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Srednicki QFT

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A full textbook by Mark Srednicki of UC Santa Barbara, filed among Phil's downloaded physics books rather than his own writing. It is organized in 97 short sections in three parts: Spin Zero, Spin One Half and Spin One. Topics include canonical quantization, path integrals, Feynman rules, renormalization, spinors, QED, nonabelian gauge theory, anomalies, the Standard Model, instantons, supersymmetry and grand unification.

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Quantum Field Theory Mark Srednicki University of California, Santa Barbara [email protected] c∝ci∇cleco√y∇t2006 by M. Srednicki All rights reserved. Please DO NOT DISTRIBUTE this document. Instead, link to http://www.physics.ucsb.edu/ ∼mark/qft.html 1 To my parents Casimir and Helen Srednicki with gratitude Contents Preface for Students 8 Preface for Instructors 12 Acknowledgments 16 I Spin Zero 18 1 Attempts at relativistic quantum mechanics 19 2 Lorentz Invariance (prerequisite: 1) 30 3 Canonical Quantization of Scalar Fields (2) 36 4 The Spin-Statistics Theorem (3) 45 5 The LSZ Reduction Formula (3) 49 6 Path Integrals in Quantum Mechanics 57 7 The Path Integral for the Harmonic Oscillator (6) 63 8 The Path Integral for Free Field Theory (3, 7) 67 9 The Path Integral for Interacting Field Theory (8) 71 10 Scattering Amplitudes and the Feynman Rules (5, 9) 87 11 Cross Sections and Decay Rates (10) 93 12 Dimensional Analysis with ¯h=c= 1(3) 104 13 The Lehmann-K¨ all´ en Form of the Exact Propagator (9) 106 14 Loop Corrections to the Propagator (10, 12, 13) 109 15 The One-Loop Correction in Lehmann-K¨ all´ en Form (14) 12 0 16 Loop Corrections to the Vertex (14) 124 17 Other 1PI Vertices (16) 127 18 Higher-Order Corrections and Renormalizability (17) 12 9 4 19 Perturbation Theory to All Orders (18) 133 20 Two-Particle Elastic Scattering at One Loop (19) 135 21 The Quantum Action (19) 139 22 Continuous Symmetries and Conserved Currents (8) 144 23 Discrete Symmetries: P,T,C, andZ(22) 152 24 Nonabelian Symmetries (22) 157 25 Unstable Particles and Resonances (14) 161 26 Infrared Divergences (20) 167 27 Other Renormalization Schemes (26) 172 28 The Renormalization Group (27) 178 29 Eﬀective Field Theory (28) 185 30 Spontaneous Symmetry Breaking (21) 196 31 Broken Symmetry and Loop Corrections (30) 200 32 Spontaneous Breaking of Continuous Symmetries (22, 30)2 05 II Spin One Half 210 33 Representations of the Lorentz Group (2) 211 34 Left- and Right-Handed Spinor Fields (3, 33) 215 35 Manipulating Spinor Indices (34) 222 36 Lagrangians for Spinor Fields (22, 35) 226 37 Canonical Quantization of Spinor Fields I (36) 236 38 Spinor Technology (37) 240 39 Canonical Quantization of Spinor Fields II (38) 246 40 Parity, Time Reversal, and Charge Conjugation (23, 39) 25 4 5 41 LSZ Reduction for Spin-One-Half Particles (5, 39) 263 42 The Free Fermion Propagator (39) 268 43 The Path Integral for Fermion Fields (9, 42) 272 44 Formal Development of Fermionic Path Integrals (43) 276 45 The Feynman Rules for Dirac Fields (10, 12, 41, 43) 282 46 Spin Sums (45) 292 47 Gamma Matrix Technology (36) 295 48 Spin-Averaged Cross Sections (46, 47) 298 49 The Feynman Rules for Majorana Fields (45) 303 50 Massless Particles and Spinor Helicity (48) 308 51 Loop Corrections in Yukawa Theory (19, 40, 48) 314 52 Beta Functions in Yukawa Theory (28, 51) 323 53 Functional Determinants (44, 45) 326 III Spin One 331 54 Maxwell’s Equations (3) 332 55 Electrodynamics in Coulomb Gauge (54) 335 56 LSZ Reduction for Photons (5, 55) 339 57 The Path Integral for Photons (8, 56) 343 58 Spinor Electrodynamics (45, 57) 345 59 Scattering in Spinor Electrodynamics (48, 58) 351 60 Spinor Helicity for Spinor Electrodynamics (50, 59) 356 61 Scalar Electrodynamics (58) 364 62 Loop Corrections in Spinor Electrodynamics (51, 59) 369 6 63 The Vertex Function in Spinor Electrodynamics (62) 378 64 The Magnetic Moment of the Electron (63) 383 65 Loop Corrections in Scalar Electrodynamics (61, 62) 386 66 Beta Functions in Quantum Electrodynamics (52, 62) 395 67 Ward Identities in Quantum Electrodynamics I (22, 59) 399 68 Ward Identities in Quantum Electrodynamics II (63, 67) 40 3 69 Nonabelian Gauge Theory (24, 58) 407 70 Group Representations (69) 412 71 The Path Integral for Nonabelian Gauge Theory (53, 69) 420 72 The Feynman Rules for Nonabelian Gauge Theory (71) 424 73 The Beta Function in Nonabelian Gauge Theory (70, 72) 427 74 BRST Symmetry (70, 71) 435 75 Chiral Gauge Theories and Anomalies (70, 72) 443 76 Anomalies in Global Symmetries (75) 455 77 Anomalies and the Path Integral for Fermions (76) 459 78 Background Field Gauge (73) 465 79 Gervais–Neveu Gauge (78) 473 80 The Feynman Rules for N×NMatrix Fields (10) 476 81 Scattering in Quantum Chromodynamics (60, 79, 80) 482 82 Wilson Loops, Lattice Theory, and Conﬁnement (29, 73) 494 83 Chiral Symmetry Breaking (76, 82) 502 84 Spontaneous Breaking of Gauge Symmetries (32, 70) 512 85 Spontaneously Broken Abelian Gauge Theory (61, 84) 517 7 86 Spontaneously Broken Nonabelian Gauge Theory (85) 523 87 The Standard Model: Gauge and Higgs Sector (84) 527 88 The Standard Model: Lepton Sector (75, 87) 532 89 The Standard Model: Quark Sector (88) 540 90 Electroweak Interactions of Hadrons (83, 89) 546 91 Neutrino Masses (89) 555 92 Solitons and Monopoles (84) 558 93 Instantons and Theta Vacua (92) 571 94 Quarks and Theta Vacua (77, 83, 93) 582 95 Supersymmetry (69) 590 96 The Minimal Supersymmetric Standard Model (89, 95) 602 97 Grand Uniﬁcation (89) 605 Bibliography 615 8 Preface for Students Quantum ﬁeld theory is the basic mathematical language that is used to describe and analyze the physics of elementary particles. T he goal of this book is to provide a concise, step-by-step introduction to t his subject, one that covers all the key concepts that are needed to understan d the Standard Model of elementary particles, and some of its proposed exte nsions. In order to be prepared to undertake the study of quantum ﬁeld theory, you should recognize and understand the following equation s: dσ dΩ=|f(θ,φ)|2 a†|n∝an}b∇acket∇i}ht=√ n+1|n+1∝an}b∇acket∇i}ht J±|j,m∝an}b∇acket∇i}ht=√ j(j+1)−m(m±1)|j,m±1∝an}b∇acket∇i}ht A(t) =e+iHt/¯hAe−iHt/¯h H=p˙q−L ct′=γ(ct−βx) E= (p2c2+m2c4)1/2 E=−˙A/c− ∇ϕ This list is not, of course, complete; but if you are familiar with these equations, you probably know enough about quantum mechanic s, classical mechanics, special relativity, and electromagnetism to ta ckle the material in this book. Quantum ﬁeld theory has a reputation as a subject that is hard to learn. The problem, I think, is not so much that its basic ingr edients are unusually diﬃcult to master (indeed, the conceptual shift n eeded to go from quantum mechanics to quantum ﬁeld theory is not nearly a s severe as the one needed to go from classical mechanics to quantum me chanics), but rather that there are a lotof these ingredients. Some are fundamental, but many are just technical aspects of an unfamiliar form of p erturbation theory. In this book, I have tried to make the subject as accessible to beginners as possible. There are three main aspects to my approach. Logical development of the basic concepts. This is, of course, very diﬀer- ent from the historical development of quantum ﬁeld theory, which, like the historical development of most worthwhile subjects, was ﬁl led with inspired guesses and brilliant extrapolations of sometimes fuzzy id eas, as well as its fair share of mistakes, misconceptions, and dead ends. None of that is in this book. From this book, you will (I hope) get the impressio n that the 9 whole subject is eﬀortlessly clear and obvious, with one ste p following the next like sunshine after a refreshing rain. Illustration of the basic concepts with the simplest exampl es.In most ﬁelds of human endeavor, newcomers are not expected to do the most de- manding tasks right away. It takes time, dedication, and lot s of practice to work up to what the accomplished masters are doing. There is n o reason to expect quantum ﬁeld theory to be any diﬀerent in this regard. Therefore, we will start oﬀ analyzing quantum ﬁeld theories that are not immediately applicable to the real world of electrons, photons, protons , etc., but that will allow us to gain familiarity with the tools we will need, and to practice using them. Then, when we do work up to “real physics”, we will be fully ready for the task. To this end, the book is divided into three parts: Spin Zero, Spin One Half, and Spin One. The technical complexitie s associated with a particular type of particle increase with its spin. We will therefore ﬁrst learn all we can about spinless particles before moving on to the more diﬃcult (and more interesting) nonzero spins. Once we get to them, we will do a good variety of calculations in (and beyond) the Sta ndard Model of elementary particles. User friendliness. Each of the three parts is divided into numerous sec- tions. Each section is intended to treat one idea or concept o r calculation, and each is written to be as self-contained as possible. For e xample, when an equation from an earlier section is needed, I usually just repeat it, rather than ask you to leaf back and ﬁnd it (a reader’s task that I’ve a lways found annoying). Furthermore, each section is labeled with its im mediate pre- requisites, so you can tell exactly what you need to have lear ned in order to proceed. This allows you to construct chains to whatever m aterial may interest you, and to get there as quickly as possible. That said, I expect that most readers of this book will encoun ter it as the textbook in a course on quantum ﬁeld theory. In that case, of course, your reading will be guided by your professor, who I hope will ﬁnd the above features useful. If, however, you are reading this boo k on your own, I have two pieces of advice. The ﬁrst (and most important) is this: ﬁnd someone else to rea d it with you. I promise that it will be far more fun and rewarding that w ay; talking about a subject to another human being will inevitably impro ve the depth of your understanding. And you will have someone to work with you on the problems. (As with all physics texts, the problems are a k ey ingredient. I will not belabor this point, because if you have gotten this far in physics, you already know it well.) The second piece of advice echoes the novelist and Nobel laur eate William Faulkner. An interviewer asked, “Mr. Faulker, some of your read- ers claim they still cannot understand your work after readi ng it two or 10 three times. What approach would you advise them to adopt?” F aulkner replied, “Read it a fourth time.” That’s my advice here as well. After the fourth attempt, thou gh, you should consider trying something else. This is, after all, n ot the only book that has ever been written on the subject. You may ﬁnd that a di ﬀerent approach (or even the same approach explained in diﬀerent wo rds) breaks the logjam in your thinking. There are a number of excellent b ooks that you could consult, some of which are listed in the Bibliograp hy. I have also listed particular books that I think could be helpful on spec iﬁc topics in Reference Notes at the end of some of the sections. This textbook (like all ﬁnite textbooks) has a number of deﬁc iencies. One of these is a rather low level of mathematical rigor. This is partly en- demic to the subject; rigorous proofs in quantum ﬁeld theory are relatively rare, and do not appear in the overwhelming majority of resea rch papers. Even some of the most basic notions lack proof; for example, c urrently you can get a million dollars from the Clay Mathematics Insti tute simply for proving that nonabelian gauge theory actually exists an d has a unique ground state. Given this general situation, and since this i s an introductory book, the proofs that we do have are only outlined. those proo fs that we do have are only outlined. Another deﬁciency of this book is that there is no discussion of the application of quantum ﬁeld theory to condensed matter phys ics, where it also plays an important role. This connection has been imp ortant in the historical development of the subject, and is especiall y useful if you already know a lot of advanced statistical mechanics. I do no t want this to be a prerequisite, however, and so I have chosen to keep the focus on applications within elementary particle physics. Yet another deﬁciency is that there are no references to the original literature . In this regard, I am following a standard trend: as the foun- dations of a branch of science retreat into history, textboo ks become more and more synthetic and reductionist. For example, it is now r are to see a new textbook on quantum mechanics that refers to the origina l papers by the famous founders of the subject. For guides to the origina l literature on quantum ﬁeld theory, there are a number of other books with extensive references that you can consult; these include Peskin & Schroeder ,Wein- berg, andSiegel. (Italicized names refer to works listed in the Bibliograph y.) Unless otherwise noted, experimental numbers are taken fro m the Review of Particle Properties, available online at http://pdg.lb l.gov. Experimen- tal numbers quoted in this book have an uncertainty of roughl y±1 in the last signiﬁciant digit. The Review should be consulted for t he most recent experimental results, and for more precise statements of th eir uncertainty. To conclude, let me say that you are about to embark on a tour of one of 11 humanity’s greatest intellectual endeavors, and certainl y the one that has produced the most precise and accurate description of the na tural world as we ﬁnd it. I hope you enjoy the ride. 12 Preface for Instructors On learning that a new text on quantum ﬁeld theory has appeare d, one is surely tempted to respond with Isidor Rabi’s famous comment about the muon: “Who ordered that?” After all, many excellent textbooks on quan- tum ﬁeld theory are already available. I, for example, would not want to be without my well-worn copies of Quantum Field Theory by Lowell S. Brown (Cambridge 1994), Aspects of Symmetry by Sidney Coleman (Cambridge 1985), Introduction to Quantum Field Theory by Michael E. Peskin and Daniel V. Schroeder (Westview 1995), Field Theory: A Modern Primer by Pierre Ramond (Addison-Wesley 1990), Fields by Warren Siegel (arXiv.org 2005), The Quantum Theory of Fields , Volumes I, II, and III, by Steven Weinberg (Cambridge 1995), and Quantum Field Theory in a Nutshell by my colleague Tony Zee (Princeton 2003), to name just a few of t he more recent texts. Nevertheless, despite the excellence of thes e and other books, I have never followed any of them very closely in my twenty yea rs of on- and-oﬀ teaching of a year-long course in relativistic quant um ﬁeld theory. As discussed in the Preface for Students, this book is based o n the no- tion that quantum ﬁeld theory is most readily learned by star ting with the simplest examples and working through their details in a log ical fashion. To this end, I have tried to set things up at the very beginning to antici- pate the eventual need for renormalization, and not be caval ier about how the ﬁelds are normalized and the parameters deﬁned. I believ e that these precautions take a lot of the “hocus pocus” (to quote Feynman ) out of the “dippy process” of renormalization. Indeed, with this appr oach, even the anharmonic oscillator is in need of renormalization; see pr oblem 14.7. A ﬁeld theory with many pedagogical virtues is ϕ3theory in six di- mensions, where its coupling constant is dimensionless. Pe rhaps because six dimensions used to seem too outre (though today’s prospe ctive string theorists don’t even blink), the only introductory textboo k I know of that treats this model is Quantum Field Theory by George Sterman (Cambridge 1993), though it is also discussed in some more advanced book s, such as Renormalization by John Collins (Cambridge 1984) and Foundations of Quantum Chromodynamics by T. Muta (World Scientiﬁc 1998). (There is also a series of lectures by Ed Witten on quantum ﬁeld theory f or math- ematicians, available online, that treat ϕ3theory.) The reason ϕ3theory in six dimensions is a nice example is that its Feynman diagra ms have a simple structure, but still exhibit the generic phenomena o f renormalizable quantum ﬁeld theory at the one-loop level. (The same cannot b e said forϕ4 theory in four dimensions, where momentum-dependent corre ctions to the propagator do not appear until the two-loop level.) Thus, in Part I of this text,ϕ3theory in six dimensions is the primary example. I use it to gi ve 13 introductory treatments of most aspects of relativistic qu antum ﬁeld theory for spin-zero particles, with a minimum of the technical com plications that arise in more realistic theories (like QED) with higher-spi n particles. Although I eventually discuss the Wilson approach to renorm alization and eﬀective ﬁeld theory (in section 29), and use eﬀective ﬁe ld theory exten- sively for the physics of hadrons in Part III, I do not feel it i s pedagogically useful to bring it in at the very beginning, as is sometimes ad vocated. The problem is that the key notion of the decoupling of physical p rocesses at dif- ferent length scales is an unfamiliar one for most students; there is nothing in typical courses on quantum mechanics or electomagnetism or classical mechanics to prepare students for this idea (which was deeme d worthy of a Nobel Prize for Ken Wilson in 1982). It also does not provide f or a simple calculational framework, since one must deal with the inﬁni te number of terms in the eﬀective lagrangian, and then explain why most o f them don’t matter after all. It’s noteworthy that Wilson himself did no t spend a lot of time computing properly normalized perturbative S-matrix elements, a skill that we certainly want our students to have; we want the m to have it because a great deal of current research still depends on i t. Indeed, the vaunted success of quantum ﬁeld theory as a description of th e real world is based almost entirely on our ability to carry out these pertu rbative calcula- tions. Studying renormalization early on has other pedagog ical advantages. With the Nobel Prizes to Gerard ’t Hooft and Tini Veltman in 19 99 and to David Gross, David Politzer, and Frank Wilczek in 2004, toda y’s students are well aware of beta functions and running couplings, and w ould like to understand them. I ﬁnd that they are generally much more exci ted about this (even in the context of toy models) than they are about le arning to reproduce the nearly century-old tree-level calculations of QED. And ϕ3 theory in six dimensions is asymptotically free, which ulti mately provides for a nice segue to the “real physics” of QCD. In general I have tried to present topics so that the more inte resting as- pects (from a present-day point of view) come ﬁrst. An exampl e is anoma- lies; the traditional approach is to start with the π0→γγdecay rate, but such a low-energy process seems like a dusty relic to most of today’s students. I therefore begin by demonstrating that anomalie s destroy the self-consistency of the great majority of chiral gauge theo ries, a fact that strikes me (and, in my experience, most students) as much mor e interest- ing and dramatic than an incorrect calculation of the π0decay rate. Then, when we do eventually get to this process (in section 90), it a ppears as a straightforward consequence of what we already learned abo ut anomalies in sections 75–77. Nevertheless, I want this book to be useful to those who disag ree with my pedagogical choices, and so I have tried to structure it to allow for 14 maximum ﬂexibility. Each section treats a particular idea o r concept or calculation, and is as self-contained as possible. Each sec tion also lists its immediate prerequisites, so that it is easy to see how to r earrange the material to suit your personal preferences. In some cases, alternative approaches are developed in the p roblems. For example, I have chosen to introduce path integrals relat ively early (though not before canonical quantization and operator met hods are ap- plied to free-ﬁeld theory), and use them to derive Dyson’s ex pansion. For those who would prefer to delay the introduction of path inte grals (but since you will have to cover them eventually, why not get it over wit h?), problem 9.5 outlines the operator-based derivation in the interact ion picture. Another point worth noting is that a textbook and lectures ar e ideally complementary. Many sections of this book contain rather te dious mathe- matical detail that I would not and do not write on the blackbo ard during a lecture. (Indeed, the earliest origins of this book are sup plementary notes that I typed up and handed out.) For example, much of the devel opment of Weyl spinors in sections 34–37 can be left to outside reading . I do encour- age you not to eliminate this material entirely, however; pe dagogically, the problem with skipping directly to four-component notation is explaining that (in four dimensions) the hermitian conjugate of a left- handed ﬁeld is right handed, a deeply important fact that is the key to solvi ng problems such as 36.5 and 83.1, which are in turn vital to understandin g the struc- ture of the Standard Model and its extensions. A related topi c is computing scattering amplitudes for Majorana ﬁelds; this is essentia l for modern re- search on massive neutrinos and supersymmetric particles, though it could be left out of a time-limited course. While I have sometimes included more mathematical detail th an is ideal for a lecture, I have also tended to omit explanations based o n “physical intuition.” For example, in section 90, we compute the π−→ℓ−¯νℓdecay amplitude (where ℓis a charged lepton) and ﬁnd that it is proportional to the lepton mass. There is a well-known heuristic explanatio n of this fact that goes something like this: “The pion has spin zero, and so the lepton and the antineutrino must emerge with opposite spin, and the refore the same helicity. An antineutrino is always right-handed, and so the lepton must be as well. But only the left-handed lepton couples to th eW−, so the decay amplitude vanishes if the left- and right-handed l eptons are not coupled by a mass term.” This is essentially correct, but the reasoning is a bit more s ubtle than it ﬁrst appears. A student may ask, “Why can’t there be orbita l angular momentum? Then the lepton and the antineutrino could have th e same spin.” The answer is that orbital angular momentum must be pe rpendicular to the linear momentum, whereas helicity is (by deﬁnition) p arallel to the 15 linear momentum; so adding orbital angular momentum cannot change the helicity assignments. (This is explored in a simpliﬁed mode l in problem 48.4.) The larger point is that intuitive explanations can a lmost always be probed more deeply. This is ﬁne in a classroom, where you are a vailable to answer questions, but a textbook author has a hard time knowi ng where to stop. Too little detail renders the explanations opaque, and too much can be overwhelming; furthermore the happy medium tends to d iﬀer from student to student. The calculation, on the other hand, is de ﬁnitive (at least within the framework being explored, and modulo the po ssibility of mathematical error). As Roger Penrose once said, “The great thing about physical intuition is that it can be adjusted to ﬁt the facts. ” So, in this book, I have tended to emphasize calculational detail at the expense of heuristic reasoning. Lectures should ideally invert this t o some extent. I should also mention that a section of the book is not intende d to coincide exactly with a lecture. The material in some sectio ns could easily be covered in less than an hour, and some would clearly take mo re. My approach in lecturing is to try to keep to a pace that allows th e students to follow the analysis, and then try to come to a more-or-less na tural stopping point when class time is up. This sometimes means ending in th e middle of a long calculation, but I feel that this is better than tryi ng to artiﬁcially speed things along to reach a predetermined destination. It would take at least three semesters of lectures to cover th is entire book, and so a year-long course must omit some. A sequence I mi ght follow is 1–23, 26–28, 33–43, 45–48, 51, 52, 54–59, 62–64, 66 –68, 24, 69, 70, 44, 53, 71–73, 75–77, 30, 32, 84, 87–89, 29, 82, 83, 90, and, if any time was left, a selection of whatever seemed of most interest to me an d the students of the remaining material. To conclude, I hope you ﬁnd this book to be a useful tool in work ing to- wards our mutual goal of bringing humanity’s understanding of the physics of elementary particles to a new audience. 16 Acknowledgments Every book is a collaborative eﬀort, even if there is only one author on the title page. Any skills I may have as a teacher were ﬁrst gle aned as a student in the classes of those who taught me. My ﬁrst and most important teachers were my parents, Casimir and Helen Srednicki, to wh om this book is dedicated. In our small town in Ohio, my excellent public- school teachers included Esta Kefauver, Marie Casher, Carol Baird, Jim Chas e, Joe Gerin, Hugh Laughlin, and Tom Murphy. In college at Cornell, Don Har till, Bruce Kusse, Bob Siemann, John Kogut, and Saul Teukolsky taught pa rticularly memorable courses. In graduate school at Stanford, Roberto Peccei gave me my ﬁrst exposure to quantum ﬁeld theory, in a superb course that required bicycling in by 8:30AM (which seemed like a major sa criﬁce at the time). Everyone in that class very much hoped that Robert o would one day turn his extensive hand-written lecture notes (which he put on reserve in the library) into a book. He never did, but I’d like to think that perhaps a bit of his consummate skill has found its way into this text. I have also used a couple of his jokes. My thesis advisor at Stanford, Lenny Susskind, taught me how to think about physics without getting bogged down in the details. Th is book in- cludes a lot of detail that Lenny would no doubt have left out, but while writing it I have tried to keep his exemplary clarity of thoug ht in mind as something to strive for. During my time in graduate school, and subsequently in postd octoral positions at Princeton and CERN, and ﬁnally as a faculty memb er at UC Santa Barbara, I was extremely fortunate to be able to intera ct with many excellent physicists, from whom I learned an enormous amoun t. These include Stuart Freedman, Eduardo Fradkin, Steve Shenker, S idney Cole- man, Savas Dimopoulos, Stuart Raby, Michael Dine, Willy Fis chler, Curt Callan, David Gross, Malcolm Perry, Sam Trieman, Arthur Wig htman, Ed Witten, Hans-Peter Nilles, Daniel Wyler, Dmitri Nanopoulo s, John Ellis, Keith Olive, Jose Fulco, Ray Sawyer, John Cardy, Frank Wilcz ek, Jim Hartle, Gary Horowitz, Andy Strominger, and Tony Zee. I am es pecially grateful to my Santa Barbara colleagues David Berenstein, S teve Giddings, Don Marolf, Joe Polchinski, and Bob Sugar, who used various d rafts of this book while teaching quantum ﬁeld theory, and made various su ggestions for improvement. I am also grateful to physicists at other institutions who re ad parts of the manuscript and also made suggestions, including Oliver de W olfe, Marcelo Gleiser, Steve Gottlieb, Arkady Tsetlyn, and Arkady Vainsh tein. I must single out for special thanks Professor Heidi Fearn of Cal St ate Fullerton, whose careful reading of Parts I and II allowed me to correct m any unclear 17 passages and outright errors that would otherwise have slip ped by. Students over the years have suﬀered through my varied attem pts to arrive at a pedagogically acceptable scheme for teaching qu antum ﬁeld theory. I thank all of them for their indulgence. I am especia lly grateful to Sam Pinansky, Tae Min Hong, and Sho Yaida for their diligen ce in ﬁnding and reporting errors, and to Brian Wignal for help wit h formatting the manuscript. Also, a number of students from around the wo rld (as well as Santa Barbara) kindly reported errors in versions of this book that were posted online; these include Omri Bahat-Treidel, Hee- Joong Chung, Yevgeny Kats, Sue Ann Koay, Peter Lee, Nikhil Jayant Joshi, K evin Weil, Dusan Simic, and Miles Stoudenmire. I thank them for their he lp, and apologize to anyone that I may have missed. Throughout this project, the assistance and support of my wi fe Elo¨ ıse and daughter Julia were invaluable. Elo¨ ıse read through th e manuscript and made suggestions that often clariﬁed the language. Juli a oﬀered advice on the cover design (a highly stylized Feynman diagram). And they both kindly indulged the amount of time I spent working on this boo k that you now hold in your hands. Part I Spin Zero 1: Attempts at relativistic quantum mechanics 19 1Attempts at relativistic quantum mechanics Prerequisite: none In order to combine quantum mechanics and relativity, we mus t ﬁrst un- derstand what we mean by “quantum mechanics” and “relativit y”. Let us begin with quantum mechanics. Somewhere in most textbooks on the subject, one can ﬁnd a list of the “axioms of quantum mechanics”. These include statements al ong the lines of The state of the system is represented by a vector in Hilbert space. Observables are represented by hermitian operators. The measurement of an observable yields one of its eigenvalu es as the result. And so on. We do not need to review these closely here. The axio m we need to focus on is the one that says that the time evolution of the state of the system is governed by the Schr¨ odinger equation, i¯h∂ ∂t|ψ,t∝an}b∇acket∇i}ht=H|ψ,t∝an}b∇acket∇i}ht, (1.1) whereHis the hamiltonian operator, representing the total energy . Let us consider a very simple system: a spinless, nonrelativ istic particle with no forces acting on it. In this case, the hamiltonian is H=1 2mP2, (1.2) wheremis the particle’s mass, and Pis the momentum operator. In the position basis, eq.(1.1) becomes i¯h∂ ∂tψ(x,t) =−¯h2 2m∇2ψ(x,t), (1.3) whereψ(x,t) =∝an}b∇acketle{tx|ψ,t∝an}b∇acket∇i}htis the position-space wave function. We would like to generalize this to relativistic motion. The obvious way to proceed is to take H= +/radicalig P2c2+m2c4, (1.4) 1: Attempts at relativistic quantum mechanics 20 which yields the correct relativistic energy-momentum rel ation. If we for- mally expand this hamiltonian in inverse powers of the speed of lightc, we get H=mc2+1 2mP2+... . (1.5) This is simply a constant (the rest energy), plus the usual no nrelativistic hamiltonian, eq.(1.2), plus higher-order corrections. Wi th the hamiltonian given by eq.(1.4), the Schr¨ odinger equation becomes i¯h∂ ∂tψ(x,t) = +/radicalig −¯h2c2∇2+m2c4ψ(x,t). (1.6) Unfortunately, this equation presents us with a number of di ﬃculties. One is that it apparently treats space and time on a diﬀerent foot ing: the time derivative appears only on the left, outside the square root , and the space derivatives appear only on the right, under the square root. This asymme- try between space and time is not what we would expect of a rela tivistic theory. Furthermore, if we expand the square root in powers o f∇2, we get an inﬁnite number of spatial derivatives acting on ψ(x,t); this implies that eq.(1.6) is not local in space. We can alleviate these problems by squaring the diﬀerential operators on each side of eq.(1.6) before applying them to the wave func tion. Then we get −¯h2∂2 ∂t2ψ(x,t) =/parenleftig −¯h2c2∇2+m2c4/parenrightig ψ(x,t). (1.7) This is the Klein-Gordon equation , and it looks a lot nicer than eq.(1.6). It is second-order in both space and time derivatives, and th ey appear in a symmetric fashion. To better understand the Klein-Gordon equation, let us cons ider in more detail what we mean by “relativity”. Special relativit y tells us that physics looks the same in allinertial frames . To explain what this means, we ﬁrst suppose that a certain spacetime coordinate system ( ct,x) represents (by ﬁat) an inertial frame. Let us deﬁne x0=ct, and write xµ, where µ= 0,1,2,3, in place of ( ct,x). It is also convenient (for reasons not at all obvious at this point) to deﬁne x0=−x0andxi=xi, wherei= 1,2,3. This can be expressed more elegantly if we ﬁrst introduce the Minkowski metric , gµν= −1 +1 +1 +1 , (1.8) where blank entries are zero. We then have xµ=gµνxν, where a repeated index is summed. 1: Attempts at relativistic quantum mechanics 21 To invert this formula, we introduce the inverse of g, which is confusingly also calledg, except with both indices up: gµν= −1 +1 +1 +1 . (1.9) We then have gµνgνρ=δµρ, whereδµρis the Kronecker delta (equal to one if its two indices take on the same value, zero otherwise). No w we can also writexµ=gµνxν. It is a general rule that any pair of repeated (and therefore s ummed) indices must consist of one superscript and one subscript; t hese indices are said to be contracted . Also, any unrepeated (and therefore unsummed) indices must match (in both name and height) on the left- and r ight-hand sides of any valid equation. Now we are ready to specify what we mean by an inertial frame. I f the coordinates xµrepresent an inertial frame (which they do, by assumption), then so do any other coordinates ¯ xµthat are related by ¯xµ= Λµνxν+aµ, (1.10) where Λµνis aLorentz transformation matrix andaµis atranslation vector . Both Λµνandaµare constant (that is, independent of xµ). Furthermore, Λµνmust obey gµνΛµρΛνσ=gρσ. (1.11) Eq.(1.11) ensures that the interval between two diﬀerent spacetime points that are labeled by xµandx′µin one inertial frame, and by ¯ xµand ¯x′µin another, is the same. This interval is deﬁned to be (x−x′)2≡gµν(x−x′)µ(x−x′)ν = (x−x′)2−c2(t−t′)2. (1.12) In the other frame, we have (¯x−¯x′)2=gµν(¯x−¯x′)µ(¯x−¯x′)ν =gµνΛµρΛνσ(x−x′)ρ(x−x′)σ =gρσ(x−x′)ρ(x−x′)σ = (x−x′)2, (1.13) as desired. When we say that physics looks the same , we mean that two observers (Alice and Bob, say) using two diﬀerent sets of coordinates ( representing 1: Attempts at relativistic quantum mechanics 22 two diﬀerent inertial frames) should agree on the predicted results of all possible experiments. In the case of quantum mechanics, thi s requires Alice and Bob to agree on the value of the wave function at a particul ar spacetime point, a point that is called xby Alice and ¯ xby Bob. Thus if Alice’s predicted wave function is ψ(x), and Bob’s is ¯ψ(¯x), then we should have ψ(x) =¯ψ(¯x). Furthermore, in order to maintain ψ(x) =¯ψ(¯x) throughout spacetime,ψ(x) and ¯ψ(¯x) should obey identical equations of motion. Thus a candidate wave equation should take the same form in any ine rtial frame. Let us see if this is true of the Klein-Gordon equation. We ﬁrs t introduce some useful notation for spacetime derivatives: ∂µ≡∂ ∂xµ=/parenleftbigg +1 c∂ ∂t,∇/parenrightbigg , (1.14) ∂µ≡∂ ∂xµ=/parenleftbigg −1 c∂ ∂t,∇/parenrightbigg . (1.15) Note that ∂µxν=gµν, (1.16) so that our matching-index-height rule is satisﬁed. If ¯xandxare related by eq.(1.10), then ¯∂and∂are related by ¯∂µ= Λµν∂ν. (1.17) To check this, we note that ¯∂ρ¯xσ= (Λρµ∂µ)(Λσνxν+aµ) = ΛρµΛσν(∂µxν) = ΛρµΛσνgµν=gρσ, (1.18) as expected. The last equality in eq.(1.18) is another form o f eq.(1.11); see section 2. We can now write eq.(1.7) as −¯h2c2∂2 0ψ(x) = (−¯h2c2∇2+m2c4)ψ(x). (1.19) After rearranging and identifying ∂2≡∂µ∂µ=−∂2 0+∇2, we have (−∂2+m2c2/¯h2)ψ(x) = 0. (1.20) This is Alice’s form of the equation. Bob would write (−¯∂2+m2c2/¯h2)¯ψ(¯x) = 0. (1.21) Is Bob’s equation equivalent to Alice’s equation? To see tha t it is, we set ¯ψ(¯x) =ψ(x), and note that ¯∂2=gµν¯∂µ¯∂ν=gµνΛµρΛµσ∂ρ∂σ=∂2. (1.22) 1: Attempts at relativistic quantum mechanics 23 Thus, eq.(1.21) is indeed equivalent to eq.(1.20). The Klei n-Gordon equa- tion is therefore manifestly consistent with relativity: i t takes the same form in every inertial frame. This is the good news. The bad news is that the Klein-Gordon eq uation violates one of the axioms of quantum mechanics: eq.