)∝/integraldisplay d3+nkei/vectork./vectorx1 /vectork2∝1 r1+n(2) to obtain a 1 /r1+nlaw. Doesn’t this immediately contradict observation?Well, no, because Newton’s law continues to hold for r/greatermuchR. In this regime, the extra coordinates are effectively zero compared to the characteristic length scale rwe are inter- ested in. The flux cannot spread far in the direction of the nextra coordinates. Think of the flux being forced to spread in only the three spatial directions we know, just as theelectromagnetic field in a wave guide is forced to propagate down the tube. Effectively weare back in (3+1)-dimensional spacetime and V( r) reverts to a 1 /rdependence. The new law of gravity (2) holds only in the opposite regime r/lessmuchR. Heuristically, when Ris much larger than the separation between the two particles, the flux does not know that the extra coordinates are finite in extent and thinks that it lives in an (n+3+1)- dimensional universe. Because of the weakness of gravity, Newton’s force law has not been tested to much accuracy at laboratory distance scales, and so there is plenty of room for theorists tospeculate in: Rcould easily be much larger than the scale of elementary particles and yet much smaller than the scale of everyday phenomena. Incredibly, the universe could have“large extra dimensions”! (The word “large” means large on the scale of particle physics.) Planck mass T o be quantitative, let us define the Planck mass MPlby writing Newton’s law more rationally as V( r)=GNm1m2(1/r)=(m1m2/M2 Pl)(1/r). Numerically, MPl/similarequal1019Gev. This enormous value obviously reflects the weakness of gravity. In fundamental units in which /planckover2piandcare set to unity, gravity defines an intrinsic mass or energy scale much higher than any scale we have yet explored experimentally. Indeed,one of the fundamental mysteries of contemporary particle physics is why this mass scaleis so high compared to anything else we know of. I will come back to this so-called hierarchyproblem in due time. For the moment, let us ask if this new picture of gravity, new in thewaning moments of the last century, can alleviate the hierarchy problem by lowering theintrinsic mass scale of gravity. Denote the mass scale (the “true scale” of gravity) characteristic of gravity in the (n+3 +1)-dimensional universe by M TGso that the gravitational potential between two objects of masses m1andm2separated by a distance r/lessmuchRis given by V( r)=m1m2 [MTG]2+n1 r1+n 42 | I. Motivation and Foundation Note that the dependence on MTGfollows from dimensional analysis: two powers to cancel m1m2andnpowers to match the nextra powers of 1 /r.F o rr/greatermuchR, as we have argued, the geometric spread of the gravitational flux is cut off by Rso that the potential becomes V( r)=m1m2 [MTG]2+n1 Rn1 r Comparing with the observed law V( r)=(m 1m2/M2 Pl)(1/r) we obtain M2 TG=M2 Pl [MTGR]n(3) IfMTGRcould be made large enough, we have the intriguing possibility that the funda- mental scale of gravity MTGmay be much lower than what we have always thought. Accelerators (such as the large Hadron Collider) could offer an exciting verification of this intriguing possibility. If the true scale of gravity MTGlies in an energy range accessible to the accelerator, there may be copious production of gravitons escaping into the higherdimensional universe. Experimentalists would see a massive amount of missing energy. Exercise I.6.1 Putting in the numbers, show that the case n=1 is already ruled out. I.7 Feynman Diagrams Feynman brought quantum field theory to the masses. —J. Schwinger Anharmonicity in field theory The free field theory we studied in the last few chapters was easy to solve because the defin-ing path integral (I.3.14) is Gaussian, so we could simply apply (I.2.15). (This correspondsto solving the harmonic oscillator in quantum mechanics.) As I noted in chapter I.3, withinthe harmonic approximation the vibrational modes on the mattress can be linearly super-posed and thus they simply pass through each other. The particles represented by wavepackets constructed out of these modes do not interact: 1hence the term free field theory. T o have the modes scatter off each other we have to include anharmonic terms in the La-grangian so that the equation of motion is no longer linear. For the sake of simplicity letus add only one anharmonic term − λ 4!ϕ4to our free field theory and, recalling (I.3.11), try to evaluate Z(J)=/integraldisplay Dϕ ei/integraltext d4x{1 2[(∂ϕ)2−m2ϕ2]−λ 4!ϕ4+Jϕ}(1) (We suppress the dependence of Zonλ.) Doing quantum field theory is no sweat, you say, it just amounts to doing the functional integral (1). But the integral is not easy! If you could do it, it would be big news. Feynman diagrams made easy As an undergraduate, I heard of these mysterious little pictures called Feynman diagramsand really wanted to learn about them. I am sure that you too have wondered about those 1A potential source of confusion: Thanks to the propagation of ϕ, the sources coupled to ϕinteract, as was seen in chapter I.4, but the particles associated with ϕdo not interact with each other. This is like saying that charged particles coupled to the photon interact, but (to leading approximation) photons do not interact with each other. 44 | I. Motivation and Foundation funny diagrams. Well, I want to show you that Feynman diagrams are not such a big deal: Indeed we have already drawn little spacetime pictures in chapters I.3 and I.4 showinghow particles can appear, propagate, and disappear. Feynman diagrams have long posed somewhat of an obstacle for first-time learners of quantum field theory. T o derive Feynman diagrams, traditional texts typically adopt thecanonical formalism (which I will introduce in the next chapter) instead of the path integralformalism used here. As we will see, in the canonical formalism fields appear as quantumoperators. T o derive Feynman diagrams, we would have to solve the equation of motionof the field operators perturbatively in λ. A formidable amount of machinery has to be developed. In the opinion of those who prefer the path integral, the path integral formalism derivation is considerably simpler (naturally!). Nevertheless, the derivation can still getrather involved and the student could easily lose sight of the forest for the trees. There isno getting around the fact that you would have to put in some effort. I will try to make it as easy as possible for you. I have hit upon the great pedagogical device of letting you discover the Feynman diagrams for yourself. My strategy is to let youtackle two problems of increasing difficulty, what I call the baby problem and the childproblem. By the time you get through these, the problem of evaluating (1) will seem muchmore tractable. A baby problem The baby problem is to evaluate the ordinary integral Z(J)=/integraldisplay+∞ −∞dqe−1 2m2q2−λ 4!q4+Jq(2) evidently a much simpler version of (1). First, a trivial point: we can always scale q→q/m so that Z=m−1F(λ m4,J m), but we won’t. Forλ=0 this is just one of the Gaussian integrals done in the appendix of chapter I.2. Well, you say, it is easy enough to calculate Z(J) as a series in λ: expand Z(J)=/integraldisplay+∞ −∞dqe−1 2m2q2+Jq/bracketleftbigg 1−λ 4!q4+1 2(λ 4!)2q8+.../bracketrightbigg and integrate term by term. You probably even know one of several tricks for computing/integraltext+∞ −∞dqe−1 2m2q2+Jqq4n: you write it as (d dJ)4n/integraltext+∞ −∞dqe−1 2m2q2+Jqand refer to (I.2.19). So Z(J)=(1−λ 4!(d dJ)4+1 2(λ 4!)2(d dJ)8+...)/integraldisplay+∞ −∞dqe−1 2m2q2+Jq(3) =e−λ 4!(d dJ)4/integraldisplay+∞ −∞dqe−1 2m2q2+Jq=(2π m2)1 2e−λ 4!(d dJ)4e1 2m2J2 (4) (There are other tricks, such as differentiating/integraltext+∞ −∞dqe−1 2m2q2+Jqwith respect to m2 repeatedly, but I want to discuss a trick that will also work for field theory.) By expanding I.7. Feynman Diagrams | 45 J J JJ JJ J J JJ JJλλ λ (a) (b) (c) ( )4λJ4 1 m2 Figure I.7.1 the two exponentials we can obtain any term in a double series expansion of Z(J) inλand J. [We will often suppress the overall factor (2π/m2)1 2=Z(J=0,λ=0)≡Z(0, 0)since it will be common to all terms. When we want to be precise, we will define ˜Z=Z(J)/Z(0, 0 ).] For example, suppose we want the term of order λandJ4in˜Z. We extract the order J8term in eJ2/2m2, namely, [1 /4!(2m2)4]J8, replace e−(λ/4! )(d/dJ)4by−(λ/4! )(d/dJ)4, and differentiate to get [8! (−λ)/(4! )3(2m2)4]J4. Another example: the term of order λ2 andJ4is [12! (−λ)2/(4!)36!2(2m2)6]J4. A third example: the term of order λ2andJ6is 1 2(λ/4!)2(d/dJ)8[1/7!(2m2)7]J14=[14!(−λ)2/(4!)26!7!2(2m2)7]J6. Finally, the term of order λandJ0is [1/2(2m2)2](−λ) . You can do this as well as I can! Do a few more and you will soon see a pattern. In fact, you will eventually realize that you can associate diagrams with each term and codifysome rules. Our four examples are associated with the diagrams in figures I.7.1–I.7.4.You can see, for a reason you will soon understand, that each term can be associated withseveral diagrams. I leave you to work out the rules carefully to get the numerical factorsright (but trust me, the “future of democracy” is not going to depend on them). The rulesgo something like this: (1) diagrams are made of lines and vertices at which four linesmeet; (2) for each vertex assign a factor of (−λ); (3) for each line assign 1 /m 2; and (4) for each external end assign J(e.g., figure I.7.3 has seven lines, two vertices, and six ends, giving ∼[(−λ)2/(m2)7]J6.) (Did you notice that twice the number of lines is equal to four times the number of vertices plus the number of ends? We will meet relations like that inchapter III.2.) In addition to the two diagrams shown in figure I.7.3, there are ten diagrams obtained by adding an unconnected straight line to each of the ten diagrams in figure I.7.2. (Do youunderstand why?) For obvious reasons, some diagrams (e.g., figure I.7.1a, I.7.3a) are known as tree 2 diagrams and others (e.g., Figs. I.7.1b and I.7.2a) as loop diagrams. Do as many examples as you need until you feel thoroughly familiar with what is going on, because we are going to do exactly the same thing in quantum field theory. It willlook much messier, but only superficially. Be sure you understand how to use diagrams to 2The Chinese character for tree (A. Zee, Swallowing Clouds ) is shown in fig. I.7.5. I leave it to you to figure out why this diagram does not appear in our Z(J) . J J J J/H9261 (a) (b) (d) (e) (i) (j) ( )6/H92612J4(f ) (g) (h)(c)/H9261 1 m2 Figure I.7.2 J J J J/H9261 (a) (b) ( )7/H92612J6 1 m2J /H9261 J Figure I.7.3 Figure I.7.4 I.7. Feynman Diagrams | 47 Figure I.7.5 represent the double series expansion of ˜Z(J) before reading on. Please. In my experience teaching, students who have not thoroughly understood the expansion of ˜Z(J) have no hope of understanding what we are going to do in the field theory context. Wick contraction It is more obvious than obvious that we can expand Z(J) in powers of J, if we please, instead of in powers of λ. As you will see, particle physicists like to classify in power of J. In our baby problem, we can write Z(J)=∞/summationdisplay s=01 s!Js/integraldisplay+∞ −∞dqe−1 2m2q2−(λ/4! )q4qs≡Z(0, 0)∞/summationdisplay s=01 s!JsG(s)(5) The coefficient G(s), whose analogs are known as “Green’s functions” in field theory, can be evaluated as a series in λwith each term determined by Wick contraction (I.2.10). For instance, the O(λ) term in G(4)is −λ 4!Z(0, 0)/integraldisplay+∞ −∞dqe−1 2m2q2q8=−7!! 4!1 m8 which of course better be equal3to what we obtained above for the λJ4term in ˜Z. Thus, there are two ways of computing Z: you expand in λfirst or you expand in Jfirst. Connected versus disconnected You will have noticed that some Feynman diagrams are connected and others are not. Thus, figure I.7.3a is connected while 3b is not. I presaged this at the end of chapter I.4and in figures I.4.2 and I.4.3. Write Z(J ,λ)=Z(J=0,λ)eW(J ,λ)=Z(J=0,λ)∞/summationdisplay N=01 N![W(J ,λ)]N(6) By definition, Z(J=0,λ)consists of those diagrams with no external source J, such as the one in figure I.7.4. The statement is that Wis a sum of connected diagrams while 3As a check on the laws of arithmetic we verify that indeed 7!! /(4!)2=8!/(4!)324. 48 | I. Motivation and Foundation Zcontains connected as well as disconnected diagrams. Thus, figure I.7.3b consists of two disconnected pieces and comes from the term (1/2!)[W(J ,λ)]2in (6), the 2! taking into account that it does not matter which of the two pieces you put “on the left or on theright.” Similarly, figure I.7.2i comes from (1/3!)[W(J ,λ)] 3. Thus, it is Wthat we want to calculate, not Z. If you’ve had a good course on statistical mechanics, you will recognize that this business of connected graphs versus disconnected graphs is just what underliesthe relation between free energy and the partition function. Propagation: from here to there All these features of the baby problem are structurally the same as the correspondingfeatures of field theory and we can take over the discussion almost immediately. But beforewe graduate to field theory, let us consider what I call a child problem, the evaluation of amultiple integral instead of a single integral: Z(J)=/integraldisplay+∞ −∞/integraldisplay+∞ −∞.../integraldisplay+∞ −∞dq1dq2...dqNe−1 2q.A.q−(λ/4! )q4+J.q(7) withq4≡/summationtext iq4 i. Generalizing the steps leading to (3) we obtain Z(J)=/bracketleftbigg(2π)N det[A]/bracketrightbigg1 2 e−(λ/4! )/summationtext i(∂/∂J i)4e1 2J.A−1.J(8) Alternatively, just as in (5) we can expand in powers of J Z(J)=∞/summationdisplay s=0N/summationdisplay i1=1...N/summationdisplay is=11 s!Ji1...Jis/integraldisplay+∞ −∞/parenleftBigg/productdisplay ldql/parenrightBigg e−1 2q.A.q−(λ/4! )q4qi1...qis =Z(0, 0)∞/summationdisplay s=0N/summationdisplay i1=1...N/summationdisplay is=11 s!Ji1...JisG(s) i1...is(9) which again we can expand in powers of λand evaluate by Wick contracting. The one feature the child problem has that the baby problem doesn’t is propagation “from here to there”. Recall the discussion of the propagator in chapter I.3. Just as in(I.3.16) we can think of the index ias labeling the sites on a lattice. Indeed, in (I.3.16) we had in effect evaluated the “2-point Green’s function” G (2) ijto zeroth order in λ(differentiate (I.3.16) with respect to Jtwice): G(2) ij(λ=0)=/bracketleftBigg/integraldisplay+∞ −∞/parenleftBigg/productdisplay ldql/parenrightBigg e−1 2q.A.qqiqj/bracketrightBigg /Z(0, 0 )=(A−1)ij (see also the appendix to chapter I.2). The matrix element ( A−1)ijdescribes propagation fromitoj. In the baby problem, each term in the expansion of Z(J) can be associated with several diagrams but that is no longer true with propagation. I.7. Feynman Diagrams | 49 Let us now evaluate the “4-point Green’s function” G(4) ijklto order λ: G(4) ijkl=/integraldisplay+∞ −∞/parenleftBigg/productdisplay mdqm/parenrightBigg e−1 2q.A.qqiqjqkql/bracketleftBigg 1−λ 4!/summationdisplay nq4 n+O(λ2)/bracketrightBigg /Z(0, 0 ) =(A−1)ij(A−1)kl+(A−1)ik(A−1)jl+(A−1)il(A−1)jk −λ/summationdisplay n(A−1)in(A−1)jn(A−1)kn(A−1)ln+...+O(λ2) (10) The first three terms describe one excitation propagating from itojand another propa- gating from ktol, plus the two possible permutations on this “history.” The order λterm tells us that four excitations, propagating from iton, from jton, from kton, and from l ton, meet at nand interact with an amplitude proportional to λ, where nis anywhere on the lattice or mattress. By the way, you also see why it is convenient to define the interac-tion(λ/4!)ϕ 4with a 1 /4! :qihas a choice of four qn’s to contract with, qjhas three qn’s to contract with, and so on, producing a factor of 4! to cancel the (1/4!). I intentionally did not display in (10) the O(λ) terms produced by Wick contracting some of the qn’s with each other. There are two types: (I) Contracting a pair of qn’s produces something like λ(A−1)ij(A−1)kn(A−1)ln(A−1)nnand (II) contracting the qn’s with each other produces the first three terms in (10) multiplied by (A−1)nn(A−1)nn. We see that (I) and (II) correspond to diagrams b and c in figure I.7.1, respectively. Evidently, the twoexcitations do not interact with each other. I will come back to (II) later in this chapter. Perturbative field theory You should now be ready for field theory! Indeed, the functional integral in (1) (which I repeat here) Z(J)=/integraldisplay Dϕ ei/integraltext d4x{1 2[(∂ϕ)2−m2ϕ2]−(λ/4! )ϕ4+Jϕ}(11) has the same form as the ordinary integral in (2) and the multiple integral in (7). There is one minor difference: there is no iin (2) and (7), but as I noted in chapter I.2 we can Wick rotate (11) and get rid of the i, but we won’t. The significant difference is that Jand ϕin (11) are functions of a continuous variable x, while Jandqin (2) are not functions of anything and in (7) are functions of a discrete variable. Aside from that, everything goesthrough the same way. As in (3) and (8) we have Z(J)=e−(i/4! )λ/integraltext d4w[δ/iδJ(w) ]4/integraldisplay Dϕei/integraltext d4x{1 2[(∂ϕ)2−m2ϕ2]+Jϕ} =Z(0, 0)e−(i/4! )λ/integraltext d4w[δ/iδJ(w) ]4e−(i/2)/integraltext/integraltext d4xd4yJ(x)D(x −y)J(y)(12) The structural similarity is total. The role of 1 /m2in (3) and of A−1(8) is now played by the propagator D(x−y)=/integraldisplayd4k (2π)4eik.(x−y) k2−m2+iε 50 | I. Motivation and Foundation Incidentally, if you go back to chapter I.3 you will see that if we were in d-dimensional spacetime, D(x−y)would be given by the same expression with d4k/(2π)4replaced by ddk/(2π)d. The ordinary integral (2) is like a field theory in 0-dimensional spacetime: if we set d=0, there is no propagating around and D(x−y)collapses to −1/m2. You see that it all makes sense. We also know that J(x) corresponds to sources and sinks. Thus, if we expand Z(J) as a series in J, the powers of Jwould indicate the number of particles involved in the process. (Note that in this nomenclature the scattering process ϕ+ϕ→ϕ+ϕcounts as a 4-particle process: we count the total number of incoming and outgoing particles.) Thus,in particle physics it often makes sense to specify the power of J. Exactly as in the baby and child problems, we can expand in Jfirst: Z(J)=Z(0, 0)∞/summationdisplay s=0is s!/integraldisplay dx1...dxsJ(x 1)...J(xs)G(s)(x1,... ,xs) =∞/summationdisplay s=0is s!/integraldisplay dx1...dxsJ(x 1)...J(xs)/integraldisplay Dϕ ei/integraltext d4x{1 2[(∂ϕ)2−m2ϕ2]−(λ/4! )ϕ4} ϕ(x1)...ϕ(xs) (13) In particular, we have the 2-point Green’s function G(x 1,x2)≡1 Z(0, 0)/integraldisplay Dϕ ei/integraltext d4x{1 2[(∂ϕ)2−m2ϕ2]−(λ/4! )ϕ4}ϕ(x1)ϕ(x 2) (14) the 4-point Green’s function, G(x1,x2,x3,x4)≡1 Z(0, 0)/integraldisplay Dϕ ei/integraltext d4x{1 2[(∂ϕ)2−m2ϕ2]−(λ/4! )ϕ4}ϕ(x 1)ϕ(x 2)ϕ(x 3)ϕ(x 4) (15) and so on. [Sometimes Z(J) is called the generating functional as it generates the Green’s functions.] Obviously, by translation invariance, G(x 1,x2)does not depend on x1andx2 separately, but only on x1−x2. Similarly, G(x1,x2,x3,x4)only depends on x1−x4,x2−x4, andx3−x4.F o rλ=0,G(x1,x2)reduces to iD(x1−x2), the propagator introduced in chapter I.3. While D(x1−x2)describes the propagation of a particle between x1andx2in the absence of interaction, G(x 1−x2)describes the propagation of a particle between x1 andx2in the presence of interaction. If you understood our discussion of G(4) ijkl, you would know that G(x 1,x2,x3,x4)describes the scattering of particles. In some sense, there are two ways of doing field theory, what I might call the Schwinger way (12) or the Wick way (13). Thus, to summarize, Feynman diagrams are just an extremely convenient way of rep- resenting the terms in a double series expansion of Z(J) inλandJ. As I said in the preface, I have no intention of turning you into a whiz at calculating Feynman diagrams. In any case, that can only come with practice. Instead, I tried to giveyou as clear an account as I can muster of the concept behind this marvellous invention of I.7. Feynman Diagrams | 51 3 4 1 2 xt Figure I.7.6 Feynman’s, which as Schwinger noted rather bitterly, enables almost anybody to become a field theorist. For the moment, don’t worry too much about factors of 4! and 2! Collision between particles As I already mentioned, I described in chapter I.4 the strategy of setting up sources andsinks to watch the propagation of a particle (which I will call a meson) associated withthe field ϕ. Let us now set up two sources and two sinks to watch two mesons scatter off each other. The setup is shown in figure I.7.6. The sources localized in regions 1 and 2both produce a meson, and the two mesons eventually disappear into the sinks localizedin regions 3 and 4. It clearly suffices to find in Za term containing J(x 1)J(x 2)J(x3)J(x4). But this is just G(x1,x2,x3,x4). Let us be content with first order in λ. Going the Wick way we have to evaluate 1 Z(0, 0)/parenleftbigg −iλ 4!/parenrightbigg/integraldisplay d4w/integraldisplay Dϕ ei/integraltext d4x{1 2[(∂ϕ)2−m2ϕ2]} ϕ(x 1)ϕ(x2)ϕ(x3)ϕ(x4)ϕ(w)4(16) Just as in (10) we Wick contract and obtain (−iλ)/integraldisplay d4wD(x 1−w)D(x 2−w)D(x 3−w)D(x 4−w) (17) As a check, let us also derive this the Schwinger way. Replace e−(i/4! )λ/integraltext d4w(δ/δJ(w))4by −(i/4! )λ/integraltext d4w(δ/δJ(w))4ande−(i/2)/integraltext/integraltext d4xd4yJ(x)D(x −y)J(y)by i4 4!24/bracketleftbigg/integraldisplay/integraldisplay d4xd4yJ(x)D(x −y)J(y)/bracketrightbigg4 52 | I. Motivation and Foundation x3 x4 x1 x2k3 k4 k1 k2 (a) (b)w Figure I.7.7 To save writing, it would be sagacious to introduce the abbreviations JaforJ(xa),/integraltext afor/integraltext d4xa, andDabforD(xa−xb). Dropping overall numerical factors, which I invite you to fill in, we obtain ∼iλ/integraldisplay w(δ δJw)4/integraldisplay/integraldisplay/integraldisplay/integraldisplay/integraldisplay/integraldisplay/integraldisplay/integraldisplay DaeDbfDcgDdhJaJbJcJdJeJfJgJh (18) The four (δ/δJw)’s hit the eight J’s in all possible combinations producing many terms, which again I invite you to write out. Two of the three terms are disconnected. Theconnected term is ∼iλ/integraldisplay w/integraldisplay/integraldisplay/integraldisplay/integraldisplay DawDbwDcwDdwJaJbJcJd (19) Evidently, this comes from the four (δ/δJ w)’s hitting Je,Jf,Jg, and Jh, thus setting xe,xf,xg, andxhtow. Compare (19) with [8! (−λ)/(4! )3(2m2)4]J4in the baby problem. Recall that we embarked on this calculation in order to produce two mesons by sources localized in regions 1 and 2, watch them scatter off each other, and then get rid of themwith sinks localized in regions 3 and 4. In other words, we set the source function J(x) equal to a set of delta functions peaked at x 1,x2,x3, and x4. This can be immediately read off from (19): the scattering amplitude is just −iλ/integraltext wD1wD2wD3wD4w, exactly as in (17). The result is very easy to understand (see figure I.7.7a). Two mesons propagate from their birthplaces at x1andx2to the spacetime point w, with amplitude D(x1−w)D(x2− w), scatter with amplitude −iλ , and then propagate from wto their graves at x3andx4, with amplitude D(w−x3)D(w −x4)[note that D(x)=D(−x) ]. The integration over w just says that the interaction point wcould have been anywhere. Everything is as in the child problem. It is really pretty simple once you get it. Still confused? It might help if you think of (12) as some kind of machine e−(i/4! )λ/integraltext d4w[δ/iδJ(w) ]4operating on Z(J ,λ=0)=e−(i/2)/integraltext/integraltext d4xd4yJ(x)D(x −y)J(y) When expanded out, Z(J ,λ=0)is a bunch of J’s thrown here and there in spacetime, with pairs of J’s connected by D’s. Think of a bunch of strings with the string ends I.7. Feynman Diagrams | 53 x3 x4 x1 x2 (a)y4 y3 y1 y2 x1 x2x3 x4 w x3 x4 x1 x2y4 y3 y1 y2 x1 x2x3 x4 w1u2 z2u1 z1 (b)w2 Figure I.7.8 corresponding to the J’s. What does the “machine” do? The machine is a sum of terms, for example, the term ∼λ2/integraldisplay d4w1/integraldisplay d4w2/bracketleftbiggδ δJ(w 1)/bracketrightbigg4/bracketleftbiggδ δJ(w 2)/bracketrightbigg4 When this term operates on a term in Z(J ,λ=0)it grabs four string ends and glues them together at the point w2; then it grabs another 4 string ends and glues them together at the point w1. The locations w1andw2are then integrated over. Two examples are shown in figure I.7.8. It is a game you can play with a child! This childish game of gluing four string ends together generates all the Feynman diagrams of our scalar field theory. Do it once and for all Now Feynman comes along and says that it is ridiculous to go through this long-winded yakkety-yak every time. Just figure out the rules once and for all. 54 | I. Motivation and Foundation For example, for the diagram in figure I.7.7a associate the factor −iλ with the scattering, the factor D(x1−w)with the propagation from x1tow, and so forth—conceptually exactly the same as in our baby problem. See, you could have invented Feynman diagrams. (Well,not quite. Maybe not, maybe yes.) Just as in going from (I.4.1) to (I.4.2), it is easier to pass to momentum space. Indeed, that is how experiments are actually done. A meson with momentum k 1and a meson with momentum k2collide and scatter, emerging with momenta k3andk4(see figure I.7.7b). Each spacetime propagator gives us D(xa−w)=/integraldisplayd4ka (2π)4e±ika(xa−w) k2 a−m2+iε Note that we have the freedom of associating with the dummy integration variable either a plus or a minus sign in the exponential. Thus in integrating over win (17) we obtain /integraldisplay d4we−i(k 1+k 2−k 3−k 4)w=(2π)4δ(4)(k1+k2−k3−k4). That the interaction could occur anywhere in spacetime translates into momentum con- servation k1+k2=k3+k4. (We put in the appropriate minus signs in two of the D’s so that we can think of k3andk4as outgoing momenta.) So there are Feynman diagrams in real spacetime and in momentum space. Spacetime Feynman diagrams are literally pictures of what happened. (A trivial remark: the orien-tation of Feynman diagrams is a matter of idiosyncratic choice. Some people draw themwith time running vertically, others with time running horizontally. We follow Feynmanin this text.) The rules We have thus derived the celebrated momentum space Feynman rules for our scalar fieldtheory: 1. Draw a Feynman diagram of the process (fig. I.7.7b for the example we just discussed). 2. Label each line with a momentum kand associate with it the propagator i/(k2−m2+iε). 3. Associate with each interaction vertex the coupling −iλ and(2π)4δ(4)(/summationtext iki−/summationtext jkj), forcing the sum of momenta/summationtext ikicoming into the vertex to be equal to the sum of momenta /summationtext jkjgoing out of the vertex. 4. Momenta associated with internal lines are to be integrated over with the measured4k (2π)4. Incidentally, this corresponds to the summation over intermediate states in ordinary per-turbation theory. 5. Finally, there is a rule about some really pesky symmetry factors. They are the analogs of those numerical factors in our baby problem. As a result, some diagrams are to be multipliedby a symmetry factor such as 1 2. These originate from various combinatorial factors counting the different ways in which the (δ/δJ )’s can hit the J’s in expressions such as (18). I will let you discover a symmetry factor in exercise I.7.2. We will illustrate by examples what these rules (and the concept of internal lines) mean. I.7. Feynman Diagrams | 55 Our first example is just the diagram (fig. I.7.7b) that we have calculated. Applying the rules we obtain −iλ( 2π)4δ(4)(k1+k2−k3−k4)4/productdisplay a=1/parenleftBigg i k2 a−m2+iε/parenrightBigg You would agree that it is silly to drag the factor/producttext4 a=1/parenleftbigg i k2a−m2+iε/parenrightbigg around, since it would be common to all diagrams in which two mesons scatter into two mesons. So we append to the Feynman rules an additional rule that we do not associate a propagator with externallines. (This is known in the trade as “amputating the external legs”.) In an actual scattering experiment, the external particles are of course real and on shell, that is, their momenta satisfy k 2 a−m2=0. Thus we better not keep the propagators of the external lines around. Arithmetically, this amounts to multiplying the Green’s functions[and what we have calculated thus far are indeed Green’s functions, see (16)] by the factor/Pi1 a(−i)(k2 a−m2)and then set k2 a=m2(“putting the particles on shell” in conversation). At this point, this procedure sounds like formal overkill. We will come back to the rationalebehind it at the end of the next chapter. Also, since there is always an overall factor for momentum conservation we should not drag the overall delta function around either. Thus, we have two more rules: 6. Do not associate a propagator with external lines. 7. A delta function for overall momentum conservation is understood. Applying these rules we obtain an amplitude which we will denote by M. For example, for the diagram in figure I.7.7b M=−iλ. The birth of particles As explained in chapter I.1 one of the motivations for constructing quantum field theory was to describe the production of particles. We are now ready to describe how two col-liding mesons can produce four mesons. The Feynman diagram in figure I.7.9 (comparefig. I.7.3a) can occur in order λ 2in perturbation theory. Amputating the external legs, we drop the factor/producttext6 a=1[i/k2 a−m2+iε] associated with the six external lines, keeping only the propagator associated with the one internal line. For each vertex we put in a factor of(−iλ) and a momentum conservation delta function (rule 3). Then we integrate over the internal momentum q(rule 4) to obtain (−iλ)2/integraldisplayd4q (2π)4i q2−m2+iε(2π)4δ(4)(k1+k2−k3−q)(2π)4δ(4)[q−(k4+k5+k6)] (20) The integral over qis a cinch, giving (−iλ)2 i (k4+k5+k6)2−m2+iε(2π)4δ(4)[k1+k2−(k3+k4+k5+k6)] (21) 56 | I. Motivation and Foundation k3k4 k1 k2k5 k6 q Figure I.7.9 We have already agreed (rule 7) not to drag the overall delta function around. This exam- ple teaches us that we didn’t have to write down the delta functions and then annihilate (allbut one of) them by integrating. In figure I.7.9 we could have simply imposed momentumconservation from the beginning and labeled the internal line as k 4+k5+k6instead of q. With some practice you could just write down the amplitude M=(−iλ)2 i (k4+k5+k6)2−m2+iε(22) directly: just remember, a coupling (−iλ) for each vertex and a propagator for each internal (but not external) line. Pretty easy once you get the hang of it. As Schwinger said, the massescould do it. The cost of not being real The physics involved is also quite clear: The internal line is associated with a virtual particle whose relativistic 4-momentum k4+k5+k6squared is not necessarily equal to m2,a si t would have to be if the particle were real. The farther the momentum of the virtual particle is from the mass shell the smaller the amplitude. You are penalized for not being real. According to the quantum rules for dealing with identical particles, to obtain the full amplitude we have to symmetrize among the four final momenta. One way of saying it isto note that the line labeled k 3in figure (I.7.9) could have been labeled k4,k5,o rk6, and we have to add all four contributions. To make sure that you understand the Feynman rules I insist that you go through the path integral calculation to obtain (21) starting with (12) and (13). I am repeating myself but I think it is worth emphasizing again that there is nothing particularly magical about Feynman diagrams. I.7. Feynman Diagrams | 57 k1 k2k3 k4 k k1 + k2 − k Figure I.7.10 Loops and a first look at divergence Just as in our baby problem, we have tree diagrams and loop diagrams. So far we have only looked at tree diagrams. Our next example is the loop diagram in figure I.7.10 (comparefig. I.7.2a.) Applying the Feynman rules, we obtain 1 2(−iλ)2/integraldisplayd4k (2π)4i k2−m2+iεi (k1+k2−k)2−m2+iε(23) As above, the physics embodied in (23) is clear: As kranges over all possible values, the integrand is large only if one or the other or both of the virtual particles associated withthe two internal lines are close to being real. Once again, there is a penalty for not beingreal (see exercise I.7.4). For large kthe integrand goes as 1 /k 4. The integral is infinite! It diverges as/integraltext d4k(1/k4). We will come back to this apparent disaster in chapter III.1. With some practice, you will be able to write down the amplitude by inspection. As another example, consider the three-loop diagram in figure I.7.11 contributing in O(λ4)to meson-meson scattering. First, for each loop pick an internal line and label the momentum it carries, p,q, andrin our example. There is considerable freedom of choice in labeling— your choice may well not agree with mine, but of course the physics should not depend on it. The momenta carried by the other internal lines are then fixed by momentumconservation, as indicated in the figure. Write down a coupling for each vertex, and apropagator for each internal line, and integrate over the internal momenta p,q, andr. 58 | I. Motivation and Foundation k1k2k3 k4 r pqp − q − rk1 + k2 − r k1 + k2 − p Figure I.7.11 Thus, without worrying about symmetry factors, we have the amplitude (−iλ)4/integraldisplayd4p (2π)4d4q (2π)4d4r (2π)4i p2−m2+iεi (k1+k2−p)2−m2+iε i q2−m2+iεi (p−q−r)2−m2+iεi r2−m2+iεi (k1+k2−r)2−m2+iε(24) Again, this triple integral also diverges: It goes as/integraltext d12P(1/P12). An assurance When I teach quantum field theory, at this point in the course some students get un- accountably very anxious about Feynman diagrams. I would like to assure the reader that the whole business is really quite simple. Feynman diagrams can be thought of simply aspictures in spacetime of the antics of particles, coming together, colliding and producingother particles, and so on. One student was puzzled that the particles do not move in I.7. Feynman Diagrams | 59 straight lines. Remember that a quantum particle propagates like a wave; D(x−y)gives us the amplitude for the particle to propagate from xtoy. Evidently, it is more convenient to think of particles in momentum space: Fourier told us so. We will see many more examplesof Feynman diagrams, and you will soon be well acquainted with them. Another studentwas concerned about evaluating the integrals in (23) and (24). I haven’t taught you how yet,but will eventually. The good news is that in contemporary research on the frontline fewtheoretical physicists actually have to calculate Feynman diagrams getting all the factorsof 2 right. In most cases, understanding the general behavior of the integral is sufficient.But of course, you should always take pride in getting everything right. In chapters II.6,III.6, and III.7 I will calculate Feynman diagrams for you in detail, getting all the factorsright so as to be able to compare with experiments. Vacuum fluctuations Let us go back to the terms I neglected in (18) and which you are supposed to havefigured out. For example, the four [ δ/δJ(w)]’s could have hit J c,Jd,Jg, and Jhin (18) thus producing something like −iλ/integraldisplay/integraldisplay/integraldisplay/integraldisplay DaeDbfJaJbJeJf(/integraldisplay wDwwDww). The coefficient of J(x1)J(x2)J(x3)J(x4)is then D13D24(−iλ/integraltext wDwwDww)plus terms obtained by permuting. The corresponding physical process is easy to describe in words and in pictures (see figure I.7.12). The source at x1produces a particle that propagates freely without any interaction to x3, where it comes to a peaceful death. The particle produced at x2leads a similarly uneventful life before being absorbed at x4. The two particles did not interact at all. Somewhere off at the point w, which could perfectly well be anywhere in the universe, there was an interaction with amplitude −iλ . This is known as a vacuum fluctuation: t xwx3 x4 x1 x2 Figure I.7.12 60 | I. Motivation and Foundation As explained in chapter I.1, quantum mechanics and special relativity inevitably cause particles to pop out of the vacuum, and they could even interact before vanishing againinto the vacuum. Look at different time slices (one of which is indicated by the dotted line)in figure I.7.12. In the far past, the universe has no particles. Then it has two particles,then four, then two again, and finally in the far future, none. We will have a lot more tosay in chapter VIII.2 about these fluctuations. Note that vacuum fluctuations occur also inour baby and child problems (see, e.g., Figs. I.7.1c, I.7.2g,h,i,j, and so forth). Two words about history I believe strongly that any self-respecting physicist should learn about the history of phys-ics, and the history of quantum field theory is among the most fascinating. Unfortunately,I do not have the space to go into it here. 4The path integral approach to field theory using sources J(x) outlined here is mainly associated with Julian Schwinger, who referred to it as “sorcery” during my graduate student days (so that I could tell people who inquired thatI was studying sorcery in graduate school.) In one of the many myths retold around tribalfires by physicists, Richard Feynman came upon his rules in a blinding flash of insight.In 1949 Freeman Dyson showed that the Feynman rules which so mystified people at thePocono conference a year earlier could actually be obtained from the more formal work ofJulian Schwinger and of Shin-Itiro Tomonaga. Exercises I.7.1 Work out the amplitude corresponding to figure I.7.11 in (24). I.7.2 Derive (23) from first principles, that is from (11). It is a bit tedious, but straightforward. You should find a symmetry factor1 2. I.7.3 Draw all the diagrams describing two mesons producing four mesons up to and including order λ2. Write down the corresponding Feynman amplitudes. I.7.4 By Lorentz invariance we can always take k1+k2=(E,/vector0)in (23). The integral can be studied as a function ofE. Show that for both internal particles to become real Emust be greater than 2 m. Interpret physically what is happening. 4An excellent sketch of the history of quantum field theory is given in chapter 1 of The Quantum Theory of Fields by S. Weinberg. For a fascinating history of Feynman diagrams, see Drawing Theories Apart by D. Kaiser. I.8 Quantizing Canonically Quantum electrodynamics is made to appear more difficult than it actually is by the very many equivalent methods bywhich it may be formulated. —R. P. Feynman Always create before we annihilate, not the other way around. —Anonymous Complementary formalisms I adopted the path integral formalism as the quickest way to quantum field theory. But Imust also discuss the canonical formalism, not only because historically it was the methodused to develop quantum field theory, but also because of its continuing importance.Interestingly, the canonical and the path integral formalisms often appear complementary,in the sense that results difficult to see in one are clear in the other. Heisenberg and Dirac Let us begin with a lightning review of Heisenberg’s approach to quantum mechanics. Given a classical Lagrangian for a single particle L=1 2˙q2−V( q) (we set the mass equal to 1), the canonical momentum is defined as p≡δL/δ˙q=˙q. The Hamiltonian is then given byH=p˙q−L=p2/2+V( q) . Heisenberg promoted position q(t) and momentum p(t) to operators by imposing the canonical commutation relation [p,q]=−i (1) Operators evolve in time according to dp dt=i[H,p]=−V/prime(q) (2) 62 | I. Motivation and Foundation and dq dt=i[H,q]=p (3) In other words, operators constructed out of pandqevolve according to O(t)= eiHtO(0)e−iHt. In (1) pandqare understood to be taken at the same time. We obtain the operator equation of motion ¨q=−V/prime(q) by combining (2) and (3). Following Dirac, we invite ourselves to consider at some instant in time the operator a≡(1/√ 2ω)(ωq +ip) with some parameter ω. From (1) we have [a,a†]=1 (4) The operator a(t) evolves according to da dt=i[H,1√ 2ω(ωq+ip)]=−i/radicalbiggω 2/parenleftbigg ip+1 ωV/prime(q)/parenrightbigg which can be written in terms of aanda†. The ground state |0/angbracketrightis defined as the state such thata|0/angbracketright=0. In the special case V/prime(q)=ω2qwe get the particularly simple resultda dt=−iωa . This is of course the harmonic oscillator L=1 2˙q2−1 2ω2q2andH=1 2(p2+ω2q2)=ω(a†a+1 2). The generalization to many particles is immediate. Starting with L=/summationdisplay a1 2˙qa2−V( q1,q2,...,qN) we have pa=δL/δ˙qaand [pa(t),qb(t)]=−iδab (5) The generalization to field theory is almost as immediate. In fact, we just use our handy- dandy substitution table (I.3.6) and see that in D−dimensional space Lgeneralizes to L=/integraldisplay dDx{1 2(˙ϕ2−(/vector∇ϕ)2−m2ϕ2)−u(ϕ)} (6) where we denote the anharmonic term (the interaction term in quantum field theory) by u(ϕ) . The canonical momentum density conjugate to the field ϕ(/vectorx,t)is then π(/vectorx,t)=δL δ˙ϕ(/vectorx,t)=∂0ϕ(/vectorx,t) (7) so that the canonical commutation relation at equal times now reads [using (I.3.6)] [π(/vectorx,t),ϕ(/vectorx/prime,t)]=[∂0ϕ(/vectorx,t),ϕ(/vectorx/prime,t)]=−iδ(D)(/vectorx−/vectorx/prime) (8) (and of course also [ π(/vectorx,t),π(/vectorx/prime,t)]=0 and [ ϕ(/vectorx,t),ϕ(/vectorx/prime,t)]=0.)Note that δabin (5) gets promoted to δ(D)(/vectorx−/vectorx/prime)in (8). You should check that (8) has the correct dimension. Turning the canonical crank we find the Hamiltonian H=/integraldisplay dDx[π(/vectorx,t)∂0ϕ(/vectorx,t)−L] =/integraldisplay dDx{1 2[π2+(/vector∇ϕ)2+m2ϕ2]+u(ϕ)} (9) I.8. Quantizing Canonically | 63 For the case u=0, corresponding to the harmonic oscillator, we have a free scalar field theory and can go considerably farther. The field equation reads (∂2+m2)ϕ=0 (10) Fourier expanding, we have ϕ(/vectorx,t)=/integraldisplaydDk/radicalbig (2π)D2ωk[a(/vectork)e−i(ω kt−/vectork./vectorx)+a†(/vectork)ei(ωkt−/vectork./vectorx)] (11) withωk=+/radicalbig /vectork2+m2so that the field equation (10) is satisfied. The peculiar looking factor (2ωk)−1 2is chosen so that the canonical relation [a(/vectork),a†(/vectork/prime)]=δ(D)(/vectork−/vectork/prime) (12) appropriate for creation and annihilation operators implies the canonical commutation [∂0ϕ(/vectorx,t),ϕ(/vectorx/prime,t)]=−iδ(D)(/vectorx−/vectorx/prime)in (8 ). You should check this but you can see why the factor (2ωk)−1 2is needed since in ∂0ϕa factor ωkis brought down from the exponential. As in quantum mechanics, the vacuum or ground state ||0/angbracketrightis defined by the condition a(/vectork)|0/angbracketright=0 for all /vectorkand the single particle state by |/vectork/angbracketright≡a†(/vectork)|0/angbracketright. Thus, for example, using (12) we have /angbracketleft0|ϕ(/vectorx,t)|/vectork/angbracketright=(1//radicalbig (2π)D2ωk)e−i(ω kt−/vectork./vectorx), which we could think of as the relativistic wave function of a single particle with momentum /vectork. For later use, we will write this more compactly as (1/ρ(k))e−ik.x, with ρ(k)≡/radicalbig (2π)D2ωka normalization factor and k0=ωk. To make contact with the path integral formalism let us calculate /angbracketleft0|ϕ(/vectorx,t)ϕ(/vector0, 0)|0/angbracketright fort> 0. Of the four terms a†a†,a†a,aa†, and aa in the product of the two fields onlyaa†survives, and thus using (12) we obtain/integraltext [dDk/(2π)D2ωk]e−i(ω kt−/vectork./vectorx). In other words, if we define the time-ordered product T[ϕ(x)ϕ(y) ]=θ(x0−y0)ϕ(x)ϕ(y) +θ(y0− x0)ϕ(y)ϕ(x),w ef i n d /angbracketleft0|T[ϕ(/vectorx,t)ϕ(/vector0, 0)]|0/angbracketright= /integraldisplaydDk (2π)D2ωk[θ(t)e−i(ω kt−/vectork./vectorx)+θ(−t)e+i(ω kt−/vectork./vectorx)] (13) Go back to (I.3.23). We discover that /angbracketleft0|T[ϕ(x)ϕ( 0)]|0/angbracketright=iD(x) , the propagator for a particle to go from 0 to xwe obtained using the path integral formalism. This further justifies the iεprescription in (I.3.22). The physical meaning is that we always create before we annihilate, not the other way around. This is a form of causality as formulatedin quantum field theory. A remark: The combination d Dk/(2ωk), even though it does not look Lorentz invariant, is in fact a Lorentz invariant measure. To see this, we use (I.2.13) to show that (exercise I.8.1) /integraldisplay d(D+1)kδ(k2−m2)θ(k0)f (k0,/vectork)=/integraldisplaydDk 2ωkf( ωk,/vectork) (14) for any function f( k) . Since Lorentz transformations cannot change the sign of k0, the step function θ(k0)is Lorentz invariant. Thus the left-hand side is manifestly Lorentz invariant, and hence the right-hand side must also be Lorentz invariant. This shows that relationssuch as (13) are Lorentz invariant; they are frame-independent statements. 64 | I. Motivation and Foundation Scattering amplitude Now that we have set up the canonical formalism it is instructive to see how the invariant amplitude Mdefined in the preceding chapter arises using this alternative formalism. Let us calculate the amplitude /angbracketleft/vectork3/vectork4|e−iHT|/vectork1/vectork2/angbracketright=/angbracketleft/vectork3/vectork4|ei/integraltext d4xL(x)|/vectork1/vectork2/angbracketrightfor meson scatter- ing/vectork1+/vectork2→/vectork3+/vectork4in order λwithu(ϕ)=λ 4!ϕ4. (We have, somewhat sloppily, turned the large transition time Tinto/integraltext dx0when going over to the Lagrangian.) Expanding in λ, we obtain (−iλ 4!)/integraltext d4x/angbracketleft/vectork3/vectork4|ϕ4(x)|/vectork1/vectork2/angbracketright. The calculation of the matrix element is not dissimilar to the one we just did for the propagator. There we have the product of two field operators between the vacuum state.Here we have the product of four field operators, all evaluated at the same spacetime pointx, sandwiched between two-particle states. There we look for a term of the form a(/vectork)a †(/vectork), Here, plugging the expansion (11) of the field into the product ϕ4(x), we look for terms of the form a†(/vectork4)a†(/vectork3)a(/vectork2)a(/vectork1), so that we could remove the two incoming particles and produce the two outgoing particles. (To avoid unnecessary complications we assume thatall four momenta are different.) We now annihilate and create. The annihilation operatora(/vectork 1)could have come from any one of the four ϕfields in ϕ4, giving a factor of 4, a(/vectork1) could have come from any one of the three remaining ϕfields, giving a factor of 3, a†(/vectork3) could have come from either of the two remaining ϕfields, giving a factor of 2, so that we end up with a factor of 4!, which cancels the factor of1 4!included in the definition of λ. (This is of course why, for the sake of convenience, λis defined the way it is. Recall from the preceding chapter an analogous step.) As you just learned and as you can see from (11), we obtain a factor of 1 /ρ(k)e−ik.xfor each incoming particle and of 1 /ρ(k)e+ik.xfor each outgoing particle, giving all together /parenleftbigg /Pi14 α=11 ρ(kα)/parenrightbigg/integraldisplay d4xei(k3+k 4−k 1−k 2).x=/parenleftbigg /Pi14 α=11 ρ(kα)/parenrightbigg (2π)4δ4(k3+k4−k1−k2) It is conventional to refer to Sfi=/angbracketleftf|e−iHT|i/angbracketright, with some initial and final state, as elements of an “S -matrix” and to define the “transition matrix” T-matrix by S=I+iT, that is, Sfi=δfi+iTfi (15) In general, for initial and final states consisting of scalar particles characterized only by their momenta, we write (using an obvious notation, for example/summationtext ikis the sum of the particle momenta in the initial state), invoking momentum conservation: iTfi=(2π)4δ4⎛ ⎝/summationdisplay fk−/summationdisplay ik⎞ ⎠/parenleftbigg /Pi1α1 ρ(kα)/parenrightbigg M(f←i) (16) In our simple example, iT(/vectork3/vectork4,/vectork1/vectork2)=(−iλ 4!)/integraltext d4x/angbracketleft/vectork3/vectork4|ϕ4(x)|/vectork1/vectork2/angbracketright, and our little cal- culation showed that M=−iλ, precisely as given in the preceding chapter. But this con- nection between TfiandM, being “merely” kinematical, should hold in general, with the invariant amplitude Mdetermined by the Feynman rules. I will not give a long boring for- I.8. Quantizing Canonically | 65 mal proof, but you should check this assertion by working out some more involved cases, such as the scattering amplitude to order λ2, recovering (I.7.23), for example. Thus, quite pleasingly, we see that the invariant amplitude M determined by the Feynman rules represents the “heart of the matter” with the momentum conservationdelta function and normalization factors stripped away. My pedagogical aim here is merely to make one more contact (we will come across more in later chapters) between the canonical and path integral formalisms, givingthe simplest possible example avoiding all subtleties and complications. Those readersinto rigor are invited to replace the plane wave states |/vectork 1/vectork2/angbracketrightwith wave packet states/integraltext d3k1/integraltext d3k2f1(/vectork1)f2(/vectork2)|/vectork1/vectork2/angbracketrightfor some appropriate functions f1andf2, starting in the far past when the wave packets were far apart, evolving into the far future, so on and soforth, all the while smiling with self-satisfaction. The entire procedure is after all no differ-ent from the treatment of scattering 1in elementary nonrelativistic quantum mechanics. Complex scalar field Thus far, we have discussed a hermitean (often called real in a minor abuse of terminology)scalar field. Consider instead (as we will in chapter I.10) a nonhermitean (usually calledcomplex in another minor abuse) scalar field governed by L=∂ϕ †∂ϕ−m2ϕ†ϕ. Again, following Heisenberg, we find the canonical momentum density conjugate to the field ϕ(/vectorx,t), namely π(/vectorx,t)=δL/ [δ˙ϕ(/vectorx,t)]=∂0ϕ†(/vectorx,t), so that [ π(/vectorx,t),ϕ(/vectorx/prime,t)]= [∂0ϕ†(/vectorx,t),ϕ(/vectorx/prime,t)]=−iδ(D)(/vectorx−/vectorx/prime). Similarly, the canonical momentum density conju- gate to the field ϕ†(/vectorx,t)is∂0ϕ(/vectorx,t). Varying ϕ†we obtain the Euler-Lagrange equation of motion (∂2+m2)ϕ=0. [Similarly, varying ϕwe obtain (∂2+m2)ϕ†=0.] Once again, we could Fourier expand, but now the nonhermiticity of ϕmeans that we have to replace (11) by ϕ(/vectorx,t)=/integraldisplaydDk/radicalbig (2π)D2ωk/bracketleftBig a(/vectork)e−i(ω kt−/vectork./vectorx)+b†(/vectork)ei(ωkt−/vectork./vectorx))/bracketrightBig (17) In (11) hermiticity fixed the second term in the square bracket in terms of the first. Here in contrast, we are forced to introduce two independent sets of creation and annihilation operators (a,a†)and(b,b†). You should verify that the canonical commutation relations imply that these indeed behave like creation and annihilation operators. Consider the current Jμ=i(ϕ†∂μϕ−∂μϕ†ϕ) (18) Using the equations of motion you should check that ∂μJμ=i(ϕ†∂2ϕ−∂2ϕ†ϕ). (This follows immediately from the fact that the equation of motion for ϕ†is the hermitean 1For example, M. L. Goldberger and K. M. Watson, Collision Theory. 66 | I. Motivation and Foundation conjugate of the equation of motion for ϕ.) The current is conserved and the corresponding time-independent charge is given by (verify this!) Q=/integraldisplay dDxJ0(x)=/integraldisplay dDk[a†(/vectork)a(/vectork)−b†(/vectork)b(/vectork)]. Thus the particle created by a†(call it the “particle”) and the particle created by b†(call it the “antiparticle”) carry opposite charges. Explicitly, using the commutation relation wehaveQa †|0/angbracketright=+a†|0/angbracketrightandQb†|0/angbracketright=−b†|0/angbracketright. We conclude that ϕ†creates a particle and annihilates an antiparticle, that is, it produces one unit of charge. The field ϕdoes the opposite. You should understand this point thoroughly, as we will need it when we come to the Dirac field for the electron and positron. The energy of the vacuum As an instructive exercise let us calculate in the free scalar field theory the expectationvalue/angbracketleft0|H|0/angbracketright=/integraltext d Dx1 2/angbracketleft0|(π2+(/vector∇ϕ)2+m2ϕ2)|0/angbracketright, which we may loosely refer to as the “energy of the vacuum.” It is merely a matter of putting together (7), (11), and (12). Let usfocus on the third term in /angbracketleft0|H|0/angbracketright, which in fact we already computed, since /angbracketleft0|ϕ(/vectorx,t)ϕ(/vectorx,t)|0/angbracketright=/angbracketleft 0|ϕ(/vector0, 0)ϕ(/vector0, 0)|0/angbracketright = lim /vectorx,t→/vector0,0/angbracketleft0|ϕ(/vectorx,t)ϕ(/vector0, 0)|0/angbracketright= lim /vectorx,t→/vector0,0/integraldisplaydDk (2π)D2ωke−i(ω kt−/vectork./vectorx)=/integraldisplaydDk (2π)D2ωk The first equality follows from translation invariance, which also implies that the factor/integraltext dDxin/angbracketleft0|H|0/angbracketrightcan be immediately replaced by V, the volume of space. The calculation of the other two terms proceeds in much the same way: for example, the two factors of /vector∇ in(/vector∇ϕ)2just bring down a factor of /vectork2. Thus /angbracketleft0|H|0/angbracketright=V/integraldisplaydDk (2π)D2ωk1 2(ω2 k+/vectork2+m2)=V/integraldisplaydDk (2π)D1 2/planckover2piωk (19) upon restoring /planckover2pi. You should find this result at once gratifying and alarming, gratifying because we recog- nize it as the zero point energy of the harmonic oscillator integrated over all momentummodes and over all space, and alarming because the integral over /vectorkclearly diverges. But we should not be alarmed: the energy of any physical configuration, for example the mass of aparticle, is to be measured relative to the “energy of the vacuum.” We ask for the differencein the energy of the world with and without the particle. In other words, we could simplydefine the correct Hamiltonian to be H−/angbracketleft0|H|0/angbracketright. We will come back to some of these issues in chapters II.5, III.1, and VIII.2. Nobody is perfect In the canonical formalism, time is treated differently from space, and so one might worry about the Lorentz invariance of the resulting field theory. In the standard treatment given I.8. Quantizing Canonically | 67 in many texts, we would go on from this point and use the Hamiltonian to generate the dynamics, developing a perturbation theory in the interaction u(ϕ). After a number of formal steps, we would manage to derive the Feynman rules, which manifestly define aLorentz-invariant theory. Historically, there was a time when people felt that quantum field theory should be defined by its collection of Feynman rules, which gives us a concrete procedure to calculatemeasurable quantities, such as scattering cross sections. An extreme view along this lineheld that fields are merely mathematically crutches used to help us arrive at the Feynmanrules and should be thrown away at the end of the day. This view became untenable starting in the 1970s, when it was realized that there is a lot more to quantum field theory than Feynman diagrams. Field theory containsnonperturbative effects invisible by definition to Feynman diagrams. Many of these effects,which we will get to in due time, are more easily seen using the path integral formalism. As I said, the canonical and the path integral formalism often appear to be complemen- tary, and I will refrain from entering into a discussion about which formalism is superior.In this book, I adopt a pragmatic attitude and use whatever formalism happens to be easierfor the problem at hand. Let me mention, however, some particularly troublesome features in each of the two formalisms. In the canonical formalism fields are quantum operators containing an in-finite number of degrees of freedom, and sages once debated such delicate questions ashow products of fields are to be defined. On the other hand, in the path integral formal-ism, plenty of sins can be swept under the rug known as the integration measure (seechapter IV .7). Appendix 1 It may seem a bit puzzling that in the canonical formalism the propagator has to be defined with time ordering, which we did not need in the path integral formalism. To resolve this apparent puzzle, it suffices to look atquantum mechanics. LetA[q(t 1)] be a function of q, evaluated at time t1. What does the path integral/integraltext Dq(t) A [q(t 1)]ei/integraltextT 0dtL(˙q,q) represent in the operator language? Well, working backward to (I.2.4) we see that we would slip A[q(t 1)] into the factor /angbracketleftqj+1|e−iHδt|qj/angbracketright, where the integer jis determined by the condition that the time t1occurs between the times jδt and(j+1)δt. In the resulting factor /angbracketleftqj+1|e−iHδtA[q(t1)]|qj/angbracketright, we could replace the c-number A[q(t1)] by the operator A[ˆq], since A[ˆq]|qj/angbracketright=A[ qj]|qj/angbracketright/similarequalA[ q(t 1)]|qj/angbracketrightto the accuracy we are working with. Note that ˆqis evidently a Schr ¨odinger operator. Thus, putting in this factor /angbracketleftqj+1|e−iHδtA[ˆq]|qj/angbracketrighttogether with all the factors of /angbracketleftqi+1|e−iHδt|qi/angbracketright, we find that the integral in question, namely/integraltext Dq(t) A [q(t 1)]ei/integraltextT 0dtL(˙q,q), actually represents /angbracketleftqF|e−iH(T −t1)A[ˆq]e−iHt 1|qI/angbracketright=/angbracketleftqF|e−iHTA[ˆq(t1)]|qI/angbracketright where ˆq(t 1)is now evidently a Heisenberg operator. [We have used the standard relation between Heisenberg and Schr ¨odinger operators, namely, OH(t)=eiHtOSe−iHt.] I find this passage back and forth between the Dirac, Schr ¨odinger, and Heisenberg pictures quite instructive, perhaps even amusing. We are now prepared to ask the more complicated question: what does the path integral /integraltext Dq(t) A [q(t1)]B[q(t2)]ei/integraltextT 0dtL(˙q,q)represent in the operator language? Here B[q(t2)] is some other function of qevaluated at time t2. So we also slip B[q(t 2)] into the appropriate factor in (I.2.4) and replace B[q(t2)]b yB[ˆq]. But now we see that we have to keep track of whether t1ort2is the earlier of the two times. If t2is earlier, the operator A[ˆq] would appear to the left of the operator B[ˆq], and if t1is earlier, to the right. Explicitly, if t2is earlier 68 | I. Motivation and Foundation thant1,we would end up with the sequence e−iH(T −t1)A[ˆq]e−iH(t 1−t2)B[ˆq]e−iHt 2=e−iHTA[ˆq(t1)]B[ˆq(t2)] (20) upon passing from the Schr ¨odinger to the Heisenberg picture, just as in the simpler situation above. Thus we define the time-ordered product T[A[ˆq(t1)]B[ˆq(t2)]]≡θ(t1−t2)A[ˆq(t1)]B[ˆq(t2)]+θ(t2−t1)B[ˆq(t2)]A[ˆq(t1)] (21) We just learned that /angbracketleftqF|e−iHTT[A[ˆq(t 1)]B[ˆq(t 2)]]|qI/angbracketright=/integraldisplay Dq(t) A [q(t 1)]B[q(t 2)]ei/integraltextT 0dtL(˙q,q)(22) The concept of time ordering does not appear on the right-hand side, but is essential on the left-hand side. Generalizing the discussion here, we see that the Green’s functions G(n)(x1,x2,...,xn)introduced in the preceding chapter [see (I.7.13–15)] is given in the canonical formalism by the vacuum expectation value of a time-ordered product of field operators /angbracketleft0|T{ϕ(x 1)ϕ(x2)...ϕ(xn)}|0/angbracketright. That (13) gives the propagator is a special case of this relationship. We could also consider /angbracketleft0|T{O 1(x1)O 2(x2)...On(xn)}|0/angbracketright, the vacuum expectation value of a time-ordered product of various operators Oi(x) [the current Jμ(x), for example] made out of the quantum field. Such objects will appear in later chapters [ for example, (VII.3.7)]. Appendix 2: Field redefinition This is perhaps a good place to reveal to the innocent reader that there does not exist an international commissionin Brussels mandating what field one is required to use. If we use ϕ, some other guy is perfectly entitled to use η, assuming that the two fields are related by some invertible function with η=f( ϕ) . (To be specific, it is often helpful to think of η=ϕ+αϕ 3with some parameter α.) This is known as a field redefinition, an often useful thing to do, as we will see repeatedly. TheS-matrix amplitudes that experimentalists measure are invariant under field redefinition. But this is tautological trivia: the scattering amplitude /angbracketleft/vectork3/vectork4|e−iHT|/vectork1/vectork2/angbracketright, for example, does not even know about ϕandη. The issue is with the formalism we use to calculate the S-matrix. In the path integral formalism, it is also trivial that we could write Z(J)=/integraltext Dη ei[S(η)+/integraltext d4xJη ]just as well asZ(J)=/integraltext Dϕ ei[S(ϕ)+/integraltext d4xJϕ ]. This result, a mere change of integration variable, was known to Newton and Leibniz. But suppose we write ˜Z(J)=/integraltext Dϕ ei[S(ϕ)+/integraltext d4xJη ]. Now of course any dolt could see that ˜Z(J)/negationslash=Z(J) , and a fortiori, the Green’s functions (I.7.14,15) obtained by differentiating ˜Z(J) andZ(J) are not equal. The nontrivial physical statement is that the S-matrix amplitudes obtained from ˜Z(J) andZ(J) are in fact the same. This better be the case, since we are claiming that the path integral formalism provides a way to actualphysics. To see how this apparent “miracle" (Green’s functions completely different, S-matrix amplitudes the same) occurs, let us think physically. We set up our sources to produce or remove one single field disturbance, as indicated in figure I.4.1. Our friend, who uses ˜Z(J) , in contrast, set up his sources to produce or remove η=ϕ+αϕ 3(we specialize for pedagogical clarity), so that once in a while (with a probability determined by α) he is producing three field disturbances instead of one, as shown in figure I.8.1. As a result, while he thinksthat he is scattering four mesons, occasionally he is actually scattering six mesons. (Perhaps he should give hisaccelerator a tune up.) But to obtain S-matrix amplitudes we are told to multiply the Green’s functions by (k 2−m2)for each external leg carrying momentum k, and then set k2tom2. When we do this, the diagram in figure I.8.1a survives, since it has a pole that goes like 1 /(k2−m2)but the extraneous diagram in figure I.8.1b is eliminated. Very simple. One point worth emphasizing is that mhere is the actual physical mass of the particle. Let’s be precise when we should. Take the single particle state |/vectork/angbracketright. Act on it with the Hamiltonian. Then H|/vectork/angbracketright=/radicalbig /vectork2+m2|/vectork/angbracketright. The mthat appears in the eigenvalue of the Hamiltonian is the actual physical mass. We will come back to the issue of thephysical mass in chapter III.3. In the canonical formalism, the field is an operator, and as we saw just now, the calculation of S-matrix amplitudes involves evaluating products of field operators between physical states. In particular, the matrix elements /angbracketleft/vectork|ϕ|0/angbracketrightand/angbracketleft0|ϕ|/vectork/angbracketright(related by hermitean conjugation) come in crucially. If we use some other field η, I.8. Quantizing Canonically | 69 /H9272/H9272/H9272 /H9272 (a) (b)JJ Figure I.8.1 what matters is merely that /angbracketleft/vectork|η|0/angbracketrightis not zero, in which case we could always write /angbracketleft/vectork|η|0/angbracketright=Z1 2/angbracketleft/vectork|ϕ|0/angbracketrightwith Zsome c-number. We simply divide the scattering amplitude by the appropriate powers of Z1 2. Exercises I.8.1 Derive (14). Then verify explicitly that dDk/(2ωk)is indeed Lorentz invariant. Some authors prefer to replace/radicalbig 2ωkin (11) by 2 ωkwhen relating the scalar field to the creation and annihilation operators. Show that the operators defined by these authors are Lorentz covariant. Work out their commutationrelation. I.8.2 Calculate /angbracketleft/vectork /prime|H|/vectork/angbracketright, where |/vectork/angbracketright=a†(/vectork)|0/angbracketright. I.8.3 For the complex scalar field discussed in the text calculate /angbracketleft0|T[ϕ(x)ϕ†(0)]|0/angbracketright. I.8.4 Show that [ Q,ϕ(x) ]=−ϕ(x) . I.9 Disturbing the Vacuum Casimir effect In the preceding chapter, we computed the energy of the vacuum /angbracketleft0|H|0/angbracketrightand obtained the gratifying result that it is given by the zero point energy of the harmonic oscillatorintegrated over all momentum modes and over space. I explained that the energy of anyphysical configuration, for example, the mass of a particle, is to be measured relative tothis vacuum energy. In effect, we simply subtract off this vacuum energy and define thecorrect Hamiltonian to be H−/angbracketleft0|H|0/angbracketright. But what if we disturb the vacuum?Physically, we could compare the energy of the vacuum before and after we introduce the disturbance, by varying the disturbance for example. Of course, it is not just ourtextbook scalar field that contributes to the energy of the vacuum. The electromagneticfield, for instance, also undergoes quantum fluctuation and would contribute, with itstwo polarization degrees of freedom, to the energy density εof the vacuum the amount 2/integraltext d 3k/(2π)31 2/planckover2piωk. In 1948 Casimir had the brilliant insight that we could disturb the vacuum and produce a shift /Delta1ε. While εis not observable, /Delta1εshould be observable since we can control how we disturb the vacuum. In particular, Casimir considered introducingtwo parallel “perfectly” conducting plates (formally of zero thickness and infinite extent)into the vacuum. The variation of /Delta1εwith the distance dbetween the plates would lead to a force between the plates, known as the Casimir force. In reality, it is the electromagneticfield that is responsible, not our silly scalar field. Call the direction perpendicular to the plates the xaxis. Because of the boundary conditions the electromagnetic field must satisfy on the conducting plates, the wave vector /vectorkcan only take on the values (πn/d ,k y,kz), with nan integer. Thus the energy per unit area between the plates is changed to/summationtext n/integraltext dkydkz/(2π)2/radicalBig (πn/d)2+k2 y+k2 z. To calculate the force, we vary d, but then we would have to worry about how the energy density outside the two plates varies. A clever trick exists for avoiding this worry: weintroduce three plates! See figure (I.9.1). We hold the two outer plates fixed and move I.9. Disturbing the Vacuum | 71 Ld L− d Figure I.9.1 only the inner plate. Now we don’t have to worry about the world outside the plates. The separation Lbetween the two outer plates can be taken to be as large as we like. In the spirit of this book (and my philosophy) of avoiding computational tedium as much as possible, I propose two simplifications: (I) do the calculation for a massless scalarfield instead of the electromagnetic field so we won’t have to worry about polarizationand stuff like that, and (II) retreat to a (1+1)-dimensional spacetime so we won’t have to integrate over k yandkz. Readers of quantum field theory texts do not need to watch their authors show off their prowess in doing integrals. As you will see, the calculation isexceedingly instructive and gives us an early taste of the art of extracting finite physicalresults from seemingly infinite expressions, known as regularization, that we will studyin chapters III.1–3. With this set-up, the energy E=f( d)+f( L−d)with f( d)=1 2∞/summationdisplay n=1ωn=π 2d∞/summationdisplay n=1n (1) since the modes are given by sin (nπx/d) (n =1,...,∞) with the corresponding energy ωn=nπ/d . Aagh! What do we do with∞/summationtext n=1n? None of the ancient Greeks from Zeno on could tell us. What they should tell us is that we are doing physics, not mathematics! Physical plates cannot keep arbitrarily high frequency waves from leaking out.1 To incorporate this piece of all-important physics, we should introduce a factor e−aωn/π with a parameter ahaving the dimension of time (or in our natural units, length) so that modes with ωn/greatermuchπ/a do not contribute: they don’t see the plates! The characteristic 1See footnote 1 in chapter III.1. 72 | I. Motivation and Foundation frequency π/a parametrizes the high frequency response of the conducting plates. Thus we have f( d)=π 2d∞/summationdisplay n=1ne−an/d=−π 2∂ ∂a∞/summationdisplay n=1e−an/d=−π 2∂ ∂a1 1−e−a/d=π 2dea/d (ea/d−1)2 Since we want a−1to be large, we take the limit asmall so that f( d)=πd 2a2−π 24d+πa2 480d3+O(a4/d5). (2) Note that f( d) blows up as a→0, as it should, since we are then back to (1). But the force between two conducting plates shouldn’t blow up. Experimentalists might have noticed itby now! Well, the force is given by F=−∂E ∂d=− {f/prime(d)−f/prime(L−d)}=−/braceleftbigg/parenleftbigg1 2πa2+π 24d2+.../parenrightbigg −(d→L−d)/bracerightbigg −→ a→0−π 24/parenleftbigg1 d2−1 (L−d)2/parenrightbigg −→ L/greatermuchd−π/planckover2pic 24d2(3) Behold, the parameter awe have introduced to make sense of the divergent sum in (1) has disappeared in our final result for the physically measurable force. In the last step, usingdimensional analysis we restored /planckover2pito underline the quantum nature of the force. The Casimir force between two plates is attractive. Notice that the 1 /d 2of the force simply follows from dimensional analysis since in natural units force has dimension ofan inverse length squared. In a tour de force, experimentalists have measured this tinyforce. The fluctuating quantum field is quite real! To obtain a sensible result we need to regularize in the ultraviolet (namely the high frequency or short time behavior parametrized by a) and in the infrared (namely the long distance cutoff represented by L). Notice how aandL“work” together in (3). This calculation foreshadows the renormalization of quantum field theories, a topic made mysterious and almost incomprehensible in many older texts. In fact, it is perfectlysensible. We will discuss renormalization in detail in chapters III.1 and III.2, but for nowlet us review what we just did. Instead of panicking when faced with the divergent sum in (1), we remind ourselves that we are proud physicists and that physics tells us that the sum should not go all theway to infinity. In a conducting plate, electrons rush about to counteract any applied tan-gential electric field. But when the incident wave oscillates at sufficiently high frequency,the electrons can’t keep up. Thus the idealization of a perfectly conducting plate fails. Weregularize (such an ugly term but that’s what field theorists use!) the sum in a mathemati-cally convenient way by introducing a damping factor. The single parameter ais supposed to summarize the unknown high frequency physics that causes the electron to fail to keepup. In reality, a −1is related to the plasma frequency of the metal making up the plate. I.9. Disturbing the Vacuum | 73 A priori, the Casimir force between the two plates could end up depending on the parameter a. In that case, the Casimir force would tell us something about the response of a conducting plate to high frequency electric fields, and it would have made for an interestingchapter in a text on solid state physics. Since this is in fact a quantum field theory text, youmight have suspected that the Casimir force represents some fundamental physics aboutfluctuating quantum fields and that awould drop out. That the Casimir force is “universal” makes it unusually interesting. Notice however, as is physically sensible, that the O(1/d 4) correction to the Casimir force does depend on whether the experimentalist used copperor aluminum plates. We might then wonder whether the leading term F=−π/(24d 2)depends on the particular regularization we used. What if we suppress the higher terms in the divergentsum with some other function? We will address this question in the appendix to thischapter. Amusingly, the 24 in (3) is the same 24 that appears in string theory! (The dimension of spacetime the quantum bosonic string must live in is determined to be 24 +2=26.) The reader who knows string theory would know what these two cryptic statements areabout (summing up the zero modes of the string). Appallingly, in an apparent attempt tomake the subject appear even more mysterious than it is, some treatments simply assert that the sum ∞/summationtext n=1nis by some mathematical sleight-of-hand equal to −1/12. Even though it would have allowed us to wormhole from (1) to (3) instantly, this assertion is manifestly absurd. What we did here, however, makes physical sense. Appendix Here we address the fundamental issue of whether a physical quantity we extract by cutting off high frequency contributions could depend on how we cut. Let me mention that in recent years the study of Casimir force foractual physical situations has grown into an active area of research, but clearly my aim here is not to give a realisticaccount of this field, but to study in an easily understood context an issue (as you will see) central to quantumfield theory. My hope is that by the time you get to actually regularize a field theory in (3+1)-dimensional spacetime, you would have amply mastered the essential physics and not have to struggle with the mechanics ofregularization. Let us first generalize a bit the regularization scheme we used and write f( d)=π 2d∞/summationdisplay n=1ng/parenleftbiggna d/parenrightbigg =π 2∂ ∂a∞/summationdisplay n=1h/parenleftbiggna d/parenrightbigg ≡π 2∂ ∂aH/parenleftbigga d/parenrightbigg (4) Hereg(v)=h/prime(v) is a rapidly decreasing function so that the sums make sense, chosen judiciously to allow ready evaluation of H(a/d) ≡∞/summationtext n=1h(na/d). [In (2), we chose g(v)=e−vand hence h(v)=−e−v.] We would like to know how the Casimir force, −F=∂f (d) ∂d−(d→L−d)=π 2∂2 ∂d∂aH(a d)−(d→L−d) (5) depends on g(v) . Let us try to get as far as we can using physical arguments and dimensional analysis. Expand Has follows: πH(a/d) =...+γ−2d2/a2+γ−1d/a+γ0+γ1a/d+γ2a2/d2+.... We might be tempted to just write a Taylor series in a/d , but nothing tells us that H(a/d) might not blow up as a→0. Indeed, the example in (3) contains a term like d/a , and so we better be cautious. 74 | I. Motivation and Foundation We will presently argue physically that the series in fact terminate in one direction. The force is given by F=/parenleftbigg ...+γ−22d a3+γ−11 2a2+γ11 2d2+γ22a d3+.../parenrightbigg −(d→L−d) (6) Look at the γ−2term: it contributes to the force a term like (d−(L−d))/a3. But as remarked earlier, the two outer plates could be taken as far apart as we like. The force could not depend on L, and thus on physical grounds γ−2must vanish. Similarly, all γ−kfork> 2 must vanish. Next, we note that the γ−1, although definitely not zero, gets subtracted away since it does not depend on d. (Theγ0term has already gone away.) At this point, notice that, furthermore, the γkterms with k> 2 all vanish asa→0. You could check that all these assertions hold for the g(v) used in the text. Remarkably, the Casimir force is determined by γ1alone: F=γ1/(2d2). As noted earlier, the force has to be proportional to 1 /d2. This fact alone shows us that in (6) we only need to keep the γ1term. In the text, we found γ1=−π/12. In exercise I.9.1 I invite you to go through an amusing calculation obtaining the same value for γ1 with an entirely different choice of g(v) . This already suggests that the Casimir force is regularization independent, that it tells us more about the vacuum than about metallic conductivity, but still it is highly instructive to study an entire class of damping or regularizing functions to watch how regularization independence emerges. Let us regularize the sum over zero point energies to f( d)=1 2∞/summationtext n=1ωnK(ωn)with K(ω) =/summationdisplay αcα/Lambda1α ω+/Lambda1α(7) Herecαis a bunch of real numbers and /Lambda1α(known as regulators or regulator frequencies) a bunch of high frequencies subject to certain conditions but otherwise chosen entirely at our discretion. For the sum∞/summationtext n=1ωnK(ωn) to converge, we need K(ωn)to vanish faster than 1 /ω2 n. In fact, for ωmuch larger than /Lambda1α,K(ω)→1 ω/summationtext αcα/Lambda1α− 1 ω2/summationtext αcα/Lambda1α2+.... The requirement that the 1 /ωand 1/ω2terms vanish gives the conditions /summationdisplay αcα/Lambda1α=0 (8) and /summationdisplay αcα/Lambda12 α=0 (9) respectively. Furthermore, low frequency physics is not to be modified, and so we want K(ω) →1 forω< </Lambda1α, thus requiring /summationdisplay αcα=1 (10) At this point, we do not even have to specify the set the index αruns over beyond the fact that the three conditions (8),(9), and (10) require that αmust take on at least three values. Note also that some of the cα’s must be negative. Incidentally, we could do with fewer regulators if we are willing to invoke some knowledge of metals, for instance,thatK(ω) =K(−ω) , but that is not the issue here. We now show that the Casimir force between the two plates does not depend on the choices of c αand/Lambda1α. First, being physicists rather than mathematicians, we freely interchange the two sums in f( d) and write f( d)=1 2/summationdisplay αcα/Lambda1α/summationdisplay nωn ωn+/Lambda1α=−1 2/summationdisplay αcα/Lambda1α/summationdisplay n/Lambda1α ωn+/Lambda1α(11) I.9. Disturbing the Vacuum | 75 where, without further ceremony, we have used condition (8). Next, keeping in mind that the sum/summationtext n/Lambda1α/(ωn+/Lambda1α)is to be put back into (11), we massage it (defining for convenience bα=π//Lambda1 α) as follows: ∞/summationdisplay n=1/Lambda1 ωn+/Lambda1=∞/summationdisplay n=1/integraldisplay∞ 0dte−t( 1+nb d)=/integraldisplay∞ 0dte−t/parenleftBigg 1 1−e−bt d−1/parenrightBigg =/integraldisplay∞ 0dte−t[d tb−1 2+tb 12d+O/parenleftBig b3/parenrightBig ] (12) (To avoid clutter we have temporarily suppressed the index α.) All these manipulations make perfect sense since the entire expression is to be inserted into the sum over αin (11) after we restore the index α. It appears that the result would depend on cαandλα. In fact, mentally restoring and inserting, we see that the 1 /bterm in (12) can be thrown away since /summationdisplay αcα/Lambda1α/bα=π/summationdisplay αcα/Lambda12 α=0 (13) [There is in fact a bit of an overkill here since this term corresponds to the γ−1term, which does not appear in the force anyway. Thus the condition (9) is, strictly speaking, not necessary. We are regularizing not merely the force, but f( d) so that it defines a sensible function.] Similarly, the b0term in (12) can be thrown away thanks to (8). Thus, keeping only the bterm in (12), we obtain f( d)=−1 24d/integraldisplay∞ 0dte−tt/summationdisplay αcα/Lambda1αbα+O/parenleftbigg1 d3/parenrightbigg =−π 24d+O/parenleftbigg1 d3/parenrightbigg (14) Indeed, f( d) , and a fortiori the Casimir force, do not depend on the cα’s and /Lambda1α’s. To the level of rigor enter- tained by physicists (but certainly not mathematicians), this amounts to a proof of regularization independencesince with enough regulators we could approximate any (reasonable) function K(ω) that actually describes real conducting plates. Again, as is physically sensible, you could check that the O(1/d 3)term in f( d) does depend on the regularization scheme. The reason that I did this calculation in detail is that we will encounter this class of regularization, known as Pauli-Villars, in chapter III.1 and especially in the calculation of the anomalous magnetic moment of the electronin chapter III.7, and it is instructive to see how regularization works in a more physical context before dealingwith all the complications of relativistic field theory. Exercises I.9.1 Choose the damping function g(v)=1/(1+v)2instead of the one in the text. Show that this re- sults in the same Casimir force. [Hint: To sum the resulting series, pass to an integral representation H(ξ)=−∞/summationtext n=11/(1+nξ)=−∞/summationtext n=1/integraltext∞ 0dte−(1+nξ)t=/integraltext∞ 0dte−t/(1−eξt). Note that the integral blows up logarithmically near the lower limit, as expected.] I.9.2 Show that with the regularization used in the appendix, the 1 /dexpansion of the force between two conducting plates contains only even powers. I.9.3 Show off your skill in doing integrals by calculating the Casimir force in (3+1)-dimensional spacetime. For help, see M. Kardar and R. Golestanian, Rev. Mod. Phys. 71: 1233, 1999; J. Feinberg, A. Mann, and M. Revzen, Ann. Phys. 288: 103, 2001. I.10 Symmetry Symmetry, transformation, and invariance The importance of symmetry in modern physics cannot be overstated.1 When a law of physics does not change upon some transformation, that law is said to exhibit a symmetry. I have already used Lorentz invariance to restrict the form of an action. Lorentz invari- ance is of course a symmetry of spacetime, but fields can also transform in what is thoughtof as an internal space. Indeed, we have already seen a simple example of this. I noted inpassing in chapter I.3 that we could require the action of a scalar field theory to be invariantunder the transformation ϕ→−ϕand so exclude terms of odd power in ϕ, such as ϕ 3, from the action. With the ϕ3term included, two mesons could scatter and go into three mesons, for example by the diagrams in (fig. I.10.1). But with this term excluded, you can easilyconvince yourself that this process is no longer allowed. You will not be able to drawa Feynman diagram with an odd number of external lines. (Think about modifying the integral in our baby problem in chapter I.7 to/integraltext +∞ −∞dqe−1 2m2q2−gq3−λq4+Jq.)Thus the simple reflection symmetry ϕ→−ϕimplies that in any scattering process the number of mesons is conserved modulo 2. Now that we understand one scalar field, let us consider a theory with two scalar fields ϕ1andϕ2satisfying the reflection symmetry ϕa→−ϕa(a=1o r2): L=1 2(∂ϕ1)2−1 2m2 1ϕ2 1−λ1 4ϕ4 1+1 2(∂ϕ2)2−1 2m2 2ϕ2 2−λ2 4ϕ4 2−ρ 2ϕ2 1ϕ2 2(1) We have two scalar particles, call them 1 and 2, with mass m1andm2. To lowest order, they scatter in the processes 1 +1→1+1, 2+2→2+2, 1+2→1+2, 1+1→2+2, and 2 +2→1+1 (convince yourself). With the five parameters m1,m2,λ1,λ2, and ρ completely arbitrary, there is no relationship between the two particles. 1A. Zee, Fearful Symmetry . I.10. Symmetry | 77 Figure I.10.1 It is almost an article of faith among theoretical physicists, enunciated forcefully by Einstein among others, that the fundamental laws should be orderly and simple, ratherthan arbitrary and complicated. This orderliness is reflected in the symmetry of the action. Suppose that m 1=m2andλ1=λ2; then the two particles would have the same mass and their interaction, with themselves and with each other, would be the same. The LagrangianLbecomes invariant under the interchange symmetry ϕ 1←→ϕ2. Next, suppose we further impose the condition ρ=λ1=λ2so that the Lagrangian becomes L=1 2/bracketleftBig (∂ϕ 1)2+(∂ϕ 2)2/bracketrightBig −1 2m2/parenleftBig ϕ2 1+ϕ2 2/parenrightBig −λ 4/parenleftBig ϕ2 1+ϕ2 2/parenrightBig2 (2) It is now invariant under the 2-dimensional rotation {ϕ1(x)→cosθϕ1(x)+sinθϕ2(x), ϕ2(x)→−sin θϕ 1(x)+cosθϕ 2(x)} withθan arbitrary angle. We say that the theory enjoys an “internal” SO( 2)symmetry, internal in the sense that the transformation has nothing to do with spacetime. In contrast to the interchange symmetry ϕ1←→ϕ2the transformation depends on the continuous parameter θ, and the corresponding symmetry is said to be continuous. We see from this simple example that symmetries exist in hierarchies. Continuous symmetries If we stare at the equations of motion (∂2+m2)ϕa=−λ/vectorϕ2ϕalong enough we see that if we define Jμ≡i(ϕ1∂μϕ2−ϕ2∂μϕ1), then ∂μJμ=i(ϕ1∂2ϕ2−ϕ2∂2ϕ1)=0 so that Jμis a conserved current. The corresponding charge Q=/integraltext dDxJ0, just like electric charge, is conserved. Historically, when Heisenberg noticed that the mass of the newly discovered neutron was almost the same as the mass of a proton, he proposed that if electromagnetism weresomehow turned off there would be an internal symmetry transforming a proton into aneutron. An internal symmetry restricts the form of the theory, just as Lorentz invariance restricts the form of the theory. Generalizing our simple example, we could construct a field theorycontaining Nscalar fields ϕ a, with a=1,...,Nsuch that the theory is invariant under the transformations ϕa→Rabϕb(repeated indices summed), where the matrix Ris an element of the rotation group SO(N) (see appendix B for a review of group theory). The fields ϕa transform as a vector /vectorϕ=(ϕ1,...,ϕN). We can form only one basic invariant, namely the 78 | I. Motivation and Foundation cd a b/H110022i/H9261 (/H9254ab/H9254cd /H11001 /H9254ac/H9254bd /H11001 /H9254ad/H9254bc) Figure I.10.2 scalar product /vectorϕ./vectorϕ=ϕaϕa=/vectorϕ2(as always, repeated indices are summed unless otherwise specified). The Lagrangian is thus restricted to have the form L=1 2/bracketleftBig (∂/vectorϕ)2−m2/vectorϕ2/bracketrightBig −λ 4(/vectorϕ2)2(3) The Feynman rules are given in fig. I.10.2. When we draw Feynman diagrams, each line carries an internal index in addition to momentum. Symmetry manifests itself in physical amplitudes. For example, imagine calculating the propagator iDab(x)=/integraltext D/vectorϕeiSϕa(x)ϕb(0). We assume that the measure D/vectorϕis in- variant under SO(N). By thinking about how Dab(x) transforms under the symmetry group SO(N) you see easily (exercise I.10.2) that it must be proportional to δab. You can check this by drawing a few Feynman diagrams or by considering an ordinary integral/integraltext d/vectorqe−S(q)qaqb. No matter how complicated a diagram you draw (fig. I.10.3, e.g.) you al- ways get this factor of δab. Similarly, scattering amplitudes must also exhibit the symmetry. Without the SO(N) symmetry, many other terms would be possible (e.g., ϕaϕbϕcϕdfor some arbitrary choice of a,b,c, andd)in (3). We can write R=eθ.Twhere θ.T=/summationtext AθATAis a real antisymmetric matrix. The group SO(N) hasN(N−1)/2 generators, which we denote by TA. [Think of the familiar case of SO( 3).] Under an infinitesimal transformation (repeated indices summed) ϕa→Rabϕb/similarequal (1+θATA)abϕb, or in other words, we have the infinitesimal change δϕa=θATAabϕb. Noether’s theorem We now come to one of the most profound observations in theoretical physics, namely Noether’s theorem, which states that a conserved current is associated with each generator ab dhg c ef ijm kl Figure I.10.3 I.10. Symmetry | 79 of a continuous symmetry. The appearance of a conserved current for (2) is not an acci- dent. As is often the case with the most important theorems, the proof of Noether’s theorem is astonishingly simple. Denote the fields in our theory generically by ϕa.Since the symmetry is continuous, we can consider an infinitesimal change δϕa. Since Ldoes not change, we have 0=δL=δL δϕaδϕa+δL δ∂μϕaδ∂μϕa =δL δϕaδϕa+δL δ∂μϕa∂μδϕa (4) We would have been stuck at this point, but if we use the equations of motion δL/δϕa= ∂μ(δL/δ∂μϕa)we can combine the two terms and obtain 0=∂μ/parenleftBigg δL δ∂μϕaδϕa/parenrightBigg (5) If we define Jμ≡δL δ∂μϕaδϕa (6) then (5) says that ∂μJμ=0. We have found a conserved current! [It is clear from the derivation that the repeated index ais summed in (6)]. Let us immediately illustrate with the simple scalar field theory in (3). Plugging δϕa= θA(TA)abϕbinto (6) and noting that θAis arbitrary, we obtain N(N−1)/2 conserved currents JA μ=∂μϕa(TA)abϕb, one for each generator of the symmetry group SO(N). In the special case N=2, we can define a complex field ϕ≡(ϕ1+iϕ2)/√ 2. The La- grangian in (3) can be written as L=∂ϕ†∂ϕ−m2ϕ†ϕ−λ(ϕ†ϕ)2, and is clearly invariant under ϕ→eiθϕandϕ†→e−iθϕ†. We find from (6) that Jμ= i(ϕ†∂μϕ−∂μϕ†ϕ), the current we met already in chapter I.8. Mathematically, this is because the groups SO( 2)andU(1)are isomorphic (see appendix B). For pedagogical clarity I have used the example of scalar fields transforming as a vector under the group SO(N) . Obviously, the preceding discussion holds for an arbitrary group Gwith the fields ϕtransforming under an arbitrary representation RofG. The conserved currents are still given by JA μ=∂μϕa(TA)abϕbwithTAthe generators evaluated in the representation R. For example, if ϕtransform as the 5-dimensional representation of SO( 3)thenTAi sa5b y5matrix. For physics to be invariant under a group of transformations it is only necessary that the action be invariant. The Lagrangian density Lcould very well change by a total divergence: δL=∂μKμ, provided that the relevant boundary term could be dropped. Then we would see immediately from (5) that all we have to do to obtain a formula for the conservedcurrent is to modify (6) to J μ≡(δL/δ∂μϕa)δϕa−Kμ. As we will see in chapter VIII.4, many supersymmetric field theories are of this type. 80 | I. Motivation and Foundation Charge as generators Using the canonical formalism of chapter I.8, we can derive an elegant result for the charge associated with the conserved current Q≡/integraldisplay d3xJ0=/integraldisplay d3xδL δ∂0ϕaδϕa Note that Qdoes not depend on the time at which the integral is evaluated: dQ dt=/integraldisplay d3x∂0J0=−/integraldisplay d3x∂iJi=0 (7) Recognizing that δL/δ∂ 0ϕais just the canonical momentum conjugate to the field ϕa,w e see that i[Q,ϕa]=δϕa (8) The charge operator generates the corresponding transformation on the fields. An impor- tant special case is for the complex field ϕinSO( 2)/similarequalU(1)theory we discussed; then [Q,ϕ]=ϕandeiθQϕe−iθQ=eiθϕ. Exercises I.10.1 Some authors prefer the following more elaborate formulation of Noether’s theorem. Suppose that the action does not change under an infinitesimal transformation δϕa(x)=θAVA a[withθAsome parameters labeled by AandVA asome function of the fields ϕb(x) and possibly also of their first derivatives with respect to x]. It is important to emphasize that when we say the action Sdoes not change we are not allowed to use the equations of motion. After all, the Euler-Lagrange equations of motion follow fromdemanding that δS=0 for any variation δϕ asubject to certain boundary conditions. Our scalar field theory example nicely illustrates this point, which is confused in some books: δS=0 merely because S is constructed using the scalar product of O(N) vectors. Now let us do something apparently a bit strange. Let us consider the infinitesimal change written above but with the parameters θAdependent on x. In other words, we now consider δϕa(x)=θA(x)VA a. Then of course there is no reason for δSto vanish; but, on the other hand, we know that since δSdoes vanish when θAis constant, δSmust have the form δS=/integraltext d4xJμ(x)∂μθA(x). In practice, this gives us a quick way of reading off the current Jμ(x); it is just the coefficient of ∂μθA(x) inδS. Show how all this works for the Lagrangian in (3). I.10.2 Show that Dab(x) must be proportional to δabas stated in the text. I.10.3 Write the Lagrangian for an SO( 3)invariant theory containing a Lorentz scalar field ϕtransforming in the 5-dimensional representation up to quartic terms. [Hint: It is convenient to write ϕas a 3 by 3 symmetric traceless matrix.] I.10.4 Add a Lorentz scalar field ηtransforming as a vector under SO( 3)to the Lagrangian in exercise I.10.3, maintaining SO( 3)invariance. Determine the Noether currents in this theory. Using the equations of motion, check that the currents are conserved. I.11 Field Theory in Curved Spacetime General coordinate transformation In Einstein’s theory of gravity, the invariant Minkowskian spacetime interval ds2= ημνdxμdxν=(dt)2−(d/vectorx)2is replaced by ds2=gμνdxμdxν, where the metric tensor gμν(x) is a function of the spacetime coordinates x. The guiding principle, known as the principle of general covariance, states that physics, as embodied in the action S, must be invariant under arbitrary coordinate transformations x→x/prime(x). More precisely, the prin- ciple1states that with suitable restrictions the effect of a gravitational field is equivalent to that of a coordinate transformation. Since ds2=g/prime λσdx/primeλdx/primeσ=g/prime λσ∂x/primeλ ∂xμ∂x/primeσ ∂xνdxμdxν=gμνdxμdxν the metric transforms as g/prime λσ(x/prime)∂x/primeλ ∂xμ∂x/primeσ ∂xν=gμν(x) (1) The inverse of the metric gμνis defined by gμνgνρ=δμ ρ. A scalar field by its very name does not transform: ϕ(x)=ϕ/prime(x/prime). The gradient of the scalar field transforms as ∂μϕ(x)=∂x/primeλ ∂xμ∂ϕ/prime(x/prime) ∂x/primeλ=∂x/primeλ ∂xμ∂/prime λϕ/prime(x/prime) By definition, a (covariant) vector field transforms as Aμ(x)=∂x/primeλ ∂xμA/prime λ(x/prime) 1For a precise statement of the principle of general covariance, see S. Weinberg, Gravitation and Cosmology , p. 92. 82 | I. Motivation and Foundation so that ∂μϕ(x) is a vector field. Given two vector fields Aμ(x) andBν(x), we can contract them with gμν(x) to form gμν(x)Aμ(x)Bν(x), which, as you can immediately check, is a scalar. In particular, gμν(x)∂μϕ(x)∂ νϕ(x) is a scalar. Thus, if we simply replace the Minkowski metric ημνin the Lagrangian L=1 2[(∂ϕ)2−m2ϕ2]=1 2(ημν∂μϕ∂νϕ−m2ϕ2)by the Einstein metric gμν, the Lagrangian is invariant under coordinate transformation. The action is obtained by integrating the Lagrangian over spacetime. Under a coordinate transformation d4x/prime=d4xdet(∂x/prime/∂x) . Taking the determinant of (1), we have g≡detgμν=detg/prime λσ∂x/primeλ ∂xμ∂x/primeσ ∂xν=g/prime/bracketleftbigg det/parenleftbigg∂x/prime ∂x/parenrightbigg/bracketrightbigg2 (2) We see that the combination d4x√−g=d4x/prime/radicalbig −g/primeis invariant under coordinate transfor- mation. Thus, given a quantum field theory we can immediately write down the theory in curved spacetime. All we have to do is promote the Minkowski metric ημνin our Lagrangian to the Einstein metric gμνand include a factor√−g in the spacetime integration measure.2The action Swould then be invariant under arbitrary coordinate transformations. For example, the action for a scalar field in curved spacetime is simply S=/integraldisplay d4x√−g1 2(gμν∂μϕ∂νϕ−m2ϕ2) (3) (There is a slight subtlety involving the spin1 2field that we will talk about in part II. We will eventually come to it in chapter VIII.1.) There is no essential difficulty in quantizing the scalar field in curved spacetime. We simply treat gμνas given [e.g., the Schwarzschild metric in spherical coordinates: g00= (1−2GM/r) ,grr=−(1−2GM/r)−1,gθθ=−r2, and gφφ=−r2sin2θ] and study the path integral/integraltext DϕeiS, which is still a Gaussian integral and thus do-able. The propagator of the scalar field D(x ,y)can be worked out and so on and so forth. (see exercise I.11.1). At this point, aside from the fact that gμν(x)carries Lorentz indices while ϕ(x) does not, the metric gμνlooks just like a field and is in fact a classical field. Write the action of the world S=Sg+SMas the sum of two terms: Sgdescribing the dynamics of the gravitational fieldgμν(which we will come to in chapter VIII.1) and SMdescribing the dynamics of all the other fields in the world [the “matter fields,” namely ϕin our simple example with SM as given in (3)]. We could quantize gravity by integrating over gμνas well, thus extending the path integral to/integraltext DgDϕeiS. Easier said than done! As you have surely heard, all attempts to integrate over gμν(x) have been beset with difficulties, eventually driving theorists to seek solace in string theory.I will explain in due time why Einstein’s theory is known as “nonrenormalizable.” 2We also have to replace ordinary derivatives ∂μby the covariant derivatives Dμof general relativity, but acting on a scalar field ϕthe covariant derivative is just the ordinary derivative Dμϕ=∂μϕ. I.11. Field Theory in Spacetime | 83 What the graviton listens to One of the most profound results to come out of Einstein’s theory of gravity is a funda- mental definition of energy and momentum. What exactly are energy and momentum anyway? Energy and momentum are what the graviton listens to. (The graviton is of coursethe particle associated with the field g μν.) The stress-energy tensor Tμνis defined as the variation of the matter action SMwith respect to the metric gμν(holding the coordinates xμfixed): Tμν(x)=−2√−gδSM δgμν(x)(4) Energy is defined as E=P0=/integraltext d3x√−gT00(x)and momentum as Pi=/integraltext d3x√−gT0i(x). Even if we are not interested in curved spacetime per se, (4) still offers us a simple (and fundamental) way to determine the stress energy of a field theory in flat spacetime. Wesimply vary around the Minkowski metric η μνby writing gμν=ημν+hμνand expand SM to first order in h. According to (4), we have3 SM(h)=SM(h=0)−/integraldisplay d4x/bracketleftBig 1 2hμνTμν+O(h2)/bracketrightBig . (5) The symmetric tensor field hμν(x) is in fact the graviton field (see chapters I.5 and VIII.1). The stress-energy tensor Tμν(x) is what the graviton field couples to, just as the electromagnetic current Jμ(x) is what the photon field couples to. Consider a general SM=/integraltext d4x√−g(A +gμνBμν+gμνgλρCμνλρ+...). Since −g= 1+ημνhμν+O(h2)andgμν=ημν−hμν+O(h2), we find Tμν=2(Bμν+2Cμνλρηλρ+...)−ημνL (6) in flat spacetime. Note T≡ημνTμν=−(4A+2ημνBμν+0.ημνηλρCμνλρ+...) (7) which we have written in a form emphasizing that Cμνλρ does not contribute to the trace of the stress-energy tensor. We now show the power of this definition of Tμνby obtaining long-familiar results about the electromagnetic field. Promoting the Lagrangian of the massive spin 1 fieldto curved spacetime we have 4L=(−1 4gμνgλρFμλFνρ+1 2m2gμνAμAν)and thus5Tμν= −FμλFλ ν+m2AμAν−ημνL. 3I use the normal convention in which indices are summed regardless of any symmetry; in other words, 1 2hμνTμν=1 2(h01T01+h10T10+...)=h01T01+.... 4Here we use the fact that the covariant curl is equal to the ordinary curl DμAν−DνAμ=∂μAν−∂νAμand soFμνdoes not involve the metric. 5Holding xμfixed means that we hold ∂μand hence Aμfixed since Aμis related to ∂μby gauge invariance. We are anticipating (see chapter II.7) here, but you have surely heard of gauge invariance in a nonrelativistic context. 84 | I. Motivation and Foundation For the electromagnetic field we set m=0. First, L=−1 4FμνFμν=−1 4(−2F2 0i+F2 ij)= 1 2(/vectorE2−/vectorB2). Thus, T00=−F0λFλ 0−1 2(/vectorE2−/vectorB2)=1 2(/vectorE2+/vectorB2). That was comforting, to see a result we knew from “childhood.” Incidentally, this also makes clear that we canthink of /vectorE 2as kinetic energy and /vectorB2as potential energy. Next, T0i=−F0λFλ i=F0jFij= εijkEjBk=(/vectorE×/vectorB)i. The Poynting vector has just emerged. Since the Maxwell Lagrangian L=−1 4gμνgλρFμλFνρinvolves only the Cterm with Cμνλρ=−1 4FμνFλρ, we see from (7) that the stress-energy tensor of the electromagnetic field is traceless, an important fact we will need in chapter VIII.1. We can of course checkdirectly that T=0 (exercise I.11.4). 6 Appendix: A concise introduction to curved spacetime General relativity is often made to seem more difficult and mysterious than need be. Here I give a concise review of some of its basic elements for later use. Denote the spacetime coordinates of a point particle by Xμ. To construct its action note that the only coordinate invariant quantity is the “length”7of the world line traced out by the particle (fig. I.11.1), namely/integraltext ds=/integraltext/radicalbiggμνdXμdXν, where gμνis evaluated at the position of the particle of course. Thus, the action for a point particle must be proportional to /integraldisplay ds=/integraldisplay/radicalBig gμνdXμdXν=/integraldisplay/radicalBigg gμν[X(ζ) ]dXμ dζdXν dζdζ where ζis any parameter that varies monotonically along the world line. The length, being geometric, is manifestly reparametrization invariant, that is, independent of our choice of ζas long as it is reasonable. This is one of those “more obvious than obvious” facts since/integraltext/radicalbiggμνdXμdXνis manifestly independent of ζ.I fw e insist, we can check the reparametrization invariance of/integraltext ds. Obviously the powers of dζmatch. Explicitly, if we write ζ=ζ(η) , then dXμ/dζ=(dη/dζ)(dXμ/dη) anddζ=(dζ/dη)dη . Let us define K≡gμν[X(ζ) ]dXμ dζdXν dζ for ease of writing. Setting the variation of/integraltext dζ√ Kequal to zero, we obtain /integraldisplay dζ1√ K(2gμνdXμ dζdδXν dζ+∂λgμνdXμ dζdXν dζδXλ)=0 which upon integration by parts (and with δXλ=0 at the endpoints as usual) gives the equation of motion √ Kd dζ/parenleftbigg1√ K2gμλdXμ dζ/parenrightbigg −∂λgμνdXμ dζdXν dζ=0 (8) To simplify (8) we exploit our freedom in choosing ζand set dζ=ds, so that K=1. We have 2gμλd2Xμ ds2+2∂σgμλdXσ dsdXμ ds−∂λgμνdXμ dsdXν ds=0 6We see that tracelessness is related to the fact that the electromagnetic field has no mass scale. Pure electromagnetism is said to be scale or dilatation invariant. For more on dilatation invariance see S. Coleman, Aspects of Symmetry ,p .6 7 . 7We put “length” in quotes because if gμνhad a Euclidean signature then/integraltext dswould indeed be the length and minimizing/integraltext dswould give the shortest path (the geodesic) between the endpoints, but here gμνhas a Minkowskian signature. I.11. Field Theory in Spacetime | 85 Xμ(ζ) Figure I.11.1 which upon multiplication by gρλbecomes d2Xρ ds2+1 2gρλ(2∂νgμλ−∂λgμν)dXμ dsdXν ds=0 that is, d2Xρ ds2+/Gamma1ρ μν[X(s) ]dXμ dsdXν ds=0 (9) if we define the Riemann-Christoffel symbol by /Gamma1ρ μν≡1 2gρλ(∂μgνλ+∂νgμλ−∂λgμν) (10) Given the initial position Xμ(s0)and velocity (dXμ/ds)(s 0)we have four second order differential equations (9) determining the geodesic followed by the particle in curved spacetime. Note that, contrary to the impressiongiven by some texts, unlike (8), (9) is not reparametrization invariant. To recover Newtonian gravity, three conditions must be met: (1) the particle moves slowly dX i/ds/lessmuchdX0/ds; (2) the gravitational field is weak, so that the metric is almost Minkowskian gμν/similarequalημν+hμν; and (3) the gravitational field does not depend on time. Condition (1) means that d2Xρ/ds2+/Gamma1ρ 00(dX0/ds)2/similarequal0, while (2) and (3) imply that /Gamma1ρ 00/similarequal−1 2ηρλ∂λh00. Thus, (9) reduces to d2X0/ds2/similarequal0 (which implies that dX0/ds is a constant) andd2Xi/ds2+1 2∂ih00(dX0/ds)2/similarequal0, which since X0is proportional to sbecomes d2Xi/dt2/similarequal−1 2∂ih00. Thus, we obtain Newton’s equationd2/vectorX dt2/similarequal−/vector∇φif we define the gravitational potential φbyh00=2φ: g00/similarequal1+2φ (11) Referring to the Schwarzschild metric, we see that far from a massive body, φ=−GM/r , as we expect. (Note also that this derivation depends neither on hijnor on h0j, as long as they are time independent.) Thus, the action of a point particle is S=−m/integraldisplay/radicalBig gμνdXμdXν=−m/integraldisplay/radicalBigg gμν[X(ζ) ]dXμ dζdXν dζdζ (12) Themfollows from dimensional analysis. A slick way of deriving S(which also allows us to see the minus sign) is to start with the nonrelativistic action of a particle in a gravitational potential φ, namely S=/integraltext Ldt=/integraltext (1 2mv2−m−mφ)dt . Note that the rest mass 86 | I. Motivation and Foundation mcomes in with a minus sign as it is part of the potential energy in nonrelativistic physics. Now force Sinto a relativistic form: For small vandφ, S=−m/integraldisplay (1−1 2v2+φ)dt/similarequal−m/integraldisplay/radicalbig 1−v2+2φdt =−m/integraldisplay/radicalbig (1+2φ)(dt)2−(d/vectorx)2 We see8that the 2 in (11) comes from the square root in the Lorentz-Fitzgerald quantity√ 1−v2. Now that we have Swe can calculate the stress energy of a point particle using (4): Tμν(x)=m√−g/integraldisplay dζK−1 2δ(4)[x−X(ζ) ]dXμ dζdXν dζ Setting ζtos(which we call the proper time τin this context) we have Tμν(x)=m√−g/integraldisplay dτδ(4)[x−X(τ) ]dXμ dτdXν dτ In particular, as we expect the 4-momentum of the particle is given by Pν=/integraldisplay d3x√−gT0ν=m/integraldisplay dτδ [x0−X0(τ)]dX0 dτdXν dτ=mdXν dτ The action (12) given here has two defects: (1) it is difficult to deal with a path integral /integraltext DXe−im/integraltext dζ√ (dXμ/dζ)(dX μ/dζ)involving a square root, and (2) Sdoes not make sense for a massless particle. To remedy these defects, note that classically, Sis equivalent to Simp=−1 2/integraldisplay dζ/parenleftbigg1 γdXμ dζdXμ dζ+γm2/parenrightbigg (13) where (dXμ/dζ)(dXμ/dζ)=gμν(X)(dXμ/dζ)(dXν/dζ). Varying with respect to γ(ζ) we obtain m2γ2= (dXμ/dζ)(dXμ/dζ) . Eliminating γinSimp we recover S. The path integral/integraltext DXeiSimphas a standard quadratic form.9Quantum mechanics of a relativistic point parti- cle is best formulated in terms of Simp, notS. Furthermore, for m=0,Simp=−1 2/integraltext dζ[γ−1(dXμ/dζ)(dX μ/dζ)] makes perfect sense. Note that varying with respect to γnow gives the well-known fact that for a massless particle gμν(X)dXμdXν=0. The action Simp will provide the starting point for our discussion on string theory in chapter VIII.5. Exercises I.11.1 Integrate by parts to obtain for the scalar field action S=−/integraldisplay d4x√−g1 2ϕ(1√−g∂μ√−ggμν∂ν+m2)ϕ 8We should not conclude from this that gij=δij.The point is that to leading order in v/c, our particle is sensitive only to g00, as we have just shown. Indeed, restoring cin the Schwarzschild metric we have ds2=(1−2GM c2r)c2dt2−(1−2GM c2r)−1dr2−r2dθ2−r2sin2θdφ2 →c2dt2−d/vectorx2−2GM rdt2+O(1/c2) 9One technical problem, which we will address in chapter III.4, is that in the integral over X(ζ) apparently different functions X(ζ) may in fact be the same physically, related by a reparametrization. I.11. Field Theory in Spacetime | 87 and write the equation of motion for ϕin curved spacetime. Discuss the propagator of the scalar field D(x ,y)(which is of course no longer translation invariant, i.e., it is no longer a function of x−y). I.11.2 Use (4) to find Efor a scalar field theory in flat spacetime. Show that the result agrees with what you would obtain using the canonical formalism of chapter I.8. I.11.3 Show that in flat spacetime Pμas derived here from the stress energy tensor Tμνwhen interpreted as an operator in the canonical formalism satisfies [ Pμ,ϕ(x) ]=−i∂μϕ(x) , and thus does exactly what you expect the energy and momentum operators to do, namely to be conjugate to time and space and hencerepresented by −i∂ μ. I.11.4 Show that for the Maxwell field Tij=−(EiEj+BiBj)+1 2δij(/vectorE2+/vectorB2)and hence T=0. I.12 Field Theory Redux What have you learned so far? Now that we have reached the end of part I, let us take stock of what you have learned. Quantum field theory is not that difficult; it just consists of doing one great big integral Z(J)=/integraldisplay Dϕ ei/integraltext dD+1x[1 2(∂ϕ)2−1 2m2ϕ2−λϕ4+Jϕ ](1) By repeatedly functionally differentiating Z(J) and then setting J=0 we obtain /integraldisplay Dϕ ϕ(x 1)ϕ(x 2)...ϕ(xn)ei/integraltext dD+1x[1 2(∂ϕ)2−1 2m2ϕ2−λϕ4](2) which tells us about the amplitude for nparticles associated with the field ϕto come into and go out of existence at the spacetime points x1,x2,...,xn, interacting with each other in between. Birth and death, with some kind of life in between. Ah, if we could only do the integral in (1)! But we can’t. So one way of going about it is to evaluate the integral as a series in λ: ∞/summationdisplay k=0(−iλ)k k!/integraldisplay Dϕ ϕ(x 1)ϕ(x 2)...ϕ(xn)[/integraldisplay dD+1yϕ(y)4]kei/integraltext dD+1x[1 2(∂ϕ)2−1 2m2ϕ2] (3) To keep track of the terms in the series we draw little diagrams. Quantum field theorists try to dream up ways to evaluate (1), and failing that, they invent tricks and methods for extracting the physics they are interested in, by hook and by crook,without actually evaluating (1). To see that quantum field theory is a straightforward generalization of quantum me- chanics, look at how (1) reduces appropriately. We have written the theory in (D+1)- dimensional spacetime, that is, Dspatial dimensions and 1 temporal dimension. Consider (1) in(0+1)-dimensional spacetime, that is, no space; it becomes Z(J)=/integraldisplay Dϕ ei/integraltext dt[1 2(dϕ dt)2−1 2m2ϕ2−λϕ4+Jϕ ](4) I.12. Field Theory Redux | 89 where we now denote the spacetime coordinate xjust by time t. We recognize this as the quantum mechanics of an anharmonic oscillator with the position of the mass point tiedto the spring denoted by ϕand with an external force Jpushing on the oscillator. In the quantum field theory (1), each term in the action makes physical sense: The first two terms generalize the harmonic oscillator to include spatial variations, the third termthe anharmonicity, and the last term an external probe. You can think of a quantum fieldtheory as an infinite collection of anharmonic oscillators, one at each point in space. We have here a scalar field ϕ. In previous and future chapters, the notion of field was and will be generalized ever so slightly: The field can transform according to a nontrivialrepresentation of the Lorentz group. We have already encountered fields transforming asa vector and a tensor and will presently encounter a field transforming as a spinor. Lorentzinvariance and whatever other symmetries we have constrain the form of the action. Theintegral will look more complicated but the approach is exactly as outlined here. That’s just about all there is to quantum field theory. This page intentionally left blank Part II Dirac and the Spinor This page intentionally left blank II.1 The Dirac Equation Staring into a fire According to a physics legend, apparently even true, Dirac was staring into a fire one evening in 1928 when he realized that he wanted, for reasons that are no longer relevant,a relativistic wave equation linear in spacetime derivatives ∂ μ≡∂/∂xμ. At that time, the Klein-Gordon equation (∂2+m2)ϕ=0, which describes a free particle of mass mand quadratic in spacetime derivatives, was already well known. This is in fact the equation ofmotion of the scalar field theory we studied earlier. At first sight, what Dirac wanted does not make sense. The equation is supposed to have the form “some linear combination of ∂ μacting on some field ψis equal to some constant times the field.” Denote the linear combination by cμ∂μ. If the cμ’s are four ordinary numbers, then the four-vector cμdefines some direction and the equation cannot be Lorentz invariant. Nevertheless, let us follow Dirac and write, using modern notation, (iγμ∂μ−m)ψ=0 (1) At this point, the four quantities iγμare just the coefficients of ∂μandmis just a constant. We have already argued that γμcannot simply be four numbers. Well, let us see what these objects have to be in order for this equation to contain the correct physics. Acting on (1) with (iγμ∂μ+m), Dirac obtained −(γμγν∂μ∂ν+m2)ψ=0. It is tradi- tional to define, in addition to the commutator [ A,B]=AB−BA familiar from quan- tum mechanics, the anticommutator {A,B}=AB +BA. Since derivatives commute, γμγν∂μ∂ν=1 2{γμ,γν}∂μ∂ν, and we have (1 2{γμ,γν}∂μ∂ν+m2)ψ=0. In a moment of inspiration Dirac realized that if {γμ,γν}=2ημν(2) withημνthe Minkowski metric he would obtain (∂2+m2)ψ=0, which describes a particle of mass m, and thus (1) would also describe a particle of mass m. 94 | II. Dirac and the Spinor Since ημνis a diagonal matrix with diagonal elements η00=1 and ηjj=− 1, (2) says that(γ0)2=1,(γj)2=− 1, and γμγν=−γνγμforμ/negationslash=ν. This last statement, that the coefficients γμanticommute with each other, implies that they indeed cannot be ordinary numbers. Dirac’s thought would make sense if we could find four such objects. Clifford algebra A set of objects γμ(clearly dof them in d-dimensional spacetime) satisfying the relation (2) is said to form a Clifford algebra. I will develop the mathematics of Clifford algebralater. Suffice it for you to check here that the following 4 by 4 matrices satisfy (2): γ0=/parenleftBiggI 0 0−I/parenrightBigg =I⊗τ3 (3) γi=/parenleftBigg0 σi −σi0/parenrightBigg =σi⊗iτ2 (4) Hereσandτdenote the standard Pauli matrices. For historical reasons the four matrices γμare known as gamma matrices—not a very imaginative name! (Our convention is such that whether an index on a Pauli matrix is upper or lower has no significance. On the otherhand, we define γ μ≡ημνγνand it does matter whether the index on a gamma matrix is upper or lower; it is to be treated just like the index on any Lorentz vector. This conventionis useful because then γ μ∂μ=γμ∂μ.) The direct product notation is convenient for computation: For example, γiγj=(σi⊗ iτ2)(σj⊗iτ2)=(σiσj⊗i2τ2τ2)=−(σiσj⊗I)and thus {γi,γj}=− { σi,σj}⊗I= −2δijas desired. You can convince yourself that the γμ’s cannot be smaller than 4 by 4 matrices. The mathematics forces the Dirac spinor ψto have 4 components! The physical content of the Dirac equation (1) is most transparent if we transform to momentum space: we plugψ(x)=/integraltext [d 4p/(2π)4]e−ipxψ(p) into (1) and obtain (γμpμ−m)ψ(p) =0 (5) Since (5) is Lorentz invariant, as we will show below, we can examine its physical content in any frame, in particular the rest frame pμ=(m,/vector0), in which it becomes (γ0−1)ψ=0 (6) As(γ0−1)2=− 2(γ0−1)we recognize (γ0−1)as a projection operator up to a trivial nor- malization. Indeed, using the explicit form in (3), we see that there is nothing mysterious to Dirac’s equation: When written out, (6) reads /parenleftBigg00 0I/parenrightBigg ψ=0 thus telling us that 2 of the 4 components in ψare zero. II.1. The Dirac Equation | 95 This makes perfect sense since we know that the electron has 2 physical degrees of freedom, not 4. Viewed in this light, the mysterious Dirac equation is no more and no lessthan a projection that gets rid of the unwanted degrees of freedom. Compare our discussionof the equation of motion of a massive spin 1 particle (chapter I.5). There also, 1 of the 4components of A μis projected out. Indeed, the Klein-Gordon equation (∂2+m2)ϕ(x) =0 just projects out those Fourier components ϕ(k) not satisfying the mass shell condition k2=m2. Our discussion provides a unified view of the equations of motion in relativistic physics: They just project out the unphysical components. A convenient notation introduced by Feynman, /negationslasha≡γμaμfor any 4-vector aμ, is now standard. The Dirac equation then reads (i/negationslash∂−m)ψ=0. Cousins of the gamma matrices Under a Lorentz transformation x/primeν=/Lambda1ν μxμ, the 4 components of the vector field Aμ transform like, well, a vector. How do the 4 components of ψtransform? Surely not in the same way as Aμsince even under rotation ψandAμtransform quite differently: one as spin1 2and the other as spin 1. Let us write ψ(x)→ψ/prime(x/prime)≡S(/Lambda1)ψ(x) and try to determine the 4 by 4 matrix S(/Lambda1) . It is a good idea to first sort out (and name) the 16 linearly independent 4 by 4 matrices. We already know five of them: the identity matrix and the γμ’s. The strategy is simply to multiply the γμ’s together, thus generating more 4 by 4 matrices until we get all 16. Since the square of a gamma matrix γμis equal to ±1 and the γμ’s anticommute with each other, we have to consider only γμγν,γμγνγλ, andγμγνγλγρwithμ,ν,λ, andρall different from one another. Thus, the only product of four gamma matrices that we haveto consider is γ5≡iγ0γ1γ2γ3(7) This combination is so important that it has its own name! (The peculiar name comes about because in some old-fashioned notation the time coordinate was called x4with a corresponding γ4.)We have γ5=i(I⊗τ3)(σ1⊗iτ2)(σ2⊗iτ2)(σ3⊗iτ2)=i4(I⊗τ3)(σ1σ2σ3⊗τ2) and so γ5=I⊗τ1=/parenleftBigg0I I 0/parenrightBigg (8) With the factor of iincluded, γ5is manifestly hermitean. An important property is that γ5 anticommutes with the γμ’s: {γ5,γμ}=0 (9) 96 | II. Dirac and the Spinor Continuing, we see that the products of three gamma matrices, all different, can be written as γμγ5(e.g.,γ1γ2γ3=−iγ0γ5). Finally, using (2) we can write the product of two gamma matrices as γμγν=ημν−iσμν, where σμν≡i 2[γμ,γν] (10) There are 4 .3/2=6 of these σμνmatrices. Count them, we got all 16. The set of 16 matrices {1, γμ,σμν,γμγ5,γ5}forms a complete basis of the space of all 4 by 4 matrices, that is, any 4 by 4 matrix can be writtenas a linear combination of these 16 matrices. It is instructive to write out σ μνexplicitly in the representation (3) and (4): σ0i=i/parenleftBigg0σi σi0/parenrightBigg (11) σij=εijk/parenleftBiggσk0 0σk/parenrightBigg (12) We see that σijare just the Pauli matrices doubly stacked, for example, σ12=/parenleftBiggσ30 0σ3/parenrightBigg Lorentz transformation Recall from a course on quantum mechanics that a general rotation can be written as ei/vectorθ/vectorJwith/vectorJthe 3 generators of rotation and /vectorθ3 rotation parameters. Recall also that the Lorentz group contains boosts in addition to rotations, with /vectorKdenoting the 3 generators of boosts. Recall from a course on electromagnetism that the 6 generators {/vectorJ,/vectorK}transform under the Lorentz group as the components of an antisymmetric tensor just like theelectromagnetic field F μνand thus can be denoted by Jμν. I will discuss these matters in more detail in chapter II.3. For the moment, suffice it to note that with this notation we can write a Lorentz transformation as /Lambda1=e−i 2ωμνJμν, with Jijgenerating rotations, J0igenerating boosts, and the antisymmetric tensor ωμν=−ωνμwith its 6 =4.3/2 components corresponding to the 3 rotation and 3 boost parameters. Given the preceding discussion and the fact that there are six matrices σμν, we suspect that up to an overall numerical factor the σμν’s must represent the 6 generators Jμνof the Lorentz group acting on a spinor. In fact, our suspicion is confirmed by thinking about what a rotation e−i 2ωijJijdoes. Referring to (12) we see that if Jijis represented by1 2σij this would correspond exactly to how a spin1 2particle transforms in quantum mechanics. More precisely, separate the 4 components of the Dirac spinor into 2 sets of 2 components: ψ=/parenleftBiggφ χ/parenrightBigg (13) II.1. The Dirac Equation | 97 From (12) we see that under a rotation around the 3rd axis, φ→e−iω 121 2σ3φandχ→ e−iω 121 2σ3χ. It is gratifying to see that φandχtransform like 2-component Pauli spinors. We have thus figured out that a Lorentz transformation /Lambda1acting on ψis represented byS(/Lambda1)=e−(i/4 )ωμνσμν, and so, acting on ψ, the generators Jμνare indeed represented by1 2σμν. Therefore we would expect that if ψ(x) satisfies the Dirac equation (1) then ψ/prime(x/prime)≡S(/Lambda1)ψ(x) would satisfy the Dirac equation in the primed frame, (iγμ∂/prime μ−m)ψ/prime(x/prime)=0 (14) where ∂/prime μ≡∂/∂x/primeμ. To show this, calculate [ σμν,γλ]=2i(γμηνλ−γνημλ)and hence forωinfinitesimal SγλS−1=γλ−(i/4)ωμν[σμν,γλ]=γλ+γμωλ μ. Building up a finite Lorentz transformation by compounding infinitesimal transformations (just as in thestandard discussion of the rotation group in quantum mechanics), we have Sγ λS−1= /Lambda1λ μγμ. Dirac bilinears The Clifford algebra tells us that (γ0)2=+ 1 and (γi)2=− 1; hence the necessity for the i in (4). One consequence of the iis that γ0is hermitean while γiis antihermitean, a fact conveniently expressed as (γμ)†=γ0γμγ0(15) Thus, contrary to what you might think, the bilinear ψ†γμψis not hermitean; rather, ¯ψγμψis hermitean with ¯ψ≡ψ†γ0. The necessity for introducing ¯ψin addition to ψ†in relativistic physics is traced back to the (+,−,−,−)signature of the Minkowski metric. It follows that (σμν)†=γ0σμνγ0. Hence, S(/Lambda1)†=γ0e(i/4)ωμνσμνγ0, (which incidentally, clearly shows that Sis not unitary, a fact we knew since σ0iis not hermitean), and so ¯ψ/prime(x/prime)=ψ(x)†S(/Lambda1)†γ0=¯ψ(x)e+(i/4 )ωμνσμν. (16) We have ¯ψ/prime(x/prime)ψ/prime(x/prime)=¯ψ(x)e+(i/4 )ωμνσμνe−(i/4 )ωμνσμνψ(x)=¯ψ(x)ψ(x) You are probably used to writing ψ†ψin nonrelativistic physics. In relativistic physics you have to get used to writing ¯ψψ .I ti s ¯ψψ , notψ†ψ, that transforms as a Lorentz scalar. There are obviously 16 Dirac bilinears ¯ψ/Gamma1ψ that we can form, corresponding to the 16 linearly independent /Gamma1’s. You can now work out how various fermion bilinears transform (exercise II.1.1). The notation is rather nice: Various objects transform the way it looks like they should transform. We simply look at the Lorentz indices they carry. Thus, ¯ψ(x)γμψ(x) transforms as a Lorentz vector. 98 | II. Dirac and the Spinor Parity An important discrete symmetry in physics is that of parity or reflection in a mirror1 xμ→x/primeμ=(x0,−/vectorx) (17) Multiply the Dirac equation (1) by γ0:γ0(iγμ∂μ−m)ψ(x) =0=(iγμ∂/prime μ−m)γ0ψ(x) , where ∂/prime μ≡∂/∂x/primeμ. Thus, ψ/prime(x/prime)≡ηγ0ψ(x) (18) satisfies the Dirac equation in the space-reflected world (where ηis an arbitrary phase that we can set to 1). Note, for example, ¯ψ/prime(x/prime)ψ/prime(x/prime)=¯ψ(x)ψ(x) but¯ψ/prime(x/prime)γ5ψ/prime(x/prime)=¯ψ(x)γ0γ5γ0ψ(x)= −¯ψ(x)γ5ψ(x) . Under a Lorentz transformation ¯ψ(x)γ5ψ(x) and¯ψ(x)ψ(x) transform in the same way but under space reflection they transform in an opposite way; in other words,while ¯ψ(x)ψ(x) transforms as a scalar, ¯ψ(x)γ 5ψ(x) transforms as a pseudoscalar. You are now ready to do the all-important exercises in this chapter. The Dirac Lagrangian An interesting question: What Lagrangian would give Dirac’s equation? The answer is L=¯ψ(i/negationslash∂−m)ψ (19) Since ψis complex we can vary ψand¯ψindependently to obtain the Euler-Lagrange equation of motion. Thus, ∂μ(δL/δ∂μψ)−δL/δψ=0 gives ∂μ(i¯ψγμ)+m¯ψ=0, which upon hermitean conjugation and multiplication by γ0gives the Dirac equation (1). The other variational equation ∂μ(δL/δ∂μ¯ψ)−δL/δ¯ψ=0 gives the Dirac equation even more directly. (If you are disturbed by the asymmetric treatment of ψand¯ψ, you can always integrate by parts in the action, have ∂μact on ¯ψin the Lagrangian, and then average the two forms of the Lagrangian. The action S=/integraltext d4xLtreats ψand¯ψsymmetrically.) Slow and fast electrons Given a set of gamma matrices it is straightforward to solve the Dirac equation (/negationslashp−m)ψ(p) =0 (20) forψ(p) : It is a simple matrix equation (see exercise II.1.3). 1Rotations consist of all linear transformations xi→Rijxjsuch that det R=+ 1. Those transformations with detR=− 1 are composed of parity followed by a rotation. In (3+1)-dimensional spacetime, parity can be defined as reversing one of the spatial coordinates or all three spatial coordinates. The two operations are related by arotation. Note that in odd dimensional spacetime, parity is not the same as space inversion, in which all spatial coordinates are reversed (see exercise II.1.12). II.1. The Dirac Equation | 99 Note that if somebody uses the gamma matrices γμ, you are free to use instead γ/primeμ= W−1γμWwithWany 4 by 4 matrix with an inverse. Obviously, γ/primeμalso satisfy the Clifford algebra. This freedom of choice corresponds to a simple change of basis. Physics cannotdepend on the choice of basis, but which basis is the most convenient depends on thephysics. For example, suppose we want to study a slowly moving electron. Let us use the basis defined by (3) and (4), and the 2-component decomposition of ψ(13). Since (6) tells us thatχ(p)=0 for an electron at rest, we expect χ(p) to be much smaller than φ(p) for a slowly moving electron. In contrast, for momentum much larger than the mass, we can approximate (20) by /negationslashpψ(p) =0. Multiplying on the left by γ 5, we see that if ψ(p) is a solution then γ5ψ(p) is also a solution since γ5anticommutes with γμ. Since (γ5)2=1, we can form two projection operators PL≡1 2(1−γ5)andPR≡1 2(1+γ5), satisfying P2 L=PL,P2 R=PR, andPLPR=0. It is extremely useful to introduce the two combinations ψL=1 2(1−γ5)ψ andψR=1 2(1+γ5)ψ. Note that γ5ψL=−ψLandγ5ψR=+ψR. Physically, a relativistic electron has two degrees of freedom known as helicities: it can spin either clockwise oranticlockwise around the direction of motion. I leave it to you as an exercise to show thatψ LandψRcorrespond to precisely these two possibilities. The subscripts LandRindicate left and right handed. Thus, for fast moving electrons, a basis known as the Weyl basis,designed so that γ 5, rather than γ0, is diagonal, is more convenient. Instead of (3), we choose γ0=/parenleftBigg0I I 0/parenrightBigg =I⊗τ1 (21) We keep γias in (4). This defines the Weyl basis. We now calculate γ5≡iγ0γ1γ2γ3=i(I⊗τ1)(σ1σ2σ3⊗i3τ2)=−(I⊗τ3)=/parenleftBigg−I 0 0I/parenrightBigg (22) which is indeed diagonal as desired. The decomposition into left and right handed fields is of course defined regardless of what basis we feel like using, but in the Weyl basis we havethe nice feature that ψ Lhas two upper components and ψRhas two lower components. The spinors ψLandψRare known as Weyl spinors. Note that in going from the Dirac to the Weyl basis γ0andγ5trade places (up to a sign): Dirac : γ0diagonal; Weyl : γ5diagonal. (23) Physics dictates which basis to use: We prefer to have γ0diagonal when we deal with slowly moving spin1 2particles, while we prefer to have γ5diagonal when we deal with fast moving spin1 2particles . I note in passing that if we define σμ≡(I,/vectorσ)and¯σμ≡(I,−/vectorσ)we can write γμ=/parenleftBigg0σμ ¯σμ0/parenrightBigg more compactly in the Weyl basis. (We develop this further in appendix E.) 100 | II. Dirac and the Spinor Chirality or handedness Regardless of whether a Dirac field ψ(x) is massive or massless, it is enormously useful to decompose ψinto left and right handed fields ψ(x)=ψL(x)+ψR(x)≡1 2(1−γ5)ψ(x) + 1 2(1+γ5)ψ(x). As an exercise, show that you can write the Dirac Lagrangian as L=¯ψ(i/negationslash∂−m)ψ=¯ψLi/negationslash∂ψL+¯ψRi/negationslash∂ψR−m(¯ψLψR+¯ψRψL) (24) The kinetic energy connects left to left and right to right, while the mass term connects left to right and right to left. The transformation ψ→eiθψleaves the Lagrangian Linvariant. Applying Noether’s theorem, we obtain the conserved current associated with this symmetry Jμ=¯ψγμψ. Projecting into left and right handed fields we see that they transform the same way:ψ L→eiθψLandψR→eiθψR. Ifm=0,Lenjoys an additional symmetry, known as a chiral symmetry, under which ψ→eiφγ5ψ. Noether’s theorem tells us that the axial current J5μ≡¯ψγμγ5ψis conserved. The left and right handed fields transform in opposite ways: ψL→e−iφψLandψR→ eiφψR. These points are particularly obvious when Lis written in terms of ψLandψR,a s in (24). In 1956 Lee and Yang proposed that the weak interaction does not preserve parity. It was eventually realized (with these four words I brush over a beautiful chapter in the historyof particle physics; I urge you to read about it!) that the weak interaction Lagrangian hasthe generic form L=G¯ψ1Lγμψ2L¯ψ3Lγμψ4L (25) where ψ1, 2, 3, 4 denotes four Dirac fields and Gthe Fermi coupling constant. This La- grangian clearly violates parity: Under a spatial reflection, left handed fields are trans-formed into right handed fields and vice versa. Incidentally, henceforth when I say a Lagrangian has a certain form, I will usually indicate only one or more of the relevant terms in the Lagrangian, as in (25). The otherterms in the Lagrangian, such as ¯ψ 1(i/negationslash∂−m1)ψ1, are understood. If the term is not hermitean, then it is understood that we also add its hermitean conjugate. Interactions As we saw in (25) given the classification of bilinears in the spinor field you worked outin an exercise it is easy to introduce interactions. As another example, we can couplea scalar field ϕto the Dirac field by adding the term gϕ¯ψψ (with gsome coupling constant) to the Lagrangian L=¯ψ(i/negationslash∂−m)ψ (and of course also adding the Lagrangian forϕ). Similarly, we can couple a vector field A μby adding the term eAμ¯ψγμψ.W e note that in this case we can introduce the covariant derivative Dμ=∂μ−ieAμand write II.1. The Dirac Equation | 101 L=¯ψ(i/negationslash∂−m)ψ+eAμ¯ψγμψ=¯ψ(iγμDμ−m)ψ . Thus, the Lagrangian for a Dirac field interacting with a vector field of mass μreads L=¯ψ(iγμDμ−m)ψ−1 4FμνFμν−1 2μ2AμAμ(26) If the mass μvanishes, this is the Lagrangian for quantum electrodynamics. Varying with respect to ¯ψ, we obtain the Dirac equation in the presence of an electromagnetic field: [iγμ(∂μ−ieAμ)−m]ψ=0 (27) Charge conjugation and antimatter With coupling to the electromagnetic field, we have the concept of charge and hence of charge conjugation. Let us try to flip the charge e. Take the complex conjugate of (27): [−iγμ∗(∂μ+ieAμ)−m]ψ∗=0. Complex conjugating (2) we see that the −γμ∗also satisfy the Clifford algebra and thus must be the γμmatrices expressed in a different basis, that is, there exists a matrix Cγ0(the notation with an explicit factor of γ0is standard; see below) such that −γμ∗=(Cγ0)−1γμ(Cγ0). Plugging in, we find that [iγμ(∂μ+ieAμ)−m]ψc=0 (28) where we have defined ψc≡Cγ0ψ∗. Thus, if ψis the field of the electron, then ψcis the field of a particle with a charge opposite to that of the electron but with the same mass,namely the positron. The discovery of antimatter was one of the most momentous in twentieth-century physics. We will discuss antimatter in more detail in the next chapter. It may be instructive to look at the specific form of the charge conjugation matrix C.W e can write the defining equation for CasCγ 0γμ∗γ0C−1=−γμ. Complex conjugating the equation (γμ)†=γ0γμγ0, we obtain (γμ)T=γ0γμ∗γ0ifγ0is real. Thus, (γμ)T=−C−1γμC (29) which explains why Cis defined with a γ0attached. In both the Dirac and the Weyl bases γ2is the only imaginary gamma matrix. Then the defining equation for Cjust says that Cγ0commutes with γ2but anticommutes with the other three γmatrices. So evidently C=γ2γ0[up to an arbitrary phase not fixed by (29)] and indeed γ2γμ∗γ2=γμ. Note that we have the simple (and satisfying) relation ψc=γ2ψ∗(30) You can easily convince yourself (exercise II.1.9) that the charge conjugate of a left handed field is right handed and vice versa. As we will see later, this fact turns out to be crucial in the construction of grand unified theory. Experimentally, it is known that theneutrino is left handed. Thus, we can now predict that the antineutrino is right handed. 102 | II. Dirac and the Spinor Furthermore, ψctransforms as a spinor. Let’s check: Under a Lorentz transformation ψ→e−(i/4 )ωμνσμνψ, complex conjugating we have ψ∗→e+(i/4 )ωμν(σμν)∗ψ∗; hence ψc→ γ2e+(i/4 )ωμν(σμν)∗ψ∗=e−(i/4 )ωμνσμνψc. [Recall from (10) that σμνis defined with an explicit i.] Note that CT=γ0γ2=−Cin both the Dirac and the Weyl bases. Majorana neutrino Since ψctransforms as a spinor, Majorana2noted that Lorentz invariance allows not only the Dirac equation i/negationslash∂ψ=mψ but also the Majorana equation i/negationslash∂ψ=mψc (31) Complex conjugating this equation and multiplying by γ2, we have −γ2iγμ∗∂μψ∗= γ2m(−γ2)ψ, that is, i/negationslash∂ψc=mψ . Thus, −∂2ψ=i/negationslash∂(i/negationslash∂ψ)=i/negationslash∂mψc=m2ψ. As we antic- ipated, mis indeed the mass, known as a Majorana mass, of the particle associated with ψ. The Majorana equation (31) can be obtained from the Lagrangian3 L=¯ψi/negationslash∂ψ−1 2m(ψTCψ+¯ψC¯ψT) (32) upon varying ¯ψ. Sinceψandψccarry opposite charge, the Majorana equation, unlike the Dirac equation, can only be applied to electrically neutral fields. However, as ψcis right handed if ψis left handed, the Majorana equation, again unlike the Dirac equation, preserves handedness.Thus, the Majorana equation is almost tailor made for the neutrino. From its conception the neutrino was assumed to be massless, but couple of years ago experimentalists established that it has a small but nonvanishing mass. As of this writing, itis not known whether the neutrino mass is Dirac or Majorana. We will see in chapter VII.7that a Majorana mass for the neutrino arises naturally in the SO( 10)grand unified theory. Finally, there is the possibility that ψ=ψ c, in which case ψis known as a Majorana spinor. Time reversal Finally, we come to time reversal,4which as you probably know, is much more confusing to discuss than parity and charge conjugation. In a famous 1932 paper Wigner showed that 2Ettore Majorana had a brilliant but tragically short career. In his early thirties, he disappeared off the coast of Sicily during a boat trip. The precise cause of his death remains a mystery. See F. Guerra and N. Robotti, Ettore Majorana: Aspects of His Scientific and Academic Activity . 3Upon recalling that Cis antisymmetric, you may have worried that ψTCψ=Cαβψαψβvanishes. In future chapters we will learn that ψhas to be treated as anticommuting “Grassmannian numbers.” 4Incidentally, I do not feel that we completely understand the implications of time-reversal invariance. See A. Zee, “Night thoughts on consciousness and time reversal,” in: Art and Symmetry in Experimental Physics : pp. 246–249 . II.1. The Dirac Equation | 103 time reversal is represented by an antiunitary operator. Since this peculiar feature already appears in nonrelativistic quantum physics, it is in some sense not the responsibility of abook on relativistic quantum field theory to explain time reversal as an antiunitary operator.Nevertheless, let me try to be as clear as possible. I adopt the approach of “letting thephysics, namely the equations, lead us.” Take the Schr ¨odinger equation i(∂/∂t)/Psi1(t) =H/Psi1(t) (and for definiteness, think of H=−(1/2m)∇ 2+V(/vectorx), just simple one particle nonrelativistic quantum mechanics.) We suppress the dependence of /Psi1on/vectorx. Consider the transformation t→t/prime=−t.W e want to find a /Psi1/prime(t/prime)such that i(∂/∂t/prime)/Psi1/prime(t/prime)=H/Psi1/prime(t/prime). Write /Psi1/prime(t/prime)=T /Psi1(t), where T is some operator to be determined (up to some arbitrary phase factor η). Plugging in, we havei[∂/∂(−t)]T /Psi1(t) =HT/Psi1(t) . Multiply by T−1, and we obtain T−1(−i)T (∂/∂t)/Psi1(t) = T−1HT/Psi1(t) . Since Hdoes not involve time in any way, we want T−1H=HT−1. Then T−1(−i)T (∂/∂t)/Psi1(t) =H/Psi1(t) . We are forced to conclude, as Wigner was, that T−1(−i)T =i (33) Speaking colloquially, we can say that in quantum physics time goes with an iand so flipping time means flipping ias well. LetT=UK , where Kcomplex conjugates everything to its right. Then T−1=KU−1 and (33) holds if U−1iU=i, that is, if U−1is just an ordinary (unitary) operator that does nothing to i. We will determine Uas we go along. The presence of Kmakes T“antiunitary.” We check that this works for a spinless particle in a plane wave state /Psi1(t)=ei(/vectork./vectorx−Et). Plugging in, we have /Psi1/prime(t/prime)=T /Psi1(t) =UK/Psi1(t) =U/Psi1∗(t)=Ue−i(/vectork./vectorx−Et); since /Psi1has only one component, Uis just a phase factor5ηthat we can choose to be 1. Rewriting, we have/Psi1/prime(t)=e−i(/vectork./vectorx+Et)=ei(−/vectork./vectorx−Et). Indeed, /Psi1/primedescribes a plane wave moving in the opposite direction. Crucially, /Psi1/prime(t)∝e−iEtand thus has positive energy as it should. Note that acting on a spinless particle T2=UKUK =UU∗K2=+ 1. Next consider a spin1 2nonrelativistic electron. Acting with Ton the spin-up state/parenleftBig1 0/parenrightBig we want to obtain the spin-down state/parenleftBig0 1/parenrightBig . Thus, we need a nontrivial matrix U=ησ2 to flip the spin: T/parenleftBigg1 0/parenrightBigg =U/parenleftBigg1 0/parenrightBigg =iη/parenleftBigg0 1/parenrightBigg Similarly, Tacting on the spin-down state produces the spin-up state. Note that acting on a spin1 2particle T2=ησ2Kησ 2K=ησ2η∗σ∗ 2KK=− 1 This is the origin of Kramer’s degeneracy: In a system with an odd number of electrons in an electric field, no matter how complicated, each energy level is twofold degenerate.The proof is very simple: Since the system is time reversal invariant, /Psi1andT/Psi1 have the same energy. Suppose they actually represent the same state. Then T/Psi1=e iα/Psi1, but then 5It is a phase factor rather than an arbitrary complex number because we require that |/Psi1/prime|2=|/Psi1|2. 104 | II. Dirac and the Spinor T2/Psi1=T( T/Psi1) =Teiα/Psi1=e−iαT/Psi1=/Psi1/negationslash=−/Psi1.S o/Psi1andT/Psi1 must represent two distinct states. All of this is beautiful stuff, which as I noted earlier you could and should have learned in a decent course on quantum mechanics. My responsibility here is to show you how itworks for the Dirac equation. Multiplying (1) by γ 0from the left, we have i(∂/∂t)ψ(t) = Hψ(t) withH=−iγ0γi∂i+γ0m. Once again, we want i(∂/∂t/prime)ψ/prime(t/prime)=Hψ/prime(t/prime)with ψ/prime(t/prime)=T ψ(t) andTsome operator to be determined. The discussion above carries over if T−1HT=H, that is, KU−1HUK =H. Thus, we require KU−1γ0UK=γ0and KU−1(iγ0γi)UK=iγ0γi. Multiplying by Kon the left and on the right, we see that we have to solve for a Usuch that U−1γ0U=γ0∗andU−1γiU=−γi∗. We now restrict ourselves to the Dirac and Weyl bases, in both of which γ2is the only imaginary guy. Okay, what flips γ1andγ3but not γ0andγ2? Well, U=ηγ1γ3(withηan arbitrary phase factor) works: ψ/prime(t/prime)=ηγ1γ3Kψ(t) (34) Since the γi’s are the same in both the Dirac and the Weyl bases, in either we have from (4) U=η(σ1⊗iτ2)(σ3⊗iτ2)=ηiσ2⊗1 As we expect, acting on the 2-component spinors contained in ψ, the time reversal operator Tinvolves multiplying by iσ2. Note also that as in the nonrelativistic case T2ψ=−ψ. It may not have escaped your notice that γ0appears in the parity operator (18), γ2in charge conjugation (30), and γ1γ3in time reversal (34). If we change a Dirac particle to its antiparticle and flip spacetime, γ5appears. CPT theorem There exists a profound theorem stating that any local Lorentz invariant field theory must be invariant under6CPT , the combined action of charge conjugation, parity, and time reversal. The pedestrian proof consists simply of checking that any Lorentz invariant localinteraction you can write down [such as (25)], while it may break charge conjugation,parity, or time reversal separately, respects CPT . The more fundamental proof involves considerable formal machinery that I will not develop here. You are urged to read aboutthe phenomenological study of charge conjugation, parity, time reversal, and CPT , surely one of the most fascinating chapters in the history of physics. 7 6A rather pedantic point, but potentially confusing to some students, is that I distinguish carefully between the action of charge conjugation Cand the matrix C: Charge conjugation Cinvolves taking the complex conjugate of ψand then scrambling the components with Cγ0. Similarly, I distinguish between the operation of time reversal Tand the matrix T. 7See, e.g., J. J. Sakurai, Invariance Principles and Elementary Particles and E. D. Commins, Weak Interactions . II.1. The Dirac Equation | 105 Two stories I end this chapter with two of my favorite physics stories—one short and one long. Paul Dirac was notoriously a man of few words. Dick Feynman told the story that when he first met Dirac at a conference, Dirac said after a long silence, “I have an equation; doyou have one too?” Enrico Fermi did not usually take notes, but during the 1948 Pocono conference (see chapter I.7) he took voluminous notes during Julian Schwinger’s lecture. When he gotback to Chicago, he assembled a group consisting of two professors, Edward Teller andGregory Wentzel, and four graduate students, Geoff Chew, Murph Goldberger, MarshallRosenbluth, and Chen-Ning Yang (all to become major figures later). The group met inFermi’s office several times a week, a couple of hours each time, to try to figure out whatSchwinger had done. After 6 weeks, everyone was exhausted. Then someone asked, “Didn’tFeynman also speak?” The three professors, who had attended the conference, said yes.But when pressed, not Fermi, nor Teller, nor Wentzel could recall what Feynman had said.All they remembered was his strange notation: pwith a funny slash through it. 8 Exercises II.1.1 Show that the following bilinears in the spinor field ¯ψψ ,¯ψγμψ,¯ψσμνψ,¯ψγμγ5ψ, and ¯ψγ5ψtrans- form under the Lorentz group and parity as a scalar, a vector, a tensor, a pseudovector or axial vector, and a pseudoscalar, respectively. [Hint: For example, ¯ψγμγ5ψ→¯ψ[1+(i/4)ωσ ]γμγ5[1−(i/4)ωσ ]ψunder an infinitesimal Lorentz transformation and →¯ψγ0γμγ5γ0ψunder parity. Work out these transforma- tion laws and show that they define an axial vector.] II.1.2 Write all the bilinears in the preceding exercise in terms of ψLandψR. II.1.3 Solve (/negationslashp−m)ψ(p) =0 explicitly (by rotational invariance it suffices to solve it for /vectorpalong the 3rd direction, say). Verify that indeed χis much smaller than φfor a slowly moving electron. What happens for a fast moving electron? II.1.4 Exploiting the fact that χis much smaller than φfor a slowly moving electron, find the approximate equation satisfied by φ. II.1.5 For a relativistic electron moving along the z-axis, perform a rotation around the z-axis. In other words, study the effect of e−(i/4 )ωσ12onψ(p) and verify the assertion in the text regarding ψLandψR. II.1.6 Solve the massless Dirac equation. II.1.7 Show explicitly that (25) violates parity. II.1.8 The defining equation for Cevidently fixes Conly up to an overall constant. Show that this constant is fixed by requiring (ψc)c=ψ. 8C. N. Yang, Lecture at the Schwinger Memorial Session of the American Physical Society meeting in Washington D. C., 1995. 106 | II. Dirac and the Spinor II.1.9 Show that the charge conjugate of a left handed field is right handed and vice versa. II.1.10 Show that ψCψ is a Lorentz scalar. II.1.11 Work out the Dirac equation in (1 +1)-dimensional spacetime. II.1.12 Work out the Dirac equation in (2 +1)-dimensional spacetime. Show that the apparently innocuous mass term violates parity and time reversal. [Hint: The three γμ’s are just the three Pauli matrices with appropriate factors of i.] II.2 Quantizing the Dirac Field Anticommutation We will use the canonical formalism of chapter I.8 to quantize the Dirac field. Long and careful study of atomic spectroscopy revealed that the wave function of two electrons had to be antisymmetric upon exchange of their quantum numbers. It followsthat we cannot put two electrons into the same energy level so that they will have thesame quantum numbers. In 1928 Jordan and Wigner showed how this requirement of anantisymmetric wave function can be formalized by having the creation and annihilationoperators for electrons satisfy anticommutation rather than commutation relations as in(I.8.12). Let us start out with a state with no electron |0/angbracketrightand denote by b † αthe operator creating an electron with the quantum numbers α. In other words, the state b† α|0/angbracketrightis the state with an electron having the quantum numbers α. Now suppose we want to have another electron with the quantum numbers β, so we construct the state b† βb† α|0/angbracketright. For this to be antisymmetric upon interchanging αandβwe must have {b† α,b† β}≡b† αb† β+b† βb† α=0 (1) Upon hermitean conjugation, we have {bα,bβ}=0. In particular, b† αb† α=0, so that we cannot create two electrons with the same quantum numbers. To this anticommutation relation we add {bα,b† β}=δ αβ (2) One way of arguing for this is to say that we would like the number operator to be N=/summationtext αb† αbα, just as in the bosonic case. Show with one line of algebra that [ AB,C]= A[B,C]+[A,C]Bor [AB,C]=A{B ,C}−{A,C}B. (A heuristic way of remembering the minus sign in the anticommuting case is that we have to move CpastBin order for Cto do its anticommuting with A.)For the desired number operator to work we need [/summationtext αb† αbα,b† β]=+b† β(so that as usual N|0/angbracketright=0, and Nb† β|0/angbracketright=b† β|0/angbracketright)and so we have (2). 108 | II. Dirac and the Spinor The Dirac field Let us now turn to the free Dirac Lagrangian L=¯ψ(i/negationslash∂−m)ψ (3) The momentum conjugate to ψisπα=δL/δ∂tψα=iψ† α. We anticipate that the correct canonical procedure requires imposing the anticommutation relation: {ψα(/vectorx,t),ψ† β(/vector0,t)}=δ(3)(/vectorx)δαβ (4) We will derive this below. The Dirac field satisfies (i/negationslash∂−m)ψ=0 (5) Plugging in plane waves u(p ,s)e−ipxandv(p ,s)eipxforψ, we have (/negationslashp−m)u(p ,s)=0 (6) and (/negationslashp+m)v(p ,s)=0 (7) The index s=± 1 reminds us that each of these two equations has two solutions, spin up and spin down. Evidently, under a Lorentz transformation the two spinors uandv transform in the same way as ψ. Thus, if we define ¯u≡u†γ0and¯v≡v†γ0, then ¯uuand ¯vvare Lorentz scalars. This subject is full of “peculiar” signs and so I will proceed very carefully and show you how every sign makes sense. First, since (6) and (7) are linear we have to fix the normalization of uandv. Since ¯u(p ,s)u(p ,s)and¯v(p ,s)v(p ,s)are Lorentz scalars, the normalization condition we impose on them in the rest frame will hold in any frame. Our strategy is to do things in the rest frame using a particular basis and then invoke Lorentz invariance and basis independence. In the rest frame, (6) and (7) reduce to (γ0−1)u=0 and (γ0+1)v=0. In particular, in the Dirac basis γ0=/parenleftBigI 0 0−I/parenrightBig , so the two independent spinors u(labeled by spin s=± 1)have the form ⎛ ⎜⎜⎜⎜⎜⎝1 000⎞ ⎟⎟⎟⎟⎟⎠and⎛ ⎜⎜⎜⎜⎜⎝0 1 00⎞ ⎟⎟⎟⎟⎟⎠ II.2. Quantizing the Dirac Field | 109 while the two independent spinors vhave the form ⎛ ⎜⎜⎜⎜⎜⎝0 0 1 0⎞ ⎟⎟⎟⎟⎟⎠and⎛ ⎜⎜⎜⎜⎜⎝0 00 1⎞ ⎟⎟⎟⎟⎟⎠ The normalization conditions we have implicitly chosen are then ¯u(p ,s)u(p ,s)=1 and ¯v(p ,s)v(p ,s)=− 1. Note the minus sign thrust upon us. Clearly, we also have the orthog- onality condition ¯uv=0 and ¯vu=0. Lorentz invariance and basis independence then tell us that these four relations hold in general. Furthermore, in the rest frame /summationdisplay suα(p,s)¯uβ(p,s)=/parenleftBiggI 0 00/parenrightBigg αβ=1 2(γ0+1)αβ and /summationdisplay svα(p,s)¯vβ(p,s)=/parenleftBigg00 0−I/parenrightBigg αβ=1 2(γ0−1)αβ Thus, in general /summationdisplay suα(p,s)¯uβ(p,s)=/parenleftbigg/negationslashp+m 2m/parenrightbigg αβ(8) and /summationdisplay svα(p,s)¯vβ(p,s)=/parenleftbigg/negationslashp−m 2m/parenrightbigg αβ(9) Another way of deriving (8) is to note that the left hand side i sa4b y4 matrix (it is like a column vector multiplied by a row vector on the right) and so must be a linear combinationof the sixteen 4 by 4 matrices we listed in chapter II.1. Argue that γ 5andγμγ5are ruled out by parity and that σμνis ruled out by Lorentz invariance and the fact that only one Lorentz vector, namely pμ, is available. Hence the right hand side must be a linear combination of /negationslashpandm. Fix the relative coefficient by acting with /negationslashp−mfrom the left. The normalization is fixed by setting α=βand summing over α. Similarly for (9). In particular, setting α=β and summing over α, we recover ¯v(p ,s)v(p ,s)=− 1. We are now ready to promote ψ(x) to an operator. In analogy with (I.8.11) we expand the field in plane waves1 ψα(x)= /integraldisplayd3p (2π)3 2(Ep/m)1 2/summationdisplay s[b(p ,s)uα(p,s)e−ipx+d†(p,s)vα(p,s)eipx] (10) (Here Ep=p0=+/radicalbig /vectorp2+m2andpx=pμxμ.)The normalization factor (Ep/m)1 2is slightly different from that in (I.8.11) for reasons we will see. Otherwise, the rationale 1The notation is standard. See e.g., J. A. Bjorken and S. D. Drell, Relativistic Quantum Mechanics . 110 | II. Dirac and the Spinor for (10) is essentially the same as in (I.8.11). We integrate over momentum /vectorp, sum over spins, expand in plane waves, and give names to the coefficients in the expansion. Because ψis complex, we have, similar to the complex scalar field in chapter I.8, a boperator and ad†operator. Just as in chapter I.8, the operators bandd†must carry the same charge. Thus, if b annihilates an electron with charge e=− |e|,d†must remove charge e; that is, it creates a positron with charge −e=|e|. A word on notation: in (10) b(p ,s),d†(p,s),u(p ,s), andv(p ,s)are written as functions of the 4-momentum pbut strictly speaking they are functions of /vectorponly, with p0always understood to be +/radicalbig /vectorp2+m2. Thus let b†(p,s)andb(p ,s)be the creation and annihilation operators for an electron of momentum pand spin s. Our introductory discussion indicates that we should impose {b(p ,s),b†(p/prime,s/prime)}=δ(3)(/vectorp−/vectorp/prime)δss/prime (11) {b(p ,s),b(p/prime,s/prime)}=0 (12) {b†(p,s),b†(p/prime,s/prime)}=0 (13) There is a corresponding set of relations for d†(p,s)andd(p ,s)the creation and annihi- lation operators for a positron, for instance, {d(p ,s),d†(p/prime,s/prime)}=δ(3)(/vectorp−/vectorp/prime)δss/prime (14) We now have to show that we indeed obtain (4). Write ¯ψ(0)=/integraldisplayd3p/prime (2π)3 2(Ep/prime/m)1 2/summationdisplay s/prime[b†(p/prime,s/prime)¯u(p/prime,s/prime)+d(p/prime,s/prime)¯v(p/prime,s/prime)] Nothing to do but to plow ahead: {ψ(/vectorx,0),¯ψ(0)} =/integraldisplayd3p (2π)3(Ep/m)/summationdisplay s[u(p ,s)¯u(p ,s)ei/vectorp./vectorx+v(p ,s)¯v(p ,s)e−i/vectorp./vectorx] if we take bandb†to anticommute with dandd†. Using (8) and (9) we obtain {ψ(/vectorx,0),¯ψ(0)}=/integraldisplayd3p (2π)3(2Ep)[(/negationslashp+m)ei/vectorp./vectorx+(/negationslashp−m)e−i/vectorp./vectorx] =/integraldisplayd3p (2π)3(2Ep)2p0γ0e−i/vectorp./vectorx=γ0δ(3)(/vectorx) which is just (4) slightly disguised. Similarly, writing schematically, we have {ψ,ψ}=0 and {ψ†,ψ†}=0. We are of course free to normalize the spinors uandvhowever we like. One alternative normalization is to define uandvas the uandvgiven here multiplied by (2m)1 2, thus changing (8) and (9) to/summationtext su¯u=/negationslashp+mand/summationtext sv¯v=/negationslashp−m. Multiplying the numerator and denominator in (10) by (2m)1 2, we see that the normalization factor (Ep/m)1 2is changed to (2Ep)1 2[thus making it the same as the normalization factor for the scalar field in (I.8.11)]. II.2. Quantizing the Dirac Field | 111 This alternative normalization (let us call it “any mass normalization”) is particularly convenient when we deal with massless spin-1 2particles: we could set m=0 everywhere without ever encountering min the denominator as in (8) and (9). The advantage of the normalization used here (let us call it “rest normalization”) is that the spinors assume simple forms in the rest frame, as we have just seen. This would proveto be advantageous when we calculate the magnetic moment of the electron in chapter III.6, for example. Of course, multiplying and dividing here and there by (2m) 1 2is a trivial operation, and there is not much sense in arguing over the relative advantages of onenormalization over another. In chapter II.6 we will calculate electron scattering at energies high compared to the massm, so that effectively we could set m=0. Actually, even then, “rest normalization” has the slight advantage of providing a (rather weak) check on the calculation. We set m=0 everywhere we can, such as in the numerator of (8) and (9), but not where we can’t, such asin the denominator. Then mmust cancel out in physical quantities such as the differential scattering cross section. Energy of the vacuum An important exercise at this point is to calculate the Hamiltonian starting with theHamiltonian density H=π∂ψ ∂t−L=¯ψ(i/vectorγ./vector∂+m)ψ (15) Inserting (10) into this expression and integrating, we have H=/integraldisplay d3xH=/integraldisplay d3x¯ψ(i/vectorγ./vector∂+m)ψ=/integraldisplay d3x¯ψiγ0∂ψ ∂t(16) which works out to be H=/integraldisplay d3p/summationdisplay sEp[b†(p,s)b(p ,s)−d(p ,s)d†(p,s)] (17) We can see the all important minus sign in (17) schematically: In (16) ¯ψgives a factor ∼(b†+d), while ∂/∂t acting on ψbrings down a relative minus sign giving ∼(b−d†), thus giving us ∼(b†+d)(b−d†)∼b†b−dd†(orthogonality between spinors vu=0 kills the cross terms). To bring the second term in (17) into the right order, we anticommute −d(p ,s)d†(p,s) =d†(p,s)d(p ,s)−δ(3)(/vector0)so that H=/integraldisplay d3p/summationdisplay sEp[b†(p,s)b(p ,s)+d†(p,s)d(p ,s)] −δ(3)(/vector0)/integraldisplay d3p/summationdisplay sEp (18) The first two terms tell us that each electron and each positron of momentum pand spin shas exactly the same energy Ep, as it should. But what about the last term? That δ(3)(/vector0) should fill us with fear and loathing. 