(1.1), t he Schr¨ odinger equation in its abstract form. The abstract Schr¨ odinger eq uation has the fundamental property of being ﬁrst order in the time derivat ive, whereas the Klein-Gordon equation is second order. This may not seem too important, but in fact it has drastic consequences. One of these is that t he norm of a state, ∝an}b∇acketle{tψ,t|ψ,t∝an}b∇acket∇i}ht=/integraldisplay d3x∝an}b∇acketle{tψ,t|x∝an}b∇acket∇i}ht∝an}b∇acketle{tx|ψ,t∝an}b∇acket∇i}ht=/integraldisplay d3xψ∗(x)ψ(x), (1.23) is not in general time independent. Thus probability is not c onserved. The Klein-Gordon equation obeys relativity, but not quantum me chanics. Dirac attempted to solve this problem (for spin-one-half pa rticles) by introducing an extra discrete label on the wave function, to account for spin:ψa(x),a= 1,2. He then tried a Schr¨ odinger equation of the form i¯h∂ ∂tψa(x) =/parenleftig −i¯hc(αj)ab∂j+mc2(β)ab/parenrightig ψb(x), (1.24) where all repeated indices are summed, and αjandβare matrices in spin- space. This equation, the Dirac equation , is consistent with the abstract Schr¨ odinger equation. The state |ψ,a,t∝an}b∇acket∇i}htcarries a spin label a, and the hamiltonian is Hab=cPj(αj)ab+mc2(β)ab, (1.25) wherePjis a component of the momentum operator. Since the Dirac equation is linear in both time and space deri vatives, it has a chance to be consistent with relativity. Note that sq uaring the hamiltonian yields (H2)ab=c2PjPk(αjαk)ab+mc3Pj(αjβ+βαj)ab+ (mc2)2(β2)ab.(1.26) SincePjPkis symmetric on exchange of jandk, we can replace αjαkby its symmetric part,1 2{αj,αk}, where {A,B}=AB+BAis the anticom- mutator. Then, if we choose matrices such that {αj,αk}ab= 2δjkδab,{αj,β}ab= 0,(β2)ab=δab, (1.27) we will get (H2)ab= (P2c2+m2c4)δab. (1.28) Thus, the eigenstates of H2are momentum eigenstates, with H2eigenvalue p2c2+m2c4. This is, of course, the correct relativistic energy-momen tum 1: Attempts at relativistic quantum mechanics 24 relation. While it is outside the scope of this section to dem onstrate it, it turns out that the Dirac equation is fully consistent with re lativity provided the Dirac matrices obey eq.(1.27). So we have apparently suc ceeded in constructing a quantum mechanical, relativistic theory! There are, however, some problems. We would like the Dirac ma trices to be 2 ×2, in order to account for electron spin. However, they must in fact be larger. To see this, note that the 2 ×2 Pauli matrices obey {σi,σj}= 2δij, and are thus candidates for the Dirac αimatrices. However, there is no fourth matrix that anticommutes with these three (easily proven by writing down the most general 2 ×2 matrix and working out the three anticommutators explicitly). Also, we can show that the Dir ac matrices must be even dimensional; see problem 1.1. Thus their minimu m size is 4×4, and it remains for us to interpret the two extra possible “s pin” states. However, these extra states cause a more severe problem than a mere overcounting. Acting on a momentum eigenstate, Hbecomes the matrix cα·p+mc2β. In problem 1.1, we ﬁnd that the trace of this matrix is zero. Thus the four eigenvalues must be + E(p), +E(p),−E(p),−E(p), where E(p) = +( p2c2+m2c4)1/2. The negative eigenvalues are the problem: they indicate that there is no ground state. In a more elabora te theory that included interactions with photons, there seems to be n o reason why a positive energy electron could not emit a photon and drop do wn into a negative energy state. This downward cascade could contin ue forever. (The same problem also arises in attempts to interpret the Kl ein-Gordon equation as a modiﬁed form of quantum mechanics.) Dirac made a wildly brilliant attempt to ﬁx this problem of ne gative energy states. His solution is based on an empirical fact abo ut electrons: they obey the Pauli exclusion principle. It is impossible to put more than one of them in the same quantum state. What if, Dirac speculat ed, all the negative energy states were already occupied ? In this case, a positive energy electron could not drop into one of these states, by Pa uli exclusion. Many questions immediately arise. Why don’t we see the negat ive elec- tric charge of this Dirac sea of electrons? Dirac’s answer: because we’re used to it. (More precisely, the physical eﬀects of a uniform charge density depend on the boundary conditions at inﬁnity that we impose o n Maxwell’s equations, and there is a choice that renders such a uniform c harge density invisible.) However, Dirac noted, if one of these negative e nergy electrons were excited into a positive energy state (by, say, a suﬃcien tly energetic photon), it would leave behind a holein the sea of negative energy elec- trons. This hole would appear to have positive charge, and po sitive energy. Dirac therefore predicted (in 1927) the existence of the positron , a particle with the same mass as the electron, but opposite charge. The p ositron was found experimentally ﬁve years later. 1: Attempts at relativistic quantum mechanics 25 However, we have now jumped from an attempt at a quantum descr ip- tion of a single relativistic particle to a theory that apparently requires aninﬁnite number of particles. Even if we accept this, we still have not solved the problem of how to describe particles like photons or pions or alpha nuclei that do notobey Pauli exclusion. At this point, it is worthwhile to stop and reﬂect on why it has proven to be so hard to ﬁnd an acceptable relativistic wave equation for a sin- gle quantum particle. Perhaps there is something wrong with our basic approach. And there is. Recall the axiom of quantum mechanics that says that “Observables are represented by hermitian operators.” Thi s is not entirely true. There is one observable in quantum mechanics that is notrepresented by a hermitian operator: time. Time enters into quantum mech anics only when we announce that the “state of the system” depends on an e xtra parametert. This parameter is not the eigenvalue of any operator. This i s in sharp contrast to the particle’s position x, which isthe eigenvalue of an operator. Thus, space and time are treated very diﬀerently, a fact that is obscured by writing the Schr¨ odinger equation in terms of th e position-space wave function ψ(x,t). Since space and time are treated asymmetrically, it is not surprising that we are having trouble incorporating a symmetry that mixes them up. So, what are we to do? In principle, the problem could be an intractable one: it mig ht beim- possible to combine quantum mechanics and relativity. In this case, t here would have to be some meta-theory, one that reduces in the non relativistic limit to quantum mechanics, and in the classical limit to rel ativistic particle dynamics, but is actually neither. This, however, turns out not to be the case. We can solve our pr oblem, but we must put space and time on an equal footing at the outset . There are two ways to do this. One is to demote position from its stat us as an operator, and render it as an extra label, like time. The othe r is to promote time to an operator. Let us discuss the second option ﬁrst. If time becomes an oper ator, what do we use as the time parameter in the Schr¨ odinger equation? Happily, in relativistic theories, there is more than one notion of time . We can use the proper time τof the particle (the time measured by a clock that moves with it) as the time parameter. The coordinate time T(the time measured by a stationary clock in an inertial frame) is then promoted t o an operator. In the Heisenberg picture (where the state of the system is ﬁx ed, but the operators are functions of time that obey the classical equa tions of motion), we would have operators Xµ(τ), whereX0=T. Relativistic quantum mechanics can indeed be developed along these lines, but it i s surprisingly 1: Attempts at relativistic quantum mechanics 26 complicated to do so. (The many times are the problem; any mon otonic function of τis just as good a candidate as τitself for the proper time, and this inﬁnite redundancy of descriptions must be understood and accounted for.) One of the advantages of considering diﬀerent formalisms is that they may suggest diﬀerent directions for generalizations. For e xample, once we haveXµ(τ), why not consider adding some more parameters? Then we would have, for example, Xµ(σ,τ). Classically, this would give us a continuous family of worldlines, what we might call a worldsheet , and so Xµ(σ,τ) would describe a propagating string. This is indeed the starting point for string theory. Thus, promoting time to an operator is a viable option, but is compli- cated in practice. Let us then turn to the other option, demot ing position to a label. The ﬁrst question is, label on what? The answer is, on oper- ators. Thus, consider assigning an operator to each point xin space; call these operators ϕ(x). A set of operators like this is called a quantum ﬁeld . In the Heisenberg picture, the operators are also time depen dent: ϕ(x,t) =eiHt/¯hϕ(x,0)e−iHt/¯h. (1.29) Thus, both position and (in the Heisenberg picture) time are now labels on operators; neither is itself the eigenvalue of an operator. So, now we have two diﬀerent approaches to relativistic quan tum theory, approaches that might, in principle, yield diﬀerent result s. This, however, is not the case: it turns out that any relativistic quantum ph ysics that can be treated in one formalism can also be treated in the other. W hich we use is a matter of convenience and taste. And, quantum ﬁeld theory , the formalism in which position and time are both labels on opera tors, is much more convenient and eﬃcient for most problems. There is another useful equivalence: ordinary nonrelativi stic quantum mechanics, for a ﬁxed number of particles, can be rewritten a s a quantum ﬁeld theory. This is an informative exercise, since the corr esponding physics is already familiar. Let us carry it out. Begin with the position-basis Schr¨ odinger equation for nparticles, all with the same mass m, moving in an external potential U(x), and inter- acting with each other via an interparticle potential V(x1−x2): i¯h∂ ∂tψ=/bracketleftiggn/summationdisplay j=1/parenleftigg −¯h2 2m∇2 j+U(xj)/parenrightigg +n/summationdisplay j=1j−1/summationdisplay k=1V(xj−xk)/bracketrightigg ψ, (1.30) whereψ=ψ(x1,...,xn;t) is the position-space wave function. The quan- tum mechanics of this system can be rewritten in the abstract form of 1: Attempts at relativistic quantum mechanics 27 eq.(1.1) by ﬁrst introducing (in, for now, the Schr¨ odinger picture) a quan- tum ﬁelda(x) and its hermitian conjugate a†(x). We take these operators to have the commutation relations [a(x),a(x′)] = 0, [a†(x),a†(x′)] = 0, [a(x),a†(x′)] =δ3(x−x′), (1.31) whereδ3(x) is the three-dimensional Dirac delta function. Thus, a†(x) and a(x) behave like harmonic-oscillator creation and annihilati on operators that are labeled by a continuous index. In terms of them, we in troduce the hamiltonian operator of our quantum ﬁeld theory, H=/integraldisplay d3xa†(x)/parenleftig −¯h2 2m∇2+U(x)/parenrightig a(x) +1 2/integraldisplay d3xd3yV(x−y)a†(x)a†(y)a(y)a(x). (1.32) Now consider a time-dependent state of the form |ψ,t∝an}b∇acket∇i}ht=/integraldisplay d3x1...d3xnψ(x1,...,xn;t)a†(x1)...a†(xn)|0∝an}b∇acket∇i}ht,(1.33) whereψ(x1,...,xn;t) is some function of the nparticle positions and time, and|0∝an}b∇acket∇i}htis the vacuum state , the state that is annihilated by all the a’s, a(x)|0∝an}b∇acket∇i}ht= 0. (1.34) It is now straightforward (though tedious) to verify that eq .(1.1), the ab- stract Schr¨ odinger equation, is obeyed if and only if the fu nctionψsatisﬁes eq.(1.30). Thus we can interpret the state |0∝an}b∇acket∇i}htas a state of “no particles”, the state a†(x1)|0∝an}b∇acket∇i}htas a state with one particle at position x1, the statea†(x1)a†(x2)|0∝an}b∇acket∇i}ht as a state with one particle at position x1and another at position x2, and so on. The operator N=/integraldisplay d3xa†(x)a(x) (1.35) counts the total number of particles. It commutes with the ha miltonian, as is easily checked; thus, if we start with a state of nparticles, we remain with a state of nparticles at all times. However, we can imagine generalizations of this version of t he theory (generalizations that would not be possible without the ﬁel d formalism) in which the number of particles is notconserved. For example, we could try adding toHa term like ∆H∝/integraldisplay d3x/bracketleftig a†(x)a2(x) + h.c./bracketrightig . (1.36) 1: Attempts at relativistic quantum mechanics 28 This term does notcommute with N, and so the number of particles would not be conserved with this addition to H. Theories in which the number of particles can change as time e volves are a good thing: they are needed for correct phenomenology. We a re already familiar with the notion that atoms can emit and absorb photo ns, and so we had better have a formalism that can incorporate this phen omenon. We are less familiar with emission and absorption (that is to sa y, creation and annihilation) of electrons, but this process also occurs in nature; it is less common because it must be accompanied by the emission or abso rption of a positron, antiparticle to the electron. There are not a lot of positrons around to facilitate electron annihilation, while e+e−pair creation requires us to have on hand at least 2 mc2of energy available for the rest-mass energy of these two particles. The photon, on the other hand, is its own antiparticle, and has zero rest mass; thus photons are easil y and copiously produced and destroyed. There is another important aspect of the quantum theory spec iﬁed by eqs.(1.32) and (1.33). Because the creation operators comm ute with each other, only the completely symmetric part of ψsurvives the integration in eq.(1.33). Therefore, without loss of generality, we can restrict our attention to ψ’s of this type: ψ(...xi...xj...;t) = +ψ(...xj...xi...;t). (1.37) This means that we have a theory of bosons , particles that (like photons or pions or alpha nuclei) obey Bose-Einstein statistics. If we want Fermi-Dirac statistics instead, we must replace eq.(1.31) with {a(x),a(x′)}= 0, {a†(x),a†(x′)}= 0, {a(x),a†(x′)}=δ3(x−x′), (1.38) where again {A,B}=AB+BAis the anticommutator. Now only the fully antisymmetric part of ψsurvives the integration in eq.(1.33), and so we can restrict our attention to ψ(...xi...xj...;t) =−ψ(...xj...xi...;t). (1.39) Thus we have a theory of fermions . It is straightforward to check that the abstract Schr¨ odinger equation, eq.(1.1), still impli es thatψobeys the diﬀerential equation (1.30).1Interestingly, there is no simple way to write 1Now, however, the ordering of the aanda†operators in the last term of eq.(1.32) becomes signiﬁcant, and must be as written. 1: Attempts at relativistic quantum mechanics 29 down a quantum ﬁeld theory with particles that obey Boltzman n statistics, corresponding to a wave function with no particular symmetr y. This is a hint of the spin-statistics theorem, which applies to relativistic quantum ﬁeld theory. It says that interacting particles with integer spin must be bosons, and interacting particles with half-integer spin must be fermions. In our nonrelativistic example, the interacting particles clearly have spin zero (because their creation operators carry no labels that could be inter- preted as corresponding to diﬀerent spin states), but can be either bosons or fermions, as we have seen. Now that we have seen how to rewrite the nonrelativistic quan tum me- chanics of multiple bosons or fermions as a quantum ﬁeld theo ry, it is time to try to construct a relativistic version. Reference Notes The history of the physics of elementary particles is recoun ted in Pais. A brief overview can be found in Weinberg I . More details on quantum ﬁeld theory for nonrelativistic particles can be found in Brown . Problems 1.1) Show that the Dirac matrices must be even dimensional. H int: show that the eigenvalues of βare all ±1, and that Tr β= 0. To show that Trβ= 0, consider, e.g., Tr α2 1β. Similarly, show that Tr αi= 0. 1.2) With the hamiltonian of eq.(1.32), show that the state d eﬁned in eq.(1.33) obeys the abstract Schr¨ odinger equation, eq.(1 .1), if and only if the wave function obeys eq.(1.30). Your demonstrati on should apply both to the case of bosons, where the particle creation and anni- hilation operators obey the commutation relations of eq.(1 .31), and to fermions, where the particle creation and annihilation o perators obey the anticommutation relations of eq.(1.38). 1.3) Show explicitly that [ N,H] = 0, where His given by eq.(1.32) and Nby eq.(1.35). 2Lorentz Invariance Prerequisite: 1 ALorentz transformation is a linear, homogeneous change of coordinates fromxµto ¯xµ, ¯xµ= Λµνxν, (2.1) that preserves the intervalx2betweenxµand the origin, where x2≡xµxµ=gµνxµxν=x2−c2t2. (2.2) This means that the matrix Λµνmust obey gµνΛµρΛνσ=gρσ, (2.3) where gµν= −1 +1 +1 +1 . (2.4) is the Minkowski metric. Note that this set of transformations includes ordinary spa tial rotations: take Λ00= 1, Λ0i= Λi0= 0, and Λij=Rij, whereRis an orthogonal rotation matrix. The set of all Lorentz transformations forms a group: the product of any two Lorentz transformations is another Lorentz transfo rmation; the product is associative; there is an identity transformatio n, Λµν=δµν; and every Lorentz transformation has an inverse. It is easy t o demonstrate these statements explicitly. For example, to ﬁnd the invers e transformation (Λ−1)µν, note that the left-hand side of eq.(2.3) can be written as Λ νρΛνσ, and that we can raise the ρindex on both sides to get Λ νρΛνσ=δρσ. On the other hand, by deﬁnition, (Λ−1)ρνΛνσ=δρσ. Therefore (Λ−1)ρν= Λνρ. (2.5) Another useful version of eq.(2.3) is gµνΛρµΛσν=gρσ. (2.6) To get eq.(2.6), start with eq.(2.3), but with the inverse tr ansformations (Λ−1)µρand (Λ−1)νσ. Then use eq.(2.5), raise all down indices, and lower all up indices. The result is eq.(2.6). For an inﬁnitesimal Lorentz transformation, we can write Λµν=δµν+δωµν. (2.7) 2: Lorentz Invariance 31 Eq.(2.3) can be used to show that δωwith both indices down (or up) is antisymmetric: δωρσ=−δωσρ. (2.8) Thus there are six independent inﬁnitesimal Lorentz transf ormations (in four spacetime dimensions). These can be divided into three rotations (δωij=−εijkˆnkδθfor a rotation by angle δθabout the unit vector ˆ n) and three boosts ( δωi0= ˆniδηfor a boost in the direction ˆ nbyrapidityδη). Not all Lorentz transformations can be reached by compoundi ng in- ﬁnitesimal ones. If we take the determinant of eq.(2.5), we g et (det Λ)−1= detΛ, which implies detΛ = ±1. Transformations with detΛ = +1 are proper , and transformations with detΛ = −1 areimproper . Note that the product of any two proper Lorentz transformations is proper , and that inﬁnitesimal transformations of the form Λ = 1 + δωare proper. There- fore, any transformation that can be reached by compounding inﬁnitesimal ones is proper. The proper transformations form a subgroup of the Lorentz group. Another subgroup is that of the orthochronous Lorentz transformations: those for which Λ00≥+1. Note that eq.(2.3) implies (Λ00)2−Λi0Λi0= 1; thus, either Λ00≥+1 or Λ00≤ −1. An inﬁnitesimal transformation is clearly orthochronous, and it is straightforward to show th at the product of two orthochronous transformations is also orthochronou s. Thus, the Lorentz transformations that can be reached by com pounding inﬁnitesimal ones are both proper and orthochronous, and th ey form a subgroup. We can introduce two discrete transformations th at take us out of this subgroup: parity andtime reversal . The parity transformation is Pµν= (P−1)µν= +1 −1 −1 −1 . (2.9) It is orthochronous, but improper. The time-reversal trans formation is Tµν= (T−1)µν= −1 +1 +1 +1 . (2.10) It is nonorthochronous and improper. Generally, when a theory is said to be Lorentz invariant , this means under the proper orthochronous subgroup only. Parity and ti me reversal are treated separately. It is possible for a quantum ﬁeld the ory to be invariant under the proper orthochronous subgroup, but not under parity and/or time-reversal. 2: Lorentz Invariance 32 From here on, in this section, we will treat the proper orthoc hronous subgroup only. Parity and time reversal will be treated in se ction 23. In quantum theory, symmetries are represented by unitary (o r antiu- nitary) operators. This means that we associate a unitary op eratorU(Λ) to each proper, orthochronous Lorentz transformation Λ. Th ese operators must obey the composition rule U(Λ′Λ) =U(Λ′)U(Λ). (2.11) For an inﬁnitesimal transformation, we can write U(1+δω) =I+i 2¯hδωµνMµν, (2.12) whereMµν=−Mνµis a set of hermitian operators called the generators of the Lorentz group . If we start with U(Λ)−1U(Λ′)U(Λ) =U(Λ−1Λ′Λ), let Λ′= 1 +δω′, and expand both sides to linear order in δω, we get δωµνU(Λ)−1MµνU(Λ) =δωµνΛµρΛνσMρσ. (2.13) Then, since δωµνis arbitrary (except for being antisymmetric), the anti- symmetric part of its coeﬃcient on each side must be the same. In this case, because Mµνis already antisymmetric (by deﬁnition), we have U(Λ)−1MµνU(Λ) = ΛµρΛνσMρσ. (2.14) We see that each vector index on Mµνundergoes its own Lorentz trans- formation. This is a general result: any operator carrying o ne or more vector indices should behave similarly. For example, consi der the energy- momentum four-vector Pµ, whereP0is the hamiltonian HandPiare the components of the total three-momentum operator. We expect U(Λ)−1PµU(Λ) = ΛµνPν. (2.15) If we now let Λ = 1 + δωin eq.(2.14), expand to linear order in δω, and equate the antisymmetric part of the coeﬃcients of δωµν, we get the commutation relations [Mµν,Mρσ] =i¯h/parenleftig gµρMνσ−(µ↔ν)/parenrightig −(ρ↔σ). (2.16) These commutation relations specify the Lie algebra of the Lorentz group. We can identify the components of the angular momentum opera torJas Ji≡1 2εijkMjk, and the components of the boost operator KasKi≡Mi0. We then ﬁnd from eq.(2.16) that [Ji,Jj] =i¯hεijkJk, [Ji,Kj] =i¯hεijkKk, [Ki,Kj] =−i¯hεijkJk. (2.17) 2: Lorentz Invariance 33 The ﬁrst of these is the usual set of commutators for angular m omentum, and the second says that Ktransforms as a three-vector under rotations. The third implies that a series of boosts can be equivalent to a rotation. Similarly, we can let Λ = 1 + δωin eq.(2.15) to get [Pµ,Mρσ] =i¯h/parenleftig gµσPρ−(ρ↔σ)/parenrightig , (2.18) which becomes [Ji,H] = 0, [Ji,Pj] =i¯hεijkPk, [Ki,H] =i¯hPi, [Ki,Pj] =i¯hδijH , (2.19) Also, the components of Pµshould commute with each other: [Pi,Pj] = 0, [Pi,H] = 0. (2.20) Together, eqs.(2.17), (2.19), and (2.20) form the Lie algeb ra of the Poincar´ e group. Let us now consider what should happen to a quantum scalar ﬁel dϕ(x) under a Lorentz transformation. We begin by recalling how ti me evolution works in the Heisenberg picture: e+iHt/¯hϕ(x,0)e−iHt/¯h=ϕ(x,t). (2.21) Obviously, this should have a relativistic generalization , e−iPx/¯hϕ(0)e+iPx/¯h=ϕ(x), (2.22) wherePx=Pµxµ=P·x−Hct. We can make this a little fancier by deﬁning the unitary spacetime translation operator T(a)≡exp(−iPµaµ/¯h). (2.23) Then we have T(a)−1ϕ(x)T(a) =ϕ(x−a). (2.24) For an inﬁnitesimal translation, T(δa) =I−i ¯hδaµPµ. (2.25) Comparing eqs.(2.12) and (2.25), we see that eq.(2.24) lead s us to expect U(Λ)−1ϕ(x)U(Λ) =ϕ(Λ−1x). (2.26) 2: Lorentz Invariance 34 Derivatives of ϕthen carry vector indices that transform in the appropriate way, e.g., U(Λ)−1∂µϕ(x)U(Λ) = Λµρ¯∂ρϕ(Λ−1x), (2.27) where the bar on a derivative means that it is with respect to t he argument ¯x= Λ−1x. Eq.(2.27) also implies U(Λ)−1∂2ϕ(x)U(Λ) = ¯∂2ϕ(Λ−1x), (2.28) so that the Klein-Gordon equation, ( −∂2+m2/¯h2c2)ϕ= 0, is Lorentz invariant, as we saw in section 1. Reference Notes A detailed discussion of quantum Lorentz transformations c an be found in Weinberg I . Problems 2.1) Verify that eq.(2.8) follows from eq.(2.3). 2.2) Verify that eq.(2.14) follows from U(Λ)−1U(Λ′)U(Λ) =U(Λ−1Λ′Λ). 2.3) Verify that eq.(2.16) follows from eq.(2.14). 2.4) Verify that eq.(2.17) follows from eq.(2.16). 2.5) Verify that eq.(2.18) follows from eq.(2.15). 2.6) Verify that eq.(2.19) follows from eq.(2.18). 2.7) What property should be attributed to the translation o peratorT(a) that could be used to prove eq.(2.20)? 2.8) a) Let Λ = 1 + δωin eq.(2.26), and show that [ϕ(x),Mµν] =Lµνϕ(x), (2.29) where Lµν≡¯h i(xµ∂ν−xν∂µ). (2.30) b) Show that [[ ϕ(x),Mµν],Mρσ] =LµνLρσϕ(x). c) Prove the Jacobi identity , [[A,B],C] + [[B,C],A] + [[C,A],B] = 0. Hint: write out all the commutators. d) Use your results from parts (b) and (c) to show that [ϕ(x),[Mµν,Mρσ]] = (LµνLρσ− LρσLµν)ϕ(x). (2.31) 2: Lorentz Invariance 35 e) Simplify the right-hand side of eq.(2.31) as much as possi ble. f) Use your results from part (e) to verify eq.(2.16), up to th e possi- bility of a term on the right-hand side that commutes with ϕ(x) and its derivatives. (Such a term, called a central charge , in fact does not arise for the Lorentz algebra.) 2.9) Let us write Λρτ=δρτ+i 2¯hδωµν(Sµν V)ρτ, (2.32) where (Sµν V)ρτ≡¯h i(gµρδντ−gνρδµτ) (2.33) are matrices which constitute the vector representation of the Lorentz generators. a) Let Λ = 1 + δωin eq.(2.27), and show that [∂ρϕ(x),Mµν] =Lµν∂ρϕ(x) + (Sµν V)ρτ∂τϕ(x). (2.34) b) Show that the matrices Sµν Vmust have the same commutation relations as the operators Mµν. Hint: see the previous problem. c) For a rotation by an angle θabout thezaxis, we have Λµν= 1 0 0 0 0 cosθ−sinθ0 0 sinθcosθ0 0 0 0 1 . (2.35) Show that Λ = exp( −iθS12 V/¯h). (2.36) d) For a boost by rapidityηin thezdirection, we have Λµν= coshη0 0 sinh η 0 1 0 0 0 0 1 0 sinhη0 0 cosh η . (2.37) Show that Λ = exp(+iηS30 V/¯h). (2.38) 3: Canonical Quantization of Scalar Fields 36 3Canonical Quantization of Scalar Fields Prerequisite: 2 Let us go back and drastically simplify the hamiltonian we co nstructed in section 1, reducing it to the hamiltonian for free particles : H=/integraldisplay d3xa†(x)/parenleftig −1 2m∇2/parenrightig a(x) =/integraldisplay d3p1 2mp2/tildewidea†(p)/tildewidea(p), (3.1) where /tildewidea(p) =/integraldisplayd3x (2π)3/2e−ip·xa(x). (3.2) Here we have simpliﬁed our notation by setting ¯h= 1. (3.3) The appropriate factors of ¯ hcan always be restored in any of our formulas via dimensional analysis. The commutation (or anticommuta tion) relations of the/tildewidea(p) and/tildewidea†(p) operators are [/tildewidea(p),/tildewidea(p′)]∓= 0, [/tildewidea†(p),/tildewidea†(p′)]∓= 0, [/tildewidea(p),/tildewidea†(p′)]∓=δ3(p−p′), (3.4) where [A,B]∓is either the commutator (if we want a theory of bosons) or the anticommutator (if we want a theory of fermions). Thus /tildewidea†(p) can be interpreted as creating a state of deﬁnite momentum p, and eq.(3.1) describes a theory of free particles. The ground state is the vacuum |0∝an}b∇acket∇i}ht; it is annihilated by /tildewidea(p), /tildewidea(p)|0∝an}b∇acket∇i}ht= 0, (3.5) and so its energy eigenvalue is zero. The other eigenstates o fHare all of the form/tildewidea†(p1).../tildewidea†(pn)|0∝an}b∇acket∇i}ht, and the corresponding energy eigenvalue is E(p1) +...+E(pn), whereE(p) =1 2mp2. It is easy to see how to generalize this theory to a relativist ic one; all we need to do is use the relativistic energy formula E(p) = +( p2c2+m2c4)1/2: H=/integraldisplay d3p(p2c2+m2c4)1/2/tildewidea†(p)/tildewidea(p). (3.6) Now we have a theory of free relativistic spin-zero particles, and they can be either bosons or fermions. 3: Canonical Quantization of Scalar Fields 37 Is this theory really Lorentz invariant? We will answer this question (in the aﬃrmative) in a very roundabout way: by constructing it a gain, from a rather diﬀerent point of view, a point of view that emphasiz es Lorentz invariance from the beginning. We will start with the classical physics of a real scalar ﬁeld ϕ(x).Real means that ϕ(x) assigns a real number to every point in spacetime. Scalar means that Alice [who uses coordinates xµand calls the ﬁeld ϕ(x)] and Bob [who uses coordinates ¯ xµ, related to Alice’s coordinates by ¯ xµ= Λµνxν+aν, and calls the ﬁeld ¯ ϕ(¯x)], agree on the numerical value of the ﬁeld: ϕ(x) = ¯ϕ(¯x). This then implies that the equation of motion for ϕ(x) must be the same as that for ¯ ϕ(¯x). We have already met an equation of this type: the Klein-Gordon equation, (−∂2+m2)ϕ(x) = 0. (3.7) Here we have simpliﬁed our notation by setting c= 1 (3.8) in addition to ¯ h= 1. As with ¯ h, factors of ccan restored, if desired, by dimensional analysis. We will adopt eq.(3.7) as the equation of motion that we would like ϕ(x) to obey. It should be emphasized at this point that we are doi ng classical physics of areal scalar ﬁeld . We are notto think of ϕ(x) as a quantum wave function. Thus, there should not be any factors of ¯hin this version of the Klein-Gordon equation. This means that the pa rameterm must have dimensions of inverse length; misnot(yet) to be thought of as a mass. The equation of motion can be derived from variation of an act ion S=/integraltextdtL, whereLis the lagrangian. Since the Klein-Gordon equation is local, we expect that the lagrangian can be written as the spa ce integral of alagrangian density L:L=/integraltextd3xL. Thus,S=/integraltextd4xL. The integration measured4xis Lorentz invariant: if we change to coordinates ¯ xµ= Λµνxν, we haved4¯x=|detΛ|d4x=d4x. Thus, for the action to be Lorentz in- variant, the lagrangian density must be a Lorentz scalar: L(x) =¯L(¯x). Then we have ¯S=/integraltextd4¯x¯L(¯x) =/integraltextd4xL(x) =S. Any simple function of ϕis a Lorentz scalar, and so are products of derivatives with a ll indices contracted, such as ∂µϕ∂µϕ. We will take for L L=−1 2∂µϕ∂µϕ−1 2m2ϕ2+ Ω0, (3.9) where Ω 0is an arbitrary constant. We ﬁnd the equation motion (also kn own as the Euler-Lagrange equation ) by making an inﬁnitesimal variation δϕ(x) 3: Canonical Quantization of Scalar Fields 38 inϕ(x), and requiring the corresponding variation of the action t o vanish: 0 =δS =/integraldisplay d4x/bracketleftig −1 2∂µδϕ∂µϕ−1 2∂µϕ∂µδϕ−m2ϕδϕ/bracketrightig =/integraldisplay d4x/bracketleftig +∂µ∂µϕ−m2ϕ/bracketrightig δϕ. (3.10) In the last line, we have integrated by parts in each of the ﬁrs t two terms, putting both derivatives on ϕ. We assume δϕ(x) vanishes at inﬁnity in any direction (spatial or temporal), so that there is no surf ace term. Since δϕhas an arbitrary xdependence, eq.(3.10) can be true if and only if (−∂2+m2)ϕ= 0. One solution of the Klein-Gordon equation is a plane wave of t he form exp(ik·x±iωt), where kis an arbitrary real wave-vector, and ω= +(k2+m2)1/2. (3.11) The general solution (assuming boundary conditions that re quireϕto re- main ﬁnite at spatial inﬁnity) is then ϕ(x,t) =/integraldisplayd3k f(k)/bracketleftig a(k)eik·x−iωt+b(k)eik·x+iωt/bracketrightig , (3.12) wherea(k) andb(k) are arbitrary functions of the wave vector k, andf(k) is a redundant function of the magnitude of kwhich we have inserted for later convenience. Note that, ifwe were attempting to interpret ϕ(x) as a quantum wave function (which we most deﬁnitely are not), then the second term would constitute the “negative energy” contrib utions to the wave function. This is because a plane-wave solution of the n onrelativistic Schr¨ odinger equation for a single particle looks like exp( ip·x−iE(p)t), withE(p) =1 2mp2; there is a minus sign in front of the positive energy. We aretrying to interpret eq.(3.12) as a realclassical ﬁeld, but this formula does not generically result in ϕbeing real. We must impose ϕ∗(x) =ϕ(x), where ϕ∗(x,t) =/integraldisplayd3k f(k)/bracketleftig a∗(k)e−ik·x+iωt+b∗(k)e−ik·x−iωt/bracketrightig =/integraldisplayd3k f(k)/bracketleftig a∗(k)e−ik·x+iωt+b∗(−k)e+ik·x−iωt/bracketrightig .(3.13) In the second term on the second line, we have changed the dumm y inte- gration variable from kto−k. Comparing eqs.(3.12) and (3.13), we see 3: Canonical Quantization of Scalar Fields 39 thatϕ∗(x) =ϕ(x) requiresb∗(−k) =a(k). Imposing this condition, we can rewrite ϕas ϕ(x,t) =/integraldisplayd3k f(k)/bracketleftig a(k)eik·x−iωt+a∗(−k)eik·x+iωt/bracketrightig =/integraldisplayd3k f(k)/bracketleftig a(k)eik·x−iωt+a∗(k)e−ik·x+iωt/bracketrightig =/integraldisplayd3k f(k)/bracketleftig a(k)eikx+a∗(k)e−ikx/bracketrightig , (3.14) wherekx=k·x−ωtis the Lorentz-invariant product of the four-vectors xµ= (t,x) andkµ= (ω,k):kx=kµxµ=gµνkµxν. Note that k2=kµkµ=k2−ω2=−m2. (3.15) A four-momentum kµthat obeysk2=−m2is said to be on the mass shell , oron shell for short. It is now convenient to choose f(k) so thatd3k/f(k) is Lorentz invariant. An integration measure that is manifestly invariant under o rthochronous Lorentz transformations is d4kδ(k2+m2)θ(k0), whereδ(x) is the Dirac delta function,θ(x) is the unit step function, and k0is treated as an independent integration variable. We then have /integraldisplay+∞ −∞dk0δ(k2+m2)θ(k0) =1 2ω. (3.16) Here we have used the rule /integraldisplay+∞ −∞dxδ(g(x)) =/summationdisplay i1 |g′(xi)|, (3.17) whereg(x) is any smooth function of xwith simple zeros at x=xi; in our case, the only zero is at k0=ω. Thus we see that if we take f(k)∝ω, thend3k/f(k) will be Lorentz invariant. We will take f(k) = (2π)32ω. It is then convenient to give the corresponding Lorentz-invariant diﬀerential its own name : /tildewiderdk≡d3k (2π)32ω. (3.18) Thus we ﬁnally have ϕ(x) =/integraldisplay /tildewiderdk/bracketleftig a(k)eikx+a∗(k)e−ikx/bracketrightig . (3.19) 3: Canonical Quantization of Scalar Fields 40 We can also invert this formula to get a(k) in terms of ϕ(x). We have /integraldisplay d3xe−ikxϕ(x) =1 2ωa(k) +1 2ωe2iωta∗(−k), /integraldisplay d3xe−ikx∂0ϕ(x) =−i 2a(k) +i 2e2iωta∗(−k). (3.20) We can combine these to get a(k) =/integraldisplay d3xe−ikx/bracketleftig i∂0ϕ(x) +ωϕ(x)/bracketrightig =i/integraldisplay d3xe−ikx↔∂0ϕ(x), (3.21) wheref↔∂µg≡f(∂µg)−(∂µf)g, and∂0ϕ=∂ϕ/∂t = ˙ϕ. Note that a(k) is time independent. Now that we have the lagrangian, we can construct the hamilto nian by the usual rules. Recall that, given a lagrangian L(qi,˙qi) as a function of some coordinates qiand their time derivatives ˙ qi, the conjugate momenta are given by pi=∂L/∂ ˙qi, and the hamiltonian by H=/summationtext ipi˙qi−L. In our case, the role of qi(t) is played by ϕ(x,t), with xplaying the role of a (continuous) index. The appropriate generalizations are t hen Π(x) =∂L ∂˙ϕ(x)(3.22) and H= Π ˙ϕ− L, (3.23) where His the hamiltonian density , and the hamiltonian itself is H=/integraltextd3xH. In our case, we have Π(x) = ˙ϕ(x) (3.24) and H=1 2Π2+1 2(∇ϕ)2+1 2m2ϕ2−Ω0. (3.25) Using eq.(3.19), we can write Hin terms of the a(k) anda∗(k) coeﬃcients: H=−Ω0V+1 2/integraldisplay /tildewiderdk/tildewiderdk′d3x/bracketleftig /parenleftig −iωa(k)eikx+iωa∗(k)e−ikx/parenrightig/parenleftig −iω′a(k′)eik′x+iω′a∗(k′)e−ik′x/parenrightig +/parenleftig +ika(k)eikx−ika∗(k)e−ikx/parenrightig ·/parenleftig +ik′a(k′)eik′x−ik′a∗(k′)e−ik′x/parenrightig +m2/parenleftig a(k)eikx+a∗(k)e−ikx/parenrightig/parenleftig a(k′)eik′x+a∗(k′)e−ik′x/parenrightig/bracketrightig 3: Canonical Quantization of Scalar Fields 41 =−Ω0V+1 2(2π)3/integraldisplay /tildewiderdk/tildewiderdk′/bracketleftig δ3(k−k′)(+ωω′+k·k′+m2) ×/parenleftig a∗(k)a(k′)e−i(ω−ω′)t+a(k)a∗(k′)e+i(ω−ω′)t/parenrightig +δ3(k+k′)(−ωω′−k·k′+m2) ×/parenleftig a(k)a(k′)e−i(ω+ω′)t+a∗(k)a∗(k′)e+i(ω+ω′)t/parenrightig =−Ω0V+1 2/integraldisplay /tildewiderdk1 2ω/bracketleftig (+ω2+k2+m2)/parenleftig a∗(k)a(k) +a(k)a∗(k)/parenrightig + (−ω2+k2+m2)/parenleftig a(k)a(−k)e−2iωt+a∗(k)a∗(−k)e+2iωt/parenrightig/bracketrightig =−Ω0V+1 2/integraldisplay /tildewiderdkω/parenleftig a∗(k)a(k) +a(k)a∗(k)/parenrightig , (3.26) whereVis the volume of space. To get the second equality, we used /integraldisplay d3xeiq·x= (2π)3δ3(q). (3.27) To get the third equality, we integrated over k′, using/tildewiderdk′=d3k′/(2π)32ω′. The last equality then follows from ω= (k2+m2)1/2. Also, we were careful to keep the ordering of a(k) anda∗(k) unchanged throughout, in anticipa- tion of passing to the quantum theory where these classical f unctions will become operators that may not commute. Let us take up the quantum theory now. We can go from classical to quantum mechanics via canonical quantization . This means that we promoteqiandpito operators, with commutation relations [ qi,qj] = 0, [pi,pj] = 0, and [ qi,pj] =i¯hδij. In the Heisenberg picture, these operators should be taken at equal times. In our case, where the “index” is continuous (and we have set ¯ h= 1), we have [ϕ(x,t),ϕ(x′,t)] = 0, [Π(x,t),Π(x′,t)] = 0, [ϕ(x,t),Π(x′,t)] =iδ3(x−x′). (3.28) From these canonical commutation relations , and from eqs.(3.21) and (3.24), we can deduce [a(k),a(k′)] = 0, [a†(k),a†(k′)] = 0, [a(k),a†(k′)] = (2π)32ωδ3(k−k′). (3.29) 3: Canonical Quantization of Scalar Fields 42 We are now denoting a∗(k) asa†(k), sincea†(k) is now the hermitian conjugate (rather than the complex conjugate) of the operat ora(k). We can now rewrite the hamiltonian as H=/integraldisplay /tildewiderdkωa†(k)a(k) + (E0−Ω0)V , (3.30) where E0=1 2(2π)−3/integraldisplay d3kω (3.31) is the total zero-point energy of all the oscillators per uni t volume, and, using eq.(3.27), we have interpreted (2 π)3δ3(0) as the volume of space V. If we integrate in eq.(3.31) over the whole range of k, the value of E0is inﬁnite. If we integrate only up to a maximum value of Λ, a numb er known as the ultraviolet cutoﬀ , we ﬁnd E0=Λ4 16π2, (3.32) where we have assumed Λ ≫m. This is physically justiﬁed if, in the real world, the formalism of quantum ﬁeld theory breaks down at so me large energy scale. For now, we simply note that the value of Ω 0is arbitrary, and so we are free to choose Ω 0=E0. With this choice, the ground state has energy eigenvalue zero. Now, if we like, we can take the limit Λ→ ∞, with no further consequences. (We will meet more of these ultraviolet divergences after we introduce interactions.) The hamiltonian of eq.(3.30) is now the same as that of eq.(3. 6), with a(k) = [(2π)32ω]1/2/tildewidea(k). The commutation relations (3.4) and (3.29) are also equivalent, if we choose commutators (rather than anti commutators) in eq.(3.4). Thus, we have re-derived the hamiltonian of free relativistic bosons by quantization of a scalar ﬁeld whose equation of mot ion is the Klein-Gordon equation. The parameter min the lagrangian is now seen to be the mass of the particle in the quantum theory. (More preci sely, since mhas dimensions of inverse length, the particle mass is ¯ hcm.) What if we want fermions? Then we should use anticommutators in eqs.(3.28) and (3.29). There is a problem, though; eq.(3.26 ) does not then become eq.(3.30). Instead, we get H=−Ω0V, a simple constant. Clearly there is something wrong with using anticommutators. This i s another hint of the spin-statistics theorem, which we will take up in sect ion 4. Next, we would like to add Lorentz-invariant interactions t o our theory. With the formalism we have developed, this is easy to do. Any l ocal func- tion ofϕ(x) is a Lorentz scalar, and so if we add a term like ϕ3orϕ4to the lagrangian density L, the resulting action will still be Lorentz invariant. Now, however, we will have interactions among the particles . Our next task is to deduce the consequences of these interactions. 3: Canonical Quantization of Scalar Fields 43 However, we already have enough tools at our disposal to prov e the spin-statistics theorem for spin-zero particles, and that is what we turn to next. Problems 3.1) Derive eq.(3.29) from eqs.(3.21), (3.24), and (3.28). 3.2) Use the commutation relations, eq.(3.29), to show expl icitly that a state of the form |k1...kn∝an}b∇acket∇i}ht ≡a†(k1)...a†(kn)|0∝an}b∇acket∇i}ht (3.33) is an eigenstate of the hamiltonian, eq.(3.30), with eigenv alueω1+ ...+ωn. The vacuum |0∝an}b∇acket∇i}htis annihilated by a(k),a(k)|0∝an}b∇acket∇i}ht= 0, and we take Ω 0=E0in eq.(3.30). 3.3) UseU(Λ)−1ϕ(x)U(Λ) =ϕ(Λ−1x) to show that U(Λ)−1a(k)U(Λ) =a(Λ−1k), U(Λ)−1a†(k)U(Λ) =a†(Λ−1k), (3.34) and hence that U(Λ)|k1...kn∝an}b∇acket∇i}ht=|Λk1...Λkn∝an}b∇acket∇i}ht, (3.35) where |k1...kn∝an}b∇acket∇i}ht=a†(k1)...a†(kn)|0∝an}b∇acket∇i}htis a state of nparticles with momentak1,...,kn. 3.4) Recall that T(a)−1ϕ(x)T(a) =ϕ(x−a), whereT(a)≡exp(−iPµaµ) is the spacetime translation operator, and P0is identiﬁed as the hamiltonian H. a) Letaµbe inﬁnitesimal, and derive an expression for [ ϕ(x),Pµ]. b) Show that the time component of your result is equivalent t o the Heisenberg equation of motion i˙ϕ= [ϕ,H]. c) For a free ﬁeld, use the Heisenberg equation to derive the K lein- Gordon equation. d) Deﬁne a spatial momentum operator P≡ −/integraldisplay d3xΠ(x)∇ϕ(x). (3.36) Use the canonical commutation relations to show that Pobeys the relation you derived in part (a). e) Express Pin terms of a(k) anda†(k). 3: Canonical Quantization of Scalar Fields 44 3.5) Consider a complex (that is, nonhermitian) scalar ﬁeld ϕwith la- grangian density L=−∂µϕ†∂µϕ−m2ϕ†ϕ+ Ω0. (3.37) a) Show that ϕobeys the Klein-Gordon equation. b) Treatϕandϕ†as independent ﬁelds, and ﬁnd the conjugate mo- mentum for each. Compute the hamiltonian density in terms of these conjugate momenta and the ﬁelds themselves (but not their ti me derivatives). c) Write the mode expansion of ϕas ϕ(x) =/integraldisplay /tildewiderdk/bracketleftig a(k)eikx+b†(k)e−ikx/bracketrightig . (3.38) Expressa(k) andb(k) in terms of ϕandϕ†and their time derivatives. d) Assuming canonical commutation relations for the ﬁelds a nd their conjugate momenta, ﬁnd the commutation relations obeyed by a(k) andb(k) and their hermitian conjugates. e) Express the hamiltonian in terms of a(k) andb(k) and their her- mitian conjugates. What value must Ω 0have in order for the ground state to have zero energy? 4: The Spin-Statistics Theorem 45 4The Spin-Statistics Theorem Prerequisite: 3 Let us consider a theory of free, spin-zero particles speciﬁ ed by the hamil- tonian H0=/integraldisplay /tildewiderdkωa†(k)a(k), (4.1) whereω= (k2+m2)1/2, and either the commutation or anticommutation relations [a(k),a(k′)]∓= 0, [a†(k),a†(k′)]∓= 0, [a(k),a†(k′)]∓= (2π)32ωδ3(k−k′). (4.2) Of course, if we want a theory of bosons, we should use commuta tors, and if we want fermions, we should use anticommutators. Now let us consider adding terms to the hamiltonian that will result in local, Lorentz invariant interactions. In order to do this, it is convenient to deﬁne a nonhermitian ﬁeld, ϕ+(x,0)≡/integraldisplay /tildewiderdkeik·xa(k), (4.3) and its hermitian conjugate ϕ−(x,0)≡/integraldisplay /tildewiderdke−ik·xa†(k). (4.4) These are then time-evolved with H0: ϕ+(x,t) =eiH0tϕ+(x,0)e−iH0t=/integraldisplay /tildewiderdkeikxa(k), ϕ−(x,t) =eiH0tϕ−(x,0)e−iH0t=/integraldisplay /tildewiderdke−ikxa†(k). (4.5) Note that the usual hermitian free ﬁeld ϕ(x) is just the sum of these: ϕ(x) =ϕ+(x) +ϕ−(x). For a proper orthochronous Lorentz transformation Λ, we hav e U(Λ)−1ϕ(x)U(Λ) =ϕ(Λ−1x). (4.6) This implies that the particle creation and annihilation op erators transform as U(Λ)−1a(k)U(Λ) =a(Λ−1k), U(Λ)−1a†(k)U(Λ) =a†(Λ−1k). (4.7) 4: The Spin-Statistics Theorem 46 This, in turn, implies that ϕ+(x) andϕ−(x) are Lorentz scalars: U(Λ)−1ϕ±(x)U(Λ) =ϕ±(Λ−1x). (4.8) We will then have local, Lorentz invariant interactions if w e take the in- teraction lagrangian density L1to be a hermitian function of ϕ+(x) and ϕ−(x). To proceed we need to recall some facts about time-dependent pertur- bation theory in quantum mechanics. The transition amplitu deTf←ito start with an initial state |i∝an}b∇acket∇i}htat timet=−∞and end with a ﬁnal state |f∝an}b∇acket∇i}ht at timet= +∞is Tf←i=∝an}b∇acketle{tf|T exp/bracketleftbigg −i/integraldisplay+∞ −∞dtHI(t)/bracketrightbigg |i∝an}b∇acket∇i}ht, (4.9) whereHI(t) is the perturbing hamiltonian in the interaction picture , HI(t) = exp(+iH0t)H1exp(−iH0t), (4.10) H1is the perturbing hamiltonian in the Schr¨ odinger picture, and T is the time ordering symbol : a product of operators to its right is to be ordered, not as written, but with operators at later times to the left o f those at earlier times. We write H1=/integraltextd3xH1(x,0), and specify H1(x,0) as a hermitian function of ϕ+(x,0) andϕ−(x,0). Then, using eqs.(4.5) and (4.10), we can see that, in the interaction picture, the perturbing ham iltonian density HI(x,t) is simply given by the same function of ϕ+(x,t) andϕ−(x,t). Now we come to the key point: for the transition amplitude Tf←ito be Lorentz invariant, the time ordering must be frame independent . The time ordering of two spacetime points xandx′is frame independent if their separation is timelike ; this means that the interval between them is negative, (x−x′)2<0. Two spacetime points whose separation is spacelike , (x−x′)2>0, can have diﬀerent temporal ordering in diﬀerent frames. I n order to avoid Tf←ibeing diﬀerent in diﬀerent frames, we must then require [HI(x),HI(x′)] = 0 whenever ( x−x′)2>0. (4.11) Obviously, [ ϕ+(x),ϕ+(x′)]∓= [ϕ−(x),ϕ−(x′)]∓= 0. However, [ϕ+(x),ϕ−(x′)]∓=/integraldisplay /tildewiderdk/tildewiderdk′ei(kx−k′x′)[a(k),a†(k′)]∓ =/integraldisplay /tildewiderdkeik(x−x′) =m 4π2rK1(mr) ≡C(r). (4.12) 4: The Spin-Statistics Theorem 47 In the next-to-last line, we have taken ( x−x′)2=r2>0, andK1(z) is a modiﬁed Bessel function. (This Lorentz-invariant integr al is most easily evaluated in the frame where t′=t.) The function C(r) isnotzero for any r>0. (Not even when m= 0; in this case, C(r) = 1/4π2r2.) On the other hand, HI(x) must involve both ϕ+(x) andϕ−(x), by hermiticity. Thus, generically, we will not be able to satisfy eq.(4.11). To resolve this problem, let us try using only particular lin ear combi- nations ofϕ+(x) andϕ−(x). Deﬁne ϕλ(x)≡ϕ+(x) +λϕ−(x), ϕ† λ(x)≡ϕ−(x) +λ∗ϕ+(x), (4.13) whereλis an arbitrary complex number. We then have [ϕλ(x),ϕ† λ(x′)]∓= [ϕ+(x),ϕ−(x′)]∓+|λ|2[ϕ−(x),ϕ+(x′)]∓ = (1∓ |λ|2)C(r) (4.14) and [ϕλ(x),ϕλ(x′)]∓=λ[ϕ+(x),ϕ−(x′)]∓+λ[ϕ−(x),ϕ+(x′)]∓ =λ(1∓1)C(r). (4.15) Thus, if we want ϕλ(x) to either commute or anticommute with both ϕλ(x′) andϕ† λ(x′) at spacelike separations, we must choose |λ|= 1,andwe must choose commutators. Then (and only then), we can build a suit ableHI(x) by making it a hermitian function of ϕλ(x). But this has simply returned us to the theory of a real scalar ﬁ eld, because, for λ=eiα,e−iα/2ϕλ(x) is hermitian. In fact, if we make the replacements a(k)→e+iα/2a(k) anda†(k)→e−iα/2a†(k), then the com- mutation relations of eq.(4.2) are unchanged, and e−iα/2ϕλ(x) =ϕ(x) = ϕ+(x)+ϕ−(x). Thus, our attempt to start with the creation and annihila- tion operators a†(k) anda(k) as the fundamental objects has simply led us back to the real, commuting, scalar ﬁeld ϕ(x) as the fundamental object. Let us return to thinking of ϕ(x) as fundamental, with a lagrangian den- sity given by some function of the Lorentz scalars ϕ(x) and∂µϕ(x)∂µϕ(x). Then, quantization will result in [ ϕ(x),ϕ(x′)]∓= 0 fort=t′. If we choose anticommutators, then [ ϕ(x)]2= 0 and [∂µϕ(x)]2= 0, resulting in a trivial Lthat is at most linear in ϕ, and independent of ˙ ϕ. This clearly does not lead to the correct physics. This situation turns out to generalize to ﬁelds of higher spi n, in any number of spacetime dimensions. One choice of quantization (commuta- tors or anticommutators) always leads to a trivial L, and so this choice 4: The Spin-Statistics Theorem 48 is disallowed. Furthermore, the allowed choice is always co mmutators for ﬁelds of integer spin, and anticommutators for ﬁelds of half -integer spin. If we try treating the particle creation and annihilation op erators as fun- damental, rather than the ﬁelds, we ﬁnd a situation similar t o that of the spin-zero case, and are led to the reconstruction of a ﬁeld th at must obey the appropriate quantization scheme. Reference Notes This discussion of the spin-statistics theorem follows tha t ofWeinberg I , which has more details. Problems 4.1) Verify eq.(4.12). Verify its limit as m→0. 5: The LSZ Reduction Formula 49 5The LSZ Reduction Formula Prerequisite: 3 Let us now consider how to construct appropriate initial and ﬁnal states for scattering experiments. In the free theory, we can creat e a state of one particle by acting on the vacuum state with a creation operat or |k∝an}b∇acket∇i}ht=a†(k)|0∝an}b∇acket∇i}ht, (5.1) where a†(k) =−i/integraldisplay d3xeikx↔∂0ϕ(x). (5.2) The vacuum state |0∝an}b∇acket∇i}htis annihilated by every a(k), a(k)|0∝an}b∇acket∇i}ht= 0, (5.3) and has unit norm, ∝an}b∇acketle{t0|0∝an}b∇acket∇i}ht= 1. (5.4) The one-particle state |k∝an}b∇acket∇i}htthen has the Lorentz-invariant normalization ∝an}b∇acketle{tk|k′∝an}b∇acket∇i}ht= (2π)32ωδ3(k−k′), (5.5) whereω= (k2+m2)1/2. Next, let us deﬁne a time-independent operator that (in the f ree theory) creates a particle localized in momentum space near k1, and localized in position space near the origin: a† 1≡/integraldisplay d3kf1(k)a†(k), (5.6) where f1(k)∝exp[−(k−k1)2/4σ2] (5.7) is an appropriate wave packet, and σis its width in momentum space. Consider the state a† 1|0∝an}b∇acket∇i}ht. If we time evolve this state in the Schr¨ odinger picture, the wave packet will propagate (and spread out). Th e particle is thus localized far from the origin as t→ ±∞ . If we consider instead a state of the form a† 1a† 2|0∝an}b∇acket∇i}ht, where k1∝ne}ationslash=k2, then the two particles are widely separated in the far past. Let us guess that this still works in the interacting theory. One compli- cation is that a†(k) will no longer be time independent, and so a† 1, eq.(5.6), becomes time dependent as well. Our guess for a suitable init ial state of a scattering experiment is then |i∝an}b∇acket∇i}ht= lim t→−∞a† 1(t)a† 2(t)|0∝an}b∇acket∇i}ht. (5.8) 5: The LSZ Reduction Formula 50 By appropriately normalizing the wave packets, we can make ∝an}b∇acketle{ti|i∝an}b∇acket∇i}ht= 1, and we will assume that this is the case. Similarly, we can consid er a ﬁnal state |f∝an}b∇acket∇i}ht= lim t→+∞a† 1′(t)a† 2′(t)|0∝an}b∇acket∇i}ht, (5.9) where k′ 1∝ne}ationslash=k′ 2, and∝an}b∇acketle{tf|f∝an}b∇acket∇i}ht= 1. This describes two widely separated par- ticles in the far future. (We could also consider acting with more creation operators, if we are interested in the production of some ext ra particles in the collision of two.) Now the scattering amplitude is simpl y given by ∝an}b∇acketle{tf|i∝an}b∇acket∇i}ht. We need to ﬁnd a more useful expression for ∝an}b∇acketle{tf|i∝an}b∇acket∇i}ht. To this end, let us note that a† 1(+∞)−a† 1(−∞) =/integraldisplay+∞ −∞dt∂0a† 1(t) =−i/integraldisplay d3kf1(k)/integraldisplay d4x∂0/parenleftig eikx↔ ∂0ϕ(x)/parenrightig =−i/integraldisplay d3kf1(k)/integraldisplay d4xeikx(∂2 0+ω2)ϕ(x) =−i/integraldisplay d3kf1(k)/integraldisplay d4xeikx(∂2 0+k2+m2)ϕ(x) =−i/integraldisplay d3kf1(k)/integraldisplay d4xeikx(∂2 0−← ∇2+m2)ϕ(x) =−i/integraldisplay d3kf1(k)/integraldisplay d4xeikx(∂2 0−→ ∇2+m2)ϕ(x) =−i/integraldisplay d3kf1(k)/integraldisplay d4xeikx(−∂2+m2)ϕ(x).(5.10) The ﬁrst equality is just the fundamental theorem of calculu s. To get the second, we substituted the deﬁnition of a† 1(t), and combined the d3xfrom this deﬁnition with the dtto getd4x. The third comes from straightforward evaluation of the time derivatives. The fourth uses ω2=k2+m2. The ﬁfth writes k2as−∇2acting oneik·x. The sixth uses integration by parts to move the ∇2onto the ﬁeld ϕ(x); here the wave packet is needed to avoid a surface term. The seventh simply identiﬁes ∂2 0− ∇2as−∂2. In free-ﬁeld theory, the right-hand side of eq.(5.10) is zer o, sinceϕ(x) obeys the Klein-Gordon equation. In an interacting theory, with (say) L1=1 6gϕ3, we have instead ( −∂2+m2)ϕ=1 2gϕ2. Thus the right-hand side of eq.(5.10) is not zero in an interacting theory. Rearranging eq.(5.10), we have a† 1(−∞) =a† 1(+∞) +i/integraldisplay d3kf1(k)/integraldisplay d4xeikx(−∂2+m2)ϕ(x).(5.11) We will also need the hermitian conjugate of this formula, wh ich (after a little more rearranging) reads a1(+∞) =a1(−∞) +i/integraldisplay d3kf1(k)/integraldisplay d4xe−ikx(−∂2+m2)ϕ(x).(5.12) 5: The LSZ Reduction Formula 51 Let us return to the scattering amplitude, ∝an}b∇acketle{tf|i∝an}b∇acket∇i}ht=∝an}b∇acketle{t0|a1′(+∞)a2′(+∞)a† 1(−∞)a† 2(−∞)|0∝an}b∇acket∇i}ht. (5.13) Note that the operators are in time order. Thus, if we feel lik e it, we can put in a time-ordering symbol without changing anything: ∝an}b∇acketle{tf|i∝an}b∇acket∇i}ht=∝an}b∇acketle{t0|Ta1′(+∞)a2′(+∞)a† 1(−∞)a† 2(−∞)|0∝an}b∇acket∇i}ht. (5.14) The symbol T means the product of operators to its right is to b e ordered, not as written, but with operators at later times to the left o f those at earlier times. Now let us use eqs.(5.11) and (5.12) in eq.(5.14). The time-o rdering symbol automatically moves all ai′(−∞)’s to the right, where they anni- hilate |0∝an}b∇acket∇i}ht. Similarly, all a† i(+∞)’s move to the left, where they annihilate ∝an}b∇acketle{t0|. The wave packets no longer play a key role, and we can take the σ→0 limit in eq.(5.7), so that f1(k) =δ3(k−k1). The initial and ﬁnal states now have a delta-function normalization, the multiparticl e generalization of eq.(5.5). We are left with ∝an}b∇acketle{tf|i∝an}b∇acket∇i}ht=in+n′/integraldisplay d4x1eik1x1(−∂2 1+m2)... d4x′ 1e−ik′ 1x′ 1(−∂2 1′+m2)... × ∝an}b∇acketle{t0|Tϕ(x1)...ϕ(x′ 1)...|0∝an}b∇acket∇i}ht. (5.15) This formula has been written to apply to the more general cas e ofn incoming particles and n′outgoing particles; the ellipses stand for similar factors for each of the other incoming and outgoing particle s. Eq.(5.15) is the Lehmann-Symanzik-Zimmermann reduction formula , or LSZ formula for short. It is one of the key equations of quan tum ﬁeld theory. However, our derivation of the LSZ formula relied on the supp osition that the creation operators of freeﬁeld theory would work comparably in theinteracting theory. This is a rather suspect assumption, and so we must review it. Let us consider what we can deduce about the energy and moment um eigenstates of the interacting theory on physical grounds. First, we assume that there is a unique ground state |0∝an}b∇acket∇i}ht, with zero energy and momentum. The ﬁrst excited state is a state of a single particle with mas sm. This state can have an arbitrary three-momentum k; its energy is then E= ω= (k2+m2)1/2. The next excited state is that of two particles. These two particles could form a bound state with energy lessthan 2m(like the 5: The LSZ Reduction Formula 52 2m m 0E P Figure 5.1: The exact energy eigenstates in the ( P,E) plane. The ground state is isolated at ( 0,0), the one-particle states form an isolated hyperbola that passes through ( 0,m), and the multi-particle continuum lies at and above the hyperbola that passes through ( 0,2m). hydrogen atom in quantum electrodynamics), but, to keep thi ngs simple, let us assume that there are no such bound states. Then the lowest possible energy of a two-particle state is 2 m. However, a two-particle state with zero total three-momentum can have anyenergy above 2 m, because the two particles could have some relative momentum that contributes to their total energy. Thus we are led to a picture of the states of theo ry as shown in ﬁg.(5.1). Now let us consider what happens when we act on the ground stat e with the ﬁeld operator ϕ(x). To this end, it is helpful to write ϕ(x) = exp( −iPµxµ)ϕ(0)exp(+iPµxµ), (5.16) wherePµis the energy-momentum four-vector. (This equation, intro duced in section 2, is just the relativistic generalization of the Heisenberg equa- tion.) Now let us sandwich ϕ(x) between the ground state (on the right), and other possible states (on the left). For example, let us p ut the ground state on the left as well. Then we have ∝an}b∇acketle{t0|ϕ(x)|0∝an}b∇acket∇i}ht=∝an}b∇acketle{t0|e−iPxϕ(0)e+iPx|0∝an}b∇acket∇i}ht =∝an}b∇acketle{t0|ϕ(0)|0∝an}b∇acket∇i}ht. (5.17) 5: The LSZ Reduction Formula 53 To get the second line, we used Pµ|0∝an}b∇acket∇i}ht= 0. The ﬁnal expression is just a Lorentz-invariant number. Since |0∝an}b∇acket∇i}htis the exact ground state of the interacting theory, we have (in general) no idea what this nu mber is. We would like ∝an}b∇acketle{t0|ϕ(0)|0∝an}b∇acket∇i}htto be zero. This is because we would like a† 1(±∞), when acting on |0∝an}b∇acket∇i}ht, to create a single particle state. We do not wanta† 1(±∞) to create a linear combination of a single particle state an d the ground state. But this is precisely what will happen if ∝an}b∇acketle{t0|ϕ(0)|0∝an}b∇acket∇i}htis not zero. So, ifv≡ ∝an}b∇acketle{t0|ϕ(0)|0∝an}b∇acket∇i}htis not zero, we will shift the ﬁeld ϕ(x) by the constantv. This means that we go back to the lagrangian, and replace ϕ(x) everywhere by ϕ(x) +v. This is just a change of the name of the operator of interest, and does not aﬀect the physics. Howeve r, the shifted ϕ(x) obeys, by construction, ∝an}b∇acketle{t0|ϕ(x)|0∝an}b∇acket∇i}ht= 0. Let us now consider ∝an}b∇acketle{tp|ϕ(x)|0∝an}b∇acket∇i}ht, where |p∝an}b∇acket∇i}htis a one-particle state with four-momentum p, normalized according to eq.(5.5). Again using eq.(5.16), we have ∝an}b∇acketle{tp|ϕ(x)|0∝an}b∇acket∇i}ht=∝an}b∇acketle{tp|e−iPxϕ(0)e+iPx|0∝an}b∇acket∇i}ht =e−ipx∝an}b∇acketle{tp|ϕ(0)|0∝an}b∇acket∇i}ht, (5.18) where ∝an}b∇acketle{tp|ϕ(0)|0∝an}b∇acket∇i}htis a Lorentz-invariant number. It is a function of p, but the only Lorentz-invariant functions of pare functions of p2, andp2is just the constant −m2. So∝an}b∇acketle{tp|ϕ(0)|0∝an}b∇acket∇i}htis just some number that depends on m and (presumably) the other parameters in the lagrangian. We would like ∝an}b∇acketle{tp|ϕ(0)|0∝an}b∇acket∇i}htto be one. That is what it is in free-ﬁeld theory, and we know that, in free-ﬁeld theory, a† 1(±∞) creates a correctly normal- ized one-particle state. Thus, for a† 1(±∞) to create a correctly normalized one-particle state in the interacting theory, we must have ∝an}b∇acketle{tp|ϕ(0)|0∝an}b∇acket∇i}ht= 1. So, if∝an}b∇acketle{tp|ϕ(0)|0∝an}b∇acket∇i}htis not equal to one, we will rescale (or, one might say, renormalize )ϕ(x) by a multiplicative constant. This is just a change of the name of the operator of interest, and does not aﬀect the physi cs. However, the rescaled ϕ(x) obeys, by construction, ∝an}b∇acketle{tp|ϕ(0)|0∝an}b∇acket∇i}ht= 1. Finally, consider ∝an}b∇acketle{tp,n|ϕ(x)|0∝an}b∇acket∇i}ht, where |p,n∝an}b∇acket∇i}htis a multiparticle state with total four-momentum p, andnis short for all other labels (such as relative momenta) needed to specify this state. We have ∝an}b∇acketle{tp,n|ϕ(x)|0∝an}b∇acket∇i}ht=∝an}b∇acketle{tp,n|e−iPxϕ(0)e+iPx|0∝an}b∇acket∇i}ht =e−ipx∝an}b∇acketle{tp,n|ϕ(0)|0∝an}b∇acket∇i}ht =e−ipxAn(p), (5.19) whereAn(p) is a function of Lorentz invariant products of the various (relative and total) four-momenta needed to specify the sta te. Note that, 5: The LSZ Reduction Formula 54 from ﬁg.(5.1), p0= (p2+M2)1/2withM≥2m. The invariant mass Mis one of the parameters included in the set n. We would like ∝an}b∇acketle{tp,n|ϕ(x)|0∝an}b∇acket∇i}htto be zero, because we would like a† 1(±∞), when acting on |0∝an}b∇acket∇i}ht, to create a single particle state. We do notwant a† 1(±∞) to create any multiparticle states. But this is precisely w hat may happen if ∝an}b∇acketle{tp,n|ϕ(x)|0∝an}b∇acket∇i}htis not zero. Actually, we are being a little too strict. We really need ∝an}b∇acketle{tp,n|a† 1(±∞)|0∝an}b∇acket∇i}ht to be zero, and perhaps it will be zero even if ∝an}b∇acketle{tp,n|ϕ(x)|0∝an}b∇acket∇i}htis not. Also, we really should test a† 1(±∞)|0∝an}b∇acket∇i}htonly against normalizable states. Mathemat- ically, non-normalizable states cause all sorts of trouble ; mathematicians don’t consider them to be states at all. In physics, this usua lly doesn’t bother us, but here we must be especially careful. So let us wr ite |ψ∝an}b∇acket∇i}ht=/summationdisplay n/integraldisplay d3pψn(p)|p,n∝an}b∇acket∇i}ht, (5.20) where theψn(p)’s are wave packets for the total three-momentum p. Note that eq.(5.20) is highly schematic; the sum over nincludes integrals over continuous parameters like relative momenta. Now we want to examine ∝an}b∇acketle{tψ|a† 1(t)|0∝an}b∇acket∇i}ht=−i/summationdisplay n/integraldisplay d3pψ∗ n(p)/integraldisplay d3kf1(k)/integraldisplay d3xeikx↔∂0∝an}b∇acketle{tp,n|ϕ(x)|0∝an}b∇acket∇i}ht. (5.21) We will take the limit t→ ±∞ in a moment. Using eq.(5.19), eq.(5.21) becomes ∝an}b∇acketle{tψ|a† 1(t)|0∝an}b∇acket∇i}ht=−i/summationdisplay n/integraldisplay d3pψ∗ n(p)/integraldisplay d3kf1(k)/integraldisplay d3x/parenleftig eikx↔ ∂0e−ipx/parenrightig An(p) =/summationdisplay n/integraldisplay d3pψ∗ n(p)/integraldisplay d3kf1(k)/integraldisplay d3x(p0+k0)ei(k−p)xAn(p). (5.22) Next we use/integraltextd3xei(k−p)·x= (2π)3δ3(k−p) to get ∝an}b∇acketle{tψ|a† 1(t)|0∝an}b∇acket∇i}ht=/summationdisplay n/integraldisplay d3p(2π)3(p0+k0)ψ∗ n(p)f1(p)An(p)ei(p0−k0)t,(5.23) wherep0= (p2+M2)1/2andk0= (p2+m2)1/2. Now comes the key point. Note that p0is strictly greater than k0, becauseM≥2m > m . Thus the integrand of eq.(5.23) contains a phase factor that oscillates more and more rapidly as t→ ±∞ . Therefore, by theRiemann-Lebesgue lemma , the right-hand side of eq.(5.23) vanishes as t→ ±∞ . 5: The LSZ Reduction Formula 55 Physically, this means that a one-particle wave packet spre ads out dif- ferently than a multiparticle wave packet, and the overlap b etween them goes to zero as the elapsed time goes to inﬁnity. Thus, even th ough our operatora† 1(t) creates some multiparticle states that we don’t want, we can “follow” the one-particle state that we do want by using a n appropri- ate wave packet. By waiting long enough, we can make the multi particle contribution to the scattering amplitude as small as we like . Let us recap. The basic formula for a scattering amplitude in terms of the ﬁelds of an interacting quantum ﬁeld theory is the LSZ for mula, which is worth writing down again: ∝an}b∇acketle{tf|i∝an}b∇acket∇i}ht=in+n′/integraldisplay d4x1eik1x1(−∂2 1+m2)... d4x1′e−ik′ 1x′ 1(−∂2 1′+m2)... × ∝an}b∇acketle{t0|Tϕ(x1)...ϕ(x′ 1)...|0∝an}b∇acket∇i}ht. (5.24) The LSZ formula is valid provided that the ﬁeld obeys ∝an}b∇acketle{t0|ϕ(x)|0∝an}b∇acket∇i}ht= 0 and ∝an}b∇acketle{tk|ϕ(x)|0∝an}b∇acket∇i}ht=e−ikx. (5.25) These normalization conditions may conﬂict with our origin al choice of ﬁeld and parameter normalization in the lagrangian. Consider, f or example, a lagrangian originally speciﬁed as L=−1 2∂µϕ∂µϕ−1 2m2ϕ2+1 6gϕ3. (5.26) After shifting and rescaling (and renaming some parameters ), we will have instead L=−1 2Zϕ∂µϕ∂µϕ−1 2Zmm2ϕ2+1 6Zggϕ3+Yϕ. (5.27) Here the three Z’s andYare as yet unknown constants. They must be chosen to ensure the validity of eq.(5.25); this gives us two conditions in four unknowns. We ﬁx the parameter mby requiring it to be equal to the actual mass of the particle (equivalently, the energy of the ﬁrst excited state relative to the ground state), and we ﬁx the parameter gby requiring some particular scattering cross section to depend on gin some particular way. (For example, in quantum electrodynamics, the parameter an alogous tog is the electron charge e. The low-energy Coulomb scattering cross section is proportional to e4, with a deﬁnite constant of proportionality and no higher-order corrections; this relationship deﬁnes e.) Thus we have four conditions in four unknowns, and it is possible to calculate Yand the three Z’s order by order in powers of g. Next, we must develop the tools needed to compute the correla tion functions ∝an}b∇acketle{t0|Tϕ(x1)...|0∝an}b∇acket∇i}htin an interacting quantum ﬁeld theory. 5: The LSZ Reduction Formula 56 Reference Notes Useful discussions of the LSZ reduction formula can be found inBrown , Itzykson & Zuber ,Peskin & Schroeder , andWeinberg I . Problems 5.1) Work out the LSZ reduction formula for the complex scala r ﬁeld that was introduced in problem 3.5. Note that we must specify the t ype (aorb) of each incoming and outgoing particle. 6: Path Integrals in Quantum Mechanics 57 6Path Integrals in Quantum Mechanics Prerequisite: none Consider the nonrelativistic quantum mechanics of one part icle in one di- mension; the hamiltonian is H(P,Q) =1 2mP2+V(Q), (6.1) wherePandQare operators obeying [ Q,P] =i. (We set ¯ h= 1 for notational convenience.) We wish to evaluate the probabili ty amplitude for the particle to start at position q′at timet′, and end at position q′′at time t′′. This amplitude is ∝an}b∇acketle{tq′′|e−iH(t′′−t′)|q′∝an}b∇acket∇i}ht, where |q′∝an}b∇acket∇i}htand|q′′∝an}b∇acket∇i}htare eigenstates of the position operator Q. We can also formulate this question in the Heisenberg pictur e, where op- erators are time dependent and the state of the system is time independent, as opposed to the more familiar Schr¨ odinger picture. In the Heisenberg pic- ture, we write Q(t) =eiHtQe−iHt. We can then deﬁne an instantaneous eigenstate of Q(t) viaQ(t)|q,t∝an}b∇acket∇i}ht=q|q,t∝an}b∇acket∇i}ht. These instantaneous eigenstates can be expressed explicitly as |q,t∝an}b∇acket∇i}ht=e+iHt|q∝an}b∇acket∇i}ht, whereQ|q∝an}b∇acket∇i}ht=q|q∝an}b∇acket∇i}ht. Then our transition amplitude can be written as ∝an}b∇acketle{tq′′,t′′|q′,t′∝an}b∇acket∇i}htin the Heisenberg picture. To evaluate ∝an}b∇acketle{tq′′,t′′|q′,t′∝an}b∇acket∇i}ht, we begin by dividing the time interval T≡ t′′−t′intoN+ 1 equal pieces of duration δt=T/(N+ 1). Then introduce Ncomplete sets of position eigenstates to get ∝an}b∇acketle{tq′′,t′′|q′,t′∝an}b∇acket∇i}ht=/integraldisplayN/productdisplay j=1dqj∝an}b∇acketle{tq′′|e−iHδt|qN∝an}b∇acket∇i}ht∝an}b∇acketle{tqN|e−iHδt|qN−1∝an}b∇acket∇i}ht...∝an}b∇acketle{tq1|e−iHδt|q′∝an}b∇acket∇i}ht. (6.2) The integrals over the q’s all run from −∞to +∞. Now consider ∝an}b∇acketle{tq2|e−iHδt|q1∝an}b∇acket∇i}ht. We can use the Campbell-Baker-Hausdorf formula exp(A+B) = exp(A)exp(B)exp(−1 2[A,B] +...) (6.3) to write exp(−iHδt) = exp[ −i(δt/2m)P2]exp[−iδtV(Q)]exp[O(δt2)].(6.4) Then, in the limit of small δt, we should be able to ignore the ﬁnal expo- nential. Inserting a complete set of momentum states then gi ves ∝an}b∇acketle{tq2|e−iHδt|q1∝an}b∇acket∇i}ht=/integraldisplay dp1∝an}b∇acketle{tq2|e−i(δt/2m)P2|p1∝an}b∇acket∇i}ht∝an}b∇acketle{tp1|e−iδtV(Q)|q1∝an}b∇acket∇i}ht =/integraldisplay dp1e−i(δt/2m)p2 1e−iδtV(q1)∝an}b∇acketle{tq2|p1∝an}b∇acket∇i}ht∝an}b∇acketle{tp1|q1∝an}b∇acket∇i}ht 6: Path Integrals in Quantum Mechanics 58 =/integraldisplaydp1 2πe−i(δt/2m)p2 1e−iδtV(q1)eip1(q2−q1). =/integraldisplaydp1 2πe−iH(p1,q1)δteip1(q2−q1). (6.5) To get the third line, we used ∝an}b∇acketle{tq|p∝an}b∇acket∇i}ht= (2π)−1/2exp(ipq). If we happen to be interested in more general hamiltonians th an eq.(6.1), then we must worry about the ordering of the PandQoperators in any term that contains both. If we adopt Weyl ordering , where the quantum hamiltonian H(P,Q) is given in terms of the classical hamiltonian H(p,q) by H(P,Q)≡/integraldisplaydx 2πdk 2πeixP+ikQ/integraldisplay dpdqe−ixp−ikqH(p,q),(6.6) then eq.(6.5) is not quite correct; in the last line, H(p1,q1) should be replaced with H(p1,¯q1), where ¯q1=1 2(q1+q2). For the hamiltonian of eq.(6.1), which is Weyl ordered, this replacement makes no d iﬀerence in the limitδt→0. Adopting Weyl ordering for the general case, we now have ∝an}b∇acketle{tq′′,t′′|q′,t′∝an}b∇acket∇i}ht=/integraldisplayN/productdisplay k=1dqkN/productdisplay j=0dpj 2πeipj(qj+1−qj)e−iH(pj,¯qj)δt, (6.7) where ¯qj=1 2(qj+qj+1),q0=q′, andqN+1=q′′. If we now deﬁne ˙ qj≡ (qj+1−qj)/δt, and take the formal limit of δt→0, we get ∝an}b∇acketle{tq′′,t′′|q′,t′∝an}b∇acket∇i}ht=/integraldisplay DqDpexp/bracketleftbigg i/integraldisplayt′′ t′dt/parenleftig p(t) ˙q(t)−H(p(t),q(t))/parenrightig/bracketrightbigg .(6.8) The integration is to be understood as over all paths in phase space that start atq(t′) =q′(with an arbitrary value of the initial momentum) and end atq(t′′) =q′′(with an arbitrary value of the ﬁnal momentum). IfH(p,q) is no more than quadratic in the momenta [as is the case for eq.(6.1)], then the integral over pis gaussian, and can be done in closed form. If the term that is quadratic in pis independent of q[as is the case for eq.(6.1)], then the prefactors generated by the gaussia n integrals are all constants, and can be absorbed into the deﬁnition of Dq. The result of integrating out pis then ∝an}b∇acketle{tq′′,t′′|q′,t′∝an}b∇acket∇i}ht=/integraldisplay Dqexp/bracketleftigg i/integraldisplayt′′ t′dtL( ˙q(t),q(t))/bracketrightigg , (6.9) whereL( ˙q,q) is computed by ﬁrst ﬁnding the stationary point of the p integral by solving 0 =∂ ∂p/parenleftig p˙q−H(p,q)/parenrightig = ˙q−∂H(p,q) ∂p(6.10) 6: Path Integrals in Quantum Mechanics 59 forpin terms of ˙ qandq, and then plugging this solution back into p˙q−H to getL. We recognize this procedure from classical mechanics: we a re passing from the hamiltonian formulation to the lagrangian formulation. Now that we have eqs.(6.8) and (6.9), what are we going to do wi th them? Let us begin by considering some generalizations; let us examine, for example, ∝an}b∇acketle{tq′′,t′′|Q(t1)|q′,t′∝an}b∇acket∇i}ht, wheret′<t1<t′′. This is given by ∝an}b∇acketle{tq′′,t′′|Q(t1)|q′,t′∝an}b∇acket∇i}ht=∝an}b∇acketle{tq′′|e−iH(t′′−t1)Qe−iH(t1−t′)|q′∝an}b∇acket∇i}ht. (6.11) In the path integral formula, the extra operator Qinserted at time t1will simply result in an extra factor of q(t1). Thus ∝an}b∇acketle{tq′′,t′′|Q(t1)|q′,t′∝an}b∇acket∇i}ht=/integraldisplay DpDqq(t1)eiS, (6.12) whereS=/integraltextt′′ t′dt(p˙q−H). Now let us go in the other direction; consider/integraltextDpDqq(t1)q(t2)eiS. This clearly requires the operators Q(t1) andQ(t2), but their order depends on whether t1<t2ort2<t1. Thus we have /integraldisplay DpDqq(t1)q(t2)eiS=∝an}b∇acketle{tq′′,t′′|TQ(t1)Q(t2)|q′,t′∝an}b∇acket∇i}ht. (6.13) where T is the time ordering symbol : a product of operators to its right is to be ordered, not as written, but with operators at later tim es to the left of those at earlier times. This is signiﬁcant, because time- ordered products enter into the LSZ formula for scattering amplitudes. To further develop these methods, we need another trick: functional derivatives . We deﬁne the functional derivative δ/δf(t) via δ δf(t1)f(t2) =δ(t1−t2), (6.14) whereδ(t) is the Dirac delta function. Also, functional derivatives are deﬁned to satisfy all the usual rules of derivatives (produc t rule, chain rule, etc). Eq.(6.14) can be thought of as the continuous gen eralization of (∂/∂xi)xj=δij. Now, consider modifying the lagrangian of our theory by incl uding ex- ternal forces acting on the particle: H(p,q)→H(p,q)−f(t)q(t)−h(t)p(t), (6.15) wheref(t) andh(t) are speciﬁed functions. In this case we will write ∝an}b∇acketle{tq′′,t′′|q′,t′∝an}b∇acket∇i}htf,h=/integraldisplay DpDqexp/bracketleftigg i/integraldisplayt′′ t′dt/parenleftig p˙q−H+fq+hp/parenrightig/bracketrightigg .(6.16) 6: Path Integrals in Quantum Mechanics 60 whereHis the original hamiltonian. Then we have 1 iδ δf(t1)∝an}b∇acketle{tq′′,t′′|q′,t′∝an}b∇acket∇i}htf,h=/integraldisplay DpDqq(t1)ei/integraltext dt[p˙q−H+fq+hp], 1 iδ δf(t1)1 iδ δf(t2)∝an}b∇acketle{tq′′,t′′|q′,t′∝an}b∇acket∇i}htf,h=/integraldisplay DpDqq(t1)q(t2)ei/integraltext dt[p˙q−H+fq+hp], 1 iδ δh(t1)∝an}b∇acketle{tq′′,t′′|q′,t′∝an}b∇acket∇i}htf,h=/integraldisplay DpDqp(t1)ei/integraltext dt[p˙q−H+fq+hp], (6.17) and so on. After we are done bringing down as many factors of q(ti) or p(ti) as we like, we can set f(t) =h(t) = 0, and return to the original hamiltonian. Thus, ∝an}b∇acketle{tq′′,t′′|TQ(t1)...P(tn)...|q′,t′∝an}b∇acket∇i}ht =1 iδ δf(t1)...1 iδ δh(tn)...∝an}b∇acketle{tq′′,t′′|q′,t′∝an}b∇acket∇i}htf,h/vextendsingle/vextendsingle/vextendsingle/vextendsingle f=h=0.(6.18) Suppose we are also interested in initial and ﬁnal states oth er than position eigenstates. Then we must multiply by the wave func tions for these states, and integrate. We will be interested, in parti cular, in the ground state as both the initial and ﬁnal state. Also, we will take the limitst′→ −∞ andt′′→+∞. The object of our attention is then ∝an}b∇acketle{t0|0∝an}b∇acket∇i}htf,h= lim t′→−∞ t′′→+∞/integraldisplay dq′′dq′ψ∗ 0(q′′)∝an}b∇acketle{tq′′,t′′|q′,t′∝an}b∇acket∇i}htf,hψ0(q′), (6.19) whereψ0(q) =∝an}b∇acketle{tq|0∝an}b∇acket∇i}htis the ground-state wave function. Eq.