112 | II. Dirac and the Spinor It is OK: Noting that δ(3)(/vectorp)=[1/(2π)3]/integraltext d3xei/vectorp/vectorx, we see that δ(3)(/vector0)=[1/(2π)3]/integraltext d3x (we encounter the same maneuver in exercise I.8.2) and so the last term contributes to H E0=−1 h3/integraldisplay d3x/integraldisplay d3p/summationdisplay s2(1 2Ep) (19) (since in natural units we have /planckover2pi=1 and hence h=2π). We have an energy −1 2Epin each unit-size phase-space cell (1/h3)d3xd3pin the sense of statistical mechanics, for each spin and for the electron and positron separately (hence the factor of 2) . This infinite additivetermE 0is precisely the analog of the zero point energy1 2/planckover2piωof the harmonic oscillator you encountered in your quantum mechanics course. But it comes in with a minus sign! The sign is bizarre and peculiar! Each mode of the Dirac field contributes −1 2/planckover2piωto the vacuum energy. In contrast, each mode of a scalar field contributes1 2/planckover2piωas we saw in chapter I.8. This fact is of crucial importance in the development of supersymmetry,which we will discuss in chapter VIII.4. Fermion propagating through spacetime In analogy with (I.8.14), the propagator for the electron is given by iSαβ(x)≡ /angbracketleft0|Tψα(x)¯ψβ(0)|0/angbracketright, where the argument of ¯ψhas been set to 0 by translation invariance. As we will see, the anticommuting character of ψrequires us to define the time-ordered product with a minus sign, namely T ψ(x) ¯ψ(0)≡θ(x0)ψ(x) ¯ψ(0)−θ(−x0)¯ψ(0)ψ(x) (20) Referring to (10), we obtain for x0>0, iS(x)=/angbracketleft0|ψ(x)¯ψ(0)|0/angbracketright=/integraldisplayd3p (2π)3(Ep/m)/summationdisplay su(p ,s)¯u(p ,s)e−ipx =/integraldisplayd3p (2π)3(Ep/m)/negationslashp+m 2me−ipx Forx0<0, we have to be a bit careful about the spinorial indices: iSαβ(x)=− /angbracketleft 0|¯ψβ(0)ψα(x)|0/angbracketright =−/integraldisplayd3p (2π)3(Ep/m)/summationdisplay s¯vβ(p,s)vα(p,s)e−ipx =−/integraldisplayd3p (2π)3(Ep/m)(/negationslashp−m 2m)αβe−ipx using the identity (9). Putting things together we obtain iS(x)=/integraldisplayd3p (2π)3(Ep/m)/bracketleftbigg θ(x0)/negationslashp+m 2me−ipx−θ(−x0)/negationslashp−m 2me+ipx/bracketrightbigg (21) We will now show that this fermion propagator can be written more elegantly as a 4-dimensional integral: II.2. Quantizing the Dirac Field | 113 iS(x)=i/integraldisplayd4p (2π)4e−ip.x/negationslashp+m p2−m2+iε=/integraldisplayd4p (2π)4e−ip.x i /negationslashp−m+iε(22) To show that (22) is indeed equivalent to (21) we go through essentially the same steps as after (I.8.14). In the complex p0plane the integrand has poles at p0=±/radicalbig /vectorp2+m2−iε/similarequal ±(Ep−iε).F o rx0>0 the factor e−ip0x0tells us to close the contour in the lower half-plane. We go around the pole at +(Ep−iε)clockwise and obtain iS(x)=(−i)i/integraldisplayd3p (2π)3e−ip.x/negationslashp+m 2Ep producing the first term in (21). For x0<0 we are now told to close the contour in the upper half-plane and thus we go around the pole at −(Ep−iε)anticlockwise. We obtain iS(x)=i2/integraldisplayd3p (2π)3e+iEpx0+i/vectorp./vectorx 1 −2Ep(−Epγ0−/vectorp/vectorγ+m) and flipping /vectorpwe have iS(x)=−/integraldisplayd3p (2π)3eip.x1 2Ep(Epγ0−/vectorp/vectorγ−m)=−/integraldisplayd3p (2π)3eip.x1 2Ep(/negationslashp−m) precisely the second term in (21) with the minus sign and all. Thus, we must define the time-ordered product with the minus sign as in (20). After all these steps, we see that in momentum space the fermion propagator has the elegant form iS(p) =i /negationslashp−m+iε(23) This makes perfect sense: S(p) comes out to be the inverse of the Dirac operator /negationslashp−m, just as the scalar boson propagator D(k)=1/(k2−m2+iε) is the inverse of the Klein- Gordon operator k2−m2. Poetic but confusing metaphors In closing this chapter let me ask you some rhetorical questions. Did I speak of an electron going backward in time? Did I mumble something about a sea of negative energyelectrons? This metaphorical language, when used by brilliant minds, the likes of Diracand Feynman, was evocative and inspirational, but unfortunately confused generationsof physics students and physicists. The presentation given here is in the modern spirit,which seeks to avoid these potentially confusing metaphors. Exercises II.2.1 Use Noether’s theorem to derive the conserved current Jμ=¯ψγμψ. Calculate [Q ,ψ], thus showing that bandd†must carry the same charge. II.2.2 Quantize the Dirac field in a box of volume of Vand show that the vacuum energy E0is indeed proportional to V. [Hint: The integral over momentum/integraltext d3pis replaced by a sum over discrete values of the momentum.] II.3 Lorentz Group and Weyl Spinors The Lorentz algebra In chapter II.1 we followed Dirac’s brilliantly idiosyncratic way of deriving his equation. We develop here a more logical and mathematical theory of the Dirac spinor. A deeperunderstanding of the Dirac spinor not only gives us a certain satisfaction, but is alsoindispensable, as we will see later, in studying supersymmetry, one of the foundationalconcepts of superstring theory; and of course, most of the fundamental particles such asthe electron and the quarks carry spin 1 2and are described by spinor fields. Let us begin by reminding ourselves how the rotation group works. The three generators Ji(i=1, 2, 3 or x,y,z) of the rotation group satisfy the commutation relation [Ji,Jj]=i/epsilon1ijkJk (1) When acting on the spacetime coordinates, written as a column vector ⎛ ⎜⎜⎜⎜⎜⎝x 0 x1 x2 x3⎞ ⎟⎟⎟⎟⎟⎠ the generators of rotations are represented by the hermitean matrices J1=⎛ ⎜⎜⎜⎜⎜⎝000 0 000 0000−i00 i 0⎞ ⎟⎟⎟⎟⎟⎠(2) withJ2andJ3obtained by cyclic permutations. You should verify by laboriously multi- plying these three matrices that (1) is satisfied. Note that the signs of Jiare fixed by the commutation relation (1). II.3. Lorentz Group and Spinors | 115 Now add the Lorentz boosts. A boost in the x≡x1direction transforms the spacetime coordinates: t/prime=(cosh ϕ) t+(sinh ϕ) x ;x/prime=(sinh ϕ) t+(cosh ϕ) x (3) or for infinitesimal ϕ t/prime=t+ϕx;x/prime=x+ϕt (4) In other words, the infinitesimal generator of a Lorentz boost in the xdirection is repre- sented by the hermitean matrix (x0≡tas usual) iK1=⎛ ⎜⎜⎜⎜⎜⎝0100 1000 00000000⎞ ⎟⎟⎟⎟⎟⎠(5) Similarly, iK 2=⎛ ⎜⎜⎜⎜⎜⎝0010 0000 1000 0000⎞ ⎟⎟⎟⎟⎟⎠(6) I leave it to you to write down K3. Note that Kiis defined to be antihermitean. Check that [ Ji,Kj]=i/epsilon1ijkKk. To see that this implies that the boost generators Ki transform as a 3-vector /vectorKunder rotation, as you would expect, apply a rotation through an infinitesimal angle θaround the 3-axis. Then (you might wish to review the material in appendix B at this point) K1→eiθJ 3K1e−iθJ 3=K1+iθ[J3,K1]+O(θ2)=K1+iθ(iK2)+ O(θ2)=cosθK1−sinθK2to the order indicated. You are now about to do one of the most significant calculations in the history of twentieth century physics. By brute force compute [ K1,K2], evidently an antisymmetric matrix. You will discover that it is equal to −iJ 3. Two Lorentz boosts produce a rotation! (You might recall from your course on electromagnetism that this mathematical fact isresponsible for the physics of the Thomas precession.) Mathematically, the generators of the Lorentz group satisfy the following algebra [known to the cognoscenti as SO( 3, 1)]: [Ji,Jj]=i/epsilon1ijkJk (7) [Ji,Kj]=i/epsilon1ijkKk (8) [Ki,Kj]=−i/epsilon1ijkJk (9) Note the all-important minus sign! 116 | II. Dirac and the Spinor How do we study this algebra? The crucial observation is that the algebra falls apart into two pieces if we form the combinations J±i≡1 2(Ji±iKi). You should check that [J+i,J+j]=i/epsilon1ijkJ+k (10) [J−i,J−j]=i/epsilon1ijkJ−k (11) and most remarkably [J+i,J−j]=0 (12) This last commutation relation tells us that J+andJ−form two separate SU( 2)algebras. (For more, see appendix B.) From algebra to representation This means that you can simply use what you have already learned about angular mo-mentum in elementary quantum mechanics and the representation of SU( 2)to deter- mine all the representations of SO( 3, 1). As you know, the representations of SU( 2) are labeled by j=0, 1 2,1 ,3 2,.... We can think of each representation as consisting of (2j+1)objects ψmwithm=−j,−j+1,...,j−1,jthat transform into each other under SU( 2). It follows immediately that the representations of SO( 3, 1)are labeled by (j+,j−)withj+andj−each taking on the values 0,1 2,1 ,3 2, . . . . Each representation consists of (2j++1)(2j−+1)objects ψm+m−withm+=−j+,−j++1 ,..., j+−1,j+ andm−=−j−,−j−+1 ,..., j−−1,j−. Thus, the representations of SO( 3, 1)are(0, 0),(1 2,0),(0,1 2),(1, 0),(0, 1),(1 2,1 2), and so on, in order of increasing dimension. We recognize the 1-dimensional representation(0, 0)as clearly the trivial one, the Lorentz scalar. By counting dimensions, we expect that the 4-dimensional representation ( 1 2,1 2)has to be the Lorentz vector, the defining representation of the Lorentz group (see exercise II.3.1). Spinor representations What about the representation (1 2,0)? Let us write the two objects as ψαwithα=1, 2. Well, what does the notation (1 2,0)mean? It says that J+i=1 2(Ji+iKi)acting on ψαis represented by1 2σiwhile J−i=1 2(Ji−iKi)acting on ψαis represented by 0. By adding and subtracting we find that Ji=1 2σi (13) and iKi=1 2σi (14) II.3. Lorentz Group and Spinors | 117 where the equal sign means “represented by” in this context. (By convention we do not distinguish between upper and lower indices on the 3-dimensional quantities Ji,Ki, and σi.)Note again that Kiis anti-hermitean. Similarly, let us denote the two objects in (0,1 2)by the peculiar symbol ¯χ˙α. I should em- phasize the trivial but potentially confusing point that unlike the bar used in chapter II.1,the bar on ¯χ ˙αis a typographical element: Think of the symbol ¯χas a letter in the Hittite alphabet if you like. Similarly, the symbol ˙αbears no relation to α: we do not obtain ˙α by operating on αin any way. The rather strange notation is known informally as “dotted and undotted” and more formally as the van der Waerden notation—a bit excessive for ourrather modest purposes at this point but I introduce it because it is the notation used insupersymmetric physics and superstring theory. (Incidentally, Dirac allegedly said that hewished he had invented the dotted and undotted notation.) Repeating the same steps asabove, you will find that on the representation (0, 1 2)we have Ji=1 2σiandiKi=−1 2σi. The minus sign is crucial. The 2-component spinors ψαand¯χ˙αare called Weyl spinors and furnish perfectly good representations of the Lorentz group. Why then does the Dirac spinor have 4 components? The reason is parity. Under parity, /vectorx→−/vectorxand/vectorp→−/vectorp, and thus /vectorJ→/vectorJand/vectorK→−/vectorK, and so /vectorJ+↔/vectorJ−. In other words, under parity the representations (1 2,0)↔(0,1 2). There- fore, to describe the electron we must use both of these 2-dimensional representations, orin mathematical notation, the 4-dimensional reducible representation ( 1 2,0)⊕(0,1 2). We thus stack two Weyl spinors together to form a Dirac spinor /Psi1=/parenleftBiggψα ¯χ˙α/parenrightBigg (15) The spinor /Psi1(p) is of course a function of 4-momentum p[and by implication also ψα(p) and¯χ˙α(p)] but we will suppress the pdependence for the time being. Referring to (13) and (14) we see that acting on /Psi1the generators of rotation /vectorJ=/parenleftBigg1 2/vectorσ 0 01 2/vectorσ/parenrightBigg where once again the equality means “represented by,” and the generators of boost i/vectorK=/parenleftBigg1 2/vectorσ 0 0−1 2/vectorσ/parenrightBigg Note once again the all-important minus sign. Parity forces us to have a 4-component spinor but we know on the other hand that the electron has only two physical degrees of freedom. Let us go to the rest frame. We mustproject out two of the components contained in /Psi1(p r)with the rest momentum pr≡ (m,/vector0). With the benefit of hindsight, we write the projection operator as P=1 2(1−γ0). You are probably guessing from the notation that γ0will turn out to be one of the gamma matrices, but at this point, logically γ0is just some 4 by 4 matrix. The condition P2=P implies that (γ0)2=1 so that the eigenvalues of γ0are±1. Since ψα↔¯χ˙αunder parity we naturally guess that ψαand¯χ˙αcorrespond to the left and right handed fields of chapter II.1. 118 | II. Dirac and the Spinor We cannot simply use the projection to set for example ¯χ˙αto 0. Parity means that we should treatψαand¯χ˙αon the same footing. We choose γ0=/parenleftBigg01 10/parenrightBigg , or more explicitly ⎛ ⎜⎜⎜⎜⎜⎝0010 0001 1000 0100⎞ ⎟⎟⎟⎟⎟⎠ (Different choices of γ0correspond to the different basis choices discussed in chapter II.1.) In other words, in the rest frame ψα−¯χ˙α=0. The projection to two degrees of freedom can be written as (γ0−1)/Psi1(p r)=0 (16) Indeed, we recognize this as just the Weyl basis introduced in chapter II.1. The Dirac equation We have derived the Dirac equation, a bit in disguise! Since our derivation is based on a step-by-step study of the spinor representation of the Lorentz group, we know how to obtain the equation satisfied by /Psi1(p) for any p: We simply boost. Writing /Psi1(p)=e−i/vectorϕ/vectorK/Psi1(pr), we have (e−i/vectorϕ/vectorKγ0ei/vectorϕ/vectorK−1)/Psi1(p) =0. Introducing the notation γμpμ/m≡e−i/vectorϕ/vectorKγ0ei/vectorϕ/vectorK, we obtain the Dirac equation (γμpμ−m)/Psi1(p) =0 (17) You can work out the details as an exercise. The derivation here represents the deep group theoretic way of looking at the Dirac equation: It is a projection boosted into an arbitrary frame. Note that this is an example of the power of symmetry, which pervades modern physics and this book: Our knowledge of how the electron field transforms under the rotationgroup, namely that it has spin 1 2, allows us to know how it transforms under the Lorentz group. Symmetry rules! In appendix E we will develop the dotted and undotted notation further for later use in the chapter on supersymmetry. In light of your deeper group theoretic understanding it is a good idea to reread chap- ter II.1 and compare it with this chapter. II.3. Lorentz Group and Spinors | 119 Exercises II.3.1 Show by explicit computation that (1 2,1 2)is indeed the Lorentz vector. II.3.2 Work out how the six objects contained in the (1, 0)and(0, 1)transform under the Lorentz group. Recall from your course on electromagnetism how the electric and magnetic fields /vectorEand/vectorBtransform. Conclude that the electromagnetic field in fact transforms as (1, 0)⊕(0, 1). Show that it is parity that once again forces us to use a reducible representation. II.3.3 Show that e−i/vectorϕ/vectorKγ0ei/vectorϕ/vectorK=/parenleftBigg0e−/vectorϕ/vectorσ e/vectorϕ/vectorσ0/parenrightBigg and e/vectorϕ/vectorσ=coshϕ+/vectorσ.ˆϕsinhϕ with the unit vector ˆϕ≡/vectorϕ/ϕ . Identifying /vectorp=mˆϕsinhϕ, derive the Dirac equation. Show that γi=/parenleftBigg0σi −σi0/parenrightBigg II.3.4 Show that a spin3 2particle can be described by a vector-spinor /Psi1αμ, namely a Dirac spinor carrying a Lorentz index. Find the corresponding equations of motion, known as the Rarita-Schwinger equations.[Hint: The object /Psi1 αμhas 16 components, which we need to cut down to 2 .3 2+1=4 components.] II.4 Spin-Statistics Connection There is no one fact in the physical world which has a greater impact on the way things are, than the Pauli exclusion principle.1 Degrees of intellectual incompleteness In a course on nonrelativistic quantum mechanics you learned about the Pauli exclusion principle2and its later generalization stating that particles with half integer spins, such as electrons, obey Fermi-Dirac statistics and want to stay apart, while in contrast particles withinteger spins, such as photons or pairs of electrons, obey Bose-Einstein statistics and loveto stick together. From the microscopic structure of atoms to the macroscopic structureof neutron stars, a dazzling wealth of physical phenomena would be incomprehensiblewithout this spin-statistics rule. Many elements of condensed matter physics, for instance,band structure, Fermi liquid theory, superfluidity, superconductivity, quantum Hall effect,and so on and so forth, are consequences of this rule. Quantum statistics, one of the most subtle concepts in physics, rests on the fact that in the quantum world, all elementary particles and hence all atoms, are absolutely identicalto, and thus indistinguishable from, one other. 3It should be recognized as a triumph of quantum field theory that it is able to explain absolute identity and indistinguishabilityeasily and naturally. Every electron in the universe is an excitation in one and the sameelectron field ψ. Otherwise, one might be able to imagine that the electrons we now 1I. Duck and E. C. G. Sudarshan, Pauli and the Spin-Statistics Theorem ,p .2 1 . 2While a student in Cambridge, E. C. Stoner came to within a hair of stating the exclusion principle. Pauli himself in his famous paper ( Zeit. f. Physik 31: 765, 1925) only claimed to “summarize and generalize Stoner’s idea.” However, later in his Nobel Prize lecture Pauli was characteristically ungenerous toward Stoner’scontribution. A detailed and fascinating history of the spin and statistics connection may be found in Duck and Sudarshan, op. cit. 3Early in life, I read in one of George Gamow’s popular physics books that he could not explain quantum statistics—all he could manage for Fermi statistics was an analogy, invoking Greta Garbo’s famous remark “Ivont to be alone.”—and that one would have to go to school to learn about it. Perhaps this spurs me, later in life, to write popular physics books also. See A. Zee, Einstein ’s Universe ,p .x . II.4. Spin-Statistics Connection | 121 know came off an assembly line somewhere in the early universe and could all be slightly different owing to some negligence in the manufacturing process. While the spin-statistics rule has such a profound impact in quantum mechanics, its explanation had to wait for the development of relativistic quantum field theory. Imaginea civilization that for some reason developed quantum mechanics but has yet to discoverspecial relativity. Physicists in this civilization eventually realize that they have to inventsome rule to account for the phenomena mentioned above, none of which involves motionfast compared to the speed of light. Physics would have been intellectually unsatisfyingand incomplete. One interesting criterion in comparing different areas of physics is their degree of intellectual incompleteness. Certainly, in physics we often accept a rule that cannot be explained until we move to the next level. For instance, in much of physics, we take as a given the fact that the charge of theproton and the charge of electron are exactly equal and opposite. Quantum electrodynamicsby itself is not capable of explaining this striking fact either. This fact, charge quantization,can only be deduced by embedding quantum electrodynamics into a larger structure, suchas a grand unified theory, as we will see in chapter VII.6. (In chapter IV .4 we will learn thatthe existence of magnetic monopoles implies charge quantization, but monopoles do notexist in pure quantum electrodynamics.) Thus, the explanation of the spin-statistics connection, by Fierz and by Pauli in the late 1930s, and by L ¨uders and Zumino and by Burgoyne in the late 1950s, ranks as one of the great triumphs of relativistic quantum field theory. I do not have the space to give a generaland rigorous proof 4here. I will merely sketch what goes terribly wrong if we violate the spin-statistics connection. The price of perversity A basic quantum principle states that if two observables commute then they are simul-taneously diagonalizable and hence observable. A basic relativistic principle states that iftwo spacetime points are spacelike with respect to each other then no signal can propagatebetween them, and hence the measurement of an observable at one of the points cannotinfluence the measurement of another observable at the other point. Consider the charge density J 0=i(ϕ†∂0ϕ−∂0ϕ†ϕ)in a charged scalar field theory. According to the two fundamental principles just enunciated, J0(/vectorx,t=0)andJ0(/vectory,t=0) should commute for /vectorx/negationslash=/vectory. In calculating the commutator of J0(/vectorx,t=0)withJ0(/vectory,t=0), we simply use the fact that ϕ(/vectorx,t=0)and∂0ϕ(/vectorx,t=0)commute with ϕ(/vectory,t=0)and ∂0ϕ(/vectory,t=0), so we just move the field at /vectorxsteadily past the field at /vectory. The commutator vanishes almost trivially. 4See I. Duck and E. C. G. Sudarshan, Pauli and the Spin-Statistics Theorem , and R. F. Streater and A. S. Wightman, PCT , Spin Statistics, and All That . 122 | II. Dirac and the Spinor Now suppose we are perverse and quantize the creation and annihilation operators in the expansion (I.8.11) ϕ(/vectorx,t=0)=/integraldisplaydDk/radicalbig (2π)D2ωk[a(/vectork)ei/vectork./vectorx+a†(/vectork)e−i/vectork./vectorx] (1) according to anticommutation rules {a(/vectork),a†(/vectorq)}=δ(D)(/vectork−/vectorq) and {a(/vectork),a(/vectorq)}=0={a†(/vectork),a†(/vectorq)} instead of the correct commutation rules. What is the price of perversity?Now when we try to move J 0(/vectory,t=0)pastJ0(/vectorx,t=0), we have to move the field at /vectory past the field at /vectorxusing the anticommutator {ϕ(/vectorx,t=0),ϕ(/vectory,t=0)} =/integraldisplay/integraldisplaydDk/radicalbig (2π)D2ωkdDq/radicalBig (2π)D2ωq{[a(/vectork)ei/vectork./vectorx+a†(/vectork)e−i/vectork./vectorx], [a(/vectorq)ei/vectorq./vectory+a†(/vectorq)e−i/vectorq./vectory]} =/integraldisplaydDk (2π)D2ωk(ei/vectork.(/vectorx−/vectory)+e−i/vectork.(/vectorx−/vectory)) (2) You see the problem? In a normal scalar-field theory that obeys the spin-statistics connection, we would have computed the commutator, and then in the last expression in (2) we would have gotten (ei/vectork.(/vectorx−/vectory)−e−i/vectork.(/vectorx−/vectory))instead of (ei/vectork.(/vectorx−/vectory)+e−i/vectork.(/vectorx−/vectory)). The integral /integraldisplaydDk (2π)D2ωk(ei/vectork.(/vectorx−/vectory)−e−i/vectork.(/vectorx−/vectory)) would obviously vanish and all would be well. With the plus sign, we get in (2) a nonvan- ishing piece of junk. A disaster if we quantize the scalar field as anticommuting! A spin 0field has to be commuting. Thus, relativity and quantum physics join hands to force thespin-statistics connection. It is sometimes said that because of electromagnetism you do not sink through the floor and because of gravity you do not float to the ceiling, and you would be sinking or floatingin total darkness were it not for the weak interaction, which regulates stellar burning.Without the spin-statistics connection, electrons would not obey Pauli exclusion. Matterwould just collapse. 5 Exercise II.4.1 Show that we would also get into trouble if we quantize the Dirac field with commutation instead of anticommutation rules. Calculate the commutator [ J0(/vectorx,0),J0(0)]. 5The proof of the stability of matter, given by Dyson and Lenard, depends crucially on Pauli exclusion. II.5Vacuum Energy, Grassmann Integrals, and Feynman Diagrams for Fermions The vacuum is a boiling sea of nothingness, full of sound and fury, signifying a great deal. —Anonymous Fermions are weird I developed the quantum field theory of a scalar field ϕ(x) first in the path integral formalism and then in the canonical formalism. In contrast, I have thus far developedthe quantum field theory of the free spin 1 2fieldψ(x) only in the canonical formalism. We learned that the spin-statistics connection forces the field operator ψ(x) to satisfy anticommutation relations. This immediately suggests something of a mystery in writingdown the path integral for the spinor field ψ. In the path integral formalism ψ(x) is not an operator but merely an integration variable. How do we express the fact that its operatorcounterpart in the canonical formalism anticommutes? We presumably cannot represent ψas a commuting variable, as we did ϕ. Indeed, we will discover that in the path integral formalism ψis to be treated not as an ordinary complex number but as a novel kind of mathematical entity known as a Grassmann number. If you thought about it, you would realize that some novel mathematical structure is needed. In chapter I.3 we promoted the coordinates of point particles q i(t)in quantum mechanics to the notion of a scalar field ϕ(/vectorx,t). But you already know from quantum mechanics that a spin1 2particle has the peculiar property that its wave function turns into minus itself when rotated through 2 π. Unlike particle coordinates, half integral spin is not an intuitive concept. Vacuum energy To motivate the introduction of Grassmann-valued fields I will discuss the notion of vacuum energy. The reason for this apparently strange strategy will become clear shortly. Quantum field theory was first developed to describe the scattering of photons and electrons, and later the scattering of particles. Recall that in chapter I.7 while studying the 124 | II. Dirac and the Spinor scattering of particles we encountered diagrams describing vacuum fluctuations, which we simply neglected (see fig. I.7.12). Quite naturally, particle physicists considered thesefluctuations to be of no importance. Experimentally, we scatter particles off each other. Whocares about fluctuations in the vacuum somewhere else? It was only in the early 1970s thatphysicists fully appreciated the importance of vacuum fluctuations. We will come back tothe importance of the vacuum 1in a later chapter. In chapters I.8 and II.2 we calculated the vacuum energy of a free scalar field and of a free spinor field using the canonical formalism. To motivate the use of Grassmann numbersto formulate the path integral for the spinor field I will adopt the following strategy. First,I use the path integral formalism to obtain the result we already have for the free scalarfield using the canonical formalism. Then we will see that in order to produce the resultwe already have for the free spinor field we must modify the path integral. By definition, vacuum fluctuations occur even when there are no sources to produce particles. Thus, let us consider the generating functional of a free scalar field theory in theabsence of sources: 2 Z=/integraldisplay Dϕei/integraltext d4x1 2[(∂ϕ)2−m2ϕ2]=C/parenleftbigg1 det[∂2+m2]/parenrightbigg1 2 =Ce−1 2Tr log(∂2+m2)(1) For the first equality we used (I.2.15) and absorbed inessential factors into the constant C. In the second equality we used the important identity detM=eTr log M(2) which you encountered in exercise I.11.2. Recall that Z=/angbracketleft0|e−iHT|0/angbracketright(with T→∞ understood so that we integrate over all of spacetime in (1)), which in this case is just e−iETwithEthe energy of the vacuum. Evaluating the trace in (1) TrO=/integraldisplay d4x/angbracketleftx|O|x/angbracketright =/integraldisplay d4x/integraldisplayd4k (2π)4/integraldisplayd4q (2π)4/angbracketleftx|k/angbracketright/angbracketleftk|O|q/angbracketright/angbracketleftq|x/angbracketright we obtain iET=1 2VT/integraldisplayd4k (2π)4log(k2−m2+iε)+A where Ais an infinite constant corresponding to the multiplicative factor Cin (1). Recall that in the derivation of the path integral we had lots of divergent multiplicative factors;this is where they can come in. The presence of Ais a good thing here since it solves a problem you might have noticed: The argument of the log is not dimensionless. Let usdefine m /primeby writing 1Indeed, we have already discussed one way to observe the effects of vacuum fluctuations in chapter I.9. 2Strictly speaking, to render the expressions here well defined we should replace m2bym2−iεas discussed earlier. II.5. Feynman Diagrams for Fermions | 125 A=−1 2VT/integraldisplayd4k (2π)4log(k2−m/prime2+iε) In other words, we do not calculate the vacuum energy as such, but only the difference between it and the vacuum energy we would have had if the particle had mass m/primeinstead ofm. The arbitrarily long time Tcancels out and Eis proportional to the volume of space V, as might be expected. Thus, the (difference in) vacuum energy density is E V=−i 2/integraldisplayd4k (2π)4log/bracketleftbiggk2−m2+iε k2−m/prime2+iε/bracketrightbigg (3) =−i 2/integraldisplayd3k (2π)3/integraldisplaydω 2πlog/bracketleftBigg ω2−ω2 k+iε ω2−ω/prime2 k+iε/bracketrightBigg where ω/prime k≡+/radicalbig /vectork2+m/prime2. We treat the (convergent) integral over ωby integrating by parts: /integraldisplaydω 2πdω dωlog/bracketleftBigg ω2−ω2 k+iε ω2−ω/prime2 k+iε/bracketrightBigg =− 2/integraldisplaydω 2πω/bracketleftBigg ω ω2−ω2 k+iε−(ωk→ω/prime k)/bracketrightBigg =−i2ω2 k(1 −2ωk)−(ωk→ω/prime k) =+i(ωk−ω/prime k) (4) Indeed, restoring /planckover2piwe get the result we want: E V=/integraldisplayd3k (2π)3(1 2/planckover2piωk−1 2/planckover2piω/prime k) (5) We had to go through a few arithmetical steps to obtain this result, but the important point is that using the path integral formalism we have managed to obtain a result previouslyobtained using the canonical formalism. A peculiar sign for fermions Our goal is to figure out the path integral for the spinor field. Recall from chapter II.2 that the vacuum energy of the spinor field comes out to have the opposite sign to the vacuum energy of the scalar field, a sign that surely ranks among the “top ten” signs of theoreticalphysics. How are we to get it using the path integral? As explained in chapter I.3, the origin of (1) lies in the simple Gaussian integration formula /integraldisplay+∞ −∞dxe−1 2ax2=/radicalbigg 2π a=√ 2πe−1 2loga Roughly speaking, we have to find a new type of integral so that the analog of the Gaussian integral would go something like e+1 2loga. 126 | II. Dirac and the Spinor Grassmann math It turns out that the mathematics we need was invented long ago by Grassmann. Let us postulate a new kind of number, called the Grassmann or anticommuting number,such that if ηandξare Grassmann numbers, then ηξ=−ξη. In particular, η 2=0. Heuristically, this mirrors the anticommutation relation satisfied by the spinor field.Grassmann assumed that any function of ηcan be expanded in a Taylor series. Since η 2=0, the most general function of ηisf( η)=a+bη, with aandbtwo ordinary numbers. How do we define integration over η? Grassmann noted that an essential property of ordinary integrals is that we can shift the dummy integration variable:/integraltext+∞ −∞dxf(x +c)=/integraltext+∞ −∞dxf(x) . Thus, we should also insist that the Grassmann integral obey the rule/integraltext dηf(η +ξ)=/integraltext dηf(η) , where ξis an arbitrary Grassmann number. Plugging into the most general function given above, we find that/integraltext dηbξ=0. Since ξis arbitrary this can only hold if we define/integraltext dηb=0 for any ordinary number b, and in particular/integraltext dη≡/integraltext dη1=0 Since given three Grassmann numbers χ,η, andξ, we have χ(ηξ) =(ηξ)χ , that is, the product (ηξ) commutes with any Grassmann number χ, we feel that the product of two anticommuting numbers should be an ordinary number. Thus, the integral/integraltext dηη is just an ordinary number that we can simply take to be 1: This fixes the normalization of dη. Thus Grassmann integration is extraordinarily simple, being defined by two rules: /integraldisplay dη=0 (6) and /integraldisplay dηη=1 (7) With these two rules we can integrate any function of η: /integraldisplay dηf(η) =/integraldisplay dη(a+bη)=b (8) ifbis an ordinary number so that f( η) is Grassmannian, and /integraldisplay dηf(η) =/integraldisplay dη(a+bη)=−b (9) ifbis Grassmannian so that f( η) is an ordinary number. Note that the concept of a range of integration does not exist for Grassmann integration. It is much easier to masterGrassmann integration than ordinary integration! Letηand¯ηbe two independent Grassmann numbers and aan ordinary number. Then the Grassmannian analog of the Gaussian integral gives /integraldisplay dη/integraldisplay d¯ηe¯ηaη=/integraldisplay dη/integraldisplay d¯η(1+¯ηaη)=/integraldisplay dηaη=a=e+loga(10) Precisely what we had wanted! We can generalize immediately: Let η=(η1,η2,...,ηN)beNGrassmann numbers, and similarly for ¯η; we then have II.5. Feynman Diagrams for Fermions | 127 /integraldisplay dη/integraldisplay d¯ηe¯ηAη=detA (11) forA={Aij}an antisymmetric NbyNmatrix. (Note that contrary to the bosonic case, the inverse of Aneed not exist.) We can further generalize to a functional integral. As we will see shortly, we now have all the mathematics we need. Grassmann path integral In analogy with the generating functional for the scalar field Z=/integraldisplay DϕeiS(ϕ)=/integraldisplay Dϕei/integraltext d4x1 2[(∂ϕ)2−(m2−iε)ϕ2] we would naturally write the generating functional for the spinor field as Z=/integraldisplay DψD ¯ψeiS(ψ ,¯ψ)=/integraldisplay Dψ/integraldisplay D¯ψei/integraltext d4x¯ψ(i/negationslash∂−m+iε)ψ Treating the integration variables ψand¯ψas Grassmann-valued Dirac spinors, we imme- diately obtain Z=/integraldisplay Dψ/integraldisplay D¯ψei/integraltext d4x¯ψ(i/negationslash∂−m+iε)ψ=C/primedet(i/negationslash∂−m+iε) =C/primeetr log(i/negationslash∂−m+iε)(12) where C/primeis some multiplicative constant. Using the cyclic property of the trace, we note that (here mis understood to be m−iε) tr log(i/negationslash∂−m)=tr logγ5(i/negationslash∂−m)γ5=tr log(−i/negationslash∂−m) =1 2[tr log(i/negationslash∂−m)+tr log(−i/negationslash∂−m)] =1 2tr log(∂2+m2). (13) Thus, Z=C/primee1 2tr log(∂2+m2−iε)[compare with (1)!]. We see that we get the same vacuum energy we obtained in chapter II.2 using the canonical formalism if we remember that the trace operation here contains a factor of4 compared to the trace operation in (1), since (i/negationslash∂−m)i sa4b y4matrix. Heuristically, we can now see the necessity for Grassmann variables. If we were to treat ψand¯ψas complex numbers in (12), we would obtain something like (1/det[i/negationslash∂−m])= e −tr log (i/negationslash∂−m)and so have the wrong sign for the vacuum energy. We want the determinant to come out in the numerator rather than in the denominator. Dirac propagator Now that we have learned that the Dirac field is to be quantized by a Grassmann path integral we can introduce Grassmannian spinor sources ηand¯η: Z(η ,¯η)=/integraldisplay DψD ¯ψei/integraltext d4x[¯ψ(i/negationslash∂−m)ψ +¯ηψ+¯ψη](14) 128 | II. Dirac and the Spinor pp + kpk Figure II.5.1 and proceed pretty much as before. Completing the square just as in the case of the scalar field, we have ¯ψKψ +¯ηψ+¯ψη=(¯ψ+¯ηK−1)K(ψ +K−1η)−¯ηK−1η (15) and thus Z(η ,¯η)=C/prime/primee−i¯η(i/negationslash∂−m)−1η(16) The propagator S(x) for the Dirac field is the inverse of the operator (i/negationslash∂−m): in other words, S(x) is determined by (i/negationslash∂−m)S(x) =δ(4)(x) (17) As you can verify, the solution is iS(x)=/integraldisplayd4p (2π)4ie−ipx /negationslashp−m+iε(18) in agreement with (II.2.22). Feynman rules for fermions We can now derive the Feynman rules for fermions in the same way that we derived the Feynman rules for a scalar field. For example, consider the theory of a scalar fieldinteracting with a Dirac field L=¯ψ(iγμ∂μ−m)ψ+1 2[(∂ϕ)2−μ2ϕ2]−λϕ4+fϕ¯ψψ (19) The generating functional Z(η ,¯η,J)=/integraldisplay DψD ¯ψDϕeiS(ψ ,¯ψ,ϕ)+i/integraltext d4x(Jϕ+¯ηψ+¯ψη)(20) can be evaluated as a double series in the couplings λandf. The Feynman rules (not repeating the rules involving only the boson) are as follows: 1. Draw a diagram with straight lines for the fermion and dotted lines for the boson, and label each line with a momentum, for example, as in figure II.5.1. II.5. Feynman Diagrams for Fermions | 129 2. Associate with each fermion line the propagator i /negationslashp−m+iε=i/negationslashp+m p2−m2+iε(21) 3. Associate with each interaction vertex the coupling factor ifand the factor (2π)4δ(4)(/summationtext inp −/summationtext outp)expressing momentum conservation (the two sums are taken over the incoming and the outgoing momenta, respectively). 4. Momenta associated with internal lines are to be integrated over with the measure/integraltext [d4p/(2π)4]. 5. External lines are to be amputated. For an incoming fermion line write u(p ,s)and for an outgoing fermion line ¯u(p/prime,s/prime). The sources and sinks have to recognize the spin polarization of the fermion being produced and absorbed. [For antifermions, we would have ¯v(p ,s)andv(p/prime,s/prime). You can see from (II.2.10) that an outgoing antifermion is associated withvrather than ¯v.] 6. A factor of (−1)is to be associated with each closed fermion line. The spinor index carried by the fermion should be summed over, thus leading to a trace for each closed fermion line.[For an example, see (II.7.7–9).] Note that rule 6 is unique to fermions, and is needed to account for their negative contribution to the vacuum energy. The Feynman diagram corresponding to vacuumfluctuation has no external line. I will discuss these points in detail in chapter IV .3. For the theory of a massive vector field interacting with a Dirac field mentioned in chapter II.1 L=¯ψ[iγμ(∂μ−ieAμ)−m]ψ−1 4FμνFμν+1 2μ2AμAμ(22) the rules differ from above as follows. The vector boson propagator is given by i k2−μ2/parenleftbiggkμkν μ2−gμν/parenrightbigg (23) and thus each vector boson line is associated not only with a momentum, but also with indices μandν. The vertex (figure II.5.2) is associated with ieγμ. p + k pk ieγμ Figure II.5.2 130 | II. Dirac and the Spinor If the vector boson line in figure II.5.2 is external and on shell, we have to specify its polarization. As discussed in chapter I.5, a massive vector boson has three degrees ofpolarizations described by the polarization vector ε (a) μfora=1, 2, 3. The amplitude for emitting or absorbing a vector boson with polarization aisieγμε(a) μ=ie/negationslashε(a) μ. In Schwinger’s sorcery, the source for producing a vector boson Jμ(x), in contrast to the source for producing a scalar meson J(x) , carries a Lorentz index. Work in momen- tum space. Current conservation kμJμ(k)=0 implies that we can decompose Jμ(k)= /Sigma13 a=1J(a)(k)ε(a) μ(k). The clever experimentalist sets up her machine, that is, chooses the functions J(a)(k), so as to produce a vector boson of the desired momentum kand polar- ization a. Current conservation requires kμε(a) μ(k)=0. For kμ=(ω(k) ,0 ,0 ,k) , we could choose ε(1) μ(k)=(0, 1, 0, 0 ),ε(2) μ(k)=(0, 0, 1, 0 ),ε(3) μ(k)=(−k ,0 ,0 ,ω(k))/m (24) In the canonical formalism, we have in analogy with the expansion of the scalar field ϕ in (I.8.11) Aμ(/vectorx,t)=/integraldisplaydDk/radicalbig (2π)D2ωk/Sigma13 a=1{a(a)(/vectork)ε(a) μ(k)e−i(ω kt−/vectork./vectorx)+a(a)†(/vectork)ε(a)∗ μ(k)ei(ωkt−/vectork./vectorx)} (25) (I trust you not to confuse the letter aused to denote annihilation and used to label polarization.) The point is that in contrast to ϕ,Aμcarries a Lorentz index, which the creation and annihilation operators have to “know about” (through the polarization label.)It is instructive to compare with the expansion of the fermion field ψin (II.2.10): the spinor index αonψis carried in the expansion by the spinors u(p ,s)andv(p ,s). In each case, an index (μ in the case of the vector and αin the case of the spinor) known to the Lorentz group is “traded” for a label specifying the spin polarization (a andsrespectively.) A minor technicality: notice that I have complex conjugated the polarization vector associated with the creation operator a (a)†(/vectork)in (25) even though the polarization vectors in (24) are real. This is because experimentalists sometimes enjoy using circularly polarizedphotons with polarization vectors ε (1) μ(k)=(0, 1,i,0)/√ 2,ε(2) μ(k)=(0, 1,−i,0)/√ 2. pp + kpk Figure II.5.3 II.5. Feynman Diagrams for Fermions | 131 Exercises II.5.1 Write down the Feynman amplitude for the diagram in figure II.5.1 for the scalar theory (19). The answer is given in chapter III.3. II.5.2 Applying the Feynman rules for the vector theory (22) show that the amplitude for the diagram in figure II.5.3 is given by (ie)2i2/integraldisplayd4k (2π)41 k2−μ2/parenleftbiggkμkν μ2−gμν/parenrightbigg ¯u(p)γν/negationslashp+ /negationslashk+m (p+k)2−m2γμu(p) (26) II.6 Electron Scattering and Gauge Invariance Electron-proton scattering We will now finally calculate a physical process that experimentalists can go out and measure. Consider scattering an electron off a proton. (For the moment let us ignore thestrong interaction that the proton also participates in. We will learn in chapter III.6 howto take this fact into account. Here we pretend that the proton, just like the electron, is astructureless spin- 1 2fermion obeying the Dirac equation.) To order e2the relevant Feynman diagram is given in figure II.6.1 in which the electron and the proton exchange a photon. But wait, from chapter I.5 we only know how to write down the propagator iDμν= i/parenleftbigkμkν μ2−ημν/parenrightbig /(k2−μ2)for a hypothetical massive photon. (Trivial notational change: the mass of the photon is now called μ, since mis reserved for the mass of the electron and Mfor the mass of the proton.) In that chapter I outlined our philosophy: we will plunge ahead and calculate with a nonzero μand hope that at the end we can set μto zero. Indeed, when we calculated the potential energy between two external charges, we find that we canletμ→0 without any signs of trouble [see (I.5.6)]. In this chapter and the next, we would like to see whether this will always be the case. Applying the Feynman rules, we obtain the amplitude for the diagram in figure II.6.1 (withk=P−pthe momentum transfer in the scattering) M(P,PN)=(−ie)(ie)i (P−p)2−μ2/parenleftbiggkμkν μ2−ημν/parenrightbigg ¯u(P)γμu(p)¯u(PN)γνu(pN) (1) We have suppressed the spin labels and used the subscript N(for nucleon) to refer to the proton. Now notice that kμ¯u(P)γμu(p)=(P−p)μ¯u(P)γμu(p)=¯u(P)(/negationslashP− /negationslashp)u(p) =¯u(P)(m −m)u(p) =0 (2) by virtue of the equations of motion satisfied by ¯u(P) andu(p) . Similarly, kμ¯u(PN)γμu(pN) =0. II.6. Scattering and Gauge Invariance | 133 kPN pNP p Figure II.6.1 This important observation implies that the kμkν/μ2term in the photon propagator does not enter. Thus M(P,PN)=−ie2 1 (P−p)2−μ2¯u(P)γμu(p)¯u(PN)γμu(pN) (3) and we can now set the photon mass μto zero with impunity and replace (P−p)2−μ2 in the denominator by (P−p)2. Note that the identity that allows us to set μto zero is just the momentum space version of electromagnetic current conservation ∂μJμ=∂μ(¯ψγμψ)=0. You would notice that this calculation is intimately related to the one we did in going from (I.5.4) to (I.5.5), with ¯u(P)γμu(p) playing the role of Jμ(k). Potential scattering That the proton mass Mis so much larger than the electron mass mallows us to make a useful approximation familiar from elementary physics. In the limit M/m tending to infinity, the proton hardly moves, and we could use, for the proton, the spinors for a particleat rest given in chapter II.2, so that ¯u(P N)γ0u(pN)≈1 and ¯u(PN)γiu(pN)≈0. Thus M=−ie2 k2¯u(P)γ0u(p) (4) We recognize that we are scattering the electron in the Coulomb potential generated by the proton. Work out the (familiar) kinematics: p=(E,0 ,0 ,|/vector p|)andP=(E,0 ,|/vectorp|sinθ, |/vectorp|cosθ). We see that k=P−pis purely spacelike and k2=−/vectork2=− 4|/vectorp|2sin2(θ/2). Recall from (I.4.7) that /integraldisplay d3xei/vectork./vectorx/parenleftbigg −e 4πr/parenrightbigg =−e /vectork2(5) We represent potential scattering by the Feynman diagram in figure II.6.2: the proton has disappeared and been replaced by a cross, which supplies the virtual photon the electroninteracts with. It is in this sense that you could think of the Coulomb potential picturesquelyas a swarm of virtual photons. 134 | II. Dirac and the Spinor k XP p Figure II.6.2 Once again, it is instructive to use the canonical formalism to derive this expression for Mfor potential scattering. We want the transition amplitude /angbracketleftP,S|e−iHT|p,s/angbracketrightwith the single electron state |p,s/angbracketright≡b†(p,s)|0/angbracketright. The term in the Lagrangian describing the elec- tron interacting with the external c-number potential Aμ(x) is given in (II.5.22) and thus to leading order we have the transition amplitude ie/integraltext d4x/angbracketleftP,S|¯ψ(x)γμψ(x)|p,s/angbracketrightAμ(x). Using (II.2.10–11) we evaluate this as ie/integraldisplay d4x(1/ρ(P))(1 /ρ(p))( ¯u(P ,S)γμu(p ,s))ei(P−p)xAμ(x) Hereρ(p) denotes the fermion normalization factor/radicalBig (2π)3Ep/m in (II.2.10). Given that the Coulomb potential has only a time component and does not depend on time, we see that integration over time gives us an energy conservation delta function, and integrationover space the Fourier transform of the potential, as in (5). Thus the above becomes (1/ρ(P))(1 /ρ(p))(2 π)δ(E P−Ep)(−ie2 /vectork2)¯u(P ,S)γ0u(p ,s). Satisfyingly, we have recovered the Feynman amplitude up to normalization factors and an energy conservation deltafunction, just as in (I.8.16) except for the substitution of boson for fermion normalizationfactors. Notice that we have energy conservation but not 3-momentum conservation, a factwe understand perfectly well when we dribble a basketball ball, for example. Electron-electron scattering Next, we graduate to two electrons scattering off each other: e−(p1)+e−(p2)→e−(P1)+ e−(P2). Here we have a new piece of physics: the two electrons are identical. A profound tenet of quantum physics states that we cannot distinguish between the two outgoing electrons. Now there are two Feynman diagrams (see fig. II.6.3) to order e2, obtained by interchanging the two outgoing electrons. The electron carrying momentum P1could have “come from” the incoming electron carrying momentum p1or the incoming electron carrying momentum p2. We have for figure II.6.3a the amplitude A(P 1,P2)=(ie2/(P 1−p1)2)¯u(P 1)γμu(p 1)¯u(P 2)γμu(p 2) II.6. Scattering and Gauge Invariance | 135 (a)P1 P2 p1 p2k (b)P1 P2 p1 p2 Figure II.6.3 as before. We have only indicated the dependence of Aon the final momenta, suppressing the other dependence. By Fermi statistics, the amplitude for the diagram in figure II.6.3is then −A(P 2,P1). Thus the invariant amplitude for two electrons of momentum p1and p2to scatter into two electrons with momentum P1andP2is M=A(P 1,P2)−A(P 2,P1) (6) To obtain the cross section we have to square the amplitude |M|2=[|A(P 1,P2)|2+(P1↔P2)]−2R eA(P 2,P1)∗A(P 1,P2) (7) 136 | II. Dirac and the Spinor At this point we have to do a fair amount of arithmetic, but keep in mind that there is nothing conceptually intricate in what follows. First, we have to learn to complex con-jugate spinor amplitudes. Using (II.1.15), note that in general (¯u(p /prime)γμ...γνu(p))∗= u(p)†γ† ν...γ† μγ0u(p/prime)=¯u(p)γν...γμu(p/prime). Here γμ...γνrepresents a product of any number of γmatrices. Complex conjugation reverses the order of the product and inter- changes the two spinors. Thus we have |A(P 1,P2)|2=e4 k4[¯u(P 1)γμu(p 1)¯u(p 1)γνu(P1)][¯u(P 2)γμu(p2)¯u(p 2)γνu(P 2)] (8) which factorizes with one factor involving spinors carrying momentum with subscript 1 and another factor involving spinors carrying momentum with subscript 2. In contrast,the interference term A(P 2,P1)∗A(P1,P2)does not factorize. In the simplest experiments, the initial electrons are unpolarized, and the polarization of the outgoing electrons is not measured. We average over initial spins and sum over finalspins using (II.2.8): /summationdisplay su(p ,s)¯u(p ,s)=/negationslashp+m 2m(9) In averaging and summing |A(P1,P2)|2we encounter the object (displaying the spin labels explicitly) τμν(P1,p1)≡1 2/summationdisplay/summationdisplay ¯u(P 1,S)γμu(p 1,s)¯u(p 1,s)γνu(P 1,S) (10) =1 2(2m)2tr(/negationslashP1+m)γμ(/negationslashp1+m)γν(11) which is to be multiplied by τμν(P2,p2). Well, we, or rather you, have to develop some technology for evaluating the trace of products of gamma matrices. The key observation is that the square of a gamma matrixis either +1o r −1, and different gamma matrices anticommute. Clearly, the trace of a product of an odd number of gamma matrices vanish. Furthermore, since there are onlyfour different gamma matrices, the trace of a product of six gamma matrices can always bereduced to the trace of a product of four gamma matrices, since there are always pairs ofgamma matrices that are equal and can be brought together by anticommuting. Similarlyfor the trace of a product of an even higher number of gamma matrices. Hence τ μν(P1,p1)=1 2(2m)2(tr(/negationslashP1γμ/negationslashp1γν)+m2tr(γμγν)). Writing tr (/negationslashP1γμ/negationslashp1γν)= P1ρp1λtr(γργμγλγν)and using the expressions for the trace of a product of an even number of gamma matrices listed in appendix D, we obtain τμν(P1,p1)=1 2(2m)24(Pμ 1pν 1− ημνP1.p1+Pν 1pμ 1+m2ημν). In averaging and summing A(P2,P1)∗A(P 1,P2)we encounter the more involved object κ≡1 22/summationdisplay/summationdisplay/summationdisplay/summationdisplay ¯u(P 1)γμu(p 1)¯u(P 2)γμu(p 2)¯u(p 1)γνu(P 2)¯u(p 2)γνu(P 1) (12) where for simplicity of notation we have suppressed the spin labels. Applying (9) we can writeκas a single trace. The evaluation of κis quite tedious, since it involves traces of products of up to eight gamma matrices. II.6. Scattering and Gauge Invariance | 137 We will be content to study electron-electron scattering in the relativistic limit in which mmay be neglected compared to the momenta. As explained in chapter II.2, while we are using the “rest normalization” for spinors we can nevertheless set mto 0 wherever possible. Then κ=1 4(2m)4tr(/negationslashP1γμ/negationslashp1γν/negationslashP2γμ/negationslashp2γν) (13) Applying the identities in appendix D to (13) we obtain tr (/negationslashP1γμ/negationslashp1γν/negationslashP2γμ/negationslashp2γν)= −2tr(/negationslashP1γμ/negationslashp1/negationslashp2γμ/negationslashP2)=− 32p1.p2P1.P2. In the same limit τμν(P1,p1)=2 (2m)2(Pμ 1pν 1+Pν 1pμ 1−ημνP1.p1)and thus τμν(P1,p1)τμν(P2,p2)=4 (2m)4(Pμ 1pν 1+Pν 1pμ 1−ημνP1.p1)(2P2μp2ν−ημνP2.p2) (14) =4.2 (2m)4(p1.p2P1.P2+p1.P2p2.P1) (15) An amusing story to break up this tedious calculation: Murph Goldberger, who was a graduate student at the University of Chicago after working on the Manhattan Projectduring the war and whom I mentioned in chapter II.1 regarding the Feynman slash, toldme that Enrico Fermi marvelled at this method of taking a trace that young people wereusing to sum over spin- 1 2polarizations. Fermi and others in the older generation had simply memorized the specific form of the spinors in the Dirac basis (which you know fromdoing exercise II.1.3) and consequently the expressions for ¯u(P ,S)γ μu(p ,s). They simply multiplied these expressions together and added up the different possibilities. Fermi wasskeptical of the fancy schmancy method the young Turks were using and challenged Murphto a race on the blackboard. Of course, with his lightning speed, Fermi won. To me, it isamazing, living in the age of string theory, that another generation once regarded thetrace as fancy math. I confessed that I was even a bit doubtful of this story until I looked atFeynman’s book Quantum Electrodynamics, but guess what, Feynman indeed constructed, on page 100 in the edition I own, a table showing the result for the amplitude squared forvarious spin polarizations. Some pages later, he mentioned that polarizations could alsobe summed using the spur (the original German word for trace). Another amusing aside:spur is cognate with the English word spoor, meaning animal droppings, and hence alsomeaning track, trail, and trace. All right, back to work! While it is not the purpose of this book to teach you to calculate cross sections for a living, it is character building to occasionally push calculations to the bitter end. Hereis a good place to introduce some useful relativistic kinematics. In calculating the crosssection for the scattering process p 1+p2→P1+P2(with the masses of the four particles all different in general) we typically encounter Lorentz invariants such as p1.P2. A priori, you might think there are six such invariants, but in fact, you know that there are only physical variables, the incident energy Eand the scattering angle θ. The cleanest way to organize these invariants is to introduce what are called Mandelstam variables: s≡(p1+p2)2=(P1+P2)2(16) t≡(P1−p1)2=(P2−p2)2(17) u≡(P2−p1)2=(P1−p2)2(18) 138 | II. Dirac and the Spinor You know that there must be an identity reducing the three variables s,t, anduto two. Show that (with an obvious notation) s+t+u=m2 1+m2 2+M2 1+M2 2(19) For our calculation here, we specialize to the center of mass frame in the relativistic limit p1=E(1, 0, 0, 1 ),p2=E(1, 0, 0, −1),P1=E(1, sinθ,0 ,c o s θ), andP2=E(1,−sinθ,0 , −cosθ). Hence p1.p2=P1.P2=2E2=1 2s (20) p1.P1=p2.P2=2E2sin2θ 2=−1 2t (21) and p1.P2=p2.P1=2E2cos2θ 2=−1 2u (22) Also, in this limit (P1−p1)4=(−2p1.P1)2=16E4sin4(θ/2)=t4. Putting it together, we obtain1 4/summationtext/summationtext/summationtext/summationtext|M|2=(e4/4m4)f (θ), where f( θ)=s2+u2 t2+2s2 tu+s2+t2 u2 =s4+t4+u4 t2u2 =/parenleftBigg 1+cos4(θ/2) sin4(θ/2)+2 sin2(θ/2)cos2(θ/2)+1+sin4(θ/2) cos4(θ/2)/parenrightBigg (23) =2/parenleftbigg1 sin4(θ/2)+1+1 cos4(θ/2)/parenrightbigg (24) The physical origin of each of the terms in (23) [before we simplify with trigonometric identities] to get to (24) is clear. The first term strongly favors forward scattering dueto the photon propagator ∼1/k 2blowing up at k∼0. The third term is required by the indistinguishability of the two outgoing electrons: the scattering must be symmetric underθ→π−θ, since experimentalists can’t tell whether a particular incoming electron has scattered forward or backward. The second term is the most interesting of all: it comes fromquantum interference. If we had mistakenly thought that electrons are bosons and takenthe plus sign in (6), the second term in f( θ) would come with a minus sign. This makes a big difference: for instance, f( π/ 2)would be 5 −8+5=2 instead of 5 +8+5=18. Since the conversion of a squared probability amplitude to a cross section is conceptually the same as in nonrelativistic quantum mechanics (divide by the incoming flux, etc.), Iwill relegate the derivation to an appendix and let you go the last few steps and obtain thedifferential cross section as an exercise: dσ d/Omega1=α2 8E2f( θ) (25) with the fine structure constant α≡e2/4π≈1/137. II.6. Scattering and Gauge Invariance | 139 An amazing subject When you think about it, theoretical physics is truly an amazing business. After the appropriate equipments are assembled and high energy electrons are scattered off eachother, experimentalists indeed would find the differential cross section given in (25). Thereis almost something magical about it. Appendix: Decay rate and cross section To make contact with experiments, we have to convert transition amplitudes into the scattering cross sections and decay rates that experimentalists actually measure. I assume that you are already familiar with the physical concepts behind these measurements from a course on nonrelativistic quantum mechanics, and thus here wefocus more on those aspects specific to quantum field theory. To be able to count states, we adopt an expedient probably already familiar to you from quantum statistical mechanics, namely that we enclose our system in a box, say a cube with length Lon each side with Lmuch larger than the characteristic size of our system. With periodic boundary conditions, the allowed plane wave states e i/vectorp./vectorx carry momentum /vectorp=2π L(nx,ny,nz) (26) where the ni’s are three integers. The allowed values of momentum form a lattice of points in momentum space with spacing 2 π/L between points. Experimentalists measure momentum with finite resolution, small but much larger than 2 π/L . Thus, an infinitesimal volume d3pin momentum space contains d3p/(2π/L)3=Vd3p/(2π)3 states with V=L3the volume of the box. We obtain the correspondence /integraldisplayd3p (2π)3f( p)↔1 V/summationdisplay pf( p) (27) In the sum the values of pranges over the discrete values in (26). The correspondence (27) between continuum normalization and the discrete box normalization implies that δ(3)(/vectorp−/vectorp/prime)↔V (2π)3δ/vectorp/vectorp/prime (28) with the Kronecker delta δ/vectorp/vectorp/primeequal to 1 if /vectorp=/vectorp/primeand 0 otherwise. One way of remembering these correspondences is simply by dimensional matching. Let us now look at the expansion (I.8.17) of a complex scalar field ϕ(/vectorx,t)=/integraldisplayd3k/radicalbig (2π)32ωk[a(/vectork)e−i(ω kt−/vectork./vectorx)+b†(/vectork)ei(ωkt−/vectork./vectorx)] (29) in terms of creation and annihilation operators. Henceforth, in order not to clutter up the page, I will abuse notation slightly, for example, dropping the arrows on vectors when there is no risk of confusion. Going over to the box normalization, we replace the commutation relation [ a(k) ,a†(k/prime)]=δ(3)(/vectork−/vectork/prime)by [a(k) ,a†(k/prime)]=V (2π)3δ/vectork/vectork/prime (30) We now normalize the creation and annihilation operators by a(k)=/parenleftbiggV (2π)3/parenrightbigg1 2 ˜a(k) (31) 140 | II. Dirac and the Spinor so that [˜a(k) ,˜a†(k/prime)]=δk,k/prime (32) Thus the state |/vectork/angbracketright≡˜a†(/vectork)|0/angbracketrightis properly normalized: /angbracketleft/vectork|/vectork/angbracketright=1. Using (27) and (31), we end up with ϕ(x)=1 V1 2/summationdisplay k1/radicalbig 2ωk(˜ae−ikx+˜b†eikx) (33) We specified a complex, rather than a real, scalar field because then, as you showed in exercise I.8.4, a conserved current can be defined with the corresponding charge Q=/integraltext d3xJ0=/integraltext d3k(a†(k)a(k) −b†(k)b(k)) →/summationtext k(˜a†(k)˜a(k)−˜b†(k)˜b(k)) . It follows immediately that /angbracketleft/vectork|Q|/vectork/angbracketright=1, so that for the state |/vectork/angbracketrightwe have one particle in the box. To derive the formula for the decay rate, we focus, for the sake of pedagogical clarity, on a toy Lagrangian L=g(η†ξ†ϕ+h.c.)describing the decay ϕ→η+ξof a meson into two other mesons. (As usual, we display only the part of the Lagrangian that is of immediate interest. In other words, we suppress the stuff you have longsince mastered: L=∂ϕ †∂ϕ−mϕϕ†ϕ+...and all the rest.) The transition amplitude /angbracketleft/vectorp,/vectorq|e−iHT|/vectork/angbracketrightis given to lowest order by A=i/angbracketleft/vectorp,/vectorq|/integraltext d4x(gη†(x)ξ†(x)ϕ(x)) |/vectork/angbracketright. Here we use the states we “carefully” normalized above, namely the ones created by the various “analogs” of ˜a†. (Just as in quantum mechanics, strictly, we should use wave packets instead of plane wave states. I assume thatyou have gone through that at least once.) Plugging in the various “analogs” of (33), we have A=ig(1 V1 2)3/summationdisplay p/prime/summationdisplay q/prime/summationdisplay k/prime1/radicalbig2ωp/prime2ωq/prime2ωk/prime/integraldisplay d4xei(p/prime+q/prime−k/prime)/angbracketleft/vectorp,/vectorq|˜a†(p/prime)˜a†(q/prime)˜a(k/prime)|/vectork/angbracketright =ig1 V3 21/radicalbig2ωp2ωq2ωk(2π)4δ(4)(p+q−k)(34) Here we have committed various minor transgressions against notational consistency. For example, since the three particles ϕ,η, andξhave different masses, the symbol ωrepresents, depending on context, different functions of its subscript (thus ωp=/radicalBig /vectorp2+m2 η, and so forth). Similarly, ˜a(k/prime)should really be written as ˜aϕ(k/prime), and so forth. Also, we confound 3- and 4-momenta. I would like to think that these all fall under the category of what the Catholic church used to call venial sins. In any case, you know full well what I am talking about. Next, we square the transition amplitude Ato find the transition probability. You might be worried, because it appears that we will have to square the Dirac delta function. But fear not, we have enclosed ourselves in a box. Furthermore, we are in reality calculating /angbracketleft/vectorp,/vectorq|e−iHT|/vectork/angbracketright, the amplitude for the state |/vectork/angbracketrightto become the state |/vectorp,/vectorq/angbracketrightafter a large but finite time T. Thus we could in all comfort write [(2π)4δ(4)(p+q−k)]2=(2π)4δ(4)(p+q−k)/integraldisplay d4xei(p+q−k)x =(2π)4δ(4)(p+q−k)/integraldisplay d4x=(2π)4δ(4)(p+q−k)VT(35) Thus the transition probability per unit time, aka the transition rate, is equal to |A|2 T=V V3/parenleftBigg 1 2ωp2ωq2ωk/parenrightBigg (2π)4δ(4)(p+q−k)g2(36) Recall that there are Vd3p/(2π)3states in the volume d3pin momentum space. Hence, multiplying the number of final states (V d3p/(2π)3)(V d3q/(2π)3)by the transition rate |A|2/T, we obtain the differential decay rate of a meson into two mesons carrying off momenta in some specified range d3pandd3q: d/Gamma1=1 2ωkV V3/parenleftBigg Vd3p (2π)32ωp/parenrightBigg/parenleftBigg Vd3q (2π)32ωq/parenrightBigg (2π)4δ(4)(p+q−k)g2(37) Yes sir, indeed, the factors of Vcancel, as they should. To obtain the total decay rate /Gamma1we integrate over d3pandd3q. Notice the factor 1 /2ωk: the decay rate for a moving particle is smaller than that of a resting particle by a factor m/ωk. We have derived time dilation, as we had better. II.6. Scattering and Gauge Invariance | 141 We are now ready to generalize to the decay of a particle carrying momentum Pintonparticles carrying momenta k1,...,kn. For definiteness, we suppose that these are all Bose particles. First, we draw all the relevant Feynman diagrams and compute the invariant amplitude M. (In our toy example, M=ig.) Second, the transition probability contains a factor 1 /Vn+1, one factor of 1 /V for each particle, but when we squared the momentum conservation delta function we also obtained a factor of VT, which converts the transition probability into a transition rate and knocks off one power of V, leaving the factor 1 /Vn. Next, when we sum over final states, we have a factor Vd3ki/((2π)32ωki)for each particle in the final state. Thus the factors of Vindeed cancel. The differential decay rate of a boson of mass Min its rest frame is thus given by d/Gamma1=1 2Md3k1 (2π)32ω(k1)...d3kn (2π)32ω(kn)(2π)4δ(4)/parenleftBigg P−n/summationdisplay i=1ki/parenrightBigg |M|2(38) At this point, we recall that, as explained in chapter II.2, in the expansion of a fermion field into creation and annihilation operators [see (II.2.10)], we have a choice of two commonly used normalizations, trivially related by a factor (2m)1 2. If you choose to use the “rest normalization" so that spinors come out nice in the rest frame, then the field expansion contains the normalization factor (Ep/m)1 2instead of the factor (2ωk)1 2for a Bose field [see (I.8.11)]. This entails the trivial replacement, for each fermion, of the factor 2 ω(k)=2/radicalbig /vectork2+m2by E(p)/m =/radicalbig /vectorp2+m2/m. In particular, for the decay rate of a fermion the factor 1 /2Mshould be removed. If you choose the “any mass renormalization,” you have to remember to normalize the spinors appearing in M correctly, but you need not touch the phase space factors derived here. We next turn to scattering cross sections. As I already said, the basic concepts involved should already be familiar to you from nonrelativistic quantum mechanics. Nevertheless, it may be helpful to review the basicnotions involved. For the sake of definiteness, consider some happy experimentalist sending a beam of haplesselectrons crashing into a stationary proton. The flux of the beam is defined as the number of electrons crossingan imagined unit area per unit time and is thus given by F=nv, where nandvdenote the density and velocity of the electrons in the beam. The measured event rate divided by the flux of the beam is defined to be the crosssection σ, which has the dimension of an area and could be thought of as the effective size of the proton as seen by the electrons. It may be more helpful to go to the rest frame of the electrons, in which the proton is plowing through the cloud of electrons like a bulldozer. In time /Delta1tthe proton moves through a distance v/Delta1t and thus sweeps through a volume σv/Delta1t , which contains nσv/Delta1t electrons. Dividing this by /Delta1tgives us the event rate nvσ . To measure the differential cross section, the experimentalist sets up, typically in the lab frame in which the target particle is at rest, a detector spanning a solid angle d/Omega1=sinθdθdφ and counts the number of events per unit time. All of this is familiar stuff. Now we could essentially take over our calculation of the differential decay rate almost in its entirety to calculate the differential cross section for the process p 1+p2→k1+k2+...+kn. With two particles in the initial state we now have a factor of (1/V)n+2in the transition probability. But as before, the square of the momentum conservation delta function produces one power of Vand counting the momentum final states gives a factor Vn, so that we are left with a factor of 1 /V. You might be worried about this remaining factor of 1 /V, but recall that we still have to divide by the flux, given by |/vectorv1−/vectorv2|n. Since we have normalized to one particle in the box the density nis 1/V. Once again, all factors of Vcancel, as they must. The procedure is thus to draw all relevant diagrams to the order desired and calculate the Feynman amplitude Mfor the process p1+p2→k1+k2+...+kn. Then the differential cross section is given by (again assuming all particles to be bosons) dσ=1 |/vectorv1−/vectorv2|2ω(p 1)2ω(p 2)d3k1 (2π)32ω(k 1)...d3kn (2π)32ω(kn)(2π)4δ(4)/parenleftBigg p1+p2−n/summationdisplay i=1ki/parenrightBigg |M|2 (39) We are implicitly working in a collinear frame in which the velocities of the incoming particles, /vectorv1and/vectorv2, point in opposite directions. This class of frames includes the familiar center of mass frame and the lab frame (inwhich /vectorv 2=0). In a collinear frame, p1=E1(1, 0, 0, v1)andp2=E2(1, 0, 0, v2), and a simple calculation shows that((p 1p2)2−m2 1m22)=(E1E2(v1−v2))2. We could write the factor |/vectorv1−/vectorv2|E1E2indσin the more invariant- looking form ((p1p2)2−m2 1m22)1 2, thus showing explicitly that the differential cross section is invariant under Lorentz boosts in the direction of the beam, as physically must be the case. 142 | II. Dirac and the Spinor An often encountered case involves two particles scattering into two particles in the center of mass frame. Let us do the phase space integral/integraltext (d3k1/2ω1)(d3k2/2ω2)δ(4)(P−k1−k2)here for easy reference. We will do it in two different ways for your edification. We could immediately integrate over d3k2thus knocking out the 3-dimensional momentum conservation delta function δ3(/vectork1+/vectork2). Writing d3k1=k2 1dk1d/Omega1, we integrate over the remaining energy conservation delta function δ(/radicalBig k2 1+m2 1+/radicalBig k2 1+m2 2−Etotal). Using (I.2.13), we find that the integral over k1givesk1ω1ω2/E total, where ω1≡/radicalBig k2 1+m2 1andω2≡/radicalBig k2 1+m2 2, withk1determined by/radicalBig k2 1+m2 1+/radicalBig k2 1+m2 2=Etotal. Thus we obtain /integraldisplayd3k1 2ω1d3k2 2ω2δ(4)(P−k1−k2)=k1 4Etotal/integraldisplay d/Omega1 (40) Once again, if you use the “rest normalization" for fermions, remember to make the replacement as explained above for the decay rate. The factor of1 4should be replaced by mf/2 for one fermion and one boson, and by m1m2for two fermions. Alternatively, we use (I.8.14) and regressing, write d3k2=/integraltext d4k2θ(k0 2)δ(k2 2−m2 2)2ω2. Integrate over d4k2and knock out the 4-dimensional delta function, leaving us with/integraltext 2ω2dk1k2 1d/Omega1/(2 ω12ω2)δ((P −k1)2−m2 2). The argument of the delta function is E2 total−2Etotalk1+m2 1−m2 2, and thus integrating over k1we get a factor of 2Etotal in the denominator, giving a result in agreement with (40). For the record, you could work out the kinematics and obtain k1=/radicalBig (E2 total−(m 1+m2)2)(E2 total−(m1−m2)2)/2Etotal Evidently, this phase space integral also applies to the decay into two particles in the rest frame of the parent particle, in which case we replace Etotal byM. In particular, for our toy example, we have /Gamma1=g2 16πM3/radicalbig (M2−(m+μ)2)(M2−(m−μ)2) (41) The differential cross section for two-into-two scattering in the center of mass frame is given by dσ d/Omega1=1 (2π)2|/vectorv1−/vectorv2|2ω(p 1)2ω(p 2)k1 EtotalF|M|2(42) In particular, in the text we calculated electron-electron scattering in the relativistic limit. As shown there, we can write |M|2=|/hatwiderM|2/(2m)4in terms of some reduced invariant amplitude /hatwiderM. The factor 1 /(2m)4transforms the factors 2 ω(p) into 2E. Things simplify enormously, with |/vectorv1−/vectorv2|=2 and k1=1 2Etotal, so that finally dσ d/Omega1=1 24(4π)2E2|/hatwiderM|2(43) Last, we come to the statistical factor Sthat must be included in calculating the total decay rate and the total cross section to avoid over-counting if there are identical particles in the final state. The factor Shas nothing to do with quantum field theory per se and should already be familiar to you from nonrelativistic quantum mechanics.The rule is that if there are n iidentical particles of type iin the final state, the total decay rate or the total cross section must be multiplied by S=/Pi1i1/ni! to account for indistinguishability. To see the necessity for this factor, it suffices to think about the simplest case of two identical Bose particles. To be specific, consider electron-positron annihilation into two photons (which we will study in chapter II.8). Forsimplicity, average and sum over all spin polarizations. Let us calculate dσ/d/Omega1 according to (43) above. This is the probability that a photon will check into a detector set up at angles θandφrelative to the beam direction. If the detector clicks, then we know that the other photon emerged at an angle π−θrelative to the beam direction. Thus the total cross section should be σ=1 2/integraldisplay d/Omega1dσ d/Omega1=1 2/integraldisplayπ 0dθdσ dθ(44) (The second equality is for all the elementary cases we will encounter in which dσ does not depend on the azimuthal angle φ.) In other words, to avoid double counting, we should divide by 2 if we integrate over the full angular range of θ. More formally, we argue as follows. In quantum mechanics, a set of states |α/angbracketrightis complete if 1 =/summationtext α|α/angbracketleft/angbracketrightα| (“decomposition of 1”). Acting with this on |β/angbracketrightwe see that these states must be normalized according to /angbracketleftα|β/angbracketright=δαβ. II.6. Scattering and Gauge Invariance | 143 Now consider the state |k1,k2/angbracketright≡1√ 2˜a†(k1)˜a†(k2)|0/angbracketright=|k2,k1/angbracketright (45) containing two identical bosons. By repeatedly using the commutation relation (32), we compute /angbracketleftq1,q2|k1,k2/angbracketright= /angbracketleft0|˜a(q 1)˜a(q 2)˜a†(k1)˜a†(k2)|0/angbracketright=1 2(δq1k1δq2k2+δq2k1δq1k2). Thus/summationtext q1/summationtext q2|q1,q2/angbracketright/angbracketleftq 1,q2|k1,k2/angbracketright =/summationtext q1/summationtext q2|q1,q2/angbracketright1 2(δq1k1δq2k2+δq2k1δq1k2)=1 2(|k1,k2/angbracketright+|k2,k1/angbracketright)=|k1,k2/angbracketright. Thus the states |k1,k2/angbracketrightare nor- malized properly. In the sum over states, we have 1 =...+/summationtext q1/summationtext q2|q1,q2/angbracketright/angbracketleftq 1,q2|+ .... In other words, if we are to sum over q1andq2independently, then we must normalize our states as in (45) with the factor of 1 /√ 2. But then this factor would appear multiplying M. In calculating the total decay rate or the total cross section, we are effectively summing over a complete set of final states. In summary, we havetwo options: either we treat the integration over d 3k1d3k2as independent in which case we have to multiply the integral by1 2, or we integrate over only half of phase space. We readily generalize from this factor of1 2to the statistical factor S. In closing, let me mention two interesting pieces of physics.To calculate the cross section σ, we have to divide by the flux, and hence σis proportional to 1 /|/vectorv 1−/vectorv2|. For exothermal processes, such as electron-positron annihilation into photons or slow neutron capture, σcould become huge as the relative velocity vrel→0. Fermi exploited this fact to great advantage in studying nuclear fission. Note that although the cross section, which has dimension of an area, formally goes to infinity, thereaction rate (the number of reactions per unit time) remains finite. Positronium decay into photons is an example of a bound state decaying in finite time. In positronium, the positron and electron are not approaching each other in plane wave states, as we assumed in our cross sectioncalculation. Rather, the probability (per unit volume) that the positron finds itself near the electron is given by|ψ(0)| 2according to elementary quantum mechanics, with ψ(x) the bound state wave function for whatever state of positronium we are interested in. In other words, |ψ(0)|2gives the volume density of positrons near the electron. Since vσis a volume divided by time, the decay rate is given by /Gamma1=vσ|ψ(0)|2. Exercises II.6.1 Show that the differential cross section for a relativistic electron scattering in a Coulomb potential is given by dσ d/Omega1=α2 4/vectorp2v2sin4(θ/2)(1−v2sin2(θ/2)). Known as the Mott cross section, it reduces to the Rutherford cross section you derived in a course on quantum mechanics in the limit the electron velocity v→0. II.6.2 To order e2the amplitude for positron scattering off a proton is just minus the amplitude (3) for electron scattering off a proton. Thus, somewhat counterintuitively, the differential cross sections for positronscattering off a proton and for electron scattering off a proton are the same to this order. Show that to the next order this is no longer true. II.6.3 Show that the trace of a product of odd number of gamma matrices vanishes. II.6.4 Prove the identity s+t+u=/summationtext am2 a. II.6.5 Verify the differential cross section for relativistic electron electron scattering given in (25). II.6.6 For those who relish long calculations, determine the differential cross section for electron-electron scattering without taking the relativistic limit. II.6.7 Show that the decay rate for one boson of mass Minto two bosons of masses mandμis given by /Gamma1=|M|2 16πM3/radicalbig (M2−(m+μ)2)(M2−(m−μ)2) II.7 Diagrammatic Proof of Gauge Invariance Gauge invariance Conceptually, rather than calculate cross sections, we have the more important task of proving that we can indeed set the photon mass μequal to zero with impunity in calculating any physical process. With μ=0, the Lagrangian given in chapter II.1 becomes the Lagrangian for quantum electrodynamics: L=¯ψ[iγμ(∂μ−ieAμ)−m]ψ−1 4FμνFμν(1) We are now ready for one of the most important observations in the history of theoretical physics. Behold, the Lagrangian is left invariant by the gauge transformation ψ(x)→ei/Lambda1(x)ψ(x) (2) and Aμ(x)→Aμ(x)+1 iee−i/Lambda1(x)∂μei/Lambda1(x)=Aμ(x)+1 e∂μ/Lambda1(x) (3) which implies Fμν(x)→Fμν(x) (4) You are of course already familiar with (3) and the invariance of Fμνfrom classical electromagnetism. In contemporary theoretical physics, gauge invariance1is regarded as fundamental and all important, as we will see later. The modern philosophy is to look at (1) as a consequence of (2) and (3). If we want to construct a gauge invariant relativistic field theory involving aspin 1 2and a spin 1 field, then we are forced to quantum electrodynamics. 1The discovery of gauge invariance was one of the most arduous in the history of physics. Read J. D. Jackson and L. B. Okun, “Historical roots of gauge invariance,” Rev. Mod. Phys. 73, 2001 and learn about the sad story of a great physicist whose misfortune in life was that his name differed from that of another physicist by only one letter. II.7. Proof of Gauge Invariance | 145 You will notice that in (3) I have carefully given two equivalent forms. While it is simpler, and commonly done in most textbooks, to write the second form, we should also keep thefirst form in mind. Note that /Lambda1(x) and/Lambda1(x)+2πgive exactly the same transformation. Mathematically speaking, the quantities e i/Lambda1(x)and∂μ/Lambda1(x) are well defined, but /Lambda1(x) is not. After these apparently formal but actually physically important remarks, we are ready to work on the proof. I will let you give the general proof, but I will show you the way byworking through some representative examples. Recall that the propagator for the hypothetical massive photon is iD μν=i(kμkν/μ2− gμν)/(k2−μ2). We can set the μ2in the denominator equal to zero without further ado and write the photon propagator effectively as iDμν=i(kμkν/μ2−gμν)/k2. The dangerous term is kμkν/μ2. We want to show that it goes away. A specific example First consider electron-electron scattering to order e4. Of the many diagrams, focus on the two in figure II.7.1.a. The Feynman amplitude is then ¯u(p/prime)/parenleftbigg γλ 1 /negationslashp+ /negationslashk−mγμ+γμ 1 /negationslashp/prime− /negationslashk−mγλ/parenrightbigg u(p)i k2/parenleftbiggkμkν μ2−δν μ/parenrightbigg /Gamma1λν (5) where /Gamma1λνis some factor whose detailed structure does not concern us. For the specific case shown in figure II.7.1a we can of course write out /Gamma1λνexplicitly if we want. Note the plus sign here from interchanging the two photons since photons obey Bose statistics. p + kλ μp’ − k λμ (a)p’ k pp’ k pk’ k’ Figure II.7.1 146 | II. Dirac and the Spinor (b)p’ k pp’ k pp’ − k λμ p + kλ μ Figure II.7.1 (continued ) Focus on the dangerous term. Contracting the ¯u(p/prime)(...)u(p) factor in (5) with kμwe have ¯u(p/prime)/parenleftbigg γλ 1 /negationslashp+ /negationslashk−m/negationslashk+ /negationslashk1 /negationslashp/prime− /negationslashk−mγλ/parenrightbigg u(p) (6) The trick is to write the /negationslashkin the numerator of the first term as (/negationslashp+ /negationslashk−m)−(/negationslashp−m), and in the numerator of the second term as (/negationslashp/prime−m)−(/negationslashp/prime− /negationslashk−m). Using (/negationslashp−m)u(p) =0 and¯u(p/prime)(/negationslashp/prime−m)=0, we see that the expression in (6) vanishes. This proves the theorem in this simple example. But since the explicit form of /Gamma1λνdid not enter, the proof would have gone through even if figure II.7.1a were replaced by the more general figure II.7.1b,where arbitrarily complicated processes could be going on under the shaded blob. Indeed we can generalize to figure II.7.1c. Apart from the photon carrying momentum kthat we are focusing on, there are already nphotons attached to the electron line. These nphotons are just “spectators” in the proof in the same way that the photon carrying momentum k /primein figure II.7.1a never came into the proof that (6) vanishes. The photon we are focusing on can attach to the electron line in n+1 different places. You can now extend the proof as an exercise. Photon landing on an internal line In the example we just considered, the photon line in question lands on an external electron line. The fact that the line is “capped at the two ends” by ¯u(p/prime)andu(p) is crucial in the proof. What if the photon line in question lands on an internal line? An example is shown in figure II.7.2, contributing to electron-electron scattering in order e8. The figure contains three distinct diagrams. The electron “on the left” emits II.7. Proof of Gauge Invariance | 147 (c)p’ k p Figure II.7.1 (continued ) three photons, which attach to an internal electron loop. The electron “on the right” emits a photon with momentum k, which can attach to the loop in three distinct ways. Since what we care about is whether the kμkρ/μ2piece in the photon propagator i(kμkρ/μ2−gμρ)/k2goes away or not, we can for our purposes replace that photon propagator by kμ. To save writing slightly, we define p1=p+q1andp2=p1+q2(see the momentum labels in figure II.7.2): Let’s focus on the relevant part of the three diagrams,referring to them as A,B, andC. A=/integraldisplayd4p (2π)4tr/parenleftbigg γν 1 /negationslashp2+ /negationslashk−mγσ 1 /negationslashp1+ /negationslashk−mγλ 1 /negationslashp+ /negationslashk−m/negationslashk1 /negationslashp−m/parenrightbigg (7) B=/integraldisplayd4p (2π)4tr/parenleftbigg γν 1 /negationslashp2+ /negationslashk−mγσ 1 /negationslashp1+ /negationslashk−m/negationslashk1 /negationslashp1−mγλ 1 /negationslashp−m/parenrightbigg (8) and C=/integraldisplayd4p (2π)4tr/parenleftbigg γν 1 /negationslashp2+ /negationslashk−m/negationslashk1 /negationslashp2−mγσ 1 /negationslashp1−mγλ 1 /negationslashp−m/parenrightbigg (9) This looks like an unholy mess, but it really isn’t. We use the same trick we used before. InCwrite/negationslashk=(/negationslashp2+ /negationslashk−m)−(/negationslashp2−m), so that C=/integraldisplayd4p (2π)4/bracketleftbigg tr/parenleftbigg γν 1 /negationslashp2−mγσ 1 /negationslashp1−mγλ 1 /negationslashp−m/parenrightbigg −tr/parenleftbigg γν 1 /negationslashp2+ /negationslashk−mγσ 1 /negationslashp1−mγλ 1 /negationslashp−m/parenrightbigg/bracketrightbigg (10) InBwrite/negationslashk=(/negationslashp1+ /negationslashk−m)−(/negationslashp1−m), so that B=/integraldisplayd4p (2π)4/bracketleftbigg tr(γν 1 /negationslashp2+ /negationslashk−mγσ 1 /negationslashp1−mγλ 1 /negationslashp−m) −tr/parenleftbigg γν 1 /negationslashp2+ /negationslashk−mγσ 1 /negationslashp1+ /negationslashk−mγλ 1 /negationslashp−m/parenrightbigg/bracketrightbigg (11) 148 | II. Dirac and the Spinor p + k k pp1 + k p2 + kq1 q2 −(q1 + q2 + k)λ σ νA k pp1 + k p2 + kq1 q2 −(q1 + q2 + k)λ σ νBp1 kpp2 p2+kq1 q2 −(q1 + q2 + k)λ σ νCp1 Figure II.7.2 II.7. Proof of Gauge Invariance | 149 Finally, in Awrite/negationslashk=(/negationslashp+ /negationslashk−m)−(/negationslashp−m) A=/integraldisplayd4p (2π)4/bracketleftbigg tr/parenleftbigg γν 1 /negationslashp2+ /negationslashk−mγσ 1 /negationslashp1+ /negationslashk−mγλ 1 /negationslashp−m/parenrightbigg −tr/parenleftbigg γν 1 /negationslashp2+ /negationslashk−mγσ 1 /negationslashp1+ /negationslashk−mγλ 1 /negationslashp+ /negationslashk−m/parenrightbigg/bracketrightbigg (12) Now you see what is happening. When we add the three diagrams together terms cancel in pairs, leaving us with A+B+C=/integraldisplayd4p (2π)4/bracketleftbigg tr/parenleftbigg γν 1 /negationslashp2−mγσ 1 /negationslashp1−mγλ 1 /negationslashp−m/parenrightbigg −tr/parenleftbigg γν 1 /negationslashp2+ /negationslashk−mγσ 1 /negationslashp1+ /negationslashk−mγλ 1 /negationslashp+ /negationslashk−m/parenrightbigg/bracketrightbigg (13) If we shift (see exercise II.7.2) the dummy integration variable p→p−kin the second term, we see that the two terms cancel. Indeed, the kμkρ/μ2piece in the photon propagator goes away and we can set μ=0. I will leave the general proof to you. We have done it for one particular process. Try it for some other process. You will see how it goes. Ward-Takahashi identity Let’s summarize. Given any physical amplitude Tμ...(k,...)with external electrons on shell [this is jargon for saying that all necessary factors u(p) and¯u(p) are included in Tμ...(k,...)] describing a process with a photon carrying momentum kcoming out of, or going into, a vertex labeled by the Lorentz index μ, we have kμTμ...(k,...)=0 (14) This is sometimes known as a Ward-Takahashi identity. The bottom line is that we can write iDμν=−igμν/k2for the photon propagator. Since we can discard the kμkν/μ2term in the photon propagator i(kμkν/μ2−gμν)/k2we can also add in a kμkν/k2term with an arbitrary coefficient. Thus, for the photon propagator we can use iDμν=i k2/bracketleftbigg (1−ξ)kμkν k2−gμν/bracketrightbigg (15) where we can choose the number ξto simplify our calculation as much as possible. Evidently, the choice of ξamounts to a choice of gauge for the electromagnetic field. In particular, the choice ξ=1 is known as the Feynman gauge, and the choice ξ=0 is known as the Landau gauge. If you find an especially nice choice, you can have a gauge named after you as well! For fairly simple calculations, it is often advisable to calculate with anarbitrary ξ. The fact that the end result must not depend on ξprovides a useful check on the arithmetic. 150 | II. Dirac and the Spinor This completes the derivation of the Feynman rules for quantum electrodynamics: They are the same rules as those given in chapter II.5 for the massive vector boson theory exceptfor the photon propagator given in (15). We have given here a diagrammatic proof of the gauge invariance of quantum electrodynamics. We will worry later (in chapter IV .7) about the possibility that the shift ofintegration momentum used in the proof may not be allowed in some cases. The longitudinal mode We now come back to the worry we had in chapter I.5. Consider a massive spin 1 mesonmoving along the z−direction. The 3 polarization vectors are fixed by the condition k λελ= 0 with kλ=(ω,0 ,0 , k)(recall chapter I.5) and the normalization ελελ=− 1, so that ε(1) λ= (0, 1, 0, 0 ),ε(2) λ=(0, 0, 1, 0 ),ε(3) λ=(−k ,0 ,0 ,ω)/μ . Note that as μ→0, the longitudinal polarization vector ε(3) λbecomes proportional to kλ=(ω,0 ,0 , −k) . The amplitude for emitting a meson with a longitudinal polarization in the process described by (14) isgiven by ε (3) λTλ...=(−kT0...+ωT3...)/μ=(−kT0...+/radicalbig k2+μ2T3...)/μ/similarequal(−kT0...+ (k+μ2 2k)T3...)/μ (forμ/lessmuchk), namely −(kλTλ.../μ)+μ 2kT3...withkλ=(k,0 ,0 ,−k) . Upon using (14) we see that the amplitude ε(3) λTλ...→μ 2kT3...→0a sμ→0. The longitudinal mode of the photon does not exist because it decouples from all physical processes. Here is an apparent paradox. Mr. Boltzmann tells us that in thermal equilibrium each degree of freedom is associated with1 2T. Thus, by measuring some thermal property (such as the specific heat) of a box of photon gas to an accuracy of 2 /3 an experimentalist could tell if the photon is truly massless rather than have a mass of a zillionth of an electron volt. The resolution is of course that as the coupling of the longitudinal mode vanishes as μ→0 the time it takes for the longitudinal mode to come to thermal equilibrium goes to infinity. Our crafty experimentalist would have to be very patient. Emission and absorption of photons According to chapter II.5, the amplitude for emitting or absorbing an external on-shell photon with momentum kand polarization a( a=1, 2)is given by ε(a) μ(k)Tμ...(k,...). Thanks to (14), we are free to vary the polarization vector ε(a) μ(k)→ε(a) μ(k)+λkμ (16) for arbitrary λ. You should recognize (16) as the momentum space version of (3). As we will see in the next chapter, by a judicious choice of ε(a) μ(k), we can simplify a given calculation considerably. In one choice, known as the “transverse gauge,” the 4-vectorsε (a) μ(k)=(0,/vectorε(k)) fora=1, 2 do not have time components. (For a photon moving in the z-direction, this is just the choice specified in the preceding section.) II.7. Proof of Gauge Invariance | 151 Exercises II.7.1 Extend the proof to cover figure II.7.1c. [Hint: To get oriented, note that figure II.7.1b corresponds to n=1.] II.7.2 You might have worried whether the shift of integration variable is allowed. Rationalizing the denomi- nators in the first integral /integraldisplayd4p (2π)4tr(γν 1 /negationslashp2−mγσ 1 /negationslashp1−mγλ 1 /negationslashp−m) in (13) and imagining doing the trace, you can convince yourself that this integral is only logarithmically divergent and hence that the shift is allowed. This issue will come up again in chapter IV .7 and we areanticipating a bit here. II.8 Photon-Electron Scattering and Crossing Photon scattering on an electron We now apply what we just learned to calculating the amplitude for the Compton scattering of a photon on an electron, namely, the process γ(k)+e(p)→γ(k/prime)+e(p/prime). First step: draw the Feynman diagrams, and notice that there are two, as indicated in figure II.8.1.The electron can either absorb the photon carrying momentum kfirst or emit the photon carrying momentum k /primefirst. Think back to the spacetime stories we talked about in chapter I.7. The plot of our biopic here is boringly simple: the electron comes along, absorbs andthen emits a photon, or emits and absorbs a photon, and then continues on its merry way.Because this is a quantum movie, the two alternate plots are shown superposed. So, apply the Feynman rules (chapter II.5) to get (just to make the writing a bit easier, we take the polarization vectors εandε /primeto be real) M=A(ε/prime,k/prime;ε,k)+(ε/prime↔ε,k/prime↔−k) (1) where A(ε/prime,k/prime;ε,k)=(−ie)2¯u(p/prime)/negationslashε/prime i /negationslashp+ /negationslashk−m/negationslashεu(p) =i(−ie)2 2pk¯u(p/prime)/negationslashε/prime(/negationslashp+ /negationslashk+m)/negationslashεu(p)(2) In either case, absorb first or emit first, the electron is penalized for not being real, by the factor of 1 /((p+k)2−m2)=1/(2pk) in one case, and 1 /((p−k/prime)2−m2)=− 1/(2pk/prime)in the other. At this point, to obtain the differential cross section, you just have take a deep breath and calculate away. I will show you, however, that we could simplify the calculation considerably by a clever choice of polarization vectors and of the frame of reference. (The calculation isstill a big mess, though!) For a change, we will be macho guys and not average and sumover the photon polarizations. II.8. Photon-Electron Scattering | 153 p + k p − k’ (b) (a)p’ /H9255, k/H9255’, k ’ pp’ /H9255, k p/H9255’, k ’ Figure II.8.1 In any case, we have εk=0 and ε/primek/prime=0. Now choose the transverse gauge introduced in the preceding chapter, so that εandε/primehave zero time components. Then calculate in the lab frame. Since p=(m,0 ,0 ,0 ), we have the additional relations εp=0 (3) and ε/primep=0 (4) Why is this a shrewd choice? Recall that /negationslasha/negationslashb=2ab−/negationslashb/negationslasha. Thus, we could move /negationslashppast /negationslashεor/negationslashε/primeat the rather small cost of flipping a sign. Notice in (1) that (/negationslashp+ /negationslashk+m)/negationslashεu(p) = /negationslashε(−/negationslashp− /negationslashk+m)u(p) =−/negationslashε/negationslashku(p) (where we have used εk=0.) Thus A(ε/prime,k/prime;ε,k)=ie2¯u(p/prime)/negationslashε/prime/negationslashε/negationslashk 2pku(p) (5) To obtain the differential cross section, we need |M|2. We will wimp out a bit and suppose, just as in chapter II.6, that the initial electron is unpolarized and the polarizationof the final electron is not measured. Then averaging over initial polarization and summingover final polarization we have [applying II.2.8] 1 2/Sigma1/Sigma1|A(ε/prime,k/prime;ε,k)|2=e4 2(2m)2(2pk)2tr(/negationslashp/prime+m)/negationslashε/prime/negationslashε/negationslashk(/negationslashp+m)/negationslashk/negationslashε/negationslashε/prime(6) In evaluating the trace, keep in mind that the trace of an odd number of gamma matrices vanishes. The term proportional to m2contains /negationslashk/negationslashk=k2=0 and hence van- ishes. We are left with tr (/negationslashp/prime/negationslashε/prime/negationslashε/negationslashk/negationslashp/negationslashk/negationslashε/negationslashε/prime)=2kptr(/negationslashp/prime/negationslashε/prime/negationslashε/negationslashk/negationslashε/negationslashε/prime)=− 2kptr(/negationslashp/prime/negationslashε/prime/negationslashε/negationslashε/negationslashk/negationslashε/prime) =2kptr(/negationslashp/prime/negationslashε/prime/negationslashk/negationslashε/prime)=8kp[2(kε/prime)2+k/primep]. 154 | II. Dirac and the Spinor Work through the steps as indicated and the strategy should be clear. We anticommute judiciously to exploit the “zero relations” εp=0,ε/primep=0,εk=0, and ε/primek/prime=0 and the normalization conditions /negationslashε/negationslashε=ε2=− 1 and /negationslashε/prime/negationslashε/prime=ε/prime2=− 1 as much as possible. We obtain 1 2/Sigma1/Sigma1|A(ε/prime,k/prime;ε,k)|2=e4 2(2m)2(2pk)28kp[2(kε/prime)2+k/primep] (7) The other term 1 2/Sigma1/Sigma1|A(ε ,−k;ε/prime,k/prime)|2=e4 2(2m)2(2pk)28(−k/primep)[2(k/primeε)2−kp] follows immediately by inspecting figure II.8.1 and interchanging (ε/prime↔ε,k/prime↔−k). Just as in chapter II.6, the interference term 1 2/Sigma1/Sigma1A(ε ,−k;ε/prime,k/prime)∗A(ε/prime,k/prime;ε,k)=e4 2(2m)2(2pk)(−2pk/prime)tr(/negationslashp/prime+m)/negationslashε/prime/negationslashε/negationslashk(/negationslashp+m)/negationslashk/prime/negationslashε/prime/negationslashε (8) is the most tedious to evaluate. Call the trace T. Clearly, it would be best to eliminate p/prime=p+k−k/prime, since we could “do more” with /negationslashpthan/negationslashp/prime. Divide and conquer: write T= P+Q1+Q2. First, massage P≡tr(/negationslashp+m)/negationslashε/prime/negationslashε/negationslashk(/negationslashp+m)/negationslashk/prime/negationslashε/prime/negationslashε=m2tr/negationslashε/prime/negationslashε/negationslashk/negationslashk/prime/negationslashε/prime/negationslashε+ tr/negationslashp/negationslashε/prime/negationslashε/negationslashk/negationslashp/negationslashk/prime/negationslashε/prime/negationslashε. In the second term, we could sail the first /negationslashppast the /negationslashεand/negationslashε/prime(ah, so nice to work in the rest frame for this problem!) to find the combination /negationslashp/negationslashk/negationslashp=2kp/negationslashp− m2/negationslashk. The m2term gives a contribution that cancels the first term in P, leaving us with P=2kptr/negationslashε/prime/negationslashε/negationslashp/negationslashk/prime/negationslashε/prime/negationslashε=2kptr/negationslashp/negationslashk/prime/negationslashε/prime(2ε/primeε−/negationslashε/prime/negationslashε)/negationslashε=8(kp)(k/primep)[2(εε/prime)2−1]. Similarly, Q1=tr/negationslashk/negationslashε/prime/negationslashε/negationslashk/negationslashp/negationslashk/prime/negationslashε/prime/negationslashε=− 2kε/primetr/negationslashk/negationslashp/negationslashk/prime/negationslashε/prime=− 8(ε/primek)2k/primepandQ2=− tr/negationslashk/prime/negationslashε/prime/negationslashε/negationslashk/negationslashp/negationslashk/prime/negationslashε/prime/negationslashε =8(εk/prime)2kp/prime. Putting it all together and writing kp/prime=k/primep=mω/primeandk/primep/prime=kp=mω,w ef i n d 1 2/Sigma1/Sigma1|M |2=e4 (2m)2/bracketleftbiggω/prime ω+ω ω/prime+4(εε/prime)2−2/bracketrightbigg (9) We calculate the differential cross section as in chapter II.6 with some minor differences since we are in the lab frame, obtaining dσ=m (2π)22ω/bracketleftBigg/integraldisplayd3k/prime 2ω/primed3p/prime Ep/primeδ(4)(k/prime+p/prime−k−p)/bracketrightBigg 1 2/summationdisplay/summationdisplay |M|2(10) As described in the appendix to chapter II.6, we could use (I.8.14) and write/integraltextd3p/prime Ep/prime(...)= /integraltext d4p/primeθ(p/prime0)δ(p/prime2−m2)(...). Doing the integral over d4p/primeto knock out the 4-dimensional delta function, we are left with a delta function enforcing the mass shell condition 0 = p/prime2−m2=(p+k−k/prime)2−m2=2p(k−k/prime)−2kk/prime=2m(ω−ω/prime)−2ωω/prime(1−cosθ), with θthe scattering angle of the photon. Thus, the frequency of the outgoing photon and of the incoming photon are related by ω/prime=ω 1+2ω msin2θ 2(11) giving the frequency shift that won Arthur Compton the Nobel Prize. You realize of course that this formula, though profound at the time, is “merely” relativistic kinematics and hasnothing to do with quantum field theory per se. II.8. Photon-Electron Scattering | 155 /H92552, k2 /H92551, k1 /H92552, k2/H92551, k1 p1 /H11002 k2p1 /H11002 k1 p2 p1p2 p1 (b) (a) Figure II.8.2 What quantum field theory gives us is the Klein-Nishina formula (1929) dσ d/Omega1=1 (2m)2(e2 4π)2(ω/prime ω)2/bracketleftbiggω/prime ω+ω ω/prime+4(εε/prime)2−2/bracketrightbigg . (12) You ought to be impressed by the year. Electron-positron annihilation Here and in chapter II.6 we calculated the cross sections for some interesting scattering processes. At the end of that chapter we marvelled at the magic of theoretical physics.Even more magical is the annihilation of matter and antimatter, a process that occursonly in relativistic quantum field theory. Specifically, an electron and a positron meet andannihilate each other, giving rise to two photons: e −(p1)+e+(p2)→γ(ε1,k1)+γ(ε2,k2). (Annihilating into one physical, that is, on-shell, photon is kinematically impossible.)This process, often featured in science fiction, is unknown in nonrelativistic quantummechanics. Without quantum field theory, you would be clueless on how to calculate, say,the angular distribution of the outgoing photons. But having come this far, you simply apply the Feynman rules to the diagrams in fig- ure II.8.2, which describe the process to order e 2. We find the amplitude M= A(k1,ε1;k2,ε2)+A(k 2,ε2;k1,ε1)(Bose statistics for the two photons!), where A(k1,ε1;k2,ε2)=(ie)(−ie) ¯v(p 2)/negationslashε2i /negationslashp1− /negationslashk1−m/negationslashε1u(p 1) (13) Students of quantum field theory are sometimes confused that while the incoming electron goes with the spinor u, the incoming positron goes with ¯v, and not with v. You could check this by inspecting the hermitean conjugate of (II.2.10). Even simpler, note that ¯v(...)u [with(...)a bunch of gamma matrices contracted with various momenta] transforms correctly under the Lorentz group, while v(...)udoes not (and does not even make sense, since they are both column spinors.) Or note that the annihilation operator dfor the positron is associated with ¯v, notv. 156 | II. Dirac and the Spinor I want to emphasize that the positron carries momentum p2=(+/radicalBig /vectorp2 2+m2,/vectorp2)on its way to that fatal rendezvous with the electron. Its energy p0 2=+/radicalBig /vectorp2 2+m2is manifestly positive. Nor is any physical particle traveling backward in time. The honest experimental- ist who arranged for the positron to be produced wouldn’t have it otherwise. Remembermy rant at the end of chapter II.2? In figure II.8.2a I have labeled the various lines with arrows indicating momentum flow. The external particles are physical and there would have been serious legal issuesif their energies were not positive. There is no such restriction on the virtual particlebeing exchanged, though. Which way we draw the arrow on the virtual particle is purelyup to us. We could reverse the arrow, and then the momentum label would becomep 2−k2=k1−p1: the time component of this “composite” 4-vector can be either positive or negative. To make the point totally clear, we could also label the lines by dotted arrows showing the flow of (electron) charge. Indeed, on the positron line, momentum and (electron) chargeflow in opposite directions. Crossing I now invite you to discover something interesting by staring at the expression in (13) fora while. Got it? Does it remind you of some other amplitude?No? How about looking at the amplitude for Compton scattering in (2)?Notice that the two amplitudes could be turned into each other (up to an irrelevant sign) by the exchange p↔p1,k↔−k1,p/prime↔−p2,k/prime↔k2,ε↔ε1,ε/prime↔ε2,u(p)↔u(p 1),u(p/prime)↔v(p 2) (14) This is known as crossing. Diagrammatically, we are effectively turning the diagrams in figures II.8.1 and II.8.2 into each other by 90◦rotations. Crossing expresses in precise terms what people who like to mumble something about negative energy traveling backward intime have in mind. Once again, it is advantageous to work in the electron rest frame and in the transverse gauge, so that we have ε 1p1=0 and ε2p1=0 as well as ε1k1=0 and ε2k1=0. Averaging over the electron and positron polarizations we obtain dσ d/Omega1=α2 8m/parenleftbiggω1 |/vectorp|/parenrightbigg/bracketleftbiggω/prime ω+ω ω/prime−4(εε/prime)2+2/bracketrightbigg (15) withω1=m(m+E)/(m +E−pcosθ),ω2=(E−m−pcosθ)ω1/m, andp=|/vectorp|and Ethe positron momentum and energy, respectively. II.8. Photon-Electron Scattering | 157 xtime y Figure II.8.3 Special relativity and quantum mechanics require antimatter The formalism in chapter II.2 makes it totally clear that antimatter is obligatory. For us to be able to add the operators bandd†in (II.2.10) they must carry the same electric charge, and thus banddcarry opposite charge. No room for argument there. Still, it would be comforting to have a physical argument that special relativity and quantum mechanicsmandate antimatter. Compton scattering offers a context for constructing a nice heuristic argument. Think of the process in spacetime. We have redrawn figure II.8.1a in figure II.8.3: the electron ishit by the photon at the point x, propagates to the point y, and emits a photon. We have assumed implicitly that (y 0−x0)>0, since we don’t know what propagating backward in time means. (If the reader knows how to build a time machine, let me know.) Butspecial relativity tells us that another observer moving by (along the 1-direction say) wouldsee the time difference (y /prime0−x/prime0)=coshϕ(y0−x0)−sinhϕ(y1−x1), which could be negative for large enough boost parameter ϕ, provided that (y1−x1)>( y0−x0), that is, if the separation between the two spacetime points xandywere spacelike. Then this observer would see the field disturbance propagating from ytox. Since we see negative electric charge propagating from xtoy, the other observer must see positive electric charge propagating from ytox. Without special relativity, as in nonrelativistic quantum mechanics, we simply write down the Sch ¨odinger equation for the electron and that is that. Special relativity allows different observers to see different time ordering and henceopposite charges flowing toward the future. Exercises II.8.1 Show that averaging and summing over photon polarizations amounts to replacing the square bracket in (9) by 2[ω/prime ω+ω ω/prime−sin2θ]. [Hint: We are working in the transverse gauge.] II.8.2 Repeat the calculation of Compton scattering for circularly polarized photons. This page intentionally left blank Part III Renormalization and Gauge Invariance This page intentionally left blank III.1 Cutting Off Our Ignorance Who is afraid of infinities? Not I, I just cut them off. —Anonymous An apparent sleight of hand The pioneers of quantum field theory were enormously puzzled by the divergent integralsthat they often encountered in their calculations, and they spent much of the 1930sand 1940s struggling with these infinities. Many leading lights of the day, driven todesperation, advocated abandoning quantum field theory altogether. Eventually, a so-calledrenormalization procedure was developed whereby the infinities were argued away andfinite physical results were obtained. But for many years, well into the late 1960s and eventhe 1970s many physicists looked upon renormalization theory suspiciously as a sleight ofhand. Jokes circulated that in quantum field theory infinity is equal to zero and that underthe rug in a field theorist’s office had been swept many infinities. Eventually, starting in the 1970s a better understanding of quantum field theory was developed through the efforts of Ken Wilson and many others. Field theorists graduallycame to realize that there is no problem of divergences in quantum field theory at all. Wenow understand quantum field theory as an effective low energy theory in a sense I willexplain briefly here and in more detail in chapter VIII.3. Field theory blowing up We have to see an infinity before we can talk about how to deal with infinities. Well, we saw one in chapter I.7. Recall that the order λ2correction (I.7.23) to the meson-meson scattering amplitude diverges. With K≡k1+k2, we have M=1 2(−iλ)2i2/integraldisplayd4k (2π)41 k2−m2+iε1 (K−k)2−m2+iε(1) As I remarked back in chapter I.7, even without doing any calculations we can see the problem that confounded the pioneers of quantum field theory. The integrand goes as 1 /k4 162 | III. Renormalization and Gauge Invariance for large kand thus the integral diverges logarithmically as/integraltext d4k/k4. (The ordinary integral/integraltext∞dr rndiverges linearly for n=0, quadratically for n=1, and so on, and/integraltext∞dr/r diverges logarithmically.) Since this divergence is associated with large values of kit is known as an ultraviolet divergence. To see how to deal with this apparent infinity, we have to distinguish between two con- ceptually separate issues, associated with the terrible names “regularization” and “renor-malization” for historical reasons. Parametrization of ignorance Suppose we are studying quantum electrodynamics instead of this artificial ϕ4theory. It would be utterly unreasonable to insist that the theory of an electron interacting with aphoton would hold to arbitrarily high energies. At the very least, with increasingly higherenergies other particles come in, and eventually electrodynamics becomes merely part ofa larger electroweak theory. Indeed, these days it is thought that as we go to higher andhigher energies the whole edifice of quantum field theory will ultimately turn out to be anapproximation to a theory whose identity we don’t yet know, but probably a string theoryaccording to some physicists. The modern view is that quantum field theory should be regarded as an effective low energy theory, valid up to some energy (or momentum in a Lorentz invariant theory) scale/Lambda1. We can imagine living in a universe described by our toy ϕ 4theory. As physicists in this universe explore physics to higher and higher momentum scales they will eventuallydiscover that their universe is a mattress constructed out of mass points and springs. Thescale/Lambda1is roughly the inverse of the lattice spacing. When I teach quantum field theory, I like to write “Ignorance is no shame” on the blackboard for emphasis when I get to this point. Every physical theory should have adomain of validity beyond which we are ignorant of the physics. Indeed were this not truephysics would not have been able to progress. It is a good thing that Feynman, Schwinger,Tomonaga, and others who developed quantum electrodynamics did not have to knowabout the charm quark for example. I emphasize that /Lambda1should be thought of as physical, parametrizing our threshold of ignorance, and not as a mathematical construct. 1Indeed, physically sensible quantum field theories should all come with an implicit /Lambda1. If anyone tries to sell you a field theory claiming that it holds up to arbitrarily high energies, you should check to see if he soldused cars for a living. (As I wrote this, a colleague who is an editor of Physical Review Letters told me that he worked as a garbage collector during high school vacations, adding jokinglythat this experience prepared him well for his present position.) 1We saw a particularly vivid example of this in chapter I.8. When we define a conducting plate as a surface on which a tangential electric field vanishes, we are ignorant of the physics of the electrons rushing about tocounter any such imposed field. At extremely high frequencies, the electrons can’t rush about fast enough andnew physics comes in, namely that high frequency modes do not see the plates. In calculating the Casimir force we parametrize our ignorance with a∼/Lambda1 −1. III.1. Cutting Off Our Ignorance | 163 Figure III.1.1 Thus, in evaluating (1) we should integrate only up to /Lambda1, known as a cutoff. We literally cut off the momentum integration (fig. III.1.1).2The integral is said to have been “regularized.” Since my philosophy in this book is to emphasize the conceptual rather than the computational, I will not actually do the integral but merely note that it is equal to2iClog(/Lambda1 2/K2)where Cis some numerical constant that you can compute if you want (see appendix 1 to this chapter). For the sake of simplicity I also assumed that m2<< K2 so that we could neglect m2in the integrand. It is convenient to use the kinematic variables s≡K2=(k1+k2)2,t≡(k1−k3)2, andu≡(k1−k4)2introduced in chapter II.6. (Writing out the kj’s explicitly in the center-of-mass frame, you see that s,t, anduare related to rather mundane quantities such as the center-of-mass energy and the scattering angle.)After all this, the meson-meson scattering amplitude reads M=−iλ+iCλ2[log/parenleftbigg/Lambda12 s/parenrightbigg +log/parenleftbigg/Lambda12 t/parenrightbigg +log/parenleftbigg/Lambda12 u/parenrightbigg ]+O(λ3) (2) This much is easy enough to understand. After regularization, we speak of cutoff- dependent quantities instead of divergent quantities, and Mdepends logarithmically on the cutoff. 2A. Zee, Einstein ’s Universe , p. 204. Cartooning schools apparently teach that physicists in general, and quantum field theorists in particular, all wear lab coats. 164 | III. Renormalization and Gauge Invariance What is actually measured Now that we have dealt with regularization, let us turn to renormalization, a terrible word because it somehow implies we are doing normalization again when in fact we haven’tyet. The key here is to imagine what we would tell an experimentalist about to measure meson-meson scattering. We tell her (or him if you insist) that we need a cutoff /Lambda1and she is not bothered at all; to an experimentalist it makes perfect sense that any given theoryhas a finite domain of validity. Our calculation is supposed to tell her how the scattering will depend on the center-of- mass energy and the scattering angle. So we show her the expression in (2). She points toλand exclaims, “What in the world is that?” We answer, “The coupling constant,” but she says, “What do you mean, coupling constant, it’s just a Greek letter!” A confused student, Confusio, who has been listening in, pipes up, “Why the fuss? I have been studying physics for years and years, and the teachers have shown us lots ofequations with Latin and Greek letters, for example, Hooke’s law F=−kx, and nobody gets upset about kbeing just a Latin letter.” Smart Experimentalist: “But that is because if you give me a spring I can go out and measure k. That’s the whole point! Mr. Egghead Theorist here has to tell me how to measure this λ.” Woah, that is a darn smart experimentalist. We now have to think more carefully what a coupling constant really means. Think about α, the coupling constant of quantum elec- trodynamics. Well, it is the coefficient of 1 /rin Coulomb’s law. Fine, Monsieur Coulomb measured it using metallic balls or something. But a modern experimentalist could justas well have measured αby scattering an electron at such and such an energy and at such and such a scattering angle off a proton. We explain all this to our experimentalistfriend. SE, nodding, agrees: “Oh yes, recently my colleague so and so measured the coupling for meson-meson interaction by scattering one meson off another at such and such an energyand at such and such a scattering angle, which correspond to your variables s,t, andu having values s 0,t0, andu0. But what does the coupling constant my colleague measured, let us call it λP, with the subscript meaning “physical,” have to do with your theoretical λ, which, as far as I am concerned, is just a Greek letter in something you call a Lagrangian!” Confusio, “Hey, if she’s going to worry about small lambda, I am going to worry about big lambda. How do I know how big the domain of validity is?” SE: “Confusio, you are not as dumb as you look! Mr. Egghead Theorist, if I use your formula (2), what is the precise value of /Lambda1that I am supposed to plug in? Does it depend on your mood, Mr. Theorist? If you wake up feeling optimistic, do you use 2 /Lambda1instead of /Lambda1? And if your girl friend left you, you use1 2/Lambda1?” We assert, “Ha, we know the answer to that one. Look at (2): Mis supposed to be an actual scattering amplitude and should not depend on /Lambda1. If someone wants to change /Lambda1 III.1. Cutting Off Our Ignorance | 165 we just shift λin such a way so that Mdoes not change. In fact, a couple lines of arithmetic will show you precisely what dλ/d/Lambda1 has to be (see exercise III.1.3).” SE: “Okay, so λis secretly a function of /Lambda1. Your notation is lousy.” We admit, “Exactly, this bad notation has confused generations of physicists.”SE: “I am still waiting to hear how the λ Pmy experimental colleague measured is related to your λ.” We say, “Aha, that’s easy. Just look at (2), which is repeated here for clarity and for your reading convenience: M=−iλ+iCλ2/bracketleftbigg log/parenleftbigg/Lambda12 s/parenrightbigg +log/parenleftbigg/Lambda12 t/parenrightbigg +log/parenleftbigg/Lambda12 u/parenrightbigg/bracketrightbigg +O(λ3) (3) According to our theory, λPis given by −iλP=−iλ+iCλ2/bracketleftbigg log/parenleftbigg/Lambda12 s0/parenrightbigg +log/parenleftbigg/Lambda12 t0/parenrightbigg +log/parenleftbigg/Lambda12 u0/parenrightbigg/bracketrightbigg +O(λ3) (4) To show you clearly what is involved, let us denote the sum of logarithms in the square bracket in (3) and in (4) by Land by L0, respectively, so that we can write (3) and (4) more compactly as M=−iλ+iCλ2L+O(λ3) (5) and −iλP=−iλ+iCλ2L0+O(λ3) (6) That is how λPandλare related.” SE: “If you give me the scattering amplitude expressed in terms of the physical coupling λPthen it’s of use to me, but it’s not of use in terms of λ. I understand what λPis, but notλ.” We answer: “Fine, it just takes two lines of algebra to eliminate λin favor of λP. Big deal. Solving (6) for λgives −iλ=−iλP−iCλ2L0+O(λ3)=−iλP−iCλ2 PL0+O(λ3 P) (7) The second equality is allowed to the order of approximation indicated. Now plug this into (5) M=−iλ+iCλ2L+O(λ3)=−iλP−iCλ2 PL0+iCλ2 PL+O(λ3 P) (8) Please check that all manipulations are legitimate up to the order of approximation indi- cated.” The “miracle” Lo and behold! The miracle of renormalization! Now in the scattering amplitude Mwe have the combination L−L0=[log(s0/s)+ log(t0/t)+log(u0/u)]. In other words, the scattering amplitude comes out as M=−iλP+iCλ2 P/bracketleftbigg log/parenleftbiggs0 s/parenrightbigg +log/parenleftbiggt0 t/parenrightbigg +log/parenleftbiggu0 u/parenrightbigg/bracketrightbigg +O(λ3 P) (9) 166 | III. Renormalization and Gauge Invariance We announce triumphantly to our experimentalist friend that when the scattering amplitude is expressed in terms of the physical coupling constant λPas she had wanted, the cutoff /Lambda1disappear completely! The answer should always be in terms of physically measurable quantities The lesson here is that we should express physical quantities not in terms of “fictitious” theoretical quantities such as λ, but in terms of physically measurable quantities such as λP. By the way, in the literature, λPis often denoted by λRand for historical reasons called the “renormalized coupling constant.” I think that the physics of “renormalization” is muchclearer with the alternative term “physical coupling constant,” hence the subscript P.W e never did have a “normalized coupling constant.” Suddenly Confusio pipes up again; we have almost forgotten him!Confusio: “You started out with an Min (2) with two unphysical quantities λand/Lambda1, and their “unphysicalness” sort of cancel each other out.” SE: “Yeah, it is reminiscent of what distinguishes the good theorists from the bad ones. The good ones always make an even number of sign errors, and the bad ones always makean odd number.” Integrating over only the slow modes In the path integral formulation, the scattering amplitude Mdiscussed here is obtained by evaluating the integral (chapter I.7) /integraldisplay Dϕ ϕ(x 1)ϕ(x 2)ϕ(x 3)ϕ(x4)ei/integraltext ddx{1 2[(∂ϕ)2−m2ϕ2]−λ 4!ϕ4} The regularization used here corresponds roughly to restricting ourselves, in the integral/integraltext Dϕ, to integrating over only those field configurations ϕ(x) whose Fourier transform ϕ(k) vanishes for k>∼/Lambda1. In other words, the fields corresponding to the internal lines in the Feynman diagrams in fig. (I.7.10) are not allowed to fluctuate too energetically. We willcome back to this path integral formulation later when we discuss the renormalizationgroup. Alternative lifestyles I might also mention that there are a number of alternative ways of regularizing Feyn- man diagrams, each with advantages and disadvantages that make them suitable for somecalculations but not others. The regularization used here, known as Pauli-Villars, has theadvantage of being physically transparent. Another often used regularization is known as III.1. Cutting Off Our Ignorance | 167 dimensional regularization. We pretend that we are calculating in d-dimensional space- time. After the Feynman integral has been beaten down to a suitable form, we do an analyticcontinuation in dand set d=4 at the end of the day. The cutoff dependences of various integrals now show up as poles as we let d→4. Just as the cutoff /Lambda1disappears when the scattering amplitude is expressed in terms of the physical coupling constant λ P, in di- mensional regularization the scattering amplitude expressed in terms of λPis free of poles. While dimensional regularization proves to be useful in certain contexts as I will note in alater chapter, it is considerably more abstract and formal than Pauli-Villars regularization.Each to his or her own taste when it comes to regularizing. Since the emphasis in this book is on the conceptual rather than the computational, I won’t discuss other regularization schemes but will merely sketch how Pauli-Villars anddimensional regularizations work in two appendixes to this chapter. Appendix 1: Pauli-Villars regularization The important message of this chapter is the conceptual point that when physical amplitudes are expressed in term of physical coupling constants the cutoff dependence disappears. The actual calculation of the Feynmanintegral is unimportant. But I will show you how to do the integral just in case you would like to do Feynmanintegrals for a living. Let us start with the convergent integral /integraldisplayd4k (2π)41 (k2−c2+iε)3=−i 32π2c2(10) The dependence on c2follows from dimensional analysis. The overall factor is calculated in appendix D. Applying the identity (D.15) 1 xy=/integraldisplay1 0dα1 [αx+(1−α)y ]2(11) to (1) we have M=1 2(−iλ)2i2/integraldisplayd4k (2π)4/integraldisplay1 0dα1 D with D=[α(K−k)2+(1−α)k2−m2+iε]2=[(k−αK)2+α(1−α)K2−m2+iε]2 Shift the integration variable k→k+αK and we meet the integral/integraltext [d4k/(2π)4][ 1/(k2−c2+iε)2], where c2=m2−α(1−α)k2. Pauli-Villars proposed replacing it by /integraldisplayd4k (2π)4/bracketleftbigg1 (k2−c2+iε)2−1 (k2−/Lambda12+iε)2/bracketrightbigg (12) with/Lambda12/greatermuchc2.F o rk much smaller than /Lambda1the added second term in the integrand is of order /Lambda1−4and is negligible compared to the first term since /Lambda1is much larger than c.F o rkmuch larger than /Lambda1, the two terms almost cancel and the integrand vanishes rapidly with increasing k, effectively cutting off the integral. Upon differentiating (12) with respect to c2and using (10) we deduce that (12) must be equal to (i/16π2)log(/Lambda12/c2). Thus, the integral /integraldisplay/Lambda1d4k (2π)41 (k2−c2+iε)2=i 16π2log/parenleftbigg/Lambda12 c2/parenrightbigg (13) is indeed logarithmically dependent on the cutoff, as anticipated in the text. 168 | III. Renormalization and Gauge Invariance For what it is worth, we obtain M=iλ2 32π2/integraldisplay1 0dαlog/parenleftbigg/Lambda12 m2−α(1−α)K2−iε/parenrightbigg (14) Appendix 2: Dimensional regularization The basic idea behind dimensional regularization is very simple. When we reach I=/integraltext [d4k/(2π)4][1/(k2−c2+iε)2] we rotate to Euclidean space and generalize to ddimensions (see appen- dix D): I(d)=i/integraldisplaydd Ek (2π)d1 (k2+c2)2=i/bracketleftbigg2πd/2 /Gamma1(d/ 2)/bracketrightbigg1 (2π)d/integraldisplay∞ 0dk kd−1 1 (k2+c2)2 As I said, I don’t want to get bogged down in computation in this book, but we’ve got to do what we’ve got to do. Changing the integration variable by setting k2+c2=c2/xwe find /integraldisplay∞ 0dk kd−1 1 (k2+c2)2=1 2cd−4/integraldisplay1 0dx(1−x)d/2−1x1−d/ 2, which we are supposed to recognize as the integral representation of the beta function. After the dust settles, we obtain i/integraldisplaydd Ek (2π)d1 (k2+c2)2=i1 (4π)d/2/Gamma1/parenleftbigg4−d 2/parenrightbigg cd−4(15) Asd→4, the right-hand side becomes i1 (4π)2/bracketleftbigg2 4−d−logc2+log(4π)−γ+O(d−4)/bracketrightbigg where γ=0.577 ...denotes the Euler-Mascheroni constant. Comparing with (13) we see that log /Lambda12in Pauli-Villars regularization has been effectively replaced by the pole 2/(4−d). As noted in the text, when physical quantities are expressed in terms of physical coupling constants, all such poles cancel. Exercises III.1.1 Work through the manipulations leading to (9) without referring to the text. III.1.2 Regard (1) as an analytic function of K2. Show that it has a cut extending from 4 m2to infinity. [Hint: If you can’t extract this result directly from (1) look at (14). An extensive discussion of this exercise will begiven in chapter III.8.] III.1.3 Change /Lambda1toe ε/Lambda1. Show that for M not to change, to the order indicated λmust change by δλ= 6εCλ2+O(λ3), that is, /Lambda1dλ d/Lambda1=6Cλ2+O(λ3) III.2 Renormalizable versus Nonrenormalizable Old view versus new view We learned that if we were to write the meson meson scattering amplitude in terms of a physically measured coupling constant λP, the dependence on the cutoff /Lambda1would disappear (at least to order λ2 P). Were we lucky or what? Well, it turns out that there are quantum field theories in which this would happen and that there are quantum field theories in which this would not happen, which gives us abinary classification of quantum field theories. Again, for historical reasons, the formerare known as “renormalizable theories” and are considered “nice.” The latter are knownas “nonrenormalizable theories,” evoking fear and loathing in theoretical physicists. Actually, with the new view of field theories as effective low energy theories to some underlying theory, physicists now look upon nonrenormalizable theories in a much moresympathetic light than a generation ago. I hope to make all these remarks clear in this anda later chapter. High school dimensional analysis Let us begin with some high school dimensional analysis. In natural units in which /planckover2pi=1 andc=1, length and time have the same dimension, the inverse of the dimension of mass (and of energy and momentum). Particle physicists tend to count dimension in terms ofmass as they are used to thinking of energy scales. Condensed matter physicists, on theother hand, usually speak of length scales. Thus, a given field operator has (equal and)opposite dimensions in particle physics and in condensed matter physics. We will use theconvention of the particle physicists. Since the action S≡/integraltext d 4xLappears in the path integral as eiS, it is clearly dimension- less, thus implying that the Lagrangian (Lagrangian density, strictly speaking) Lhas the same dimension as the 4th power of a mass. We will use the notation [ L]=4 to indicate 170 | III. Renormalization and Gauge Invariance that Lhas dimension 4. In this notation [ x]=− 1 and [ ∂]=1. Consider the scalar field theory L=1 2[(∂ϕ)2−m2ϕ2]−λϕ4. For the term (∂ϕ)2to have dimension 4, we see that [ϕ]=1 (since 2 (1+[ϕ])=4). This then implies that [ λ]=0, that is, the coupling λis dimen- sionless. The rule is simply that for each term in L, the dimensions of the various pieces, including the coupling constant and mass, have to add up to 4 (thus, e.g., [ λ]+4[ϕ]=4). How about the fermion field ψ? Applying this rule to the Lagrangian L=¯ψiγμ∂μψ+ ... we see that [ ψ]=3 2. (Henceforth we will suppress the ...; it is understood that we are looking at a piece of the Lagrangian. Furthermore, since we are doing dimensionalanalysis we will often suppress various irrelevant factors, such as numerical factors andthe gamma matrices in the Fermi interaction that we will come to presently.) Looking atthe coupling fϕ¯ψψ we see that the Yukawa coupling fis dimensionless. In contrast, in the theory of the weak interaction with L=G¯ψψ¯ψψ we see that the Fermi coupling G has dimension −2 (since −2+4( 3 2)=4; got that?). From the Maxwell Lagrangian −1 4FμνFμνwe see that [ Aμ]=1 and hence Aμhas the same dimension as ∂μ: The vector field has the same dimension as the scalar field. The electromagnetic coupling eAμ¯ψγμψtells us that eis dimensionless, which we can also deduce from Coulomb’s law written in natural units V( r)=α/r , with the fine structure constant α=e2/4π. Scattering amplitude blows up We are now ready for a heuristic argument regarding the nonrenormalizability of a theory. Consider Fermi’s theory of the weak interaction. Imagine calculating the amplitude M for a four-fermion interaction, say neutrino-neutrino scattering at an energy much smallerthan/Lambda1. In lowest order, M∼G. Let us try to write down the amplitude to the next order: M∼G+G 2(?), where we will try to guess what (?) is. Since all masses and energies are by definition small compared to the cutoff /Lambda1, we can simply set them equal to zero. Since [ G]=− 2, by high school dimensional analysis the unknown factor (?)must have dimension +2. The only possibility for (?)is/Lambda12. Hence, the amplitude to the next order must have the form M∼G+G2/Lambda12. We can also check this conclusion by looking at the Feynman diagram in figure III.2.1: Indeed it goes as G2/integraltext/Lambda1d4p(1/p)(1/p)∼G2/Lambda12. Without a cutoff on the theory, or equivalently with /Lambda1=∞ , theorists realized that the theory was sick: Infinity was the predicted value for a physical quantity. Fermi’s weakinteraction theory was said to be nonrenormalizable. Furthermore, if we try to calculate tohigher order, each power of Gis accompanied by another factor of /Lambda1 2. In desperation, some theorists advocated abandoning quantum field theory altogether. Others expended an enormous amount of effort trying to “cure” weak interaction theory.For instance, one approach was to speculate that the series (with coefficients suppressed) M∼G[1+G/Lambda1 2+(G/Lambda12)2+(G/Lambda12)4+...] summed to Gf (G/Lambda12), where the unknown function fmight have the property that f(∞) was finite. In hindsight, we now know that this is not a fruitful approach. III.2. Renormalization Issues | 171 ν ννν νν Figure III.2.1 Instead, what happened was that toward the late 1960s S. Glashow, A. Salam, and S. Weinberg, building on the efforts of many others, succeeded in constructing an elec-troweak theory unifying the electromagnetic and weak interactions, as I will discuss inchapter VII.2. Fermi’s weak interaction theory emerges within electroweak theory as alow energy effective theory. Fermi’s theory cried out In modern terms, we think of the cutoff /Lambda1as really being there and we hear the cutoff dependence of the four-fermion interaction amplitude M∼G+G2/Lambda12as the sound of the theory crying out that something dramatic has to happen at the energy scale /Lambda1∼(1/G)1 2. The second term in the perturbation series becomes comparable to the first, so at the veryleast perturbation theory fails. Here is another way of making the same point. Suppose that we don’t know anything about cutoff and all that. With Ghaving mass dimension −2, just by high school dimen- sional analysis we see that the neutrino-neutrino scattering amplitude at center-of-mass energy Ehas to go as M∼G+G 2E2+.... When Ereaches the scale ∼(1/G)1 2the am- plitude reaches order unity and some new physics must take over just because the cross section is going to violate the unitarity bound from basic quantum mechanics. (Rememberphase shift and all that?) In fact, what that something is goes back to Yukawa, who at the same time that he suggested the meson theory for the nuclear forces also suggested that an intermediatevector boson could account for the Fermi theory of the weak interaction. (In the 1930s thedistinction between the strong and the weak interactions was far from clear.) Schematically,consider a theory of a vector boson of mass Mcoupled to a fermion field via a dimensionless coupling constant g: L=¯ψ(iγμ∂μ−m)ψ−1 4FμνFμν+M2AμAμ+gAμ¯ψγμψ (1) 172 | III. Renormalization and Gauge Invariance k Figure III.2.2 Let’s calculate fermion-fermion scattering. The Feynman diagram in figure III.2.2 gen- erates an amplitude (−ig)2(¯uγμu)[i/(k2−M2+iε)](¯uγμu), which when the momentum transfer kis much less than Mbecomes i(g2/M2)(¯uγμu)(¯uγμu). But this is just as if the fermions are interacting via a Fermi theory of the form G(¯ψγμψ)(¯ψγμψ)withG=g2/M2. If we blithely calculate with the low energy effective theory G(¯ψγμψ)(¯ψγμψ), it cries out that it is going to fail. Yes sir indeed, at the energy scale (1/G)1 2=M/g , the vector boson is produced. New physics appears. I find it sobering and extremely appealing that theories in physics have the ability to announce their own eventual failure and hence their domains of validity, in contrast totheories in some other areas of human thought. Einstein’s theory is now crying out The theory of gravity is also notoriously nonrenormalizable. Simply comparing Newton’slawV( r)=G NM1M2/rwith Coulomb’s V( r)=α/r we see that Newton’s gravitational constant GNhas mass dimension −2. No more need be said. We come to the same morose conclusion that the theory of gravity, just like Fermi’s theory of weak interaction,is nonrenormalizable. To repeat the argument, if we calculate graviton-graviton scatteringat energy E, we encounter the series ∼[1+G NE2+(GNE2)2+...]. Just as in our discussion of the Fermi theory, the nonrenormalizability of quantum grav- ity tells us that at the Planck energy scale (1/GN)1 2≡MPlanck∼1019mproton new physics must appear. Fermi’s theory cried out, and the new physics turned out to be the elec- troweak theory. Einstein’s theory is now crying out. Will the new physics turn out to bestring theory? 1 Exercise III.2.1 Consider the d-dimensional scalar field theory S=/integraltext ddx(1 2(∂ϕ)2+1 2m2ϕ2+λϕ4+...+λnϕn+...). Show that [ ϕ]=(d−2)/2 and [ λn]=n(2−d)/2+d. Note that ϕis dimensionless for d=2. 1J. Polchinski, String Theory . III.3 Counterterms and Physical Perturbation Theory Renormalizability The heuristic argument of the previous chapter indicates that theories whose coupling has negative mass dimension are nonrenormalizable. What about theories with dimensionlesscouplings, such as quantum electrodynamics and the ϕ 4theory? As a matter of fact, both of these theories have been proved to be renormalizable. But it is much more difficult toprove that a theory is renormalizable than to prove that it is nonrenormalizable. Indeed,the proof that nonabelian gauge theory (about which more later) is renormalizable tookthe efforts of many eminent physicists, culminating in the work of ’t Hooft, Veltman, B.Lee, Zinn-Justin, and many others. Consider again the simple ϕ 4theory. First, a trivial remark: The physical coupling constant λPis a function of s0,t0, andu0[see (III.1.4)]. For theoretical purposes it is much less cumbersome to set s0,t0, and u0equal to μ2and thus use, instead of (III.1.4), the simpler definition −iλP=−iλ+3iCλ2log/parenleftbigg/Lambda12 μ2/parenrightbigg +O(λ3) (1) This is purely for theoretical convenience.1 We saw that to order λ2the meson-meson scattering amplitude when expressed in terms of the physical coupling λPis independent of the cutoff /Lambda1. How do we prove that this is true to all orders in λ? Dimensional analysis only tells us that to any order in λthe dependence of the meson scattering amplitude on the cutoff must be a sum of terms going as [log (/Lambda1/μ)]p with some power p. The meson-meson scattering amplitude is certainly not the only quantity that depends on the cutoff. Consider the inverse of the ϕpropagator to order λ2as shown in figure III.3.1. 1In fact, the kinematic point s0=t0=u0=μ2cannot be reached experimentally, but that’s of concern to theorists. 174 | III. Renormalization and Gauge Invariance kkq kk qp (a) (b) Figure III.3.1 The Feynman diagram in figure III.3.1a gives something like −iλ/integraldisplay/Lambda1/bracketleftbiggd4q (2π)4/bracketrightbigg/bracketleftbiggi q2−m2+iε/bracketrightbigg The precise value does not concern us; we merely note that it depends quadratically on the cutoff /Lambda1but not on k2. The diagram in figure III.3.1b involves a double integral I(k,m,/Lambda1;λ)≡(−iλ)2/integraldisplay/Lambda1/integraldisplay/Lambda1d4p (2π)4d4q (2π)4 i p2−m2+iεi q2−m2+iεi (p+q+k)2−m2+iε(2) Counting powers of pandqwe see that the integral ∼/integraltext (d8P/P6)and so Idepends quadratically on the cutoff /Lambda1. By Lorentz invariance Iis a function of k2, which we can expand in a series D+Ek2+ Fk4+.... The quantity Dis just Iwith the external momentum kset equal to zero and so depends quadratically on the cutoff /Lambda1. Next, we can obtain Eby differentiating Iwith respect to ktwice and then setting kequal to zero. This clearly decreases the powers of pand qin the integrand by 2 and so Edepends only logarithmically on the cutoff /Lambda1. Similarly, we can obtain Fby differentiating Iwith respect to kfour times and then setting kequal to zero. This decreases the powers of pandqin the integrand by 4 and thus Fis given by an integral that goes as ∼/integraltext d8P/P10for large P. The integral is convergent and hence cutoff independent. We can clearly repeat the argument ad infinitum. Thus Fand the terms in (...)are cutoff independent as the cutoff goes to infinity and we don’t have to worry about them. Putting it altogether, we have the inverse propagator k2−m2+a+bk2up toO(k2)with aandb, respectively, quadratically and logarithmically cutoff dependent. The propagator is changed to 1 k2−m2→1 (1+b)k2−(m2−a)(3) The pole in k2is shifted to m2 P≡m2+δm2≡(m2−a)(1+b)−1, which we identify as the physical mass. This shift is known as mass renormalization. Physically, it is quitereasonable that quantum fluctuations will shift the mass. III.3. Physical Perturbation Theory | 175 What about the fact that the residue of the pole in the propagator is no longer 1 but (1+b)−1? To understand this shift in the residue, recall that we blithely normalized the field ϕso that L=1 2(∂ϕ)2+.... That the coefficient of k2in the lowest order inverse propagator k2−m2is equal to 1 reflects the fact that the coefficient of1 2(∂ϕ)2inLis equal to 1. There is certainly no guarantee that with higher order corrections included the coefficientof 1 2(∂ϕ)2in an effective Lwill stay at 1. Indeed, we see that it is shifted to (1+b).F o r historical reasons, this is known as “wave function renormalization” even though there isno wave function anywhere in sight. A more modern term would be field renormalization.(The word renormalization makes some sense in this case, as we did normalize the fieldwithout thinking too much about it.) Incidentally, it is much easier to say “logarithmic divergent” than to say “logarithmically dependent on the cutoff /Lambda1,” so we will often slip into this more historical and less accurate jargon and use the word divergent. In ϕ 4theory, the wave function renormalization and the coupling renormalization are logarithmically divergent, while the mass renormalizationis quadratically divergent. Bare versus physical perturbation theory What we have been doing thus far is known as bare perturbation theory. We should haveput the subscript 0 on what we have been calling ϕ,m, andλ. The field ϕ 0is known as the bare field, and m0andλ0are known as the bare mass and bare coupling, respectively. I did not put on the subscript 0 way back in part I because I did not want to clutter up thenotation before you, the student, even knew what a field was. Seen in this light, using bare perturbation theory seems like a really stupid thing to do, and it is. Shouldn’t we start out with a zeroth order theory already written in terms ofthe physical mass m Pand physical coupling λPthat experimentalists actually measure, and perturb around that theory? Yes, indeed, and this way of calculating is known asrenormalized or dressed perturbation theory, or as I prefer to call it, physical perturbationtheory. We write L=1 2[(∂ϕ)2−m2 Pϕ2]−λP 4!ϕ4+A(∂ϕ)2+Bϕ2+Cϕ4(4) (A word on notation: The pedantic would probably want to put a subscript Pon the field ϕ, but let us clutter up the notation as little as possible.) Physical perturbation theory works as follows. The Feynman rules are as before, but with the crucial difference thatfor the coupling we use λ Pand for the propagator we write i/(k2−m2 P+iε) with the physical mass already in place. The last three terms in (4) are known as counterterms. The coefficients A,B, andCare determined iteratively (see later) as we go to higher and higher order in perturbation theory. They are represented as crosses in Feynman diagrams,as indicated in figure III.3.2, with the corresponding Feynman rules. All momentumintegrals are cut off. 176 | III. Renormalization and Gauge Invariance k+2i(Ak2 + B)i k2−mp2k k−iλp 4!iC Figure III.3.2 Let me now explain how A,B, and Care determined iteratively. Suppose we have determined them to order λN P. Call their values to this order AN,BN,andCN. Draw all the diagrams that appear in order λN+1 P. We determine AN+1,BN+1, andCN+1by requiring that the propagator calculated to the order λN+1 Phas a pole at mPwith a residue equal to 1, and that the meson-meson scattering amplitude evaluated at some specified values ofthe kinematic variables has the value −iλ P. In other words, the counterterms are fixed by the condition that mPandλPare what we say they are. Of course, A,B, andCwill be cutoff dependent. Note that there are precisely three conditions to determine the threeunknowns A N+1,BN+1, andCN+1. Explained in this way, you can see that it is almost obvious that physical perturbation theory works, that is, it works in the sense that all the physical quantities that we calculatewill be cutoff independent. Imagine, for example, that you labor long and hard to calculatethe meson-meson scattering amplitude to order λ 17 P. It would contain some cutoff depen- dent and some cutoff independent terms. Then you simply add a contribution given byC 17and adjust C17to cancel the cutoff dependent terms. But ah, you start to worry. You say, “What if I calculate the amplitude for two mesons to go into four mesons, that is, diagrams with six external legs? If I get a cutoff dependentanswer, then I am up the creek, as there is no counterterm of the form Dϕ 6in (4) to soak up the cutoff dependence.” Very astute of you, but this worry is covered by the following power counting theorem. Degree of divergence Consider a diagram with BEexternal ϕlines. First, a definition: A diagram is said to have a superficial degree of divergence Dif it diverges as /Lambda1D. (A logarithmic divergence log /Lambda1 counts as D=0.)The theorem says that Dis given by D=4−BE (5) III.3. Physical Perturbation Theory | 177 Figure III.3.3 I will give a proof later, but I will first illustrate what (5) means. For the inverse propagator, which has BE=2, we are told that D=2. Indeed, we encountered a quadratic divergence. For the meson-meson scattering amplitude, BE=4, and so D=0, and indeed, it is logarithmically divergent. According to the theorem, if you calculate a diagram with six external legs (what is technically sometimes known as the six-point function), BE=6 andD=− 2. The theorem says that your diagram is convergent or cutoff independent (i.e., the cutoff dependencedisappears as /Lambda1 −2). You didn’t have to worry. You should draw a few diagrams to check this point. Diagrams with more external legs are even more convergent. The proof of the theorem follows from simple power counting. In addition to BEand D, let us define BIas the number of internal lines, Vas the number of vertices, and L as the number of loops. (It is helpful to focus on a specific diagram such as the one infigure III.3.3 with B E=6,D=− 2,BI=5,V=4, and L=2.) The number of loops is just the number of/integraltext [d4k/(2π)4] we have to do. Each internal line carries with it a momentum to be integrated over, so we seem to have BIintegrals to do. But the actual number of integrals to be done is of course decreased by the momentumconservation delta functions associated with the vertices, one to each vertex. There are thusVdelta functions, but one of them is associated with overall momentum conservation of the entire diagram. Thus, the number of loops is L=BI−(V−1) (6) [If you have trouble following this argument, you should work things out for the diagram in figure III.3.3 for which this equation reads 2 =5−(4−1).] For each vertex there are four lines coming out (or going in, depending on how you look at it). Each external line comes out of (or goes into) one vertex. Each internal line connectstwo vertices. Thus, 4V=BE+2BI (7) (For figure III.3.3 this reads 4 .4=6+2.5.) 178 | III. Renormalization and Gauge Invariance Finally, for each loop there is a/integraltext d4kwhile for each internal line there is a i/(k2−m2+iε), bringing the powers of momentum down by 2. Hence, D=4L−2BI (8) (For figure III.3.3 this reads −2=4.2−2.5.) Putting (6), (7), and (8) together, we obtain the theorem (5).2As you can plainly see, this formalizes the power counting we have been doing all along. Degree of divergence with fermions To test if you understood the reasoning leading to (5), consider the Yukawa theory we metin (II.5.19). (We suppress the counterterms for typographical clarity.) L=¯ψ(iγμ∂μ−mP)ψ+1 2[(∂ϕ)2−μ2 Pϕ2]−λPϕ4+fPϕ¯ψψ (9) Now we have to count FIandFE, the number of internal and external fermion lines, respectively, and keep track of VfandVλ, the number of vertices with the coupling fand λ, respectively. We have five equations altogether. For instance, (7) splits into two equationsbecause we now have to count fermion lines as well as boson lines. For example, we nowhave Vf+4Vλ=BE+2BI (10) We (that is, you) finally obtain D=4−BE−3 2FE (11) So the divergent amplitudes, that is, those classes of diagrams with D≥0, have (BE,FE)= (0, 2),(2, 0),(1, 2), and(4, 0). We see that these correspond to precisely the six terms in the Lagrangian (9), and thus we need six counterterms. Note that this counting of superficial powers of divergence shows that all terms with mass dimension ≤4 are generated. For example, suppose that in writing down the La- grangian (9) we forgot to include the λPϕ4term. The theory would demand that we include this term: We have to introduce it as a counterterm [the term with (BE,FE)=(4, 0)in the list above]. A common feature of (5) and (11) is that they both depend only on the number of external lines and not on the number of vertices V. Thus, for a given number of external lines, no matter to what order of perturbation theory we go, the superficial degree of divergenceremains the same. Further thought reveals that we are merely formalizing the dimension-counting argument of the preceding chapter. [Recall that the mass dimension of a Bosefield [ϕ] is 1 and of a Fermi field [ ψ]i s 3 2. Hence the coefficients 1 and3 2in (11).] 2The superficial degree of divergence measures the divergence of the Feynman diagram as all internal momenta are scaled uniformly by k→akwithatending to infinity. In a more rigorous treatment we have to worry about the momenta in some subdiagram (a piece of the full diagram) going to infinity with other momenta held fixed. III.3. Physical Perturbation Theory | 179 Our discussion hardly amounts to a rigorous proof that theories such as the Yukawa theory are renormalizable. If you demand rigor, you should consult the many field theorytomes on renormalization theory, and I do mean tomes, in which such arcane topics asoverlapping divergences and Zimmerman’s forest formula are discussed in exhaustive andexhausting detail. Nonrenormalizable field theories It is instructive to see how nonrenormalizable theories reveal their unpleasant personali-ties when viewed in the context of this discussion. Consider the Fermi theory of the weakinteraction written in a simplified form: L=¯ψ(iγμ∂μ−mP)ψ+G(¯ψψ)2. The analogs of (6), (7), and (8) now read L=FI−(V−1),4V=FE+2FI, and D= 4L−FI. Solving for the superficial degree of divergence in terms of the external number of fermion lines, we find D=4−3 2FE+2V (12) Compared to the corresponding equations for renormalizable theories (5) and (11), D now depends on V. Thus if we calculate fermion-fermion scattering (FE=4), for example, the divergence gets worse and worse as we go to higher and higher order in the perturbationseries. This confirms the discussion of the previous chapter. But the really bad news isthat for any F E, we would start running into divergent diagrams when Vgets sufficiently large, so we would have to include an unending stream of counterterms (¯ψψ)3,(¯ψψ)4, (¯ψψ)5,..., each with an arbitrary coupling constant to be determined by an experimental measurement. The theory is severely limited in predictive power. At one time nonrenormalizable theories were considered hopeless, but they are accepted in the modern view based on the effective field theory approach, which I will discuss inchapter VIII.3. Dependence on dimension The superficial degree of divergence clearly depends on the dimension dof spacetime since each loop is associated with/integraltext ddk. For example, consider the Fermi interaction G(¯ψψ)2in (1+1)-dimensional spacetime. Of the three equations that went into (12) one is changed toD=2L−FI, giving D=2−1 2FE (13) the analog of (12) for 2-dimensional spacetime. In contrast to (12), Vno longer enters, and the only superficial diagrams have FE=2 and 4, which we can cancel by the appropriate 180 | III. Renormalization and Gauge Invariance counterterms. The Fermi interaction is renormalizable in (1+1)-dimensional spacetime. I will come back to it in chapter VII.4. The Weisskopf phenomenon I conclude by pointing out that the mass correction to a Bose field and to a Fermi fielddiverge differently. Since this phenomenon was first discovered by Weisskopf, I refer toit as the Weisskopf phenomenon. To see this go back to (11) and observe that B EandFE contribute differently to the superficial degree of divergence D.F o rBE=2,FE=0, we haveD=2, and thus the mass correction to a Bose field diverges quadratically, as we have already seen explicitly [the quantity ain (3)]. But for FE=2,BE=0, we have D=1 and it looks like the fermion mass is linearly divergent. Actually, in 4-dimensional field theorywe cannot possibly get a linear dependence on the cutoff. To see this, it is easiest to look at,as an example, the Feynman integral you wrote down for exercise II.5.1, for the diagramin figure II.5.1: (if )2i2/integraldisplayd4k (2π)41 k2−μ2/negationslashp+ /negationslashk+m (p+k)2−m2≡A(p2)/negationslashp+B(p2) (14) where I define the two unknown functions A(p2)andB(p2)for convenience. (For the purpose of this discussion it doesn’t matter whether we are doing bare or physical pertur-bation theory. If the latter, then I have suppressed the subscript Pfor the sake of notational clarity.) Look at the integrand for large k. You see that the integral goes as/integraltext /Lambda1d4k(/negationslashk/k4)and looks linearly divergent, but by reflection symmetry k→−kthe integral to leading order vanishes. The integral in (14) is merely logarithmically divergent. The superficial degreeof divergence Doften gives an exaggerated estimate of how bad the divergence can be (hence the adjective “superficial”). In fact, staring at (14) we can prove more, for instance,thatB(p 2)must be proportional to m. As an exercise you can show, using the Feynman rules given in chapter II.5, that the same conclusion holds in quantum electrodynamics. For a boson, quantum fluctuations give δμ∝/Lambda12/μ, while for a fermion such as the electron, quantum correction to its mass δm∝mlog(/Lambda1/m) is much more benign. It is interesting to note that in the early twentieth century physicists thought of the electronas a ball of charge of radius a. The electrostatic energy of such a ball, of the order e 2/a, was identified as the electron mass. Interpreting 1 /aas/Lambda1, we could say that in classical physics the electron mass is proportional to /Lambda1and diverges linearly. Thus, one way of stating the Weisskopf phenomenon is that “bosons behave worse than a classical charge,but fermions behave better.” As Weisskopf explained in 1939, the difference in the degree of the divergence can be understood heuristically in terms of quantum statistics. The “bad” behavior of bosons has to do with their gregariousness. A fermion would push away the virtual fermions fluctuat-ing in the vacuum, thus creating a cavity in the vacuum charge distribution surrounding III.3. Physical Perturbation Theory | 181 it. Hence its self-energy is less singular than would be the case were quantum statistics not taken into account. A boson does the opposite. The “bad” behavior of bosons will come back to haunt us later. Power of /planckover2picounts the number of loops This is a convenient place to make a useful observation, though unrelated to divergences and cut-off dependence. Suppose we restore Planck’s unit of action /planckover2pi. In the path integral, the integrand becomes eiS//planckover2pi(recall chapter I.2) so that effectively L→L//planckover2pi. Consider L=−1 2ϕ(∂2+m2)ϕ−λ 4!ϕ4, just to be definite. The coupling λ→λ//planckover2pi, so that each vertex is now associated with a factor of 1 //planckover2pi. Recall that the propagator is essentially the inverse of the operator (∂2+m2), and so in momentum space 1 /(k2−m2)→/planckover2pi/(k2−m2). Thus the powers of /planckover2piare given by the number of internal lines minus the number of vertices P=BI−V=L−1, where we used (6). You can check that this holds in general and not just for ϕ4theory. This observation shows that organizing Feynman diagrams by the number of loops amounts to an expansion in Planck’s constant (sometimes called a semi-classical expan-sion), with the tree diagrams providing the leading term. We will come across this againin chapter IV .3. Exercises III.3.1 Show that in (1+1)-dimensional spacetime the Dirac field ψhas mass dimension1 2, and hence the Fermi coupling is dimensionless. III.3.2 Derive (11) and (13). III.3.3 Show that B(p2)in (14) vanishes when we set m=0. Show that the same behavior holds in quantum electrodynamics. III.3.4 We showed that the specific contribution (14) to δmis logarithmically divergent. Convince yourself that this is actually true to any finite order in perturbation theory. III.3.5 Show that the result P=L−1 holds for all the theories we have studied. III.4 Gauge Invariance: A Photon Can Find No Rest When the central identity blows up I explained in chapter I.7 that the path integral for a generic field theory can be formally evaluated in what deserves to be called the Central Identity of Quantum Field Theory: /integraldisplay Dϕe−1 2ϕ.K.