(6.19) is a rather cumbersome formula, however. We will, therefore, em ploy a trick to simplify it. Let|n∝an}b∇acket∇i}htdenote an eigenstate of Hwith eigenvalue En. We will suppose thatE0= 0; if this is not the case, we will shift Hby an appropriate constant. Next we write |q′,t′∝an}b∇acket∇i}ht=eiHt′|q′∝an}b∇acket∇i}ht =∞/summationdisplay n=0eiHt′|n∝an}b∇acket∇i}ht∝an}b∇acketle{tn|q′∝an}b∇acket∇i}ht =∞/summationdisplay n=0ψ∗ n(q′)eiEnt′|n∝an}b∇acket∇i}ht, (6.20) whereψn(q) =∝an}b∇acketle{tq|n∝an}b∇acket∇i}htis the wave function of the nth eigenstate. Now, replaceHwith (1 −iǫ)Hin eq.(6.20), where ǫis a small positive inﬁnites- imal. Then, take the limit t′→ −∞ of eq.(6.20) with ǫheld ﬁxed. Every 6: Path Integrals in Quantum Mechanics 61 state except the ground state is then multiplied by a vanishi ng exponential factor, and so the limit is simply ψ∗ 0(q′)|0∝an}b∇acket∇i}ht. Next, multiply by an arbi- trary function χ(q′), and integrate over q′. The only requirement is that ∝an}b∇acketle{t0|χ∝an}b∇acket∇i}ht ∝ne}ationslash= 0. We then have a constant times |0∝an}b∇acket∇i}ht, and this constant can be absorbed into the normalization of the path integral. A simi lar analysis of ∝an}b∇acketle{tq′′,t′′|=∝an}b∇acketle{tq′′|e−iHt′′shows that the replacement H→(1−iǫ)Halso picks out the ground state as the ﬁnal state in the t′′→+∞limit. What all this means is that if we use (1 −iǫ)Hinstead ofH, we can be cavalier about the boundary conditions on the endpoints of t he path. Any reasonable boundary conditions will result in the ground st ate as both the initial and ﬁnal state. Thus we have ∝an}b∇acketle{t0|0∝an}b∇acket∇i}htf,h=/integraldisplay DpDqexp/bracketleftbigg i/integraldisplay+∞ −∞dt/parenleftig p˙q−(1−iǫ)H+fq+hp/parenrightig/bracketrightbigg .(6.21) Now let us suppose that H=H0+H1, where we can solve for the eigenstates and eigenvalues of H0, andH1can be treated as a perturbation. Suppressing the iǫ, eq.(6.21) can be written as ∝an}b∇acketle{t0|0∝an}b∇acket∇i}htf,h=/integraldisplay DpDqexp/bracketleftbigg i/integraldisplay+∞ −∞dt/parenleftig p˙q−H0(p,q)−H1(p,q) +fq+hp/parenrightig/bracketrightbigg = exp/bracketleftbigg −i/integraldisplay+∞ −∞dtH1/parenleftbigg1 iδ δh(t),1 iδ δf(t)/parenrightbigg/bracketrightbigg ×/integraldisplay DpDqexp/bracketleftbigg i/integraldisplay+∞ −∞dt/parenleftig p˙q−H0(p,q) +fq+hp/parenrightig/bracketrightbigg .(6.22) To understand the second line of this equation, take the expo nential prefac- tor inside the path integral. Then the functional derivativ es (that appear as the arguments of H1) just pull out appropriate factors of p(t) andq(t), generating the right-hand side of the ﬁrst line. We assume th at we can compute the functional integral in the second line, since it involves only the solvable hamiltonian H0. The exponential prefactor can then be ex- panded in powers of H1to generate a perturbation series. IfH1depends only on q(and not on p), and if we are only interested in time-ordered products of Q’s (and not P’s), and ifHis no more than quadratic in P, and if the term quadratic in Pdoes not involve Q,then eq.(6.22) can be simpliﬁed to ∝an}b∇acketle{t0|0∝an}b∇acket∇i}htf= exp/bracketleftbigg i/integraldisplay+∞ −∞dtL1/parenleftbigg1 iδ δf(t)/parenrightbigg/bracketrightbigg ×/integraldisplay Dqexp/bracketleftbigg i/integraldisplay+∞ −∞dt/parenleftig L0( ˙q,q) +fq/parenrightig/bracketrightbigg . (6.23) whereL1(q) =−H1(q). 6: Path Integrals in Quantum Mechanics 62 Reference Notes Brown andRamond I have especially clear treatments of various aspects of path integrals. For a careful derivation of the midpoint rule of eq.(6.7), seeBerry & Mount . Problems 6.1) a) Find an explicit formula for Dqin eq.(6.9). Your formula should be of the form Dq=C/producttextN j=1dqj, whereCis a constant that you should compute. b) For the case of a free particle, V(Q) = 0, evaluate the path integral of eq.(6.9) explicitly. Hint: integrate over q1, thenq2, etc, and look for a pattern. Express you ﬁnal answer in terms of q′,t′,q′′,t′′, and m. Restore ¯hby dimensional analysis. c) Compute ∝an}b∇acketle{tq′′,t′′|q′,t′∝an}b∇acket∇i}ht=∝an}b∇acketle{tq′′|e−iH(t′′−t′)|q′∝an}b∇acket∇i}htby inserting a complete set of momentum eigenstates, and performing the integral ov er the momentum. Compare with your result in part (b). 7: The Path Integral for the Harmonic Oscillator 63 7The Path Integral for the Harmonic Oscillator Prerequisite: 6 Consider a harmonic oscillator with hamiltonian H(P,Q) =1 2mP2+1 2mω2Q2. (7.1) We begin with the formula from section 6 for the ground state t o ground state transition amplitude in the presence of an external fo rce, specialized to the case of a harmonic oscillator: ∝an}b∇acketle{t0|0∝an}b∇acket∇i}htf=/integraldisplay DpDqexpi/integraldisplay+∞ −∞dt/bracketleftig p˙q−(1−iǫ)H+fq/bracketrightig . (7.2) Looking at eq.(7.1), we see that multiplying Hby 1−iǫis equivalent to the replacements m−1→(1−iǫ)m−1[or, equivalently, m→(1+iǫ)m] and mω2→(1−iǫ)mω2. Passing to the lagrangian formulation then gives ∝an}b∇acketle{t0|0∝an}b∇acket∇i}htf=/integraldisplay Dqexpi/integraldisplay+∞ −∞dt/bracketleftig 1 2(1+iǫ)m˙q2−1 2(1−iǫ)mω2q2+fq/bracketrightig .(7.3) From now on, we will simplify the notation by setting m= 1. Next, let us use Fourier-transformed variables, /tildewideq(E) =/integraldisplay+∞ −∞dteiEtq(t), q (t) =/integraldisplay+∞ −∞dE 2πe−iEt/tildewideq(E). (7.4) The expression in square brackets in eq.(7.3) becomes /bracketleftig ···/bracketrightig =1 2/integraldisplay+∞ −∞dE 2πdE′ 2πe−i(E+E′)t/bracketleftig/parenleftig −(1+iǫ)EE′−(1−iǫ)ω2/parenrightig /tildewideq(E)/tildewideq(E′) +/tildewidef(E)/tildewideq(E′) +/tildewidef(E′)/tildewideq(E)/bracketrightig . (7.5) Note that the only tdependence is now in the prefactor. Integrating over t then generates a factor of 2 πδ(E+E′). Then we can easily integrate over E′to get S=/integraldisplay+∞ −∞dt/bracketleftig ···/bracketrightig =1 2/integraldisplay+∞ −∞dE 2π/bracketleftig/parenleftig (1+iǫ)E2−(1−iǫ)ω2/parenrightig /tildewideq(E)/tildewideq(−E) +/tildewidef(E)/tildewideq(−E) +/tildewidef(−E)/tildewideq(E)/bracketrightig . (7.6) 7: The Path Integral for the Harmonic Oscillator 64 The factor in large parentheses is equal to E2−ω2+i(E2+ω2)ǫ, and we can absorb the positive coeﬃcient into ǫto getE2−ω2+iǫ. Now it is convenient to change integration variables to /tildewidex(E) =/tildewideq(E) +/tildewidef(E) E2−ω2+iǫ. (7.7) Then we get S=1 2/integraldisplay+∞ −∞dE 2π/bracketleftigg /tildewidex(E)(E2−ω2+iǫ)/tildewidex(−E)−/tildewidef(E)/tildewidef(−E) E2−ω2+iǫ/bracketrightigg .(7.8) Furthermore, because eq.(7.7) is just a shift by a constant, Dq=Dx. Now we have ∝an}b∇acketle{t0|0∝an}b∇acket∇i}htf= exp/bracketleftigg i 2/integraldisplay+∞ −∞dE 2π/tildewidef(E)/tildewidef(−E) −E2+ω2−iǫ/bracketrightigg ×/integraldisplay Dxexp/bracketleftbiggi 2/integraldisplay+∞ −∞dE 2π/tildewidex(E)(E2−ω2+iǫ)/tildewidex(−E)/bracketrightbigg .(7.9) Now comes the key point. The path integral on the second line o f eq.(7.9) is what we get for ∝an}b∇acketle{t0|0∝an}b∇acket∇i}htfin the case f= 0. On the other hand, if there is no external force, a system in its ground state wil l remain in its ground state, and so ∝an}b∇acketle{t0|0∝an}b∇acket∇i}htf=0= 1. Thus ∝an}b∇acketle{t0|0∝an}b∇acket∇i}htfis given by the ﬁrst line of eq.(7.9), ∝an}b∇acketle{t0|0∝an}b∇acket∇i}htf= exp/bracketleftigg i 2/integraldisplay+∞ −∞dE 2π/tildewidef(E)/tildewidef(−E) −E2+ω2−iǫ/bracketrightigg . (7.10) We can also rewrite ∝an}b∇acketle{t0|0∝an}b∇acket∇i}htfin terms of time-domain variables as ∝an}b∇acketle{t0|0∝an}b∇acket∇i}htf= exp/bracketleftbiggi 2/integraldisplay+∞ −∞dtdt′f(t)G(t−t′)f(t′)/bracketrightbigg , (7.11) where G(t−t′) =/integraldisplay+∞ −∞dE 2πe−iE(t−t′) −E2+ω2−iǫ. (7.12) Note thatG(t−t′) is a Green’s function for the oscillator equation of motion : /parenleftigg ∂2 ∂t2+ω2/parenrightigg G(t−t′) =δ(t−t′). (7.13) This can be seen directly by plugging eq.(7.12) into eq.(7.1 3) and then taking theǫ→0 limit. We can also evaluate G(t−t′) explicitly by treating the integral over Eon the right-hand side of eq.(7.12) as a contour integral 7: The Path Integral for the Harmonic Oscillator 65 in the complex Eplane, and then evaluating it via the residue theorem. The result is G(t−t′) =i 2ωexp/parenleftig −iω|t−t′|/parenrightig . (7.14) Consider now the formula from section 6 for the time-ordered product of operators. In the case of initial and ﬁnal ground states, i t becomes ∝an}b∇acketle{t0|TQ(t1)...|0∝an}b∇acket∇i}ht=1 iδ δf(t1)...∝an}b∇acketle{t0|0∝an}b∇acket∇i}htf/vextendsingle/vextendsingle/vextendsingle f=0. (7.15) Using our explicit formula, eq.(7.11), we have ∝an}b∇acketle{t0|TQ(t1)Q(t2)|0∝an}b∇acket∇i}ht=1 iδ δf(t1)1 iδ δf(t2)∝an}b∇acketle{t0|0∝an}b∇acket∇i}htf/vextendsingle/vextendsingle/vextendsingle f=0 =1 iδ δf(t1)/bracketleftbigg/integraldisplay+∞ −∞dt′G(t2−t′)f(t′)/bracketrightbigg ∝an}b∇acketle{t0|0∝an}b∇acket∇i}htf/vextendsingle/vextendsingle/vextendsingle f=0 =/bracketleftig 1 iG(t2−t1) + (term with f’s)/bracketrightig ∝an}b∇acketle{t0|0∝an}b∇acket∇i}htf/vextendsingle/vextendsingle/vextendsingle f=0 =1 iG(t2−t1). (7.16) We can continue in this way to compute the ground-state expec tation value of the time-ordered product of more Q(t)’s. If the number of Q(t)’s is odd, then there is always a left-over f(t) in the prefactor, and so the result is zero. If the number of Q(t)’s is even, then we must pair up the functional derivatives in an appropriate way to get a nonzero result. Th us, for exam- ple, ∝an}b∇acketle{t0|TQ(t1)Q(t2)Q(t3)Q(t4)|0∝an}b∇acket∇i}ht=1 i2/bracketleftig G(t1−t2)G(t3−t4) +G(t1−t3)G(t2−t4) +G(t1−t4)G(t2−t3)/bracketrightig .(7.17) More generally, ∝an}b∇acketle{t0|TQ(t1)...Q(t2n)|0∝an}b∇acket∇i}ht=1 in/summationdisplay pairingsG(ti1−ti2)...G(ti2n−1−ti2n).(7.18) Problems 7.1) Starting with eq.(7.12), do the contour integral to ver ify eq.(7.14). 7.2) Starting with eq.(7.14), verify eq.(7.13). 7: The Path Integral for the Harmonic Oscillator 66 7.3) a) Use the Heisenberg equation of motion, ˙A=i[H,A], to ﬁnd explicit expressions for ˙Qand˙P. Solve these to get the Heisenberg-picture operatorsQ(t) andP(t) in terms of the Schr¨ odinger picture operators QandP. b) Write the Schr¨ odinger picture operators QandPin terms of the creation and annihilation operators aanda†, whereH= ¯hω(a†a+1 2). Then, using your result from part (a), write the Heisenberg- picture operatorsQ(t) andP(t) in terms of aanda†. c) Using your result from part (b), and a|0∝an}b∇acket∇i}ht=∝an}b∇acketle{t0|a†= 0, verify eqs.(7.16) and (7.17). 7.4) Consider a harmonic oscillator in its ground state at t=−∞. It is then then subjected to an external force f(t). Compute the probabil- ity|∝an}b∇acketle{t0|0∝an}b∇acket∇i}htf|2that the oscillator is still in its ground state at t= +∞. Write your answer as a manifestly real expression, and in ter ms of the Fourier transform/tildewidef(E) =/integraltext+∞ −∞dteiEtf(t). Your answer should not involve any other unevaluated integrals. 8: The Path Integral for Free Field Theory 67 8The Path Integral for Free Field Theory Prerequisite: 3, 7 Our results for the harmonic oscillator can be straightforw ardly generalized to a free ﬁeld theory with hamiltonian density H0=1 2Π2+1 2(∇ϕ)2+1 2m2ϕ2. (8.1) The dictionary we need is q(t)−→ϕ(x,t) (classical ﬁeld) Q(t)−→ϕ(x,t) (operator ﬁeld) f(t)−→J(x,t) (classical source ) (8.2) The distinction between the classical ﬁeld ϕ(x) and the corresponding op- erator ﬁeld should be clear from context. To employ the ǫtrick, we multiply H0by 1−iǫ. The results are equiv- alent to replacing m2inH0withm2−iǫ. From now on, for notational simplicity, we will write m2when we really mean m2−iǫ. Let us write down the path integral (also called the functional integral ) for our free ﬁeld theory: Z0(J)≡ ∝an}b∇acketle{t0|0∝an}b∇acket∇i}htJ=/integraldisplay Dϕei/integraltext d4x[L0+Jϕ], (8.3) where L0=−1 2∂µϕ∂µϕ−1 2m2ϕ2(8.4) is the lagrangian density, and Dϕ∝/productdisplay xdϕ(x) (8.5) is the functional measure . Note that when we say path integral , we now mean a path in the space of ﬁeld conﬁgurations. We can evaluate Z0(J) by mimicking what we did for the harmonic oscillator in section 7. We introduce four-dimensional Fou rier transforms, /tildewideϕ(k) =/integraldisplay d4xe−ikxϕ(x), ϕ (x) =/integraldisplayd4k (2π)4eikx/tildewideϕ(k), (8.6) wherekx=−k0t+k·x, andk0is an integration variable. Then, starting withS0=/integraltextd4x[L0+Jϕ], we get S0=1 2/integraldisplayd4k (2π)4/bracketleftig −/tildewideϕ(k)(k2+m2)/tildewideϕ(−k)+/tildewideJ(k)/tildewideϕ(−k)+/tildewideJ(−k)/tildewideϕ(k)/bracketrightig ,(8.7) 8: The Path Integral for Free Field Theory 68 wherek2=k2−(k0)2. We now change path integration variables to /tildewideχ(k) =/tildewideϕ(k)−/tildewideJ(k) k2+m2. (8.8) Since this is merely a shift by a constant, we have Dϕ=Dχ. The action becomes S0=1 2/integraldisplayd4k (2π)4/bracketleftigg/tildewideJ(k)/tildewideJ(−k) k2+m2−/tildewideχ(k)(k2+m2)/tildewideχ(−k)/bracketrightigg . (8.9) Just as for the harmonic oscillator, the integral over χsimply yields a factor ofZ0(0) =∝an}b∇acketle{t0|0∝an}b∇acket∇i}htJ=0= 1. Therefore Z0(J) = exp/bracketleftigg i 2/integraldisplayd4k (2π)4/tildewideJ(k)/tildewideJ(−k) k2+m2−iǫ/bracketrightigg = exp/bracketleftbiggi 2/integraldisplay d4xd4x′J(x)∆(x−x′)J(x′)/bracketrightbigg . (8.10) Here we have deﬁned the Feynman propagator , ∆(x−x′) =/integraldisplayd4k (2π)4eik(x−x′) k2+m2−iǫ. (8.11) The Feynman propagator is a Green’s function for the Klein-G ordon equa- tion, (−∂2 x+m2)∆(x−x′) =δ4(x−x′). (8.12) This can be seen directly by plugging eq.(8.11) into eq.(8.1 2) and then taking the ǫ→0 limit. We can also evaluate ∆( x−x′) explicitly by treating the k0integral on the right-hand side of eq.(8.11) as a contour integral in the complex k0plane, and then evaluating it via the residue theorem. The result is ∆(x−x′) =i/integraldisplay /tildewiderdkeik·(x−x′)−iω|t−t′| =iθ(t−t′)/integraldisplay /tildewiderdkeik(x−x′)+iθ(t′−t)/integraldisplay /tildewiderdke−ik(x−x′),(8.13) whereθ(t) is the unit step function. The integral over/tildewiderdkcan also be performed in terms of Bessel functions; see section 4. Now, by analogy with the formula for the ground-state expect ation value of a time-ordered product of operators for the harmoni c oscillator, we have ∝an}b∇acketle{t0|Tϕ(x1)...|0∝an}b∇acket∇i}ht=1 iδ δJ(x1)...Z 0(J)/vextendsingle/vextendsingle/vextendsingle J=0. (8.14) 8: The Path Integral for Free Field Theory 69 Using our explicit formula, eq.(8.10), we have ∝an}b∇acketle{t0|Tϕ(x1)ϕ(x2)|0∝an}b∇acket∇i}ht=1 iδ δJ(x1)1 iδ δJ(x2)Z0(J)/vextendsingle/vextendsingle/vextendsingle J=0 =1 iδ δJ(x1)/bracketleftbigg/integraldisplay d4x′∆(x2−x′)J(x′)/bracketrightbigg Z0(J)/vextendsingle/vextendsingle/vextendsingle J=0 =/bracketleftig 1 i∆(x2−x1) + (term with J’s)/bracketrightig Z0(J)/vextendsingle/vextendsingle/vextendsingle J=0 =1 i∆(x2−x1). (8.15) We can continue in this way to compute the ground-state expec tation value of the time-ordered product of more ϕ’s. If the number of ϕ’s is odd, then there is always a left-over Jin the prefactor, and so the result is zero. If the number of ϕ’s is even, then we must pair up the functional derivatives in an appropriate way to get a nonzero result. Thus, for examp le, ∝an}b∇acketle{t0|Tϕ(x1)ϕ(x2)ϕ(x3)ϕ(x4)|0∝an}b∇acket∇i}ht=1 i2/bracketleftig ∆(x1−x2)∆(x3−x4) + ∆(x1−x3)∆(x2−x4) + ∆(x1−x4)∆(x2−x3)/bracketrightig .(8.16) More generally, ∝an}b∇acketle{t0|Tϕ(x1)...ϕ(x2n)|0∝an}b∇acket∇i}ht=1 in/summationdisplay pairings∆(xi1−xi2)...∆(xi2n−1−xi2n).(8.17) This result is known as Wick’s theorem . Problems 8.1) Starting with eq.(8.11), verify eq.(8.12). 8.2) Starting with eq.(8.11), verify eq.(8.13). 8.3) Starting with eq.(8.13), verify eq.(8.12). Note that t he time deriva- tives in the Klein-Gordon wave operator can act on either the ﬁeld (which obeys the Klein-Gordon equation) or the time-orderi ng step functions. 8.4) Use eqs.(3.19), (3.29), and (5.3) (and its hermitian co njugate) to verify the last line of eq.(8.15). 8.5) The retarded and advanced Green’s functions for the Kle in-Gordon wave operator satisfy ∆ ret(x−y) = 0 forx0≥y0and ∆ adv(x−y) = 0 forx0≤y0. Find the pole prescriptions on the right-hand side of eq.(8.11) that yield these Green’s functions. 8: The Path Integral for Free Field Theory 70 8.6) LetZ0(J) = expiW0(J), and evaluate the real and imaginary parts ofW0(J). 8.7) Repeat the analysis of this section for the complex scal ar ﬁeld that was introduced in problem 3.5, and further studied in problem 5. 1. Write your source term in the form J†ϕ+Jϕ†, and ﬁnd an explicit formula, analogous to eq.(8.10), for Z0(J†,J). Write down the appropriate generalization of eq.(8.14), and use it to compute ∝an}b∇acketle{t0|Tϕ(x1)ϕ(x2)|0∝an}b∇acket∇i}ht, ∝an}b∇acketle{t0|Tϕ†(x1)ϕ(x2)|0∝an}b∇acket∇i}ht, and ∝an}b∇acketle{t0|Tϕ†(x1)ϕ†(x2)|0∝an}b∇acket∇i}ht. Then verify your re- sults by using the method of problem 8.4. Finally, give the ap propri- ate generalization of eq.(8.17). 8.8) A harmonic oscillator (in units with m= ¯h= 1) has a ground-state wave function ∝an}b∇acketle{tq|0∝an}b∇acket∇i}ht ∝e−ωq2/2. Now consider a real scalar ﬁeld ϕ(x), and deﬁne a ﬁeld eigenstate |A∝an}b∇acket∇i}htthat obeys ϕ(x,0)|A∝an}b∇acket∇i}ht=A(x)|A∝an}b∇acket∇i}ht, (8.18) where the function A(x) is everywhere real. For a free-ﬁeld theory speciﬁed by the hamiltonian of eq.(8.1), Show that the ground-state wave functional is ∝an}b∇acketle{tA|0∝an}b∇acket∇i}ht ∝exp/bracketleftigg −1 2/integraldisplayd3k (2π)3ω(k)˜A(k)˜A(−k)/bracketrightigg , (8.19) where ˜A(k)≡/integraltextd3xe−ik·xA(x) andω(k)≡(k2+m2)1/2. 9: The Path Integral for Interacting Field Theory 71 9The Path Integral for Interacting Field Theory Prerequisite: 8 Let us consider an interacting quantum ﬁeld theory speciﬁed by a la- grangian of the form L=−1 2Zϕ∂µϕ∂µϕ−1 2Zmm2ϕ2+1 6Zggϕ3+Yϕ. (9.1) As we discussed at the end of section 5, we ﬁx the parameter mby requiring it to be equal to the actual mass of the particle (equivalentl y, the energy of the ﬁrst excited state relative to the ground state), and w e ﬁx the pa- rametergby requiring some particular scattering cross section to de pend ongin some particular way. (We will have more to say about this af ter we have learned to calculate cross sections.) We also assume th at the ﬁeld is normalized by ∝an}b∇acketle{t0|ϕ(x)|0∝an}b∇acket∇i}ht= 0 and ∝an}b∇acketle{tk|ϕ(x)|0∝an}b∇acket∇i}ht=e−ikx. (9.2) Here|0∝an}b∇acket∇i}htis the ground state, normalized via ∝an}b∇acketle{t0|0∝an}b∇acket∇i}ht= 1, and |k∝an}b∇acket∇i}htis a state of one particle with four-momentum kµ, wherek2=kµkµ=−m2, normalized via ∝an}b∇acketle{tk′|k∝an}b∇acket∇i}ht= (2π)32k0δ3(k′−k). (9.3) Thus we have four conditions (the speciﬁed values of m,g,∝an}b∇acketle{t0|ϕ|0∝an}b∇acket∇i}ht, and ∝an}b∇acketle{tk|ϕ|0∝an}b∇acket∇i}ht), and we will use these four conditions to determine the valu es of the four remaining parameters ( Yand the three Z’s) that appear in L. Before going further, we should note that this theory (known asϕ3 theory, pronounced “phi-cubed”) actually has a fatal ﬂaw. T he hamiltonian density is H=1 2Z−1 ϕΠ2−Yϕ+1 2Zmm2ϕ2−1 6Zggϕ3. (9.4) Classically, we can make this arbitrarily negative by choos ing an arbitrarily large value for ϕ. Quantum mechanically, this means that this hamiltonian has no ground state. If we start oﬀ near ϕ= 0, we can tunnel through the potential barrier to large ϕ, and then “roll down the hill”. However, this process is invisible in perturbation theory in g. The situation is exactly analogous to the problem of a harmonic oscillator perturbed by aq3term. This system also has no ground state, but perturbation theor y (both time dependent and time independent) does not “know” this. We wil l be inter- ested in eq.(9.1) only as an example of how to do perturbation expansions in a simple context, and so we will overlook this problem. We would like to evaluate the path integral for this theory, Z(J)≡ ∝an}b∇acketle{t0|0∝an}b∇acket∇i}htJ=/integraldisplay Dϕei/integraltext d4x[L0+L1+Jϕ]. (9.5) 9: The Path Integral for Interacting Field Theory 72 We can evaluate Z(J) by mimicking what we did for quantum mechanics at the end of section 6. Speciﬁcally, we can rewrite eq.(9.5) as Z(J) =ei/integraltext d4xL1/parenleftbig1 iδ δJ(x)/parenrightbig/integraldisplay Dϕei/integraltext d4x[L0+Jϕ]. ∝ei/integraltext d4xL1/parenleftbig1 iδ δJ(x)/parenrightbig Z0(J), (9.6) whereZ0(J) is the result in free-ﬁeld theory, Z0(J) = exp/bracketleftbiggi 2/integraldisplay d4xd4x′J(x)∆(x−x′)J(x′)/bracketrightbigg . (9.7) We have written Z(J) as proportional to (rather than equal to) the right- hand side of eq.(9.6) because the ǫtrick does not give us the correct overall normalization; instead, we must require Z(0) = 1, and enforce this by hand. Note that, in eq.(9.7), we have implicitly assumed that L0=−1 2∂µϕ∂µϕ−1 2m2ϕ2, (9.8) since this is the L0that gives us eq.(9.7). Therefore, the rest of Lmust be included in L1. We write L1=1 6Zggϕ3+Lct, Lct=−1 2(Zϕ−1)∂µϕ∂µϕ−1 2(Zm−1)m2ϕ2+Yϕ, (9.9) where Lctis called the counterterm lagrangian. We expect that, as g→0, Y→0 andZi→1. In fact, as we will see, Y=O(g) andZi= 1 +O(g2). In order to make use of eq.(9.7), we will have to compute lots a nd lots of functional derivatives of Z0(J). Let us begin by ignoring the counterterms. We deﬁne Z1(J)∝exp/bracketleftigg i 6Zgg/integraldisplay d4x/parenleftbigg1 iδ δJ(x)/parenrightbigg3/bracketrightigg Z0(J), (9.10) where the constant of proportionality is ﬁxed by Z1(0) = 1. We now make a dual Taylor expansion in powers of gandJto get Z1(J)∝∞/summationdisplay V=01 V!/bracketleftigg iZgg 6/integraldisplay d4x/parenleftbigg1 iδ δJ(x)/parenrightbigg3/bracketrightiggV ×∞/summationdisplay P=01 P!/bracketleftbiggi 2/integraldisplay d4yd4zJ(y)∆(y−z)J(z)/bracketrightbiggP . (9.11) If we focus on a term in eq.(9.11) with particular values of VandP, then the number of surviving sources (after we take all the functi onal derivatives) 9: The Path Integral for Interacting Field Theory 73 3 S = 2 S = 2 x 3! Figure 9.1: All connected diagrams with E= 0 andV= 2. 4 3 S = 24 S = 4! S = 2 x 3!3S = 2 S = 2 Figure 9.2: All connected diagrams with E= 0 andV= 4. isE= 2P−3V. (HereEstands for external , a terminology that should become clear by the end of the next section; Vstands for vertex andPfor propagator .) The overall phase factor of such a term is then iV(1/i)3ViP= iV+E−P, and the 3 Vfunctional derivatives can act on the 2 Psources in (2P)!/(2P−3V)! diﬀerent combinations. However, many of the resulting expressions are algebraically identical. To organize them, we introduce Feynman diagrams . In these diagrams, a line segment (straight or curved) stands for a propagator1 i∆(x−y), a ﬁlled circle at one end of a line segment for a source i/integraltextd4xJ(x), and a vertex joining three line segments for iZgg/integraltextd4x. Sets of diagrams with diﬀerent values of EandVare shown in ﬁgs.(9.1–9.11). To count the number of terms on the right-hand side of eq.(9.1 1) that result in a particular diagram, we ﬁrst note that, in each dia gram, the num- ber of lines is Pand the number of vertices is V. We can rearrange the three functional derivatives from a particular vertex with out changing the resulting diagram; this yields a counting factor of 3! for ea ch vertex. Also, we can rearrange the vertices themselves; this yields a coun ting factor of V!. Similarly, we can rearrange the two sources at the ends of a particular propagator without changing the resulting diagram; this yi elds a counting 9: The Path Integral for Interacting Field Theory 74 factor of 2! for each propagator. Also, we can rearrange the p ropagators themselves; this yields a counting factor of P!. All together, these count- ing factors neatly cancel the numbers from the dual Taylor ex pansions in eq.(9.11). However, this procedure generally results in an overcounti ng of the num- ber of terms that give identical results. This happens when s ome rearrange- ment of derivatives gives the same match-up to sources as some rearrange- ment of sources. This possibility is always connected to som e symmetry property of the diagram, and so the factor by which we have ove rcounted is called the symmetry factor . The ﬁgures show the symmetry factor Sof each diagram. Consider, for example, the second diagram of ﬁg.(9.1). The t hree prop- agators can be rearranged in 3! ways, and all these rearrange ments can be duplicated by exchanging the derivatives at the vertices . Furthermore the endpoints of each propagator can be simultaneously swap ped, and the eﬀect duplicated by swapping the two vertices. Thus, S= 2×3! = 12. Let us consider two more examples. In the ﬁrst diagram of ﬁg.( 9.6), the exchange of the two external propagators (along with the ir attached sources) can be duplicated by exchanging all the derivative s at one vertex for those at the other, and simultaneously swapping the endp oints of each semicircular propagator. Also, the eﬀect of swapping the to p and bottom semicircular propagators can be duplicated by swapping the corresponding derivatives at each vertex. Thus, the symmetry factor is S= 2×2 = 4. In the diagram of ﬁg.(9.10), we can exchange derivatives to m atch swaps of the top and bottom external propagators on the left, or the top and bottom external propagators on the right, or the set of exter nal propagators on the left with the set of external propagators on the right. Thus, the symmetry factor is S= 2×2×2 = 8. The diagrams in ﬁgs.(9.1–9.11) are all connected : we can trace a path through the diagram between any two points on it. However, th ese are not the only contributions to Z(J). The most general diagram consists of a product of several connected diagrams. Let CIstand for a particular connected diagram, including its symmetry factor. A genera l diagramD can then be expressed as D=1 SD/productdisplay I(CI)nI, (9.12) wherenIis an integer that counts the number of CI’s inD, andSDis the additional symmetry factor for D(that is, the part of the symmetry factor that is not already accounted for by the symmetry factors alr eady included in each of the connected diagrams). We now need to determine SD. 9: The Path Integral for Interacting Field Theory 75 S = 2 Figure 9.3: All connected diagrams with E= 1 andV= 1. S = 23 S = 22 S = 22 Figure 9.4: All connected diagrams with E= 1 andV= 3. S = 2 Figure 9.5: All connected diagrams with E= 2 andV= 0. S = 22 S = 22 Figure 9.6: All connected diagrams with E= 2 andV= 2. 9: The Path Integral for Interacting Field Theory 76 Since we have already accounted for propagator and vertex re arrange- ments within eachCI, we need to consider only exchanges of propagators and vertices among diﬀerent connected diagrams. These can leave the total diagramDunchanged only if (1) the exchanges are made among diﬀerent butidentical connected diagrams, and only if (2) the exchanges involve all of the propagators and vertices in a given connected diagram . If there are nIfactors ofCIinD, there are nI! ways to make these rearrangements. Overall, then, we have SD=/productdisplay InI!. (9.13) NowZ1(J) is given (up to an overall normalization) by summing all dia - gramsD, and eachDis labeled by the integers nI. Therefore Z1(J)∝/summationdisplay {nI}D ∝/summationdisplay {nI}/productdisplay I1 nI!(CI)nI ∝/productdisplay I∞/summationdisplay nI=01 nI!(CI)nI ∝/productdisplay Iexp (CI) ∝exp (/summationtext ICI). (9.14) Thus we have a remarkable result: Z1(J) is given by the exponential of the sum of connected diagrams. This makes it easy to impose the normalization Z1(0) = 1: we simply omit the vacuum diagrams (those with no sources), like those of ﬁgs.(9.1) and (9.2). We then have Z1(J) = exp[iW1(J)], (9.15) where we have deﬁned iW1(J)≡/summationdisplay I/ne}ationslash={0}CI, (9.16) and the notation I∝ne}ationslash={0}means that the vacuum diagrams are omitted from the sum, so that W1(0) = 0.1 Were it not for the counterterms in L1, we would have Z(J) =Z1(J). Let us see what we would get if this was, in fact, the case. In pa rticular, let us compute the vacuum expectation value of the ﬁeld ϕ(x), which is given 1We have included a factor of ion the left-hand side of eq.(9.16) because then W1(J) is real in free-ﬁeld theory; see problem 8.6. 9: The Path Integral for Interacting Field Theory 77 S = 23 S = 23 3 S = 2 S = 24 S = 22 S = 22 S = 2 S = 22 3 S = 2 2 Figure 9.7: All connected diagrams with E= 2 andV= 4. S = 3! Figure 9.8: All connected diagrams with E= 3 andV= 1. 9: The Path Integral for Interacting Field Theory 78 S = 22S = 22 S = 3! Figure 9.9: All connected diagrams with E= 3 andV= 3. S = 23 Figure 9.10: All connected diagrams with E= 4 andV= 2. 2 S = 23S = 24 S = 24 S = 2 S = 22 S = 22 Figure 9.11: All connected diagrams with E= 4 andV= 4. 9: The Path Integral for Interacting Field Theory 79 S = 2 S = 2 S = 2 S = 1 Figure 9.12: All connected diagrams with E= 1,X≥1 (whereXis the number of one-point vertices from the linear counterterm), andV+X≤3. by ∝an}b∇acketle{t0|ϕ(x)|0∝an}b∇acket∇i}ht=1 iδ δJ(x)Z1(J)/vextendsingle/vextendsingle/vextendsingle/vextendsingle J=0 =δ δJ(x)W1(J)/vextendsingle/vextendsingle/vextendsingle/vextendsingle J=0. (9.17) This expression is then the sum of all diagrams [such as those in ﬁgs.(9.3) and (9.4)] that have a single source, with the source removed : ∝an}b∇acketle{t0|ϕ(x)|0∝an}b∇acket∇i}ht=1 2ig/integraldisplay d4y1 i∆(x−y)1 i∆(y−y) +O(g3). (9.18) Here we have set Zg= 1 in the ﬁrst term, since Zg= 1+O(g2). We see the vacuum-expectation value of ϕ(x) is not zero, as is required for the validity of the LSZ formula. To ﬁx this, we must introduce the countert ermYϕ. Including this term in the interaction lagrangian L1introduces a new kind of vertex, one where a single line segment ends; the correspo nding vertex factor isiY/integraltextd4y. The simplest diagrams including this new vertex are shown in ﬁg.(9.12), with a cross symbolizing the vertex. AssumingY=O(g), only the ﬁrst diagram in ﬁg.(9.12) contributes at O(g), and we have ∝an}b∇acketle{t0|ϕ(x)|0∝an}b∇acket∇i}ht=/parenleftig iY+1 2(ig)1 i∆(0)/parenrightig/integraldisplay d4y1 i∆(x−y) +O(g3). (9.19) Thus, in order to have ∝an}b∇acketle{t0|ϕ(x)|0∝an}b∇acket∇i}ht= 0, we should choose Y=1 2ig∆(0) +O(g3). (9.20) The factor of iis disturbing, because Ymust be a real number: it is the coeﬃcient of a hermitian operator in the hamiltonian, as see n in eq.(9.4). Therefore, ∆(0) must be purely imaginary, or we are in troubl e. We have ∆(0) =/integraldisplayd4k (2π)41 k2+m2−iǫ. (9.21) 9: The Path Integral for Interacting Field Theory 80 From eq.(9.21), it is not immediately obvious whether or not ∆(0) is purely imaginary, but eq.(9.21) does reveal another problem: the i ntegral diverges at largek. This is another example of an ultraviolet divergence , similar to the one we encountered in section 3 when we computed the zero- point energy of the ﬁeld. To make some progress, we introduce an ultraviolet cutoﬀ Λ, which we assume is much larger than mand any other energy of physical interest. Modiﬁcations to the propagator above some cutoﬀ may be well j ustiﬁed physically; for example, quantum ﬂuctuations in spacetime itself should become important above the Planck scale , which is given by the inverse square root of Newton’s constant, and has the numerical valu e of 1019GeV (compared to, say, the proton mass, which is 1GeV). In order to retain the Lorentz-transformation properties o f the propa- gator, we implement the ultraviolet cutoﬀ in a more subtle wa y than we did in section 3; specﬁcally, we make the replacement ∆(x−y)→/integraldisplayd4k (2π)4eik(x−y) k2+m2−iǫ/parenleftigg Λ2 k2+ Λ2−iǫ/parenrightigg2 . (9.22) The integral is now convergent, and we can evaluate the modiﬁ ed ∆(0) with the methods of section 14; for Λ ≫m, the result is ∆(0) =i 16π2Λ2. (9.23) ThusYis real, as required. If we like, we can now formally take the l imit Λ→ ∞. The parameter Ybecomes inﬁnite, but ∝an}b∇acketle{t0|ϕ(x)|0∝an}b∇acket∇i}htremains zero, at least to this order in g. It may be disturbing to have a parameter in the lagrangian tha t is formally inﬁnite. However, such parameters are not directl y measurable, and so need not obey our preconceptions about their magnitud es. Also, it is important to remember that Yincludes a factor of g; this means that we can expand in powers of Yas part of our general expansion in powers of g. When we compute something measurable (like a scattering cro ss section), all the formally inﬁnite numbers will cancel in a well-deﬁne d way, leaving behind ﬁnite coeﬃcients for the various powers of g. We will see how this works in detail in sections 14–20. As we go to higher orders in g, things become more complicated, but in principle the procedure is the same. Thus, at O(g3), we sum up the diagrams of ﬁgs.(9.4) and (9.12), and then add to YwhateverO(g3) term is needed to maintain ∝an}b∇acketle{t0|ϕ(x)|0∝an}b∇acket∇i}ht= 0. In this way we can determine the value ofYorder by order in powers of g. Once this is done, there is a remarkable simpliﬁcation. Our a djustment ofYto keep ∝an}b∇acketle{t0|ϕ(x)|0∝an}b∇acket∇i}ht= 0 means that the sum of all connected diagrams 9: The Path Integral for Interacting Field Theory 81 Figure 9.13: All connected diagrams without tadpoles with E≤4 and V≤4. with a single source is zero. Consider now that same inﬁnite s et of diagrams, but replace the single source in each of them with some other s ubdiagram. Here is the point: no matter what this replacement subdiagram is, the sum of all these diagrams is still zero. Therefore, we need not bother to compute any of them! The rule is this: ignore any diagram that, when a s ingle line is cut, falls into two parts, one of which has no sources . All of these diagrams (known as tadpoles ) are canceled by the Ycounterterm, no matter what subdiagram they are attached to. The diagrams that remain (a nd need to be computed!) are shown in ﬁg.(9.13). We turn next to the remaining two counterterms. For notation al sim- plicity we deﬁne A=Zϕ−1, B =Zm−1, (9.24) 9: The Path Integral for Interacting Field Theory 82 and recall that we expect each of these to be O(g2). We now have Z(J) = exp/bracketleftbigg −i 2/integraldisplay d4x/parenleftbigg1 iδ δJ(x)/parenrightbigg/parenleftig −A∂2 x+Bm2/parenrightig/parenleftbigg1 iδ δJ(x)/parenrightbigg/bracketrightbigg Z1(J). (9.25) We have integrated by parts to put both ∂x’s onto one δ/δJ(x). (Note that the time derivatives in this interaction should really be tr eated by including an extra source term for the conjugate momentum Π = ˙ ϕ. However, the space derivatives are correctly treated, and then the time d erivatives must work out comparably by Lorentz invariance.) Eq.(9.25) results in a new vertex at which two lines meet. The corre- sponding vertex factor is ( −i)/integraltextd4x(−A∂2 x+Bm2); the∂2 xacts on the xin one or the other (but not both) propagators. (Which one does n ot matter, and can be changed via integration by parts.) Diagramatical ly, all we need do is sprinkle these new vertices onto the propagators in our existing dia- grams. How many of these vertices we need to add depends on the order ingwe are working to achieve. This completes our calculation of Z(J) inϕ3theory. We express it as Z(J) = exp[iW(J)], (9.26) whereW(J) is given by the sum of all connected diagrams with no tad- poles and at least two sources, and including the counterter m vertices just discussed. Now that we have Z(J), we must ﬁnd out what we can do with it. Problems 9.1) Compute the symmetry factor for each diagram in ﬁg.(9.1 3). (You can then check your answers by consulting the earlier ﬁgures .) 9.2) Consider a real scalar ﬁeld with L=L0+L1, where L0=−1 2∂µϕ∂µϕ−1 2m2ϕ2, L1=−1 24Zλλϕ4+Lct, Lct=−1 2(Zϕ−1)∂µϕ∂µϕ−1 2(Zm−1)m2ϕ2. a) What kind of vertex appears in the diagrams for this theory (that is, how many line segments does it join?), and what is the asso ciated vertex factor? b) Ignoring the counterterms, draw all the connected diagra ms with 1≤E≤4 and 0 ≤V≤2, and ﬁnd their symmetry factors. c) Explain why we did not have to include a counterterm linear inϕ to cancel tadpoles. 