ϕ−V( ϕ) +J.ϕ=e−V( δ / δJ)e1 2J.K−1.J(1) For any field theory we can always gather up all the fields, put them into one giant column vector, and call the vector ϕ. We then single out the term quadratic in ϕwrite it as1 2ϕ.K.ϕ, and call the rest V( ϕ) . I am using a compact notation in which spacetime coordinates and any indices on the field, including Lorentz indices, are included in the indices of the formalmatrix K. We will often use (1) with V=0: /integraldisplay Dϕe−1 2ϕ.K.ϕ+J.ϕ=e1 2J.K−1.J(2) But what if Kdoes not have an inverse? This is not an esoteric phenomenon that occurs in some pathological field theory, but in one of the most basic actions of physics, the Maxwell action S(A)=/integraldisplay d4xL=/integraldisplay d4x/bracketleftBig 1 2Aμ(∂2gμν−∂μ∂ν)Aν+AμJμ/bracketrightBig . (3) The formal matrix Kin (2) is proportional to the differential operator (∂2gμν−∂μ∂ν)≡ Qμν. A matrix does not have an inverse if some of its eigenvalues are zero, that is, if when acting on some vector, the matrix annihilates that vector. Well, observe that Qμνannihilates vectors of the form ∂ν/Lambda1(x) :Qμν∂ν/Lambda1(x)=0. Thus Qμνhas no inverse. There is absolutely nothing mysterious about this phenomenon; we have already en- countered it in classical physics. Indeed, when we first learned the concepts of electricity, we were told that only the “voltage drop” between two points has physical meaning. Ata more sophisticated level, we learned that we can always add any constant (or indeedany function of time) to the electrostatic potential (which is of course just “voltage”) since III.4. Gauge Invariance | 183 by definition its gradient is the electric field. At an even more sophisticated level, we see that solving Maxwell’s equation (which of course comes from just extremizing the action)amounts to finding the inverse Q −1. [In the notation I am using here Maxwell’s equation ∂μFμν=Jνis written as QμνAν=Jμ, and the solution is Aν=(Q−1)νμJμ.] Well,Q−1does not exist! What do we do? We learned that we must impose an additional constraint on the gauge potential Aμ, known as “fixing a gauge.” A mundane nonmystery To emphasize the rather mundane nature of this gauge fixing problem (which some older texts tend to make into something rather mysterious and almost hopelessly dif-ficult to understand), consider just an ordinary integral/integraltext +∞ −∞dAe−A.K.A, with A= (a,b)a 2-component vector and K=/parenleftBig10 00/parenrightBig , a matrix without an inverse. Of course you realize what the problem is: We have/integraltext+∞ −∞/integraltext+∞ −∞da db e−a2and the integral over bdoes not exist. To define the integral we insert into it a delta function δ(b−ξ). The integral becomes defined and actually does not depend on the arbitrary num-berξ. More generally, we can insert δ[f( b) ] with fsome function of our choice. In the context of an ordinary integral, this procedure is of course ludicrous overkill, butwe will use the analog of this procedure in what follows. In this baby problem, wecould regard the variable b, and thus the integral over it, as “redundant.” As we will see, gauge invariance is also a redundancy in our description of massless spin 1 parti-cles. A massless spin 1 field is intrinsically different from a massive spin 1 field—that’s the crux of the problem. The photon has only two polarization degrees of freedom. (You alreadylearned in classical electrodynamics that an electromagnetic wave has two transversedegrees of freedom.) This is the true physical origin of gauge invariance. In this sense, gauge invariance is, strictly speaking, not a “real” symmetry but merely a reflection of the fact that we used a redundant description: a Lorentz vector field to describetwo physical degrees of freedom. Restricting the functional integral I will now discuss the method for dealing with this redundancy invented by Faddeev and Popov. As you will see presently, it is the analog of the method we used in our baby problem above. Even in the context of electromagnetism this method is a bit of overkill, but it willprove to be essential for nonabelian gauge theories (as we will see in chapter VII.1) andfor gravity. I will describe the method using a completely general and somewhat abstract language. In the next section, I will then apply the discussion here to a specific example.If you have some trouble with this section, you might find it helpful to go back and forthbetween the two sections. 184 | III. Renormalization and Gauge Invariance Suppose we have to do the integral I≡/integraltext DAeiS(A); this can be an ordinary integral or a path integral. Suppose that under the transformation A→Agthe integrand and the measure do not change, that is, S(A)=S(Ag)andDA=DAg. The transformations obviously form a group, since if we transform again with g/prime, the integrand and the measure do not change under the combined effect of gandg/primeandAg→(Ag)g/prime=Agg/prime. We would like to write the integral Iin the form I=(/integraltext Dg)J , with Jindependent of g. In other words, we want to factor out the redundant integration over g. Note that Dg is the invariant measure over the group of transformations and/integraltext Dg is the volume of the group. Be aware of the compactness of the notation in the case of a path integral: Aandgare both functions of the spacetime coordinates x. I want to emphasize that this hardly represents anything profound or mysterious. If you have to do the integral I=/integraltext dx dy eiS(x ,y)withS(x ,y)some function of x2+y2, you know perfectly well to go to polar coordinates I=(/integraltext dθ)J=(2π)J , where J=/integraltext dr reiS(r) is an integral over the radial coordinate ronly. The factor 2 πis precisely the volume of the group of rotations in 2 dimensions. Faddeev and Popov showed how to do this “going over to polar coordinates” in a unified and elegant way. Following them, we first write the numeral “one” as1=/Delta1(A)/integraltext Dgδ [f( A g)], an equality that merely defines /Delta1(A) . Here fis some function of our choice and /Delta1(A) , known as the Faddeev-Popov determinant, of course depends on f. Next, note that [ /Delta1(Ag/prime)]−1=/integraltext Dgδ [f( Ag/primeg)]=/integraltext Dg/prime/primeδ[f( Ag/prime/prime)]=[/Delta1(A)]−1, where the second equality follows upon defining g/prime/prime=g/primegand noting that Dg/prime/prime=Dg. In other words, we showed that /Delta1(A) =/Delta1(A g): the Faddeev-Popov determinant is gauge invariant. We now insert 1 into the integral Iwe have to do: I=/integraldisplay DAeiS(A) =/integraldisplay DAeiS(A)/Delta1(A)/integraldisplay Dgδ [f( Ag)] =/integraldisplay Dg/integraldisplay DAeiS(A)/Delta1(A)δ [f( Ag)] (4) As physicists and not mathematicians, we have merrily interchanged the order of integration. At the physicist’s level of rigor, we are always allowed to change integration variables until proven guilty. So let us change AtoAg−1; then I=/parenleftbigg/integraldisplay Dg/parenrightbigg/integraldisplay DAeiS(A)/Delta1(A)δ [f (A)] (5) where we have used the fact that DA ,S(A) , and/Delta1(A) are all invariant under A→Ag−1. That’s it. We’ve done it. The group integration (/integraltext Dg) has been factored out. The volume of a compact group is finite, but in gauge theories there is a separate group at every point in spacetime, and hence (/integraltext Dg) is an infinite factor. (This also explains why there is no gauge fixing problem in theories with global symmetries introduced in III.4. Gauge Invariance | 185 chapter I.10.) Fortunately, in the path integral Zfor field theory we do not care about overall factors in Z, as was explained in chapter I.3, and thus the factor (/integraltext Dg) can simply be thrown away. Fixing the electromagnetic gauge Let us now apply the Faddeev-Popov method to electromagnetism. The transformationleaving the action invariant is of course A μ→Aμ−∂μ/Lambda1,s ogin the present context is denoted by /Lambda1andAg≡Aμ−∂μ/Lambda1. Note also that since the integral Iwe started with is independent of fit is still independent of fin spite of its appearance in (5). Choose f (A)=∂A−σ, where σis a function of x. In particular, Iis independent of σand so we can integrate Iwith an arbitrary functional of σ, in particular, the functional e−(i/2ξ)/integraltext d4xσ(x)2. We now turn the crank. First, we calculate [/Delta1(A)]−1≡/integraldisplay Dgδ [f( Ag)]=/integraldisplay D/Lambda1δ(∂A −∂2/Lambda1−σ) (6) Next we note that in (5) /Delta1(A) appears multiplied by δ[f (A)] and so in evaluating [/Delta1(A)]−1 in (6) we can effectively set f (A)=∂A−σto zero. Thus from (6) we have /Delta1(A) “=” [/integraltext D/Lambda1δ(∂2/Lambda1)]−1. But this object does not even depend on A, so we can throw it away. Thus, up to irrelevant overall factors that could be thrown away Iis just/integraltext DAeiS(A)δ(∂A−σ). Integrating over σ(x) as we said we were going to do, we finally obtain Z=/integraldisplay Dσe−(i/2ξ)/integraltext d4xσ(x)2/integraldisplay DAeiS(A)δ(∂A−σ) =/integraldisplay DAeiS(A)−(i/2 ξ)/integraltext d4x(∂A)2(7) Nifty trick by Faddeev and Popov, eh? Thus, S(A) in (3) is effectively replaced by Seff(A)=S(A)−1 2ξ/integraldisplay d4x(∂A)2 =/integraldisplay d4x/braceleftbigg1 2Aμ/bracketleftbigg ∂2gμν−/parenleftbigg 1−1 ξ/parenrightbigg ∂μ∂ν/bracketrightbigg Aν+AμJμ/bracerightbigg (8) andQμνbyQμν eff=∂2gμν−(1−1/ξ)∂μ∂νor in momentum space Qμν eff=−k2gμν+(1− 1/ξ)kμkν, which does have an inverse. Indeed, you can check that Qμν eff/bracketleftbigg −gνλ+(1−ξ)kνkλ k2/bracketrightbigg1 k2=δμ λ Thus, the photon propagator can be chosen to be (−i) k2/bracketleftbigg gνλ−(1−ξ)kνkλ k2/bracketrightbigg (9) in agreement with the conclusion in chapter II.7. 186 | III. Renormalization and Gauge Invariance While the Faddeev-Popov argument is a lot slicker, many physicists still prefer the explicit Feynman argument given in chapter II.7. I do. When we deal with the Yang-Mills theoryand the Einstein theory, however, the Faddeev-Popov method is indispensable, as I havealready noted. A photon can find no rest Let us understand the physics behind the necessity for imposing by hand a (gauge fixing)constraint in gauge theories. In chapter I.5 we sidestepped this whole issue of fixing thegauge by treating the massive vector meson instead of the photon. In effect, we changedQ μνto(∂2+m2)gμν−∂μ∂ν, which does have an inverse (in fact we even found the inverse explicitly). We then showed that we could set the mass mto 0 in physical calculations. There is, however, a huge and intrinsic difference between massive and massless particles. Consider a massive particle moving along. We can always boost it to its restframe, or in more mathematical terms, we can always Lorentz transform the momentumof a massive particle to the reference momentum q μ=m(1, 0, 0, 0 ). (As is the case elsewhere in this book, if there is no risk of confusion, we write column vectors as rowvectors for typographical convenience.) To study the spin degrees of freedom, we shouldevidently sit in the rest frame of the particle and study how its states respond to rotation.The fancy pants way of saying this is we should study how the states of the particletransform under that particular subgroup of the Lorentz group (known as the little group)consisting of those Lorentz transformations /Lambda1that leave q μinvariant, namely /Lambda1μ νqν=qμ. Forqμ=m(1, 0, 0, 0 ), the little group is obviously the rotation group SO( 3). We then apply what we learned in nonrelativistic quantum mechanics and conclude that a spin jparticle has(2j+1)spin states (or polarizations in classical physics), as already noted back in chapter I.5. But if the particle is massless, we can no longer find a Lorentz boost that would bring us to its rest frame. A photon can find no rest! For a massless particle, the best we can do is to transform the particle’s momentum to the reference momentum qμ=ω(1, 0, 0, 1 )for some arbitrarily chosen ω. Again, this is just a fancy way of saying that we can always call the direction of motion the third axis. What isthe little group that leaves q μinvariant? Obviously, rotations around the third axis, forming the group O(2), leave qμinvariant. The spin states of a massless particle of any spin around its direction of motion are known as helicity states, as was already mentioned in chapter II.1. For a particle of spin j, the helicities ±j are transformed into each other by parity and time reversal, and thus both helicities must be present if the interactions the particleparticipates in respect these discrete symmetries, as is the case with the photon and thegraviton. 1In particular, the photon, as we have seen repeatedly, has only two polarization degrees of freedom, instead of three, since we no longer have the full rotation group SO( 3). 1But not with the neutrino. III.4. Gauge Invariance | 187 (You already learned in classical electrodynamics that an electromagnetic wave has two transverse degrees of freedom.) For more on this, see appendix B. In this sense, gauge invariance is strictly speaking not a “real” symmetry but merely a reflection of the fact that we used a redundant description: we used a vector field Aμwith its four degrees of freedom to describe two physical degrees of freedom. This is the truephysical origin of gauge invariance. The condition /Lambda1 μ νqν=qμshould leave us with a 3-parameter subgroup. To find the other transformations, it suffices to look in the neighborhood of the identity, that is, at Lorentztransformations of the form /Lambda1(α ,β)=I+αA+βB+.... By inspection, we see that A=⎛ ⎜⎜⎜⎜⎜⎝010 0 100 −1 000 0 010 0⎞ ⎟⎟⎟⎟⎟⎠=i(K 1+J2), B=⎛ ⎜⎜⎜⎜⎜⎝001 0 000 0 100 −1 001 0⎞ ⎟⎟⎟⎟⎟⎠=i(K 2−J1) (10) where we used the notation for the generators of the Lorentz group from chapter II.3. Note thatAandBare to a large extent determined by the fact that JandKare symmetric and antisymmetric, respectively. By direct computation or by invoking the celebrated minus sign in (II.3.9), we find that [A,B]=0. Also, [J3,A]=Band [J3,B]=−Aso that, as expected, (A,B)form a 2-component vector under O(2)rotations around the third axis. (For those who must know, the generators A,B, andJ3generate the group ISO( 2), the invariance group of the Euclidean 2-plane, consisting of two translations and one rotation.) The preceding paragraph establishing the little group for massless particles applies for any spin, including zero. Now specialize to a spin 1 massless particle with the twopolarization vectors /epsilon1 ±(q)=(1/√ 2)(0, 1, ±i,0). The polarization vectors are defined by how they transform under rotation eiφJ 3. So it is natural to ask how /epsilon1±(q)transform under /Lambda1(α ,β). Inspecting (10), we see that /epsilon1±(q)→/epsilon1±(q)+1√ 2(α±iβ)q (11) We recognize (11) as a gauge transformation (as was explained in chapter II.7). For a mass- less spin 1 particle, the gauge transformation is contained in the Lorentz transformations! Suppose we construct the corresponding spin 1 field as in chapter II.5 [and in analogy to (I.8.11) and (II.2.10)]: Aμ(x)=/integraldisplayd3k/radicalbig (2π)32ωk/summationdisplay α=1, 2[a(α)(/vectork)ε(α) μ(k)e−i(ω kt−/vectork./vectorx)+a†(α)(/vectork)ε∗(α) μ(k)ei(ωkt−/vectork./vectorx))] (12) withωk=|/vectork|. The polarization vectors ε(α) μ(k)are of coursed determined by the condition kμε(α) μ(k)=0, which we could easily satisfy by defining ε(α)(k)=/Lambda1(q→k)ε(α)(q), where /Lambda1(q→k)denotes a Lorentz transformation that brings the reference momentum qtok. Note that the ε(α)(k) thus constructed has a vanishing time component. (To see this, first boost ε(α) μ(q) along the third axis and then rotate, for example.) Hence, kμε(α) μ(k)= −/vectork./vectorε(α)(k)=0. These properties of ε(α) μ(k) translate into A0(x)=0 and− →∇./vectorA(x)=0. 188 | III. Renormalization and Gauge Invariance These two constraints cut the four degrees of freedom contained in Aμ(x) down to two and fix what is known as the Coulomb or radiation gauge.2 Given the enormous importance of gauge invariance, it might be instructive to review the logic underlying the “poor man’s approach” to gauge invariance (which, as I mentionedin chapter I.5, I learned from Coleman) adopted in this book for pedagogical reasons. Youcould have fun faking Feynman’s mannerism and accent, saying, “Aw shucks, all that fancytalk about little groups! Who needs it? Those experimentalists won’t ever be able to provethat the photon mass is mathematically zero anyway.” So start, as in chapter I.5, with the two equations needed for describing a spin 1 massive particle: (∂2+m2)Aμ=0 (13) and ∂μAμ=0 (14) Equation (14) is needed to cut the number of degrees of freedom contained in Aμdown from four to three. Lo and behold, (13) and (14) are equivalent to the single equation ∂μ(∂μAν−∂νAμ)+m2Aν=0 (15) Obviously, (13) and (14) together imply (15). To verify that (15) implies (13) and (14), we act with ∂νon (15) and obtain m2∂A=0 (16) which for m/negationslash=0 requires ∂A=0, namely (14). Plugging this into (15) we obtain (13). Having packaged two equations into one, we note that we can derive this single equation (15) by varying the Lagrangian L=−1 4FμνFμν+1 2m2A2(17) withFμν≡∂μAν−∂νAμ. Next, suppose we include a source Jμfor this particle by changing the Lagrangian to L=−1 4FμνFμν+1 2m2A2+AμJμ(18) with the resulting equation of motion ∂μ(∂μAν−∂νAμ)+m2Aν=−Jν (19) But now observe that when we act with ∂νon (19) we obtain m2∂A=−∂J (20) 2For a much more detailed and leisurely discussion, see S. Weinberg, Quantum Theory of Fields, pp. 69–74 and 246–255. III.4. Gauge Invariance | 189 We recover (14) only if ∂μJμ=0, that is, if the source producing the particle, commonly know as the current, is conserved. Put more vividly, suppose the experimentalists who constructed the accelerator (or whatever) to produce the spin 1 particle messed up and failed to insure that ∂μJμ=0; then∂A/negationslash=0 and a spin 0 excitation would also be produced. To make sure that the beam of spin 1 particles is not contaminated with spin 0 particles, the accelerator builders mustassure us that the source J μin the Lagrangian (18) is indeed conserved. Now, if we want to study massless spin 1 particles, we simply set m=0 in (18). The “poor man” ends up (just like the “rich man”) using the Lagrangian L=−1 4FμνFμν+AμJμ(21) to describe the photon. Lo and behold (as we exclaimed in chapter II.7), Lis left invariant by the gauge transformation Aμ→Aμ−∂μ/Lambda1for any /Lambda1(x). (As was also explained in that chapter, the third polarization decouples in the limit m→0.) The “poor man” has thus discovered gauge invariance! However, as I warned in chapter I.5, depending on his or her personality, the poor man could also wake up in the middle of the night worrying that physics might be discontinuousin the limit m→0. Thus the little group discussion is needed to remove that nightmare. But then a “real” physicist in the Feynman mode could always counter that for any physicalmeasurement everything must be okay as long as the duration of the experiment is shortcompared to the characteristic time 1 /m. More on this issue in chapter VIII.1. A reflection on gauge symmetry As we will see later and as you might have heard, much of the world beyond electro- magnetism is also described by gauge theories. But as we saw here, gauge theories arealso deeply disturbing and unsatisfying in some sense: They are built on a redundancyof description. The electromagnetic gauge transformation A μ→Aμ−∂μ/Lambda1is not truly a symmetry stating that two physical states have the same properties. Rather, it tells us thatthe two gauge potentials A μandAμ−∂μ/Lambda1describe the same physical state. In your or- derly study of physics, the first place where Aμbecomes indispensable is the Schr ¨odinger equation, as I will explain in chapter IV .4. Within classical physics, you got along perfectlywell with just /vectorEand/vectorB. Some physicists are looking for a formulation of quantum elec- trodynamics without using A μ, but so far have failed to turn up an attractive alternative to what we have. It is conceivable that a truly deep advance in theoretical physics wouldinvolve writing down quantum electrodynamics without writing A μ. III.5 Field Theory without Relativity Slower in its maturity Quantum field theory at its birth was relativistic. Later in its maturity, it found applications in condensed matter physics. We will have a lot more to say about the role of quantum fieldtheory in condensed matter, but for now, we have the more modest goal of learning howto take the nonrelativistic limit of a quantum field theory. The Lorentz invariant scalar field theory L=(∂/Phi1†)(∂/Phi1) −m2/Phi1†/Phi1−λ(/Phi1†/Phi1)2(1) (withλ>0 as always) describes a bunch of interacting bosons. It should certainly contain the physics of slowly moving bosons. For clarity consider first the relativistic Klein-Gordonequation (∂2+m2)/Phi1=0 (2) for a free scalar field. A mode with energy E=m+εwould oscillate in time as /Phi1∝e−iEt. In the nonrelativistic limit, the kinetic energy εis much smaller than the rest mass m. It makes sense to write /Phi1(/vectorx,t)=e−imtϕ(/vectorx,t), with the field ϕoscillating in time much more slowly than e−imt. Plugging into (2) and using the identity (∂/∂t)e−imt(...)= e−imt(−im +∂/∂t)( ...)twice, we obtain (−im +∂/∂t)2ϕ−/vector∇2ϕ+m2ϕ=0. Dropping the term(∂2/∂t2)ϕas small compared to −2im(∂/∂t)ϕ , we find Schr ¨odinger’s equation, as we had better: i∂ ∂tϕ=−/vector∇2 2mϕ (3) By the way, the Klein-Gordon equation was actually discovered before Schr ¨odinger’s equation. III.5. Nonrelativistic Field Theory | 191 Having absorbed this, you can now easily take the nonrelativistic limit of a quantum field theory. Simply plug /Phi1(/vectorx,t)=1√ 2me−imtϕ(/vectorx,t) (4) into (1) . (The factor 1 /√ 2mis for later convenience.) For example, ∂/Phi1† ∂t∂/Phi1 ∂t−m2/Phi1†/Phi1→1 2m/braceleftbigg/bracketleftbigg/parenleftbigg im+∂ ∂t/parenrightbigg ϕ†/bracketrightbigg/bracketleftbigg /parenleftbigg −im+∂ ∂t/parenrightbigg ϕ/bracketrightbigg −m2ϕ†ϕ/bracerightbigg /similarequal1 2i/parenleftBigg ϕ†∂ϕ ∂t−∂ϕ† ∂tϕ/parenrightBigg (5) After an integration by parts we arrive at L=iϕ†∂0ϕ−1 2m∂iϕ†∂iϕ−g2(ϕ†ϕ)2(6) where g2=λ/4m2. As we saw in chapter I.10 the theory (1) enjoys a conserved Noether current Jμ= i(/Phi1†∂μ/Phi1−∂μ/Phi1†/Phi1). The density J0reduces to ϕ†ϕ, precisely as you would expect, while Jireduces to (i/2m)(ϕ†∂iϕ−∂iϕ†ϕ). When you first took a course in quantum mechanics, didn’t you wonder why the density ρ≡ϕ†ϕand the current Ji=(i/2m)(ϕ†∂iϕ−∂iϕ†ϕ) look so different? As to be expected, various expressions inevitably become uglier whenreduced from a more symmetric to a less symmetric theory. Number is conjugate to phase angle Let me point out some differences between the relativistic and nonrelativistic case. The most striking is that the relativistic theory is quadratic in time derivative, while the nonrelativistic theory is linear in time derivative. Thus, in the nonrelativistic theorythe momentum density conjugate to the field ϕ, namely δL/δ∂ 0ϕ, is just iϕ†, so that [ϕ†(/vectorx,t),ϕ(/vectorx/prime,t)]=−δ(D)(/vectorx−/vectorx/prime). In condensed matter physics it is often illuminating to writeϕ=√ρeiθso that L=i 2∂0ρ−ρ∂0θ−1 2m/bracketleftbigg ρ(∂iθ)2+1 4ρ(∂iρ)2/bracketrightbigg −g2ρ2(7) The first term is a total divergence. The second term tells us something of great impor- tance1in condensed matter physics: in the canonical formalism (chapter I.8), the momen- tum density conjugate to the phase field θ(x) isδL/δ∂ 0θ=−ρand thus Heisenberg tells us that [ρ(/vectorx,t),θ(/vectorx/prime,t)]=iδ(D)(/vectorx−/vectorx/prime) (8) 1See P. Anderson, Basic Notions of Condensed Matter Physics , p. 235. 192 | III. Renormalization and Gauge Invariance Integrating and defining N≡/integraltext dDxρ(/vectorx,t)=the total number of bosons, we find one of the most important relations in condensed matter physics [N,θ]=i (9) Number is conjugate to phase angle, just as momentum is conjugate to position. Marvel at the elegance of this! You would learn in a condensed matter course that this fundamentalrelation underlies the physics of the Josephson junction. You may know that a system of bosons with a “hard core” repulsion between them is a superfluid at zero temperature. In particular, Bogoliubov showed that the system containsan elementary excitation obeying a linear dispersion relation. 2I will discuss superfluidity in chapter V .1. In the path integral formalism, going from the complex field ϕ=ϕ1+iϕ2toρand θamounts to a change of integration variables, as I remarked back in chapter I.8. In the canonical formalism, since one deals with operators, one has to tread with somewhat morefinesse. The sign of repulsion In the nonrelativistic theory (7) it is clear that the bosons repel each other: Piling particlesinto a high density region would cost you an energy density g 2ρ2. But it is less clear in the relativistic theory that λ(/Phi1†/Phi1)2withλpositive corresponds to repulsion. I outline one method in exercise III.5.3, but here let’s just take a flying heuristic guess. The Hamiltonian(density) involves the negative of the Lagrangian and hence goes as λ(/Phi1 †/Phi1)2for large /Phi1 and would thus be unbounded below for λ<0. We know physically that a free Bose gas tends to condense and clump, and with an attractive interaction it surely might want tocollapse. We naturally guess that λ>0 corresponds to repulsion. I next give you a more foolproof method. Using the central identity of quantum field theory we can rewrite the path integral for the theory in (1) as Z=/integraldisplay D/Phi1Dσei/integraltext d4x[(∂/Phi1†)(∂/Phi1)−m2/Phi1†/Phi1+2σ/Phi1†/Phi1+(1/λ)σ2](10) Condensed matter physicists call the transformation from (1) to the Lagrangian L= (∂/Phi1†)(∂/Phi1) −m2/Phi1†/Phi1+2σ/Phi1†/Phi1+(1/λ)σ2the Hubbard-Stratonovich transformation. In field theory, a field that does not have kinetic energy, such as σ, is known as an auxiliary field and can be integrated out in the path integral. When we come to the superfield formalismin chapter VIII.4, auxiliary fields will play an important role. Indeed, you might recall from chapter III.2 how a theory with an intermediate vector boson could generate Fermi’s theory of the weak interaction. The same physics is involvedhere: The theory (10) in which the /Phi1field is coupled to an “intermediate σboson” can generate the theory (1). 2For example, L.D. Landau and E. M. Lifschitz, Statisical Physics , p. 238. III.5. Nonrelativistic Field Theory | 193 Ifσwere a “normal scalar field” of the type we have studied, that is, if the terms quadratic inσin the Lagrangian had the form1 2(∂σ)2−1 2M2σ2, then its propagator would be i/(k2−M2+iε). The scattering amplitude between two /Phi1bosons would be proportional to this propagator. We learned in chapter I.4 that the exchange of a scalar field leads to anattractive force. Butσis not a normal field as evidenced by the fact that the Lagrangian contains only the quadratic term +(1/λ)σ 2. Thus its propagator is simply i/(1/λ)=iλ, which (for λ>0)has a sign opposite to the normal propagator evaluated at low-momentum transfer i/(k2−M2+iε)/similarequal−i/M2. We conclude that σexchange leads to a repulsive force. Incidentally, this argument also shows that the repulsion is infinitely short ranged, like a delta function interaction. Normally, as we learned in chapter I.4 the range is determinedby the interplay between the k 2and the M2terms. Here the situation is as if the M2term is infinitely large. We can also argue that the interaction λ(/Phi1†/Phi1)2involves creating two bosons and then annihilating them both at the same spacetime point. Finite density One final point of physics that people trained as particle physicists do not always remem-ber: Condensed matter physicists are not interested in empty space, but want to have afinite density ¯ρof bosons around. We learned in statistical mechanics to add a chemical potential term μϕ †ϕto the Lagrangian (6). Up to an irrelevant (in this context!) additive constant, we can rewrite the resulting Lagrangian as L=iϕ†∂0ϕ−1 2m∂iϕ†∂iϕ−g2(ϕ†ϕ−¯ρ)2(11) Amusingly, mass appears in different places in relativistic and nonrelativistic field theories. To proceed further, I have to develop the concept of spontaneous symmetrybreaking. Thus, adios for now. We will come back to superfluidity in due time. Exercises III.5.1 Obtain the Klein-Gordon equation for a particle in an electrostatic potential (such as that of the nucleus) by the gauge principle of replacing (∂/∂t) in (2) by ∂/∂t−ieA 0. Show that in the nonrelativistic limit this reduces to the Schr ¨odinger’s equation for a particle in an external potential. III.5.2 Take the nonrelativistic limit of the Dirac Lagrangian. III.5.3 Given a field theory we can compute the scattering amplitude of two particles in the nonrelativistic limit. We then postulate an interaction potential U(/vectorx)between the two particles and use nonrelativistic quan- tum mechanics to calculate the scattering amplitude, for example in Born approximation. Comparingthe two scattering amplitudes we can determine U(/vectorx). Derive the Yukawa and the Coulomb potentials this way. The application of this method to the λ(/Phi1 †/Phi1)2interaction is slightly problematic since the delta function interaction is a bit singular, but it should be all right for determining whether the force isrepulsive or attractive. III.6 The Magnetic Moment of the Electron Dirac’s triumph I said in the preface that the emphasis in this book is not on computation, but how can I not tell you about the greatest triumph of quantum field theory? After Dirac wrote down his equation, the next step was to study how the electron interacts with the electromagnetic field. According to the gauge principle already used to writethe Schr ¨odinger’s equation in an electromagnetic field, to obtain the Dirac equation for an electron in an external electromagnetic field we merely have to replace the ordinaryderivative ∂ μby the covariant derivative Dμ=∂μ−ieAμ: (iγμDμ−m)ψ=0 (1) Recall (II.1.27). Acting on this equation with (iγμDμ+m), we obtain −(γμγνDμDν+m2)ψ=0. We haveγμγνDμDν=1 2({γμ,γν}+[γμ,γν])DμDν=DμDμ−iσμνDμDνandiσμνDμDν= (i/2)σμν[Dμ,Dν]=(e/2)σμνFμν. Thus /parenleftbigg DμDμ−e 2σμνFμν+m2/parenrightbigg ψ=0 (2) Now consider a weak constant magnetic field pointing in the 3rd direction for definite- ness, weak so that we can ignore the ( Ai)2term in ( Di)2. By gauge invariance, we can choose A0=0,A1=−1 2Bx2, andA2=1 2Bx1(so that F12=∂1A2−∂2A1=B). As we will see, this is one calculation in which we really have to keep track of factors of 2. Then (Di)2=(∂i)2−ie(∂iAi+Ai∂i)+O(A2 i) =(∂i)2−2ie 2B(x1∂2−x2∂1)+O(A2 i) =/vector∇2−e/vectorB./vectorx×/vectorp+O(A2 i) (3) Note that we used ∂iAi+Ai∂i=(∂iAi)+2Ai∂i=2Ai∂i, where in (∂iAi)the partial deriva- tive acts only on Ai. You may have recognized /vectorL≡/vectorx×/vectorpas the orbital angular momentum III.6. Magnetic Moment of Electron | 195 operator. Thus, the orbital angular momentum generates an orbital magnetic moment that interacts with the magnetic field. This calculation makes good physical sense. If we were studying the interaction of a charged scalar field /Phi1with an external electromagnetic field we would start with (DμDμ+m2)/Phi1=0 (4) obtained by replacing the ordinary derivative in the Klein-Gordon equation by covariant derivatives. We would then go through the same calculation as in (3). Comparing (4) with(2) we see that the spin of the electron contributes the additional term (e/2)σ μνFμν. As in chapter II.1 we write ψ=/parenleftBigφ χ/parenrightBig in the Dirac basis and focus on φsince in the nonrelativistic limit it dominates χ. Recall that in that basis σij=εijk/parenleftBigσk0 0σk/parenrightBig . Thus (e/2)σμνFμνacting on φis effectively equal to (e/2)σ3(F12−F21)=(e/2)2σ3B=2e/vectorB./vectorS since /vectorS=(/vectorσ/2). Make sure you understand all the factors of 2! Meanwhile, according to what I told you in chapter II.1, we should write φ=e−imt/Psi1, where /Psi1oscillates much more slowly than e−imtso that (∂2 0+m2)e−imt/Psi1/similarequale−imt[−2im(∂/∂t)/Psi1 ]. Putting it all together, we have /bracketleftbigg −2im∂ ∂t−/vector∇2−e/vectorB.(/vectorL+2/vectorS)/bracketrightbigg /Psi1=0 (5) There you have it! As if by magic, Dirac’s equation tells us that a unit of spin angular momentum interacts with a magnetic field twice as much as a unit of orbital angularmomentum, an observational fact that had puzzled physicists deeply at the time. Thecalculation leading to (5) is justly celebrated as one of the greatest in the history of physics. The story is that Dirac did not do this calculation until a day after he discovered his equation, so sure was he that the equation had to be right. Another version is that hedreaded the possibility that the magnetic moment would come out wrong and that Naturewould not take advantage of his beautiful equation. Another way of seeing that the Dirac equation contains a magnetic moment is by the Gordon decomposition, the proof of which is given in an exercise: ¯u(p/prime)γμu(p)=¯u(p/prime)/bracketleftbigg(p/prime+p)μ 2m+iσμν(p/prime−p)ν 2m/bracketrightbigg u(p) (6) Looking at the interaction with an electromagnetic field ¯u(p/prime)γμu(p)Aμ(p/prime−p), we see that the first term in (6) only depends on the momentum (p/prime+p)μand would have been there even if we were treating the interaction of a charged scalar particle with the electromagnetic field to first order. The second term involves spin and gives the mag-netic moment. One way of saying this is that ¯u(p /prime)γμu(p) contains a magnetic moment component. 196 | III. Renormalization and Gauge Invariance The anomalous magnetic moment With improvements in experimental techniques, it became clear by the late 1940’s that the magnetic moment of the electron was larger than the value calculated by Dirac by a factorof 1.00118 ±0.00003. The challenge to any theory of quantum electrodynamics was to calculate this so-called anomalous magnetic moment. As you probably know, Schwinger’sspectacular success in meeting this challenge established the correctness of relativisticquantum field theory, at least in dealing with electromagnetic phenomena, beyond anydoubt. Before we plunge into the calculation, note that Lorentz invariance and current conser- vation tell us (see exercise III.6.3) that the matrix element of the electromagnetic currentmust have the form (here |p,s/angbracketrightdenotes a state with an electron of momentum pand polarization s) /angbracketleftp/prime,s/prime|Jμ(0)|p,s/angbracketright=¯u(p/prime,s/prime)/bracketleftbigg γμF1(q2)+iσμνqν 2mF2(q2)/bracketrightbigg u(p ,s) (7) where q≡(p/prime−p). The functions F1(q2)andF2(q2), about which Lorentz invariance can tell us nothing, are known as form factors. To leading order in momentum transfer q, (7) becomes ¯u(p/prime,s/prime)/braceleftbigg(p/prime+p)μ 2mF1(0)+iσμνqν 2m[F1(0)+F2(0)]/bracerightbigg u(p ,s) by the Gordon decomposition. The coefficient of the first term is the electric charge observed by experimentalists and is by definition equal to 1. (To see this, think of potentialscattering, for example. See chapter II.6.) Thus F 1(0)=1. The magnetic moment of the electron is shifted from the Dirac value by a factor 1 +F2(0). Schwinger’s triumph Let us now calculate F2(0)to order α=e2/4π. First draw all the relevant Feynman dia- grams to this order (fig. III.6.1). Except for figure 1b, all the Feynman diagrams are clearlyproportional to ¯u(p /prime,s/prime)γμu(p ,s)and thus contribute to F1(q2), which we don’t care about. Happy are we! We only have to calculate one Feynman diagram. pqp’ pqp’ p’ + k p + kk (a) (b) (c) (d) (e) Figure III.6.1 III.6. Magnetic Moment of Electron | 197 It is convenient to normalize the contribution of figure 1b by comparing it to the lowest order contribution of figure 1a and write the sum of the two contributions as ¯u(γμ+/Gamma1μ)u. Applying the Feynman rules, we find /Gamma1μ=/integraldisplayd4k (2π)4−i k2/parenleftbigg ieγν i /negationslashp/prime+ /negationslashk−mγμ i /negationslashp+ /negationslashk−mieγν/parenrightbigg (8) I will now go through the calculation in some detail not only because it is important, but also because we will be using a variety of neat tricks springing from the brilliant minds ofSchwinger and Feynman. You should verify all the steps of course. Simplifying somewhat we obtain /Gamma1 μ=−ie2/integraltext [d4k/(2π)4](Nμ/D), where Nμ=γν(/negationslashp/prime+ /negationslashk+m)γμ(/negationslashp+ /negationslashk+m)γν (9) and 1 D=1 (p/prime+k)2−m21 (p+k)2−m21 k2=2/integraldisplay dα dβ1 D. (10) We have used the identity (D.16). The integral is evaluated over the triangle in the ( α-β) plane bounded by α=0,β=0, and α+β=1, and D=[k2+2k(αp/prime+βp) ]3=[l2−(α+β)2m2]3+O(q2) (11) where we completed a square by defining k=l−(αp/prime+βp) . The momentum integration is now over d4l. Our strategy is to massage Nμinto a form consisting of a linear combination of γμ,pμ, andp/primeμ. Invoking the Gordon decomposition (6) we can write (7) as ¯u/braceleftbigg γμ[F1(q2)+F2(q2)]−1 2m(p/prime+p)μF2(q2)/bracerightbigg u Thus, to extract F2(0)we can throw away without ceremony any term proportional to γμ that we encounter while massaging Nμ. So, let’s proceed. Eliminating kin favor of lin (9) we obtain Nμ=γν[/negationslashl+ /negationslashP/prime+m]γμ[/negationslashl+ /negationslashP+m]γν (12) where P/primeμ≡(1−α)p/primeμ−βpμandPμ≡(1−β)pμ−αp/primeμ. I will use the identities in appendix D repeatedly, without alerting you every time I use one. It is convenient toorganize the terms in N μby powers of m. (Here I give up writing in complete grammatical sentences.) 1. The m2term: a γμterm, throw away. 2. The mterms: organize by powers of l. The term linear in lintegrates to 0 by symmetry. Thus, we are left with the term independent of l: m(γν/negationslashP/primeγμγν+γνγμ/negationslashPγν)=4m[(1−2α)p/primeμ+(1−2β)pμ] →4m(1−α−β)(p/prime+p)μ(13) In the last step I used a handy trick; since Dis symmetric under α←→β, we can sym- metrize the terms we get in Nμ. 198 | III. Renormalization and Gauge Invariance 3. Finally, the most complicated m0term. The term quadratic in l: note that we can effectively replace lσlτinside/integraltext d4l/(2π)4by1 4ηστl2by Lorentz invariance (this step is possible because we have shifted the integration variable so that Dis a Lorentz invariant function of l2.) Thus, the term quadratic in lgives rise to a γμterm. Throw it away. Again we throw away the term linear in l, leaving [use (D.6) here!] γν/negationslashP/primeγμ/negationslashPγν=− 2/negationslashPγμ/negationslashP/prime →− 2[(1−β)/negationslashp−αm]γμ[(1−α)/negationslashp/prime−βm] (14) where in the last step we remembered that /Gamma1μis to be sandwiched between ¯u(p/prime)andu(p) . Again, it is convenient to organize the terms in (14) by powers of m. With the various tricks we have already used, we find that the m2term can be thrown away, the mterm gives 2m(p/prime+p)μ[α(1−α)+β(1−β)], and the m0term gives 2 m(p/prime+p)μ[−2(1−α)(1−β)]. Putting it altogether, we find that Nμ→2m(p/prime+p)μ(α+β)(1−α−β) We can now do the integral/integraltext [d4l/(2π)4](1/D)using (D.11). Finally, we obtain /Gamma1μ=− 2ie2/integraldisplay dα dβ(−i 32π2)1 (α+β)2m2Nμ =−e2 8π21 2m(p/prime+p)μ(15) and thus, trumpets please: F2(0)=e2 8π2=α 2π(16) Schwinger’s announcement of this result in 1948 had an electrifying impact on the theo- retical physics community. I gave you in this chapter not one, but two, of the great triumphs of twentieth century physics, although admittedly the first is not a result of field theory per se. Exercises III.6.1 Evaluate ¯u(p/prime)(/negationslashp/primeγμ+γμ/negationslashp)u(p) in two different ways and thus prove Gordon decomposition. III.6.2 Check that (7) is consistent with current conservation. [Hint: By translation invariance (we suppress the spin variable) /angbracketleftp/prime|Jμ(x)|p/angbracketright=/angbracketleftp/prime|Jμ(0)|p/angbracketrightei(p/prime−p)x and hence /angbracketleftp/prime|∂μJμ(x)|p/angbracketright=i(p/prime−p)μ/angbracketleftp/prime|Jμ(0)|p/angbracketrightei(p/prime−p)x Thus current conservation implies that qμ/angbracketleftp/prime|Jμ(0)|p/angbracketright=0.] III.6.3 By Lorentz invariance the right hand side of (7) has to be a vector. The only possibilities are ¯uγμu, (p+p/prime)μ¯uu, and(p−p/prime)μ¯uu. The last term is ruled out because it would not be consistent with current conservation. Show that the form given in (7) is in fact the most general allowed. III.6. Magnetic Moment of Electron | 199 III.6.4 In chapter II.6, when discussing electron-proton scattering, we ignored the strong interaction that the proton participates in. Argue that the effects of the strong interaction could be included phenomenolog-ically by replacing the vertex ¯u(P ,S)γ μu(p ,s)in (II.6.1) by /angbracketleftP,S|Jμ(0)|p,s/angbracketright=¯u(P ,S)/bracketleftbigg γμF1(q2)+iσμνqν 2mF2(q2)/bracketrightbigg u(p ,s) (17) Careful measurements of electron-proton scattering, thus determining the two proton form factors F1(q2)andF2(q2), earned R. Hofstadter the 1961 Nobel Prize. While we could account for the general behavior of these two form factors, we are still unable to calculate them from first principles (in contrastto the corresponding form factors for the electron.) See chapters IV .2 and VII.3. III.7Polarizing the Vacuum and Renormalizing the Charge A photon can fluctuate into an electron and a positron One early triumph of quantum electrodynamics is the understanding of how quantum fluctuations affect the way the photon propagates. A photon can always metamorphose intoan electron and a positron that, after a short time mandated by the uncertainty principle,annihilate each other becoming a photon again. The process, which keeps on repeatingitself, is depicted in figure III.7.1. Quantum fluctuations are not limited to what we just described. The electron and positron can interact by exchanging a photon, which in turn can change into an electronand a positron, and so on and so forth. The full process is shown in figure III.7.2, where theshaded object, denoted by i/Pi1 μν(q) and known as the vacuum polarization tensor, is given by an infinite number of Feynman diagrams, as shown in figure III.7.3. Figure III.7.1 isobtained from figure III.7.2 by approximating i/Pi1 μν(q) by its lowest order diagram. It is convenient to rewrite the Lagrangian L=¯ψ[iγμ(∂μ−ieAμ)−m]ψ−1 4FμνFμνby letting A→(1/e)A, which we are always allowed to do, so that L=¯ψ[iγμ(∂μ−iAμ)−m]ψ−1 4e2FμνFμν(1) Note that the gauge transformation leaving Linvariant is given by ψ→eiαψandAμ→ Aμ+∂μα. The photon propagator (chapter III.4), obtained roughly speaking by inverting (1/4e2)FμνFμν, is now proportional to e2: iDμν(q)=−ie2 q2/bracketleftbigg gμν−(1−ξ)qμqν q2/bracketrightbigg (2) Every time a photon is exchanged, the amplitude gets a factor of e2. This is just a trivial but convenient change and does not affect the physics in the slightest. For example, in the Feynman diagram we calculated in chapter II.6 for electron-electron scattering, the factore 2can be thought of as being associated with the photon propagator rather than as coming from the interaction vertices. In this interpretation e2measures the ease with which the III.7. Polarizing the Vacuum | 201 + + + Figure III.7.1 + + + Figure III.7.2 = + ++ ++ pp + q Figure III.7.3 photon propagates through spacetime. The smaller e2, the more action it takes to have the photon propagate, and the harder for the photon to propagate, the weaker the effect ofelectromagnetism. The diagrammatic proof of gauge invariance given in chapter II.7 implies that q μ/Pi1μν(q)=0. T ogether with Lorentz invariance, this requires that /Pi1μν(q)=(qμqν−gμνq2)/Pi1(q2) (3) The physical or renormalized photon propagator as shown in figure III.7.2 is then given by the geometric series iDP μν(q)=iDμν(q)+iDμλ(q)i/Pi1λρ(q)iD ρν(q) +iDμλ(q)i/Pi1λρ(q)iDρσ(q)i/Pi1σκ(q)iDκν(q)+... =−ie2 q2gμν{1−e2/Pi1(q2)+[e2/Pi1(q2)]2+...}+q μqνterm =−ie2 q2gμν1 1+e2/Pi1(q2)+qμqνterm (4) Because of (3) the (1−ξ)(qμqλ/q2)part of Dμλ(q) is annihilated when it encounters /Pi1λρ(q). Thus, in iDP μν(q) the gauge parameter ξenters only into the qμqνterm and drops out in physical amplitudes, as explained in chapter II.7. 202 | III. Renormalization and Gauge Invariance The residue of the pole in iDP μν(q) is the physical or renormalized charge squared: e2 R=e2 1 1+e2/Pi1(0)(5) Respect for gauge invariance In order to determine eRin terms of e, let us calculate to lowest order i/Pi1μν(q)=(−)/integraldisplayd4p (2π)4tr/parenleftbigg iγν i /negationslashp+ /negationslashq−miγμi /negationslashp−m/parenrightbigg (6) For large pthe integrand goes as 1 /p2with a subleading term going as m2/p4causing the integral to have a quadratically divergent and a logarithmically divergent piece. (Yousee, it is easy to slip into bad language.) Not a conceptual problem at all, as I explainedin chapter III.1. We simply regularize. But now there is a delicate point: Since gaugeinvariance plays a crucial role, we must make sure that our regularization respects gaugeinvariance. In the Pauli-Villars regularization (III.1.13) we replace (6) by i/Pi1μν(q)=(−)/integraldisplayd4p (2π)4/bracketleftbigg tr/parenleftbigg iγν i /negationslashp+ /negationslashq−miγμi /negationslashp−m/parenrightbigg −/summationdisplay acatr/parenleftbigg iγν i /negationslashp+ /negationslashq−maiγμi /negationslashp−ma/parenrightbigg/bracketrightBigg (7) Now the integrand goes as (1−/summationtext aca)(1/p2)with a subleading term going as (m2−/summationtext acam2 a)(1/p4), and thus the integral would converge if we choose caandmasuch that /summationdisplay aca=1 (8) and /summationdisplay acam2 a=m2(9) Clearly, we have to introduce at least two regulator masses. We are confessing to ignorance of the physics above the mass scale ma. The integral in (7) is effectively cut off when the momentum pexceeds ma. Does a bell ring for you? It should, as this discussion conceptually parallels that in the appendix to chapter I.9. The gauge invariant form (3) we expect to get actually suggests that we need fewer regulator terms than we think. Imagine expanding (6) in powers of q. Since /Pi1μν(q)=(qμqν−gμνq2)[/Pi1(0)+...] we are only interested in terms of O(q2)and higher in the Feynman integral. If we expand the integrand in (6), we see that the term of O(q2)goes as 1 /p4for large p, thus giving a logarithmically divergent (speaking bad language again!) contribution. (Incidentally, you III.7. Polarizing the Vacuum | 203 may recall that this sort of argument was also used in chapter III.3.) It seems that we need only one regulator. This argument is not rigorous because we have not proved that /Pi1(q2) has a power series expansion in q2, but instead of worrying about it let us proceed with the calculation. Once the integral is convergent, the proof of gauge invariance given in chapter II.7 now goes through. Let us recall briefly how the proof went. In computing qμ/Pi1μν(q) we use the identity 1 /negationslashp+ /negationslashq−m/negationslashq1 /negationslashp−m=1 /negationslashp−m−1 /negationslashp+ /negationslashq−m to split the integrand into two pieces that cancel upon shifting the integration variable p→p+q. Recall from exercise (II.7.2) that we were concerned that in some cases the shift may not be allowed, but it is allowed if the integral is sufficiently convergent, as isindeed the case now that we have regularized. In any event, the proof is in the eating ofthe pudding, and we will see by explicit calculation that /Pi1 μν(q) indeed has the form in (3). Having learned various computational tricks in the previous chapter you are now ready to tackle the calculation. I will help by walking you through it. In order not to clutter upthe page I will suppress the regulator terms in (7) in the intermediate steps and restorethem toward the end. After a few steps you should obtain i/Pi1μν(q)=−/integraldisplayd4p (2π)4Nμν D where Nμν=tr[γν(/negationslashp+ /negationslashq+m)γμ(/negationslashp+m)] and 1 D=/integraldisplay1 0dα1 D withD=[l2+α(1−α)q2−m2+iε]2, where l=p+αq. Eliminating pin favor of land beating on Nμνyou will find that Nμνis effectively equal to −4/parenleftbigg1 2gμνl2+α(1−α)(2qμqν−gμνq2)−m2gμν/parenrightbigg Integrate over lusing (D.12) and (D.13) and, writing the contribution from the regulators explicitly, obtain /Pi1μν(q)=−1 4π2/integraldisplay1 0dα/bracketleftBigg Fμν(m)−/summationdisplay acaFμν(ma)/bracketrightBigg (10) where Fμν(m) =1 2gμν/braceleftbigg /Lambda12−2[m2−α(1−α)q2] log/Lambda12 m2−α(1−α)q2+m2−α(1−α)q2/bracerightbigg −[α(1−α)(2qμqν−gμνq2)−m2gμν]/bracketleftbigg log/Lambda12 m2−α(1−α)q2−1/bracketrightbigg (11) Remember, you are doing the calculation; I am just pointing the way. In appendix D, /Lambda1was introduced to give meaning to various divergent integrals. Since our integral is convergent, 204 | III. Renormalization and Gauge Invariance we should not need /Lambda1, and indeed, it is gratifying to see that in (10) /Lambda1drops out thanks to the conditions (8) and (9). Some other terms drop out as well, and we end up with /Pi1μν(q)=−1 2π2(qμqν−gμνq2)/integraldisplay1 0dα α( 1−α) {log[m2−α(1−α)q2]−/summationdisplay acalog[m2 a−α(1−α)q2]} (12) Lo and behold! The vacuum polarization tensor indeed has the form /Pi1μν(q)=(qμqν− gμνq2)/Pi1(q2). Our regularization scheme does respect gauge invariance. Forq2/lessmuchm2 a(the kinematic regime we are interested in had better be much lower than our threshold of ignorance) we simply define log M2≡/summationtext acalogm2 ain (12) and obtain /Pi1(q2)=1 2π2/integraldisplay1 0dα α( 1−α)logM2 m2−α(1−α)q2(13) Note that our heuristic argument is indeed correct. In the end, effectively we need only one regulator, but in the intermediate steps we needed two. Actually, this bickering overthe number of regulators is beside the point. In chapter III.1 I mentioned dimensional regularization as an alternative to Pauli-Villars regularization. Historically, dimensional regularization was invented to preserve gaugeinvariance in nonabelian gauge theories (which I will discuss in a later chapter). It isinstructive to calculate /Pi1using dimensional regularization (exercise III.7.1). Electric charge Physically, we end up with a result for /Pi1(q2)containing a parameter M2expressing our threshold of ignorance. We conclude that e2 R=e2 1 1+(e2/12π2)log(M2/m2)/similarequale2/parenleftbigg 1−e2 12π2logM2 m2/parenrightbigg (14) Quantum fluctuations effectively diminish the charge. I will explain the physical origin of this effect in a later chapter on renormalization group flow. You might argue that physically charge is measured by how strongly one electron scatters off another electron. T o order e4, in addition to the diagrams in chapter II.7, we also have, among others, the diagrams shown in figure III.7.4a,b,c. We have computed 4a, but whatabout 4b and 4c? In many texts, it is shown that contributions of III.7.4b and III.7.4c tocharge renormalization cancel. The advantage of using the Lagrangian in (1) is that thisfact becomes self-evident: Charge is a measure of how the photon propagates. T o belabor a more or less self-evident point let us imagine doing physical or renormalized perturbation theory as explained in chapter III.3. The Lagrangian is written in terms ofphysical or renormalized fields (and as before we drop the subscript Pon the fields) L=¯ψ(iγμ(∂μ−iAμ)−mP)ψ−1 4e2 PFμνFμν +A¯ψiγμ(∂μ−iAμ)ψ+B¯ψψ−CFμνFμν(15) III.7. Polarizing the Vacuum | 205 + (a) (b) (c) Figure III.7.4 where the coefficients of the counterterms A,B, and Care determined iteratively. The point is that gauge invariance guarantees that ¯ψiγμ∂μψand¯ψγμAμψalways occur in the combination ¯ψiγμ(∂μ−iAμ)ψ: The strength of the coupling of Aμto¯ψγμψcannot change. What can change is the ease with which the photon propagates through spacetime. This statement has profound physical implications. Experimentally, it is known to a high degree of accuracy that the charges of the electron and the proton are opposite andexactly equal. If the charges were not exactly equal, there would be a residual electrostaticforce between macroscopic objects. Suppose we discovered a principle that tells us thatthe bare charges of the electron and the proton are exactly equal (indeed, as we will see,in grand unification theories, this fact follows from group theory). How do we knowthat quantum fluctuations would not make the charges slightly unequal? After all, theproton participates in the strong interaction and the electron does not and thus manymore diagrams would contribute to the long range electromagnetic scattering betweentwo protons. The discussion here makes clear that this equality will be maintained for theobvious reason that charge renormalization has to do with the photon. In the end, it is alldue to gauge invariance. Modifying the Coulomb potential We have focused on charge renormalization, which is determined completely by /Pi1(0), but in (13) we obtained the complete function /Pi1(q2), which tells us how the qdependence of the photon propagator is modified. According to the discussion in chapter I.5, the Coulombpotential is just the Fourier transform of the photon propagator (see also exercise III.5.3).Thus, the Coulomb interaction is modified from the venerable 1 /rlaw at a distance scale of the order of (2m) −1, namely the inverse of the characteristic value of qin/Pi1(q2). This modification was experimentally verified as part of the Lamb shift in atomic spectroscopy,another great triumph of quantum electrodynamics. 206 | III. Renormalization and Gauge Invariance Exercises III.7.1 Calculate /Pi1μν(q) using dimensional regularization. The procedure is to start with (6), evaluate the trace inNμν, shift the integration momentum from ptol, and so forth, proceeding exactly as in the text, until you have to integrate over the loop momentum l. At that point you “pretend” that you are living ind-dimensional spacetime, so that the term like lμlνinNμν, for example, is to be effectively replaced by(1/d)g μνl2. The integration is to be performed using (III.1.15) and various generalizations thereof. Show that the form (3) automatically emerges when you continue to d=4. III.7.2 Study the modified Coulomb’s law as determined by the Fourier integral/integraltext d3q{1//vectorq2[1+e2/Pi1(/vectorq2)]}ei/vectorq/vectorx. III.8 Becoming Imaginary and Conserving Probability When Feynman amplitudes go imaginary Let us admire the polarized vacuum, viz (III.7.13): /Pi1(q2)=1 2π2/integraldisplay1 0dαα( 1−α)log/Lambda12 m2−α(1−α)q2−iε(1) Dear reader, you have come a long way in quantum field theory, to be able to calculate such an amazing effect. Quantum fluctuations alter the way a photon propagates! For a spacelike photon, with q2negative, /Pi1is real and positive for momentum small compared to the threshold of our ignorance /Lambda1. For a timelike photon, we see that if q2>0 is large enough, the argument of the logarithm may go negative, and thus /Pi1becomes complex. As you know, the logarithmic function log zcould be defined in the complex zplane with a cut that can be taken conventionally to go along the negative real axis, so that for wreal and positive, log (−w±iε)=log(w)±iπ[since in polar coordinates log(ρeiθ)=log(ρ)+iθ]. We now invite ourselves to define a function in the complex plane: /Pi1(z)≡ 1/(2π2)/integraltext1 0dαα( 1−α)log/Lambda12/(m2−α(1−α)z) . The integrand has a cut on the positive realzaxis extending from z=m2/(α(1−α)) to infinity (fig. III.8.1). Since the maximum value of α(1−α)in the integration range is1 4,/Pi1(z) is an analytic function in the complex zplane with a cut along the real axis starting at zc=4m2. The integral over αsmears all those cuts of the integrand into one single cut. For timelike photons with large enough q2, a mathematician might be paralyzed won- dering which side of the cut to go to, but we as physicists know, as per the iεprescription from chapter I.3, that we should approach the cut from above, namely that we should take/Pi1(q 2+iε) (withε, as always, a positive infinitesimal) as the physical value. Ultimately, causality tells us which side of the cut we should be on. That the imaginary part of /Pi1starts at/radicalbig q2>2mprovides a strong hint of the physics behind amplitudes going complex. We began the preceding chapter talking about how 208 | III. Renormalization and Gauge Invariance z 4m2 Figure III.8.1 a photon merrily propagating along could always metamorphose into a virtual electron- positron pair that, after a short time dictated by the uncertainly principle, annihilate eachother to become a photon again. For/radicalbig q2>2mthe pair is no longer condemned to be virtual and to fluctuate out of existence almost immediately. The pair has enough energyto get real. (If you did the exercises religiously, you would recognize that these points werealready developed in exercises I.7.4 and III.1.2.) Physically, we could argue more forcefully as follows. Imagine a gauge boson of mass M coupling to electrons just like the photon. (Indeed, in this book we started out supposingthat the photon has a mass.) The vacuum polarization diagram then provides a one-loopcorrection to the vector boson propagator. For M> 2m, the vector boson becomes unstable against decay into an electron-positron pair. At the same time, /Pi1acquires an imaginary part. You might suspect that Im /Pi1might have something to do with the decay rate. We will verify these suspicions later and show that, hey, your physical intuition is pretty good. When we ended the preceding chapter talking about the modifications to the Coulomb potential, we thought of a spacelike virtual photon being exchanged between two charges asin electron-electron or electron-proton scattering (chapter II.6). Use crossing (chapter II.8)to map electron-electron scattering into electron-positron scattering. The vacuum polariza-tion diagram then appears (fig. III.8.2) as a correction to electron-positron scattering. Onefunction /Pi1covers two different physical situations. Incidentally, the title of this section should, strictly speaking, have the word “complex,” but it is more dramatic to say “When Feynman amplitudes go imaginary,” if only to echocertain movie titles. Dispersion relations and high frequency behavior One of the most remarkable discoveries in elementary particle physics has been that of the existence of thecomplex plane. —J. Schwinger Considering that amplitudes are calculated in quantum field theory as integrals over products of propagators, it is more or less clear that amplitudes are analytic functions III.8. Becoming Imaginary | 209 e/H11001e/H11002 e/H11001e/H11002 Figure III.8.2 of the external kinematic variables. Another example is the scattering amplitude Min chapter III.1: it is manifestly an analytic function of s,t, anduwith various cuts. From the late 1950s until the early 1960s, considerable effort was devoted to studying analyticity inquantum field theory, resulting in a vast literature. Here we merely touch upon some elementary aspects. Let us start with an embarrass- ingly simple baby example: f( z)=/integraltext 1 0dα1/(z−α)=log((z−1)/z). The integrand has a pole at z=α, which got smeared by the integral over αinto a cut stretching from 0 to 1. At the level of physicist rigor, we may think of a cut as a lot of poles mashed together anda pole as an infinitesimally short cut. We will mostly encounter real analytic functions, namely functions satisfying f ∗(z)= f( z∗)(such as log z). Furthermore, we focus on functions that have cuts along the real axis, as exemplified by /Pi1(z) . For the class of analytic functions specified here, the discontinuity of the function across the cut is given by disc f( x)≡f( x+iε)−f( x−iε)=f( x+iε)− f( x+iε)∗=2iImf( x+iε). Define, for σreal,ρ(σ)=Imf( σ+iε). Using Cauchy’s theorem with a contour Cthat goes around the cut as indicated in figure III.8.3, we could write f( z)=/contintegraldisplay Cdz/prime 2πif( z/prime) z/prime−z. (2) Assuming that f( z) vanishes faster than 1 /zasz→∞ , we can drop the contribution from infinity and write f( z)=1 π/integraldisplay dσρ(σ) σ−z(3) where the integral ranges over the cut. Note that we can check this equation using the identity (I.2.14): Imf( x+iε)=1 π/integraldisplay dσρ(σ) Im1 σ−x−iε=1 π/integraldisplay dσρ(σ)πδ(σ −x) 210 | III. Renormalization and Gauge Invariance z C Figure III.8.3 This relation tells us that knowing the imaginary part of falong the cut allows us to construct fin its entirety, including a fortiori its real part on and away from the cut. Relations of this type, known collectively as dispersion relations, go back at least to thework of Kramers and Kr ¨onig on optics and are enormously useful in many areas of physics. We will use it in, for example, chapter VII.4. We implicitly assumed that the integral over σconverges. If not, we can always (formally) subtract f(0)= 1 π/integraltext dσρ(σ)/σ fromf( z) as given above and write f( z)=f(0)+z π/integraldisplay dσρ(σ) σ(σ−z)(4) The integral over σnow enjoys an additional factor of 1 /σand hence is more convergent. In this case, to reconstruct f( z) , we need, in addition to knowledge of the imaginary part offon the cut, an unknown constant f(0). Evidently, we could repeat this process until we obtain a convergent integral. A bell rings, and you, the astute reader, see the connection with the renormalization procedure of introducing counterterms. In the dispersion weltanschauung, divergentFeynman integrals correspond to integrals over σthat do not converge. Once again, divergent integrals do not bend real physicists out of shape: we simply admit to ignoranceof the high σregime. During the height of the dispersion program, it was jokingly said that particle theorists either group or disperse, depending on whether you like group theory or complex analysisbetter. III.8. Becoming Imaginary | 211 Imaginary part of Feynman integrals Going back to the calculation of vacuum polarization in the preceding chapter, we see that the numerator Nμν, which comes from the spin of the photon and of the electron, is irrelevant in determining the analytic structure of the Feynman diagram. It is thedenominator Dthat counts. Thus, to get at a conceptual understanding of analyticity in quantum field theory, we could dispense with spins and study the analog of vacuumpolarization in the scalar field theory with the interaction term L=g(η †ξ†ϕ+h.c.), introduced in the appendix to chapter II.6. The ϕpropagator is corrected by the analog of the diagrams in figure III.7.1 to [compare with (4)] iDP(q)=i q2−M2+i/epsilon1+i q2−M2+i/epsilon1i/Pi1(q2)i q2−M2+i/epsilon1+... =i q2−M2+/Pi1(q2)+i/epsilon1(5) T o order g2we have i/Pi1(q2)=i4g2/integraldisplayd4k (2π)41 k2−μ2+iε1 (q−k)2−m2+iε(6) As in the preceding chapter we need to regulate the integral, but we will leave that implicit. Having practiced with the spinful calculation of the preceding chapter, you can now whiz through this spinless calculation and obtain /Pi1(z)=g2 16π2/integraldisplay1 0dα log/Lambda12 αm2+(1−α)μ2−α(1−α)z(7) with/Lambda1some cutoff. (Please do whiz and not imagine that you could whiz.) We use the same Greek letter /Pi1and allow the two particles in the loop to have different masses, in contrast to the situation in quantum electrodynamics. As before, for zreal and negative, the argument of the log is real and positive, and /Pi1is real. By the same token, for zreal and positive enough, the argument of the log becomes negative for some value of α, and/Pi1(z) goes complex. Indeed, Im/Pi1(σ+iε)=−g2 16π2/integraldisplay1 0dα(−π)θ[ α(1−α)σ−αm2−(1−α)μ2] =g2 16π/integraldisplayα+ α−dα =g2 16πσ/radicalbig (σ−(m+μ)2)(σ−(m−μ)2) (8) withα±the two roots of the quadratic equation obtained by setting the argument of the step function to zero. 212 | III. Renormalization and Gauge Invariance Decay and distintegration At this point you might already be flipping back to the expression given in chapter II.6 for the decay rate of a particle. Earlier we entertained the suspicion that the imaginary part of/Pi1(z) corresponds to decay. T o confirm our suspicion, let us first go back to elementary quantum mechanics. The higher energy levels in a hydrogen atom, say, are, strictlyspeaking, not eigenstates of the Hamiltonian: an electron in a higher energy level will, in afinite time, emit a photon and jump to a lower energy level. Phenomenologically, however,the level could be assigned a complex energy E−i 1 2/Gamma1. The probability of staying in this level then goes with time like |ψ(t)|2∝|e−i(E−i1 2/Gamma1)t|2=e−/Gamma1t. (Note that in elementary quantum mechanics, the Coulomb and radiation components of the electromagnetic fieldare treated separately: the former is included in the Schr ¨odinger equation but not the latter. One of the aims of quantum field theory is to remedy this artificial split.) We now go back to (5) and field theory: note that /Pi1(q 2)effectively shifts M2→M2− /Pi1(q2). Recall from (III.3.3) that we have counter terms available to, well, counter two cutoff-dependent pieces of /Pi1(q2). But we have nothing to counter the imaginary part of /Pi1(q2)with, and so it better be cutoff independent. Indeed it is! The cutoff only appears in the real part in (7). We conclude that the effect of /Pi1going imaginary is to shift the mass of the ϕmeson by a cutoff-independent amount from Mto/radicalbig M2−iIm/Pi1(M2)≈M−iIm/Pi1(M2)/(2M). Note that to order g2it suffices to evaluate /Pi1at the unshifted mass squared M2, since the shift in mass is itself of order g2. Thus /Gamma1=Im/Pi1(M2)/M gives the decay rate, as we suspected. We obtain (g has dimension of mass and so the dimension is correct) /Gamma1=g2 16πM3/radicalBig [M2−(m+μ)2][M2−(m−μ)2] (9) precisely what we had in (II.6.7). You and I could both take a bow for getting all the factors exactly right! Note that both the treatment given in elementary quantum mechanics and here are in the spirit of treating the decay as a small perturbation. As the width becomes large, at somepoint it no longer makes good sense to talk of the field associated with the particle ϕ. Taking the imaginary part directly We ought to be able to take the imaginary part of the Feynman integral in (6) directly, rather than having to first calculate it as an integral over the Feynman parameter α. I will now show you how to do this using a trick. For clarity and convenience, change notation from(6), label the momentum carried by the two internal lines in figure III.8.4a separately, andrestore the momentum conservation delta function, so that i/Pi1(q) =(ig)2i2/integraldisplayd4kη (2π)4d4kξ (2π)4(2π)4δ4(kη+kξ−q)/braceleftBigg 1 k2 η−m2 η+i/epsilon11 k2 ξ−m2 ξ+i/epsilon1/bracerightBigg (10) III.8. Becoming Imaginary | 213 qk/H9264k/H9257 /H9272ox /H9264/H9257 (b) (a) Figure III.8.4 Write the propagator as 1 /(k2−m2+i/epsilon1)=P(1/(k2−m2))−iπδ(k2−m2)and, noting an explicit overall factor of i, take the real part of the curly bracket above, thus obtaining Im/Pi1(q)=−g2/integraldisplay d/Phi1(PηPξ−/Delta1η/Delta1ξ) (11) For the sake of compactness, we have introduced the notation d/Phi1=d4kη (2π)4d4kξ (2π)4(2π)4δ4(kη+kξ−q),Pη=P1 k2 η−m2 η,/Delta1η=πδ(k2 η−m2 η) and so on. We welcome the product of two delta functions; they are what we want, restricting the two particles ηandξon shell. But yuck, what do we do with the product of the two principal values? They don’t correspond to anything too physical that we know of. T o get rid of the two principal values, we use a trick.1First, we regress and recall that we started out with Feynman diagrams as spacetime diagrams (for example, fig. I.7.6) ofthe process under study. Here (fig. III.8.4b) a ϕexcitation turns into an ηand a ξwith amplitude igat some spacetime point, which by translation invariance we could take to be the origin; the ηand the ξexcitations propagate to some point xwith amplitude iD η(x) andiDξ(x), respectively, and then recombine into ϕwith amplitude ig(note: not −ig ). Fourier transforming this product of spacetime amplitudes gives i/Pi1(q) =(ig)2i2/integraldisplay dxe−iqxDη(x)D ξ(x) (12) T o see that this is indeed the same as (10), all you have to do is to plug in the expression (I.3.22) for Dη(x) andDξ(x). Incidentally, while many “professors of Feynman diagrams” think almost exclusively in momentum space, Feynman titled his 1949 paper “Space-Time Approach to QuantumElectrodynamics,” and on occasions it is useful to think of the spacetime roots of a givenFeynman diagram. Now is one of those occasions. 1C. Itzkyson and J.-B. Zuber, Quantum Field Theory, p. 367. 214 | III. Renormalization and Gauge Invariance Next, go back to exercise I.3.3 and recall that the advanced propagator Dadv(x) and retarded propagator Dret(x)vanish for x0<0 andx0>0, respectively, and thus the product Dadv(k)D ret(k) manifestly vanishes for all x. Also recall that the advanced and retarded propagators Dadv(k)andDret(k)differ from the Feynman propagator D(k) by simply, but crucially, having their poles in different half-planes in the complex k0plane. Thus 0=−ig2/integraldisplay dxe−iqxDη, adv(x)D ξ, ret(x) =/integraldisplayd4kη (2π)4d4kξ (2π)4(2π)4δ4(kη+kξ−q)1 k2 η−m2 η−iση/epsilon11 k2 ξ−m2 ξ+iσξ/epsilon1(13) where we used the shorthand ση=sgn(k0 η)andσξ=sgn(k0 ξ). (The sign function is defined by sgn (x)=± 1 according to whether x> 0o r<0.) T aking the imaginary part of 0, we obtain [compare (11)] 0=−g2/integraldisplay d/Phi1(PηPξ+σησξ/Delta1η/Delta1ξ) (14) Subtracting (14) from (11) to get rid of the rather unpleasant term PηPξ, we find finally Im/Pi1(q)=+g2/integraldisplay d/Phi1( 1+σησξ)(/Delta1η/Delta1ξ) =g2π2/integraldisplayd4kη (2π)4d4kξ (2π)4(2π)4δ4(kη+kξ−q)θ(k0 η)δ(k2 η−m2 η)θ(k0 ξ)δ(k2 ξ−m2 ξ)(1+σησξ) (15) Thus/Pi1(q) develops an imaginary part only when the three delta functions can be satisfied simultaneously. T o see what these three conditions imply, we can, since /Pi1(q) is a function of q2,g o to a frame in which q=(Q,/vector0)withQ>0 with no loss of generality. Since k0 η+k0 ξ= Q>0 and since (1+σησξ)vanishes unless k0 ηandk0 ξhave the same sign, k0 ηandk0 ξ must be both positive if Im /Pi1(q) is to be nonzero, but that is already mandated by the two step functions. Furthermore, we need to solve the conservation of energy condition Q=/radicalBig /vectork2+m2 η+/radicalBig /vectork2+m2 ξfor some 3-vector /vectork. This is possible only if Q>m η+mξ,i n which case, using the identity (I.8.14) θ(k0)δ(k2−μ2)=θ(k0)δ(k0−εk) 2εk, (16) we obtain Im/Pi1(q)=1 2g2/integraldisplayd3kη (2π)32ωηd3kξ (2π)32ωξ(2π)4δ4(kη+kξ−q) (17) We see that (and as we will see more generally) Im /Pi1(q) works out to be a finite integral over delta functions. Indeed, no counter term is needed. Some readers might feel that this trick of invoking the advanced and retarded propa- gators is perhaps a bit “too tricky.” For them, I will show a more brute force method inappendix 1. III.8. Becoming Imaginary | 215 Unitarity and the Cutkosky cutting rule The simple example we just went through in detail illustrates what is known as the Cutkosky cutting rule, which states that to calculate the imaginary part of a Feynmanamplitude we first “cut” through a diagram (as indicated by the dotted line in figure II.8.4a).For each internal line cut, replace the propagator 1 /(k 2−m2+i/epsilon1)byδ(k2−m2); in other words, put the virtual excitation propagating through the cut onto the mass shell. Thus,in our example, we could jump from (10) to (15) directly. This validates our intuition thatFeynman amplitudes go imaginary when virtual particles can “get real.” For a precise statement of the cutting rule, see below. (The Cutkosky cut is not be confused with the Cauchy cut in the complex plane, of course.) The Cutkosky cutting rule in fact follows in all generality from unitarity. A basic postulate of quantum mechanics is that the time evolution operator e −iHTis unitary and hence preserves probability. Recall from chapter I.8 that it is convenient to split off from theS matrix S fi=/angbracketleftf|e−iHT|i/angbracketrightthe piece corresponding to “nothing is happening”: S=I+ iT. Unitarity S†S=Ithen implies 2 Im T=i(T†−T)=T†T. Sandwiching this between initial and final states and inserting a complete set of intermediate states (1 =/summationtext n|n/angbracketright/angbracketleftn| ) we have 2I mTfi=/summationdisplay nT† fnTni (18) which some readers might recognize as a generalization of the optical theorem from elementary quantum mechanics. It is convenient to introduce F=−iM. (We are merely taking out an explicit factor of iinM: in our simple example, Mcorresponds to i/Pi1,Fto/Pi1.) Then the relation (I.8.16) between TandMbecomes Tfi=(2π)4δ(4)(Pfi)(/Pi1fi1/ρ)F(f←i), where for the sake of compactness we have introduced some obvious notations [thus (/Pi1fi1 ρ)denotes the product of the normalization factors 1 /ρ(see chapter I.8), one for each of the particle in the state iand in the state f, andPfithe sum of the momenta in fminus the sum of the momenta in i.] With this notation, the left-hand side of the generalized optical theorem becomes 2ImTfi=2(2π)4δ(4)(Pfi)(/Pi1fi1/ρ)Im F(f←i)and the right-hand side /summationdisplay nT† fnTni=/summationdisplay n(2π)4δ(4)(Pfn)(2π)4δ(4)(Pni)/parenleftbigg /Pi1fn1 ρ/parenrightbigg/parenleftbigg /Pi1ni1 ρ/parenrightbigg (F(n←f) )∗F(n←i) The product of two delta functions δ(4)(Pfn)δ(4)(Pni)=δ(4)(Pfi)δ(4)(Pni), and thus we could cancel off δ(4)(Pfi). Also (/Pi1fn1/ρ)(/Pi1 ni1/ρ)/(/Pi1 fi1/ρ)=(/Pi1n1/ρ2), and we happily recover the more familiar factor ρ2[namely (2π)32ωfor bosons]. Thus finally, the gener- alized optical theorem tells that 2ImF(f←i)=/summationdisplay n(2π)4δ(4)(Pni)/parenleftbigg /Pi1n1 ρ2/parenrightbigg (F(n←f) )∗F(n←i), (19) 216 | III. Renormalization and Gauge Invariance namely, that the imaginary part of the Feynman amplitude F(f←i)is given by a sum of(F(n←f) )∗F(n←i)over intermediate states |n/angbracketright. The particles in the intermediate state are of course physical and on shell. We are to sum over all possible states |n/angbracketrightallowed by quantum numbers and by the kinematics. According to Cutkosky, given a Feynman diagram, to obtain its imaginary part, we simply cut through it in the several different ways allowed, corresponding to the differentpossible intermediate state |n/angbracketright. Particles in the state |n/angbracketrightare manifestly real, not virtual. Note that unitarity and hence the optical theorem are nonlinear in the transition am- plitude. This has proved to be enormously useful in actual computation. Suppose we areperturbing in some coupling gand we know F(n←i)andF(n←f)to order g N. The op- tical theorem gives us Im F(f←i)to order gN+1, and we could then construct F(f←i) to order gN+1using a dispersion relation. The application of the Cutkosky rule to the vacuum polarization function discussed in this chapter is particularly simple: there is only one possible way of cutting the Feynmandiagram for /Pi1. Here the initial and final states |i/angbracketrightand|f/angbracketrightboth consist of a single ϕmeson, while the intermediate state |n> consists of an ηand a ξmeson. Referring to the appendix to chapter II.6, we recall that/summationtext ncorresponds to/integraltextd3kη (2π)3d3kξ (2π)3. Thus the optical theorem as stated in (19) says that Im/Pi1(q)=1 2g2/integraldisplayd3kη (2π)32ωηd3kξ (2π)32ωξ(2π)4δ4(kη+kξ−q) (20) precisely what we obtained in (17). You and I could take another bow, since we even get the factor of 2 correctly (as we must!). We should also say, in concluding this chapter, “Vive Cauchy!” Appendix 1: Taking the imaginary part by brute force For those readers who like brute force, we will extract the imaginary part of /Pi1=−ig2/integraldisplayd4k (2π)41 k2−μ2+iε1 (q−k)2−m2+iε(21) by a more staightforward method, as promised in the text. Since /Pi1depends only on q2, we have the luxury of setting q=(M,/vector0). We already know that in the complex M2plane, /Pi1has a cut on the axis starting at M2=(m+μ)2. Let us verify this by brute force. We could restrict ourselves to M> 0. Factorizing, we find that the denominator of the integrand is a product of four factors, k0−(εk−iε),k0+(εk−iε),k0−(M+Ek−iε), andk0−(M−Ek+iε), and thus the integrand has four poles in the complex k0plane. (Evidently, εk=/radicalBig /vectork2+μ2,Ek=/radicalbig /vectork2+m2, and if you are perplexed over the difference between εkandε, then you are hopelessly confused.) We now integrate over k0, choosing to close the contour in the lower half plane and going around picking up poles. Picking up the pole at εk−iε, we obtain /Pi11=−g2/integraltext (d3k/(2π)3)(1/(2εk(εk−M−Ek)(εk−M+Ek))). Picking up the pole at M+Ek−iε, we obtain /Pi12=−g2/integraltext (d3k/(2π)3)(1/((M+Ek−εk)(M+Ek+εk)(2Ek))). We now regard /Pi1=/Pi11+/Pi12as a function of M: /Pi1=−g2/integraldisplayd3k (2π)31 (M+Ek−εk)/bracketleftbigg1 2εk(M−εk−Ek+iε)+1 2Ek(M+Ek+εk−iε)/bracketrightbigg (22) III.8. Becoming Imaginary | 217 In spite of appearances, there is no pole at M≈εk−Ek. (Since this pole would lead to a cut at μ−m, there better not be!) For M> 0 we only care about the pole at M≈εk+Ek=/radicalBig /vectork2+μ2+/radicalbig /vectork2+m2. When we integrate over /vectork, this pole gets smeared into a cut starting at m+μ. So far so good. T o calculate the discontinuity across the cut, we use the identity (16) once again Restoring the iε’s and throwing away the term we don’t care about, we have effectively /Pi1=−g2/integraldisplayd3k (2π)31 2εk(εk−M−Ek+iε)(ε k−M+Ek−iε) =− 2πg2/integraldisplayd4k (2π)4θ(k0)δ(k2−μ2)1 (M−εk+Ek−iε)(M −(εk+Ek)+iε) The discontinuity of /Pi1across the cut just specified is determined by applying (I.2.13) to the factor 1/(M−(εk+Ek)+iε), giving Im /Pi1=2π2g2/integraltext (d4k/(2π)4)θ(k0)δ(k2−μ2)δ(M−(εk+Ek))/2Ek. Use the identity (16) again in the form θ(q0−k0)θ((q−k)2−m2)=θ(q0−k0)θ((q0−k0)−Ek) 2Ek(23) and we obtain Im/Pi1=2π2g2/integraldisplayd4k (2π)4θ(k0)δ(k2−μ2)θ(q0−k0)δ((q−k)2−m2) (24) Remarkably, as Cutkosky taught us, to obtain the imaginary part we simply replace the propagators in (21) by delta functions. Appendix 2: A dispersion representation for the two-point amplitude I would like to give you a bit more flavor of the dispersion program once active and now being revived (seepart N). Consider the two-point amplitude iD(x)≡/angbracketleft0|T(O(x)O(0))|0/angbracketright, with O(x)some operator in the canonical formalism. For example, for O(x) equal to the field ϕ(x) ,D(x) would be the propagator. In chapter I.8, we were able to evaluate D(x) for a free field theory, because then we could solve the field equation of motion and expand ϕ(x) in terms of creation and annihilation operators. But what can we do in a fully interacting field theory? There is no hope of solving the operator field equation of motion. The goal of the dispersion program of the 1950s and 1960s is to say as much as possible about D(x) based on general considerations such as analyticity. OK, so first write iD(x)=θ(x 0)/angbracketleft0|eiPxO(0)e−iPxO(0)|0/angbracketright+θ( −x0)/angbracketleft0|O(0)eiPxO(0)e−iPx|0/angbracketright, where we used spacetime translation O(x)=eiP.xO(0)e−iP.x. By the way, if you are not totally sure of this relation, dif- ferentiate it to obtain ∂μO(x)=i[Pμ,O(x)] which you should recognize as the relativistic version of the usual Heisenberg equations (I.8.2, 3). Now insert 1 =/summationtext n|n/angbracketright/angbracketleftn| , with |n/angbracketrighta complete set of intermediate states, to obtain /angbracketleft0|eiPxO(0)e−iPxO(0)|0/angbracketright=/angbracketleft 0|O(0)e−iPxO(0)|0/angbracketright=/summationtext n/angbracketleft0|O(0)|n/angbracketright/angbracketleftn| e−iPxO(0)|0/angbracketright=/summationtext ne−iPnx|O0n|2, where we used Pμ|0/angbracketright=0 and Pμ|n/angbracketright=Pμ n|n/angbracketright and defined O0n≡/angbracketleft0|O(0)|n/angbracketright. Next, use the integral representations for the step function θ(t)=−i/integraltext (dω/2 π)eiωt/(ω−iε)andθ(−t)=i/integraltext (dω/2 π)eiωt/(ω+iε). Again, if you are not sure of this, simply differentiated dtθ(t)=−id dt/integraltext (dω/2 π)eiωt/(ω−iε)=/integraltext (dω/2 π)eiωt, which you recog- nize from (I.2.12) as indeed the integral representation of the delta function δ(t)=d dtθ(t) . In other words, the representation used here is the integral of the representation in (I.2.12). Putting it all together, we obtain iD(q)=/integraldisplay d4xeiq.xiD(x)=−i(2π)3/summationdisplay n|O0n|2/braceleftBigg δ(3)(/vectorq−/vectorPn) P0 n−q0−iε+δ(3)(/vectorq+/vectorPn) P0 n+q0−iε/bracerightBigg . (25) The integral over d3xproduced the 3-dimensional delta function, while the integral over dx0=dtpicked up the denominator in the integral representation for the step function. Now take the imaginary part using Im1 /(P0 n−q0−iε)=πδ(q0−P0 n). We thus obtain Im(i/integraldisplay d4xeiqx/angbracketleft0|T(O(x)O(0))|0/angbracketright)=π(2π)3/summationdisplay n|O0n|2(δ(4)(q−Pn)+δ(4)(q+Pn)) (26) 218 | III. Renormalization and Gauge Invariance with the more pleasing 4-dimensional delta function. Note that for q0>0 the term involving δ(4)(q+Pn)drops out, since the energies of physical states must be positive. What have we accomplished? Even though we are totally incapable of calculating D(q), we have managed to represent its imaginary part in terms of physical quantities that are measurable in principle, namely the absolutesquare |O 0n|2of the matrix element of O(0)between the vacuum state and the state |n/angbracketright. For example, if O(x) is the meson field ϕ(x) in aϕ4theory, the state |n/angbracketrightwould consist of the single-meson state, the three-meson state, and so on. The general hope during the dispersion era was that by keeping a few states we could obtain a decentapproximation to D(q). Note that the result does not depend on perturbing in some coupling constant. The contribution of the single meson state |/vectork/angbracketrighthas a particularly simple form, as you might expect. With our normalization of single-particle states (as in chapter I.8), Lorentz invariance implies /angbracketleft/vectork|O(0)|0/angbracketright= Z1 2//radicalbig (2π)32ωk, with ωk=/radicalbig /vectork2+m2andZ1 2an unknown constant, measuring the “strength” with which Ois capable of producing the single meson from the vacuum. Putting this into (25) and recognizing that the sum over single-meson states is now given by/integraltext d3k|/vectork/angbracketright/angbracketleft/vectork|[with the normalization /angbracketleft/vectork/prime|/vectork/angbracketright=δ3(/vectork/prime−/vectork)], we find that the single-meson contribution to iD(q) is given by −i( 2π)3/integraldisplay d3kZ (2π)32ωk/braceleftBigg δ(3)(/vectorq−/vectork) ωk−q0−iε+(q→−q)/bracerightBigg =−iZ 2ωq/braceleftBigg 1 ωq−q0−iε+(q→−q)/bracerightBigg =iZ q2−m2+iε(27) This is a very satisfying result: even though we cannot calculate iD(q), we know that it has a pole at a position determined by the meson mass with a residue that depends on how Ois normalized. As a check, we can also easily calculate the contribution of the single-meson state to −ImD(q). Plugging into (26), we find, for q0>0,πZ/integraltext (d3k/2ωk)δ4(q−k)=(πZ/2 ωq)δ(q0−ωq)=πZδ(q2−m2), where we used (I.8.14, 16) in the last step. Given our experience with the vacuum polarization function, we would expect D(q) (which by Lorentz invariance is a function of q2) to have a cut starting at q2=(3m)2. T o verify this, simply look at (26) and choose /vectorq= 0. The contribution of the three-meson state occurs at/radicalbig q2=q0=P0 “3”=/radicalBig /vectork2 1+m2+/radicalBig /vectork2 2+m2+/radicalBig /vectork2 3+m2≥ 3m. The sum over states is now a triple integration over /vectork1,/vectork2, and/vectork3, subject to the constraint /vectork1+/vectork2+/vectork3=0. Knowing the imaginary part of D(q) we can now write a dispersion relation of the kind in (3). Finally, if you stare at (26) long enough (see exercise III.8.3) you will discover the relation Im(i/integraldisplay d4xeiqx/angbracketleft0|T(O(x)O(0))|0/angbracketright)=1 2/integraldisplay d4xeiqx/angbracketleft0|[O(x),O(0)]|0/angbracketright (28) The discussion here is relevant to the discussion of field redefinition in chapter I.8. Suppose our friend uses η=Z1 2ϕ+αϕ3instead of ϕ; then the present discussion shows that his propagator/integraltext d4xeiqx/angbracketleft0|T(η(x)η( 0))|0/angbracketright still has a pole at q2−m2. The important point is that physics fixes the pole to be at the same location. Here we have taken Oto be a Lorentz scalar. In applications (see chapter VII.3) the role of Ois often played by the electromagnetic current Jμ(x) (treated as an operator). The same discussion holds except that we have to keep track of some Lorentz indices. Indeed, we recognize that the vacuum polarization function /Pi1μνthen corresponds to the function Din this discussion. Exercises III.8.1 Evaluate the imaginary part of the vacuum polarization function, and by explicit calculation verify that it is related to the decay rate of a vector particle into an electron and a positron. III.8.2 Suppose we add a term gϕ3to our scalar ϕ4theory. Show that to order g4there is a “box diagram” contributing to meson scattering p1+p2→p3+p4with the amplitude I=g4/integraldisplayd4k (2π)41 (k2−m2−iε)((k +p2)2−m2−iε)((k −p1)2−m2−iε)((k +p2−p3)2−m2−iε) III.8. Becoming Imaginary | 219 Calculate the integral explicitly as a function of s=(p1+p2)2andt=(p3−p2)2. Study the analyticity property of Ias a function of sfor fixed t. Evaluate the discontinuity of Iacross the cut and verify Cutkosky’s cutting rule. Check that the optical theorem works. What about the analyticity property of I as a function of tfor fixed s? And as a function of u=(p3−p1)2? III.8.3 Prove (28). [Hint: Do unto/integraltext d4xeiqx/angbracketleft0|[O(x),O(0)]|0/angbracketrightwhat we did to/integraltext d4xeiqx/angbracketleft0||T(O(x)O(0))|0/angbracketright, namely, insert 1 =/summationtext n|n/angbracketright/angbracketleftn| (with |n/angbracketrighta complete set of states) between O(x) andO(0)in the commu- tator. Now we don’t have to bother with representing the step function.] This page intentionally left blank Part IV Symmetry and Symmetry Breaking This page intentionally left blank IV.1 Symmetry Breaking A symmetric world would be dull While we would like to believe that the fundamental laws of Nature are symmetric, a completely symmetric world would be rather dull, and as a matter of fact, the realworld is not perfectly symmetric. More precisely, we want the Lagrangian, but not theworld described by the Lagrangian, to be symmetric. Indeed, a central theme of modernphysics is the study of how symmetries of the Lagrangian can be broken. We will see insubsequent chapters that our present understanding of the fundamental laws is built uponan understanding of symmetry breaking. Consider the Lagrangian studied in chapter I.10: L=1 2/bracketleftBig (∂/vectorϕ)2−μ2/vectorϕ2/bracketrightBig −λ 4(/vectorϕ2)2(1) where /vectorϕ=(ϕ1,ϕ2,... ,ϕN). This Lagrangian exhibits an O(N) symmetry under which /vectorϕ transforms as an N-component vector. We can easily add terms that do not respect the symmetry. For instance, add terms such as ϕ2 1,ϕ4 1andϕ2 1/vectorϕ2and break the O(N) symmetry down to O(N−1), under which ϕ2,..., ϕNrotate as an (N −1)-component vector. This way of breaking the symmetry, “by hand” as it were, is known as explicit breaking. We can break the symmetry in stages. Obviously, if we want to, we can break it down to O(N−M) by hand, for any M<N . Note that in this example, with the terms we added, the reflection symmetry ϕa→−ϕa (anya)still holds. It is easy enough to break this symmetry as well, by adding a term such asϕ3 a, for example. Breaking the symmetry by hand is not very interesting. Indeed, we might as well start with a nonsymmetric Lagrangian in the first place. 224 | IV . Symmetry and Symmetry Breaking V(q) q Figure IV .1.1 Spontaneous symmetry breaking A more subtle and interesting way is to let the system “break the symmetry itself,” a phenomenon known as spontaneous symmetry breaking. I will explain by way of anexample. Let us flip the sign of the /vectorϕ 2term in (1) and write L=1 2/bracketleftBig (∂/vectorϕ)2+μ2/vectorϕ2/bracketrightBig −λ 4(/vectorϕ2)2(2) Naively, we would conclude that for small λthe field ϕcreates a particle of mass/radicalbig −μ2= iμ. Something is obviously wrong. The essential physics is analogous to what would happen if we give the spring constant in an anharmonic oscillator the wrong sign and write L=1 2(˙q2+kq2)−(λ/4)q4. We all know what to do in classical mechanics. The potential energy V( q)=−1 2kq2+(λ/4)q4 [known as the double-well potential (figure IV .1.1)] has two minima at q=±v, where v≡(k/λ)1 2. At low energies, we choose either one of the two minima and study small oscillations around that minimum. Committing to one or the other of the two minimabreaks the reflection symmetry q→−qof the system. In quantum mechanics, however, the particle can tunnel between the two minima, the tunneling barrier being V(0)−V(±v) . The probability of being in one or the other of the two minima must be equal, thus respecting the reflection symmetry q→−qof the Hamiltonian. In particular, the ground state wave function ψ(q)=ψ(−q) is even. Let us try to extend the same reasoning to quantum field theory. For a generic scalar field Lagrangian L= 1 2(∂0ϕ)2−1 2(∂iϕ)2−V( ϕ) we again have to find the minimum of the potential energy/integraltext dDx[1 2(∂iϕ)2+V( ϕ) ], where Dis the dimension of space. Clearly, any spatial variation in ϕonly increases the energy, and so we set ϕ(x) to equal a spacetime independent quantity ϕand look for the minimum of V( ϕ) . In particular, for the example in (2), we have V( ϕ)=−1 2μ2/vectorϕ2+λ 4(/vectorϕ2)2(3) As we will see, the N=1 case is dramatically different from the N≥2 cases. IV .1. Symmetry Breaking | 225 Difference between quantum mechanics and quantum field theory Study the N=1 case first. The potential V( ϕ) looks exactly the same as the potential in figure IV .1.1 with the horizontal axis relabeled as ϕ. There are two minima at ϕ=±v= ±(μ2/λ)1 2. But some thought reveals a crucial difference between quantum field theory and quan- tum mechanics. The tunneling barrier is now [ V(0)−V(±v) ]/integraltext dDx(where Ddenotes the dimension of space) and hence infinite (or more precisely, extensive with the volume ofthe system)! T unneling is shut down, and the ground state wave function is concentratedaround either +v or−v. We have to commit to one or the other of the two possibilities for the ground state and build perturbation theory around it. It does not matter whichone we choose: The physics is equivalent. But by making a choice, we break the reflectionsymmetry ϕ→−ϕof the Lagrangian. The reflection symmetry is broken spontaneously! We did not put symmetry breaking terms into the Lagrangian by hand but yet the reflection symmetry is broken. 1 Let’s choose the ground state at +v and write ϕ=v+ϕ/prime. Expanding in ϕ/primewe find after a bit of arithmetic that L=μ4 4λ+1 2(∂ϕ/prime)2−μ2ϕ/prime2−O(ϕ/prime3) (4) The physical particle created by the shifted field ϕ/primehas mass√ 2μ. The physical mass squared has to come out positive since, after all, it is just −V/prime/prime(ϕ)|ϕ=v, as you can see after a moment’s thought. Similarly, you would recognize that the first term in (4) is just −V( ϕ) |ϕ=v. If we are only interested in the scattering of the mesons associated with ϕ/primethis term does not enter at all. Indeed, we are always free to add an arbitrary constant to Lto begin with. We had quite arbitrarily set V( ϕ=0)equal to 0. The same situation appears in quantum mechanics: In the discussion of the harmonic oscillator the zero point energy1 2/planckover2piωis not observable; only transitions between energy levels are physical. We will return to this point in chapter VIII.2. Yet another way of looking at (2) is that quantum field theory amounts to doing the Euclidean functional integral Z=/integraldisplay Dϕe−/integraltext ddx{1 2[(∂ϕ)2−μ2ϕ2]+λ 4(ϕ2)2} and perturbation theory just corresponds to studying the small oscillations around a minimum of the Euclidean action. Normally, with μ2positive, we expand around the minimum ϕ=0. With μ2negative, ϕ=0 is a local maximum and not a minimum. In quantum field theory what is called the ground state is also known as the vacuum, since it is literally the state in which the field is “at rest,” with no particles present. Here we 1An insignificant technical aside for the nitpickers: Strictly speaking, in field theory the ground state wave function should be called a wave functional, since /Psi1[ϕ(/vectorx)] is a functional of the function ϕ(/vectorx). 226 | IV . Symmetry and Symmetry Breaking have two physically equivalent vacua from which we are to choose one. The value assumed byϕin the ground state, either vor−vin our example, is known as the vacuum expectation value of ϕ. The field ϕis said to have acquired a vacuum expectation value. Continuous symmetry Let us now turn to (2) with N≥2. The potential (3) is shown in figure IV .1.2 for N=2. The shape of the potential has been variously compared to the bottom of a punted wine bottleor a Mexican hat. The potential is minimized at /vectorϕ 2=μ2/λ. Something interesting is going on: We have an infinite number of vacua characterized by the direction of /vectorϕin that vacuum. Because of the O(2)symmetry of the Lagrangian they are all physically equivalent. The result had better not depend on our choice. So let us choose /vectorϕto point in the 1 direction, that is, ϕ1=v≡+/radicalbig μ2/λandϕ2=0. Now consider fluctuations around this field configuration, in other words, write ϕ1= v+ϕ/prime 1andϕ2=ϕ/prime 2, plug into (2) for N=2, and expand Lout. I invite you to do the arithmetic. You should find (after dropping the primes on the fields; why clutter thenotation, right?) L=μ4 4λ+1 2/bracketleftBig (∂ϕ 1)2+(∂ϕ2)2/bracketrightBig −μ2ϕ2 1+O(ϕ3) (5) The constant term is exactly as in (4), and just like the field ϕ/primein (4), the field ϕ1has mass√ 2μ. But now note the remarkable feature of (5): the absence of a ϕ2 2term. The field ϕ2is massless! Emergence of massless boson Thatϕ2comes out massless is not an accident. I will now explain that the masslessness is a general and exact phenomenon. Referring back to figure IV .1.2 we can easily understand the particle spectrum. Excitation in the ϕ1field corresponds to fluctuation in the radial direction, “climbing the wall” so to speak, while excitation in the ϕ2field corresponds to fluctuation in the angular direction, “rolling along the gutter” so to speak. It costs no energy for a marble to roll along theminima of the potential energy, going from one minimum to another. Another way ofsaying this is to picture a long wavelength excitation of the form ϕ 2=asin(ωt−/vectork/vectorx)with asmall. In a region of length scale small compared to |/vectork|−1, the field ϕ2is essentially constant and thus the field /vectorϕis just rotated slightly away from the 1 direction, which by theO(2)symmetry is equivalent to the vacuum. It is only when we look at regions of length scale large compared to |/vectork|−1that we realize that the excitation costs energy. Thus, as|/vectork|→0, we expect the energy of the excitation to vanish. We now understand the crucial difference between the N=1 and the N=2 cases: In the former we have a reflection symmetry, which is discrete, while in the latter we have anO(2)symmetry, which is continuous. IV .1. Symmetry Breaking | 227 Figure IV .1.2 We have worked out the N=2 case in detail. You should now be able to generalize our discussion to arbitrary N≥2 (see exercise IV .1.1). Meanwhile, it is worth looking at N=2 from another point of view. Many field theories can be written in more than one form and it is important to know them under differentguises. Construct the complex field ϕ=(1/√ 2)(ϕ1+iϕ2); we have ϕ†ϕ=1 2(ϕ2 1+ϕ2 2)and so can write (2) as L=∂ϕ†∂ϕ+μ2ϕ†ϕ−λ(ϕ†ϕ)2(6) which is manifestly invariant under the U(1)transformation ϕ→eiαϕ(recall chapter I.10). You may recognize that this amounts to saying that the groups O(2)andU(1)are locally isomorphic. Just as we can write a vector in Cartesian or polar coordinates we are freeto parametrize the field by ϕ(x)=ρ(x)e iθ(x)(as in chapter III.5) so that ∂μϕ=(∂μρ+ iρ∂μθ)eiθ. We obtain L=ρ2(∂θ)2+(∂ρ)2+μ2ρ2−λρ4. Spontaneous symmetry breaking means setting ρ=v+χwithv=+/radicalbig μ2/2λ, whereupon L=v2(∂θ)2+⎡ ⎣(∂χ)2−2μ2χ2−4/radicalBigg μ2λ 2χ3−λχ4⎤ ⎦+⎛ ⎝/radicalBigg 2μ2 λχ+χ2⎞ ⎠(∂θ)2(7) We recognize the phase θ(x) as the massless field. We have arranged the terms in the Lagrangian in three groups: the kinetic energy of the massless field θ, the kinetic and potential energy of the massive field χ, and the interaction between θandχ. (The additive constant in (5) has been dropped to minimize clutter.) 228 | IV . Symmetry and Symmetry Breaking Goldstone’s theorem We will now prove Goldstone’s theorem, which states that whenever a continuous sym- metry is spontaneously broken, massless fields, known as Nambu2-Goldstone bosons, emerge. Recall that associated with every continuous symmetry is a conserved charge Q. That Q generates a symmetry is stated as [H,Q]=0 (8) Let the vacuum (or ground state in quantum mechanics) be denoted by |0/angbracketright. By adding an appropriate constant to the Hamiltonian H→H+cwe can always write H|0/angbracketright=0. Normally, the vacuum is invariant under the symmetry transformation, eiθQ|0/angbracketright=| 0/angbracketright,o r in other words Q|0/angbracketright=0. But suppose the symmetry is spontaneously broken, so that the vacuum is not invariant under the symmetry transformation; in other words, Q|0/angbracketright /negationslash= 0. Consider the state Q|0/angbracketright. What is its energy? Well, HQ|0/angbracketright=[H,Q]|0/angbracketright=0 (9) [The first equality follows from H|0/angbracketright=0 and the second from (8).] Thus, we have found another state Q|0/angbracketrightwith the same energy as |0/angbracketright. Note that the proof makes no reference to either relativity or fields. You can also see that it merely formalizes the picture of the marble rolling along the gutter. In quantum field theory, we have local currents, and so Q=/integraldisplay dDxJ0(/vectorx,t) where Ddenotes the dimension of space and conservation of Qsays that the integral can be evaluated at any time. Consider the state |s/angbracketright=/integraldisplay dDxe−i/vectork/vectorxJ0(/vectorx,t)|0/angbracketright which has3spatial momentum /vectork.A s/vectorkgoes to zero it goes over to Q|0/angbracketright, which as we learned in (9) has zero energy. Thus, as the momentum of the state |s/angbracketrightgoes to zero, its energy goes to zero. In a relativistic theory, this means precisely that |s/angbracketrightdescribes a massless particle. 2Y . Nambu, quite deservedly, received the 2008 physics Nobel Prize for his profound contribution to our understanding of spontaneous symmetry breaking. 3Acting on it with Pi(exercise I.11.3) and using Pi|0/angbracketright=0, we have Pi|s/angbracketright=/integraldisplay dDxe−i/vectork/vectorx[Pi,J0(/vectorx,t)]|0/angbracketright=−i/integraldisplay dDe−i/vectork./vectorx∂iJ0(/vectorx,t)|0/angbracketright=ki|s/angbracketright upon integrating by parts. IV .1. Symmetry Breaking | 229 The proof makes clear that the theorem practically exudes generality: It applies to any spontaneously broken continuous symmetry. Counting Nambu-Goldstone bosons From our proof, we see that the number of Nambu-Goldstone bosons is clearly equal tothe number of conserved charges that do not leave the vacuum invariant, that is, do notannihilate |0/angbracketright. For each such charge Q α, we can construct a zero-energy state Qα|0/angbracketright. In our example, we have only one current Jμ=i(ϕ1∂μϕ2−ϕ2∂μϕ1)and hence one Nambu-Goldstone boson. In general, if the Lagrangian is left invariant by a symmetrygroup Gwithn(G) generators, but the vacuum is left invariant by only a subgroup HofG withn(H) generators, then there are n(G)− n(H) Nambu-Goldstone bosons. If you want to show off your mastery of mathematical jargon you can say that the Nambu-Goldstonebosons live in the coset space G/H . Ferromagnet and spin wave The generality of the proof suggests that the usefulness of Goldstone’s theorem is not restricted to particle physics. In fact, it originated in condensed matter physics, the classicexample there being the ferromagnet. The Hamiltonian, being composed of just theinteraction of nonrelativistic electrons with the ions in the solid, is of course invariantunder the rotation group SO( 3), but the magnetization /vectorMpicks out a direction, and the ferromagnet is left invariant only under the subgroup SO( 2)consisting of rotations about the axis defined by /vectorM. The Nambu-Goldstone theorem is easy to visualize physically. Consider a “spin wave” in which the local magnetization /vectorM(/vectorx)varies slowly from point to point. A physicist living in a region small compared to the wavelength does not evenrealize that he or she is no longer in the “vacuum.” Thus, the frequency of the wave mustgo to zero as the wavelength goes to infinity. This is of course exactly the same heuristicargument given earlier. Note that quantum mechanics is needed only to translate the wavevector /vectorkinto momentum and the frequency ωinto energy. I will come back to magnets and spin wave in chapters V .3 and VI.5. Quantum fluctuations and the dimension of spacetime Our discussion of spontaneous symmetry breaking is essentially classical. What happenswhen quantum fluctuations are included? I will address this question in detail in chap-ter IV .3, but for now let us go back to (5). In the ground state, ϕ 1=vandϕ2=0. Recall that in the mattress model of a scalar field theory the mass term comes from the springsholding the mattress to its equilibrium position. The term −μ 2ϕ/prime2 1(note the prime) in (5) 230 | IV . Symmetry and Symmetry Breaking tells us that it costs action for ϕ/prime 1to wander away from its ground state value ϕ/prime 1=0. But now we are worried: ϕ2is massless. Can it wander away from its ground state value? T o answer this question let us calculate the mean square fluctuation /angbracketleft(ϕ2(0))2/angbracketright=1 Z/integraldisplay DϕeiS(ϕ)[ϕ2(0)]2 =lim x→01 Z/integraldisplay DϕeiS(ϕ)ϕ2(x)ϕ 2(0) =lim x→0/integraldisplayddk (2π)deikx k2(10) (We recognized the functional integral that defines the propagator; recall chapter I.7.) The upper limit of the integral in (10) is cut off at some /Lambda1(which would correspond to the inverse of the lattice spacing when applying these ideas to a ferromagnet) and so asexplained in chapter III.1 (and as you will see in chapter VIII.3) we are not particularlyworried about the ultraviolet divergence for large k. But we do have to worry about a possible infrared divergence for small k.(Note that for a massive field 1 /k 2in (10) would have been replaced by 1 /(k2+μ2)and there would be no infrared divergence.) We see that there is no infrared divergence for d> 2. Our picture of spontaneously breaking a continuous symmetry is valid in our (3+1)-dimensional world. However, for d≤2 the mean square fluctuation of ϕ2comes out infinite, so our naive picture is totally off. We have arrived at the Coleman-Mermin-Wagner theorem (provedindependently by a particle theorist and two condensed matter theorists), which statesthat spontaneous breaking of a continuous symmetry is impossible for d=2. Note that while our discussion is given for O(2)symmetry the conclusion applies to any continuous symmetry since the argument depends only on the presence of Nambu-Goldstone fields. In our examples, symmetry is spontaneously broken by a scalar field ϕ, but nothing says that the field ϕmust be elementary. In many condensed matter systems, superconductors, for example, symmetries are spontaneously broken, but we know that the system consistsof electrons and atomic nuclei. The field ϕis generated dynamically, for example as a bound state of two electrons in superconductors. More on this in chapter V .4. The spontaneousbreaking of a symmetry by a dynamically generated field is sometimes referred to asdynamical symmetry breaking. 4 Exercises IV .1.1 Show explicitly that there are N−1 Nambu-Goldstone bosons in the G=O(N) example (2). IV .1.2 Construct the analog of (2) with Ncomplex scalar fields and invariant under SU(N) . Count the number of Nambu-Goldstone bosons when one of the scalar fields acquires a vacuum expectation value. 4This chapter is dedicated to the memory of the late Jorge Swieca. IV.2 The Pion as a Nambu-Goldstone Boson Crisis for field theory After the spectacular triumphs of quantum field theory in the electromagnetic interaction, physicists in the 1950s and 1960s were naturally eager to apply it to the strong and weakinteractions. As we have already seen, field theory when applied to the weak interactionappeared not to be renormalizable. As for the strong interaction, field theory appeared to-tally untenable for other reasons. For one thing, as the number of experimentally observedhadrons (namely strongly interacting particles) proliferated, it became clear that were weto associate a field with each hadron the resulting field theory would be quite a mess, withnumerous arbitrary coupling constants. But even if we were to restrict ourselves to nucle-ons and pions, the known coupling constant of the interaction between pions and nucleonsis a large number. (Hence the term strong interaction in the first place!) The perturbativeapproach that worked so spectacularly well in quantum electrodynamics was doomed tofailure. Many eminent physicists at the time advocated abandoning quantum field theory al- together, and at certain graduate schools, quantum field theory was even dropped fromthe curriculum. It was not until the early 1970s that quantum field theory made a tri-umphant comeback. A field theory for the strong interaction was formulated, not in termsof hadrons, but in terms of quarks and gluons. I will get to that in chapter VII.3. Pion weak decay T o understand the crisis facing field theory, let us go back in time and imagine what a fieldtheorist might be trying to do in the late 1950s. Since this is not a book on particle physics,I will merely sketch the relevant facts. You are urged to consult one of the texts on the subject. 1By that time, many semileptonic decays such as n→p+e−+ν,π−→e−+ν, 1See, e.g., E. Commins and P. H. Bucksbaum, Weak Interactions of Leptons and Quarks. 232 | IV . Symmetry and Symmetry Breaking andπ−→π0+e−+νhad been measured. Neutron βdecay n→p+e−+νwas of course the process for which Fermi invented his theory, which by that time had assumedthe form L=G[ eγμ(1−γ5)ν][pγμ(1−γ5)n], where nis a neutron field annihilating a neutron, pa proton field annihilating a proton, νa neutrino field annihilating a neutrino (or creating an antineutrino as in βdecay), and ean electron field annihilating an electron. It became clear that to write down a field for each hadron and a Lagrangian for each decay process, as theorists were in fact doing for a while, was a losing battle. Instead, weshould write L=G[eγμ(1−γ5)ν](Jμ−J5μ) (1) withJμandJ5μtwo currents transforming as a Lorentz vector, and axial vector respectively. We think of JμandJ5μas quantum operators in a canonical formulation of field theory. Our task would then be to calculate the matrix elements between hadron states, /angbracketleftp|(Jμ− J5μ)|n/angbracketright,/angbracketleft0|(Jμ−J5μ)|π−/angbracketright,/angbracketleftπ0|(Jμ−J5μ)|π−/angbracketright, and so on, corresponding to the three decay processes I listed above. (I should make clear that although I am talking about weakdecays, the calculation of these matrix elements is a problem in the strong interaction. Inother words, in understanding these decays, we have to treat the strong interaction to allorders in the strong coupling, but it suffices to treat the weak interaction to lowest order inthe weak coupling G.) Actually, there is a precedent for the attitude we are adopting here. T o account for nuclear βdecay (Z,A)→(Z+1,A)+e −+ν, Fermi certainly did not write a separate Lagrangian for each nucleus. Rather, it was the task of the nuclear theoristto calculate the matrix element /angbracketleftZ+1,A|[ pγμ(1−γ5)n]|Z,A/angbracketright. Similarly, it is the task of the strong interaction theorist to calculate matrix elements such as /angbracketleftp|(Jμ−J5μ)|n/angbracketright. For the story I am telling, let me focus on trying to calculate the matrix element of the axial vector current Jμ 5between a neutron and a proton. Here we make a trivial change in notation: We no longer indicate that we have a neutron in the initial state and aproton in the final state, but instead we specify the momentum pof the neutron and the momentum p /primeof the proton. Incidentally, in (1) the fields and the currents are of course all functions of the spacetime coordinates x. Thus, we want to calculate /angbracketleftp/prime|Jμ 5(x)|p/angbracketright, but by translation invariance this is equal to /angbracketleftp/prime|Jμ 5(0)|p/angbracketrighte−i(p/prime−p).x. Henceforth, we simply calculate /angbracketleftp/prime|Jμ 5(0)|p/angbracketrightand suppress the 0. Note that spin labels have already been suppressed. Lorentz invariance and parity can take us some distance: They imply that2 /angbracketleftp/prime|Jμ 5|p/angbracketright=¯u(p/prime)[γμγ5F(q2)+qμγ5G(q2)]u(p) (2) withq≡p/prime−p[compare with (III.6.7)]. But Lorentz invariance and parity can only take us so far: We know nothing about the “form factors” F(q2)andG(q2). 2Another possible term of the form (p/prime+p)μγ5can be shown to vanish by charge conjugation and isospin symmetries. IV .2. Pion as Nambu-Goldstone Boson | 233 (a) (b) Figure IV .2.1 Similarly, for the matrix element /angbracketleft0|Jμ 5|π−/angbracketrightLorentz invariance tells us that /angbracketleft0|Jμ 5|k/angbracketright=fkμ(3) I have again labeled the initial state by the momentum kof the pion. The right-hand side of (3) has to be a vector but since kis the only vector available it has to be proportional tok. Just like F(q2)andG(q2), the constant fis a strong interaction quantity that we don’t know how to calculate. On the other hand, F(q2),G(q2), and fcan all be measured experimentally. For instance, the rate for the decay π−→e−+νclearly depends on f2. Too many diagrams Let us look over the shoulder of a field theorist trying to calculate /angbracketleftp/prime|Jμ 5|p/angbracketrightand/angbracketleft0|Jμ 5|k/angbracketright in (2)in the late 1950s. He would draw Feynman diagrams such as the ones in figures IV .2.1 and IV .2.2 and soon realize that it would be hopeless. Because of the strong coupling, hewould have to calculate an infinite number of diagrams, even if the strong interaction weredescribed by a field theory, a notion already rejected by many luminaries of the time. Figure IV .2.2 234 | IV . Symmetry and Symmetry Breaking In telling the story of the breakthrough I am not going to follow the absolutely fascinating history of the subject, full of total confusion and blind alleys. Instead, with the benefit ofhindsight, I am going to tell the story using what I regard as the best pedagogical approach. The pion is very light The breakthrough originated in the observation that the mass of the π−at 139 Mev was considerably less than the mass of the proton at 938 Mev. For a long time this was simplytaken as a fact not in any particular need of an explanation. But eventually some theoristswondered why one hadron should be so much lighter than another. Finally, some theorists took the bold step of imagining an “ideal world” in which the π −is massless. The idea was that this ideal world would be a good approximation of our world, to an accuracy of about 15% (∼139/938). Do you remember one circumstance in which a massless spinless particle would emerge naturally? Yes, spontaneous symmetry breaking! In one of the blinding insights that havecharacterized the history of particle physics, some theorists proposed that the πmesons are the Nambu-Goldstone bosons of some spontaneous broken symmetry. Indeed, let’s multiply (3) by k μ: kμ/angbracketleft0|Jμ 5|k/angbracketright=fk2=fm2 π(4) which is equal to zero in the ideal world. Recall from our earlier discussion on translation invariance that /angbracketleft0|Jμ 5(x)|k/angbracketright=/angbracketleft 0|Jμ 5(0)|k/angbracketrighte−ik.x and hence /angbracketleft0|∂μJμ 5(x)|k/angbracketright=−ikμ/angbracketleft0|Jμ 5(0)|k/angbracketrighte−ik.x Thus, if the axial current is conserved, ∂μJμ 5(x)=0, in the ideal world, kμ/angbracketleft0|Jμ 5|k/angbracketright=0 and (4) would indeed imply m2 π=0. The ideal world we are discussing enjoys a symmetry known as the chiral symmetry of the strong interaction. The symmetry is spontaneously broken in the ground state weinhabit, with the πmeson as the Nambu-Goldstone boson. The Noether current associated with this symmetry is the conserved J μ 5. In fact, you should recognize that the manipulation here is closely related to the proof of the Nambu-Goldstone theorem given in chapter IV .1. Goldberger-Treiman relation Now comes the punchline. Multiply (2 )by(p/prime−p)μ. By the same translation invariance argument we just used, (p/prime−p)μ/angbracketleftp/prime|Jμ 5(0)|p/angbracketright=i/angbracketleftp/prime|∂μJμ 5(x)|p/angbracketrightei(p/prime−p).x IV .2. Pion as Nambu-Goldstone Boson | 235 and hence vanishes if ∂μJμ 5=0. On the other hand, multiplying the right-hand side of (2)by(p/prime−p)μwe obtain ¯u(p/prime)[(/negationslashp/prime− /negationslashp)γ5F(q2)+q2γ5G(q2)]u(p) . Using the Dirac equation (do it!) we conclude that 0=2mNF(q2)+q2G(q2) (5) withmNthe nucleon mass. The form factors F(q2)andG(q2)are each determined by an infinite number of Feynman diagrams we have no hope of calculating, but yet we have managed to relatethem! This represents a common strategy in many areas of physics: When faced withvarious quantities we don’t know how to calculate, we can nevertheless try to relate them. We can go farther by letting q→0 in (5). Referring to (2) we see that F(0)is measured experimentally in n→p+e −+ν(the momentum transfer is negligible on the scale of the strong interaction). But oops, we seem to have a problem: We predict the nucleon massm N=0! In fact, we are saved by examining figure IV .2.1b: There are an infinite number of diagrams exhibiting a pole due to none other than the massless πmeson, which you can see gives fqμ1 q2gπNN¯u(p/prime)γ5u(p) (6) When the πpropagator joins onto the nucleon line, an infinite number of diagrams summed together gives the experimentally measured pion-nucleon coupling constantg πNN . Thus, referring to (2), we see that for q∼0 the form factor G(q2)∼f(1/q2)gπNN . Plugging into (5), we obtain the celebrated Goldberger-T reiman relation 2mNF(0)+fgπNN=0 (7) relating four experimentally measured quantities. As might be expected, it holds with about a 15% error, consistent with our not living in a world with an exactly massless πmeson. Toward a theory of the strong interaction The art of relating infinite sets of Feynman diagrams without calculating them, and it is an art form involving a great deal of cleverness, was developed into a subject called dispersionrelations and S-matrix theory, which we mentioned briefly in chapter III.8. Our present understanding of the strong interaction was built on this foundation. You could see fromthis example that an important component of dispersion relations was the study of theanalyticity properties of Feynman diagrams as described in chapter III.8. The essence ofthe Goldberger-T reiman argument is separating the infinite number of diagrams into thosewith a pole in the complex q 2-plane and those without a pole (but with a cut.) The discovery that the strong interaction contains a spontaneously broken symmetry provided a crucial clue to the underlying theory of the strong interaction and ultimatelyled to the concepts of quarks and gluons. 236 | IV . Symmetry and Symmetry Breaking A note for the historian of science: Whether theoretical physicists regard a quantity as small or large depends (obviously) on the cultural and mental framework they grew upin. T reiman once told me that the notion of setting 138 Mev to zero, when the energyreleased per nucleon in nuclear fission is of order 10 Mev, struck the generation that grewup with the atomic bomb (as T reiman did—he was with the armed forces in the Pacific) assurely the height of absurdity. Now of course a new generation of young string theoristsis perfectly comfortable in regarding anything less than the Planck energy 10 19Gev as essentially zero. IV.3 Effective Potential Quantum fluctuations and symmetry breaking The important phenomenon of spontaneous symmetry breaking was based on minimizing the classical potential energy V( ϕ) of a quantum field theory. It is natural to wonder how quantum fluctuations would change this picture. T o motivate the discussion, consider once again (III.3.3) L=1 2(∂ϕ)2−1 2μ2ϕ2−1 4!λϕ4+A(∂ϕ)2+Bϕ2+Cϕ4(1) (Speaking of quantum fluctuations, we have to include counterterms as indicated.) What have you learned about this theory? For μ2>0, the action is extremized at ϕ=0, and quantizing the small fluctuations around ϕ=0 we obtain scalar particles that scatter off each other. For μ2<0, the action is extremized at some ϕmin, and the discrete symmetry ϕ→−ϕis spontaneously broken, as you learned in chapter IV .1. What happens when μ=0? T o break or not to break, that is the question. A quick guess is that quantum fluctuations would break the symmetry. The μ=0 theory is posed on the edge of symmetry breaking, and quantum fluctuations ought to push itover the brink. Think of a classical pencil perfectly balanced on its tip. Then “switch on”quantum mechanics. Wisdom of the son-in-law Let us follow Schwinger and Jona-Lasinio and develop the formalism that enables us toanswer this question. Consider a scalar field theory defined by Z=eiW(J)=/integraldisplay Dϕei[S(ϕ)+Jϕ ](2) [with the convenient shorthand Jϕ=/integraltext d4xJ(x)ϕ(x)]. If we can do the functional integral, we obtain the generating functional W(J) . As explained in chapter I.7, by differentiating 238 | IV . Symmetry and Symmetry Breaking Wwith respect to the source J(x) repeatedly, we can obtain any Green’s function and hence any scattering amplitude we want. In particular, ϕc(x)≡δW δJ(x)=1 Z/integraldisplay Dϕei[S(ϕ)+Jϕ ]ϕ(x) (3) The subscript cis used traditionally to remind us (see appendix 2 in chapter I.8) that in a canonical formalism ϕc(x) is the expectation value /angbracketleft0|ˆϕ|0/angbracketrightof the quantum operator ˆϕ.I t is certainly not to be confused with the integration dummy variable ϕin (3). The relation (3) determines ϕc(x) as a functional of J. Given a functional WofJwe can perform a Legendre transform to obtain a functional /Gamma1ofϕc. Legendre transform is just the fancy term for the simple relation /Gamma1(ϕc)=W(J) −/integraldisplay d4xJ(x)ϕ c(x) (4) The relation is simple, but be careful about what it says: It defines a functional of ϕc(x) through the implicit dependence of Jonϕc. On the right-hand side of (4) Jis to be eliminated in favor of ϕcby solving (3). We expand the functional /Gamma1(ϕc)in the form /Gamma1(ϕc)=/integraldisplay d4x[−V eff(ϕc)+Z(ϕc)(∂ϕc)2+...] (5) where (...)indicates terms with higher and higher powers of ∂. We will soon see the wisdom of the notation Veff(ϕc). The point of the Legendre transform is that the functional derivative of /Gamma1is nice and simple: δ/Gamma1(ϕc) δϕc(y)=/integraldisplay d4xδJ(x) δϕc(y)δW(J) δJ(x)−/integraldisplay d4xδJ(x) δϕc(y)ϕc(x)−J(y) =−J(y) (6) a relation we can think of as the “dual” of δW(J)/δJ(x) =ϕc(x). If you vaguely feel that you have seen this sort of manipulation before in your physics eduction, you are quite right! It was in a course on thermodynamics, where you learnedabout the Legendre transform relating the free energy to the energy: F=E−TS with Fa function of the temperature TandEa function of the entropy S. Thus Jand ϕare “conjugate” pairs just like TandS(or even more clearly magnetic field Hand magnetization M). Convince yourself that this is far more than a mere coincidence. ForJandϕ cindependent of xwe see from (5) that the condition (6) reduces to V/prime eff(ϕc)=J (7) This relation makes clear what the effective potential Veff(ϕc)is good for. Let’s ask what happens when there is no external source J. The answer is immediate: (7) tells us that V/prime eff(ϕc)=0, (8) IV .3. Effective Potential | 239 In other words, the vacuum expectation value of ˆϕin the absence of an external source is determined by minimizing Veff(ϕc). First order in quantum fluctuations All of these formal manipulations are not worth much if we cannot evaluate W(J) .I n fact, in most cases we can only evaluate eiW(J)=/integraltext Dϕei[S(ϕ)+Jϕ ]in the steepest descent approximation (see chapter I.2). Let us turn the crank and find the steepest descent “point”ϕ s(x), namely the solution of (henceforth I will drop the subscript c as there is little risk of confusion) δ[S(ϕ)+/integraltext d4yJ(y)ϕ(y)] δϕ(x)/vextendsingle/vextendsingle/vextendsingle/vextendsingle/vextendsingle ϕs=0 (9) or more explicitly, ∂2ϕs(x)+V/prime[ϕs(x)]=J(x) (10) Write the dummy integration variable in (2) as ϕ=ϕs+/tildewideϕand expand to quadratic order in/tildewideϕto obtain Z=e(i//planckover2pi)W(J)=/integraldisplay Dϕe(i//planckover2pi)[S(ϕ)+Jϕ ] /similarequale(i//planckover2pi)[S(ϕs)+Jϕ s]/integraldisplay D/tildewideϕe(i//planckover2pi)/integraltext d4x1 2[(∂/tildewideϕ)2−V/prime/prime(ϕs)/tildewideϕ2] =e(i//planckover2pi)[S(ϕs)+Jϕ s]−1 2tr log[∂2+V/prime/prime(ϕs)](11) We have used (II.5.2) to represent the determinant we get upon integrating over /tildewideϕ. Note that I have put back Planck’s constant /planckover2pi. Here ϕs, as a solution of (10), is to be regarded as a function of J. Now that we have determined W(J) =[S(ϕs)+Jϕs]+i/planckover2pi 2tr log[∂2+V/prime/prime(ϕs)]+O(/planckover2pi2) it is straightforward to Legendre transform. I will go painfully slowly here: ϕ=δW δJ=δ[S(ϕs)+Jϕs] δϕsδϕs δJ+ϕs+O(/planckover2pi)=ϕs+O(/planckover2pi) T o leading order in /planckover2pi,ϕ(namely the object formerly known as ϕc)is equal to ϕs. Thus, from (4) we obtain /Gamma1(ϕ)=S(ϕ)+i/planckover2pi 2tr log[∂2+V/prime/prime(ϕ)]+O(/planckover2pi2) (12) Nice though this formula looks, in practice it is impossible to evaluate the trace for arbitrary ϕ(x) : We have to find all the eigenvalues of the operator ∂2+V/prime/prime(ϕ), take their log, and sum. Our task simplifies drastically if we are content with studying /Gamma1(ϕ) for 240 | IV . Symmetry and Symmetry Breaking ϕindependent of x, in which case V/prime/prime(ϕ) is a constant and the operator ∂2+V/prime/prime(ϕ) is translation invariant and easily treated in momentum space: tr log[∂2+V/prime/prime(ϕ)]=/integraldisplay d4x/angbracketleftx|log[∂2+V/prime/prime(ϕ)]|x/angbracketright =/integraldisplay d4x/integraldisplayd4k (2π)4/angbracketleftx|k/angbracketright/angbracketleftk|log[∂2+V/prime/prime(ϕ)]|k/angbracketright/angbracketleftk|x/angbracketright =/integraldisplay d4x/integraldisplayd4k (2π)4log[−k2+V/prime/prime(ϕ)] (13) Referring to (5), we obtain Veff(ϕ)=V( ϕ)−i/planckover2pi 2/integraldisplayd4k (2π)4log/bracketleftbiggk2−V/prime/prime(ϕ) k2/bracketrightbigg +O(/planckover2pi2) (14) known as the Coleman-Weinberg effective potential. What we computed is the order /planckover2pi correction to the classical potential V( ϕ) . Note that we have added a ϕindependent constant to make the argument of the logarithm dimensionless. We can give a nice physical interpretation of (14). Let the universe be suffused with the scalar field ϕ(x) taking on the value ϕ, a background field so to speak. For V( ϕ)= 1 2μ2ϕ2+(1/4!)λϕ4, we have V/prime/prime(ϕ)=μ2+1 2λϕ2≡μ(ϕ)2, which, as the notation μ(ϕ)2 suggests, we recognize as the ϕ-dependent effective mass squared of a scalar particle propagating in the background field ϕ. The mass squared μ2in the Lagrangian is corrected by a term1 2λϕ2due to the interaction of the particle with the background field ϕ. Now we see clearly what (14) tells us: The first term V( ϕ) is the classical energy density contained in the background ϕ, while the second term is the vacuum energy density of a scalar field with mass squared equal to V/prime/prime(ϕ) [see (II.5.3) and exercise IV .3.4]. Your renormalization theory at work The integral in (14) is quadratically divergent, or more correctly, quadratically dependent on the cutoff. But no sweat, we were instructed to introduce three counterterms (of whichonly two are relevant here since ϕis independent of x). Thus, we actually have Veff(ϕ)=V( ϕ)+/planckover2pi 2/integraldisplayd4kE (2π)4log/bracketleftBigg k2 E+V/prime/prime(ϕ) k2 E/bracketrightBigg +Bϕ2+Cϕ4+O(/planckover2pi2) (15) where we have Wick rotated to a Euclidean integral (see appendix D). Using (D.9) and integrating up to k2 E=/Lambda12, we obtain (suppressing /planckover2pi) Veff(ϕ)=V( ϕ)+/Lambda12 32π2V/prime/prime(ϕ)−[V/prime/prime(ϕ)]2 64π2loge1 2/Lambda12 V/prime/prime(ϕ)+Bϕ2+Cϕ4(16) As expected, since the integrand in (15) goes as 1 /k2 Efor large k2 Ethe integral depends quadratically and logarithmically on the cutoff /Lambda12. IV .3. Effective Potential | 241 Watch renormalization theory at work! Since Vis a quartic polynomial in ϕ,V/prime/prime(ϕ) is a quadratic polynomial and [ V/prime/prime(ϕ)]2a quartic polynomial. Thus, we have just enough coun- terterms Bϕ2+Cϕ4to absorb the cutoff dependence. This is a particularly transparent example of how the method of adding counterterms works. T o see how bad things can happen in a nonrenormalizable theory, suppose in contrast thatVis a polynomial of degree 6 in ϕ. Then we are allowed to have three counterterms Bϕ2+Cϕ4+Dϕ6, but that is not enough since [ V/prime/prime(ϕ)]2is now a polynomial of degree 8. This means that we should have started out with Va polynomial of degree 8, but then [V/prime/prime(ϕ)]2would be a polynomial of degree 12. Clearly, the process escalates into an infinite- degree polynomial. We see the hallmark of a nonrenormalizable theory: its insatiableappetite for counterterms. Imposing renormalization conditions Waking up from the nightmare of an infinite number of counterterms chasing us, let usgo back to the sweetly renormalizable ϕ 4theory. In chapter III.3 we fix the counterterms by imposing conditions on various scattering amplitudes. Here we would have to fix thecoefficients BandCby imposing two conditions on V eff(ϕ) at appropriate values of ϕ.W e are working in field space, so to speak, rather than momentum space, but the conceptualframework is the same. We could proceed with the general quartic polynomial V( ϕ) , but instead let us try to answer the motivating question of this chapter: What happens when μ=0, that is, when V( ϕ)=(1/4!)λϕ 4? The arithmetic is also simpler. Evaluating (16) we get Veff(ϕ)=(/Lambda12 64π2λ+B)ϕ2+(1 4!λ+λ2 (16π)2logϕ2 /Lambda12+C)ϕ4+O(λ3) (after absorbing some ϕ-independent constants into C). We see explicitly that the /Lambda1 dependence can be absorbed into BandC. We started out with a purely quartic V( ϕ) . Quantum fluctuations generate a quadrati- cally divergent ϕ2term that we can cancel with the Bcounterterm. What does μ=0 mean? It means that (d2V/ dϕ2)|ϕ=0vanishes. T o say that we have a μ=0 theory means that we have to maintain a vanishing renormalized mass squared, defined here as the coefficient ofϕ2. Thus, we impose our first condition d2Veff dϕ2/vextendsingle/vextendsingle/vextendsingle/vextendsingle ϕ=0=0 (17) This is a somewhat long-winded way of saying that we want B=−(/Lambda12/64π2)λto this order. Similarly, we might think that the second condition would be to set (d4Veff/dϕ4)|ϕ=0 equal to some coupling, but differentiating the ϕ4logϕterm in Vefffour times we are going to get a term like log ϕ, which is not defined at ϕ=0. We are forced to impose our 242 | IV . Symmetry and Symmetry Breaking condition on d4Veff/dϕ4not at ϕ=0 but at ϕequal to some arbitrarily chosen mass M. (Recall that ϕhas the dimension of mass.) Thus, the second condition reads d4Veff dϕ4/vextendsingle/vextendsingle/vextendsingle/vextendsingle ϕ=M=λ(M) (18) where λ(M) is a coupling manifestly dependent on M. Plugging Veff(ϕ)=(1 4!λ+λ2 (16π)2logϕ2 /Lambda12+C)ϕ4+O(λ3) into (18) we see that λ(M) is equal to λplusO(λ2)corrections, among which is a term like λ2logM. We can get a clean relation by differentiating λ(M) : Mdλ(M) dM=3 16π2λ2+O(λ3) =3 16π2λ(M)2+O[λ(M)3] (19) where the second equality is correct to the order indicated. This interesting relation tells us how the coupling λ(M) depends on the mass scale Mat which it is defined. Recall exercise III.1.3. We will come back to this relation in chapter VI.7 on the renormalization group. Meanwhile, let us press on. Using (18) to determine Cand plugging it into Veffwe obtain Veff(ϕ)=1 4!λ(M)ϕ4+λ(M)2 (16π)2ϕ4/parenleftbigg logϕ2 M2−25 6/parenrightbigg +O[λ(M)3] (20) You are no longer surprised, I suppose, that Cand the cutoff /Lambda1have both disappeared. That’s a renormalizable theory for you! The fact that Veffdoes not depend on the arbitrarily chosen M, namely, M(dV eff/dM) = 0, reproduces (19) to the order indicated. Breaking by quantum fluctuations Now we can answer the motivating question: T o break or not to break? Quantum fluctuations generate a correction to the potential of the form +ϕ4logϕ2, but log ϕ2is whopping big and negative for small ϕ! The O(/planckover2pi)correction overwhelms the classical O(/planckover2pi0)potential +ϕ4nearϕ=0. Quantum fluctuations break the discrete symmetry ϕ→−ϕ. It is easy enough to determine the minima ±ϕmin ofVeff(ϕ) (which you should plot as a function of ϕto get a feeling for). But closer inspection shows us that we cannot take the precise value of ϕmin seriously; Veffhas the form λϕ4(1+λlogϕ+...)suggesting that the expansion parameter is actually λlogϕrather than λ. [T ry to convince yourself that (...)starts with (λlogϕ)2.] The minima ϕmin ofVeffclearly occurs when the expansion parameter is of order unity. In an exercise in chapter IV .7 you will see a clever way of gettingaround this problem. IV .3. Effective Potential | 243 Fermions In (11) ϕsplays the role of an external field while /tildewideϕcorresponds to a quantum field we integrate over. The role of /tildewideϕcan also be played by a fermion field ψ. Consider adding ¯ψ(i/negationslash∂−m−fϕ) ψ to the Lagrangian. In the path integral Z=/integraldisplay DϕD ¯ψDψei/integraltext d4x[1 2(∂ϕ)2−V( ϕ) +¯ψ(i/negationslash∂−m−fϕ) ψ ](21) we can always choose to integrate over ψfirst, obtaining Z=/integraldisplay Dϕei/integraltext d4x[1 2(∂ϕ)2−V( ϕ) ]+tr log (i/negationslash∂−m−fϕ)(22) Repeating the steps in (13) we find that the fermion field contributes VF(ϕ)=+i/integraldisplayd4p (2π)4tr log/negationslashp−m−fϕ /negationslashp(23) toVeff(ϕ). (The trace in (23) is taken over the gamma matrices.) Again from chapter II.5, we see that physically VF(ϕ) represents the vacuum energy of a fermion with the effective massm(ϕ)≡m+fϕ. We can massage the trace of the logarithm using tr log M=log det M(II.5.12) and cyclically permuting factors in a determinant): tr log(/negationslashp−a)=tr logγ5(/negationslashp−a)γ5=tr log(− /negationslashp−a) =1 2tr(log(/negationslashp−a)+log(/negationslashp+a))+1 2tr log(−1) =1 2tr log(−1)(p2−a2). (24) Hence, tr log(/negationslashp−a) /negationslashp=1 2tr logp2−a2 p2=2 logp2−a2 p2(25) and so VF(ϕ)=2i/integraldisplayd4p (2π)4logp2−m(ϕ)2 p2(26) Contrast the overall sign with the sign in (14): the difference in sign between fermionic and bosonic loops was explained in chapter II.5. Thus, in the end the effective potential generated by the quantum fluctuations has a pleasing interpretation: It is just the energy density due to the fluctuating energy, entirelyanalogous to the zero point energy of the harmonic oscillator, of quantum fields living inthe background ϕ(see exercise IV .3.5). Exercises IV .3.1 Consider the effective potential in (0+1)-dimensional spacetime: Veff(ϕ)=V( ϕ)+/planckover2pi 2/integraldisplaydkE (2π)logk2 E+V/prime/prime(ϕ) k2 E+O(/planckover2pi2) 244 | IV . Symmetry and Symmetry Breaking No counterterm is needed since the integral is perfectly convergent. But (0+1)-dimensional field theory is just quantum mechanics. Evaluate the integral and show that Veffis in complete accord with your knowledge of quantum mechanics. IV .3.2 Study Veffin(1+1)−dimensional spacetime. IV .3.3 Consider a massless fermion field ψcoupled to a scalar field ϕbyfϕ¯ψψ in(1+1)-dimensional spacetime. Show that VF=1 2π(f ϕ)2logϕ2 M2(27) after a suitable counterterm has been added. This result is important in condensed matter physics, as we will see in chapter V .5 on the Peierls instability. IV .3.4 Understand (14) using Feynman diagrams. Show that Veffis generated by an infinite number of di- agrams. [Hint: Expand the logarithm in (14) as a series in V/prime/prime(ϕ)/k2and try to associate a Feynman diagram with each term in the series.] IV .3.5 Consider the electrodynamics of a complex scalar field L=−1 4FμνFμν+/bracketleftBig (∂μ+ieAμ)ϕ†/bracketrightBig/bracketleftbig (∂μ−ieAμ)ϕ/bracketrightbig +μ2ϕ†ϕ−λ(ϕ†ϕ)2(28) In a universe suffused with the scalar field ϕ(x) taking on the value ϕindependent of xas in the text, the Lagrangian will contain a term (e2ϕ†ϕ)AμAμso that the effective mass squared of the photon field becomes M(ϕ)2≡e2ϕ†ϕ. Show that its contribution to Veff(ϕ) has the form /integraldisplayd4k (2π)4logk2−M(ϕ)2 k2(29) Compare with (14) and (26). [Hint: Use the Landau gauge to simplify the calculation.] If you need help, I strongly urge you to read S. Coleman and E. Weinberg, Phys. Rev. D7: 1883, 1973, a paragon of clarity in exposition. IV.4 Magnetic Monopole Quantum mechanics and magnetic monopoles Curiously enough, while electric charges are commonplace nobody has ever seen a mag- netic charge or monopole. Within classical physics we can perfectly well modify one ofMaxwell’s equations to /vector∇./vectorB=ρ M, with ρMdenoting the density of magnetic monopoles. The only price we have to pay is that the magnetic field /vectorBcan no longer be represented as/vectorB=/vector∇×/vectorAsince otherwise /vector∇./vectorB=/vector∇./vector∇×/vectorA=εijk∂i∂jAk=0 identically. Newton and Leibniz told us that derivatives commute with each other. So what, you say. Indeed, who cares that /vectorBcannot be written as /vector∇×/vectorA? The vector po- tential /vectorAwas introduced into physics only as a mathematical crutch, and indeed that is still how students are often taught in a course on classical electromagnetism. As the dis-tinguished nineteenth-century physicist Heaviside thundered, “Physics should be purgedof such rubbish as the scalar and vector potentials; only the fields /vectorEand/vectorBare physical.” With the advent of quantum mechanics, however, Heaviside was proved to be quite wrong. Recall, for example, the nonrelativistic Schr ¨odinger equation for a charged particle in an electromagnetic field: /bracketleftbigg −1 2m(/vector∇−ie/vectorA)2+eφ/bracketrightbigg ψ=Eψ (1) Charged particles couple directly to the vector and scalar potentials /vectorAandφ, which are thus seen as being more fundamental, in some sense, than the electromagnetic fields /vectorE and/vectorB, as I alluded to in chapter III.4. Quantum physics demands the vector potential. Dirac noted brilliantly that these remarks imply an intrinsic conflict between quantum mechanics and the concept of magnetic monopoles. Upon closer analysis, he found thatquantum mechanics does not actually forbid the existence of magnetic monopoles. Itallows magnetic monopoles, but only those carrying a specific amount of magnetic charge. 246 | IV . Symmetry and Symmetry Breaking Differential forms For the following discussion and for the next chapter on Y ang-Mills theory, it is highly convenient to use the language of differential forms. Fear not, we will need only a fewelementary concepts. Let x μbeDreal variables (thus, the index μtakes on Dvalues) andAμ(not necessarily the electromagnetic gauge potential in this purely mathematical section) be Dfunctions of the x’s. In our applications, xμrepresent coordinates and, as we will see, differential forms have natural geometric interpretations. We call the object A≡Aμdxμa 1-form. The differentials dxμare treated following New- ton and Leibniz. If we change coordinates x→x/prime, then as usual dxμ=(∂xμ/∂x/primeν)dx/primeνso thatA≡Aμdxμ=Aμ(∂xμ/∂x/primeν)dx/primeν≡A/prime νdx/primeν. This reproduces the standard transforma- tion law of vectors under coordinate transformation A/prime ν=Aμ(∂xμ/∂x/primeν). As an example, consider A=cosθd ϕ . Regarding θandϕas angular coordinates on a 2-sphere (namely the surface of a 3-ball), we have Aθ=0 and Aϕ=cosθ. Similarly, we define a p-form as H=(1/p!)Hμ1μ2...μpdxμ1dxμ2...dxμp. (Repeated indices are summed, as always.) The “degenerate” example is that of a 0-form, call it /Lambda1, which is just a scalar function of the coordinates xμ. An example of a 2-form is F=(1/2!)Fμνdxμdxν. We now face the question of how to think about products of differentials. In an ele- mentary course on calculus we learned that dx dy represents the area of an infinitesimal rectangle with length dxand width dy. At that level, we more or less automatically regard dy dx as the same as dx dy . The order of writing the differentials does not matter. However, think about making a coordinate transformation so that x=x(x/prime,y/prime)andy=y(x/prime,y/prime)are now functions of the new coordinates x/primeandy/prime. Now look at dx dy =/parenleftbigg∂x ∂x/primedx/prime+∂x ∂y/primedy/prime/parenrightbigg/parenleftbigg∂y ∂x/primedx/prime+∂y ∂y/primedy/prime/parenrightbigg (2) Note that the coefficient of dx/primedy/primeis(∂x/∂x/prime)(∂y/∂y/prime)and that the coefficient of dy/primedx/prime is(∂x/∂y/prime)(∂y/∂x/prime). We see that it is much better if we regard the differentials dxμ as anticommuting objects [what mathematicians would call Grassmann variables (recall chapter (II.5)] so that dy/primedx/prime=−dx/primedy/primeanddx/primedx/prime=0=dy/primedy/prime. Then (2) simplifies neatly to dx dy =/parenleftbigg∂x ∂x/prime∂y ∂y/prime−∂x ∂y/prime∂y ∂x/prime/parenrightbigg dx/primedy/prime≡J(x ,y;x/prime,y/prime)dx/primedy/prime(3) We obtain the correct Jacobian J(x ,y;x/prime,y/prime)for transforming the area element dx dy to the area element dx/primedy/prime. In many texts, dx dy is written as dx^dy . We will omit the wedge—no reason to clutter up the page. This little exercise tells us that we should define dxμdxν=−dxνdxμand regard the area element dxμdxνas directional. The area elements dxμdxνanddxνdxμhave the same magnitude but point in opposite directions. IV .4. Magnetic Monopole | 247 We now define a differential operation dto act on any form. Acting on a p-form H,i t gives by definition dH=1 p!∂νHμ1μ2...μpdxνdxμ1dxμ2...dxμp Thus, d/Lambda1=∂ν/Lambda1dxνand dA=∂νAμdxνdxμ=1 2(∂νAμ−∂μAν)dxνdxμ In the last step, we used dxμdxν=−dxνdxμ. We see that this mathematical formalism is almost tailor made to describe electromag- netism. If we call A≡Aμdxμthe potential 1-form and think of Aμas the electromag- netic potential, then F=dA is in fact the field 2-form. If we write Fout in terms of its components F=(1/2!)Fμνdxμdxν, then Fμνis indeed equal to the electromagnetic field (∂μAν−∂νAμ). Note that xμis not a form, and dxμis not dacting on a form. If you like, you can think of differential forms as “merely” an elegantly compact notation. The point is to think of physical objects such as AandFas entities, without having to commit to any particular coordinate system. This is particularly convenient when one hasto deal with objects more complicated than AandF, for example in string theory. By using differential forms, we avoid drowning in a sea of indices. An important identity is dd=0 (4) which says that acting with don any form twice gives zero. Verify this as an exercise. In particular dF=ddA=0. If you write this out in components you will recognize it as a standard identity (the “Bianchi identity”) in electromagnetism. Closed is not necessarily globally exact It is convenient here to introduce some jargon. A p-form αis said to be closed if dα=0. It is said to be exact if there exists a (p−1)-form βsuch that α=dβ. T alking the talk, we say that (4) tells us that exact forms are closed.Is the converse of (4) true? Kind of. The Poincar ´e lemma states that a closed form is locally exact. In other words, if dH=0 with Hsome p-form, then locally H=dK (5) for some (p−1)-form K. However, it may or may not be the case that H=dK globally, that is, everywhere. Actually, whether you know it or not, you are already familiar with the Poincar ´e lemma. For example, surely you learned somewhere that if the curl of a vector field vanishes, the vector field is locally the gradient of some scalar field. Forms are ready made to be integrated over. For example, given the 2-form F= (1/2!)Fμνdxμdxν, we can write/integraltext MFfor any 2-manifold M. Note that the measure is 248 | IV . Symmetry and Symmetry Breaking already included and there is no need to specify a coordinate choice. Again, whether you know it or not, you are already familiar with the important theorem /integraldisplay MdH=/integraldisplay ∂MH (6) withHap-form and ∂M the boundary of a (p+1)-dimensional manifold M. Dirac quantization of magnetic charge After this dose of mathematics, we are ready to do some physics. Consider a sphere surrounding a magnetic monopole with magnetic charge g. Then the electromagnetic field 2-form is given by F=(g/4π)d cosθd ϕ . This is almost a definition of what we mean by a magnetic monopole (see exercise IV .4.3.) In particular, calculate the magnetic flux byintegrating Fover the sphere S 2 /integraldisplay S2F=g (7) As I have already noted, the area element is automatically included. Indeed, you might have recognized dcosθd ϕ=−sin θd θd ϕ as precisely the area element on a unit sphere. Note that in “ordinary notation” (7) implies the magnetic field /vectorB=(g/4πr2)ˆr, with ˆrthe unit vector in the radial direction. I will now give a rather mathematical, but rigorous, derivation, originally developed by Wu and Y ang, of Dirac’s quantization of the magnetic charge g. First, let us recall how gauge invariance works, from, for example, (II.7.3). Under a transformation of the electron field ψ(x)→ei/Lambda1(x)ψ(x) , the electromagnetic gauge poten- tial changes by Aμ(x)→Aμ(x)+1 iee−i/Lambda1(x)∂μei/Lambda1(x) or in the language of forms, A→A+1 iee−i/Lambda1dei/Lambda1(8) Differentiating, we can of course write Aμ(x)→Aμ(x)+1 e∂μ/Lambda1(x) as is commonly done. The form given in (8) reminds us that gauge transformation is defined as multiplication by a phase factor ei/Lambda1(x), so that /Lambda1(x) and/Lambda1(x)+2πdescribe exactly the same transformation. In quantum mechanics Ais physical, pace Heaviside, and so we should ask what A would give rise to F=(g/4π)d cosθd ϕ . Easy, you say; clearly A=(g/4π)cosθd ϕ . (In checking this by calculating dA, remember that dd=0.) But not so fast; your mathematician friend says that dϕis not defined at the north and south poles. Put his objection into everyday language: If you are standing on the north pole,what is your longitude? So strictly speaking it is forbidden to write A=(g/4π)cosθd ϕ . IV .4. Magnetic Monopole | 249 But, you are smart enough to counter, then what about AN=(g/4π)(cosθ−1)dϕ , eh? When you act with donANyou obtain the desired F; the added piece (g/4π)(−1)dϕ gets annihilated by dthanks once again to the identity (4). At the north pole, cos θ=1,AN vanishes, and is thus perfectly well defined. OK, but your mathematician friend points out that your ANis not defined at the south pole, where it is equal to (g/4π)(−2)dϕ . Right, you respond, I anticipated that by adding the subscript N. I am now also forced to define AS=(g/4π)(cosθ+1)dϕ . Note that dacting on ASagain gives the desired F. But now ASis defined everywhere except at the north pole. In mathematical jargon, we say that the gauge potential Ais defined locally, but not globally. The gauge potential ANis defined on a “coordinate patch” covering the northern hemisphere and extending past the equator as far south as we want as long as we donot include the south pole. Similarly, A Sis defined on a “coordinate patch” covering the southern hemisphere and extending past the equator as far north as we want as long aswe do not include the north pole. But what happens where the two coordinate patches overlap, for example, along the equator. The gauge potentials A NandASare not the same: AS−AN=2g 4πdϕ (9) Now what? Aha, but this is a gauge theory: If ASandANare related by a gauge transforma- tion, then all is well. Thus, referring to (8) we require that 2 (g/4π)dϕ =(1/ie)e−i/Lambda1dei/Lambda1 for some phase function ei/Lambda1. By inspection we have ei/Lambda1=ei2(eg/4π)ϕ. Butϕ=0 and ϕ=2πdescribe exactly the same point. In order for ei/Lambda1to make sense, we must have ei2(eg/4π)(2π)=ei2(eg/4π)(0)=1; in other words, eieg=1, or g=2π en (10) where ndenotes an integer. This is Dirac’s famous discovery that the magnetic charge on a magnetic monopole is quantized in units of 2 π/e . A “dual” way of putting this is that if the monopole exists then electric charge is quantized in units of 2 π/g . Note that the whole point is that Fis locally but not globally exact; otherwise by (6) the magnetic charge g=/integraltext S2Fwould be zero. I show you this rigorous mathematical derivation partly to cut through a lot of the confusion typical of the derivations in elementary texts and partly because this type ofargument is used repeatedly in more advanced areas of physics, such as string theory. Electromagnetic duality That a duality may exist between electric and magnetic fields has tantalized theoretical physicists for a century and a half. By the way, if you read Maxwell, you will discover that heoften talked about magnetic charges. You can check that Maxwell’s equations are invariantunder the elegant transformation (/vectorE+i/vectorB)→e iθ(/vectorE+i/vectorB)if magnetic charges exist. 250 | IV . Symmetry and Symmetry Breaking (a) (b)xμ(σ,τ) xμ(τ) Figure IV .4.1 One intriguing feature of (10)is that if eis small, then gis large, and vice versa. What would magnetic charges look like if they exist? They wouldn’t look any different fromelectric charges: They too interact with a 1 /rpotential, with likes repelling and opposites attracting. In principle, we could have perfectly formulated electromagnetism in terms ofmagnetic charges, with magnetic and electric fields exchanging their roles, but the theorywould be strongly coupled, with the coupling grather than e. Theoretical physicists are interested in duality because it allows them a glimpse into field theories in the strongly coupled regime. Under duality, a weakly coupled field theoryis mapped into a strongly coupled field theory. This is exactly the reason why the discoverysome years ago that certain string theories are dual to others caused such enormousexcitement in the string theory community: We get to know how string theories behave inthe strongly coupled regime. More on duality in chapter VI.3. Forms and geometry The geometric character of differential forms is further clarified by thinking about theelectromagnetic current of a charged particle tracing out the world line X μ(τ) inD- dimensional spacetime (see figure IV .4.1a): Jμ(x)=/integraldisplay dτdXμ dτδ(D)[x−X(τ) ] (11) The interpretation of this elementary formula from electromagnetism is clear: dXμ/dτ is the 4-velocity at a given value of the parameter τ(“proper time”) and the delta function ensures that the current at xvanishes unless the particle passes through x. Note that Jμ(x) is invariant under the reparametrization τ→τ/prime(τ). The generalization to an extended object is more or less obvious. Consider a string. It traces out a world sheet Xμ(τ,σ)in spacetime (see figure 1b), where σis a parameter IV .4. Magnetic Monopole | 251 telling us where we are along the length of the string. [For example, for a closed string, σis conventionally taken to range between 0 and 2 πwithXμ(τ,0)=Xμ(τ,2π).] The current associated with the string is evidently given by Jμν(x)=/integraldisplay dτdσ det/parenleftBigg∂τXμ∂τXν ∂σXμ∂σXν/parenrightBigg δ(D)[x−X(τ ,σ)] (12) where ∂τ≡∂/∂τ and so forth. The determinant is forced on us by the requirement of invariance under reparametrization τ→τ/prime(τ,σ),σ→σ/prime(τ,σ). It follows that Jμνis an antisymmetric tensor. Hence, the analog of the electromagnetic potential Aμcoupling to the current Jμis an antisymmetric tensor field Bμνcoupling to the current Jμν. Thus, string theory contains a 2-form potential B=1 2Bμνdxμdxνand the corresponding 3-form fieldH=dB. In fact, string theory typically contains numerous p-forms. Aharonov-Bohm effect The reality of the gauge potential Awas brought home forcefully in 1959 by Aharonov and Bohm. Consider a magnetic field Bconfined to a region /Omega1as illustrated in figure IV .4.2. The quantum physics of an electron is described by solving the Schr ¨odinger equation (1). In Feynman’s path integral formalism the amplitude associated with a path Pis modified by a multiplicative factor eie/integraltext P/vectorA.d/vectorx, where the line integral is evaluated along the path P. Thus, in the path integral calculation of the probability for an electron to propagate fromatob(fig. IV .4.2), there will be interference between the contributions from path 1 and path 2 of the form /parenleftbigg eie/integraltext P1/vectorA.d/vectorx/parenrightbigg/parenleftbigg eie/integraltext P2/vectorA.d/vectorx/parenrightbigg∗ =/parenleftbigg eie/contintegraltext/vectorA.d/vectorx/parenrightbigg ab B = 0B = 0 Ω P2P1 Figure IV .4.2 252 | IV . Symmetry and Symmetry Breaking but/contintegraltext/vectorA.d/vectorx=/integraltext/vectorB.d/vectorSis precisely the flux enclosed by the closed curve ( P1−P2), namely the curve going from atobalong P1and then returning from btoaalong (−P2)since complex conjugation in effect reverses the direction of the path P2. Remarkably, the electron feels the effect of the magnetic field even though it never wanders into a regionwith a magnetic field present. When the Aharonov-Bohm paper was first published, no less an authority than Niels Bohr was deeply disturbed. The effect has since been conclusively demonstrated in a seriesof beautiful experiments by T onomura and collaborators. Coleman once told of a gedanken prank that connects the Aharonov-Bohm effect to Dirac quantization of magnetic charge. Let us fabricate an extremely thin solenoid so thatit is essentially invisible and thread it into the lab of an unsuspecting experimentalist,perhaps our friend from chapter III.1. We turn on a current and generate a magnetic fieldthrough the solenoid. When the experimentalist suddenly sees the magnetic flux comingout of apparently nowhere, she gets so excited that she starts planning to go to Stockholm. What is the condition that prevents the experimentalist from discovering the prank? A careful experimentalist might start scattering electrons around to see if she can detect asolenoid. The condition that she does not see an Aharonov-Bohm effect and thus doesnot discover the prank is precisely that the flux going through the solenoid is an integertimes 2 π/e . This implies that the apparent magnetic monopole has precisely the magnetic charge predicted by Dirac! Exercises IV .4.1 Prove dd=0. IV .4.2 Show by writing out the components explicitly that dF=0 expresses something that you are familiar with but disguised in a compact notation. IV .4.3 Consider F=(g/4π) d cosθd ϕ . By transforming to Cartesian coordinates show that this describes a magnetic field pointing outward along the radial direction. IV .4.4 Restore the factors of /planckover2piandcin Dirac’s quantization condition. IV .4.5 Write down the reparametrization-invariant current Jμνλof a membrane. IV .4.6 Letg(x) be the element of a group G. The 1-form v=gdg†is known as the Cartan-Maurer form. Then tr vNis trivially closed on an N-dimensional manifold since it is already an N-form. Consider Q=/integraltext SNtrvNwithSNtheN-dimensional sphere. Discuss the topological meaning of Q. These con- siderations will become important later when we discuss topology in field theory in chapter V .7. [Hint:Study the case N=3 and G=SU( 2).] IV.5 Nonabelian Gauge Theory Most such ideas are eventually discarded or shelved. But some persist and may become obsessions. Occasionally an obsessiondoes finally turn out to be something good. —C. N. Y ang talking about an idea that he first had as a student and that he kept coming back to year after year. 1 Local transformation It was quite a nice little idea. T o explain the idea Y ang was talking about, recall our discussion of symmetry in chapter I.10. For the sake of definiteness let ϕ(x)={ϕ1(x),ϕ2(x),... ,ϕN(x)} be an N- component complex scalar field transforming as ϕ(x)→Uϕ(x), with Uan element of SU(N) . Since ϕ†→ϕ†U†andU†U=1, we have ϕ†ϕ→ϕ†ϕand∂ϕ†∂ϕ→∂ϕ†∂ϕ. The invariance of the Lagrangian L=∂ϕ†∂ϕ−V( ϕ†ϕ)under SU(N) is obvious for any poly- nomial V. In the theoretical physics community there are many more people who can answer well- posed questions than there are people who can pose the truly important questions. Thelatter type of physicist can invariably also do much of what the former type can do, but thereverse is certainly not true. In 1954 C.N. Y ang and R. Mills asked what will happen if the transformation varies from place to place in spacetime, or in other words, if U=U(x) is a function of x. Clearly, ϕ †ϕis still invariant. But in contrast ∂ϕ†∂ϕis no longer invariant. Indeed, ∂μϕ→∂μ(Uϕ)=U∂μϕ+(∂μU)ϕ=U[∂μϕ+(U†∂μU)ϕ ] T o cancel the unwanted term (U†∂μU)ϕ , we generalize the ordinary derivative ∂μto a covariant derivative Dμ, which when acting on ϕ, gives Dμϕ(x)=∂μϕ(x)−iAμ(x)ϕ(x) (1) The field Aμis called a gauge potential in direct analogy with electromagnetism. 