9: The Path Integral for Interacting Field Theory 83 9.3) Consider a complex scalar ﬁeld (see problems 3.5, 5.1, a nd 8.7) with L=L0+L1, where L0=−∂µϕ†∂µϕ−m2ϕ†ϕ, L1=−1 4Zλλ(ϕ†ϕ)2+Lct, Lct=−(Zϕ−1)∂µϕ†∂µϕ−(Zm−1)m2ϕ†ϕ. This theory has two kinds of sources, JandJ†, and so we need a way to tell which is which when we draw the diagrams. Rather th an labeling the source blobs with a JorJ†, we will indicate which is which by putting an arrow on the attached propagator that poi nts towards the source if it is a J†, andaway from the source if it is a J. a) What kind of vertex appears in the diagrams for this theory , and what is the associated vertex factor? Hint: your answer shou ld involve those arrows! b) Ignoring the counterterms, draw all the connected diagra ms with 1≤E≤4 and 0 ≤V≤2, and ﬁnd their symmetry factors. Hint: the arrows are important! 9.4) Consider the integral expW(g,J)≡1√ 2π/integraldisplay+∞ −∞dxexp/bracketleftig −1 2x2+1 6gx3+Jx/bracketrightig .(9.27) This integral does not converge, but it can be used to generat e a joint power series in gandJ, W(g,J) =∞/summationdisplay V=0∞/summationdisplay E=0CV,EgVJE. (9.28) a) Show that CV,E=/summationdisplay I1 SI, (9.29) where the sum is over all connected Feynman diagrams with Esources andVthree-point vertices, and SIis the symmetry factor for each diagram. b) Use eqs.(9.27) and (9.28) to compute CV,EforV≤4 andE≤5. (This is most easily done with a symbolic manipulation progr am like Mathematica.) Verify that the symmetry factors given in ﬁgs .(9.1– 9.11) satisfy the sum rule of eq.(9.29). 9: The Path Integral for Interacting Field Theory 84 c) Now consider W(g,J+Y), withYﬁxed by the “no tadpole” con- dition∂ ∂JW(g,J+Y)/vextendsingle/vextendsingle/vextendsingle/vextendsingle J=0= 0. (9.30) Then write W(g,J+Y) =∞/summationdisplay V=0∞/summationdisplay E=0/tildewideCV,EgVJE. (9.31) Show that /tildewideCV,E=/summationdisplay I1 SI, (9.32) where the sum is over all connected Feynman diagrams with Esources andVthree-point vertices and no tadpoles , andSIis the symmetry factor for each diagram. d) LetY=a1g+a3g3+..., and use eq.(9.30) to determine a1and a3. Compute/tildewideCV,EforV≤4 andE≤4. Verify that the symmetry factors for the diagrams in ﬁg.(9.13) satisfy the sum rule of eq.(9.32). 9.5)The interaction picture. In this problem, we will derive a formula for ∝an}b∇acketle{t0|Tϕ(xn)...ϕ(x1)|0∝an}b∇acket∇i}htwithout using path integrals. Suppose we have a hamiltonian density H=H0+H1, where H0=1 2Π2+1 2(∇ϕ)2+ 1 2m2ϕ2, andH1is a function of Π( x,0) andϕ(x,0) and their spatial derivatives. (It should be chosen to preserve Lorentz invar iance, but we will not be concerned with this issue.) We add a constant to H so thatH|0∝an}b∇acket∇i}ht= 0. Let |∅∝an}b∇acket∇i}htbe the ground state of H0, with a constant added toH0so thatH0|∅∝an}b∇acket∇i}ht= 0. (H1is then deﬁned as H−H0.) The Heisenberg-picture ﬁeld is ϕ(x,t)≡eiHtϕ(x,0)e−iHt. (9.33) We now deﬁne the interaction-picture ﬁeld ϕI(x,t)≡eiH0tϕ(x,0)e−iH0t. (9.34) a) Show that ϕI(x) obeys the Klein-Gordon equation, and hence is a free ﬁeld. b) Show that ϕ(x) =U†(t)ϕI(x)U(t), whereU(t)≡eiH0te−iHtis unitary. c) Show that U(t) obeys the diﬀerential equation id dtU(t) =HI(t)U(t), whereHI(t) =eiH0tH1e−iH0tis the interaction hamiltonian in the in- teraction picture, and the boundary condition U(0) = 1. 9: The Path Integral for Interacting Field Theory 85 d) IfH1is speciﬁed by a particular function of the Schr¨ odinger-pi cture ﬁelds Π( x,0) andϕ(x,0), show that HI(t) is given by the same func- tion of the interaction-picture ﬁelds Π I(x,t) andϕI(x,t). e) Show that, for t>0, U(t) = T exp/bracketleftbigg −i/integraldisplayt 0dt′HI(t′)/bracketrightbigg (9.35) obeys the diﬀerential equation and boundary condition of pa rt (c). What is the comparable expression for t<0? Hint: you may need to deﬁne a new ordering symbol. f) DeﬁneU(t2,t1)≡U(t2)U†(t1). Show that, for t2>t1, U(t2,t1) = T exp/bracketleftbigg −i/integraldisplayt2 t1dt′HI(t′)/bracketrightbigg . (9.36) What is the comparable expression for t1>t2? g) For any time ordering, show that U(t3,t1) =U(t3,t2)U(t2,t1) and thatU†(t1,t2) =U(t2,t1). h) Show that ϕ(xn)...ϕ(x1) =U†(tn,0)ϕI(xn)U(tn,tn−1)ϕI(xn−1) ...U(t2,t1)ϕI(x1)U(t1,0). (9.37) i) Show that U†(tn,0) =U†(∞,0)U(∞,tn) and also that U(t1,0) = U(t1,−∞)U(−∞,0). j) ReplaceH0with (1 −iǫ)H0, and show that ∝an}b∇acketle{t0|U†(∞,0) =∝an}b∇acketle{t0|∅∝an}b∇acket∇i}ht∝an}b∇acketle{t∅| and thatU(−∞,0)|0∝an}b∇acket∇i}ht=|∅∝an}b∇acket∇i}ht∝an}b∇acketle{t∅|0∝an}b∇acket∇i}ht. k) Show that ∝an}b∇acketle{t0|ϕ(xn)...ϕ(x1)|0∝an}b∇acket∇i}ht=∝an}b∇acketle{t∅|U(∞,tn)ϕI(xn)U(tn,tn−1)ϕI(xn−1)... U(t2,t1)ϕI(x1)U(t1,−∞)|∅∝an}b∇acket∇i}ht × |∝an}b∇acketle{t∅|0∝an}b∇acket∇i}ht|2. (9.38) l) Show that ∝an}b∇acketle{t0|Tϕ(xn)...ϕ(x1)|0∝an}b∇acket∇i}ht=∝an}b∇acketle{t∅|TϕI(xn)...ϕI(x1)e−i/integraltext d4xHI(x)|∅∝an}b∇acket∇i}ht × |∝an}b∇acketle{t∅|0∝an}b∇acket∇i}ht|2. (9.39) m) Show that |∝an}b∇acketle{t∅|0∝an}b∇acket∇i}ht|2= 1/∝an}b∇acketle{t∅|Te−i/integraltext d4xHI(x)|∅∝an}b∇acket∇i}ht. (9.40) 9: The Path Integral for Interacting Field Theory 86 Thus we have ∝an}b∇acketle{t0|Tϕ(xn)...ϕ(x1)|0∝an}b∇acket∇i}ht=∝an}b∇acketle{t∅|TϕI(xn)...ϕI(x1)e−i/integraltext d4xHI(x)|∅∝an}b∇acket∇i}ht ∝an}b∇acketle{t∅|Te−i/integraltext d4xHI(x)|∅∝an}b∇acket∇i}ht. (9.41) We can now Taylor expand the exponentials on the right-hand s ide of eq.(9.41), and use free-ﬁeld theory to compute the result ing corre- lation functions. 10: Scattering Amplitudes and the Feynman Rules 87 10Scattering Amplitudes and the Feynman Rules Prerequisite: 5, 9 Now that we have an expression for Z(J) = expiW(J), we can take func- tional derivatives to compute vacuum expectation values of time-ordered products of ﬁelds. Consider the case of two ﬁelds; we deﬁne th e exact propagator via 1 i∆(x1−x2)≡ ∝an}b∇acketle{t0|Tϕ(x1)ϕ(x2)|0∝an}b∇acket∇i}ht. (10.1) For notational simplicity let us deﬁne δj≡1 iδ δJ(xj). (10.2) Then we have ∝an}b∇acketle{t0|Tϕ(x1)ϕ(x2)|0∝an}b∇acket∇i}ht=δ1δ2Z(J)/vextendsingle/vextendsingle/vextendsingle J=0 =δ1δ2iW(J)/vextendsingle/vextendsingle/vextendsingle J=0−δ1iW(J)/vextendsingle/vextendsingle/vextendsingle J=0δ2iW(J)/vextendsingle/vextendsingle/vextendsingle J=0 =δ1δ2iW(J)/vextendsingle/vextendsingle/vextendsingle J=0. (10.3) To get the last line we used δjW(J)|J=0=∝an}b∇acketle{t0|ϕ(xj)|0∝an}b∇acket∇i}ht= 0. Diagramat- ically,δ1removes a source, and labels the propagator endpoint x1. Thus 1 i∆(x1−x2) is given by the sum of diagrams with two sources, with those sources removed and the endpoints labeled x1andx2. (The labels must be applied in both ways. If the diagram was originally symmetri c on exchange of the two sources, the associated symmetry factor of 2 is the n canceled by the double labeling.) At lowest order, the only contributio n is the “barbell” diagram of ﬁg.(9.5) with the sources removed. Thus we recove r the obvious fact that1 i∆(x1−x2) =1 i∆(x1−x2) +O(g2). We will take up the subject of theO(g2) corrections in section 14. For now, let us go on to compute ∝an}b∇acketle{t0|Tϕ(x1)ϕ(x2)ϕ(x3)ϕ(x4)|0∝an}b∇acket∇i}ht=δ1δ2δ3δ4Z(J) =/bracketleftig δ1δ2δ3δ4iW + (δ1δ2iW)(δ3δ4iW) + (δ1δ3iW)(δ2δ4iW) + (δ1δ4iW)(δ2δ3iW)/bracketrightig J=0.(10.4) We have dropped terms that contain a factor of ∝an}b∇acketle{t0|ϕ(x)|0∝an}b∇acket∇i}ht= 0. According to eq.(10.3), the last three terms in eq.(10.4) simply give p roducts of the exact propagators. 10: Scattering Amplitudes and the Feynman Rules 88 Let us see what happens when these terms are inserted into the LSZ formula for two incoming and two outgoing particles, ∝an}b∇acketle{tf|i∝an}b∇acket∇i}ht=i4/integraldisplay d4x1d4x2d4x′ 1d4x′ 2ei(k1x1+k2x2−k′ 1x′ 1−k′ 2x′ 2) ×(−∂2 1+m2)(−∂2 2+m2)(−∂2 1′+m2)(−∂2 2′+m2) ×∝an}b∇acketle{t0|Tϕ(x1)ϕ(x2)ϕ(x′ 1)ϕ(x′ 2)|0∝an}b∇acket∇i}ht. (10.5) If we consider, for example,1 i∆(x1−x′ 1)1 i∆(x2−x′ 2) as one term in the correlation function in eq.(10.5), we get from this term /integraldisplay d4x1d4x2d4x′ 1d4x′ 2ei(k1x1+k2x2−k′ 1x′ 1−k′ 2x′ 2)F(x11′)F(x22′) = (2π)4δ4(k1−k′ 1)(2π)4δ4(k2−k′ 2)/tildewideF(¯k11′)/tildewideF(¯k22′),(10.6) whereF(xij)≡(−∂2 i+m2)(−∂2 j+m2)∆(xij),/tildewideF(k) is its Fourier transform, xij′≡xi−x′ j, and ¯kij′≡(ki+k′ j)/2. The important point is the two delta functions: these tell us that the four-momenta of the two out going particles (1′and 2′) are equal to the four-momenta of the two incoming particles (1 and 2). In other words, no scattering has occurred. This is not the event whose probability we wish to compute! The other two sim ilar terms in eq.(10.4) either contribute to “no scattering” events, o r vanish due to factors like δ4(k1+k2) (which is zero because k0 1+k0 2≥2m>0). In general, the diagrams that contribute to the scattering process of in terest are only those that are fully connected : every endpoint can be reached from every other endpoint by tracing through the diagram. These are the diagrams that arise from all the δ’s acting on a single factor of W. Therefore, from here on, we restrict our attention to those diagrams alone. W e deﬁne the connected correlation functions via ∝an}b∇acketle{t0|Tϕ(x1)...ϕ(xE)|0∝an}b∇acket∇i}htC≡δ1...δEiW(J)/vextendsingle/vextendsingle/vextendsingle J=0, (10.7) and use these instead of ∝an}b∇acketle{t0|Tϕ(x1)...ϕ(xE)|0∝an}b∇acket∇i}htin the LSZ formula. Returning to eq.(10.4), we have ∝an}b∇acketle{t0|Tϕ(x1)ϕ(x2)ϕ(x′ 1)ϕ(x′ 2)|0∝an}b∇acket∇i}htC=δ1δ2δ1′δ2′iW/vextendsingle/vextendsingle/vextendsingle J=0. (10.8) The lowest-order (in g) nonzero contribution to this comes from the diagram of ﬁg.(9.10), which has four sources and two vertices. The fo urδ’s remove the four sources; there are 4! ways of matching up the δ’s to the sources. These 24 diagrams can then be collected into 3 groups of 8 diag rams each; the 8 diagrams in each group are identical. The 3 distinct dia grams are shown in ﬁg.(10.1). Note that the factor of 8 neatly cancels t he symmetry factorS= 8 of the diagram with sources. 10: Scattering Amplitudes and the Feynman Rules 89 11 2 2 11 212 1 2 2 Figure 10.1: The three tree-level Feynman diagrams that con tribute to the connected correlation function ∝an}b∇acketle{t0|Tϕ(x1)ϕ(x2)ϕ(x′ 1)ϕ(x′ 2)|0∝an}b∇acket∇i}htC. This is a general result for tree diagrams (those with no closed loops): once the sources have been stripped oﬀ and the endpoints labe led, each diagram with a distinct endpoint labeling has an overall sym metry factor of one. The tree diagrams for a given process represent the lo west-order (in g) nonzero contribution to that process. We now have ∝an}b∇acketle{t0|Tϕ(x1)ϕ(x2)ϕ(x′ 1)ϕ(x′ 2)|0∝an}b∇acket∇i}htC = (ig)2/parenleftig 1 i/parenrightig5/integraldisplay d4yd4z∆(y−z) ×/bracketleftig ∆(x1−y)∆(x2−y)∆(x′ 1−z)∆(x′ 2−z) + ∆(x1−y)∆(x′ 1−y)∆(x2−z)∆(x′ 2−z) + ∆(x1−y)∆(x′ 2−y)∆(x2−z)∆(x′ 1−z)/bracketrightig +O(g4). (10.9) Next, we use eq.(10.9) in the LSZ formula, eq.(10.5). Each Kl ein-Gordon wave operator acts on a propagator to give (−∂2 i+m2)∆(xi−y) =δ4(xi−y). (10.10) The integrals over the external spacetime labels x1,2,1′,2′are then trivial, and we get ∝an}b∇acketle{tf|i∝an}b∇acket∇i}ht= (ig)2/parenleftig 1 i/parenrightig/integraldisplay d4yd4z∆(y−z)/bracketleftig ei(k1y+k2y−k′ 1z−k′ 2z) +ei(k1y+k2z−k′ 1y−k′ 2z) +ei(k1y+k2z−k′ 1z−k′ 2y)/bracketrightig +O(g4).(10.11) This can be simpliﬁed by substituting ∆(y−z) =/integraldisplayd4k (2π)4eik(y−z) k2+m2−iǫ(10.12) 10: Scattering Amplitudes and the Feynman Rules 90 into eq.(10.9). Then the spacetime arguments appear only in phase factors, and we can integrate them to get delta functions: ∝an}b∇acketle{tf|i∝an}b∇acket∇i}ht=ig2/integraldisplayd4k (2π)41 k2+m2−iǫ ×/bracketleftig (2π)4δ4(k1+k2+k)(2π)4δ4(k′ 1+k′ 2+k) + (2π)4δ4(k1−k′ 1+k)(2π)4δ4(k′ 2−k2+k) + (2π)4δ4(k1−k′ 2+k)(2π)4δ4(k′ 1−k2+k)/bracketrightig +O(g4) =ig2(2π)4δ4(k1+k2−k′ 1−k′ 2) ×/bracketleftbigg1 (k1+k2)2+m2+1 (k1−k′ 1)2+m2+1 (k1−k′ 2)2+m2/bracketrightbigg +O(g4). (10.13) In eq.(10.13), we have left out the iǫ’s for notational convenience only; m2 is reallym2−iǫ. The overall delta function in eq.(10.13) tells that that four-momentum is conserved in the scattering process, whic h we should, of course, expect. For a general scattering process, it is then convenient to deﬁne a scattering matrix element Tvia ∝an}b∇acketle{tf|i∝an}b∇acket∇i}ht= (2π)4δ4(kin−kout)iT, (10.14) wherekinandkoutare the total four-momenta of the incoming and outgoing particles, respectively. Examining the calculation which led to eq.(10.13), we can ta ke away some universal features that lead to a simple set of Feynman rules for computing contributions to iTfor a given scattering process. The Feynman rules are: 1. Draw lines (called external lines ) for each incoming and each outgoing particle. 2. Leave one end of each external line free, and attach the oth er to a vertex at which exactly three lines meet. Include extra internal lines in order to do this. In this way, draw all possible diagrams th at are topologically inequivalent . 3. On each incoming line, draw an arrow pointing towards the v ertex. On each outgoing line, draw an arrow pointing away from the ve rtex. On each internal line, draw an arrow with an arbitrary direct ion. 4. Assign each line its own four-momentum. The four-momentu m of an external line should be the four-momentum of the correspo nding particle. 10: Scattering Amplitudes and the Feynman Rules 91 2 k2k1 k1k2+k1 k2k1 k2k1k1k k12k1 k2k1 kk k1 2 Figure 10.2: The tree-level s-,t-, andu-channel diagrams contributing to iTfor two particle scattering. 5. Think of the four-momenta as ﬂowing along the arrows, and c onserve four-momentum at each vertex. For a tree diagram, this ﬁxes t he momenta on all the internal lines. 6. The value of a diagram consists of the following factors: for each external line, 1; for each internal line with momentum k,−i/(k2+m2−iǫ); for each vertex, iZgg. 7. A diagram with Lclosed loops will have Linternal momenta that are not ﬁxed by rule #5. Integrate over each of these momenta ℓiwith measured4ℓi/(2π)4. 8. A loop diagram may have some leftover symmetry factors if t here are exchanges of internal propagators and vertices that leave the diagram unchanged; in this case, divide the value of the diagram by th e sym- metry factor associated with exchanges of internal propaga tors and vertices. 9. Include diagrams with the counterterm vertex that connects two prop- agators, each with the same four-momentum k. The value of this vertex is −i(Ak2+Bm2), whereA=Zϕ−1 andB=Zm−1, and each isO(g2). 10. The value of iTis given by a sum over the values of all these diagrams. For the two-particle scattering process, the tree diagrams resulting from these rules are shown in ﬁg.(10.2). Now that we have our procedure for computing the scattering a mplitude T, we must see how to relate it to a measurable cross section. Problems 10: Scattering Amplitudes and the Feynman Rules 92 10.1) Use eq.(9.41) of problem 9.5 to rederive eq.(10.9). 10.2) Write down the Feynman rules for the complex scalar ﬁel d of prob- lem 9.3. Remember that there are two kinds of particles now (w hich we can think of as positively and negatively charged), and th at your rules must have a way of distinguishing them. Hint: the most d irect approach requires two kinds of arrows: momentum arrows (as d is- cussed in this section) and what we might call “charge” arrow s (as discussed in problem 9.3). Try to ﬁnd a more elegant approach that requires only one kind of arrow. 10.3) Consider a complex scalar ﬁeld ϕthat interacts with a real scalar ﬁeldχviaL1=gχϕ†ϕ. Use a solid line for the ϕpropagator and a dashed line for the χpropagator. Draw the vertex (remember the arrows!), and ﬁnd the associated vertex factor. 10.4) Consider a real scalar ﬁeld with L1=1 2gϕ∂µϕ∂µϕ. Find the associ- ated vertex factor. 10.5) The scattering amplitudes should be unchanged if we ma ke aﬁeld redeﬁnition . Suppose, for example, we have L=−1 2∂µϕ∂µϕ−1 2m2ϕ2, (10.15) and we make the ﬁeld redeﬁnition ϕ→ϕ+λϕ2. (10.16) Work out the lagrangian in terms of the redeﬁned ﬁeld, and the cor- responding Feynman rules. Compute (at tree level) the ϕϕ→ϕϕ scattering amplitude. You should get zero, because this is a free-ﬁeld theory in disguise. (At the loop level, we also have to take in to ac- count the transformation of the functional measure Dϕ; see section 85.) 11: Cross Sections and Decay Rates 93 11Cross Sections and Decay Rates Prerequisite: 10 Now that we have a method for computing the scattering amplit udeT, we must convert it into something that could be measured in an ex periment. In practice, we are almost always concerned with one of two ge neric cases: one incoming particle, for which we compute a decay rate , or two incoming particles, for which we compute a cross section . We begin with the latter. Let us also specialize, for now, to the case of two outgoing pa rticles as well as two incoming particles. In ϕ3theory, we found in section 10 that in this case we have T=g2/bracketleftbigg1 (k1+k2)2+m2+1 (k1−k′ 1)2+m2+1 (k1−k′ 2)2+m2/bracketrightbigg +O(g4), (11.1) wherek1andk2are the four-momenta of the two incoming particles, k′ 1and k′ 2are the four-momenta of the two outgoing particles, and k1+k2=k′ 1+k′ 2. Also, these particles are all on shell :k2 i=−m2 i. (Here, for later use, we allow for the possibility that the particles all have diﬀere nt masses.) Let us think about the kinematics of this process. In the center-of- mass frame , orCM frame for short, we take k1+k2=0, and choose k1 to be in the + zdirection. Now the only variable left to specify about the initial state is the magnitude of k1. Equivalently, we could specify the total energy in the CM frame, E1+E2. However, it is even more convenient to deﬁne a Lorentz scalar s≡ −(k1+k2)2. In the CM frame, sreduces to (E1+E2)2;sis therefore called the center-of-mass energy squared . Then, sinceE1= (k2 1+m2 1)1/2andE2= (k2 1+m2 2)1/2, we can solve for |k1|in terms ofs, with the result |k1|=1 2√s/radicalig s2−2(m2 1+m2 2)s+ (m2 1−m2 2)2(CM frame) .(11.2) Now consider the two outgoing particles. Since momentum is c onserved, we must have k′ 1+k′ 2=0, and since energy is conserved, we must also have (E′ 1+E′ 2)2=s. Then we ﬁnd |k′ 1|=1 2√s/radicalig s2−2(m2 1′+m2 2′)s+ (m2 1′−m2 2′)2(CM frame) .(11.3) Now the only variable left to specify about the ﬁnal state is t he angleθ between k1andk′ 1. However, it is often more convenient to work with the Lorentz scalar t≡ −(k1−k′ 1)2, which is related to θby t=m2 1+m2 1′−2E1E′ 1+ 2|k1||k′ 1|cosθ. (11.4) 11: Cross Sections and Decay Rates 94 This formula is valid in any frame. The Lorentz scalars sandtare two of the three Mandelstam variables , deﬁned as s≡ −(k1+k2)2=−(k′ 1+k′ 2)2, t≡ −(k1−k′ 1)2=−(k2−k′ 2)2, u≡ −(k1−k′ 2)2=−(k2−k′ 1)2. (11.5) The three Mandelstam variables are not independent; they sa tisfy the linear relation s+t+u=m2 1+m2 2+m2 1′+m2 2′. (11.6) In terms of s,t, andu, we can rewrite eq.(11.1) as T=g2/bracketleftbigg1 m2−s+1 m2−t+1 m2−u/bracketrightbigg +O(g4), (11.7) which demonstrates the notational utility of the Mandelsta m variables. Now let us consider a diﬀerent frame, the ﬁxed target orFT frame (also sometimes called the lab frame ), in which particle #2 is initially at rest: k2=0. In this case we have |k1|=1 2m2/radicalig s2−2(m2 1+m2 2)s+ (m2 1−m2 2)2(FT frame) .(11.8) Note that, from eqs.(11.8) and (11.2), m2|k1|FT=√s|k1|CM. (11.9) This will be useful later. We would now like to derive a formula for the diﬀerential scat tering cross section. In order to do so, we assume that the whole expe riment is taking place in a big box of volume V, and lasts for a large time T. We should really think about wave packets coming together, but we will use some simple shortcuts instead. Also, to get a more general an swer, we will let the number of outgoing particles be arbitrary. Recall from section 10 that the overlap between the initial a nd ﬁnal states is given by ∝an}b∇acketle{tf|i∝an}b∇acket∇i}ht= (2π)4δ4(kin−kout)iT. (11.10) To get a probability, we must square ∝an}b∇acketle{tf|i∝an}b∇acket∇i}ht, and divide by the norms of the initial and ﬁnal states: P=|∝an}b∇acketle{tf|i∝an}b∇acket∇i}ht|2 ∝an}b∇acketle{tf|f∝an}b∇acket∇i}ht∝an}b∇acketle{ti|i∝an}b∇acket∇i}ht. (11.11) 11: Cross Sections and Decay Rates 95 The numerator of this expression is |∝an}b∇acketle{tf|i∝an}b∇acket∇i}ht|2= [(2π)4δ4(kin−kout)]2|T |2. (11.12) We write the square of the delta function as [(2π)4δ4(kin−kout)]2= (2π)4δ4(kin−kout)×(2π)4δ4(0), (11.13) and note that (2π)4δ4(0) =/integraldisplay d4xei0·x=VT . (11.14) Also, the norm of a single particle state is given by ∝an}b∇acketle{tk|k∝an}b∇acket∇i}ht= (2π)32k0δ3(0) = 2k0V . (11.15) Thus we have ∝an}b∇acketle{ti|i∝an}b∇acket∇i}ht= 4E1E2V2, (11.16) ∝an}b∇acketle{tf|f∝an}b∇acket∇i}ht=n′/productdisplay j=12k′ j0V , (11.17) wheren′is the number of outgoing particles. If we now divide eq.(11.11) by the elapsed time T, we get a probability per unit time ˙P=(2π)4δ4(kin−kout)V|T |2 4E1E2V2/producttextn′ j=12k′0 jV. (11.18) This is the probability per unit time to scatter into a set of o utgoing par- ticles with precise momenta. To get something measurable, w e should sum each outgoing three-momentum k′ jover some small range. Due to the box, all three-momenta are quantized: k′ j= (2π/L)n′ j, whereV=L3, andn′ jis a three-vector with integer entries. (Here we have assumed p eriodic bound- ary conditions, but this choice does not aﬀect the ﬁnal resul t.) In the limit of largeL, we have /summationdisplay n′ j→V (2π)3/integraldisplay d3k′ j. (11.19) Thus we should multiply ˙Pby a factor of Vd3k′ j/(2π)3for each outgoing particle. Then we get ˙P=(2π)4δ4(kin−kout) 4E1E2V|T |2n′/productdisplay j=1/tildewiderdk′ j, (11.20) 11: Cross Sections and Decay Rates 96 where we have identiﬁed the Lorentz-invariant phase-space diﬀerential /tildewiderdk≡d3k (2π)32k0(11.21) that we ﬁrst introduced in section 3. To convert ˙Pto a diﬀerential cross section dσ, we must divide by the incident ﬂux. Let us see how this works in the FT frame, where p article #2 is at rest. The incident ﬂux is the number of particles per u nit volume that are striking the target particle (#2), times their spee d. We have one incident particle (#1) in a volume Vwith speed v=|k1|/E1, and so the incident ﬂux is |k1|/E1V. Dividing eq.(11.20) by this ﬂux cancels the last factor ofV, and replaces E1in the denominator with |k1|. We also set E2=m2and note that eq.(11.8) gives |k1|m2as a function of s;dσwill be Lorentz invariant if, in other frames, we simply use this f unction as the value of |k1|m2. Adopting this convention, and using eq.(11.9), we have dσ=1 4|k1|CM√s|T |2dLIPSn′(k1+k2), (11.22) where |k1|CMis given as a function of sby eq.(11.2), and we have deﬁned then′-body Lorentz-invariant phase-space measure dLIPSn′(k)≡(2π)4δ4(k−/summationtextn′ j=1k′ i)n′/productdisplay j=1/tildewiderdk′ j. (11.23) Eq.(11.22) is our ﬁnal result for the diﬀerential cross sect ion for the scat- tering of two incoming particles into n′outgoing particles. Let us now specialize to the case of two outgoing particles. W e need to evaluate dLIPS 2(k) = (2π)4δ4(k−k′ 1−k′ 2)/tildewiderdk′ 1/tildewiderdk′ 2, (11.24) wherek=k1+k2. SincedLIPS 2(k) is Lorentz invariant, we can compute it in any convenient frame. Let us work in the CM frame, where k= k1+k2=0andk0=E1+E2=√s; then we have dLIPS 2(k) =1 4(2π)2E′ 1E′ 2δ(E′ 1+E′ 2−√s)δ3(k′ 1+k′ 2)d3k′ 1d3k′ 2.(11.25) We can use the spatial part of the delta function to integrate overd3k′ 2, with the result dLIPS 2(k) =1 4(2π)2E′ 1E′ 2δ(E′ 1+E′ 2−√s)d3k′ 1, (11.26) 11: Cross Sections and Decay Rates 97 where now E′ 1=/radicalig k′ 12+m2 1′andE′ 2=/radicalig k′ 12+m2 2′. (11.27) Next, let us write d3k′ 1=|k′ 1|2d|k′ 1|dΩCM, (11.28) wheredΩCM= sinθdθdφ is the diﬀerential solid angle, and θis the angle between k1andk′ 1in the CM frame. We can carry out the integral over the magnitude of k′ 1in eq.(11.26) using/integraltextdxδ(f(x)) =/summationtext i|f′(xi)|−1, wherexi satisﬁesf(xi) = 0. In our case, the argument of the delta function vanishes at just one value of |k′ 1|, the value given by eq.(11.3). Also, the derivative of that argument with respect to |k′ 1|is ∂ ∂|k′ 1|/parenleftig E′ 1+E′ 2−√s/parenrightig =|k′ 1| E′ 1+|k′ 1| E′ 2 =|k′ 1|/parenleftbiggE′ 1+E′ 2 E′ 1E′ 2/parenrightbigg =|k′ 1|√s E′ 1E′ 2. (11.29) Putting all of this together, we get dLIPS 2(k) =|k′ 1| 16π2√sdΩCM. (11.30) Combining this with eq.(11.22), we have dσ dΩCM=1 64π2s|k′ 1| |k1||T |2, (11.31) where |k1|and|k′ 1|are the functions of sgiven by eqs.(11.2) and (11.3), anddΩCMis the diﬀerential solid angle in the CM frame. The diﬀerential cross section can also be expressed in a fram e-independent manner by noting that, in the CM frame, we can take the diﬀeren tial of eq.(11.4) at ﬁxed sto get dt= 2|k1||k′ 1|dcosθ (11.32) = 2|k1||k′ 1|dΩCM 2π. (11.33) Now we can rewrite eq.(11.31) as dσ dt=1 64πs|k1|2|T |2, (11.34) 11: Cross Sections and Decay Rates 98 where |k1|is given as a function of sby eq.(11.2). We can now transform dσ/dt intodσ/dΩ in any frame we might like (such as the FT frame) by taking the diﬀerential of eq.(11.4) in that frame. In general, though, |k′ 1|depends on θas well ass, so the result is more complicated than it is in eq.(11.32) for the CM frame. Returning to the general case of n′outgoing particles, we can deﬁne a Lorentz invariant total cross section by integrating completely over all the outgoing momenta, and dividing by an appropriate symmetry factor S. If there aren′ iidentical outgoing particles of type i, then S=/productdisplay in′ i!, (11.35) and σ=1 S/integraldisplay dσ, (11.36) wheredσis given by eq.(11.22). We need the symmetry factor because merely integrating over all the outgoing momenta in dLIPSn′treats the ﬁnal state as being labeled by an ordered list of these momenta. But if some outgoing particles are identical, this is not correct; the momenta of the identical particles should be speciﬁed by an unordered list (because, for example, the state a† 1a† 2|0∝an}b∇acket∇i}htis identical to the state a† 2a† 1|0∝an}b∇acket∇i}ht). The symmetry factor provides the appropriate correction. In the case of two outgoing particles, eq.(11.36) becomes σ=1 S/integraldisplay dΩCMdσ dΩCM(11.37) =2π S/integraldisplay+1 −1dcosθdσ dΩCM, (11.38) whereS= 2 if the two outgoing particles are identical, and S= 1 if they are distinguishable. Equivalently, we can compute σfrom eq.(11.34) via σ=1 S/integraldisplaytmax tmindtdσ dt, (11.39) wheretminandtmaxare given by eq.(11.4) in the CM frame with cos θ=−1 and +1, respectively. To compute σwith eq.(11.38), we should ﬁrst express tanduin terms of sandθvia eqs.(11.4) and (11.6), and then integrate overθat ﬁxeds. To compute σwith eq.(11.39), we should ﬁrst express u in terms of sandtvia eq.(11.6), and then integrate over tat ﬁxeds. Let us see how all this works for the scattering amplitude of ϕ3theory, eq.(11.7). In this case, all the masses are equal, and so, in t he CM frame, 11: Cross Sections and Decay Rates 99 E=1 2√sfor all four particles, and |k′ 1|=|k1|=1 2(s−4m2)1/2. Then eq.(11.4) becomes t=−1 2(s−4m2)(1−cosθ). (11.40) From eq.(11.6), we also have u=−1 2(s−4m2)(1 + cosθ). (11.41) Thus|T |2is quite a complicated function of sandθ. In the nonrelativistic limit,|k1| ≪mor equivalently s−4m2≪m2, we have T=5g2 3m2/bracketleftigg 1−8 15/parenleftigg s−4m2 m2/parenrightigg +5 18/parenleftbigg 1 +27 25cos2θ/parenrightbigg/parenleftigg s−4m2 m2/parenrightigg2 +.../bracketrightigg +O(g4). (11.42) Thus the diﬀerential cross section is nearly isotropic. In t he extreme rela- tivistic limit, |k1| ≫mor equivalently s≫m2, we have T=g2 ssin2θ/bracketleftigg 3 + cos2θ−/parenleftigg (3 + cos2θ)2 sin2θ−16/parenrightigg m2 s+.../bracketrightigg +O(g4). (11.43) Now the diﬀerential cross section is sharply peaked in the fo rward (θ= 0) and backward ( θ=π) directions. We can compute the total cross section σfrom eq.(11.39). We have in this casetmin=−(s−4m2) andtmax= 0. Since the two outgoing particles are identical, the symmetry factor is S= 2. Then setting u= 4m2−s−t, and performing the integral in eq.(11.39) over tat ﬁxeds, we get σ=g4 32πs(s−4m2)/bracketleftigg 2 m2+s−4m2 (s−m2)2−2 s−3m2 +4m2 (s−m2)(s−2m2)ln/parenleftigg s−3m2 m2/parenrightigg/bracketrightigg +O(g6).(11.44) In the nonrelativistic limit, this becomes σ=25g4 1152πm6/bracketleftigg 1−79 60/parenleftigg s−4m2 m2/parenrightigg +.../bracketrightigg +O(g6). (11.45) In the extreme relativistic limit, we get σ=g4 16πm2s2/bracketleftigg 1 +7 2m2 s+.../bracketrightigg +O(g6). (11.46) 11: Cross Sections and Decay Rates 100 These results illustrate how even a very simple quantum ﬁeld theory can yield speciﬁc predictions for cross sections that could be t ested experimen- tally. Let us now turn to the other basic problem mentioned at the beg inning of this section: the case of a single incoming particle that decays ton′other particles. We have an immediate conceptual problem. According to our de velop- ment of the LSZ formula in section 5, each incoming and outgoi ng particle should correspond to a single-particle state that is an exac t eigenstate of the exact hamiltonian. This is clearly not the case for a part icle that can decay. Referring to ﬁg.(5.1), the hyperbola of such a partic le must lie above the continuum threshold. Strictly speaking, then, the LSZ f ormula is not applicable. A proper understanding of this issue requires a study of loop corrections that we will undertake in section 25. For now, we will simply a ssume that the LSZ formula continues to hold for a single incoming parti cle. Then we can retrace the steps from eq.(11.11) to eq.(11.20); the onl y change is that the norm of the initial state is now ∝an}b∇acketle{ti|i∝an}b∇acket∇i}ht= 2E1V (11.47) instead of eq.(11.16). Identifying the diﬀerential decay r atedΓ with ˙Pthen gives dΓ =1 2E1|T |2dLIPSn′(k1), (11.48) where nows=−k2 1=m2 1. In the CM frame (which is now the rest frame of the initial particle), we have E1=m1; in other frames, the relative factor ofE1/m1indΓ accounts for relativistic time dilation of the decay rate. We can also deﬁne a total decay rate by integrating over all th e outgoing momenta, and dividing by the symmetry factor of eq.(11.35): Γ =1 S/integraldisplay dΓ. (11.49) We will compute a decay rate in problem 11.1 Reference Notes For a derivation with wave packets, see Brown ,Itzykson & Zuber , orPeskin & Schroeder . Problems 11: Cross Sections and Decay Rates 101 11.1) a) Consider a theory of a two real scalar ﬁelds AandBwith an interaction L1=gAB2. Assuming that mA>2mB, compute the total decay rate of the Aparticle at tree level. b) Consider a theory of a real scalar ﬁeld ϕand a complex scalar ﬁeld χwithL1=gϕχ†χ. Assuming that mϕ>2mχ, compute the total decay rate of the ϕparticle at tree level. 11.2) Consider Compton scattering , in which a massless photon is scattered by an electron, initially at rest. (This is the FT frame.) In p roblem 59.1, we will compute |T |2for this process (summed over the possible spin states of the scattered photon and electron, and averag ed over the possible spin states of the initial photon and electron) , with the result |T |2= 32π2α2/bracketleftigg m4+m2(3s+u)−su (m2−s)2+m4+m2(3u+s)−su (m2−u)2 +2m2(s+u+ 2m2) (m2−s)(m2−u)/bracketrightigg +O(α4) (11.50) whereα= 1/137.036 is the ﬁne-structure constant. a) Express the Mandelstam variables sanduin terms of the initial and ﬁnal photon energies ωandω′. b) Express the scattering angle θFTbetween the initial and ﬁnal pho- ton three-momenta in terms of ωandω′. c) Express the diﬀerential scattering cross section dσ/dΩFTin terms ofωandω′. Show that your result is equivalent to the Klein-Nishina formula dσ dΩFT=α2 2m2ω′2 ω2/bracketleftbiggω ω′+ω′ ω−sin2θFT/bracketrightbigg . (11.51) 11.3) Consider the process of muon decay ,µ−→e−νeνµ. In section 88, we will compute |T |2for this process (summed over the possible spin states of the decay products, and averaged over the possible spin states of the initial muon), with the result |T |2= 64G2 F(k1·k′ 2)(k′ 1·k′ 3), (11.52) whereGFis the Fermi constant ,k1is the four-momentum of the muon, and k′ 1,2,3are the four-momenta of the νe,νµ, ande−, respec- tively. In the rest frame of the muon, its decay rate is theref ore Γ =32G2 F m/integraldisplay (k1·k′ 2)(k′ 1·k′ 3)dLIPS 3(k1), (11.53) 11: Cross Sections and Decay Rates 102 wherek1= (m,0), andmis the muon mass. The neutrinos are massless, and the electron mass is 200 times less than the muo n mass, so we can take the electron to be massless as well. To evaluate Γ, we perform the following analysis. a) Show that Γ =32G2 F m/integraldisplay /tildewiderdk′ 3k1µk′ 3ν/integraldisplay k′ 2µk′ 1νdLIPS 2(k1−k′ 3). (11.54) b) Use Lorentz invariance to argue that, for m1′=m2′= 0, /integraldisplay k′ 1µk′ 2νdLIPS 2(k) =Ak2gµν+Bkµkν, (11.55) whereAandBare numerical constants. c) Show that, for m1′=m2′= 0, /integraldisplay dLIPS 2(k) =1 8π. (11.56) d) By contracting both sides of eq.(11.55) with gµνand withkµkν, and using eq.(11.56), evaluate AandB. e) Use the results of parts (b) and (d) in eq.(11.54). Set k1= (m,0), and compute dΓ/dEe; hereEe≡E′ 3is the energy of the electron. Note that the maximum value of Eeis reached when the electron is emitted in one direction, and the two neutrinos in the oppo site direction; what is this maximum value? f) Perform the integral over Eeto obtain the muon decay rate Γ. g) The measured lifetime of the muon is 2 .197×10−6s. The muon mass is 105 .66MeV. Determine the value of GFin GeV−2. (Your answer is too low by about 0.2%, due to loop corrections to the decay rate.) h) Deﬁne the energy spectrum of the electron P(Ee)≡Γ−1dΓ/dEe. Note thatP(Ee)dEeis the probability for the electron to be emit- ted with energy between EeandEe+dEe. Draw a graph of P(Ee) vs.Ee/mµ. 11.4) Consider a theory of three real scalar ﬁelds ( A,B, andC) with L=−1 2∂µA∂µA−1 2m2 AA2 −1 2∂µB∂µB−1 2m2 BB2 −1 2∂µC∂µC−1 2m2 CC2 +gABC . (11.57) 11: Cross Sections and Decay Rates 103 Write down the tree-level scattering amplitude (given by th e sum of the contributing tree diagrams) for each of the following pr ocesses: AA→AA, AA→AB , AA→BB, AA→BC , AB→AB , AB→AC . (11.58) Your answers should take the form T=g2/bracketleftbiggcs m2s−s+ct m2 t−t+cu m2u−u/bracketrightbigg , (11.59) where, in each case, each ciis a positive integer, and each m2 iism2 A orm2 Borm2 C. Hint: Tmay be zero for some processes. 12: Dimensional Analysis with ¯h=c= 1 104 12Dimensional Analysis with ¯h=c= 1 Prerequisite: 3 We have set ¯ h=c= 1. This allows us to convert a time Tto a length L viaT=c−1L, and a length Lto an inverse mass M−1viaL= ¯hc−1M−1. Thus any quantity Acan be thought of as having units of mass to some power (positive, negative, or zero) that we will call [ A]. For example, [m] = +1, (12.1) [xµ] =−1, (12.2) [∂µ] = +1, (12.3) [ddx] =−d. (12.4) In the last line, we have generalized our considerations to t heories ind spacetime dimensions. Let us now consider a scalar ﬁeld in dspacetime dimensions with la- grangian density L=−1 2∂µϕ∂µϕ−1 2m2ϕ2−N/summationdisplay n=31 n!gnϕn. (12.5) The action is S=/integraldisplay ddxL, (12.6) and the path integral is Z(J) =/integraldisplay Dϕexp/bracketleftbigg i/integraldisplay ddx(L+Jϕ)/bracketrightbigg . (12.7) From eq.(12.7), we see that the action Smust be dimensionless, because it appears as the argument of the exponential function. Ther efore [S] = 0. (12.8) Combining eqs.(12.4) and (12.8) yields [L] =d. (12.9) Then, from eqs.(12.9) and (12.3), and the fact that ∂µϕ∂µϕis a term in L, we see that we must have [ϕ] =1 2(d−2). (12.10) Then, since gnϕnis also a term in L, we must have [gn] =d−1 2n(d−2). (12.11) 12: Dimensional Analysis with ¯h=c= 1 105 In particular, for the ϕ3theory we have been working with, we have [g3] =1 2(6−d). (12.12) Thus we see that the coupling constant of ϕ3theory is dimensionless in d= 6 spacetime dimensions. Theories with dimensionless couplings tend to be more inter esting than theories with dimensionful couplings. This is because any n ontrivial de- pendence of a scattering amplitude on a coupling must be expr essed as a function of a dimensionless parameter. If the coupling is it self dimension- ful, this parameter must be the ratio of the coupling to the ap propriate power of either the particle mass m(if it isn’t zero) or, in the high-energy regimes≫m2, the Mandelstam variable s. Thus the relevant parame- ter isgs−[g]/2. If [g] is negative [and it usually is: see eq.