1C. N. Y ang, Selected Papers 1945–1980 with Commentary ,p .1 9 . 254 | IV . Symmetry and Symmetry Breaking How must Aμtransform, so that Dμϕ(x)→U(x)D μϕ(x) ? In other words, we would likeDμϕ(x) to transform the way ∂μϕ(x) transformed when Udid not depend on x.I f so, then [ Dμϕ(x) ]†Dμϕ(x)→[Dμϕ(x) ]†Dμϕ(x) and can be used as an invariant kinetic energy term for the field ϕ. Working backward, we see that Dμϕ(x)→U(x)D μϕ(x) if ( and it goes without saying that you should be checking this) Aμ→UAμU†−i(∂μU)U†=UAμU†+iU∂μU†(2) (The equality follows from UU†=1.)We refer to Aμas the nonabelian gauge potential and to (2) as a nonabelian gauge transformation. Let us now make a series of simple observations. 1. Clearly, Aμhave to be NbyNmatrices. Work out the transformation law for A† μusing (2) and show that the condition Aμ−A† μ=0 is preserved by the gauge transformation. Thus, it is consistent to take Aμto be hermitean. Specifically, you should work out what this means for the group SU( 2)so that U=eiθ.τ/2where θ.τ=θaτa, with τathe familiar Pauli matrices. 2. Writing U=eiθ.TwithTathe generators of SU(N), we have Aμ→Aμ+iθa[Ta,Aμ]+∂μθaTa(3) under an infinitesimal transformation U/similarequal1+iθ.T. For most purposes, the infinitesimal form (3) suffices. 3. T aking the trace of (3) we see that the trace of Aμdoes not transform and so we can take Aμ to be traceless as well as hermitean. This means that we can always write Aμ=Aa μTaand thus decompose the matrix field Aμinto component fields Aa μ. There are as many Aa μ’s as there are generators in the group [3 for SU( 2), 8 for SU( 3), and so forth.] 4. You are reminded in appendix B that the Lie algebra of the group is defined by [ Ta,Tb]= ifabcTc, where the numbers fabcare called structure constants. For example, fabc=εabc forSU( 2). Thus, (3) can be written as Aa μ→Aa μ−fabcθbAc μ+∂μθa(4) Note that if θdoes not depend on x, theAa μ’s transform as the adjoint representation of the group. 5. IfU(x)=eiθ(x)is just an element of the abelian group U(1), all these expressions simplify andAμis just the abelian gauge potential familiar from electromagnetism, with (2 ) the usual abelian gauge transformation. Hence, Aμis known as the nonabelian gauge potential. A transformation Uthat depends on the spacetime coordinates xis known as a gauge transformation or local transformation. A Lagrangian Linvariant under a gauge transfor- mation is said to be gauge invariant. IV .5. Nonabelian Gauge Theory | 255 Construction of the field strength We can now immediately write a gauge invariant Lagrangian, namely L=(D μϕ)†(Dμϕ)−V( ϕ†ϕ) (5) but the gauge potential Aμdoes not yet have dynamics of its own. In the familiar example ofU(1)gauge invariance, we have written the coupling of the electromagnetic potential Aμto the matter field ϕ, but we have yet to write the Maxwell term −1 4FμνFμνin the Lagrangian. Our first task is to construct a field strength Fμνout of Aμ. How do we do that? Y ang and Mills apparently did it by trial and error. As an exercise you might alsowant to try that before reading on. At this point the language of differential forms introduced in chapter IV .4 proves to be of use. It is convenient to absorb a factor of −i by defining A M μ≡−iAP μ, where AP μ denotes the gauge potential we have been using all along. Until further notice, when we write Aμwe mean AM μ. Referring to (1) we see that the covariant derivative has the cleaner form Dμ=∂μ+Aμ. (Incidentally, the superscripts MandPindicate the potential appearing in the mathematical and physical literature, respectively.) As before, let usintroduce A=A μdxμ, now a matrix 1-form, that is, a form that also happens to be a matrix in the defining representation of the Lie algebra [e.g., an NbyNtraceless hermitean matrix forSU(N).] Note that A2=AμAνdxμdxν=1 2[Aμ,Aν]dxμdxν is not zero for a nonabelian gauge potential. (Obviously, there is no such object in electro- magnetism.) Our task is to construct a 2-form F=1 2Fμνdxμdxνout of the 1-form A. We adopt a direct approach. Out of Awe can construct only two possible 2-forms: dA andA2.S oFmust be a linear combination of the two. In the notation we are using the transformation law (2) reads A→UAU†+UdU†(6) withUa 0-form (and so dU†=∂μU†dxμ.)Applying dto (6) we have dA→UdAU†+dUAU†−UAdU†+dUdU†(7) Note the minus sign in the third term, from moving the 1-form dpast the 1-form A.O n the other hand, squaring (6) we have A2→UA2U†+UAdU†+UdU†UAU†+UdU†UdU†(8) Applying dtoUU†=1 we have UdU†=−dUU†. Thus, we can rewrite (8)as A2→UA2U†+UAdU†−dUAU†−dUdU†(9) 256 | IV . Symmetry and Symmetry Breaking Lo and behold! If we add (7) and (9), six terms knock each other off, leaving us with something nice and clean: dA+A2→U(dA +A2)U†(10) The mathematical structure thus led Y ang and Mills to define the field strength F=dA+A2(11) Unlike A, the field strength 2-form Ftransforms homogeneously (10): F→UFU†(12) In the abelian case A2vanishes and Freduces to the usual electromagnetic form. In the nonabelian case, Fis not gauge invariant, but gauge covariant. Of course, you can also construct Fa μνwithout using differential forms. As an exercise you should do it starting with (4). The exercise will make you appreciate differential forms! At the very least, we can regard differential forms as an elegantly compact notation thatsuppresses the indices aandμin (4). At the same time, the fact that (11) emerges so smoothly clearly indicates a profound underlying mathematical structure. Indeed, thereis a one-to-one translation between the physicist’s language of gauge theory and themathematician’s language of fiber bundles. Let me show you another route to (11). In analogy to d, define D=d+A, understood as an operator acting on a form to its right. Let us calculate D2=(d+A)(d+A)=d2+dA+Ad+A2 The first term vanishes, the second can be written as dA=(dA)−Ad; the parenthesis emphasizes that dacts only on A. Thus, D2=(dA)+A2=F (13) Pretty slick? I leave it as an exercise for you to show that D2transforms homogeneously and hence so does F. Elegant though differential forms are, in physics it is often desirable to write more explicit formulas. We can write (11) out as F=(∂μAν+AμAν)dxμdxν=1 2(∂μAν−∂νAμ+[Aμ,Aν])dxμdxν(14) With the definition F≡1 2Fμνdxμdxνwe have Fμν=∂μAν−∂νAμ+[Aμ,Aν] (15) At this point, we might also want to switch back to physicist’s notation. Recall that Aμ in (15) is actually AM μ≡−iAP μand so by analogy define FM μν=−iFP μν. Thus, Fμν=∂μAν−∂νAμ−i[Aμ,Aν] (16) where, until further notice, Aμstands for AP μ. (One way to see the necessity for the iin (16) is to remember that physicists like to take Aμto be a hermitean matrix and the commutator of two hermitean matrices is antihermitean.) IV .5. Nonabelian Gauge Theory | 257 As long as we are being explicit we might as well go all the way and exhibit the group indices as well as the Lorentz indices. We already wrote Aμ=Aa μTaand so we naturally writeFμν=Fa μνTa. Then (16) becomes Fa μν=∂μAa ν−∂νAa μ+fabcAb μAcν (17) I mention in passing that for SU( 2)Aand Ftransform as vectors and the structure constant fabcis just εabc, so the vector notation /vectorFμν=∂μ/vectorAν−∂ν/vectorAμ+/vectorAμ×/vectorAνis often used. The Y ang-Mills Lagrangian Given that Ftransforms homogeneously (12) we can immediately write down the analog of the Maxwell Lagrangian, namely the Y ang-Mills Lagrangian L=−1 2g2trFμνFμν(18) We are normalizing Taby trTaTb=1 2δabso that L=−(1/4g2)Fa μνFaμν. The theory described by this Lagrangian is known as pure Y ang-Mills theory or nonabelian gaugetheory. Apart from the quadratic term (∂ μAa ν−∂νAa μ)2, the Lagrangian L=−(1/4g2)Fa μνFaμν also contains a cubic term fabcAbμAcν(∂μAa ν−∂νAa μ)and a quartic term (fabcAb μAcν)2. As in electromagnetism the quadratic term describes the propagation of a massless vector boson carrying an internal index a, known as the nonabelian gauge boson or the Y ang- Mills boson. The cubic and quartic terms are not present in electromagnetism and describethe self-interaction of the nonabelian gauge boson. The corresponding Feynman rules aregiven in figure IV .5.1a, 1b, and c. The physics behind this self-interaction of the Y ang-Mills bosons is not hard to under- stand. The photon couples to charged fields but is not charged itself. Just as the charge (c) (b)(a) Figure IV .5.1 258 | IV . Symmetry and Symmetry Breaking of a field tells us how the field transforms under the U(1)gauge group, the analog of the charge of a field in a nonabelian gauge theory is the representation the field belongs to. TheY ang-Mills bosons couple to all fields transforming nontrivially under the gauge group. Butthe Y ang-Mills bosons themselves transform nontrivially: In fact, as we have noted, theytransform under the adjoint representation. Thus, they must couple to themselves. Pure Maxwell theory is free and so essentially trivial. It contains a noninteracting photon. In contrast, pure Y ang-Mills theory contains self-interaction and is highly nontrivial. Notethat the structure coefficients f abcare completely fixed by group theory, and thus in contrast to a scalar field theory, the cubic and quartic self-interactions of the gauge bosons,including their relative strengths, are totally fixed by symmetry. If any 4-dimensional fieldtheory can be solved exactly, pure Y ang-Mills theory may be it, but in spite of the enormousamount of theoretical work devoted to it, it remains unsolved (see chapters VII.3 and VII.4). ’t Hooft’s double-line formalism While it is convenient to use the component fields Aa μfor many purposes, the matrix fieldAμ=Aa μTaembodies the mathematical structure of nonabelian gauge theory more elegantly. The propagator for the components of the matrix field in a U(N) gauge theory has the form /angbracketleft0|TAμ(x)i jAν(0)k l|0/angbracketright =/angbracketleft0|TAa μ(x)Abν(0)|0/angbracketright(Ta)i j(Tb)kl (19) ∝δab(Ta)i j(Tb)kl∝δi lδk j [We have gone from an SU(N) to aU(N) theory for the sake of simplicity. The generators ofSU(N) satisfy a traceless condition T r Ta=0, as a result of which we would have to subtract1 Nδi jδk lfrom the right-hand side.] The matrix structure Ai μjnaturally suggests that we, following ’t Hooft, introduce a double-line formalism, in which the gauge potential isdescribed by two lines, each associated with one of the two indices iandj. We choose the convention that the upper index flows into the diagram, while the lower index flows outof the diagram. The propagator in (19) is represented in figure IV .5.2a. The double-lineformalism allows us to reproduce the index structure δ i lδk jnaturally. The cubic and quartic couplings are represented in figure IV .5.2b and c. The constant gintroduced in (18) is known as the Y ang-Mills coupling constant. We can always write the quadratic term in (18) in the convention commonly used in electromag-netism by a trivial rescaling A→gA. After this rescaling, the cubic and quartic couplings of the Y ang-Mills boson go as gandg 2, respectively. The covariant derivative in (1) be- comes Dμϕ=∂μϕ−igAμϕ, showing that galso measures the coupling of the Y ang-Mills boson to matter. The convention we used, however, brings out the mathematical structure more clearly. As written in (18), g2measures the ease with which the Y ang-Mills boson can propagate. Recall that in chapter III.7 we also found this way of defining the coupling as IV .5. Nonabelian Gauge Theory | 259 i jl k (a) (b) (c) Figure IV .5.2 a measure of propagation useful in electromagnetism. We will see in chapter VIII.1 that Newton’s coupling appears in the same way in the Einstein-Hilbert action for gravity. Theθterm Besides tr FμνFμν, we can also form the dimension-4 term εμνλρtrFμνFλρ. Clearly, this term violates time reversal invariance Tand parity Psince it involves one time index and three space indices. We will see later that the strong interaction is describedby a nonabelian gauge theory, with the Lagrangian containing the so-called θterm (θ/32π 2)εμνλρtrFμνFλρ. As you will show in exercise IV .5.3 this term is a total diver- gence and does not contribute to the equation of motion. Nevertheless, it induces anelectric dipole moment for the neutron. The experimental upper bound on the electricdipole moment for the neutron translates into an upper bound on θof the order 10 −9.I will not go into how particle physicists resolve the problem of making sure that θis small enough or vanishes outright. Coupling to matter fields We took the scalar field ϕto transform in the fundamental representation of the group. In general, ϕcan transform in an arbitrary representation Rof the gauge group G.W e merely have to write the covariant derivative more generally as Dμϕ=(∂μ−iAa μTa (R))ϕ (20) where Ta (R)represents the ath generator in the representation R(see exercise IV .5.1). Clearly, the prescription to turn a globally symmetric theory into a locally symmetric theory is to replace the ordinary derivative ∂μacting on any field, boson or fermion, 260 | IV . Symmetry and Symmetry Breaking belonging to the representation Rby the covariant derivative Dμ=(∂μ−iAa μTa (R)). Thus, the coupling of the nonabelian gauge potential to a fermion field is given by L=¯ψ(iγμDμ−m)ψ=¯ψ(iγμ∂μ+γμAa μTa (R)−m)ψ . (21) Fields listen to the Y ang-Mills gauge bosons according to the representation Rthat they belong to, and those that belong to the trivial identity representation do not hearthe call of the gauge bosons. In the special case of a U(1)gauge theory, also known as electromagnetism, Rcorresponds to the electric charge of the field. Those fields that transform trivially under U(1)are electrically neutral. Appendix Let me show you another context, somewhat surprising at first sight, in which the Y ang-Mills structure pops up.2 Consider Schr ¨odinger’s equation i∂ ∂t/Psi1(t)=H(t)/Psi1(t) (22) with a time dependent Hamiltonian H(t) . The setup is completely general: For instance, we could be talking about spin states in a magnetic field or about a single particle nonrelativistic Hamiltonian with the wave function/Psi1(/vectorx,t). We suppress the dependence of Hand/Psi1on variables other than time t. First, solve the eigenvalue problem of H(t) . Suppose that because of symmetry or some other reason the spectrum of H(t ) contains an n-fold degeneracy, in other words, there exist ndistinct solutions of the equations H(t)ψ a(t)=E(t)ψa(t), with a=1,... ,n. Note that E(t) can vary with time and that we are assuming that the degeneracy persists with time, that is, the degeneracy does not occur “accidentally” at one instant in time. We canalways replace H(t) byH(t)−E(t) so that henceforth we have H(t)ψ a(t)=0. Also, the states can be chosen to be orthogonal so that /angbracketleftψb(t)|ψ a(t)/angbracketright=δ ba. (For notational reasons it is convenient to jump back and forth between the Schr ¨odinger and the Dirac notation. T o make it absolutely clear, we have /angbracketleftψb(t)|ψ a(t)/angbracketright=/integraldisplay d/vectorxψ∗ b(/vectorx,t)ψa(/vectorx,t) if we are talking about single particle quantum mechanics.) Let us now study (22) in the adiabatic limit, that is, we assume that the time scale over which H(t) varies is much longer than 1 //Delta1E , where /Delta1E denotes the energy gap separating the states ψa(t)from neighboring states. In that case, if /Psi1(t) starts out in the subspace spanned by {ψa(t)} it will stay in that subspace and we can write /Psi1(t)=/summationtext aca(t)ψa(t). Plugging this into (22) we obtain immediately/summationtext a[(dca/dt)ψ a(t)+ca(t)(∂ψa/∂t)]=0. T aking the scalar product with ψb(t), we obtain dcb dt=−/summationdisplay aAbaca (23) with the nbynmatrix Aba(t)≡i/angbracketleftψb(t)|∂ψa ∂t/angbracketright (24) Now suppose somebody else decides to use a different basis, ψ/prime a(t)=U∗ ac(t)ψc(t), related to ours by a unitary transformation. (The complex conjugate on the unitary matrix Uis just a notational choice so that our final equation will come out looking the same as a celebrated equation in the text; see below.) I have also passed 2F. Wilczek and A. Zee, “Appearance of Gauge Structure in Simple Dynamical Systems,” Phys. Rev. Lett. 52:2111, 1984. IV .5. Nonabelian Gauge Theory | 261 to the repeated indices summed notation. Differentiate to obtain (∂ψ/prime a/∂t)=U∗ ac(t)(∂ψ c/∂t)+(dU∗ ac/dt)ψc(t). Contracting this with ψ/prime∗ b(t)=Ubd(t)ψ∗ d(t)and multiplying by i,w ef i n d A/prime=UAU†+iU∂U† ∂t(25) Suppose the Hamiltonian H(t) depends on dparameters λ1,... ,λd. We vary the parameters, thus tracing out a path defined by {λμ(t),μ=1,... ,d}in the d-dimensional parameter space. For example, for a spin Hamiltonian, {λμ}could represent an external magnetic field. Now (23) becomes dcb dt=−/summationdisplay a(Aμ)bacadλμ dt(26) if we define (Aμ)ba≡i/angbracketleftψb|∂μψa/angbracketright, where ∂μ≡∂/∂λμ, and (25) generalizes to A/prime μ=UAμU†+iU∂μU†(27) We have recovered (IV .5.2). Lo and behold, a Y ang-Mills gauge potential Aμhas popped up in front of our very eyes! The “transport” equation (26) can be formally solved by writing c(λ)=Pe−/integraltext Aμdλμ, where the line integral is over a path connecting an initial point in the parameter space to some final point λandPdenotes a path ordering operation. We break the path into infinitesimal segments and multiply together the noncommuting contribution e−Aμ/Delta1λμfrom each segment, ordered along the path. In particular, if the path is a closed curve, by the time we return to the initial values of the parameters, the wave function will have acquired a matrix phasefactor, known as the nonabelian Berry’s phase. This discussion is clearly intimately related to the discussion ofthe Aharonov-Bohm phase in the preceding chapter. T o see this nonabelian phase, all we have to do is to find some quantum system with degeneracy in its spectrum and vary some external parameter such as a magnetic field. 3In their paper, Y ang and Mills spoke of the degeneracy of the proton and neutron under isospin in an idealized world and imagined transporting a proton from one pointin the universe to another. That a proton at one point can be interpreted as a neutron at another necessitates theintroduction of a nonabelian gauge potential. I find it amusing that this imagined transport can now be realizedanalogously in the laboratory. You will realize that the discussion here parallels the discussion in the text leading up to (IV .5.2). The spacetime dependent symmetry transformation corresponds to a parameter dependent change of basis. When I discussgravity in chapter VIII.1 it will become clear that moving the basis {ψ a}around in the parameter space is the precise analog of parallel transporting a local coordinate frame in differential geometry and general relativity. We will also encounter the quantity Pe−/integraltext Aμdλμagain in chapter VII.1 in the guise of a Wilson loop. Exercises IV .5.1 Write down the Lagrangian of an SU( 2)gauge theory with a scalar field in the I=2 representation. IV .5.2 Prove the Bianchi identity DF≡dF+[A,F]=0. Write this out explicitly with indices and show that in the abelian case it reduces to half of Maxwell’s equations. IV .5.3 In 4-dimensions εμνλρtrFμνFλρcan be written as tr F2. Show that dtrF2=0 in any dimensions. IV .5.4 Invoking the Poincar ´e lemma (IV .4.5) and the result of exercise IV .5.3 show that tr F2=dtr(AdA +2 3A3). Write this last equation out explicitly with indices. Identify these quantities in the case of electromag-netism. 3A. Zee, “On the Non-Abelian Gauge Structure in Nuclear Quadrupole Resonance,” Phys. Rev. A38:1, 1988. The proposed experiment was later done by A. Pines. 262 | IV . Symmetry and Symmetry Breaking IV .5.5 For a challenge show that tr Fn, which appears in higher dimensional theories such as string theory, are all total divergences. In other words, there exists a (2n−1)-form ω2n−1(A) such that tr Fn=dω 2n−1(A) . [Hint: A compact representation of the form ω2n−1(A)=/integraltext1 0dt f 2n−1(t,A) exists.] Work out ω5(A) explicitly and try to generalize knowing ω3andω5. Determine the (2n−1)-form f2n−1(t,A). For help, see B. Zumino et al., Nucl. Phys. B239:477, 1984. IV .5.6 Write down the Lagrangian of an SU( 3)gauge theory with a fermion field in the fundamental or defining triplet representation. IV.6 The Anderson-Higgs Mechanism The gauge potential eats the Nambu-Goldstone boson As I noted earlier the ability to ask good questions is of crucial importance in physics. Here is an excellent question: How does spontaneous symmetry breaking manifest itselfin gauge theories? Going back to chapter IV .1, we gauge the U(1)theory in (IV .1.6) by replacing ∂ μϕwith Dμϕ=(∂μ−ieAμ)ϕso that L=−1 4FμνFμν+(Dϕ)†Dϕ+μ2ϕ†ϕ−λ(ϕ†ϕ)2(1) Now when we go to polar coordinates ϕ=ρeiθwe have Dμϕ=[∂μρ+iρ(∂μθ−eAμ)]eiθ and thus L=−1 4FμνFμν+ρ2(∂μθ−eAμ)2+(∂ρ)2+μ2ρ2−λρ4(2) (Compare this with L=ρ2(∂μθ)2+(∂ρ)2+μ2ρ2−λρ4in the absence of the gauge field.) Under a gauge transformation ϕ→eiαϕ(so that θ→θ+α)andeAμ→eAμ+∂μα, and thus the combination Bμ≡Aμ−(1/e)∂μθis gauge invariant. The first two terms in L thus become −1 4FμνFμν+e2ρ2B2 μ. Note that Fμν=∂μAν−∂νAμ=∂μBν−∂νBμhas the same form in terms of the potential Bμ. Upon spontaneous symmetry breaking, we write ρ=(1/√ 2)(v+χ), with v=/radicalbig μ2/λ. Hence L=−1 4FμνFμν+1 2M2B2 μ+e2vχB2 μ+1 2e2χ2B2 μ +1 2(∂χ)2−μ2χ2−√ λμχ3−λ 4χ4+μ4 4λ(3) 264 | IV . Symmetry and Symmetry Breaking The theory now consists of a vector field Bμwith mass M=ev (4) interacting with a scalar field χwith mass√ 2μ. The phase field θ, which would have been the Nambu-Goldstone boson in the ungauged theory, has disappeared. We say that thegauge field A μhas eaten the Nambu-Goldstone boson; it has gained weight and changed its name to Bμ. Recall that a massless gauge field has only 2 degrees of freedom, while a massive gauge field has 3 degrees of freedom. A massless gauge field has to eat a Nambu-Goldstone bosonin order to have the requisite number of degrees of freedom. The Nambu-Goldstoneboson becomes the longitudinal degree of freedom of the massive gauge field. We do notlose any degrees of freedom, as we had better not. This phenomenon of a massless gauge field becoming massive by eating a Nambu- Goldstone boson was discovered by numerous particle physicists 1and is known as the Higgs mechanism. People variously call ϕ, or more restrictively χ, the Higgs field. The same phenomenon was discovered in the context of condensed matter physics by Landau,Ginzburg, and Anderson, and is known as the Anderson mechanism. Let us give a slightly more involved example, an O(3)gauge theory with a Higgs field ϕ a (a=1, 2, 3)transforming in the vector representation. The Lagrangian contains the kinetic energy term1 2(Dμϕa)2, with Dμϕa=∂μϕa+gεabcAb μϕcas indicated in (IV .5.20). Upon spontaneous symmetry breaking, /vectorϕacquires a vacuum expectation value which without loss of generality we can choose to point in the 3-direction, so that /angbracketleftϕa/angbracketright=vδa3. We set ϕ3=vand see that 1 2(Dμϕa)2→1 2(gv)2(A1 μAμ1+A2 μAμ2) (5) The gauge potential A1 μandA2 μacquires mass gv[compare with (4)] while A3 μremains massless. A more elaborate example is that of an SU( 5)gauge theory with ϕtransforming as the 24- dimensional adjoint representation. (See appendix B for the necessary group theory.) Thefieldϕi sa5b y5hermitean traceless matrix. Since the adjoint representation transforms asϕ→ϕ+iθ a[Ta,ϕ], we have Dμϕ=∂μϕ−igAa μ[Ta,ϕ] with a=1,... , 24 running over the 24 generators of SU( 5). By a symmetry transformation the vacuum expectation value of ϕcan be taken to be diagonal /angbracketleftϕi j/angbracketright=vjδi j(i,j=1,... ,5), with/summationtext jvj=0. (This is the analog of our choosing /angbracketleft/vectorϕ/angbracketrightto point in the 3-direction in the preceding example.) We have in the Lagrangian tr(Dμϕ)(Dμϕ)→g2tr[Ta,/angbracketleftϕ/angbracketright][/angbracketleftϕ/angbracketright,Tb]Aa μAμb(6) The gauge boson masses squared are given by the eigenvalues of the 24 by 24 matrix g2tr[Ta,/angbracketleftϕ/angbracketright][/angbracketleftϕ/angbracketright,Tb], which we can compute laboriously for any given /angbracketleftϕ/angbracketright. 1Including P. Higgs, F. Englert, R. Brout, G. Guralnik, C. Hagen, and T . Kibble. IV .6. Anderson-Higgs Mechanism | 265 It is easy to see, however, which gauge bosons remain massless. As a specific example (which will be of interest to us in chapter VII.6), suppose /angbracketleftϕ/angbracketright=v⎛ ⎜⎜⎜⎜⎜⎜⎜⎜⎝200 0 0 020 0 0002 0 0000−30000 0 −3⎞ ⎟⎟⎟⎟⎟⎟⎟⎟⎠(7) Which generators Tacommute with /angbracketleftϕ/angbracketright? Clearly, generators of the form/parenleftBig A0 00/parenrightBig and of the form/parenleftBig 00 0B/parenrightBig . Here Arepresents 3 by 3 hermitean traceless matrices (of which there are 32−1=8, the so-called Gell-Mann matrices) and Brepresents 2 by 2 hermitean traceless matrices (of which there are 22−1=3, namely the Pauli matrices). Furthermore, the generator ⎛ ⎜⎜⎜⎜⎜⎜⎜⎜⎝200 0 0 020 0 0002 0 0000−30000 0 −3⎞ ⎟⎟⎟⎟⎟⎟⎟⎟⎠(8) being proportional to /angbracketleftϕ/angbracketright, obviously commutes with /angbracketleftϕ/angbracketright. Clearly, these generators gen- erateSU( 3),SU( 2), and U(1), respectively. Thus, in the 24 by 24 mass-squared matrix g2tr[Ta,/angbracketleftϕ/angbracketright][/angbracketleftϕ/angbracketright,Tb] there are blocks of submatrices that vanish, namely, an 8 by 8 block, a 3 by 3 block, anda1b y1block. We have 8 +3+1=12 massless gauge bosons. The remaining 24 −12=12 gauge bosons acquire mass. Counting massless gauge bosons In general, consider a theory with the global symmetry group Gspontaneously broken to a subgroup H. As we learned in chapter IV .1, n(G)−n(H) Nambu-Goldstone bosons appear. Now suppose the symmetry group Gis gauged. We start with n(G) massless gauge bosons, one for each generator. Upon spontaneous symmetry breaking, the n(G)− n(H) Nambu-Goldstone bosons are eaten by n(G)−n(H) gauge bosons, leaving n(H) massless gauge bosons, exactly the right number since the gauge bosons associated withthe surviving gauge group Hshould remain massless. In our simple example, G=U(1),H=nothing: n(G)=1 andn(H)=0. In our second example, G=O(3),H=O(2)/similarequalU(1):n(G)=3 and n(H)=1, and so we end up with one massless gauge boson. In the third example, G=SU( 5),H=SU( 3)⊗SU( 2)⊗U(1) so that n(G)=24 and n(H)=12. Further examples and generalizations are worked out in the exercises. 266 | IV . Symmetry and Symmetry Breaking Gauge boson mass spectrum It is easy enough to work out the mass spectrum explicitly. The covariant derivative of a Higgs field is Dμϕ=∂μϕ+gAa μTaϕ, where gis the gauge coupling, Taare the generators of the group Gwhen acting on ϕ, andAa μthe gauge potential corresponding to the ath generator. Upon spontaneous symmetry breaking we replace ϕby its vacuum expectation value/angbracketleftϕ/angbracketright=v. Hence Dμϕis replaced by gAa μTav. The kinetic term1 2(Dμϕ.Dμϕ)[here (.)denotes the scalar product in the group G] in the Lagrangian thus becomes 1 2g2(Tav.Tbv)AμaAb μ≡1 2Aμa(μ2)abAbμ where we have introduced the mass-squared matrix (μ2)ab=g2(Tav.Tbv) (9) for the gauge bosons. [You will recognize (9) as the generalization of (4); also compare (5) and (6).] We diagonalize (μ2)abto obtain the masses of the gauge bosons. The eigenvectors tell us which linear combinations of Aa μcorrespond to mass eigenstates. Note that μ2is ann(G) byn(G) matrix with n(H) zero eigenvalues, whose existence can also be seen explicitly. Let Tcbe a generator of H. The statement that Hremains unbroken by the vacuum expectation value vmeans that the symmetry transformation generated by Tcleaves vinvariant; in other words, Tcv=0, and hence the gauge boson associated with Tcremains massless, as it should. All these points are particularly evident in the SU( 5) example we worked out. Feynman rules in spontaneously broken gauge theories It is easy enough to derive the Feynman rules for spontaneously broken gauge theories.T ake, for example, (3). As usual, we look at the terms quadratic in the fields, Fouriertransform, and invert. We see that the gauge boson propagator is given by −i k2−M2+iε(gμν−kμkν M2) (10) and the χpropagator by i k2−2μ2+iε(11) I leave it to you to work out the rules for the interaction vertices. As I said in another context, field theories often exist in several equivalent forms. T ake the U(1)theory in (1) and instead of polar coordinates go to Cartesian coordinates ϕ=(1/√ 2)(ϕ1+iϕ2)so that Dμϕ=∂μϕ−ieAμϕ=1√ 2[(∂μϕ1+eAμϕ2)+i(∂μϕ2−eAμϕ1)] IV .6. Anderson-Higgs Mechanism | 267 Then (1) becomes L=−1 4FμνFμν+1 2[(∂μϕ1+eAμϕ2)2+(∂μϕ2−eAμϕ1)2] (12) +1 2μ2(ϕ2 1+ϕ2 2)−1 4λ(ϕ2 1+ϕ2 2)2 Spontaneous symmetry breaking means setting ϕ1→v+ϕ/prime 1withv=/radicalbig μ2/λ. The physical content of (12) and (3) should be the same. Indeed, expand the Lagrangian (12) to quadratic order in the fields: L=μ4 4λ−1 4FμνFμν+1 2M2A2 μ−MAμ∂μϕ2+1 2[(∂μϕ/prime 1)2−2μ2ϕ/prime2 1] +1 2(∂μϕ2)2+... (13) The spectrum, a gauge boson Awith mass M=evand a scalar boson ϕ/prime 1with mass√ 2μ, is identical to the spectrum in (3). (The particles there were named Bandχ.) But oops, you may have noticed something strange: the term −MAμ∂μϕ2which mixes the fields Aμandϕ2. Besides, why is ϕ2still hanging around? Isn’t he supposed to have been eaten? What to do? We can of course diagonalize but it is more convenient to get rid of this mixing term. Referring to the Fadeev-Popov quantization of gauge theories discussed in chapter III.4 wenote that the gauge fixing term generates a term to be added to L. We can cancel the unde- sirable mixing term by choosing the gauge function to be f (A)=∂A+ξevϕ 2−σ. Going through the steps, we obtain the effective Lagrangian Leff=L−(1/2ξ)(∂A +ξMϕ 2)2 [compare with (III.4.7)]. The undesirable cross term −MAμ∂μϕ2inLis now canceled upon integration by parts. In Leffthe terms quadratic in Anow read −1 4FμνFμν+1 2M2A2 μ− (1/2ξ)(∂A)2while the terms quadratic in ϕ2read1 2[(∂μϕ2)2−ξM2ϕ22], immediately giving us the gauge boson propagator −i k2−M2+iε/bracketleftbigg gμν−(1−ξ)kμkν k2−ξM2+iε/bracketrightbigg (14) and the ϕ2propagator i k2−ξM2+iε(15) This one-parameter class of gauge choices is known as the Rξgauge. Note that the would- be Goldstone field ϕ2remains in the Lagrangian, but the very fact that its mass depends on the gauge parameter ξbrands it as unphysical. In any physical process, the ξdependence in theϕ2andApropagators must cancel out so as to leave physical amplitudes ξindependent. In exercise IV .6.9 you will verify that this is indeed the case in a simple example. Different gauges have different advantages You might wonder why we would bother with the Rξgauge. Why not just use the equivalent formulation of the theory in (3), known as the unitary gauge, in which the gauge boson 268 | IV . Symmetry and Symmetry Breaking propagator (10) looks much simpler than (14) and in which we don’t have to deal with the unphysical ϕ2field? The reason is that the Rξgauge and the unitary gauge complement each other. In the Rξgauge, the gauge boson propagator (14) goes as 1 /k2for large kand so renormalizability can be proved rather easily. On the other hand, in the unitary gaugeall fields are physical (hence the name “unitary”) but the gauge boson propagator (10)apparently goes as k μkν/k2for large k; to prove renormalizability we must show that the kμkνpiece of the propagator does not contribute. Using both gauges, we can easily prove that the theory is both renormalizable and unitary. By the way, note that in the limit ξ→∞ (14) goes over to (10) and ϕ2disappear, at least formally. In practical calculations, there are typically many diagrams to evaluate. In the Rξgauge, the parameter ξdarn well better disappears when we add everything up to form the physical mass shell amplitude. The Rξgauge is attractive precisely because this requirement provides a powerful check on practical calculations. I remarked earlier that strictly speaking, gauge invariance is not so much a symmetry as the reflection of a redundancy in the degrees of freedom used. (The photon has only 2degrees of freedom but we use a field A μwith 4 components.) A purist would insist, in the same vein, that there is no such thing as spontaneously breaking a gauge symmetry.T o understand this remark, note that spontaneous breaking amounts to setting ρ≡|ϕ|to vandθto 0 in (2). The statement |ϕ|=v is perfectly U(1)invariant: It defines a circle in ϕ space. By picking out the point θ=0 on the circle in a globally symmetric theory we break the symmetry. In contrast, in a gauge theory, we can use the gauge freedom to fix θ=0 everywhere in spacetime. Hence the purists. I will refrain from such hair-splitting in thisbook and continue to use the convenient language of symmetry breaking even in a gaugetheory. Exercises IV .6.1 Consider an SU( 5)gauge theory with a Higgs field ϕtransforming as the 5-dimensional representation: ϕi,i=1, 2, ... , 5. Show that a vacuum expectation value of ϕbreaks SU( 5)toSU( 4). Now add another Higgs field ϕ/prime, also transforming as the 5-dimensional representation. Show that the symmetry can either remain at SU( 4)or be broken to SU( 3). IV .6.2 In general, there may be several Higgs fields belonging to various representations labeled by α. Show that the mass squared matrix for the gauge bosons generalize immediately to (μ2)ab=/summationtext αg2(Ta αvα.Tb αvα), where vαis the vacuum expectation value of ϕαandTa αis theath generator represented on ϕα. Combine the situations described in exercises IV .6.1 and IV .6.2 and work out the mass spectrum of the gaugebosons. IV .6.3 The gauge group Gdoes not have to be simple; it could be of the form G 1⊗G2⊗...⊗Gk, with coupling constants g1,g2,... ,gk. Consider, for example, the case G=SU( 2)⊗U(1)and a Higgs fieldϕtransforming like the doublet under SU( 2)and like a field with charge1 2under U(1), so that Dμϕ=∂μϕ−i[gAa μ(τa/2)+g/primeBμ1 2]ϕ. Let/angbracketleftϕ/angbracketright=/parenleftBig 0 v/parenrightBig . Determine which linear combinations of the gauge bosons Aa μandBμacquire mass. IV .6.4 In chapter IV .5 you worked out an SU( 2)gauge theory with a scalar field ϕin the I=2 representation. Write down the most general quartic potential V( ϕ) and study the possible symmetry breaking pattern. IV .6. Anderson-Higgs Mechanism | 269 IV .6.5 Complete the derivation of the Feynman rules for the theory in (3) and compute the amplitude for the physical process χ+χ→B+B. IV .6.6 Derive (14). [Hint: The procedure is exactly the same as that used to obtain (III.4.9).] Write L=1 2AμQμνAν withQμν=(∂2+M2)gμν−[1−(1/ξ)]∂μ∂νor in momentum space Qμν=−(k2−M2)gμν+[1− (1/ξ)]kμkν. The propagator is the inverse of Qμν. IV .6.7 Work out the ( ...)in(13)and the Feynman rules for the various interaction vertices. IV .6.8 Using the Feynman rules derived in exercise IV .6.7 calculate the amplitude for the physical process ϕ/prime 1+ ϕ/prime 1→A+Aand show that the dependence on ξcancels out. Compare with the result in exercise IV .6.5. [Hint: There are two diagrams, one with Aexchange and the other with ϕ2exchange.] IV .6.9 Consider the theory defined in (12) with μ=0. Using the result of exercise IV .3.5 show that Veff(ϕ)=1 4λϕ4+1 64π2(10λ2+3e4)ϕ4/parenleftbigg logϕ2 M2−25 6/parenrightbigg +... (16) where ϕ2=ϕ2 1+ϕ2 2. This potential has a minimum away from ϕ=0 and thus the gauge symmetry is spontaneously broken by quantum fluctuations. In chapter IV .3 we did not have the e4term and argued that the minimum we got there was not to be trusted. But here we can balance the λϕ4against e4ϕ4log(ϕ2/M2)forλof the same order of magnitude as e4. The minimum can be trusted. Show that the spectrum of this theory consists of a massive scalar boson and a massive vector boson, with m2(scalar) m2(vector)=3 2πe2 4π(17) For help, see S. Coleman and E. Weinberg, Phys. Rev. D7: 1888, 1973. IV.7 Chiral Anomaly Classical versus quantum symmetry I have emphasized the importance of asking good questions. Here is another good one: Is a symmetry of classical physics necessarily a symmetry of quantum physics? We have a symmetry of classical physics if a transformation ϕ→ϕ+δϕleaves the action S(ϕ) invariant. We have a symmetry of quantum physics if the transformation leaves the path integral/integraltext DϕeiS(ϕ)invariant. When our question is phrased in this path integral language, the answer seems obvious: Not necessarily. Indeed, the measure Dϕ may or may not be invariant. Yet historically, field theorists took as almost self-evident the notion that any symmetry of classical physics is necessarily a symmetry of quantum physics, and indeed, almostall the symmetries they encountered in the early days of field theory had the property ofbeing symmetries of both classical and quantum physics. For instance, we certainly expectquantum mechanics to be rotational invariant. It would be very odd indeed if quantumfluctuations were to favor a particular direction. You have to appreciate the frame of mind that field theorists operated in to understand their shock when they discovered in the late 1960s that quantum fluctuations can indeedbreak classical symmetries. Indeed, they were so shocked as to give this phenomenon therather misleading name “anomaly,” as if it were some kind of sickness of field theory. Withthe benefits of hindsight, we now understand the anomaly as being no less conceptuallyinnocuous as the elementary fact that when we change integration variables in an integralwe better not forget the Jacobian. With the passing of time, field theorists have developed many different ways of looking at the all important subject of anomaly. They are all instructive and shed different lights on how the anomaly comes about. For this introductory text I choose to show the existenceof anomaly by an explicit Feynman diagram calculation. The diagram method is certainlymore laborious and less slick than other methods, but the advantage is that you will see IV .7. Chiral Anomaly | 271 a classical symmetry vanishing in front of your very eyes! No smooth formal argument for us. The lesser of two evils Consider the theory of a single massless fermion L=¯ψiγμ∂μψ. You can hardly ask for a simpler theory! Recall from chapter II.1 that Lis manifestly invariant under the separate transformations ψ→eiθψandψ→eiθγ5ψ, corresponding to the conserved vector current Jμ=¯ψγμψand the conserved axial current Jμ 5=¯ψγμγ5ψrespectively. You should verify that ∂μJμ=0 and ∂μJμ 5=0 follow immediately from the classical equation of motion iγμ∂μψ=0. Let us now calculate the amplitude for a spacetime history in which a fermion-anti- fermion pair is created at x1and another such pair is created at x2by the vector current, with the fermion from one pair annihilating the antifermion from the other pair and theremaining fermion-antifermion pair being subsequently annihilated by the axial current.This is a long-winded way of describing the amplitude /angbracketleft0|TJ λ 5(0)Jμ(x1)Jν(x2)|0/angbracketrightin words, but I want to make sure that you know what I am talking about. Feynman tellsus that the Fourier transform of this amplitude is given by the two “triangle” diagrams infigure IV .7.1a and b. /Delta1λμν(k1,k2)=(−1)i3/integraldisplayd4p (2π)4 tr/parenleftbigg γλγ5 1 /negationslashp− /negationslashqγν 1 /negationslashp− /negationslashk1γμ1 /negationslashp+γλγ5 1 /negationslashp− /negationslashqγμ 1 /negationslashp− /negationslashk2γν1 /negationslashp/parenrightbigg (1) withq=k1+k2. Note that the two terms are required by Bose statistics. The overall factor of(−1)comes from the closed fermion loop. Classically, we have two symmetries implying ∂μJμ=0 and ∂μJμ 5=0. In the quantum theory, if ∂μJμ=0 continues to hold, then we should have k1μ/Delta1λμν=0 and k2ν/Delta1λμν=0, and if ∂μJμ 5=0 continues to hold, then qλ/Delta1λμν=0. Now that we have things all set up, we merely have to calculate /Delta1λμνto see if the two symmetries hold up under quantum fluctuations. No big deal. p p − q p − k1γλγ5 γμγν (a)p p − qγλγ5 (b)γμγν p − k2 Figure IV .7.1 272 | IV . Symmetry and Symmetry Breaking Before we blindly calculate, however, let us ask ourselves how sad we would be if either of the two currents JμandJμ 5fails to be conserved. Well, we would be very upset if the vector current is not conserved. The corresponding charge Q=/integraltext d3xJ0counts the number of fermions. We wouldn’t want our fermions to disappear into thin air or pop out of nowhere.Furthermore, it may please us to couple the photon to the fermion field ψ. In that case, you would recall from chapter II.7 that we need ∂ μJμ=0 to prove gauge invariance and hence show that the photon has only two degrees of polarization. More explicitly, imaginea photon line coming into the vertex labeled by μin figure IV .7.1a and b with propagator (i/k 2 1)[ξ(k 1μk1ρ/k2 1)−gμρ]. The gauge dependent term ξ(k 1μk1ρ/k2 1)would not go away if the vector current is not conserved, that is, if k1μ/Delta1λμνfails to vanish. On the other hand, quite frankly, just between us friends, we won’t get too upset if quantum fluctuation violates axial current conservation. Who cares if the axial chargeQ 5=/integraltext d3xJ0 5is not constant in time? Shifting integration variable So, do k1μ/Delta1λμνandk2ν/Delta1λμνvanish? We will look over Professor Confusio’s shoulders as he calculates k1μ/Delta1λμν. (We are now in the 1960s, long after the development of renor- malization theory as described in chapter III.1 and Confusio has managed to get a tenuretrack assistant professorship.) He hits /Delta1 λμνas written in (1) with k1μand using what he learned in chapter II.7 writes /negationslashk1in the first term as /negationslashp−(/negationslashp− /negationslashk1)and in the second term as(/negationslashp− /negationslashk2)−(/negationslashp− /negationslashq), thus obtaining k1μ/Delta1λμν(k1,k2) =i/integraldisplayd4p (2π)4tr(γλγ5 1 /negationslashp− /negationslashqγν 1 /negationslashp− /negationslashk1−γλγ5 1 /negationslashp− /negationslashk2γν1 /negationslashp) (2) Just as in chapter II.7, Confusio recognizes that in the integrand the first term is just the second term with the shift of the integration variable p→p−k1. The two terms cancel and Professor Confusio publishes a paper saying k1μ/Delta1λμν=0, as we all expect. Remember back in chapter II.7 I said we were going to worry later about whether it is legitimate to shift integration variables. Now is the time to worry! You could have asked your calculus teacher long ago when it is legitimate to shift integration variables. When is/integraltext+∞ −∞dpf (p +a)equal to/integraltext+∞ −∞dpf (p) ? The difference between these two integrals is /integraldisplay+∞ −∞dp(ad dpf( p)+...)=a(f(+∞)−f(−∞)) +... Clearly, if f(+∞) andf(−∞) are two different constants, then it is not okay to shift. But if the integral/integraltext+∞ −∞dpf (p) is convergent, or even logarithmically divergent, it is certainly okay. It was okay in chapter II.7 but definitely not here in (2)! IV .7. Chiral Anomaly | 273 As usual, we rotate the Feynman integrand to Euclidean space. Generalizing our obser- vation above to d-dimensional Euclidean space, we have /integraldisplay dd Ep[f( p+a)−f( p) ]=/integraldisplay dd Ep[aμ∂μf( p)+...] which by Gauss’s theorem is given by a surface integral over an infinitely large sphere enclosing all of Euclidean spacetime and hence equal to lim P→∞aμ/parenleftbiggPμ P/parenrightbigg f( P) Sd−1(P) where Sd−1(P) is the area of a (d−1)-dimensional sphere (see appendix D) and where an average over the surface of the sphere is understood. (Recall from our experience evaluatingFeynman diagrams that the average of P μPν/P2is equal to1 4ημνby a symmetry argument, with the normalization1 4fixed by contracting with ημν.)Rotating back, we have for a 4- dimensional Minkowskian integral /integraldisplay d4p[f( p+a)−f( p) ]=lim P→∞iaμ/parenleftbiggPμ P/parenrightbigg f( P) ( 2π2P3) (3) Note the ifrom Wick rotating back. Applying (3) with f( p)=tr/parenleftbigg γλγ5 1 /negationslashp− /negationslashk2γν1 /negationslashp/parenrightbigg =tr[γ5(/negationslashp− /negationslashk2)γν/negationslashpγλ] (p−k2)2p2=4iετνσλk2τpσ (p−k2)2p2 we obtain k1μ/Delta1λμν=i (2π)4lim P→∞i(−k 1)μPμ P4iετνσλk2τPσ P42π2P3=i 8π2ελντσk1τk2σ Contrary to what Confusio said, k1μ/Delta1λμν/negationslash=0. As I have already said, this would be a disaster. Fermion number is not conserved and matter would be disintegrating all around us! What is the way out? In fact, we are only marginally smarter than Professor Confusio. We did not notice that the integral defining /Delta1λμνin (1) is linearly divergent and is thus not well defined. Oops, even before we worry about calculating k1μ/Delta1λμνandk2ν/Delta1λμνwe better worry about whether or not /Delta1λμνdepends on the physicist doing the calculation. In other words, suppose another physicist chooses1to shift the integration variable pin the linearly divergent integral in (1) by an arbitrary 4-vector aand define /Delta1λμν(a,k1,k2) =(−1)i3/integraldisplayd4p (2π)4tr(γλγ5 1 /negationslashp+ /negationslasha− /negationslashqγν 1 /negationslashp+ /negationslasha− /negationslashk1γμ 1 /negationslashp+ /negationslasha) +{μ,k1↔ν,k2} (4) There can be as many results for the Feynman diagrams in figure IV .7.1a and b as there are physicists! That would be the end of physics, or at least quantum field theory, for sure. 1This is the freedom of choice in labeling internal momenta mentioned in chapter I.7. 274 | IV . Symmetry and Symmetry Breaking Well, whose result should we declare to be correct? The only sensible answer is that we trust the person who chooses an asuch that k1μ/Delta1λμν(a,k1,k2)andk2ν/Delta1λμν(a,k1,k2)vanish, so that the photon will have the right number of degrees of freedom should we introduce a photon into the theory. Let us compute /Delta1λμν(a,k1,k2)−/Delta1λμν(k1,k2)by applying (3) to f( p)= tr(γλγ51 /negationslashp−/negationslashqγν 1 /negationslashp−/negationslashk1γμ1 /negationslashp). Noting that f( P)=lim P→∞tr(γλγ5/negationslashPγν/negationslashPγμ/negationslashP) P6 =2Pμtr(γλγ5/negationslashPγν/negationslashP)−P2tr(γλγ5/negationslashPγνγμ) P6=+4iP2Pσεσνμλ P6 we see that /Delta1λμν(a,k1,k2)−/Delta1λμν(k1,k2)=4i 8π2lim P→∞aωPωPσ P2εσνμλ+{μ,k1↔ν,k2} =i 8π2εσνμλaσ+{μ,k1↔ν,k2} (5) There are two independent momenta k1andk2in the problem, so we can take a= α(k1+k2)+β(k1−k2). Plugging into (5), we obtain /Delta1λμν(a,k1,k2)=/Delta1λμν(k1,k2)+iβ 4π2ελμνσ(k1−k2)σ (6) Note that αdrops out. As expected, /Delta1λμν(a,k1,k2)depends on β, and hence on a. Our unshakable desire to have a conserved vector current, that is, k1μ/Delta1λμν(a,k1,k2)=0, now fixes the parameter β upon recalling k1μ/Delta1λμν(k1,k2)=i 8π2ελντσk1τk2σ Hence, we must choose to deal with /Delta1λμν(a,k1,k2)withβ=−1 2. One way of viewing all this is to say that the Feynman rules do not suffice in determining /angbracketleft0|TJλ 5(0)Jμ(x1)Jν(x2)|0/angbracketright. They have to be supplemented by vector current conservation. The amplitude /angbracketleft0|TJλ 5(0)Jμ(x1)Jν(x2)|0/angbracketrightis defined by /Delta1λμν(a,k1,k2)withβ=−1 2. Quantum fluctuation violates axial current conservation Now we come to the punchline of the story. We insisted that the vector current be conserved. Is the axial current also conserved? T o answer this question, we merely have to compute qλ/Delta1λμν(a,k1,k2)=qλ/Delta1λμν(k1,k2)+i 4π2εμνλσk1λk2σ (7) IV .7. Chiral Anomaly | 275 By now, you know how to do this: qλ/Delta1λμν(k1,k2)=i/integraldisplayd4p (2π)4tr/parenleftbigg γ5 1 /negationslashp− /negationslashqγν 1 /negationslashp− /negationslashk1γμ −γ5 1 /negationslashp− /negationslashk2γν1 /negationslashpγμ/parenrightbigg +{μ,k1↔ν,k2} =i 4π2εμνλσk1λk2σ (8) Indeed, you recognize that the integration has already been done in (2). We finally obtain qλ/Delta1λμν(a,k1,k2)=i 2π2εμνλσk1λk2σ (9) The axial current is not conserved! In summary, in the simple theory L=¯ψiγμ∂μψwhile the vector and axial currents are both conserved classically, quantum fluctuation destroys axial current conservation. Thisphenomenon is known variously as the anomaly, the axial anomaly, or the chiral anomaly. Consequences of the anomaly As I said, the anomaly is an extraordinarily rich subject. I will content myself with a seriesof remarks, the details of which you should work out as exercises. 1. Suppose we gauge our simple theory L=¯ψiγμ(∂μ−ieAμ)ψand speak of Aμas the photon field. Then in figure IV .7.1 we can think of two photon lines coming out of the vertices labeledμandν. Our central result (9) can then be written elegantly as two operator equations: Classical physics: ∂ μJμ 5=0 (10) Qu antum physics: ∂μJμ 5=e2 (4π)2εμνλσFμνFλσ (11) The divergence of the axial current ∂μJμ 5is not zero, but is an operator capable of producing two photons. 2. Applying the same type of argument as in chapter IV .2 we can calculate the rate of the decay π0→γ+γ. Indeed, historically people used the erroneous result (10) to deduce that this experimentally observed decay cannot occur! See exercise IV .7.2. The resolution of thisapparent paradox led to the correct result (11). 3. Writing the Lagrangian in terms of left and right handed fields ψ RandψLand introducing the left and right handed currents Jμ R≡¯ψRγμψRandJμ L≡¯ψLγμψL, we can repackage the anomaly as ∂μJμ R=1 2e2 (4π)2εμνλσFμνFλσ and ∂μJμ L=−1 2e2 (4π)2εμνλσFμνFλσ (12) 276 | IV . Symmetry and Symmetry Breaking (Hence the name chiral!) We can think of left handed and right handed fermions running around the loop in figure IV .7.1, contributing oppositely to the anomaly. 4. Consider the theory L=¯ψ(iγμ∂μ−m)ψ . Then invariance under the transformation ψ→ eiθγ5ψis spoiled by the mass term. Classically, ∂μJμ 5=2m¯ψiγ5ψ: The axial current is explicitly not conserved. The anomaly now says that quantum fluctuation produces anadditional term. In the theory L=¯ψ[iγ μ(∂μ−ieAμ)−m]ψ, we have ∂μJμ 5=2m¯ψiγ5ψ+e2 (4π)2εμνλσFμνFλσ (13) 5. Recall that in chapter III.7 we introduced Pauli-Villars regulators to calculate vacuum polarization. We subtract from the integrand what the integrand would have been if theelectron mass were replaced by some regulator mass. The analog of electron mass in (1) isin fact 0 and so we subtract from the integrand what the integrand would have been if 0were replaced by a regulator mass M. In other words, we now define /Delta1 λμν(k1,k2)=(−1)i3/integraldisplayd4p (2π)4tr/parenleftbigg γλγ5 1 /negationslashp− /negationslashqγν 1 /negationslashp− /negationslashk1γμ1 /negationslashp −γλγ5 1 /negationslashp− /negationslashq−Mγν 1 /negationslashp− /negationslashk1−Mγμ 1 /negationslashp−M/parenrightbigg +{μ,k1↔ν,k2}. (14) Note that as p→∞ the integrand now vanishes faster than 1 /p3. This is in accordance with the philosophy of regularization outlined in chapters III.1 and III.7: For p/lessmuchM, the threshold of ignorance, the integrand is unchanged. But for p/greatermuchM, the integrand is cut off. Now the integral in (14) is superficially logarithmically divergent and we can shift theintegration variable pat will. So how does the chiral anomaly arise? By including the regulator mass Mwe have broken axial current conservation explicitly. The anomaly is the statement that this breaking persistseven when we let Mtend to infinity. It is extremely instructive (see exercise IV .7.4) to work this out. 6. Consider the nonabelian theory L=¯ψiγ μ(∂μ−igAa μTa)ψ. We merely have to include in the Feynman amplitude a factor of Taat the vertex labeled by μand a factor of Tbat the vertex labeled by ν. Everything goes through as before except that in summing over all the different fermions that run around the loop we obtain a factor tr TaTb. Thus, we see instantly that in a nonabelian gauge theory ∂μJμ 5=g2 (4π)2εμνλσtrFμνFλσ (15) where Fμν=Fa μνTais the matrix field strength defined in chapter IV .5. Nonabelian sym- metry tells us something remarkable: The object εμνλσtrFμνFλσcontains not only a term quadratic in A, but also terms cubic and quartic in A, and hence there is also a chiral anomaly with three and four gauge bosons coming in, as indicated in figure IV .7.2a andb. Some people refer to the anomaly produced in figures IV .7.1 and IV .7.2 as the triangle,square, and pentagon anomaly. Historically, after the triangle anomaly was discovered, therewas a controversy as to whether the square and pentagon anomaly existed. The nonabelian IV .7. Chiral Anomaly | 277 γλγ5 (a)γλγ5 (b)TcTa Tb TcTbTaTd Figure IV .7.2 w2 w1 Figure IV .7.3 symmetry argument given here makes things totally obvious, but at the time people calcu- lated Feynman diagrams explicitly and, as we just saw, there are subtleties lying in wait forthe unwary. 7. We will see in chapter V .7 that the anomaly has deep connections to topology.8. We computed the chiral anomaly in the free theory L=¯ψ(iγ μ∂μ−m)ψ . Suppose we couple the fermion to a scalar field by adding fϕ¯ψψ or to the electromagnetic field for that matter. Now we have to calculate higher order diagrams such as the three-loop diagramin figure IV .7.3. You would expect that the right-hand side of (9) would be multiplied by1+h(f ,e,...), where his some unknown function of all the couplings in the theory. Surprise! Adler and Bardeen proved that h=0. This apparently miraculous fact, known as the nonrenormalization of the anomaly, can be understood heuristically as follows. Beforewe integrate over the momenta of the scalar propagators in figure IV .7.3 (labeled by w 1 andw2)the Feynman integrand has seven fermion propagators and thus is more than sufficiently convergent that we can shift integration variables with impunity. Thus, beforewe integrate over w 1andw2all the appropriate Ward identities are satisfied, for instance, qλ/Delta1λμν 3 loops(k1,k2;w1,w2)=0. You can easily complete the proof. You will give a proof2based on topology in exercise V .7.13. 2For a simple proof not involving topology, see J. Collins, Renormalization , p. 352. 278 | IV . Symmetry and Symmetry Breaking (a)π° γγ (b)π° γ γ Figure IV .7.4 9. The preceding point was of great importance in the history of particle physics as it led directly to the notion of color, as we will discuss in chapter VII.3. The nonrenormalization of theanomaly allowed the decay amplitude for π 0→γ+γto be calculated with confidence in the late 1960s. In the quark model of the time, the amplitude is given by an infinite number ofFeynman diagrams, as indicated in figure IV .7.4 (with a quark running around the fermionloop), but the nonrenormalization of the anomaly tells us that only figure IV .7.4a contributes.In other words, the amplitude does not depend on the details of the strong interaction. Thatit came out a factor of 3 too small suggested that quarks come in 3 copies, as we will see inchapter VII.3. 10. It is natural to speculate as to whether quarks and leptons are composites of yet more fundamental fermions known as preons. The nonrenormalization of the chiral anomalyprovides a powerful tool for this sort of theoretical speculation. No matter how complicatedthe relevant interactions might be, as long as they are described by field theory as we knowit, the anomaly at the preon level must be the same as the anomaly at the quark-lepton level.This so-called anomaly matching condition 3severely constrains the possible preon theories. 11. Historically, field theorists were deeply suspicious of the path integral, preferring the canonical approach. When the chiral anomaly was discovered, some people even argued thatthe existence of the anomaly proved that the path integral was wrong. Look, these peoplesaid, the path integral /integraldisplay D¯ψDψ e i/integraltext d4x¯ψiγμ(∂μ−iAμ)ψ(16) is too stupid to tell us that it is not invariant under the chiral transformation ψ→eiθγ5ψ. Fujikawa resolved the controversy by showing that the path integral did know about theanomaly: Under the chiral transformation the measure D¯ψDψ changes by a Jacobian. Recall that this was how I motivated this chapter: The action may be invariant but not thepath integral. 3G. ’t Hooft, in: G. ’t Hooft et al., eds., Recent Developments in Gauge Theories ; A. Zee, Phys. Lett. 95B:290, 1980. IV .7. Chiral Anomaly | 279 Exercises IV .7.1 Derive (11) from (9). The momentum factors k1λandk2σin (9) become the two derivatives in FμνFλσin (11). IV .7.2 Following the reasoning in chapter IV .2 and using the erroneous (10) show that the decay amplitude for the decay π0→γ+γwould vanish in the ideal world in which the π0is massless. Since the π0does decay and since our world is close to the ideal world, this provided the first indication historically that(10) cannot possibly be valid. IV .7.3 Repeat all the calculations in the text for the theory L=¯ψ(iγ μ∂μ−m)ψ . IV .7.4 T ake the Pauli-Villars regulated /Delta1λμν(k1,k2)and contract it with qλ. The analog of the trick in chapter II.7 is to write /negationslashqγ5in the second term as [2 M+(/negationslashp−M)−(/negationslashp− /negationslashq+M)]γ5. Now you can freely shift integration variables. Show that qλ/Delta1λμν(k1,k2)=− 2M/Delta1μν(k1,k2) (17) where /Delta1μν(k1,k2)≡(−1)i3/integraldisplayd4p (2π)4 tr/parenleftbigg γ5 1 /negationslashp− /negationslashq−Mγν 1 /negationslashp− /negationslashk1−Mγμ 1 /negationslashp−M/parenrightbigg +{μ,k1↔ν,k2} Evaluate /Delta1μνand show that /Delta1μνgoes as 1 /M in the limit M→∞ and so the right hand side of (17) goes to a finite limit. The anomaly is what the regulator leaves behind as it disappears from the lowenergy spectrum: It is like the smile of the Cheshire cat. [We can actually argue that /Delta1 μνgoes as 1 /M without doing a detailed calculation. By Lorentz invariance and because of the presence of γ5,/Delta1μν must be proportional to εμνλρk1λk2ρ, but by dimensional analysis, /Delta1μνmust be some constant times εμνλρk1λk2ρ/M . You might ask why we can’t use something like 1 /(k2 1)1 2instead of 1 /M to make the dimension come out right. The answer is that from your experience in evaluating Feynman diagrams in (3+1)-dimensional spacetime you can never get a factor like 1 /(k2 1)1 2.] IV .7.5 There are literally Nways of deriving the anomaly. Here is another. Evaluate /Delta1λμν(k1,k2)=(−1)i3/integraldisplayd4p (2π)4 tr/parenleftbigg γλγ5 1 /negationslashp− /negationslashq−mγν 1 /negationslashp− /negationslashk1−mγμ 1 /negationslashp−m/parenrightbigg +{μ,k1↔ν,k2} in the massive fermion case not by brute force but by first using Lorentz invariance to write /Delta1λμν(k1,k2)=ελμνσk1σA1+...+εμνστk1σk2τkλ 2A8 where Ai≡Ai(k2 1,k2 2,q2)are eight functions of the three Lorentz scalars in the problem. You are supposed to fill in the dots. By counting powers as in chapters III.3 and III.7 show that two of thesefunctions are given by superficially logarithmically divergent integrals while the other six are given byperfectly convergent integrals. Next, impose Bose statistics and vector current conservation k 1μ/Delta1λμν= 0=k2ν/Delta1λμνto show that we can avoid calculating the superficially logarithmically divergent integrals. Compute the convergent integrals and then evaluate qλ/Delta1λμν(k1,k2). IV .7.6 Discuss the anomaly by studying the amplitude /angbracketleft0|TJλ 5(0)Jμ 5(x1)Jν 5(x2)|0/angbracketright 280 | IV . Symmetry and Symmetry Breaking given in lowest orders by triangle diagrams with axial currents at each vertex. [Hint: Call the momentum space amplitude /Delta1λμν 5(k1,k2).] Show by using (γ5)2=1 and Bose symmetry that /Delta1λμν 5(k1,k2)=1 3[/Delta1λμν(a,k1,k2)+/Delta1μνλ(a,k2,−q)+/Delta1νλμ(a,−q,k1)] Now use (9) to evaluate qλ/Delta1λμν 5(k1,k2). IV .7.7 Define the fermionic measure Dψ in (16) carefully by going to Euclidean space. Calculate the Jacobian upon a chiral transformation and derive the anomaly. [Hint: For help, see K. Fujikawa, Phys. Rev. Lett. 42: 1195, 1979.] IV .7.8 Compute the pentagon anomaly by Feynman diagrams in order to check remark 6 in the text. In other words, determine the coefficient cin∂μJμ 5=...+cεμνλσtrAμAνAλAσ. Part V Field Theory and Collective Phenomena I mentioned in the introduction that one of the more intellectually satisfying developments in the last two or three decades has been the increasingly important role played by fieldtheoretic methods in condensed matter physics. This is a rich and diverse subject; in thisand subsequent chapters I can barely describe the tip of the iceberg and will have to contentmyself with a few selected topics. Historically, field theory was introduced into condensed matter physics in a rather direct and straightforward fashion. The nonrelativistic electrons in a condensed matter systemcan be described by a field ψ, along the lines discussed in chapter III.5. Field theoretic Lagrangians may then be written down, Feynman diagrams and rules developed, and soon and so forth. This is done in a number of specialized texts. What we present here is to alarge extent the more modern view of an effective field theoretic description of a condensedmatter system, valid at low energy and momentum. One of the fascinations of condensedmatter physics is that due to highly nontrivial many body effects the low energy degreesof freedom might be totally different from the electrons we started out with. A particularlystriking example (to be discussed in chapter VI.2) is the quantum Hall system, in whichthe low energy effective degree of freedom carries fractional charge and statistics. Another advantage of devoting a considerable portion of a field theory book to condensed matter physics is that historically and pedagogically it is much easier to understand therenormalization group in condensed matter physics than in particle physics. I will defiantly not stick to a legalistic separation between condensed matter and particle physics. Some of the topics treated in Parts V and VI actually belong to particle physics.And of course I cannot be responsible for explaining condensed matter physics, any morethan I could be responsible for explaining particle physics in chapter IV .2. This page intentionally left blank V.1 Superfluids Repulsive bosons Consider a finite density ¯ρof nonrelativistic bosons interacting with a short ranged repul- sion. Return to (III.5.11): L=iϕ†∂0ϕ−1 2m∂iϕ†∂iϕ−g2(ϕ†ϕ−¯ρ)2(1) The last term is exactly the Mexican well potential of chapter IV .1, forcing the magnitude ofϕto be close to√¯ρ, thus suggesting that we use polar variables ϕ≡√ρeiθas we did in (III.5.7). Plugging in and dropping the total derivative (i/2)∂0ρ, we obtain L=−ρ∂0θ−1 2m/bracketleftbigg1 4ρ(∂iρ)2+ρ(∂iθ)2/bracketrightbigg −g2(ρ−¯ρ)2(2) Spontaneous symmetry breaking As in chapter IV .1 write√ρ=√¯ρ+h(the vacuum expectation value of ϕis√¯ρ), assume h/lessmuch√¯ρ, and expand1: L=− 2/radicalbig ¯ρh∂ 0θ−¯ρ 2m(∂iθ)2−1 2m(∂ih)2−4g2¯ρh2+... (3) Picking out the terms up to quadratic in hin (3) we use the “central identity of quantum field theory” (see appendix A) to integrate out h, obtaining L=¯ρ∂0θ1 4g2¯ρ−(1/2m)∂2 i∂0θ−¯ρ 2m(∂iθ)2+... =1 4g2(∂0θ)2−¯ρ 2m(∂iθ)2+... (4) 1Note that we have dropped the (potentially interesting) term −¯ρ∂0θbecause it is a total divergence. 284 | V . Field Theory and Collective Phenomena In the second equality we assumed that we are looking at processes with wave number k small compared to/radicalbig 8g2¯ρm so that (1/2m)∂2 iis negligible compared to 4 g2¯ρ. Thus, we see that there exists in this fluid of bosons a gapless mode (often referred to as the phonon)with the dispersion ω2=2g2¯ρ m/vectork2(5) The learned among you will have realized that we have obtained Bogoliubov’s classic result without ever doing a Bogoliubov rotation.2 Let me briefly remind you of Landau’s idealized argument3that a linearly dispersing mode (that is, ωis linear in k)implies superfluidity. Consider a mass Mof fluid flowing down a tube with velocity v. It could lose momentum and slow down to velocity v/prime by creating an excitation of momentum k:Mv=Mv/prime+/planckover2pik. This is only possible with sufficient energy to spare if1 2Mv2≥1 2Mv/prime2+/planckover2piω(k) . Eliminating v/primewe obtain for M macroscopic v≥ω/k . For a linearly dispersing mode this gives a critical velocity vc≡ω/k below which the fluid cannot lose momentum and is hence super. [Thus, from (5) theidealized v c=g√2¯ρ/m .] Suitably scaling the distance variable, we can summarize the low energy physics of superfluidity in the compact Lagrangian L=1 4g2(∂μθ)2(6) which we recognize as the massless version of the scalar field theory we studied in part I, but with the important proviso that θis a phase angle field, that is, θ(x) andθ(x)+2πare really the same. This gapless mode is evidently the Nambu-Goldstone boson associatedwith the spontaneous breaking of the global U(1)symmetry ϕ→e iαϕ. Linearly dispersing gapless mode The physics here becomes particularly clear if we think about a gas of free bosons. We can give a momentum /planckover2pi/vectorkto any given boson at the cost of only (/planckover2pi/vectork)2/2min energy. There exist many low energy excitations in a free boson system. But as soon as a short rangedrepulsion is turned on between the bosons, a boson moving with momentum /vectorkwould affect all the other bosons. A density wave is set up as a result, with energy proportionaltokas we have shown in (5). The gapless mode has gone from quadratically dispersing to linearly dispersing. There are far fewer low energy excitations. Specifically, recall that thedensity of states is given by N(E) ∝k D−1(dk/dE) . For example, for D=2 the density of states goes from N(E) ∝constant (in the presence of quadratically dispersing modes) to N(E) ∝E(in the presence of linearly dispersing modes) at low energies. 2L. D. Landau and E. M. Lifshitz, Statistical Physics , p. 238. 3Ibid., p. 192. V .1. Superfluids | 285 As was emphasized by Feynman4among others, the physics of superfluidity lies not in the presence of gapless excitations, but in the paucity of gapless excitations. (After all,the Fermi liquid has a continuum of gapless modes.) There are too few modes that thesuperfluid can lose energy and momentum to. Relativistic versus nonrelativistic This is a good place to discuss one subtle difference between spontaneous symmetrybreaking in relativistic and nonrelativistic theories. Consider the relativistic theory studiedin chapter IV .1: L=(∂/Phi1 †)(∂/Phi1) −λ(/Phi1†/Phi1−v2)2. It is often convenient to take the λ→∞ limit holding vfixed. In the language used in chapter IV .1 “climbing the wall” costs infinitely more energy than “rolling along the gutter.” The resulting theory is defined by L=(∂/Phi1†)(∂/Phi1) (7) with the constraint /Phi1†/Phi1=v2. This is known as a nonlinear σmodel, about which much more in chapter VI.4. The existence of a Nambu-Goldstone boson is particularly easy to see in the nonlinear σ model. The constraint is solved by /Phi1=veiθ, which when plugged into Lgives L=v2(∂θ)2. There it is: the Nambu-Goldstone boson θ. Let’s repeat this in the nonrelativistic domain. T ake the limit g2→∞ with¯ρheld fixed so that (1) becomes L=iϕ†∂0ϕ−1 2m∂iϕ†∂iϕ (8) with the constraint ϕ†ϕ=¯ρ. But now if we plug the solution of the constraint ϕ=√¯ρeiθ into L(and drop the total derivative −¯ρ∂0θ), we get L=−(¯ρ/2m)(∂iθ)2with the equation of motion ∂2 iθ=0. Oops, what is this? It’s not even a propagating degree of freedom? Where is the Nambu-Goldstone boson? Knowing what I already told you, you are not going to be puzzled by this apparent paradox5for long, but believe me, I have stumped quite a few excellent relativistic minds with this one. The Nambu-Goldstone boson is still there, but as we can see from (5) itspropagation velocity ω/k scales to infinity as gand thus it disappears from the spectrum for any nonzero /vectork. Why is it that we are allowed to go to this “nonlinear” limit in the relativistic case? Because we have Lorentz invariance! The velocity of a linearly dispersing mode, if such amode exists, is guaranteed to be equal to 1. 4R. P. Feynman, Statistical Mechanics . 5This apparent paradox was discussed by A. Zee, “From Semionics to T opological Fluids” in O. J. P. ´Ebolic et al., eds., Particle Physics , p. 415. 286 | V . Field Theory and Collective Phenomena Exercises V .1.1 Verify that the approximation used to reach (3) is consistent. V .1.2 T o confine the superfluid in an external potential W(/vectorx)we would add the term −W(/vectorx)ϕ†(/vectorx,t)ϕ(/vectorx,t) to (1). Derive the corresponding equation of motion for ϕ. The equation, known as the Gross-Pitaevski equation, has been much studied in recent years in connection with the Bose-Einstein condensate. V.2Euclid, Boltzmann, Hawking, and Field Theory at Finite T emperature Statistical mechanics and Euclidean field theory I mentioned in chapter I.2 that to define the path integral more rigorously we should perform a Wick rotation t=−itE. The scalar field theory, instead of being defined by the Minkowskian path integral Z=/integraldisplay Dϕe(i//planckover2pi)/integraltext ddx[1 2(∂ϕ)2−V( ϕ) ](1) is then defined by the Euclidean functional integral Z=/integraldisplay Dϕe−(1//planckover2pi)/integraltext dd Ex[1 2(∂ϕ)2+V( ϕ) ]=/integraldisplay Dϕe−(1//planckover2pi)E(ϕ)(2) where ddx=−idd Ex, with dd Ex≡dtEd(d−1)x.I n( 1 )(∂ϕ)2=(∂ϕ/∂t)2−(/vector∇ϕ)2, while in (2) (∂ϕ)2=(∂ϕ/∂t E)2+(/vector∇ϕ)2: The notation is a tad confusing but I am trying not to introduce too many extraneous symbols. You may or may not find it helpful to think of (/vector∇ϕ)2+V( ϕ) as one unit, untouched by Wick rotation. I have introduced E(ϕ)≡/integraltext dd Ex[1 2(∂ϕ)2+V( ϕ) ], which may naturally be regarded as a static energy functional of the field ϕ(x) . Thus, given a configuration ϕ(x) ind-dimensional space, the more it varies, the less likely it is to contribute to the Euclidean functional integral Z. The Euclidean functional integral (2) may remind you of statistical mechanics. Indeed, Herr Boltzmann taught us that in thermal equilibrium at temperature T=1/β, the probability for a configuration to occur in a classical system or the probability for a state tooccur in a quantum system is just the Boltzmann factor e −βEsuitably normalized, where E is to be interpreted as the energy of the configuration in a classical system or as the energyeigenvalue of the state in a quantum system. In particular, recall the classical statisticalmechanics of an N-particle system for which E(p ,q)=/summationdisplay i1 2mp2 i+V( q 1,q2,... ,qN) 288 | V . Field Theory and Collective Phenomena The partition function is given (up to some overall constant) by Z=/productdisplay i/integraldisplay dpidqie−βE(p ,q) After doing the integrals over pwe are left with the (reduced) partition function Z=/productdisplay i/integraldisplay dqie−βV(q 1,q2,...,qN) Promoting this to a field theory as in chapter I.3, letting i→xandqi→ϕ(x) as before, we see that the partition function of a classical field theory with the static energy functionalE(ϕ) has precisely the form in (2), upon identifying the symbol /planckover2pias the temperature T=1/β. Thus, Euclidean quantum field theory in d-dimensional spacetime ∼Classical statistical mechanics in d-dimensional space(3) Functional integral representation of the quantum partition function More interestingly, we move on to quantum statistical mechanics. The integration over phase space {p,q}is replaced by a trace, that is, a sum over states: Thus the partition function of a quantum mechanical system (say of a single particle to be definite) with theHamiltonian His given by Z=tre−βH=/summationdisplay n/angbracketleftn|e−βH|n/angbracketright In chapter I.2 we worked out the integral representation of /angbracketleftF|e−iHT|I/angbracketright. (You should not confuse the time Twith the temperature Tof course.) Suppose we want an integral representation of the partition function. No need to do any more work! We simply replacethe time Tby−iβ , set|I/angbracketright=|F/angbracketright=|n/angbracketrightand sum over |n/angbracketrightto obtain Z=tre−βH=/integraldisplay PBCDqe−/integraltextβ 0dτL(q)(4) T racing the steps from (I.2.3) to (I.2.5) you can verify that here L(q)=1 2(dq/dτ)2+V( q) is precisely the Lagrangian corresponding to Hin the Euclidean time τ. The integral over τruns from 0 to β. The trace operation sets the initial and final states equal and so the functional integral should be done over all paths q(τ) with the boundary condition q(0)= q(β) . The subscript PBC reminds us of this all important periodic boundary condition. The extension to field theory is immediate. If His the Hamiltonian of a quantum field theory in D-dimensional space [and hence d=(D+1)-dimensional spacetime], then the partition function (4) is Z=tre−βH=/integraldisplay PBCDϕe−/integraltextβ 0dτ/integraltext dDxL(ϕ)(5) V .2. Finite T emperatures | 289 with the integral evaluated over all paths ϕ(/vectorx,τ)such that ϕ(/vectorx,0)=ϕ(/vectorx,β) (6) (Here ϕrepresents all the Bose fields in the theory.) A remarkable result indeed! T o study a field theory at finite temperature all we have to do is rotate it to Euclidean space and impose the boundary condition (6). Thus, Euclidean quantum field theory in (D+1)-dimensional spacetime, 0 ≤τ<β ∼Quantum statistical mechanics in D-dimensional space(7) In the zero temperature limit β→∞ we recover from (5) the standard Wick-rotated quantum field theory over an infinite spacetime, as we should. Surely you would hit it big with mystical types if you were to tell them that temperature is equivalent to cyclic imaginary time. At the arithmetic level this connection comes merelyfrom the fact that the central objects in quantum physics e −iHTand in thermal physics e−βHare formally related by analytic continuation. Some physicists, myself included, feel that there may be something profound here that we have not quite understood. Finite temperature Feynman diagrams If we so desire, we can develop the finite temperature perturbation theory of (5), workingout the Feynman rules and so forth. Everything goes through as before with one majordifference stemming from the condition (6) ϕ(/vectorx,τ=0)=ϕ(/vectorx,τ=β). Clearly, when we Fourier transform with the factor e iωτ, the Euclidean frequency ωcan take on only discrete values ωn≡(2π/β)n , withnan integer. The propagator of the scalar field becomes 1/(k2 4+/vectork2)→1/(ω2 n+/vectork2). Thus, to evaluate the partition function, we simply write the relevant Euclidean Feynman diagrams and instead of integrating over frequency we sumover a discrete set of frequencies ω n=(2πT)n ,n=− ∞ ,... ,+∞ . In other words, after you beat a Feynman integral down to the form/integraltext dd EkF(k2 E), all you have to do is replace it by 2πT/summationtext n/integraltext dDkF[(2πT)2n2+/vectork2]. It is instructive to see what happens in the high-temperature T→∞ limit. In summing overωn, then=0 term dominates since the combination (2πT)2n2+/vectork2occurs in the denominator. Hence, the diagrams are evaluated effectively in D-dimensional space. We lose a dimension! Thus, Euclidean quantum field theory in D-dimensional spacetime ∼High-temperature quantum statistical mechanics in D-dimensional space(8) This is just the statement that at high-temperature quantum statistical mechanics goes classical [compare (3)]. 290 | V . Field Theory and Collective Phenomena An important application of quantum field theory at finite temperature is to cosmology: The early universe may be described as a soup of elementary particles at some hightemperature. Hawking radiation Hawking radiation from black holes is surely the most striking prediction of gravitationalphysics of the last few decades. The notion of black holes goes all the way back to Michelland Laplace, who noted that the escape velocity from a sufficiently massive object mayexceed the speed of light. Classically, things fall into black holes and that’s that. But withquantum physics a black hole can in fact radiate like a black body at a characteristictemperature T. Remarkably, with what little we learned in chapter I.11 and here, we can actually determine the black hole temperature. I hasten to add that a systematic development wouldbe quite involved and fraught with subtleties; indeed, entire books are devoted to thissubject. However, what we need to do is more or less clear. Starting with chapter I.11, wewould have to develop quantum field theory (for instance, that of a scalar field ϕ)in curved spacetime, in particular in the presence of a black hole, and ask what a vacuum state (i.e.,a state devoid of ϕquanta) in the far past evolves into in the far future. We would find a state filled with a thermal distribution of ϕquanta. We will not do this here. In hindsight, people have given numerous heuristic arguments for Hawking radiation. Here is one. Let us look at the Schwarzschild solution (see chapter I.11) ds2=/parenleftbigg 1−2GM r/parenrightbigg dt2−/parenleftbigg 1−2GM r/parenrightbigg−1 dr2−r2dθ2−r2sin2θd φ2(9) At the horizon r=2GM , the coefficients of dt2anddr2change sign, indicating that time and space, and hence energy and momentum, are interchanged. Clearly, somethingstrange must occur. With quantum fluctuations, particle and antiparticle pairs are alwayspopping in and out of the vacuum, but normally, as we had discussed earlier, the uncer-tainty principle limits the amount of time /Delta1tthe pairs can exist to ∼1//Delta1E . Near the black hole horizon, the situation is different. A pair can fluctuate out of the vacuum right at thehorizon, with the particle just outside the horizon and the antiparticle just inside; heuris-tically the Heisenberg restriction on /Delta1t may be evaded since what is meant by energy changes as we cross the horizon. The antiparticle falls in while the particle escapes to spa-tial infinity. Of course, a hand-waving argument like this has to be backed up by detailedcalculations. If black holes do indeed radiate at a definite temperature T, and that is far from obvious a priori, we can estimate Teasily by dimensional analysis. From (9) we see that only the combination GM , which evidently has the dimension of a length, can come in. Since T has the dimension of mass, that is, length inverse, we can only have T∝1/GM . T o determine Tprecisely, we resort to a rather slick argument. I warn you from the outset that the argument will be slick and should be taken with a grain of salt. It is onlymeant to whet your appetite for a more correct treatment. V .2. Finite T emperatures | 291 Imagine quantizing a scalar field theory in the Schwarzschild metric, along the line described in chapter I.11. If upon Wick rotation the field “feels” that time is periodic withperiod β, then according to what we have learned in this chapter the quanta of the scalar field would think that they are living in a heat bath with temperature T=1/β. Setting t→−iτ, we rotate the metric to ds2=−/bracketleftBigg/parenleftbigg 1−2GM r/parenrightbigg dτ2+/parenleftbigg 1−2GM r/parenrightbigg−1 dr2+r2dθ2+r2sin2θdφ2/bracketrightBigg (10) In the region just outside the horizon r>∼2GM , we perform the general coordinate trans- formation (τ,r)→(α,R)so that the first two terms in ds2become R2dα2+dR2, namely the length element squared of flat 2-dimensional Euclidean space in polar coordinates. T o leading order, we can write the Schwarzchild factor (1−2GM/r) as (r−2GM)/(2 GM)≡γ2R2with the constant γto be determined. Then the second term becomes dr2/(γ2R2)=(4GM)2γ2dR2, and thus we set γ=1/(4GM) to get the desired dR2. The first two terms in −ds2are then given by R2(dτ/(4 GM))2+dR2. Thus the Eu- clidean time is related to the polar angle by τ=4GMα and so has a period of 8πGM =β. We obtain thus the Hawking temperature T=1 8πGM=/planckover2pic3 8πGM(11) Restoring /planckover2piby dimensional analysis, we see that Hawking radiation is indeed a quantum effect. It is interesting to note that the Wick rotated geometry just outside the horizon is given by the direct product of a plane with a 2-sphere of radius 2 GM , although, this observation is not needed for the calculation we just did. Exercises V .2.1 Study the free field theory L=1 2(∂ϕ)2−1 2m2ϕ2at finite temperature and derive the Bose-Einstein distribution. V .2.2 It probably does not surprise you that for fermionic fields the periodic boundary condition (6) is replaced by an antiperiodic boundary condition ψ(/vectorx,0)=−ψ(/vectorx,β)in order to reproduce the results of chap- ter II.5. Prove this by looking at the simplest fermionic functional integral. [Hint: The clearest expositionof this satisfying fact may be found in appendix A of R. Dashen, B. Hasslacher, and A. Neveu, Phys. Rev. D12: 2443, 1975.] V .2.3 It is interesting to consider quantum field theory at finite density, as may occur in dense astrophysical objects or in heavy ion collisions. (In the previous chapter we studied a system of bosons at finite densityand zero temperature.) In statistical mechanics we learned to go from the partition function to the grand partition function Z=tre −β(H −μN), where a chemical potential μis introduced for every conserved particle number N. For example, for noninteracting relativistic fermions, the Lagrangian is modified toL=¯ψ(i/negationslash∂−m)ψ+μ¯ψγ0ψ. Note that finite density, as well as finite temperature, breaks Lorentz invariance. Develop the subject of quantum field theory at finite density as far as you can. V.3 Landau-Ginzburg Theory of Critical Phenomena The emergence of nonanalyticity Historically, the notion of spontaneous symmetry breaking, originating in the work of Landau and Ginzburg on second-order phase transitions, came into particle physics fromcondensed matter physics. Consider a ferromagnetic material in thermal equilibrium at temperature T. The mag- netization /vectorM(x) is defined as the average of the atomic magnetic moments taken over a region of a size much larger than the length scale characteristic of the relevant micro-scopic physics. (In this chapter, we are discussing a nonrelativistic theory and xdenotes the spatial coordinates only.) We know that at low temperatures, rotational invariance is spon-taneously broken and that the material exhibits a bulk magnetization pointing in somedirection. As the temperature is raised past some critical temperature T cthe bulk mag- netization suddenly disappears. We understand that with increased thermal agitation theatomic magnetic moments point in increasingly random directions, canceling each otherout. More precisely, it was found experimentally that just below T cthe magnetization |/vectorM| vanishes as ∼(Tc−T)β, where the so-called critical exponent β/similarequal0.37. This sudden change is known as a second order phase transition, an example of a critical phenomenon. Historically, critical phenomena presented a challenge to theoreticalphysicists. In principle, we are to compute the partition function Z=tre −H/Twith the microscopic Hamiltonian H, but Zis apparently smooth in Texcept possibly at T=0. Some physicists went as far as saying that nonanalytic behavior such as (Tc− T)βis impossible and that within experimental error |/vectorM|actually vanishes as a smooth function of T. Part of the importance of Onsager’s famous exact solution in 1944 of the 2- dimensional Ising model is that it settled this question definitively. The secret is that aninfinite sum of terms each of which may be analytic in some variable need not be analyticin that variable. The trace in tr e −H/Tsums over an infinite number of terms. V .3. Theory of Critical Phenomena | 293 Arguing from symmetry In most situations, it is essentially impossible to calculate Zstarting with the microscopic Hamiltonian. Landau and Ginzburg had the brilliant insight that the form of the freeenergy Gas a function of /vectorMfor a system with volume Vcould be argued from general principles. First, for /vectorMconstant in x, we have by rotational invariance G=V[a/vectorM2+b(/vectorM2)2+...] (1) where a,b,... are unknown (but expected to be smooth) functions of T. Landau and Ginzburg supposed that avanishes at some temperature Tc. Unless there is some special reason, we would expect that for TnearTcwe have a=a1(T−Tc)+...[rather than, say, a=a2(T−Tc)2+...]. But you already learned in chapter IV .1 what would happen. For T> T c,Gis minimized at /vectorM=0, but as Tdrops below Tc, new minima suddenly develop at|/vectorM|=/radicalbig (−a/ 2b)∼(Tc−T)1 2. Rotational symmetry is spontaneously broken, and the mysterious nonanalytic behavior pops out easily. T o include the possibility of /vectorMvarying in space, Landau and Ginzburg argued that G must have the form G=/integraldisplay d3x{∂i/vectorM∂i/vectorM+a/vectorM2+b(/vectorM2)2+...} (2) where the coefficient of the ( ∂i/vectorM)2term has been set to 1 by rescaling /vectorM. You would recognize (2) as the Euclidean version of the scalar field theory we have been studying. Bydimensional analysis we see that 1 /√ asets the length scale. More precisely, for T> Tc, let us turn on a perturbing external magnetic field /vectorH(x) by adding the term −/vectorH./vectorM. Assuming /vectorMsmall and minimizing Gwe obtain (−∂2+a)/vectorM/similarequal/vectorH, with the solution /vectorM(x) =/integraldisplay d3y/integraldisplayd3k (2π)3ei/vectork.(/vectorx−/vectory) /vectork2+a/vectorH(y) =/integraldisplay d3y1 4π|/vectorx−/vectory|e−√a|/vectorx−/vectory|/vectorH(y) (3) [Recall that we did the integral in (I.4.7)—admire the unity of physics!] It is standard to define a correlation function </vectorM(x) /vectorM(0)> by asking what the mag- netization /vectorM(x) will be if we use a magnetic field sharply localized at the origin to create a magnetization /vectorM(0)there. We expect the correlation function to die off as e−|/vectorx|/ξover some correlation length ξthat goes to infinity as Tapproaches Tcfrom above. The critical exponent νis traditionally defined by ξ∼1/(T−Tc)ν. In Landau-Ginzburg theory, also known as mean field theory, we obtain ξ=1/√aand hence ν=1 2. The important point is not how well the predicted critical exponents such as βandν agree with experiment but how easily they emerge from Landau-Ginzburg theory. The theory provides a starting point for a complete theory of critical phenomena, which waseventually developed by Kadanoff, Fisher, Wilson, and others using the renormalizationgroup (to be discussed in chapter VI.8). 294 | V . Field Theory and Collective Phenomena The story goes that Landau had a logarithmic scale with which he ranked theoretical physicists, with Einstein on top, and that after working out Landau-Ginzburg theory hemoved himself up by half a notch. Exercise V .3.1 Another important critical exponent γis defined by saying that the susceptibility χ≡(∂M/∂H) |H=0 diverges ∼1/|T−Tc|γasTapproaches Tc. Determine γin Landau-Ginzburg theory. [Hint: Instructively, there are two ways of doing it: (a) Add −/vectorH./vectorMto (1) for /vectorMand/vectorHconstant in space and solve for /vectorM(/vectorH). (b) Calculate the susceptibility function χij(x−y)≡[∂Mi(x)/∂H j(y)]|H=0and integrate over space.] V.4 Superconductivity Pairing and condensation When certain materials are cooled below a certain critical temperature Tc, they suddenly be- come superconducting. Historically, physicists had long suspected that the superconduct-ing transition, just like the superfluid transition, has something to do with Bose-Einsteincondensation. But electrons are fermions, not bosons, and thus they first have to pairinto bosons, which then condense. We now know that this general picture is substantiallycorrect: Electrons form Cooper pairs, whose condensation is responsible for superconduc-tivity. With brilliant insight, Landau and Ginzburg realized that without having to know the detailed mechanism driving the pairing of electrons into bosons, they could understanda great deal about superconductivity by studying the field ϕ(x) associated with these con- densing bosons. In analogy with the ferromagnetic transition in which the magnetization /vectorM(x) in a ferromagnet suddenly changes from zero to a nonzero value when the temper- ature drops below some critical temperature, they proposed that ϕ(x) becomes nonzero for temperatures below T c. (In this chapter xdenotes spatial coordinates only.) In statisti- cal physics, quantities such as /vectorM(x) andϕ(x) that change through a phase transition are known as order parameters. The field ϕ(x) carries two units of electric charge and is therefore complex. The dis- cussion now unfolds much as in chapter V .3 except that ∂iϕshould be replaced by Diϕ≡ (∂i−i2eAi)ϕsinceϕis charged. Following Landau and Ginzburg and including the energy of the external magnetic field, we write the free energy as F=1 4F2 ij+|Diϕ|2+a|ϕ|2+b 2|ϕ|4+... (1) which is clearly invariant under the U(1)gauge transformation ϕ→ei2e/Lambda1ϕandAi→Ai+ ∂i/Lambda1. As before, setting the coefficient of |Diϕ|2equal to 1 just amounts to a normalization choice for ϕ. The similarity between (1) and (IV .6.1) should be evident. 296 | V . Field Theory and Collective Phenomena Meissner effect A hallmark of superconductivity is the Meissner effect, in which an external magnetic field/vectorBpermeating the material is expelled from it as the temperature drops below Tc. This indicates that a constant magnetic field inside the material is not favored energetically. Theeffective laws of electromagnetism in the material must somehow change at T c. Normally, a constant magnetic field would cost an energy of the order ∼/vectorB2V, where Vis the volume of the material. Suppose that the energy density is changed from the standard /vectorB2to/vectorA2 (where as usual /vector∇×/vectorA=/vectorB). For a constant magnetic field /vectorB,/vectorAgrows as the distance and hence the total energy would grow faster than V. After the material goes superconducting, we have to pay an unacceptably large amount of extra energy to maintain the constantmagnetic field and so it is more favorable to expel the magnetic field. Note that a term like /vectorA 2in the effective energy density preserves rotational and transla- tional invariance but violates electromagnetic gauge invariance. But we already know howto break gauge invariance from chapter IV .6. Indeed, the U(1)gauge theory described there and the theory of superconductivity described here are essentially the same, related by aWick rotation. As in chapter V .3 we suppose that for temperature T/similarequalT c,a/similarequala1(T−Tc)while b remains positive. The free energy Fis minimized by ϕ=0 above Tc, and by |ϕ|=√−a/b ≡vbelow Tc. All this is old hat to you, who have learned that upon symmetry breaking in a gauge theory the gauge field gains a mass. We simply read off from (1) that F=1 4F2 ij+(2ev)2A2 i+... (2) which is precisely what we need to explain the Meissner effect. London penetration length and coherence length Physically, the magnetic field does not drop precipitously from some nonzero value outside the superconductor to zero inside, but drops over some characteristic length scale, calledthe London penetration length. The magnetic field leaks into the superconductor a bit overa length scale l, determined by the competition between the energy in the magnetic field F 2 ij∼(∂A)2∼A2/l2and the Meissner term (2ev)2A2in (2). Thus, Landau and Ginzburg obtained the London penetration length lL∼(1/ev)=(1/e)√b/−a. Similarly, the characteristic length scale over which the order parameter ϕvaries is known as the coherence length lϕ, which can be estimated by balancing the second and third terms in (1), roughly (∂ϕ)2∼ϕ2/l2 ϕandaϕ2against each other, giving a coherence length of order lϕ∼1/√−a. Putting things together, we have lL lϕ∼√ b e(3) V .4. Superconductivity | 297 You might recognize from chapter IV .6 that this is just the ratio of the mass of the scalar field to the mass of the vector field. As I remarked earlier, the concept of spontaneous symmetry breaking went from con- densed matter physics to particle physics. After hearing a talk at the University of Chicagoon the Bardeen-Cooper-Schrieffer theory of superconductivity by the young Schrieffer,Nambu played an influential role in bringing spontaneous symmetry breaking to the par-ticle physics community. Exercises V .4.1 Vary (1) to obtain the equation for Aand determine the London penetration length more carefully. V .4.2 Determine the coherence length more carefully. V.5 Peierls Instability Noninteracting hopping electrons The appearance of the Dirac equation and a relativistic field theory in a solid would be surprising indeed, but yes, it is possible. Consider the Hamiltonian H=−t/summationdisplay j(c† j+1cj+c† jcj+1) (1) describing noninteracting electrons hopping on a 1-dimensional lattice (figure V .5.1). Here cjannihilates an electron on site j. Thus, the first term describes an electron hopping from sitejto site j+1 with amplitude t. We have suppressed the spin labels. This is just about the simplest solid state model; a good place to read about it is in Feynman’s “Freshmanlectures.” Fourier transforming c j=/summationtext keikajc(k) (where ais the spacing between sites), we immediately find the energy spectrum ε(k)=− 2tcoska(fig. V .5.2). Imposing a periodic boundary condition on a lattice with Nsites, we have k=(2π/Na)n withnan integer from−1 2Nto1 2N.A sN→∞ ,kbecomes a continuous rather than a discrete variable. As usual, the Brillouin zone is defined by −π/a < k ≤π/a . There is absolutely nothing relativistic about any of this. Indeed, at the bottom of the spectrum the energy (up to an irrelevant additive constant) goes as ε(k)/similarequal2t1 2(ka)2≡ k2/2meff. The electron disperses nonrelativistically with an effective mass meff. jj + 1 j − 1 Figure V .5.1 V .5. Peierls Instability | 299 ε(k) εF kπ a_ +π a Figure V .5.2 But now let us fill the system with electrons up to some Fermi energy εF(see fig. V .5.2). Focus on an electron near the Fermi surface and measure its energy from εF and momentum from +kF. Suppose we are interested in electrons with energy small compared to εF, that is, E≡ε−εF/lessmuchεF, and momentum small compared to kF, that is, p≡k−kF/lessmuchkF. These electrons obey a linear energy-momentum dispersion E=vFp with the Fermi velocity vF=(∂ε/∂k)| k=kF. We will call the field associated with these electrons ψR, where the subscript indicates that they are “right moving.” It satisfies the equation of motion (∂/∂t+vF∂/∂x)ψR=0. Similarly, the electrons with momentum around −kFobey the dispersion E=−vFp. We will call the field associated with these electrons ψLwithLfor “left moving,” satisfying (∂/∂t−vF∂/∂x)ψL=0. Emergence of the Dirac equation The Lagrangian summarizing all this is simply L=iψ† R/parenleftbigg∂ ∂t+vF∂ ∂x/parenrightbigg ψR+iψ† L/parenleftbigg∂ ∂t−vF∂ ∂x/parenrightbigg ψL (2) Introducing a 2-component field ψ=/parenleftBig ψL ψR/parenrightBig ,¯ψ≡ψ†γ0≡ψ†σ2, and choosing units so thatvF=1, we may write Lmore compactly as L=iψ†/parenleftbigg∂ ∂t−σ3∂ ∂x/parenrightbigg ψ=¯ψiγμ∂μψ (3) withγ0=σ2andγ1=iσ1satisfying the Clifford algebra {γμ,γν}=2gμν. 300 | V . Field Theory and Collective Phenomena Amazingly enough, the (1+1)-dimensional Dirac Lagrangian emerges in a totally nonrelativistic situation! An instability I will now go on to discuss an important phenomenon known as Peierls’s instability. I willnecessarily have to be a bit sketchy. I don’t have to tell you again that this is not a text onsolid state physics, but in any case you will not find it difficult to fill in the gaps. Peierls considered a distortion in the lattice, with the ion at site jdisplaced from its equilibrium position by cos[ q(ja) ]. (Shades of our mattress from chapter I.1!) A lattice distortion with wave vector q=2k Fwill connect electrons with momentum kFwith electrons with momentum −kF. In other words, it connects right moving ones with left moving electrons, or in our field theoretic language ψRwithψL. Since the right moving electrons and the left moving electrons on the surface of the Fermi sea have thesame energy (namely ε F, duh!) we have the always interesting situation of degenerate perturbation theory:/parenleftBig εF 0 0εF/parenrightBig +/parenleftBig 0δ δ0/parenrightBig with eigenvalues εF±δ. A gap opens at the surface of the Fermi sea. Here δrepresents the perturbation. Thus, Peierls concluded that the spectrum changes drastically and the system is unstable under a perturbation with wavevector 2 k F. A particularly interesting situation occurs when the system is half filled with electrons (so that the density is one electron per site—recall that electrons have up and down spin).In other words, k F=π/2aand thus 2 kF=π/a . A lattice distortion of the form shown in figure V .5.3 has precisely this wave vector. Peierls showed that a half-filled system wouldwant to distort the lattice in this way, doubling the unit cell. It is instructive to see how thisphysical phenomenon emerges in a field theoretic formulation. Denote the displacement of the ion at site jbyd j. In the continuum limit, we should be able to replace djby a scalar field. Show that a perturbation connecting ψRandψLcouples to¯ψψ and¯ψγ5ψ, and that a linear combination ¯ψψ and¯ψγ5ψcan always be rotated to ¯ψψ by a chiral transformation (see exercise V .5.1.) Thus, we extend (3) to L=¯ψiγμ∂μψ+1 2[(∂tϕ)2−v2(∂xϕ)2]−1 2μ2ϕ2+gϕ¯ψψ+... (4) Remember that you worked out the effective potential Veff(ϕ) of this (1+1)-dimensional field theory in exercise IV .3.2: Veff(ϕ) goes as ϕ2logϕ2for small ϕ, which overwhelms the 1 2μ2ϕ2term. Thus, the symmetry ϕ→−ϕis dynamically broken. The field ϕacquires a vacuum expectation value and ψbecomes massive. In other words, the electron spectrum develops a gap. Figure V .5.3 V .5. Peierls Instability | 301 Exercise V .5.1 Parallel to the discussion in chapter II.1 you can see easily that the space of 2 by 2 matrices is spanned by the four matrices I,γμ, andγ5≡γ0γ1=σ3. (Note the peculiar but standard notation of γ5.)Convince yourself that1 2(I±γ5)projects out right- and left handed fields just as in (3 +1)-dimensional spacetime. Show that in the bilinear ¯ψγμψleft handed fields are connected to left handed fields and right handed to right handed and that in the scalar ¯ψψ and the pseudoscalar ¯ψγ5ψright handed is connected to left handed and vice versa. Finally, note that under the transformation ψ→eiθγ5ψthe scalar and the pseudoscalar rotate into each other. Check that this transformation leaves the massless Dirac Lagrangian(3) invariant. V.6 Solitons Breaking the shackles of Feynman diagrams When I teach quantum field theory I like to tell the students that by the mid-1970s field theorists were breaking the shackles of Feynman diagrams. A bit melodramatic, yes,but by that time Feynman diagrams, because of their spectacular successes in quantumelectrodynamics, were dominating the thinking of many field theorists, perhaps to excess.As a student I was even told that Feynman diagrams define quantum field theory, thatquantum fields were merely the “slices of venison” 1used to derive the Feynman rules, and should be discarded once the rules were obtained. The prevailing view was that it barelymade sense to write down ϕ(x) . This view was forever shattered with the discovery of topological solitons, as we will now discuss. Small oscillations versus lumps Consider once again our favorite toy model L=1 2(∂ϕ)2−V( ϕ) with the infamous double- well potential V( ϕ)=(λ/4)(ϕ2−v2)2in(1+1)-dimensional spacetime. In chapter IV .1 we learned that of the two vacua ϕ=±vwe are to pick one and study small oscillations around it. So, pick one and write ϕ=v+χ, expand Linχ, and study the dynamics of theχmeson with mass μ=(λv2)1 2. Physics then consists of suitably quantized waves oscillating about the vacuum v. But that is not the whole story. We can also have a time independent field configuration withϕ(x) (in this and the next chapter xwill denote only space unless it is clearly meant to be otherwise from the context) taking on the value −v asx→− ∞ and+v asx→+ ∞ , and changing from −v to+v around some point x0over some length scale las shown in 1Gell-Mann used to speak about how pheasant meat is cooked in France between two slices of venison which are then discarded. He forcefully advocated a program to extract and study the algebraic structure of quantum field theories which are then discarded. V .6. Solitons | 303 (a) (b)+v −vϕ x x0 ε x x0 Figure V .6.1 figure V .6.1a. [Note that if we consider the Euclidean version of the field theory, identify the time coordinate as the ycoordinate, and think of ϕ(x ,y)as the magnetization (as in chapter V .3), then the configuration here describes a “domain wall” in a 2-dimensionalmagnetic system.] Think about the energy per unit length ε(x)=1 2/parenleftbiggdϕ dx/parenrightbigg2 +λ 4(ϕ2−v2)2(1) for this configuration, which I plot in figure V .6.1b. Far away from x0we are in one of the two vacua and there is no energy density. Near x0, the two terms in ε(x) both contribute to the energy or mass M=/integraltext dx ε(x) : the “spatial variation” (in a slight abuse of terminology often called the “kinetic energy”) term/integraltext dx1 2(dϕ/dx)2∼l(v/l)2∼v2/l, and the “potential energy” term/integraltext dxλ(ϕ2−v2)2∼lλv4. T o minimize the total energy the spatial variation term wants lto be large, while the potential term wants lto be small. The competition dM/dl =0 gives v2/l∼lλv4, thus fixing l∼(λv2)−1 2∼1/μ. The mass comes out to be ∼μv2∼μ(μ2/λ). We have a lump of energy spread over a region of length lof the order of the Compton wavelength of the χmeson. By translation invariance, the center of the lump x0can be anywhere. Furthermore, since the theory is Lorentz invariant, we can always boost to 304 | V . Field Theory and Collective Phenomena send the lump moving at any velocity we like. Recalling a famous retort in the annals of American politics (“It walks like a duck, quacks like a duck, so Mr. Senator, why don’tyou want to call it a duck?”) we have here a particle, known as a kink or a soliton, withmass ∼μ(μ 2/λ) and size ∼l. Perhaps because of the way the soliton was discovered, many physicists think of it as a big lumbering object, but as we have seen, the size of asoliton l∼1/μcan be made as small as we like by increasing μ. So a soliton could look like a point particle. We will come back to this point in chapter VI.3 when we discuss duality. T opological stability While the kink and the meson are the same size, for small λthe kink is much more massive than the meson. Nevertheless, the kink cannot decay into mesons because it costsan infinite amount of energy to undo the kink [by “lifting” ϕ(x) over the potential energy barrier to change it from +v to−v forxfrom some point >∼x 0to+∞, for example]. The kink is said to be topologically stable. The stability is formally guaranteed by the conserved current Jμ=1 2vεμν∂νϕ (2) with the charge Q=/integraldisplay+∞ −∞dxJ0(x)=1 2v[ϕ(+∞)−ϕ(−∞)] Mesons, which are small localized packets of oscillations in the field clearly have Q=0, while the kink has Q=1. Thus, the kink cannot decay into a bunch of mesons. Incidentally, the charge density J0=(1/2v)(dϕ/dx) is concentrated at x0where ϕchanges most rapidly, as you would expect. Note that ∂μJμ=0 follows immediately from the antisymmetric symbol εμνand does not depend on the equation of motion. The current Jμis known as a “topological current.” Its existence does not follow from Noether’s theorem (chapter I.10) but from topology. Our discussion also makes clear the existence of an antikink with Q=− 1 and described by a configuration with ϕ(−∞)=+vandϕ(+∞)=−v. The name is justified by consider- ing the configuration pictured in figure V .6.2 containing a kink and an antikink far apart.As the kink and the antikink move closer to each other, they clearly can annihilate intomesons, since the configuration shown in figure V .6.2 and the vacuum configuration withϕ(x)=+veverywhere are separated by a finite amount of energy. A nonperturbative phenomenon That the mass of the kink comes out inversely proportional to the coupling λis a clear sign that field theorists could have done perturbation theory in λtill they were blue in the face without ever discovering the kink. Feynman diagrams could not have told us about it. V .6. Solitons | 305 Antikink Kink Figure V .6.2 You can calculate the mass of a kink by minimizing M=/integraldisplay dx/bracketleftBigg 1 2/parenleftbiggdϕ dx/parenrightbigg2 +λ 4/parenleftBig ϕ2−v2/parenrightBig2/bracketrightBigg =/parenleftbiggμ2 λ/parenrightbigg μ/integraldisplay dy/bracketleftBigg 1 2/parenleftbiggdf dy/parenrightbigg2 +1 4/parenleftBig f2−1/parenrightBig2/bracketrightBigg where in the last step we performed the obvious scaling ϕ(x)→vf (y) andy=μx. This scaling argument immediately showed that the mass of the kink M=a(μ2/λ)μ with aa pure number: The heuristic estimate of the mass proved to be highly trustworthy. The actual function ϕ(x) and hence acan be computed straightforwardly with standard variational methods. Bogomol’nyi inequality More cleverly, observe that the energy density (1) is the sum of two squares. Usinga 2+b2≥2|ab|we obtain M≥/integraldisplay dx/parenleftbiggλ 2/parenrightbigg1 2/vextendsingle/vextendsingle/vextendsingle/vextendsingle/parenleftbiggdϕ dx/parenrightbigg/parenleftBig ϕ2−v2/parenrightBig/vextendsingle/vextendsingle/vextendsingle/vextendsingle≥/parenleftbiggλ 2/parenrightbigg1 2/vextendsingle/vextendsingle/vextendsingle/vextendsingle/vextendsingle/bracketleftbigg1 3ϕ3−v2ϕ2/bracketrightbigg+∞ −∞/vextendsingle/vextendsingle/vextendsingle/vextendsingle/vextendsingle=/vextendsingle/vextendsingle/vextendsingle/vextendsingle4 3√ 2μ/parenleftbiggμ2 λ/parenrightbigg Q/vextendsingle/vextendsingle/vextendsingle/vextendsingle We have the elegant result M≥|Q| (3) with mass Mmeasured in units of (4/3√ 2)μ(μ2/λ). This is an example of a Bogomol’nyi inequality, which plays an important role in string theory. Exercises V .6.1 Show that if ϕ(x) is a solution of the equation of motion, then so is ϕ[(x−vt)/√ 1−v2]. V .6.2 Discuss the solitons in the so-called sine-Gordon theory L=1 2(∂ϕ)2−gcos(βϕ) . Find the topological current. Is the Q=2 soliton stable or not? V .6.3 Compute the mass of the kink by the brute force method and check the result from the Bogomol’nyi inequality. V.7 Vortices, Monopoles, and Instantons Vortices The kink is merely the simplest example of a large class of topological objects in quantum field theory. Consider the theory of a complex scalar field in (2+1)-dimensional spacetime L= ∂ϕ†∂ϕ−λ(ϕ†ϕ−v2)2with the now familiar Mexican hat potential. With some minor changes in notation, this is the theory we used to describe interacting bosons and su-perfluids. (We choose to study the relativistic rather than the nonrelativistic version but asyou will see the issue does not enter for the questions I want to discuss here.) Are there solitons, that is, objects like the kink, in this theory?Given some time-independent configuration ϕ(x) let us look at its mass or energy M=/integraldisplay d2x[∂iϕ†∂iϕ+λ(ϕ†ϕ−v2)2]. (1) The integrand is a sum of two squares, each of which must give a finite contribution. In particular, for the contribution of the second term to be finite the magnitude of ϕmust approach vat spatial infinity. This finite energy requirement does not fix the phase of ϕhowever. Using polar coordi- nates(r,θ)we will consider the Ansatz ϕ−→ r→∞veiθ. Writing ϕ=ϕ1+iϕ2, we see that the vector (ϕ1,ϕ2)=v(cosθ, sinθ)points radially outward at infinity. Recall the definition of the current Ji=i(∂iϕ†ϕ−ϕ†∂iϕ)in a bosonic fluid given in chapter III.5. The flow whirls about at spatial infinity, and thus this configuration is known as the vortex. By explicit differentiation or dimensional analysis, we have ∂iϕ∼v(1/r) asr→∞ .N o w look at the first term in M. Oops, the energy diverges logarithmically as v2/integraltext d2x(1/r2). Is there a way out? Not unless we change the theory. V .7. Monopoles and Instantons | 307 Vortex into flux tubes Suppose we gauge the theory by replacing ∂iϕbyDiϕ=∂iϕ−ieAiϕ. Now we can achieve finite energy by requiring that the two terms in Diϕknock each other out so that Diϕ−→r→∞0 faster than 1 /r. In other words, Ai−→ r→∞−(i/e)(1 /|ϕ|2)ϕ†∂iϕ=(1/e)∂iθ. Immediately, we have Flux≡/integraldisplay d2xF12=/contintegraldisplay CdxiAi=2π e(2) where Cis an infinitely large circle at spatial infinity and we have used Stokes’ theorem. Thus, in a gauged U(1)theory the vortex carries a magnetic flux inversely proportional to the charge. When I say magnetic, I am presuming that Arepresents the electromagnetic gauge potential. The vortex discussed here appears as a flux tube in so-called type IIsuperconductors. It is worth remarking that this fundamental unit of flux (2) is normallywritten in the condensed matter physics literature in unnatural units as /Phi10=hc e(3) very pleasingly uniting three fundamental constants of Nature. Homotopy groups Since spatial infinity in 2 dimensional space is topologically a unit circle S1and since the field configuration with |ϕ|=v also forms a circle S1, this boundary condition can be characterized as a map S1→S1. Since this map cannot be smoothly deformed into the trivial map in which S1is mapped onto a point in S1, the corresponding field configuration is indeed topologically stable. (Think of wrapping a loop of string around a ring.) Mathematically, maps of Sninto a manifold Mare classified by the homotopy group /Pi1n(M), which counts the number of topologically inequivalent maps. You can look up the homotopy groups for various manifolds in tables.1In particular, for n≥1,/Pi1n(Sn)=Z, where Zis the mathematical notation for the set of all integers. The simplest example /Pi11(S1)=Zis proved almost immediately by exhibiting the maps ϕ−→r→∞veimθ, with m any integer (positive or negative), using the context and notation of our discussion for convenience. Clearly, this map wraps one circle around the other mtimes. The language of homotopy groups is not just to impress people, but gives us a unifying language to discuss topological solitons. Indeed, looking back you can now see that thekink is a physical manifestation of /Pi1 0(S0)=Z2, where Z2denotes the multiplicative group consisting of {+1,−1}(since the 0-dimensional sphere S0={ + 1,−1}consists of just two points and is topologically equivalent to the spatial infinity in 1-dimensional space). 1See tables 6.V and 6.VI, S. Iyanaga and Y . Kawada, eds., Encyclopedic Dictionary of Mathematics , p. 1415. 308 | V . Field Theory and Collective Phenomena Hedgehogs and monopoles If you absorbed all this, you are ready to move up to (3 +1)-dimensional spacetime. Spatial infinity is now topologically S2. By now you realize that if the scalar field lives on the manifold M, then we have at infinity the map of S2→M. The simplest choice is thus to takeS2forM. Hence, we are led to scalar fields ϕa(a=1, 2, 3)transforming as a vector /vectorϕ under an internal symmetry group O(3)and governed by L=1 2∂/vectorϕ.∂/vectorϕ−V(/vectorϕ./vectorϕ). (There should be no confusion in using the arrow to indicate a vector in the internal symmetrygroup.) Let us choose V=λ(/vectorϕ 2−v2)2. The story unfolds much as the story of the vortex. The requirement that the mass of a time independent configuration M=/integraldisplay d3x[1 2(∂/vectorϕ)2+λ(/vectorϕ2−v2)2] (4) be finite forces |/vectorϕ|=v at spatial infinity so that /vectorϕ(r=∞)indeed lives on S2. The identity map S2→S2indicates that we should consider a configuration such that ϕa−→ r→∞vxa r(5) This equation looks a bit strange at first sight since it mixes the index of the internal symmetry group with the index of the spatial coordinates (but in fact we have alreadyencountered this phenomenon in the vortex). At spatial infinity, the field /vectorϕis pointing radially outward, so this configuration is known picturesquely as a hedgehog. Draw apicture if you don’t get it! As in the vortex story, the requirement that the first term in (4) be finite forces us to introduce an O(3)gauge potential A b μso that we can replace the ordinary derivative ∂iϕa by the covariant derivative Diϕa=∂iϕa+eεabcAb iϕc. We can then arrange Diϕato vanish at infinity. Simple arithmetic shows that with (5) the gauge potential has to go as Ab i−→r→∞1 eεbijxj r2(6) Imagine yourself in a lab at spatial infinity. Inside a small enough lab, the /vectorϕfield at different points are all pointing in approximately the same direction. The gauge groupO(3)is broken down to O(2)/similarequalU(1). The experimentalists in this lab observe a massless gauge field associated with the U(1), which they might as well call the electromagnetic field (“quacks like a duck”). Indeed, the gauge invariant tensor field Fμν≡Fa μνϕa |ϕ|−εabcϕa(Dμϕ)b(Dνϕ)c e|ϕ|3(7) can be identified as the electromagnetic field (see exercise V .7.5). There is no electric field since the configuration is time independent and Ab 0=0. We can only have a magnetic field /vectorBwhich you can immediately calculate since you know Ab i, but by symmetry we already see that /vectorBcan point only in the radial direction. This is the fabled magnetic monopole first postulated by Dirac! V .7. Monopoles and Instantons | 309 The presence of magnetic monopoles in spontaneously broken gauge theory was dis- covered by ’t Hooft and Polyakov. If you calculate the total magnetic flux coming out of themonopole/integraltext d/vectorS./vectorB, where as usual d/vectorSdenotes a small surface element at infinity pointing radially outward, you will find that it is quantized in suitable units, exactly as Dirac hadstated, as it must (recall chapter IV .4). We can once again write a Bogomol’nyi inequality for the mass of the monopole M=/integraldisplay d3x/bracketleftBig 1 4(/vectorFij)2+1 2(Di/vectorϕ)2+V(/vectorϕ)/bracketrightBig (8) [/vectorFijtransforms as a vector under O(3); recall (IV .5.17).] Observe that 1 4(/vectorFij)2+1 2(Di/vectorϕ)2=1 4(/vectorFij±εijkDk/vectorϕ)2∓1 2εijk/vectorFij.Dk/vectorϕ Thus M≥/integraldisplay d3x/bracketleftBig ∓1 2εijk/vectorFij.Dk/vectorϕ+V(/vectorϕ)/bracketrightBig (9) We next note that /integraldisplay d3x1 2εijk/vectorFij.Dk/vectorϕ=/integraldisplay d3x1 2εijk∂k(/vectorFij./vectorϕ)=v/integraldisplay d/vectorS./vectorB=4πvg has an elegant interpretation in terms of the magnetic charge gof the monopole. Further- more, if we can throw away V(/vectorϕ)while keeping |/vectorϕ|− →r→∞v, then the inequality M≥4πv|g| is saturated by /vectorFij=±εijkDk/vectorϕ. The solutions of this equation are known as Bogomol’nyi- Prasad-Sommerfeld or BPS states. It is not difficult to construct an electrically charged magnetic monopole, known as a dyon. We simply take Ab 0=(xb/r)f (r) with some suitable function f( r) . One nice feature of the topological monopole is that its mass comes out to be ∼MW/α∼ 137MW(exercise V .7.11), where MWdenotes the mass of the intermediate vector boson of the weak interaction. We are anticipating chapter VII.2 a bit in that the gauge boson thatbecomes massive by the Anderson-Higgs mechanism of chapter IV .6 may be identifiedwith the intermediate vector boson. This explains naturally why the monopole has not yetbeen discovered. Instanton Consider a nonabelian gauge theory, and rotate the path integral to 4-dimensional Eu- clidean space. We might wish to evaluate Z=/integraltext DAe−S(A)in the steepest descent approx- imation, in which case we would have to find the extrema of S(A)=/integraldisplay d4x1 2g2trFμνFμν with finite action. This implies that at infinity |x|=∞ ,Fμνmust vanish faster than 1 /|x|2, and so the gauge potential Aμmust be a pure gauge: A=gdg†forgan element of the gauge group [see (IV .5.6)]. Configurations for which this is true are known as instantons. 310 | V . Field Theory and Collective Phenomena We see that the instanton is yet one more link in the “great chain of being”: kink-vortex- monopole-instanton. Choose the gauge group SU( 2)to be definite. In the parametrization g=x4+i/vectorx./vectorσ we have by definition g†g=1 and det g=1, thus implying x2 4+/vectorx2=1. We learn that the group manifold of SU( 2)isS3. Thus, in an instanton, the gauge potential at infinity A−→ |x|→∞gdg†+O(1/|x|2)defines a map S3→S3. Sound familiar? Indeed, you’ve already seenS0→S0,S1→S1,S2→S2playing a role in field theory. Recall from chapter IV .5 that tr F2=dtr(AdA +2 3A3). Thus /integraldisplay trF2=/integraldisplay S3tr(AdA +2 3A3)=/integraldisplay S3tr(AF−1 3A3)=−1 3/integraldisplay S3tr(gdg†)3(10) where we used the fact that Fvanishes at infinity. This shows explicitly that/integraltext trF2 depends only on the homotopy of the map S3→S3defined by gand is thus a topological quantity. Incidentally,/integraltext S3tr(gdg†)3is known to mathematicians as the Pontryagin index (see exercise V .7.12). I mentioned in chapter IV .7 that the chiral anomaly is not affected by higher-order quantum fluctuations. You are now in position to give an elegant topological proof of thisfact (exercise V .7.13). Kosterlitz-Thouless transition We were a bit hasty in dismissing the vortex in the nongauged theory L=∂ϕ†∂ϕ−λ(ϕ†ϕ− v2)2in(2+1)-dimensional spacetime. Around a vortex ϕ∼veiθand so it is true, as we have noted, that the energy of a single vortex diverges logarithmically. But what about avortex paired with an antivortex? Picture a vortex and an antivortex separated by a distance Rlarge compared to the distance scales in the theory. Around an antivortex ϕ∼ve −iθ. The field ϕwinds around the vortex one way and around the antivortex the other way. Convince yourself by drawinga picture that at spatial infinity ϕdoes not wind at all: It just goes to a fixed value. The winding one way cancels the winding the other way. Thus, a configuration consisting of a vortex-antivortex pair does not cost infinite energy. But it does cost a finite amount of energy: In the region between the vortex and theantivortex ϕis winding around, in fact roughly twice as fast (as you can see by drawing a picture). A rough estimate of the energy is thus v 2/integraltext d2x(1/r2)∼v2log(R/a) , where we integrated over a region of size R, the relevant physical scale in the problem. (T o make sense of the problem we divide Rby the size aof the vortex.) The vortex and the antivortex attract each other with a logarithmic potential. In other words, the configuration cannotbe static: The vortex and the antivortex want to get together and annihilate each other ina fiery embrace and release that finite amount of energy v 2log(R/a). (Hence the term antivortex.) All of this is at zero temperature, but in condensed matter physics we are interested in the free energy F=E−TS (withSthe entropy) at some temperature Trather than the V .7. Monopoles and Instantons | 311 energy E. Appreciating this elementary point, Kosterlitz and Thouless discovered a phase transition as the temperature is raised. Consider a gas of vortices and antivortices at somenonzero temperature. Because of thermal agitation, the vortices and antivortices movingaround may or may not find each other to annihilate. How high do we have to crank upthe temperature for this to happen? Let us do a heuristic estimate. Consider a single vortex. Herr Boltzmann tells us that the entropy is the logarithm of the “number” of ways in which we can put the vortex insidea box of size L(which we will let tend to infinity). Thus, S∼log(L/a) . The entropy Sis to battle the energy E∼v 2log(L/a) . We see that the free energy F∼(v2−T)log(L/a) goes to infinity if T<∼v2, which we identify as essentially the critical temperature Tc. A single vortex cannot exist below Tc. Vortices and antivortices are tightly bound below Tcbut are liberated above Tc. Black hole The discovery in the 1970s of these topological objects that cannot be seen in perturbation theory came as a shock to the generation of physicists raised on Feynman diagrams andcanonical quantization. People (including yours truly) were taught that the field operatorϕ(x) is a highly singular quantum operator and has no physical meaning as such, and that quantum field theory is defined perturbatively by Feynman diagrams. Even quiteeminent physicists asked in puzzlement what a statement such as ϕ−→ r→∞veiθwould mean. Learned discussions that in hindsight are totally irrelevant ensued. As I said in introducing chapter V .6, I like to refer to this historical process as “field theorists breaking the shacklesof Feynman diagrams.” It is worth mentioning one argument physicists at that time used to convince themselves that solitons do exist. After all, the Schwarzschild black hole, defined by the metric g μν(x) (see chapter I.11), had been known since 1916. Just what are the components of the metricg μν(x)? They are fields in exactly the same way that our scalar field ϕ(x) and our gauge potential Aμ(x)are fields, and in a quantum theory of gravity gμνwould have to be replaced by a quantum operator just like ϕandAμ. So the objects discovered in the 1970s are conceptually no different from the black hole known in the 1910s. But in the early 1970s most particle theorists were not particularly aware of quantum gravity. Exercises V .7.1 Explain the relation between the mathematical statement /Pi10(S0)=Z2and the physical result that there are no kinks with |Q|≥2. V .7.2 In the vortex, study the length scales characterizing the variation of the fields ϕandA. Estimate the mass of the vortex. 312 | V . Field Theory and Collective Phenomena V .7.3 Consider the vortex configuration in which ϕ−→r→∞veiνθ, with νan integer. Calculate the magnetic flux. Show that the magnetic flux coming out of an antivortex (for which ν=− 1)is opposite to the magnetic flux coming out of a vortex. V .7.4 Mathematically, since g(θ)≡eiνθmay be regarded as an element of the group U(1), we can speak of a map of S1, the circle at spatial infinity, onto the group U(1). Calculate (i/2π)/integraltext S1gdg†, thus showing that the winding number is given by this integral of a 1-form. V .7.5 Show that within a region in which ϕais constant, Fμνas defined in the text is the electromagnetic field strength. Compute /vectorBfar from the center of a magnetic monopole and show that Dirac quantization holds. V .7.6 Display explicitly the map S2→S2, which wraps one sphere around the other twice. Verify that this map corresponds to a magnetic monopole with magnetic charge 2. V .7.7 Write down the variational equations that minimize (8). V .7.8 Find the BPS solution explicitly. V .7.9 Discuss the dyon solution. Work it out in the BPS limit. V .7.10 Verify explicitly that the magnetic monopole is rotation invariant in spite of appearances. By this is meant that all physical gauge invariant quantities such as /vectorBare covariant under rotation. Gauge variant quantities such as Ab ican and do vary under rotation. Write down the generators of rotation. V .7.11 Show that the mass of the magnetic monopole is about 137 MW. V .7.12 Evaluate n≡−(1/24π2)/integraltext S3tr(gdg†)3for the map g=ei/vectorθ./vectorσ. [Hint: By symmetry, you need calculate the integrand only in a small neighborhood of the identity element of the group or equivalently the north pole of S3. Next, consider g=ei(θ1σ1+θ 2σ2+mθ 3σ3)forman integer and convince yourself that mmeasures the number of times S3wraps around S3.] Compare with exercise V .7.4 and admire the elegance of mathematics. V .7.13 Prove that higher order corrections do not change the chiral anomaly ∂μJμ 5=[1/(4π)2]εμνλσtrFμνFλσ (I have rescaled A→(1/g)A). [Hint: Integrate over spacetime and show that the left hand side is given by the number of right-handed fermion quanta minus the number of left-handed fermion quanta, sothat both sides are given by integers.] Part VI Field Theory and Condensed Matter This page intentionally left blank VI.1Fractional Statistics, Chern-Simons T erm, and T opological Field Theory Fractional statistics The existence of bosons and fermions represents one of the most profound features of quantum physics. When we interchange two identical quantum particles, the wavefunction acquires a factor of either +1o r −1. Leinaas and Myrheim, and later Wilczek independently, had the insight to recognize that in (2+1)-dimensional spacetime particles can also obey statistics other than Bose or Fermi statistics, a statistics now known asfractional or anyon statistics. These particles are now known as anyons. T o interchange two particles, we can move one of them half-way around the other and then translate both of them appropriately. When you take one anyon half-way aroundanother anyon going anticlockwise, the wave function acquires a factor of e iθwhere θ is a real number characteristic of the particle. For θ=0, we have bosons and for θ=π, fermions. Particles half-way between bosons and fermions, with θ=π/2, are known as semions. After Wilczek’s paper came out, a number of distinguished senior physicists were thor- oughly confused. Thinking in terms of Schr ¨odinger’s wave function, they got into endless arguments about whether the wave function must be single valued. Indeed, anyon statis-tics provides a striking example of the fact that the path integral formalism is sometimessignificantly more transparent. The concept of anyon statistics can be formulated in termsof wave functions but it requires thinking clearly about the configuration space over whichthe wave function is defined. Consider two indistinguishable particles at positions x i 1andxi 2at some initial time that end up at positions xf 1andxf 2a time Tlater. In the path integral representation for/angbracketleftxf 1,xf 2|e−iHT|xi 1,xi 2/angbracketrightwe have to sum over all paths. In spacetime, the worldlines of the two particles braid around each other (see fig. VI.1.1). (We are implicitly assuming that the particles cannot go through each other, which is the case if there is a hard corerepulsion between them.) Clearly, the paths can be divided into topologically distinctclasses, characterized by an integer nequal to the number of times the worldlines of the 316 | VI. Field Theory and Condensed Matter t i 1xx i 2xf 2xf 1x Figure VI.1.1 two particles braid around each other. Since the classes cannot be deformed into each other, the corresponding amplitudes cannot interfere quantum mechanically, and withthe amplitudes in each class we are allowed to associate an additional phase factor e iαn beyond the usual factor coming from the action. The dependence of αnonnis determined by how quantum amplitudes are to be combined. Suppose one particle goes around the other through an angle /Delta1ϕ1, a history to which we assign an additional phase factor eif (/Delta1ϕ 1)withfsome as yet unknown function. Suppose this history is followed by another history in which our particle goes around theother by an additional angle /Delta1ϕ 2. The phase factor eif (/Delta1ϕ 1+/Delta1ϕ 2)we assign to the combined history clearly has to satisfy the composition law eif (/Delta1ϕ 1+/Delta1ϕ 2)=eif (/Delta1ϕ 1)eif (/Delta1ϕ 2). In other words, f (/Delta1ϕ) has to be a linear function of its argument. We conclude that in (2+1)-dimensional spacetime we can associate with the quantum amplitude corresponding to paths in which one particle goes around the other anti-clockwise through an angle /Delta1ϕ a phase factor e i(θ/π)/Delta1ϕ, with θan arbitrary real parameter. Note that when one particle goes around the other clockwise through an angle /Delta1ϕ the quantum amplitude acquires a phase factor e−iθ π/Delta1ϕ. When we interchange two anyons, we have to be careful to specify whether we do it “anticlockwise” or “clockwise,” producing factors eiθande−iθ, respectively. This indicates immediately that parity Pand time reversal invariance Tare violated. Chern-Simons theory The next important question is whether all this can be incorporated in a local quantum field theory. The answer was given by Wilczek and Zee, who showed that the notion of fractionalstatistics can result from the effect of coupling to a gauge potential. The significance of VI.1. T opological Field Theory | 317 a field theoretic formulation is that it demonstrates conclusively that the idea of anyon statistics is fully compatible with the cherished principles that we hold dear and that gointo the construction of quantum field theory. Given a Lagrangian L 0with a conserved current jμ, construct the Lagrangian L=L0+γεμνλaμ∂νaλ+aμjμ(1) Hereεμνλdenotes the totally antisymmetric symbol in (2+1)-dimensional spacetime and γis an arbitrary real parameter. Under a gauge transformation aμ→aμ+∂μ/Lambda1, the term εμνλaμ∂νaλ, known as the Chern-Simons term, changes by εμνλaμ∂νaλ→εμνλaμ∂νaλ+ εμνλ∂μ/Lambda1∂νaλ. The action changes by δS=γ/integraltext d3xεμνλ∂μ(/Lambda1∂νaλ)and thus, if we are allowed to drop boundary terms, as we assume to be the case here, the Chern-Simonsaction is gauge invariant. Note incidentally that in the language of differential form youlearned in chapters IV .5 and IV .6 the Chern-Simons term can be written compactly as ada . Let us solve the equation of motion derived from (1): 2γεμνλ∂νaλ=−jμ(2) for a particle sitting at rest (so that ji=0). Integrating the μ=0 component of (2), we obtain /integraldisplay d2x(∂ 1a2−∂2a1)=−1 2γ/integraldisplay d2xj0(3) Thus, the Chern-Simons term has the effect of endowing the charged particles in the theory with flux. (Here the term charged particles simply means particles that couple tothe gauge potential a μ. In this context, when we refer to charge and flux, we are obviously not referring to the charge and flux associated with the ordinary electromagnetic field. Weare simply borrowing a useful terminology.) By the Aharonov-Bohm effect (chapter IV .4), when one of our particles moves around another, the wave function acquires a phase, thus endowing the particles with anyonstatistics with angle θ=1/4γ(see exercise VI.1.5). Strictly speaking, the term “fractional statistics” is somewhat misleading. First, a trivial remark: The statistics parameter θdoes not have to be a fraction. Second, statistics is not directly related to counting how many particles we can put into a state. The statisticsbetween anyons is perhaps better thought of as a long ranged phase interaction betweenthem, mediated by the gauge potential a. The appearance of ε μνλin (1) signals the violation of parity Pand time reversal invari- anceT, something we already know. Hopf term An alternative treatment is to integrate out ain (1). As explained in chapter III.4, and as in any gauge theory, the inverse of the differential operator ε∂is not defined: It has a zero mode since (εμνλ∂ν)(∂λF(x)) =0 for any smooth function F(x) . Let us choose the Lorenz 318 | VI. Field Theory and Condensed Matter gauge ∂μaμ=0. Then, using the fundamental identity of field theory (see appendix A) we obtain the nonlocal Lagrangian LHopf=1 4γ/parenleftbigg jμεμνλ∂ν ∂2jλ/parenrightbigg (4) known as the Hopf term. T o determine the statistics parameter θconsider a history in which one particle moves half-way around another sitting at rest. The current jis then equal to the sum of two terms describing the two particles. Plugging into (4) we evaluate the quantum phase eiS=ei/integraltext d3xLHopfand obtain θ=1/(4γ). T opological field theory There is something conceptually new about the pure Chern-Simons theory S=γ/integraldisplay Md3xεμνλaμ∂νaλ (5) It is topological. Recall from chapter I.11 that a field theory written in flat spacetime can be immediately promoted to a field theory in curved spacetime by replacing the Minkowski metric ημνby the Einstein metric gμνand including a factor√−g in the spacetime integration measure. But in the Chern-Simons theory ημνdoes not appear! Lorentz indices are contracted with the totally antisymmetric symbol εμνλ. Furthermore, we don’t need the factor√−g, as I will now show. Recall also from chapter I.11 that a vector field transforms as aμ(x)= (∂x/primeλ/∂xμ)a/prime λ(x/prime)and so for three vector fields εμνλaμ(x)bν(x)cλ(x)=εμνλ∂x/primeσ ∂xμ∂x/primeτ ∂xν∂x/primeρ ∂xλa/prime σ(x/prime)b/prime τ(x/prime)c/prime ρ(x/prime) =det/parenleftbigg∂x/prime ∂x/parenrightbigg εστρa/prime σ(x/prime)b/prime τ(x/prime)c/prime ρ(x/prime) On the other hand, d3x/prime=d3xdet(∂x/prime/∂x) . Observe, then, d3xεμνλaμ(x)bν(x)cλ(x)=d3x/primeεστρa/prime σ(x/prime)b/prime τ(x/prime)c/prime ρ(x/prime) which is invariant without the benefit of√−g. So, the Chern-Simons action in (5) is invariant under general coordinate transforma- tion—it is already written for curved spacetime. The metric gμνdoes not enter anywhere. The Chern-Simons theory does not know about clocks and rulers! It only knows about thetopology of spacetime and is rightly known as a topological field theory. In other words,when the integral in (5) is evaluated over a closed manifold Mthe property of the field theory/integraltext Dae iS(a)depends only on the topology of the manifold, and not on whatever metric we might put on the manifold. VI.1. T opological Field Theory | 319 Ground state degeneracy Recall from chapter I.11 the fundamental definition of energy and momentum. The energy-momentum tensor is defined by the variation of the action with respect to gμν, but hey, the action here does not depend on gμν. The energy-momentum tensor and hence the Hamiltonian is identically zero! One way of saying this is that to define the Hamiltonianwe need clocks and rulers. What does it mean for a quantum system to have a Hamiltonian H=0? Well, when we took a course on quantum mechanics, if the professor assigned an exam problem to findthe spectrum of the Hamiltonian 0, we could do it easily! All states have energy E=0. We are ready to hand it in. But the nontrivial problem is to find how many states there are. This number is known as the ground state degeneracy and depends only on the topology of the manifold M. Massive Dirac fermions and the Chern-Simons term Consider a gauge potential aμcoupled to a massive Dirac fermion in (2+1)-dimensional spacetime: L=¯ψ(i/negationslash∂+ /negationslasha−m)ψ . You did an exercise way back in chapter II.1 discovering the rather surprising phenomenon that in (2+1)-dimensional spacetime the Dirac mass term violates PandT. (What? You didn’t do it? You have to go back.) Thus, we would expect to generate the PandTviolating the Chern-Simons term εμνλaμ∂λaνif we integrate out the fermion to get the term tr log (i/negationslash∂+ /negationslasha−m)in the effective action, along the lines discussed in chapter IV .3. In one-loop order we have the vacuum polarization diagram (diagrammatically exactly the same as in chapter III.7 but in a spacetime with one less dimension) with a Feynmanintegral proportional to /integraldisplayd3p (2π)3tr/parenleftbigg γν 1 /negationslashp+ /negationslashq−mγμ 1 /negationslashp−m/parenrightbigg (6) As we will see, the change 4 →3 makes all the difference in the world. I leave it to you to evaluate (6) in detail (exercise VI.1.7) but let me point out the salient features here. Since the ∂λin the Chern-Simons term corresponds to qλin momentum space, in order to identify the coefficient of the Chern-Simons term we need only differentiate (6) withrespect to q λand set q→0: /integraldisplayd3p (2π)3tr(γν 1 /negationslashp−mγλ 1 /negationslashp−mγμ 1 /negationslashp−m) =/integraldisplayd3p (2π)3tr[γν(/negationslashp+m)γλ(/negationslashp+m)γμ(/negationslashp+m)] (p2−m2)3(7) 320 | VI. Field Theory and Condensed Matter I will simply focus on one piece of the integral, the piece coming from the term in the trace proportional to m3: εμνλm3/integraldisplayd3p (p2−m2)3(8) As I remarked in exercise II.1.12, in (2 +1)-dimensional spacetime the γμ’s are just the three Pauli matrices and thus tr (γνγλγμ)is proportional to εμνλ: The antisymmetric symbol appears as we expect from PandTviolation. By dimensional analysis, we see that the integral in (8) is up to a numerical constant equal to 1 /m3and so mcancels. But be careful! The integral depends only on m2and doesn’t know about the sign of m. The correct answer is proportional to 1 /|m|3, not 1 /m3. Thus, the coefficient of the Chern- Simons term is equal to m3/|m|3=m/|m|= sign of m, up to a numerical constant. An instructive example of an important sign! This makes sense since under P(orT)a Dirac field with mass mis transformed into a Dirac field with mass −m . In a parity-invariant theory, with a doublet of Dirac fields with masses mand−m a Chern-Simons term should not be generated. Exercises VI.1.1 In a nonrelativistic theory you might think that there are two separate Chern-Simons terms, εijai∂0ajand εija0∂iaj. Show that gauge invariance forces the two terms to combine into a single Chern-Simons term εμνλaμ∂νaλ. For the Chern-Simons term, gauge invariance implies Lorentz invariance. In contrast, the Maxwell term would in general be nonrelativistic, consisting of two terms, f2 0iandf2 ij, with an arbitrary relative coefficient between them (with fμν=∂μaν−∂νaμas usual). VI.1.2 By thinking about mass dimensions, convince yourself that the Chern-Simons term dominates the Maxwell term at long distances. This is one reason that relativistic field theorists find anyon fluids soappealing. As long as they are interested only in long distance physics they can ignore the Maxwellterm and play with a relativistic theory (see exercise VI.1.1). Note that this picks out (2+1)-dimensional spacetime as special. In (3+1)-dimensional spacetime the generalization of the Chern-Simons term ε μνλσfμνfλσhas the same mass dimension as the Maxwell term f2.I n(4 +1)-dimensional space the termερμνλσaρfμνfλσis less important at long distances than the Maxwell term f2. VI.1.3 There is a generalization of the Chern-Simons term to higher dimensional spacetime different from that given in exercise IV .1.2. We can introduce a p-form gauge potential (see chapter IV .4). Write the generalized Chern-Simons term in (2p+1)-dimensional spacetime and discuss the resulting theory. VI.1.4 Consider L=γaε∂a −(1/4g2)f2. Calculate the propagator and show that the gauge boson is massive. Some physicists puzzled by fractional statistics have reasoned that since in the presence of the Maxwellterm the gauge boson is massive and hence short ranged, it can’t possibly generate fractional statistics,which is manifestly an infinite ranged interaction. (No matter how far apart the two particles we areinterchanging are, the wave function still acquires a phase.) The resolution is that the information is in fact propagated over an infinite range by a q=0 pole associated with a gauge degree of freedom. This apparent paradox is intimately connected with the puzzlement many physicists felt when they firstheard of the Aharonov-Bohm effect. How can a particle in a region with no magnetic field whatsoeverand arbitrarily far from the magnetic flux know about the existence of the magnetic flux? VI.1. T opological Field Theory | 321 VI.1.5 Show that θ=1/4γ. There is a somewhat tricky factor1of 2. So if you are off by a factor of 2, don’t despair. T ry again. VI.1.6 Find the nonabelian version of the Chern-Simons term ada. [Hint: As in chapter IV .6 it might be easier to use differential forms.] VI.1.7 Using the canonical formalism of chapter I.8 show that the Chern-Simons Lagrangian leads to the Hamiltonian H=0. VI.1.8 Evaluate (6). 1X.G. Wen and A. Zee, J. de Physique, 50: 1623, 1989. VI.2 Quantum Hall Fluids Interplay between two pieces of physics Over the last decade or so, the study of topological quantum fluids (of which the Hall fluid is an example) has emerged as an interesting subject. The quantum Hall system consistsof a bunch of electrons moving in a plane in the presence of an external magnetic fieldBperpendicular to the plane. The magnetic field is assumed to be sufficiently strong so that the electrons all have spin up, say, so they may be treated as spinless fermions. As iswell known, this seemingly innocuous and simple physical situation contains a wealth ofphysics, the elucidation of which has led to two Nobel prizes. This remarkable richnessfollows from the interplay between two basic pieces of physics. 1. Even though the electron is pointlike, it takes up a finite amount of room.Classically, a charged particle in a magnetic field moves in a Larmor circle of radius rdetermined by evB=mv 2/r. Classically, the radius is not fixed, with more energetic particles moving in larger circles, but if we quantize the angular momentum mvr to be h=2π(in units in which /planckover2piis equal to unity) we obtain eBr2∼2π. A quantum electron takes up an area of order πr2∼2π2/eB . 2. Electrons are fermions and want to stay out of each other’s way. Not only does each electron insist on taking up a finite amount of room, each has to have its own room. Thus, the quantum Hall problem may be described as a sort of housingcrisis, or as the problem of assigning office space at an Institute for Theoretical Physics tovisitors who do not want to share offices. Already at this stage, we would expect that when the number of electrons N eis just right to fill out space completely, namely when Neπr2∼Ne(2π2/eB)∼A, the area of the system, something special happens. VI.2. Quantum Hall Fluids | 323 Landau levels and the integer Hall effect These heuristic considerations could be made precise, of course. The textbook problem of a single spinless electron in a magnetic field −[(∂x−ieAx)2+(∂y−ieAy)2]ψ=2mEψ was solved by Landau decades ago. The states occur in degenerate sets with energy En=/parenleftbig n+1 2/parenrightbigeB m,n=0, 1, 2, ... , known as the nth Landau level. Each Landau level has degeneracy BA/ 2π, where Ais the area of the system, reflecting the fact that the Larmor circles may be placed anywhere. Note that one Landau level is separated from the next bya finite amount of energy (eB/m) . Imagine putting in noninteracting electrons one by one. By the Pauli exclusion principle, each succeeding electron we put in has to go into a different state in the Landau level. Sinceeach Landau level can hold BA/ 2πelectrons it is natural (see exercise VI.2.1) to define a filling factor ν≡N e/(BA/2 π). When νis equal to an integer, the first νLandau levels are filled. If we want to put in one more electron, it would have to go into the (ν+1)st Landau level, costing us more energy than what we spent for the preceding electron. Thus, for νequal to an integer the Hall fluid is incompressible. Any attempt to compress it lessens the degeneracy of the Landau levels (the effective area Adecreases and so the degeneracy BA/ 2πdecreases) and forces some of the electrons to the next level, costing us lots of energy. An electric field Eyimposed on the Hall fluid in the ydirection produces a current Jx=σxyEyin the xdirection with σxy=ν(in units of e2/h) . This is easily understood in terms of the Lorentz force law obeyed by electrons in the presence of a magnetic field. Thesurprising experimental discovery was that the Hall conductance σ xywhen plotted against Bgoes through a series of plateaus, which you might have heard about. T o understand these plateaus we would have to discuss the effect of impurities. I will touch upon thefascinating subject of impurities and disorder in chapter VI.8. So, the integer quantum Hall effect is relatively easy to understand. Fractional Hall effect After the integer Hall effect, the experimental discovery of the fractional Hall effect, namely that the Hall fluid is also incompressible for filling factor νequal to simple odd- denominator fractions such as1 3and1 5, took theorists completely by surprise. For ν=1 3, only one-third of the states in the first Landau level are filled. It would seem that throwing in a few more electrons would not have that much effect on the system. Why should theν= 1 3Hall fluid be incompressible? Interaction between electrons turns out to be crucial. The point is that saying the first Landau level is one-third filled with noninteracting spinless electrons does not define aunique many-body state: there is an enormous degeneracy since each of the electrons can 324 | VI. Field Theory and Condensed Matter go into any of the BA/ 2πstates available subject only to Pauli exclusion. But as soon as we turn on a repulsive interaction between the electrons, a presumably unique ground stateis picked out within the vast space of degenerate states. Wen has described the fractionalHall state as an intricate dance of electrons: Not only does each electron occupy a finiteamount of room on the dance floor, but due to the mutual repulsion, it has to be carefulnot to bump into another electron. The dance has to be carefully choreographed, possibleonly for certain special values of ν. Impurities also play an essential role, but we will postpone the discussion of impurities to chapter VI.6. In trying to understand the fractional Hall effect, we have an important clue. You will remember from chapter V .7 that the fundamental unit of flux is given by 2 π, and thus the number of flux quanta penetrating the plane is equal to N φ=BA/ 2π. Thus, the puzzle is that something special happens when the number of flux quanta per electron Nφ/Ne=ν−1 is an odd integer. I arranged the chapters so that what you learned in the previous chapter is relevant to solving the puzzle. Suppose that ν−1flux quanta are somehow bound to each electron. When we interchange two such bound systems there is an additional Aharonov-Bohmphase in addition to the ( −1)from the Fermi statistics of the electrons. For ν −1odd these bound systems effectively obey Bose statistics and can be described by a complex scalarfieldϕ. The condensation of ϕturns out to be responsible for the physics of the quantum Hall fluid. Effective field theory of the Hall fluid We would like to derive an effective field theory of the quantum Hall fluid, first obtainedby Kivelson, Hansson, and Zhang. There are two alternative derivations, a long way and ashort way. In the long way, we start with the Lagrangian describing spinless electrons in a magnetic field in the second quantized formalism (we will absorb the electric charge eintoA μ), L=ψ+i(∂ 0−iA 0)ψ+1 2mψ†(∂i−iAi)2ψ+V( ψ†ψ) (1) and massage it into the form we want. In the previous chapter, we learned that by intro- ducing a Chern-Simons gauge field we can transform ψinto a scalar field. We then invoke duality, which we will learn about in the next chapter, to represent the phase degree offreedom of the scalar field as a gauge field. After a number of steps, we will discover thatthe effective theory of the Hall fluid turns out to be a Chern-Simons theory. Instead, I will follow the short way. We will argue by the “what else can it be” method or, to put it more elegantly, by invoking general principles. Let us start by listing what we know about the Hall system. 1. We live in (2+1)-dimensional spacetime (because the electrons are restricted to a plane.) 2. The electromagnetic current Jμis conserved: ∂μJμ=0. VI.2. Quantum Hall Fluids | 325 These two statements are certainly indisputable; when combined they tell us that the current can be written as the curl of a vector potential Jμ=1 2π/epsilon1μνλ∂νaλ (2) The factor of 1 /(2π)defines the normalization of aμ. We learned in school that in 3- dimensional spacetime, if the divergence of something is zero, then that something isthe curl of something else. That is precisely what (2) says. The only sophistication here isthat what we learned in school works in Minkowskian space as well as Euclidean space—itis just a matter of a few signs here and there. The gauge potential comes looking for us Observe that when we transform aμbyaμ→aμ−∂μ/Lambda1, the current is unchanged. In other words, aμis a gauge potential. We did not go looking for a gauge potential; the gauge potential came looking for us! There is no place to hide. The existence of a gauge potential follows from completelygeneral considerations. 3. We want to describe the system field theoretically by an effective local Lagrangian.4. We are only interested in the physics at long distance and large time, that is, at small wave number and low frequency. Indeed, a field theoretic description of a physical system may be regarded as a means of organizing various aspects of the relevant physics in a systematic way according to theirrelative importance at long distances and according to symmetries. We classify terms in afield theoretic Lagrangian according to powers of derivatives, powers of the fields, and soforth. A general scheme for classifying terms is according to their mass dimensions, asexplained in chapter III.2. The gauge potential a μhas dimension 1, as is always the case for any gauge potential coupled to matter fields according to the gauge principle, and thus (2)is consistent with the fact that the current has mass dimension 2 in (2+1)-dimensional spacetime. 5. Parity and time reversal are broken by the external magnetic field.This last statement is just as indisputable as statements 1 and 2. The experimentalist produces the magnetic field by driving a current through a coil with the current flowingeither clockwise or anticlockwise. Given these five general statements we can deduce the form of the effective Lagrangian.Since gauge invariance forbids the dimension-2 term a μaμin the Lagrangian, the simplest possible term is in fact the dimension-3 Chern-Simons term /epsilon1μνλaμ∂νaλ. Thus, the Lagrangian is simply L=k 4πa/epsilon1∂a+... (3) where kis a dimensionless parameter to be determined. 326 | VI. Field Theory and Condensed Matter We have introduced and will use henceforth the compact notation /epsilon1a∂b≡/epsilon1μνλaμ∂νbλ= /epsilon1b∂a for two vector fields aμandbμ. The terms indicated by ( ...) in (3) include the dimension-4 Maxwell term (1/g2)(f2 0i− βf2 ij)and other terms with higher dimensions. (Here βis some constant; see exer- cise VI.1.1.) The important observation is that these higher dimensional terms are lessimportant at long distances. The long distance physics is determined purely by the Chern-Simons term. In general the coefficient kmay well be zero, in which case the physics is determined by the short distance terms represented by the (...)in (3). Put differently, a Hall fluid may be defined as a 2-dimensional electron system for which the coefficient ofthe Chern-Simons term does not vanish, and consequently is such that its long distancephysics is largely independent of the microscopic details that define the system. Indeed,we can classify 2-dimensional electron systems according to whether kis zero or not. Coupling the system to an “external” or “additional” electromagnetic gauge potential A μ and using (2) we obtain (after integrating by parts and dropping a surface term) L=k 4π/epsilon1μνλaμ∂νaλ−1 2π/epsilon1μνλAμ∂νaλ=k 4π/epsilon1μνλaμ∂νaλ−1 2π/epsilon1μνλaμ∂νAλ (4) Note that the gauge potential of the magnetic field responsible for the Hall effect should not be included in Aμ; it is implicitly contained already in the coefficient k. The notion of quasiparticles or “elementary” excitations is basic to condensed matter physics. The effects of a many-body interaction may be such that the quasiparticles in thesystem are no longer electrons. Here we define the quasiparticles as the entities that coupleto the gauge potential and thus write L=k 4πa/epsilon1∂a+aμjμ−1 2π/epsilon1μνλaμ∂νAλ... (5) Defining ˜jμ≡jμ−(1/2π)/epsilon1μνλ∂νAλand integrating out the gauge field we obtain (see VI.1.4) L=π k˜jμ/parenleftbigg/epsilon1μνλ∂ν ∂2/parenrightbigg ˜jλ (6) Fractional charge and statistics We can now simply read off the physics from (6). The Lagrangian contains three types of terms: AA ,Aj, andjj. TheAA term has the schematic form A(/epsilon1∂/epsilon1∂/epsilon1∂/∂2)A. Using /epsilon1∂/epsilon1∂∼∂2and canceling between numerator and denominator, we obtain L=1 4πkA/epsilon1∂A (7) Varying with respect to Awe determine the electromagnetic current Jμ em=1 4πk/epsilon1μνλ∂νAλ (8) VI.2. Quantum Hall Fluids | 327 We learn from the μ=0 component of this equation that an excess density δnof electrons is related to a local fluctuation of the magnetic field by δn=(1/2πk)δB ; thus we can identify the filling factor νas 1/k, and from the μ=icomponents that an electric field produces a current in the orthogonal direction with σxy=(1/k)=ν. TheAjterm has the schematic form A(/epsilon1∂/epsilon1∂/∂2)j. Canceling the differential operators, we find L=1 kAμjμ(9) Thus, the quasiparticle carries electric charge 1 /k. Finally, the quasiparticles interact with each other via L=π kjμ/epsilon1μνλ∂ ∂2jλ(10) We simply remove the twiddle sign in (6). Recalling chapter VI.1 we see that quasiparticles obey fractional statistics with θ π=1 k(11) By now, you may well be wondering that while all this is fine and good, what would actually tell us that ν−1has to be an odd integer? We now argue that the electron or hole should appear somewhere in the excitation spectrum. After all, the theory is supposed to describe a system of electrons and thusfar our rather general Lagrangian does not contain any reference to the electron! Let us look for the hole (or electron). We note from (9) that a bound object made up of kquasiparticles would have charge equal to 1. This is perhaps the hole! For this to work, we see that khas to be an integer. So far so good, but kdoesn’t have to be odd yet. What is the statistics of this bound object? Let us move one of these bound objects half- way around another such bound object, thus effectively interchanging them. When onequasiparticle moves around another we pick up a phase given by θ/π=1/kaccording to (11). But here we have kquasiparticles going around kquasiparticles and so we pick up a phase θ π=1 kk2=k (12) For the hole to be a fermion we must require θ/π to be an odd integer. This fixes kto be an odd integer. Sinceν=1/k, we have here the classic Laughlin odd-denominator Hall fluids with filling factor ν=1 3,1 5,1 7,.... The famous result that the quasiparticles carry fractional charge and statistics just pops out [see (9) and (11)]. This is truly dramatic: a bunch of electrons moving around in a plane with a magnetic field corresponding to ν=1 3, and lo and behold, each electron has fragmented into three pieces, each piece with charge1 3and fractional statistics1 3! 328 | VI. Field Theory and Condensed Matter A new kind of order The goal of condensed matter physics is to understand the various states of matter. States of matter are characterized by the presence (or absence) of order: a ferromagnet becomesordered below the transition temperature. In the Landau-Ginzburg theory, as we saw inchapter V .3, order is associated with spontaneous symmetry breaking, described naturallywith group theory. Girvin and MacDonald first noted that the order in Hall fluids does notreally fit into the Landau-Ginzburg scheme: We have not broken any obvious symmetry.The topological property of the Hall fluids provides a clue to what is going on. As explainedin the preceding chapter, the ground state degeneracy of a Hall fluid depends on thetopology of the manifold it lives on, a dependence group theory is incapable of accountingfor. Wen has forcefully emphasized that the study of topological order, or more generallyquantum order, may open up a vast new vista on the possible states of matter. 1 Comments and generalization Let me conclude with several comments that might spur you to explore the wealth ofliterature on the Hall fluid. 1. The appearance of integers implies that our result is robust. A slick argument can be made based on the remark in the previous chapter that the Chern-Simons term doesnot know about clocks and rulers and hence can’t possibly depend on microphysics suchas the scattering of electrons off impurities which cannot be defined without clocks andrulers. In contrast, the physics that is not part of the topological field theory and describedby ( ...) in (3) would certainly depend on detailed microphysics. 2. If we had followed the long way to derive the effective field theory of the Hall fluid, we would have seen that the quasiparticle is actually a vortex constructed (as in chapter V .7)out of the scalar field representing the electron. Given that the Hall fluid is incompressible,just about the only excitation you can think of is a vortex with electrons coherently whirlingaround. 3. In the previous chapter we remarked that the Chern-Simons term is gauge invariant only upon dropping a boundary term. But real Hall fluids in the laboratory live in sampleswith boundaries. So how can (3) be correct? Remarkably, this apparent “defect” of the theoryactually represents one of its virtues! Suppose the theory (3) is defined on a bounded 2-dimensional manifold, a disk for example. Then as first argued by Wen there must bephysical degrees of freedom living on the boundary and represented by an action whosechange under a gauge transformation cancels the change of/integraltext d 3x(k/ 4π)a/epsilon1 ∂a . Physically, it is clear that an incompressible fluid would have edge excitations2corresponding to waves on its boundary. 1X. G. Wen, Quantum Field Theory of Many-Body Systems. 2The existence of edge currents in the integer Hall fluid was first pointed out by Halperin. VI.2. Quantum Hall Fluids | 329 4. What if we refuse to introduce gauge potentials? Since the current Jμhas dimension 2, the simplest term constructed out of the currents, JμJμ, is already of dimension 4; indeed, this is just the Maxwell term. There is no way of constructing a dimension 3 localinteraction out of the currents directly. T o lower the dimension we are forced to introducethe inverse of the derivative and write schematically J(1//epsilon1∂)J , which is of course just the non-local Hopf term. Thus, the question “why gauge field?” that people often ask can beanswered in part by saying that the introduction of gauge fields allows us to avoid dealingwith nonlocal interactions. 5. Experimentalists have constructed double-layered quantum Hall systems with an infinitesimally small tunneling amplitude for electrons to go from one layer to the other.Assuming that the current J μ I(I=1, 2)in each layer is separately conserved, we introduce two gauge potentials by writing Jμ I=1 2π/epsilon1μνλ∂νaIλas in (2). We can repeat our general argument and arrive at the effective Lagrangian L=/summationdisplay I,JKIJ 4πaI/epsilon1∂aJ+... (13) The integer khas been promoted to a matrix K. As an exercise, you can derive the Hall conductance, the fractional charge, and the statistics of the quasiparticles. You would notbe surprised that everywhere 1 /kappears we now have the matrix inverse K −1instead. An interesting question is what happens when Khas a zero eigenvalue. For example, we could have K=/parenleftBig 11 11/parenrightBig . Then the low energy dynamics of the gauge potential a−≡a1−a2 is not governed by the Chern-Simons term, but by the Maxwell term in the ( ...)in (13). We have a linearly dispersing mode and thus a superfluid! This striking prediction3was verified experimentally. 6. Finally, an amusing remark: In this formalism electron tunneling corresponds to the nonconservation of the current Jμ −≡Jμ 1−Jμ 2=(1/2π)/epsilon1μνλ∂νa−λ. The difference N1−N2 of the number of electrons in the two layers is not conserved. But how can ∂μJμ −/negationslash=0 even though Jμ −is the curl of a−λ(as I have indicated explicitly)? Recalling chapter IV .4, you the astute reader say, aha, magnetic monopoles! T unneling in a double-layered Hall system inEuclidean spacetime can be described as a gas of monopoles and antimonopoles. 4(Think, why monopoles and antimonopoles?) Note of course that these are not monopoles in theusual electromagnetic gauge potential but in the gauge potential a −λ. What we have given in this section is certainly a very slick derivation of the effective long distance theory of the Hall fluid. Some would say too slick. Let us go back to ourfive general statements or principles. Of these five, four are absolutely indisputable. Infact, the most questionable is the statement that looks the most innocuous to the casualreader, namely statement 3. In general, the effective Lagrangian for a condensed mattersystem would be nonlocal. We are implicitly assuming that the system does not contain amassless field, the exchange of which would lead to a nonlocal interaction. 5Also implicit 3X. G. Wen and A. Zee, Phys. Rev. Lett. 69: 1811, 1992. 4X. G. Wen and A. Zee, Phys. Rev. B47: 2265, 1993. 5A technical remark: Vortices (i.e., quasiparticles) pinned to impurities in the Hall fluid can generate an interaction nonlocal in time. 330 | VI. Field Theory and Condensed Matter in (3) is the assumption that the Lagrangian can be expressed completely in terms of the gauge potential a. A priori, we certainly do not know that there might not be other relevant degrees of freedom. The point is that as long as these degrees of freedom are not gaplessthey can be safely integrated out. Exercises VI.2.1 T o define filling factor precisely, we have to discuss the quantum Hall system on a sphere rather than on a plane. Put a magnetic monopole of strength G(which according to Dirac can be only a half-integer or an integer) at the center of a unit sphere. The flux through the sphere is equal to Nφ=2G. Show that the single electron energy is given by El=(1 2/planckover2piωc)/bracketleftbig l(l+1)−G2/bracketrightbig /G with the Landau levels corresponding tol=G,G+1,G+2, . . . , and that the degeneracy of the lth level is 2 l+1. With LLandau levels filled with noninteracting electrons ( ν=L)show that Nφ=ν−1Ne−S, where the topological quantity Sis known as the shift. VI.2.2 For a challenge, derive the effective field theory for Hall fluids with filling factor ν=m/k withkan odd integer, such as ν=2 5. [Hint: You have to introduce mgauge potentials aIλand generalize (2) to Jμ=(1/2π)/epsilon1μνλ∂ν/summationtextm I=1aIλ. The effective theory turns out to be L=1 4πm/summationdisplay I,J=1aIKIJ/epsilon1∂aJ+m/summationdisplay I=1aIμ˜jIμ+... with the integer kreplaced by a matrix K. Compare with (13).] VI.2.3 For the Lagrangian in (13), derive the analogs of (8), (9), and (11). VI.3 Duality A far reaching concept Duality is a profound and far reaching concept1in theoretical physics, with origins in electromagnetism and statistical mechanics. The emergence of duality in recent years inseveral areas of modern physics, ranging from the quantum Hall fluids to string theory,represents a major development in our understanding of quantum field theory. Here Itouch upon one particular example just to give you a flavor of this vast subject. My plan is to treat a relativistic theory first, and after you get the hang of the subject, I will go on to discuss the nonrelativistic theory. It makes sense that some of the interestingphysics of the nonrelativistic theory is absent in the relativistic formulation: A largersymmetry is more constraining. By the same token, the relativistic theory is actually mucheasier to understand if only because of notational simplicity. Vortices Couple a scalar field in (2+1)-dimensions to an external electromagnetic gauge potential, with the electric charge qindicated explicitly for later convenience: L=1 2|(∂μ−iqAμ)ϕ|2−V( ϕ†ϕ) (1) We have already studied this theory many times, most recently in chapter V .7 in connection with vortices. As usual, write ϕ=|ϕ|eiθ. Minimizing the potential Vat|ϕ|=v gives the ground state field configuration. Setting ϕ=veiθin (1) we obtain L=1 2v2(∂μθ−qAμ)2(2) 1For a first introduction to duality, I highly recommend J. M. Figueroa-O’Farrill, Electromagnetic Duality for Children , http://www.maths.ed.ac.uk/~jmf/T eaching/Lectures/EDC.html. 332 | VI. Field Theory and Condensed Matter which upon absorbing θintoAby a gauge transformation we recognize as the Meissner Lagrangian. For later convenience we also introduce the alternative form L=−1 2v2ξ2 μ+ξμ(∂μθ−qAμ) (3) We recover (2) upon eliminating the auxiliary field ξμ(see appendix A and chapter III.5). In chapter V .7 we learned that the excitation spectrum includes vortices and anti- vortices, located where |ϕ|vanishes. If, around the zero of |ϕ|,θchanges by 2 π, we have a vortex. Around an antivortex, /Delta1θ=− 2π. Recall that around a vortex sitting at rest, the electromagnetic gauge potential has to go as qAi→∂iθ (4) at spatial infinity in order for the energy of the vortex to be finite, as we can see from (2). The magnetic flux /integraldisplay d2xεij∂iAj=/contintegraldisplay d/vectorx./vectorA=/Delta1θvortex q=2π q(5) is quantized in units of 2 π/q . Let us pause to think physically for a minute. On a distance scale large compared to the size of the vortex, vortices and antivortices appear as points. As discussed in chapter V .7,the interaction energy of a vortex and an antivortex separated by a distance Ris given by simply plugging into (2). Ignoring the probe field A μ, which we can take to be as weak as possible, we obtain ∼/integraltextR adr r(∇θ)2∼log(R/a) where ais some short distance cutoff. But recall that the Coulomb interaction in 2-dimensional space is logarithmic since by dimensional analysis/integraltext d2k(ei/vectork./vectorx/k2)∼log(|/vectorx|/a) (with a−1some ultraviolet cutoff). Thus, a gas of vortices and antivortices appears as a gas of point “charges” with a Coulombinteraction between them. Vortex as charge in a dual theory Duality is often made out by some theorists to be a branch of higher mathematics butin fact it derives from an entirely physical idea. In view of the last paragraph, can we notrewrite the theory so that vortices appear as point “charges” of some as yet unknown gaugefield? In other words, we want a dual theory in which the fundamental field creates andannihilates vortices rather than ϕquanta. We will explain the word “dual” in due time. Remarkably, the rewriting can be accomplished in just a few simple steps. Proceeding physically and heuristically, we picture the phase field θas smoothly fluctuating, except that here and there it winds around 2 π. Write ∂ μθ=∂μθsmooth +∂μθvortex . Plugging into (3) we write L=−1 2v2ξ2 μ+ξμ(∂μθsmooth +∂μθvortex−qAμ) (6) Integrate over θsmooth and obtain the constraint ∂μξμ=0, which can be solved by writing ξμ=εμνλ∂νaλ (7) VI.3. Duality | 333 a trick we used earlier in chapter VI.2. As in that chapter, a gauge potential comes looking for us, since the change aλ→aλ+∂λ/Lambda1does not change ξμ. Plugging into (6), we find L=−1 4v2f2 μν+εμνλ∂νaλ(∂μθvortex−qAμ) (8) where fμν=∂μaν−∂νaμ. Our treatment is heuristic because we ignore the fact that |ϕ|vanishes at the vortices. Physically, we think of the vortices as almost pointlike so that |ϕ|=v “essentially” every- where. As mentioned in chapter V .6, by appropriate choice of parameters we can makesolitons, vortices, and so on as small as we like. In other words, we neglect the couplingbetween θ vortex and|ϕ|. A rigorous treatment would require a proper short distance cutoff by putting the system on a lattice.2But as long as we capture the essential physics, as we assuredly will, we will ignore such niceties. Note for later use that the electromagnetic current Jμ, defined as the coefficient of −Aμ in (8), is determined in terms of the gauge potential aλto be Jμ=qεμνλ∂νaλ (9) Let us integrate the term εμνλ∂νaλ∂μθvortex in (8) by parts to obtain aλελμν∂μ∂νθvortex . According to Newton and Leibniz, ∂μcommutes with ∂ν, and so apparently we get zero. But∂μand∂νcommute only when acting on a globally defined function, and, heavens to Betsy, θvortex is not globally defined since it changes by 2 πwhen we go around a vortex. In particular, consider a vortex at rest and look at the quantity a0couples to in (8) namely εij∂i∂jθvortex=/vector∇×(/vector∇θvortex)in the notation of elementary physics. Integrating this over a region containing the vortex gives/integraltext d2x/vector∇×(/vector∇θvortex)=/contintegraltext d/vectorx./vector∇θvortex=2π. Thus, we recognize (1/2π)εij∂i∂jθvortex as the density of vortices, the time component of some vortex current jλ vortex. By Lorentz invariance, jλ vortex=(1/2π)ελμν∂μ∂νθvortex . Thus, we can now write (8) as L=−1 4v2f2 μν+(2π)aμjμ vortex−Aμ(qεμνλ∂νaλ) (10) Lo and behold, we have accomplished what we set out to do. We have rewritten the theory so that the vortex appears as an “electric charge” for the gauge potential aμ. Sometimes this is called a dual theory, but strictly speaking, it is more accurate to refer to it as the dualrepresentation of the original theory (1). Let us introduce a complex scalar field /Phi1, which we will refer to as the vortex field, to create and annihilate the vortices and antivortices. In other words, we “elaborate” thedescription in (10) to L=−1 4v2f2 μν+1 2|(∂μ−i(2π)aμ)/Phi1|2−W(/Phi1) −Aμ(qεμνλ∂νaλ) (11) 2For example, M. P. A. Fisher, “Mott Insulators, Spin Liquids, and Quantum Disordered Superconductivity,” cond-mat/9806164, appendix A. 334 | VI. Field Theory and Condensed Matter The potential W(/Phi1) contains terms such as λ(/Phi1†/Phi1)2describing the short distance inter- action of two vortices (or a vortex and an antivortex.) In principle, if we master all the shortdistance physics contained in the original theory (1) then these terms are all determinedby the original theory. Vortex of a vortex Now we come to the most fascinating aspect of the duality representation and the reasonwhy the word “dual” is used in the first place. The vortex field /Phi1is a complex scalar field, just like the field ϕwe started with. Thus we can perfectly well form a vortex out of /Phi1, namely a place where /Phi1vanishes and around which the phase of /Phi1goes through 2 π. Amusingly, we are forming a vortex of a vortex, so to speak. So, what is a vortex of a vortex?The duality theorem states that the vortex of a vortex is nothing but the original charge, described by the field ϕwe started out with! Hence the word duality. The proof is remarkably simple. The vortex in the theory (11) carries “magnetic flux.” Referring to (11) we see that 2 πa i→∂iθat spatial infinity. By exactly the same manipulation as in (5), we have 2π/integraldisplay d2xεij∂iaj=2π/contintegraldisplay d/vectorx./vectora=2π (12) Note that I put quotation marks around the term “magnetic flux” since as is evident I am talking about the flux associated with the gauge potential aμand not the flux associated with the electromagnetic potential Aμ. But remember that from (9) the electromagnetic current Jμ=qεμνλ∂νaλand in particular J0=qεij∂iaj. Hence, the electric charge (note no quotation marks) of this vortex of a vortex is equal to/integraltext d2xJ0=q, precisely the charge of the original complex scalar field ϕ. This proves the assertion. Here we have studied vortices, but the same sort of duality also applies to monopoles. As I remarked in chapter IV .4, duality allows us a glimpse into field theories in the stronglycoupled regime. We learned in chapter V .7 that certain spontaneously broken nonabeliangauge theories in (3+1)-dimensional spacetime contains magnetic monopoles. We can write a dual theory in terms of the monopole field out of which we can construct mono-poles. The monopole of a monopole turns out be none other than the charged fields of theoriginal gauge theory. This duality was first conjectured many years ago by Olive and Mon-tonen and later shown to be realized in certain supersymmetric gauge theories by Seibergand Witten. The understanding of this duality was a “hot” topic a few years ago as it ledto deep new insight about how certain string theories are dual to each other. 3In contrast, according to one of my distinguished condensed matter colleagues, the important notion of duality is still underappreciated in the condensed matter physics community. 3For example, D. I. Olive and P. C. West, eds., Duality and Supersymmetric Theories . VI.3. Duality | 335 Meissner begets Maxwell and so on We will close by elaborating slightly on duality in (2+1)-dimensional spacetime and how it might be relevant to the physics of 2-dimensional materials. Consider a Lagrangian L(a) quadratic in a vector field aμ. Couple an external electromagnetic gauge potential Aμto the conserved current εμνλ∂νaλ: L=L(a)+Aμ(εμνλ∂νaλ) (13) Let us ask: For various choices of L(a), if we integrate out awhat is the effective Lagrangian L(A) describing the dynamics of A? If you have gotten this far in the book, you can easily do the integration. The central identity of quantum field theory again! Given L(a)∼aKa (14) we have L(A)∼(ε∂A)1 K(ε∂A) ∼A(ε∂1 Kε∂)A (15) We have three choices for L(a) to which I attach various illustrious names: L(a)∼a2Meissner L(a)∼aε∂a Chern-Simons (16) L(a)∼f2∼a∂2aMaxwell Since we are after conceptual understanding, I won’t bother to keep track of indices and irrelevant overall constants. (You can fill them in as an exercise.) For example, givenL(a)=f μνfμνwithfμν=∂μaν−∂νaμwe can write L(a)∼a∂2aand so K=∂2. Thus, the effective dynamics of the external electromagnetic gauge potential is given by (15) asL(A)∼A[ε∂(1/∂ 2)ε∂]A∼A2, the Meissner Lagrangian! In this “quick and dirty” way of making a living, we simply set εε∼1 and cancel factors of ∂in the numerator against those in the denominator. Proceeding in this way, we construct the following table: Dynamics of a K Effective Lagrangian L(A)∼A[ε∂(1/K)ε∂ ]ADynamics of the external probe A Meissner a21A(ε∂ε∂)A ∼A∂2A Maxwell F2 Chern-Simons aε∂a ε∂A(ε∂1 ε∂ε∂)A∼Aε∂A Chern-Simons Aε∂A Maxwell f2∼a∂2a∂2A(ε∂1 ∂2ε∂)A∼AA Meissner A2 (17) 336 | VI. Field Theory and Condensed Matter Meissner begets Maxwell, Chern-Simons begets Chern-Simons, and Maxwell begets Meissner. I find this beautiful and fundamental result, which represents a form of duality,very striking. Chern-Simons is self-dual: It begets itself. Going nonrelativistic It is instructive to compare the nonrelativistic treatment of duality.4Go back to the super- fluid Lagrangian of chapter V .1: L=iϕ†∂0ϕ−1 2m∂iϕ†∂iϕ−g2(ϕ†ϕ−¯ρ)2(18) As before, substitute ϕ≡√ρeiθto obtain L=−ρ∂0θ−ρ 2m(∂iθ)2−g2(ρ−¯ρ)2+... (19) which we rewrite as L=−ξμ∂μθ+m 2ρξ2 i−g2(ρ−¯ρ)2+... (20) In (19) we have dropped a term ∼(∂iρ1/2)2. In (20) we have defined ξ0≡ρ. Integrating outξiin (20) we recover (19). All proceeds as before. Writing θ=θsmooth +θvortex and integrating out θsmooth ,w e obtain the constraint ∂μξμ=0, solved by writing ξμ=/epsilon1μνλ∂νˆaλ. The hat on ˆaλis for later convenience. Note that the density ξ0≡ρ=/epsilon1ij∂iˆaj≡ˆf (21) is the “magnetic” field strength while ξi=/epsilon1ij(∂0ˆaj−∂jˆa0)≡/epsilon1ijˆf0j (22) is the “electric” field strength. Putting all of this into (20) we have L=m 2ρˆf2 0i−g2(ˆf−¯ρ)2−2πˆaμjμ vortex+... (23) T o “subtract out” the background “magnetic” field ¯ρ, an obviously sensible move is to write ˆaμ=¯aμ+aμ (24) where we define the background gauge potential by ¯a0=0,∂0¯aj=0 (no background “electric” field) and /epsilon1ij∂i¯aj=¯ρ (25) 4The treatment given here follows essentially that given by M. P. A. Fisher and D. H. Lee. VI.3. Duality | 337 The Lagrangian (23) then takes on the cleaner form L=/parenleftbiggm 2¯ρf2 0i−g2f2/parenrightbigg −2πaμjμ vortex−2π¯aiji vortex+... (26) We have expanded ρ∼¯ρin the first term. As in (10) the first two terms form the Maxwell Lagrangian, and the ratio of their coefficients determines the speed of propagation c=/parenleftbigg2g2¯ρ m/parenrightbigg1/2 (27) In suitable units in which c=1, we have L=−m 4¯ρfμνfμν−2πaμjμ vortex−2π¯aiji vortex+... (28) Compare this with (10). The one thing we missed with our relativistic treatment is the last term in (28), for the simple reason that we didn’t put in a background. Recall that the term like AiJiin ordinary electromagnetism means that a moving particle associated with the current Ji sees a magnetic field /vector∇×/vectorA. Thus, a moving vortex will see a “magnetic field” /epsilon1ij∂i(¯a+a)j=¯ρ+/epsilon1ij∂iaj (29) equal to the sum of ¯ρ, the density of the original bosons, and a fluctuating field. In the Coulomb gauge ∂iai=0 we have (f0i)2=(∂0ai)2+(∂ia0)2, where the cross term (∂0ai)(∂ia0)effectively vanishes upon integration by parts. Integrating out the Coulomb fielda0, we obtain L=−¯ρ 2m(2π)2/integraldisplay/integraldisplay d2xd2y/bracketleftbigg j0(/vectorx)log|/vectorx−/vectory| aj0(/vectory)/bracketrightbigg +m 2¯ρ(∂0ai)2−g2f2+2π(ai+¯ai)jvortex i(30) The vortices repel each other by a logarithmic interaction/integraltext d2k(ei/vectork./vectorx/k2)∼log(|/vectorx|/a) as we have known all along. A self-dual theory Interestingly, the spatial part f2of the Maxwell Lagrangian comes from the short ranged repulsion between the original bosons. If we had taken the bosons to interact by an arbitrary potential V( x) we would have, instead of the last term in (20), /integraldisplay/integraldisplay d2xd2y[ρ(x)−¯ρ]V( x−y)[ρ(y)−¯ρ] (31) It is easy to see that all the steps go through essentially as before, but now the second term in (26) becomes /integraldisplay/integraldisplay d2xd2yf (x)V (x −y)f(y) (32) 338 | VI. Field Theory and Condensed Matter Thus, the gauge field propagates according to the dispersion relation ω2=(2¯ρ/m)V (k) /vectork2(33) where V( k) is the Fourier transformation of V( x) . In the special case V( x)=g2δ(2)(x) we recover the linear dispersion given in (27). Indeed, we have a linear dispersion ω∝|/vectork|as long as V( x) is sufficiently short ranged for V(/vectork=0)to be finite. An interesting case is when V( x) is logarithmic. Then V( k) goes as 1 /k2and so ω∼ constant: The gauge field aibecomes massive and drops out. The low energy effective theory consists of a bunch of vortices with a logarithmic interaction between them. Thus,a theory of bosons with a logarithmic repulsion between them is self dual in the low energylimit. The dance of vortices and antivortices Having gone through this nonrelativistic discussion of duality, let us reward ourselves byderiving the motion of vortices in a fluid. Let the bulk of the fluid be at rest. According to (28)the vortex behaves like a charged particle in a background magnetic field ¯bproportional to the mean density of the fluid ¯ρ. Thus, the force acting on a vortex is the usual Lorentz force /vectorv×/vectorB, and the equation of motion of the vortex in the presence of a force Fis then just ¯ρ/epsilon1ij˙xj=Fi (34) This is the well-known result that a vortex, when pushed, moves in a direction perpendic- ular to the force. Consider two vortices. According to (30) they repel each other by a logarithmic inter- action. They move perpendicular to the force. Thus, they end up circling each other. Incontrast, consider a vortex and an antivortex, which attract each other. As a result of thisattraction, they both move in the same direction, perpendicular to the straight line joiningthem (see fig. VI.3.1). The vortex and antivortex move along in step, maintaining the dis-tance between them. This in fact accounts for the famous motion of a smoke ring. If wecut a smoke ring through its center and perpendicular to the plane it lies in, we have just ++_ (a) (b)+ Figure VI.3.1 VI.3. Duality | 339 a vortex with an antivortex for each section. Thus, the entire smoke ring moves along in a direction perpendicular to the plane it lies in. All of this can be understood by elementary physics, as it should be. The key observation is simply that vortices and antivortices produce circular flows in the fluid around them, sayclockwise for vortices and anticlockwise for antivortices. Another basic observation is thatif there is a local flow in the fluid, then any object, be it a vortex or an antivortex, caught init would just flow along in the same direction as the local flow. This is a consequence ofGalilean invariance. By drawing a simple picture you can see that this produces the samepattern of motion as discussed above. VI.4 TheσModels as Effective Field Theories The Lagrangian as a mnemonic Our beloved quantum field theory has had two near death experiences. The first started around the mid-1930s when physical quantities came out infinite. But it roared back tolife in the late 1940s and early 1950s, thanks to the work of the generation that includedFeynman, Schwinger, Dyson, and others. The second occurred toward the late 1950s. Aswe have already discussed, quantum field theory seemed totally incapable of addressingthe strong interaction: The coupling was far too strong for perturbation theory to be of anyuse. Many physicists—known collectively as the S-matrix school—felt that field theory was irrelevant for studying the strong interaction and advocated a program of trying toderive results from general principles without using field theory. For example, in derivingthe Goldberger-T reiman relation, we could have foregone any mention of field theory andFeynman diagrams. Eventually, in a reaction against this trend, people realized that if some results could be obtained from general considerations such as notions of spontaneous symmetry breakingand so forth, any Lagrangian incorporating these general properties had to produce thesame results. At the very least, the Lagrangian provides a mnemonic for any physical resultderived without using quantum field theory. Thus was born the notion of long distanceor low energy effective field theory, which would prove enormously useful in both particleand condensed matter physics (as we have already seen and as we will discuss further inchapter VIII.3). The strong interaction at low energies One of the earliest examples is the σmodel of Gell-Mann and L ´evy, which describes the interaction of nucleons and pions. We now know that the strong interaction has to bedescribed in terms of quarks and gluons. Nevertheless, at long distances, the degrees of VI.4.σModels | 341 freedom are the two nucleons and the three pions. The proton and the neutron transform as a spinor ψ≡/parenleftbigp n/parenrightbig under the SU( 2)of isospin. Consider the kinetic energy term ¯ψiγ∂ψ = ¯ψLiγ∂ψ L+¯ψRiγ∂ψ R. We note that this term has the larger symmetry SU( 2)L×SU( 2)R, with the left handed field ψLand the right handed field ψRtransforming as a doublet under SU( 2)LandSU( 2)R, respectively. [The SU( 2)of isospin is the diagonal subgroup ofSU( 2)L×SU( 2)R.] We can write ψL∼(1 2,0)andψR∼(0,1 2). Now we see a problem immediately: The mass term m¯ψψ=m(¯ψLψR+h.c.) is not allowed since ¯ψLψR∼(1 2,1 2), a 4-dimensional representation of SU( 2)L×SU( 2)R,a group locally isomorphic to SO( 4). At this point, lesser physicists would have said, what is the problem, we knew all along that the strong interaction is invariant only under the SU( 2)of isospin, which we will write as SU( 2)I. Under SU( 2)Ithe bilinears constructed out of ¯ψLandψRtransform as 1 2×1 2=0+1, the singlet being ¯ψψ and the triplet ¯ψiγ5τaψ. With only SU( 2)Isymmetry, we can certainly include the mass term ¯ψψ . T o say it somewhat differently, to fully couple to the four bilinears we can construct out of¯ψLandψR, namely ¯ψψ and¯ψiγ5τaψ, we need four meson fields transforming as the vector representation under SO( 4). But only the three pion fields are known. It seems clear that we only have SU( 2)Isymmetry. Nevertheless, Gell-Mann and L ´evy boldly insisted on the larger symmetry SU( 2)L× SU( 2)R/similarequalSO( 4)and simply postulated an additional meson field, which they called σ, so that (σ,/vectorπ)form the 4-dimensional representation. I leave it to you to verify that ¯ψL(σ+i/vectorτ./vectorπ)ψR+h.c.=¯ψ(σ+i/vectorτ./vectorπγ 5)ψis invariant. Hence, we can write down the invariant Lagrangian L=¯ψ[iγ∂+g(σ+i/vectorτ./vectorπγ 5)]ψ+L(σ,/vectorπ) (1) where the part not involving the nucleons reads L(σ,/vectorπ)=1 2/parenleftBig (∂σ)2+(∂/vectorπ)2/parenrightBig +μ2 2(σ2+/vectorπ2)−λ 4(σ2+/vectorπ2)2(2) This is known as the linear σmodel. Theσmodel would have struck most physicists as rather strange at the time it was introduced: The nucleon does not have a mass and there is an extra meson field. Aha, butyou would recognize (2) as precisely the Lagrangian (IV .1.2) (for N=4)that we studied, which exhibits spontaneous symmetry breaking. The four scalar fields (ϕ 4,ϕ1,ϕ2,ϕ3) in (IV .1.2) correspond to (σ,/vectorπ). With no loss of generality, we can choose the vacuum expectation value of ϕto point in the 4th direction, namely the vacuum in which /angbracketleft0|σ|0/angbracketright=/radicalbig μ2/λ≡vand/angbracketleft0|/vectorπ|0/angbracketright=0. Expanding σ=v+σ/primewe see immediately that the nucleon has a mass M=gv. You should not be surprised that the pion comes out massless. The meson associated with the field σ/prime, which we will call the σmeson, has no reason to be massless and indeed is not. Can the all-important parameter vbe related to a measurable quantity? Indeed. From chapter I.10 you will recall that the axial current is given by Noether’s theorem asJ a μ5=¯ψγμγ5(τa/2)ψ+πa∂μσ−σ∂μπa. After σacquires a vacuum expectation value, 342 | VI. Field Theory and Condensed Matter Ja μ5contains a term −v∂μπa. This term implies that the matrix element /angbracketleft0|Ja μ5|πb/angbracketright=ivk μ, where kdenotes the momentum of the pion, and thus vis proportional to the fdefined in chapter IV .2. Indeed, we recognize the mass relation M=gvas precisely the Goldberger- T reiman relation (IV .2.7) with F(0)=1 (see exercise VI.4.4). The nonlinear σmodel It was eventually realized that the main purpose in life of the potential in L(σ,/vectorπ)is to force the vacuum expectation values of the fields to be what they are, so the potential canbe replaced by a constraint σ 2+/vectorπ2=v2. A more physical way of thinking about this point is by realizing that the σmeson, if it exists at all, must be very broad since it can decay via the strong interaction into two pions. We might as well force it out of the low energyspectrum by making its mass large. By now, you have learned from chapters IV .1 and V .1that the mass of the σmeson, namely√ 2μ, can be taken to infinity while keeping vfixed by letting μ2andλtend to infinity, keeping their ratio fixed. We will now focus on L(σ,/vectorπ). Instead of thinking abut L(σ,/vectorπ)=1 2[(∂σ)2+(∂/vectorπ)2] with the constraint σ2+/vectorπ2=v2, we can simply solve the constraint and plug the solution σ=√ v2−/vectorπ2into the Lagrangian, thus obtaining what is known as the nonlinear σmodel: L=1 2/bracketleftbigg (∂/vectorπ)2+(/vectorπ.∂/vectorπ)2 f2−/vectorπ2/bracketrightbigg =1 2(∂/vectorπ)2+1 2f2(/vectorπ.∂/vectorπ)2+... (3) Note that Lcan be written in the form L=(∂πa)Gab(/vectorπ)(∂πb); some people like to think of Gabas a “metric” in field space. [Incidentally, recall that way back in chapter I.3 we restricted ourselves to the simplest possible kinetic energy term1 2(∂ϕ)2, rejecting possibilities such asU(ϕ)(∂ϕ)2. But recall also that in chapter IV .3 we noted that such a term would arise by quantum fluctuations.] In accordance with the philosophy that introduced this chapter, any Lagrangian that captures the correct symmetry properties should describe the same low energy physics.1 This means that anybody, including you, can introduce his or her own parametrization ofthe fields. The nonlinear σmodel is actually an example of a broad class of field theories whose Lagrangian has a simple form but with the fields appearing in it subject to some nontrivialconstraint. An example is the theory defined by L(U)=f2 4tr(∂μU†.∂μU) (4) withU(x) a matrix-valued field and an element of SU( 2). Indeed, if we write U=e(i/f )/vectorπ./vectorτ we see that L(U)=1 2(∂/vectorπ)2+(1/2f2)(/vectorπ.∂/vectorπ)2+... , identical to (3) up to the terms indicated. The /vectorπfield here is related to the one in (3) by a field redefinition. There is considerably more we can say about the nonlinear σmodels and their applica- tions in particle and condensed matter physics, but a thorough discussion would take us 1S. Weinberg, Physica 96A: 327, 1979. VI.4.σModels | 343 far beyond the scope of this book. Instead, I will develop some of their properties in the exercises and in the next chapter will sketch how they can arise in one class of condensedmatter systems. Exercises VI.4.1 Show that the vacuum expectation value of (σ,/vectorπ)can indeed point in any direction without changing the physics. At first sight, this statement seems strange since, by virtue of its γ5coupling to the nucleon, the pion is a pseudoscalar field and cannot have a vacuum expectation without breaking parity. But (σ,/vectorπ) are just Greek letters. Show that by a suitable transformation of the nucleon field parity is conserved, asit should be in the strong interaction. VI.4.2 Calculate the pion-pion scattering amplitude up to quadratic order in the external momenta, using the nonlinear σmodel (3). [Hint: For help, see S. Weinberg, Phys. Rev. Lett. 17: 616, 1966.] VI.4.3 Calculate the pion-pion scattering amplitude up to quadratic order in the external momenta, using the linear σmodel (2). Don’t forget the Feynman diagram involving σmeson exchange. You should get the same result as in exercise VI.4.2. VI.4.4 Show that the mass relation M=gvamounts to the Goldberger-T reiman relation. VI.5 Ferromagnets and Antiferromagnets Magnetic moments In chapters IV .1 and V .3 I discussed how the concept of the Nambu-Goldstone boson originated as the spin wave in a ferromagnetic or an antiferromagnetic material. A cartoondescription of such materials consists of a regular lattice on each site of which sits alocal magnetic moment, which we denote by a unit vector /vectorn jwithjlabeling the site. In a ferromagnetic material the magnetic moments on neighboring sites want to pointin the same direction, while in an antiferromagnetic material the magnetic momentson neighboring sites want to point in opposite directions. In other words, the energy isH=J/summationtext <ij>/vectorni./vectornj, where iandjlabel neighboring sites. For antiferromagnets J> 0, and for ferromagnets J< 0. I will merely allude to the fully quantum description formulated in terms of a spin /vectorSjoperator on each site j; the subject lies far beyond the scope of this text. In a more microscopic treatment, we would start with a Hamiltonian (such as the Hubbard Hamiltonian) describing the hopping of electrons and the interaction betweenthem. Within some approximate mean field treatment the classical variable /vectorn jwould then emerge as the unit vector pointing in the direction of /angbracketleftc† j/vectorσcj/angbracketrightwithc† jandcjthe electron creation and annihilation operators, respectively. But this is not a text on solid state physics. First versus second order in time Here we would like to derive an effective low energy description of the ferromagnet and antiferromagnet in the spirit of the σmodel description of the preceding chapter. Our treatment will be significantly longer than the standard discussion given in some fieldtheory texts, but has the slight advantage of being correct. The somewhat subtle issue is what kinetic energy term we have to add to −H to form the Lagrangian L. Since for a unit vector /vectornwe have /vectorn.(d/vectorn/dt) =(d(/vectorn./vectorn)/dt) =0, we cannot VI.5. Magnetic Systems | 345 make do with one time derivative. With two derivatives we can form (d/vectorn/dt) .(d/vectorn/dt) and so Lwrong=1 2g2/summationdisplay j∂/vectornj ∂t.∂/vectornj ∂t−J/summationdisplay <ij>/vectorni./vectornj (1) A typical field theory text would then pass to the continuum limit and arrive at the Lagrangian density L=1 2g2(∂/vectorn ∂t.∂/vectorn ∂t−c2 s/summationdisplay l∂/vectorn ∂xl.∂/vectorn ∂xl) (2) with the constraint [ /vectorn(x ,t)]2=1. This is another example of a nonlinear σmodel. Just as in the nonlinear σmodel discussed in chapter VI.4, the Lagrangian looks free, but the nontrivial dynamics comes from the constraint. The constant cs(which is determined in terms of the microscopic variable J)is the spin wave velocity, as you can see by writing down the equation of motion (∂2/∂t2)/vectorn−c2 s∇2/vectorn=0. But you can feel that something is wrong. You learned in a quantum mechanics course that the dynamics of a spin variable /vectorSis first order in time. Consider the most basic example of a spin in a constant magnetic field described by H=μ/vectorS./vectorB. Then d/vectorS/dt= i[H,/vectorS]=μ/vectorB×/vectorS. Besides, you might remember from a solid state physics course that in a ferromagnet the dispersion relation of the spin wave has the nonrelativistic form ω∝k2 and not the relativistic form ω2∝k2implied by (2). The resolution of this apparent paradox is based on the Pauli-Hopf identity: Given a unit vector /vectornwe can always write /vectorn=z†/vectorσz, where z=/parenleftbigz1 z2/parenrightbig consists of two complex numbers such that z†z≡z† 1z1+z† 2z2=1. Verify this! (A mathematical aside: Writing z1andz2out in terms of real numbers we see that this defines the so-called Hopf map S3→S2.)While we cannot form a term quadratic in /vectornand linear in time derivative, we can write a term quadratic in the complex doublet zand linear in time derivative. Can you figure it out before looking at the next line? The correct version of (1) is Lcorrect=i/summationdisplay jz† j∂zj ∂t+1 2g2/summationdisplay j∂/vectornj ∂t.∂/vectornj ∂t−J/summationdisplay <ij>/vectorni./vectornj (3) The added term is known as the Berry’s phase term and has deep topological meaning. You should derive the equation of motion using the identity /integraldisplay dt δ/parenleftbigg z† j∂zj ∂t/parenrightbigg =1 2i/integraldisplay dt δ/vectornj./parenleftbigg /vectornj×∂/vectornj ∂t/parenrightbigg (4) Remarkably, although z† j(∂zj/∂t) cannot be written simply in terms of /vectornj, its variation can be. Low energy modes in the ferromagnet and the antiferromagnet In the ground state of a ferromagnet, the magnetic moments all point in the same direction, which we can choose to be the z-direction. Expanding the equation of motion in 346 | VI. Field Theory and Condensed Matter small fluctuations around this ground state /vectornj=ˆez+δ/vectornj(where evidently ˆezdenotes the appropriate unit vector) and Fourier transforming, we obtain /parenleftBigg−ω2 g2+h(k) −1 2iω 1 2iω −ω2 g2+h(k)/parenrightBigg/parenleftBiggδnx(k) δny(k)/parenrightBigg =0 (5) linking the two components δnx(k)andδny(k)ofδ/vectorn(k) . The condition /vectornj./vectornj=1 says that δnz(k)=0. Here ais the lattice spacing and h(k)≡4J[2−cos(kxa)−cos(kya)]/similarequal2Ja2k2 for small k. (I am implicitly working in two spatial dimensions as evidenced by kxandky.) At low frequency the Berry term iωdominates the naive term ω2/g2, which we can therefore throw away. Setting the determinant of the matrix equal to zero, we see that weget the correct quadratic dispersion relation ω∝k 2. The treatment of the antiferromagnet is interestingly different. The so-called N ´eel state1 for an antiferromagnet is defined by /vectornj=(−1)jˆez. Writing /vectornj=(−1)jˆez+δ/vectornj, we obtain /parenleftBigg−ω2 g2+f( k) −1 2iω 1 2iω −ω2 g2+f( k)/parenrightBigg/parenleftBiggδnx(k) δny(k+Q)/parenrightBigg =0 (6) linking δnx and δny evaluated at different momenta. Here f( k)=4J [2+cos(kxa)+cos(kya)] andQ=[π/a ,π/a ]. The appearance of Qis due to (−1)j=eiQaj. (I will let you figure out the somewhat overly compact notation.) The antiferromagneticfactor (−1) jexplicitly breaks translation invariance and kicks in the momentum Qwhen- ever it occurs. A similar equation links δny(k) andδnx(k+Q). Solving these equations, you will find that there is a high frequency branch that we are not interested in and a lowfrequency branch with the linear dispersion ω∝k. Thus, the low frequency dynamics of the antiferromagnet can be described by the nonlinear σmodel (2), which when the spin wave velocity is normalized to 1 can be written in the relativistic form: L=1 2g2∂μ/vectorn.∂μ/vectorn (7) Exercises VI.5.1 Work out the two branches of the spin wave spectrum in the ferromagnetic case, paying particular attention to the polarization. VI.5.2 Verify that in the antiferromagnetic case the Berry’s phase term merely changes the spin wave velocity and does not affect the spectrum qualitatively as in the ferromagnetic case. 1Note that while the N ´eel state describes the lowest energy configuration for a classical antiferromagnet, it does not describe the ground state of a quantum antiferromagnet. The terms S+ iS− j+S− iS+ jin the Hamiltonian J/summationtext <ij>/vectorSi./vectorSjflip the spins up and down. VI.6 Surface Growth and Field Theory In this chapter I will discuss a topic, rather unusual for a field theory text, taken from non- equilibrium statistical mechanics, one of the hottest growth fields in theoretical physicsover the last few years. I want to introduce you to yet another area in which field theoreticconcepts are of use. Imagine atoms being deposited randomly on some surface. This is literally how some novel materials are grown. The height h(x ,t)of the surface as it grows is governed by the Kardar-Parisi-Zhang equation ∂h ∂t=ν∇2h+λ 2(∇h)2+η(/vectorx,t) (1) This equation describes a deceptively simple prototype of nonequilibrium dynamics and has a remarkably wide range of applicability. T o understand (1), consider the various terms on the right-hand side. The term ν∇2h (withν> 0)is easy to understand: Positive in the valleys of hand negative on the peaks, it tends to smooth out the surface. With only this term the problem would be linear andhence trivial. The nonlinear term (λ/2)(∇h) 2renders the problem highly nontrivial and interesting; I leave it to you as an exercise to convince yourself of the geometric origin ofthis term. The third term describes the random arrival of atoms, with the random variableη(/vectorx,t)usually assumed to be Gaussian distributed, with zero mean, 1and correlations /angbracketleftbig η(/vectorx,t)η(/vectorx/prime,t/prime)/angbracketrightbig =2σ2δD(/vectorx−/vectorx/prime)δ(t−t/prime) (2) In other words, the probability distribution for a particular η(/vectorx,t)is given by P(η)∝e−1 2σ2/integraltext dDxdt η( /vectorx,t)2 Here /vectorxrepresents coordinates in D-dimensional space. Experimentally, D=2 for the situation I described, but theoretically we are free to investigate the problem for any D. 1There is no loss of generality here since an additive constant in ηcan be absorbed by the shift h→h+ct. 348 | VI. Field Theory and Condensed Matter Typically, condensed matter physicists are interested in calculating the correlation be- tween the height of the surface at two different positions in space and time: /angbracketleft[h(/vectorx,t)−h(/vectorx/prime,t/prime)]2/angbracketright=|/vectorx−/vectorx/prime|2χf/parenleftbigg|/vectorx−/vectorx/prime|z |t−t/prime|/parenrightbigg , (3) The bracket /angbracketleft.../angbracketrighthere and in (2) denotes averaging over different realizations of the random variable η(/vectorx,t). On the right-hand side of (3) I have written the dynamic scaling form typically postulated in condensed matter physics, where χandzare the so-called roughness and dynamic exponents. The challenge is then to show that the scaling form iscorrect and to calculate χandz. Note that the dynamic exponent z(which in general is not an integer) tells us, roughly speaking, how many powers of space is worth one power oftime. (For λ=0 we have simple diffusion for which z=2.)Herefdenotes an unknown function. I will not go into more technical details. Our interest here is to see how this problem, which does not even involve quantum mechanics, can be converted into a quantum fieldtheory. Start with Z≡/integraldisplay Dh/integraldisplay Dηe−1 2σ2/integraltext dDxdt η( /vectorx,t)2 δ/bracketleftbigg∂h ∂t−ν∇2h−λ 2(∇h)2−η(/vectorx,t)/bracketrightbigg (4) Integrating over η, we obtain Z=/integraltext Dhe−S(h)with the action S(h)=1 2σ2/integraldisplay dD/vectorxd t/bracketleftbigg∂h ∂t−ν∇2h−λ 2(∇h)2/bracketrightbigg2 (5) You will recognize that this describes a nonrelativistic field theory of a scalar field h(/vectorx,t). The physical quantity we are interested in is then given by /angbracketleft/bracketleftbig h(/vectorx,t)−h(/vectorx/prime,t/prime)/bracketrightbig2/angbracketright=1 Z/integraldisplay Dhe−S(h)[h(/vectorx,t)−h(/vectorx/prime,t/prime)]2(6) Thus, the challenge of determing the roughness and dynamic exponents in statistical physics is equivalent to the problem of determining the propagator D(/vectorx,t)≡1 Z/integraldisplay Dhe−S(h)h(/vectorx,t)h(/vector0, 0) of the scalar field h. Incidentally, by scaling t→t/ν andh→/radicalbig σ2/ν h, we can write the action as S(h)=1 2/integraldisplay dD/vectorxd t/bracketleftbigg/parenleftbigg∂ ∂t−∇2/parenrightbigg h−g 2(∇h)2/bracketrightbigg2 (7) withg2≡λ2σ2/ν3. Expanding the action in powers of has usual S(h)=1 2/integraldisplay dD/vectorxd t/braceleftBigg/bracketleftbigg/parenleftbigg∂ ∂t−∇2/parenrightbigg h/bracketrightbigg2 −g(∇h)2/parenleftbigg∂ ∂t−∇2/parenrightbigg h+g2 4(∇h)4/bracerightBigg , (8) we recognize the quadratic term as giving us the rather unusual propagator 1 /(ω2+k4)for the scalar field h, and the cubic and quartic term as describing the interaction. As always, VI.6. Surface Growth | 349 h h(x) θdcos θ x Figure VI.6.1 to calculate the desired physical quantity we evaluate the functional or “path” integral Z=/integraldisplay Dhe−S(h)+/integraltext dDxdtJ(x ,t)h(x ,t) and then functionally differentiate repeatedly with respect to J. My intent here is not so much to teach you nonequilibrium statistical mechanics as to show you that quantum field theory can emerge in a variety of physical situations, includingthose involving only purely classical physics. Note that the “quantum fluctuations” herearise from the random driving term. Evidently, there is a close methodological connectionbetween random dynamics and quantum physics. Exercises VI.6.1 An exercise in elementary geometry: Draw a straight line tilted at an angle θwith respect to the horizontal. The line represents a small segment of the surface at time t. Now draw a number of circles of diameter dtangent to and on top of this line. Next draw another line tilted at angle θwith respect to the horizontal and lying on top of the circles, namely tangent to them. This new line represents the segment of thesurface some time later (see fig. VI.6.1). Note that /Delta1h=d/cosθ/similarequald(1+ 1 2θ2). Show that this generates the nonlinear term (λ/2)(∇h)2in the KPZ equation (1). For applications of the KPZ equation, see for example, T . Halpin–Healy and Y .-C. Zhang, Phys. Rep. 254: 215, 1995; A. L. Barabasi and H. E. Stanley, Fractal Concepts in Surface Growth . VI.6.2 Show that the scalar field hhas the propagator 1 /(ω2+k4). VI.6.3 Field theory can often be cast into apparently rather different forms by a change of variable. Show that by writing U=e1 2ghwe can change the action (7) to S=2 g2/integraldisplay dD/vectorxd t/parenleftbigg U−1∂ ∂tU−U−1∇2U/parenrightbigg2 (9) a kind of nonlinear σmodel. VI.7 Disorder: Replicas and Grassmannian Symmetry Impurities and random potential An important area in condensed matter physics involves the study of disordered systems, a subject that has been the focus of a tremendous amount of theoretical work over thelast few decades. Electrons in real materials scatter off the impurities inevitably presentand effectively move in a random potential. In the spirit of this book I will give you a briefintroduction to this fascinating subject, showing how the problem can be mapped into aquantum field theory. The prototype problem is that of a quantum particle obeying the Schr ¨odinger equa- tionHψ=[−∇ 2+V( x) ]ψ=Eψ , where V( x) is a random potential (representing the impurities) generated with the Gaussian white noise probability distribution P(V) = Ne−/integraltext dDx(1/2g2)V (x)2with the normalization factor Ndetermined by/integraltext DVP(V) =1. The parameter gmeasures the strength of the impurities: the larger g, the more disordered the system. This of course represents an idealization in which interaction between electronsand a number of other physical effects are neglected. As in statistical mechanics we think of an ensemble of systems each of which is charac- terized by a particular function V( x) taken from the distribution P(V) . We study the aver- age or typical properties of the system. In particular, we might want to know the averageddensity of states defined by ρ(E)=/angbracketlefttrδ(E−H)/angbracketright=/angbracketleft/summationtext iδ(E−Ei)/angbracketright, where the sum runs over the ith eigenstate of Hwith corresponding eigenvalue Ei. We denote by /angbracketleftO(V) /angbracketright≡/integraltext DVP( V) O( V) the average of any functional O(V) ofV( x) . Clearly,/integraltextE∗+δE E∗dE ρ(E) counts the number of states in the interval from E∗toE∗+δE, an important quantity in, for example, tunneling experiments. VI.7. Disorder | 351 Anderson localization Another important physical question is whether the wave functions at a particular energy Eextend over the entire system or are localized within a characteristic length scale ξ(E) . Clearly, this issue determines whether the material is a conductor or an insulator. At firstsight, you might think that we should study S(x ,y;E)≡/angbracketleftBig/summationdisplay iδ(E−Ei)ψ∗ i(x)ψi(y)/angbracketrightBig which might tell us how the wave function at xis correlated with the wave function at some other point y, butSis unsuitable because ψ∗ i(x)ψ i(y) has a phase that depends on V. Thus, Swould vanish when averaged over disorder. Instead, the correct quantity to study is K(x−y;E)≡/angbracketleftBig/summationdisplay iδ(E−Ei)ψ∗ i(x)ψi(y)ψ∗ i(y)ψi(x)/angbracketrightBig sinceψ∗ i(x)ψi(y)ψ∗ i(y)ψi(x)is manifestly positive. Note that upon averaging over all possi- bleV( x) we recover translation invariance so that Kdoes not depend on xandyseparately, but only on the separation |x−y|.A s|x −y|→∞ ,i fK(x−y;E)∼e−|x−y|/ξ(E)decreases exponentially the wave functions around the energy Eare localized over the so-called local- ization length ξ(E) . On the other hand, if K(x−y;E)decreases as a power law of |x−y|, the wave functions are said to be extended. Anderson and his collaborators made the surprising discovery that localization prop- erties depend on D, the dimension of space, but not on the detailed form of P(V) (an example of the notion of universality). For D=1 and 2 all wave functions are localized, regardless of how weak the impurity potential might be. This is a highly nontrivial state-ment since a priori you might think, as eminent physicists did at the time, that whether thewave functions are localized or not depends on the strength of the potential. In contrast, forD=3, the wave functions are extended for Ein the range (−E c,Ec).A sEapproaches the energy Ec(known as the mobility edge) from above, the localization length ξ(E) diverges asξ(E)∼1/(E−Ec)μwith some critical exponent1μ. Anderson received the Nobel Prize for this work and for other contributions to condensed matter theory. Physically, localization is due to destructive interference between the quantum waves scattering off the random potential. When a magnetic field is turned on perpendicular to the plane of a D=2 electron gas the situation changes dramatically: An extended wave function appears at E=0. For non- zeroE, all wave functions are still localized, but with the localization length diverging as ξ(E)∼1/|E|ν. This accounts for one of the most striking features of the quantum Hall effect (see chapter VI.2): The Hall conductivity stays constant as the Fermi energy increases but then suddenly jumps by a discrete amount due to the contribution of the extended state 1This is an example of a quantum phase transition. The entire discussion is at zero temperature. In contrast to the phase transition discussed in chapter V .3, here we vary Einstead of the temperature. 352 | VI. Field Theory and Condensed Matter as the Fermi energy passes through E=0. Understanding this behavior quantitatively poses a major challenge for condensed matter theorists. Indeed, many consider an analyticcalculation of the critical exponent νas one of the “Holy Grails” of condensed matter theory. Green’s function formalism So much for a lightning glimpse of localization theory. Fascinating though the localization transition might be, what does quantum field theory have to do with it? This is after all afield theory text. Before proceeding we need a bit of formalism. Consider the so-calledGreen’s function G(z)≡/angbracketlefttr[1/(z−H)]/angbracketrightin the complex z-plane. Since tr[1 /(z−H)]=/summationtext i1/(z−Ei), this function consists of a sum of poles at the eigenvalues Ei. Upon averaging, the poles merge into a cut. Using the identity (I.2.13) lim ε→0Im[1/(x+iε)]= −πδ(x), we see that ρ(E)=−1 πlim ε→0ImG(E+iε) (1) So if we know G(z) we know the density of states. The infamous denominator I can now explain how quantum field theory enters into the problem. We start by taking the logarithm of the identity (A.15) J†.K−1.J=log(/integraldisplay Dϕ†Dϕe−ϕ†.K.ϕ+J†.ϕ+ϕ†.J) (where as usual we have dropped an irrelevant term). Differentiating with respect to J† andJand then setting J†andJequal to 0 we obtain an integral representation for the inverse of a hermitean matrix: (K−1)ij=/integraltext Dϕ†Dϕe−ϕ†.K.ϕϕiϕ† j/integraltext Dϕ†Dϕe−ϕ†.K.ϕ(2) (Incidentally, you may recognize this as essentially related to the formula (I.7.14) for the propagator of a scalar field.) Now that we know how to represent 1 /(z−H)we have to take its trace, which means setting i=jin (2) and summing. In our problem, H=− ∇2+V( x) and the index icorresponds to the continuous variable xand the summation to an integration over space. Replacing Kbyi(z−H) (and taking care of the appropriate delta function) we obtain tr−i z−H=/integraldisplay dDy⎧ ⎨ ⎩/integraltext Dϕ†Dϕei/integraltext dDx{∂ϕ†∂ϕ+[V( x)−z]ϕ†ϕ}ϕ(y)ϕ†(y) /integraltext Dϕ†Dϕei/integraltext dDx{∂ϕ†∂ϕ+[V( x)−z]ϕ†ϕ}⎫ ⎬ ⎭(3) VI.7. Disorder | 353 This is starting to look like a scalar field theory in D-dimensional Euclidean space with the action S=/integraltext dDx{∂ϕ†∂ϕ+[V( x)−z]ϕ†ϕ}. [Note that for (3) to be well defined zhas to be in the lower half-plane.] But now we have to average over V( x) , that is, integrate over Vwith the probability distribution P(V) . We immediately run into the difficulty that confounded theorists for a long time. The denominator in (3) stops us cold: If that denominator were not there,then the functional integration over V( x) would just be the Gaussian integral you have learned to do over and over again. Can we somehow lift this infamous denominator intothe numerator, so to speak? Clever minds have come up with two tricks, known as thereplica method and the supersymmetric method, respectively. If you can come up withanother trick, fame and fortune might be yours. Replicas The replica trick is based on the well-known identity (1/x)=lim n→0xn−1, which allows us to write that much disliked denominator as lim n→0/parenleftBig/integraldisplay Dϕ†Dϕei/integraltext dDx{∂ϕ†∂ϕ+[V( x)−z]ϕ†ϕ}/parenrightBign−1 =lim n→0/integraldisplayn/productdisplay a=2Dϕ† aDϕaei/integraltext dDx/summationtextn a=2{∂ϕ† a∂ϕa+[V( x)−z]ϕ† aϕa} Thus (3) becomes tr1 z−H=lim n→0i/integraldisplay dDy/integraldisplay/parenleftBiggn/productdisplay a=1Dϕ† aDϕa/parenrightBigg ei/integraltext dDx/summationtextn a=1{∂ϕ† a∂ϕa+[V( x)−z]ϕ† aϕa}ϕ1(y)ϕ† 1(y) (4) Note that the functional integral is now over ncomplex scalar fields ϕa. The field ϕhas been replicated. For positive integers, the integrals in (4) are well defined. We hope thatthe limit n→0 will not blow up in our face. Averaging over the random potential, we recover translation invariance; thus the in- tegrand for/integraltext d Dydoes not depend on yand/integraltext dDyjust produces the volume Vof the system. Using (A.13) we obtain /angbracketleftbigg tr1 z−H/angbracketrightbigg =iVlim n→0/integraldisplay/parenleftBiggn/productdisplay a=1Dϕ† aDϕa/parenrightBigg ei/integraltext dDxLϕ1(0)ϕ† 1(0) (5) where L(ϕ)≡n/summationdisplay a=1(∂ϕ† a∂ϕa−zϕ† aϕa)+ig2 2/parenleftBiggn/summationdisplay a=1ϕ† aϕa/parenrightBigg2 (6) We obtain a field theory (with a peculiar factor of i)ofnscalar fields with a good old ϕ4interaction invariant under O(n) (known as the replica symmetry.) Note the wisdom of 354 | VI. Field Theory and Condensed Matter replacing Kbyi(z−H); if we didn’t include the ithe functional integral would diverge at large ϕ, as you can easily check. For zin the upper half-plane we would replace Kby −i(z−H). The quantity from which we can extract the desired averaged density of states is given by the propagator of the scalar field. Incidentally, we can replace ϕ1(0)ϕ† 1(0)in (5) by the more symmetric expression (1/n)/summationtextn b=1ϕ† bϕb. Absorbing Vso that we are calculating the density of states per unit volume, we find G(z)=ilim n→0/integraldisplay/parenleftBiggn/productdisplay a=1Dϕ† aDϕa/parenrightBigg eiS(ϕ)/parenleftBigg 1 nn/summationdisplay b=1ϕ† b(0)ϕb(0)/parenrightBigg (7) For positive integer nthe field theory is perfectly well defined, so the delicate step in the replica approach is in taking the n→0 limit. There is a fascinating literature on this limit. (Consult a book devoted to spin glasses.) Some particle theorists used to speak disparagingly of condensed matter physics as dirt physics, and indeed the influence of impurities and disorder on matter is one of the centralconcerns of modern condensed matter physics. But as we see from this example, in manyrespects there is no mathematical difference between averaging over randomness andsumming over quantum fluctuations. We end up with a ϕ 4field theory of the type that many particle theorists have devoted considerable effort to studying in the past. Furthermore,Anderson’s surprising result that for D=2 any amount of disorder, no matter how small, localizes all states means that we have to understand the field theory defined by (6) ina highly nontrivial way. The strength of the disorder shows up as the coupling g 2,s o no amount of perturbation theory in g2can help us understand localization. Anderson localization is an intrinsically nonperturbative effect. Grassmannian approach As I mentioned earlier, people have dreamed up not one, but two, tricks in dealing withthe nasty denominator. The second trick is based on what we learned in chapter II.5 onintegration over Grassmann variables: Let η(x) and¯η(x) be Grassmann fields, then /integraldisplay DηD¯ηe−/integraltext d4x¯ηKη=CdetK=/tildewideC/parenleftbigg/integraldisplay DϕDϕ†e−/integraltext d4xϕ†Kϕ/parenrightbigg−1 where Cand/tildewideCare two uninteresting constants that we can absorb into the definition of DηD¯η. With this identity we can write (3) as tr1 z−H=i/integraldisplay dDy/integraldisplay Dϕ†DϕDηD ¯ηei/integraltext dDx{{∂ϕ†∂ϕ+[V( x)−z]ϕ†ϕ}+{∂¯η∂η+[ V( x)−z]¯ηη}}ϕ(y)ϕ†(y) (8) and then easily average over the disorder to obtain (per unit volume) /angbracketleftbigg tr1 z−H/angbracketrightbigg =i/integraldisplay Dϕ†DϕDηD ¯ηei/integraltext dDxL(¯η,η,ϕ†,ϕ)ϕ(0)ϕ†(0) (9) with L(¯η,η,ϕ†,ϕ)=∂ϕ†∂ϕ+∂¯η∂η−z(ϕ†ϕ+¯ηη)+ig2 2(ϕ†ϕ+¯ηη)2(10) VI.7. Disorder | 355 We end up with a field theory with bosonic (commuting) fields ϕ†andϕand fermionic (anticommuting) fields ¯ηandηinteracting with a strength determined by the disorder. The action Sexhibits an obvious symmetry rotating bosonic fields into fermionic fields and vice versa, and hence this approach is known in the condensed matter physics communityas the supersymmetric method. (It is perhaps worth emphasizing that ¯ηandηare not spinor fields, which we underline by not writing them as ¯ψandψ. The supersymmetry here, perhaps better referred to as Grassmannian symmetry, is quite different from thesupersymmetry in particle physics to be discussed in chapter VIII.4.) Both the replica and the supersymmetry approaches have their difficulties, and I was not kidding when I said that if you manage to invent a new approach without some of thesedifficulties it will be met with considerable excitement by condensed matter physicists. Probing localization I have shown you how to calculate the averaged density of states ρ(E) . How do we study localization? I will let you develop the answer in an exercise. From our earlier discussionit should be clear that we have to study an object obtained from (3) by replacing ϕ(y)ϕ †(y) byϕ(x)ϕ†(y)ϕ(y)ϕ†(x). If we choose to think of the replica field theory in the language of particle physics as describing the interaction of some scalar meson, then rather pleasingly,we see that the density of states is determined by the meson propagator and localizationis determined by meson-meson scattering. Exercises VI.7.1 Work out the field theory that will allow you to study Anderson localization. [Hint: Consider the object /angbracketleftbigg/parenleftbigg1 z−H/parenrightbigg (x,y)/parenleftbigg1 w−H/parenrightbigg (y,x)/angbracketrightbigg for two complex numbers zandw. You will have to introduce two sets of replica fields, commonly denoted byϕ+ aandϕ− a.] {Notation: [1 /(z−H)](x,y)denotes the xyelement of the matrix or operator [1 /(z−H)].} VI.7.2 As another example from the literature on disorder, consider the following problem. Place Npoints randomly in a D-dimensional Euclidean space of volume V. Denote the locations of the points by /vectorxi(i=1 ,..., N). Let f(/vectorx)=(−)/integraldisplaydDk (2π)Dei/vectork/vectorx k2+m2 Consider the NbyNmatrix Hij=f/parenleftbig /vectorxi−/vectorxj/parenrightbig . Calculate ρ(E) , the density of eigenvalues of Has we average over the ensemble of matrices, in the limit N→∞ ,V→∞ , with the density of points ρ≡N/V (not to be confused with ρ(E) of course) held fixed. [Hint: Use the replica method and arrive at the field theory action S(ϕ)=/integraldisplay dDx/bracketleftBiggn/summationdisplay a=1(|∇ϕa|2+m2|ϕa|2)−ρe−(1/z)/summationtextn a=1|ϕa|2/bracketrightBigg This problem is not entirely trivial; if you need help consult M. M ´ezard et al., Nucl. Phy. B559: 689, 2000, cond-mat/9906135. VI.8Renormalization Group Flow as a Natural Concept in High Energy and Condensed Matter Physics Therefore, conclusions based on the renormalization group arguments...a r e dangerous and must be viewed with due caution. So is it with all conclusions from localrelativistic field theories. —J. Bjorken and S. Drell, 1965 It is not dangerous The renormalization group represents the most important conceptual advance in quantumfield theory over the last three or four decades. The basic ideas were developed simultane-ously in both the high energy and condensed matter physics communities, and in someareas of research renormalization group flow has become part of the working language. As you can easily imagine, this is an immensely rich and multifaceted subject, which we can discuss from many different points of view, and a full exposition would require a bookin itself. Unfortunately, there has never been a completely satisfactory and comprehensivetreatment of the subject. The discussions in some of the older books are downrightmisleading and confused, such as the well-known text from which I learned quantum fieldtheory and from which the quote above was taken. In the limited space available here, Iwill attempt to give you a flavor of the subject rather than all the possible technical details.I will first approach it from the point of view of high energy physics and then from thatof condensed matter physics. As ever, the emphasis will be on the conceptual rather thanthe computational. As you will see, in spite of the order of my presentation, it is easierto grasp the role of the renormalization group in condensed matter physics than in highenergy physics. I laid the foundation for the renormalization group in chapter III.1—I do plan ahead! Let us go back to our experimentalist friend with whom we were discussing λϕ 4theory. We will continue to pretend that our world is described by a simple λϕ4theory and that an approximation to order λ2suffices. VI.8. Renormalization Group Flow | 357 What experimentalists insist on Our experimentalist friend was not interested in the coupling constant λwe wrote down on a piece of paper, a mere Greek letter to her. She insisted that she would accept onlyquantities she and her experimental colleagues can actually measure, even if only inprinciple. As a result of our discussion with her we sharpened our understanding of whata coupling constant is and learned that we should define a physical coupling constant by[see (III.1.4)] λP(μ)=λ−3Cλ2log/parenleftbigg/Lambda12 μ2/parenrightbigg +O(λ3) (1) At her insistence, we learned to express our result for physical amplitudes in terms of λP(μ), and not in terms of the theoretical construct λ. In particular, we should write the meson-meson scattering amplitude as M=−iλP(μ)+iCλP(μ)2/bracketleftbigg log/parenleftbiggμ2 s/parenrightbigg +log/parenleftbiggμ2 t/parenrightbigg +log/parenleftbiggμ2 u/parenrightbigg/bracketrightbigg +O[λP(μ)3] (2) What is the physical significance of λP(μ)? T o be sure, it measures the strength of the interaction between mesons as reflected in (2). But why one particular choice of μ? Clearly, from (2) we see that λP(μ) is particularly convenient for studying physics in the regime in which the kinematic variables s,t, and uare all of order μ2. The scattering amplitude is given by −iλP(μ) plus small logarithmic corrections. (Recall from a footnote in chapter III.3 that the renormalization point s0=t0=u0=μ2is adopted purely for theoretical convenience and cannot be reached in actual experiments. For our conceptualunderstanding here this is not a relevant issue.) In short, λ P(μ) is known as the coupling constant appropriate for physics at the energy scale μ. In contrast, if we were so idiotic as to use the coupling constant λP(μ/prime)while exploring physics in the regime with s,t, anduof order μ2, with μ/primevastly different from μ, then we would have a scattering amplitude M=−iλP(μ/prime)+iCλP(μ/prime)2/bracketleftbigg log/parenleftbiggμ/prime2 s/parenrightbigg +log/parenleftbiggμ/prime2 t/parenrightbigg +log/parenleftbiggμ/prime2 u/parenrightbigg/bracketrightbigg +O[λP(μ/prime)3] (3) in which the second term [with log (μ/prime2/μ2)large] can be comparable to or larger than the first term. The coupling constant λP(μ/prime)is not a convenient choice. Thus, for each energy scaleμthere is an “appropriate” coupling constant λP(μ). Subtracting (2) from (3) we can easily relate λP(μ) andλP(μ/prime)forμ∼μ/prime: λP(μ/prime)=λP(μ)+3CλP(μ)2log/parenleftbiggμ/prime2 μ2/parenrightbigg +O[λP(μ)3] (4) We can express this as a differential “flow equation” μd dμλP(μ)=6CλP(μ)2+O(λ3 P) (5) 358 | VI. Field Theory and Condensed Matter As you have already seen repeatedly, quantum field theory is full of historical misnomers. The description of how λP(μ) changes with μis known as the renormalization group. The only appearance of a group concept here is the additive group of transformationμ→μ+δμ. For the conceptual discussion in chapter III.1 and here, we don’t need to know what the constant Chappens to be. If Chappens to be negative, then the coupling λ P(μ) will decrease as the energy scale μincreases, and the opposite will occur if Chappens to be positive. (In fact, the sign is positive, so that as we increase the energy scale, λPflows away from the origin.) Flow of the electromagnetic coupling The behavior of λis typical of coupling constants in 4-dimensional quantum field theories. For example, in quantum electrodynamics, the coupling eor equivalently α=e2/4π, measures the strength of the electromagnetic interaction. The story is exactly as that toldfor the λϕ 4theory: Our experimentalist friend is not interested in the Latin letter e, but wants to know the actual interaction when the relevant momenta squared are of the orderμ 2. Happily, we have already done the computation: We can read off the effective coupling at momentum transferred squared q2=μ2from (III.7.14): eP(μ)2=e2 1 1+e2/Pi1(μ2)/similarequale2[1−e2/Pi1(μ2)+O(e4)] T akeμmuch larger than the electron mass mbut much smaller than the cutoff mass M. Then from (III.7.13) μd dμeP(μ)=−1 2e3μd dμ/Pi1(μ2)+O(e5)=+1 12π2e3 P+O(e5 P) (6) We learn that the electromagnetic coupling increases as the energy scale increases. Electromagnetism becomes stronger as we go to higher energies, or equivalently shorterdistances. Physically, the origin of this phenomenon is closely related to the physics of dielectrics. Consider a photon interacting with an electron, which we will call the test electron toavoid confusion in what follows. Due to quantum fluctuations, as described way back inchapter I.1, spacetime is full of electron-positron pairs, popping in and out of existence.Near the test electron, the electrons in these virtual pairs are repelled by the test electronand thus tend to move away from the test electron while the positrons tend to move towardthe test electron. Thus, at long distances, the charge of the test electron is shielded to someextent by the cloud of positrons, causing a weaker coupling to the photon, while at shortdistances the coupling to the photon becomes stronger. The quantum vacuum is just asmuch a dielectric as a lump of actual material. You may have noticed by now that the very name “coupling constant” is a terrible misnomer due to the fact that historically much of physics was done at essentially oneenergy scale, namely “almost zero”! In particular, people speak of the fine structure VI.8. Renormalization Group Flow | 359 “constant” α=1/137 and crackpots continue to try to “derive” the number 137 from numerology or some fancier method. In fact, αis merely the coupling “constant” of the electromagnetic interaction at very low energies. It is an experimental fact that α, more properly written as αP(μ)≡e2 P(μ)/4π, varies with the energy scale μwe are exploring. But alas, we are probably stuck with the name “coupling constant.” Renormalization group flow In general, in a quantum field theory with a coupling constant g, we have the renormal- ization group flow equation μdg dμ=β(g) (7) which is sometimes written as dg/dt =β(g) upon defining t≡log(μ/μ 0). I will now suppress the subscript Pon physical coupling constants. If the theory happens to have several coupling constants gi,i=1,... ,N, then we have dgi dt=βi(g1,... ,gN) (8) We can think of (g1,... ,gN)as the coordinate of a particle in N-dimensional space, tas time, and βi(g1,... ,gN)a position dependent velocity field. As we increase μort we would like to study how the particle moves or flows. For notational simplicity, we willnow denote (g 1,... ,gN)collectively as g. Clearly, those couplings at which βi(g∗)(for all i)happen to vanish are of particular interest: g∗is known as a fixed point. If the velocity field around a fixed point g∗is such that the particle moves toward that point (and once reaching it stays there since its velocity is now zero) the fixed point is known as attractive orstable. Thus, to study the asymptotic behavior of a quantum field theory at high energieswe “merely” have to find all its attractive fixed points under the renormalization groupflow. In a given theory, we can typically see that some couplings are flowing toward largervalues while others are flowing toward zero. Unfortunately, this wonderful theoretical picture is difficult to implement in practice because we essentially have no way of calculating the functions β i(g). In particular, g∗could well be quite large, associated with what is known as a strong coupling fixed point, and perturbation theory and Feynman diagrams are of no use in determining the propertiesof the theory there. Indeed, we know the fixed point structure of very few theories. Happily, we know of one particularly simple fixed point, namely g ∗=0, at which perturbation theory is certainly applicable. We can always evaluate (8) perturbatively: dgi/dt=cjk igjgk+djkl igjgkgl+... . (In some theories, the series starts with quadratic terms and in others, with cubic terms. Sometimes there is also a linear term.) Thus, as we have already seen in a couple of examples, the asymptotic or high energy behavior of the theory depends on the sign of βiin (8). Let us now join the film “Physics History” already in progress. In the late 1960s, experimentalists studying the so-called deep inelastic scattering of electrons on protons 360 | VI. Field Theory and Condensed Matter discovered that their data seemed to indicate that after being hit by a highly energetic electron, one of the quarks inside the proton would propagate freely without interactingstrongly with the other quarks. Normally, of course, the three quarks inside the proton arestrongly bound to each other to form the proton. Eventually, a few theorists realized thatthis puzzling state of affairs could be explained if the theory of strong interaction is suchthat the coupling flows toward the fixed point g ∗=0. If so, then the strong interaction between quarks would actually weaken at higher and higher energy scales. All of this is of course now “obvious” with the benefit of hindsight, but dear students, remember that at that time field theory was pronounced as possibly unsuitable for youngminds and the renormalizable group was considered “dangerous” even in a field theorytext! The theory of strong interaction was unknown. But if we were so bold as to accept the dangerous renormalization group ideas then we might even find the theory of the stronginteraction by searching for asymptotically free theories, which is what theories with anattractive fixed point at g ∗=0 became known as. Asymptotically free theories are clearly wonderful. Their behavior at high energies can be studied using perturbative methods. And so in this way the fundamental theory of thestrong interaction, now known as quantum chromodynamics, about which more later, wasfound. Looking at physics on different length scales The need for renormalization groups is really transparent in condensed matter physics.Instead of generalities, let me focus on a particularly clear example, namely surfacegrowth. Indeed, that was why I chose to introduce the Kardar-Parisi-Zhang equation inchapter VI.6. We learned that to study surface growth we have to evaluate the functionalor path integral Z(/Lambda1) =/integraldisplay /Lambda1Dhe−S(h). (9) with, you will recall, S(h)=1 2/integraldisplay dD/vectorxd t/parenleftbigg∂h ∂t−∇2h−g 2(∇h)2/parenrightbigg2 . (10) This defines a field theory. As with any field theory, and as I indicate, a cutoff /Lambda1has to be introduced. We integrate over only those field configurations h(/vectorx,t)that do not contain Fourier components with /vectorkandωlarger than /Lambda1. (In principle, since this is a nonrelativistic theory we should have different cutoffs for /vectorkand for ω, but for simplicity of exposition let us just refer to them together generically as /Lambda1.)The appearance of the cutoff is completely physical and necessary. At the very least, on length scales comparable to the size of therelevant molecules, the continuum description in terms of the field h(/vectorx,t)has long since broken down. Physically, since the random driving term η(/vectorx,t)is a white noise, that is, ηat/vectorxand at/vectorx /prime(and also at different times) are not correlated at all, we expect the surface to look VI.8. Renormalization Group Flow | 361 (a) (b) Figure VI.8.1 very uneven on a microscopic scale, as depicted in figure VI.8.1a. But suppose we are not interested in the detailed microscopic structure, but more in how the surface behaves ona larger scale. In other words, we are content to put on blurry glasses so that the surfaceappears as in figure VI.8.1b. This is a completely natural way to study a physical system,one that we are totally familiar with from day one in studying physics. We may be interestedin physics over some length scale Land do not care about what happens on length scales much less than L. The renormalization group is the formalism that allows us to relate the physics on differ- ent length scales or, equivalently, physics on different energy scales. In condensed matterphysics, one tends to think of length scales, and in particle physics, of energy scales. Themodern approach to renormalization groups came out of the study of critical phenomenaby Kadanoff, Fisher, Wilson, and others, as mentioned in chapter V .3. Consider, for exam-ple, the Ising model, with the spin at each site either up or down and with a ferromagneticinteraction between neighboring spins. At high temperatures, the spins point randomlyup and down. As the temperature drops toward the ferromagnetic transition point, islandsof up spins (we say up spins to be definite, we could just as easily talk of down spins) startto appear. They grow ever larger in size until the critical temperature T cat which all the spins in the entire system point up. The characteristic length scale of the physics at any particular temperature is given by the typical size of the islands. The physically motivatedblock spin method of Kadanoff et al. treats blocks of up spin as one single effective upspin, and similarly blocks of down spins. The notion of a renormalization group is thenthe natural one for describing these effective spins by an effective Hamiltonian appropriateto that length scale. It is more or less clear how to implement this physical idea of changing length scales in the functional integral (9). We are supposed to integrate over those h(/vectork,ω), with /vectorkandω less than /Lambda1. Suppose we do only a fraction of what we are supposed to do. Let us integrate 362 | VI. Field Theory and Condensed Matter over those h(/vectork,ω)with/vectorkandωlarger than /Lambda1−δ/Lambda1 but smaller than /Lambda1. This is precisely what we mean when we say that we don’t care about the fluctuations of h(/vectorx,t)on length and time scales less than (/Lambda1 −δ/Lambda1)−1. Putting on blurry glasses For the sake of simplicity, let us go back to our favorite, the λϕ4theory, instead of the surface growth problem. Recall from the preceding chapters the importance of the Euclidean λϕ4 theory in modern condensed matter theory. So, continue the λϕ4theory to Euclidean space and stare at the integral Z(/Lambda1) =/integraldisplay /Lambda1Dϕe−/integraltext ddxL(ϕ)(11) The notation/integraltext /Lambda1instructs us to include only those field configurations ϕ(x)=/integraltext [ddk/(2π)d]eikxϕ(k) such that ϕ(k)=0 for|k|≡(/summationtextd i=1k2 i)1 2larger than /Lambda1. As explained in the text this amounts to putting on blurry glasses with resolution L=1//Lambda1:W ed on o t admit or see fluctuations with length scales less than L. Evidently, the O(d) invariance, namely the Euclidean equivalent of Lorentz invariance, will make our lives considerably easier. In contrast, for the surface growth problem we willneed special glasses that blur space and time differently. 1 We are now ready to make our glasses blurrier by letting /Lambda1→/Lambda1−δ/Lambda1 (withδ/Lambda1 > 0). Write ϕ=ϕs+ϕw(sfor “smooth” and wfor “wriggly”), defined such that the Fourier components ϕs(k) andϕw(k) are nonzero only for |k|≤(/Lambda1 −δ/Lambda1) and (/Lambda1 −δ/Lambda1)≤|k|≤ /Lambda1, respectively. (Obviously, the designation “smooth” and “wriggly” is for convenience.)Plugging into (11) we can write Z(/Lambda1) =/integraldisplay /Lambda1−δ/Lambda1Dϕse−/integraltext ddxL(ϕs)/integraldisplay Dϕwe−/integraltext ddxL1(ϕs,ϕw)(12) where all the terms in L1(ϕs,ϕw)depend on ϕw. (What we are doing here is somewhat reminiscent of what we did in chapter IV .3.) Imagine doing the integral over ϕw. Call the result e−/integraltext ddxδL(ϕs)≡/integraldisplay Dϕwe−/integraltext ddxL1(ϕs,ϕw) and thus we have Z(/Lambda1) =/integraldisplay /Lambda1−δ/Lambda1Dϕse−/integraltext ddx[L(ϕs)+δL(ϕs)](13) There, we have done it! We have rewritten the theory in terms of the “smooth” field ϕs. Of course, this is all formal, since in practice the integral over ϕwcan only be done perturbatively assuming that the relevant couplings are small. If we could do the integral 1In condensed matter physics, the so-called dynamical exponent zmeasures this difference. More precisely, in the context of the surface growth problem, the correlator (introduced in chapter VI.6) satisfies the dynamicscaling form given in (VI.6.3). Naively, the dynamical exponent zshould be 2. (For a brief review of all this, see M. Kardar and A. Zee, Nucl. Phys. B464[FS]: 449, 1996, cond-mat/9507112.) VI.8. Renormalization Group Flow | 363 overϕwexactly, we might as well just do the integral over ϕand then we would have no need for all this renormalization group stuff. For pedagogical purposes, consider more generally L=1 2(∂ϕ)2+/summationtext nλnϕn+... (so thatλ2is the usual1 2m2andλ4the usual λ.)Since terms such as ∂ϕs∂ϕwintegrate to zero, we have /integraldisplay ddxL1(ϕs,ϕw)=/integraldisplay ddx/parenleftbigg1 2(∂ϕw)2+1 2m2ϕ2 w+.../parenrightbigg withϕshiding in the (...). This describes a field ϕwinteracting with both itself and a background field ϕs(x). By symmetry considerations δL(ϕs)has the same form as L(ϕs) but with different coefficients. Adding δL(ϕs)toL(ϕs)thus shifts2the couplings λn[and the coefficient of1 2(∂ϕs)2.] These shifts generate the flow in the space of couplings I described earlier. We could have perfectly well left (13) as our end result. But suppose we want to compare (13) with (11). Then we would like to change the/integraltext /Lambda1−δ/Lambda1in (13) to/integraltext /Lambda1. For convenience, introduce the real number b< 1b y/Lambda1−δ/Lambda1=b/Lambda1.I n/integraltext /Lambda1−δ/Lambda1we are told to integrate over fields with |k|≤b/Lambda1 . So all we have to do is make a trivial change of variable: Let k=bk/prime so that |k/prime|≤/Lambda1 . But then correspondingly we have to change x=x/prime/bso that eikx=eik/primex/prime. Plugging in, we obtain /integraldisplay ddxL(ϕs)=/integraldisplay ddx/primeb−d/bracketleftBigg 1 2b2(∂/primeϕs)2+/summationdisplay nλnϕn s+.../bracketrightBigg (14) where ∂/prime=∂/∂x/prime=(1/b)∂/∂x . Define ϕ/primebyb2−d(∂/primeϕs)2=(∂/primeϕ/prime)2or in other words ϕ/prime= b1 2(2−d)ϕs. Then (14) becomes /integraldisplay ddx/prime/bracketleftBigg 1 2(∂/primeϕ/prime)2+/summationdisplay nλnb−d+(n/2 )(d−2)ϕ/primen+.../bracketrightBigg Thus, if we define the coefficient of ϕ/primenasλ/prime nwe have λ/prime n=b(n/2)(d−2)−dλn (15) an important result in renormalization group theory. Relevant, irrelevant, and marginal Let us absorb what this means. (For the time being, let us ignore δL(ϕs)to keep the discussion simple.) As we put on blurrier glasses, in other words, as we become interestedin physics over longer distance scales, we can once again write Z(/Lambda1) as in (11) except that the couplings λ nhave to be replaced by λ/prime n. Since b< 1 we see from (15) that the λn’s with (n/2)(d−2)−d> 0 get smaller and smaller and can eventually be neglected. A dose of jargon here: The corresponding operators ϕn(for historical reasons we revert for an instant 2T erms such as (∂ϕ)4can also be generated and that is why I wrote L(ϕ) with the (...)under which terms such as these can be swept. You can check later that for most applications these terms are irrelevant in the technicalsense to be defined below. 364 | VI. Field Theory and Condensed Matter from the functional integral language to the operator language) are called irrelevant. They are the losers. Conversely, the winners, namely the ϕn’s for which (n/2)(d−2)−d< 0, are called relevant. Operators for which (n/2)(d−2)−d=0 are called marginal. For example, take n=2:m/prime2=b−2m2and the mass term is always relevant in any dimension. On the other hand, take n=4, and we see that λ/prime=bd−4λandϕ4is relevant ford< 4, irrelevant for d> 4, and marginal at d=4. Similarly, λ/prime 6=b2d−6λandϕ6is marginal at d=3 and becomes irrelevant for d> 3. We also see that d=2 is special: All the ϕn’s are relevant. Now all this may ring a bell if you did the exercises religiously. In exercise III.2.1 you showed that the coupling λnhas mass dimension [ λn]=(n/2)(2−d)+d. Thus, the quantity (n/2)(d−2)−dis just the length dimension of λn. For example, for d=4, λ6has mass dimension −2 and thus as explained in chapter III.2 the ϕ6interaction is nonrenormalizable, namely that it has nasty behavior at high energy. But condensed matterphysicists are interested in the long distance limit, the opposite limit from the one thatinterests particle physicsts. Thus, it is the nasty guys like ϕ 6that become irrelevant in the long distance limit. One more piece of jargon: Given a scalar field theory, the dimension dat which the most relevant interaction becomes marginal is known as the critical dimension in condensedmatter physics. For example, the critical dimension for a ϕ 6theory is 3. It is now just a matter of “high school arithmetic” to translate (15) into differential form. Write λ/prime n= λn+δλn; then from b=1−(δ/Lambda1//Lambda1) we have δλn=− [n 2(d−2)−d]λn(δ/Lambda1//Lambda1) . Let us now be extra careful about signs. As I have already remarked, for (n/2)(d−2)− d> 0 the coupling λn(which, for definiteness, we will think of as positive) get smaller, as is evident from (15). But since we are decreasing /Lambda1to/Lambda1−δ/Lambda1, a positive δ/Lambda1 actually corresponds to the resolution of our blurry glasses L=/Lambda1−1changing to L+L(δ/Lambda1//Lambda1) . Thus we obtain Ldλn dL=−/bracketleftbiggn 2(d−2)−d/bracketrightbigg λn, (16) so that for (n/2)(d−2)−d> 0 a positive λnwould decrease as Lincreases.3 In particular, for n=4,L(dλ/dL) =(4−d)λ . In most condensed matter physics appli- cations, d≤3 and so λincreases as the length scale of the physics under study increases. Theϕ4coupling is relevant as noted above. TheδL(ϕs), which we provisionally neglected, contributes an additional term, which we call dynamical in contrast to the geometrical or “trivial” term displayed, to the right-hand side of (16). Thus, in general L(dλn/dL)=− [(n/2)(d−2)−d]λn+K(d ,n,...,λj,...), with the dynamical term Kdepending not only on dandn, but also on all the other couplings. [For example, in (5) the “trivial” term vanishes since we are in 4-dimensionalspacetime; there is only a dynamical contribution.] 3Note that what appears on the right-hand side is minus the length dimension of λn, not the length dimension (n/2)(d−2)−das one might have guessed naively. VI.8. Renormalization Group Flow | 365 As you can see from this discussion, a more descriptive name for the renormalization group might be “the trick of doing an integral a little bit at a time.” Exploiting symmetry T o determine the renormalization group flow of the coupling gin the surface growth problem we can repeat the same type of computation we did to determine the flow of thecoupling λand of ein our two previous examples, namely we would calculate, to use the language of particle physics, the amplitude for h-h scattering to one loop order. But instead, let us follow the physical picture of Kadanoff et al. In Z(/Lambda1)=/integraltext Dhe −S(h)we integrate over only those h(/vectork,ω)with/vectorkandωlarger than /Lambda1−δ/Lambda1 but less than /Lambda1. I will now show you how to exploit the symmetry of the problem to minimize our labor. The important thing is not necessarily to learn about the dynamics of surface growth,but to learn the methodology that will serve you well in other situations. I have pickeda particularly “difficult” nonrelativistic problem whose symmetries are not manifest, sothat if you master the renormalization group for this problem you will be ready for almostanything. Imagine having done this partial integration and call the result/integraltext Dhe −˜S(h). At this point you should work out the symmetries of the problem as indicated in the exercises. Thenyou can argue that ˜S(h) must have the form ˜S(h)=1 2/integraldisplay dDxd t/bracketleftbigg/parenleftbigg α∂ ∂t−β∇2/parenrightbigg h−αg 2(∇h)2/bracketrightbigg2 +... , (17) depending on two parameters αandβ. The ( ...)indicates terms involving higher powers ofhand its derivatives. The simplifying observation is that the same coefficient αmultiplies both∂h/∂t and(g/2)(∇h)2. Once we know αandβthen by suitable rescaling we can bring the action ˜S(h) back into the same form as S(h) and thus find out how gchanges. Therefore, it suffices to look at the (∂h/∂t)2and(∇2h)2terms in the action, or equivalently at the propagator, which is considerably simpler to calculate. As Rudolf Peierls once said4 to the young Hans Bethe, “Erst kommt das Denken, dann das Integral.” (Roughly, “Firstthink, then do the integral.”) We will not do the computation here. Suffice it to note that g has the high school dimension of (length) 1 2(D−2)(see exercise VI.8.5). Thus, according to the preceding discussion we should have Ldg dL=1 2(2−D)g+cDg3+... (18) A detailed calculation is needed to determine the coefficient cD, which obviously depends on the dimension of space Dsince the Feynman integrals depend on D. The equation tells us how g, an effective measure of nonlinearity in the physics of 4John Wheeler gave me similar advice when I was a student: “Never calculate without first knowing the answer.” 366 | VI. Field Theory and Condensed Matter surface growth, changes when we change the length scale L. For the record, cD= [S(D)/ 4(2π)D](2D−3)/D , with S(D) theD-dimensional solid angle. The interesting factor is of course (2D−3), changing sign between5D=1 and 2. Localization As I said earlier, renormalization group flow has literally become part of the language of condensed matter and high energy physics. Let me give you another example of the powerof the renormalization group. Go back to Anderson localization (chapter VI.7), which soastonished the community at the time. People were surprised that the localization behaviordepends so drastically on the dimension of space D, and perhaps even more so, that for D=2 all states are localized no matter how weak the strength of the disorder. Our usual physical intuition would say that there is a critical strength. As we will now see, bothfeatures are quite naturally accounted for in the renormalization group language. Already,you see in (18) that Denters in an essential way. I now offer you a heuristic but beautiful (at least to me) argument given by Abra- hams, Anderson, Licciardello, and Ramakrishnan, who as a result became known to thecondensed matter community as the “Gang of Four.” First, you have to understand thedifference between conductivity σand conductance Gin solid state physics lingo. Conduc- tivity 6is defined by /vectorJ=σ/vectorE, where /vectorJmeasures the number of electrons passing through a unit area per unit time. Conductance Gis the inverse of resistance (the mnemonic: the two words rhyme). Resistance Ris the property of a lump of material and defined in high school physics by V=IR, where the current Imeasures the number of electrons passing by per unit time. T o relate σandG, consider a lump of material, taken to be a cube of size L, with a voltage drop Vacross it. Then I=JL2=σEL2=σ(V/L)L2=σLV and thus7 G(L)=1/R=I/V=σL. Next, let us go to two dimensions. Consider a thin sheet of ma- terial of length and width Land thickness a/lessmuchL. (We are doing real high school physics, not talking about some sophisticated field theorist’s idea of two dimensional space!) Again,apply a voltage drop Vover the length L:I=J(aL) =σEaL =σ(V/L)aL =σVa and so G(L)=1/R=I/V=σa. I will let you go on to one dimension: Consider a wire of length Land width and thickness a. In this way, we obtain G(L)∝L D−2. Incidentally, condensed matter physicists customarily define a dimensionless conductance g(L)≡/planckover2piG(L)/e2. 5Incidentally, the theory is exactly solvable for D=1 (with methods not discussed in this book). 6Over the years I have asked a number of high energy theorists how is it possible to obtain /vectorJ=σ/vectorE, which manifestly violates time reversal invariance, if the microscopic physics of an electron scattering on an impurityatom perfectly well respects time reversal invariance. Very few knew the answer. The resolution of this apparentparadox is in the order of limits! Condensed matter theorists calculate a frequency and wave vector dependent conductivity σ(ω ,/vectork)and then take the limit ω,/vectork→0 and /vectork 2/ω→0. Before the limit is taken, time reversal invariance holds. The time it takes the particle to find out that it is in a box of size of order 1 /kis of order 1 /(Dk2) (withDthe diffusion constant). The physics is that this time has to be much longer than the observation time ∼1/ω. 7Sam T reiman told me that when he joined the U.S. Army as a radio operator he was taught that there were three forms of Ohm’s law: V=IR,I=V/R , andR=V/I . In the second equality here we use the fourth form. VI.8. Renormalization Group Flow | 367 1 −1β(g) gcgD = 3 D = 2 D = 1 Figure VI.8.2 We also know the behavior of g(L) when g(L) is small or, in other words, when the material is an insulator for which we expect g(L)∼ce−L/ξ, with ξsome length charac- teristic of the material and determined by the microscopic physics. Thus, for g(L) small, L(dg/dL) =−(L/ξ)g(L) =g(L)[log g(L)−logc], where the constant log cis negligible in the regime under consideration. Putting things together, we obtain β(g)≡L gdg dL=/braceleftBigg(D−2)+... for large g logg+... for small g(19) First, a trivial note: in different subjects, people define β(g) differently (without affecting the physics of course). In localization theory, β(g) is traditionally defined as dlogg/d logL as indicated here. Given (19) we can now make a “most plausible” plot of β(g) as shown in figure VI.8.2. You see that for D=2 (and D=1)the conductance g(L) always flows toward 0 as we go to long distances (macroscopic measurements on macroscopic materials)regardless of where we start. In contrast, for D=3, ifg 0the initial value of gis greater than a critical gctheng(L) flows to infinity (presumably cut off by physics we haven’t included) and the material is a metal, while if g0<gc, the material is an insulator. Incidentally, condensed matter theorists often speak of a critical dimension Dcat which the long distance behavior of a system changes drastically; in this case Dc=2. Effective description In a sense, the renormalization group goes back to a basic notion of physics, that the effective description can and should change as we move from one length scale to another.For example, in hydrodynamics we do not have to keep track of the detailed interaction 368 | VI. Field Theory and Condensed Matter among water molecules. Similarly, when we apply the renormalization group flow to the strong interaction, starting at high energies and moving toward low energies, the effectivedescription goes from a theory of quarks and gluons to a theory of nucleons and mesons.In this more general picture then, we no longer think of flowing in a space of couplingconstants, but in “the space of Hamiltonians” that some condensed matter physicists liketo talk about. Exercises VI.8.1 Show that the solution of dg/dt =−bg3+...is given by 1 α(t)=1 α(0)+8πbt+... (20) where we defined α(t)=g(t)2/4π. VI.8.2 In our discussion of the renormalization group, in λϕ4theory or in QED, for the sake of simplicity we assumed that the mass mof the particle is much smaller than μand thus set mequal to zero. But nothing in the renormalization group idea tells us that we can’t flow to a mass scale below m. Indeed, in particle physics many orders of magnitude separate the top quark mass mtfrom the up quark mass mu. We might want to study how the strong interaction coupling flows from some mass scale far above mt down to some mass scale μbelow mtbut still large compared to mu. As a crude approximation, people often set any mass mbelow μequal to zero and any mabove μto infinity (i.e., not contributing to the renormalization group flow). In reality, as μapproaches mfrom above the particle starts to contribute less and drops out as μbecomes much less than m. T aking either the λϕ4theory or QED study this so-called threshold effect. VI.8.3 Show that (10) is invariant under the so-called Galilean transformation h(/vectorx,t)→h/prime(/vectorx,t)=h/parenleftbig /vectorx+g/vectorut,t/parenrightbig +/vectoru./vectorx+g 2u2t (21) Show that because of this symmetry only two parameters αandβappear in (17). VI.8.4 In˜S(h) only derivatives of the field hcan appear and not the field itself. (Since the transformation h(/vectorx,t)→h(/vectorx,t)+cwithca constant corresponds to a trivial shift of where we measure the surface height from, the physics must be invariant under this transformation.) T erms involving only one power ofhcannot appear since they are all total divergences. Thus, ˜S(h) must start with terms quadratic in h. Verify that the ˜S(h) given in (17) is indeed the most general. A term proportional to (∇h)2is also allowed by symmetries and is in fact generated. However, such a term can be eliminated by transforming to amoving coordinate frame h→h+ct. VI.8.5 Show that ghas the high school dimension of (length)1 2(D−2). [Hint: The form of S(h) implies that thas the dimension of length squared and so hhas the dimension (length)1 2(2−D).] Comparing the terms ∇2h andg(∇h)2we determine the dimension of g.] VI.8.6 Calculate the hpropagator to one loop order. Extract the coefficients of the ω2andk4terms in a low frequency and wave number expansion of the inverse propagator and determine αandβ. VI.8.7 Study the renormalization group flow of gforD=1, 2, 3. Part VII Grand Unification This page intentionally left blank VII.1Quantizing Yang-Mills Theory and Lattice Gauge Theory One reason that Y ang-Mills theory was not immediately taken up by physicists is that people did not know how to calculate with it. At the very least, we should be able towrite down the Feynman rules and calculate perturbatively. Feynman himself took up thechallenge and concluded, after looking at various diagrams, that extra fields with ghostlikeproperties had to be introduced for the theory to be consistent. Nowadays we know howto derive this result more systematically. The story goes that Feynman wanted to quantize gravity but Gell-Mann suggested to him to first quantize Y ang-Mills theory as a warm-up exercise. Consider pure Y ang-Mills theory—it will be easy to add matter fields later. Follow what we have learned. Split the Lagrangian L=L 0+L1as usual into two pieces (we also choose to scale A→gA) : L0=−1 4(∂μAa ν−∂νAa μ)2(1) and L1=−1 2g(∂μAaν−∂νAa μ)fabcAbμAcν−1 4g2fabcfadeAbμAcνAdμAeν(2) Then invert the differential operator in the quadratic piece (1) to obtain the propagator. This part looks the same as the corresponding procedure for quantum electrodynamics,except for the occurrence of the index a. Just as in electrodynamics, the inverse does not exist and we have to fix a gauge. I built up the elaborate Faddeev-Popov method to quantize quantum electrodynamics and as I noted, it was a bit of overkill in that context. But here comes the payoff: We can nowturn the crank. Recall from chapter III.4 that the Faddeev-Popov method would give us Z=/integraldisplay DAeiS(A)/Delta1(A)δ [f (A)] (3) with/Delta1(A) ≡{/integraltext Dgδ [f( Ag)]}−1andS(A)=/integraltext d4xLthe Y ang-Mills action. (As in chap- ter III.4, Ag≡gAg−1−i(∂g)g−1denotes the gauge transform of A. Here g≡g(x) denotes 372 | VII. Grand Unification the group element that defines the gauge transformation at xand is obviously not to be confused with the coupling constant.) Since /Delta1(A) appears in (3) multiplied by δ[f (A)], in the integral over gwe expect, for a reasonable choice of f (A) , only infinitesimal gto be relevant. Let us choose f (A)= ∂A−σ. Under an infinitesimal transformation, Aa μ→Aa μ−fabcθbAc μ+∂μθaand thus /Delta1(A) ={/integraldisplay Dθδ[∂Aa−σa−∂μ(fabcθbAc μ−∂μθa)]}−1(4) “=”{/integraldisplay Dθδ[∂μ(fabcθbAc μ−∂μθa)]}−1. where the “effectively equal sign” follows since /Delta1(A) is to be multiplied later by δ[f (A)]. Let us write formally ∂μ(fabcθbAc μ−∂μθa)=/integraldisplay d4yKab(x,y)θb(y) (5) thus defining the operator Kab(x,y)=∂μ(fabcAc μ−∂μδab)δ(4)(x−y). Note that in con- trast to electromagnetism here Kdepends on the gauge potential. The elementary result/integraltext dθδ(Kθ) =1/K forθandKreal numbers can be generalized to/integraltext dθδ(Kθ) =1/detK forθa real vector and Ka nonsingular matrix. Regarding Kab(x,y)as a matrix, we ob- tain/Delta1(A) =detK, but we know from chapter II.5 how to represent the determinant as a functional integral over Grassmann variables: Write /Delta1(A) =/integraltext DcDc†eiSghost(c†,c), with Sghost(c†,c)=/integraldisplay d4xd4yc† a(x)Kab(x,y)cb(y) =/integraldisplay d4x[∂c† a(x)∂c a(x)−∂μc† a(x)fabcAcμ(x)cb(x)] =/integraldisplay d4x∂c† a(x)Dc a(x) (6) and with Dthe covariant derivative for the adjoint representation, to which the fields ca andc† abelong just like Aa μ. The fields caandc† aare known as ghost fields because they violate the spin-statistics connection: Though scalar, they are treated as anticommuting.This “violation” is acceptable because they are not associated with physical particles andare introduced merely to represent /Delta1(A) in a convenient form. This takes care of the /Delta1(A) factor in (3). As for the δ[f (A)] factor, we use the same trick as in chapter III.4 and integrate Zoverσ a(x) with a Gaussian weight e−(i/2ξ)/integraltext d4xσa(x)2 so that δ[f (A)] gets replaced by e−(i/2ξ)/integraltext d4x(∂Aa)2. Putting it all together, we obtain Z=/integraldisplay DADcDc†eiS(A)−(i/2 ξ)/integraltext d4x(∂A)2+iS ghost(c†,c)(7) withξa gauge parameter. Comparing with the corresponding expression for an abelian gauge theory in chapter III.4, we see that in nonabelian gauge theories we have a ghost action Sghost in addition to the Y ang-Mills action. Thus, L0and L1are changed to L0=−1 4(∂μAa ν−∂νAa μ)2−1 2ξ(∂μAa μ)2+∂c† a∂ca (8) VII.1. Quantizing Yang-Mills Theory | 373 a,μ c,λ b,νk1 k3k2a,μb,ν d,ρc,λ c,μ abp(a) (b) (c) Figure VII.1.1 and L1=−1 2g(∂μAa ν−∂νAa μ)fabcAbμAcν+1 4g2fabcfadeAbμAcνAdμAeν−∂μc† agfabcAcμcb(x) (9) We can now read off the propagators for the gauge boson and for the ghost field imme- diately from (8). In particular, we see that except for the group index athe terms quadratic in the gauge potential are exactly the same as the terms quadratic in the electromagneticgauge potential in (III.4.8). Thus, the gauge boson propagator is (−i) k2/bracketleftbigg gνλ−(1−ξ)kνkλ k2/bracketrightbigg δab (10) Compare with (III.4.9). From the term ∂c† a∂cain (8) we find the ghost propagator to be (i/k2)δab. From L1we see that there is a cubic and a quartic interaction between the gauge bosons, and an interaction between the gauge boson and the ghost field, as illustrated in figure VII.1.1. The cubic and the quartic couplings can be easily read off as gfabc[gμν(k1−k2)λ+gνλ(k2−k3)μ+gλμ(k3−k1)ν] (11) and −ig2[fabefcde(gμλgνρ−gμρgνλ)+fadefcbe(gμλgνρ−gμνgρλ) +facefbde(gμνgλρ−gμρgνλ)] (12) respectively. The coupling to the ghost field is gfabcpμ(13) 374 | VII. Grand Unification Obviously, we can exploit various permutation symmetries in writing these down. For instance, in (12) the second term is obtained from the first by the interchange {c,λ}↔ {d,ρ}, and the third and fourth terms are obtained from the first and second by the interchange {a,μ}↔{ c,λ}. Unnatural act In a highly symmetric theory such as Y ang-Mills, perturbating is clearly an unnatural act as it involves brutally splitting Linto two parts: a part quadratic in the fields and the rest. Consider, for example, an exactly soluble single particle quantum mechanicsproblem, such as the Schr ¨odinger equation with V( x)=1−(1/coshx) 2. Imagine writing V( x)=1 2x2+W(x) and treating W(x) as a perturbation on the harmonic oscillator. You would have a hard time reproducing the exact spectrum, but this is exactly how we brutalizeY ang-Mills theory in the perturbative approach: We took the “holistic entity” tr F μνFμνand split it up into the “harmonic oscillator” piece tr (∂μAν−∂νAμ)2and a “perturbation.” If Y ang-Mills theory ever proves to be exactly soluble, the perturbative approach with its mangling of gauge invariance is clearly not the way to do it. Lattice gauge theory Wilson proposed a way out: Do violence to Lorentz invariance rather than to gauge in-variance. Let us formulate Y ang-Mills theory on a hypercubic lattice in 4-dimensionalEuclidean spacetime. As the lattice spacing a→0 we expect to recover 4-dimensional rotational invariance and (by a Wick rotation) Lorentz invariance. Wilson’s formulation,known as lattice gauge theory, is easy to understand, but the notation is a bit awkward, dueto the lack of rotational invariance. Denote the location of the lattice sites by the vector x i. On each link, say the one going from xito one of its nearest neighbors xj, we associate an NbyNsimple unitary matrix Uij. Consider the square, known as a plaquette, bounded by the four corners xi,xj,xk, and xl(with these nearest neighbors to each other.) See figure VII.1.2. For each plaquette Pwe associate the quantity S(P)=Re trUijUjkUklUli, constructed to be invariant under the local transformation Uij→V† iUijVj (14) The symmetry is local because for each site xiwe can associate an independent Vi. Wilson defined Y ang-Mills theory by Z=/integraldisplay /Pi1dUe(1/2f2)/summationtext PS(P)(15) where the sum is taken over all the plaquettes in the lattice. The coupling strength f controls how wildly the unitary matrices Uij’s fluctuate. For small f, large values of S(P) are favored, and so the Uij’s are all approximately equal to the unit matrix (up to an irrelevant global transformation.) VII.1. Quantizing Yang-Mills Theory | 375 xi xjxl xk Uli UjkUkl Uij Figure VII.1.2 Without doing any arithmetic, we can argue by symmetry that in the continuum limit a→0, Y ang-Mills theory as we know it must emerge: The action is manifestly invariant under local SU(N) transformation. T o actually see this, define a field Aμ(x) withμ= 1, 2, 3, 4, permeating the 4-dimensional Euclidean space the lattice lives in, by Uij=V† ieiaAμ(x)Vj (16) where x=1 2(xi+xj)(namely the midpoint of the link Uijlives on) and μis the direction connecting xitoxj(namely ˆμ≡(xj−xi)/a is the unit vector in the μdirection.) The V’s just reflect the gauge freedom in (14) and obviously do not enter into the plaquette actionS(P) by construction. I will let you show in an exercise that trUijUjkUklUli=treia2Fμν+O(a3)(17) withFμνthe Y ang-Mills field strength evaluated at the center of the plaquette. Indeed, we could have discovered the Y ang-Mills field strength in this way. I hope that you start tosee the deep geometric significance of F μν. Continuing the exercise you will find that the action on each plaquette comes out to be S(P)=Re treia2Fμν+O(a3) =Re tr[1 +ia2Fμν−1 2a4FμνFμν+O(a5)]=tr 1−1 2a4trFμνFμν+... (18) and so up to an irrelevant additive constant we recover in (15) the Y ang-Mills action in the continuum limit. Again, it is worth emphasizing that without going through any arithmeticwe could have fixed the a 4term in (18) (up to an overall constant) by dimensional analysis and gauge invariance.1 1The sign can be easily checked against the abelian case. 376 | VII. Grand Unification The Wilson formulation is beautiful in that none of the hand-wringing over gauge fixing, Faddeev-Popov determinant, ghost fields, and so forth is necessary for (15) to makesense. Recalling chapter V .3 you see that (15) defines a statistical mechanics problemlike any other. Instead of integrating over some spin variables say, we integrate over thegroup SU(N) for each link. Most importantly, the lattice gauge formulation opens up the possibility of computing the properties of a highly nontrivial quantum field theorynumerically. Lattice gauge theory is a thriving area of research. For a challenge, try toincorporate fermions into lattice gauge theory: This is a difficult and ongoing problembecause fermions and spinor fields are naturally associated with SO( 4), which does not sit well on a lattice. Wilson loop Field theorists usually deal with local observables, that is, observables defined at a space-time point x, such as J μ(x) or trFμν(x)Fμν(x), but of course we can also deal with nonlocal observables, such as ei/contintegraltext CdxμAμin electromagnetism, where the line integral is evaluated over a closed curve C. The gauge invariant quantity in the exponential is equal to the electromagnetic flux going through the surface bounded by C. (Indeed, recall chap- ter IV .4.) Wilson pointed out that lattice gauge theory contains a natural gauge invariant but nonlocal observable W(C) ≡trUijUjk...UnmUmi, where the set of links connecting xi toxjtoxket cetera and eventually to xmand back to xitraces out a loop called C. Referring to (16) we see that W(C) , known as the Wilson loop, is the trace of a product of many factors of eiaAμ. Thus, in the continuum limit a→0, we have evidently W(C) ≡trPei/contintegraltext CdxμAμ(19) withCnow an arbitrary curve in Euclidean spacetime. Here Pdenotes path ordering, clearly necessary since the Aμ’s associated with different segments of C, being matrices, do not commute with each other. [Indeed, Pis defined by the lattice definition of W(C) .] T o understand the physical meaning of the Wilson loop, Recall chapters I.4 and I.5. T o obtain the potential energy Ebetween two oppositely charged lumps we have to compute lim T→∞1 Z/integraldisplay DAeiSMaxwell (A)+i/integraltext d4xAμJμ=e−iET For two lumps held at a distance Rapart we plug in Jμ(x)=ημ0{δ(3)(/vectorx)−δ(3)[/vectorx−(R,0 ,0)]} and see that we are actually computing the expectation value /angbracketleftei(/integraltext C1dxμAμ−/integraltext C2dxμAμ)/angbracketrightin a fluctuating electromagnetic field, where C1andC2denote two straight line segments at/vectorx=(0, 0, 0 )and/vectorx=(R,0 ,0), respectively. It is convenient to imagine bringing the two lumps together in the far future (and similarly in the far past). Then we deal instead with the manifestly gauge invariant quantity /angbracketleftei/contintegraltext CdxμAμ/angbracketright, where Cis the rectangle shown VII.1. Quantizing Yang-Mills Theory | 377 T RTime Space C Figure VII.1.3 in figure VII.1.3. Note that for Tlarge log /angbracketleftei/contintegraltext CdxμAμ/angbracketright∼−iE(R)T , which is essentially proportional to the perimeter length of the rectangle C. As we will discuss in chapter VII.3 and as you have undoubtedly heard, the currently accepted theory of the strong interaction involves quarks coupled to a nonabelian Y ang-Mills gauge potential A μ. Thus, to determine the potential energy E(R) between a quark and an antiquark held fixed at a distance Rfrom each other we “merely” have to compute the expectation value of the Wilson loop /angbracketleftW(C) /angbracketright=1 Z/integraldisplay /Pi1dUe−(1/2f2)/summationtext PS(P)W(C) (20) In lattice gauge theory we could compute log /angbracketleftW(C) /angbracketrightforCthe large rectangle in fig- ure VII.1.3 numerically, and extract E(R) . (We lost the ibecause we are living in Euclidean spacetime for the purpose of this discussion.) Quark confinement You have also undoubtedly heard that since free quarks have not been observed, quarks aregenerally believed to be permanently confined. In particular, it is believed that the potentialenergy between a quark and an antiquark grows linearly with separation E(R)∼σR. One imagines a string tying the quark to the antiquark with a string tension σ. If this conjecture is correct, then log /angbracketleftW(C) /angbracketright∼σRT should go as the area RT enclosed by C. Wilson calls this behavior the area law, in contrast to the perimeter law characteristic offamiliar theories such as electromagnetism. T o prove the area law in Y ang-Mills theory isone of the outstanding challenges of theoretical physics. 378 | VII. Grand Unification Exercises VII.1.1 The gauge choice in the text preserves Lorentz invariance. It is often useful to choose a gauge that breaks Lorentz invariance, for example, f (A)=nμAμ(x)withnsome fixed 4-vector. This class of gauge choices, known as the axial gauge, contains various popular gauges, each of which corresponds to a particularchoice of n. For instance, in light-cone gauge, n=(1, 0, 0, 1 ), in space-cone gauge, n=(0, 1,i,0). Show that for any given A(x) we can find a gauge transformation so that n.A /prime(x)=0. VII.1.2 Derive (17) and relate fto the coupling gin the continuum formulation of Y ang-Mills theory. [Hint: Use the Baker-Campbell-Hausdorff formula eAeB=eA+B+1 2[A,B]+1 12([A,[A,B]]+[B,[B,A]])+... VII.1.3 Consider a lattice gauge theory in ( D+1)-dimensional space with the lattice spacing ainD-dimensional space and bin the extra dimension. Obtain the continuum D-dimensional field theory in the limit a→0 withbkept fixed. VII.1.4 Study in (2) the alternative limit b→0 with akept fixed so that you obtain a theory on a spatial lattice but with continuous time. VII.1.5 Show that for lattice gauge theory the Wilson area law holds in the limit of strong coupling. [Hint: Expand (20) in powers of f−2.] VII.2 Electroweak Unification The scourge of massless spin 1 particles With the benefit of hindsight, we now know that Nature likes Y ang-Mills theory. In the late 1960s and early 1970s, the electromagnetic and weak interactions were unified intoan electroweak interaction, described by a nonabelian gauge theory based on the groupSU( 2)⊗U(1). Somewhat later, in the early 1970s, it was realized that the strong interaction can be described by a nonabelian gauge theory based on the group SU( 3). Nature literally consists of a web of interacting Y ang-Mills fields. But when the theory was first proposed in 1954, it seemed to be totally inconsistent with observations as they were interpreted at that time. As Y ang and Mills themselves pointedout in their paper, the theory contains massless spin 1 particles, which were certainly notknown experimentally. Thus, except for interest on the part of a few theorists (Schwinger,Glashow, Bludman, and others) who found the mathematical structure elegantly attractiveand felt that nonabelian gauge theory must somehow be relevant for the weak interaction,the theory gradually sank into oblivion and was not part of the standard graduate curricu-lum in particle physics in the 1960s. Again with the benefit of hindsight, it would seem that there are only two logical solutions to the difficulty that experimentalists do not see any massless spin 1 particlesexcept for the photon: (1) the Y ang-Mills particles somehow acquire mass, or (2) the Y ang-Mills particles are in fact massless but are somehow not observed. We now know that thefirst possibility was realized in the electroweak interaction and the second in the stronginteraction. Constructing the electroweak theory We now discuss electroweak unification. It is perhaps pedagogically clearest to motivate how we would go about constructing such a theory. As I have said before, this is not a 380 | VII. Grand Unification textbook on particle physics and I necessarily will have to keep the discussion of particle physics to the bare minimum. I gave you a brief introduction to the structure of the weakinteraction in chapter IV .2. The other salient fact is that weak interaction violates parity,as mentioned in chapter II.1. In particular, the left handed electron field e Land the right handed electron field eR, which transform into each other under parity, enter into the weak interaction quite differently. Let us start with the weak decay of the muon, μ−→e−+¯ν+ν/prime, with νandν/primethe electron neutrino and muon neutrino, respectively. The relevant term in the Lagrangianis¯ν /prime LγμμL¯eLγμνL, with the left hand electron field eL, the electron neutrino field (which is left handed) νL, and so forth. The field μLannihilates a muon, the field ¯eLcreates an electron, and so on. (Henceforth, we will suppress the word field.) As you probably know,the elementary constituents of matter form three families, with the first family consistingofν,e, and the up uand down dquarks, the second of ν /prime,μ, and the charm cand strange squarks, and so on. For our purposes here, we will restrict our attention to the first family. Thus, we start with ¯νLγμeL¯eLγμνL. As I remarked in chapter III.2, a Fermi interaction of this type can be generated by the exchange of an intermediate vector boson W+with the coupling W+ μ¯νLγμeL+W− μ¯eLγμνL. The idea is then to consider an SU( 2)gauge theory with a triplet of gauge bosons denoted byWa μ, with a=1, 2, 3. Put νLandeLinto the doublet representation and the right handed electron field eRinto a singlet representation, thus ψL≡/parenleftBiggν e/parenrightBigg L,eR (1) (The notation is such that the upper component of ψLisνLand the lower component is eL.) The fields νLandeL, but not eR, listen to the gauge bosons Wa μ. Indeed, according to (IV .5.21) the Lagrangian contains Wa μ¯ψLτaγμψL=(W1−i2 μ¯ψL1 2τ1+i2γμψL+h.c.)+W3 μ¯ψLτ3γμψL where W1−i2 μ≡W1 μ−iW2 μand so forth. We recognize τ1+i2≡τ1+iτ2as the raising operator and the first two terms as (W1−i2 μ¯νLγμeL+h.c.), precisely what we want. By design, the exchange of W± μgenerates the desired term ¯νLγμeL¯eLγμνL. We need more room We would hope that the boson W3we were forced to introduce would turn out to be the pho- ton so that electromagnetism is included. But alas, W3couples to the current ¯ψLτ3γμψL= (¯νLγμνL−¯eLγμeL), not the electromagnetic current −(¯eLγμeL+¯eRγμeR). Oops! Another problem lurks. T o generate a mass term for the electron, we need a doublet Higgs field ϕ≡/parenleftbigϕ+ ϕ0/parenrightbig in order to construct the SU( 2)invariant term f¯ψLϕeRin the Lagrangian so that when ϕacquires the vacuum expectation value/parenleftbig0 v/parenrightbig we will have VII.2. Electroweak Unification | 381 f¯ψLϕeR→f(¯ν,¯e)L/parenleftBigg0 v/parenrightBigg eR=fv¯eLeR (2) But none of the SU( 2)transformations leaves/parenleftbig0 v/parenrightbig invariant: The vacuum expectation value ofϕspontaneously breaks the entire SU( 2)symmetry, leaving all three Wbosons massive. There is no room for the photon in this failed theory. Aagh! We need more room. Remarkably, we can avoid both the oops and the aagh by extending the gauge symmetry to SU( 2)⊗U(1). Denoting the generator of U(1)by1 2Y(called the hypercharge) and the associated gauge potential by Bμ[and their counterparts TaandWa μ forSU( 2)] we have the covariant derivative Dμ=∂μ−igWa μTa−ig/primeBμY 2. With four gauge bosons, we dare to hope that one of them might turn out to be the photon. The gauge potentials are normalized by the corresponding kinetic energy terms, L=−1 4(Bμν)2−1 4(Wa μν)2+...with the abelian Bμν=∂μBν−∂νBμand nonabelian field strength Wa μν=∂μWa ν−∂νWa μ+εabcWb μWc ν. The generators Taare of course normalized by the commutation relations that define SU( 2). In contrast, there is no commutation relation in the abelian algebra U(1)to fix the normalization of the generator1 2Y. Until this is fixed, the normalization of the U(1)gauge coupling g/primeis not fixed. How do we fix the normalization of the generator1 2Y? By construction, we want spon- taneous symmetry breaking to leave a linear combination of T3and1 2Yinvariant, to be identified as the generator the massless photon couples to, namely the charge operator Q. Thus, we write Q=T3+1 2Y (3) Once we know T3and1 2Yof any field, this equation tells us its charge. For example, Q(νL)=1 2+1 2Y(νL)andQ(eL)=−1 2+1 2Y(eL). In particular, we see that the coefficient ofT3in (3) must be 1 since the charges of νLandeLdiffer by 1. The relation (3) fixes the normalization of1 2Y. Determining the hypercharge The next step is to determine the hypercharge of various multiplets in the theory, which in turn determines how Bμcouples to these multiplets. Consider ψL.F o reLto have charge −1, the doublet ψLmust have1 2Y=−1 2. In contrast, the field eRhas1 2Y=− 1 since T3=0 oneR. Given the hypercharge of ψLandeRwe see that the invariance of the term f¯ψLϕeR under SU( 2)⊗U(1)forces the Higgs field ϕto have1 2Y=+1 2. Thus, according to (3) the upper component of ϕhas electric charge Q=+1 2+1 2=+ 1 and the lower component Q=−1 2+1 2=0. Thus, we write ϕ=/parenleftbigϕ+ ϕ0/parenrightbig . Recall that ϕhas the vacuum expectation value/parenleftbig0 v/parenrightbig . The fact that the electrically neutral field ϕ0acquires a vacuum expectation value but the charged field ϕ+does not provide a consistency check. 382 | VII. Grand Unification The theory works itself out Now that the couplings of the gauge bosons to the various fields, in particular, the Higgs field, are determined, we can easily work out the mass spectrum of the gauge bosons, asindeed, let me remind you, you have already done in exercise IV .6.3! Upon spontaneous symmetry breaking ϕ→(1/√ 2)/parenleftbig0 v/parenrightbig (the normalization is conven- tional): We simply plug in L=(Dμϕ)†(Dμϕ)→g2v2 4W+ μW−μ+v2 8(gW3 μ−g/primeBμ)2(4) I trust that this is what you got! Thus, the linear combination gW3 μ−g/primeBμbecomes massive while the orthogonal combination remains massless and is identified with the photon. Itis clearly convenient to define the angle θby tan θ=g /prime/g. Then, Zμ=cosθW3 μ−sinθBμ (5) describes a massive gauge boson known as the Zboson, while the electromagnetic po- tential is given by Aμ=sinθW3 μ+cosθBμ. Combine (4) and (5) and verify that the mass squared of the Zboson is M2 Z=v2(g2+g/prime2)/4, and thus by elementary trigonometry ob- tain the relation MW=MZcosθ (6) The exchange of the Wboson generates the Fermi weak interaction L=−g2 2M2 W¯νLγμeL¯eLγμνL=−4G√ 2¯νLγμeL¯eLγμνL where the second equality merely gives the historical definition of the Fermi coupling G. Thus, G√ 2=g2 8M2 W(7) Next, we write the relevant piece of the covariant derivative gW3 μT3+g/primeBμY 2=g(cosθZμ+sinθAμ)T3+g/prime(−sinθZμ+cosθAμ)Y 2 in terms of the physically observed ZandA. The coefficient of Aμworks out to be gsinθT3+g/primecosθ(Y/ 2)=gsinθ(T3+Y/2); the fact that the combination Q=T3+ Y/2 emerges provides a nice check on the formalism. Furthermore, we obtain e=gsinθ (8) Meanwhile, it is convenient to write gcosθT3−g/primesinθ(Y/ 2), the coefficient of Zμin the covariant derivative, in terms of the physically familiar electric charge Qrather than the theoretical hypercharge Y: Thus, gcosθT3−g/primesinθ(Q−T3)=g cosθ(T3−sin2θQ) VII.2. Electroweak Unification | 383 In other words, we have determined the coupling of the Zboson to an arbitrary fermion field/Psi1in the theory: L=g cosθZμ¯/Psi1γμ(T3−sin2θQ)/Psi1 (9) For example, using (9) we can immediately write the coupling of Zto leptons: L=g cosθZμ[1 2(¯νLγμνL−¯eLγμeL)+sin2θ¯eγμe] (10) Including quarks How to include the hadrons is now almost self evident. Given that only left handed fields participate in the weak interaction, we put the quarks of the first generation intoSU( 2)⊗U(1)multiplets as follows: qα L≡/parenleftBigguα dα/parenrightBigg L,uα R,dα R(11) where α=1, 2, 3 denotes the color index, which I will discuss in the next chapter. The right handed quarks uα Randdα Rare put into singlets so that they do not hear the weak bosons Wa. Recall that the up quark uand the down quark dhave electric charges2 3and −1 3respectively. Referring to (3) we see1 2Y=1 6,2 3, and−1 3forqα L,uα R, anddα R, respectively. From (9) we can immediately read off the coupling of the Zboson to the quarks: L=g cosθZμ[1 2(¯uLγμuL−¯dLγμdL)−sin2θJμ em] (12) Finally, I leave it to you to verify that of the four degrees of freedom contained in ϕ (since ϕ+andϕ0are complex) three are eaten by the WandZbosons, leaving one physical degree of freedom Hcorresponding to the elusive Higgs particle that experimenters are still searching for as of this writing. The neutral current By virtue of its elegantly economical gauge group structure, this SU( 2)⊗U(1)electroweak theory of Glashow, Salam, and Weinberg ushered in the last great predictive era of theo- retical particle physics. Writing (10) and (12) as L=g cosθZμ(Jμ leptons+Jμ quarks) and using (6) we see that Zboson exchange generates a hitherto unknown neutral current interaction Lneutral current =−g2 2M2 W(Jleptons +Jquarks)μ(Jleptons +Jquarks)μ between leptons and quarks. By studying various processes described by Lneutral current we can determine the weak angle θ. Once θis determined, we can predict gfrom (8). Once g 384 | VII. Grand Unification is determined, we can predict MWfrom (7). Once MWis determined, we can predict MZ from (6). Concluding remarks As I mentioned, there are three families of leptons and quarks in Nature, consisting of(νe,e,u,d),(νμ,μ,c,s), and (ντ,τ,t,b). The appearance of this repetitive family structure, about which the SU( 2)⊗U(1)theory has nothing to say, represents one of the great unsolved puzzles of particle physics. The three families, with the appropriaterotation angles between them, are simply incorporated into the theory by repeating whatwe wrote above. A more logical approach than the one given here would be to start with an SU( 2)⊗U(1) theory with a doublet Higgs field with some hypercharge, and to say, “Behold, uponspontaneous symmetry breaking, one linear combination of generators remains unbrokenwith a corresponding massless gauge field.” I think that our quasi-historical approach isclearer. As I have mentioned on several occasions, Fermi’s theory of the weak interaction is nonrenormalizable. In 1999, ’t Hooft and Veltman were awarded the Nobel Prize forshowing that the SU( 2)⊗U(1)electroweak theory is renormalizable, thus paving the way for the triumph of nonabelian gauge theories in describing the strong, electromagnetic,and weak interactions. I cannot go into the details of their proof here, but I would liketo mention that the key is to start with the nonabelian analog of the unitary gauge (recallchapter IV .6) and proceed to the R ξgauge. At large momenta, the massive gauge boson propagators go as ∼(kμkν/k2)in the unitary gauge, but as ∼(1/k2)in the Rξgauge. The theory is then renormalizable by power counting. Exercises VII.2.1 Unfortunately, the mass of the elusive Higgs particle Hdepends on the parameters in the double well potential V=−μ2ϕ†ϕ+λ(ϕ†ϕ)2responsible for the spontaneous symmetry breaking. Assuming that His massive enough to decay into W++W−andZ+Z, determine the rates for Hto decay into various modes. VII.2.2 Show that it is possible to stay with the SU( 2)gauge group and to identify W3as the photon A, but at the cost of inventing some experimentally unobserved lepton fields. This theory does not describe ourworld: For one thing, it is essentially impossible to incorporate the quarks. Show this! [Hint: We have toput the leptons into a triplet of SU( 2)instead of a doublet.] VII.3 Quantum Chromodynamics Quarks Quarks come in six flavors, known as up, down, strange, charm, bottom, and top, denoted byu,d,s,c,b, andt. The proton, for example, is made of two up quarks and a down quark ∼(uud), while the neutral pion corresponds to ∼(u¯u−d¯d)/√ 2. Please consult any text on particle physics for details. By the late 1960s the notion of quarks was gaining wide acceptance, but two separate lines of evidence indicated that a crucial element was missing. In studying how hadronsare made of quarks, people realized that the wave function of the quarks in a nucleon doesnot come out to be antisymmetric under the interchange of any pair of quarks, as requiredby the Pauli exclusion principle. At around the same time, it was realized that in the idealworld we used to derive the Goldberger-T reiman relation and in which the pion is masslesswe can calculate the decay rate for the process π 0→γ+γ, as mentioned in chapter IV .7. Puzzlingly enough, the calculated rate came out smaller than the observed rate by a factor of 9=32. Both puzzles could be resolved in one stroke by having quarks carry a hitherto unknown internal degree of freedom that Gell-Mann called color. For any specified flavor, a quarkcomes in one of three colors. Thus, the up quark can be red, blue, or yellow. In a nucleon,the wave function of the three quarks will then contain a factor referring to color, besidesthe factors referring to orbital motion, spin, and so on. We merely have to make the colorpart of the wave function antisymmetric; in fact, we simply take it to be ε αβγ, where α,β, andγdenote the colors carried by the three quarks. With quarks in three colors, we have to multiply the amplitude for π0decay by a factor of 3, thus neatly resolving the discrepancy between theory and experiment. 386 | VII. Grand Unification Asymptotic freedom As I mentioned in chapter VI.6, the essential clue came from studying deep inelastic scat- tering of electrons off nucleons. Experimentalists made the intriguing discovery that whenhit hard the quarks in the nucleons act as if they hardly interact with each other, in otherwords, as if they are free. On the other hand, since quarks are never seen as isolated enti-ties, they appear to be tightly bound to each other within the nucleon. As I have explained,this puzzling and apparently contradictory behavior of the quarks can be understood if thestrong interaction coupling flows to zero in the large momentum (ultraviolet) limit and toinfinity or at least to some large value in the small momentum (infrared) limit. A numberof theorists proposed searching for theories whose couplings would flow to zero in theultraviolet limit, now known as asymptotic free theories. Eventually, Gross, Wilczek, andPolitzer discovered that Y ang-Mills theory is asymptotically free. This result dovetails perfectly with the realization that quarks carry color. The nonabelian gauge transformation would take a quark of one color into a quark of another color. Thus, towrite down the theory of the strong interaction we simply take the result of exercise IV .5.6, L=−1 4g2Fa μνFaμν+¯q(iγμDμ−m)q (1) with the covariant derivative Dμ=∂μ−iAμ. The gauge group is SU( 3)with the quark field qin the fundamental representation. In other words, the gauge fields Aμ=Aa μTa, where Ta(a=1 ,...,8 )are traceless hermitean 3 by 3 matrices. Explicitly, ( Aμq)α=Aa μ(Ta)αβqβ, where α,β=1, 2, 3. The theory is known as quantum chromodynamics, or QCD for short, and the nonabelian gauge bosons are known as gluons. T o incorporate flavor, we simplywrite/summationtext f j=1¯qj(iγμDμ−mj)qjfor the second term in (1), where the index jgoes over the fflavors. Note that quarks of different flavors have different masses. Infrared slavery The flip side of asymptotic freedom is infrared slavery. We cannot follow the renormaliza- tion group flow all the way down to the low momentum scale characteristic of the quarksbound inside hadrons since the coupling gbecomes ever stronger and our perturbative cal- culation of β(g) is no longer adequate. Nevertheless, it is plausible although never proven thatggoes to infinity and that the gluons keep the quarks and themselves in permanent confinement. The Wilson loop introduced in chapter VII.1 provides the order parameterfor confinement. In elementary physics forces decrease with the separation between interacting objects, so permanent confinement is a rather bizarre concept. Are there any other instances ofpermanent confinement? Consider a magnetic monopole in a superconductor. We get to combine what we learned in chapters IV .4 and V .4 (and even VI.2)! A quantized amount of magnetic flux comes outof the monopole, but according to the Meissner effect a superconductor expels magneticflux. Thus, a single magnetic monopole cannot live inside a superconductor. VII.3. Quantum Chromodynamics | 387 M M R Figure VII.3.1 Now consider an antimonopole a distance Raway (figure VII.3.1). The magnetic flux coming out of the monopole can go into the antimonopole, forming a tube connectingthe monopole and the antimonopole and obliging the superconductor to give up being asuperconductor in the region of the flux tube. In the language of chapter V .4, it is no longerenergetically favorable for the field or order parameter ϕto be constant everywhere; instead it vanishes in the region of the flux tube. The energy cost of this arrangement evidentlygrows as R(consistent with Wilson’s area law). In other words, an experimentalist living inside a superconductor would find that it costs more and more energy to pull a monopole and an antimonopole apart. Thisconfinement of monopoles inside a superconductor is often taken to be a model of the yet-to-be-proven confinement of quarks. Invoking electromagnetic duality we can imaginea magnetic superconductor in contrast to the usual electric superconductor. Inside amagnetic superconductor, electric charges would be permanently confined. Our universemay be likened to a color magnetic superconductor in which quarks (the analog of electriccharges) are confined. On distance scales large compared to the radius of the color flux tube connecting a quark to an antiquark, the tube can be thought of as a string. Historically, that was how stringtheory originated. The challenge, boys and girls, is to prove that the ground state or vacuumof (1) is a color magnetic superconductor. Symmetries of the strong interaction Now that we have a theory of the strong interaction, we can understand the origin of thesymmetries of the strong interaction, namely the isospin symmetry of Heisenberg and thechiral symmetry that when spontaneously broken leads to the appearance of the pion as aNambu-Goldstone boson (as discussed in chapters IV .2 and VI.4). Consider a world with two flavors, which is all that is relevant for a discussion of the pion. Introduce the notation u≡q 1,d≡q2, andq=/parenleftbigu d/parenrightbig so that we can write the Lagrangian as L=−1 4g2Fa μνFaμν+¯q(iγμDμ−m)q with m=/parenleftBiggmu 0 0md/parenrightBigg 388 | VII. Grand Unification where muandmdare the masses of the up and down quarks, respectively. If mu=md, the Lagrangian is invariant under q→eiθ.τq, corresponding to Heisenberg’s isospin symmetry. In the limit in which muandmdvanish, the Lagrangian is invariant under q→eiϕ.τγ5q, known as the chiral SU( 2)symmetry, chiral because the right handed quarks qRand the left handed quarks qLtransform differently. T o the extent that muandmdare both much smaller than the energy scale of the strong interaction, chiral SU( 2)is an approximate symmetry. The pion is the Nambu-Goldstone boson associated with the spontaneous breaking of the chiral SU( 2). Indeed, this is an example of dynamical symmetry breaking since there is no elementary scalar field around to acquire a vacuum expectation. Instead, thestrong interaction dynamics is supposed to drive the composite scalar fields ¯uuand¯ddto “condense into the vacuum” so that /angbracketleft0|¯uu|0/angbracketright=/angbracketleft 0|¯dd|0/angbracketrightbecome nonvanishing, where the equality between the two vacuum expectation values ensures that Heisenberg’s isospin isnot spontaneously broken, an experimental fact since there are no corresponding Nambu-Goldstone bosons. In terms of the doublet field q, the QCD vacuum is supposed to be such that/angbracketleft0|¯qq|0/angbracketright /negationslash= 0 while /angbracketleft0|¯q/vectorτq|0/angbracketright=0. Renormalization group flow The renormalization group flow of the QCD coupling is governed by dg dt=β(g)=−11 3T2(G)g3 16π2(2) with the all-crucial minus sign. Here T2(G)δab=facdfbcd(3) I will not go through the calculation of β(g) here, but having mastered chapters VI.8 and VII.1 you should feel that you can do it if you want to.1At the very least, you should understand the factor g3andT2(G) by drawing the relevant Feynman diagrams. When fermions are included, dg dt=β(g)=/bracketleftBig −11 3T2(G)+4 3T2(F)/bracketrightBigg3 16π2(4) where T2(F)δab=tr[Ta(F)Tb(F)] (5) I do expect you to derive (4) given (2). For SU(N) T2(F)=1 2for each fermion in the fundamental representation. Note that asymptotic freedom is lost when there are too many fermions. 1For a detailed calculation, see, e.g., S. Weinberg, The Quantum Theory of Fields , Vol. 2, sec. 18.7. VII.3. Quantum Chromodynamics | 389 γ e–e+ Figure VII.3.2 You already solved an equation like (4) in exercise VI.8.1. Let us define, in analogy to quantum electrodynamics, αS(μ)≡g(μ)2/4π, the strong coupling at the momentum scale μ. From (4) we obtain2 αS(Q)=αS(μ) 1+(1/4π)(11−2 3nf)αS(μ) log(Q2/μ2)(6) showing explicitly that αS(Q)→0 logarithmically as Q→∞ . Electron-positron annihilation I have space to show you only one physical application. Experimentalists have measured the cross section σofe+e−annihilation into hadrons as a function of the total center-of- mass energy E. The amplitude is shown in figure VII.3.2. T o calculate the cross section in terms of the amplitude, we have to go through what some people call “boring kinematics,”such as normalizing everything correctly, dividing by the flux of the two beams, and soforth (see the appendix to chapter II.6). For the good of your soul, you should certainly gothrough this type of calculation at least once. Believe me, I did it more times than I careto remember. But happily, as I will now show you, we can avoid most of this grunge labor.First, consider the ratio R(E)≡σ(e+e−→hadrons) σ(e+e−→μ+μ−) The kinematic stuff cancels out. In figure VII.3.2 the half of the diagram involving the electron positron lines and the photon propagator also appears in the Feynman diagram e+e−→μ+μ−(figure VII.3.3) and so cancels out in R(E) . The blob in figure VII.3.2, which hides all the complexity of the strong interaction, is given by /angbracketleft0|Jμ(0)|h/angbracketright, where Jμis the 2For the accumulated experimental evidence on αS(Q) , see figure 14.3 in F. Wilczek, in: V . Fitch et al., eds., Critical Problems in Physics , p. 281. 390 | VII. Grand Unification γ e−e+μ−μ+ Figure VII.3.3 electromagnetic current and the state |h/angbracketrightcan contain any number of hadrons. T o obtain the cross section we have to square the amplitude, include a δ-function for momentum conservation, and sum over all |h/angbracketright, thus arriving at /summationdisplay h(2π)4δ4(ph−pe+−pe−)/angbracketleft0|Jμ(0)|h/angbracketright/angbracketlefth| Jν(0)|0/angbracketright (7) [withq≡pe++pe−=(E,/vector0)]. This quantity can be written as /integraldisplay d4xeiqx/angbracketleft0|Jμ(x)Jν(0)|0/angbracketright=/integraldisplay d4xeiqx/angbracketleft0|[Jμ(x),Jν(0)]|0/angbracketright =2I m(i/integraldisplay d4xeiqx/angbracketleft0|TJμ(x)Jν(0)|0/angbracketright) (The first equality follows from E> 0 and the second was explained in chapter III.8.) T o determine this quantity, we would have to calculate an infinite number of Feynmandiagrams involving lots of quarks and gluons. A typical diagram is shown in figure VII.3.4.Completely hopeless! This is where asymptotic freedom rides to the rescue! From chapter VI.7 you learned that for a process at energy Ethe appropriate coupling strength to use is g(E) . But as we crank up E,g(E) gets smaller and smaller. Thus diagrams such as figure VII.3.4 involving many powers of g(E) all fall away, leaving us with the diagrams with no power of g(E) (fig. VII.3.5a) and two powers of g(E) (figs. VII.3.5b,c,d). No calculation is necessary to obtain the leading term in R(E) , since the diagram in figure VII.3.5a is the same one that enters into e +e−→μ+μ−: We merely replace the quark propagator by the muon propagator (quark and muon masses are negligible compared to E). At high energy, the quarks are free and R(E) merely counts the square of the charge Qaof the various quarks contributing at that energy. We predict R(E) −→ E→∞3/summationdisplay aQ2 a(8) The factor of 3 accounts for color. VII.3. Quantum Chromodynamics | 391 γ γ Figure VII.3.4 Not only does QCD turn itself off at high energies, it tells us how fast it is turning itself off. Thus, we can determine how the limit in (8) is approached: R(E)=/parenleftBigg 3/summationdisplay aQ2 a/parenrightBigg/parenleftBigg 1+C2 (11−2 3nf)log(E/μ)+.../parenrightBigg (9) I will leave it to you to calculate C. Dreams of exact solubility An analytic solution of quantum chromodynamics is something of a “Holy Grail” for field theorists (a grail that now carries a prize of one million dollars: see www.ams.org/claymath/). Many field theorists have dreamed that at least “pure” QCD, that is QCDwithout quarks, might be exactly soluble. After all, if any 4-dimensional quantum field (a) (b) (c) (d)γ γ Figure VII.3.5 392 | VII. Grand Unification 1 QCD μg(μ)2 4π Figure VII.3.6 theory turns out to be exactly soluble, pure Y ang-Mills, with all its fabulous symmetries, is the most likely possibility. (Perhaps an even more likely candidate for solubility issupersymmetric Y ang-Mills theory. We will touch on supersymmetry in chapter VIII.4.) Let me be specific about what it means to solve QCD. Consider a world with only up and down quarks with m uandmdboth set equal to zero, namely a world described by L=−1 4g2Fa μνFaμν+¯qiγμDμq (10) The goal would be to calculate something like the ratio of the mass of the ρmeson mρto the mass of the proton mP. T o make progress, theoretical physicists typically need to have a small parameter to expand in, but in trying to solve (10) we are confronted with the immediate difficultythat there is no such parameter. You might think that gis a parameter, but you would be mistaken. The renormalization group analysis taught us that g(μ) is a function of the energy scale μat which it is measured. Thus, there is no particular dimensionless number we can point to and say that it measures the strength of QCD. Instead, the best we can dois to point to the value of μat which (g(μ) 2/4π)becomes of order 1. This is the energy, known as /Lambda1QCD , at which the strong interaction becomes strong as we come down from high energy (fig. VII.3.6). But /Lambda1QCD merely sets the scale against which other quantities are to be measured. In other words, if you manage to calculate mPit better come out proportional to /Lambda1QCD since/Lambda1QCD is the only quantity with dimension of mass around. Similarly for mρ. Put in precise terms, if you publish a paper with a formula giving mρ/mP in terms of pure numbers such as 2 and π, the field theory community will hail you as a conquering hero who has solved QCD exactly. The apparent trade of a dimensionless coupling gfor a dimensional mass scale /Lambda1QCD is known as dimensional transmutation, of which we will see another example in the next chapter. VII.3. Quantum Chromodynamics | 393 Exercises VII.3.1 Calculate Cin (9). [Hint: If you need help, consult T . Appelquist and H. Georgi, Phys. Rev. D8: 4000, 1973; and A. Zee, Phys. Rev. D8: 4038, 1973.] VII.3.2 Calculate (2). VII.4 LargeNExpansion Inventing an expansion parameter Quantum chromodynamics is a zero-parameter theory, so it is difficult to give even a first approximation. In desperation, field theorists invented a parameter in which to expandQCD. Suppose instead of three colors we have Ncolors. ’t Hooft 1noticed that as N→∞ remarkable simplifications occur. The idea is that if we can calculate mρ/mP, for example, in the large Nlimit the result may be close to the actual value. People sometimes joke that particle physicists regard 3 as a large number, but actually the correction to the large N limit is typically of order 1 /N2, about 10% in the real world. Particle physicists would be more than happy to be able to calculate hadron masses to this degree of accuracy. As with spontaneous symmetry breaking and a number of other important concepts, the largeNexpansion came out of condensed matter physics but nowadays is used routinely in all sorts of contexts. For example, people have tried a large Napproach to solve high- temperature superconductivity and to fold RNA.2 Scaling the QCD coupling So, let the color group be U(N) and write L=−Na 2g2trFμνFμν+¯ψ[i(/negationslash∂−i/negationslashA)−m]ψ (1) Note that we have replaced g2byg2/Na. For finite Nthis change has no essential signifi- cance. The point is to choose the power aso that interesting simplifications occur in the limitN→∞ withg2held fixed. The cubic and quartic interaction vertices of the gluons 1G. ’t Hooft, Under the Spell of the Gauge Principle , p. 378. 