(12.11)], then gs−[g]/2blows up at high energies, and the perturbative expansion br eaks down. This behavior is connected to the nonrenormalizability of theories with couplings with negative mass dimension, a subject we wi ll take up in section 18. It turns out that such theories require an inﬁnit e number of input parameters to make sense; see section 29. In the opposi te case, [g] positive, the theory becomes trivial at high energy, becaus egs−[g]/2goes rapidly to zero. Thus the case of [ g] = 0 is just right: scattering amplitudes can have a nontrivial dependence on gat all energies. Therefore, from here on, we will be primarily interested in ϕ3theory in d= 6 spacetime dimensions, where [ g3] = 0. Problems 12.1) Express ¯ hcin GeVfm, where 1fm = 1 Fermi = 10−13cm. 12.2) Express the masses of the proton, neutron, pion, elect ron, muon, and tau in GeV. 12.3) The proton is a strongly interacting blob of quarks and gluons. It has a nonzero charge radius rp, given byr2 p=/integraltextd3xρ(r)r2, whereρ(r) is the quantum expectation value of the electric charge dist ribution inside the proton. Estimate the value of rp, and then look up its measured value. How accurate was your estimate? 13: The Lehmann-K¨ all´ en Form of the Exact Propagator 106 13The Lehmann-K ¨all´en Form of the Exact Propagator Prerequisite: 9 Before turning to the subject of loop corrections to scatter ing amplitudes, it will be helpful to consider what we can learn about the exac t propagator ∆(x−y) from general principles. We deﬁne the exact propagator via ∆(x−y)≡i∝an}b∇acketle{t0|Tϕ(x)ϕ(y)|0∝an}b∇acket∇i}ht. (13.1) We take the ﬁeld ϕ(x) to be normalized so that ∝an}b∇acketle{t0|ϕ(x)|0∝an}b∇acket∇i}ht= 0 and ∝an}b∇acketle{tk|ϕ(x)|0∝an}b∇acket∇i}ht=e−ikx. (13.2) Indspacetime dimensions, the one-particle state |k∝an}b∇acket∇i}hthas the normalization ∝an}b∇acketle{tk|k′∝an}b∇acket∇i}ht= (2π)d−12ωδd−1(k−k′), (13.3) withω= (k2+m2)1/2. The corresponding completeness statement is /integraldisplay /tildewiderdk|k∝an}b∇acket∇i}ht∝an}b∇acketle{tk|=I1, (13.4) whereI1is the identity operator in the one-particle subspace, and /tildewiderdk≡dd−1k (2π)d−12ω(13.5) is the Lorentz invariant phase-space diﬀerential. We also d eﬁne the exact momentum-space propagator ˜∆(k2) via ∆(x−y)≡/integraldisplayddk (2π)deik(x−y)˜∆(k2). (13.6) In free-ﬁeld theory, the momentum-space propagator is ˜∆(k2) =1 k2+m2−iǫ. (13.7) It has an isolated pole at k2=−m2with residue one; mis the actual, phys- ical mass of the particle, the mass that enters into the energ y-momentum relation. We begin our analysis with eq.(13.1). We take x0> y0, and insert a complete set of energy eigenstates between the two ﬁelds. R ecall from section 5 that there are three general classes of energy eige nstates: 13: The Lehmann-K¨ all´ en Form of the Exact Propagator 107 1. The ground state or vacuum |0∝an}b∇acket∇i}ht, which is a single state with zero energy and momentum. 2. The one particle states |k∝an}b∇acket∇i}ht, speciﬁed by a three-momentum kand with energy ω= (k2+m2)1/2. 3. States in the multiparticle continuum |k,n∝an}b∇acket∇i}ht, speciﬁed by a three- momentum kand other parameters (such as relative momenta among the diﬀerent particles) that we will collectively denote as n. The energy of one of these states is ω= (k2+M2)1/2, whereM≥2m;M is one of the parameters in the set n. Thus we get ∝an}b∇acketle{t0|ϕ(x)ϕ(y)|0∝an}b∇acket∇i}ht=∝an}b∇acketle{t0|ϕ(x)|0∝an}b∇acket∇i}ht∝an}b∇acketle{t0|ϕ(y)|0∝an}b∇acket∇i}ht +/integraldisplay /tildewiderdk∝an}b∇acketle{t0|ϕ(x)|k∝an}b∇acket∇i}ht∝an}b∇acketle{tk|ϕ(y)|0∝an}b∇acket∇i}ht +/summationdisplay n/integraldisplay /tildewiderdk∝an}b∇acketle{t0|ϕ(x)|k,n∝an}b∇acket∇i}ht∝an}b∇acketle{tk,n|ϕ(y)|0∝an}b∇acket∇i}ht.(13.8) The sum over nis schematic, and includes integrals over continuous pa- rameters like relative momenta. The ﬁrst two terms in eq.(13.8) can be simpliﬁed via eq.(13.2 ). Also, writing the ﬁeld as ϕ(x) = exp( −iPµxµ)ϕ(0)exp(+iPµxµ), wherePµis the energy-momentum operator, gives us ∝an}b∇acketle{tk,n|ϕ(x)|0∝an}b∇acket∇i}ht=e−ikx∝an}b∇acketle{tk,n|ϕ(0)|0∝an}b∇acket∇i}ht, (13.9) wherek0= (k2+M2)1/2. We now have ∝an}b∇acketle{t0|ϕ(x)ϕ(y)|0∝an}b∇acket∇i}ht=/integraldisplay /tildewiderdkeik(x−y)+/summationdisplay n/integraldisplay /tildewiderdkeik(x−y)|∝an}b∇acketle{tk,n|ϕ(0)|0∝an}b∇acket∇i}ht|2.(13.10) Next, we deﬁne the spectral density ρ(s)≡/summationdisplay n|∝an}b∇acketle{tk,n|ϕ(0)|0∝an}b∇acket∇i}ht|2δ(s−M2). (13.11) Obviously, ρ(s)≥0 fors≥4m2, andρ(s) = 0 fors<4m2. Now we have ∝an}b∇acketle{t0|ϕ(x)ϕ(y)|0∝an}b∇acket∇i}ht=/integraldisplay /tildewiderdkeik(x−y)+/integraldisplay∞ 4m2dsρ(s)/integraldisplay /tildewiderdkeik(x−y).(13.12) In the ﬁrst term, k0= (k2+m2)1/2, and in the second term, k0= (k2+s)1/2. Clearly we can also swap xandyto get ∝an}b∇acketle{t0|ϕ(y)ϕ(x)|0∝an}b∇acket∇i}ht=/integraldisplay /tildewiderdke−ik(x−y)+/integraldisplay∞ 4m2dsρ(s)/integraldisplay /tildewiderdke−ik(x−y)(13.13) 13: The Lehmann-K¨ all´ en Form of the Exact Propagator 108 as well. We can then combine eqs.(13.12) and (13.13) into a fo rmula for the time-ordered product ∝an}b∇acketle{t0|Tϕ(x)ϕ(y)|0∝an}b∇acket∇i}ht=θ(x0−y0)∝an}b∇acketle{t0|ϕ(x)ϕ(y)|0∝an}b∇acket∇i}ht+θ(y0−x0)∝an}b∇acketle{t0|ϕ(y)ϕ(x)|0∝an}b∇acket∇i}ht, (13.14) whereθ(t) is the unit step function, by means of the identity /integraldisplayddk (2π)deik(x−y) k2+m2−iǫ=iθ(x0−y0)/integraldisplay /tildewiderdkeik(x−y) +iθ(y0−x0)/integraldisplay /tildewiderdke−ik(x−y); (13.15) the derivation of eq.(13.15) was sketched in section 8. Comb ining eqs.(13.12– 13.15), we get i∝an}b∇acketle{t0|Tϕ(x)ϕ(y)|0∝an}b∇acket∇i}ht=/integraldisplayddk (2π)deik(x−y)/bracketleftigg 1 k2+m2−iǫ +/integraldisplay∞ 4m2dsρ(s)1 k2+s−iǫ/bracketrightigg .(13.16) Comparing eqs.(13.1), (13.6), and (13.16), we see that ˜∆(k2) =1 k2+m2−iǫ+/integraldisplay∞ 4m2dsρ(s)1 k2+s−iǫ. (13.17) This is the Lehmann-K¨ all´ en form of the exact momentum-space propagator ˜∆(k2). We note in particular that ˜∆(k2)has an isolated pole at k2=−m2 with residue one , just like the propagator in free-ﬁeld theory. Problems 13.1) Consider an interacting scalar ﬁeld theory in dspacetime dimensions, L=−1 2Zϕ∂µϕ∂µϕ−1 2Zmm2ϕ2− L1(ϕ), (13.18) where L1(ϕ) is a function of ϕ(and not its derivatives). The exact momentum-space propagator for ϕcan be expressed in Lehmann- K¨ all´ en form by eq.(13.17). Find a formula for the renormal izing fac- torZϕin terms of ρ(s). Hint: consider the commutator [ ϕ(x),˙ϕ(y)]. 14: Loop Corrections to the Propagator 109 14Loop Corrections to the Propagator Prerequisite: 10, 12, 13 In section 10, we wrote the exact propagator as 1 i∆(x1−x2)≡ ∝an}b∇acketle{t0|Tϕ(x1)ϕ(x2)|0∝an}b∇acket∇i}ht=δ1δ2iW(J)/vextendsingle/vextendsingle/vextendsingle J=0, (14.1) whereiW(J) is the sum of connected diagrams, and δiacts to remove a source from a diagram and label the corresponding propagato r endpoint xi. Inϕ3theory, the O(g2) corrections to1 i∆(x1−x2) come from the di- agrams of ﬁg.(14.1). To compute them, it is simplest to work d irectly in momentum space, following the Feynman rules of section 10. A n appro- priate assignment of momenta to the lines is shown in ﬁg.(14. 1); we then have 1 i˜∆(k2) =1 i˜∆(k2) +1 i˜∆(k2)/bracketleftig iΠ(k2)/bracketrightig 1 i˜∆(k2) +O(g4), (14.2) where ˜∆(k2) =1 k2+m2−iǫ(14.3) is the free-ﬁeld propagator, and iΠ(k2) =1 2(ig)2/parenleftig 1 i/parenrightig2/integraldisplayddℓ (2π)d˜∆((ℓ+k)2)˜∆(ℓ2) −i(Ak2+Bm2) +O(g4) (14.4) is the self-energy . Here we have written the integral appropriate for d spacetime dimensions; for now we will leave darbitrary, but later we will want to focus on d= 6, where the coupling gis dimensionless. In the ﬁrst term in eq.(14.4), the factor of one-half is the sy mmetry factor associated with exchanging the top and bottom semici rcular prop- agators. Also, we have written the vertex factor as igrather than iZgg because we expect Zg= 1 +O(g2), and so the Zg−1 contribution can be lumped into the O(g4) term. In the second term, A=Zϕ−1 and B=Zm−1 are both expected to be O(g2). It will prove convenient to deﬁne Π( k2) to all orders via the geometric series 1 i˜∆(k2) =1 i˜∆(k2) +1 i˜∆(k2)/bracketleftig iΠ(k2)/bracketrightig 1 i˜∆(k2) +1 i˜∆(k2)/bracketleftig iΠ(k2)/bracketrightig 1 i˜∆(k2)/bracketleftig iΠ(k2)/bracketrightig 1 i˜∆(k2) +... . (14.5) 14: Loop Corrections to the Propagator 110 k k kkk k+l l Figure 14.1: The O(g2) corrections to the propagator. Figure 14.2: The geometric series for the exact propagator. This is illustrated in ﬁg.(14.2). The sum in eq.(14.5) will i nclude allthe diagrams that contribute to ˜∆(k2) if we take iΠ(k2) to be given by the sum of all diagrams that are one-particle irreducible , or 1PI for short. A diagram is 1PI if it is still connected after any one line is cut. The 1P I diagrams that make an O(g4) contribution to iΠ(k2) are shown in ﬁg.(14.3). When writing down the value of one of these diagrams, we omit the tw o external propagators. If we sum up the series in eq.(14.5), we get ˜∆(k2) =1 k2+m2−iǫ−Π(k2). (14.6) In section 13, we learned that the exact propagator has a pole atk2=−m2 with residue one. This is consistent with eq.(14.6) if and on ly if Π(−m2) = 0, (14.7) Π′(−m2) = 0, (14.8) where the prime denotes a derivative with respect to k2.We will use eqs.(14.7) and (14.8) to ﬁx the values of AandB. Figure 14.3: The O(g4) contributions to iΠ(k2). 14: Loop Corrections to the Propagator 111 Next we turn to the evaluation of the O(g2) contribution to iΠ(k2) in eq.(14.4). We have the immediate problem that the integral o n the right- hand side diverges at large ℓford≥4. We faced a similar situation in section 9 when we evaluated the lowest-order tadpole diagra m. There we introduced an ultraviolet cutoﬀ Λ that modiﬁed the behavior of ˜∆(ℓ2) at largeℓ2. Here, for now, we will simply restrict our attention to d <4, where the integral in eq.(14.4) is ﬁnite. Later we will see wh at we can say about larger values of d. We will evaluate the integral in eq.(14.4) with a series of tr icks. We ﬁrst use Feynman’s formula to combine denominators, 1 A1...An=/integraldisplay dFn(x1A1+...+xnAn)−n, (14.9) where the integration measure over the Feynman parameters xiis /integraldisplay dFn= (n−1)!/integraldisplay1 0dx1...dxnδ(x1+...+xn−1). (14.10) This measure is normalized so that /integraldisplay dFn1 = 1. (14.11) We will prove eq.(14.9) in problem 14.1. In the case at hand, we have ˜∆((k+ℓ)2)˜∆(ℓ2) =1 (ℓ2+m2)((ℓ+k)2+m2) =/integraldisplay1 0dx/bracketleftig x((ℓ+k)2+m2) + (1−x)(ℓ2+m2)/bracketrightig−2 =/integraldisplay1 0dx/bracketleftig ℓ2+ 2xℓ·k+xk2+m2/bracketrightig−2 =/integraldisplay1 0dx/bracketleftig (ℓ+xk)2+x(1−x)k2+m2/bracketrightig−2 =/integraldisplay1 0dx/bracketleftig q2+D/bracketrightig−2, (14.12) where we have suppressed the iǫ’s for notational convenience; they can be restored via the replacement m2→m2−iǫ. In the last line we have deﬁned q≡ℓ+xk (14.13) and D≡x(1−x)k2+m2. (14.14) 14: Loop Corrections to the Propagator 112 Req0Im q0 Figure 14.4: The q0integration contour along the real axis can be rotated to the imaginary axis without passing through the poles at q0=−ω+iǫ andq0= +ω−iǫ. We then change the integration variable in eq.(14.4) from ℓtoq; the jaco- bian is trivial, and we have ddℓ=ddq. Next, think of the integral over q0from−∞to +∞as a contour integral in the complex q0plane. If the integrand vanishes fast enough as |q0| → ∞ , we can rotate this contour clockwise by 90◦, as shown in ﬁg.(14.4), so that it runs from −i∞to +i∞. In making this Wick rotation , the contour does not pass over any poles. (The iǫ’s are needed to make this statement unambiguous.) Thus the value of the integral is unchanged. I t is now convenient to deﬁne a euclidean d-dimensional vector ¯ qviaq0=i¯qdand qj= ¯qj; thenq2= ¯q2, where ¯q2= ¯q2 1+...+ ¯q2 d. (14.15) Also,ddq=idd¯q. Therefore, in general, /integraldisplay ddqf(q2−iǫ) =i/integraldisplay dd¯qf(¯q2) (14.16) as long asf(¯q2)→0 faster than 1 /¯qdas ¯q→ ∞. Now we can write Π(k2) =1 2g2I(k2)−Ak2−Bm2+O(g4), (14.17) where I(k2)≡/integraldisplay1 0dx/integraldisplaydd¯q (2π)d1 (¯q2+D)2. (14.18) It is now straightforward to evaluate the d-dimensional integral over ¯ qin spherical coordinates. Before we perform this calculation, however, let us introdu ce another trick, one that can simplify the task of ﬁxing AandBthrough the impo- sition of eqs.(14.7) and (14.8). Here is the trick: diﬀerent iate Π(k2) twice with respect to k2to get Π′′(k2) =1 2g2I′′(k2) +O(g4), (14.19) 14: Loop Corrections to the Propagator 113 where, from eqs.(14.18) and (14.14), I′′(k2) =/integraldisplay1 0dx6x2(1−x)2/integraldisplaydd¯q (2π)d1 (¯q2+D)4. (14.20) Then, after we evaluate these integrals, we can get Π( k2) by integrating with respect tok2, subject to the boundary conditions of eqs.(14.7) and (14.8 ). In this way we can construct Π( k2) without ever explicitly computing A andB. Notice that this trick does something else for us as well. The integral over ¯qin eq.(14.20) is ﬁnite for any d<8, whereas the original integral in eq.(14.18) is ﬁnite only for d<4. This expanded range of dnow includes the value of greatest interest, d= 6. How did this happen? We can gain some insight by making a Taylo r expansion of Π( k2) aboutk2=−m2: Π(k2) =/bracketleftig 1 2g2I(−m2) + (A−B)m2/bracketrightig +/bracketleftig 1 2g2I′(−m2) +A/bracketrightig (k2+m2) +1 2!/bracketleftig 1 2g2I′′(−m2)/bracketrightig (k2+m2)2+... +O(g4). (14.21) From eqs.(14.18) and (14.14), it is straightforward to see t hatI(−m2) is divergent for d≥4,I′(−m2) is divergent for d≥6, and, in general, I(n)(−m2) is divergent for d≥4 + 2n. We can use the O(g2) terms in AandBto cancel oﬀ the1 2g2I(−m2) and1 2g2I′(−m2) terms in Π( k2), whether or not they are divergent. But if we are to end up with a ﬁnite Π(k2), all of the remaining terms must be ﬁnite, since we have no mo re free parameters left to adjust. This is the case for d<8. Of course, for 4 ≤d <8, the values of AandB(and hence the la- grangian coeﬃcients Z= 1 +AandZm= 1 +B) are formally inﬁnite, and this may be disturbing. However, these coeﬃcients are not di rectly mea- surable, and so need not obey our preconceptions about their magnitudes. Also, it is important to remember that AandBeach includes a factor of g2; this means that we can expand in powers of AandBas part of our general expansion in powers of g. When we compute Π( k2) (which enters into observable cross sections), all the formally inﬁnite n umbers cancel in a well-deﬁned way, provided d<8. Ford≥8, this procedure breaks down, and we do not obtain a ﬁnite expression for Π( k2). In this case, we say that the theory is nonrenormaliz- able. We will discuss the criteria for renormalizability of a the ory in detail in section 18. It turns out that ϕ3theory is renormalizable for d≤6. (The 14: Loop Corrections to the Propagator 114 problem with 6 <d< 8 arises from higher-order corrections, as we will see in section 18.) Now let us return to the calculation of Π( k2). Rather than using the trick of ﬁrst computing Π′′(k2), we will instead evaluate Π( k2) directly from eq.(14.18) as a function of dford <4. Then we will analytically continue the result to arbitrary d. This procedure is known as dimensional regularization . Then we will ﬁx AandBby imposing eqs.(14.7) and (14.8), and ﬁnally take the limit d→6. We could just as well use the method of section 9. Making the re place- ment ˜∆(p2)→1 p2+m2−iǫΛ2 p2+ Λ2−iǫ, (14.22) where Λ is the ultraviolet cutoﬀ, renders the O(g2) term in Π( k2) ﬁnite ford <8; This procedure is known as Pauli–Villars regularization . We then evaluate Π( k2) as a function of Λ, ﬁx AandBby imposing eqs.(14.7) and (14.8), and take the Λ → ∞ limit. Calculations with Pauli-Villars regularization are generally much more cumbersome than the y are with dimensional regularization. However, the ﬁnal result for Π (k2) is the same. Eq.(14.21) demonstrates that anyregularization scheme will give the same result ford <8, at least as long as it preserves the Lorentz invariance of the integrals. We therefore turn to the evaluation of I(k2), eq.(14.18). The angu- lar part of the integral over ¯ qyields the area Ω dof the unit sphere in d dimensions, which is Ωd=2πd/2 Γ(1 2d); (14.23) this is most easily veriﬁed by computing the gaussian integr al/integraltextdd¯qe−¯q2in both cartesian and spherical coordinates. Here Γ( x) is the Euler gamma function; for a nonnegative integer nand smallx, Γ(n+1) =n!, (14.24) Γ(n+1 2) =(2n)! n!2n√π, (14.25) Γ(−n+x) =(−1)n n!/bracketleftbigg1 x−γ+/summationdisplayn k=1k−1+O(x)/bracketrightbigg ,(14.26) whereγ= 0.5772...is the Euler-Mascheroni constant. The radial part of the ¯ qintegral can also be evaluated in terms of gamma functions. The overall result (generalized slightly) is /integraldisplaydd¯q (2π)d(¯q2)a (¯q2+D)b=Γ(b−a−1 2d)Γ(a+1 2d) (4π)d/2Γ(b)Γ(1 2d)D−(b−a−d/2). (14.27) 14: Loop Corrections to the Propagator 115 We will make frequent use of this formula throughout this boo k. In the case of interest, eq.(14.18), we have a= 0 andb= 2. There is one more complication to deal with. Recall that we wa nt to focus ond= 6 because in that case gis dimensionless. However, for general d,ghas mass dimension ε/2, where ε≡6−d. (14.28) To account for this, we introduce a new parameter ˜ µwith dimensions of mass, and make the replacement g→g˜µε/2. (14.29) In this way gremains dimensionless for all ε. Of course, ˜ µis not an actual parameter of the d= 6 theory. Therefore, nothing measurable (like a cross section) can depend on it. This seemingly innocuous statement is actually quite power ful, and will eventually serve as the foundation of the renormalization g roup. We now return to eq.(14.18), use eq.(14.26), and set d= 6−ε; we get I(k2) =Γ(−1+ε 2) (4π)3/integraldisplay1 0dxD/parenleftbigg4π D/parenrightbiggε/2 . (14.30) Hence, with the substitution of eq.(14.29), and deﬁning α≡g2 (4π)3(14.31) for notational convenience, we have Π(k2) =1 2αΓ(−1+ε 2)/integraldisplay1 0dxD/parenleftigg 4π˜µ2 D/parenrightiggε/2 −Ak2−Bm2+O(α2). (14.32) Now we can take the ε→0 limit, using eq.(14.26) and Aε/2= 1 +ε 2lnA+O(ε2). (14.33) The result is Π(k2) =−1 2α/bracketleftigg/parenleftig 2ε+ 1/parenrightig/parenleftig 1 6k2+m2/parenrightig +/integraldisplay1 0dxDln/parenleftigg 4π˜µ2 eγD/parenrightigg/bracketrightigg −Ak2−Bm2+O(α2). (14.34) 14: Loop Corrections to the Propagator 116 Here we have used/integraltext1 0dxD=1 6k2+m2. It is now convenient to deﬁne µ≡√ 4πe−γ/2˜µ, (14.35) and rearrange things to get Π(k2) =1 2α/integraldisplay1 0dxDln(D/m2) −/braceleftig 1 6α/bracketleftig 1ε+ ln(µ/m) +1 2/bracketrightig +A/bracerightig k2 −/braceleftig α/bracketleftig 1ε+ ln(µ/m) +1 2/bracketrightig +B/bracerightig m2+O(α2).(14.36) If we takeAandBto have the form A=−1 6α/bracketleftig 1ε+ ln(µ/m) +1 2+κA/bracketrightig +O(α2), (14.37) B=−α/bracketleftig 1ε+ ln(µ/m) +1 2+κB/bracketrightig +O(α2), (14.38) whereκAandκBare purely numerical constants, then we get Π(k2) =1 2α/integraldisplay1 0dxDln(D/m2) +α/parenleftig 1 6κAk2+κBm2/parenrightig +O(α2).(14.39) Thus this choice of AandBrenders Π(k2) ﬁnite and independent of µ, as required. To ﬁxκAandκB, we must still impose the conditions Π( −m2) = 0 and Π′(−m2) = 0. The easiest way to do this is to ﬁrst note that, schematic ally, Π(k2) =1 2α/integraldisplay1 0dxDlnD+ linear in k2andm2+O(α2).(14.40) We can then impose Π( −m2) = 0 via Π(k2) =1 2α/integraldisplay1 0dxDln(D/D 0) + linear in ( k2+m2) +O(α2).(14.41) where D0≡D/vextendsingle/vextendsingle/vextendsingle k2=−m2= [1−x(1−x)]m2. (14.42) Now it is straightforward to diﬀerentiate eq.(14.41) with r espect tok2, and ﬁnd that Π′(−m2) vanishes for Π(k2) =1 2α/integraldisplay1 0dxDln(D/D 0)−1 12α(k2+m2) +O(α2). (14.43) The integral over xcan be done in closed form; the result is Π(k2) =1 12α/bracketleftig c1k2+c2m2+ 2k2f(r)/bracketrightig +O(α2), (14.44) 14: Loop Corrections to the Propagator 117 -30 -20 -10 10 20 30k2 /FrΑctionBΑrExt/FrΑctionBΑrExt/FrΑctionBΑrExt/FrΑctionBΑrExt/FrΑctionBΑrExt/FrΑctionBΑrExt/FrΑctionBΑrExtm2 -0.2-0.10.1Α Figure 14.5: The real and imaginary parts of Π( k2)/(k2+m2) in units of α. wherec1= 3−π√ 3,c2= 3−2π√ 3, and f(r) =r3tanh−1(1/r), (14.45) r= (1 + 4m2/k2)1/2. (14.46) There is a branch point at k2=−4m2, and Π(k2) acquires an imaginary part fork2<−4m2; we will discuss this further in the next section. We can write the exact propagator as ˜∆(k2) =/parenleftbigg1 1−Π(k2)/(k2+m2)/parenrightbigg1 k2+m2−iǫ. (14.47) In ﬁg.(14.5), we plot the real and imaginary parts of Π( k2)/(k2+m2) in units ofα. We see that its values are quite modest for the plotted range . For much larger values of |k2|, we have Π(k2) k2+m2≃1 12α/bracketleftig ln(k2/m2) +c1/bracketrightig +O(α2). (14.48) If we had kept track of the iǫ’s,k2would bek2−iǫ; whenk2is negative, we have ln( k2−iǫ) = ln |k2| −iπ. The imaginary part of Π( k2)/(k2+m2) therefore approaches the asymptotic value of −1 12πα+O(α2) whenk2is large and negative. The real part of Π( k2)/(k2+m2), however, continues to increase logarithmically with |k2|when|k2|is large. We will begin to address the meaning of this in section 26. 14: Loop Corrections to the Propagator 118 Problems 14.1) Derive a generalization of Feynman’s formula, 1 Aα1 1...Aαn n=Γ(/summationtext iαi)/producttext iΓ(αi)1 (n−1)!/integraldisplay dFn/producttext ixαi−1 i (/summationtext ixiAi)/summationtext iαi.(14.49) Hint: start with Γ(α) Aα=/integraldisplay∞ 0dttα−1e−At, (14.50) which deﬁnes the gamma function. Put an index on A,α, andt, and take the product. Then multiply on the right-hand side by 1 =/integraldisplay∞ 0dsδ(s−/summationtext iti). (14.51) Make the change of variable ti=sxi, and carry out the integral over s. 14.2) Verify eq.(14.23). 14.3) a) Show that /integraldisplay ddqqµf(q2) = 0, (14.52) /integraldisplay ddqqµqνf(q2) =C2gµν/integraldisplay ddqq2f(q2), (14.53) and evaluate the constant C2in terms ofd. Hint: use Lorentz symme- try to argue for the general structure, and evaluate C2by contracting withgµν. b) Similarly evaluate/integraltextddqqµqνqρqσf(q2). 14.4) Compute the values of κAandκB. 14.5) Compute the O(λ) correction to the propagator in ϕ4theory (see problem 9.2) in d= 4−εspacetime dimensions, and compute the O(λ) terms inAandB. 14.6) Repeat problem 14.5 for the theory of problem 9.3. 14.7)Renormalization of the anharmonic oscillator. Consider an anhar- monic oscillator, speciﬁed by the lagrangian L=1 2Z˙q2−1 2Zωω2q2−Zλλω3q4. (14.54) We set ¯h= 1 andm= 1;λis then dimensionless. 14: Loop Corrections to the Propagator 119 a) Find the hamiltonian Hcorresponding to L. Write it as H= H0+H1, whereH0=1 2P2+1 2ω2Q2, and [Q,P] =i. b) Let |0∝an}b∇acket∇i}htand|1∝an}b∇acket∇i}htbe the ground and ﬁrst excited states of H0, and let|Ω∝an}b∇acket∇i}htand|I∝an}b∇acket∇i}htbe the ground and ﬁrst excited states of H. (We take all these eigenstates to have unit norm.) We deﬁne ωto be the excitation energy of H,ω≡EI−EΩ. We normalize the position operatorQby setting ∝an}b∇acketle{tI|Q|Ω∝an}b∇acket∇i}ht=∝an}b∇acketle{t1|Q|0∝an}b∇acket∇i}ht= (2ω)−1/2. Finally, to make things mathematically simpler, we set Zλequal to one, rather than using a more physically motivated deﬁnition. Write Z= 1 +A andZω= 1 +B, whereA=κAλ+O(λ2) andB=κBλ+O(λ2). Use Rayleigh–Schr¨ odinger perturbation theory to compute theO(λ) corrections to the unperturbed energy eigenvalues and eige nstates. c) Find the numerical values of κAandκBthat yieldω=EI−EΩ and∝an}b∇acketle{tI|Q|Ω∝an}b∇acket∇i}ht= (2ω)−1/2. d) Now think of the lagrangian of eq.(14.54) as specifying a q uantum ﬁeld theory in d= 1 dimensions. Compute the O(λ) correction to the propagator. Fix κAandκBby requiring the propagator to have a pole atk2=−ω2with residue one. Do your results agree with those of part (c)? Should they? 15: The One-Loop Correction in Lehmann-K¨ all´ en Form 120 15The One-Loop Correction in Lehmann-K ¨all´en Form Prerequisite: 14 In section 13, we found that the exact propagator could be wri tten in Lehmann-K¨ all´ en form as ˜∆(k2) =1 k2+m2−iǫ+/integraldisplay∞ 4m2dsρ(s)1 k2+s−iǫ, (15.1) where the spectral density ρ(s) is real and nonnegative. In section 14, on the other hand, we found that the exact propagator could be wr itten as ˜∆(k2) =1 k2+m2−iǫ−Π(k2), (15.2) and that, to O(g2) inϕ3theory in six dimensions, Π(k2) =1 2α/integraldisplay1 0dxDln(D/D 0)−1 12α(k2+m2) +O(α2), (15.3) where α≡g2/(4π)3, (15.4) D=x(1−x)k2+m2−iǫ, (15.5) D0= [1−x(1−x)]m2. (15.6) In this section, we will attempt to reconcile eqs.(15.2) and (15.3) with eq.(15.1). Let us begin by considering the imaginary part of the propaga tor. We will always take k2andm2to be real, and explicitly include the appropriate factors ofiǫwhenever they are needed. We can use eq.(15.1) and the identity 1 x−iǫ=x x2+ǫ2+iǫ x2+ǫ2 =P1 x+iπδ(x), (15.7) wherePmeans the principal part, to write Im˜∆(k2) =πδ(k2+m2) +/integraldisplay∞ 4m2dsρ(s)πδ(k2+s) =πδ(k2+m2) +πρ(−k2), (15.8) 15: The One-Loop Correction in Lehmann-K¨ all´ en Form 121 whereρ(s)≡0 fors<4m2. Thus we have πρ(s) = Im ˜∆(−s) fors≥4m2. (15.9) Let us now suppose that Im Π( k2) = 0 for some range of k2. (In section 14, we saw that the O(α) contribution to Π( k2) is purely real for k2> −4m2.) Then, from eqs.(15.2) and (15.7), we get Im˜∆(k2) =πδ(k2+m2−Π(k2)) for ImΠ( k2) = 0. (15.10) From Π( −m2) = 0, we know that the argument of the delta function van- ishes atk2=−m2, and from Π′(−m2) = 0, we know that the derivative of this argument with respect to k2equals one at k2=−m2. Therefore Im˜∆(k2) =πδ(k2+m2) for ImΠ( k2) = 0. (15.11) Comparing this with eq.(15.8), we see that ρ(−k2) = 0 if Im Π( k2) = 0. Now suppose Im Π( k2) isnotzero for some range of k2. (In section 14, we saw that the O(α) contribution to Π( k2) has a nonzero imaginary part fork2<−4m2.) Then we can ignore the iǫin eq.(15.2), and Im˜∆(k2) =Im Π(k2) (k2+m2+ ReΠ(k2))2+ (Im Π(k2))2for ImΠ(k2)∝ne}ationslash= 0. (15.12) Comparing this with eq.(15.8) we see that πρ(s) =ImΠ(−s) (−s+m2+ Re Π( −s))2+ (Im Π( −s))2. (15.13) Since we know ρ(s) = 0 fors <4m2, this tells us that we must also have Im Π(−s) = 0 fors <4m2, or equivalently Im Π( k2) = 0 fork2>−4m2. This is just what we found for the O(α) contribution to Π( k2) in section 14. We can also see this directly from eq.(15.3), without doing t he integral overx. The integrand in this formula is real as long as the argument of the logarithm is real and positive. From eq.(15.5), we see that Dis real and positive if and only if x(1−x)k2>−m2. The maximum value of x(1−x) is 1/4, and so the argument of the logarithm is real and positive fo r the whole integration range 0 ≤x≤1 if and only if k2>−4m2. In this regime, ImΠ( k2) = 0. On the other hand, for k2<−4m2, the argument of the logarithm becomes negative for some of the integratio n range, and so ImΠ(k2)∝ne}ationslash= 0 fork2<−4m2. This is exactly what we need to reconcile eqs.(15.2) and (15.3) with eq.(15.1). Problems 15: The One-Loop Correction in Lehmann-K¨ all´ en Form 122 15.1) In this problem we will verify the result of problem 13. 1 toO(α). a) Let Π loop(k2) be given by the ﬁrst line of eq.(14.32), with ε >0. Show that, up to O(α2) corrections, A= Π′ loop(−m2). (15.14) Then use Cauchy’s integral formula to write this as A=/contintegraldisplaydw 2πiΠloop(w) (w+m2)2, (15.15) where the contour of integration is a small counterclockwis e circle around −m2in the complex wplane. b) By examining eq.(14.32), show that the only singularity o f Πloop(k2) is a branch point at k2=−4m2. Take the cut to run along the neg- ative real axis. c) Distort the contour in eq.(15.15) to a circle at inﬁnity wi th a detour around the branch cut. Examine eq.(14.32) to show that, for ε>0, the circle at inﬁnity does not contribute. The contour aroun d the branch cut then yields A=/integraldisplay−4m2 −∞dw 2πi1 (w+m2)2/bracketleftig Πloop(w+iǫ)−Πloop(w−iǫ)/bracketrightig ,(15.16) whereǫis inﬁnitesimal (and is not to be confused with ε= 6−d). d) Examine eq.(14.32) to show that the real part of Π loop(w) is con- tinuous across the branch cut, and that the imaginary part ch anges sign, so that Πloop(w+iǫ)−Πloop(w−iǫ) =−2iIm Π loop(w−iǫ). (15.17) e) Letw=−sin eq.(15.16) and use eq.(15.17) to get A=−1 π/integraldisplay∞ 4m2dsIm Π loop(−s−iǫ) (s−m2)2. (15.18) Use this to verify the result of problem 13.1 to O(α). 15.2)Dispersion relations. Consider the exact Π( k2), withε= 0. Assume that its only singularity is a branch point at k2=−4m2, that it obeys eq.(15.17), and that Π( k2) grows more slowly than |k2|2at large |k2|. By recapitulating the analysis in the previous problem, sho w that Π′′(k2) =2 π/integraldisplay∞ 4m2dsIm Π(−s−iε) (k2+s)3. (15.19) 15: The One-Loop Correction in Lehmann-K¨ all´ en Form 123 This is a twice subtracted dispersion relation . It gives Π′′(k2) through- out the complex k2plane in terms of the values of the imaginary part of Π(k2) along the branch cut. 16: Loop Corrections to the Vertex 124 16Loop Corrections to the Vertex Prerequisite: 14 Consider the O(g3) diagram of ﬁg.(16.1), which corrects the ϕ3vertex. In this section we will evaluate this diagram. We can deﬁne an exact three-point vertex function iV3(k1,k2,k3) as the sum of one-particle irreducible diagrams with three extern al lines carrying momentak1,k2, andk3, all incoming, with k1+k2+k3= 0 by momentum conservation. (In adopting this convention, we allow k0 ito have either sign; ifkiis the momentum of an external particle, then the sign of k0 iis positive if the particle is incoming, and negative if it is outgoing.) The original vertexiZggis the ﬁrst term in this sum, and the diagram of ﬁg.(16.1) is the second. Thus we have iV3(k1,k2,k3) =iZgg+ (ig)3/parenleftig 1 i/parenrightig3/integraldisplayddℓ (2π)d˜∆((ℓ−k1)2)˜∆((ℓ+k2)2)˜∆(ℓ2) +O(g5). (16.1) In the second term, we have set Zg= 1 +O(g2). We proceed immediately to the evaluation of this integral, using the series of trick s from section 14. First we use Feynman’s formula to write ˜∆((ℓ−k1)2)˜∆((ℓ+k2)2)˜∆(ℓ2) =/integraldisplay dF3/bracketleftig x1(ℓ−k1)2+x2(ℓ+k2)2+x3ℓ2+m2/bracketrightig−3,(16.2) where /integraldisplay dF3= 2/integraldisplay1 0dx1dx2dx3δ(x1+x2+x3−1). (16.3) We manipulate the right-hand side of eq.(16.2) to get ˜∆((ℓ−k1)2)˜∆((ℓ+k2)2)˜∆(ℓ2) =/integraldisplay dF3/bracketleftig ℓ2−2ℓ·(x1k1−x2k2) +x1k2 1+x2k2 2+m2/bracketrightig−3 =/integraldisplay dF3/bracketleftig (ℓ−x1k1+x2k2)2+x1(1−x1)k2 1+x2(1−x2)k2 2 + 2x1x2k1·k2+m2/bracketrightig−3 =/integraldisplay dF3/bracketleftig q2+D/bracketrightig−3. (16.4) In the last line, we have deﬁned q≡ℓ−x1k1+x2k2, and D≡x1(1−x1)k2 1+x2(1−x2)k2 2+ 2x1x2k1·k2+m2 =x3x1k2 1+x3x2k2 2+x1x2k2 3+m2, (16.5) 16: Loop Corrections to the Vertex 125 k1 k k2 3 l + k2l k l 1 Figure 16.1: The O(g3) correction to the vertex iV3(k1,k2,k3). where we used k2 3= (k1+k2)2andx1+x2+x3= 1 to simplify the second line. After making a Wick rotation of the q0contour, we have V3(k1,k2,k3)/g=Zg+g2/integraldisplay dF3/integraldisplaydd¯q (2π)d1 (¯q2+D)3+O(g4),(16.6) where ¯qis a euclidean vector. This integral diverges for d≥6. We therefore evaluate it for d<6, using the general formula from section 14; the result is/integraldisplaydd¯q (2π)d1 (¯q2+D)3=Γ(3−1 2d) 2(4π)d/2D−(3−d/2). (16.7) Now we set d= 6−ε. To keepgdimensionless, we make the replacement g→g˜µε/2. Then we have V3(k1,k2,k3)/g=Zg+1 2αΓ(ε 2)/integraldisplay dF3/parenleftigg 4π˜µ2 D/parenrightiggε/2 +O(α2),(16.8) whereα=g2/(4π)3. Now we can take the ε→0 limit. The result is V3(k1,k2,k3)/g=Zg+1 2α/bracketleftigg 2 ε+/integraldisplay dF3ln/parenleftigg 4π˜µ2 eγD/parenrightigg/bracketrightigg +O(α2),(16.9) where we have used/integraltextdF3= 1. We now let µ2= 4πe−γ˜µ2, set Zg= 1 +C , (16.10) and rearrange to get V3(k1,k2,k3)/g= 1 +/braceleftig α/bracketleftig 1ε+ ln(µ/m)/bracketrightig +C/bracerightig −1 2α/integraldisplay dF3ln(D/m2) +O(α2). (16.11) 16: Loop Corrections to the Vertex 126 If we takeCto have the form C=−α/bracketleftig 1ε+ ln(µ/m) +κC/bracketrightig +O(α2), (16.12) whereκCis a purely numerical constant, we get V3(k1,k2,k3)/g= 1−1 2α/integraldisplay dF3ln(D/m2)−κCα+O(α2).(16.13) Thus this choice of Crenders V3(k1,k2,k3) ﬁnite and independent of µ, as required. We now need a condition, analogous to Π( −m2) = 0 and Π′(−m2) = 0, to ﬁx the value of κC. These conditions on Π( k2) were mandated by known properties of the exact propagator, but there is nothing dir ectly comparable for the vertex. Diﬀerent choices of κCcorrespond to diﬀerent deﬁnitions of the coupling g. This is because, in order to measure g, we would measure a cross section that depends on g; these cross sections also depend on κC. Thus we can use any value for κCthat we might fancy, as long as we all agree on that value when we compare our calculations with exp erimental measurements. It is then most convenient to simply set κC= 0. This corresponds to the condition V3(0,0,0) =g. (16.14) This condition can then also be used to ﬁx the higher-order (i ng) terms in Zg. The integrals over the Feynman parameters in eq.(16.13) can not be done in closed form, but it is easy to see that if (for example) |k2 1| ≫m2, then V3(k1,k2,k3)/g≃1−1 2α/bracketleftig ln(k2 1/m2) +O(1)/bracketrightig +O(α2). (16.15) Thus the magnitude of the one-loop correction to the vertex f unction in- creases logarithmically with |k2 i|when|k2 i| ≫m2. This is the same behavior that we found for Π( k2)/(k2+m2) in section 14. Problems 16.1) Compute the O(λ2) correction to V4inϕ4theory (see problem 9.2) in d= 4−εspacetime dimensions. Take V4=λwhen all four external momenta are on shell, and s= 4m2. What is the O(λ) contribution toC? 16.2) Repeat problem 16.1 for the theory of problem 9.3. 17: Other 1PI Vertices 127 17Other 1PI Vertices Prerequisite: 16 In section 16, we deﬁned the three-point vertex function iV3(k1,k2,k3) as the sum of all one-particle irreducible diagrams with three external lines, with the external propagators removed. We can extend this de ﬁnition to then-point vertex iVn(k1,...,kn). There are two key diﬀerences between Vn>3andV3inϕ3theory. The ﬁrst is that there is no tree-level contribution to Vn>3. The second is that the one-loop contribution to Vn>3is ﬁnite for d <2n. In particular, the one-loop contribution to Vn>3is ﬁnite for d= 6. Let us see how this works for the case n= 4. We treat all the external momenta as incoming, so that k1+k2+k3+k4= 0. One of the three contributing one-loop diagrams is shown in ﬁg.(17.1); in th is diagram, the k3vertex is opposite to the k1vertex. Two other inequivalent diagrams are then obtained by swapping k3↔k2andk3↔k4. We then have iV4=g4/integraldisplayd6ℓ (2π)6˜∆((ℓ−k1)2)˜∆((ℓ+k2)2)˜∆((ℓ+k2+k3)2)˜∆(ℓ2) + (k3↔k2) + (k3↔k4) +O(g6). (17.1) Feynman’s formula gives ˜∆((ℓ−k1)2)˜∆((ℓ+k2)2)˜∆((ℓ+k2+k3)2)˜∆(ℓ2) =/integraldisplay dF4/bracketleftig x1(ℓ−k1)2+x2(ℓ+k2)2+x3(ℓ+k2+k3)2+x4ℓ2+m2/bracketrightig−4 =/integraldisplay dF4/bracketleftig q2+D1234/bracketrightig−4, (17.2) whereq=ℓ−x1k1+x2k2+x3(k2+k3) and, after making repeated use of x1+x2+x3+x4= 1 andk1+k2+k3+k4= 0, D1234=x1x4k2 1+x2x4k2 2+x2x3k2 3+x1x3k2 4 +x1x2(k1+k2)2+x3x4(k2+k3)2+m2. (17.3) We see that the integral over qis ﬁnite for d <8, and in particular for d= 6. After a Wick rotation of the q0contour and applying the general formula of section 14, we ﬁnd /integraldisplayd6q (2π)61 (q2+D)4=i 6(4π)3D. (17.4) 17: Other 1PI Vertices 128 k1 k2k4 k3k2k2k3k1 l + + l l + l Figure 17.1: One of the three one-loop Feynman diagrams cont ributing to the four-point vertex iV4(k1,k2,k3,k4); the other two are obtained by swappingk3↔k2andk3↔k4. Thus we get V4=g4 6(4π)3/integraldisplay dF4/parenleftbigg1 D1234+1 D1324+1 D1243/parenrightbigg +O(g6). (17.5) This expression is ﬁnite and well-deﬁned; the same is true fo r the one-loop contribution to Vnfor alln>3. Problems 17.1) Verify eq.(17.3). 