2M. Bon, G. Vernizzi, H. Orland, and A. Zee, “T opological classification of RNA structures,” J. Mol. Biol. 379:900, 2008. VII.4. Large NExpansion | 395 γγ γγ (a) (b) Figure VII.4.1 are proportional to Na. On the other hand, since the gluon propagator goes as the inverse of the quadratic terms in L, it is proportional to 1 /Na. The coupling of the gluon to the quark does not depend on N. To fi x a, let us focus on a specific application, the calculation of σ(e+e−→hadrons) discussed in the last chapter. Suppose we want to calculate this cross section at lowenergies. Consider the two-gluon exchange diagrams shown in figures VII.4.1a and b.The two diagrams are of order g 4and we would have to calculate both. Note that 1b is nonplanar: Since one gluon crosses over the other, the diagram cannot be drawn on theplane if we insist that lines cannot go through each other. Now the double-line formalism introduced in chapter IV .5 shines. In this formalism the diagrams figure VII.4.1a and b are redrawn as in figure VII.4.2a and b. The two gluonpropagators common to both diagrams give a factor 1 /N 2a. Now comes the punchline. We sum over three independent color indices in 2a, thus getting a factor N3. Grab some crayons and try to color each line in 2a with a different color: you will need three crayons.In contrast, we sum over only one independent index in 2b, getting only a factor of N.I n other words, 2a dominates 2b by a factor N 2. In the large Nlimit we can throw 2b away. Clearly, the rule is to associate one factor of Nwith each loop. Thus, the lowest order diagram, shown in 2c, with Ndifferent colors circulating in it, scales as N; 2a scales as N3/N2a. We want 2a and 2c to scale in the same way and thus we choose a=1. (a) (b) (c) (d) Figure VII.4.2 396 | VII. Grand Unification By drawing more diagrams [e.g., 2d scales as N(1/N4)N4, with the three factors coming from the quartic coupling, the propagators, and the sum over colors, respectively], you canconvince yourself that planar diagrams dominate in the large Nlimit, all scaling as N.F o r a challenge, try to prove it. Evidently, there is a topological flavor to all this. The reduction to planar diagrams is a vast simplification but there are still an infinite number of diagrams. At this stage in our mastery of field theory, we still can’t solve largeNQCD. (As I started writing this book, there were tantalizing clues, based on insight and techniques developed in string theory, that a solution of large NQCD might be within sight. As I now go through the final revision, that hope has faded.) The double-line formalism has a natural interpretation. Group theoretically, the matrix gauge potential A i jtransforms just like ¯qiqj(but assuredly we are not saying that the gluon is a quark-antiquark bound state) and the two lines may be thought of as describing a quarkand an antiquark propagating along, with the arrows showing the direction in which coloris flowing. Random matrix theory There is a much simpler theory, structurally similar to large NQCD, that actually can be solved. I am referring to random matrix theory. Exaggerating a bit, we can say that quantum mechanics consists of writing down a matrix known as the Hamiltonian and then finding its eigenvalues and eigenvectors. In theearly 1950s, when confronted with the problem of studying the properties of complicatedatomic nuclei, Eugene Wigner proposed that instead of solving the true Hamiltonian insome dubious approximation we might generate large matrices randomly and study thedistribution of the eigenvalues—a sort of statistical quantum mechanics. Random matrixtheory has since become a rich and flourishing subject, with an enormous and growingliterature and applications to numerous areas of theoretical physics and even to puremathematics (such as operator algebra and number theory.) 3It has obvious applications to disordered condensed matter systems and less obvious applications to random surfacesand hence even to string theory. Here I will content myself with showing how ’t Hooft’sobservation about planar diagrams works in the context of random matrix theory. Let us generate NbyNhermitean matrices ϕrandomly according to the probability P(ϕ)=1 Ze−N trV( ϕ)(2) withV( ϕ) a polynomial in ϕ. For example, let V( ϕ)=1 2m2ϕ2+gϕ4. The normalization/integraltext dϕP(ϕ) =1 fixes Z=/integraldisplay dϕe−N trV( ϕ)(3) The limit N→∞ is always understood. 3For a glimpse of the mathematical literature, see D. Voiculescu, ed., Free Probability Theory . VII.4. Large NExpansion | 397 As in chapter VI.7 we are interested in ρ(E) , the density of eigenvalues of ϕ. T o make sure that you understand what is actually meant, let me describe what we would do werewe to evaluate ρ(E) numerically. For some large integer N, we would ask the computer to generate a hermitean matrix ϕwith the probability P(ϕ) and then to solve the eigenvalue equation ϕv=Ev. After this procedure had been repeated many times, the computer could plot the distribution of eigenvalues in a histogram that eventually approaches a smoothcurve, called the density of eigenvalues ρ(E) . We already developed the formalism to compute ρ(E) in (VI.7.1): Compute the real analytic function G(z)≡/angbracketleft(1/N) tr[1/(z−ϕ)]/angbracketrightandρ(E)=−(1/π) lim ε→0ImG(E+iε). The average /angbracketleft.../angbracketrightis taken with the probability P(ϕ) : /angbracketleftO(ϕ) /angbracketright=1 Z/integraldisplay Dϕe−N trV( ϕ)O(ϕ) You see that my choice of notation, ϕfor the matrix and V( ϕ)=1 2m2ϕ2+gϕ4as an example, is meant to be provocative. The evaluation of Zis just like the evaluation of a path integral, but for an action S(ϕ)=NtrV( ϕ) that does not involve/integraltext ddx. Random matrix theory can be thought of as a quantum field theory in (0+0)-dimensional spacetime! Various field theoretic methods, such as Feynman diagrams, can all be applied to random matrix theory. But life is sweet in (0+0)-dimensional spacetime: There is no space, no time, no energy, and no momentum and hence no integral to do in evaluatingFeynman diagrams. The Wigner semicircle law Let us see how this works for the simple case V( ϕ)=1 2m2ϕ2(we can always absorb minto ϕbut we won’t). Instead of G(z), it is slightly easier to calculate Gi j(z)≡/angbracketleftBigg/parenleftbigg1 z−ϕ/parenrightbiggi j/angbracketrightBigg =δi jG(z) The last equality follows from invariance under unitary transformations: P(ϕ)=P(U†ϕU) (4) Expand Gi j(z)=∞/summationdisplay n=01 z2n+1/angbracketleft(ϕ2n)i j/angbracketright (5) Do the Gaussian integral 1 Z/integraldisplay dϕe−N tr1 2m2ϕ2ϕi kϕl j=1 Z/integraldisplay dϕe−N1 2m2/summationtext p,qϕp qϕq pϕi kϕl j=δi jδl k1 Nm2(6) Setting k=land summing, we find the n=1 term in (5) is equal to (1/z3)δi j(1/m2). Just as in any field theory we can associate a Feynman diagram with each of the terms in (5). For the n=1 term, we have figure VII.4.3. The matrix character of ϕlends itself naturally to ’t Hooft’s double-line formalism and thus we can speak of quark and gluon 398 | VII. Grand Unification jl k i Figure VII.4.3 propagators with a good deal of ease. The Feynman rules are given in figure VII.4.4. We recognize ϕas the gluon field and (5) as the gluon propagator. Indeed, we can formulate our problem as follows: Given the bare quark propagator 1 /z, compute the true quark propagator G(z) with all interaction effects taken into account. Let us now look at the n=2 term in (5) 1 /z5<ϕi hϕh kϕk lϕl j>, which we represent in figure VII.4.5a. With a bit of thought you can see that the index ican be contracted with k,l,o rj, thus giving rise to figures VII.4.5b, c, d. Summing over color indices, just as in QCD, we see that the planar diagrams in 5b and 5d dominate the diagram in 5c by a factorN 2. We can take over ’t Hooft’s observation that planar diagrams dominate. Incidentally, in this example, you see how large Nis essential, allowing us to get rid of nonplanar diagrams. After all, if I ask you to calculate the density of eigenvalues for sayN=7 you would of course protest saying that the general formula for solving a degree-7 polynomial equation is not even known. The simple example in figure VII.4.5 already indicates how all possible diagrams could be constructed. In 5b the same “unit” is repeated, while in 5d the same “unit” is nestedinside a more basic diagram. A more complicated example is shown in 5e. You can convinceyourself that for N=∞ all diagrams contributing to G(z) can be generated by either “nesting” existing diagrams inside an overarching gluon propagator or “repeating” an i kj l1 z 1 Nm2δi j δl k 1 Figure VII.4.4 VII.4. Large NExpansion | 399 i h jl k (a) (b) (c) (d) (e) Figure VII.4.5 existing structure over and over again. T ranslate the preceding sentence into two equations: “Repeat” (see figure VII.4.6a), G(z)=1 z+1 z/Sigma1(z)1 z+1 z/Sigma1(z)1 z/Sigma1(z)1 z+... =1 z−/Sigma1(z)(7) and “nest” (see figure VII.4.6b), /Sigma1(z)=1 m2G(z) (8) 400 | VII. Grand Unification + = + +GΣ Σ Σ ΣG(a) (b)= Figure VII.4.6 Combining these two equations we obtain a simple quadratic equation for G(z) that we can immediately solve to obtain G(z)=m2 2/parenleftBigg z−/radicalbigg z2−4 m2/parenrightBigg (9) (From the definition of G(z) we see that G(z)→1/zfor large zand thus we choose the negative root.) We immediately deduce that ρ(E)=2 πa2/radicalbig a2−E2 (10) where a2=4/m2. This is a famous result known as Wigner’s semicircle law. The Dyson gas I hope that you are struck by the elegance of the large Nplanar diagram approach. But you might have also noticed that the gluons do not interact. It is as if we have solved quantum electrodynamics while we have to solve quantum chromodynamics. What if we have todeal with V( ϕ)= 1 2m2ϕ2+gϕ4? The gϕ4term causes the gluons to interact with each other, generating horrible diagrams such as the one in figure VII.4.7. Clearly, diagrams proliferate and as far as I know nobody has ever been able to calculate G(z) using the Feynman diagram approach. Happily, G(z) can be evaluated using another method known as the Dyson gas approach. The key is to write ϕ=U†/Lambda1U (11) VII.4. Large NExpansion | 401 Figure VII.4.7 where /Lambda1denotes the NbyNdiagonal matrix with diagonal elements equal to λi,i= 1 ,..., N. Change the integration variable in (3) from ϕtoUand/Lambda1: Z=/integraldisplay dU/integraldisplay/parenleftbig /Pi1idλi/parenrightbig Je−N/summationtext kV( λk)(12) withJthe Jacobian. Since the integrand does not depend on Uwe can throw away the integral over U. It just gives the volume of the group SU(N) . Does this remind you of chapter VII.1? Indeed, in (11) Ucorresponds to the unphysical gauge degrees of freedom— the relevant degrees of freedom are the eigenvalues {λi}. As an exercise you can use the Faddeev-Popov method to calculate J. Instead, we will follow the more elegant tack of determining Jby arguing from general principles. The change of integration variables in (11) is ill defined when any two of the λi’s are equal, at which point Jmust vanish. (Recall that the change from Cartesian coordinates to spherical coordinates is ill defined at the north and south poles and indeed the Jacobianin sin θdθdϕ vanishes at θ=0 and π.) Since the λ i’s are created equal, interchange symmetry dictates that J=[/Pi1m>n(λm−λn)]β. The power βcan be fixed by dimensional analysis. With N2matrix elements dϕobviously has dimension λN2while (/Pi1idλi)Jhas dimension λNλβN(N −1)/2; thus β=2. Having determined J, let us rewrite (12) as Z=/integraldisplay (/Pi1idλi)[/Pi1m>n(λm−λn)]2e−N/summationtext kV( λk) =/integraldisplay (/Pi1idλi)e−N/summationtext kV( λk)+1 2/summationtext m/negationslash=nlog(λm−λn)2 (13) Dyson pointed out that in this form Z=/integraltext (/Pi1idλi)e−NE(λ 1,...,λN)is just the partition function of a classical 1-dimensional gas (recall chapter V .2). Think of λi, a real number, as the position of the ith molecule. The energy of a configuration E(λ 1,..., λN)=/summationdisplay kV( λk)−1 2N/summationdisplay m/negationslash=nlog(λm−λn)2(14) consists of two terms with obvious physical interpretations. The gas is confined in a potential well V( x) and the molecules repel4each other with the two-body potential −(1/N) log(x−y)2. Note that the two terms in Eare of the same order in Nsince each 4Note that this corresponds to the repulsion between energy levels in quantum mechanics. 402 | VII. Grand Unification sum counts for a power of N. In the large Nlimit (we can think of Nas the inverse temperature), we evaluate Zby steepest descent and minimize E, obtaining V/prime(λk)=2 N/summationdisplay n/negationslash=k1 λk−λn(15) which in the continuum limit, as the poles in (15) merge into a cut, becomes V/prime(λ)= 2P/integraltext dμ[ρ(μ)/(λ −μ)], where ρ(μ) is the unknown function we want to solve for and P denotes principal value. Defining as before G(z)=/integraltext dμ[ρ(μ)/(z −μ)] we see that our equation for ρ(μ) can be written as Re G(λ+iε)=1 2V/prime(λ). In other words, G(z) is a real analytic function with cuts along the real axis. We are given the real part of G(z) on the cut and are to solve for the imaginary part. Br ´ezin, Itzykson, Parisi, and Zuber have given an elegant solution of this problem. Assume for simplicity that V( z ) is an even polynomial and that there is only one cut (see exercise VII.4.7). Invoke symmetry and, incorporating what we know, postulatethe form G(z)=1 2/bracketleftBig V/prime(z)−P(z)/radicalbig z2−a2/bracketrightBig withP(z) an unknown even polynomial. Remarkably, the requirement G(z)→1/zfor largezcompletely determines P(z) . Pedagogically, it is clearest to go to a specific example, sayV( z )=1 2m2z2+gz4. Since V/prime(z)is a cubic polynomial in z,P(z) has to be a quadratic (even) polynomial in z. T aking the limit z→∞ and requiring the coefficients of z3and ofzinG(z) to vanish and the coefficient of 1 /zto be 1 gives us three equations for three unknowns [namely aand the two unknowns in P(z) ]. The density of eigenvalues is then determined to be ρ(E)=(1/π)P(E)√ a2−E2. I think the lesson to take away here is that Feynman diagrams, in spite of their historical importance in quantum electrodynamics and their usefulness in helping us visualize whatis going on, are vastly overrated. Surely, nobody imagines that QCD, even large NQCD, will one day be solved by summing Feynman diagrams. What is needed is the analog ofthe Dyson gas approach for large NQCD. Conversely, if a reader of this book manages to calculate G(z) by summing planar diagrams (after all, the answer is known!), the insight he or she gains might conceivably be useful in seeing how to deal with planar diagramsin large NQCD. Field theories in the large Nlimit A number of field theories have also been solved in the large Nexpansion. I will tell you about one example, the Gross-Neveu model, partly because it has some of the flavor ofQCD. The model is defined by S(ψ)=/integraldisplay d2x⎡ ⎣N/summationdisplay a=1¯ψai/negationslash∂ψa+g2 2N/parenleftBiggN/summationdisplay a=1¯ψaψa/parenrightBigg2⎤ ⎦ (16) Recall from chapter III.3 that this theory should be renormalizable in (1+1)-dimensional spacetime. For some finite N, sayN=3, this theory certainly appears no easier to solve VII.4. Large NExpansion | 403 than any other fully interacting field theory. But as we will see, as N→∞ we can extract a lot of interesting physics. Using the identity (A.14) we can rewrite the theory as S(ψ ,σ)=/integraldisplay d2x/bracketleftBiggN/summationdisplay a=1¯ψa(i/negationslash∂−σ)ψa−N 2g2σ2/bracketrightBigg (17) By introducing the scalar field σ(x) we have “undone” the four-fermion interaction. (Recall that we used the same trick in chapter III.5.) You will note that the physics involved issimilar to that behind the introduction of the weak boson to generate the Fermi interaction.Using what we learned in chapters II.5 and IV .3 we can immediately integrate out thefermion fields to obtain an action written purely in terms of the σfield S(σ)=−/integraldisplay d2xN 2g2σ2−iN tr log(i/negationslash∂−σ) (18) Note the factor of Nin front of the tr log term coming from the integration over Nfermion fields. With the malice of forethought we, or rather Gross and Neveu, have introduced anexplicit factor of 1 /Nin the coupling strength in (16), so that the two terms in (18) both scale asN. Thus, the path integral Z=/integraltext Dσe iS(σ)may be evaluated by the steepest descent or stationary phase method in the large Nlimit. We simply extremize S(σ) . Incidentally, we can see the judiciousness of the choice a=1 in large NQCD in the same way. Integrating out the quarks in (1) we get S=−/integraldisplay d4xN 2g2trFμνFμν+Ntr log(i(/negationslash∂−i/negationslashA)−m) and thus the two terms both scale as Nand can balance each other. The increase in the number of degrees of freedom has to be offset by a weakening of the coupling. T o study the ground state behavior of the theory, we restrict our attention to field configurations σ(x) that do not depend on x. (In other words, we are not expecting translation symmetry to be spontaneously broken.) We can immediately take over the resultyou got in exercise IV .3.3 and write the effective potential 1 NV( σ)=1 2g(μ)2σ2+1 4πσ2/parenleftbigg logσ2 μ2−3/parenrightbigg (19) We have imposed the condition (1/N)[d2V (σ)/dσ2]|σ=μ=1/g(μ)2as the definition of the mass scale dependent coupling g(μ) (compare IV .3.18). The statement that V( σ) is independent of μimmediately gives 1 g(μ)2−1 g(μ/prime)2=1 πlogμ μ/prime(20) Asμ→∞ ,g(μ)→0. Remarkably, this theory is asymptotically free, just like QCD. If we want to, we can work backward to find the flow equation μd dμg(μ)=−1 2πg(μ)3+... (21) The theory in its different incarnations, (16), (17), and (18), enjoys a discrete Z2symme- try under which ψa→γ5ψaandσ→−σ. As in chapter IV .3, this symmetry is dynamically 404 | VII. Grand Unification broken by quantum fluctuations. The minimum of V( σ) occurs at σmin=μe1−π/g(μ)2and so according to (17) the fermions acquire a mass mF=σmin=μe1−π/g(μ)2(22) Note that this highly nontrivial result can hardly be seen by staring at (16) and we have no way of proving it for finite N. In the spirit of the large Napproach, however, we expect that the fermion mass might be given by mF=μe1−π/g(μ)2+O(1/N2)so that (22) would be a decent approximation even for say, N=3. Since mFis physically measurable, it better not depend on μ. You can check that. This theory also exhibits dimensional transmutation as described in the previous chap- ter. We start out with a theory with a dimensionless coupling gand end up with a dimen- sional fermion mass mF. Indeed, any other quantity with dimension of mass would have to be equal to mFtimes a pure number. Dynamically generated kinks I discuss the existence of kinks and solitons in chapter V .6. You clearly understood that the existence of such objects follows from general considerations of symmetry and topology,rather than from detailed dynamics. Here we have a (1+1)-dimensional theory with a discrete Z 2symmetry, so we certainly expect a kink, namely a time independent configu- ration σ(x) (henceforth xwill denote only the spatial coordinate and will no longer label a generic point in spacetime) such that σ(−∞)=−σmin andσ(+∞)=σmin. [Obviously, there is also the antikink with σ(−∞)=σmin andσ(+∞)=−σmin.] At first sight, it would seem almost impossible to determine the precise shape of the kink. In principle, we have to evaluate tr log[ i/negationslash∂−σ(x) ] for an arbitrary function σ(x) such thatσ(+∞)=−σ(−∞) (and as I explained in chapter IV .3, this involves finding all the eigenvalues of the operator i/negationslash∂−σ(x) , summing over the logarithm of the eigenvalues), and then varying this functional of σ(x) to find the optimal shape of the kink. Remarkably, the shape can actually be determined thanks to a clever observation.5In analogy with the steps leading to (IV .3.24) we note that tr log[i/negationslash∂−σ(x) ]=tr logγ5[i/negationslash∂−σ(x) ]γ5=tr log(−1)[i/negationslash∂+σ(x) ] and thus up to an irrelevant additive constant tr log(i/negationslash∂−σ(x)) =1 2tr log[i/negationslash∂−σ(x) ][i/negationslash∂+σ(x) ] =1 2tr log/braceleftBig −∂2+iγ1σ/prime(x)−[σ(x) ]2/bracerightBig (23) 5C. Callan, S. Coleman, D. Gross, and A. Zee, (unpublished). See D. J. Gross, “Applications of the Renormal- ization Group to High-Energy Physics,” in: R. Balian and J. Zinn-Justin, eds., Methods in Field Theory , p. 247. By the way, I recommend this book to students of field theory. VII.4. Large NExpansion | 405 Since γ1has eigenvalues ±i, this is equal to 1 2/braceleftBig tr log{−∂2+σ/prime(x)−[σ(x) ]2}+tr log {−∂2−σ/prime(x)−[σ(x) ]2}/bracerightBig but these two terms are equal by parity (space reflection) and hence tr log[i/negationslash∂−σ(x) ]=tr log{−∂2−σ/prime(x)−[σ(x) ]2} Referring to (18) we see that S(σ) is the sum of two terms, a term quadratic in σ(x) and a term that depends only on the combination σ/prime(x)+[σ(x) ]2. But we know that σmin minimizes S(σ) . Thus, the soliton is given by the solution of the ordinary differential equation σ/prime(x)+[σ(x) ]2=σ2 min(24) namely σ(x)=σmin tanhσminx. The soliton would be observed as an object of size 1/σmin=1/mF. I leave it to you to show that its mass is given by mS=N πmF (25) Precisely as theorized in the last chapter, the ratio mS/mFcomes out to be a pure number, N/π , as it must. By an even more clever method that I do not have space to describe, Dashen, Hasslacher, and Neveu were able to study time dependent configurations of σand determine the mass spectrum of this model. Exercises VII.4.1 Since the number of gluons only differs by one, it is generally argued that it does not make any difference whether we choose to study the U(N) theory or the SU(N) theory. Discuss how the gluon propagator in aU(N) theory differs from the gluon propagator in an SU(N) theory and decide which one is easier. VII.4.2 As a challenge, solve large NQCD in (1+1)-dimensional spacetime. [Hint: The key is that in (1+1)- dimensional spacetime with a suitable gauge choice we can integrate out the gauge potential Aμ.] For help, see ’t Hooft, Under the Spell of the Gauge Principle , p. 443. VII.4.3 Show that if we had chosen to calculate G(z)≡/angbracketleft(1/N) tr(1/z−ϕ)/angbracketright, we would have to connect the two open ends of the quark propagator. We see that figures VII.4.5b and d lead to the same diagram. Completethe calculation of G(z) in this way. VII.4.4 Suppose the random matrix ϕis real symmetric rather than hermitean. Show that the Feynman rules are more complicated. Calculate the density of eigenvalues. [Hint: The double-line propagator can twist.] VII.4.5 For hermitean random matrices ϕ, calculate Gc(z,w)≡/angbracketleftbigg1 Ntr1 z−ϕ1 Ntr1 w−ϕ/angbracketrightbigg −/angbracketleftbigg1 Ntr1 z−ϕ/angbracketrightbigg/angbracketleftbigg1 Ntr1 w−ϕ/angbracketrightbigg forV( ϕ)=1 2m2ϕ2using Feynman diagrams. [Note that this is a much simpler object to study than the object we need to study in order to learn about localization (see exercise VI.6.1).] Show that by takingsuitable imaginary parts we can extract the correlation of the density of eigenvalues with itself. For help,see E. Br ´ezin and A. Zee, Phys. Rev. E51: p. 5442, 1995. 406 | VII. Grand Unification VII.4.6 Use the Faddeev-Popov method to calculate Jin the Dyson gas approach. VII.4.7 ForV( ϕ)=1 2m2ϕ2+gϕ4, determine ρ(E) .F o rm2sufficiently negative (the double well potential again) we expect the density of eigenvalues to split into two pieces. This is evident from the Dyson gas picture.Find the critical value m 2 c.F o rm2<m2 cthe assumption of G(z) having only one cut used in the text fails. Show how to calculate ρ(E) in this regime. VII.4.8 Calculate the mass of the soliton (25). VII.5 Grand Unification Crying out for unification A gauge theory is specified by a group and the representations the matter fields belong to. Let us go back to chapter VII.2 and make a catalogue for the SU( 3)⊗SU( 2)⊗U(1) theory. For example, the left handed up and down quarks are in a doublet/parenleftbiguα dα/parenrightbig Lwith hypercharge1 2Y=1 6. Let us denote this by (3, 2,1 6)L, with the three numbers indicating how these fields transform under SU( 3)⊗SU( 2)⊗U(1). Similarly, the right handed up quark is (3, 1,2 3)R. The leptons are (1, 2,−1 2)Land(1, 1,−1)R, where the “1” in the first entry indicates that these fields do not participate in the strong interaction. Writing it alldown, we see that the quarks and leptons of each family are placed in (3, 2,1 6)L,(3, 1,2 3)R,(3, 1,−1 3)R,(1, 2,−1 2)L, and (1, 1,−1)R (1) This motley collection of representations practically cries out for further unification. Who would have constructed the universe by throwing this strange looking list down? What we would like to have is a larger gauge group Gcontaining SU( 3)⊗SU( 2)⊗ U(1), such that this laundry list of representations is unified into (ideally) one great big representation. The gauge bosons in G[but not in SU( 3)⊗SU( 2)⊗U(1)of course] would couple the representations in (1) to each other. Before we start searching for G, note that since gauge transformations commute with the Lorentz group, these desired gauge transformations cannot change left handed fieldsto right handed fields. So let us change all the fields in (1) to left handed fields. Recall fromexercise II.1.9 that charge conjugation changes left handed fields to right handed fieldsand vice versa. Thus, instead of (1) we can write (3, 2,1 6),(3∗,1 ,−2 3),(3∗,1 ,1 3),(1, 2,−1 2), and (1, 1, 1 ) (2) We now omit the subscripts LandR: everybody is left handed. 408 | VII. Grand Unification A perfect fit The smallest group that contains SU( 3)⊗SU( 2)⊗U(1)isSU( 5). (If you are shaky about group theory, study appendix B now.) Recall that SU( 5)has 52−1=24 generators. Explicitly, the generators are represented by 5 by 5 hermitean traceless matrices acting onfive objects we denote by ψ μwithμ=1, 2, . . . , 5. [These five objects form the fundamental or defining representation of SU( 5).] It is now obvious how we can fit SU( 3)andSU( 2)intoSU( 5). Of the 24 matrices that generate SU( 5), eight have the form/parenleftbigA0 00/parenrightbig and three the form/parenleftbig00 0B/parenrightbig , where Arepresents 3 by 3 hermitean traceless matrices (of which there are 32−1=8, the so-called Gell- Mann matrices) and Brepresents 2 by 2 hermitean traceless matrices (of which there are 22−1=3, namely the Pauli matrices). Clearly, the former generate an SU( 3)and the latter an SU( 2). Furthermore, the 5 by 5 hermitean traceless matrix 1 2Y=⎛ ⎜⎜⎜⎜⎜⎜⎜⎜⎝− 1 300 0 0 0−1 300 0 00 −1 300 0001 20 000 01 2⎞ ⎟⎟⎟⎟⎟⎟⎟⎟⎠(3) generates a U(1). Without being coy about it, we have already called this matrix the hypercharge1 2Y. In other words, if we separate the index μ={α,i}withα=1, 2, 3 and i=4, 5, then theSU( 3)acts on the index αand the SU( 2)acts on the index i. Thus, the three objects ψαtransform as a 3-dimensional representation under SU( 3)and hence could be a 3 o ra3∗. Let us choose ψαas transforming as 3; we will see shortly that this is the right choice with Y/2 given as in (3). The three objects ψαdo not transform under SU( 2) and hence each of them belongs to the singlet 1 representation. Furthermore, they carryhypercharge − 1 3as we can read off from (3). T o sum up, ψαtransform as (3, 1,−1 3) under SU( 3)⊗SU( 2)⊗U(1). On the other hand, the two objects ψitransform as 1 under SU( 3)and 2 under SU( 2), and carry hypercharge1 2; thus they transform as (1, 2,1 2).I n other words, we embed SU( 3)⊗SU( 2)⊗U(1)intoSU( 5)by specifying how the defining representation of SU( 5)decomposes into representations of SU( 3)⊗SU( 2)⊗U(1) 5→(3, 1,−1 3)⊕(1, 2,1 2) (4) T aking the conjugate we see that 5∗→(3∗,1 ,1 3)⊕(1, 2,−1 2) (5) Inspecting (2), we see that (3∗,1 ,1 3)and(1, 2,−1 2)appear on the list. We are on the right track! The fields in these two representations fit snugly into 5∗. This accounts for five of the fields contained in (2); we still have the ten fields (3, 2,1 6),(3∗,1 ,−2 3), and(1, 1, 1 ) (6) VII.5. Grand Unification | 409 Consider the next representation of SU( 5)in order of size, namely the antisymmetric tensor representation ψμν. Its dimension is (5×4)/2=10, precisely the number we want, if only the quantum numbers under SU( 3)⊗SU( 2)⊗U(1)work out! Since we know that 5 →(3, 1,−1 3)⊕(1, 2,1 2), we simply (again, see appendix B!) have to work out the antisymmetric product of (3, 1,−1 3)⊕(1, 2,1 2)with itself, namely the direct sum of (where ⊗Adenotes the antisymmetric product) (3, 1,−1 3)⊗A(3, 1,−1 3)=(3∗,1 ,−2 3) (7) (3, 1,−1 3)⊗A(1, 2,1 2)=(3, 2,−1 3+1 2)=(3, 2,1 6) (8) and (1, 2,1 2)⊗A(1, 2,1 2)=(1, 1, 1 ) (9) [I will walk you through (7): In SU( 3)3⊗A3=3∗(remember εijkfrom appendix B?), in SU( 2)1⊗A1=1, and in U(1)the hypercharges simply add −1 3−1 3=−2 3.] Lo and behold, these SU( 3)⊗SU( 2)⊗U(1)representations form exactly the collection of representations in (6). In other words, 10→(3, 2,1 6)⊕(3∗,1 ,−2 3)⊕(1, 1, 1 ) (10) The known quark and lepton fields in a given family fit perfectly into the 5∗and 10 representations of SU( 5)! I have just described the SU( 5)grand unified theory of Georgi and Glashow. In spite of the fact that the theory has not been directly verified by experiment, it is extremely difficultfor me and for many other physicists not to believe that SU( 5)is at least structurally correct, in view of the perfect group theoretic fit. It is often convenient to display the contents of the representation 5 ∗and 10, using the names given to the various fields historically. We write 5∗as a column vector ψμ=/parenleftBiggψα ψi/parenrightBigg =⎛ ⎜⎜⎝¯dα ν e⎞ ⎟⎟⎠(11) and the 10 as an antisymmetric matrix ψμν={ψαβ,ψαi,ψij} =⎛ ⎜⎜⎜⎜⎜⎜⎜⎜⎝0¯u−¯udu −¯u 0¯ud u ¯u−¯u 0du −d−d−d 0¯e −u−u−u−¯e 0⎞ ⎟⎟⎟⎟⎟⎟⎟⎟⎠(12) (I suppressed the color indices.) 410 | VII. Grand Unification Deepening our understanding of physics Aside from its esthetic appeal, grand unification deepens our understanding of physics enormously. 1. Ever wondered why electric charge is quantized? Why don’t we see particles with charge equal to√πtimes the electron’s charge? In quantum electrodynamics, you could perfectly well write down L=¯ψ[i(/negationslash∂−i/negationslashA)−m]ψ+¯ψ/prime[i(/negationslash∂−i√π/negationslashA)−m/prime]ψ/prime+... (13) In contrast, in grand unified theory Aμcouples to a generator of the grand unifying gauge group, and you know that the generators of any group such as SU(N) (that is not given by the direct product of U(1)with other groups) are forced by the nontrivial commutation relations [ Ta,Tb]=ifabcTcto assume quantized values. For example, the eigenvalues of T3inSU( 2), which depend on the representation of course, must be multiples of1 2. Within SU( 3)×SU( 2)×U(1), we cannot understand charge quantization: The generator of U(1)is not quantized. But upon grand unification into SU( 5)[or more generally any group without U(1)factors] electric charge is quantized. The result here is deeply connected to Dirac’s remark (chapter IV .4) that electric charge is quantized if the magnetic monopole exists. We know from chapter V .7 that spontaneouslybroken nonabelian gauge theories such as the SU( 5)theory contain the monopole. 2. Ever wondered why the proton charge is exactly equal and opposite to the electron charge? This important fact allows us to construct the universe as we know it. Atoms mustbe electrically neutral to some fantastic degree of accuracy for standard cosmology to work;otherwise, electrostatic forces between macroscopic matter would tear the universe apart. This remarkable fact is nicely incorporated into SU( 5). It is fun to see how it goes. Evaluating tr Q=0 over the 5 ∗implies that 3 Q¯d=−Qe−. I have used the fact that the strong interaction commutes with electromagnetism and hence quarks with different colorhave the same charge. Now let us calculate the proton charge Q P: QP=2Qu+Qd=2(Qd+1)+Qd=3Qd+2=Qe−+2 (14) IfQe−=− 1, then QP=−Qe−, as is indeed the case! 3. Recall that in electroweak theory we defined tan θ=g1/g2, with the coupling of the gauge bosons g2Aa μTa+g1Bμ(Y/2). Since the normalization of Aa μandBμis fixed by their respective kinetic energy term, the relative strength of g2andg1is determined by the normalization of Y/2 relative to T3. Let us evaluate tr T2 3and tr (Y/2)2on the defining representation 5 : tr T2 3=(1 2)2+(1 2)2=1 2and tr (Y/2)2=(1 3)23+(1 2)22=5 6. Thus, T3and/radicalbig 3/5(Y/2)are normalized equally. So the correct grand unified combina- tion is Aa μTa+Bμ/radicalbig 3/5(Y/2), and therefore tan θ=g1/g2=/radicalbig 3/5o r sin2θ=3 8(15) VII.5. Grand Unification | 411 at the grand unification scale. T o compare with the experimental value of sin2θwe would have to study how the couplings g2andg1flow under the renormalization group down to low energies. We will postpone this discussion until the next chapter. Freedom from anomaly Recall from chapter VII.2 that the key to proving renormalizability of nonabelian gaugetheory is the ability to pass freely between the unitary gauge and the R ξgauge. The crucial ingredient is gauge invariance and the resulting Ward-T akahashi identities (seechapter II.7). Suddenly you start to worry. What about the chiral anomaly? The existence of the anomaly means that some Ward-T akahashi identities fail to hold. For our theories to makesense, they had better be free from anomalies. I remarked in chapter IV .7 that the historicalname “anomaly” makes it sound like some kind of sickness. Well, in a way, it is. We should have already checked the SU( 3)⊗SU( 2)⊗U(1)theory for anomalies, but we didn’t. I will let you do it as an exercise. Here I will show that the SU( 5)theory is healthy. If theSU( 5)theory is anomaly-free, then a fortiori so is the SU( 3)⊗SU( 2)⊗U(1)theory. In chapter IV .7 I computed the anomaly in an abelian theory but as I remarked there clearly all we have to do to generalize to a nonabelian theory is to insert a generator T a of the gauge group at each vertex of the triangle diagram in figure IV .7.1. Summing over the various fermions running around the loop, we see that the anomaly is proportionaltoA abc(R)≡tr(Ta{Tb,Tc}), where Rdenotes the representation to which the fermions belong. We have to sum Aabc(R) over all the representations in the theory, remembering to associate opposite signs to left handed and right handed fermion fields. (It may behelpful to remind yourself of remark 3 in chapter IV .7 and exercise IV .7.6.) We are now ready to give the SU( 5)theory a health check. First, all fermion fields in (2) are left handed. Second, convince yourself (simply imagine calculating A abcfor all possible abc) that it suffices to set Ta,Tb, andTcall equal to T≡⎛ ⎜⎜⎜⎜⎜⎜⎜⎜⎝2 0 000 0 2 000002 0 0000 −30 000 0 −3⎞ ⎟⎟⎟⎟⎟⎟⎟⎟⎠ a multiple of the hypercharge. Let us now evaluate tr T3on the 5∗representation, trT3|5∗=3(−2)3+2(+3)3=30 (16) and on the 10, trT3|10=3(+4)3+6(−1)3+(−6)3=− 30 (17) An apparent miracle! The anomaly cancels. 412 | VII. Grand Unification This remarkable cancellation between sums of cubes of a strange list of numbers suggests strongly, to say the least, that SU( 5)is not the end of the story. Besides, it would be nice if the 5∗and 10 could be unified into a single representation. Exercises VII.5.1 Write down the charge operator Qacting on 5, the defining representation ψμ. Work out the charge content of the 10 =ψμνand identify the various fields contained therein. VII.5.2 Show that for any grand unified theory, as long as it is based on a simple group, we have at the unification scale sin2θ=/summationtextT2 3/summationtextQ2(18) where the sum is taken over all fermions. VII.5.3 Check that the SU( 3)⊗SU( 2)⊗U(1)theory is anomaly-free. [Hint: The calculation is more involved than in SU( 5)since there are more independent generators. First show that you only have to evaluate trY{Ta,Tb}and tr Y3, with TaandYthe generators of SU( 2)andU(1), respectively.] VII.5.4 Construct grand unified theories based on SU( 6),SU( 7),SU( 8), . . . , until you get tired of the game. People used to get tenure doing this. [Hint: You would have to invent fermions yet to be experimentallydiscovered.] VII.6 Protons Are Not Forever Proton decay Charge conservation guarantees the stability of the electron, but what about the stability of the proton? Charge conservation allows p→π0+e+. No fundamental principle says that the proton lives forever, but yet the proton is known for its longevity: It has been aroundessentially since the universe began. The stability of the proton had to be decreed by an authority figure: Eugene Wigner was the first to proclaim the law of baryon number conservation. The story goes that whenWigner was asked how he knew that the proton lives forever he quipped, “I can feel it inmy bones.” I take the remark to mean that just from the fact that we do not glow in thedark we can set a fairly good lower bound on the proton’s life span. As soon as we start grand unifying, we better start worrying. Generically, when we grand unify we put quarks and leptons into the same representation of some gauge group [see(VII.5.11 and VII.5.12)]. This miscegenation immediately implies that there are gaugebosons transforming quarks into leptons and vice versa. The bag of three quarks knownas the proton could very well get turned into leptons upon the exchange of these gaugebosons. In other words, the proton, the rock on which our world is built upon, may not beforever! Thus, grand unification runs the risk of being immediately falsified. LetM Xdenote generically the masses of those gauge bosons transforming quarks into leptons and vice versa. Then the amplitude for proton decay is of order g2/M2 X, with g the coupling strength of the grand unifying gauge group, and the proton decay rate /Gamma1is given by (g2/M2 X)2times a phase space factor controlled essentially by the proton mass mPsince the pion and positron masses are negligible compared to the proton mass. By dimensional analysis, we determine that /Gamma1∼(g2/M2 X)2m5P. Since the proton is known to live for something like at least 1031years, MXhad better be huge compared to the kind of energy scales we can reach experimentally. The mass MXis of the same order as the mass scale MGUT at which the grand unified theory is spontaneously broken down to SU( 3)⊗SU( 2)⊗U(1). Specifically, in the SU( 5) 414 | VII. Grand Unification μ3 5( )4πg 2 4πg 2 4πg 24πg 2 MGUT1 2 3 Figure VII.6.1 theory, a Higgs field Hμ νtransforming as the adjoint 24, with its vacuum expectation value /angbracketleftHμ ν/angbracketrightequal to the diagonal matrix with elements (−1 3,−1 3,−1 3,1 2,1 2)times some v, can do the job, as was discussed in chapter IV .6. The gauge bosons in SU( 3)⊗SU( 2)⊗U(1) remain massless while the other gauge bosons acquire mass MXof order gv. T o determine MGUT , we apply renormalization group flow to g3,g2, andg1, the cou- plings of SU( 3),SU( 2), and U(1), respectively. The idea is that as we move up in the mass or energy scale μthe two asymptotically free couplings g3(μ) andg2(μ) decrease while g1(μ) increases. Thus, at some mass scale MGUT they will meet and that is where SU( 3)⊗SU( 2)⊗U(1)is unified into SU( 5)(see figure VII.6.1). Because of the extremely slow logarithmic running (it should be called walking or even crawling but again for his-torical reasons we are stuck with running) of the coupling constant, we anticipate that theunification mass scale M GUT will come out to be much larger than any scale we were used to in particle physics prior to grand unification. In fact, MGUT will turn out to have an enormous value of the order 1014−15Gev and the idea of grand unification passes its first hurdle. Stability of the world implies the weakness of electromagnetism Using the result of exercise VI.8.1 we obtain (here αS≡g2 3/4πandαGUT≡g2/4πdenote the strong interaction and grand unification analog of the fine structure constant α, respectively, with Fthe number of families) 4π [g3(μ)]2≡1 αS(μ)=1 αGUT+1 6π(4F−33)logMGUT μ(1) 4π [g2(μ)]2≡sin2θ(μ) α(μ)=1 αGUT+1 6π(4F−22)logMGUT μ(2) 3 54π [g1(μ)]2≡3 5cos2θ(μ) α(μ)=1 αGUT+1 6π4FlogMGUT μ(3) VII.6. Protons Are Not Forever | 415 Byθ(μ) we mean the value of θat the scale μ.A tμ=MGUT , the three couplings are related through SU( 5). We evaluate these equations for some experimentally accessible value of μ, plugging in measured values of αSandα. With three equations, we not only manage to determine the unification scale MGUT and coupling αGUT , but we can predict θ. In other words, unless the ratio g1tog2is precisely right, the three lines in figure VII.6.1 will not meet at one point. Note that the number of fermion families Fcontributes equally to (1), (2), and (3). This is as it should be since the fermions are effectively massless for the purpose of this calculationand do not “know” that the unifying group has been broken into SU( 3)⊗SU( 2)⊗U(1). These equations are derived assuming that all fermion masses are small compared to μ. Rearranging these equations somewhat, we find sin2θ=1 6+5α(μ) 9αS(μ)(4) sin2θ α(μ)=1 αS(μ)+1 6π11 logMGUT μ(5) 1 α(μ)=8 31 αGUT+1 6π/parenleftbigg32 3F−22/parenrightbigg logMGUT μ(6) We obtain in (4) a prediction for sin2θ(μ) independent of MGUT and of the number of families. Note that (5) gives the bound 1 α(μ)≥1 6π11 logMGUT μ(7) A lower bound on the proton lifetime (and hence on MGUT)translates into an upper bound on the fine structure constant. Amusingly, the stability of the world implies the weaknessof electromagnetism. As I noted earlier, plugging in the measured value of α S, we obtain a huge value for MGUT . I regard this as a triumph of grand unification: MGUT could have come out to have a much lower scale, leading to an immediate contradiction with the observed stability of theproton, but it didn’t. Another way of looking at it is that if we are somehow given M GUT and αGUT , grand unification fixes the couplings of all three nongravitational interactions! The point is not that this simplest try at grand unification doesn’t quite agree with experiment: The miracle is that it works at all. It is beyond the scope of this book to discuss in detail the comparison of (4), (5), and (6) with experiment. T o do serious phenomenology, one has to include threshold effects(see exercise VII.6.1), higher order corrections, and so on. T o make a long story short,after grand unified theory came out there was enormous excitement over the possibilityof proton decay. Alas, the experimental lower bound on the proton lifetime was eventuallypushed above the prediction. This certainly does not mean the demise of the notion of grand unification. Indeed, as I mentioned earlier, the perfect fit is enough to convincemost particle theorists of the essential correctness of the idea. Over the years people haveproposed adding various hypothetical particles to the theory to promote proton longevity. 416 | VII. Grand Unification The idea is that these particles would affect the renormalization group flow and hence MGUT . The proton lifetime is actually not the most critical issue. With more accurate measurements of αSand of θ, it was found that the three couplings do not quite meet at a point. Indeed, for believers in low energy supersymmetry, part of their faith is foundedon the fact that with supersymmetric particles included, the three coupling constants domeet. 1But skeptics of course can point to the extra freedom to maneuver. Branching ratios You may have realized that (1), (2), and (3) are not specific for SU( 5): they hold as long asSU( 3)⊗SU( 2)⊗U(1)is unified into some simple group (simple so that there is only one gauge coupling g). Let us now focus on SU( 5). Recall that we decompose the SU( 5)index μ, which can take on five values, into two types. In other words, the index μis labeled by {α,i}, where α takes on three values and itakes on two values. The gauge bosons in SU( 5)correspond to the 24 independent components of the traceless hermitean field Aμ ν(μ,ν=1 ,2 ,...,5 ) transforming as the adjoint representation. Focusing on the group theory of SU( 5),w e will suppress Lorentz indices, spinor indices, etc. Clearly, the eight gauge bosons in SU( 3) transform an index of type αinto an index of type α, while the three gauge bosons in SU( 2) transform an index of type iinto an index of type i. Then there is the U(1)gauge boson that couples to the hypercharge1 2Y. (Of course, you know what I mean by my somewhat loose language: The SU( 3)gauge bosons transform fields carrying a color index into a field carrying a color index.) The fun comes with the gauge bosons Aα iandAi α, which transform the index αinto the index iand vice versa. Since αtakes on three values, and itakes on two values, there are 6+6=12 such gauge bosons, thus accounting for all the gauge bosons in SU( 5).I n other words, 24 →(8, 1)+(1, 3)+(1, 1)+(3, 2)+(3∗,2). We will now see explicitly that the exchange of these bosons between quarks and leptons leads to proton decay. We merely have to write down the terms in the Lagrangian involving the coupling of the bosons Aα iandAi αto fermions and draw the appropriate Feynman diagrams. I will go through part of the group theoretic analysis, leaving you to work out the rest. Simplyby contracting indices we see that the boson A μ νacting on ψμtakes it to ψνand acting on ψνρtakes it to ψμρ. Let us look at what A5 αdoes, using your result from exercise VII.5.1. It takes ψ5=e−→ψα=¯d (8) ψαβ=¯u→ψ5β=u (9) and ψα4=d→ψ54=e+(10) 1See, e.g., F. Wilczek, in: V . Fitch et al., eds., Critical Problems in Physics , p. 297. VII.6. Protons Are Not Forever | 417 de+ u u de+ u u u u Figure VII.6.2 Thus, the exchange of A5 αgenerates the process (figure VII.6.2) u+d→¯u+e+, leading to proton decay p(uud) →π0(u¯u)+e+. Observe that while the decay p→π0+e+violates both baryon number Band lepton number L, it conserves the combination B−L. In exercise VII.6.2 you will work out the branching ratios for various decay modes. T oo bad experimentalists have not yet measured them. Fermion masses We might hope that with grand unification we would gain new understanding of quarkand lepton masses. Unfortunately, the situation on fermion masses in SU( 5)is muddled, and to this day nobody understands the origin of quark and lepton masses. Introducing a Higgs field ϕ μtransforming as the 5 (as indicated by the notation) we can write the coupling ψμCψμνϕν (11) and ψμνCψλρϕσεμνλρσ (12) (withϕνthe conjugate 5∗), reflecting the group theoretic fact (see appendix C) that 5∗⊗10 contains the 5 and 10 ⊗10 contains the 5∗. Since 5 →(3, 1,−1 3)⊕(1, 2,1 2)we see that this Higgs field is just the natural extension of the SU( 2)⊗U(1)Higgs doublet (1, 2,1 2). Not wanting to break electromagnetism, we allow only the electrically neutral fourth component of ϕto acquire a vacuum expectation value. Setting /angbracketleftϕ4/angbracketright=v , we obtain (up to uninteresting overall constants) ψαCψα4+ψ5Cψ54/equal1⇒md=me (13) and ψαβCψγ5εαβγ/equal1⇒mu/negationslash=0 (14) 418 | VII. Grand Unification The larger symmetry yields a mass relation md=meat the unification scale; we again have to apply the renormalization group flow. It is worth noting that the mass relationm d=mecomes about because as far as the fermions are concerned, SU( 5)has been only broken down to SU( 4)byϕ. The trouble is that we obtain more or less the same relation for each of the three families, since most of the running occurs between the unificationscaleM GUT and the top quark mass so that threshold effects give only a small correction. Putting in numbers one gets something like mb mτ∼ms mμ∼md me∼3 (15) Let us use this to predict the down sector quark masses in terms of the lepton masses. The formula mb∼3mτworks rather well and provides indirect evidence that there can only be three families since the renormalization group flow depends on F. The formula ms∼3mμis more or less in the ballpark, depending on what “experimental” value one takes for ms. The formula for md, on the other hand, is downright embarrassing. People mumble something about the first family being so light and hence other effects, such asone-loop corrections might be important. At the cost of making the theory uglier, peoplealso concoct various schemes by introducing more Higgs fields, such as the 45, to givemass to fermions. Note that in one respect SU( 5)is not as “economical” as SU( 2)⊗U(1), in which the same Higgs field that gives mass to the gauge bosons also gives mass to the fermions. The universe is not empty, but almost I mention in passing another triumph of grand unification: its ability to explain the originof matter in our universe. It has long behooved physicists to understand two fundamentalfacts about the universe: (1) the universe is not empty, and (2) the universe is almost empty.T o physicists, (1) means that the universe is not symmetric between matter and antimatter,that is, the net baryon number N Bis nonzero; and (2) is quantified by the strikingly small observed value NB/Nγ∼10−10of the ratio of the number of baryons to the number of photons. Suppose we start with a universe with equal quantities of matter and antimatter. For the universe to evolve into the observed matter dominated universe, three conditions must besatisfied: (1) The laws of the universe must be asymmetric between matter and antimatter.(2) The relevant physical processes had to be out of equilibrium so that there was an arrowof time. (3) Baryon number must be violated. We know for a fact that conditions (1) and (2) indeed hold in the world: There is CP violation in the weak interaction and the early universe expanded rapidly. As for (3), grand unification naturally violates baryon number. Furthermore, while proton decay(suppressed by a factor of 1 /M 2 GUTin amplitude) proceeds at an agonizingly slow rate (for those involved in the proton decay experiment!), in the early universe, when the Xand Ybosons are produced in abundance, their fast decays could easily drive baryon number VII.6. Protons Are Not Forever | 419 violation. The suppression factor 1 /M2 GUTdoes not come in. I have no doubt that eventually the number 10−10measuring “the amount of dirt in the universe” will be calculated in some grand unified theory. Hierarchy I promised you that the Weisskopf phenomenon would come back to haunt us. That thegrand unification mass scale M GUT naturally comes out so large counts as a triumph, but it also leads to a problem known as the hierarchy problem. The hierarchy refers tothe enormous ratio M GUT/MEW, where MEWdenotes the electroweak unification scale, of order 102Gev. I will sketch this rather murky subject. Look at the Higgs field ϕrespon- sible for breaking electroweak theory. We don’t know its renormalized or physical massprecisely, but we do know that it is of order M EW. Imagine calculating the bare pertur- bation series in some grand unified theory—the precise theory does not enter into thediscussion—starting with some bare mass μ 0forϕ. The Weisskopf phenomenon tells us that quantum correction shifts μ2 0by a huge quadratically cutoff dependent amount δμ2 0∼f2/Lambda12∼f2M2 GUT, where we have substituted for /Lambda1the only natural mass scale around, namely MGUT , and where fdenotes some dimensionless coupling. T o have the physical mass squared μ2=μ2 0+δμ2 0come out to be of order M2 EW, something like 28 or- ders of magnitude smaller than M2 GUT, would require an extremely fine-tuned and highly unnatural cancellation between μ2 0andδμ2 0. How this could happen “naturally” poses a severe challenge to theoretical physicists. Naturalness The hierarchy problem is closely connected with the notion of naturalness dear to the the-oretical physics community. We naturally expect that dimensionless ratios of parametersin our theories should be of order unity, where the phrase “order unity” is interpreted lib-erally between friends, say anywhere from 10 −2or 10−3to 102or 103. Following ’t Hooft, we can formulate a technical definition of naturalness: The smallness of a dimensionlessparameter ηwould be considered natural only if a symmetry emerges in the limit η→0. Thus, fermion masses could be naturally small, since, as you will recall from chapter II.1,a chiral symmetry emerges when a fermion mass is set equal to zero. On the other hand,no particular symmetry emerges when we set either the bare or renormalized mass of ascalar field equal to zero. This represents the essence of the hierarchy problem. Exercises VII.6.1 Suppose there are F/primenew families of quarks and leptons with masses of order M/prime. Adopting the crude approximation described in exercise VI.8.2 of ignoring these families for μbelow M/primeand of treating M/prime 420 | VII. Grand Unification as negligible for μabove M/prime, run the renormalization group flow and discuss how various predictions, such as proton lifetime, are changed. VII.6.2 Work out proton decay in detail. Derive relations between the following decay rates: /Gamma1(p→π0e+), /Gamma1(p→π+¯ν),/Gamma1(n→π−e+), and/Gamma1(n→π0¯ν). VII.6.3 Show that SU( 5)conserves the combination B−L. For a challenge, invent a grand unified theory that violates B−L. VII.7 SO(10) Unification Each family into a single representation At the end of chapter VII.5 we felt we had good reason to think that SU( 5)unification is not the end of the story. Let us ask if we might be able to fit the 5 and 10∗into a single representation of a bigger group Gcontaining SU( 5). It turns out that there is a natural embedding of SU( 5)into the orthogonal SO( 10) that works,1but to explain that I have to teach you some group theory. The starting point is perhaps somewhat surprising: We go back to chapter II.3, where we learned that theLorentz group SO( 3, 1), or its Euclidean cousin SO( 4), has spinor representations. We will now generalize the concept of spinors to d-dimensional Euclidean space. I will work out the details for deven and leave the odd dimensions as an exercise for you. You might also want to review appendix B now. Clifford algebra and spinor representations Start with an assertion. For any integer nwe claim that we can find 2 nhermitean matrices γi(i=1, 2, ... ,2n)that satisfy the Clifford algebra {γi,γj}=2δij (1) In other words, to prove our claim we have to produce 2 nhermitean matrices γithat anticommute with each other and square to the identity matrix. We will refer to the γi’s as theγmatrices for SO( 2n). Forn=1, it is a breeze: γ1=τ1andγ2=τ2. There you are. 1Howard Georgi told me that he actually found SO( 10)before SU( 5). 422 | VII. Grand Unification Now iterate. Given the 2 nγ matrices for SO( 2n)we construct the (2 n+2)γmatrices forSO( 2n+2)as follows γ(n+1) j=γ(n) j⊗τ3=/parenleftBiggγ(n) j0 0−γ(n) j/parenrightBigg ,j=1, 2, ... ,2n (2) γ(n+1) 2n+1=1⊗τ1=/parenleftBigg01 10/parenrightBigg (3) γ(n+1) 2n+2=1⊗τ2=/parenleftBigg0−i i 0/parenrightBigg (4) (Throughout this book 1 denotes a unit matrix of the appropriate size.) The superscript in parentheses is obviously for us to keep track of which set of γmatrices we are talking about. Verify that if the γ(n)’s satisfy the Clifford algebra, the γ(n+1)’s do as well. For example, {γ(n+1) j,γ(n+1) 2n+1}=(γ(n) j⊗τ3).(1⊗τ1)+(1⊗τ1).(γ(n) j⊗τ3) =γ(n) j⊗{τ3,τ1}=0 This iterative construction yields for SO( 2n)theγmatrices γ2k−1=1⊗1⊗...⊗1⊗τ1⊗τ3⊗τ3⊗...⊗τ3 (5) and γ2k=1⊗1⊗...⊗1⊗τ2⊗τ3⊗τ3⊗...⊗τ3 (6) with 1 appearing k−1 times and τ3appearing n−ktimes. The γ’s are evidently 2nby 2n matrices. When and if you feel confused at any point in this discussion you should work things out explicitly for SO( 4),SO( 6), and so on. In analogy with the Lorentz group, we define 2 n(2n−1)/2=n(2n−1)hermitean matrices σij≡i 2[γi,γj] (7) Note that σijis equal to iγiγjfori/negationslash=jand vanishes for i=j. The commutation of the σ’s with each other is thus easy to work out. For example, [σ12,σ23]=− [γ1γ2,γ2γ3]=−γ1γ2γ2γ3+γ2γ3γ1γ2=− [γ1,γ3]=2iσ13 Roughly speaking, the γ2’s inσ12andσ23knock each other out. Thus, you see that the 1 2σij’s satisfy the same commutation relations as the generators Jij’s ofSO( 2n)(as given in appendix B). The1 2σij’s represent the Jij’s. As 2nby 2nmatrices, the σ’s act on an object ψwith 2ncomponents that we will call the spinor ψ. Consider the unitary transformation ψ→eiωijσijψwithωij=−ωjia set of real numbers. Then ψ†γkψ→ψ†e−iωijσijγkeiωijσijψ=ψ†γkψ−iωijψ†[σij,γk]ψ+... forωijinfinitesimal. Using the Clifford algebra we easily evaluate the commutator as [σij,γk]=− 2i(δikγj−δjkγi). (Ifkis not equal to either iorjthenγkclearly commutes VII.7. SO(10) Unification | 423 withσij, and if kis equal to either iorj, then we use γ2 k=1.)We see that the set of objects vk≡ψ†γkψ,k=1,... ,2ntransforms as a vector in 2 n-dimensional space, with 4 ωijthe infinitesimal rotation angle in the ijplane: vk→vk−2(ωkjvj−ωikvi)=vk−4ωkjvj (8) (in complete analogy to ¯ψγμψtransforming as a vector under the Lorentz group.) This gives an alternative proof that1 2σijrepresents the generators of SO( 2n). We define the matrix γFIVE=(−i)nγ1γ2...γ2n, which in the basis we are using has the explicit form γFIVE=τ3⊗τ3⊗...⊗τ3 (9) withτ3appearing ntimes. By analogy with the Lorentz group we define the “left handed” spinor ψL≡1 2(1−γFIVE)ψand the “right handed” spinor ψR≡1 2(1+γFIVE)ψ, such that γFIVEψL=−ψLandγFIVEψR=ψR. Under ψ→eiωijσijψ, we have ψL→eiωijσijψLand ψR→eiωijσijψRsinceγFIVEcommutes with σij. The projection into left and right handed spinors cut the number of components into halves and thus we arrive at the importantconclusion that the two irreducible spinor representations of SO( 2n)have dimension 2 n−1. (Convince yourself that the representation cannot be reduced further.) In particular, thespinor representation of SO( 10)is 2 10/2−1=24=16−dimensional. We will see that the 5∗and 10 of SU( 5)can be fit into the 16 of SO( 10). Embedding unitary groups into orthogonal groups The unitary group SU( 5)can be naturally embedded into the orthogonal group SO( 10). In fact, I will now show you that embedding SU(n) intoSO( 2n)is as easy as z=x+iy. Consider the 2 n-dimensional real vectors x=(x1,... ,xn,y1,... ,yn)and x/prime=(x/prime 1,... ,x/prime n,y/prime 1,... ,y/prime n). By definition, SO( 2n)consists of linear transformations on these two real vectors leaving their scalar product x/primex=/summationtextn j=1(x/prime jxj+y/prime jyj)invariant. Now out of these two real vectors we can construct two n-dimensional complex vectors z=(x1+iy1,... ,xn+iyn)andz/prime=(x/prime 1+iy/prime 1,... ,x/prime n+iy/prime n). The group U(n) consists of transformations on the two n-dimensional complex vectors zandz/primeleaving invariant their scalar product (z/prime)∗z=n/summationdisplay j=1(x/prime j+iy/prime j)∗(xj+iyj) =n/summationdisplay j=1(x/prime jxj+y/prime jyj)+in/summationdisplay j=1(x/prime jyj−y/prime jxj) In other words, SO( 2n)leaves/summationtextn j=1(x/prime jxj+y/prime jyj)invariant, but U(n) consists of the subset of those transformations in SO( 2n)that leave invariant not only/summationtextn j=1(x/prime jxj+y/prime jyj) but also/summationtextn j=1(x/prime jyj−y/prime jxj). Now that we understand this natural embedding of U(n) intoSO( 2n), we see that the defining or vector representation of SO( 2n), which we will call simply 2 n, decomposes 424 | VII. Grand Unification upon restriction to U(n) into the two defining representations of U(n) ,nandn∗; thus 2n→n⊕n∗(10) In other words, (x1,... ,xn,y1,... ,yn)can be written as (x1+iy1,... ,xn+iyn)and (x1−iy1,... ,xn−iyn)Note that this is the analog of (VII.5.4) indicating that the defining representation of SU( 5)decomposes into representations of SU( 3)⊗SU( 2)⊗U(1): 5→(3∗,1 ,1 3)⊕(1, 2,−1 2). (11) Given the decomposition law (10), we can now figure out how other representations of SO( 2n)decompose when restricted to the natural subgroup U(n) . The tensor representa- tions of SO( 2n)are easy, since they are constructed out of the vector representation. [This is precisely what we did in going from (VII.5.4) to (VII.5.7, 8, and 9).] For example, the adjointrepresentation of SO( 2n), which has dimension 2 n(2n−1)/2=n(2n−1), transforms as an antisymmetric 2-index tensor 2 n⊗ A2nand so decomposes into 2n⊗A2n→(n⊕n∗)⊗A(n⊕n∗) (12) according to (10). The antisymmetric product ⊗Aon the right hand side is, of course, to be evaluated within U(n). For instance, n⊗Anis the n(n−1)/2 representation of U(n) . In this way, we see that n(2n−1)→n2−1 (the adjoint) ⊕1 (the singlet) ⊕n(n−1)/2 ⊕(n(n−1)/2)∗(13) As a check, the total dimension of the representations of U(n) on the right hand side adds up to(n2−1)+1+2n(n−1)/2=n(2n−1). In particular, for SO( 10)⊃SU( 5), we have 45→24⊕1⊕10⊕10∗and of course 24 +1+10+10=45. Decomposing the spinor It is more difficult to figure out how the spinor representation of SO( 2n)decompose upon restriction to U(n) . I give here a heuristic argument that satisfies most physicists, but certainly not mathematicians. I will just do SO( 10)⊃SU( 5)and let you work out the general case. The question is how the 16 falls apart. Just from numerology and fromknowing the dimensions of the smaller representations of SU( 5)(1, 5, 10, 15) we see there are only so many possibilities, some of them rather unlikely, for example, the 16 fallingapart into 16 1’s. Picture the spinor 16 of SO( 10)breaking up into a bunch of representations of SU( 5). By definition, the 45 generators of SO( 10)scramble all these representations together. Let us ask what the various pieces of 45, namely 24 ⊕1⊕10⊕10 ∗, do to these representations. The 24 transform each of the representations of SU( 5)into itself, of course, because they are the 24 generators of SU( 5)and that is what generators were born to do. The generator VII.7. SO(10) Unification | 425 1 can only multiply each of these representations by a real number. (In other words, the corresponding group element multiplies each of the representations by a phase factor.) What does the 10, which as you recall from chapter VII.5 is represented as an antisym- metric tensor with two upper indices and hence also known as [2], do to these represen-tations? Suppose the bunch of representations that Sbreaks up into contains the singlet [0]=1o fSU( 5). The 10 =[2] acting on [0] gives the [2] =10. (Almost too obvious for words! An antisymmetric tensor of two indices combined with a tensor with no indices is an anti-symmetric tensor of two indices.) What about 10 =[2] acting on [2]? The result is a tensor with four upper indices. It certainly contains the [4], which is equivalent to [1] ∗=5∗.B u t look, 1 ⊕10⊕5∗already add up to 16. Thus, we have accounted for everybody. There can’t be more. So we conclude S+→[0]⊕[2]⊕[4]=1⊕10⊕5∗(14) The 5∗and the 10 of SU( 5)fit inside the 16+ofSO( 10)! We will learn later that the two spinor representations of SO( 10)are conjugate to each other. Indeed, you may have noticed that I snuck a superscript plus on the letter S. The conjugate spinor S−breaks up into the conjugate of the representations in (14): S−→[1]⊕[3]⊕[5]=5⊕10∗⊕1∗(15) The long lost antineutrino The fit would be perfect if we introduce one more field transforming as a 1, that is, a singlet under SU( 5)and hence a fortiori a singlet under SU( 3)⊗SU( 2)⊗U(1). In other words, this field does not participate in the strong, weak, and electromagnetic interactions, or inplain English, it describes a lepton with no electric charge and is not involved in the knownweak interaction. Thus, this field can be identified as the “long lost” antineutrino field ν c L. This guy does not listen to any of the known gauge bosons. Recall that we are using a convention in which all fermion fields are left handed, and hence we have written νc L. By a conjugate transformation, as explained earlier, this is equivalent to the right handed neutrino field νR. SinceνRis anSU( 5)singlet, we can give it a Majorana mass Mwithout breaking SU( 5). Hence we expect Mto be larger than or of the same order of magnitude as the mass scale at which SU( 5)is broken, which as we saw in chapter VII.5 is much higher than the mass scales that have been explored experimentally. This explains why νRhas not been seen. On the other hand, with the presence of νRwe can have a Dirac mass term m(¯νLνR+ h.c.). Since this term breaks SU( 2)⊗U(1)just like the mass terms for the quarks and leptons we know, we expect mto be of the same order of magnitude2as the known quark and lepton masses (which for reasons unknown span an enormous range). 2Explicitly, with νRnow available we can add to the SU( 2)⊗U(1)theory of chapter VII.2 the term f/prime˜ϕ¯νRψL, where ˜ϕ≡τ2ϕ†. In the absence of any indication to the contrary, we might suppose that f/primeis of the same order of magntiude as the coupling fthat leads to the electron mass. 426 | VII. Grand Unification Thus, in the space spanned by (ν,νc)we have the (Majorana) mass matrix M=/parenleftBigg0m mM/parenrightBigg (16) withM/greatermuchm. Since the trace and determinant of MareMand−m2, respectively, Mhas a large eigenvalue ∼Mand a small eigenvalue ∼m2/M . A tiny mass ∼m2/M, suppressed relative to the usual quark and lepton masses by the factor m/M , is naturally generated for the (observed) left handed neutrino. This rather attractive scenario, known as theseesaw mechanism for obvious reason, was discovered independently by Minkowski andby Glashow, and somewhat later by Y anagida and by Gell-Mann, Ramond, and Slansky. Again, the tight fit of the 5 ∗and the 10 of SU( 5)inside the 16+ofSO( 10)has convinced many physicists that it is surely right. A binary code for the world Given the product form of the γmatrices in (5) and (6), and hence of σij, we can write the states of the spinor representations as |ε1ε2...εn/angbracketright (17) where each of the ε’s takes on the values ±1. For example, for n=1,τ1|+/angbracketright=|−/angbracketright and τ1|−/angbracketright=|+/angbracketright , while τ2|+/angbracketright=i |−/angbracketright andτ2|−/angbracketright=− i|+/angbracketright . From (9) we see that γFIVE|ε1ε2...εn/angbracketright=(/Pi1n j=1εj)|ε1ε2...εn/angbracketright (18) The right handed spinor S+consists of those states |ε1ε2...εn/angbracketrightwith(/Pi1n j=1εj)=+ 1, and the left handed spinor S−those states with (/Pi1n j=1εj)=− 1. Indeed, the spinor represen- tations have dimension 2n−1. Thus, in SO( 10)unification the fundamental quarks and leptons are described by a five- bit binary code, with states such as |++−−+/angbracketright and|−+−−−/angbracketright . Personally, I find this a rather pleasing picture of the world. Let us work out the states explicitly. This also gives me a chance to make sure that you understand the group theory presented in this chapter. Start with the much simplercase of SO( 4). The spinor S +consists of |++ /angbracketright and|−− /angbracketright while the spinor S−consists of|+− /angbracketright and|−+ /angbracketright . As discussed in chapter II.3, SO( 4)contains two distinct SU( 2) subgroups. Removing a few factors of ifrom the discussion in chapter II.3 we see that the third generator of SU( 2), call it σ3, can be taken to be either σ12−σ34orσ12+σ34. The two choices correspond to the two distinct SU( 2)subgroups. We choose (arbitrarily) σ3=1 2(σ12−σ34). From (5) and (6) we have σ12=iγ1γ2=i(τ 1⊗τ3)(τ2⊗τ3)=−τ3⊗1 and σ34=− 1⊗τ3, and so σ3=1 2(−τ 3⊗1+1⊗τ3). T o figure out how the four states |++ /angbracketright , |−− /angbracketright ,|+− /angbracketright , and|−+ /angbracketright transform under our chosen SU( 2), let us act on them with σ3. For example, σ3|++ /angbracketright=1 2(−τ3⊗1+1⊗τ3)|++ /angbracketright=1 2(−1+1)|++ /angbracketright=0 VII.7. SO(10) Unification | 427 and σ3|−+ /angbracketright=1 2(−τ3⊗1+1⊗τ3)|−+ /angbracketright=1 2(1+1)|−+ /angbracketright=|−+ /angbracketright Aha, under SU( 2)|++ /angbracketright and|−− /angbracketright are two singlets while |+− /angbracketright and|−+ /angbracketright make up a doublet. Note that this is consistent with the generalization of (14) and (15), namely that upon the restriction of SO( 2n)toU(n) the spinors decompose as S+→[0]⊕[2]⊕... (19) and S−→[1]⊕[3]⊕... (20) I have not indicated the end of the two sequences: A moment’s reflection indicates that it depends on whether nis even or odd. In our example, n=2,and thus 2+→[0]⊕[2]=1⊕1 and 2−→[1]=2. Similarly, for n=3, upon the restriction of SO( 6)toU(3),4+→ [0]⊕[2]=1⊕3∗and 4−→[1]⊕[3]=3⊕1. (Our choice of which triplet representation ofU(3)to call 3 or 3∗is made to conform to common usage, as we will see presently.) We are now ready to figure out the identity of each of the 16 states such as |++−−+/angbracketright inSO( 10)unification. First of all, (18) tells us that under the subgroup SO( 4)⊗SO( 6)of SO( 10)the spinor 16+decomposes as (since /Pi15 j=1εj=+ 1 implies ε1ε2=ε3ε4ε5) 16+→(2+,4+)⊕(2−,4−) (21) We identify the natural SU( 2)subgroup of SO( 4)as the SU( 2)of the electroweak interac- tion and the natural SU( 3)subgroup of SO( 6)as the color SU( 3)of the strong interaction. Thus, according to the preceding discussion, (2+,4+)are the SU( 2)singlets of the stan- dardU(1)⊗SU( 2)⊗SU( 3)model, while (2−,4−)are the SU( 2)doublets. Here is the lineup (all fields being left handed as usual): SU( 2)doublets: ν=|−+−−−/angbracketright e−=|+−−−−/angbracketright u=|−+++−/angbracketright ,|−++−+/angbracketright , and |−+−++/angbracketright d=|+−++−/angbracketright ,|+−+−+/angbracketright , and |+−−++/angbracketright SU( 2)singlets: νc=|+++++/angbracketright e+=|−−+++/angbracketright uc=|+++−−/angbracketright ,|++−+−/angbracketright , and |++−−+/angbracketright dc=|−−+−−/angbracketright ,|−−−+−/angbracketright , and |−−−−+/angbracketright . I assure you that this is a lot of fun to work out and I urge you to reconstruct this table without looking at it. Here are a few hints if you need help. From our discussion 428 | VII. Grand Unification ofSU( 2)I know that ν=|−+ ε3ε4ε5/angbracketrightande−=|+− ε3ε4ε5/angbracketright, but how do I know that ε3=ε4=ε5=− 1? First, I know that ε3ε4ε5=− 1. I also know that 4−→3⊕1 upon restricting SO( 6)to color SU( 3). Well, of the four states |−−−/angbracketright ,|++−/angbracketright ,|+−+/angbracketright , and|−++/angbracketright the “odd man out” is clearly |−−−/angbracketright . By the same heuristic argument, among the 16 possible states |+++++/angbracketright is the “odd man out” and so must be νc. There are lots of consistency checks. For example, once I identify ν=|−+−−−/angbracketright , e−=|+−−−−/angbracketright , and νc=|+++++/angbracketright , I can figure out the electric charge Q, which, since it transforms as a singlet under color SU( 3), must have the value Q= aε1+bε2+c(ε 3+ε4+ε5)when acting on the state |ε1ε2ε3ε4ε5/angbracketright. The constants a,b, and ccan be determined from the three equations Q(ν)=−a+b−3c=0,Q(e−)=− 1, and Q(νc)=0. Thus, Q=−1 2ε1+1 6(ε3+ε4+ε5). Living in the computer age, I find it intriguing that the fundamental constituents of matter are coded by five bits. You can tell your condensed matter colleagues thattheir beloved electron is composed of the binary strings +−−−− and−−+++ .A n intriguing possibility 3suggests itself, that quarks and leptons may be composed of five different species of fundamental fermionic objects. We construct composites, writing a+if that species is present, and a −if it is absent. For example, from the expression for Qgiven above, we see that species 1 carries electric charge − 1 2, species 2 is neutral, and species 3, 4, and 5 carry charge1 6. A more or less concrete model can even be imagined by binding these fundamental fermionic objects to a magnetic monopole. I emphasize that particles transforming in 16−, such as |+−+++/angbracketright , have not been observed experimentally. A speculation on the origin of families One of the great unsolved puzzles in particle physics is the family problem. Why do quarksand leptons come in three generations {ν e,e,u,d},{νμ,μ,c,s}, and{ντ,τ,t,b}? The way we incorporate this experimental fact into our present day theory can only be describedas pathetic: We repeat the fermionic sector of the Lagrangian three times without anyunderstanding whatsoever. Three generations living together gives rise to a nagging familyproblem. Our binary code view of the world suggests a wildly speculative (perhaps too speculative to mention in a textbook?) approach to the family problem: We add more bits. T o me,a reasonable possibility is to “hyperunify” into an SO( 18)theory, putting all fermions into a single spinorial representation S +=256+, which upon the breaking of SO( 18)to SO( 10)⊗SO( 8)decomposes as 256+→(16+,8+)⊕(16−,8−) (22) We have a lot of 16+’s. Unhappily, we see that group theory [see also (21)] dictates that we also get a bunch of unwanted 16−’s. One suggestion is that Nature might repeat the trick 3For further details, see F. Wilczek and A. Zee, Phys. Rev. D25: 553, 1982, Section IV . VII.7. SO(10) Unification | 429 She uses with color SU( 3), whose strong force confines fields that are not color singlets (chapter VII.3). Interestingly, we can exploit a striking feature of SO( 8), which some people regard as the most beautiful of all groups. In particular, the two spinorial representations8 ±have the same dimension as the vectorial representation 8v(the equation 2n−1=2n has the unique solution n=4). There is a transformation that cyclically rotates these three representations 8+,8−, and 8vinto each other (in the jargon, the group SO( 8)admits an outer automorphism). Thus, there exists a subgroup SO( 5)ofSO( 8)such that when we break SO( 8)into that SO( 5)8+behaves like 8vwhile 8−behaves like a spinor, namely 8+→5⊕1⊕1⊕1 and 8−→4⊕4∗(23) If we call this4SO( 5)hypercolor and assumes that the strong force associated with it con- fines all fields that are not hypercolor singlets, then only three 16+’s remain! Unfortunately, as the relevant physics occurs in the energy regime above grand unification, our knowl-edge of the dynamics of symmetry breaking is far too paltry for us to make any furtherstatements. Charge conjugation The product ⊗notation we use here allows us to construct the conjugation matrix Cexplic- itly. By definition C−1σ∗ ijC=−σij(so that Cchanges eiθijσijinto its complex conjugate.) From (2), (3), and (4) we see that we can construct C(n+1)={C(n)⊗τ1ifnodd C(n)⊗τ2 ifneven(24) You can check that this gives C−1γ∗ jC=(−1)nγjand hence the desired result. Explicitly, Cis a direct product of an alternating sequence of τ1andτ2and so we deduce an important property. Acting on |ε1ε2...εn/angbracketright,Cflips the sign of all the ε’s. Thus C changes the sign of (/Pi1n j=1εj)fornodd, and does not for neven. For nodd, the two spinor representations S+andS−are conjugates of each other, while for neven, they are conjugates of themselves, or in other words, they are real. This can also be seendirectly from C −1γFIVEC=(−1)nγFIVE. You can check this with all the explicit examples we have encountered: SO( 2),SO( 4),SO( 6),SO( 8),SO( 10), and SO( 18). See also exercise VII.7.3. Anomalies What about anomalies in SO( 2n)grand unification? According to the discussion in chap- ter VII.5 we have to evaluate Aijklmn≡tr(Jij{Jkl,Jmn})over the fermion representation. 4The reader savvy with group theory would recognize that SO( 5)is isomorphic with the symplectic group Sp( 4)and that the Dynkin diagram of SO( 8)is the most symmetric of all. 430 | VII. Grand Unification Applying an SO( 2n)transformation Jij→OTJijOwe see easily that Aijklmnis an invari- ant tensor. Can we construct an invariant 6-index tensor with the appropriate symmetryproperties (e.g., A ijklmn=−Ajiklmn)inSO( 2n)? We can’t, except in SO( 6), for which we haveεijklmn. Thus, Aijklmnvanishes except in SO( 6), where it is proportional to εijklmn. An elegant one line proof that any grand unified theory based on SO( 2n)forn/negationslash=3 is free from anomaly! The cancellation of the anomaly between 5∗and 10 at the end of chapter VII.5 doesn’t seem so miraculous any more. Miracles tend to fade away as we gain deeper understanding. Amusingly, by discussing a physics question, namely whether a gauge theory is renor- malizable or not, we have discovered a mathematical fact. What is so special about SO( 6)? See exercise VII.7.5. Exercises VII.7.1 Work out the Clifford algebra in d-dimensional space for dodd. VII.7.2 Work out the Clifford algebra in d-dimensional Minkowski space. VII.7.3 Show that the Clifford algebra for d=4kand for d=4k+2 have somewhat different properties. (If you need help with this and the two preceding exercises, look up F. Wilczek and A. Zee, Phys. Rev. D25: 553, 1982.) VII.7.4 Discuss the Higgs sector of the SO( 10). What do you need to give mass to the quarks and leptons? VII.7.5 The group SO( 6)has 6(6−1)/2=15 generators. Notice that the group SU( 4)also has 42−1=15 generators. Substantiate your suspicion that SO( 6)andSU( 4)are isomorphic. Identify some low dimensional representations. VII.7.6 Show that (unfortunately) the number of families we get in SO( 18)depends on which subgroup of SO( 8) we take to be hypercolor. VII.7.7 If you want to grow up to be a string theorist, you need to be familiar with the Dirac equation in various dimensions but especially in 10. As a warm up, study the Dirac equation in 2-dimensional spacetime.Then proceed to study the Dirac equation in 10-dimensional spacetime. Part VIII Gravity and Beyond This page intentionally left blank VIII.1Gravity as a Field Theory and the Kaluza-Klein Picture Including gravity Field theory texts written a generation ago typically do not even mention gravity. The gravitational interaction, being so much weaker than the other three interactions, wassimply not included in the education of particle physicists. The situation has changed witha vengeance: The main drive of theoretical high energy physics today is the unification ofgravity with the other three interactions, with string theory the main candidate for a unifiedtheory. From a course on general relativity you would have learned about the Einstein-Hilbert action for gravity S=1 16πG/integraldisplay d4x√−gR≡/integraldisplay d4x√−gM2 PR (1) where g=detgμνdenotes the determinant of the curved metric gμνof spacetime, Ris the scalar curvature, and Gis Newton’s constant. Let me remind you that the Riemann curvature tensor Rλ μνκ=∂ν/Gamma1λ μκ−∂κ/Gamma1λ μν+/Gamma1σ μκ/Gamma1λ νσ−/Gamma1σ μν/Gamma1λ κσ(2) is constructed out of the Riemann-Christoffel symbol (recall chapter I.11): /Gamma1λ μν=1 2gλρ(∂νgρμ+∂μgρν−∂ρgμν) (3) The Ricci tensor is defined by Rμκ=Rν μνκand the scalar curvature by R=gμνRμν. Varying Sgives us1the Einstein field equation Rμν−1 2gμνR=− 8πGTμν (4) The Einstein-Hilbert action is uniquely determined if we require the action to be coor- dinate invariant and to involve two powers of spacetime derivative. As you can see from 1See, e.g., S. Weinberg, Gravitation and Cosmology , p. 364. 434 | VIII. Gravity and Beyond (2) and (3) the scalar curvature Rinvolves two powers of derivative and the dimensionless fieldgμνand thus has mass dimension 2. Hence G−1must have mass dimension 2. The second form in (1) emphasizes this point and is often preferred in modern work on grav-ity. (The modified Planck mass M P≡1/√ 16πG differs from the usual Planck mass by a trivial factor, much like the relation between hand/planckover2pi.) The theory sprang from Einstein’s profound intuition regarding the curvature of space- time and is manifestly formulated in terms of geometric concepts. In many textbooks,Einstein’s theory is developed, and rightly so, in purely geometric terms. On the other hand, as I hinted back in chapter I.6, gravity can be treated on the same footing as the other interactions. After all, the graviton may be regarded as just anotherelementary particle like the photon. The action (1), however, does not look anything likethe field theories we have studied thus far. I will now show you that in fact it does have thesame kind of structure. Gravity as a field theory Let us write gμν=ημν+hμν, where ημνdenotes the flat Minkowski metric and hμνthe deviation from the flat metric. Expand the action in powers of hμν. In order not to drown in a sea of Lorentz indices, let us suppress them for a first go-around. Merely from thefact that the scalar curvature Rinvolves two derivatives ∂in its definition, we see that the expansion must have the schematic form S=/integraldisplay d4x1 16πG(∂h∂h +h∂h∂h +h2∂h∂h+...) (5) after dropping total divergences. As I remarked in chapter I.11, the field hμν(x) describes a graviton in flat space and is to be treated like any other field. The first term ∂h∂h , which governs how the graviton propagates, is conceptually no different than the first term in theaction for a scalar field ∂ϕ∂ϕ or for the photon field ∂A∂A . The terms cubic and higher in hdetermine the interaction of the graviton with itself. The Einstein-Hilbert action in the weak field expansion is structurally reminiscent of the Y ang-Mills action, which may be written in schematic form as S=/integraltext d 4x(1/g2)(∂A∂A + A2∂A+A4). As I explained in chapter IV .5, we understand the self interaction of the Y ang- Mills bosons physically: The bosons themselves carry the charge to which they couple. We can understand the self interaction of the graviton similarly: The graviton couples toanything carrying energy and momentum, and it certainly carries energy and momentum.In contrast, the photon does not couple to itself. We say that Y ang-Mills and Einstein theories are nonlinear, while Maxwell theory is linear. The former are hard, the latter easy. But while the Y ang-Mills action terminates, the Einstein-Hilbert action, because of the presence of√ −g and of the inverse of gμν, is an infinite series in the graviton field hμν. The other major difference is that while Y ang-Mills theory is renormalizable, gravity is notoriously nonrenormalizable, as we argued by dimensional analysis in chapter III.2. Weare now in a position to see this explicitly. Consider the self energy correction to the graviton VIII.1. Gravity as a Field Theory | 435 (a) (b) Figure VIII.1.1 propagator shown in figure VIII.1.1a. We see from the second term in (5) that the three- graviton coupling involves two powers of momentum. Thus the Feynman integral goes as/integraltext d4k(kkkk/k2k2), with four powers of kin the numerator from the two vertices and four powers in the denominator from the propagators. T aking out two powers of momentum toextract the coefficient of ∂h∂h , we see that the correction to 1 /G is quadratically divergent. Because of the explicit powers of momentum in the coupling, the divergence gets worseand worse as we go to higher and higher order. Compare figure VIII.1.1b to 1a: We havethree more propagators, worth ∼1/k 6, and one more loop integration/integraltext d4k, but two more vertices ∼k4. The degree of divergence goes up by 2. Of course we already knew all this by dimensional analysis. As mentioned in chapter I.11 the fundamental definition Tμν(x)=−2√−gδSM δgμν(x) tells us that coupling of the graviton to matter (in the weak field limit) can be included by adding the term −/integraldisplay d4x1 2hμνTμν(6) to the action, where Tμνstands for the (flat spacetime) stress-energy tensor of all the matter fields of the world, a matter field being any field that is not the graviton field. Thus, withthe inclusion of matter (5) is modified schematically 2to S=/integraldisplay d4x[1 16πG(∂h∂h +h∂h∂h +h2∂h∂h+...)+(hT+...)] (7) In chapter IV .5 I noted that we can bring Y ang-Mills theory into the same convention commonly used in Maxwell theory by a trivial rescaling A→gA. Similarly, we can also bring Einstein theory into the same convention by rescaling the graviton field hμν→√ Ghμνso that the action becomes (to ease writing we absorb 16 πintoGwhenever we feel like it) S=/integraldisplay d4x (∂h∂h +√ Gh∂h∂h +Gh2∂h∂h+...+√ GhT ) 2If this is to represent an expansion of Sin powers of h, then strictly speaking, if we display the terms cubic and quartic in the Einstein-Hilbert action, we should also display the contribution coming from the terms of higher order in hcontained in Tμν(x)=−(2/√−g)δSM/δgμν(x). 436 | VIII. Gravity and Beyond We see explicitly that√ 16πG=1/MPmeasures the strength of the graviton coupling to itself and to all other fields. Once again, the enormity of MP(compared to the scale of the strong interaction, say) indicates the feebleness of gravity. Here we expanded gμνaround a flat metric but we could just as well expand gμν= ¯gμν+hμν, with ¯gμνa curved metric, that of a black hole (see chapter V .7) for instance. Determining the weak field action After this index free survey we are ready to tackle the indices. We would like to determine the first term ∂h∂h in (7) so that we can obtain the graviton propagator. Thus, we have to expand the action S≡M2 P/integraltext d4x√−ggμνRμνup to and including order h2. From (2) and (3) we see that the Ricci tensor Rμνstarts in O(h) so that it suffices to evaluate√−ggμνto O(h) . That’s easy: As we have already seen in chapter I.11, g=− [1+ημνhμν+O(h2)] and gμν=ημν−hμν+O(h2)so that√−ggμν=ημν−hμν+1 2ημνh+O(h2), where we have defined h≡ημνhμν. We now must calculate RμνtoO(h2), a straightforward but tedious task starting from (2) and (3). In line with the spirit of this book, which is to avoid tedious calculation whenever possible, I will now show you how to get around this. We invoke symmetry considerations! Under a general coordinate transformation xμ→x/primeμ=xμ−εμ(x) the metric changes tog/primeμν=(∂x/primeμ/∂xσ)(∂x/primeν/∂xτ)gστ. Plugging in gμν=ημν−hμν+..., lowering the in- dices (with ημνto this order), and using (∂x/primeμ/∂xσ)=δμ σ−∂σεμ, we find, treating ∂μενas of the same order as hμν: h/prime μν=hμν+∂μεν+∂νεμ (8) Note the structural similarity to the electromagnetic gauge transformation A/prime μ=Aμ− ∂μ/Lambda1. Very nice! We will explore the sense in which gravity can be regarded as a gauge theory in more detail later. We are looking for the terms in the action quadratic in hand quadratic in ∂. Lorentz invariance tells us that there are four possible terms (T o see this, first write down termswith the indices on the two ∂matching, then the terms with the index on a ∂matching an index on an h, and so on): S=/integraldisplay d4x(a∂ λhμν∂λhμν+b∂λhμ μ∂λhνν+c∂λhλν∂μhμν+dhλ λ∂μ∂νhμν) with four unknown constants a,b,c, andd. Now vary Swithδhμν=∂μεν+∂νεμ, inte- grating by parts freely. For example, δ(∂λhμν∂λhμν)=2[∂λ(2∂μεν)](∂λhμν)“=”4εν∂2∂μhμν Since there are three objects linear in h, linear in ε, and cubic in ∂(namely εν∂2∂νhand εν∂ν∂λ∂μhλμin addition to the one already shown) the condition δS=0 gives three equa- tions, just enough to fix the action up to an overall constant, corresponding to Newton’sconstant. The invariant combination turns out to be I≡1 2∂λhμν∂λhμν−1 2∂λhμ μ∂λhνν−∂λhλν∂μhμν+∂νhλ λ∂μhμν (9) VIII.1. Gravity as a Field Theory | 437 Thus, even if we had never heard of the Einstein-Hilbert action we could still determine the action for gravity in the weak field limit by requiring that the action be invariant underthe transformation (8). This is hardly surprising since coordinate invariance determinesthe Einstein-Hilbert action. Still, it is nice to construct gravity “from scratch.” Referring to (6), we can now write the weak field expansion of Sas Swfg=/integraldisplay d4x/parenleftbigg1 32πGI−1 2hμνTμν/parenrightbigg without having to expand RtoO(h2). The coefficient of Iis fixed by the requirement that we reproduce the usual Newtonian gravity (see later). The graviton propagator As we anticipated in (5) the action Swfg indeed has the same quadratic structure of all the field theories we have studied, and so as usual the graviton propagator is just theinverse of a differential operator. But just as in Maxwell and Y ang-Mills theories the relevantdifferential operator in Einstein-Hilbert theory does not have an inverse because of the“gauge invariance” in (8). No problem. We have already developed the Faddeev-Popov method to deal with this difficulty. In fact, for my limited purposes here, to derive the graviton propagator in flatspacetime, I don’t even need the full-blown Faddeev-Popov formalism with ghosts and all. 3 Indeed, recall from chapter III.4 that for the Feynman gauge (ξ =1)we simply add (∂A)2 to the invariant 1 2FμνFμν=∂μAν(∂μAν−∂νAμ)“=”−Aμημν∂2Aν−(∂A)2 thus canceling the last term. Inverting the differential operator −ημν∂2we obtain the photon propagator in the Feynman gauge −iημν/k2. We play the same “trick” for gravity. After staring at I=1 2∂λhμν∂λhμν−1 2∂λhμ μ∂λhνν−∂λhλν∂μhμν+∂νhλ λ∂μhμν for a while, we see that by adding (∂μhμν−1 2∂νhλ λ)2we can knock off the last two terms in Iso that Swfgeffectively becomes Swfg=/integraldisplay d4x1 2/bracketleftbigg1 32πG/parenleftbigg ∂λhμν∂λhμν−1 2∂λh∂λh/parenrightbigg −hμνTμν/bracketrightbigg (10) In other words, the freedom in choosing hμνin (8) allows us to impose the so-called harmonic gauge condition ∂μhμ ν=1 2∂νhλλ (11) (the linearized version of ∂μ(√−ggμν)=0.) 3This is because (8) does not involve the field hμν, just as in the Maxwell case but unlike the Y ang-Mills case. Since we do not intend to calculate loop diagrams in quantum gravity, we do not need the full power of the Faddeev-Popov method. 438 | VIII. Gravity and Beyond Writing (10) in the form S=1 32πG/integraldisplay d4x/bracketleftBig hμνKμν;λσ(−∂2)hλσ+O(h3)/bracketrightBig we see that we have to invert the matrix Kμν;λσ≡1 2(ημληνσ+ημσηνλ−ημνηλσ) regarding μν andλσ as the two indices. Note that we have to maintain the symmetry of hμν. In other words, we are dealing with matrices acting in a linear space spanned by symmetric two-index tensors. Thus, the identity matrix is actually Iμν;λσ≡1 2(ημληνσ+ημσηνλ) You can check that Kμν;λσKλσ ;ρω=Iμν;ρωso that K−1=K. Thus, in the harmonic gauge the graviton propagator in flat spacetime is given by (scaling out Newton’s constant) Dμν,λσ(k)=1 2ημληνσ+ημσηνλ−ημνηλσ k2+iε(12) Newton from Einstein Varying (10) with respect to hμνwe obtain the Euler-Lagrange equation of motion4 1 32πG(−2∂2hμν+ημν∂2h)−Tμν=0. T aking the trace, we find ∂2h=16πGT (withT≡ ημνTμν)and so we obtain5 ∂2hμν=− 16πG(Tμν−1 2ημνT) (13) In the static limit, T00is the dominant component6of the stress-energy tensor and (13) reduces to /vector∇2φ=4πGT00upon recalling from chapter I.5 that the Newtonian gravitational potential φ≡1 2h00. We have just derived Poisson’s equation for φ. Incidentally, this suggests another way of avoiding the tedious task of expanding the Einstein-Hilbert action (and hence R)toO(h2)if you are willing to accept the Einstein field equation (4) as given. You need expand Rμνonly to O(h) to obtain (13) from (4), and from (13) you can reconstruct the action to O(h2). Indeed, from (2) and (3) you easily get Rμν=1 2(−∂2hμν+∂μ∂λhλ ν+∂ν∂λhλ μ−∂μ∂νhλ λ)+O(h2)→−1 2∂2hμν+O(h2) with the further simplification in harmonic gauge. But this is not quite fair since con- siderable technology7(Palatini identity and all the rest) is needed to derive (4) from (1). 4Note that the flat spacetime energy momentum conservation ∂μTμν=0 together with the equation of motion implies ∂2(∂μhμν−1 2∂νh)=0. 5Thus, the Einstein equation in vacuum Rμν=0 reduces to ∂2hμν=0; hence the name “harmonic.” 6Note that, in contrast to T00,h00does not dominate the other components of hμν. 7See S. Weinberg, Gravitation and Cosmology , pp. 290 and 364. VIII.1. Gravity as a Field Theory | 439 Einstein’s theory and the deflection of light Consider two particles with stress-energy tensors Tμν (1)andTμν (2)respectively interacting via the exchange of a graviton. The scattering amplitude is then (up to some overall constantnot essential for our purposes here) given by GTμν (1)Dμν,λσ(k)Tλσ (2)=G 2k2(2Tμν (1)T(2)μν−T(1)T(2)) For nonrelativistic matter T00is much larger than the other components T0jandTij(as I have just remarked), so the scattering amplitude between two lumps of nonrelativisticmatter (say, the earth and you) is proportional to G 2k2(2T00 (1)T00 (2)−T00 (1)T00 (2))=G 2k2T00 (1)T00 (2) As explained way back in chapters I.4 and I.5, the interaction potential is given by the Fourier transform of the scattering amplitude, namely G/integraldisplay/integraldisplay d3xd3x/primeT(1)00(x)T(2)00(x/prime)/integraldisplay d3kei/vectork.(/vectorx−/vectorx/prime)1 /vectork2 and thus for two well-separated objects we recover the Newtonian potential GM(1)M(2)/r. We are now able to address the issue raised at the end of chapter I.5. Suppose a particle theorist, Dr. Gravity, wants to propose a theory of gravity to rival Einstein’s theory. Dr. Gclaims that gravity is due to the exchange of a spin 2 particle with a teeny mass m Gcoupled to the stress-energy tensor Tμν. In chapter I.5 we worked out the propagator of a massive spin 2 particle, namely Dspin 2 μν,λσ(k)=1 2(GμλGνσ+GμσGνλ−2 3GμνGλσ)/(k2−m2 G+iε) withGμν=ημν−kμkν/m2 G(after a trivial notational adjustment). Since the particle is coupled to a conserved source kμTμν=0 we can replace Gμνbyημν. Thus, in the limit mG→0 we have the propagator Dspin 2 μν,λσ(k)=1 2ημληνσ+ημσηνλ−2 3ημνηλσ k2+iε(14) Compare this with (12). Dr. G’s propagator differs from Einstein’s:2 3versus 1. Remarkably, gravity is not generated by an almost massless spin 2 particle. The “2 3discontinuity” between (12) and (14) was discovered in 1970 independently by Iwasaki, by van Dam and Veltman, and by Zakharov. In Dr. G’s theory (with his own gravitational coupling GG), the interaction between two particles is given by GGTμν (1)Dμν,λσ(k)Tλσ (2)=GG 2k2(2Tμν (1)T(2)μν−2 3T(1)T(2)) For two lumps of nonrelativistic matter this becomes GG 2k2(2T00 (1)T00 (2)−2 3T00 (1)T00 (2))=4 3GG 2k2T00 (1)T00 (2) Dr. G simply takes his GG=3 4Gand his theory passes all experimental tests. 440 | VIII. Gravity and Beyond But wait! There is also the famous 1919 observation of the deflection of starlight by the sun, and the photon is definitely not a lump of nonrelativistic matter. Indeed, recallfrom chapter I.11 (or from your course on electromagnetism) that T≡T μ μvanishes for the photon. Thus, taking Tμν (1)andTμν (2)to be the stress-energy tensor of the sun and of the photon respectively, Einstein would have for the scattering amplitude (G/2k2)2Tμν (1)T(2)μν while Dr. G would have (GG/2k2)2Tμν (1)T(2)μν=3 4(G/2k2)2Tμν (1)T(2)μν. Dr. G would have predicted a deflection angle of 3 GM/R instead of 4 GM/R (with MandRthe mass and radius of the sun). On the Brazilian island of Sobral in 1919 Einstein triumphedover Dr. G. As explained in chapter I.5, while a massive spin 2 particle has 5 degrees of freedom the massless graviton has only 2. (I give an analysis of the helicity ±2 structure of one graviton exchange in appendix 2.) The 5 degrees of freedom may be thought of as consisting of thehelicity ±2 degrees of freedom we want plus 2 helicity ±1 and a helicity 0 degrees of freedom. The coupling of the helicity ±1 degrees of freedom vanishes because k μTμν=0. Thus, effectively, we are left with an extra scalar coupling to the trace T≡ημνTμνof the stress-energy tensor; as we can see plainly the discrepancy indeed resides in the last termof (12) and (14). You should be disturbed that a measurement of the deflection of starlight can show that a physical quantity, the graviton mass m G, is mathematically zero rather than less than some extremely small value. This apparent paradox was resolved by A. Vainshtein in1972. 8He found that Dr. G’s theory contains a distance scale rV=/parenleftBigg GM m4 G/parenrightBigg1 5 in the gravitational field around a body of mass M. The helicity 0 degree of freedom becomes effective only on the distance scales r/greatermuchrV. Inside the Vainshtein radius rV, the gravitational field is the same as in Einstein’s theory and experiments cannot distinguishbetween Einstein’s and Dr. G’s theories. With the current astrophysical bound m G/lessmuch (1024cm)−1andMthe mass of the sun, rVcomes out to be much larger than the size of the solar system. In other words, the apparent paradox arose because of an interchangeof limits: We can take either the characteristic distance of the measurement r obs(the radius of the sun in the deflection of starlight) or the Vainshtein radius rVto infinity first. So all is well: Dr. G’s theory is consistent with current measurements provided that he takes mGsmall enough. What he is not allowed to do is use the one graviton exchange approximation. Instead, he should solve the massive analog of Einstein’s field equation (4)around a massive body such as the sun, as Vainshtein did. This is equivalent to expandingto all orders in the graviton field hand resumming: In Feynman diagram language we 8A. I. Vainshtein, Phys. Lett. 39B:393, 1972; see also C. Deffayet, G. Dvali, G. Gabadadze, and A. I. Vainshtein, Phys. Rev. D65:044026, 2002. VIII.1. Gravity as a Field Theory | 441 q k1 k2p1 p2 Figure VIII.1.2 have an infinite number of diagrams corresponding to the sun emitting 1, 2, 3, ... ,∞ gravitons respectively. The paradox is formally resolved by noting that the higher ordersare increasingly singular as m G→0. The gravity of light At this point, you are ready to do perturbative quantum gravity: You have the graviton propagator (12), and you can read off the interaction between gravitons from the detailedversion of (7) and the interaction between the graviton and any other field from the term− 1 2hμνTμν. The only trouble is that you might “drown in a sea of indices” if you don’t watch out, as I have already warned you. I know of one calculation (in fact one of my favorites in theoretical physics) in which we can beat the indices down easily. An interesting question: Einstein said that lightis deflected by a massive object, but is light deflected gravitationally by light? T olman,Ehrenfest, and Podolsky discovered that in the weak field limit two light beams movingin the same direction do not interact gravitationally, but two light beams moving in theopposite directions do. Surprising, eh? The scattering of two photons k 1+k2→p1+p2via the exchange of a graviton is given by the Feynman diagram in figure VIII.1.2, with the momentum transfer q≡p1−k1, plus another diagram with p1andp2interchanged. The Feynman rule for coupling a graviton to two photons can be read off from hμνTμν=−hμν(FμλFνλ−1 4ημνFρλFρλ) but all we need is that the interaction involve two powers of spacetime derivatives ∂acting on the electromagnetic potential Aμso that the graviton-photon-photon vertex involves 2 powers of momenta, one from each photon. Hence the scattering amplitude (with all Lorentz indices suppressed) has the schematic form ∼(k1p1)D(k 2p2). The η’s in the graviton propagator Dtie the indices on (k1p1)and(k2p2)together. (We have suppressed the polarization vectors of the photons, imagining that they are to be averaged over in the amplitude squared.) Referring to (12), we see that the amplitude is the sum of three terms such as ∼(k1.p1)(k 2.p2)/q2,∼(k1.k2)(p 1.p2)/q2, and ∼(k1.p2)(k 2.p1)/q2. Since according to Fourier the long distance part of the interaction potential is given by 442 | VIII. Gravity and Beyond the small qbehavior of the scattering amplitude, we need only evaluate these terms in the limitq→0. We can throw almost everything away! For example, k1.p1→k1.k1=0,k1.p2=k1.(k1+k2−p1)→k1.k2 Just imagining contracting all those indices in our heads is good enough: We obtain the amplitude ∼(k1.k2)(p1.p2)/q2. Ifk1andk2point in the same direction, k1.k2∝k1.k1=0. Two photons moving in the same direction do not interact gravitationally. Of course, this result is not of any practical importance since electromagnetic effects are far more important, but this is not an engineering text. In appendix 1 I give an alternativederivation of this amusing result. Kaluza-Klein compactification You have probably read about how excited Einstein was when he heard of the proposal ofKaluza and of Klein to extend the dimension of spacetime to 5 and thus unify electromag-netism and gravity. The 5th dimension is supposed to be compactified into a tiny circle ofradius afar smaller than what experimentalists can see; in other words, x 5is an angular variable with x5=x5+2πa. You have surely heard that string theory, at least in some ver- sion, is based on the Kaluza-Klein idea. Strings live in 10-dimensional spacetime, with 6of the dimensions compactified. I can now show you how the Kaluza-Klein mechanism works. Start with the action S=1 16πG 5/integraldisplay d5x/radicalbig −g 5R5 (15) in 5-dimensional spacetime. The subscript 5 serves to indicate the 5-dimensional quanti- ties. We denote the 5-dimensional metric by gABwith the indices AandBrunning over 0, 1, 2, 3, 5. Assume that gABdoes not depend on x5. Plug into S, integrate over x5, and compute the effective 4-dimensional action. Since R5and the 4-dimensional scalar curvature R both involve two powers of ∂andgAB contains gμν, we must have (exercise VIII.1.5) R5=R+.... Thus, (15) contains the Einstein-Hilbert action with Newton’s gravitational constant G∼G5/a. What else do we get? We don’t even have to work through the arithmetic. We can argue by symmetry. Under the 5-dimensional coordinate transformation xA→x/primeA= xA+εA(x), we have [see (8)] h/prime AB=hAB−∂AεB−∂BεA. Let us choose εμ=0 andε5(x) to be independent of x5: We go around and rotate each of the tiny circles attached to every point in our spacetime a tiny bit. Well, we have h/prime μν=hμνandh/prime 55=h55, buth/prime μ5=hμ5−∂με5. But if we give the Lorentz 4-vector hμ5and 4-scalar ε5new names, call them Aμand/Lambda1, this just says A/prime μ=Aμ−∂μ/Lambda1, the usual electromagnetic gauge transformation! Since we know that the 5-dimensional action (15) is invariant under xA→x/primeA=xA+ εA(x), the resulting 4-dimensional action must be invariant under Aμ→A/prime μ=Aμ−∂μ/Lambda1 VIII.1. Gravity as a Field Theory | 443 and hence must contain the Maxwell action. Note once again the power of symmetry considerations. No need to do tedious calculations. Electromagnetism comes out of gravity! Differential geometry of Riemannian manifolds I hinted earlier at a deep connection between general coordinate transformation and gaugetransformation. Let us flesh this out by looking at differential geometry and gravity. Forthis sketch we will consider locally Euclidean (rather than Minkowskian) spaces. The differential geometry of Riemannian manifolds can be elegantly summarized in the language of differential forms. Consider a Riemannian manifold (such as a sphere)with the metric g μν(x). Locally, the manifold is Euclidean by definition, which means gμν(x)=ea μ(x)δabeb ν(x) (16) where the matrix e(x) may be thought of as a similarity transformation that diagonalizes gμνand scales it to the unit matrix. Thus, for a D-dimensional manifold there exist D “world vectors” ea μ(x) obviously dependent on xand labeled by the index a=1, 2, ... ,D. The functions ea μ(x) are known as “vielbeins” (meaning “many legs” in German, vierbeins =four legs for D=4, dreibeins =three legs for D=3, and so on.) In some sense, the vielbeins can be thought as the “square root” of the metric. Let us clarify by a simple example. The familiar two-sphere (of unit radius) has the line element9ds2=dθ2+sin2θdϕ2. From the metric (g θθ=1,gϕϕ=sin2θ)we can read off e1 θ=1 and e2 ϕ=sinθ(all other components are zero). We are invited to define D1-forms ea=ea μdxμ. (In our example, e1=dθ,e2=sinθdϕ .) On a curved manifold, when we parallel transport a vector, the vector changes when expressed in terms of the locally Euclidean coordinate frame. (This is just the familiarstatement that on a curved manifold such as the surface of the earth the notion of a vectorpointing straight north is a local concept: When we move infinitesimally away keeping our“north vector” pointing in the same direction, it will end up being infinitesimally rotatedaway from the “north vector” defined at the point we have just moved to.) This infinitesimalrotation of the vielbeins is described by dea=−ωabeb(17) Note that since ωgenerates an infinitesimal rotation it is an antisymmetric matrix: ωab= −ωba. Since deais a 2-form, ωis a 1-form, known as the connection: It “connects” the locally Euclidean frames at nearby points. (Since the indices a,b, etc. are associated with the Euclidean metric δabwe do not have to distinguish between upper and lower indices. When we do write upper or lower indices a,b, etc. it is for typographical convenience.) In 9Note that this represents the square of an infinitesimal distance element and not an area element, and so a quantity such as dθ2is literally the square of dθand not the wedge product dθdθ (of chapter IV .4), which would have been identically zero. 444 | VIII. Gravity and Beyond the simple example of the sphere, de1=0 and de2=cosθd θd ϕ and so the connection has only one nonvanishing component ω12=−ω21=−cos θd ϕ . At any point, we are free to rotate the vielbeins: If you use the vielbeins ea μI am free to use some other vielbeins e/primea μinstead, as long as mine are related to yours by a rotation ea μ(x)= Oa b(x)e/primeb μ(x). [You can check that gμν(x)=ea μ(x)δabeb ν(x)=e/primea μ(x)δabe/primeb ν(x) ifOTO=1.] The connection ω/primeis defined by de/primea=−ω/primeabe/primeb. You can readily work out that (suppressing indices) ω=Oω/primeOT−(dO)OT(18) The local curvature of the manifold is a measure of how the connection varies from point to point. We would like the curvature to be invariant under the local rotation O(or at least to transform as a tensor so that by contracting it with vectors we can form a scalar).The desired object is the 2-form R ab=dωab+ωacωcb. You can check that R=OR/primeOT. (For the sphere, R12=dω12+ω1cωc2=sinθd θd ϕ .)Written out in components, Rab= Rab μνdxμdxν. I leave it to you to verify that Rab μνeλ aeσ bis the usual Riemann curvature tensor Rλσ μν, where eλ ais the inverse of the matrix ea λ. In particular, Rab μνeμ aeν bis the scalar curvature, which in our convention works out to be +1 for the sphere. Thus, Riemannian geometry can be elegantly summarized by the two statements (again suppressing indices) de+ωe=0 (19) and R=dω+ω2(20) Look familiar? You should be struck by the similarity between (20) and the expression for the field strength in nonabelian gauge theories F=dA+A2. Note ωtransforms [see (18)] exactly the same way as the gauge potential A. But one nagging difference, namely the lack of an analog of ein gauge theory, has long bothered some theoretical physicists (but is shrugged off by most as inconsequential). Also, note that Einstein theory is linearinRwhile Y ang-Mills theory is quadratic in F. Gravity and Y ang-Mills We can make the connection between gravity and Y ang-Mills theory more explicit by looking at the derivative of a vector field. Y ang-Mills theory was born of the requirement that a field ϕand its derivative ∂μϕtransform in the same way under a spacetime- dependent internal symmetry transformation (IV .5.1). In Einstein gravity a vector field Wμ(x) transforms as W/primeμ(x/prime)=Sμ ν(x)Wν(x) withSμ ν(x)=∂x/primeμ/∂xν. Since the matrix S depends on the spacetime coordinate x, we see that ∂λWμcould not possibly transform like a tensor with one upper and one lower index, as we would like naively just by lookingat indices. We would have to introduce a covariant derivative. Not surprisingly, this closely VIII.1. Gravity as a Field Theory | 445 parallels the discussion in chapter IV .5. Historically, Y ang and Mills were inspired by Einstein gravity. Using the chain rule and the product rule, we have ∂/prime λW/primeμ(x/prime)=∂W/primeμ(x/prime) ∂x/primeλ=∂xρ ∂x/primeλ∂ ∂xρ[Sμ ν(x)Wν(x)]=(S−1)ρ λSμ ν∂ρWν+[(S−1)ρ λ∂ρSμ ν]Wν(21) Were the second term in (21), which comes from differentiating S, not there, the naive guess, that ∂λWμtransforms like a tensor, would be valid. The fact that the transformation Svaries from place to place has negated the naive guess. What is happening is quite clear: as the vector Wvaries from a given point to a neighbor- ing point, the coordinate axes that define the components of Walso change. This suggests that we could define a more suitable derivative, called the covariant derivative and writtenasD λWμ, to take this effect into account, so that DλWμwould indeed transform like a tensor. Exactly as in Y ang-Mills theory (IV .5.1), we have to add an extra term to knock outthe second term in (21). Just the way the indices hang together immediately suggests the correct construction. The factor (S −1)ρ λ∂ρSμ νin the unwanted second term in (21) has one upper index and two lower indices, so we need an object with this set of indices. Lo, the Riemann-Christoffelsymbol /Gamma1 μ λνin (3) (and introduced in chapter I.11) fits the bill perfectly. I will let you have the fun of verifying that the covariant derivative defined by DλWμ≡∂λWμ+/Gamma1μ λνWν(22) indeed transforms like a tensor (note that /Gamma1was normalized correctly for this purpose). I end with a technical remark about the coupling of gravity to spin1 2fields. First, we of course have to Wick rotate so that the vierbein ea μerects a locally Minkowskian rather than a Euclidean coordinate frame. The indices a,b, etc. are now contracted with the Minkowskian metric ηab. The slight subtlety is that the Dirac gamma matrices γa are associated with the Lorentz rotation of the vierbein ea μ(x)=Oa b(x)e/primeb μ(x/prime)and thus carry the Lorentz index arather than the “world” index μ. Similarly, the Dirac spinor ψ(x) is defined relative to the local Lorentz frame specified by the vierbein, and thus its covariant derivative has to be defined in terms of the connection ωrather than the symbol /Gamma1. Hence the flat space Dirac action/integraltext d4x¯ψ(iγμ∂μ−m)ψ must be general- ized to/integraltext d4x√−g¯ψ(iγaηabebμDμ−m)ψ , where the covariant derivative Dμψ=∂μψ− i 4ωμabσabψexpresses the rotation of the local Lorentz frame as we move from a point xto a neighboring point. In contrast to the action for integer spin fields in curved spacetime (see chapter I.11), the Dirac action in curved spacetime involves the vierbein explicitly. Appendix 1: Light on light again The stress-energy tensor Tμνof a light beam moving in the x-direction has four nonzero components: the energy density T00of course, then T0x=T00since photons carry the same energy and momentum, next Tx0=T0xby symmetry, and finally Txx=T00since the stress-energy tensor of the electromagnetic field is traceless (chapter I.11). Without having to solve Einstein’s equations in the weak field limit (13) explicitly weknow immediately that h 00=h0x=hx0=hxx≡h. The metric around the light beam is given by g00=1+h, 446 | VIII. Gravity and Beyond g0x=gx0=−h, and gxx=− 1+h(and of course gyy=gzz=− 1 plus a bunch of vanishing components). Consider a photon moving parallel to the light beam. Its worldline is determined by (recall chapter I.11) d2xρ dζ2=−/Gamma1ρ μνdxμ dζdxν dζ Let’s calculate d2y/dζ2andd2z/dζ2with(dy/dζ) ,(dz/dζ) /lessmuch(dt/dζ) ,(dx/dζ). Using (3) we find (with μ,ν restricted to 0, x) d2y dζ2=1 2(∂νgyμ+∂μgyν−∂ygμν)dxμ dζdxν dζ =−1 2(∂yh)/bracketleftbigg (dt dζ)2+(dx dζ)2−2dt dζdx dζ/bracketrightbigg =−1 2(∂yh)(dt dζ−dx dζ)2 For a photon moving in the same direction as the light beam dt=dxandd2y/dζ2=d2z/dζ2=0. We have once again derived the T olman-Ehrenfest-Podolsky effect. Note we never had to solve for h. Incidentally, if you are a bit unsure of dt=dx, the condition ds=0 for a light beam moving in the x-direction amounts to (1+h)dt2−2hdtdx −(1−h)dx2=0. Upon division by dt2we obtain −(1+h)+2hv+(1−h)v2= 0, with v≡dx/dt . The quadratic equation has two roots v=∓(1±h)/(1−h). The negative root gives v=1, and thus for a photon moving in the same direction as the light beam dx/dt =1. In contrast, the positive root v=−(1+h)/(1−h)describes a photon moving in the opposite direction. Appendix 2: The helicity structure of gravity T o gain a deeper understanding of the difference between Einstein’s and Dr. G’s theories let us look at the helicity structure of the interaction in the two cases. T o warm up, consider the interaction between two conserved currents due to the exchange of a spin 1 particle of momentum kand mass m:Jμ (1)J(2)μ=J0 (1)J0 (2)−Ji (1)Ji (2). Use current conservation kμJμ=0 to eliminate J0=kiJi/ω(withω≡k0). We obtain (kikj/ω2−δij)Ji (1)Jj (2). Let/vectork point in the 3rd direction and use /vectork2=ω2−m2to write this as −[(m2/ω2)J3 (1)J3 (2)+J1 (1)J1 (2)+J2 (1)J2 (2)]. We see that as m→0 the longitudinal component of the current J3indeed decouples as explained in chapter II.7 and we obtain −1 2(J1+i2 (1)J1−i2 (2)+J1−i2 (1)J1+i2 (2)), showing explicitly that the photon has helicity ±1. (Obvious notation: J1+i2≡J1+iJ2etc.) Onward to gravity. Consider the interaction Tμν (1)T(2)μν−ξT(1)T(2), where ξ=1 2for Einstein and1 3for Dr. G. For ease of writing I will now omit the subscripts (1) and (2). Conservation kμTμν=0 allows us to eliminate T0i=kjTji/ωandT00=kjklTjl/ω2. Again taking /vectorkto point in the 3rd direction we obtain the mess /parenleftbiggm ω/parenrightbigg4 T33T33+2/parenleftbiggm ω/parenrightbigg2 (T13T13+T23T23)+T11T11+T22T22+2T12T12 −ξ/bracketleftBigg/parenleftbiggm ω/parenrightbigg2 T33+T11+T22/bracketrightBigg/bracketleftBigg/parenleftbiggm ω/parenrightbigg2 T33+T11+T22/bracketrightBigg which simplifies in the limit m→0t o T11T11+T22T22+2T12T12−ξ(T11+T22)(T11+T22) In Einstein’s theory, ξ=1 2and this becomes 1 2(T11−T22)(T11−T22)+2T12T12 which lo and behold is equal to1 2(T1+i2,1+i2T1−i2,1−i2+T1−i2,1−i2T1+i2,1+i2), showing that indeed the graviton carries helicity ±2. In Dr. G’s theory, this would not be the case. VIII.1. Gravity as a Field Theory | 447 Exercises VIII.1.1 Work out Tμνfor a scalar field. Draw the Feynman diagram for the contribution of one-graviton exchange to the scattering of two scalar mesons. Calculate the amplitude and extract the interaction energy betweentwo mesons sitting at rest, thus deriving Newton’s law of gravity. VIII.1.2 Work out T μνfor the Y ang-Mills field. VIII.1.3 Show that if hμνdoes not satisfy the harmonic gauge, we can always make a gauge transformation with ενdetermined by ∂2εν=∂μhμ ν−1 2∂νhλλso that it does. All of this should be conceptually familiar from your study of electromagnetism. VIII.1.4 Count the number of degrees of polarization of a graviton. [Hint: Consider a plane wave hμν(x)= hμν(k)eikxjust because it is a bit easier to work in momentum space. A symmetric tensor has 10 components and the harmonic gauge kμhμ ν=1 2kνhλλimposes 4 conditions. Oops, we are left with 6 degrees of freedom. What is going on?] [Hint: You can make a further gauge transformation and stillstay in the harmonic gauge. The graviton should have only 2 degrees of polarization.] VIII.1.5 The Kaluza-Klein result that we argued by symmetry considerations can of course be derived explicitly. Let me sketch the calculation for you. Consider the metric ds2=gμνdxμdxν−a2[dθ+Aμ(x)dxμ]2 where θdenotes an angular variable 0 ≤θ< 2π. With Aμ=0, this is just the metric of a curved spacetime, which has a circle of radius aattached at every point. The transformation θ→θ+/Lambda1(x) leaves dsinvariant provided that we also transform Aμ(x)→Aμ(x)−∂μ/Lambda1(x) . Calculate the 5-dimensional scalar curvature R5and show that R5=R4−1 4a2FμνFμν. Except for the precise coefficient1 4this result follows entirely from symmetry considerations and from the fact that R5involves two derivatives on the 5-dimensional metric, as explained in the text. After some suitable rescaling this is the usual action for gravity pluselectromagnetism. Note that the 5-dimensional metric has the explicit form g5 AB=/parenleftBigggμν−a2AμAν−a2Aμ −a2Aν −a2/parenrightBigg (23) VIII.1.6 Generalize the Kaluza-Klein construction by replacing the circles by higher dimensional spheres. Show that Y ang-Mills fields emerge. VIII.1.7 Starting with the connection 1-form ω12=−cos θdϕ for the sphere, show that the scalar curvature is a constant independent of θandϕ. VIII.1.8 The vielbeins for a spacetime with Minkowski metric is defined by gμν(x)=ea μ(x)ηabeb ν(x), where the Minkowski metric ηabreplaces the Euclidean metric δab. The indices aandbare to be contracted with ηab. For example, Rab=dωab+ωacηcdωdb. Show that everything goes through as expected. VIII.2The Cosmological Constant Problem and the Cosmic Coincidence Problems The force that knows too much The word paradox has been debased by loose usage in the physics literature. A real paradox should involve a major and clear-cut discrepancy between theoretical expectation andexperimental measurement. The ultraviolet catastrophe, for example, is a paradox, theresolution of which around the dawn of the twentieth century ushered in quantum physics.I now come to the most egregious paradox of present day physics. The electromagnetic force knows about the particles carrying charge, and the strong force knows about the particles carrying color. And the gravitational force? It knowseverybody! More precisely, anybody carrying energy and momentum. Within a particle physics frame of mind, which is the only frame of mind we have in exploring the fundamental structure of physics, the graviton can be regarded as justanother particle. Indeed, given that a massless spin 2 particle couples to the stress-energytensor, one can reconstruct Einstein’s theory. Nevertheless, there is an uncomfortable feel to this whole picture. Gravity has to do with the curvature of spacetime, the arena in which all fields and particles live. The graviton isnot just another particle. This in essence is the root origin 1of the paradox of the cosmological constant. The graviton is not just another particle—it knows too much! The cosmological constant In the absence of gravity, the addition of a constant /Lambda1to the Lagrangian L→L−/Lambda1has no effect whatsoever. In classical physics the Euler-Lagrange equations of motion depend 1For more along this line, see A. Zee, hep-th/0805.2183 in Proceedings of the Conference in Honor of C. N. Y ang’s 85th Birthday, World Scientific, Singapore 2008, p. 131. VIII.2. Cosmic Coincidence Problem | 449 only on the variation of the Lagrangian. In quantum field theory we have to evaluate the functional integral Z=/integraltext Dϕei/integraltext d4xL(x), which upon the inclusion of /Lambda1merely acquires a multiplicative factor. As we have seen repeatedly, a multiplicative factor in Zdoes not enter into the calculation of Green’s function and scattering amplitudes. Gravity, however, knows about /Lambda1. Physically, the inclusion of /Lambda1corresponds to a shift in the Hamiltonian H→H+/integraltext d3x/Lambda1. Thus, the “cosmological constant” /Lambda1describes a constant energy or mass per unit volume permeating the universe, and of course gravityknows about it. More technically, the term in the action −/integraltext d 4x/Lambda1 is not invariant under a coordinate transformation x→x/prime(x). In the presence of gravity, general coordinate invariance re- quires that the term −/integraltext d4x/Lambda1 in the action Sbe modified to −/integraltext d4x√g/Lambda1, as I explained way back in chapter I.11. Thus, the gravitational field gμνknows about /Lambda1, the infamous cosmological constant introduced by Einstein and lamented by him as his biggest mistake.This often quoted lament is itself a mistake. The introduction of the cosmological constantis not a mistake: It should be there. Symmetry breaking generates vacuum energy In our discussion on spontaneous symmetry breaking, we repeatedly ignored an additivetermμ 4/4λthat appears in L. Particle physics is built on a series of spontaneous symmetry breaking. As the universe cools, grand unified symmetry is spontaneously broken, followed by electroweak symme-try breaking, then chiral symmetry breaking, just to mention a few that we have discussed.At every stage a term like μ 4/4λappears in the Lagrangian, and gravity duly takes note. How large do we expect the cosmological constant /Lambda1to be? As we will see, for our purposes the roughest order of magnitude estimate suffices. Let us take λto be of order 1. As for μ, for the three kinds of symmetry breaking I just mentioned, μis of order 1017, 102, and 1 Gev, respectively. We thus expect the cosmological constant /Lambda1to be roughly μ4=μ/(μ−1)3, where the last form of writing μ4reminds us that /Lambda1is a mass or energy density: An energy of order μpacked into a cube of size μ−1. But this is outrageous even if we take the smallest value for μ: We know that the universe is not permeated with a mass density of the order of 1 Gev in every cube of size 1 (Gev)−1. We don’t have to put in actual numbers to see that there is a humongous discrepancy be- tween theoretical expectation and observational reality. If you want numbers, the currentobservational bound on the cosmological constant is <∼(10 −3ev)4. With the grand unifi- cation energy scale, we are off by (17+9+3)×4=116 orders of magnitude. This is the mother of all discrepancies! With the Planck mass MPl∼1019Gev the natural scale of gravity, we would expect /Lambda1∼M4 Plif it is of gravitational origin. We are then off by 124 orders of magnitude. We are not talking about the crummy calculation of some pitiful theorist not fitting someexperimental curve by a factor of 2. 450 | VIII. Gravity and Beyond We can imagine the universe starting out with a negative cosmological constant, fined tuned to cancel the cosmological constant generated by the various episodes of sponta-neous symmetry breaking. Or there must be a dynamical mechanism that adjusts thecosmological constant to zero. Notice I say zero, because the cosmological constant problem is basically an enormous mismatch between the units natural to particle physics and natural to cosmology. Measuredin units of Gev 4the cosmological constant is so incredibly tiny that particle physicists have traditionally assumed that it must be zero and have looked in vain for a plausiblemechanism to drive it to zero. One of the disappointments of string theory is its inabilityto resolve the cosmological constant problem. As of the writing of this chapter around theturn of the millennium, the brane world scenarios (chapter I.6) have generated a great dealof excitement by offering a glimmer of a hope. Roughly, the idea is that the gravitationaldynamics of the larger space that our universe is embedded in may cancel the effect of thecosmological constant. Cosmic coincidence But Nature has a big surprise for us. While theorists racked their brains trying to come upwith a convincing argument that /Lambda1=0, observational cosmologists steadily refined their measurements and discovered dark energy. The “cleanest” explanation of dark energy byfar is that it represents the cosmological constant. Assuming that this is the case (andwho knows?), the upper bound on the cosmological constant would be changed to anapproximate equality /Lambda1∼(10−3ev)4!!! (1) The cosmological constant paradox deepens. Theoretically, it is easier to explain why some quantity is mathematically 0 than why it happens to be ∼10−124in the units natural (?) to the problem. T o make things worse, (10−3ev)4happens to be the same order of magnitude as the present matter density of the universe ρM. More precisely, dark energy accounts for ∼74% of the mass content of the universe, dark matter for ∼22%, and ordinary matter for ∼4%. First, the ordinary matter we know and love is reduced to an almost negligibly smallcomponent of the universe. Second, why should ρ Mbe comparable to /Lambda1to within a factor of 3? This is sometimes referred to as the cosmic coincidence problem. Now the cosmological constant /Lambda1is, within our present understanding, a parameter in the Lagrangian. On the other hand, since most of the mass density of the universe residesin rest mass, as the universe expands ρ M(t)decreases as [1 /R(t)]3, where R(t) denotes the scale size of the universe.2In the far past, ρMwas much larger than /Lambda1, and in the 2For an easy introduction to cosmology, see A. Zee, Unity of Forces in the Universe , vol. II, chap. 10. VIII.2. Cosmic Coincidence Problem | 451 far future, it will be much smaller. It just so happens that, in this particular epoch of the universe, when you and I are around, ρM∼/Lambda1. Or to be less anthropocentric, the epoch when ρM∼/Lambda1happens to be when galaxy formation has been largely completed. Very bizarre!In their desperation, some theorists have even been driven to invoke anthropic selection. 3 3For a recent review, see A. Vilenkin, hep-th/0106083. VIII.3Effective Field Theory Approach to Understanding Nature Low energy manifestation The pioneers of quantum field theory, Dirac for example, tended to regard field theory as a fundamental description of Nature, complete in itself. As I have mentioned severaltimes, in the 1950s, after the success of quantum electrodynamics many leading particlephysicists rejected quantum field theory as incapable of dealing with the strong and weakinteractions, not to mention gravity. Then came the great triumph of field theory in the early1970s. But after particle physicists retrieved field theory from the dust bin of theoreticalphysics, they realized that the field theories they were studying might be “merely” the lowenergy manifestation of a deeper structure, a structure first identified as a grand unifiedtheory and later as a string theory. Thus was developed an outlook known as the effectivefield theory approach, pace Dirac. The general idea is that we can use field theory to say something about physics at low energies or equivalently long distances even if we don’t know anything about the ultimatetheory, be it a theory built on strings or some as yet undreamed of structure. An importantconsequence of this paradigm shift was that nonrenormalizable field theories becameacceptable. I will illuminate these remarks with specific examples. The emergence of this effective field theory philosophy, championed especially by Wil- son, marks another example of cross fertilization between condensed matter and particlephysics. T oward the late 1960s, Wilson and others developed a powerful effective field the-ory approach to understanding critical phenomena, culminating in his Nobel Prize. Thesituation in condensed matter physics is in many ways the opposite of that in particle phys-ics at least as particle physics was understood in the 1960s. Condensed matter physicistsknow the short distance physics, namely the quantum mechanics of electrons and ions.But it certainly doesn’t help in most cases to write down the Schr ¨odinger equation for the electrons and ions. Rather, what one would like to have is an effective description of howa system would respond when probed at low frequency and small wave vector. A strikingexample is the effective theory of the quantum Hall fluid as described in chapter VI.2: The VIII.3. Effective Field Theory | 453 relevant degree of freedom is a gauge field, certainly a far cry from the underlying elec- tron. As in the σmodel description (chapter VI.4) of quantum chromodynamics, it is fair to say that without experimental guidance theorists would have a terribly hard time de-ciding what the relevant low energy long distance degrees of freedom might be. You haveseen numerous other examples in condensed matter physics, from the Landau-Ginzburgtheory of superconductivity to Peierls instability. The threshold of ignorance In our discussion of renormalization, I espouse the philosophy that a quantum fieldtheory provides an effective description of physics up to a certain energy scale /Lambda1,a threshold of ignorance beyond which physics not included in the theory comes intoplay. In a nonrenormalizable theory, various physical quantities that we might wish tocalculate will come out dependent on /Lambda1, thus indicating that the physics at or beyond the scale /Lambda1is essential for understanding the low energy physics we are interested in. Nonrenormalizable theories suffer from not being totally predictive, but nevertheless theymay be useful. After all, the Fermi theory of the weak interaction described experimentsand even foretold its own demise. In a renormalizable theory, various physical quantities come out independent of /Lambda1, provided that the calculated results are expressed in terms of physical coupling constantsand masses, rather than in terms of some not particularly meaningful bare couplingconstants and masses. Low energy physics is not sensitive to what happens at highenergies, and we are able to parametrize our ignorance of high energy physics in terms ofa few physical constants. From the late 1960s to the 1970s, one main thrust of fundamental physics was to classify and study renormalizable theories. As we know, this program was “more thanspectacularly successful.” It allowed us to pin down the theory of the strong, the weak, andthe electromagnetic interactions. Renormalization group flow and dimensional analysis The effective field theory philosophy is intrinsically tied to renormalization group flow.In a given field theory, as we flow toward low energies, some couplings may tend to zerowhile others do not (and if they tend to infinity as in QCD, then we are unable to figureout the effective theory without experimental input). Thus, the first step is to calculate therenormalization group flow. A simple example is given in exercise VIII.3.1. In many cases, we can simply use dimensional analysis. As I explained in our earlier discussion on renormalization theory, couplings with negative dimensions of mass are not important at low energies. T o be specific, suppose we add a gϕ 6term to a λϕ4theory. The coupling ghas the dimension of inverse mass squared. Let us define M2≡1/g.A t low energies, the effect of the gϕ6term is suppressed by (E/M)2. 454 | VIII. Gravity and Beyond How do we understand Schwinger’s spectacular calculation of the anomalous magnetic moment of the electron in the effective field theory philosophy? Let me first tell the traditional (i.e., pre-Wilsonian) version of the story. A student could have asked, “Professor Schwinger, why didn’t you include the term (1/M)¯ψσμνψFμνin the Lagrangian?” The answer is that we better not. Otherwise, we would lose our prediction for the anomalous magnetic moment; it would depend on M. Recall that [ ψ]=3 2and [Aμ]=1, and hence ¯ψσμνψFμνhas mass dimension3 2+3 2+1+1=5>4. The requirement of renormalizability, that the Lagrangian be restricted to contain operators of dimension 4 orless, provides the rationale for excluding this term. Actually, the “real” punchline of my story is that Schwinger probably would not have answered the question. When I took Schwinger’s field theory class, it was well knownamong the students that it was forbidden to ask questions. Schwinger would simplyignore any raised hands. There was no opportunity to ask questions after class either:As he uttered his last sentence of an invariably beautifully prepared lecture, he would sailmajestically out of the room. Dirac dealt with questions differently. I was too young to havewitnessed it, but the story goes that when a student asked, “Professor Dirac, I did not under-stand..., ” Dirac replied, “That is an assertion, not a question.” The modern retelling of the magnetic moment story turns it around. We now regard the Lagrangian of quantum electrodynamics as an effective Lagrangian which should includean infinite sequence of terms of ever higher dimensions, with coefficients parametrizingour threshold of ignorance. The physics of electrons and photons is now described by L=¯ψ(iγμ(∂μ−ieAμ)−m)ψ−1 4FμνFμν+1 M¯ψσμνψFμν+... Yes, the term (1/M)¯ψσμνψFμνis there, with some unknown Mhaving the dimension of a mass. Schwinger’s result, that quantum fluctuations generate a term(α/2π)(1/2m e)¯ψσμνψFμν, should then be interpreted as saying that the anomalous magnetic moment of the electron is predicted to be [ (α/2π)(1/2me)+1/M]. The close agreement of (α/2π)(1/2me)with the experimental value of the anomalous magnetic moment can then be turned around to set a lower bound on M/greatermuch(4π/α)me. Equivalently, Schwinger’s result predicts the anomalous magnetic moment of the elec- tron if we have independent evidence that Mis much larger than [ (α/2π)(1/2me)]−1.I want to emphasize that all of this makes total physical sense. For example, if you speculatethat the electron has some finite size a, then you would expect M∼1/a. The anomalous magnetic moment calculation gives an upper bound for a, telling us that the electron must be pointlike down to some small scale. Alternatively, we could have had independent evi-dence, from electron scattering for example, that ahas to be smaller than a certain length, thus giving us a lower bound on M. T o underscore this point, imagine that in 1948 we followed Schwinger and quickly calculated the anomalous magnetic moment of the proton. We could literally have doneit in 3 seconds, since all we have to do is replace m ebympin the Lagrangian, thus obtaining (α/2π)(1/2mp)¯ψσμνψFμν, which would of course disagree resoundingly with VIII.3. Effective Field Theory | 455 experiment. The disagreement tells us that we had not included all the relevant physics, namely that the proton interacts strongly and is not pointlike. Indeed, we now know thatthe anomalous magnetic moment of the proton gets contributions from the anomalousmagnetic moments and the orbital motion of the quarks inside the proton. Effective theory of proton decay It may seem that with the effective field theory approach we lose some predictive power. Buteffective field theories can also be surprisingly predictive. Let me give a specific example.Suppose we had never heard of grand unified theory. All we know is the SU( 3)⊗SU( 2)⊗ U(1)theory. An experimentalist tells us that he is planning to see if the proton would decay. Without the foggiest notion about what would cause the proton to decay we can still write down a field theory to describe proton decay. The Lagrangian Lis to be constructed out of quark qand lepton lfields and must satisfy the symmetries that we know. Three quarks disappear, so we write down schematically qqq , but three spinors do not a Lorentz scalar make. We have to include a lepton field and write qqql . Since four fermion fields are involved, the terms qqql have mass dimension 6 and so in Lthey have to appear as (1/M 2)qqql with some mass M, corresponding to the mass scale of the physics responsible for proton decay. The experimental lower bound on the lifetimeof the proton sets a lower bound on M. It is instructive to contrast this analysis with an (imagined) effective field theory analysis of proton decay long before the concept of quarks was invented, say around 1950. We wouldconstruct an effective Lagrangian out of the available fields, namely the proton field p, the electron field e, and the pion field π, and thus write down the dimension 4 operator f¯pe +π0with some dimensionless constant f. T o estimate f, we would naively compare this operator with the one describing pion-nucleon coupling (chapter IV .2) g¯pnπ+in the effective Lagrangian. Since f¯pe+π0violates isospin invariance, we might expect f∼αg, namely the same order as gmultiplied by some measure of isospin breaking, say the fine structure constant. But this would give an unacceptably short lifetime to the proton. Weare forced to set fto a ridiculously small number, which seems highly unnatural. Thus, at least in hindsight, we can say that the extremely long lifetime of the proton almost pointsto the existence of quarks. The key, as we saw above, is to promote of the mass dimensionof the term in the effective Lagrangian responsible for proton decay from 4 to 6. (Can thecosmological constant puzzle be solved in the same way?) Another way of saying this is that SU( 3)⊗SU( 2)⊗U(1)plus renormalizability predicts one of the most striking facts of the universe, the stability of the proton. In contrast, theold pion-nucleon theory glaringly failed to explain this experimental fact. In accordance with our philosophy, Lmust be invariant under SU( 3)⊗SU( 2)⊗U(1), under which quark and lepton fields transform rather idiosyncratically, as we saw inchapter VII.5. T o construct Lwe have to sit down and list all Lorentz invariant SU( 3)⊗ SU( 2)⊗U(1)terms of the form qqql . 456 | VIII. Gravity and Beyond Sitting down, we would find that, assuming only one family of quarks and leptons for simplicity, there are only four terms we can write down for proton decay, which I list herefor the sake of completeness: (/tildewidel LCqL)(uRCdR),(eRCuR)(/tildewideqLCqL),(/tildewidelLCqL)(/tildewideqLCqL), and (eRCuR)(uRCdR). Here lL=/parenleftbigν e/parenrightbig LandqL=/parenleftbigu d/parenrightbig Ldenote the lepton and quark doublet ofSU( 2)⊗U(1), the twiddle is defined by /tildewidelj=liεijwithSU( 2)indices i,j=1, 2 (see appendix B), and Cdenotes the charge conjugation matrix. Color indices on the quark fields are contracted in the only possible way. The effective Lagrangian is then given by thesum of these four terms, with four unknown coefficients. The effective field theory tells us that all possible baryon number violating decay pro- cesses can be determined in terms of four unknowns. We expect that these predictions willhold to an accuracy of order (M W/M)2. (IfMWwere zero, SU( 3)⊗SU( 2)⊗U(1)would be exact.) Of course, we can increase our predictive power by making further assumptions. For example, if we think that proton decay is mediated by a vector particle, as in a generic grandunified theory, then only the first two terms in the above list are allowed. In a specific grandunified theory, such as the SU( 5)theory, the two unknown coefficients are determined in terms of the grand unified coupling and the mass of the Xboson. T o appreciate the predictive power of the effective field theory approach, inspect the list of the four possible operators. We can immediately predict that while proton decayviolates both baryon number Band lepton number L, it conserves the combination B−L. I emphasize that this is not at all obvious before doing the analysis. Could you have toldthe experimentalist which of the two possible modes n→e +π−orn→e−π+he should expect? A priori, it could well be that B+Lis conserved. Note that Fermi’s theory of the weak interaction would be called an effective field theory these days. Of course, in contrast to proton decay, beta decay was actually seen, and theprediction from this sort of symmetry analysis, namely the existence of the neutrino, wastriumphantly confirmed. Along the same line, we could construct an effective field theory of neutrino masses. Surely one of the most exciting experimental discoveries in particle physics of recentyears was that neutrinos are not massless. Let us construct an SU( 2)⊗U(1)invariant effective theory. Since ν Lresides inside lL, without doing any detailed analysis we can see that a dimension-5 operator is required: schematically lLlLcontains the desired neutrino bilinear but it carries hypercharge Y/2=− 1; on the other hand, the Higgs doublet ϕ carries hypercharge +1 2, and so the lowest dimensional operator we can form is of the formllϕϕ with dimension3 2+3 2+1+1=5. Thus, the effective Lmust contain a term (1/M)llϕϕ , with Mthe mass scale of the new physics responsible for the neutrino mass. Thus, by dimensional analysis we can estimate mν∼m2 l/M, with mlsome typical charged lepton mass. If we take mlto be the muon mass ∼102Mev and mν∼10−1ev, we find M∼(102Mev)2/10−1(10−6Mev)=108Gev. The philosophy of effective field theories valid up to a certain energy scale /Lambda1seems so obvious by now that it is almost difficult to imagine that at one time many eminentphysicists demanded much more of quantum field theory: that it be fundamental up toarbitrarily high energy scales. VIII.3. Effective Field Theory | 457 Indeed, we now regard all quantum field theories as effective field theory. For all we know, spacetime on some short distance does consist of a lattice, and so the Y ang-Millsaction is but the leading term in an expansion of the Wilson lattice action. The Einstein-Hilbert Lagrangian, being nonrenormalizable, is a fortiori “merely" the leading term inan effective field theory L=√−g(M4 /Lambda1+M2 PR+c1R2+c2RμνRμν+c3RμνσρRμνσρ+1 M2(d1R3+...)+...) Herec1, 2, 3 andd1are dimensionless numbers presumably of order 1. The three terms quadratic in the curvature involve four powers of derivatives versus the two powers inthe Einstein-Hilbert term, and hence their effects relative to the leading terms are sup-pressed by (E/M P)2withEan energy scale characteristic of the process we are studying. Thus, these so-called Weyl-Eddington terms could be safely ignored in any conceivableexperiment. [A technical aside: The Gauss-Bonnet theorem implies that the combination(R 2−4RμνRμν+RμνσρRμνσρ)is a total derivative, so c3can be effectively set to 0, but that is besides the point here.] We have indicated only one representative dimension 6 termR 3(out of many). Its coefficient, in accordance with high school dimensional analysis, is suppressed by two powers of some mass M. What do we expect the mass scale Mto be? Suppose we live in a universe with only gravity (and of course we don’t, actually) then once again, we could risk being presumptuous andtakeMto be the intrinsic mass scale of gravity, namely the Planck mass M P, but we have not yet recovered from our third-degree burn from supposing that M/Lambda1∼MP. If we could ignore the cosmological constant problem for a moment, then the standard (but quitepossibly wrong!) consensus is that in a universe of pure gravity our theory of gravity is aneffective expansion in powers of (E/M P)2. Alternatively, we could treat Las the effective theory of gravity after we integrate out all the matter degree of freedom. In that case, Mwould be of order me(imagine gravitons coupled to an electron loop; see exercise VIII.3.5), or perhaps even mν(generated by a neutrino loop). Effective field theory of the blue sky As another application of the effective field theory philosophy, consider the scattering of electromagnetic waves on an electrically neutral spinless particle described by a scalar field /Phi1. Since /Phi1is neutral, the lowest dimension gauge invariant term that can be added to L=∂/Phi1†∂/Phi1+m2/Phi1†/Phi1+...is (1/M2)/Phi1†/Phi1FμνFμν. A factor of 1 /M2, with Msome mass scale, has to be included with the dimension 1 +1+2+2=6 operator to bring the high school dimension down to 4. The two powers of derivative in FμνFμνtell us immediately that the amplitude for photon scattering on this neutral particle goes like M∝ω2, with ωthe frequency of the electromagnetic wave. Thus we conclude that the scattering cross section varies like σ(ω)∝ω4. 458 | VIII. Gravity and Beyond We have arrived at Rayleigh’s celebrated explanation of the color of the sky. In passing through the atmosphere red light scatters less than blue light on air molecules and hencethe sky is blue. For application to spinless atoms or molecules, we can pass to the nonrelativistic limit as described in chapter III.5, setting /Phi1=(1/√ 2m)e−imtϕ, so that the effective Lagrangian now reads L=ϕ†i∂0ϕ−1 2m∂iϕ†∂iϕ+1 mM2ϕ†ϕ(c 1/vectorE2−c2/vectorB2)+... In this case, since we understand the microscopic physics governing atoms and molecules, we know perfectly well what the mass scale Mrepresents. The coupling of a photon to an electrically neutral system such as an atom or a molecule must vanish likethe characteristic size dof the system, since as d→0 the positive and negative charges are on top of each other, giving a vanishing net coupling to the photon. Rotational invarianceimplies that the coupling ∼/vectork./vectord. The scattering amplitude then goes like M∝(ωd) 2, since the coupling has to act twice, once for the incoming photon and once for the out-going photon. (Note that by rotational invariance the expectation value of the operator /vectord vanishes, but we are doing second order perturbation theory so that we have to evalu-ate the expectation value of a quantity quadratic in /vectord.) 1Squaring Mand invoking some elementary quantum mechanics and dimensional analysis, we obtain the cross sectionσ(ω)∼d 6ω4. Appendix: Reshuffling terms in effective field theory The Lagrangian of an effective field theory consists of an infinite sequence of terms arranged in an orderly progression of higher and higher mass dimension, constrained only by the assumed symmetries of the theory.In fact, some terms could be effectively eliminated. T o explain this, we focus on a toy example: L=1 2(∂ϕ)2−λϕ4+1 M2(aϕ6+bϕ3∂2ϕ+c(∂2ϕ)2)+O/parenleftbigg1 M4/parenrightbigg (1) We are secretly dealing with the action and thus we freely integrate by parts. For arithmetical simplicity, we did not include a mass term, so that to leading order in 1 /M the equation of motion reads simply ∂2ϕ=0. The three possible dimension 6 terms are shown explicitly [we integrate by parts to get rid of the term ϕ2(∂ϕ)2]. Are we allowed to use the equation of motion to eliminate the two dimension 6 terms that are proportional to∂2ϕ? We know that we could make a field redefinition without changing the on shell amplitudes, so let us rede- fineϕ→ϕ+(1/M2)F. Then1 2(∂ϕ)2→1 2(∂ϕ)2−(1/M2)F∂2ϕ+O(1/M4)andλϕ4→λ(ϕ4+(1/M2)ϕ3F+ O(1/M4)).S e tF=pϕ3+q∂2ϕ. We see that with an appropriate choice of pandqwe can cancel off bandc. Notice that in the process we also change ato some other value. The answer to the question is yes, but the naive statement that the equation of motion ∂2ϕ=0 empowers us to simply set ∂2ϕto zero in the nonleading terms in the effective field theory is, legalistically speaking, incorrect, or at least misleading. We see that we actually generated O(1/M4)terms and changed the ϕ6term. Thus, more correctly, a field redefinition allows us to shuffle terms around and to higher order. The net effect, however, isthe same as if we trusted the naive statement and set ∂ 2ϕto zero in the nonleading terms. 1For details, see, for example, J. J. Sakurai, Advanced Quantum Mechanics, Addison-Wesley, New York, 1967, p. 47. VIII.3. Effective Field Theory | 459 This procedure works for fermions also. As an example, consider the effective Lagrangian L=¯ψ(iγμ∂μ− m)ψ+(1/M3)¯ψ(iγμ∂μ−m)ψ( ¯ψψ)+.... Then the field redefinition ψ→ψ−(1/2M3)ψ(¯ψψ) gets rid of the dimension 7 term shown. We could also apply what we just learned to the effective theory of gravity if without any understanding we set the cosmological constant to zero. Also, use the Gauss-Bonnet theorem to get rid of the RμνσρRμνσρterm, so that we have L=√−g/parenleftbigg M2 PR+c1R2+c2RμνRμν+1 M2(d1R3+...)+.../parenrightbigg (2) Make a field redefinition gμν→gμν+δgμνand use δ/integraldisplay d4x√−gR=−/integraldisplay d4x√−g(Rμν−1 2gμνR)δgμν Setδgμν=pRμν+qgμνR. Then we can cancel off c1andc2with a judicious choice of pandq. I emphasize that this works only if we set the cosmological constant to zero without any ado. Exercises VIII.3.1 Consider L=1 2/bracketleftBig (∂ϕ 1)2+(∂ϕ2)2/bracketrightBig −λ(ϕ4 1+ϕ4 2)−gϕ2 1ϕ2 2(3) We have taken the O(2)theory from chapter I.10 and broken the symmetry explicitly. Work out the renormalization group flow in the (λ−g)plane and draw your own conclusions. VIII.3.2 Assuming the nonexistence of the right handed neutrino field νR(i.e., assuming the minimal particle content of the standard model) write down all SU( 2)⊗U(1)invariant terms that violate lepton number Lby 2 and hence construct an effective field theory of the neutrino mass. Of course, by constructing a specific theory one can be much more predictive. Out of the product lLlLwe can form a Lorentz scalar transforming as either a singlet or triplet under SU( 2). T ake the singlet case and construct a theory. [Hint: For help, see A. Zee, Phys. Lett. 93B : p. 389, 1980.] VIII.3.3 LetA,B,C,Ddenote four spin1 2fields and label their handedness by a subscript: γ5Ah=hAhwith h=± 1. Thus, A+is right handed, A−left handed, and so on. Show that (AhBh)(C−hD−h)=−1 2(AhγμD−h)(C−hγμBh) (4) This is an example of a broad class of identities known as Fierz identities (some of which we will need in discussing supersymmetry.) Argue that if proton decay proceeds in lowest order from the exchange of avector particle then only the terms (/tildewidel LCqL)(uRCdR)and(eRCuR)(/tildewideqLCqL)are allowed in the Lagrangian. VIII.3.4 Given the conclusion of the previous exercise show that the decay rate for the processes p→π++¯ν, p→π0+e+,n→π0+¯ν, andn→π−+e+are proportional to each other, with the proportionality factors determined by a single unknown constant [the ratio of the coefficients of (/tildewidelLCqL)(uRCdR)and (eRCuR)(/tildewideqLCqL)]. For help on these last three exercises see S. Weinberg, Phys. Rev. Lett. 43: 1566, 1979; F. Wilczek and A. Zee, ibid. p. 1571; H. A. Weldon and A. Zee, Nucl. Phys. B173: 269, 1980. VIII.3.5 Imagine a mythical (and presumably impossible) race of physicists who only understand physics at energies less than the electron mass me. They manage to write down the effective field theory for the one particle they know, the photon, L=−1 4FμνFμν+1 m4 e{a(FμνFμν)2+b(Fμν˜Fμν)2}+ ... (5) 460 | VIII. Gravity and Beyond with ˜Fμν=1 2εμνρσFρσthe dual field strength as usual and aandbtwo dimensionless constants presumably of order unity.(a) Show that Lrespects charge conjugation ( A→−Ain this context), parity, and time reversal, (and of course gauge invariance.) (b) Draw the Feynman diagrams that give rise to the two dimension 8 terms shown. The coefficients aandbwere calculated by Euler and Kockel in 1935 and by Heisenberg and Euler in 1936, quite a feat since they did not know about Feynman diagrams and any of the modern quantum field theoryset up. (c) Explain why dimensional 6 terms are absent in L. [Hint: One possible term is ∂ λFμν∂λFμν.] (d) Our mythical physicists do not know about the electron, but they are getting excited. They are going to start doing photon-photon scattering experiments with a machine called LPC that could producephotons with energy greater than m e. Discuss what they will see. Apply unitarity and the Cutkosky rules. VIII.3.6 Use the effective field theory approach to show that the scattering cross section of light on an electrically neutral spin1 2particle (such as the neutron) goes like σ∝ω2to leading order, not ω4. Argue further that the constant of proportionality can be fixed in terms of the magnetic moment μof the particle. [Historical note: This result was first obtained in 1954 by F. Low ( Phys. Rev. 96: 1428) and by M. Gell-Mann and Murph L. Goldberger (Phys. Rev. 96: 1433) using much more elaborate arguments.] VIII.4 Supersymmetry: A Very Brief Introduction Unifying bosons and fermions Let me start with a few of the motivations for supersymmetry. (1) All experimentally known symmetries relate bosons to bosons and fermions to fermions. We would like to have asymmetry, supersymmetry, relating bosons and fermions. (2) It is natural for fermions tobe massless (recall chapter VII.6), but not for bosons. Perhaps by pairing the Higgs fieldwith a fermion field we can resolve the hierarchy problem mentioned in chapter VII.6.