18: Higher-Order Corrections and Renormalizability 129 18Higher-Order Corrections and Renormalizability Prerequisite: 17 In sections 14–17, we computed the one-loop diagrams with tw o, three, and four external lines for ϕ3theory in six dimensions. We found that the ﬁrst two involved divergent momentum integrals, but that these d ivergences could be absorbed into the coeﬃcients of terms in the lagrang ian. If this is true for all higher-order (in g) contributions to the propagator and to the one-particle irreducible vertex functions (with n≥3 external lines), then we say that the theory is renormalizable . If this is not the case, and further divergences arise, it may be possible to absorb them by addin g some new terms to the lagrangian. If a ﬁnite number of such new terms is required, the theory is still said to be renormalizable. However, if an inﬁnite number of new terms is required, then the theory is said to be nonrenormalizable . Despite the inﬁnite number of parameters needed to specify i t, a nonrenor- malizable theory is generally able to make useful predictio ns at energies below some ultraviolet cutoﬀ Λ; we will discuss this in section 29. In this section, we will deduce the necessary conditions for renormaliz- ability. As an example, we will analyze a scalar ﬁeld theory i ndspacetime dimensions of the form L=−1 2Zϕ∂µϕ∂µϕ−1 2Zmm2ϕ2−∞/summationdisplay n=31 n!Zngnϕn. (18.1) Consider a Feynman diagram with Eexternal lines, Iinternal lines, L closed loops, and Vnvertices that connect nlines. (Here Vnis just a num- ber, not to be confused with the vertex function Vn.) Do the momentum integrals associated with this diagram diverge? We begin by noting that each closed loop gives a factor of ddℓi, and each internal propagator gives a factor of 1 /(p2+m2), wherepis some linear combination of external momenta kiand loop momenta ℓi. The diagram would then appear to have an ultraviolet divergence at large ℓiif there are moreℓ’s in the numerator than there are in the denominator. The num ber ofℓ’s in the numerator minus the number of ℓ’s in the denominator is the diagram’s superﬁcial degree of divergence D≡dL−2I , (18.2) and the diagram appears to be divergent if D≥0. (18.3) 18: Higher-Order Corrections and Renormalizability 130 Next we derive a more useful formula for D. The diagram has Eexternal lines, so another contributing diagram is the tree diagram w here all the lines are joined by a single vertex, with vertex factor −iZEgE; this is, in fact, the value of this entire diagram, which then has mass dimensi on [gE]. (The Z’s are all dimensionless, by deﬁnition.) Therefore, the ori ginal diagram also has mass dimension [ gE], since both are contributions to the same scattering amplitude: [diagram] = [ gE]. (18.4) On the other hand, the mass dimension of any diagram is given b y the sum of the mass dimensions of its components, namely [diagram] = dL−2I+∞/summationdisplay n=3Vn[gn]. (18.5) From eqs.(18.2), (18.4), and (18.5), we get D= [gE]−∞/summationdisplay n=3Vn[gn]. (18.6) This is the formula we need. From eq.(18.6), it is immediately clear that if any [ gn]<0, we expect uncontrollable divergences, since Dincreases with every added vertex of this type. Therefore, a theory with any [gn]<0is nonrenormalizable . According to our results in section 12, the coupling constan ts have mass dimension [gn] =d−1 2n(d−2), (18.7) and so we have [gn]<0 ifn>2d d−2. (18.8) Thus we are limited to powers no higher than ϕ4in four dimensions, and no higher than ϕ3in six dimensions. The same criterion applies to more complicated theories as w ell:a the- ory is nonrenormalizable if any coeﬃcient of any term in the l agrangian has negative mass dimension . What about theories with couplings with only positive or zer o mass dimension? We see from eq.(18.6) that the only dangerous dia grams (those withD≥0) are those for which [ gE]≥0. But in this case, we can absorb the divergence simply by adjusting the value of ZE. This discussion also applies to the propagator; we can think of Π( k2) as representing the loop- corrected counterterm vertex Ak2+Bm2, withAandBm2playing the roles of two couplings. We have [ A] = 0 and [Bm2] = 2, so the contributing 18: Higher-Order Corrections and Renormalizability 131 Figure 18.1: The one-loop contribution to V4. Figure 18.2: A two-loop contribution to V4, and the corresponding coun- terterm insertion. diagrams are expected to be divergent (as we have already see n in detail), and the divergences must be absorbed into AandBm2. Dis called the superﬁcial degree of divergence because a diagram might diverge even if D <0, or might be ﬁnite even if D≥0. The latter can happen if there are cancellations among ℓ’s in the numerator. Quantum electrodynamics provides an example of this phenomenon tha t we will en- counter in Part III; see problem 62.3. For now we turn our atte ntion to the case of diagrams with D<0 that nevertheless diverge. Consider, for example, the diagrams of ﬁgs.(18.1) and (18.2 ). The one- loop diagram of ﬁg.(18.1) with E= 4 is ﬁnite, but the two-loop correction from the ﬁrst diagram of ﬁg.(18.2) is not: the bubble on the up per prop- agator diverges. This is an example of a divergent subdiagram . However, this is not a problem in this case, because this divergence is canceled by the second diagram of ﬁg.(18.2), which has a counterterm ver tex in place of the bubble. This is the generic situation: divergent subdiagrams are di agrams that, considered in isolation, have D≥0. These are precisely the diagrams whose divergences can be canceled by adjusting the Zfactor of the corresponding tree diagram (in theories where [ gn]≥0 for all nonzero gn). Thus, we expect that theories with couplings whose mass dime nsions are all positive or zero will be renormalizable. A detailed s tudy of the 18: Higher-Order Corrections and Renormalizability 132 properties of the momentum integrals in Feynman diagrams is necessary to give a complete proof of this. It turns out to be true withou t further restrictions for theories that have spin-zero and spin-one -half ﬁelds only. Theories with spin-one ﬁelds are renormalizable for d= 4 if and only if these spin-one ﬁelds are associated with a gauge symmetry . We will study this in Part III. Theories of ﬁelds with spin greater than one are never renorm alizable ford≥4. Reference Notes Explicit two-loop calculations in ϕ3theory can be found in Collins ,Muta, andSterman . Problems 18.1) In any number dof spacetime dimensions, a Dirac ﬁeld Ψα(x) car- ries a spin index α, and has a kinetic term of the form iΨγµ∂µΨ, where we have suppressed the spin indices; the gamma matrices γµ are dimensionless, and Ψ = Ψ†γ0. a) What is the mass dimension [Ψ] of the ﬁeld Ψ? b) Consider interactions of the form gn(ΨΨ)n, wheren≥2 is an integer. What is the mass dimension [ gn] ofgn? c) Consider interactions of the form gm,nϕm(ΨΨ)n, whereϕis a scalar ﬁeld, andm≥1 andn≥1 are integers. What is the mass dimension [gm,n] ofgm,n? d) Ind= 4 spacetime dimensions, which of these interactions are allowed in a renormalizable theory? 19: Perturbation Theory to All Orders 133 19Perturbation Theory to All Orders Prerequisite: 18 In section 18, we found that, generally, a theory is renormal izable if all of its lagrangian coeﬃcients have positive or zero mass dime nsion. In this section, using ϕ3theory in six dimensions as our example, we will see how to construct a ﬁnite expression for a scattering amplitude t o arbitrarily high order in the ϕ3couplingg. We begin by summing all one-particle irreducible diagrams w ith two external lines; this gives us the self-energy Π( k2). We next sum all 1PI diagrams with three external lines; this gives us the three- point vertex function V3(k1,k2,k3). Order by order in g, we must adjust the value of the lagrangian coeﬃcients Zϕ,Zm, andZgto maintain the conditions Π(−m2) = 0, Π′(−m2) = 0, and V3(0,0,0) =g. Next we will construct the n-point vertex functions Vn(k1,...,kn) with 4≤n≤E, whereEis the number of external lines in the process of interest. We compute these using a skeleton expansion . This means that we draw all the contributing 1PI diagrams, but omit diagrams that include either propagator or three-point vertex corrections. That is, we include only diagrams that are not only 1PI, but also 2PI and 3PI: they remain connected when any one, two, or three lines are cut. (Cutting three lines may isolate a single tree-level vertex, but nothing more com plicated.) Then we take the propagators and vertices in these diagrams to be g iven by the exact propagator ˜∆(k2) = (k2+m2−Π(k2))−1and vertex V3(k1,k2,k3), rather than by the tree-level propagator ˜∆(k2) = (k2+m2)−1and vertex g. We then sum these skeleton diagrams to get Vnfor 4≤n≤E. Order by order in g, this procedure is equivalent to computing Vnby summing the usual set of contributing 1PI diagrams. Next we draw all tree-level diagrams that contribute to the process of interest (which has Eexternal lines), including not only three-point vertices, but also n-point vertices for n= 3,4,...,E . Then we evaluate these diagrams using the exact propagator ˜∆(k2) for internal lines, and the exact 1PI vertices Vn; external lines are assigned a factor of one.1We sum these tree diagrams to get the scattering amplitude; loo p corrections have all been accounted for already in ˜∆(k2) andVn. Order by order in g, this procedure is equivalent to computing the scattering a mplitude by summing the usual set of contributing diagrams. Thus we now know how to compute an arbitrary scattering ampli tude 1This is because, in the LSZ formula, each Klein-Gordon wave o perator becomes (in momentum space) a factor of k2 i+m2that multiplies each external propagator, leaving behind only the residue of the pole in that propagator at k2 i=−m2; by construction, this residue is one. 19: Perturbation Theory to All Orders 134 to arbitrarily high order. The procedure is the same in any qu antum ﬁeld theory; only the form of the propagators and vertices change , depending on the spins of the ﬁelds. The tree-level diagrams of the ﬁnal step can be thought of as t he Feyn- man diagrams of a quantum action (oreﬀective action , orquantum eﬀective action ) Γ(ϕ). There is a simple and interesting relationship between th e ef- fective action Γ( ϕ) and the sum of connected diagrams with sources iW(J). We derive it in section 21. Reference Notes The detailed procedure for renormalization at higher order s is discussed in Coleman ,Collins ,Muta, andSterman . 20: Two-Particle Elastic Scattering at One Loop 135 20Two-Particle Elastic Scattering at One Loop Prerequisite: 19 We now illustrate the general rules of section 19 by computin g the two- particle elastic scattering amplitude, including all one- loop corrections, in ϕ3theory in six dimensions. Elastic means that the number of outgoing particles (of each species, in more general contexts) is the same as the number of incoming particles (of each species). We computed the amplitude for this process at tree level in se ction 10, with the result iTtree=1 i(ig)2/bracketleftig˜∆(−s) +˜∆(−t) +˜∆(−u)/bracketrightig , (20.1) where ˜∆(−s) = 1/(−s+m2−iǫ) is the free-ﬁeld propagator, and s,t, and uare the Mandelstam variables. Later we will need to remember thatsis positive, that tanduare negative, and that s+t+u= 4m2. The exact scattering amplitude is given by the diagrams of ﬁg .(20.1), with all propagators and vertices interpreted as exact propagators and ver- tices. (Recall, however, that each external propagator con tributes only the residue of the pole at k2=−m2, and that this residue is one; thus the fac- tor associated with each external line is simply one.) We get the one-loop approximation to the exact amplitude by using the one-loop e xpressions for the internal propagators and vertices. We thus have iT1−loop=1 i/parenleftig [iV3(s)]2˜∆(−s) + [iV3(t)]2˜∆(−t) + [iV3(u)]2˜∆(−u)/parenrightig +iV4(s,t,u), (20.2) where, suppressing the iǫ’s, ˜∆(−s) =1 −s+m2−Π(−s), (20.3) Π(−s) =1 2α/integraldisplay1 0dxD 2(s)ln/parenleftig D2(s)/D0/parenrightig −1 12α(−s+m2),(20.4) V3(s)/g= 1−1 2α/integraldisplay dF3ln/parenleftig D3(s)/m2/parenrightig , (20.5) V4(s,t,u) =1 6g2α/integraldisplay dF4/bracketleftbigg1 D4(s,t)+1 D4(t,u)+1 D4(u,s)/bracketrightbigg . (20.6) Hereα=g2/(4π)3, the Feynman integration measure is /integraldisplay dFnf(x) = (n−1)!/integraldisplay1 0dx1...dxnδ(x1+...+xn−1)f(x) 20: Two-Particle Elastic Scattering at One Loop 136 = (n−1)!/integraldisplay1 0dx1/integraldisplay1−x1 0dx2.../integraldisplay1−x1−...−xn−2 0dxn−1 ×f(x)/vextendsingle/vextendsingle/vextendsingle xn=1−x1−...−xn−1, (20.7) and we have deﬁned D2(s) =−x(1−x)s+m2, (20.8) D0= +[1−x(1−x)]m2, (20.9) D3(s) =−x1x2s+ [1−(x1+x2)x3]m2, (20.10) D4(s,t) =−x1x2s−x3x4t+ [1−(x1+x2)(x3+x4)]m2.(20.11) We obtain V3(s) from the general three-point function V3(k1,k2,k3) by set- ting two of the three k2 ito−m2, and the third to −s. We obtain V4(s,t,u) from the general four-point function V4(k1,...,k 4) by setting all four k2 i to−m2, (k1+k2)2to−s, (k1+k3)2to−t, and (k1+k4)2to−u. (Recall that the vertex functions are deﬁned with all momenta treate d as incoming; here we have identiﬁed −k3and−k4as the outgoing momenta.) Eqs.(20.2–20.11) are formidable expressions. To gain some intuition about them, let us consider the limit of high-energy, ﬁxed an gle scattering, where we take s,|t|, and|u|all much larger than m2. Equivalently, we are considering the amplitude in the limit of zero particle mass . We can then set m2= 0 inD2(s),D3(s), andD4(s,t). For the self- energy, we get Π(−s) =−1 2αs/integraldisplay1 0dxx(1−x)/bracketleftbigg ln/parenleftbigg−s m2/parenrightbigg + ln/parenleftbiggx(1−x) 1−x(1−x)/parenrightbigg/bracketrightbigg +1 12αs =−1 12αs/bracketleftig ln(−s/m2) + 3−π√ 3/bracketrightig . (20.12) Thus, ˜∆(−s) =1 −s−Π(−s) =−1 s/parenleftig 1 +1 12α/bracketleftig ln(−s/m2) + 3−π√ 3/bracketrightig/parenrightig +O(α2).(20.13) The appropriate branch of the logarithm is found by replacin gsbys+iǫ. Forsreal and positive, −slies just below the negative real axis, and so ln(−s) = lns−iπ. (20.14) Fort(oru), which is negative, we have instead ln(−t) = ln |t|, lnt= ln|t|+iπ. (20.15) 20: Two-Particle Elastic Scattering at One Loop 137 kk1 k1 k2 k2 k1 k2k1 k2k1k1k2k1k1 k2k1k2+ k1 k2k1k21 k2 Figure 20.1: The Feynman diagrams contributing to the two-p article elastic scattering amplitude; a double line stands for the exact pro pagator1 i˜∆(k), a circle for the exact three-point vertex V3(k1,k2,k3), and a square for the exact four-point vertex V4(k1,k2,k3,k4). An external line stands for the unit residue of the pole at k2=−m2. 20: Two-Particle Elastic Scattering at One Loop 138 For the three-point vertex, we get V3(s)/g= 1−1 2α/integraldisplay dF3/bracketleftig ln(−s/m2) + ln(x1x2)/bracketrightig , = 1−1 2α/bracketleftig ln(−s/m2)−3/bracketrightig , (20.16) where the same comments about the appropriate branch apply. For the four-point vertex, the integral over the Feynman par ameters can be done in closed form, with the result /integraldisplaydF4 D4(s,t)=−3 s+t/parenleftbigg π2+/bracketleftig ln(s/t)/bracketrightig2/parenrightbigg = +3 u/parenleftbigg π2+/bracketleftig ln(s/t)/bracketrightig2/parenrightbigg , (20.17) where the second line follows from s+t+u= 0. Putting all of this together, we have T1−loop=g2/bracketleftig F(s,t,u) +F(t,u,s) +F(u,s,t)/bracketrightig , (20.18) where F(s,t,u)≡ −1 s/parenleftbigg 1−11 12α/bracketleftig ln(−s/m2) +c/bracketrightig −1 2α/bracketleftig ln(t/u)/bracketrightig2/parenrightbigg ,(20.19) andc= (6π2+π√ 3−39)/11 = 2.33. This is a typical result of a loop calculation: the original tree-level amplitude is correct ed by powers of log- arithms of kinematic variables. Problems 20.1) Verify eq.(20.17). 20.2) Compute the O(α) correction to the two-particle scattering amplitude at threshold , that is, for s= 4m2andt=u= 0, corresponding to zero three-momentum for both the incoming and outgoing part icles. 21: The Quantum Action 139 21The Quantum Action Prerequisite: 19 In section 19, we saw how to compute (in ϕ3theory ind= 6 dimensions) the 1PI vertex functions Vn(k1,...,kn) forn≥4 via the skeleton expansion : draw all Feynman diagrams with nexternal lines that are one-, two-, and three-particle irreducible, and compute them using the exa ct propagator ˜∆(k2) and three-point vertex function V3(k1,k2,k3). We now deﬁne the quantum action (oreﬀective action , orquantum eﬀective action ) Γ(ϕ)≡1 2/integraldisplayddk (2π)dx/tildewideϕ(−k)/parenleftig k2+m2−Π(k2)/parenrightig /tildewideϕ(k) +∞/summationdisplay n=31 n!/integraldisplayddk1 (2π)d...ddkn (2π)d(2π)dδd(k1+...+kn) ×Vn(k1,...,kn)/tildewideϕ(k1).../tildewideϕ(kn), (21.1) where/tildewideϕ(k) =/integraltextddxe−ikxϕ(x). The quantum action has the property that thetree-level Feynman diagrams it generates give the complete scattering amplitude of the original theory. In this section, we will determine the relationship between Γ(ϕ) and the sum of connected diagrams with sources, iW(J), introduced in section 9. Recall that W(J) is related to the path integral Z(J) =/integraldisplay Dϕexp/bracketleftbigg iS(ϕ) +i/integraldisplay ddxJϕ/bracketrightbigg , (21.2) whereS=/integraltextddxLis the action, via Z(J) = exp[iW(J)]. (21.3) Consider now the path integral ZΓ(J)≡/integraldisplay Dϕexp/bracketleftbigg iΓ(ϕ) +i/integraldisplay ddxJϕ/bracketrightbigg (21.4) = exp[iWΓ(J)]. (21.5) WΓ(J) is given by the sum of connected diagrams (with sources) in w hich each line represents the exact propagator, and each n-point vertex rep- resents the exact 1PI vertex Vn.WΓ(J) would be equal to W(J) if we included only tree diagrams in WΓ(J). 21: The Quantum Action 140 We can isolate the tree-level contribution to a path integra l by means of the following trick. Introduce a dimensionless parameter t hat we will call ¯h, and the path integral ZΓ,¯h(J)≡/integraldisplay Dϕexp/bracketleftbiggi ¯h/parenleftbigg Γ(ϕ) +/integraldisplay ddxJϕ/parenrightbigg/bracketrightbigg (21.6) = exp[iWΓ,¯h(J)]. (21.7) In a given connected diagram with sources, every propagator (including those that connect to sources) is multiplied by ¯ h, every source by 1 /¯h, and every vertex by 1 /¯h. The overall factor of ¯ his then ¯hP−E−V, whereVis the number of vertices, Eis the number of sources (equivalently, the number of external lines after we remove the sources), and Pis the number of prop- agators (external and internal). We next note that P−E−Vis equal to L−1, whereLis the number of closed loops. This can be seen by counting the number of internal momenta and the constraints among the m. Specif- ically, assign an unﬁxed momentum to each internal line; the re areP−E of these momenta. Then the Vvertices provide Vconstraints. One lin- ear combination of these constraints gives overall momentu m conservation, and so does not constrain the internal momenta. Therefore, t he number of internal momenta left unﬁxed by the vertex constraints is ( P−E)−(V−1), and the number of unﬁxed momenta is the same as the number of lo opsL. So,WΓ,¯h(J) can be expressed as a power series in ¯ hof the form WΓ,¯h(J) =∞/summationdisplay L=0¯hL−1WΓ,L(J). (21.8) If we take the formal limit of ¯ h→0, the dominant term is the one with L= 0, which is given by the sum of tree diagrams only. This is jus t what we want. We conclude that W(J) =WΓ,L=0(J). (21.9) Next we perform the path integral in eq.(21.6) by the method o f station- ary phase. We ﬁnd the point (actually, the ﬁeld conﬁguration ) at which the exponent is stationary; this is given by the solution of t hequantum equation of motion δ δϕ(x)Γ(ϕ) =−J(x). (21.10) LetϕJ(x) denote the solution of eq.(21.10) with a speciﬁed source fu nction J(x). Then the stationary-phase approximation to ZΓ,¯h(J) is ZΓ,¯h(J) = exp/bracketleftbiggi ¯h/parenleftbigg Γ(ϕJ) +/integraldisplay ddxJϕJ/parenrightbigg +O(¯h0)/bracketrightbigg . (21.11) 21: The Quantum Action 141 Combining the results of eqs.(21.7), (21.8), (21.9), and (2 1.11), we ﬁnd W(J) = Γ(ϕJ) +/integraldisplay ddxJϕJ. (21.12) This is the main result of this section. Let us explore it further. Recall from section 9 that the vacu um expec- tation value of the ﬁeld operator ϕ(x) is given by ∝an}b∇acketle{t0|ϕ(x)|0∝an}b∇acket∇i}ht=δ δJ(x)W(J)/vextendsingle/vextendsingle/vextendsingle/vextendsingle J=0. (21.13) Now consider what we get if we do not set J= 0 after taking the derivative: ∝an}b∇acketle{t0|ϕ(x)|0∝an}b∇acket∇i}htJ≡δ δJ(x)W(J). (21.14) This is the vacuum expectation value of ϕ(x) in the presence of a nonzero source function J(x). We can get some more information about it by using eq.(21.12) for W(J). Making use of the product rule for derivatives, we have ∝an}b∇acketle{t0|ϕ(x)|0∝an}b∇acket∇i}htJ=δ δJ(x)Γ(ϕJ) +ϕJ(x) +/integraldisplay d6yJ(y)δϕJ(y) δJ(x). (21.15) We can evaluate the ﬁrst term on the right-hand side by using t he chain rule, δ δJ(x)Γ(ϕJ) =/integraldisplay d6yδΓ(ϕJ) δϕJ(y)δϕJ(y) δJ(x). (21.16) Then we can combine the ﬁrst and third terms on the right-hand side of eq.(21.15) to get ∝an}b∇acketle{t0|ϕ(x)|0∝an}b∇acket∇i}htJ=/integraldisplay d6y/bracketleftbiggδΓ(ϕJ) δϕJ(y)+J(y)/bracketrightbiggδϕJ(y) δJ(x)+ϕJ(x). (21.17) Now we note from eq.(21.10) that the factor in large brackets on the right- hand side of eq.(21.17) vanishes, and so ∝an}b∇acketle{t0|ϕ(x)|0∝an}b∇acket∇i}htJ=ϕJ(x). (21.18) That is, the vacuum expectation value of the ﬁeld operator ϕ(x) in the presence of a nonzero source function is also the solution to the quantum equation of motion, eq.(21.10). We can also write the quantum action in terms of a derivative expansion , Γ(ϕ) =/integraldisplay ddx/bracketleftig − U(ϕ)−1 2Z(ϕ)∂µϕ∂µϕ+.../bracketrightig , (21.19) 21: The Quantum Action 142 where the ellipses stand for an inﬁnite number of terms with m ore and more derivatives, and U(ϕ) and Z(ϕ) are ordinary functions (not functionals) ofϕ(x).U(ϕ) is called the quantum potential (oreﬀective potential , or quantum eﬀective potential ), and it plays an important conceptual role in theories with spontaneous symmetry breaking; see section 3 1. However, it is rarely necessary to compute it explicitly, except in thos e cases where we are unable to do so. Reference Notes Construction of the quantum action is discussed in Coleman ,Itzykson & Zuber,Peskin & Schroeder , and Weinberg II. Problems 21.1) Show that Γ(ϕ) =W(Jϕ)−/integraldisplay ddxJϕϕ, (21.20) whereJϕ(x) is the solution of δ δJ(x)W(J) =ϕ(x) (21.21) for a speciﬁed ϕ(x). 21.2)Symmetries of the quantum action. Suppose that we have a set of ﬁeldsϕa(x), and that both the classical action S(ϕ) and the integra- tion measure Dϕare invariant under ϕa(x)→/integraldisplay ddyRab(x,y)ϕb(y) (21.22) for some particular function Rab(x,y). Typically Rab(x,y) is a con- stant matrix times δd(x−y), or a ﬁnite number of derivatives of δd(x−y); see sections 22, 23, and 24 for some examples. a) Show that W(J) is invariant under Ja(x)→/integraldisplay ddyJb(y)Rba(y,x). (21.23) b) Use eqs.(21.20) and (21.23) to show that the quantum actio n Γ(ϕ) is invariant under eq.(21.22). This is an important result t hat we will use frequently. 21: The Quantum Action 143 21.3) Consider performing the path integral in the presence of abackground ﬁeld¯ϕ(x); we deﬁne exp[iW(J; ¯ϕ)]≡/integraldisplay Dϕexp/bracketleftbigg iS(ϕ+ ¯ϕ) +i/integraldisplay ddxJϕ/bracketrightbigg .(21.24) ThenW(J;0) is the original W(J) of eq.(21.3). We also deﬁne the quantum action in the presence of the background ﬁeld, Γ(ϕ; ¯ϕ)≡W(Jϕ; ¯ϕ)−/integraldisplay ddxJϕϕ, (21.25) whereJϕ(x) is the solution of δ δJ(x)W(J; ¯ϕ) =ϕ(x) (21.26) for a speciﬁed ϕ(x). Show that Γ(ϕ; ¯ϕ) = Γ(ϕ+ ¯ϕ;0), (21.27) where Γ(ϕ;0) is the original quantum action of eq.(21.1). 22: Continuous Symmetries and Conserved Currents 144 22Continuous Symmetries and Conserved Currents Prerequisite: 8 Suppose we have a set of scalar ﬁelds ϕa(x), and a lagrangian density L(x) =L(ϕa(x),∂µϕa(x)). Consider what happens to L(x) if we make an inﬁnitesimal change ϕa(x)→ϕa(x) +δϕa(x) in each ﬁeld. We have L(x)→ L(x) +δL(x), whereδL(x) is given by the chain rule, δL(x) =∂L ∂ϕa(x)δϕa(x) +∂L ∂(∂µϕa(x))∂µδϕa(x). (22.1) Next consider the classical equations of motion (also known as the Euler- Lagrange equations, or the ﬁeld equations ), given by the action principle δS δϕa(x)= 0, (22.2) whereS=/integraltextd4yL(y) is the action, and δ/δϕa(x) is a functional derivative. (For deﬁniteness, we work in four spacetime dimensions, tho ugh our re- sults will apply in any number.) We have (with repeated indic es implicitly summed) δS δϕa(x)=/integraldisplay d4yδL(y) δϕa(x) =/integraldisplay d4y/bracketleftigg ∂L(y) ∂ϕb(y)δϕb(y) δϕa(x)+∂L(y) ∂(∂µϕb(y))δ(∂µϕb(y)) δϕa(x)/bracketrightigg =/integraldisplay d4y/bracketleftigg ∂L(y) ∂ϕb(y)δbaδ4(y−x) +∂L(y) ∂(∂µϕb(y))δba∂µδ4(y−x)/bracketrightigg =∂L(x) ∂ϕa(x)−∂µ∂L(x) ∂(∂µϕa(x)). (22.3) We can use this result to make the replacement ∂L(x) ∂ϕa(x)→∂µ∂L(x) ∂(∂µϕa(x))+δS δϕa(x)(22.4) in eq.(22.1). Then, combining two of the terms, we get δL(x) =∂µ/parenleftigg ∂L(x) ∂(∂µϕa(x))δϕa(x)/parenrightigg +δS δϕa(x)δϕa(x). (22.5) 22: Continuous Symmetries and Conserved Currents 145 Next we identify the object in large parentheses in eq.(22.5 ) as the Noether current jµ(x)≡∂L(x) ∂(∂µϕa(x))δϕa(x). (22.6) Eq.(22.5) then implies ∂µjµ(x) =δL(x)−δS δϕa(x)δϕa(x). (22.7) If the classical ﬁeld equations are satisﬁed, then the secon d term on the right-hand side of eq.(22.7) vanishes. The Noether current plays a special role if we can ﬁnd a set of i n- ﬁnitesimal ﬁeld transformations that leaves the lagrangia n unchanged, or invariant . In this case, we have δL= 0, and we say that the lagrangian has acontinuous symmetry . From eq.(22.7), we then have ∂µjµ= 0 whenever the ﬁeld equations are satisﬁed, and we say that the Noether c urrent is conserved . In terms of its space and time components, this means that ∂ ∂tj0(x) +∇ ·j(x) = 0. (22.8) If we interpret j0(x) as acharge density , andj(x) as the corresponding cur- rent density , then eq.(22.8) expresses the local conservation of this ch arge. Let us see an example of this. Consider a theory of a complex scalar ﬁeld with lagrangian L=−∂µϕ†∂µϕ−m2ϕ†ϕ−1 4λ(ϕ†ϕ)2. (22.9) We can also rewrite Lin terms of two real scalar ﬁelds by setting ϕ= (ϕ1+iϕ2)/√ 2 to get L=−1 2∂µϕ1∂µϕ1−1 2∂µϕ2∂µϕ2−1 2m2(ϕ2 1+ϕ2 2)−1 16λ(ϕ2 1+ϕ2 2)2.(22.10) In the form of eq.(22.9), it is obvious that Lis left invariant by the trans- formation ϕ(x)→e−iαϕ(x), (22.11) whereαis a real number. This is called a U(1) transformation , a transfor- mation by a unitary 1 ×1 matrix. In terms of ϕ1andϕ2, this transformation reads /parenleftiggϕ1(x) ϕ2(x)/parenrightigg →/parenleftiggcosαsinα −sinαcosα/parenrightigg/parenleftiggϕ1(x) ϕ2(x)/parenrightigg . (22.12) If we think of ( ϕ1,ϕ2) as a two-component vector, then eq.(22.12) is just a rotation of this vector in the plane by angle α. Eq.(22.12) is called an 22: Continuous Symmetries and Conserved Currents 146 SO(2) transformation , a transformation by an orthogonal 2 ×2 matrix with a special value of the determinant (namely +1, as opposed to −1, the only other possibility for an orthogonal matrix). We have learne d that a U(1) transformation can be mapped into an SO(2) transformation. The inﬁnitesimal form of eq.(22.11) is ϕ(x)→ϕ(x)−iαϕ(x), ϕ†(x)→ϕ†(x) +iαϕ†(x), (22.13) whereαis now inﬁnitesimal. In eq.(22.6), we should treat ϕandϕ†as in- dependent ﬁelds. It is also conventional to scale the inﬁnit esimal parameter out of the current, so that we have αjµ=∂L ∂(∂µϕ)δϕ+∂L ∂(∂µϕ†)δϕ† = (−∂µϕ†)(−iαϕ) + (−∂µϕ)(+iαϕ†) =αIm(ϕ†↔ ∂µϕ), (22.14) whereA↔∂µB≡A∂µB−(∂µA)B. Canceling out α, we ﬁnd that the Noether current is jµ= Im(ϕ†↔ ∂µϕ). (22.15) We can also repeat this exercise using the SO(2) form of the tr ans- formation. For inﬁnitesimal α, eq.(22.12) becomes δϕ1= +αϕ2and δϕ2=−αϕ1. Then the Noether current is given by αjµ=∂L ∂(∂µϕ1)δϕ1+∂L ∂(∂µϕ2)δϕ2 = (−∂µϕ1)(+αϕ2) + (−∂µϕ2)(−αϕ1) =α(ϕ1↔ ∂µϕ2), (22.16) which is (hearteningly) equivalent to eq.(22.14). Let us deﬁne the Noether charge Q≡/integraldisplay d3xj0(x) =/integraldisplay d3xIm(ϕ†↔ ∂0ϕ), (22.17) and investigate its properties. If we integrate eq.(22.8) o verd3x, use Gauss’s law to write the volume integral of ∇·jas a surface integral, and assume that the boundary conditions at inﬁnity ﬁx j(x) = 0 on that surface, then 22: Continuous Symmetries and Conserved Currents 147 we ﬁnd that Qis constant in time. To get a better idea of the physical implications of this, let us rewrite Qusing the free-ﬁeld expansions ϕ(x) =/integraldisplay /tildewiderdk/bracketleftig a(k)eikx+b∗(k)e−ikx/bracketrightig , ϕ†(x) =/integraldisplay /tildewiderdk/bracketleftig b(k)eikx+a∗(k)e−ikx/bracketrightig . (22.18) We have written a∗(k) andb∗(k) rather than a†(k) andb†(k) because so far our discussion has been about the classical ﬁeld theory. In a theory with interactions, these formulae (and their ﬁrst time deri vatives) are valid at any one particular time (say, t=−∞). Then, we can plug them into eq.(22.17), and ﬁnd (after some manipulation similar to wha t we did for the hamiltonian in section 3) Q=/integraldisplay /tildewiderdk/bracketleftig a∗(k)a(k)−b(k)b∗(k)]. (22.19) In the quantum theory, this becomes an operator that counts t he number ofaparticles minus the number of bparticles. This number is then time- independent, and so the scattering amplitude vanishes iden tically for any process that changes the value of Q. This can be seen directly from the Feynman rules, which conserve Qat every vertex. To better understand the implications of the Noether curren t in the quantum theory, we begin by considering the path integral, Z(J) =/integraldisplay Dϕei[S+/integraltext d4yJaϕa]. (22.20) The value of Z(J) is unchanged if we make the change of variable ϕa(x)→ ϕa(x) +δϕa(x), withδϕa(x) an arbitrary inﬁnitesimal shift that (we as- sume) leaves the measure Dϕinvariant. Thus we have 0 =δZ(J) =i/integraldisplay Dϕei[S+/integraltext d4yJbϕb]/integraldisplay d4x/parenleftbiggδS δϕa(x)+Ja(x)/parenrightbigg δϕa(x).(22.21) We can now take nfunctional derivatives with respect to Jaj(xj), and then setJ= 0, to get 0 =/integraldisplay DϕeiS/integraldisplay d4x/bracketleftigg iδS δϕa(x)ϕa1(x1)...ϕan(xn) +n/summationdisplay j=1ϕa1(x1)...δaajδ4(x−xj)...ϕan(xn)/bracketrightigg δϕa(x).(22.22) 22: Continuous Symmetries and Conserved Currents 148 Sinceδϕa(x) is arbitrary, we can drop it (and the integral over d4x). Then, since the path integral computes the vacuum expectation val ue of the time- ordered product, we have 0 =i∝an}b∇acketle{t0|TδS δϕa(x)ϕa1(x1)...ϕan(xn)|0∝an}b∇acket∇i}ht +n/summationdisplay j=1∝an}b∇acketle{t0|Tϕa1(x1)...δaajδ4(x−xj)...ϕan(xn)|0∝an}b∇acket∇i}ht.(22.23) These are the Schwinger-Dyson equations for the theory. To get a feel for them, let us look at free-ﬁeld theory for a sin gle real scalar ﬁeld, for which δS/δϕ (x) = (∂2 x−m2)ϕ(x). Forn= 1 we get (−∂2 x+m2)i∝an}b∇acketle{t0|Tϕ(x)ϕ(x1)|0∝an}b∇acket∇i}ht=δ4(x−x1). (22.24) That the Klein-Gordon wave operator should sit outside the t ime-ordered product (and hence act on the time-ordering step functions) is clear from the path integral form of eq.(22.22). We see from eq.(22.24) that the free- ﬁeld propagator, ∆( x−x1) =i∝an}b∇acketle{t0|Tϕ(x)ϕ(x1)|0∝an}b∇acket∇i}ht, is a Green’s function for the Klein-Gordon wave operator, a fact we ﬁrst learned in sec tion 8. More generally, we can write ∝an}b∇acketle{t0|TδS δϕa(x)ϕa1(x1)...ϕan(xn)|0∝an}b∇acket∇i}ht= 0 forx∝ne}ationslash=x1,...,n. (22.25) We see that the classical equation of motion is satisﬁed by a q uantum ﬁeld inside a correlation function, as long as its spacetime argu ment diﬀers from those of all the other ﬁelds. When this is not the case, we get e xtracontact terms. Let us now consider a theory that has a continuous symmetry an d a corresponding Noether current. Take eq.(22.22), and set δϕa(x) to be the inﬁnitesimal change in ϕa(x) that results in δL(x) = 0. Now sum over the indexa, and use eq.(22.7). The result is the Ward (orWard-Takahashi ) identity 0 =∂µ∝an}b∇acketle{t0|Tjµ(x)ϕa1(x1)...ϕan(xn)|0∝an}b∇acket∇i}ht +in/summationdisplay j=1∝an}b∇acketle{t0|Tϕa1(x1)...δϕaj(x)δ4(x−xj)...ϕan(xn)|0∝an}b∇acket∇i}ht.(22.26) Thus, conservation of the Noether current holds in the quant um theory, with the current inside a correlation function, up to contac t terms with a speciﬁc form that depends on the details of the inﬁnitesimal transformation that leaves Linvariant. 22: Continuous Symmetries and Conserved Currents 149 The Noether current is also useful in a slightly more general context. Suppose we have a transformation of the ﬁelds such that δL(x) is not zero, but instead is a total divergence: δL(x) =∂µKµ(x) for someKµ(x). Then there is still a conserved current, now given by jµ(x) =∂L(x) ∂(∂µϕa(x))δϕa(x)−Kµ(x). (22.27) An example of this is provided by the symmetry of spacetime translations . We transform the ﬁelds via ϕa(x)→ϕa(x−a), whereaµis a constant four- vector. The inﬁnitesimal version of this is ϕa(x)→ϕa(x)−aν∂νϕa(x), and so we haveδϕa(x) =−aν∂νϕa(x). Under this transformation, we obviously haveL(x)→ L(x−a), and soδL(x) =−aν∂νL(x) =−∂ν(aνL(x)). Thus in this case Kν(x) =−aνL(x), and the conserved current is jµ(x) =∂L(x) ∂(∂µϕa(x))(−aν∂νϕa(x))−aµL(x) =aνTµν(x), (22.28) where we have deﬁned the stress-energy orenergy-momentum tensor Tµν(x)≡ −∂L(x) ∂(∂µϕa(x))∂νϕa(x) +gµνL(x). (22.29) For a renormalizable theory of a set of real scalar ﬁelds ϕa(x), the lagrangian takes the form L=−1 2∂µϕa∂µϕa−V(ϕ), (22.30) whereV(ϕ) is a polynomial in the ϕa’s. In this case Tµν=∂µϕa∂νϕa+gµνL. (22.31) In particular, T00=1 2Π2 a+1 2(∇ϕa)2+V(ϕ), (22.32) where Πa=∂0ϕais the canonical momentum conjugate to the ﬁeld ϕa. We recognize T00as the hamiltonian density Hthat corresponds to the lagrangian density of eq.(22.30). Then, by Lorentz symmetr y,T0jmust be the corresponding momentum density. We have T0j=∂0ϕa∂jϕa=−Πa∇jϕa. (22.33) To check that this is a sensible result, we use the free-ﬁeld e xpansion for a set of real scalar ﬁelds [the same as eq.(22.18) but with b(k) =a(k) for each ﬁeld]; then we ﬁnd that the momentum operator is given by Pj=/integraldisplay d3xT0j(x) =/integraldisplay /tildewiderdkkja† a(k)aa(k), (22.34) 22: Continuous Symmetries and Conserved Currents 150 which is just what we would expect. We therefore identify the energy- momentum four-vector as Pµ=/integraldisplay d3xT0µ(x). (22.35) Recall that in section 2 we deﬁned the spacetime translation operator as T(a)≡exp(−iPµaµ), (22.36) and announced that it had the property that T(a)−1ϕa(x)T(a) =ϕa(x−a). (22.37) Now that we have an explicit formula for Pµ, we can check this. This is easiest to do for inﬁnitesimal aµ; then eq.(22.37) becomes [ϕa(x),Pµ] =1 i∂µϕa(x). (22.38) This can indeed be veriﬁed by using the canonical commutatio n relations forϕa(x) and Πa(x). One more symmetry we can investigate is Lorentz symmetry. If we make an inﬁnitesimal Lorentz transformation, we have ϕa(x)→ϕa(x+δω·x), whereδω·xis shorthand for δωνρxρ. This case is very similar to that of spacetime translations; the only diﬀerence is that the tran slation parameter aνis nowxdependent, aν→ −δωνρxρ. The resulting conserved current is Mµνρ(x) =xνTµρ(x)−xρTµν(x), (22.39) and it obeys ∂µMµνρ= 0, with the derivative contracted with the ﬁrst index. Mµνρis antisymmetric on its second two indices; this comes about becauseδωνρis antisymmetric. The conserved charges associated with th is current are Mνρ=/integraldisplay d3xM0νρ(x), (22.40) and these are the generators of the Lorentz group that were introduced in section 2. Again, we can use the canonical commutation rel ations for the ﬁelds to check that the Lorentz generators have the right commutation relations, both with the ﬁelds and with each other. Reference Notes The path-integral approach to Ward identities is treated in more de- tail in Peskin & Schroeder . An operator-based derivation can be found in Weinberg I . 22: Continuous Symmetries and Conserved Currents 151 Problems 22.1) For the Noether current of eq.(22.6), and assuming tha tδϕadoes not involve time derivatives, use the canonical commutation re lations to show that [ϕa,Q] =iδϕa, (22.41) whereQis the Noether charge. 22.2) Use the canonical commutation relations to verify eq. (22.38). 22.3) a) With Tµνgiven by eq.(22.31), compute the equal-time ( x0=y0) commutators [ T00(x),T00(y)], [T0i(x),T00(y)], and [T0i(x),T0j(y)]. b) Use your results to verify eqs.(2.17), (2.19), and (2.20) . 