(3) Recalling from chapter II.5 that fermions contribute negatively to the vacuum energy,you might be tempted to speculate that the cosmological constant problem could be solvedif we could get the fermion contribution to cancel the boson contribution. Disappointingly, it has been more than 30 years 1since the conception of supersymmetry (Golfand and Likhtman constructed the first supersymmetric field theory in 1971) anddirect experimental evidence is still lacking. All existing supersymmetric theories pairknown bosons with unknown fermions and known fermions with unknown bosons.Supersymmetry has to be broken at some mass scale Mbeyond the regime already explored experimentally, but then (as explained in chapter VIII.2) we might expect a cosmologicalconstant of order M 4. Be that as it may, supersymmetric field theories have many nice properties (hardly surprising since the relevant symmetry is much larger). Supersymmetry has thus attracteda multitude of devotees. I give you here as brief an introduction to supersymmetry as I canwrite. In the spirit of a first exposure, I will avoid mentioning any subtleties and caveats,hoping that this brief introduction will be helpful to students before they tackle the tomesout there. 1For a fascinating account of the early history of supersymmetry, see G. Kane and M. Shifman, eds., The Supersymmetric World: The Beginning of the Theory . 462 | VIII. Gravity and Beyond Inventing supersymmetry Suppose one day you woke up wanting to invent a field theory with a symmetry relating bosons to fermions. The first thing you would need is the same number of fermionic andbosonic degrees of freedom. The simplest fermion field is the two-component Weyl spinorψ. You would now have one complex degree of freedom, 2so you would have to throw in a complex scalar field ϕ. You could proceed by trial and error: Write down a Lagrangian including all terms with dimension up to four and then adjust the various parameters inthe Lagrangian until the desired symmetry appears. For instance, you might adjust μin the mass terms μ 2ϕ†ϕ+m(ψψ +¯ψ¯ψ) until the theory becomes more symmetrical so that the boson and the fermion have the same mass. If you were to try to play the game by using a Dirac spinor /Psi1and a complex scalar ϕyou would be doomed to failure from the very start since there would be twice as many fermionic degrees of freedom as bosonic degrees of freedom. I believe that thedevelopment of supersymmetry was very much retarded by the fact that until the early1970s most field theorists, having grown up with Dirac spinors, had little knowledge ofWeyl spinors. That was a hint that now is the time for you to get thoroughly familiar withthe dotted and undotted notation of appendix E. T o read this chapter, you need to be fluentwith that notation. Supersymmetric algebra It is perfectly feasible to construct this supersymmetric field theory, known as the Wess-Zumino model, by trial and error, but instead I will show you an elegant but more abstractapproach known as the superspace and superfield formalism, invented by Salam andStrathdee. We will have to develop a considerable amount of formal machinery. Everythingis very super here. Write the supersymmetry generator taking us from ϕtoψ αasQα(known as the supercharge). The statement that Qαtransforms as a Weyl spinor means [ Jμν,Qα]= −i(σμν)αβQβ, where Jμνdenotes the generators of the Lorentz group. Of course, since Qαis independent of the spacetime coordinates [ Pμ,Qα]=0. From appendix E we denote the conjugate of Qαby¯Q˙αand [Jμν,¯Q˙α]=−i(¯σμν)˙α˙β¯Q˙β. We have to write down the anticommutation relation between the Grassman objects Qα and¯Q˙βand now the work we did in appendix E really pays off. The supersymmetry algebra is given by {Qα,¯Q˙β}=2(σμ)α˙βPμ (1) 2One complex degree of freedom on mass shell and two complex degrees of freedom off mass shell. See the superfield formalism below. VIII.4. Supersymmetry | 463 We argue by the “what else can it be?” method. The right-hand side must carry the indices αand˙βand we know that the only object that carries these indices is σμ. The Lorentz index μhas to be contracted and the only vector around is Pμ. The factor of 2 fixes the normalization of Q. By the same kind of argument we must have {Qα,Qβ}=c1(σμν)αβJμν+c2δβ α. Com- muting with Pλwe see that the constant c1must vanish. Recalling that Qγ=εγβQβ,w e have{Qα,Qγ}=c 2εγα; but since the left-hand side is symmetric in αandγwe have c2=0. Thus, {Qα,Qβ}=0 and {¯Q˙α,¯Q˙β}=0 (see exercise VIII.4.2). A basic theorem An important physical fact follows immediately from (1). Contracting with (¯σν)˙βαwe obtain 4Pν=(¯σν)˙βα{Qα,¯Q˙β} (2) In particular the time component tells us about the Hamiltonian 4H=/summationdisplay α{Qα,¯Q˙α}=/summationdisplay α{Qα,Q† α}=/summationdisplay α(QαQ† α+Q† αQα) (3) We obtain the important theorem that in a supersymmetric field theory any physical state |S/angbracketrightmust have nonnegative energy: /angbracketleftS|H|S/angbracketright=1 2/summationdisplay α/summationdisplay S/prime|/angbracketleftS/prime|Qα|S/angbracketright|2≥0 (4) Superspace Now that we have constructed the supersymmetric algebra let us keep in mind our goal of constructing supersymmetric field theories. T o do that, we need to figure out and classifyhow fields transform under this supersymmetric algebra. We have to go through a lot offormalism, the necessity for which will become clear in due course. Imagine that you are trying to invent the superspace formalism. Let us motivate it by staring at the basic relation (1) {Q α,¯Q˙β}=2(σμ)α˙βPμ. A supersymmetric transformation Qfollowed by its conjugate ¯Q˙βgenerates a translation Pμ. Hmm, let’s see, Pμ≡i(∂/∂xμ) generates translation in xμ, so perhaps Qα, being Grassmannian, would generate transla- tion in some abstract Grassmannian coordinate θα? (Similarly, ¯Q˙βwould generate trans- lation in ¯θ˙β.) Salam and Strathdee invented the notion of a superspace with bosonic and fermionic coordinates {xμ,θα,¯θ˙β}with the supersymmetry algebra represented by translations in this space. So let us try Qαand¯Q˙βbeing something like ∂/∂θαand∂/∂¯θ˙β, respectively. But then {Qα,¯Q˙β}=0 and we don’t get (1). We have to keep playing around modifying Qαand¯Q˙β. You may already see what we need. If we add a term such as θσμ∂μto¯Q˙β, then the ∂/∂θα 464 | VIII. Gravity and Beyond inQαacting on θσμ∂μwill produce something like the right-hand side of (1). Similarly, we will want to add a term such as ¯θσμ∂μtoQα. (Once again, the dotted and undotted notation we worked hard to develop fixes what we must write, namely (σμ)α˙α¯θ˙α∂μso that the indices match and obey the “southwest to northeast” rule.) Thus, we represent thesupercharges as Qα=∂ ∂θα−i(σμ)α˙α¯θ˙α∂μ (5) and ¯Q˙β=−∂ ∂¯θ˙β+iθβ(σμ)β˙β∂μ (6) You see that (1) is now satisfied. Interestingly, when we translate in the fermionic direction we have to translate a bit in the bosonic direction as well. Superfield A superfield /Phi1(xμ,θα,¯θ˙β), as the name suggests, is just a field living in superspace. An infinitesimal supersymmetry transformation takes /Phi1→/Phi1/prime=(1+iξαQα+i¯ξ˙α¯Q˙α)/Phi1 (7) withξand¯ξtwo Grassmannian parameters. It turns out that we can impose some condition on /Phi1and restrict this rather broad definition a bit. After staring at (5) and (6) for a while, you may realize that there are twoother objects, Dα=∂ ∂θα+i(σμ)α˙α¯θ˙α∂μ and ¯D˙β=−/bracketleftbigg∂ ∂¯θ˙β+iθβ(σμ)β˙β∂μ/bracketrightbigg that we can define, sort of the combinations orthogonal to Qαand¯Q˙β. Clearly, Dαand ¯D˙βanticommute with Qαand¯Q˙β. The significance of this fact is that if we impose the condition ¯D˙β/Phi1=0 on the superfield /Phi1, then according to (7) its transform /Phi1/primealso satisfies the condition. A superfield /Phi1satisfying the condition ¯D˙β/Phi1=0 is known as a chiral superfield. The con- dition is actually easy to implement:3Observe that if we define yμ≡(xμ+iθα(σμ)α˙α¯θ˙α) (note we are adding two bosonic quantities here), then ¯D˙βyμ=−/bracketleftbigg∂ ∂¯θ˙β+iθβ(σν)β˙β∂ν/bracketrightbigg yμ=−/bracketleftBig −iθα(σμ)α˙β+iθβ(σμ)β˙β/bracketrightBig =0 Thus, a superfield /Phi1(y ,θ)that depends on yandθonly is a chiral superfield. 3This is analogous to the problem of constructing a function f( x ,y)satisfying the condition Lf=0 with L≡[x(∂/∂y) −y(∂/∂x) ]. We define r≡(x2+y2)1 2and observe that Lr=0. Then any fthat only depends on rsatisfies the desired condition. VIII.4. Supersymmetry | 465 Let us expand /Phi1in powers of θholding yfixed. Remember that θcontains two compo- nents (θ1,θ2). Thus, we can form an object with at most two powers of θ, namely θθ, which you worked out in exercise E.3. Thus, as usual, power series in Grassmannian variablesterminate, and we have /Phi1(y ,θ)=ϕ(y)+√ 2θψ(y) +θθF(y) withϕ(y) ,ψ(y) , andF(y) merely coefficients in the series at this stage. We can T aylor expand once more around x: /Phi1(y ,θ)=ϕ(x)+√ 2θψ(x) +θθF(x) +iθσμ¯θ∂μϕ(x)−1 2θσμ¯θθσν¯θ∂μ∂νϕ(x)+√ 2θiθσμ¯θ∂μψ(x)(8) We see that a chiral superfield /Phi1contains a Weyl fermion field ψ, and two complex scalar fields ϕandF. Finding a total divergence Let’s do a bit of dimensional analysis for fun and profit. Given that Pμhas the dimension of mass, which we write as [Pμ]=1 using the same notation as in chapter III.2, then (1), (5), and (6) tell us that [ Q]=[¯Q]=1 2and [θ]=[¯θ]=−1 2. Given [ ϕ]=1, then (8) tells us [ψ]=3 2, which we know already, and [ F]=2, which we didn’t know. In fact, we have never met a Lorentz scalar field with mass dimension 2. How can we have a kinetic energy termforFinLwith dimension 4? We can’t. The term F †Falready has dimension 4, and any derivative is going to make the dimension even higher. Also, didn’t we say that with ϕand ψwe balance the same number of bosonic and fermionic degrees of freedom? The field F(x) definitely has something strange about him. What is he doing in our theory? Under an infinitesimal supersymmetric transformation the superfield changes by δ/Phi1= i(ξQ+¯ξ¯Q)/Phi1 . Referring to (8), (5), and (6), you can work out how the component fields ϕ, ψ, andFtransform (see exercise VIII.4.5). But we can go a long way invoking symmetry and dimensional analysis. For example, δF is linear in ξor¯ξ, which by dimensional analysis must multiply something with dimension [5 2] since [ F]=2 and [ ξ]=[¯ξ]=−1 2. The only thing around with dimension [5 2]i s∂μψ, which carries an undotted index. Note it can’t be ∂μ¯ψsince/Phi1does not contain ¯ψ. By Lorentz invariance we have to find something carrying the index μ, and that can only be (σμ)α˙α. The dotted index on (σμ)α˙αcan only be contracted with ¯ξ. So everything is fixed except for an overall constant: δF∼∂μψα(σμ)α˙α¯ξ˙α(9) Arguing along the same lines you can easily show that δϕ∼ξψ andδψ∼ξF+∂μϕσμ¯ξ. The important point here is not the overall constant in (9) but that δF is a total diver- gence. Given any superfield /Phi1let us denote by [ /Phi1]Fthe coefficient of θθin an expansion of /Phi1 [as in (8)]. What we have learned is that under a supersymmetric transformation δ([/Phi1]F) is a total divergence and thus/integraltext d4x[/Phi1]Fis invariant under supersymmetry. 466 | VIII. Gravity and Beyond Our next observation is that if ¯D˙β/Phi1=0, then ¯D˙β/Phi12=0 also. In other words, if /Phi1is a chiral superfield, then so is /Phi12(and by extension, /Phi13,/Phi14, and so forth). Supersymmetric action What do we want to achieve anyway? We want to construct an action invariant under supersymmetry. Finally, after all this formalism we are ready. In fact, it is almost staring us in the face:/integraltext d4x[1 2m/Phi12+1 3g/Phi13+...]Fis invariant under supersymmetry by virtue of the last two paragraphs. Squaring (8) and extracting the coefficient of θθwe see by inspection that [/Phi12]F=(2Fϕ−ψψ) . Similarly, [ /Phi13]F=3(Fϕ2−ϕψψ) . Now do exercise VIII.4.6. Looks like we have generated a mass term for the Weyl fermion ψand its coupling to the scalar field ϕ, but where are the kinetic energy terms, such as ¯ψ˙α(¯σμ)˙αα∂μψα? Vector superfield The kinetic energy terms contain ¯ψ˙α, which does not appear in /Phi1. T o get the conjugate field¯ψ˙α, we obviously have to use /Phi1†, and so we are led to consider /Phi1†/Phi1. More formalism here! We call a superfield V( x ,θ,¯θ)a vector superfield if V=V†. For example, /Phi1†/Phi1is a vector superfield. Imagine expanding /Phi1†/Phi1=ϕ†ϕ+... or any vector superfield Vin powers of θand¯θ. The highest power is uniquely ¯θ¯θθθ since by the properties of Grassmannian variables the only object we can form is ¯θ˙1¯θ˙2θ1θ2. Any object quadratic in θand quadratic in ¯θ, such as (θσμ¯θ)(θσ μ¯θ), can be beaten down to ¯θ¯θθθ by using the kind of identities you discovered in the exercises in appendix E. Let [ V]Ddenote the coefficient of ¯θ¯θθθ in the expansion ofV. Again, dimensional analysis can carry us a long way. If Vhas mass dimension n, then [ V]Dhas mass dimension n+2 since θand¯θeach has mass dimension −1 2. Let us study how [ V]Dchanges under an infinitesimal supersymmetry transformation δV= i(ξQ+¯ξ¯Q)V . We use the same kind of argument as before: δ([V]D)is linear in ξor¯ξ, which by dimensional analysis must multiply something with dimension n+5 2since [ ξ]= [¯ξ]=−1 2. This can only be the derivative ∂of something with dimension n+3 2, namely the coefficients of ¯θ¯θθand¯θθθ in the expansion of V. We conclude that δ([V]D)has to have the form ∂μ(...), namely that δ([V]D)is a total divergence. This is the same type of argument that allows us to conclude that δ([/Phi1]F)is a total divergence. Thus, the action/integraltext d4x[/Phi1†/Phi1]Dis invariant under supersymmetry. Staring at (8), which I repeat for your convenience, /Phi1(y ,θ)=ϕ(x)+√ 2θψ(x) +θθF(x) (10) +iθσμ¯θ∂μϕ(x)−1 2θσμ¯θθσν¯θ∂μ∂νϕ(x)+√ 2θiθσμ¯θ∂μψ(x) VIII.4. Supersymmetry | 467 we see that/integraltext d4x[/Phi1†/Phi1]Dcontains/integraltext d4xϕ†∂2ϕ(from multiplying the first term in /Phi1†with the fifth term in /Phi1),/integraltext d4x∂ϕ†∂ϕ(from multiplying the fourth term in /Phi1†with the fourth term in /Phi1),/integraltext d4x¯ψ¯σμ∂μψ(from multiplying the second term in /Phi1†with the sixth term in/Phi1), and finally/integraltext d4xF†F(from multiplying the third term in /Phi1†with the third term in/Phi1). It is quite amusing how derivatives of fields arise in supersymmetric field theories: Note that the action/integraltext d4x[/Phi1†/Phi1]Ddoes not contain derivatives explicitly. T o summarize, given a chiral superfield /Phi1we have constructed the supersymmetric action S=/integraldisplay d4x{[/Phi1†/Phi1]D−([W(/Phi1)]F+h.c.)} (11) Explicitly, with the choice W(/Phi1) =1 2m/Phi12+1 3g/Phi13, we have S=/integraldisplay d4x{∂ϕ†∂ϕ+i¯ψ¯σμ∂μψ+F†F−(mFϕ −1 2mψψ +gFϕ2−gϕψψ +h.c.)} (12) An auxiliary field From the very beginning the field Fseemed strange. Since [ F]=2 we anticipated that it cannot have a kinetic energy term with mass dimension 4 and indeed it doesn’t. We see thatit is not a dynamical field that propagates—it is an auxiliary field (just like σin chapter III.5 andξ μin chapter VI.3) and can be integrated out in the path integral/integraltext DF†DFeiS. Indeed, collect the terms that depend on FinS, namely F†F−F(mϕ +gϕ2)−F†(mϕ†+gϕ†2)=|F−(mϕ+gϕ2)†|2−|mϕ+gϕ2|2 So, integrate over FandF†and get S=/integraldisplay d4x{∂ϕ†∂ϕ+i¯ψ¯σμ∂μψ−|mϕ+gϕ2|2+(1 2mψψ −gϕψψ +h.c.)} (13) Note that the scalar potential V( ϕ†,ϕ)=|mϕ+gϕ2|2≥0 in accordance with (4) and vanishes at its minimum, giving a zero cosmological constant. Note that we are no longerfree to add an arbitrary constant to V( ϕ †,ϕ)as we could in a nonsupersymmetric field theory. As expected, supersymmetric field theories are much more restrictive than ordinary field theories, and, duh, also much more symmetric. The formalism described here canbe extended to construct supersymmetric Y ang-Mills theory. Another important generalization is to introduce, instead of one supercharge Q α,Nsu- percharges QI α, with I=1,... ,N(exercise VIII.4.2). Since each charge QI αtransforms like the Sz=1 2component of a spin1 2operator, it takes a state with Sz=min a super- multiplet to a state with Sz=m+1 2. Thus the integer Nis bounded from above. For supersymmetric Y ang-Mills theory, the maximum number of supersymmetry generatorsisN=4 if we do not want to introduce fields with spin ≥1. Similarly, the most supersym- metric supergravity theory we could construct (exercise VIII.4.3) has N=8. 468 | VIII. Gravity and Beyond As mentioned in chapter VII.3, if any nontrivial 4-dimensional quantum field theory turned out to be exactly soluble, the supersymmetric N=4 Y ang-Mills theory is probably our best bet. In all likelihood, the first relativistic quantum field theory to be solved exactlywould be N=4 Y ang-Mills in the planar large Nlimit of chapter VII.4. I hope that this brief introduction gave you a flavor of supersymmetry and will enable you to go on to specialized treatises. Exercises VIII.4.1 Construct the Wess-Zumino Lagrangian by the trial and error approach. VIII.4.2 In general there may be Nsupercharges QI α, with I=1 ,..., N. Show that we can have {QI α,QJ β}= εαβZIJ, where ZIJdenotes c-numbers known as central charges. VIII.4.3 From the fact that we do not know how to write consistent quantum field theories with fields having spin greater than 2 show that the Nin the previous exercise cannot exceed 8. Theories with N=8 supersymmetry are said to be maximally supersymmetric. Show that if we do not want to include gravity,Ncannot be greater than 4. Supersymmetric N=4 Y ang-Mills theory has many remarkable properties. VIII.4.4 Show that ∂θ α/∂θβ=εαβ. VIII.4.5 Work out δϕ,δψ, andδFprecisely by computing δ/Phi1=i(ξαQα+¯ξ˙α¯Q˙α)/Phi1. VIII.4.6 For any polynomial W(/Phi1) show that [W(/Phi1)]F=F[dW(ϕ)/dϕ ]+terms not involving F. Show that for the theory (11) the potential energy is given by V( ϕ†,ϕ)=|∂W(ϕ)/∂ϕ |2. VIII.4.7 Construct a field theory in which supersymmetry is spontaneously broken. [Hint: You need at least three chiral superfields.] VIII.4.8 If we can construct supersymmetric quantum field theory, surely we can construct supersymmetric quantum mechanics. Indeed, consider Q1≡1 2[σ1P+σ2W(x) ] andQ2≡1 2[σ2P−σ1W(x) ], where the momentum operator P=−i(d/dx) as usual. Define Q≡Q1+iQ2. Study the properties of the Hamil- tonian Hdefined by {Q,Q†}=2H. VIII.5A Glimpse of String Theory as a 2-Dimensional Field Theory Geometrical action for the bosonic string In this chapter, I will try to give you a tiny glimpse into string theory. Needless to say, you can get only the merest whiff of the subject here, but fortunately excellent texts do exist andI believe that this book has prepared you for them. My main purpose is to show you thatperhaps surprisingly the basic formulation of string theory is naturally phrased in termsof a 2-dimensional field theory. In chapter I.11 I described a point particle tracing out a world line given by X μ(τ) in D−dimensional spacetime. Recall that the action is given geometrically by the length of the world line S=−m/integraldisplay dτ/radicalBigg dXμ dτdXμ dτ(1) and remains unchanged under reparametrization τ→τ/prime(τ). Recall also that classically, S is equivalent to Simp=−1 2/integraldisplay dτ/parenleftbigg1 γdXμ dτdXμ dτ+γm2/parenrightbigg (2) Now consider a string sweeping out a world sheet given by Xμ(τ,σ)inD-dimensional spacetime, which we have already encountered in chapter IV .4 in connection with differ- ential forms. In analogy with (1), Nambu and Goto proposed an action given geometricallyby the area of the world sheet SNG=T/integraldisplay dτdσ/radicalBig det(∂aXμ∂bXμ) (3) where ∂1Xμ≡∂Xμ/∂τ ,∂2Xμ≡∂Xμ/∂σ , and(∂aXμ∂bXμ)denotes the abelement of a 2 by 2 matrix. Here, as in (1), μranges over Dvalues: 0, 1, . . . , D−1. The constant T(≡1/2πα/prime withα/primethe slope of the Regge trajectory in particle phenomenology) corresponds to the string tension since stretching the string to enlarge the world sheet costs an extra amountof action proportional to T. 470 | VIII. Gravity and Beyond In a precise parallel with the discussion for the point particle, it is preferable to avoid the square root and instead use the action S=1 2T/integraldisplay dτdσγ1 2γab(∂aXμ∂bXμ) (4) withγ=detγabin the path integral to quantize the string. We will now show that Sis equivalent classically to SNG. As in (2), we vary Swith respect to the auxiliary variable γab, which we then eliminate. For a matrix M,δM−1=−M−1(δM)M−1andδdetM=δetr logM=etr logMtrM−1δM= (detM)trM−1δM. Thus, δγab=−γacδγcdγdbandδγ=γγbaδγab. For ease of writing, define hab≡∂aXμ∂bXμ. The variation of the integrand in (4) thus gives δ[γ1 2γabhab]=γ1 2[1 2γdcδγcd(γabhab)−γacδγcdγdbhab] Setting the coefficient of δγcdequal to 0 we obtain hcd=1 2γcd(γabhab) (5) where the indices on hare raised and lowered by the metric γ. Multiplying (5) by hdc(and summing over repeated indices) we find γabhab=2 and thus γcd=hcd. Plugging this into (4) we find that S=T/integraltext dτdσ( deth)1 2. Thus, SandSNGare indeed equivalent classically. The action (4), first discovered by Brink, Di Vecchia, and Howe and by Deser and Zumino,is known as the Polyakov action. Note that (5) determines γ abonly up to an arbitrary local rescaling known as a Weyl transformation: γab(τ,σ)→e2ω(τ ,σ)γab(τ,σ) (6) Thus, the action (4) must be invariant under the Weyl transformation. Staring at the string action (4), you will recognize that it is just the action for a quantum field theory of Dmassless scalar fields Xμ(τ,σ)in 2-dimensional spacetime with coor- dinates (τ,σ), albeit with some unusual signs. The index μplays the role of an internal index, and Poincar ´e invariance in our original D-dimensional spacetime now appears as an internal symmetry. Indeed, a good deal of string theory is devoted to the study of quan-tum field theories in 2-dimensional spacetime! It is amusing how quantum field theorymanages to stay on the stage. T o this bosonic string theory we can add fermionic variables in such a way as to make the action supersymmetric. The result, as you surely have heard, is superstring theory,thought by some to be the theory of everything. 1 1“T o understand macroscopic properties of matter based on understanding these microscopic laws is just unrealistic. Even though the microscopic laws are, in a strict sense, controlling what happens at the larger scale,they are not the right way to understand that. And that is why this phrase, “theory of everything,” sounds sleazy.”— J. Schwarz, one of the founders of string theory. VIII.5. A Glimpse of String Theory | 471 This infinitesimal introduction to string theory is all I can give you here, but I hope that this book has prepared you adequately to begin studying various specialized texts on stringtheory. 2 2For a brief but authoritative introduction, see E. Witten, “Reflections on the Fate of Spacetime,” Physics T oday , April 1996, p. 24. This page intentionally left blank Closing Words As I confessed in the preface, I started out intending to write a concise introduction to quantum field theory, but the book grew and grew. The subject is simply too rich. As Imentioned, after a period of almost being abandoned, quantum field theory came roaringback. T o quote my thesis advisor Sidney Coleman, the triumph of quantum field theorywas veritably “a victory parade” that made “the spectator gasp with awe and laugh withjoy.” String theory is beautiful and marvellous, but until it is verified, quantum field theory remains the true theory of everything. All of physics can now be said to be derivable fromfield theory. T o start with, quantum field theory contains quantum mechanics as a (0 +1)- dimensional field theory, and to end (perhaps) with, string theory may be formulated as a(1+1)-dimensional field theory. Quantum field theory can arguably be regarded as the pinnacle of human thought. (Hush, you hear the distant howls of the mathematicians, English professors, philoso-phers, and perhaps even a few stuck-up musicologists?) It is a distillation of basic notionsfrom the very beginning of the physics: Newton’s realization that energy is the square ofmomentum appears in field theory as the two powers of spatial derivative. But yet—youknew that was coming, didn’t you, with field theory set up as the pinnacle et cetera?—but yet, field theory in its present form is in my opinion still incomplete and surely somebright young minds will see how to develop it further. For one thing, field theory has not progressed much beyond the harmonic paradigm, as I presaged in the first chapter. The discovery of the soliton and instanton opened up a newvista, showing in no uncertain terms that Feynman diagrams ain’t everything, contraryto what some field theorists thought. Duality offers one way of linking perturbative weak coupling theory to strong coupling, but as yet practically nothing is known of the strongcoupling regime. When speaking of renormalization groups, we bravely speak of flowingto a strong coupling fixed point, but we merely have the boat ticket: We have little idea of 474 | Closing Words what the destination looks like. Perhaps in the not too distant future, lattice field theorists can extract the field configurations that dominate. Another restriction is to two powers of the derivative, a restriction going back to Newton as I remarked above. In modern applications of field theory to problems far beyond particlephysics, there is no reason at all to impose this restriction. For example, in studying visualperception, one encounters field theories much more involved than those we have studiedin this book. (See the appendix for a brief description.) These field theories are Euclideanin any case and the corresponding functional integral with higher derivatives certainlymakes sense: It is only in Minkowskian theories that we do not know how to handlehigher derivatives. Newton again—certainly economists consider the rate of change ofthe acceleration as well as acceleration. Another innovative application is the formulation 1 of a class of problems in nonequilibrium statistical mechanics as field theories. Typically,various objects wander around and react when they meet. This class of problems appearsin areas ranging from chemical reactions to population biology. We can go far beyond the restriction on the number of derivatives in the Lagrangian. Who said that we can only have integrands of the form “exponential of a spacetimeintegral”? Most modifications you can think of might immediately run afoul of some basic principles (for example,/integraltext Dϕe −/integraltext d4xL(ϕ)−[/integraltext d4x˜L(ϕ)]2would violate locality), but surely others might not. Another speculative thought I like to entertain goes along the followingline: Classical and quantum physics are formulated in terms of differential equationsand functional integrals, respectively. But how are differential equations contained in integrals? The answer is that the integrals/integraltext Dϕe −(1//planckover2pi)/integraltext d4xL(ϕ)contain a parameter /planckover2piso that in the limit /planckover2pigoing to zero the evaluation of the integrals amounts to solving partial differential equations. Can we go beyond quantum field theory by finding a mathematicaloperation that in the limit of some parameter ¯kgoing to zero reduces to doing the integral/integraltext Dϕe−(1//planckover2pi)/integraltext d4xL(ϕ)? The arena of local field theory has always been restricted to the set of dreal numbers xμ. The recent excitement over noncommutative field theory promises to take us beyond. (I was tempted to discuss noncommutative field theory too, but then the nutshell wouldtruly burst.) But perhaps the most unsatisfying feature of field theory is the present formulation of gauge theories. Gauge “symmetry” does not relate two different physical states, but twodescriptions of the same physical state. We have this strange language full of redundancywe can’t live without. We start with unneeded baggage that we then gauge-fix away. We evenknow how to avoid this redundancy from the start but at the price of discretizing spacetime.This redundancy of description is particularly glaring in the manufactured gauge theoriesnow fashionable in condensed matter physics, in which the gauge symmetry is not thereto begin with. Also, surely the way we calculate in nonabelian gauge theories by cuttingthe Y ang-Mills action up into pieces and doing violence to gauge invariance will be held 1By M. Doi, L. Peliti, J. C. Cardy, and others. See for example J. C. Cardy, cond-mat/9607163, “Renormalisation Group Approach to Reaction-Diffusion Problems,” in: J.-B. Zuber, ed., Mathematical Beauty of Physics , p. 113. Closing Words | 475 up to ridicule a hundred years from now. I would not be surprised if a brilliant reader of this book finds a more elegant formulation of what we now call gauge theories. Look at the development of the very first field theory, namely Maxwell’s theory of electromagnetism. By the end of the nineteenth century it had been thoroughly studied andthe overwhelming consensus was that at least the mathematical structure was completelyunderstood. Yet the big news of the early twentieth century was that the theory, surprisesurprise, contains two hidden symmetries, Lorentz invariance and gauge invariance: twosymmetries that, as we now know, literally hold the key to the secrets of the universe. Mightnot our present day theory also contain some unknown hidden symmetries, symmetrieseven more lovely than Lorentz and gauge invariance? I think that most physicists wouldsay that the nineteenth-century greats missed these two crucial symmetries because oftheir lousy notation 2and tendency to use equations of motion instead of the action. Some of these same people would doubt that we could significantly improve our notation andformalism, but the dotted-undotted notation looks clunky to me and I have a naggingfeeling 3that a more powerful formalism will one day replace the path integral formalism. Since the point of good pedagogy is to make things look easy, students sometimes do not fully appreciate that symmetries do not literally leap out at you. If someone had writtena supersymmetric Y ang-Mills theory in the mid-1950s, it would certainly have been a longtime before people realized that it contained a hidden symmetry. So it is entirely possiblethat an insightful reader could find a hitherto unknown symmetry hidden in our well-studied field theories. It is not just a matter of clearer notation and formalism that caused the nineteenth- century greats to miss two important symmetries; it is also that they did not possessthe mind set for symmetry. The old paradigm “experiments →action →symmetry” had to be replaced 4in fundamental physics by the new paradigm “symmetry →action → experiments,” the new paradigm being typified by grand unified theory and later by stringtheory. Surely, some future physicists will remark archly that we of the early twenty-firstcentury did not possess the right mind set. In physics textbooks, many subjects have a finished completed feel to them, but not quantum field theory. Some people say to me, what else is there to say about field theory?I would like to remind those people that a large portion of the material in this book wasunknown 30 years ago. Of course, while I feel that further developments are possible, Ihave no idea what—otherwise I would have published it—so I can’t tell you what. But let memention two recent developments that I find extremely intriguing. (1) Some field theoriesmay be dual to string theories. (2) In dimensional deconstruction a d-dimensional field theory may look (d+1)-dimensional in some range of the energy scale: the field theory can literally generate a spatial dimension. These developments suggest that quantum field 2It is said, and I agree, that one of Einstein’s great contributions is the repeated indices summation convention. T ry to read Maxwell’s treatises and you will appreciate the importance of good notation. 3I once asked Feynman how he would solve the finite square well using the path integral. 4A. Zee, Fearful Symmetry, chap. 6. 476 | Closing Words theories contain considerable hidden structures waiting to be uncovered. Perhaps another golden age is in store for quantum field theory. So boys and girls, the parade is over, and now it’s up to you to get another parade going.5 Appendix An image presented to the visual system can be described as a 2-dimensional Euclidean field ϕ0(x), with ϕ0 representing the gray scale from black ( ϕ0=− ∞ )to white ( ϕ0=+ ∞ ). [You can see that color might be included by going to a field /vectorϕtransforming under some internal SO( 2)group for example.] The image actually perceived, ϕ(x) , is the actual image ϕ0(x) distorted to ϕ0[y(x) ] plus some noise η(x) . Distortion is described by a map x→y(x) of the 2-dimensional Euclidean plane. Your brain’s task is to decide whether the actual image is ϕ0(x) or some other ϕ1(x). Your ability to discriminate between images depends on the functional integral Z=/integraldisplay Dy(x)/integraldisplay Dη(x)e−W [y(x) ]−(1/2C)/integraltext d2xη(x)2δ{ϕ0[y(x) ]+η(x)−ϕ(x)} (1) =/integraldisplay Dy(x)e−W [y(x) ]−(1/2C)/integraltext d2x{ϕ(x)−ϕ 0[y(x) ]}2 where for simplicity I have taken the noise, measured by the parameter C, to be Gaussian and white. The weighting function W[y(x) ] is presumably hard wired by evolution into our visual system, telling us that certain distortions (translations, rotations, and dilations) are much more likely than others. Writing y(x)=x+A(x) we note that Zdefines a field theory of the 2-component field Ai(x), which can always be written as Ai=∂iη+εij∂jχ. Note that the field Ai(x) appears “inside” an “external” field ϕ0. From symmetry considerations we might argue that W=−/integraldisplay d2x/parenleftbigg1 g2η∂6η+1 f2χ∂6χ/parenrightbigg with two coupling constants fandg. I can give here only the briefest of sketches and refer the interested reader to the literature.6Clearly, one can think of other examples. This particular example serves only to show that there are many more field theories than those described in standard texts. 5As the Beatles said, quantum fields forever! 6W . Bialek and A. Zee, Statistical mechanics and invariant perception, Phys. Rev. Lett. 58: 741, 1987; Under- standing the efficiency of human perception, Phys. Rev. Lett. 61: 1512, 1988. Part N While quantum field theory was discovered and developed in the twentieth century, I will introduce in this part, added in the second edition of this text, some topics that have beenworked out in the twenty-first century. At the rate these topics are rapidly evolving, I maybe quite foolish to include them here. But I am taking the plunge as I think that I wouldserve my readers better by letting them have a taste of the twenty-first century ratherthan expanding on the twentieth. More likely than not, by the time this second editionis published, there will be better ways of treating the material contained here. You shouldread part N in this spirit and regard what is given here as an entry key to a fast growing 1 research literature. 1Indeed, by the time the manuscript was copyedited (April 2009) it had been discovered that the amplitudes discussed in chapters N.2–4 could be written even more simply using a twistor and dual twistor formalism. See p. 494 and N. Arkani-Hamed, P. Cachazo, C. Cheung, and J. Kaplan, arXiv:09032110. This page intentionally left blank N.1 Gravitational Waves and Effective Field Theory An unfinished symphony One astounding prediction of Einstein gravity is the existence of ripples crisscrossing the fabric of spacetime, what one writer refers to as Einstein’s unfinished symphony.1Massive detectors have been built, with more to come, in a human “curious George” effort to tunein to the song of the cosmos. Consider a black hole of size r S=2Gm (its Schwarzschild radius—see chapter I.10, with mits mass) a distance rOfrom another black hole, moving with velocity v. As the black holes spiral into each other they emit gravitational waves, with a characteristic wavelengthdetermined by the orbital period λ=2πr O/v. Thus the physics contains three distance scales: rS,rO, andλ. We will stay within the simple post Newtonian regime rS/lessmuchrO/lessmuchλ. T oward the end, as rO/similarequalrSandv/similarequal1, relativistic effects rear their nasty heads, from which we will prudently stay away. In the Closing Words to the first edition of this text, I mention that one intriguing development over the last few decades has been the use of effective field theory to describesituations involving more than one energy scale (or equivalently, length and time scales).The physics at the high energy scale Mis then represented in the low energy effective Lagrangian by higher dimensional terms, suppressed by powers of Mbut constrained by the symmetries we know. Examples abound in this book, from the quantum Hall effectto surface growth to proton decay. The latter provides a classic example: while we professignorance of the physics responsible for proton decay, we can nevertheless make usefulpredictions by adding 4-fermion interactions invariant under the low energy gauge groupSU( 3)⊗SU( 2)⊗U(1), as shown in chapter VIII.3. An interesting recent development is an elegant description of the emission of gravi- tational waves by inspiraling black holes using effective field theory. Here, in contrast toproton decay, we actually know the short distance physics involved. Effective field theory 1M. Bartusiak, Einstein ’s Unfinished Symphony. 480 | Part N nevertheless offers an efficient and sensible way to organize and compartmentalize physics on the various distance scales. We will merely touch upon one aspect of this approach. Finite size objects in general relativity SincerS/lessmuchrO, the leading approximation would be to treat the black hole as a point particle using the action (I.11.12) Spp=−m/integraltext dτ=−m/integraltext/radicalbiggμνdXμdXν=−m/integraltext dτ/radicalBig gμν˙Xμ˙Xν, where ˙Xμ=dXμ/dτ . Let us now include the corrections due to the finite size of the black hole. As you will see, the following discussion actually applies not only to black holes butto any finite sized object, including you. In the spirit of effective field theory, we add to S pphigher dimensional terms to be formed out of the point particle degree of freedom ˙Xμand the ambient gμνthe particle moves in, subject to local coordinate invariance, of course. The invariant tensors we can form out ofg μνare, to leading order, the scalar curvature R, the Ricci curvature Rμν, and the Riemann curvature tensor Rμλνρ . You might start with the scalar curvature and the Ricci curvature, and add to Sppthe terms Sdrop=/integraltext dτ(cSR(X) +cRRμν(X)˙Xμ˙Xν). The curvatures R(X) andRμν(X) are evaluated on the worldline Xμ(τ) of the particle of course. Einstein’s equation of motion Rμν−1 2gμνR=0 implies Rμν(X)=0 and thus also R(X) =0. Following the discussion in chapter VIII.3, we now show that, as we might intuitively feel, we are allowed to drop Sdrop. For our problem we have the total action S=SEH+Spp+Sdrop with the Einstein-Hilbert action (VIII.1.1) SEH=/integraltext d4x√−gM2 PR. Under a field redefinition gμν→gμν+δgμν, δSEH=/integraldisplay d4x√−gM2 P(Rμν−1 2gμνR)δg μν (1) Note that was how we would have derived the equation of motion for gravity, by varying gμν. Here we are not making an arbitrary variation, but rather our goal is to choose a specificδg μνso that the resulting δSEHnegates Sdrop. Since Sdrop consists of an integral over the worldline of the particle while δSEHis given by an integral over spacetime, we need a delta function in δgμνto switch from one kind of integral to another. The choice δgμν(x)=1/radicalbig −g(x)M2 P/integraldisplay dτδ4(x−X(τ)) [agμν(X)+b˙Xμ˙Xν] (2) gives δS=/integraldisplay dτ(−aR(X) +b/bracketleftbigg Rμν(X)−1 2gμνR(X)/bracketrightbigg ˙Xμ˙Xν) (3) So, with some appropriate values of aandb, we can indeed cancel off2Sdrop, thus vindicating our intuition that the particle does not feel the Ricci and scalar curvaturesfor the obvious reason that they vanish. 2A technicality: field redefinition also induces a contact interaction of the form/integraltext dτ( 1/√−g)δ4(X1(τ)− X2(τ)) between the two massive objects. Going from field theory to the point particle description represents a conceptual step backward, so that we should expect delta function effects at the location of the point particles. N.1. Gravitational Waves | 481 How about terms we can construct out of the Riemann curvature tensor Rμλνρ ? For your convenience, I list its symmetry properties3here:Rτρμν=−Rτρνμ=−Rρτμν ,Rτρμν= Rμντρ , andRτρμν+Rτμνρ+Rτνρμ=0. Thus, due to the antisymmetry, we are not able to contract all four indices of Rμλνρ with˙Xμ. We can contract at most two indices, to form the two objects Eμν(X)≡Rμλνρ(X)˙Xλ˙XρandBμν(X)≡˜Rμλνρ(X)˙Xλ˙Xρ, where ˜Rμλνρ(x)≡1 2√−gεμλσηRση νρ(x) denotes the dual of the curvature tensor. These are 2-indexed tensors and we need to square them to form scalars to put into the action. Hence, to next order the particle action becomes Sp=/integraldisplay dτ(−m+cEEμνEμν+cBBμνBμν+...) (4) Note that the unknown constants cEandcBhave dimension of inverse mass cubed. Before we explore the physical content of this effective action, let us understand the meaning of EandBby retreating to the more familiar case of a point particle mov- ing in an electromagnetic field Fμν(in flat space). Form Eμ≡Fμν˙XνandBμ≡˜Fμν˙Xν, where ˜Fμν(x)≡1 2εμνσηFση. Going to the rest frame of the particle, where ˙X0=1 and ˙Xi=0, we see that, as the notation suggests, this is just the familiar decomposition of electromagnetism into electricity and magnetism. Similarly, EμνandBμνrepresent the decomposition of curvature into its “electric” and “magnetic” components. In chapter I.11 we varied the first term in (4) to obtain the standard geodesic equation that is at the heart of Einstein’s theory. Here we obtain d2Xρ dτ2+/Gamma1ρ μν(X(τ))dXμ dτdXν dτ=fρ(X(τ)) (5) where fμ(X(τ)) comes from varying the EandBterms in (4). A finite sized body experiences a tidal force fμdue to the varying gravitational force acting on it. It no longer follows a geodesic. The fact that we had to square the electric and magnetic components of the curvature to form the effective action (4) means that the effects of these correction terms are highlysuppressed. Since Riemann curvature contains two derivatives, the correction terms in-volve four derivatives. T o estimate the magnitude of c EandcBwe exploit a rather cute argument as follows. Consider the scattering of a graviton off this point particle (which, remember, is a black hole in the problem we are studying) generated by the couplings in (4): iM∼ ...+icE, Bω4/M2 P+... where ωdenotes the energy of the graviton. The powers of ω follows from the four derivatives just mentioned. (If you don’t understand the powersofM Pyou need to read chapter VIII.1 again.) Here cE, B denotes the two unknown couplings cE∼cBgenerically. Imagine calculating the total scattering cross section for a graviton on a black hole. Squaring the amplitude Metc., we would end up with σ(ω)∼ ...+c2 E, Bω8/M4 P+.... 3S. Weinberg, Gravitation and Cosmology, p. 141. 482 | Part N This treatment of the black hole as a point particle is only valid for ωrS/lessmuch1 of course. The (...)iniMrepresents diagrams we have not included, for example, the one originating from the first term in (4) (namely the term responsible for keeping us down to earth!). Anice feature of the argument I am about to give is that we don’t even need to know whatthe terms in (...)are. On the other hand, we argue that by dimensional analysis the cross section must have the form σ(ω)=r 2 Sf( ωr S)since the only length scale in the Schwarzschild metric is rS. Expanding the unknown function f( ωr S)in powers of its argument we have σ(ω)= ...+αω8r10 S+...withαsome constant. (A technical aside: the massless graviton could produce infrared factors like log ωrS, which we ignore for our purposes.) Requiring that the two expressions agree, we obtain cE, B∼M2 Pr5 S. Indeed, as expected, the couplings cE, Bare highly suppressed as rS→0. Exercise N.1.1 Using considerations similar to those in the text, show that the scattering cross section for a photon of frequency ωon an atom or a molecule vanishes like ω4asω→0, a result which, as mentioned in chapter VIII.3, underlies the well-known explanation of why the sky is blue. N.2 Gluon Scattering in Pure Yang-Mills Theory Boil and toil with Feynman diagrams You might think that after some 50 years, there could not possibly be any novelty in calculating Feynman amplitudes. But you would be wrong. Over the last dozen years or so,and largely since the first edition of this text, a group of intrepid searchers have found someamazingly powerful methods of tackling Feynman diagrams. As I said at the beginning ofchapters VIII.4 and 5, I can only give you an introduction to this subject, telling you justenough for you to explore this fast-growing literature. T o best appreciate this new development, you should do a little calculation before reading further. Consider pure Y ang-Mills theory, by consensus the nicest field theory we have,simple to write down and perfumed with symmetries. Not even any fermions around tomess things up. Call the gauge bosons gluons for convenience. Now calculate 5-gluonscattering at tree level as shown in figure N.2.1. No loops, just trees. The Feynman rulesare given in chapter VII.1 and also in appendix C. You really must calculate before reading on. I will wait for you. You think to yourself, this is easy, just a bunch of tree diagrams. In fact, to make it easier, put all the externalgluons on-shell, that is, set p 2 i=0,i=1, 2, ... ,5 . This calculation is not merely an idle exercise, but is in fact phenomenologically im- portant. At an accelerator such as the soon-to-be-operational Large Hadron Collider, two /H11001 /H11001 /H11001... Figure N.2.1 484 | Part N Result of a brute force calculation (actually only a small part of it): k1 /H11080 k4/H92552 /H11080 k1/H92551 /H11080 /H9255 3/H92554 /H11080 /H9255 5 Figure N.2.2 protons are smashed together at high energies. Two gluons, one from each proton, collide and produce three gluons, which then materialize into three jets of hadrons. Because ofasymptotic freedom, at high energies the effective coupling gbecomes small enough for perturbative field theory to be relevant, and the tree amplitude you are busily calculatingprovides a key ingredient for the phenomenological models used to study the experimentalmeasurements. Time’s up! A small part of the answer is shown in figure N.2.2, taken from a lecture by Zvi Bern. 1You really should take a look in order to appreciate, to be grateful for even, the formalism to be explained in this chapter. You know that the amplitude is linear in eachof the five polarization vectors /epsilon1 i. The 3-gluon vertex (VII.1.11) is linear in momentum and there are three of them in a typical diagram. Thus a typical term in the numerator of the Feynman amplitude would be, as shown in the figure, p1.p4/epsilon12.p1/epsilon11./epsilon13/epsilon14./epsilon15.A rough estimate shows that there are almost 10,000 such terms. That’s why, in spite of myadmonition, you didn’t finish the calculation before reading ahead. Incidentally, you couldsee that even 4-gluon scattering at tree level, though doable by hand, is rather involved. 1Z. Bern, “Magic T ricks for Scattering Amplitudes,” http://online.itp.ucsb.edu/online/colloq/bern1/pdf/ Bern1.pdf. N.2. Gluon Scattering | 485 New technology for Feynman diagrams In practice, phenomenologists studying jet production have developed elaborate computer codes based on numerical recursion and these prove to be quite efficient. In this introduc-tory text, however, we are not after numerical efficiency but a deeper understanding of thestructure of multi-gluon amplitudes. I have set you up so that, surely, after your abortiveattempt to calculate the 5-gluon amplitude you now fully appreciate the need for new waysof approaching Feynman diagrams. I will now explain some of the novel methods peoplehave invented over the last 15 years or so. A relatively simple first step is to strip the color off the amplitude. Evidently, it is much better to use the matrix notation of (IV .5.16) than the index notation of (IV .5.17). Insteadof the structure constants f abcand their products in the “indexed” Feynman rules in (VII.1.11–13) we have colored Feynman rules (read appendix 1 now) with objects liketr(T a[Tb,Tc])and tr ([Ta,Tb][Tc,Td])for the cubic and quartic coupling vertex, respec- tively, where Tadenotes the matrix representing the suitably normalized generators of the gauge group. Denote the color matrices carried by the external gluons by Ta1,Ta2,...Tan (withn=5 in the example you failed to do). In calculating a multi-gluon scattering amplitude in tree approximation, you would find each term multiplied by the product of a bunch of color traces, such as tr (TeA)tr(TeB) where AandBdenote products of T’s. Here the index eis carried by a virtual gluon and hence is summed over. We now use the group theoretic identity (with esummed over) for the gauge group SU(N) [recall (IV .5.19)] (Te)i j(Te)kl=1 2/parenleftbigg δi lδk j−1 Nδi jδk l/parenrightbigg (1) (Here e=1,...,N2−1 and the indices i,j,k,l=1,...,N, of course.) The second term takes care of the traceless condition tr Te=0. However, we can drop it, since if we extend the gauge group to U(N) , that extra gluon does not couple to the other gluons anyway. Thus tr(TeA)tr(TeB)=1 2tr(AB). Repeating this procedure, we reduce the product of traces to a single trace of nTa’s multiplied together in some specific order. Indeed, the astute reader will have noted that had we used the double-line formalism of figure IV .5.2, this entire discussion would not even have been necessary. As we also sawin chapter VII.4, the double-line formalism does offer many advantages. The other simplifying step is to specify the helicity of the gluon instead of writing the amplitude in terms of polarization vectors. You recall, from way back when, that amassless spin 1 particle moving along the third-direction k=ω(1, 0, 0, 1 )can have helicity h=+ , corresponding to the polarization vector /epsilon1=1/(√ 2)(0, 1, i,0), or helicity h=− , corresponding to the polarization vector /epsilon1=(1/√ 2)(0, 1, −i,0). We specify the external gluons by momentum, helicity, and color: (p1,h1,a1,p2,h2,a2,... ,pn,hn,an). Thus we can write the n-gluon amplitude as M=i/summationdisplay permutationstr(Ta1Ta2...Tan)A(1, 2, ... ,n) (2) 486 | Part N Following the literature, we have compressed the notation further and denote {pi,hi}byi. The sum is over all possible permutations of the ngluons. We can now focus on the “color stripped” amplitude A(1, 2, ... ,n). First, a triviality. It is convenient to treat the gluons as all outgoing (or if you prefer, as all incoming), so that/summationtext ipi=0 and the time component of some of the momenta can be negative. We can then obtain the physically desired amplitude by crossing. Keep in mindthat under crossing p→−pand/epsilon1→/epsilon1 ∗, that is, the helicity flips. The spinor helicity formalism Now we are ready to return to the expression in figure N.2.2. The technical term for this expression is an “unholy mess.” It turns out that the key to unraveling this hopeless morasscan be found in exercise II.3.1: that the Lorentz vector sits in the representation ( 1 2,1 2)and thus can be constructed as a product of two spinors, one from the representation (1 2,0), the other from (0,1 2). You did the exercise, didn’t you? So you know how to write, for example, the momentum vector pμas a product of two spinors. T o go on, you should also read appendices B and E. I am now ready to explain the spinor helicity formalism designed to exploit this peculiar property of the Lorentz vector. Or, to say it a bit more mysteriously, I am going to showyou how to take the square root of the momentum. Now you appreciate the power of the undotted-dotted notation introduced in appendix E. The undotted index goes with ( 1 2,0), and the dotted with (0,1 2). We are looking for an object transforming like (1 2,0)⊗(0,1 2)to represent a vector. The problem can then be stated as follows: instead of writing momentum as pμ, we want to write it as pα˙α, an object carrying an undotted and a dotted index, namelya2b y2matrix in cruder language. We merely have to flip through appendix E and look for an object carrying the desired indices. There it is, (σμ)α˙α, and indeed, its μindex is begging to be contracted with pμ. Thus, with no further work, we can write [since σμ=(I,/vectorσ)] pα˙α≡pμ(σμ)α˙α=(p0I−piσi)α˙α=/parenleftBigg(p0−p3)−(p1−ip2) −(p1+ip2)( p0+p3)/parenrightBigg α˙α(3) We have succeeded in writing the momentum a sa2b y2matrix. You may recognize this as nothing but the matrix XM(with some trivial change in notation) used in appendix B to construct the covering of SO( 3, 1)bySL( 2,C). Given two vectors pandq, their scalar product is given by p.q=εαβε˙α˙βpα˙αqβ˙β (4) which you can check explicitly, writing the right-hand side as a trace and once again using σ2σT i=−σiσ2, as we did in appendix E. For q=p, this reduces to p.p=εαβε˙α˙βpα˙αpβ˙β= detp; here we recognized a definition of the determinant. [Of course, you could also eval- uate the determinant of (3) by inspection, or recall that this was also used in appendix B.] N.2. Gluon Scattering | 487 Clearly, there is an unavoidable notational overload: the single letter pdenotes both the vector and the matrix, but you should be able to tell from the context which object is beingreferred to. Here we are going to apply this formalism to massless gluons with lightlike momenta. Things simplify considerably: for plightlike, det p=0 and thus the matrix pgenerically has one 0 eigenvalue. (In fancy talk, the matrix has rank 1 rather than 2.) From elementarylinear algebra we recall thata2b y2matrix mof rank 1 can always be written as m ij=viwj, withvandwtwo 2-component vectors, (obviously since the vector orthogonal to wprovides the 0 eigenvector.) Thus, for a lightlike vector, we can write pα˙α=λα˜λ˙α (5) in terms of two 2-component spinors λand˜λ. For physical momentum, the components pμare real, of course. I invite you to verify, however, that everything we just did from (3) to (5) goes through even if pμare complex. It turns out that in the next chapters we will find it convenient to consider complexmomentum. Upon first exposure, the formalism appears quite opaque, but actually, like a lot of formalisms, it is fairly simple or perhaps even trivial. If you are confused at any point inthe following exposition, just work things out explicitly. For example, consider a physicalmomentum with p 0=E> 0. With no loss of generality, you can choose /vectorpto point along the third direction, so that (with a trivial abuse of notation p=|/vectorp|) p=/parenleftBiggE−p 0 0 E+p/parenrightBigg which for plightlike collapses to the rank 1 matrix p=2E/parenleftBigg00 01/parenrightBigg =2E/parenleftBigg0 1/parenrightBigg (01) Thus, in this case, λand˜λare both equal to √ 2E/parenleftBigg0 1/parenrightBigg numerically. (T o make sure you get it, work this out for /vectorppointing in some other direction.) You can think of the Pauli spinors λand˜λas the “square root” of the Lorentz vector pμ. Note how the group theory discussion in chapter II.3 foreordained this rather nontrivialpossibility. After all, there we saw how a Lorentz vector can be constructed out of two Diracspinors uandu /prime. Interestingly, in discussing ferromagnets and antiferromagnets in chapter VI.5, we used a poor man’s version of (3), namely /vectorn=z†/vectorσz. You learned in school that the ordinary square root has a sign ambiguity. Analogously, in (5)pdoes not determine λand˜λuniquely. We can always rescale λ→uλand˜λ→1 u˜λ for any complex number u. (You might have wondered what fixed the overall constant in λand˜λin the simple example above: I made an arbitrary choice.) 488 | Part N For real momentum, the matrix pα˙α=pμ(σμ)α˙αis hermitean, which implies that ˜λ=λ∗ is the complex conjugate of λ. The spinor ˜λis not independent of λ, and so the rescaling parameter uis restricted to be a phase factor eiγ. [Also, recall from appendix B how XM transforms under SL( 2,C) and you will see that it is all consistent.] In this case, the condition that phas rank 1 allows for two solutions: pα˙α=±λα˜λ˙α, with the two possible signs corresponding to whether p0>0 or not. A side remark at this point: We will see that it is useful to consider the group SO( 2, 2) instead of the Lorentz group SO( 3, 1). Thus, as the discussion in appendix B indicates, you can also take the square root of an SO( 2, 2)vector and write pα˙α=λα˜λ˙α, but with λ and˜λtwo independent real spinors, as is consistent with the local isomorphism between SO( 2, 2)andSL( 2,R)⊗SL( 2,R). The rescaling mentioned above is now restricted to u being a real number. It is instructive to count the number of real degrees of freedom for these different cases. A complex lightlike momentum depends on 4 ×2−2=6 real numbers, since the condition p2now amounts to two real conditions, while λand˜λeach contains 2 complex numbers, but with rescaling we are left with 2 ×2−1=3 complex numbers, that is, 6 real numbers. A real lightlike momentum depends on 4 −1=3 real numbers, but now ˜λis tied to λcontaining 2 complex numbers, which get reduced to 3 real numbers after rescaling by a phase factor. For a (real) lightlike vector transforming under SO( 2, 2),w e have 2 real spinors, which after rescaling contains 3 real numbers. So it all works out, ofcourse. I mention all this here for future use. It should be evident to you, for the rest of this chapter, which statements hold for complex momenta and which hold only for realmomenta. At the end of the day, when we arrive at a physical quantity, such as theamplitude, we will of course set the momenta contained therein to be real. For two lightlike vectors pandq, write p α˙α=λα˜λ˙αandqα˙α=μα˜μ˙α, then we have p.q=(εαβλαμβ)(ε˙α˙β˜λ˙α˜μ˙β)≡/angbracketleftλ,μ/angbracketright[˜λ,˜μ] (6) Here we have defined the two Lorentz invariants /angbracketleftλ,μ/angbracketright≡εαβλαμβ=− /angbracketleftμ,λ/angbracketright (7) and [˜λ,˜μ]≡ε˙α˙β˜λ˙α˜μ˙β=− [˜μ,˜λ] (8) (treating the spinors as c-number objects.) Note in passing that with our convention, λ1=λ2andλ2=−λ1, and so /angbracketleftλ,μ/angbracketright=− λ1μ2+λ2μ1=−εαβλαμβ. We have already verified in (E.13) that /angbracketleftλ,μ/angbracketrightis invariant, but for the sake of total pedagogical clarity let us check it once more, this time using infinitesimal transformations.Write (E.4) more compactly as δλ α=σβ αλβ, where σdenotes some linear combination of Pauli matrices. Noting that /angbracketleftλ,μ/angbracketrightis nothing but λσ2μup to some irrelevant overall constant, we have indeed δ(λσ2μ)=(λσTσ2μ+λσ2σμ)=0. A notational remark: the twiddles in [ ˜λ,˜μ] are redundant. The square bracket is defined only for spinors transforming like (0,1 2). Henceforth, we will write [ λ,μ]≡ε˙α˙β˜λ˙α˜μ˙β. N.2. Gluon Scattering | 489 For real physical momenta, ˜λ=λ∗so that /angbracketleftλ,μ/angbracketright=[ λ,μ]∗. Then p.q=/angbracketleftλ,μ/angbracketright[λ,μ] implies that /angbracketleftλ,μ/angbracketright=√p.qeiφand [λ,μ]=√p.qe−iφ, with some phase factor eiφ.W e thus conclude that the two spinorial products may be regarded as the (two) square rootsof the Lorentz dot product p.qup to a phase factor. You could now raise an interesting question: how do we write the polarization vectors /epsilon1(p) of a massless gluon? The requirement that /epsilon1(p) .p=0 can be satisfied, according to (4), by setting /epsilon1 α˙α= d−1λα˜μ˙α, for an arbitrary ˜μ˙αand with the factor ddetermined as follows. We require that, for an arbitrary complex number w, scaling ˜μ→w˜μdoes not change /epsilon1(since ˜μis arbitrary after all). Thus dhas to be linear in ˜μ. The further requirement that dbe Lorentz invariant implies, as we just learned, that d=[x,μ], where ˜xis some (0,1 2)spinor. The only spinor available is ˜λand hence we obtain /epsilon1− α˙α=λα˜μ˙α [λ,μ](9) By convention, we will call this polarization negative helicity. The arbitrary choice of ˜μ˙αrepresents the freedom inherent in a gauge theory. Indeed, we see that gauge transformation corresponds to the spinorial shift ˜μ→˜μ+y˜λ(for some arbitrary number y) under which /epsilon1α˙α→/epsilon1α˙α+yλα˜λ˙α, which translates into the usual shift of/epsilon1by some multiple of p. The positive helicity polarization is given by the other possible choice /epsilon1+ α˙α=μα˜λ˙α /angbracketleftμ,λ/angbracketright(10) Check that it works. Gauge transformation now corresponds to the shift μ→μ+yλ. Note that the polarization vectors are normalized as /epsilon1+./epsilon1−=/angbracketleftμλ/angbracketright[μλ]/(/angbracketleftμλ/angbracketright[ μλ])=1. Taming the unholy mess Consider the tree-level scattering amplitude with n≥4 outgoing massless gluons. (In this and the next sections, we can take all momenta to be real.) The color-stripped amplitudeis then characterized by a string of helicities (h 1,...,h n). T ake for example the amplitude with(+++ ...++). Upon crossing, it describes two gluons, each with helicity −, going inton−2 gluons all with helicity +. Both incoming gluons flip their helicity and thus this amplitude is said to be maximal helicity violating. Your intuition may tell you thatthis amplitude ought to be suppressed, since highly energetic massless particles tend tomaintain their helicities. If you try to verify this using traditional Feynman diagrams, youwould once again encounter a big mess. The spinor helicity formalism rides to the rescue. Consider the amplitude A(h 1,...,hn). For each of the ngluons, we have piα˙α=λiα˜λi˙α, and an arbitrary spinor that we are free to choose (subject to some conditions), namely either μiαor˜μi˙α, depending on whether the corresponding helicity is +or−, respectively. There are quite a few indices, but fortunately, in computing amplitudes, we encounter only Lorentz invariants, such as 490 | Part N /epsilon1i./epsilon1j=εαβε˙α˙β/epsilon1iα˙α/epsilon1jβ˙β(be sure to distinguish between the two varieties of epsilon here!), and thus the spinor indices will be contracted over and disappear. In particular, we have(omitting the comma in the angled and square brackets) /epsilon1+ i./epsilon1+ j=/angbracketleftμiμj/angbracketright[λiλj] /angbracketleftμiλi/angbracketright/angbracketleftμjλj/angbracketright(11) /epsilon1− i./epsilon1− j=/angbracketleftλiλj/angbracketright[μiμj] [λiμi][λjμj](12) /epsilon1− i./epsilon1+ j=/angbracketleftλiμj/angbracketright[μiλj] [λiμi]/angbracketleftμjλj/angbracketright(13) We also list for convenience /epsilon1+ i.pj=/angbracketleftμiλj/angbracketright[λiλj] /angbracketleftμiλi/angbracketright(14) and /epsilon1− i.pj=/angbracketleftλiλj/angbracketright[μiλj] [λiμi](15) Evidently, in this formalism, flipping helicity corresponds to interchanging the brackets /angbracketleft.../angbracketrightand [ ...]. We need one more important observation. Obviously, in a tree-level diagram for n-gluon scattering, you cannot have as many 3-gluon vertices as you like. Draw the tree diagramsforn=4 for example (see figure N.2.3). The number of 3-gluon vertices could be either 0 or 2. In general, the number of 3-gluon vertices can be at most n−2. You are asked to verify this in exercise N.2.2. As remarked earlier, while the 4-gluon vertex does not involvemomentum, the 3-gluon vertex is linear in momentum. Thus, in the numerator of theFeynman amplitude, we have npolarization vectors /epsilon1 ibut at most n−2 momenta. We are to form a scalar out of these Lorentz vectors by taking dot products. Clearly, there are atleast two polarization vectors who have to dance with each other. Therefore we concludethat the tree amplitude must contain at least one power of /epsilon1 i./epsilon1j. (In the n=5 case that power was actually 2, as we saw.) Now we are ready to rock. For the amplitude A(++ ...+)(suppressing the momentum labels), we simply choose the spinors μirepresenting the gauge degrees of freedom to all be equal. Then all dot products /epsilon1+ i./epsilon1+ jbetween polarization vectors vanish according to (11). But we just argued that the tree amplitude must contain at least one power of /epsilon1i./epsilon1j. Remarkably, we have shown that the maximal helicity-violating amplitude vanishes for any n! Our intuition suggested that these amplitudes are suppressed, but in fact they vanish. What about the next-to-maximal helicity-violating amplitudes with one negative helicity, namely A(−++ ...++)? Label the gluon with negative helicity as 1. Once again, for i=2,...n, choose μiall equal to λ1. Then /epsilon1+ i./epsilon1+ j∝/angbracketleftμiμj/angbracketright=0, for i,j/negationslash=1. Furthermore, /epsilon1− 1./epsilon1+ i∝/angbracketleftλ1μi/angbracketright=/angbracketleftλ1λ1/angbracketright=0 for i/negationslash=1. The amplitude A(−++ ...++)also vanishes! Clearly, this “cheap” trick of exploiting gauge freedom no longer works for the next amplitude with two negative helicities. T o see why the trick does not work any more, lookatA(−−+ ...++)for instance. Once again we could, for i=3,...n, choose μ iall equal N.2. Gluon Scattering | 491 (a)3 2 4 13 2 4 13 2 4 1 (b) (c) Figure N.2.3 so that /epsilon1+ i./epsilon1+ j=0 fori,j≥3, but then we don’t have enough freedom to make all the other polarization dot products vanish. In fact, at some point, we better have some nonvanishingamplitudes. In the literature, these amplitudes with two negative helicities are calledmaximal helicity-violating amplitudes. Upon crossing two of the gluons, they describetwo gluons producing n−2 gluons, with helicities + +→++ ...+,− +→−+ ...+, and− −→−−+ ...+. Explicit calculation of A(1,2,3,4) Then=4 case is the simplest. T ake a deep breath and try to calculate A(1−,2−,3+,4+) andA(1−,2+,3−,4+). For 4-gluon scattering these two are the only nonvanishing tree amplitudes, since by parity the amplitudes with three minuses are related to the amplitudeswith three pluses (which we know vanish), and so on. The bad news is that the calculation is fairly involved. The good news is that we can still exploit gauge freedom mercilessly and that the final answer is surprisingly simple. T ackle A(1 −,2−,3+,4+)first. The relevant diagrams are shown in figure N.2.3. Let us simplify the notation as much as possible: write /angbracketleft12/angbracketright=/angbracketleftλ1λ2/angbracketright, [12] =[λ1λ2], and so forth. Now we need the colored Feynman rules in the form given in appendix 1. In line with the preceding discussion let us choose ˜μ1=˜μ2=˜λ3andμ3=μ4=λ2. Then all but one of the polarization dot products vanish. For instance, /epsilon1− 2./epsilon1+ 3∝/angbracketleftλ2μ3/angbracketright[μ2λ3]∝/angbracketleftλ2λ2/angbracketright=0. The only nonzero product is /epsilon1− 1./epsilon1+ 4=/angbracketleftλ1μ4/angbracketright[μ1λ4]/([λ1μ1]/angbracketleftμ4λ4/angbracketright)=/angbracketleft12/angbracketright[34]/([13]/angbracketleft24/angbracketright), where the second equality follows from our gauge choice. This implies that the quarticdiagram N.2.3a vanishes, since it involves the product of two polarization dot products. We notice that there are only two more diagrams (fig. N.2.3b,c) rather than three. With the traditional Feynman rules there is a diagram with 1 and 3 on the same cubic vertex.Here we see another advantage of color stripping. We are looking at the coefficient oftr(T a1Ta2Ta3Ta4). The diagram we just described has Ta1next to Ta3and so does not contribute to this particular color ordering. Next, the diagram in figure N.2.3b vanishes. Look at the cubic vertex involving 2, 3, and v(for the virtual gluon): (/epsilon12./epsilon13/epsilon1v.p2+/epsilon13./epsilon1v/epsilon12.p3+/epsilon1v./epsilon12/epsilon13.pv), with /epsilon1vunderstood as a “placeholder” to be contracted with the /epsilon1∗ vfrom the other cubic vertex. The first term 492 | Part N vanishes because /epsilon12./epsilon13=0, the second term because /epsilon12.p3∝[μ2λ3]=[λ3λ3]=0, and the third term because /epsilon13.pv=−/epsilon13.(p2+p3)=−/epsilon13.p2∝/angbracketleftμ3λ2/angbracketright=/angbracketleftλ2λ2/angbracketright=0. Our gauge choice was wise indeed! Only one diagram (fig. N.2.3c) left to calculate. The cubic vertex (/epsilon11./epsilon12/epsilon1v.p1+/epsilon12./epsilon1v/epsilon11. p2+/epsilon1v./epsilon11/epsilon12.pv)is to be contracted with the other cubic vertex (/epsilon13./epsilon14/epsilon1∗ v.p3+/epsilon14./epsilon1∗ v/epsilon13. p4+/epsilon1∗ v./epsilon13/epsilon14.(−pv)). In each of these vertices, the first term vanishes, since the only nonzero polarization product is /epsilon11./epsilon14. T o obtain the amplitude we replace the polarization product /epsilon1ρ v/epsilon1ω∗ vfor the placeholder by the propagator −igρω/(p1+p2)2=−igρω/(2p1.p2). Again, since all but one of the polarization dot products vanish, only one term survives the contraction with gρω. We obtain A(1−,2−,3+,4+)=/epsilon11./epsilon14/epsilon12.p1/epsilon13.p4/p1.p2. Since we are after conceptual understanding more than anything else, now and hence- forth, in this and the next two chapters, we will suppress overall factors to keep variousexpressions as uncluttered as possible. We have already calculated /epsilon1 1./epsilon14, so it remains to evaluate /epsilon12.p1=/angbracketleft21/angbracketright[31]/[23], /epsilon13. p4=/angbracketleft24/angbracketright[34]//angbracketleft23/angbracketright, and p1.p2=/angbracketleft12/angbracketright[12]. Thus A=/angbracketleft12/angbracketright[34]2/([12][23] /angbracketleft23/angbracketright). We can now use various identities to write this in a more symmetric form. First, momen- tum conservation gives/summationtext ip(i) α˙α=/summationtext iλ(i) α˜λ(i) ˙α=0. Multiplying this by εβαε˙α˙γλ(j) β˜λ(k) ˙γwe ob- tain/summationtext i/angbracketleftji/angbracketright[ik]=0 for any jandk. Second, we have /angbracketleft34/angbracketright[34]=p3.p4=p1.p2=/angbracketleft12/angbracketright[12]. Finally, the spinors can be regarded as 2-dimensional vectors and so any two spinors μand νspan the space. Thus a third spinor λcan always be expanded as a linear combination of the other two, viz, λ=(/angbracketleftλν/angbracketrightμ−/angbracketleftλμ/angbracketrightν)/ /angbracketleftμν/angbracketright, with the coefficients determined easily by contracting with μandν. Contracting with a fourth spinor ηthen yields /angbracketleftλη/angbracketright/angbracketleftμν/angbracketright=/angbracketleftλν/angbracketright/angbracketleftμη/angbracketright−/angbracketleftλμ/angbracketright/angbracketleftνη /angbracketright (16) known as the Schouten identity. Using these identities we now massage Ainto shape. Multiply the numerator and denominator of Aby/angbracketleft34/angbracketrightto obtain /angbracketleft12/angbracketright2[34]/(/angbracketleft23/angbracketright/angbracketleft34/angbracketright[23]). Next, multiply the numerator and denominator by /angbracketleft12/angbracketright2. In the denominator write /angbracketleft12/angbracketright[23]=− /angbracketleft 14/angbracketright[43]. Finally, we obtain (suppressing overall phase factors, as promised) A(1−,2−,3+,4+)=/angbracketleft12/angbracketright4 /angbracketleft12/angbracketright/angbracketleft23/angbracketright/angbracketleft34/angbracketright/angbracketleft41/angbracketright=p1.p2 p2.p3(17) Compare this with figure N.2.2. You should be impressed, even though here we are doing then=4 rather than the n=5 case. Recall that we have another amplitude A(1−,2+,3−,4+)yet to calculate, in which the two negative-helicity gluons are not adjacent in color. You should work this out as anexercise, but it turns out that we can use a trick. Write the analog of (2) for 4-gluon scattering M=i/summationdisplay permutationstr(Ta1Ta2Ta3Ta4)A(1−,2+,3−,4+) (18) We have already remarked that if we extend the gauge group from SU(N) toU(N) , the extra gluon (known in the literature, perhaps confusingly, as the “photon”) does not coupleto the other gluons (because the couplings in Y ang-Mills theory all involve commutators;see appendix 1.) Thus if we replace, say T a2, by the identity matrix, the entire sum should vanish. The six terms in the sum then break up into two groups, multipled by either N.2. Gluon Scattering | 493 tr(Ta1Ta3Ta4)or tr(Ta1Ta4Ta3). Since the two traces are independent, the two groups vanish separately. The traces in the three terms tr (Ta1Ta2Ta3Ta4)A(1−,2+,3−,4+)+ tr(Ta1Ta3Ta2Ta4)A(1−,3−,2+,4+)+tr(Ta1Ta3Ta4Ta2)A(1−,3−,4+,2+)all become tr(Ta1Ta3Ta4). We thus obtain the so-called photon decoupling identity A(1−,2+,3−,4+) +A(1−,3−,2+,4+)+A(1−,3−,4+,2+)=0, relating the desired amplitude to two ampli- tudes already known from (17). Thus A(1−,2+,3−,4+)=−(A(1−,3−,2+,4+)+A(1−,3−,4+,2+)) =− /angbracketleft 13/angbracketright4/parenleftbigg1 /angbracketleft13/angbracketright/angbracketleft32/angbracketright/angbracketleft24/angbracketright/angbracketleft41/angbracketright+1 /angbracketleft13/angbracketright/angbracketleft34/angbracketright/angbracketleft42/angbracketright/angbracketleft21/angbracketright/parenrightbigg =/angbracketleft13/angbracketright4 /angbracketleft12/angbracketright/angbracketleft23/angbracketright/angbracketleft34/angbracketright/angbracketleft41/angbracketright(19) where we used the Schouten identity. Remarkably, the two amplitudes A(1−,2−,3+,4+)andA(1−,2+,3−,4+)have the same form. It is tempting to conjecture that for n-gluon scattering, the maximal helicity-violating amplitudes in which two of the gluons carry negative helicity and the rest positive helicityis given by the elegant expression (for n≥4) A(1+,2+,...j−,... ,k−...n+)=/angbracketleftjk/angbracketright4 /angbracketleft12/angbracketright/angbracketleft23/angbracketright/angbracketleft34/angbracketright.../angbracketleft(n−1)n/angbracketright/angbracketleftn1 /angbracketright(20) This conjecture was first put forward by Parke and T aylor and proved by Berends and Giele (using an off-shell recursion method and a precursor to the on-shell recursion method tobe explained in the next chapter.) We will prove it in the next chapter. Meanwhile, we note that one way of arguing for the conjecture’s validity is to verify that the proposed amplitude satisfies all the symmetry requirements. Besides Lorentzinvariance (obviously satisfied), amplitudes at tree level in a massless theory like pureY ang-Mills should also satisfy scale and conformal invariance. One interesting check is to count, for each i, the powers of λ iminus the powers of ˜λi. Call this quantity /Lambda1i. Then since momentum has the form ∼λ˜λ, it contributes 0 to /Lambda1i.I n contrast, for negative helicity /epsilon1− α˙α=λα˜μ˙α/[λ,μ]∼λ/˜λ. For positive helicity we have the opposite: /epsilon1+ α˙α=μα˜λ˙α//angbracketleftμ ,λ/angbracketright∼˜λ/λ. Thus we have /Lambda1i=− 2hi. We checked that indeed, in (20), we have /Lambda1i=2 fori=j,kand/Lambda1i=− 2 fori/negationslash=j,k. Keeping track of /Lambda1iduring the calculation also provides us with a useful check. Note that the n=5 scattering amplitude, which we started this chapter with, is completely determined, since there are only two independent nonzero amplitudes:A(1 −,2−,3+,4+,5+)andA(1−,2+,3−,4+,5+). Further developments The astonishing simplicity of (20) has sparked a surge of interest and further develop- ments. Here I will be content to mention some of them. Once the tree amplitudes are done, one can calculate loop amplitudes by using a more sophisticated version of the unitarity methods and of the Cutkosky cutting rulesof chapter II.8. Proceeding in this way, various authors have bootstrapped their way up to 494 | Part N multiloop amplitudes. While the actual computational labor can quickly get out of hand, it is still enormously less than the labor needed with traditional Feynman methods. What about the basic cubic vertex of the theory? We will work it out in appendix 2 and show that it fits nicely into the form in (20) with one important caveat. Surely you, the astute reader, feel that there must be some deep reason for the aston- ishing simplification from the mess in figure (1) to the elegant expression in (20). Indeed,tree amplitudes in gauge theories (and in gravity) turn out to be even simpler when writ-ten in terms of the twistors studied by Penrose decades ago. As this exciting development 2 occurred while this book was going to press, I have to balance my desire to make the bookas up-to-date as possible against pagination constraints. Thus I can provide here only anultra-concise (and hence perhaps somewhat cryptic) key to the literature, giving you nomore than a flavor of what is involved. Include the momentum conservation delta function with the amplitudes of the type studied here and define M(... ,λ i,˜λi,...)≡A(λ ,˜λ)δ(4)(/summationtextn j=1λj˜λj). Due to space con- straints, I will suppress the kinematic dependence of Mon all but the particle iand write simply M(λi,˜λi). Let us Fourier transform Min two possible ways (and overuse the letter Msomewhat): M(W i)=/integraldisplay d2λiexp(i˜μα iλiα)M(λ i,˜λi). (21a) and M(Z i)=/integraldisplay d2˜λiexp(iμ˙α i˜λi˙α)M(λ i,˜λi). (21b) where W≡(˜μ,˜λ)andZ≡(λ,μ)denote two 4 −component objects which may be re- garded for the time being as column “vectors.” The intent here is to transform Mse- quentially for i=1, 2, ...nusing either (21a) or (21b). Consider SO( 2, 2)here instead of SO( 3, 1), so that the spinors λand˜λare real, and hence we can take μand˜μto be real as well. Thus, these integral transforms are no more and no less than the Fourier trans-forms you have long been familiar with, and the variable μis conjugate to the variable ˜λ in the same sense that pis conjugate to qin quantum mechanics. The objects WandZ, known as a twistor and a dual twistor and conjugate to each other, each consisting of 4real components, naturally transform under the group SL( 4,R)(namely the set of all 4 by 4 matrices with real entries and unit determinant), with the invariant W.Z=˜μλ+˜λμ. Given more than one W’s and Z’s we also have the Lorentz invariants Z 1IZ2≡<λ1,λ2> andW1IW2≡[λ1,λ2]. (Here I, in a slightly abused notation used in the literature, evidently denotes the 4 by 4 matrix containing the 2 by 2 identity matrix either in its upper left corneror in its lower right corner depending on whether it acts on WorZ, with all other entries equal to zero.) We have (displaying the helicity hof particle iwhile suppressing the index i)M(tW ,h)=/integraltext d 2λexp(it˜μλ)M(λ ,t˜λ,h)=t−2/integraltext d2λ/primeexp(i˜μλ/prime)M(t−1λ/prime,t˜λ,h)=t2(h−1)M(W ,h) where we used the observation earlier that /Lambda1=− 2h, namely that M(t−1λ,t˜λ)= t2hM(λ ,˜λ). Similarly, M(tZ ,h)=t−2(h+1)M(Z ,h). This scaling result, which you realize comes from the little group (see p. 186), indicates that we should favor a mixed or am- 2The literature on twistors could be traced starting with the paper mentioned on p. 477. N.2. Gluon Scattering | 495 bitwistor representation for the scattering amplitude, using Wwhen the particle carries +helicity and Zwhen the particle carries −helicity. For example, for the basic Y ang-Mills cubic vertex with helicities (++− )(see appendix 2) we write M(W+ 1,W+ 2,Z− 3). The scaling relation just derived imposes powerful con- straints on this amplitude, namely M(W1,W2,Z3)=M(tW 1,W2,Z3)=M(W 1,tW 2,Z3)= M(W 1,W2,tZ3), which implies that in the ambitwistor representation the defining ver- tex for Y ang-Mills theory is apparently, up to an irrelevant overall constant, just 1! Moreprecisely, M(W 1,W2,Z3)depends on the three possible invariants W1.Z3,W2.Z3, and W1IW 2. The scaling relations (note that tcould be either positive or negative) then force Mto have the amazingly simple form M(W+ 1,W+ 2,Z− 3)=sign(W1.Z3)sign(W2.Z3)sign(W1IW2) In different kinematic regions, the basic Y ang-Mills vertex is numerically equal to ±1. T ree amplitudes live naturally in ambitwistor space. As another example, the 4-gluon scattering amplitude (19) we worked hard to get becomes simply M(W+ 1,Z− 2,W+ 3,Z− 4)=sign(W 1.Z2)sign(Z2.W3)sign(W 3.Z4)sign(Z4.W1) Let’s anticipate a bit and write the basic cubic vertex for gravity to be given in (N.3.20) in this ambitwistor representation. Indeed, the scaling relations derived above couldbe immediately applied to the graviton, for which h=± 2. We obtain M(tW ,++)= t 2M(W ,++) andM(tZ ,−−)=t2M(Z ,−−), thus immediately fixing the cubic vertex for gravity to be M(W++ 1,W++ 2,Z−− 3)=|(W1.Z3)(W2.Z3)(W1IW2)| Going from Y ang-Mills to Einstein-Hilbert, we merely have to replace the sign function by the absolute value! Clearly, the take-home message is that quantum field theory possesses hidden structures that the traditional Feynman diagram approach would likely have no hope of uncovering. Appendix 1: Colored Feynman rules for Yang-Mills theory Using the double line formalism of chapter IV .5, we can draw the cubic and quartic vertices in Y ang-Mills theory as in figure IV .5.2. Our conventions for the generators of SU(N) are [Ta,Tb]=ifabcTcand tr (TaTb)=1 2δab. Thusfabc=− 2itr([Ta,Tb]Tc). Start with the Feynman rule for the quartic vertex given in chapter VII.1 and appendix C. First, fabefcde=− 4tr([Ta,Tb][Tc,Td]). Next we multiply by polarization vectors and obtain the colored rule for the quartic vertex: 4ig2tr(TaTbTcTd)(/epsilon1 1./epsilon12/epsilon13./epsilon14−/epsilon14./epsilon11/epsilon12./epsilon13) (22) The two other terms are obtained by permutation. Similarly, the cubic vertex in (C.18) becomes (with a trivial change k→p) −4igtr(TaTbTc)(/epsilon1 1./epsilon12/epsilon13.p1+/epsilon12./epsilon13/epsilon11.p2+/epsilon13./epsilon11/epsilon12.p3) (23) As described in the text, we can now strip off the color factors tr (TaTbTc)and tr (TaTbTcTd). Color stripped amplitudes satisfy a number of useful identities. For example, the color stripped amplitude for n-gluon scattering satisfies the reflection identity A(1, 2, ... ,n)=(−1)nA(n ,... ,2 ,1). T o show this, note that the stripped quartic vertex (/epsilon11./epsilon12/epsilon13./epsilon14−/epsilon14./epsilon11/epsilon12./epsilon13)does not change sign under the reflection 1234 →4321, while the stripped cubic vertex changes sign under 123 →321. From exercise N.2.2, V3+2V4=n−2, and thus V3is odd or even according to whether nis odd or even. 496 | Part N Appendix 2: The cubic vertex in the spinor helicity formalism A rather natural question to ask is what the cubic vertex (23) looks like in the spinor helicity formalism. The first observation is that if we put all momenta on shell, p2 1=p2 2=p2 3=0, then the cubic vertex actually vanishes. By momentum conservation, we have p2 1=(p2+p3)2=p2.p3=0. The conditions pi.pj=0 then imply all three lightlike momenta point in the same direction, so that pi=Ei(1, 0, 0, 1 ),i=1, 2, 3. But this means that, for example, /epsilon13.p1∝/epsilon13.p3=0, and thus the cubic vertex (23) vanishes. Now you see the motivation for allowing the momenta to be complex. Then the conditions pi.pj=0n o longer force all three lightlike momenta to point in the same direction, and we can have a nonvanishing cubic vertex on shell. As explained in the text, to complexify momentum, we simply remove the constraint ˜λ=λ∗.B y the way, by complexifying the momenta here, we are anticipating the discussion in the next chapter a bit. As always, we are free to choose the μspinors to our advantage. A good choice here is μ1=μ2andμ3= λ1. Referring to (12) and (13), we then have /epsilon1− 1./epsilon1− 2∝[μ1μ2]=0 and /epsilon1− 1./epsilon1+ 3∝/angbracketleftλ1μ3/angbracketright=0. The cubic vertex collapses to A(1−,2−,3+)=/epsilon1− 2./epsilon1+ 3/epsilon1− 1.p2=/parenleftbigg/angbracketleftλ2μ3/angbracketright[μ2λ3] [λ2μ2]/angbracketleftμ3λ3/angbracketright/parenrightbigg/parenleftbigg/angbracketleftλ1λ2/angbracketright[μ1λ2] [λ1μ1]/parenrightbigg =/angbracketleft12/angbracketright2 /angbracketleft13/angbracketright[μ1λ3] [μ1λ1](24) As in the text, we are ignoring all overall factors. T o get rid of the unphysical μ1, we need a variant of the momentum conservation identity given in the text. Multiplying/summationtext ip(i) α˙α=/summationtext iλ(i) α˜λ(i) ˙α=0b yεβαε˙α˙γλ(j) β˜μ˙γ, we obtain/summationtext i[μλi]/angbracketleftλiλj/angbracketright=0 for any j, which for j=2 implies [ μ1λ3]/angbracketleftλ3λ2/angbracketright=− [μ1λ1]/angbracketleftλ1λ2/angbracketright. Multiplying (24) by /angbracketleftλ3λ2/angbracketright//angbracketleftλ3λ2/angbracketrightand applying the identity just derived we finally obtain the “mostly minus” cubic vertex A(1−,2−,3+)=/angbracketleft12/angbracketright4 /angbracketleft12/angbracketright/angbracketleft23/angbracketright/angbracketleft31/angbracketright(25) Satisfyingly, we have obtained an expression consistent with (20) (which we have not yet proven) but keep in mind that (25) holds only for complex momenta. I leave it to you to obtain the “mostly plus” cubic vertex A(1+,2+,3−)=[12]4 [12][23][31](26) which also follows from the rule about flipping helicities stated in the text. What about the “all plus” and “all minus” vertices? By now you should be able to determine them as a simple exercise. Exercises N.2.1 Work out the two polarization vectors for general μand˜μfor a gluon moving along the third direction. N.2.2 Show that the number of cubic vertices in tree-level n-gluon scattering can be at most n−2. N.2.3 Show that the result in (17) satisfies the reflection identity A(1−,2−,3+,4+)=A(4+,3+,2−,1−). N.2.4 Show that the “all plus” and “all minus” cubic Y ang-Mills vertices (see appendix 2) vanish. [Hint: Choose theμspinors wisely.] N.2.5 Why doesn’t the argument in the text that A(−++ ...++)vanish apply to A(−++ )? N.2.6 Insert the expression for the cubic vertex into (21) and derive M(W+ 1,W+ 2,Z− 3). N.2.7 Show that M(W+ 1,Z− 2,W+ 3,Z− 4)reproduces (19). N.2.8 Show that SL( 4,R)is locally isomorphic to the conformal group. [Hint: Identify the 15 =42−1 genera- tors of the conformal group (3 rotations Ji, 3 boosts Ki, 1 dilation D, 4 translations Pμ, and 4 conformal transformations Kμ) with the 15 traceless real 4 by 4 matrices.] N.3 Subterranean Connections in Gauge Theories Excess baggage This text, like all texts on field theory, sings the praise of gauge theories—hey, Nature loves them regardless of what physicists like—but, unlike many texts, emphasizes repeatedlythat gauge symmetry is strictly speaking not a symmetry, but a redundancy in description.Extra degrees of freedom are introduced only to be gauge fixed away. In the first edition ofthis book, I expressed in the Closing Words the hope that in the future physics will finda more elegant way of formulating this peculiar concept of local invariance. Perhaps thathope is being realized sooner rather than later! In our current formulation of gauge theories, for a process involving nmassless gauge bosons (photons or gluons) we are instructed to laboriously calculate an off-shell amplitudeM μ1μ2...μn. But experimentalists don’t know about amplitudes carrying Lorentz indices! SE from chapter III.1 speaks up again. “My gauge bosons are specified by their helicities hi,i= 1,...n, not a Lorentz index.” Come to think of it, we theorists do go through a strange two-step procedure involving a lot of excess baggage. After toiling to obtain Mμ1μ2...μnwith external momenta off shell, we then set external momenta on shell and contract with polarization vectors to determine the scattering amplitude for gluons in specified polarization states Mλ1λ2...λn≡ /epsilon1λ1μ1/epsilon1λ2μ2.../epsilon1λnμnMμ1μ2...μn|onshell . In effect, in step 2 we wash away much of the unnecessary information in Mμ1μ2...μnwe worked hard to get in step 1. The cancellation in the 5-gluon scattering in the preceding chapter, with ∼10,000 terms boiling down to a single term, should have convinced you that the traditional Feynmanway may not be so good. In your study of physics, you surely have had the pleasure ofwatching terms canceling against each other toward the end of a calculation, but 10,000 terms down to 1, that was the mother of all cancellations. The key is that in gauge theories there is a kind of secret subterranean connection between different Feynman diagrams, and cancellations are routine. Gauge invariance tells 498 | Part N us that p1 μ1(/epsilon12 μ2.../epsilon1n μnMμ1μ2...μn|onshell), for example, vanishes. Thus, the many diagrams that go into Mμ1μ2...μnmust know about each other in some intricate way. (We saw a glimpse of that way back in chapter II.7 when we proved gauge invariance.) TheS-matrix reloaded In spite of the tremendous difficulties lying ahead, I feel thatS-matrix theory is far from dead and that . . . much new interesting mathematics will be created by attemptingto formalize it. —T . Regge 1 The garbage of the past often becomes the treasure of the present (and vice versa). —A. Polyakov2 As discussed in chapter III.8, back in the 1950s and 1960s, dispersion theorists3tried to forge ahead by studying the analytic properties of various amplitudes as functionsof their external Lorentz invariants, namely the Mandelstam variables sandtfor 2-to- 2 scattering and q 2in our simple vacuum polarization example. But once one gets past 2-to-2 scattering, the analytic structure becomes unwieldy. The program failed and wasswept into the dustbin of physics history. (However, you might know that, through a ratherconvoluted process, this massive effort eventually gave birth to string theory.) Remarkably, some features of this program are being revived. In particular, in this chapter we will discuss the notion of complexifying physical variables. In an interestingtwist, it turns out to be better to complexify the external momenta (to be explained below)rather than invariants like sandt. A historical aside: Landau apparently suggested on one occasion that it might be useful to consider complex momenta. Consider the amplitude M(p i,hi)for tree-level scattering of nmassless particles with momentum and helicity (pi,hi),i=1 ,..., n, with p2 i=0. (For a gauge theory, we will define the amplitude with the color factors already stripped away. Also, suppress the trivialmultiplicative coupling constant dependence and drop all such overall factors as we movealong.) The novel idea is to pick two external momenta p r,ps, complexifying them while keeping them on shell and maintaining momentum conservation. We take all momentaas incoming. At this stage we can keep the discussion general and not even specify thetheory except to stipulate that it contains only massless particles. But to fix ideas, you canimagine a gauge theory. For some complex number z, replace p randpsby pr(z)=pr+zq and ps(z)=ps−zq (1) 1T . Regge, Publ. RIMS, Kyoto University, 12 suppl.: pp. 367–375, 1977. 2A. Polyakov, Gauge Fields and Strings, CRC, 1987, p. 1. 3See, for example, G. Barton, Dispersion T echniques in Field Theory, W . A. Benjamin, 1965. N.3. Connections in Gauge Theories | 499 Rps(z) Lpr(z) PL(z) Figure N.3.1 T o keep pr(z)2=0 and ps(z)2=0, we need q.pr,s=0 and q2=0, which is possible only if we make qcomplex. T o be explicit, go to a frame in which pr+pshas only a time component and use units so that the time component is equal to 2. Then pr=(1, 0, 0, 1 ), ps=(1, 0, 0, −1), q=(0, 1,i,0) (2) A technical aside. This is why I mentioned SO( 2, 2)in appendix B and in the preceding chapter: with a (++− − )signature one could satisfy the on-shell constraint without having a complex q. Here I will stick with the more physical SO( 3, 1)and consider complex momenta as explained in the preceding chapter. Another side remark: As you will see,the discussion goes through for any spacetime dimension d≥4. With this set up the scattering amplitude M(z)becomes an analytic function of z. Think of the complex momentum zqflowing into the diagram with p r(z), cruising through some of the internal lines, and then flowing out with −ps(z). Let us turn on our pole detector. At tree level, a pole can arise only from a propagator carrying momentum zq+.... Thus the tree amplitude M(z)has only simple poles, coming from diagrams of the type shown in figure N.3.1. Divide piinto two sets LandR, with those in Lflowing into a blob on the left-hand side and those in Rflowing into a blob on the right-hand side. The two blobs are connected by a single propagator carrying momentum PL(z), which by arbitrary convention we choose to flow into blob L. The two blobs are themselves tree amplitudes in the theory. Let nLandnRbe the number of external momenta in sets LandR, respectively (withnL+nR=n, of course, and nL≥2,nR≥2). Then the left-hand blob represents tree scattering of nL+1 particles, with nLparticles on shell and one particle with momentum PL(z)off shell, with an amplitude ML(z). Similarly, the right-hand blob represents tree scattering of nR+1 particles, with nRparticles on shell and one particle with momentum −PL(z)off shell, with an amplitude MR(z). 500 | Part N Clearly, the momentum PL(z) depends on zonly if pr(z) andps(z) do not appear in the same set. With no loss of generality let pr(z)belong to the set Landps(z)to the set R. Then PL(z)=−((/summationtext i/epsilon1Lpi)+zq)=PL(0)−zqandPL(z)2=PL(0)2−2zq.PL(0)= −2q.PL(0)(z−zL), where zL=PL(0)2/(2q.PL(0)). Thus Mhas a pole at z=zL, which, sinceqis complex, is in general complex. The amplitude M has poles all over the complex z-plane, at z=zL, one for each valid partition of the external momenta into L+R. The residue has the factorized form RL=ML(zL)MR(zL)/(2q.PL(0)), where ML(zL)andMR(zL)are now both on-shell amplitudes, since the particle carrying momentum PL(zL)is now on shell. As always, we suppress all inessential overall factors. If, and that is a crucial if, M(z)→0a sz→∞ , then/contintegraltext C(dz/z) M(z)=0, where the contour Cis a circle of infinite radius running along infinity. We then shrink the contour, picking up the pole at z=0, which contributes M(0)to the contour integral, and a bunch of poles at z=zL, contributing a sum of terms consisting of the residue at each pole, multiplied by 1 /zL. We thus determine the scattering amplitude to be M(0)=−/summationdisplay L,hRL zL=−/summationdisplay L,hML(zL)MR(zL) PL(0)2(3) Note that the sum also runs over the helicity hcarried by the intermediate particle PL. The notation has been a bit compact, but suffices to get the essential point across without cluttering the page with bloated expressions. But let us now make the notationa bit more precise. T o start with, z Lof course depends on the specific partition Lthrough the momentum PL. T o make sure you follow, let us describe ML(zL)more explicitly. It is an on-shell amplitude with (nL+1)particles coming in, respectively carrying momentum and helicity (pr(zL),hr),(pi,hi)fori/epsilon1L ,i/negationslash=r, and(PL(zL),h). Two of the momenta are complex, namely pr(zL)andPL(zL). Let us emphasize that by construction PL(zL)2=0 and so all particles are on shell. Similarly, MR(zL)is an on-shell amplitude with (nR+1) particles coming in, respectively carrying momentum and helicity (ps(zL),hs),(pi,hi)for i/epsilon1R ,i/negationslash=s, and(−PL(zL),−h). The crucial point is that, amazingly, as was discovered by Britto, Cachazo, Feng, and Witten, we can determine the n-point tree amplitude M(z)in terms of lower point on-shell tree amplitudes, specifically (3) as a sum over products of the (nL+1)-point amplitude ML(zL)and(nR+1)-point amplitude MR(zL). Note that n−1≥nL+1≥3 (similarly fornR+1), and thus by applying these so-called BCFW recursion relations repeatedly, we can calculate any on-shell tree amplitude in Y ang-Mills theory and in gravity in terms of an irreducible 3-point amplitude. Furthermore, in the primitive 3-point on-shell amplitude, allLorentz invariants constructed out of the momenta vanish, since p i.pj=1 2(pi+pj)2=0. T o determine the loop amplitudes, Bern, Dixon, and Kosower have generalized the unitarity methods sketched earlier in this text and alluded to in the preceding chapter. With these methods, one can calculate all amplitudes, trees and loops, and thus determine thetheory completely in terms of the helicity dependence of the 3-point on-shell amplitude. N.3. Connections in Gauge Theories | 501 Amazingly, the old dream of the S-matrix school comes true! Everything within pertur- bation theory is determined without our ever having to refer to a Lagrangian. Note that to obtain physical amplitudes we need only M(z=0)but to recurse to higher point amplitudes we need to know M(z/negationslash=0). As we will see, once we have M(z=0)we can obtain M(z/negationslash=0)by analytic continuation. As emphasized in the preceding chapter, the decomposition of a lightlike vector in terms of spinors pα˙α≡pμ(σμ)α˙α=λα˜λ˙α (4) works equally well for complex lightlike vectors. In that case, as already explained in chapter N.2, the two spinors λand˜λare independent of each other. Another side remark: The deformation (1) considered here has a nice form in the helicity spinor formalism of the preceding chapter. Let pr=λr˜λrandps=λs˜λs(with spinor indices suppressed). Then the spinor deformation ˜λr→˜λr+z˜λsandλs→λs−zλr (leaving λrand˜λsunchanged) gives the desired momentum deformation with q=λr˜λs, which we see is not hermitean and hence corresponds to a complex momentum. This isconsistent with the discussion in the preceding chapter, since the deformation obviouslydoes not respect the equality between ˜λandλ ∗necessary for real momenta. The naive person about to recurse Imagine that you woke up one morning and had the wonderful idea of complexifying momentum. Then suppose you had enough wits, after a bow to Cauchy, to discover thesemarvellous recursion relations. But after you calmed down, you wanted to try the recursionout on some theory. Naturally, you first chose a scalar field theory, say a ϕ 3or aϕ4theory. Your enthusiasm dies immediately. In these theories, the basic vertex is just a number, the coupling. For an n-point amplitude, there are always some Feynman diagrams in which pr(z)andps(z)meet at one of the basic vertices, and the entire diagram does not even depend on z. The crucial assumption that M(z)→0a sz→∞ is simply not true. Most physicists might give up at this point, but suppose you were possessed of strength of character and decided to take a look at Yang-Mills theory, thinking that, after all, it seemedmuch more fundamental than some dumb scalar field theory. But a quick look convincesyou that things are even worse. Consider the diagram in figure N.3.2a contributing to the n- gluon amplitude. Put p r(z)andps(z)as “far apart” as possible to maximize the number of propagators between them. There are (n−3)propagators, contributing a factor of 1 /zn−3 to the amplitude as z→∞ . But alas, this is overwhelmed by (n−2)cubic vertices with each vertex linear in momentum, thus contributing a factor of zn−2. You have not yet included the polarization vectors, which for z=0 are given by /epsilon1− r= q,/epsilon1+ r=q∗and/epsilon1− s=q∗,/epsilon1+ s=q(note that q↔q∗under r↔s, since the two momenta 502 | Part N (a) (b) (c)ps(z) pr(z) ps(z) pr(z)ps(z) pr(z)n/H110022 Figure N.3.2 N.3. Connections in Gauge Theories | 503 point in opposite directions). We also have to deform them to maintain their orthogonality with the corresponding momentum vectors: /epsilon1− r(z)=q,/epsilon1+ r(z)=q∗+zps (5) and /epsilon1− s(z)=q∗−zpr,/epsilon1+ s(z)=q (6) You should check that all conditions are satisfied, for example, /epsilon1+ r(z).pr(z)=(q∗+ zps)(pr+zq)=z(q∗q+pspr)=0. [In the notation of (N.2.9, 10), the polarization vectors here correspond to the choice μr(z)=μr(0)=λs,˜μr(z)=˜μr(0)=˜λs,μs(z)=μs(0)= λr,˜μs(z)=˜μs(0)=˜λr. The first equal sign in each relation simply emphasizes that we choose not to deform the μ’s and ˜μ’s.] Note the peculiar asymmetry between randsafter deformation: in particular, two of the polarization vectors, /epsilon1+ rand/epsilon1− s, grow with zand thus worsen the large zbehavior. Putting it all together and referring to (5) and (6), you would conclude [with the notation Mhrhs(z)] that M−+ naive(z)→zn−2 zn−3=z,M−−or++ naive(z)→z2,M+− naive(z)→z3(7) which most certainly do not →0. A seemingly unimportant comment that will become important later: Of course, some of the gluons other than randscould first interact among themselves as shown in fig- ure N.3.2b. This merely reduces the effective nin the discussion above for those particular diagrams, and we reach the same naive estimates. Reality more benign than expectation Reality turns out to be much more benign than our naive expectation! Actually, amplitudesin Yang-Mills theory behave better than amplitudes in scalar field theory, the opposite ofwhat we thought. This fact is either astonishing or not so astonishing, depending on how jaded you are. I have to admit that it sounds a bit less amazing after learning in the preceding chapter that∼10,000 terms can cancel down to a single term. We can even concoct a heuristic physical argument. Go back to Yang-Mills theory and call the particles gluons, as before. Apply crossing to gluon s, so that we have an incoming gluon rwith a huge momentum p r(z)∼zqin the large zlimit, emerging as a gluon with the huge momentum −ps(z)∼zq. The other (n−2)gluons have fixed momenta and are thus soft. We have a hard gluon blasting through a soft gluon background, something like a high energy gamma ray blasting through a magnetic field, and thus we do not expectmuch scattering as z→∞ , and even less scattering that would flip the helicity of the hard gluon. (The situation is conceptually similar to electron scattering in an external Coulombpotential, discussed in chapter II.6, except that here the field excitation being scattered isof the same type as the background field.) 504 | Part N But not so fast! Even though you and I have studied physics for years, we haven’t built up much intuition about complex momenta. At least I speak for myself. Alternatively, wecould go to SO( 2, 2)and deal with real momenta, but we haven’t much experience with signature (++− − )spacetime, either. Background field method Nevertheless, the picture of a hard gluon blasting through a soft background turns out to be helpful in guiding us toward an elegant formulation of the problem. We splitthe Yang-Mills gauge potential (which we write as Aon this occasion) into two pieces, A(x)=A(x)+a(x) , a background potential Awithout high momentum components in its Fourier transform and a fluctuating potential awith high momentum components. (You would do exactly the same split when studying a laser beam passing through a laboratorymagnetic field.) To develop this so-called background field method (which is useful forother problems besides this one), it pays to use the differential form notation used inchapter IV .5. We split the transformation law A→UAU †+UdU†intoA→UAU†+UdU†and a→UaU†. In other words, the background Atransforms like a Yang-Mills potential, while the fluctuating atransforms like a matter field in the adjoint representation. Plugging into the field strength F=dA+A2=d(A+a)+(A+a)2, we find Fequal to the sum of the background field strength F=dA+A2and the 2-form da+Aa+aA+a2=(∂μaν+ [Aμ,aν]+aμaν)dxμdxν≡1 2(Dμaν−Dνaμ+[aμ,aν])dxμdxν. Switching back from math to physics notation and defining the shorthand notation D[μaν]≡Dμaν−Dνaμ, we have Fμν=Fμν+D[μaν]−i[aμ,aν]. Here Dμaν=∂μaν−i[Aμ,aν] is the covariant derivative (with respect to the background potential A) of the adjoint field a. Since we have only two hard gluons interacting with the soft background, it suffices to expand the Yang-Mills Lagrangian to quadratic order in a: L=−1 2g2trFμνFμν =−1 2g2tr/parenleftBig FμνFμν+D[μaν]D[μaν]+2FμνD[μaν]−2iFμν[aμ,aν]/parenrightBig +O(a3) (8) Since in the action we integrate Lover spacetime, we are effectively allowed to inte- grate by parts. Thus the third term in the parenthesis, tr (FμνDμaν)=tr(Fμν(∂μaν− i[Aμ,aν]))“=”t r ((DμFμν)aν). Since the background field satisfies the field equation DμFμν=0, this term vanishes. (You should not be surprised that the term linear in a in the action is linear in the field equation.) Thus, to study the propagation of athrough the background A, we can focus on the Lagrangian quadratic in a:Lquad=−(1/g2)tr/parenleftbig (Dμaν−Dνaμ)Dμaν−iFμν[aμ,aν]/parenrightbig As always, we need to fix the gauge. Upon integration by parts we have trDνaμDμaν=tr(DμaμDνaν+iFμν[aμ,aν]) (9) N.3. Connections in Gauge Theories | 505 Note that, unlike ordinary derivatives, when the gauge derivatives DνandDμpass each other, they produce the field strength Fμ. [Verify this! You might recall the more mathe- matical form (IV .5.13).] Thus a convenient way of fixing the gauge is to add tr (DμaμDνaν) so that the gauge-fixed Lagrangian becomes Lquad=−1 g2tr(DμaνDμaν−2iFμν[aμ,aν]) (10) [Incidentally, this parallels precisely what was done in (III.4.8) to obtain the Feynman gauge withξ=1.] Return now to our problem of studying the large-z behavior of the scattering amplitude Mλρ. Recall from (7) that we obtain Mλρ→zn−2/zn−3=z. (Also recall that polarization vectors have not yet been included, and they could multiply this behavior by z0,z1,o rz2.) The culprit is the derivative in the cubic vertex ∼Aa∂a sitting inside the first term in (10). In contrast, the ∼AaAa piece in the first term and the second term tr Fμν[aμ,aν]i n (10) insert quartic vertices that do not grow with z. The situation confronting us is now best discussed by the clever trick of renaming indices. First, understand that Lorentz invariance is broken by the presence of the back-ground field A μ, to be regarded as given and fixed. (This is the same as in chapter VI.2: the presence of a background magnetic field means that parity and time reversal are broken.)But now suppose we simply relabel indices and write Lquad=−1 g2tr(ηabDμaaDμab−2iFab[aa,ab]) (11) where ηabis nothing but the humble Minkowski metric. The first term by itself enjoys a hidden “enhanced Lorentz” symmetry: an SO( 3, 1) transformation on the indices a,bleaves the Lagrangian invariant. We now exploit this hidden symmetry. Since the leading behavior of Mabfor large zcomes from repeated insertion of the cubic vertex ηabaaAμ∂μabcontained in the first term in (11), we conclude that the leading behavior must be proportional to ηab. In contrast, with one insertion of the quartic vertex from the second term in (11), we decrease the power of zby one, since it does not contain a derivative on the field a.B u t we also break the hidden “enhanced Lorentz” symmetry, since Fabis fixed. On the other hand, there is an extra bit of information: we know that it is antisymmetric in (ab). [Note that an insertion of the quartic vertex ηabaaAμAμabcontained in the first term in (11) also decreases the power of zby one, but its contribution is proportional to ηab.] Thus the hidden “enhanced Lorentz” symmetry tells us that the amplitude expanded in powers of zmust have the form Mab=(cz+...)ηab+Aab+1 zBab+... (12) withcsome unknown constant. The only thing we know about the matrix Aabis that it is antisymmetric in (ab). (I am following the notation in the literature. If you are confused between this matrix Aand the background gauge potential A(x) you need to go back to square 1.) 506 | Part N We still have gauge invariance in the form pra(z)Mab(z)εsb(z)=0 and εra(z)Mab(z)psb(z)=0, giving us valuable information. For example, looking up the form pr(z)=pr+zq, we obtain qaMab(z)εsb(z)=−(1/z)p raMab(z)εsb(z), but since from (5) /epsilon1− r(z)=q, this means that /epsilon1− ra(z)Mab(z)εsb(z)=−(1/z)p raMab(z)εsb(z). Let us now look at the specific helicity combinations for which we had naive expectations in (7). Recall that we expected M−+(z)→z. In fact, since /epsilon1+ s(z)=qandpr.q=0, we have M−+(z)=/epsilon1− ra(z)Mab(z)ε+ sb(z)=−1 zpra{(cz+...)ηab+Aab+1 zBab+...}qb =−1 zpraAabqb+O(1 z2)→1 z(13) This amplitude behaves better than naive expectation by two powers of (1/z)! Next, M−−(z)→z2naively, but in fact M−−(z)=/epsilon1− ra(z)Mab(z)ε− sb(z)=−1 zpra{(cz+...)ηab+Aab+1 zBab+...}(q∗ b−zprb) =−1 z(praAabq∗ b+praBabprb)+O(1 z2)→1 z, (14) three powers better than naive expectation. Similarly, M++(z)→1/z. Note that these conclusions hold for any n. If you have the strength, you might want to witness the cancellations by explicitly calculating the various M’s for low values of n. But not all helicity amplitudes behave better than naive expectation. We finally come toM+−(z), which →z3naively. Looking at (5) and (6) we already see trouble, since both /epsilon1+ ra(z)andε− sb(z)grow like z. Now we have M+−(z)=/epsilon1+ ra(z)Mab(z)ε− sb(z)=(q∗ a+zpsa){(cz+...)ηab+Aab+1 zBab+...}(q∗ b−zprb) =−cps.prz3+O(z2)→z3(15) Incidentally, note that our intuition about complex momenta is a bit shaky. The helicity- conserving amplitude (+→+ )[namely M+−(z)by crossing; recall that Mwas defined with all momenta going in] behaves worse than the (+→− )amplitude M++(z), the (−→+ )amplitude M−−(z), and the (−→− )amplitude M−+(z). The polarization vectors are continued for complex momentum in a nonsymmetric fashion. Confusio suddenly speaks up! “You haven’t yet exploited the gauge invariance of the background field,” he says. We forgot that he often appears in the company of SE. Indeed, he is right. Very good— Confusio did not become an assistant professor for nothing. Indeed, let us look at the cubic vertex in figure N.3.2a more carefully: we have a hard gluon carrying momentum zq+...scattering off a background gluon carrying some small momentum pinto a hard gluon with momentum zq+.... The coupling comes from the term tr ∂μaν[Aμ,aν] in the Lagrangian, and thus to leading order in zthe vertex is proportional to zqμ.Aμ(p). According to exercise VII.1.1, we can choose a gauge in which qμ.Aμ(p)=A2+i3(p)=0, known as the Chalmers-Siegel space cone gauge. We should check to see if this is possible, but to streamline the exposition let us merely do the abelian case. With Aμ(x)→Aμ(x)−∂μ/Lambda1(x), the desired gauge choice N.3. Connections in Gauge Theories | 507 requires q.A(p)=iq.p/Lambda1(p), and thus we can solve for /Lambda1(p) as long as q.p/negationslash=0. While q.pr,s=0 by construction, generically there is no reason for q.pito vanish for i/negationslash=r,s. So we conclude that indeed we can get rid of the offending cubic vertex. But not so fast! What about figure N.3.2c, in which all the soft gluons interact with each other to form one single soft gluon carrying momentum/summationtext i/negationslash=r,spi=−(pr+ps)? Since q.(pr+ps)=0, we cannot set q.A(pr+ps)=0 and the diagram in figure N.3.2c remains. Thus, even though we managed to get rid of the cubic vertices in figure N.3.2a, b,our previous conclusion about the large-z behavior of Mstill stands. “Wait! What about the color factor?” Confusio yells. Let us look at the color structure we stripped off. From figure IV .5.2b we see that the cubic vertex in figure N.3.2c requiresthat the two hard gluons be adjacent in color. It is easiest to explain the terminology byan example: a red-green gluon and a blue-yellow gluon are not adjacent in color, but theyare both adjacent to a red-yellow gluon (and to a blue-green gluon). Note that the couplingtrF μν[aμ,aν] also requires that the two hard gluons be color adjacent. Thus, if the two hard gluons are not color adjacent, the large-z behavior of Mis somewhat better, since now c=0 and Aab=O(1/z). Then M−+→1/z2instead of 1 /z, M+−→z2instead of z3, while M−−andM++are not improved. Confusio deserves credit for his partial triumph, and perhaps eventually should be given tenure. The bottom line is that, contrary to naive expectation, amplitudes in gauge theory behave well enough for the BCFW recursion program to work. We don’t even mind that M+− behaves badly; it suffices for the program that M−,any helicityvanishes for large z.I n particular, in appendix 1 we will show how to complete the calculation started in thepreceding chapter. As indicated earlier, once we determine the tree amplitudes, we can in principle obtain all loop amplitudes by using unitarity. In this modern revival of the S-matrix spirit, we deal with only on-shell amplitudes. The message here is that traditional Feynman diagramscarry around an enormous amount of unnecessary off-shell baggage. A dramatic exampleis furnished by this innocuous looking Feynman integral /integraldisplayd4l (2π)4lμlνlρlλ l2(l−k)2(l−p)2(l−q)2(16) which you can evaluate most conveniently using dimensional regularization. Try it. The in- tegral looks similar to the integrals we did back in chapters III.6,7, but looks are deceptive.The answer, if printed on a page, is a total black smudge (see http://online.kitp.ucsb.edu/online/colloq/bern2/oh/05.html). After all, this integral is just one piece of a physical am-plitude and by itself does not possess any nice qualities, such as gauge invariance. All possible Lorentz invariant theories Remarkably, not only does BCFW recursion allow us to determine all n-point on-shell amplitudes in terms of a primitive 3-point on-shell amplitude, it also restricts all possibletheories for which the recursion works. Let us sketch how this is possible. We anticipate 508 | Part N here, as we will explain in the next chapter, that the recursion program works for massless spin 2 as well as for spin 1 particles. Consider a 4-point on-shell amplitude M. The point is that we are free to deform different pairs (r,s)to determine M. Suppose we pick (r,s)= (1, 4). Then Mis the sum of two pieces, one with a pole in s=(p1+p2)2=(p3+p4)2and another with a pole in t=(p1+p3)2=(p2+p4)2. But we could have also picked (1, 2)for example. That the physical 4-point on-shell amplitude M(z=0)constructed in different ways must agree imposes powerful self-consistency conditions on the primitive 3-pointon-shell amplitude. Perhaps not surprisingly, for spin 2 massless particles, Einstein gravity is the only possible theory, while for spin 1 massless particles, Yang-Mills gauge theory. Indeed, thisresult was proven long ago by Weinberg using rather general arguments. But it is stillinstructive to see how the same result emerges from a strikingly different formalism.These self-consistency conditions also allow one to explore and search for other possibletheories. It is crucial that the primitive 3-point on-shell amplitude M 3is evaluated for complex momenta, which allow more freedom than garden-variety everyday real lightlike momenta.(As already noted in appendix 1 to chapter N.2, the Yang-Mills cubic vertex vanishesfor real lightlike momenta.) I remind you again that for complex momenta p i=λi˜λi, the two spinors λiand˜λiare independent of each other. Recall from chapter N.2 that /angbracketleftij/angbracketright=/angbracketleftλiλj/angbracketright≡εαβλiαλjβand [ij]=[λiλj]=[˜λi˜λj]≡ε˙α˙β˜λi˙α˜λj˙β. Also, pi.pj=/angbracketleftij/angbracketright[ij]. The on-mass shell conditions pi.pj=0 then become /angbracketleft12/angbracketright[12]=0,/angbracketleft23/angbracketright[23]=0, and /angbracketleft31/angbracketright[31]=0. Apparently there are several possible solutions. For example, we could have all three square brackets vanish with all three angled brackets nonzero, or we could havetwo square brackets vanish, say [12] =[23]=0, with /angbracketleft31/angbracketright=0. But there are only two independent 2-component spinors, so three spinors cannot be linearly independent [take ˜λ 1∝(0, 1)and˜λ2∝(1,w), then the third spinor ˜λ3is necessarily a linear combination of the other two]. Thus, [12] =0 and [23] =0 mean that ˜λ1∝˜λ2and˜λ2∝˜λ3, respectively, which implies that ˜λ3∝˜λ1and [31] =0. Of course, the discussion can be repeated with square and angled brackets interchanged. Thus we conclude that either /angbracketleft12/angbracketright=/angbracketleft 23/angbracketright=/angbracketleft 31/angbracketright=0 or [12] =[23]=[31]=0 (17) (For example, if [12] =[23]=[31]=0, then ˜λ2=α2˜λ1and˜λ3=α3˜λ1, and momentum conservation/summationtext ipi=/summationtext iλi˜λi=0 implies λ1+α2λ2+α3λ3=0. The information here is in the coefficients, since three 2-component spinors are always linearly dependent.) Thus, depending on the helicities, either M3=MH(/angbracketleft12/angbracketright,/angbracketleft23/angbracketright,/angbracketleft31/angbracketright)orM3= MA([12], [23], [31] ). Recall from the preceding chapter that /Lambda1i=− 2hi, where /Lambda1icounts the powers of λi minus the powers of ˜λi. But acting on MH, this just counts the powers of λi. Write MH=/angbracketleft12/angbracketrightd3/angbracketleft23/angbracketrightd1/angbracketleft31/angbracketrightd2and solve for the unknown d’s using /Lambda11=d2+d3=− 2hi, etc. Thend1=h1−h2−h3,d2=h2−h3−h1, andd3=h3−h1−h2. For example, suppose N.3. Connections in Gauge Theories | 509 the theory contains spin 1 massless particles, with different varieties labeled by an index awhose range we need not specify. Then we have, for example, M3(1− a,2− b,3+ c)=fabc/parenleftbigg/angbracketleft12/angbracketright3 /angbracketleft23/angbracketright/angbracketleft31/angbracketright/parenrightbigg (18) since the helicities h1=h2=− 1 and h3=+ 1 imply that d1=d2=− 1 and d3=3. At this stagefabcis some unknown coefficient that depends on the particle variety. As required, we have two positive powers of λ1andλ2and two negative powers of λ3. This confirms what we obtained in appendix 1 to the preceding chapter. Several remarks follow. 1. From pi=λi˜λithe spinors λand˜λhave mass dimension1 2. Thus M3has mass dimension 1, as expected (recall that the cubic coupling in gauge theory has the form ∼/epsilon1./epsilon1/epsilon1.p). 2. We obtain the 3-point amplitude M3(1+ a,2+ b,3− c)=fabc/parenleftBigg [12]3 [23][31]/parenrightBigg (19) by flipping helicities, which, as we have learned in the preceding chapter, amounts to interchanging the roles played by λand˜λ, so that it given by square instead of angled brackets. 3. Note the power of the spinor helicity formalism. We can immediately generalize to higher integer spin sby scaling the /Lambda1i’s and hence the d’s up by a factor of s. Thus we simply raise the round parenthesis in M3(1− a,2− b,3+ c)to power s. The cubic vertex for spin 2 is thus given by M3(1−− a,2−− b,3++ c)=fabc(/angbracketleft12/angbracketright3/(/angbracketleft23/angbracketright/angbracketleft31/angbracketright))2. 4. Interchanging 1and 2, we see that fabc=−fbacforsodd. Thus for sodd (s =1, for example), we cannot have a theory with only one variety of particles. We are compelled to introducethe index a(and call it color!). 5. For seven (s =2, for example) we can get away with only one variety. Call it the graviton. The coefficient f abccan be omitted and one of the two basic cubic vertices for Einstein gravity is simply given by M3(1−−,2−−,3++)=/parenleftbigg/angbracketleft12/angbracketright3 /angbracketleft23/angbracketright/angbracketleft31/angbracketright/parenrightbigg2 (20) (The other vertex is of course obtained by replacing angled brackets by square brackets.) More on the 3-graviton vertex in appendix 2. 6. Check out the power of the self-consistency argument sketched above. Consider the 4-point amplitude M(1a,2b,3c,4d)in a theory with a variety of spin 1 massless particles. Apply the recursion to construct Mas the sum of an amplitude with an schannel pole, evidently proportional to fabefcdewith an implicit sum over the label eof the intermediate particle, and an amplitude with a tchannel pole proportional to facefbde. Requiring Mconstructed with different choices of (r,s)in the recursion to be the same then gives the constraint fabefcde+facefbde+fadefbce=0 (21) 510 | Part N K/H11002j/H11002 2/H110013/H11001n/H11002 4/H110011/H11001 Figure N.3.3 But we recognize this as just the defining relation (B.19) for the generators of a Lie algebra [Ta,Tb]=ifabcTcwritten out in the adjoint representation! The coefficients fabcthat appear in the primitive 3-point on-shell amplitude are the structure constants of the algebra. If thisis too abstract for you, verify it for SU( 2). The recursion program produces Einstein gravity and Yang-Mills theory as the unique low energy theory for massless spin 2 and spin 1 particles, respectively, with sufficientlygood large-z behavior for the recursion relations to be valid. Of course, we also know that in the Lagrangian formalism, the powerful constraints of local coordinate invariance andlocal gauge invariance fix the actions for Einstein gravity and Yang-Mills completely. Appendix 1 Here, as promised, we use the recursion approach to prove the result conjectured in the preceding chapter, that forn-gluon scattering, the maximal helicity-violating amplitude is given by A(1+,2+,...j−,...,n−)=/angbracketleftjn/angbracketright4 /angbracketleft12/angbracketright/angbracketleft23/angbracketright/angbracketleft34/angbracketright.../angbracketleft(n−1)n/angbracketright/angbracketleftn1 /angbracketright(22) (Using the cyclicity of the amplitude, we have with no loss of generality let gluon ncarry negative helicity.) We take r=nands=1, and deform ˜λn→˜λn+z˜λ1andλ1→λ1−zλn(leaving λnand˜λ1unchanged), in other words, pn→pn+zqandp1→p1−zqwithq=λn˜λ1. Write (3) as (see fig. N.3.3) A(1+,2+,...j−,...,n−)=A3(ˆ1+,2+,ˆK−)An−1(−ˆK+,3+,...,j−,...,ˆn−)/PL(0)2(23) Here we define K(z)=PL(z)to simplify writing. We use a hat to indicate that the corresponding momentum has been complexified. Thus ˆ1,ˆn,ˆKremind us that p1(z),pn(z), and K(z)=−(p1(z)+p2)(evaluated at z=zL) are the three complex momenta in the problem. In the spirit of recursion, we are supposing that A3andAn−1 are given by (22) (and the corresponding expression with all helicities flipped and angled brackets replaced by square brackets). Note that (22) does not refer to any possible relation between the untwiddled λand twiddled ˜λspinors and thus makes sense for both complex and real momenta. Notice that the sum over partition Land over hin (3) collapses to one term in (23). We have used the result A(++ ...+)=0 and A(−+ ...+)=0, and the second half of (17) to eliminate a diagram similar to that in fig. N.3.3 but with particles (n−1)andnparticipating in the cubic vertex instead of 1 and 2. N.3. Connections in Gauge Theories | 511 Recursing and using PL(0)2=2p1.p2=2/angbracketleft12/angbracketright[12], we obtain [see (19)] A(1+,2+,...j−,...,n−)=[ˆ12]3 [2,ˆK][ˆK,ˆ1]/angbracketleftjˆn/angbracketright4 /angbracketleftˆK3/angbracketright/angbracketleft34/angbracketright.../angbracketleft(n−1)ˆn/angbracketright/angbracketleftˆnˆK/angbracketright1 /angbracketleft12/angbracketright[12](24) As before, we suppress overall constants. The trick consists of taking various hats off or leaving them on. Since ˜λ1is unchanged, we can remove the hat on ˆ1 when it appears in a square bracket. Similarly, since λnis unchanged, we can remove the hat on ˆnwhen it appears in an angled bracket. On the other hand, we should leave the hat on ˆK. Instead, we use momentum conservation −λK˜λK=λ1˜λ1+λ2˜λ2so that /angbracketleftˆK,3/angbracketright[ˆK,1 ]=− /angbracketleft 3,ˆK/angbracketright[ˆK,1 ]=/angbracketleft32/angbracketright[21] since [11] =0. (You might note that this is the same sort of manipulation used to derive the first identity we needed to massage A4into shape in the preceding chapter.) Similarly, /angbracketleftnˆK/angbracketright[2,ˆK]=− /angbracketleftnˆK/angbracketright[ˆK,2 ]=/angbracketleftn1/angbracketright[12]. Doing all this to (24) we obtain A(1+,2+,...j−,...,n−)=[12]3/angbracketleftjn/angbracketright4 /angbracketleft12/angbracketright[12]/angbracketleftn1/angbracketright[12]/angbracketleft32/angbracketright[21]/angbracketleft34/angbracketright.../angbracketleft(n−1)n/angbracketright =/angbracketleftjn/angbracketright4 /angbracketleft12/angbracketright/angbracketleft23/angbracketright/angbracketleft34/angbracketright.../angbracketleft(n−1)n/angbracketright/angbracketleftn1 /angbracketright, (25) precisely the conjectured result. Note how much more powerful the recursion approach is compared to the explicit spinor helicity calculation we did to obtain A(1, 2, 3, 4 )in the preceding chapter, which in turn is so much more powerful than the traditional Feynman diagram calculation. Thus theoretical physics marches on. You might be puzzled that the quartic vertex (figure IV .5.1c) of Yang-Mills theory is not needed in the recursion program. Does this nonparticipation in the program mean that we can multiply the quartic term in the Lagrangianby an arbitrary coefficient (including 0)? The resolution of this apparent paradox can be traced to the fact thatthe (perturbative) physical states of the gluon are built into the recursion relations. The quartic term is neededto guarantee gauge invariance and hence the two helicity states of the gluon. Appendix 2 By showing you the mess in figure N.2.2 I have already plenty impressed upon you that the traditional Feynmandiagram approach is almost hopeless when it comes to gluons. The situation with gravity is far worse. Considerthe 3-graviton vertex. Conceptually it is easy to understand: we write g μν=ημν+hμνand expand the Einstein- Hilbert action (VIII.1.1) to O(h3). There it is, with indices suppressed, the cubic term h∂h∂h in (VIII.1.5). Of course, this actually represents many terms with the eight indices contracted every which way, but which youcan readily work out. Next, pick the harmonic gauge for example, and derive the Feynman rule for the 3-gravitonvertex G μα,νβ,σγ(p1,p2,p3), namely the analog of the 3-gluon vertex in (C.18). Each of the three gravitons, say the one carrying momentum p1, can be created by any one of the three h’s in h∂h∂h, and thus many terms are generated simply by permuting. The two derivatives give two powers of momentum. Thus, a typical term hasthe form p 1βp2μηανησγ. Keep working! In all, Gμα,νβ,σγ(p1,p2,p3)contains about 100 terms. Now imagine calculating the one-loop contribution to graviton-graviton scattering. You get the point. By now, you fully appreciate that the traditional Feynman approach carries an enormous amount of unneces- sary off-shell information. Already, if we put p1,p2, andp3on shell and contract Gμα,νβ,σγ(p1,p2,p3)with the polarization vectors /epsilon1μα 1,/epsilon1νβ 2, and/epsilon1σγ 3, the 3-graviton vertex simplifies enormously to G(p 1,p2,p3)=/epsilon1μα 1/epsilon1νβ 2/epsilon1σγ 3(p1σημν+cyclic)(p 1γηαβ+cyclic) (26) Quite naturally, we can write the polarization vector for a spin 2 massless particle in terms of the polarization vector for a spin 1 massless particle: /epsilon1μα(p)=/epsilon1μ(p)/epsilon1α(p). This form satisfies all that is required of a polarization vector for spin 2: /epsilon1μα(p)p μ=0,/epsilon1μα(p)=/epsilon1αμ(p), and ημα/epsilon1μα(p)=0. Thus, indeed, the 3-graviton vertex G(p 1,p2,p3)=[/epsilon1μ 1/epsilon1ν 2/epsilon1σ 3(p1σημν+cyclic)]2is the square of the 3-gluon vertex (N.2.23), in confirmation of (20), which of course is just the same statement couched in another notation. 512 | Part N Exercises N.3.1 Show that the structure of Lie algebra (21) emerges naturally. N.3.2 In appendix 1 we recursed by complexifying the momenta of two external lines with helicity +and−. In the derivation of the recursion relation (3) we could have picked any two external lines to complexify.Determine the amplitude calculated directly in chapter N.2, namely A(1 −,2−,3+,4+), by complexifying lines 1 and 2. This is an example of the self-consistency argument sketched in the text. Particle physics experimentalists are fond of saying that yesterday’s spectacular discovery is today’s calibration and tomorrow’s annoying background. The canonical example is the Nobel-winning discoveryof the CP-violating decay of the K Lmeson into two pions. In theoretical physics, yesterday’s discovery is today’s homework exercise and tomorrow’s trivium. N.3.3 Using the explicit forms given for A(1−,2−,3+,4+)andA(1−,2+,3−,4+)in the preceding chapter, check the estimated large zbehavior in (13–15). N.3.4 Worry about the sloppy handling of factors of 2 in appendix 1. [Hint: The final result is correct because the polarization vectors in (5–6) are normalized to |ε|2=2 for convenience.] N.4Is Einstein Gravity Secretly the Square of Yang-Mills Theory? Gravity and gauge theory Quantum gravity has baffled generations of theoretical physicists, as you have no doubt heard. One aspect of this puzzle is the relationship between gravity and gauge theory,which describes the other fundamental interactions. While gravity and gauge theory areboth born of local invariance, the Einstein-Hilbert action/integraltext d 4x√−gR and the Yang-Mills action/integraltext d4xtr(FμνFμν)look completely different. Perturbatively, gravity is afflicted with an infinite number of interaction terms, as was explained in chapter VIII.1, and hence gravity is not renormalizable, in stark contrast togauge theory. On the other hand, the two field theories enjoy many conceptual similaritiesbetween them. Yang-Mills theory is the unique low energy effective theory of a spin 1massless field, just as Einstein gravity is the unique low energy effective theory of a spin 2massless field. String theory unifies gravity and gauge theory. This remarkable fact alone points to a deep connection between gravity and gauge theory, even though within field theory theconnection is totally obscure. One important clue is that the oscillator spectrum of the openstring contains only the gauge field but not the graviton, which appears in the spectrumof the closed string. However, the closed string spectrum could be described as two copiesof an open string spectrum, thus leading Kawai, Lewellen, and Tye to discover relationsbetween graviton scattering and gauge boson scattering. 1In the limit of the string energy scale going to infinity, we know that string theory reduces to field theory and thus a shadow of these KLT relations should survive in field theory. (As you might know, not all theoristsare convinced that string theory corresponds to reality. If string theory eventually fails,its ultimate value might well turn out to be the light it sheds on the hidden structure ofquantum field theory.) 1It is definitely beyond the scope of this book to explain these statements. See, for example, J. Polchinski, String Theory , p. 27. 514 | Part N In any case, the bottom line is that string theory strongly hints that graviton ampli- tudes can be expressed as products of Yang-Mills amplitudes, schematically Mgravitons ∼ Mgauge×Mgauge . The first reaction of many theoretical physicists when first told this is puzzled skepticism. How is this possible, they ask quite reasonably, since Yang-Millscontains an internal symmetry group while gravity doesn’t? Now that we have learned to strip color, a connection between amplitudes no longer strikes us as so implausible, particularly if we stick to on-shell scattering amplitudes forgluons in specified polarization states M λ1λ2...λn, namely amplitudes that experimentalists can measure, rather than amplitudes Mμ1μ2...μncarrying Lorentz indices that theorists using traditional methods play with. As we saw in the preceding chapter, the color strippedtree-level on-shell helicity amplitude for gauge boson scattering boils down to the /angbracketleft.../angbracketright and [ ...] products of two component spinors. We did not do the analogous calculation of the tree-level on-shell helicity amplitude for graviton scattering, but we could anticipatethat the result would again be expressed in terms of the /angbracketleft.../angbracketrightand [ ...] products. The spinor helicity formalism is intrinsic to the Lorentz group SO( 3, 1), not tied to a specific theory. In particular, the interaction vertices in Einstein gravity are again given in termsof scalar products of momenta and polarization vectors. Quite suggestively, the gravitonpolarization vectors can be written, as mentioned in appendix 1 to the preceding chapter,as/epsilon1 μν=/epsilon1μ/epsilon1ν, a product of the gauge theory polarization vectors. Indeed, I have already given part of the mystery away in the preceding chapter. We saw that the basic cubic interaction vertex of three gravitons (with complex momenta) is givenby the square of the corresponding quantity for three gluons. In summary, thanks to our string theory friends, we now know that there exists a secret structural connection between gravity and gauge theory that is totally opaque atthe Lagrangian level. Deformed graviton polarizations In this closing chapter, I give a brief introduction to the exciting quest for this secretconnection. I will be content to look at one specific calculation. Go back to the BCFW recursion (chapter N.3). It would work for gravity if the com- plexified scattering amplitude M(z)vanishes as z→∞ . But naively, it would seem that the situation for gravity is even worse than the situation for gauge theory, since the cubicgraviton vertex is quadratic in momentum and thus goes like z 2. (Recall the two powers of derivative in the scalar curvature; see chapter VIII.1.) Repeat the calculation in the pre- ceding chapter for n-graviton on shell scattering. Go back to figure N.3.2 and interpret the lines as gravitons. The (n−2)cubic vertices give a factor of z2(n−2)for large z, easily over- whelming the factor of 1 /zn−3from the (n−3)propagators. This nasty behavior occurs even before we include the polarization of the two hard gravitons. The graviton carries helicity ±2 (appendix 2 of chapter VIII.1) and hence a polarization “vector” /epsilon1μν, given by a symmetric and traceless tensor. We can naturally construct /epsilon1μν= N.4. Gravity and Yang-Mills Theory | 515 /epsilon1μ/epsilon1νout of the polarization vectors for a massless spin 1 particle (as already explained in the preceding chapter). Thus, after deformation, /epsilon1++μν r(z)=/epsilon1+μ r/epsilon1+ν r=(q∗+zps)μ(q∗+zps)ν,/epsilon1−−μν r(z)=/epsilon1−μ r/epsilon1−ν r=qμqν(1) and /epsilon1++μν s(z)=/epsilon1+μ s/epsilon1+ν s=qμqν,/epsilon1−−μν s(z)=/epsilon1−μ s/epsilon1−ν s=(q∗−zpr)μ(q∗−zpr)ν(2) Note that the /epsilon1μν(z)’s are in fact traceless and could go as either z0orz2for large z. Putting it together, we obtain the naive estimate M−−,++ naive(z)→z2(n−2) zn−3=zn−1,M−−,−− or++,++ naive(z)→zn+1,M++,−− naive(z)→zn+3(3) The escalating behavior as nincreases is the hallmark of a nonrenormalizable theory, as explained in chapter III.2. Hard graviton in a soft spacetime Once again, we hope that real life is cushier than naive expectation. By the same reasoningused for gauge theory, we study a hard graviton blasting through a gravitational field, thatis, a background of soft gravitons. So write the metric of spacetime as G μν=gμν+hμν. Plug this into the Einstein-Hilbert action (VIII.1.1) and extract the terms quadratic in h. While the calculation is straightforward, it does involve some heavy lifting. To avoid thelabor, we note that using the harmonic gauge, we did this calculation in (VIII.1.10) butonly for the special case g μν=ημν(in other words, we expanded around flat Minkowski spacetime rather than a general curved spacetime). We had L=1 64πG(ημνηλρηστ∂μhλσ∂νhρτ−1 2ημν∂μh∂νh) (4) with the trace degree of freedom h≡ημνhμν. Henceforth, we set 64 πG=1. Armed with symmetry considerations and our knowledge of gravity (chapter VIII.1), we can almost immediately guess that when we go from a flat ημνto a curved gμνbackground, this quadratic Lagrangian generalizes to L=√−g(gμνgλρgστDμhλσDνhρτ−1 2gμνDμhDνh−2Rλρσ τhλσhρτ) (5) withhnow defined as h≡gμνhμν. Here Ddenotes the covariant derivative with respect to the curved metric gμνintroduced in chapter VIII.1 and Rλρσ τthe Riemann curvature tensor constructed out of gμν. I trust you not to confuse this Dassociated with the curved background with the covariant derivative in Yang-Mills theory used in the preceding chapter and mentioned below in passing. Let us go through the various features of (5). The√−ggoes with the spacetime volume and is common to any Lagrangian in curved spacetime, as we learned way back in (I.11.2).We also learned there to promote any Lagrangian from flat to curved spacetime by replacingη μνwithgμνand the ordinary derivative by a covariant derivative (see chapter IV .5). As you 516 | Part N can see, everything pretty much works out in parallel with how things work out for gauge theory. The new feature is the term involving the Riemann curvature tensor Rλρσ τ, which vanishes upon restriction to flat spacetime. But you are not surprised that such a termcould pop up, given tr F μν[aμ,aν] in (N.3.8). Indeed, the only thing we can’t determine without doing the actual calculation is the numerical coefficient (−2)of this term. That particular number will play no role in the following discussion. Also, in the preceding chapter we dropped a term linear in aμbecause of the equation of motion DμFμν=0. Here, analogously, we dropped a term linear in hμνbecause of Einstein’s equation of motion Rμν=0. This also explains why terms involving the Ricci tensor Rμνand the scalar curvature Rdo not appear in (5). We want to calculate the large- zbehavior of various scattering amplitudes M−−,++, etc., and compare with the naive expectation (3). We hope that the same trick we usedfor the gauge theory case would also work for gravity. Now the string theory hint, thatM gravitons ∼Mgauge×Mgauge , suggests a factorized structure in the graviton amplitude, and so more or less naturally leads to the guess that the first index λand the second index σofhλσare somehow associated respectively with the two copies of Mgauge . The key to breaking the problem apart is the Bern transformation unlinking these two indices. Examining (5), we see that the only term that links the first index with the secondindex of h λσappears in the term gμνDμhDνh, since h=gμνhμνdoes precisely that. How to get rid of this term? The trick, following Bern and Grant, is to introduce a scalar field φ and add the term 2 gμν∂μφ∂νφ. We are allowed to do this, since φdoes not appear in the tree level graviton scattering amplitude we are studying. (Of course, the theory is changed frompure gravity, and φdoes circulate in loop diagrams for graviton scattering. Some readers may also know that in string theory the graviton appears with a scalar φ, the experimentally unobserved dilaton.) For pedagogical clarity in explaining what we are going to do next, it is best to retreat to the case of the flat background. Focus on the parenthesis in (4), now modified to(∂ μhλσ∂μhλσ−1 2∂μh∂μh+2∂μφ∂μφ). (The normalization of φis, in this context, just chosen for convenience.) Since we can always make a field redefinition (see the appendix tochapter VIII.3) without affecting on-shell scattering amplitudes, we let h λσ→hλσ+ηλσφ (and hence h→h+4φ) andφ→φ+1 2h. You can verify that our parenthesis changes to (∂μhλσ∂μhλσ−2∂μφ∂μφ). Since in this manipulation the role of ηλσis merely to convert hλσintoh, the same transformation works when ηλσis promoted to gλσ. The upshot is that we can effectively rewrite (5) as L=√−g(gμνgλρgστDμhλσDνhρτ−2Rλρσ τhλσhρτ) (6) Now that φhas done his job, we have unceremoniously thrown him out since he doesn’t contribute to the on-shell tree amplitudes we are interested in. We have thus dropped thetermg μν∂μφ∂μφ. There has been quite a bit of formal development and perhaps the reader has lost sight of what we are trying to do. Recall that we want to study the amplitude of a hard gravitonblasting through spacetime. Although a multitude of indices have appeared, as is alwaysthe case with gravity, you should recognize that this Lagrangian is conceptually simple: it N.4. Gravity and Yang-Mills Theory | 517 is quadratic in the quantum field hdescribing the hard graviton and contains some given c-number tensors gλρ(x)andRλρσ τ(x)pertaining to the background. Unlinked melody The important point is that the two indices carried by hλσare now unlinked from each other in the first term in (6). In chapter VIII.1, we learned to trade a “world index” likeλfor a locally flat Lorentz index aby using the vierbein e a λ(x). Here we are invited to introduce two sets of vierbein, eand˜e, with their associated connections ωand˜ω, and writehλσ≡ea λ˜e˜a σha˜a. In reality, of course e=˜eandω=˜ω, but this notation keeps track of the fact that the two sets of indices carried by hλσare unlinked. Note that hλσis treated in our quadratic Lagrangian as just some tensor field living in a curved spacetime specifiedbyg λσ≡ea λ˜e˜a σηa˜a. Also in chapter VIII.1 we emphasized that the covariant derivative acting on vectors carrying a world index and on vectors carrying a locally flat Lorentz index assumes differentforms, D μVν=∂μVν−/Gamma1λ μνVλandDμVa=∂μVa−ωb μaVb, respectively. For pedagogical clarity, I will use two different symbols DandDto denote what is conceptually the same operation. For our problem we have Dλhμν=ea μ˜e˜a νDλha˜a, with Dλha˜a=∂λha˜a−ωb λahb˜a− ˜ω˜b λ˜aha˜b. With this notation, the relevant Lagrangian becomes L=√−g(gμνηab˜η˜a˜bDμha˜aDνhb˜b−2Rab˜a˜bha˜ahb˜b) (7) We are now ready to study the large zbehavior of the scattering amplitude of a hard graviton carrying momentum zq+...blasting through a curved background spacetime gμν. The analysis proceeds much as in the Yang-Mills case discussed in the preceding chap- ter. Focus on the first term:√−ggμνηab˜η˜a˜b(∂μha˜a−ωc μahc˜a−˜ω˜c μ˜aha˜c)(∂νhb˜b−ωd νbhd˜b− ˜ω˜d ν˜bhb˜d). The leading O(z2)behavior comes from the piece containing two derivatives in the first term, namely Llead≡√−ggμνηab˜η˜a˜b∂μha˜a∂νhb˜b, and thus contributes to the am- plitude a term proportional to ηab˜η˜a˜b. In the Yang-Mills case, the Lagrangian contains a hidden “enhanced Lorentz” symmetry. Here the situation is even better: we have not one,but two hidden “enhanced Lorentz” symmetries. The term L leadis evidently left invariant by two separate SO( 3, 1)Lorentz transformations, one operating on the a,bindices, the other on the ˜a,˜bindices. The subleading O(z) behavior comes from the pieces in the first term containing one derivative and one factor of either ωor˜ω, for example√−ggμνηab˜η˜a˜b∂μha˜a(ωd νbhd˜b+ ˜ω˜d ν˜bhb˜d). In this way we find that Mab,˜a˜b→cz2ηab˜η˜a˜b+z(ηab˜A˜a˜b+Aab˜η˜a˜b)+..., with Aaband˜A˜a˜btwo matrices antisymmetric in their indices. To see this, consider for example the piece involving ω(after some relabeling of indices):√−ggμνηac˜η˜a˜b(∂μha˜a)ωb νchb˜b. This gives rise to the term Aab˜η˜a˜binMab,˜a˜b. Note that since the matrix Aabdepends on the spin connection ωab νof the background, all we can say is that it is antisymmetric in its two indices ab. Recall that this is quite analogous to what we did in the Yang-Mills case. 518 | Part N Before reading further, you could now flex your mental muscle and push ahead to obtain the sub-subleading O(z0)behavior. This comes from the pieces in the first term containing two factors of ωand˜ω, for example (again, after some relabeling of indices)√−ggμνηcd˜η˜a˜bωa μcha˜aωb νdhb˜b. All we can now conclude is that this contributes to the scat- tering amplitude a term of the form Bab˜η˜a˜b, with Baban arbitrary matrix. To this order, the second term in (7) also contributes, breaking the “enhanced Lorentz” symmetries com-pletely. Nevertheless, we can still exploit the known symmetry properties of the Riemanncurvature tensor under interchange of its indices to say something about its contributionto the scattering amplitude. Putting it all together we conclude that Mab,˜a˜b→cz2ηab˜η˜a˜b+z(ηab˜A˜a˜b+Aab˜η˜a˜b)+Aab˜a˜b+(ηab˜B˜a˜b+Bab˜η˜a˜b)+O/parenleftbigg1 z/parenrightbigg (8) Compare this with (N.3.12), which states that the amplitude for the scattering of a hard gluon off a background of soft gluons goes like Mab=(cz+...)ηab+Aab+1 zBab+.... Amazingly, you can see that the large- zbehavior for the scattering of a hard graviton off a background of soft gravitons can be obtained by “squaring” the large-z behavior for the scattering of a hard gluon off a background of soft gluons! In other words, Mab,˜a˜b∼ MabM˜a˜b, as far as the large-z behavior is concerned. Just as in the Yang-Mills case, by exploiting gauge identities like pra(z)Ma˜a,b˜b/epsilon1sb˜b(z)= 0, we can determine the large- zbehavior of various helicity amplitudes. For your conve- nience I remind you that (from the preceding chapter) pr(z)=pr+zqandps(z)=ps− zq. Thus the gauge identity just displayed says that qaMa˜a,b˜b/epsilon1sb˜b(z)= −(1/z)p raMa˜a,b˜b/epsilon1sb˜b(z). Recalling [see (1)] that /epsilon1−−μν r(z)=qμqνwe see that in calcu- lating the amplitude M−−,h(z)we can effectively replace /epsilon1−−μν r(z)by(1/z2)pμ rpν r. Thus we can immediately conclude, since /epsilon1++μν s(z)∼z0for large z[see (2], that for example M−−,++(z)→1 z2(9) which is far better than the naive expectation M−−,++ naive(z)→zn−1. Indeed, the horrible ever-escalating behavior with increasing nhas disappeared. Even more remarkably, the large-z behavior of graviton scattering amplitudes is consistent with the string-inspired notion that gravity is “the square of Yang-Mills.” Recall that in gauge theory M−+(z)→1/z. Thus, for large z, indeed M−−,++(z)∼(M−+(z))2. The bottom line here is that the large- zbehavior of gravity is surprisingly benign and vanishes fast enough for the recursion program to work. Gravity is a square? So, is Einstein gravity secretly the square of Yang-Mills theory? Already we have seen in the preceding chapter, anticipating that the recursion pro- gram works, that the primitive 3-point amplitude for gravity (for one helicity configura-tion)(/angbracketleft12/angbracketright 3//angbracketleft23/angbracketright/angbracketleft31/angbracketright)2is the square of the primitive 3-point amplitude for gauge theory N.4. Gravity and Yang-Mills Theory | 519 (/angbracketleft12/angbracketright3//angbracketleft23/angbracketright/angbracketleft31/angbracketright), something that you could have never suspected by staring at√−gR and trFμνFμνuntill you are blue in the face. The calculations in the previous section show that the large-z behavior for the scattering of a hard graviton off a background of soft gravitons could be obtained by “squaring” thelargezbehavior for the scattering of a hard gluon off a background of soft gluons, certainly something that nobody could have anticipated by looking at Lagrangians. Further evidence that the answer to the title of this chapter is “yes” comes from a recent calculation by Bern, Carrasco, and Johansson. Interestingly, they do not strip the colorfrom a Yang-Mills theory, but instead show that they can write the “color-dressed” treeamplitudes in the form Atree(1, 2, ...,n)=/summationdisplay anaca (/Pi1jp2 j)a(10) It is beyond the scope of this book to explain in detail how this expression is obtained. I merely state that the index alabels an individual diagram. For each diagram, the amplitude may be written as the product of a kinematic function naof the momenta and a color factor ca, divided by the product of the momenta pjcarried by the internal lines. (I do not explain here how naandcaare defined.) The tree amplitude is then given by a sum over all tree diagrams. Bern et al. then conjecture that the n-graviton scattering amplitude at the tree level is given, amazingly, by Mtree gravity(1, 2, ...,n)=/summationdisplay anana (/Pi1jp2 j)a(11) They have checked by explicit computation that their conjecture in fact holds up to n=8. Furthermore, they have also verified that the conjecture, suitably generalized, also holdsfor the various supercousins of Einstein gravity and Yang-Mills theory. Thus the evidence is extremely strong that, yes indeed, Einstein gravity is secretly the square of Yang-Mills theory, at least at the level of tree amplitudes. However, as of thiswriting (February 2009), there is no definitive understanding within field theory. The finalword on the subject has yet to be said, and it is not even clear what the final path to the finalword might be. I would be foolish indeed to discuss this further in a textbook when theentire subject is being rapidly developed. By the time this book is published, the conjecturethat Einstein gravity is the square of Yang-Mills theory may well have been proved. If not,then nothing would please me more than if a reader of this textbook could go on and proveit, hopefully not just at the tree level, but to all orders. What is the simplest field theory? The uninitiated would likely answer ϕ4theory. Indeed, field theory texts almost all start with some kind of scalar field theory. Even I am not able to do any better. But the sophisticated,namely you, now that you have reached the end of this text, realize that the more symmetrythe theory has the better. To theoretical physicists, simplicity actually secretly means 520 | Part N symmetry. Incidentally, I have always hated scalar field theories, and have ventured to say so publicly. It is hard to like the action L=1 2(∂ϕ)2−λϕ4, so barren of color and flavor. Some of the major problems facing particle physics, such as the hierarchy problem, mayeventually turn out to stem from our not having mastered scalar field theory. Of course, scalar field theory is the simplest in the superficial sense that you need to know the least to approach it. As I said in chapter I.12, once one is familiar with scalarfield theory the rest consists of “merely” decorating the field with various indices describingspacetime or internal symmetries. But the symmetry and the resulting structure provideus with handles to grab on to. Both Yang-Mills theory and Einstein gravity have an internallogic sorely lacking in scalar field theory. As I mentioned in chapters VII.3 and VIII.4,the consensus view is that the first exactly soluble field theory would almost certainly beN=4 supersymmetric Yang-Mills theory, the supercousin of pure Yang-Mills theory. The remarkable recent developments described in the last three chapters have only reinforcedthis view. Almost beyond belief, even gravity may be simpler than we had long thought.For large complexified momentum, graviton scattering for some helicity arrangementsactually behaves better than gluon scattering. The evidence is mounting that Einsteingravity may in some sense be the square of Yang-Mills theory. So now we are left with theamusing thought that the simplest field theory may well end up being gravity or N=8 supergravity with its maximal supersymmetry. (At this point, a friend of mine who workswithN=8 supergravity pipes up, “It sure doesn’t look simpler if you are the guy doing the calculation!” It is clear from the simple dimensional argument of chapter III.2 thatas one goes to higher order, the numerator of the Feynman integrand quickly becomesextremely involved.) Only time will tell who will win the simplest field theory contest, but we do have two convincing candidates. More Closing Words In the closing words to the first edition of this book, I wrote that Yang-Mills theory was almost begging for a better notation that would lay bare the deeper structure of thetheory. Oy, the excess baggage we have to carry! Ten thousand terms instead of one. Insome respects, the spinor helicity formalism and the recursion program explained inchapters N.2–N.4 provide a partial answer to that pious wish. Imagine some theorist idly wondering, after 1865, if there were a better notation to describe the six fields E x,Ey,Ez,Bx,By,Bzfor which Maxwell had written 20 equations (since he did not use vector notation). We can even fantasize that by fooling around withnumerology (“Look, 4 .3/2=6!”), this “crackpot” came up with an antisymmetric 4 by 4 matrix he called F. Shoehorning Maxwell’s equations in vacuum (some of them stating that the time variation of EandBis related to the space variation of EandB) into this strange notation, this guy could even stumble on a secret connection between space andtime. The spinor helicity formalism and the recursion program, though elegant, are still rooted in the perturbative expansion of the 1940s. Can they be pushed into the nonpertur-bative regime? There have been attempts in that direction. In all previous revolutions in physics, a formerly cherished concept has to be jettisoned. If we are poised before another conceptual shift, something else might have to go. Lorentzinvariance, perhaps? More likely, we may have to abandon strict locality. Again, in closingwords I mumble something (from steepest descent to integral to what?) about modifyingthe form of the path integral. The recursion program and the resuscitated S-matrix ap- proach might be a step in this direction, formulating field theory while avoiding mention of a local Lagrangian. But we need analyticity, and of course analyticity follows from local-ity and causality, as far as we understand. We know also that even local field theory couldspawn non-local constructs, most notoriously the horizon of a black hole. But there the 522 | Closing Words to Part N dynamics bends the causal structure of spacetime out of whack. The lack of strict locality is not built into the laws of physics. Of course, we also know how to imbue physics with non-locality right from the start. We have Wilson’s lattice formulation of gauge theory, and more recently, Wen’s intriguinglattice formulation of gravity. When I showed the last three chapters to our friend SE, she mused, after some reflection, “Now I see what theorists could always do when in doubt: enhance the symmetry and makeit local, complexify and bow to Cauchy, and take a square root when possible!” I nodded, “These are the three ways of the warrior theorist: I call them the Einstein way, the Heisenberg way, and the Dirac way. They were wildly successful in the past, andperhaps they will work in the future as well.” With this edition of my textbook, I can no doubt count on a new group of readers to come up with fresh insights into field theory. As these new chapters suggest, there maystill be plenty of secret structures to uncover. And thus field theory marches on. Finally, I reveal the origin of the quote at the start of the preface to the second edition. As a kid, Feynman came across a calculus book 1that proclaimed “What one fool can do, another can.” He was thus inspired to master calculus. Now that you have mastered quantum field theory, you can switch from the “understand” in the preface to the “do” in these closingwords. 1Silvanus P. Thompson (1851–1916), Calculus Made Easy , 1910, updated by Martin Gardner, St. Martin’s Press, (1998). I am kind of trying to do for quantum field theory what Thompson did for calculus. Appendix AGaussian Integration and the Central Identity of Quantum Field Theory The basic Gaussian: /integraldisplay+∞ −∞dxe−1 2x2=√ 2π (1) The scaled Gaussian: /integraldisplay+∞ −∞dxe−1 2ax2=/parenleftbigg2π a/parenrightbigg1 2 (2) Moments: /integraldisplay+∞ −∞dxe−1 2ax2x2n=/parenleftbigg2π a/parenrightbigg1 21 an(2n−1)(2n−3)...5.3.1,n≥1 (3) Gaussian with source: /integraldisplay+∞ −∞dxe−1 2ax2+Jx=/parenleftbigg2π a/parenrightbigg1 2 eJ2/2a(4) /integraldisplay+∞ −∞dxe−1 2ax2+iJx=/parenleftbigg2π a/parenrightbigg1 2 e−J2/2a(5) /integraldisplay+∞ −∞dxe1 2iax2+iJx=/parenleftbigg2πi a/parenrightbigg1 2 e−iJ2/2a(6) /integraldisplay+∞ −∞/integraldisplay+∞ −∞.../integraldisplay+∞ −∞dx1dx2...dxNei 2x.A.x+iJ.x=/parenleftbigg(2πi)N det[A]/parenrightbigg1 2 e−(i/2)J.A−1.J(7) /integraldisplay+∞ −∞/integraldisplay+∞ −∞.../integraldisplay+∞ −∞dx1dx2...dxNe−1 2x.A.x+J.x=/parenleftbigg(2π)N det[A]/parenrightbigg1 2 e1 2J.A−1.J(8) In what follows, we omit an overall factor. Central identity of quantum field theory: /integraldisplay Dϕe−1 2ϕ.K.ϕ−V( ϕ) +J.ϕ=e−V( δ / δJ)e1 2J.K−1.J(9) 524 | Appendix A. Gaussian Integration A trivial variation: /integraldisplay Dϕe−1 2ϕ.K.ϕ+J.ϕ=e1 2J.K−1.J(10) Variations: /integraldisplay Dϕe(i/2)ϕ.K.ϕ+iJ.ϕ=e−(i/2)J.K−1.J(11) /integraldisplay Dϕei/integraltext ddx[1 2ϕ(x)Kϕ(x)+J(x)ϕ(x) ]=ei/integraltext ddx[−1 2J(x)K−1J(x) ](12) /integraldisplay Dϕe−/integraltext ddx[1 2ϕ(x)Kϕ(x)+J(x)ϕ(x) ]=e/integraltext ddx[1 2J(x)K−1J(x) ](13) (where KorK−1or both may be nonlocal) A specific example: /integraldisplay Dϕei/integraltext ddx[(λ/2)ϕ2+ϕ¯ψψ]=ei/integraltext ddx[−(1/2λ)(¯ψψ)2](14) ForKhermitean with ϕcomplex: /integraldisplay Dϕ†Dϕe−ϕ†.K.ϕ+J†.ϕ+ϕ†.J=eJ†.K−1.J(15) As noted earlier, various numerical factors have been swept under the integration measure. In applying these formulas, be sure that these factors are not relevant for your purposes. Appendix B A Brief Review of Group Theory I give here a brief review of the group theory I will need in the text. I assume that you have been exposed to some group theory, otherwise this instant review might not be intelligible. Most of the concepts are illustrated withexamples, and it goes without saying that you should work out all the examples and verify the assertions madewithout proof. SO(N) The special orthogonal group SO(N) consists of all NbyNreal matrices Othat are orthogonal OTO=1 (1) and have unit determinant detO=1 (2) We denote the element in the ith row and jth column by Oij. The group SO(N) consists of rotations in N- dimensional Euclidean space and its defining or fundamental representation is given by the Ncomponent vector /vectorv={vj,j=1 ,..., N}, which transforms under the action of the group element Oaccording to (as always, all repeated indices are summed over) vi→v/primei=Oijvj(3) We define tensors as objects that transform as if they are equal to the product of vectors. For example, the tensor Tijktransforms according to Tijk→T/primeijk=OilOjmOknTlmn(4) as if it is equal to the product vivjvk. The emphasis is on the phrase “as if”: Tijkis not to be thought of as being equal to vivjvk. It is important to develop some “feel” or intuition for groups and their representations. Some people find it helpful to picture a certain number of objects being acted upon by the group and transformed into linearcombinations of each other. Thus, picture T ijkasN3objects being scrambled together. Tensors furnish representations of the group. In our particular example, each group element is represented by an N3byN3matrix acting on the N3objects Tijk. The number of objects in a tensor is called the dimension of the representation. It may well be that any given object in a representation does not transform, under all the elements of the group, into a linear combination of all the other objects, but only into a subset of them. Let me illustrate with an example. 526 | Appendix B. Brief Review of Group Theory Consider Tij→T/primeij=OilOjmTlm. Form the symmetric Sij≡1 2(Tij+Tji)and antisymmetric combinations Aij≡1 2(Tij−Tji). The symmetric combination Sijtransforms into OilOjmSlm, which is obviously symmetric. Similarly, Aijtransforms into OilOjmAlm, which is obviously antisymmetric . In other words, the set of N2 objects contained in Tijsplit into two sets:1 2N(N+1)objects contained in Sijand1 2N(N−1)objects contained inAij. TheSij’s transform among themselves and the Aij’s transform among themselves. The representation furnished by Tijis said to be reducible: It breaks apart into two representations. Obviously, representations that do not break apart are called irreducible. We just exploited the obvious fact that the symmetry properties of a tensor under permutation of its indices is not changed by the group transformation, namely that the indices on a tensor transform independently, as in (4).The various possible symmetry properties may be classified with Young tableaux, which is useful in a generaltreatment of group theory. Fortunately, in the field theory literature one rarely encounters a tensor with suchcomplex symmetry properties that one has to learn about Young tableaux. Another way of saying this is that we can restrict our attention to tensors with definite symmetry properties under permutation of their indices. In our specific example, we can always take T ijto be either symmetric or antisymmetric under the exchange of iandj. We have yet to use the properties (1) and (2). Given a symmetric tensor Tijconsider the combination T≡δijTij, known as the trace. Then T→δijT/primeij=δijOilOjmTlm=(OT)liδijOjmTlm=δlmTlm=T, where we used (1). In other words, Ttransforms into itself. We can subtract the trace from Tijforming the traceless tensor Qij≡Tij−(1/N)δijT. The1 2N(N+1)−1 objects contained in Qijtransform among themselves. To summarize, given two vectors vandw, we can form a tensor, and decompose the tensor into a symmetric traceless combination, a trace, and an antisymmetric tensor. This process is written as N⊗N=[1 2N(N+1)−1]⊕1⊕1 2N(N−1) (5) In particular, for SO( 3),3⊗3=5⊕1⊕3, a relation you should be familiar with from courses on mechanics and electromagnetism. There are two conventions for naming representations. We can simply give the dimension of the representa- tion. (This can occasionally be ambiguous: Two distinct representations may happen to have the same dimension.)Alternatively, we can specify the symmetry properties of the tensor furnishing the representation. For instance,the representation furnished by a totally antisymmetric tensor of nindices is often denoted by [ n] and the rep- resentation furnished by a totally symmetric traceless tensor of nindices by {n}. Obviously, [1] ={1}. In this notation, the decomposition in (5) can be written as {1}⊗{ 1}={ 2}⊕{ 0}⊕[2]. For the group SO( 3), with its long standing in physics, the confusion over names is almost worse than in reading Russian novels: For instance,{1}is also known as pand{2}asd. We have yet to use (2). Using the antisymmetric symbol ε 123. . . N, we write (2) as εi1i2...iNOi11Oi22...OiNN=1 (6) or equivalently εi1i2...iNOi1j1Oi2j2...OiNjN=εj1j2...jN (7) By multiplying (7) by OTrepeatedly, we can obviously generate more identities. Instead of drowning in a sea of indices, let me explain this point by specializing to say N=3. Thus, multiplying (7) by (OT)jNkN, we obtain εi1i2i3Oi1j1Oi2j2=εj1j2j3(OT)j3i3 Speaking loosely, we can think of moving some of the O’s on the left hand side of (7) to the right hand side, where they become OT’s. Using these identities, you can easily show that [ n] is equivalent to [N −n], that is, these two representations transform in the same way. For example, as is well known, in SO( 3)the antisymmetric 2-index tensor is equivalent to the vector. (The cross product of two vectors is a vector.) Any orthogonal matrix can be written as O=eA. The conditions (1) and (2) imply that Ais real and antisymmetric, so that Amay be expressed as a linear combination of N(N−1)/2 antisymmetric matrices denoted by iJij:O=eiθijJij(with repeated indices summed over). We have defined Jijas imaginary and antisymmetric and hence hermitean. Since the commutator [ Jij,Jkl] is antihermitean, it can be written as a linear combination of the iJ’s. Ironically, some students are confused at this point because of their familiarity with SO( 3), which has special properties that do not generalize to SO(N). In speaking about rotations in 3-dimensional space we can specify a rotation as either around say the third axis, with the corresponding generator J3, or as in the (1-2)-plane, with the corresponding generator J12=−J21. Appendix B. Brief Review of Group Theory | 527 In higher dimensions, for example 10-dimensional space, we can speak of a rotation in the (6-7)-plane, with the corresponding generator J67=−J76, but it is nonsense to speak of a rotation around the fifth axis. Thus, to generalize to higher dimensions we should write the standard commutation relation [ J1,J2]=iJ3forSO( 3)as [J23,J31]=iJ12, which can be generalized immediately to [Jij,Jkl]=i(δikJjl−δjkJil+δjlJik−δilJjk) (8) The right hand side reflects the antisymmetric character of Jij=−Jji. A potential confusion some students may have about the notation: Jijdenotes a matrix generating rotation in the (i -j)-plane, a matrix with element (Jij)klin the k-th row and l-th column. The indices i,j,k, andlall run from 1 to N, but in (Jij)klthe set {ij} and the set {kl}should be distinguished conceptually: The former labels the generator and the latter are matricial indices when the generator is regarded as a matrix. As an exercise, write down (Jij)klexplicitly and obtain (8) by direct computation. In studying group theory, as I have already remarked, one source of confusion comes from the fact that some of the smaller groups, which we tend to encounter first in our studies, have special properties that do notgeneralize. The special property of SO( 3)we just noted is due to the fact that the antisymmetric symbol ε ijkcarries three indices and thus Jijmay be written as Jk≡1 2εijkJij.F o rSO( 4)the antisymmetric symbol εijklcarries four indices and we can form the combinations1 2(Jij±1 2εijklJkl). Define J1 ±≡1 2(J23±J14),J2 ±≡1 2(J31±J24), and J3 ±≡1 2(J12±J34). By explicit computation, show that [ Ji +,Jj +]=iεijkJk +,[Ji −,Jj −]=iεijkJk −, and [ Ji +,Jj −]=0. This proves the well-known theorem that SO( 4)is locally isomorphic to SO( 3)⊗SO( 3). I assume that you know that SO( 3)is locally isomorphic to SU( 2). If you don’t, I give a brief review below. With a few i’s included here and there, these two results prove the statement that the Lorentz group SO( 3, 1) is locally isomorphic to SU( 2)⊗SU( 2), which we proved explicitly in chapter II.3. The Lorentz group can be thought of as an “analytic continuation” of the rotation group SO( 4). See below for a more precise statement. One highly non-obvious result of group theory is that SO(N) contains representations other than vector and tensor. I develop the relevant group theory for the spinor representations in chapter VII.7. SU(N) We next turn to the special unitary group SU(N) consisting of all NbyNmatrices Uthat are unitary U†U=1 (9) and have unit determinant detU=1 (10) The story of SU(N) has more or less the same plot as the story of SO(N) with the crucial difference that the tensors of the unitary groups can carry both upper and lower indices. We denote the element in the ith row and jth column by Ui j; the wisdom of this notation will soon become apparent. The defining or fundamental representation of SU(N) consists of Nobjects ϕj,j=1 ,..., N, that transform under the action of the group element Uaccording to ϕi→ϕ/primei=Ui jϕj(11) Taking the complex conjugate of (11) we have ϕ∗i→(Ui j)∗ϕ∗j=(U†)j iϕ∗j(12) We invite ourselves to define an object we write as ϕithat transforms in the same way as ϕ∗i; thus ϕi→ϕ/prime i=(U†)j iϕj (13) Note that we did not say that ϕiis equal to ϕ∗i; we merely said that ϕiandϕ∗itransform in the same way. As before, we can have tensors. The tensor ϕij k, for example, transforms as if it is equal to the product ϕiϕjϕk: ϕij k→ϕ/primeij k=Ui lUj m(U†)nkϕlm n(14) 528 | Appendix B. Brief Review of Group Theory Again, we emphasize that we did not say that ϕij kis equal to ϕiϕjϕk. (In some books ϕiis called a covariant vector andϕia contravariant vector. A tensor ϕ......... withmupper indices and nlower indices is defined to transform as if it is equal to the product of mcovariant vectors and ncontravariant vectors.) The possibility of complex conjugation in SU(N) leads naturally to having indices “upstairs” and “downstairs.” Note that (9) can be written out explicitly as (U†)k iUj k=δj iand thus the Kronecker delta in SU(N) carries one upper and one lower index. It is important when taking traces that we set an upper index equal to a lower index and sum over them: for example, we can consider δk jϕij k≡ϕij j, which transforms as ϕij j→Ui lUj m(U†)njϕlm n=Ui lϕlm m(15) where we have used (9). In other words, ϕij j, the trace of ϕij k, denote Nobjects that transform into linear combinations of each other in the same way as ϕi. Thus, given a tensor, we can always subtract out its trace. As in the discussion for SO(N), tensors furnish representations of the group. The discussion proceeds as before. The symmetry properties of a tensor under permutation of its indices are not changed by the grouptransformation. Another way of saying this is that given a tensor we can always take it to have definite symmetry properties under permutation of its upper indices and under permutation of its lower indices. In our specific example, we can always take ϕij kto be either symmetric or antisymmetric under the exchange of iandjand to be traceless. Thus, the symmetric traceless tensor ϕij kfurnishes a representation with dimension1 2N2(N+1)−Nand the antisymmetric traceless tensor ϕij ka representation with dimension1 2N2(N−1)−N. Thus, in summary, the irreducible representations of SU(N) are realized by traceless tensors with definite symmetry properties under permutation of indices. For example, in SU( 5), some commonly encountered representations are ϕi,ϕij(antisymmetric), ϕij(symmetric), ϕi j,ϕij k(antisymmetric in the upper indices and traceless) with dimensions 5, 10, 15, 24, and 45, respectively. Convince yourself that for SU(N) the dimensions of the representations defined by these tensors are N,N(N−1)/2,N(N+1)/2,N2−1, and1 2N2(N−1)−N, respectively. The representation defined by the traceless tensor ϕi jis known as the adjoint representation. By definition, it transforms according to ϕi j→ϕ/primei j=Ui l(U†)njϕl n=Ui lϕl n(U†)n j. We are thus invited to regard ϕi jas a matrix transforming according to ϕ→ϕ/prime=UϕU†(16) Note that if ϕis hermitean it stays hermitean, and thus we can take ϕto be a hermitean traceless matrix. (If ϕ is antihermitean we can always multiply it by i.)Another way of saying this is that given a hermitean traceless matrix X,UXU†is also hermitean and traceless if Uis an element of SU(N) . As in the SO(N) story, representations of SU(N) have many names. For example, we can refer to the representation furnished by a tensor with mupper and nlower indices as (m,n). Alternatively, we can refer to them by their dimensions, with an asterisk to distinguish representations with mostly lower indices from therepresentations with mostly upper indices. For example, an alias for (1, 0)isNand for (0, 1)isN ∗. A square bracket is used to indicate that the indices are antisymmetric and a curly bracket indicate that the indices aresymmetric. Thus, the 10 of SU( 5)is also known as [2, 0] =[2], where as indicated the 0 (no lower index) is suppressed. Similarly, 10 ∗is also known as [0, 2] =[2]∗. The condition (10) can be written as either εi1i2...iNUi1 1Ui2 2...UiN N=1 (17) or εi1i2...iNU1 i1U2 i2...UN iN=1 (18) Thus, we have two antisymmetric symbols εi1i2...iNandεi1i2...iNthat we can use to raise and lower indices. Again, we can immediately generalize (17) to εi1i2...iNUi1 j1Ui2 j2...UiN jN=εj1j2...jN and multiplying this identity by (U†)jNpNand summing over jNwe obtain εi1i2...pNUi1 j1Ui2 j2...UiN−1 jN−1=εj1j2...jN(U†)jNpN Appendix B. Brief Review of Group Theory | 529 Clearly, by repeating this process, we can peel off the U’s on the left hand side and put them back as U†’s on the right hand side. We can play a similar game with (18). To avoid drowning in a sea of indices, let me show you how to raise and lower indices in a specific example rather than in general. Consider the tensor ϕij kinSU( 4). We expect that the tensor ϕkpq≡ϕij kεijpq will transform as a tensor with three lower indices. Indeed, ϕkpq≡ϕij kεijpq→εijpqUi lUj m(U†)nkϕlm n=εlmst(U†)s p(U†)tq(U†)nkϕlm n=(U†)n k(U†)sp(U†)tqϕnst As in SO(N) we can look at the generators of SU(N) by noting that any unitary matrix can be written as U=eiH, with Hhermitean and traceless as required by (9) and (10). There are (N2−1)linearly independent NbyNhermitean traceless matrices Ta(a=1 ,2 ,..., N2−1). Any NbyNhermitean traceless matrix can be written as a linear combination of the Ta’s and thus we can write U=eiθaTa, where θaare real numbers and the index ais summed over. Since the commutator [ Ta,Tb] is antihermitean and traceless, it can also be written as a linear combination of the Ta’s: [Ta,Tb]=ifabcTc(19) (with the index csummed over.) The commutation relations (19) define the Lie algebra of SU(N), and fabcare known as the structure constants. For SU( 2)the structure constants fabcare simply given by the antisymmetric symbol εabc. Sometimes students are confused by how the generators act. Consider an infinitesimal transformation U/similarequal1+iθaTa. On the defining representation, ϕi→Ui jϕj/similarequalϕi+iθa(Ta)i jϕj. Thus, the ath generator acting on the defining representation gives Taϕ. Now consider the adjoint representation (16) ϕ→ϕ/prime/similarequal(1+iθaTa)ϕ(1+iθaTa)†/similarequalϕ+iθaTaϕ−ϕiθaTa=ϕ+iθa[Ta,ϕ] (20) In other words, the ath generator acting on the adjoint representation gives [ Ta,ϕ]. Perhaps some students are confused by the fact that ϕis used as a generic symbol to denote different objects. Since the adjoint representation ϕis hermitean and traceless it can also be written as a linear combination of the generators, thus ϕ=ϕbTb. Using (19) we can thus also write (20) as ϕc→ϕ/primec/similarequalϕc−fabcθaϕb. In particular forSU( 2), the three objects ϕatransform as a 3-vector. (Note the notation: ϕais not to be confused with ϕi:i n SU( 2)the index a=1, 2, 3 while i=1, 2.) This last remark essentially amounts to a proof that SU( 2)is locally isomorphic to SO( 3). I will now give a somewhat more formal proof. Any 2 by 2 hermitean traceless matrix Xcan be written as a linear combination of the three Pauli matrices X=/vectorx./vectorσwith three real coefficients (x1,x2,x3), which we regard as the components of a 3-vector /vectorx. For any element UofSU( 2),X/prime≡U†XU is hermitean and traceless, so that we can write X/prime=/vectorx/prime./vectorσ. Note that we have implicitly used the first defining property of an SU( 2)matrix (9). By explicit computation, we find det X=−/vectorx2. Invoking the second defining property of an SU( 2)matrix (10), we obtain det X/prime=detX and thus /vectorx/prime2=/vectorx2. The 3-vector /vectorxis rotated into the 3-vector /vectorx/prime. Thus we can associate a rotation with any given U. Since Uand−U are associated with the same rotation, this gives a double covering of SO( 3)bySU( 2).A physicist would just say that when a spin1 2particle is rotated through 2 π, its wave function changes sign. The map clearly preserves group multiplication: if two elements U1andU2ofSU( 2)are mapped to the rotations R1 andR2respectively, then the element U1U2is mapped to the rotation R1R2. Alternatively, noting that tr X2=/vectorx2 and tr X/prime2=trX2, we obtain the same conclusion. Once again, the two special unitary groups that most students learn first, namely SU( 2)andSU( 3), have special properties that do not generalize to SU(N), just as SO( 3)has special properties that do not generalize to SO(N) , possibly leading to confusion. ForSU( 2), because the antisymmetric symbol εijandεijcarry two indices, it suffices to consider only tensors with upper indices, all symmetrized: We can raise all lower indices of any tensor by contracting with εijrepeatedly. After this is done, we can remove any pair of indices in which the tensor is antisymmetric by contracting withε ij. In particular, ϕi=εijϕj, which can be stated equivalently in terms of a special property of the Pauli matrices σ2σ∗ aσ2=−σa (21) 530 | Appendix B. Brief Review of Group Theory so that σ2(ei/vectorθ/vectorσ)∗σ2=ei/vectorθ/vectorσ(22) ForSU( 2)(11) becomes ϕi→ϕ/primei=(ei/vectorθ/vectorσ)i jϕj Complex conjugating, we obtain ϕ∗i→[(ei/vectorθ/vectorσ)i j]∗ϕ∗j=[(−iσ 2)ei/vectorθ/vectorσ(iσ 2)]i jϕ∗j and so iσ2ϕ∗→ei/vectorθ/vectorσ(iσ2ϕ∗) We learn that iσ2ϕ∗transform in the same way as ϕ. Recall that we define ϕito transform in the same way as ϕ∗i. Thus, εijϕjtransforms in the same way as ϕi. In the jargon, SU( 2)is said to have only real and pseudoreal representations, but not complex representations. A pseudoreal representation is equivalent to its complexconjugate upon a similarity transformation. Recall that (21) figures into our discussion of charge conjugation inchapter II.1 and of the Higgs doublet in chapter VII.2. ForSU( 3)it suffices to consider only tensors with all their upper indices symmetrized and all their lower indices symmetrized. Thus, the representations of SU( 3)are uniquely labeled by two integers (m,n), where m andndenote the number of upper and lower indices. The reason is that the antisymmetric symbols ε ijkand εijkcarry three indices. We can always trade a pair of lower indices in which the tensor is antisymmetric for one upper index, and similarly for upper indices. You can see easily that these special properties do not generalize beyond SU( 2)andSU( 3). Multiplying representations together In a course on quantum mechanics you learn how to combine angular momentum. We have already encountered this concept in (5), which when specialized to SO( 3), tells us that 3 ⊗3=5⊕1⊕3, as we noted. This is sometimes described by saying that when we combine two angular momentum L=1 states we obtain L=0, 1, 2. Students are justifiably confused when this procedure is also known as addition of angular momentum. Given two tensors ϕandηofSU(N) , with mupper and nlower indices and with m/primeupper and n/primelower indices, respectively, we can consider a tensor Twith(m+m/prime)upper and (n+n/prime)lower indices that transforms in the same way as the product ϕη. We can then reduce Tby the various operations described above. This operation of multiplying two representations together is of course of fundamental importance in physics. In quantum fieldtheory, for example, we multiply fields together to construct the Lagrangian. As an example, multiply 5 ∗and 10 in SU( 5). To reduce Tij k=ϕkηijwe separate out the trace ϕkηkj(which transforms as a 5 )after which there is nothing more we can do. Thus, 5∗⊗10=5⊕45 (23) As another example, consider 10 ⊗10:ϕijηkl. It is easiest to write ηklequivalently as a tensor with three lower indices εmnhklηkl. The product 10 ⊗10 then carries two upper and three lower indices and we will write it as Tij mnh. Taking traces, we separate out Tij mij, which we recognize as 5∗, and the traceless part of Tij mnj, which we recognize as 45∗(see above), thus obtaining: 10⊗10=5∗⊕45∗⊕50∗(24) As exercises you can work out 5⊗5=10⊕15 (25) and 5⊗5∗=1⊕24 (26) You should recognize the 24 as the adjoint. Appendix B. Brief Review of Group Theory | 531 In physics we are often called upon to multiply a tensor by itself. Statistics then plays a role. For instance, SU( 5)grand unification contains a scalar field ϕitransforming as 5. Because of Bose statistics, the product ϕiϕj contains only the 15. Restriction to subgroup To explain the next group theoretic concept, let me take a physical example. The SU(3) of Gell-Mann and Ne’eman transforms the three quarks u,d, andsinto linear combinations of each other. It contains as a subgroup the isospin SU( 2)of Heisenberg, which transforms uandd, but leaves salone. In other words, upon restriction to the subgroup SU( 2)the irreducible representation 3 of SU( 3)decomposes as 3→2⊕1 (27) Consider an irreducible representation with dimension dof some group G. When we restrict our attention to a subgroup H, the set of dobjects will in general decompose into nsubsets, containing d1,d2,...,d nobjects, such that the objects of each subset only transform among themselves under the action of H. This makes obvious sense since there are fewer transformations in Hthan in G. The decomposition of the fundamental or defining representation specifies how the subgroup His embedded inG. Since all representations may be built up as products of the fundamental representation, once we know how the fundamental representation decomposes, we know how all representations decompose. For example,inSU( 3) 3⊗3∗=8⊕1 (28) while in SU( 2) (2⊕1)⊗(2⊕1)=(3⊕1)⊕2⊕2⊕1 (29) Comparing (28) and (29) we learn that 8→3⊕1⊕2⊕2. (30) Alternatively, we can simply look at the tensors involved. Consider ϕiofSU( 3)where the index itakes on the value 1, 2, 3. Let the index μtakes on the value 1, 2. Obviously, ϕi={ϕμ,ϕ3}corresponds to an explicit display of (27). Then ϕi j={¯ϕμ ν,ϕμ 3,ϕ3 μ,ϕ3 3}, where the bar on ¯ϕμ νis to remind us that it is traceless. This corresponds precisely to (30). Actually, SU( 3)also contains the larger subgroup SU( 2)⊗U(1), where the U(1)is generated by the traceless hermitean matrix ⎛ ⎜⎜⎝−100 0−10 00 2⎞ ⎟⎟⎠ We can then write (27) as 3 →(2,−1)⊕(1, 2), where the notation is almost self-explanatory. Thus, (2,−1) denotes a 2 under SU( 2)with “charge” −1 under U(1). In the text, we will decompose various representations of SU( 5)andSO( 10). Everything we do there will simply be somewhat more elaborate versions of what we did here. More on SO(4),SO(3,1), and SO(2,2) In chapter II.3 you learned that acting on the two objects ψαwithα=1, 2 in the spinor representation (1 2,0), the generators of rotation and boost are represented by Ji=1 2σiandiKi=1 2σi, respectively. I remind you that the equal sign means “represented by.” For most purposes (for example, classifying quantum fields) and at thelevel of rigor of this book, it suffices to think of the Lie algebra generated by commuting J iandKi. Occasionally, however, it is useful to contemplate the actual group with group elements ei/vectorθ/vectorJandei/vectorϕ/vectorK. In the spinor representation (1 2,0)the group elements are represented by ei/vectorθ/vectorσ 2ande/vectorϕ/vectorσ 2. While ei/vectorθ/vectorσ 2is special unitary, the 2 by 2 matrix e/vectorϕ/vectorσ 2, bereft of the i, is merely special but not unitary. (Incidentally, to verify these and subsequent statements, since you understand rotation thoroughly, you could, without loss of generality, choose 532 | Appendix B. Brief Review of Group Theory /vectorϕto point along the third axis, in which case e/vectorϕ/vectorσ 2is diagonal with elements eϕ 2ande−ϕ 2. Thus, while the matrix is not unitary, its determinant is manifestly equal to 1.) This set of matrices defines the multiplicative groupSL( 2,C), consisting of all 2 by 2 complex-valued matrices with unit determinant. Let us count the number of generators of this group. Two conditions on the determinant (real part =1, imaginary part =0) cut the four complex entries containing eight real numbers down to six numbers, which accounts for the six generators of the Lorentz group SO( 3, 1). To exhibit the map explicitly, we extend the earlier discussion showing that SU( 2)covers SO( 3). Consider the most general 2 by 2 hermitean matrix XM=x0I−/vectorx./vectorσ=/parenleftBiggx0−x3x1−ix2 x1+ix2x0+x3/parenrightBigg (31) By explicit computation, det XM=(x0)2−/vectorx2. (To see this instantly, choose /vectorxto point along the third axis and invoke rotational invariance.) Now consider X/prime M=L†XML, with Lan element of SL( 2,C). Manifestly, detX/prime M=detXMand thus the transformation preserves (x0)2−/vectorx2and hence corresponds to Lorentz transfor- mations. Since Land−L give the same transformation x→x/prime, we see that SL( 2,C)double covers SO( 3, 1). Mathematicians say that SO( 3, 1)=SL( 2,C)/Z 2.I fL is also unitary, then x0/prime=x0and the transformation is a rotation. The SU( 2)subgroup of SL( 2,C)double covers the rotation subgroup SO( 3)of the Lorentz group SO( 3, 1), that is, SO( 3)=SU( 2)/Z 2. Incidentally, if we introduce an iat a strategic location and define the 2 by 2 matrix XE=x4I+i/vectorx./vectorσ, regarding (/vectorx,x4)as a 4-dimensional vector, we have det XE=(x4)2+/vectorx2, the Euclidean length squared of the 4-vector. (Once again, choose /vectorxto point along the third axis so that XEis a diagonal matrix with elements x4±ix3.) Since ei/vectorθ/vectorσ 2=cosθ 2+isinθ 2(ˆθ.σ)withˆθa unit vector in the θdirection (to see this, once again choose /vectorθto point along the 3rdaxis), we see that XE/((x4)2+/vectorx2)1 2is an element of SU( 2). (We will come back to this observation in the next section.) Thus, for any two elements UandVofSU( 2), the matrix X/prime E=V†XEUcan also be decomposed in the form X/prime E=x/prime4I+i/vectorx/prime./vectorσ. Evidently, det X/prime E=detXE. Thus the transformation preserves (x4)2+/vectorx2and describes an element of SO( 4). This shows explicitly that SO( 4)is locally isomorphic to SU( 2)⊗SU( 2).I f V=U, we have a rotation, and if V†=U, the Euclidean analog of a boost. Note that while the rotation group SO( 3)is compact, the Lorentz group SO( 3, 1)is not, since the range of the boost parameters /vectorϕis unbounded. In contrast, the group SO( 4)is compact and thus can be covered by a compact group, namely, SU( 2)⊗SU( 2), but the noncompact group SO( 3, 1)cannot be. At this point, having done SO( 4)andSO( 3, 1), I might as well (with a wink toward the nuts who complained that this book is not encyclopedic enough) throw in the group SO( 2, 2)for use in part N. Let us strip the Pauli matrix σ2(kind of a “troublemaker” or at least an odd man out) of his iand define (just for this paragraph) σ2≡/parenleftBigg0−1 10/parenrightBigg Any real 2 by 2 matrix XHcould be decomposed as XH=x4I+/vectorx./vectorσ. Now det XH=(x4)2+(x2)2−(x3)2−(x1)2, the quadratic form of a spacetime with two time and two space coordinates. The set of all linear transformations(with unit determinant) on (x 1,x2,x3,x4)that preserve this quadratic form defines the group SO( 2, 2). Introduce the multiplicative group SL( 2,R)consisting of all 2 by 2 real-valued matrices with unit determinant. For any two elements LlandLrof this group, consider the transformation X/prime H=LlXHLr. Evidently, det X/prime H= detXH. This shows explicitly that the group SO( 2, 2)is locally isomorphic to SL( 2,R)⊗SL( 2,R). Although two- timing theories are bound to be trouble, we could use SO( 2, 2)formally in computing scattering amplitudes, as we will see in chapter N.3. Topological quantization of helicity As promised, let us go back to the observation in the previous section that the matrix XE/((x4)2+/vectorx2)1 2is an element of SU( 2). Define wA≡xA/((x4)2+/vectorx2)1 2forA=1, 2, 3, 4. An arbitrary element of SU( 2)can be written asU=w4I+i/vectorw./vectorσ, with det U=1=(w4)2+/vectorw2. The 4-dimensional unit vector w=(w4,/vectorw)traces out the 3-sphere S3, the surface of the 4-ball B4living in 4-dimensional Euclidean space. Thus the group manifold of SU( 2)isS3. Next, recall that SU( 2)double covers the rotation group SO( 3), or in plain talk, two elements Uand−U of SU( 2)corresponds to the same rotation. Thus the group manifold of SO( 3)isS3/Z2, that is, the 3-sphere with antipodal points identified. Appendix B. Brief Review of Group Theory | 533 Consider closed paths in SO( 3). Starting at some point PonS3, wander off a bit and come back to P. The path you traced can evidently be continuously shrunk to a point. But suppose you go off to the other side ofthe world and arrive at −P, the antipodal point of P. You also trace a closed path in SO( 3)since Pand−P correspond to the same element of SO( 3), but this closed path obviously cannot be shrunk to a point. On the other hand, if after arriving at −P you keep going and eventually return to P, then the entire path you traced can be continuously shrunk to a point. Using the language of homotopy groups introduced in chapter V .7, wesay that /Pi1 1(SO(3 ))=Z2: there are two topologically inequivalent classes of paths in the 3-dimensional rotation group. Now we can go back and tie up a loose end in chapter III.4. Back in school you learned that the nonlinear algebraic structure of the Lie algebra [ Ji,Jj]=i/epsilon1ijkJkenforces quantization of angular momentum. But the little group for a massless particle is merely O(2). In the “rich man’s approach” to gauge invariance, how do we get the helicity of the photon and the graviton quantized? The answer is that we invoke topological, rather than algebraic, quantization. A rotation through 4 πis represented by ei4πhon the helicity hstate of the massless particle, but the path traced out by this rotation can be continuously shrunk to a point. Hence, we must have ei4πh=1 and h=0,±1 2,±1 ,... . Appendix C Feynman Rules Here we gather the Feynman rules given in various chapters. Draw all possible diagrams. Label each line with a momentum. If applicable, also label each line with an incoming and an outgoing Lorentz index (for a line describing a vector field), with an incoming and an outgoinginternal index (for a line describing a field transforming under an internal symmetry), so on and so forth.Momentum is conserved at each vertex. Momenta associated with internal lines are to be integrated over withthe measure/integraltext [d 4p/(2π)4]. A factor of (−1)is to be associated with each closed fermion loop. External lines are to be amputated. For an incoming fermion line write u(p,s)and for an outgoing fermion line ¯u(p/prime,s/prime). For an incoming antifermion, write ¯v(p,s), and for an outgoing antifermion, v(p/prime,s/prime). If there are symmetry transformations leaving the diagram invariant, then we have to worry about the infamous symmetry factors.Since I don’t trust the compilations in various textbooks I work out the symmetry factors from scratch, and thatis what I advise you to do. Scalar field interacting with Dirac field L=¯ψ(iγμ∂μ−m)ψ+1 2[(∂ϕ)2−μ2ϕ2]−λ 4!ϕ4+fϕ¯ψψ (1) Scalar propagator: ki k2−μ2+iε(2) Appendix C. Feynman Rules | 535 Scalar vertex: −iλ (3) Fermion propagator: p i /negationslashp−m+iε=i/negationslashp+m p2−m2+iε(4) Scalar fermion vertex: if (5) Initial external fermion: u(p,s) (6) Final external fermion: ¯u(p,s) (7) Initial external antifermion: ¯v(p,s) (8) Final external antifermion: v(p,s) (9) Vector field interacting with Dirac field L=¯ψ(iγμ(∂μ−ieAμ)−m)ψ−1 4FμνFμν−1 2μ2AμAμ(10) Vector boson propagator: k i k2−μ2/parenleftbiggkμkν μ2−gμν/parenrightbigg (11) Photon propagator (with ξan arbitrary gauge parameter): k i k2/bracketleftbigg (1−ξ)kμkν k2−gμν/bracketrightbigg (12) 536 | Appendix C. Feynman Rules Vector boson fermion vertex: μ ieγμ(13) Initial external vector boson: εμ(k) (14) Final external vector boson: εμ(k)∗(15) Nonabelian gauge theory Gauge boson propagator: k i k2/bracketleftbigg (1−ξ)kμkν k2−gμν/bracketrightbigg δab (16) Ghost propagator: k i k2δab (17) Cubic interaction between the gauge bosons: a,μ c,λ b,νk1 k3k2gfabc[gμν(k1−k2)λ+gνλ(k2−k3)μ+gλμ(k3−k1)ν] (18) Quartic interaction between the gauge bosons: a,μb,ν d,ρc,λ−ig2[fabefcde(gμλgνρ−gμρgνλ) +fadefcbe(gμλgνρ−gμνgρλ) +facefbde(gμνgλρ−gμρgνλ)](19) Appendix C. Feynman Rules | 537 Gauge boson coupling to the ghost field: c,μ abpgfabcpμ(20) Cross sections and decay rates Given the Feynman amplitude Mfor a process p1+p2→k1+k2+...+knthe differential cross section is given by dσ=1 |/vectorv1−/vectorv2|E(p1)E(p2)d3k1 (2π)3E(k1)...d3kn (2π)3E(kn)(2π)4δ(4)(p1+p2−n/summationdisplay i=1ki)|M|2(21) Here/vectorv1and/vectorv2denote the velocities of the incoming particles. The energy factor E(p)=2/radicalbig /vectorp2+m2for bosons andE(p)=/radicalbig /vectorp2+m2/mfor fermions come from the different normalization of the creation and annhilation operators in chapters I.8 and II.2. For a decay of a particle of mass Mthe differential decay rate in its rest frame is given by d/Gamma1=1 2Md3k1 (2π)3E(k1)...d3kn (2π)3E(kn)(2π)4δ(4)(P−n/summationdisplay i=1ki)|M|2(22) Appendix D Various Identities and Feynman Integrals Gamma matrices Identities for the trace of a product of an even number of gamma matrices: trγμγν=4ημν(1) trγμγνγλγσ=4(ημνηλσ−ημληνσ+ημσηνλ) (2) We define the totally antisymmetric symbol εμνλσbyε0123=+ 1 (note ε0123=− 1). Then with our definition γ5≡iγ0γ1γ2γ3, we have trγ5γμγνγλγσ=− 4iεμνλσ(3) Identities that follow from the basic Clifford identity: γμ/negationslashpγμ=− 2/negationslashp (4) γμ/negationslashp/negationslashqγμ=4p.q (5) γμ/negationslashp/negationslashq/negationslashrγμ=− 2/negationslashr/negationslashq/negationslashp (6) I leave it to you to derive these identities. For example, to obtain (4) keep moving γμto the right in the expression γμ/negationslashpγμ=(2pμ− /negationslashpγμ)γμ=2/negationslashp−4/negationslashp=− 2/negationslashp. Evaluating Feynman diagrams Over the years, a number of tricks and identities have been developed for evaluating the integrals associated with Feynman diagrams. Let us evaluate I=/integraldisplayd4k (2π)41 (k2−m2+iε)3=/integraldisplayd3k (2π)3/integraldisplaydk0 2π1 [k2 0−(/vectork2+m2)+iε]3 Focus on the k0integral. Draw where the poles are in the complex k0-plane and you will see that the integration contour can be rotated anticlockwise so that [we denote the integrand by f( k 0)] /integraldisplay+∞ −∞dk0f( k 0)=/integraldisplay+i∞ −i∞dk0f( k 0)=i/integraldisplay+∞ −∞dk4f( ik4) (7) Appendix D. Identities and Feynman Integrals | 539 where in the last step we define k0=ik4(corresponding to the Wick rotation mentioned in chapters I.2 and V .2.) Thus, I=i(−1)3/integraldisplayd4 Ek (2π)41 (k2 E+m2)3 where d4 Ekis the integration element in Euclidean 4-dimensional space and k2 E≡k2 4+/vectork2the square of a Euclidean 4-vector. The infinitesimal εcan now be set equal to zero. We can integrate immediately over the three angles since the integrand does not depend on them. You can look up the angular element in Euclidean space in a book,but we will use a neat trick instead. I will do the more general d-dimensional integral H=/integraltext d dkF(k2), where k2=k2 1+k2 2+...+k2 dandFcan be any function as long as the integral converges. (I now drop the subscript E; the context makes clear that we are in Euclidean space.) We can of course set dequal to 4 at the end. The result for arbitrary dwill be useful to us in regularizing dimensionally (chapter III.1). We imagine integrating over the (d−1)angular variables to obtain H=C(d)/integraltext∞ 0dk kd−1F(k2). To determine C(d) we will do the integral J=/integraltext ddke−1 2k2in two different ways. Using (I.2.8) we haveJ=(√ 2π)d. Alternatively, J=C(d)/integraldisplay∞ 0dk kd−1e−1 2k2=C(d) 2d 2−1/integraldisplay∞ 0dx xd 2−1e−x=C(d) 2d 2−1/Gamma1(d 2) where we changed integration variables and recognized the integral representation of the gamma function /Gamma1(z+1)≡/integraltext∞ 0dx xze−x. (Recall that upon integration by parts we obtain /Gamma1(z+1)=z/Gamma1(z), so that /Gamma1(n)=(n−1)! fornan integer.) Therefore C(d)=2πd/2//Gamma1(d/2 )and /integraldisplay ddkF(k2)=2πd/2 /Gamma1(d/ 2)/integraldisplay∞ 0dk kd−1F(k2) (8) Setting d=1 in (8) we determine /Gamma1(1 2)=π1 2, and setting F(k2)=δ(k−1)we see that the area of the (d−1)- dimensional sphere is equal to C(d) , thus recovering various results you learned in school about circles and spheres: C(2)=2πandC(3)=4π. The new result you need as a budding field theorist is for d=4: /integraldisplay d4kF(k2)=π2/integraldisplay∞ 0dk2k2F(k2) (9) So finally we have I=−i 16π2/integraldisplay∞ 0dk2k2 1 (k2+m2)3=−i 16π21 2m2(10) We have derived the basic formula for doing Feynman integrals: /integraldisplayd4k (2π)41 (k2−m2+iε)3=−i 32π2m2(11) (With the telltale iεwe have evidently moved back to Minkowski space.) As an exercise you can go through the same steps to find /integraldisplay/Lambda1d4k (2π)41 (k2−m2+iε)2=i 16π2/bracketleftbigg log/parenleftbigg/Lambda12 m2/parenrightbigg −1+.../bracketrightbigg (12) Here a cutoff is needed, which we introduce by setting the upper limit in the integral over k2in the analog of (10) to /Lambda12. As a check, differentiate (12) with respect to m2to recover (11). As another exercise show that /integraldisplay/Lambda1d4k (2π)4k2 (k2−m2+iε)2=−i 16π2/bracketleftbigg /Lambda12−2m2log/parenleftbigg/Lambda12 m2/parenrightbigg +m2+.../bracketrightbigg (13) In (12) and (13) (...)denote terms that vanish for /Lambda12/greatermuchm2. In some texts, the (−1)in (12) is dropped by absorbing it into /Lambda12. But then we have to be careful to adjust (13) accordingly if it appears in the same calculation. 540 | Appendix D. Identities and Feynman Integrals A useful identity in combining denominators is 1 x1x2...xn=(n−1)!/integraldisplay1 0/integraldisplay1 0.../integraldisplay1 0dα1dα2...dαn δ⎛ ⎝1−n/summationdisplay jαj⎞ ⎠1 (α1x1+α2x2+...+αnxn)n(14) Forn=2, 1 xy=/integraldisplay1 0dα1 [αx+(1−α)y]2(15) and for n=3, 1 xyz=2/integraldisplay1 0/integraldisplay1 0/integraldisplay1 0dαdβdγδ(α +β+γ−1)1 (αx+βy+γz)3(16) =2/integraldisplay/integraldisplay triangledαdβ1 [z+α(x−z)+β(y−z)]3 where the integration region is the triangle in the α-βplane bounded by 0 ≤β≤1−αand 0 ≤α≤1. Appendix E Dotted and Undotted Indices and the Majorana Spinor We develop the dotted and undotted notation introduced in chapter II.3 for further use in discussing supersym- metry in chapter VIII.4 and in part N. In essence, the appearance of undotted and dotted indices can be traced back to the fact that the algebra of the Lorentz group SO( 3, 1), with the generators /vectorJ+i/vectorKand/vectorJ−i/vectorK, breaks up into two pieces, each isomorphic to the algebra of SU( 2). The absence or presence of the dot allows us to keep track of which SU( 2)we are talking about. Here I will use extensively results from chapter II.3 and from the exercises (do them!) there without bothering to write them down again here. In the Weyl basis of chapter II.1 γμ=/parenleftBigg0σμ ¯σμ0/parenrightBigg (1) where σμ=(I,/vectorσ)and¯σμ=(I,−/vectorσ). Knowing that γμacts on /Psi1=/parenleftBiggψα ¯χ˙α/parenrightBigg we see that σμand¯σμcarry indices as follows: (σμ)α˙αand(¯σμ)˙αα(2) This is consistent with what you know: the Lorentz vector transforms like (1 2,1 2)and thus straddles the two SU( 2)’s. The matrices σμand¯σμmix dotted and undotted indices. We will make good use of this observation later. Let us check that the Lorentz transformation property of the Dirac spinor /Psi1is consistent with what was discussed in chapter II.1. There we learned that /Psi1→e−i 4ωμν/Sigma1μν/Psi1, where /Sigma1μν≡i 2[γμ,γν]. (We want to use the symbol σμνfor some other quantity, hence the change of notation.) Using (1) we obtain /Sigma1μν=2i/parenleftBiggσμν0 0¯σμν/parenrightBigg where σμν≡1 4(σμ¯σν−σν¯σμ)and¯σμν≡1 4(¯σμσν−¯σνσμ). From (2) we see that these two matrices carry indices as follows: (σμν)β αand(¯σμν)˙α ˙β(3) 542 | Appendix E. Indices and the Majorana Spinor Again, this reflects the fact that the antisymmetric tensor (such as the electromagnetic field Fμν)transforms like (1, 0)+(0, 1). The matrices σμνand¯σμνmay seem alien, but recall that they are manufactured out of the familiar Pauli matrices and so they are simply Pauli matrices (what else could they be?) themselves. In particular, σ0i=−¯σ0i=−1 2σiand σij=¯σij=−i 2εijkσk Note that these relations are consistent with (σμν)†=−(¯σμν), which in turn follows from (/Sigma1μν)†=γ0/Sigma1μνγ0. Mother Nature is kind to the students of quantum field theory. The relativistic spinor /Psi1breaks up into two 2-component spinors acted on by the Pauli matrices. What you learned in nonrelativistic quantum mechanicscontinues to be relevant here. Thus under an infinitesimal Lorentz transformation ψα→/parenleftBig I+1 2ωμνσμν/parenrightBigβ αψβ (4) and ¯χ˙α→/parenleftBig I+1 2ωμν¯σμν/parenrightBig˙α ˙β¯χ˙β(5) You should check that it all works out according to plan. Everything is consistent with what we learned in chapter II.3, in particular, that boosts act oppositely on (1 2,0)and(0,1 2), but rotations act the same. Thus far, on the spinor fields ψαand¯χ˙α, the dotted indices always live upstairs and the undotted indices downstairs. What would get them to change floors? Charge conjugation. Recall from chapter II.1 that the charge conjugated field is defined by /Psi1c≡C¯/Psi1T[where Tdenotes transpose, ¯/Psi1means /Psi1†γ0, andC−1γμC=−(γμ)T.] In the Weyl basis, we can choose C=ζγ0γ2=ζ/parenleftBigg−σ20 0σ2/parenrightBigg (6) The condition (/Psi1c)c=/Psi1implies |ζ|=1. We choose ζ=−i. Explicitly, /Psi1c=/parenleftBiggiσ2¯χ∗ −iσ2ψ∗/parenrightBigg We now introduce some notation, the wisdom of which will soon become clear. Given ψαand¯χ˙α, define ¯ψ˙α≡(ψα)∗andχα≡(¯χ˙α)∗(7) Weird, complex conjugation puts on a dot and a bar. We raise and lower undotted indices as follows: ψα=εαβψβandψβ=εβγψγwhich implies that εαβεβγ=δγ α. Thus, if we choose εαβ=/parenleftBigg01 −10/parenrightBigg =(iσ 2)αβ then εβγ=/parenleftBigg0−1 10/parenrightBigg =(−iσ 2)βγ We are forced to define ε12=+ 1 and ε12=− 1 to have opposite signs, a fact to keep in mind. You should realize by now that what we are doing can again be traced back to that peculiar fact about Pauli matrices (appendix B): (iσ 2)σ∗ i(−iσ 2)=−σi (8) or equivalently σ2σT iσ2=−σi (9) Appendix E. Indices and the Majorana Spinor | 543 an identity in one guise or another familiar from quantum mechanics. We have used it again and again, in appendix B and in the text (for example, in connection with Majorana masses and with the Higgs field). From(8) we have (iσ 2)σμ∗(−iσ 2)=¯σμand hence (iσ 2)(σμν)∗(−iσ 2)=¯σμν(10) Analogously, we raise and lower dotted indices as follows: ¯ψ˙α=ε˙α˙β¯ψ˙βand¯ψ˙β=ε˙β˙γ¯ψ˙γ. Referring to (7) we see that ε˙α˙βis numerically the same as εαβ, andε˙β˙γis numerically the same as εβγ. You now see the rationale of these apparently capricious choices: we can now write /Psi1c=/parenleftBiggχα ¯ψ˙α/parenrightBigg (11) Referring to /Psi1=/parenleftBiggψα ¯χ˙α/parenrightBigg (12) we see that the point of the notation is that ψαandχαtransform in the same way and are the same kind of creature (and similarly for ¯χ˙αand¯ψ˙α.) We now come to the all-important concept of a Majorana spinor. Ettore Majorana, a brilliant physicist, mysteriously disappeared early in his career. Fermi supposedly described Majorana as “a towering giant withoutany common sense.” 1 Given a Dirac spinor /Psi1,i f/Psi1=/Psi1c, then /Psi1is said to be a Majorana spinor. Comparing (12) and (11), we see that a Majorana spinor has the form /Psi1M=/parenleftBiggψα ¯ψ˙α/parenrightBigg (13) An obvious remark but a handy mnemonic: Given a Weyl spinor ψαwe can construct a Majorana spinor, and given two Weyl spinors we can construct a Dirac spinor: one Weyl equals one Majorana, and two Weyls equalone Dirac. Incidentally, another way of seeing that complex conjugation puts on a dot is that (see chapter II.3) conjugation interchanges /vectorJ+i/vectorKand/vectorJ−i/vectorK. The point to remember is simply that given a spinor λ α, then λ˙αtransforms like (λα)∗. You should verify this, keeping in mind (10). The utility of the notation is similar to that of the covariant and contravariant (or upper and lower) indices in special and general relativity. We always contract an upper index with a lower index. Here we have the additionalrule that an undotted upper index can only be contracted with an undotted lower index, but never with a dottedlower index, (obviously, since they belong to different algebras.) It is easy to verify these rules. For example, letus show that η αψαis invariant. Using (4) we proceed with laboriously careful pedagogy: ηα→η/primeα=εαβη/prime β=εαβ(e1 2ωσ)γ βηγ=εαβ(e1 2ωσ)γ βεγρηρ=(e−1 2ωσT)α ρηρ(14) where we used once again the identity (9). Then ηαψα→η(e−1 2ωσT)T(e1 2ωσ)ψ=ηψ, which is indeed an invariant. In special and general relativity we raise and lower indices with the metric, which is of course symmetric. Here we raise and lower indices with the antisymmetric εsymbol and as a result signs pop up here and there. For example, ηαψα=εαβηβψα=ηβ(−εβα)ψα=−ηβψβ. Contrast this with the scalar product of two vectors vμwμ=vμwμ. If we want to suppress indices and write ηψ, we must decide once and for all what that means. The standard convention is to define ηψ≡ηαψα (15) and not ηβψβ. This rule is sometimes stated by saying that in contracting undotted indices we always go from the northwest to the southeast, and never from southeast to northwest. As we learned in chapter II.5, spinorfields are to be treated as anticommuting Grassman variables under the path integral, so that −η βψβ=ψβηβ. We end up with the nice rule ηψ=ψη. 1M. Gell-Mann, private communication. Incidentally, the name Ettore corresponds to Hector in English. 544 | Appendix E. Indices and the Majorana Spinor Similarly, we define ¯χ¯ξ≡¯χ˙α¯ξ˙α=¯ξ¯χ (16) In contracting dotted indices we always go from southwest to northeast. Of course, none of this “Santa Barbara to Cambridge” convention is needed if the indices are displayed explicitly. Just as in special and general relativity, where the upper and lower indices are very useful in telling us whether expressions we write down make sense, the undotted and dotted upper and lower indices allow usto see immediately that ηψandησ μνψmake sense, but that ησμψdoes not. [Look at (2) and (3) and notice the kind of indices that appear.] The notation of course just codifies in a convenient way the group theory fact that (1 2,0)⊗(1 2,0)=(0, 0)⊕(1, 0), namely, that out of two Weyl spinors we can make a scalar and a tensor but not a vector. As always, notation should be driven by physics and computational convenience (which is intimately con- nected to elegance). To gain familiarity with the dotted and undotted 2-component notation, you should work out some of the identities in the exercises. These identities are useful when working with supersymmetric field theories. Exercises E.1 Show that ησμνψ=−ψσμνηand¯χ¯σμψ=−ψσμ¯χ. E.2 Show that (θϕ)(¯χ¯ξ)=−1 2(θσμ¯ξ)(¯χ¯σμϕ). E.3 Show that θαθβ=1 2(θθ)δα β. [Hint: simply evaluate the two sides for all possible cases.] Solutions to Selected Exercises Part I I.3.1 From the text we have for x0=0, D(x)=−i/integraldisplayd3k (2π)32/radicalbig /vectork2+m2e−i/vectork./vectorx =−i 2(2π)2/integraldisplay∞ 0dkk2 /radicalbig /vectork2+m2/integraldisplay+1 −1d(cosθ)e−ikr cosθ =−1 2(2π)2r/integraldisplay∞ 0dkk/radicalbig /vectork2+m2(eikr−e−ikr)=−1 8π2r/integraldisplay∞ −∞dkk/radicalbig /vectork2+m2eikr =i 8π2r∂ ∂r/integraldisplay∞ −∞dk/radicalbig /vectork2+m2eikr The integrand in I≡/integraltext∞ −∞(dk//radicalbig /vectork2+m2)eikrhas a cut along the imaginary axis going from imtoi∞(and another cut we don’t care about.) So fold the contour around the cut and change variable to k=i(m+y): I=2/integraldisplay∞ 0dye−(y+m)r 1/radicalbig (y+m)2−m2 =2/integraldisplay∞ 1due−mru 1√ u2−1 =2/integraldisplay∞ 0dte−mr cosht. 546 | Solutions to Selected Exercises At this point you can look in a table and find that this is some Bessel function and read off the large r behavior, but it is more stylish to press on and descend steeply: we obtain D(x)=−im 4π2r/integraldisplay∞ 0dt(cosht)e−mr cosht =−im 4π2r/integraldisplay∞ 0d(sinht)e−mr cosht =−im 4π2r/integraldisplay∞ 0dse−mr√ s2+1/similarequal−im 4π2r/integraldisplay∞ 0dse−mr(1 +1 2s2) =−im2 4π2(π 2(mr)3)1 2e−mr, using the Gaussian integral from the appendix of chapter I.2. I.3.2 We evaluate D(x)=/integraldisplayd2k (2π)2eikx k2−m2+iε by contours as in the text and obtain D(x)=−i/integraldisplaydk (2π)2ωk[e−i(ω kt−kx)θ(x0)+ei(ωkt−kx)θ(−x0)] Forx0=0, we recognize the integral D(x)=−i/integraldisplay+∞ −∞dk (2π)2/radicalbig /vectork2+m2e−ikx as a Bessel function from exercise I.3.1: D(x)=−i 2πK0(m|x|)→−i 2π/radicalbiggπ 2m|x|e−m|x| with the expected exponential decay for large x. I.7.2 Expanding and keeping only the desired terms Z(J)→C/braceleftBigg 1+1 2!/parenleftbigg −i 4!λ/parenrightbigg2/integraldisplay/integraldisplay d4w1d4w2/bracketleftbiggδ iδJ(w1)/bracketrightbigg4/bracketleftbiggδ iδJ(w2)/bracketrightbigg4 1 6!/bracketleftbigg −i 2/integraldisplay/integraldisplay d4xd4yJ(x)D(x −y)J(y)/bracketrightbigg6/bracerightBigg Just keep on differentiating. I.7.4 Write k1=(√ k2+m2,0 ,0 ,k) andk2=(√ k2+m2,0 ,0 ,−k) . Then E=2√ k2+m2≥2m. Physically, a pair of mesons can be produced when E≥2m. I.8.1 Do the k0integral on the left-hand side of (I.8.14):/integraltext dk0δ((k0)2−ω2 k)θ(k0)f (k0,/vectork), where ωk≡ +/radicalbig /vectork+m2. Using (I.2.12) and picking up the positive root because of the step function, we obtain/integraltext∞ 0dk0(δ(k0−ωk)/(2k0))f (k0,/vectork)=f( ωk,/vectork)/(2ωk). To verify the invariance explicitly, boost in the xdirection and drop the subscript on ωk:kx→ sinhφω+coshφkxandω→coshφω+sinhφkx. Then, using ω2=(kx)2+...and hence ωdω= kxdkx, we have dkx→(sinhφ( kx/ω)+coshφ)dkx. Hence dkx/ω→dkx/ω. Solutions to Selected Exercises | 547 I.8.2 Clearly, only the terms aa†anda†ainHcontributes to </vectork/prime|H|/vectork> . Extract these two types of terms in /integraldisplay dDxϕ(x)2 =/integraldisplay dDx/integraldisplay/integraldisplaydDq/radicalBig (2π)D2ωqdDq/prime /radicalBig (2π)D2ωq/prime[a(/vectorq)a†(/vectorq/prime)e−i(ω qt−/vectorq./vectorx)ei(ωq/primet−/vectorq/prime./vectorx)+h.c.] =/integraldisplaydDq 2ωq[a(/vectorq)a†(/vectorq)+a†(/vectorq)a(/vectorq)] and so His for our purposes effectively equal to/integraltext dDqωq 2[a(/vectorq)a†(/vectorq)+a†(/vectorq)a(/vectorq)], which upon using the commutation relation is equal to/integraltext dDqωq 2[δ(D)(/vector0)+2a†(/vectorq)a(/vectorq)]. We recognize the first term as the vacuum calculated in the text. Note that the definition of the delta function (2π)Dδ(D)(/vectork)=/integraltext dDxei/vectork./vectorx implies δ(D)(/vector0)=[1/(2π)D]/integraltext dDx=V/(2π)D. Thus, subtracting off the vacuum energy, we have H effectively equal to/integraltext dDqωqa†(/vectorq)a(/vectorq), which just says that a mode of momentum /vectorqcarries energy ωq. In particular, using the commutation relation twice we have </vectork/prime|H|/vectork>=δ(D)(/vectork/prime−/vectork)ωk. The energy of a particle of momentum /vectorkisωkrelative to the vacuum. I.8.4 Q=/integraltext dDxJ0(x)=/integraltext dDx(ϕ†i∂0ϕ−i(∂0ϕ†)ϕ). Focus on the first term: /integraldisplay dDx/integraldisplay/integraldisplaydDk/prime /radicalbig (2π)D2ωk/primedDk/radicalbig (2π)D2ωk [a†(/vectork/prime)ei(ωk/primet−/vectork/prime./vectorx)+b(/vectork/prime)e−i(ωk/primet−/vectork/prime./vectorx)]ωk[a(/vectork)e−i(ω kt−/vectork./vectorx)−b†(/vectork)ei(ωkt−/vectork./vectorx)] Note that i∂0brings down a factor of ωkand produces a relative sign between aandb†. As in exercise I.8.2 the integral over xproduces a delta function that collapses the two kintegrals into one, giving /integraldisplay dDk1 2(a†(/vectork)a(/vectork)−b(/vectork)b†(/vectork)−a†(−/vectork)b†(/vectork)e2iωkt+b(−/vectork)a(/vectork)e−2iωkt) The second term in −i(∂ 0ϕ†)ϕinJ0(x)is just the hermitean conjugate of the first term ϕ†i∂0ϕ. Thus, adding the hermitean conjugate of what we have just obtained, we find Q=/integraldisplay dDk[a†(/vectork)a(/vectork)−b(/vectork)b†(/vectork)] =/integraldisplay dDk[a†(/vectork)a(/vectork)−b†(/vectork)b(/vectork)]+δ(D)(/vector0)/integraldisplay dDk The infinite additive constant is to be subtracted out much like the vacuum energy. In some texts a normal ordering operation, denoted by a pair of colons, is defined as follows: If you see : (...): you are instructed to move all the creation operators in the expression (...)to the left of the annihilation operators. In other words, by fiat : b(/vectork)b†(/vectork):≡b†(/vectork)b(/vectork). The current is then defined by Jμ(x)≡:(ϕ†i∂μϕ−i(∂μϕ†)ϕ) :. Since the normal ordered current differs from the naively defined current by a c-number the most crucial property of the current, namely current conservation ∂μJμ=0, is not affected. This is of course just a formal way of saying that the value of the charge in the vacuum state is to be subtracted. In any case, the result Q=/integraltext dDk[a†(/vectork)a(/vectork)−b†(/vectork)b(/vectork)] shows that aandbannihilate positive and negative charges, respectively. I.10.2 We have (repeated indices summed) Raa/primeRbb/primeiDa/primeb/prime(x)=/integraldisplay DϕR aa/primeϕa/prime(x)R bb/primeϕb/prime(0)eiS But we can change the integration variable from ϕtoRϕ. Since the action Sand the measure Dϕ are both invariant under SO(N) rotations, this is equal to/integraltext Dϕϕ a(x)ϕb(0)eiS=iDab(x). Thus, we obtain Dab=Raa/primeRbb/primeDa/primeb/prime. The properties of the rotation group are such that the only solution of this equation isDabproportional to δab. 548 | Solutions to Selected Exercises I.10.3 The field ϕtransforms as a symmetric traceless tensor (see appendix B) under SO( 3), that is, with all indices displayed, ϕab→Raa/primeRbb/primeϕa/primeb/prime=Raa/primeϕa/primeb/primeRT b/primeb=(RϕRT)ab. As suggested in the hint, writing ϕ as a 3 by 3 symmetric traceless matrix we have ϕ→RϕRTand thus the invariants are (up to quartic order in ϕ)tr(∂μϕ)2,t rϕ2,t rϕ4, and (tr ϕ2)2. Remarkably, you can prove that tr ϕ4and (tr ϕ2)2actually amount to only one invariant by diagonalizing ϕ=⎛ ⎜⎜⎝α 00 0β 0 00 −(α+β)⎞ ⎟⎟⎠ You can see by computation tr ϕ4and (tr ϕ2)2are both proportional to [ α2+β2+(α+β)2]2. Thus, if we restrict ourselves to quartic terms the Lagrangian L=1 2tr(∂μϕ)2−1 2m2trϕ2−λ(trϕ2)2actually has anSO( 5)symmetry (since ϕhas 5 components.) This is an example of what is known as “accidental symmetry.” Convince yourself that this holds only to quartic order in ϕ. I.11.2 Varying gμρgρλ=δμ λwe have ( δgμρ)gρλ=−gμρ(δgρλ), which upon multiplication by gλνbecomes δgμν=−gμρ(δgρλ)gλν. You may recognize this as just the statement δM−1=−M−1(δM)M−1for a matrix M. To evaluate δgwe use the important identity det M=eTr log M, which you can prove easily by diagonalizing Mwith a similarity transformation. The left hand side is equal to the product of the eigenvalues of M, while the right hand side is equal to the exponential of the sum of logarithms of the eigenvalues. {You can define the logarithm of a matrix by expanding log[ I+(M−I)] in a power series in(M−I).} Thus, δdetM=(detM)trM−1δM and so δg=ggνμδgμν. We are now ready to vary S=/integraldisplay d4x√−g1 2(gμν∂μϕ∂νϕ−m2ϕ2)≡/integraldisplay d4x√−gL Plugging in, we have δS=/integraldisplay d4x√−g[1 2gνμδgμνL−gμρ(δgρλ)gλν1 2∂μϕ∂νϕ] Thus, Tμν=−2√−gδS δgμν=gμρgνλ∂ρϕ∂λϕ−gμνL In the flat spacetime limit T00=(∂0ϕ)2−L=1 2((∂0ϕ)2+(/vector∇ϕ)2+m2ϕ2) precisely the energy density as promised. I.11.3 Using the expression for Tμνfrom the preceding exercise, we have Pi=/integraldisplay d3xT0i=−/integraltext d3x∂0ϕ∂iϕ and [Pi,ϕ(x) ]=−/integraldisplay d3y[∂0ϕ(y) ,ϕ(x) ]∂iϕ(y)=i∂iϕ(x) Thus, combined with the fact that P0=H, we have [Pμ,ϕ(x) ]=−i∂μϕ(x) , which just reflects the fact thatPμandxνare conjugate variables. I.11.4 Evaluating Tμν=−FμλFλ ν−ημνL, we have Tij=−FiλFλ j+1 2δij(/vectorE2−/vectorB2)=−EiEj+FikFjk+1 2δij(/vectorE2−/vectorB2) Solutions to Selected Exercises | 549 SinceFikFjk=εikmεjknBmBn=δij/vectorB2−BiBj, we obtain the announced result. Note that δijTij=1 2(/vectorE2+ /vectorB2)=T00and hence T=0. Part II II.1.1 Continuing the hint, we have δ(¯ψγμγ5ψ)=¯ψi 4ωλρ[σλρ,γμγ5]ψ=¯ψi 4ωλρ[σλρ,γμ]γ5ψ since γ5anticommutes with gamma matrices and hence commutes with the product of two gamma matrices. Inserting [ σλρ,γμ] as given in the text we have δ(¯ψγμγ5ψ)=ωμ λ¯ψγλγ5ψ, which is precisely how a vector transforms. Under parity ¯ψγμγ5ψ→¯ψγ0γμγ5γ0ψwhich equals ¯ψγ5γ0ψ=−¯ψγ0γ5ψ forμ=0, and ¯ψγ0γiγ5γ0ψ=¯ψγiγ5ψforμ=i. The time component flips sign while the spatial components do not. Thus, the behavior under parity is opposite to that of a normal vector: ¯ψγμγ5ψ is an axial vector. The other cases proceed similarly. II.1.2 FromψL=1 2(1−γ5)ψandψR=1 2(1+γ5)ψ, we find ¯ψL=ψ† Lγ0=ψ†1 2(1−γ5)γ0=¯ψ1 2(1+γ5)and ¯ψR=¯ψ1 2(1−γ5). We then just use the properties of PLandPRrepeatedly. For example, ¯ψLψR=¯ψ1 2(1+ γ5)ψand¯ψRψL=¯ψ1 2(1−γ5)ψ, or equivalently, ¯ψψ=¯ψLψR+¯ψRψLand¯ψγ5ψ=¯ψLψR−¯ψRψL.A s another example, ¯ψLγμψL=¯ψ1 2(1+γ5)γμ1 2(1−γ5)ψ=¯ψγμ1 2(1−γ5)ψand¯ψRγμψR=¯ψγμ1 2(1+ γ5)ψ. Note that various combinations vanish, for example, ¯ψLψL=0,¯ψLγμψR=0, and so on. Complete the exercise. II.1.3–4 In the appropriate basis the Dirac equation becomes /parenleftBiggE−mp σ3 −pσ 3−E−m/parenrightBigg/parenleftBiggφ χ/parenrightBigg =0 that is, (E−m)φ+pσ3χ=0 and −pσ 3φ−(E+m)χ=0. The second equation informs us that χ= −[p/(E+m)]σ3φ. For a slow electron χ/similarequal−(p/2m)σ 3φ, so that χis smaller than φby the factor p/2m. The first equation then reduces to (E−m−p2/2m)φ=0, which just reminds us of the relation between energy and momentum in the nonrelativistic limit. II.1.5 In the Weyl basis the Dirac equation for a relativistic electron moving along the 3-axis E(γ0−γ3)ψ=0 becomes /parenleftBigg0 I−σ3 I+σ30/parenrightBigg/parenleftBiggψL ψR/parenrightBigg =0 Since σ12≡i 2[γ1,γ2]=−i 2[σ1,σ2]⊗I=σ3⊗I=/parenleftBiggσ30 0σ3/parenrightBigg under a rotation around the 3-axis, ψL→e−(i/4 )ωσ3ψL=e+(i/4 )ωψLwhile ψR→e−(i/4 )ωσ3ψR= e−(i/4 )ωψR. Indeed, the left and right handed fields rotate in opposite directions. II.1.6 In the Weyl basis, the Dirac equation γ.pu=0 becomes σμpμη=0, and ¯σμpμχ=0, with u=/parenleftBiggχ η/parenrightBigg The solutions are η=/parenleftBiggp1−ip2 p0−p3/parenrightBigg and η=/parenleftBiggp0+p3 p1+ip2/parenrightBigg 550 | Solutions to Selected Exercises for the two possible helicities. The corresponding solutions for χmay be obtained by /vectorp↔−/vectorp. We have ¯uu=p.p=0. For a particle moving in the +3 direction, η=0 and χ=2E/parenleftBigg0 1/parenrightBigg and η=2E/parenleftBigg1 0/parenrightBigg andχ=0. The Lorentz vector ¯uγμu=(2E)2(1, 0, 0, 1 ). (What other direction could it point in?) For a particle moving in the −3 direction, ηandχexchange roles. This exercise shows explicitly that for massless particles we can use 2-component spinors. (What happens if parity is broken?) II.1.8 In either the Dirac or the Weyl basis, (ψc)c=γ2(γ2ψ∗)∗=ψ. II.1.9 It is easiest to work in either the Dirac or the Weyl basis. Let ψbe left handed, that is (1+γ5)ψ=0. Then(1−γ5)ψc=(1−γ5)γ2ψ∗=γ2(1+γ5)ψ∗=0 since γ5is real. II.1.10 ψCψ →ψe−i 4ωλρ(σλρ)TCe−i 4ωμνσμνψ=ψCψ since(σλρ)TC=−Cσλρ. II.1.12 Under parity or reflection in a mirror, x1→x1andx2→−x2. Choose γ0=σ3,γ0γ1=σ1, andγ0γ2= σ2. Multiply the Dirac equation (iγμ∂μ−m)ψ=0b yγ0and write [ i(∂0+γ0γi∂i)−γ0m]ψ=0. Then multiplying by σ1reverses the sign of the ∂2term, but also the mass term. I leave it to you to discuss time reversal. II.2.1 Apply Noether for the transformation ψ→eiθψ=(1+iθ)ψ Then δL δ(∂μψ)δψ+δL δ(∂μ¯ψ)δ¯ψ=¯ψiγμ(iθψ) Note that formally Ldoes not depend on ∂μ¯ψ. Thus, up to overall factors we can choose Jμ=¯ψγμψ with the corresponding charge Q=/integraltext d3x¯ψγ0ψinto which we plug (II.2.10) ψ(x)=/integraldisplayd3p (2π)3/2(Ep/m)1/2/summationdisplay s[b(p,s)u(p ,s)e−ipx+d†(p,s)v(p ,s)eipx] At this point, the calculation pretty much parallels what you did in exercise I.8.4. The integration/integraltext d3x over space produces a delta function that sets the momentum variables in ψand in ¯ψequal to one another. The new feature here is that we encounter objects such as ¯uγ0u. Invoking Lorentz invariance and referring to the rest frame form of uandvwe have ¯u(p,s)γμu(p,s/prime)=δss/primepμ/m,¯u(p,s)γμv(p,s/prime)=0, and so on. We obtain Q=/integraldisplayd3p (2π)3(Ep/m)/summationdisplay s[b†(p,s)b(p ,s)+d(p ,s)d†(p,s)] As in exercise I.8.4 we have to move the creation operator d†to the left of the annihilation operator d and subtract off an infinite constant. Thus, finally Q=/integraldisplayd3p (2π)3(Ep/m)/summationdisplay s[b†(p,s)b(p ,s)−d†(p,s)d(p ,s)] showing clearly that bannihilates a negative charge and da positive charge. Solutions to Selected Exercises | 551 To calculate [ Q,ψ(0)]=/integraltext d3x[¯ψ(x)γ0ψ(x) ,ψ(0)] we use the identity [ AB ,C]=A{B ,C}−{A,C}B and the canonical anticommutation relation (II.2.4). We find [ Q,ψ(0)]=−ψ(0), thus showing that b andd†must carry the same charge. II.3.4 The desired equations are γμ/Psi1αμ=0 (this takes out 4 components since αtakes on 4 values) and (/negationslashp−m)β α/Psi1βμ=0 (for each μthis takes out 2 components and so altogether 4 ×2=8 components.) Thus, 16 −4−8=4 components as desired. Another way of saying this is that γμ/Psi1αμis a Dirac spinor and hence the spin1 2part of the vector-spinor /Psi1αμ. II.6.4 It is good practice to be as symmetrical as one can in calculations. So define p3≡−P1andp4≡−P2and add the 6 (not 3) combinations appearing in the definitions of s,t, andu, thus obtaining 2(s+t+u)=(p1+p2)2+(p3+p4)2+(p3+p1)2+(p4+p2)2+(p4+p1)2+(p3+p2)2 =34/summationdisplay i=1m2 i+2(p1.p2+p3.p4+p1.p3+p2.p4+p1.p4+p2.p3) The second group of terms on the right-hand side collect into (/summationtext4 i=1pi)2−/summationtext4 i=1m2 i. (Obviously, we have for convenience changed notation slightly, setting m3=M1andm4=M2.) II.6.5 Referring to (C.11) we see that in dσthe factor 1 |/vectorv1−/vectorv2|E(p1)E(p2)1 (2π)3E(k1)...1 (2π)3E(kn)(2π)4 reduces to1 2(m/E)4[1/(2π)2]. Integrating the factor d3P1d3P2δ(4)(p1+p2−P1−P2)over/vectorP2we knock off 3 of the delta functions, leaving us with d/Omega1dP1P2 1δ(2E−E1), and so the integral over P1givesd/Omega11 2E2. Finally, the factor containing the “real physics” is1 2/summationtext s/summationtext S|M|2=(e4/4m4)f (θ) . Multiplying the three factors together and dividing by d/Omega1, we obtain dσ/d/Omega1 =(1 2)5[e4/(2π)2](1/E2)f (θ), as given in the text. Note that mcancels out as expected. We should be able to take the limit m→0 compared to the energies without the cross section either blowing up or vanishing. II.6.7/Gamma1=|M|2 2M/integraldisplayd3k (2π)32ωd3k/prime (2π)32ω/prime(2π)4δ4(k+k/prime−q) Knock off the /vectork/primeintegral and do the angular part of the /vectorkintegral to obtain /Gamma1=|M|2 8πM/integraldisplaydkk2 ωω/primeδ(/radicalbig k2+m2+/radicalbig k/prime2+m2−M) Using (I.2.12), we evaluate the integral as (k2/ωω/prime)(1/(k ω+k ω/prime))=k/M . Solving√ k2+m2+√ k/prime2+m2= Mforkwe obtain the stated result. Part III III.1.2 The amplitude should become nonanalytic when both denominators of the integrand (k2−m2+iε)((K −k)2−m2+iε) vanish, namely when k2=m2and(K−k)2=m2. But we found the condition in exercise I.7.4, namely thatK2≥4m2. Referring to (III.1.14) M=iλ2 32π2/integraldisplay1 0dαlog/bracketleftbigg/Lambda12 α(1−α)K2−m2+iε/bracketrightbigg we see that the log has a cut starting at K2=m2/α(1−α).A sα ranges from 0 to 1, the minimum value ofm2/α(1−α)is attained at α=1 2. So indeed, the cut starts at K2=4m2. 552 | Solutions to Selected Exercises III.1.3 Under the indicated change, log /Lambda1→logeε/Lambda1=log/Lambda1+ε, and so δM=−iδλ+iCλ23(2ε)+O(λ3). Thus, δM=0 implies δλ=6Cλ2ε+O(λ3)=6Cλ2δlog/Lambda1+O(λ3)giving the stated result for /Lambda1(dλ/d/Lambda1). III.2.1 For/integraltext ddx(∂ϕ)2to be dimensionless, we need [ϕ ]=(d−2)/2. Thus [ ϕn]=n(d−2)/2 and so in order for/integraltext ddxλnϕnto be dimensionless, we must have [ λn]=n(2−d)/2+d. III.3.3 When we set m=0, the integrand is manifestly a linear combination of γmatrices. The integral cannot produce a term independent of the γmatrices, which is what Bis. For electrodynamics, the integral is changed to (ie)2i2/integraldisplayd4k (2π)41 k2[(1−ξ)kμkν k2−gμν]γμ/negationslashp+ /negationslashk+m (p+k)2−m2γν ≡A(p2)/negationslashp+B(p2) When m=0, the integrand is a linear combination of the product of three γmatrices, which can only reduce to one γmatrix, not to none. Incidentally, an alternative way of seeing the results stated here is to recall from chapter II.1 that with m=0 the Lagrangian is invariant under the chiral transformation ψ→eiθγ5ψ. III.3.4 This essentially follows from D=4−BE−3 2FEforBE=0 and FE=2. Then D=1 but the linear divergence is reduced to logarithmic divergence by the symmetry argument given in the text. III.5.2 Basically, you have already done this problem in exercises II.1.3 and II.1.4. You merely have to replace Eand/vectorpby∂/∂t and/vector∇(see also chapter III.6). III.5.3 In nonrelativistic quantum mechanics, the scattering amplitude is given in the Born approximation by itimes the Fourier transform of the potential: i/integraltext d3xei/vectork./vectorxU(/vectorx). The scattering amplitude owing to the exchange of a scalar meson of mass mis just i/(k2−m2)/similarequal−i/(/vectork2+m2). Thus, we just repeat the calculation in chapter I.4, obtaining U(/vectorx)=−/integraldisplayd3k (2π)3ei/vectork./vectorx /vectork2+m2=−1 4πre−mr III.6.1 ¯u(p/prime)(/negationslashp/primeγμ+γμ/negationslashp)u(p) =2m¯u(p/prime)γμu(p) by the equation of motion, but using γμγν=1 2{γμ,γν}+ 1 2[γμ,γν]=ημν−iσμνwe can also write (/negationslashp/primeγμ+γμ/negationslashp)=(p/prime+p)μ+iσμν(p/prime−p)ν. We thus obtain the Gordon decomposition. III.6.2 We compute qμ¯u(p/prime)[γμF1(q2)+iσμνqν 2mF2(q2)]u(p)=¯u(p/prime)/negationslashqu(p)F 1(q2) =¯u(p/prime)(/negationslashp/prime− /negationslashp)u(p)F 1(q2) =¯u(p/prime)(m−m)u(p)F 1(q2)=0 where the first and third equality follows from the antisymmetry of σand the equation of motion, respectively. III.7.1 Proceeding as in the text but living in d−dimensional spacetime, we obtain i/Pi1μν(q)=−i/integraltextddl (2π)dNμν D where1 D=/integraltext1 0dα1 Dwith D=(l2+α(1−α)q2−m2+iε)2as before but with Nμνnow effectively equal to−d(( 1−2 d)gμνl2+α(1−α)(2qμqν−gμνq2)−m2gμν). Rotating to Euclidean space we see that we have to do the integrals (with c2≡m2−α(1−α)q2)/integraltextdd El (2π)d1 (l2+c2)2and/integraltextdd El (2π)dl2 (l2+c2)2=/integraltextdd El (2π)d1 (l2+c2)− c2/integraltextdd El (2π)d1 (l2+c2)2. I did the first of these integrals for you in appendix II in chapter III.1. Generalizing slightly we have/integraltext∞ 0dlld−1 1 (l2+c2)a=1 2cd−2a/integraltext1 0dx( 1−x)d 2−1xa−1−d 2. I will let you carry on from here. Solutions to Selected Exercises | 553 Part IV IV .1.1 Write /vectorϕ=(ϕ1,ϕ2,..., v+ϕ/prime N). We compute1 2μ2/vectorϕ2−(λ/4)(/vectorϕ2)2up toO(ϕ3)and find (upon dropping the/prime) 1 2μ2(v2+2vϕN+/vectorϕ2)−λ 4(v4+4v2ϕ2 N+4v3ϕN+2v2/vectorϕ2) The condition of no linear term in ϕNfixesv2=μ2/λand so the coefficient of /vectorϕ2is equal to1 2μ2− λ/4(2v2)=0. The (N −1)fields ϕ1,ϕ2,..., ϕN−1are massless. IV .3.1 We have/integraltext∞ 0dklog[(k2+a2)/k2]=πaand so Veff(ϕ)=V( ϕ)+/planckover2pi√V/prime/prime(ϕ)/2 +O(/planckover2pi2).F o r L=1 2(∂ϕ)2− 1 2ω2ϕ2, the quantum oscillator with ϕidentified as position, we have Veff(0)=1 2/planckover2piω. IV .3.3 We have m(ϕ)=fϕ inVF(ϕ)=2i/integraltextd2p (2π)2logp2−m(ϕ)2 p2 which after Wick rotation becomes −2/integraldisplayd2pE (2π)2logp2 E+m(ϕ)2 p2 E=−1 2π/integraldisplay∞ 0dxlogx+m(ϕ)2 x After cutting off the integral at /Lambda12and adding a counterterm Bϕ2we obtain VF=1 2π(f ϕ)2logϕ2 M2 IV .3.4 Veff=i/summationtext∞ n=1/integraltext d4k/(2π)4(1/2n)[V/prime/prime(ϕ)/k2]n.F o rV/prime/prime(ϕ)=1 2λϕ2the corresponding Feynman diagrams consist of a circle with nV ’s attached to the circumference, where the 2 nis the infamous symmetry factor that I tried to avoid talking about in chapter I.7. IV .4.1 WithHan arbitrary p-form, ddH=1 (p+1)!1 p!∂λ∂νHμ1μ2...μpdxλdxνdxμ1dxμ2...dxμp =1 21 (p+1)!1 p![∂λ,∂ν]Hμ1μ2...μpdxλdxνdxμ1dxμ2...dxμp=0 IV .5.1 If you have done all the exercises thus far (see exercise I.10.3), you have already made the acquaintance of theI=2 scalar field transforming as ϕab→RacRbdϕcd=Racϕcd(RT)db=(RϕRT)ab, which thus can be written as a traceless 3 by 3 symmetric matrix ϕ→RϕRT. Now you merely have to write out the covariant derivative Dμϕ(see IV .5.20) explicitly. [Hint: The action of the generators on ϕis similar to what is shown in (B.20).] IV .5.2 dF=d(dA+A2)=dAA−AdA and [A,F]=AdA−dAA and so dF+[A,F]=0. Explicitly with indices, this reads εμνλσ(∂νFλσ+[Aν,Fλσ])=0. In the abelian case, we have, for μ=0,εijk∂iFjk= /vector∇./vectorB=0 (recall chapter IV .4!), and for μ=i,εijk(−∂ 0Fjk+∂jF0k−∂jFk0)=−∂0Bi+(/vector∇×/vectorE)i=0. IV .5.4 From the general arguments mentioned in the problem we know that tr F2must be the “d of some- thing.” Now dtrAdA=trdAdA anddtr2 3A3=2 3tr(dAA2−AdAA +A2dA)=2t rdAA2but on the other hand tr F2=tr(dA+A2)(dA+A2)=tr(dAdA +2dAA2)since tr A4=trA3A=−tr AA3= −trA4=0. In electromagnetism, tr A3=0, and dtrAdA when written out in elementary notation is just ∂μ(εμνλσAν∂λAσ)=1 4εμνλσFμνFλσ. 554 | Solutions to Selected Exercises IV .5.6 We simply plug in the general expression in the text and obtain L=−1 4g2Fa μνFaμν+¯q(iγμDμ−m)q with the covariant derivative Dμ=∂μ−iAμ=∂μ−iAa μTa, where Ta(a=1 ,...,8 )are traceless her- mitean 3 by 3 matrices. Explicitly, (A μq)α=Aa μ(Ta)αβqβ, with α,β=1, 2, 3 (see chapter VII.3). IV .6.3 Observe Aa μτa=/parenleftBiggA3 μA1−i2 μ A1+i2 μ−A3μ/parenrightBigg with the obvious notation A1±i2 μ≡A1 μ±iA2 μ. Let/angbracketleftϕ/angbracketright=/parenleftBig 0 v/parenrightBig so that Dμϕ=∂μϕ−i(gAa μτa 2+g/primeBμ1 2)ϕ→−i 2v/parenleftBigggA1−i2 μ −gA3 μ+g/primeBμ/parenrightBigg Thus, |Dμϕ|2contains v2[g2A1+i2 μA1−i2 μ+(−gA3 μ+g/primeBμ)2]. The combinations A1+i2 μ,A1−i2 μ, and (−gA3 μ+g/primeBμ)acquire mass while (g/primeA3 μ+gBμ)remain massless. IV .7.4 We have /Delta1μν(k1,k2)=i/integraldisplayd4p (2π)4Nμν D+{μ,k1↔ν,k2} where Nμν≡trγ5(/negationslashp− /negationslashq+M)γν(/negationslashp− /negationslashk1+M)γμ(/negationslashp+M) Only the term linear in MinNμνdoes not vanish, giving Nμν=4iMεμνστk1σk2τ. Since we are interested only in terms of O(k1k2)we can set D→(p2−M2)3so that /Delta1μν(k1,k2)=− 8Mεμνστk1σk2τ/integraldisplayd4p (2π)41 (p2−M2)3=i 4π2Mεμνστk1σk2τ with a dependence on Mas stated in the problem. The effect of the regulator, like some unsavory acquaintance, remains even after we have sent him to infinity. IV .7.5 We will sketch the solution. The details may be found in the lectures given by S. Adler at the 1970 Brandeis Summer School. The point is to imagine a regularization scheme that preserves the variousrelevant symmetries, namely Lorentz invariance, vector current conservation, and Bose statistics. As you will see, we don’t actually have to specify the regularization. By Lorentz invariance, we have /Delta1λμν(k1,k2)=ελμνσk1σA1+ελμνσk2σA2+ελμστk1σk2τkν 1A3 +ελμστk1σk2τkν 2A4+ελνστk1σk2τkμ 1A5 +ελνστk1σk2τkμ 2A6+εμνστk1σk2τkλ 1A7 +εμνστk1σk2τkλ 2A8 Since the Feynman integral representing /Delta1λμνis superficially linearly divergent, we see that A3,...,A8 are all convergent since we have to pull out three powers of momentum to extract them. In contrast, A1 andA2are logarithmically divergent. But we can relate them to A3,...,A8by vector current conservation since 0 =k1μ/Delta1λμν=ελνστk1σk2τ(−A2+k2 1A5+k1.k2A6)and thus A2=k2 1A5+k1.k2A6. Similarly for A1. Rationalizing the Feynman integrand and evaluating the trace in the numerator, we can systematically ignore terms that contribute only to A1andA2. Furthermore, Bose statistics gives us relations such as A3(k2 1,k2 2,q2)=−A6(k2 2,k2 1,q2). Solutions to Selected Exercises | 555 Part V V .1.1 We dropped the term h2∂0θbut kept the term 4 g2¯ρh2. This requires ∂0θ/lessmuchg2¯ρ, that is, ω/lessmuchg2¯ρ, but since in our solution ω∼g√¯ρ/mk this requires k/lessmuchg√m¯ρ, which is consistent with what we assumed about k. Looking at the terms −2√¯ρh∂ 0θ−4g2¯ρh2inLwe see that h∼∂0θ/(g2√¯ρ)/lessmuch√¯ρ, which is also consistent. V .5.1 Withγ5=σ3,1 2(I±γ5)clearly projects out the top and bottom component of ψ=/parenleftBig ψLψR/parenrightBig , respectively. Everything is formally the same as in chapter II.1, but we can also work things out explicitly in the specific representation given here. Thus, ¯ψψ=ψ†σ2ψ=i(ψ† RψL−ψ† LψR)and¯ψγ5ψ=ψ†σ2σ3ψ= i(ψ† RψL+ψ† LψR). Under the transformation ψ→eiθγ5ψ,ψL→eiθψLandψR→e−iθψR, and the massless Dirac Lagrangian L=iψ† R(∂ ∂t+vF∂ ∂x)ψR+iψ† L(∂ ∂t−vF∂ ∂x)ψL clearly does not change. V .6.1 This of course just follows from Lorentz invariance. We have ∂tϕ(x−vt√ 1−v2)=−v√ 1−v2ϕ/prime(x−vt√ 1−v2) and ∂xϕ(x−vt√ 1−v2)=1√ 1−v2ϕ/prime(x−vt√ 1−v2) and thus the equation (∂2 t−∂2 x)ϕ(x−vt√ 1−v2)+V/prime[ϕ(x−vt√ 1−v2)]=0 becomes ϕ/prime/prime(x−vt√ 1−v2)−V/prime[ϕ(x−vt√ 1−v2)]=0 Note that this does not depend on the form of V. For any relativistic theory, the soliton moves like a relativistic particle (obviously!). V .6.2 The sine-Gordon theory has an infinite number of vacua occurring at ϕ=(2n+1)π/β . Thus, there exists a whole spectrum of solitons, such that ϕ(±∞)=(2n±+1)π/β . The topological current is Jμ=(β/2π)εμν∂νϕwith the corresponding charge Q=(n+−n−). TheQ=2 soliton decays into two Q=1 solitons. V .7.4 (i/2π)/integraltext S1gdg†=(i/2π)/integraltext S1eiνθde−iνθ=(i/2π)/integraltext S1(−iνdθ) =(ν/2π)/integraltext2π 0dθ=ν, which indeed counts the number of times eiνθwinds around the circle. What mathematicians call the winding number is indeed just the magnetic flux of the physicist. V .7.5 Within a region small enough so that we can treat ϕa=vδa3as constant, using (Dμϕ)b=∂μϕb+ eεbcdAc μϕdwe have (Dμϕ)1=evA2 μand(Dμϕ)2=−evA1 μand thus Fμν≡Fa μνϕa |ϕ|−(1/e)εabcϕa(Dμϕ)b(Dνϕ)c |ϕ|3 →F3 μν+e(A2 μA1ν−A2 νA1μ)=∂μA3 ν−∂νA3 μ precisely the electromagnetic field strength since A3 μis the massless component of the Yang-Mills field. Let us compute Bk=εijkFijfar from the magnetic monopole. To calculate the magnetic charge we are 556 | Solutions to Selected Exercises interested only in the term of order 1 /r2in/vectorB. Since Dμϕ→O(1/r2)by construction we can drop the second term in Fij. Thus, we merely have to compute Fa ij≡∂iAa j−∂jAa i+eεabcAb iAcj. Since Fa ijwill eventually be contracted with the unit vector ϕa/|ϕ|=xa/r, we can effectively drop some of the terms inFa ij, thus simplifying the computation. We have ∂iAa j=∂i(1 eεajlxl r2)“="1 eεaji1 r2 and eεabcAb iAcj=(1/e)εabcεbimεcj nxmxn r4 =(1/e)(δciδam−δcmδai)εcj nxmxn r4=1 er4εijnxaxn so that Fa ijϕa |ϕ|=Fa ijxa r=1 er3(−2+1)εaijxa=−1 er3εaijxa and hence Bk=−(1/er2)ˆxk. The magnetic charge g=− 4π/e . Our result appears to differ from Dirac’s quantization condition (IV .4.10) by a factor of 2. The resolution of this apparent paradox is instructive. In fact, we can always introduce into this theory a field/Psi1(which could be a Bose or a Fermi field) transforming in the I=1 2representation with the corresponding covariant derivative Dμ/Psi1=∂μ/Psi1−ie(1 2τa)Aa μ/Psi1. The field /Psi1carries electric charge1 2e. Thus, the fundamental unit of electric charge is actually1 2e, note, and our result g=− 4π/e=− 2π/(e/2 ) is actually nothing but the Dirac quanization condition. (The sign is trivial: just a question of which onewe call the monopole and which the antimonopole.) V .7.7 Plugging in the Ansatz ϕ a=(H(r)/er)(xa/r) andAb i=[1−K(r) ]εbij(xj/er2)[so that H(r)−→ r→∞evr and K(r)−→ r→∞0 in accordance with the asymptotic behavior (V .7.5) and (V .7.6)] into M=/integraltext d3x{1 4(/vectorFij)2+ 1 2(Di/vectorϕ)2+V(/vectorϕ)}we get Mas a functional of HandK. Minimizing Mgives (with H/prime=dH/dr etc) the equations r2H/prime/prime=2HK2+(λ/e2)[H3−(ev)2r2H] andr2K/prime/prime=K(K2−1)+KH2. For help, see M. K. Prasad and C. M. Sommerfeld, Phys. Rev. Lett. 35: 760, 1975. V .7.8 The BPS solution corresponds to setting λ=0 in the two equations in exercise V .7.7, rendering the equations soluble, with the solution H(r)=evr( cothevr)−1 andK(r)=evr/( sinhevr) . Ask yourself whyH(r) andK(r) approach their asymptotic values exponentially. What determines the length scale? V .7.9 For help, see B. Julia and A. Zee, Phys. Rev. D11: 2227, 1975. V .7.11 We derived the lower bound for the mass of the magnetic monopole 4 πv|g|∼4π(ev)/e2∼MW/α. V .7.12 Near the identity element g=ei/vectorθ./vectorσ/similarequal1+i/vectorθ./vectorσand thus gdg†/similarequal−id/vectorθ./vectorσ. In a small neighborhood of the identity element the group manifold is locally Euclidean and so tr(gdg†)3=itr(σiσjσk)dθidθjdθk=− 12dθ1dθ2dθ3 is manifestly proportional to the volume element on S3.F o r g=ei(θ1σ1+θ 2σ2+mθ 3σ3),t r(gdg†)3= −12mdθ1dθ2dθ3. V .7.13/integraltext d4x(∂μJμ 5)=/integraltext d3xJ0 5|t=+∞−/integraltext d3xJ0 5|t=−∞ . Recalling that J0 5=ψ† RψR−ψ† LψL, we see that the two spatial integrals just count the number of right moving fermion quanta minus the number of left moving fermion quanta at t=± ∞ respectively. So/integraltext d4x(∂μJμ 5)is an integer. On the other hand, in the text we proved that/integraltext trF2is a topological invariant. In other words, with suitable normalization, evidently 1/(4π)2,the integral [1 /(4π)2]/integraltext d4xεμνλσtrFμνFλσis an integer. Thus, the coefficient 1 /(4π)2cannot be shifted even a little bit by quantum fluctuations. Solutions to Selected Exercises | 557 Part VI VI.4.2 The quartic interaction term (1/2f2)(/vectorπ.∂/vectorπ)2inLgives the amplitude i(1/2f2)i2δabδcd(k1k3+k1k4+ k2k3+k2k4)+permutations =(i/2f2)δabδcd(k1+k2)2+permutations for the 4-pion interaction vertex (where for convenience we have labeled all the momenta as going outward so that k1+k2+k3+k4=0). VI.4.3 After writing σ=v+σ/prime, we find, as in chapter IV .1, that L=−1 2(2μ2)σ/prime2−λvσ/prime/vectorπ2−1 4λ(/vectorπ2)2+ ... , where we have displayed only terms relevant for our purposes. The diagrams contributing to four-pion interaction are of two types, those involving the λ(/vectorπ2)2term and those invoking σ/primeex- change. The former gives for the amplitude (−i 4λ)2 .2(δabδcd+δacδbd+δadδbc)while the latter gives (−iλv)2{2i/[(k1+k2)2−m2 σ/prime]}δabδcd. Thus, expanding to first order in momenta squared we find the coefficient of δabδcd: −iλ−2iλ2v2/bracketleftBigg −1 m2 σ/prime/bracketrightBigg (1+(k1+k2)2 m2 σ/prime)=−iλ+2iλ2v2 2μ2/bracketleftbigg 1+(k1+k2)2 2μ2/bracketrightbigg =iλ 2μ2(k1+k2)2 To compare with exercise VI.4.2 we remember that f2=v2=μ2/λso that the amplitude here is also equal to (i/2f2)δabδcd(k1+k2)2+permutations, as we had anticipated in the text. VI.4.4 We will track down factors of 2 carefully but not factors of iand−1. Let us go back to the chiral transformations ψ→[1+i/vectorθ.(/vectorτ/2)γ5]ψand¯ψ→¯ψ[1+i/vectorθ.(/vectorτ/2)γ5]. Thus, δ(¯ψψ)=θa¯ψiγ5τaψand δ(¯ψiγ5τaψ)=−θa¯ψψ . Hence, for L=¯ψ{iγ∂+g(σ+i/vectorτ./vectorπγ 5)}ψ+L(σ,/vectorπ)to be invariant we must haveδσ=θaπaandδπa=−θaσ. Applying Noether’s theorem Jμ=(δL/δ∂μϕ)δϕ, we obtain the current Ja μ5=¯ψiγμγ5(τa/2)ψ+πa∂μσ−σ∂μπawritten in the text. Comparing the term ¯piγμγ5ncontained in J1+i2 μ5≡J1 μ5+iJ2 μ5with the current J5μdefined in chapter IV .2, we see that J5μ=−iJ1+i2 μ5. The normal- ized state |π−/angbracketright=(1/√ 2)(|π1/angbracketright−i|π2/angbracketright)so that /angbracketleft0|π1+i2|π−/angbracketright= 2/√ 2. The current J1+i2 μ5contains the term−v∂μπ1+i2and thus f=√ 2v. Next, we have to work out the pion-nucleon coupling gπNN as de- fined in chapter IV .2. Here Lcontains g¯ψi/vectorτ./vectorπγ5ψ, which contains√ 2g¯piγ 5nπ−sinceπ1−i2=√ 2π−. Thus, gπNN=√ 2g. Putting it together, we see that M=gvtranslates to 2 M=fgπNN in agreement with chapter IV .2. VI.6.1 See figure VI.6.1. From /Delta1h=(d/cos θ)/similarequald(1+1 2θ2)and(∂h/∂x) =tanθ/similarequalθ, we have (∂h/∂t) ∝θ2∝ (∂h/∂x)2, thus giving rise to the term (λ/2)(/vector∇h)2. It all goes back to Mr. Pythagoras. VI.6.2 We integrate the term1 2/integraltext dD/vectorxd t [((∂/∂t) −/vector∇2)h]2inS(h) by parts to obtain −1 2/integraltext dD/vectorxd t [h((∂/∂t) + /vector∇2)((∂/∂t) −/vector∇2)h]. Thus, the propagator is the inverse of the operator (∂/∂t +/vector∇2)(∂/∂t −/vector∇2)= ∂2/∂t2−(/vector∇2)2, the Fourier transform of which is −(ω2+k4). VI.8.3 Withh/prime(/vectorx,t)=h/parenleftbig /vectorx+g/vectorut,t/parenrightbig +/vectoru./vectorx+g 2u2t, we have ∂h/prime/∂t=∂h/∂t +g/vectoru./vector∇h+(g/2)u2and/vector∇h/prime= /vector∇h+/vectoru. Thus the combination (∂h/∂t) −g 2(/vector∇h)2is invariant, as is (obviously) /vector∇2h. In other words, ˜S(h) must be constructed out of these two invariant combinations. VI.8.5 Look at the action S(h)=1 2/integraltext dD/vectorxd t [(∂h/∂t) −∇2h−g(∇h)2/2]2. Comparing ∂h/∂t −∇2hwe see that time has the dimension of length squared: T∼L2. From the term/integraltext dD/vectorx dt (∂h/∂t)2and the fact that Sis dimensionless, we have [h]2∼T2/(LDT)∼1/LD−2and so hhas the dimension of (1/LD−2)1 2. Comparing /vector∇2hwithg(/vector∇h)2we see that ghas the dimension of 1 /h, that is, L(D−2)/2. VI.8.7 We are told that L(dg/dL) =(2−D)g/ 2+(2D−3)fDg3+... . We are assuming that the terms (...)can be neglected. Thus (in what follows a2andb2are two generic positive numbers) for D=1, L(dg/dL) =a2g−b2g3andgflows toward the fixed point g∗=a/b . (Incidentally, the KPZ equation is soluble for D=1 by methods not explained in this text and both zandχare known exactly.) For D=2,L(dg/dL) =b2g3andgflows toward some unknown strong (presumably) coupling fixed point. ForD=3,L(dg/dL) =−a2g+b2g3. The fixed point g∗=a/b is unstable. For g<g∗,gflows toward the trivial (i.e., free, or Gaussian) fixed point. Since the theory at the fixed point is free we know the 558 | Solutions to Selected Exercises critical exponents exactly: z=2 and χ=(2−D)/ 2. For g>g∗,gflows toward some unknown strong (presumably) coupling fixed point. Part VII VII.1.1. SetnμA/prime μ(x)=0 with A/prime μ=U†AμU+iU†∂μU, so that n.∂U(x) =in.A(x)U(x). Define λ(x)= r.x/(r .n)for any 4-vector rand write x=λ(x)n +x⊥, so that r.x⊥=0. Then U(x)=Pei/integraltextλ(x) 0dσn.A(σn+x ⊥) (with a path ordering) solves the differential equation, since n.∂λ=1 by construction. VII.1.2 Using the BHC formula given, we have (the V’s are clearly irrelevant) UijUjk=eiaAμeiaAν=eia(Aμ+Aν)−1 2a2[Aμ,Aν]+a3C+a4D+O(a5) Similarly, UklUli=e−iaA/prime μe−iaA/primeν=e−ia(A/prime μ+A/primeν)−1 2a2[Aμ,Aν]+a3E+a4F+O(a5): the prime reminding us that the AμandAνin this expression is to be evaluated on the “north” and “west” side of the plaquette in figure VII.1.2, respectively, in contrast to the AμandAνinUijUjkwhich are evaluated on the “south” and “east” side, respectively. Here C,D,E, andFdenote various commutators, which we drag along merely to show that they eventually drop out in what interests us. (Note how thedifferent terms are associated with different powers of a, as indicated. Note also that in some places we have dropped the prime on Aand absorbed the “error” in doing so into terms of higher order in a.) Thus, UklUli=e−ia(A μ+Aν)−ia2(∂νAμ−∂μAν−1 2i[Aμ,Aν])+a3G+a4H+O(a5) where GandHdenote sums of commutators and terms such as ∂ν∂νAμand∂ν∂ν∂νAμ. Applying the BHC formula again to the order indicated we have UijUjkUklUli=eia2(∂μAν−∂νAμ)−a2[Aμ,Aν]+O(a4)=eia2Fμν+a3I+a4J+O(a5) withFμν=∂μAν−∂νAμ+i[Aμ,Aν]. The same remarks on GandHapply to IandJ. The Yang-Mills field strength emerges naturally, as we would anticipate. Since the traces of commutators and of Avanish, when we apply the trace all the junk drops out to O(a5)and we have S(P)=Re tr[1 −1 2a4FμνFμν+O(a5)] By gauge invariance, the corrections must be of even order in abut for our purposes we don’t care about them anyway. Evidently, fandgare related by some uninteresting factors of a. Part VIII VIII.1.7 R12=dω12=d(−cosθdϕ)=sinθdθdϕ =2R12 θϕdθdϕ /equal1⇒R12 θϕ=1 2sinθ. Since eθ 1=1,eϕ 2=1/sinθ, we obtain R≡Rab μνeμ aeν b=2R12 θϕeθ 1eϕ 2=1, independent of θandϕas expected. Part N N.1.1. The effective action for an electrically neutral system is given in the point particle limit by S=/integraltext dτ(−m+ bEEμEμ+bBBμBμ+...), with EμandBμdefined in the text. The interaction terms involve two powers of derivatives, which translate into two powers of ωin the scattering amplitude and hence four powers ofωin the scattering cross section. (Note that a possible term like/integraltext dτFμνFμνcan be absorbed into the two terms already displayed.) N.3.2. As in (III.3.7) we have 3 V3+4V4=2I+nwhere Idenotes the number of internal lines. The number of loops (III.3.6) L=I−(V3+V4−1)is 0 in a tree diagram. Thus V3=n−2−2V4≤n−2. Further Reading Books on field theory This is a list of field theory textbooks that I know about. I do not necessarily recommend them all. In food as in books, each has his or her own taste. T . Banks, Modern Quantum Field Theory, Cambridge University Press, New York, 2008. J. D. Bjorken and S. D. Drell, Relativistic Quantum Mechanics , McGraw-Hill, New York, 1964. ———, Relativistic Quantum Fields , McGraw-Hill, New York, 1965. L. S. Brown, Quantum Field Theory , Cambridge University Press, New York, 1992. S. J. Chang, Introduction to Quantum Field Theory , World Scientific, Singapore, 1990. T . P. Cheng and L. F. Li, Gauge Theory of Elementary Particle Physics , Clarendon Press, Oxford, 1984. F. Dyson and D. Derbes, Advanced Quantum Mechanics, World Scientific, Singapore, 2007. R. P. Feynman, Quantum Electrodynamics, W . A. Benjamin, New York, 1962. K. Huang, Quantum Field Theory , John Wiley & Sons, New York, 1998. C. Itzykson and J-B. Zuber, Quantum Field Theory , McGraw-Hill, New York,1980. T . D. Lee, Particle Physics and Introduction to Field Theory , Taylor & Francis, New York, 1981. V . P. Nair, Quantum Field Theory, Springer, New York, 2005. M. E. Peskin and D. V . Schroeder, An Introduction to Quantum Field Theory , Perseus, Reading MA, 1995. L. H. Ryder, Quantum Field Theory , 2nd Ed., Cambridge University Press, New York, 1996. M. Stednicki, Quantum Field Theory, Cambridge University Press, New York, 2007. G. Sterman, An Introduction to Quantum Field Theory, Cambridge University Press, New York, 1993. S. Weinberg, Quantum Theory of Fields ,V o l s .1&2 ,C ambridge University Press, New York, 1996. X. G. Wen, Quantum Field Theory of Many-Body Systems, Oxford University Press, New York, 2007. and finally, of course, F. Mandl, Introduction to Quantum Field Theory , Interscience, New York, 1959. 560 | Further Reading Books on various topics mentioned A. A. Abrikosov, L. Gorkov, and A. Dzyaloshinski, Methods of Quantum Field Theory in Statistical Physics , Prentice Hall, Englewood Cliffs, NJ, 1963. S. L. Adler, “Perturbation Theory Anomalies,” in: Lectures on Elementary Particles and Quantum Field Theory , 1970, Brandeis University Summer Institute in Theoretical Physics, S. Deser et al, ed., MIT Press, Cambridge, 1970. P. Anderson, Basic Notions of Condensed Matter Physics , Benjamin-Cummings, Menlo Park, CA 1984. D. Bailin and A. Love, Supersymmetric Gauge Field Theory and String Theory , IOP Publishing, Bristol and Philadelphia, 1994. R. Balian and J. Zinn-Justin, eds., Methods in Field Theory , North Holland Publishing, Amsterdam, and World Scientific, Singapore, 1981. A. L. Barabasi and H. E. Stanley, Fractal Concepts in Surface Growth , Cambridge University Press, Cambridge, 1995. D. Budker, S. J. Freedman, and P. H. Bucksbaum, eds., Art and Symmetry in Experimental Physics: Festschrift for Eugene D. Commins , American Institute of Physics, New York, 2001. J. Cardy, Scaling and Renormalization in Statistical Physics , Cambridge University Press, New York, 1996. S. Coleman, Aspects of Symmetry , Cambridge University Press, Cambridge, 1985. J. Collins, Renormalization , Cambridge University Press, Cambridge, 1985. E. D. Commins, Weak Interactions , McGraw-Hill, New York, 1973. E. D. Commins and P. H. Bucksbaum, Weak Interactions of Leptons and Quarks , Cambridge University Press, Cambridge, 2000. M. Creutz, Quarks, Gluons and Lattices , Cambridge University Press, Cambridge, 1983. P. A. M. Dirac, The Principles of Quantum Mechanics , Oxford University Press, Oxford, 1935. (On p. 253 he explained why he wanted the equation of motion for the electron to be first order in timederivative.) A. Dobado et al., Effective Lagrangians for the Standard Model , Springer-Verlag, Berlin, 1997. O.J.P. ´Eboli et al., Particle Physics , World Scientific, Singapore 1992. R. P. Feynman, Statistical Mechanics , Perseus Publishing, Reading, MA, 1998. R. P. Feynman and A. R. Hibbs, Quantum Mechanics and Path Integrals , McGraw-Hill, New York, 1965. J. M. Figueroa-O’ Farrill, Electromagnetic Duality for Children , on the World Wide Web 1998. V . Fitch et al., eds., Critical Problems in Physics , Princeton University Press, Princeton, 1997. M. Gell-Mann and Y . Ne’eman, The Eightfold Way , W . A. Benjamin, New York, 1964. H. B. Geyer, ed., Field Theory, T opology and Condensed Matter Physics , Springer, 1995 (A. Zee, “Quantum Hall Fluids.”) M. L. Goldberger and K. M. Watson, Collision Theory, Dover, New York, 2004. N. Goldenfeld, Lectures on Phase T ransitions and the Renormalization Group , Addison-Wesley, Read- ing, MA, 1992. F. Guerra and N. Robotti, Ettore Majorana: Aspects of His Scientific and Academic Activity, Springer, New York, 2008. C. Itzykson and J-M. Drouffe, Statistical Field Theory , Cambridge University Press, Cambridge, 1989. S. Iyanaga and Y . Kawada, eds., Encyclopedic Dictionary of Mathematics , MIT Press, Cambridge, 1980. L. Kadanoff, Statistical Physics , World Scientific, Singapore, 2000. G. Kane and M. Shifman, eds., The Supersymmetric World: The Beginning of the Theory ,World Scientific, Singapore, 2000. J. I. Kapusta, Finite-T emperature Field Theory , Cambridge University Press, Cambridge, 1989. L. D. Landau and E. M. Lifschitz, Statistical Physics , Addison-Wesley, Reading, MA, 1974. S. K. Ma, Modern Theory of Critical Phenomena , Benjamin/Cummings, Reading, MA, 1976. Further Reading | 561 H. J. W . M ¨uller-Kirsten and A. Wiedemann, Supersymmetry , World Scientific, Singapore 1987. T . Muta, Foundations of Quantum Chromodynamics , World Scientific, Singapore, 1998. D. I. Olive and P. C. West, eds., Duality and Supersymmetric Theories , Cambridge University Press, Cambridge, 1999. J. Polchinski, String Theory , Cambridge University Press, Cambridge, 1998. J. J. Sakurai, Invariance Principles and Elementary Particles , Princeton University Press, Princeton, 1964. L. Schulman, T echniques and Applications of Path Integrals , John Wiley & Sons, New York, 1981. R. F. Streater and A. S. Wightman, PCT , Spin Statistics, and All That , W . B. Benjamin, New York, 1968. G. ’t Hooft, Under the Spell of the Gauge Principle , Word Scientific, Singapore, 1994. G. ’t Hooft et al., eds. Recent Developments in Gauge Theories , Plenum, New York, 1980. D. Voiculescu, ed., Free Probability Theory , American Mathematical Society, Providence, R.I., 1997. S. Weinberg, Gravitation and Cosmology , John Wiley & Sons, New York, 1972. C. N. Yang, Selected Papers 1945–1980 with Commentary , W . H. Freeman, San Francisco, 1983. A. Zee, Unity of Forces in the Universe , World Scientific, Singapore, 1982. J.-B. Zuber, ed., Mathematical Beauty of Physics , World Scientific, Singapore, 1997. Some popular books and books on the history of quantum field theory M. Bartusiak, Einstein ’s Unfinished Symphony, Joseph Henry Press, Washington, D.C., 2000. I. Duck and E. C. G. Sudarshan, Pauli and the Spin-Statistics Theorem , World Scientific, Singapore 1997. R. P. Feynman, QED: The Strange Theory of Light and Matter, Princeton University Press, Princeton, 2006. D. Kaiser, Drawing Theories Apart, University of Chicago Press, Chicago, 2005. A. I. Miller, Early Quantum Electrodynamics , Cambridge University Press, Cambridge, 1994. L. O’Raifeartaigh, The Dawning of Gauge Theory , Princeton University Press, Princeton, 1997. S. S. Schweber, QED and the Men Who Made It: Dyson, Feynman, Schwinger, and T omonaga , Princeton University Press, Princeton, 1994. A. Zee, Fearful Symmetry , Princeton University Press, Princeton, 1999. ———, Einstein ’s Universe , Oxford University Press, New York, 2001. ———, Swallowing Clouds , University of Washington Press, Seattle, 2002. Further Reading for Part N In writing a textbook, the author has the luxury of not preparing a detailed scholarly bibliography (unless he or she chooses to follow the example of S. Weinberg, who is, in my opinion, mostadmirable in this regard). Even more extravagant is the freedom accorded to authors of popularbooks who in most cases give their unsuspecting and gullible readers the impression that the physicsof an entire era was done by two or three greats, individuals worthy of their own personality cults.Presenting recent developments still in flux, I am faced with the dilemma of whether to give propercredit. In scholarly publications, conscientious referencing is of course ethically mandated, but thisis a textbook. Fortunately, in this age of omniscient search engines, the reader could easily compilea bibliography more exhaustive than even a myopic humanist used to be able to muster in half alifetime. I could do the same, but it is of little help to you for me to merely list the names of those 562 | Further Reading responsible for, say, the new way of computing amplitudes using the spinor helicity formalism.1 Instead, I can best serve the typical reader by listing a few papers and review articles starting from which you can track down the literature to your scholarly heart’s desire. To those who feel that theyshould be mentioned, I apologize and refer you to Glashow’s description of a tapestry in the preface. W . Goldberger and I. Z. Rothstein, arXiv: hep-th/0409156v2. Z. Bern, L. J. Dixon, D. C. Dunbar, D. A. Kosower, arXiv: hep-ph/9602280.N. Arkani-Hamed and J. Kaplan, arXiv: hep-th/0801.2385.E. Witten, arXiv: hep-th/0312171. 1F. A. Berends, Z. Bern, L. Chang, P. De Causmaecker, L. J. Dixon, D. C. Dunbar, R. Gastmans, W . Giele, J. F. Gunion, R. Kleiss, D. A. Kosower, Z. Kunszt, M. Mangano, A. G. Morgan, S. J. Parke, W . J. Stirling, T . R.Taylor, W . Troost, T . T . Wu, Z. Xu, D. H. Zhang, and many many others. I know how to copy and paste also! Please forgive me if I inadvertently left you off this list. Index Page numbers followed by letters f and n refer to figures and notes, respectively. Abrahams, Elihu, 366 accelerators, 42, 483Adler, Steve, 277Aharonov-Bohm effect, 251–252, 261, 317, 320amplitudes: as analytic functions, 208–209; symmetry in, 78. See also meson-meson scattering amplitude “amputating the external legs,” 55analyticity, in quantum field theory, 207–209, 211, 217, 219. See also nonanalyticity Anderson, Phil: in “Gang of Four,” 366; Nobel Prize for, 351 Anderson localization, 351, 354; in renormalization group language, 366–367 Anderson mechanism, 264angular momentum, addition of, 530anharmonicity, in field theory, 43, 89anomaly (axial anomaly/chiral anomaly), 270, 275; alternative ways of deriving, 279; consequences of,275–278; Feynman diagram calculation revealing,270–274; grand unification and freedom from,411–412, 429; higher-order quantum fluctuationsand, 310; in nonabelian gauge theory, 276;nonrenormalization of, 277; and path integralformalism, 278 anthropic selection, 451anticommutation: spin-statistics connection and, 122, 123; wave function of electrons and, 107 antiferromagnet(s): effective low energy descriptionof, 344–345; magnetic moments in, 344; N ´eel state for, 346 antikink(s), 304, 305fantimatter: discovery of, 101; requirement of, 157antineutrino field, in SO (10) unification, 425–426 antiunitary operator, time reversal as, 103anyon(s), 315; interchanging, 316; statistics between, 317 approximation, steepest-descent, 16area law, 377, 387asymptotically free theories, 360, 386, 390; Gross-Neveu model as, 403 atom(s), interaction with radiation, 3attraction: quantum field theory on, 35–36; spin 1 particle and, 36–37; spin 2 particle and, 36 auxiliary field, 192, 467–468axial anomaly. See anomaly axial current conservation, quantum fluctuations destroying, 274–275 axial gauge, 378 background field method, 504–507 Bardeen, Bill, 277bare perturbation theory, 175baryon number conservation, law of, 413; grand unification and violation of, 418 BCFW recursion, 500, 507, 514 Berends, F. A., 493Bern, Zvi, 484, 484f, 516, 519 564 | Index Bern transformation, 516 Berry’s phase, nonabelian, 261, 346Berry’s phase term, in ferromagnets and antiferromagnets, 345, 346 beta decay, 456Bethe, Hans, 365Bianchi identity, 247, 261Bjorken, James, 356black hole(s): gravitational waves in, 479–482; Hawking radiation from, 290–291; Schwarzschild,311 Bludman, Sid, Yang-Mills theory and, 379blue sky, effective field theory of, 457–458Bogoliubov, N. N., 192Bogoliubov calculation, of gapless mode, 284Bogomol’nyi inequality, 305; for mass of monopole, 309 Bogomol’nyi-Prasad-Sommerfeld (BPS) states, 309Bohr, Niels, 252Boltzmann, Ludwig, 150, 287; on entropy, 311Bose-Einstein condensation, 295Bose-Einstein statistics, 120Bose field, mass correction to, divergence of, 180boson(s): “bad” behavior of, 180; electron pairing into, 295; and fermions, unification of, 461;gapless mode in, 284; gauge (see gauge boson[s]); intermediate vector, and Fermi theory of theweak interaction, 171–172, 309; Lorentz invariantscalar field theory on, 190; mass correction for,divergence of, 180; massless, emergence of, 226–227; Nambu-Goldstone ( seeNambu-Goldstone boson[s]); nonabelian gauge (Yang-Mills), 257,386, 434; in nonrelativistic theory, 192–193;repulsion of, 283, 337 BPS (Bogomol’nyi-Prasad-Sommerfeld) states, 309brane world scenarios, 40–42, 450Br´ezin, Edouard, 402 Brillouin zone, 298Brink, Lars, 470Britto, Ruth, 500Burgoyne, N., 121 Cachazo, F. A., 500 canonical formalism, 61–69; and degrees of freedom, 67; and Feynman diagrams, 43; vs. path integralformalism, 44, 61, 67; propagator in, 67–68; timeordering in, 67–68 Carrasco, J. J., 519Casimir force, between two plates, 70–75, 71fCauchy’s theorem, 209central identity of quantum field theory, 182, 524 chain rule, 445charge: in dual theory, vortices as, 332–334; asgenerator, 80; of quasiparticles, 327. See also electric charge; magnetic charge charge conjugation, 101–102; in grand unification, 429 charge quantization, deducing, 121Chern-Simons term, 317; gauge invariance of, 328; for Hall fluid, 326; massive Dirac fermions and,319–320; and Maxwell term, 320; in nonrelativistictheory, 320 Chern-Simons theory, 318–319; effective theory of Hall fluid as, 324 Chew, Geoff, 105chiral anomaly. See anomaly chiral superfield, 464, 466chiral symmetry: condition for, 419; conserved current associated with, 100; of strong interaction, 234, 387–388 classical limit, path integral formalism for taking, 19classical physics, symmetry of, 270Clifford algebra: and Dirac bilinears, 97; and Dirac equation, 94–95 closed forms, 247coherence length, 296Coleman, Sidney, 252, 473Coleman-Mermin-Wagner theorem, 230Coleman-Weinberg effective potential, 240color, quark, 385, 386complex plane, 207, 208Compton, Arthur, 154Compton scattering, 152–157condensation, and superconductivity, 295condensed matter physics: critical dimension in, 364, 367; disordered systems studied by, 350, 354;goal of, 328; Goldstone’s theorem in, 229–230;impurities studied by, 354; length scales in, 169;momentum density in, 191; number conjugateto phase angle in, 192; particle physics and, 281,452–453; quantum field theory and, 5, 190, 281;quantum Hall effect and, 351–352; quasiparticlesin, 326; renormalization group in, 360–363;spin-statistics rule and, 120 conductivity, vs. conductance, 366–367connected graphs, vs. disconnected graphs, 29, 47conserved current: charge associated with, 80; and chiral symmetry, 100; and continuous symmetry,78–79; momentum space version of, 133 continuous symmetry, 77–78, 226; conserved current and, 78–79 continuous symmetry breaking, 226; Coleman- Mermin-Wagner theorem on, 230; and masslessfields, 228–229 Cooper pairs of electrons, 295coordinate transformations, 81–82 Index | 565 cosmic coincidence problem, 450 cosmological constant, 448–449; measured in units of Gev4, 449–450; order of magnitude, 449 cosmological constant problem, 450; approach to, 455; root of, 448; string theory’s inability toresolve, 450; supersymmetry as solution to, 461 Coulomb potential, 133–134, 143; modification of, 205 Coulomb’s electric force: and Newton’s gravitational force, comparison of, 29; quantum field theory on,32–33 counterterms: cutoff dependence absorbed by, 241; in Feynman diagrams, 175–176, 176f;nonrenormalizable theories and, 179, 241 coupling, electromagnetic, 358coupling constant(s), 164, 173; dimensionless, 170; of electromagnetic interaction, 359; hadronproliferation and, 231; as misnomer, 358–359;pion-nucleon, 235; renormalized, 166; Yang-Mills,258–259 coupling renormalization, 173–174CPT theorem, 104critical dimension, in condensed matter physics, 364, 367 critical phenomena, 292; complete theory of, 293; Landau-Ginzburg theory of, 293–294 crossing, 156cubic vertex, in spinor helicity formalism, 496current conservation. See conserved current curved spacetime: Dirac action in, 445; introduction to, 84–86; quantum field theory in, 82, 290 Cutkosky cutting rule, 215–216, 217, 219, 493cutoff dependence, 163; avoiding in physical perturbation theory, 176; counterterms absorbing,241; disappearance of, 166, 167; of meson-mesonscattering amplitude, 173 dark energy, 450 dark matter, 450Dashen, Roger, 405decay rate, 139–141, 212Deser, Stanley, 470differential forms, 246–247; use in nonabelian gauge theory, 255, 256 differential operator, propagator as inverse of, 23dilatation invariance, 84ndimensional analysis, 169–170, 453; on meson- meson scattering amplitude, 173 dimensional regularization, 167, 168, 204Dirac, Paul, 105; on electric charge, quantizing of, 410; on magnetic monopole, 308; metaphorical language of, 113; on path integral formalism,10–13; on positron, 5; on quantum mechanicsand magnetic monopoles, 245; on spinor representation, 117; teaching style of, 454 Dirac basis vs. Weyl basis, 98–99Dirac bilinears, 97Dirac equation, 93–105; Clifford algebra and, 94– 95; in curved spacetime, 444; and degrees offreedom, reduction in, 95; derivation of, 118;electromagnetic field and, 101; handedness and,100; Lorentz transformation and, 96–97; magneticmoment of electron in, 194–195; origins of, 93–94; parity and, 98; in solid state physics, 298, 299;solving, 98; time reversal and, 104 Dirac field: interacting with scalar field, Feynman rules for, 53–54, 534–535; interacting with vectorfield, 100; interacting with vector field, Feynmanrules for, 129, 129f, 535–536; propagator for, 127; quantizing, 107–113, 122; quantizing byGrassmann path integral, 127; vacuum energy of,111–112, 125 Dirac operator, 113Dirac spinor, 94, 96; components of, 117; and supersymmetry, 114 disorder: Anderson localization of, 351, 354; condensed matter physics and study of, 350, 354;Grassmannian approach to, 354 dispersion relations, 208–210, 217–218, 235Di Vecchia, Paolo, 470divergence(s): degree of, 176–178; dependence on dimension of spacetime, 179; with fermions, 178–179; logarithmic, 175, 176–177; in quantum fieldtheory, 57–58, 161–162; superficial degree of,176, 179; total, supersymmetric transformation,465–466 dotted and undotted notation, 116–117, 541–544; replacing, 475 double-line formalism, 395, 396double-slit experiment, 7, 8f; expansion of, 7–9, 8f, 9fdouble-well potential, 224, 224fDrell, Sid, 356duality: in (2+1)-dimensional spacetime, 335; concept of, 331, 332; electromagnetic, 249; andlinking of perturbative weak coupling to strongcoupling, 473; of monopoles, 334; nonrelativistictreatment of, 336–337; relativistic treatment of,335–336; of string theories, 334; vortex, 334 dual theory, vortices as charges in, 332–334dynamical symmetry breaking, 230; example of, 388dynamical variable, in quantum field theory, 19dyon, 309Dyson, Freeman, 60Dyson gas approach, 400–402 effective field theory: of blue sky, 457–458; 566 | Index effective field theory (continued) development of, 452; Fermi theory of the weakinteraction as, 456; gravitational waves and, 479–482; of Hall fluid, 324–325, 452–453; of neutrinomasses, 456; predictive power of, 456–459; ofproton decay, 455–457; recent developments in,479–482; and renormalization group flow, 453;reshuffling terms in, 458–459 effective potential, 238–239; Coleman-Weinberg, 240; generated by quantum fluctuations, 243 Ehrenfest, P., 441Einstein, Albert: and cosmological constant, 449; and repeated indices summation convention, 475n Einstein-Hilbert action for gravity, 433–434; Newtonian gravity derived from, 438; Yang-Millsaction compared with, 434–435 Einstein Lagrangian, linearized, 34Einstein’s theory of gravity, 81, 83; and deflection of light, 439–440; gravitational waves in, 479–481;nonrenormalizability of, 172; Yang-Mills theorycompared with, 444–445, 513–520 electric charge: quantized, grand unification on, 410; quantum fluctuations and, 204, 205;renormalization of, 205 electric force: and gravitational force, comparison of, 29; quantum field theory on, 32–33 electromagnetic coupling, flow of, 358electromagnetic duality, 249electromagnetic field: Dirac equation in presence of, 101; as quantum field, 3–4; stress-energy tensorof, 83–84 electromagnetic force: between like charges, 33; knowledge of, 448 electromagnetic wave, degrees of freedom of, 38electromagnetism: Faddeev-Popov method applied to, 185–186; and gravity, unification of, 442; Maxwellon (see Maxwell theory of electromagnetism); weakness of, 414–415 electron(s): absolute identity of, 120–121, 134; binary strings in, 428; Bose-Einstein statistics for, 120;in condensed matter system, 281; Cooper pairsof, 295; degrees of freedom of, 95, 99; effectof magnetic field on, 251–252; energy levelsavailable to, 5; Fermi-Dirac statistics for, 120;as fermions, 322; fractional Hall state of, 324;magnetic moment of ( seemagnetic moment of electron); mass of, in classical physics, 180;noninteractive hopping, 298–299, 298f, 299f;pairing into bosons, 295; photon fluctuation into,200–202, 201f; photon scattering on, 152–157,153f, 157f; requirements of antisymmetric wave function, 107; stability of, 413 electron-positron annihilation, 155–156, 389–391electron scattering, 132–143; cross sections for, 137–143; off electrons, 134–138, 135f; off nucleons, deep inelastic, 386; off protons, 132–134, 133f,199; off protons, deep inelastic, 359; off protons,Schr ¨odinger equation for, 3; to order e 4, 145–149, 145–147f, 148f; potential, 133–134, 134f electroweak theory, 170–171, 379; construction of, 379–383; renormalizability of, 384 energy: dark, 450; fundamental definition of, 83; of mass, 35; quantum mechanics and specialrelativity on, 3; of vacuum (see vacuum energy) energy density, 35energy-momentum tensor, 319energy scales: in particle physics, 169; renormalization group and, 361 entropy, Boltzmann on, 311Euclidean path integral, 12 Euclidean quantum field theory, 287–288; and high- temperature quantum statistical mechanics, 289;and quantum statistical mechanics, 289 Euler, Leonhard, 460Euler-Lagrange equation, 12, 80, 438, 448exact forms, 247 Fadeev-Popov method, 183–185, 267, 371; applying to electromagnetism, 185–186; and derivation ofgraviton propagator, 437 Feng, Bo, 500Fermi, Enrico, 105, 137Fermi coupling, 170Fermi-Dirac statistics, 120Fermi field, mass correction to, divergence of, 180Fermi liquid, gapless modes in, 285fermion(s): and bosons, unification of, 461; degree of divergence with, 178–179; electrons as, 322;Feynman rules for, 128–131, 128f; in lattice gaugetheory, 376; mass correction for, divergence of,180; massive Dirac, and Chern-Simons term,319–320 fermion-fermion scattering, Feynman diagram for, 172, 172f fermion masses: in grand unification, 417–418; naturally small, 419 fermion normalization factors, 134fermion propagator, 112Fermi theory of the weak interaction, 232; as effective field theory, 456; intermediate vector boson and,171–172; nonrenormalizability of, 170, 179, 384;predictive power of, 453; within electroweaktheory, 171 ferromagnet(s), 229; effective low energy description of, 344–345; low energy modes in, 345–346; magnetic moments in, 344; order in, 328 ferromagnetic transition, 295Feynman, Richard: contribution of, 43; on difficulty Index | 567 of quantum electrodynamics, 61; on Dirac, 105; metaphorical language of, 113; on path integralformalism, 7–10; study of calculus by, 522; ontrace products of gamma matrices, 137; Yang-Millstheory and, 371 Feynman diagrams: beginning of, 29, 30f; breaking shackles of, 311; canonical formalism and, 43;childish game generating, 53, 53f; connectedvs. disconnected, 29, 47; counterterms in, 175–176, 176f; Cutkosky cutting rule for, 215–216,217, 219; discovering, 43–51, 45f, 46f; dominanceof, 302; for electron scattering, 132–134, 133f,134f, 135f; evaluating, 538–539; for fermion-fermion scattering, 172, 172f; finite temperature,289; function of, 50; imaginary part of, 207–219,208f, 213f; limitations of, 67; loop, 45, 57–58, 57f, 58f, 181, 494; in momentum space, 54; newapproaches to, 483–486; orientation of, 54; pathintegral formalism and, 44; in perturbation theory,55, 56f; for photon scattering, 152, 153f, 155, 155f;regularization of, alternative ways of, 166; relatinginfinite sets of, 234–235; in spacetime, 54, 58, 213 Feynman gauge, 149Feynman rules, 534–537; colored, 485, 491, 495; discovery of, 60; for fermions, 128–131, 128f; innonabelian gauge theory, 536–537; in physicalperturbation theory, 175–176, 176f; for quantumelectrodynamics, derivation of, 144–150; inrandom matrix theory, 397, 398f; for scalar field,54–55, 534–535; in spontaneously broken gaugetheories, 266–267; for vector field, 129, 130f,535–536; in Yang-Mills theory, 257, 257f, 494–495 field redefinition, 68–69, 218, 342field renormalization, 175field strength, construction of, 255–257Fierz, M., 121Fierz identities, 459Fisher, Matthew P. A., 336nFisher, Michael, 293; and renormalization groups, 361 fixed point(s), strong coupling, 359flux: fundamental unit of, 324; gauge potential and, 334 force: origin of, 29; particle and, 27–29. See also specific force forms: closed vs. exact, 247–248; geometric character of, 250–251, 250f fractional Hall effect, 323–324fractional (anyon) statistics, 315; coupling to gauge potential, 316–317; gauge boson and, 320;misleading nature of term, 317; and quasiparticles, 327 freedom, degrees of, 37–38; canonical formalism and, 67; Dirac equation and reduction in, 95; ofelectron, 95, 99; gauge invariance as redundancy in, 268; longitudinal, in massive gauge field, 264;of photon, 186–187 free field theory (Gaussian theory), 21–23, 43; in terms of Fourier transform, 26 Fujikawa, Kazuo, 278 gamma matrices, 94, 117, 538; products of, 95– 96; trace products of, evaluating, 136–137,153–154 Gamow, George, 120n“Gang of Four,” 366gapless mode, 284; Bogoliubov calculation of, 284; linearly dispersing, 284–285 gauge boson(s): and fractional statistics, 320; and intermediate vector boson, 309; mass spectrum of, 266 gauge fixing, 183gauge invariance, 83n, 144, 475; of Chern-Simons term, 328; and Dirac quantization of magneticcharge, 248; discovery of, 144n; in latticegauge theory, 376; in nonabelian gauge theory,preserving, 204; origin of, 183; proof of, 145–150,203–204; as redundancy in degrees of freedom,268; regularization respecting, 202–204; andrenormalizability, 411 gauge potential, 251; flux associated with, 334; fractional statistics and, 316–317; in Hall fluid,325–326, 329; nonabelian, 254, 255 gauge theory(ies): Faddeev-Popov quantization of, 183–185, 267; and fiber bundles, correspondencebetween, 256; gravity, as, 436; lattice, 374–376;recent developments in, 497–512; redundancyin, 183–185, 189; S-matrix theory and, 498– 501; spontaneously broken, Feynman rulesfor, 266–267; spontaneously broken, magneticmonopoles in, 309; and superconductivitytheory, 296; symmetry breaking in, 263–265,268, 296; unsatisfactory formulation of, 474,497; vortex in, 307. See also nonabelian gauge theory(ies) gauge transformation (local transformation), 187, 254; and general coordinate transformation,443 Gauss-Bonnet theorem, 457, 459Gaussian integration, 14, 523Gaussian theory (free field theory), 21–23, 43; in terms of Fourier transform, 26 Gell-Mann, Murray: and effective field theory, 460; on quark color, 385; and seesaw mechanism, 426;σmodel of, 340–341; SU (3) of, 531; Yang-Mills theory and, 371 Gell-Mann matrices, 265general coordinate invariance, 36 568 | Index general coordinate transformation, and gauge transformation, connection between, 443 general covariance, principle of, 81general relativity: finite size objects in, 480–482; and quantum mechanics marriage of, 6; review of,84–86 generator, charge as, 80Georgi, Howard, 421n; grand unification theory of, 407–409 ghost fields, 372–374Giele, W ., 493Ginzburg, V ., 264; on London penetration length, 296; on second-order transitions, 292; onsuperconductivity, 295 Girvin, Steve, 328Glashow, Sheldon, 171; electroweak theory of, 383; grand unification theory of, 407–409; and seesawmechanism, 426; Yang-Mills theory and, 379 gluon(s), 386; origins of concept, 235gluon scattering, 483–496; approaches to calculation of, 483–493, 484f, 491f; spinor helicity formalismand, 486–491, 496 Goldberger, Murph L., 105, 137, 460Goldberger-Treiman relation, 235, 342Goldstone’s theorem, 228–229; in condensed matter physics, 229 Golfand, Yu. A., 461Gordon decomposition, 195Goto, T ., 469grand unification, 452; binary code in, 426– 428; charge conjugation in, 429; and deeperunderstanding of physics, 410–411; fermionmasses in, 417–418; and freedom from anomaly,411–412, 429; and hierarchy problem, 419;need for, 407; and origin of matter, explanationfor, 418; and proton decay, 413–414, 415,456; SO (10): antineutrino field in, 425–426; SO (18), 428; spinor representation of, 421– 423, 424, 426; SU (5), 531; SU (5), Georgi and Glashow theory of, 407–409; triumph of,415–416 Grant, A. K., 516Grassmannian symmetry, 355Grassmann integration, 126–127Grassmann number(s), 123, 126; in path integral for spinor field, 124 Grassmann variables, 246gravitational force. See gravity gravitational interaction, 36gravitational waves, and effective field theory, 479–482 gravition propagator, 437–438graviton: coupling to matter, 435; definition of, 83; deformed polarizations of, 514–515; as elementaryparticle, 434, 448; force associated with, 29; in (n+3+1)-dimensional universe, 41–42; recent developments on, 513–520; self-interaction of,434; in spacetime, 515–517; spin of, 35, 39; instring theory, 513–514, 516 gravity: Einstein-Hilbert action for, 433–434; Einstein on ( seeEinstein’s theory of gravity); and electromagnetism, unification of, 442;as field theory, 434–436; as gauge theory,436; helicity structure of, 446; of light, 441;Newton on (see Newton’s gravitational force); nonrenormalizability of, 434; weak field actionfor, 436–437 Green’s function(s), 47, 55, 352; generating, 50; propagator related to, 23 Gross, David, 386 Gross-Neveu model, 402–404ground state, in quantum field theory, 37, 225ground state degeneracy, 319group theory, review of, 525–533. See also special orthogonal group SO (N); special unitary group SU (N) hadron(s): electron-positron annihilation into, 389– 391; in electroweak unification, 383; experimentalobservation of, 231; quarks as components of, 385 Hall effect, 351–352; fractional, 323–324; integer, 323Hall fluid(s), 322–330; Chern-Simons term for, 326; effective field theory of, 324–325, 452–453; electron tunneling in, 329; five generalstatements/principles of, 325, 329; gauge potentialin, 325–326, 329; incompressibility of, 323, 328;Laughlin odd-denominator, 327; order in, 328 handedness, field, 100; charge conjugation and, 101Hansson, T ., 324harmonic paradigm, 5Hasslacher, Brosl, 405Hawking radiation, 290–291Heaviside, O., 24, 245hedgehog, 308Heisenberg, Werner: approach to quantum mechanics, 61–62; and effective field theory, 460;isospin SU (2) of, 531; isospin symmetry of, 387, 388; on neutron and proton, symmetry of, 77 helicity, topological quantization of, 532–533helicity formalism, spinor, 486–491, 496, 501, 521hierarchy problem, grand unification and, 419Higgs field, covariant derivative of, 266Higgs particle, mass of, 384high energy physics: renormalization group in, 359–360 high frequency behavior, 208–210Hofstadter, R., 199homotopy groups, 307 Index | 569 Hopf term, 318; non-local, 329 Howe, P., 470Hubbard-Stratonovich transformation, 192 identity, absolute, 120–121, 134 imaginary part, of Feynman diagrams, 207–219, 208f, 213f impurities, 323; condensed matter physics and study of, 354; and random potential, 350 infinities, in quantum field theory, 161–162instanton(s), 309; discovery of, 473integer Hall effect, 323integration measure, in path integral formalism, 67integration variables, shifting, 272interchange symmetry, 77internal symmetry, 77 inverse square law, 40irrelevant operators, 363Ising model, 361isospin symmetry of Heisenberg, 387, 388Itzykson, Claude, 402Iwasaki, Y ., 439 Johansson, H., 519 Jona-Lasinio, G., 237Jordan, P., 107Josephson junction, fundamental relation underlying, 192 Kadanoff, Leo, 361; and renormalization groups, 361 Kaluza-Klein compactification, 442–443; derivation of, 447 Kardar-Parisi-Zhang equation, 347Kawai, H., 513kinks. See solitons Kivelson, Steve, 324Klein-Gordon equation, 21, 93, 95, 190; Schr ¨odinger equation derived from, 190 Klein-Gordon operator, 113Klein-Nishina formula, 155Kockel, B., 460Kosterlitz-Thouless transition, 310Kramer’s degeneracy, 103 Lagrangian: Dirac (see Dirac equation); gauge invariant, 253–254; Maxwell (see Maxwell Lagrangian); Meissner, 332, 335; as mnemonic,340, 342; for quantum electrodynamics, 101, 144;symmetries of breaking, 223; weak interaction,100; Yang-Mills, 257 Lamb shift in atomic spectroscopy, 205Landau, L. D., 264; on complex momenta, 498; on London penetration length, 296; on second-ordertransitions, 292; on superconductivity, 295; on superfluidity, 284 Landau gauge, 149Landau-Ginzburg approach to quantum field theory, 18 Landau-Ginzburg theory (mean field theory), 292–294; order in, 328 Landau levels, 323Laplace, P.-S., 290Large Hadron Collider, 483large Nexpansion, 394–396; Dyson gas approach to, 400–402; field theories in, 402–404 Larmor circle, 322, 323lattice gauge theory, 374–376; Wilson loop in, 376–377, 457 Laughlin odd-denominator Hall fluids, 327 Lee, B., 173Lee, D. H., 336nLee, T sung-dao, 100Legendre transform, 238–239Leinaas, J., 315length scales: in condensed matter physics, 169; renormalization group and, 361–362 leptons: families of, 384; generations of, 428; and quarks, neutral current interaction between, 383 L´evy, M., σmodel of, 340–341 Lewellen, David C., 513Licciardello, D., 366light, gravity of, 441light beam, stress-energy tensor of, 445Likhtman, E. P., 461linearly dispersing mode, 284; velocity of, 285local field theory, 474, 521–522localization: Anderson, 351, 354; Anderson, in renormalization group language, 366–367; studyof, 355 local transformation (see gauge transformation) logarithmic divergence, 175, 176–177London penetration length, 296loop diagrams, 45, 57–58, 57f, 58f, 181, 494Lorentz algebra, 114–116Lorentz boosts, 114–115Lorentz group: defining representation of, 116; generators of, algebra for, 115–116; spinorrepresentation of, 116–118 Lorentz invariance, 475; canonical formalism and, 63, 66–67; Euclidean equivalent of, 362; inquantum field theory, 18, 24; recent developmentsin, 507–510 Lorentz transformation: and Dirac equation, 96–97Low, F., 460 L¨uders, G., 121 MacDonald, Alan, 328 570 | Index magnetic charge (monopole), 249, 309; confinement in superconductor, 386–387; Dirac quantizationof, 248–249, 252; duality of, 334; electricallycharged (dyon), 309; mass of, 309; and Maxwell’sequations, 249; quantum mechanics and, 245; inspontaneously broken gauge theory, 309 magnetic moment of electron: anomaly in, 196; calculation of, 454; in Dirac equation, 194–195;Schwinger on, 196–198, 454 magnetic moment of ferromagnet and antiferromagnet, 344 magnetic moment of proton, anomaly in, 454–455Majorana, Ettore, 102n, 543Majorana equation, 102Majorana mass, 102; for neutrino, 102Majorana spinor, 102, 543 Mandelstam variables, 137–138, 498marginal operators, 364mass(es): attraction between, 35–36; of electron, 180; energy of, 35; of gauge boson, 266; of Higgsparticle, 384; of magnetic charge (monopole), 309;Majorana, 102; of neutrino, 102; of nucleon, 341;Planck, 41–42, 434; of soliton (kink), 305 massive gauge field, Nambu-Goldstone boson and, 264–265 massive spin 1 field, vs. massless spin 1 field, 183massive spin 1 particle: degrees of freedom of, 38; degrees of polarization of, 34; field theory of,32–33; propagator for, 34; in Yang-Mills theory,379 massive spin 2 particle: degrees of polarization of, 35; propagator for, 35, 439 mass renormalization, 174matrix (matrices): gamma ( seegamma matrices); Gell-Mann, 265; Pauli, 265; without inverse,182–183 matter: dark, 450; origin of, explanation for, 418; states of, 328 mattress model of scalar field theory, 4–5, 4f; disturbing, 21, 21f; path integral description of,17–19 Maxwell, James Clerk, 521Maxwell action, 182Maxwell equations, magnetic charges and, 249Maxwell Lagrangian, 32, 34, 84; bypassing, 33–34; derivation of, 38 Maxwell term, 320, 329Maxwell theory of electromagnetism: development of, 474–476; Yang-Mills theory compared with,257 mean field theory (Landau-Ginzburg theory), 293–294; order in, 328 Meissner effect, 296, 386Meissner Lagrangian, 332, 335meson(s): birth of, quantum field theory on, 55–56, 56f;π(see pion[s]); σ, 341, 342; soliton compared with, 304; vector, field theory of, 32–33. See also massive spin 1 particle meson-meson scattering amplitude, 357; canonical formalism and, 64–65; cutoff dependence of,173; dimensional analysis on, 173; divergenceof, 161–162; path integral formulation of, 166;regularization and, 163; renormalization and,164–166 Michell, John, 290Mills, Robert, and nonabelian gauge theory, 253, 255 Minkowski, Peter, and seesaw mechanism, 426Minkowskian path integral, 287Minkowskian spacetime, 36 momentum: complex, 498–501, 499f; fundamental definition of, 83; orbital angular, Dirac equationon, 194–195; spin angular, Dirac equation on, 195;square root of, 486–489 momentum density, in nonrelativistic theory, 191momentum space, 26; fermion propagator in, 113; Feynman diagrams in, 54 monopole. See magnetic charge Montonen, J., 334muon, weak decay of, 380Myrheim, J., 315 Nambu, Yoichiro, 297, 469; Nobel prize for, 228n Nambu-Goldstone boson(s), 228–229; gapless mode as, 284; in massive gauge field, 264–265; π mesons (pions) as, 234, 387, 388; in relativistic vs.nonrelativistic theories, 285 naturalness, notion of, 419N´eel state, for antiferromagnet, 346 Ne’eman, Y ., SU (3) of, 531 neutral current interaction, 383neutrino(s): handedness of, 101; mass of, 102neutrino masses, effective field theory of, 456neutron(s): βdecay of, 231–232; electric dipole moment for, 259; and proton, internal symmetryof, 77 Neveu, Andr ´e, 402, 405 Newton’s gravitational force: and Coulomb’s electric force, comparison of, 29; derived from Einstein-Hilbert action, 438; quantum field theory on, 32,33–36 Noether current, 191, 234Noether’s theorem, 78–79, 100, 341; elaborate formulation of, 80 nonabelian Berry’s phase, 261, 346 nonabelian gauge potential, 254, 255; coupling to a fermion field, 260 nonabelian gauge theory(ies), 253–260; chiral Index | 571 anomaly in, 276; differential forms in, 255, 256; Feynman rules in, 536–537; gaugeinvariance in, preserving, 204; ghost actionin, 372; redundancy of, Faddeev-Popovapproach to, 183; renormalizability of, 173,411; strong interaction described by, 259, 379;’t Hooft double-line formalism and, 258–259;unsatisfactory features of, 474. See also Yang-Mills theory nonanalyticity: emergence of, 292; symmetry breaking and, 293 noncommutative field theory, 474nonrenormalizable theory(ies), 169, 179, 453; counterterms in, 179, 241; Einstein’s theoryof gravity as, 172; Fermi’s theory of the weakinteraction, 170, 179 nonrenormalization of the anomaly, 277notation, dotted and undotted, 116–117, 541–544; replacing, 475 nucleon(s): attraction between, 28; electron scattering off of, deep inelastic, 386; mass of, 341; and pions,interaction between, 340–341; wave function ofquarks in, 385 Olive, D. I., 334 optical theorem, 215–216, 219orbital angular momentum, Dirac equation on, 194–195 order parameters, 295orthogonal groups, embedding unitary groups into, 423–424 Parisi, Giorgio, 402 parity, 98; Dirac equation and, 98; and Dirac spinor, 117; weak interaction and, 100, 379–380 Parke, S. J., 493particle(s): birth and death of, 4–5; birth of, quantum field theory on, 55–56; field associated with, 26–27; force associated with, 27–29; interchanging,315–316, 316f; propagation of, describing, 48–49, 50; scattering of (see scattering of particles); sources and sinks for, 20. See also specific particles particle physics: and condensed matter physics, 281, 452–453; energy scales in, 169; family problem in,428; spontaneous symmetry breaking in, 292, 297,449 partition function, in quantum statistical mechanics, 288–289 path integral formalism: vs. canonical formalism, 44, 61, 67; chiral anomaly and, 278; and classicallimit, 19; derivation of, 44; description of mattress model, 17–19; Dirac on, 10–13; Feynman on, 7–10; Grassmann math and, 127; history of, 60;integration measure in, 67; replacing, 475; forspinor field, 124; and vacuum energy, calculation of, 123–125 Pauli, Wolfgang, on spin-statistics connection, 121Pauli exclusion principle, 120, 323; history of, 120nPauli-Hopf identity, 345Pauli matrices, 265Pauli-Villars regularization, 75, 166–168Peierls, Rudolf, 365Peierls instability, 300pentagon anomaly, 276, 277fperturbation theory, 49–51; bare, 175; Feynman diagrams in, 55, 56f; finite temperature, 289;physical (renormalized/dressed), 175–176, 176f perturbative quantum gravity, 441ϕ 4theory, renormalizability of, 173, 175 phonon(s), 5, 284 photon(s): absence of rest frame for, 186–189; birth and death of, 4; Bose-Einstein statistics for, 120;degrees of freedom of, 186–187; electron-positronannihilation into, 155; emission and absorption of,150; fluctuation into electron and positron, 200–202, 201f; force associated with, 29; longitudinalmode of, 150; spin of, 36, 39 photon propagation: charge as measure of, 204; quantum fluctuations and, 200–202, 201f photon propagator, 149–150; Fourier transform of, 205; physical (renormalized), 201, 201f photon scattering, 152–157; cross sections for, 152–155; on electrons, 152–157, 153f, 157f physical perturbation theory, 175–176, 176fpion(s) (π meson): massless, 235, 341; Nambu- Goldstone boson, 388; as Nambu-Goldstoneboson, 234, 387; and nucleons, interactionbetween, 340–341; prediction regarding, 29;quarks as components of, 385; weak decay of,231–233 pion-nucleon coupling constant, 235Planck mass: modified, 434; for (n+3+1)- dimensional universe, 41–42 Planck’s constant, 181Podolsky, B., 441Poincar ´e lemma, 247 point particle: action of, constructing, 84–86; stress energy of, calculating, 86; world line traced out by,length of, 84, 85f Poisson equation, 438polarization, degrees of, 34Politzer, H. D., on Yang-Mills theory, 386Polyakov, Alexander (Sasha), 498; on magnetic monopoles, 309 Polyakov action, 470 Pontryagin index, 310positron(s): Dirac’s conception of, 5; photon fluctuation into, 200–202, 201f 572 | Index potential energy, double-well, 224, 224f power counting theorem, 176–178preons, theories about, 278product rule, 445propagation of particles, describing, 48–49, 50propagator, 23–25; in canonical formalism, 67–68; for Dirac field, 127; fermion, 112; graviton, 437–438; for massive spin 1 particle, 34; for massivespin 2 particle, 35, 439; photon, 149–150 proton(s): charge of, grand unification on, 410; electron scattering off of, 132–134, 133f, 199;electron scattering off of, deep inelastic, 359;electron scattering off of, Schr ¨odinger equation for, 3; magnetic moment of, anomaly in, 454–455;and neutron, internal symmetry of, 77; quarks ascomponents of, 385; stability of, 413 proton decay: branching ratios for, 416–417; effective theory of, 455–457; grand unification and,413–414, 415, 456; slow rate of, 418 quantum chromodynamics (QCD), 360, 386; analytic solution of, search for, 391; at high energies, 391;largeNexpansion of, 394–396; renormalization group flow of, 388–389 quantum electrodynamics (QED), 32; coupling constant of, 164; coupling in, 358; electromagneticgauge transformation in, 189; Feynman ondifficulty of, 61; Feynman rules for, derivationof, 144–150; intellectual incompleteness of, 121;Lagrangian for, 101, 144; renormalizability of,173 quantum field theory(ies): in (0+0)-dimensional spacetime, 397; in 2-dimensional spacetime,470; anharmonicity in, 43, 89; asymptoticbehavior of, study of, 359–360; central identityof, 182, 523; and condensed matter physics,5, 190, 281; crisis of, 231, 340, 452; in curvedspacetime, 82, 290; divergences in, 57–58, 161–162; Euclidean, 287–288, 289, 290; at finitedensity, 291; at finite temperature, 289–290;gravity as, 434–436; ground state in, 37, 225;harmonic paradigm and, 5; hidden structuresin, 476; history of, 60; infinities in, 161–162;innovative applications of, 473–474, 476; integralof, 88–89; low energy manifestation of, 162, 169,452; mattress model and, 17–19; motivationfor constructing, 55; need for, 3–5, 6, 123;nonrelativistic limit of, 190–191; relativisticvs. nonrelativistic, 191–193; renormalizablevs. nonrenormalizable, 169; on repulsion andattraction, 32–36; restrictions within, 474; steps toward, 235; strong and weak interactionsapplied to, 231; of strong interaction, 340;supersymmetric, 461, 467–468; surface growthand, 347–349; symmetry breaking in, 225–226;theories subsumed by, 473; threshold of ignorance in, 162–163, 453; triumph of, 452, 473; vacuumin, 20 quantum fluctuations: axial current conservation destroyed by, 274–275; effective potentialgenerated by, 243; and electric charge, 204, 205;first order in, 239–240; higher order, and chiralanomaly, 310; and photon propagation, 200–202,201f; and symmetry breaking, 229, 237, 242, 270 quantum Hall fluid. See Hall fluid(s) quantum Hall system, 281quantum mechanics: antimatter as requirement in, 157; and general relativity, marriage of, 6;harmonic oscillator in, solving, 43; Heisenberg’sapproach to, 61–62; and magnetic monopoles,245; partition function in, 288–289; path integral formalism of, 7–12; quantum field theory asgeneralization of, 88–89, 473; and relativisticphysics, joining in spin-statistics connection,122; and special relativity, marriage of, 3, 6,121; symmetry breaking in, 225–226; symmetryof, 270; time reversal in, 102–104; and vectorpotential, need for, 245 quantum statistics, 120quantum vacuum, 358quark(s): color of, 385, 386; confinement of, 377, 386–387; in electroweak unification, 383; familiesof, 384; flavors of, 385; generations of, 428; andleptons, neutral current interaction between,383; origins of concept, 235; strong interactionbetween, weakening of, 360 quasiparticle(s), 326; charge of, 327; fractional statistics and, 327; as vortex, 328 radiation: and atoms, interaction between, 3; Hawking radiation, 290–291 Ramakrishnan, T . V ., 366Ramond, Pierre, and seesaw mechanism, 426random dynamics, and quantum physics, 349random matrix theory, 396–397; Feynman rules in, 397, 398f random potential, impurities and, 350Rarita-Schwinger equations, 119Rayleigh, Lord, 458recursion, 501–503, 507–512, 521; BCFW, 500, 507, 514 redundancy, Faddeev-Popov approach to, 183–185reflection symmetry, 76, 226; breaking, 223, 224, 225 Regge, T ., 498regularization, 163; Casimir force and, 71–75; dimensional, 167, 168, 204; gauge invariancerespected by, 202–204; Pauli-Villars, 75, 166–167 relativistic physics: equations of motion in, unified view of, 95; language of, 26; and quantum physics, Index | 573 joining in spin-statistics connection, 122. See also general relativity; special relativity relativistic quantum field theory: correctness of, establishment of, 196; vs. nonrelativistic quantumfield theory, 191–193 relevant operators, 363renormalizable conditions, imposing, 241–242renormalizable theory(ies), 169, 173, 453; electroweak theory as, 384; nonabelian gauge theory as, 173,411;ϕ 4theory as, 173, 175, 178–179; Yukawa theory as, 179 renormalization, 161, 164–166; coupling, 173–174; of electric charge, 205; field, 175; mass, 174; wavefunction, 175 renormalization group, 356, 358; and Anderson localization, 366–367; in condensed matter physics, 360–363; and effective description,367; effective field theory philosophy and, 453;in high energy physics, 359–360; in quantumchromodynamics, 388–389 renormalization theory, application of, 240–241renormalized coupling constant, 166renormalized (dressed) perturbation theory, 175–176, 176f reparametrization invariance, 84replica method, 353–354representations: conventions for naming, 526; multiplying, 530–531 repulsion: of bosons, 192–193, 283, 337; quantum field theory on, 32–33; spin 1 particle and, 36–37;of vortices, 338 rest frames, for photons, absence of, 186–189Ricci tensor, 433Riemann-Christoffel symbol, 85, 445Riemann curvature tensor, 433, 480–481, 515–516Riemannian manifolds, differential geometry of, 443–444 Rosenbluth, Marshall, 105rotation group, 114; and Lorentz group, symmetry of, 118 R ξgauge, 267, 268 Salam, Abdus, 171; electroweak theory of, 383; superspace and superfield formalism of, 462, 463 scalar boson operator, 113scalar field: complex, 65–66; Feynman rules for, 54– 55, 534–535; quantizing in curved spacetime, 82;and vacuum energy, 66 scalar field theory: classical field equation in, 20; Euclidean functional integral and, 287; Euclideanversion of, 293; massless version of, 284; simplicity of, 519–520 scalar potentials, 245scattering of particles: describing, 51–53, 51f, 52f; fermion-fermion, Feynman diagram for, 172;meson-meson (see meson-meson scattering amplitude); reflection symmetry in, 76; andvacuum fluctuations, 124. See also electron scattering Schouten identity, 492, 493Schrieffer, Bob, 297Schr ¨odinger equation: electromagnetic gauge transformation in, 189; Klein-Gordon equationand, 21n, 190; limitations of, 3; Yang-Millsstructure in, 261 Schwarz, John, on string theory, 470nSchwarzschild black hole, 311Schwarzschild solution, for Hawking radiation, 290Schwinger, Julian: on complex plane, 208; and effective potential, 237; on Feynman’scontribution, 43, 50, 56; on magnetic moment of electron, 196–198, 454; on path integralformalism, 60; at Pocono conference (1948), 105;teaching style of, 454; Yang-Mills theory and,379 second-order phase transitions, 292seesaw mechanism, 37, 426Seiberg, Nathan, 334self-dual theory, 337semions, 315σmeson, 341, 342 σmodel, 340–341; for ferromagnets and antiferromagnets, 345, 346; nonlinear, 342, 346 sky color, effective field theory of, 457–458Slansky, Dick, and seesaw mechanism, 426S-matrix theory, 68, 235, 340, 498–501 solid state physics, Dirac equation in, 298, 299solitons (kinks): discovery of, 302–304, 473; dynamically generated, 400–405; mass of, 304;topological stability of, 304; unifying language fordiscussing, 307 SO (N).See special orthogonal group sources and sinks, creating, 20, 51, 51fspacetime: curved (see curved spacetime); dimension of, and symmetry breaking, 229; discretizing, 22;Feynman diagrams in, 54, 58, 213; gravitationalwaves in, 479–482; graviton in, 515–517; symmetryof, Lorentz invariance as, 76 special orthogonal group SO(N), 525–527; binary code in, 427–428; review of, 531–532; SO(3), 526– 527; SO(10) grand unification, antineutrino field in, 425–426; SO(18), 428; spinor representation of, 421–423, 424, 426 special relativity: antimatter as requirement in, 157; and quantum mechanics, marriage of, 3, 6, 121 special unitary group SU (N), 527–530; decomposing representations of, 531; of Heisenberg, 531; SU (2), 529–530; SU (3), 529, 530; SU (3), of Gell- Mann and Ne’eman, 531; SU (5), 531; SU (5), Georgi and Glashow theory of, 407–409 574 | Index spin angular momentum, Dirac equation on, 195 spinor(s): Dirac, 94, 96, 114, 117; Majorana, 102; representations of, 116–117; Weyl, 117, 462 spinor field: deriving, 125–127; path integral for, 123; path integral for, Grassmann numbers in, 124;vacuum energy of, 111–112, 125 spinor helicity formalism, 486–491, 496, 501, 521spin-statistics rule, 120–121; and anticommutation relations, 122, 123; price of violating, 121–122 spin wave, 229spontaneous symmetry breaking, 224, 225, 227; continuous: and massless fields, 228–229; ingauge theories, 263–265; in particle physics,292, 297, 449; quantum fluctuations and, 229;of reflection symmetry, 225; in relativistic vs.nonrelativistic theories, 285; second-order phase transitions and, 292; and superfluidity, 283–284 square anomaly, 276, 277fsquare root of momentum, 486–489steepest-descent approximation, 16Stokes’ theorem, 307Stoner, E. C., 120nStrathdee, J., superspace and superfield formalism of, 462, 463 stress-energy tensor, 35; definition of, 83; of light beam, 445; properties of, 84 string theory: 2-dimensional field theory, 469– 470; as candidate for unified theory, 433, 452;and cosmological constant problem, inabilityto resolve, 450; duality of, 334; future of, 513;graviton in, 515–517; Kaluza-Klein idea and, 442;origins of, 6, 387; p-forms in, 251; in quantum field theory, 473; Schwarz on, 470n strong coupling: fixed point in, 359; linking to perturbative weak coupling, 473 strong interaction: chiral symmetry of, 234; currently accepted theory of, 379; fundamental theoryof, 360; hadronic, 36; at low energies, 340–341;nonabelian gauge theory on, 259, 379; quantumfield theory of, 235, 340; renormalization groupflow applied to, 368; symmetries of, 234, 387–388 SU (N).See special unitary group supercharges, 464superconductivity, 295–297; and Meissner effect, 296superconductor(s): monopole confinement in, 386–387; type II, flux tube in, 307 superfield, 464–465; chiral, 464, 466; vector, 466–467superfluidity, 192; gapless excitations and, 284–285; Lagrangian summarizing, 284; linearly dispersingmode of, 284; spontaneous symmetry breakingand, 283–284 superspace and superfield formalism, 462–463superstring theory, 470supersymmetric action, 466supersymmetric algebra, 462–463 supersymmetric field theories, 461, 467–468; Yang-Mills, 392, 467–468 supersymmetric method, 355supersymmetric transformation, total divergence under, 465–466 supersymmetry, 112; Dirac spinor and, 114; inventing, 462; motivations for, 461 surface growth, 347, 360; and quantum field theory, 348–349 Swieca, Jorge, 230nsymmetry, 76–80; in amplitudes, 78; breaking, 226; chiral, 234, 387, 388, 419; classical vs.quantum, 270–271; conserved current and, 78–79; continuous, 77–78, 226; in field theories,475; Grassmannian, 355; Heisenberg isospin, 387, 388; interchange, 77; internal, 77; powerof, 18, 76, 118; reflection, 76, 226; replica, 353;of spacetime, Lorentz invariance as, 76; stronginteraction, 234, 387–388; tensors and, 526, 528.See also supersymmetry symmetry breaking, 223–230; continuous symmetry and, 226; dimension of spacetime and, 229;dynamical, 230, 388; in gauge theories, 263–265, 268, 296; and nonanalyticity, 293; quantumfluctuations and, 229, 237, 242, 270; in quantummechanics vs. quantum field theory, 225–226;reflection symmetry and, 223, 224, 225; andsuperfluidity, 283–284; and vacuum energy, 449.See also spontaneous symmetry breaking Taylor, T . R., 493 Teller, Edward, 105temperature: black hole, 290; and cyclic imaginary time, 289; finite, quantum field theory at, 289–290 tensor(s): energy-momentum, 319; of light beam, 445; of orthogonal group, 525–526; Ricci, 433;Riemann curvature, 433; stress-energy, 35, 83–84; symmetry properties of, 526, 528; of unitarygroup, 527–528; vacuum polarization, 200, 201f,204, 208, 209f, 211, 216, 218 tensor field, 35, 83θterm, 259 Thomas precession, 115’t Hooft, Gerardus, 173; on electroweak theory, 384; on large Nexpansion, 394; on magnetic monopoles, 309 ’t Hooft double-line formalism, 258–2593-brane, 40–42time ordering in canonical formalism, 67–68time reversal, 102–104; and Dirac equation, 104 Tolman, R., 441Tolman-Ehrenfest-Podolsky effect, 441, 446Tomonaga, Shin-Itiro, 60 Index | 575 topological current, 304 topological field theory, 318topological objects, 306; discovery of, shock of, 311. See also specific objects topological order, 328topological quantum fluids, 322. See also Hall fluid total divergence, under supersymmetric transformation, 465–466 trace, 526tree diagrams, 45, 483–484, 491–494Treiman, Sam, 236. See also Goldberger-Treiman relation triangle anomaly, 271f, 276twistor space, 494–495Tye, Henry, 513 ultraviolet catastrophe, 448 ultraviolet divergence, 162uncertainty principle, 3, 290unification. See grand unification unitarity, 215–216, 500unitary gauge, 267unitary groups, embedding into orthogonal groups, 423–424 universe: 3-brane, 40–42; early, 290; formation of structure in, 36 vacuum: disturbing of, 20, 21f, 70–73; quantum, 20, 358 vacuum energy: calculation of, using path integral fomalism, 123–125; disturbance of vacuum and,70–73; of free scalar field, 66; of free spinor field(Dirac field), 111–112, 125; Grassmann pathintegral for, 127; symmetry breaking and, 449 vacuum expectation value, 226vacuum fluctuations, 59–60, 59f; Feynmann diagram corresponding to, 129; scattering of particles and,123 vacuum polarization tensor, 200, 201, 204, 208, 209f, 211, 216, 218 van Dam, H., 439van der Waerden notation. See dotted and undotted notation vector field, interacting with Dirac field, 100; Feynman rules for, 129, 129f, 535–536 vector meson (massive spin 1 meson): field theory of, 32–33. See also massive spin 1 particle vector potential, 245vector superfield, 466–467Veltman, Tini, 173, 439; on electroweak theory, 384vielbeins, 443 visual perception, application of field theory to, 476vortex (vortices), 306, 331–332; as charges in dual theory, 332–334; density of, 333; duality of, 334;as flux tube, 307; motion in fluid, 338–339, 338f; paired with antivortex, 310–311, 339; quasiparticleas, 328; repulsion of, 337 Ward-Takahashi identity, 149, 411 wave function(s), Anderson localization of, 351, 354wave function renormalization, 175wave packets, in mattress model, 4–5, 4fweak interaction, 37; intermediate vector boson of, 171–172, 309; and parity, 100, 379–380; quantumfield theory applied to, 231. See also Fermi theory of the weak interaction weak interaction Lagrangian, 100Weinberg, Steve, 171, 508; electroweak theory of, 383Weisskopf phenomenon, 180–181; grand unification and, 419 Wen, Xiao-gang, 324, 328; and topological order, 328Wentzel, Gregory, 105Wess-Zumino model, 462Weyl basis, 98–99, 118Weyl-Eddington terms, 457Weyl spinors, 117; and supersymmetry, 462Wheeler, John, 365nWick, Gian Carlo, 14Wick contractions, 14–16, 47Wick rotation, 12, 287Wick theorem, 14Wigner, Eugene: on antisymmetric wave function of electron, 107; and law of baryon numberconservation, 413; and random matrix theory, 396;on time reversal, 102 Wigner semicircle law, 397–400Wilczek, Frank, 315, 316; on Yang-Mills theory, 386Wilson, Ken, 161; and complete theory of critical phenomena, 293; and effective field theoryapproach, 452; and lattice gauge theory, 374–376;and renormalization groups, 361 Wilson loop, 261; in lattice gauge theory, 376–377, 457; and quark confinement, 386 Witten, Ed, 334, 500Wu, Tai-tsun, 248 Yanagida, T ., and seesaw mechanism, 426 Yang, Chen-Ning, 100, 105, 248; and nonabelian gauge theory, 253, 255 Yang-Mills bosons, 257, 386; self-interaction of, 434 Yang-Mills coupling constant, 258–259Yang-Mills Lagrangian, 257Yang-Mills theory, 257–258; area law in, 377; asymptotically free, 386; Einstein-Hilbert action compared with, 434–435; Einstein’s theoryof gravity compared with, 444–445, 513–520;Feynman rules in, 257, 257f, 494–495; 576 | Index Yang-Mills theory (continued) gluon scattering in, 483–496; original responseto, 371, 379; perturbative approach to, 374;quantizing, 371–373; recent developmentsin, 483–496, 501–504, 513–520; recursion in,501–503; Schr ¨odinger equation and, 260– 261; supersymmetric, 392, 467–468; Wilsonformulation of, 374–376 Young tableaux, 526Yukawa, H., 28–29, 171Yukawa coupling, 170 Yukawa theory, renormalizability of, 178– 179 Zakharov, V ., 439 Zee, A., 316Zhang, Shou-cheng, 324Zinn-Justin, Jean, 173Zuber, Jean-Bernard, 402Zumino, Bruno, 121, 470