23: Discrete Symmetries: P,T,C, andZ 152 23Discrete Symmetries: P,T,C, and Z Prerequisite: 22 In section 2, we studied the proper orthochronous Lorentz transformations, which are continuously connected to the identity. In this se ction, we will consider the eﬀects of parity, Pµν= (P−1)µν= +1 −1 −1 −1 . (23.1) andtime reversal , Tµν= (T−1)µν= −1 +1 +1 +1 . (23.2) We will also consider certain other discrete transformatio ns such as charge conjugation . Recall from section 2 that for every proper orthochronous Lo rentz trans- formation Λµνthere is an associated unitary operator U(Λ) with the prop- erty that U(Λ)−1ϕ(x)U(Λ) =ϕ(Λ−1x). (23.3) Thus for parity and time-reversal, we expect that there are c orresponding unitary operators P≡U(P), (23.4) T≡U(T), (23.5) such that P−1ϕ(x)P=ϕ(Px), (23.6) T−1ϕ(x)T=ϕ(Tx). (23.7) There is, however, an extra possible complication. Since th ePand Tmatrices are their own inverses, a second parity or time-rev ersal trans- formation should transform all observables back into thems elves. Using eqs.(23.6) and (23.7), along with P2= 1 and T2= 1, we see that P−2ϕ(x)P2=ϕ(x), (23.8) T−2ϕ(x)T2=ϕ(x). (23.9) 23: Discrete Symmetries: P,T,C, andZ 153 Sinceϕ(x) is a hermitian operator, it is in principle an observable, a nd so eqs.(23.8) and (23.9) are just what we expect. However, anot her possibility for the parity transformation of the ﬁeld, diﬀerent from eqs .(23.6) and (23.7) but nevertheless consistent with eqs.(23.8) and (23 .9), is P−1ϕ(x)P=−ϕ(Px), (23.10) T−1ϕ(x)T=−ϕ(Tx). (23.11) This possible extra minus sign cannot arise for proper ortho chronous Lorentz transformations, because they are continuously connected to the identity, and for the identity transformation (that is, no transforma tion at all), we must obviously have the plus sign. If the minus sign appears on the right-hand side, we say that t he ﬁeld isodd under parity (or time reversal). If a scalar ﬁeld is odd under parity, we sometimes say that it is a pseudoscalar .1 So, how do we know which is right, eqs.(23.6) and (23.7), or eq s.(23.10) and (23.11)? The general answer is that we get to choose, but t here is a key principle to guide our choice: if at all possible, we want to deﬁneP andTso that the lagrangian density is even, P−1L(x)P= +L(Px), (23.12) T−1L(x)T= +L(Tx). (23.13) Then, after we integrate over d4xto get the action S, the action will be invariant. This means that parity and time-reversal are conserved . For theories with spin-zero ﬁelds only, it is clear that the c hoice of eqs.(23.6) and (23.7) always leads to eqs.(23.12) and (23.1 3), and so there is no reason to ﬂirt with eqs.(23.10) and (23.11). For theories that also include spin-one-half ﬁelds, certain scalar bilinears in these ﬁel ds are necessarily odd under parity and time reversal, as we will see in section 40. I f a scalar ﬁeld couples to such a bilinear, then eqs.(23.12) and (23.13) wil l hold if and only if we choose eqs.(23.10) and (23.11) for that scalar, and so t hat is what we must do. There is one more interesting fact about the time-reversal o peratorT: it isantiunitary , rather than unitary. Antiunitary means that T−1iT=−i. To see why this must be the case, consider a Lorentz transform ation of the energy-momentum four-vector, U(Λ)−1PµU(Λ) = ΛµνPν. (23.14) 1It is still a scalar under proper orthochronous Lorentz tran sformations; that is, eq.(23.3) still holds. Thus the appellation scalar often means eq.(23.3), and either eq.(23.6) oreq.(23.10), and that is how we will use the term. 23: Discrete Symmetries: P,T,C, andZ 154 For parity and time-reversal, we therefore expect P−1PµP=PµνPν, (23.15) T−1PµT=TµνPν. (23.16) In particular, for µ= 0, we expect P−1HP= +HandT−1HT=−H. The ﬁrst of these is ﬁne; it says the hamiltonian is invariant under parity, which is what we want.2However, eq.(23.16) is a disaster: it says that the hamiltonian is invariant under time-reversal if and only if H=−H, which is possible only if H= 0. Can we just put an extra minus sign on the right-hand side of eq .(23.16), as we did for eq.(23.11)? The answer is no. We constructed Pµexplicitly in terms of the ﬁelds in section 22, and it is easy to check that choosing eq.(23.11) for the ﬁelds does not yield an extra minus sign in eq.(23.16) for the energy-momentum four-vector. Let us reconsider the origin of eq.(23.14). We ﬁrst recall th at the space- time translation operator T(a) = exp( −iP·a). (23.17) (which should not be confused with the time-reversal operat orT) trans- forms a scalar ﬁeld according to T(a)−1ϕ(x)T(a) =ϕ(x−a). (23.18) The spacetime translation operator is a scalar with a spacet ime coordinate as a label; by analogy with eq.(23.3), we should have U(Λ)−1T(a)U(Λ) =T(Λ−1a). (23.19) Now, treat aµas inﬁnitesimal in eq.(23.19) to get U(Λ)−1(I−iaµPµ)U(Λ) =I−i(Λ−1)νµaµPν =I−iΛµνaµPν. (23.20) For time-reversal, this becomes T−1(I−iaµPµ)T=I−iTµνaµPν. (23.21) If we now identify the coeﬃcients of −iaµon each side, we get eq.(23.16), which is bad. In order to get the extra minus sign that we need, we must impose the antiunitary condition T−1iT=−i. (23.22) 2When spin-one-half ﬁelds are present, it may be that no opera tor exists that satisﬁes either eq.(23.6) or eq.(23.10) and also eq.(23.15); in this case we say that parity is explicitly broken . 23: Discrete Symmetries: P,T,C, andZ 155 We then ﬁnd T−1PµT=−TµνPν(23.23) instead of eq.(23.16). This yields T−1HT= +H, which is the correct expression of time-reversal invariance. We turn now to other unitary operators that change the signs o f scalar ﬁelds, but do nothing to their spacetime arguments. Suppose we have a theory with real scalar ﬁelds ϕa(x), and a unitary operator Zthat obeys Z−1ϕa(x)Z=ηaϕa(x), (23.24) whereηais either +1 or −1 for each ﬁeld. We will call ZaZ2operator , be- cause Z 2is the additive group of the integers modulo 2, which is equiv alent to the multiplicative group of +1 and −1. This also implies that Z2= 1, and soZ−1=Z. (For theories with spin-zero ﬁelds only, the same is also true ofPandT, but things are more subtle for higher spin, as we will see in Part II.) Consider the theory of a complex scalar ﬁeld ϕ= (ϕ1+iϕ2)/√ 2 that was introduced in section 22, with lagrangian L=−∂µϕ†∂µϕ−m2ϕ†ϕ−1 4λ(ϕ†ϕ)2(23.25) =−1 2∂µϕ1∂µϕ1−1 2∂µϕ2∂µϕ2−1 2m2(ϕ2 1+ϕ2 2)−1 16λ(ϕ2 1+ϕ2 2)2.(23.26) In the form of eq.(23.25), Lis obviously invariant under the U(1) transfor- mation ϕ(x)→e−iαϕ(x). (23.27) In the form of eq.(23.26), Lis obviously invariant under the equivalent SO(2) transformation, /parenleftiggϕ1(x) ϕ2(x)/parenrightigg →/parenleftiggcosαsinα −sinαcosα/parenrightigg/parenleftiggϕ1(x) ϕ2(x)/parenrightigg . (23.28) However, it is also obvious that Lhas an additional discrete symmetry, ϕ(x)↔ϕ†(x) (23.29) in the form of eq.(23.25), or equivalently /parenleftiggϕ1(x) ϕ2(x)/parenrightigg →/parenleftigg+1 0 0−1/parenrightigg/parenleftiggϕ1(x) ϕ2(x)/parenrightigg . (23.30) in the form of eq.(23.26). This discrete symmetry is called charge conju- gation . It always occurs as a companion to a continuous U(1) symmetr y. In terms of the two real ﬁelds, it enlarges the group from SO(2 ) (the group 23: Discrete Symmetries: P,T,C, andZ 156 of 2×2 orthogonal matrices with determinant +1) to O(2) (the grou p of 2×2 orthogonal matrices). We can implement charge conjugation by means of a particular Z2op- eratorCthat obeys C−1ϕ(x)C=ϕ†(x), (23.31) or equivalently C−1ϕ1(x)C= +ϕ1(x), (23.32) C−1ϕ2(x)C=−ϕ2(x). (23.33) We then have C−1L(x)C=L(x), (23.34) and so charge conjugation is a symmetry of the theory. Physic ally, it implies that the scattering amplitudes are unchanged if we exchange all thea-type particles (which have charge +1) with all the b-type particles (which have charge −1). This means, in particular, that the aandbparticles must have exactly the same mass. We say that bisa’santiparticle . More generally, we can also have Z 2symmetries that are not related to antiparticles. Consider, for example, ϕ4theory, where ϕis a real scalar ﬁeld with lagrangian L=−1 2∂µϕ∂µϕ−1 2m2ϕ2−1 24λϕ4. (23.35) If we deﬁne the Z 2operatorZvia Z−1ϕ(x)Z=−ϕ(x), (23.36) thenLis obviously invariant. We therefore have Z−1HZ=H, or equiva- lently [Z,H] = 0, where His the hamiltonian. If we assume that (as usual) the ground state is unique, then, since Zcommutes with H, the ground state must also be an eigenstate of Z. We can ﬁx the phase of Z[which is undetermined by eq.(23.36)] via Z|0∝an}b∇acket∇i}ht=Z−1|0∝an}b∇acket∇i}ht= +|0∝an}b∇acket∇i}ht. (23.37) Then, using eqs.(23.36) and (23.37), we have ∝an}b∇acketle{t0|ϕ(x)|0∝an}b∇acket∇i}ht=∝an}b∇acketle{t0|ZZ−1ϕ(x)ZZ−1|0∝an}b∇acket∇i}ht =−∝an}b∇acketle{t0|ϕ(x)|0∝an}b∇acket∇i}ht. (23.38) Since∝an}b∇acketle{t0|ϕ(x)|0∝an}b∇acket∇i}htis equal to minus itself, it must be zero. Thus, as long as the ground state is unique, the Z 2symmetry of ϕ4theory guarantees that the ﬁeld has zero vacuum expectation value. We therefore do n ot need to enforce this condition with a counterterm Yϕ, as we did in ϕ3theory. (The assumption of a unique ground state does not necessarily hol d, however, as we will see in section 30.) 24: Nonabelian Symmetries 157 24Nonabelian Symmetries Prerequisite: 22 Consider the theory (introduced in section 22) of a two real s calar ﬁelds ϕ1 andϕ2with L=−1 2∂µϕ1∂µϕ1−1 2∂µϕ2∂µϕ2−1 2m2(ϕ2 1+ϕ2 2)−1 16λ(ϕ2 1+ϕ2 2)2.(24.1) We can generalize this to the case of Nreal scalar ﬁelds ϕiwith L=−1 2∂µϕi∂µϕi−1 2m2ϕiϕi−1 16λ(ϕiϕi)2, (24.2) where a repeated index is summed. This lagrangian is clearly invariant under the SO( N) transformation ϕi(x)→Rijϕj(x), (24.3) whereRis an orthogonal matrix with a positive determinant: RT=R−1, detR= +1. This largrangian is also clearly invariant under the Z 2trans- formationϕi(x)→ −ϕi(x), which enlarges SO( N) to O(N); see section 23. However, in this section we will be concerned only with th e continuous SO(N) part of the symmetry. Next we will need some results from group theory. Consider an inﬁnites- imal SO(N) transformation, Rij=δij+θij+O(θ2). (24.4) Orthogonality of Rijimplies that θijis real and antisymmetric. It is con- venient to express θijin terms of a basis set of hermitian matrices ( Ta)ij. The indexaruns from 1 to1 2N(N−1), the number of linearly independent, hermitian, antisymmetric, N×Nmatrices. We can, for example, choose eachTato have a single nonzero entry −iabove the main diagonal, and a corresponding + ibelow the main diagonal. These matrices obey the normalization condition Tr(TaTb) = 2δab. (24.5) In terms of them, we can write θjk=−iθa(Ta)jk, (24.6) whereθais a set of1 2N(N−1) real, inﬁnitesimal parameters. TheTa’s are the generator matrices of SO(N). The product of any two SO(N) transformations is another SO( N) transformation; this implies (see 24: Nonabelian Symmetries 158 problem 24.2) that the commutator of any two generator matri ces must be a linear combination of generator matrices, [Ta,Tb] =ifabcTc. (24.7) The numerical factors fabcare the structure coeﬃcients of the group, and eq.(24.7) speciﬁes its Lie algebra . Iffabc= 0, the group is abelian . Other- wise, it is nonabelian . Thus, U(1) and SO(2) are abelian groups (since they each have only one generator that obviously must commute wit h itself), and SO(N) forN≥3 is nonabelian. If we multiply eq.(24.7) on the right by Td, take the trace, and use eq.(24.5), we ﬁnd fabd=−1 2iTr/parenleftig [Ta,Tb]Td/parenrightig . (24.8) Using the cyclic property of the trace, we ﬁnd that fabdmust be completely antisymmetric. Taking the complex conjugate of eq.(24.8) ( and remember- ing that the Ta’s are hermitian matrices), we ﬁnd that fabdmust be real. The simplest nonabelian group is SO(3). In this case, we can c hoose (Ta)ij=−iεaij, whereεijkis the completely antisymmetric Levi-Civita symbol, with ε123= +1. The commutation relations become [Ta,Tb] =iεabcTc. (24.9) That is, the structure coeﬃcients of SO(3) are given by fabc=εabc. Consider now a theory with Ncomplex scalar ﬁelds ϕi, and a lagrangian L=−∂µϕ† i∂µϕi−m2ϕ† iϕi−1 4λ(ϕ† iϕi)2, (24.10) where a repeated index is summed. This lagrangian is clearly invariant under the U( N) transformation ϕi(x)→Uijϕj(x), (24.11) whereUis a unitary matrix: U†=U−1. We can write Uij=e−iθ/tildewideUij, whereθis a real parameter and det/tildewideUij= +1;/tildewideUijis called a special unitary matrix. Clearly the product of two special unitary matrices is another special unitary matrix; the N×Nspecial unitary matrices form the group SU(N). The group U( N) is the direct product of the group U(1) and the group SU(N); we write U( N) = U(1) ×SU(N). Consider an inﬁnitesimal SU( N) transformation, /tildewideUij=δij−iθa(Ta)ij+O(θ2), (24.12) whereθais a set of real, inﬁnitesimal parameters. Unitarity of/tildewideUimplies that the generator matrices Taare hermitian, and det/tildewideU= +1 implies 24: Nonabelian Symmetries 159 that eachTais traceless. (This follows from the general matrix formula ln detA= TrlnA.) The index aruns from 1 to N2−1, the number of lin- early independent, hermitian, traceless, N×Nmatrices. We can choose these matrices to obey the normalization condition of eq.(2 4.5). For SU(2), the generators can be chosen to be the Pauli matrices; the str ucture coeﬃ- cients of SU(2) then turn out to be fabc= 2εabc, the same as those of SO(3), up to an irrelevant overall factor [which could be removed by changing the numerical factor on right-hand side of eq.(24.5) from 2 to1 2]. For SU(N), we can choose the Ta’s in the following way. First, there are the SO( N) generators, with one −iabove the main diagonal a corre- sponding + ibelow; there are1 2N(N−1) of these. Next, we get another set by putting one +1 above the main diagonal and a corresponding +1 below; there are1 2N(N−1) of these. Finally, there are diagonal matrices with n 1’s along the main diagonal, followed a single entry −n, followed by zeros [times an overall normalization constant to enforce eq.(24 .5)]; the are N−1 of these. The total is N2−1, as required. However, if we examine the lagrangian of eq.(24.10) more clo sely, we ﬁnd that it is actually invariant under a larger symmetry gro up, namely SO(2N). To see this, write each complex scalar ﬁeld in terms of two r eal scalar ﬁelds, ϕj= (ϕj1+iϕj2)/√ 2. Then ϕ† jϕj=1 2(ϕ2 11+ϕ2 12+...+ϕ2 N1+ϕ2 N2). (24.13) Thus, we have 2 Nreal scalar ﬁelds that enter Lsymmetrically, and so the actual symmetry group of eq.(24.10) is SO(2 N), rather than just the obvious subgroup U( N). We will, however, meet the SU( N) groups again in Parts II and III, where they will play a more important role. Problems 24.1) Show that θijin eq.(24.4) must be antisymmetric if Ris orthogonal. 24.2) By considering the SO( N) transformation R′−1R−1R′R, whereR andR′are independent inﬁnitesimal SO( N) transformations, prove eq.(24.7). 24.3) a) Find the Noether current jaµfor the transformation of eq.(24.6). b) Show that [ ϕi,Qa] = (Ta)ijϕj, whereQais the Noether charge. c) Use this result, eq.(24.7), and the Jacobi identity (see p roblem 2.8) to show that [ Qa,Qb] =ifabcQc. 24: Nonabelian Symmetries 160 24.4) The elements of the group SO( N) can be deﬁned as N×Nmatrices Rthat satisfy Rii′Rjj′δi′j′=δij. (24.14) The elements of the symplectic group Sp(2N) can be deﬁned as 2 N× 2NmatricesSthat satisfy Sii′Sjj′ηi′j′=ηij, (24.15) where the symplectic metric ηijis antisymmetric, ηij=−ηji, and squares to minus the identity: η2=−I. One way to write ηis η=/parenleftigg0I −I0/parenrightigg , (24.16) whereIis theN×Nidentity matrix. Find the number of generators of Sp(2N). 25: Unstable Particles and Resonances 161 25Unstable Particles and Resonances Prerequisite: 14 Consider a theory of two real scalar ﬁelds, ϕandχ, with lagrangian L=−1 2∂µϕ∂µϕ−1 2m2 ϕϕ2−1 2∂µχ∂µχ−1 2m2 χχ2+1 2gϕχ2+1 6hϕ3.(25.1) This theory is renormalizable in six dimensions, where gandhare dimen- sionless coupling constants. Let us assume that mϕ>2mχ. Then it is kinematically possible for theϕparticle to decay into two χparticles. The amplitude for this process is given at tree level by the Feynman diagram of ﬁg.(25.1), an d is simply T=g. We can also choose to deﬁne gas the value of the exact ϕχ2vertex function V3(k,k′ 1,k′ 2) when all three particles are on shell: k2=−m2 ϕ, k′ 12=k′ 22=−m2 χ. This implies that T=g (25.2) exactly. According to the formulae of section 11, the diﬀerential dec ay rate (in the rest frame of the initial ϕparticle) is dΓ =1 2mϕdLIPS 2|T |2, (25.3) wheredLIPS 2is the Lorentz invariant phase space diﬀerential for two out - going particles, introduced in section 11. We must make a sli ght adaptation for six dimensions: dLIPS 2≡(2π)6δ6(k′ 1+k′ 2−k)/tildewiderdk′ 1/tildewiderdk′ 2. (25.4) Herek= (mϕ,0) is the energy-momentum of the decaying particle, and /tildewiderdk=d5k (2π)52ω(25.5) is the Lorentz-invariant phase-space diﬀerential for one p article. Recall that we can also write it as /tildewiderdk=d6k (2π)62πδ(k2+m2 χ)θ(k0), (25.6) whereθ(x) is the unit step function. Performing the integral over k0turns eq.(25.6) into eq.(25.5). 25: Unstable Particles and Resonances 162 kk1 k2 Figure 25.1: The tree-level Feynman diagram for the decay of aϕparticle (dashed line) into two χparticles (solid lines). Repeating for six dimensions what we did in section 11 for fou r dimen- sions, we ﬁnd dLIPS 2=|k′ 1|3 4(2π)4mϕdΩ, (25.7) where |k′ 1|=1 2(m2 ϕ−4m2 χ)1/2is the magnitude of the spatial momentum of one of the outgoing particles. We can now plug this into eq. (25.3), and use/integraltextdΩ = Ω 5= 2π5/2/Γ(5 2) =8 3π2. We also need to divide by a symmetry factor of two, due to the presence of two identical particles in the ﬁnal state. The result is Γ =1 2·1 2mϕ/integraldisplay dLIPS 2|T |2(25.8) =1 12πα(1−4m2 χ/m2 ϕ)3/2mϕ, (25.9) whereα=g2/(4π)3. However, as we discussed in section 11, we have a conceptual p roblem. According to our development of the LSZ formula in section 5, each incom- ing and outgoing particle should correspond to a single-par ticle state that is an exact eigenstate of the exact hamiltonian. This is clea rly not the case for a particle that can decay. Let us, then, compute something else instead: the correctio n to theϕ propagator from a loop of χparticles, as shown in ﬁg.(25.2). The diagram is the same as the one we already analyzed in section 14, excep t that the internal propagators contain mχinstead ofmϕ. (There is also a contri- bution from a loop of ϕparticles, but we can ignore it if we assume that h≪g.) We have Π(k2) =1 2α/integraldisplay1 0dxDlnD−A′k2−B′m2 ϕ, (25.10) where D=x(1−x)k2+m2 χ−iǫ, (25.11) 25: Unstable Particles and Resonances 163 Figure 25.2: A loop of χparticles correcting the ϕpropagator. andA′andB′are the ﬁnite counterterm coeﬃcients that remain after the inﬁnities have been absorbed. We now try to ﬁx A′andB′by imposing the usual on-shell conditions Π( −m2 ϕ) = 0 and Π′(−m2 ϕ) = 0. But, we have a problem. For k2=−m2 ϕandmϕ>2mχ,Dis negative for part of the range of x. Therefore ln Dhas an imaginary part. This imaginary part cannot be canceled by A′andB′, sinceA′andB′must be real: they are coeﬃcients of hermitian operators in the lagr angian. The best we can do is Re Π( −m2 ϕ) = 0 and Re Π′(−m2 ϕ) = 0. Imposing these gives Π(k2) =1 2α/integraldisplay1 0dxDln(D/|D0|)−1 12α(k2+m2 ϕ), (25.12) where D0=−x(1−x)m2 ϕ+m2 χ. (25.13) Now let us compute the imaginary part of Π( k2). This arises from the integration range x−< x < x +, wherex±=1 2±1 2(1 +m2 χ/k2)1/2are the roots of D= 0 whenk2<−4m2 χ. In this range, Im ln D=−iπ; the minus sign arises because, according to eq.(25.11), Dhas a small negative imaginary part. Now we have Im Π(k2) =−1 2πα/integraldisplayx+ x−dxD =−1 12πα(1 + 4m2 χ/k2)3/2k2(25.14) whenk2<−4m2 χ. Evaluating eq.(25.14) at k2=−m2 ϕ, we get Im Π(−m2 ϕ) =1 12πα(1−4m2 χ/m2 ϕ)3/2m2 ϕ. (25.15) From this and eq.(25.9), we see that Im Π(−m2 ϕ) =mϕΓ. (25.16) This is not an accident. Instead, it is a general rule. We will argue this in two ways: ﬁrst, from the mathematics of Feynman diagr ams, and second, from the physics of resonant scattering in quantum m echanics. 25: Unstable Particles and Resonances 164 We begin with the mathematics of Feynman diagrams. Return to the diagrammatic expression for Π( k2), before we evaluated any of the integrals: Π(k2) =−1 2ig2/integraldisplayd6ℓ1 (2π)6d6ℓ2 (2π)6(2π)6δ6(ℓ1+ℓ2−k) ×1 ℓ2 1+m2χ−iǫ1 ℓ2 2+m2χ−iǫ −(Ak2+Bm2 ϕ). (25.17) Here, for later convenience, we have assigned the internal l ines momenta ℓ1andℓ2, and explicitly included the momentum-conserving delta fu nction that ﬁxes one of them. We can take the imaginary part of Π( k2) by using the identity 1 x−iǫ=P1 x+iπδ(x), (25.18) wherePmeans the principal part. We then get, in a shorthand notatio n, Im Π(k2) =−1 2g2/integraldisplay/parenleftig P1P2−π2δ1δ2/parenrightig . (25.19) Next, we notice that the integral in eq.(25.17) is the Fourie r transform of [∆(x−y)]2, where ∆(x−y) =/integraldisplayd6k (2π)6eik(x−y) k2+m2χ−iǫ(25.20) is the Feynman propagator. Recall (from problem 8.5) that we can get the retarded or advanced propagator (rather than the Feynman pr opagator) by replacing the ǫin eq.(25.20) with, respectively, −sǫor +sǫ, wheres≡ sign(k0). Therefore, in eq.(25.19), replacing δ1with−s1δ1andδ2with +s2δ2yields an integral that is the real part of the Fourier transf orm of ∆ret(x−y)∆adv(x−y). But this product is zero, because the ﬁrst factor vanishes when x0≥y0, and the second when x0≤y0. So we can subtract the modiﬁed integrand from the original without changing th e value of the integral. Thus we have Im Π(k2) =1 2g2π2/integraldisplay (1 +s1s2)δ1δ2. (25.21) The factor of 1+ s1s2vanishes ifℓ0 1andℓ0 2have opposite signs, and equals 2 if they have the same sign. Because the delta function in eq.( 25.17) enforces ℓ0 1+ℓ0 2=k0, andk0=mϕis positive, both ℓ0 1andℓ0 2must be positive. 25: Unstable Particles and Resonances 165 So we can replace the factor of 1 + s1s2in eq.(25.21) with 2 θ(ℓ0 1)θ(ℓ0 2). Rearranging the numerical factors, we have ImΠ(k2) =1 4g2/integraldisplayd6ℓ1 (2π)6d6ℓ2 (2π)6(2π)6δ6(ℓ1+ℓ2−k) ×2πδ(ℓ2 1+m2 χ)θ(ℓ0 1)2πδ(ℓ2 2+m2 χ)θ(ℓ0 2).(25.22) If we now set k2=−m2 ϕ, use eqs.(25.4) and (25.6), and recall that T=g is the decay amplitude, we can rewrite eq.(25.22) as Im Π(−m2 ϕ) =1 4/integraldisplay dLIPS 2|T |2. (25.23) Comparing eqs.(25.8) and (25.23), we see that we indeed have Im Π(−m2 ϕ) =mϕΓ. (25.24) This relation persists at higher orders in perturbation the ory. Our anal- ysis can be generalized to give the Cutkosky rules for computing the imag- inary part of any Feynman diagram, but this is beyond the scop e of our current interest. To get a more physical understanding of this result, recall t hat in non- relativistic quantum mechanics, a metastable state with en ergyE0and angular momentum quantum number ℓshows up as a resonance in the partial-wave scattering amplitude, fℓ(E)∼1 E−E0+iΓ/2. (25.25) If we imagine convolving this amplitude with a wave packet/tildewideψ(E)e−iEt, we will ﬁnd a time dependence ψ(t)∼/integraldisplay dE1 E−E0+iΓ/2/tildewideψ(E)e−iEt ∼e−iE0t−Γt/2. (25.26) Therefore |ψ(t)|2∼e−Γt, and we identify Γ as the inverse lifetime of the metastable state. In the relativistic case, consider the scattering process χχ→χχ. The contributing diagrams from the eﬀective action are those of ﬁg.(20.1), where the exact internal propagator can be either ϕorχ. Suppose that the center-of-mass energy squared sis close tom2 ϕ. Since the ϕprogator has a pole near s=m2 ϕ,s-channelϕexchange, shown in ﬁg.(25.3), makes the dominant contribution to the χχscattering amplitude. We then have T ≃g2 −s+m2ϕ−Π(−s). (25.27) 25: Unstable Particles and Resonances 166 Figure 25.3: For snearm2 ϕ, the dominant contribution to χχscattering is s-channelϕexchange. Here we have used the fact that the exact ϕχχvertex has the value gwhen all three particles are on-shell. Now let us write s= (mϕ+ε)2≃m2 ϕ+ 2mϕε, (25.28) whereε≪mϕis the amount of energy by which our incoming particles are oﬀ resonance . We ﬁnd T ≃−g2/2mϕ ε+ Π(−m2ϕ)/2mϕ. (25.29) Recalling that ReΠ( −m2 ϕ) = 0, and comparing with eq.(25.25), we see that we should make the identiﬁcation of eq.(25.24). Reference Notes The Cutkosky rules are discussed in more detail in Peskin & Schroeder . More details on resonances can be found in Weinberg I . 26: Infrared Divergences 167 26Infrared Divergences Prerequisite: 20 In section 20, we computed the ϕϕ→ϕϕscattering amplitude in ϕ3theory in six dimensions in the high-energy limit ( s,|t|, and |u|all much larger thanm2). We found that T=T0/bracketleftig 1−11 12α/parenleftig ln(s/m2) +O(m0)/parenrightig +O(α2)/bracketrightig , (26.1) where T0=−g2(s−1+t−1+u−1) is the tree-level result, and the O(m0) term includes everything without a large logarithm that blows up in the limitm→0.1 Suppose we are interested in the limit of massless particles . The large log is then problematic, since it blows up in this limit. What does this mean? It means we have made a mistake. Actually, two mistakes. In th is section, we will remedy one of them. Throughout the physical sciences, it is necessary to make va rious ide- alizations in order to make progress. (Recall the “massless springs” and “frictionless planes” of freshman mechanics.) Sometimes t hese idealiza- tions can lead us into trouble, and that is one of the things th at has gone wrong here. We have assumed that we can isolate individual particles. Th e reasoning behind this was explained in section 5, and it depends on the e xistence of an energy gap between the one-particle states and the multipar ticle continuum. However, this gap vanishes if the theory includes massless p articles. In this case, it is possible that the scattering process involved th e creation of some extra very low energy (or soft) particles that escaped detection. Or, there may have been some extra soft particles hiding in the initial state that discreetly participated in the scattering process. Or, wha t was seen as a single high-energy particle may actually have been two or mo re particles that were moving colinearly and sharing the energy. Let us, then, correct our idealization of a perfect detector and account for these possibilities. We will work with ϕ3theory, initially in dspacetime dimensions. LetTbe the amplitude for some scattering process in ϕ3theory. Now consider the possibility that one of the outgoing particles in this process splits into two, as shown in ﬁg.(26.1). The amplitude for thi s new process 1In writing Tin this form, we have traded factors of ln tand ln ufor ln sby ﬁrst using lnt= lns+ ln(t/s), and then hiding the ln( t/s) terms in the O(m0) catchall. 26: Infrared Divergences 168 k2k1 k Figure 26.1: An outgoing particle splits into two. The gray c ircle stands for the sum of all diagrams contributing to the original ampl itudeiT. is given in terms of Tby Tsplit=ig−i k2+m2T, (26.2) wherek=k1+k2, andk1andk2are the on-shell four-momenta of the two particles produced by the split. (For notational conven ience, we drop our usual primes on the outgoing momenta.) The key point is th is: in the massless limit, it is possible for 1 /(k2+m2) to diverge. To understand the physical consequences of this possibilit y, we should compute an appropriate cross-section. To get the cross sect ion for the original process (without the split), we multiply |T |2by/tildewiderdk(as well as by similar diﬀerentials for other outgoing particles, and by a n overall energy- momentum delta function). For the process with the split, we multiply |Tsplit|2by1 2/tildewiderdk1/tildewiderdk2instead of/tildewiderdk. (The factor of one-half is for counting of identical particles.) If we assume that (due to some imper fection) our detector cannot tell whether or not the one particle actuall y split into two, then we should (according to the usual rules of quantum mecha nics) add the probabilities for the two events, which are distinguish able in principle. We can therefore deﬁne an eﬀectively observable squared-am plitude via |T |2 obs/tildewiderdk=|T |2/tildewiderdk+|Tsplit|21 2/tildewiderdk1/tildewiderdk2+... . (26.3) Here the ellipses stand for all other similar processes invo lving emission of one or more extra particles in the ﬁnal state, or absorption o f one or more extra particles in the initial state. We can simplify eq.(26.3) by including a factor of 1 = (2π)d−12ωδd−1(k1+k2−k)/tildewiderdk (26.4) 26: Infrared Divergences 169 in the second term. Now all terms in eq.(26.3) include a facto r of/tildewiderdk, so we can drop it. Then, using eq.(26.2), we get |T |2 obs≡ |T |2/bracketleftigg 1 +g2 (k2+m2)2(2π)d−12ωδd−1(k1+k2−k)1 2/tildewiderdk1/tildewiderdk2+.../bracketrightigg . (26.5) Now we come to the point: in the massless limit, the phase spac e integral in the second term in eq.(26.5) can diverge. This is because, form= 0, k2= (k1+k2)2=−4ω1ω2sin2(θ/2), (26.6) whereθis the angle between the spatial momenta k1andk2, andω1,2= |k1,2|. Also, form= 0, /tildewiderdk1/tildewiderdk2∼(ωd−3 1dω1)(ωd−3 2dω2)(sind−3θdθ). (26.7) Therefore, for small θ, /tildewiderdk1/tildewiderdk2 (k2)2∼dω1 ω5−d 1dω2 ω5−d 2dθ θ7−d. (26.8) Thus the integral over each ωdiverges at the low end for d≤4, and the integral over θdiverges at the low end for d≤6. These divergent integrals would be cut oﬀ (and rendered ﬁnite) if we kept the mass mnonzero, as we will see below. Our discussion leads us to expect that the m→0 divergence in the second term of eq.(26.5) should cancel them→0 divergence in the loop correction to |T |2. We will now see how this works (or fails to work) in detail for the familiar case of two-particle scattering in six spac etime dimensions, where Tis given by eq.(26.1). For d= 6, there is no problem with soft particles (corresponding to the small- ωdivergence), but there is a problem with collinear particles (corresponding to the small- θdivergence). Let us assume that our imperfect detector cannot tell one par ticle from two nearly collinear particles if the angle θbetween their spatial momenta is less than some small angle δ. Since we ultimately want to take the m→0 limit, we will evaluate eq.(26.5) with m2/k2≪δ2≪1. We can immediately integrate over d5k2using the delta function, which results in setting k2=k−k1everywhere. Let βthen be the angle between k1(which is still to be integrated over) and k(which is ﬁxed). For two- particle scattering, |k|=1 2√sin the limit m→0. We then have (2π)52ωδ5(k1+k2−k)1 2/tildewiderdk1/tildewiderdk2→Ω4 4(2π)5ω ω1ω2|k1|4d|k1|sin3βdβ, (26.9) 26: Infrared Divergences 170 where Ω 4= 2π2is the area of the unit four-sphere. Now let γbe the angle between k2andk. The geometry of this trio of vectors implies θ=β+γ, |k1|= (sinγ/sinθ)|k|, and|k2|= (sinβ/sinθ)|k|. All three of the angles are small and positive, and it then is useful to write β=xθandγ= (1−x)θ, with 0 ≤x≤1 andθ≤δ≪1. In the low mass limit, we can safely set m= 0 everywhere in eq.(26.5) except in the propagator, 1 /(k2+m2). Then, expanding to leading order in bothθandm, we ﬁnd (after some algebra) k2+m2≃ −x(1−x)k2/bracketleftig θ2+ (m2/k2)f(x)/bracketrightig , (26.10) wheref(x) = (1 −x+x2)/(x−x2)2. Everywhere else in eq.(26.5), we can safely setω1=|k1|= (1−x)|k|andω2=|k2|=x|k|. Then, changing the integration variables in eq.(26.9) from |k1|andβtoxandθ, we get |T |2 obs=|T |2/bracketleftigg 1 +g2Ω4 4(2π)5/integraldisplay1 0x(1−x)dx/integraldisplayδ 0θ3dθ [θ2+ (m2/k2)f(x)]2+.../bracketrightigg . (26.11) Performing the integral over θyields 1 2ln(δ2k2/m2)−1 2lnf(x)−1 2. (26.12) Then, performing the integral over xand using Ω 4= 2π2andα=g2/(4π)3, we get |T |2 obs=|T |2/bracketleftig 1 +1 12α/parenleftig ln(δ2k2/m2) +c/parenrightig +.../bracketrightig , (26.13) wherec= (4−3√ 3π)/3 =−4.11. The displayed correction term accounts for the possible spl itting of one of the two outgoing particles. Obviously, there is an identi cal correction for the other outgoing particle. Less obviously (but still t rue), there is an identical correction for each of the two incoming particl es. (A glib explanation is that we are computing an eﬀective amplitude- squared, and this is the same for the reverse process, with in and outgoing particles switched. So in and out particles should be treated symmetri cally.) Then, since we have a total of four in and out particles (before acco unting for any splitting), |T |2 obs=|T |2/bracketleftig 1 +4 12α/parenleftig ln(δ2k2/m2) +c/parenrightig +O(α2)/bracketrightig . (26.14) We have now accounted for the O(α) corrections due to the failure of our detector to separate two particles whose spatial moment a are nearly 26: Infrared Divergences 171 parallel. Combining this with eq.(26.1), and recalling tha tk2=1 4s, we get |T |2 obs=|T0|2/bracketleftig 1−11 6α/parenleftig ln(s/m2) +O(m0)/parenrightig +O(α2)/bracketrightig ×/bracketleftig 1 +1 3α/parenleftig ln(δ2s/m2) +O(m0)/parenrightig +O(α2)/bracketrightig =|T0|2/bracketleftig 1−α/parenleftig 3 2ln(s/m2) +1 3ln(1/δ2) +O(m0)/parenrightig +O(α2)/bracketrightig . (26.15) We now have two kinds of large logs. One is ln(1 /δ2); this factor depends on the properties of our detector. If we build a very good dete ctor, one for whichαln(1/δ2) is not small, then we will have to do more work, and calculate higher-order corrections to eq.(26.15). The other large log is our original nemesis ln( s/m2). This factor blows up in the massless limit. This means that there is still a mist ake hidden somewhere in our analysis. Reference Notes Infrared divergences in quantum electrodynamics are discu ssed in Brown andPeskin & Schroeder . More general treatments can be found in Sterman andWeinberg I . 27: Other Renormalization Schemes 172 27Other Renormalization Schemes Prerequisite: 26 To ﬁnd the remaining mistake in eq.(26.15), we must review ou r renor- malization procedure. Recall our result from section 14 for the one-loop correction to the propagator, Π(k2) =−/bracketleftig A+1 6α/parenleftig 1ε+1 2/parenrightig/bracketrightig k2−/bracketleftig B+α/parenleftig 1ε+1 2/parenrightig/bracketrightig m2 +1 2α/integraldisplay1 0dxDln(D/µ2) +O(α2), (27.1) whereα=g2/(4π)3andD=x(1−x)k2+m2. The derivative of Π( k2) with respect tok2is Π′(k2) =−/bracketleftig A+1 6α/parenleftig 1ε+1 2/parenrightig/bracketrightig +1 2α/integraldisplay1 0dxx(1−x)/bracketleftig ln(D/µ2) + 1/bracketrightig +O(α2).(27.2) We previously determined AandBvia the requirements Π( −m2) = 0 and Π′(−m2) = 0. The ﬁrst condition ensures that the exact propagator ˜∆(k2) has a pole at k2=−m2, and the second ensures that the residue of this pole is one. Recall that the ﬁeld must be normalized in this wa y for the validity of the LSZ formula. We now consider the massless limit. We have D=x(1−x)k2, and we should apparently try to impose Π(0) = Π′(0) = 0. However, Π(0) is now automatically zero for any values of AandB, while Π′(0) is ill deﬁned. Physically, the problem is that the one-particle states are no longer sep- arated from the multiparticle continuum by a ﬁnite gap in ene rgy. Mathe- matically, the pole in ˜∆(k2) atk2=−m2merges with the branch point at k2=−4m2, and is no longer a simple pole. The only way out of this diﬃculty is to change the renormalization scheme . Let us ﬁrst see what this means in the case m∝ne}ationslash= 0, where we know what we are doing. Let us try making a diﬀerent choice of AandB. Speciﬁcally, let A=−1 6α1ε+O(α2), B=−α1ε+O(α2). (27.3) Here we have chosen AandBto cancel the inﬁnities, and nothing more; we say that AandBhave no ﬁnite parts . This choice represents a diﬀerent renormalization scheme . Our original choice (which, up until now, we have pretended was inescapable!) is called the on-shell or OS scheme. The choice 27: Other Renormalization Schemes 173 of eq.(27.3) is called the modiﬁed minimal-subtraction orMS (pronounced “emm-ess-bar”) scheme. [“Modiﬁed” because we introduced µviag→ g˜µε/2, withµ2= 4πe−γ˜µ2; had we set µ= ˜µinstead, the scheme would be just plain minimal subtraction or MS.] Now we have ΠMS(k2) =−1 12α(k2+ 6m2) +1 2α/integraldisplay1 0dxDln(D/µ2) +O(α2),(27.4) as compared to our old result in the on-shell scheme, ΠOS(k2) =−1 12α(k2+m2) +1 2α/integraldisplay1 0dxDln(D/D 0) +O(α2),(27.5) where again D=x(1−x)k2+m2, andD0= [−x(1−x)+1]m2. Notice that ΠMS(k2) has a well-deﬁned m→0 limit, whereas Π OS(k2) does not. On the other hand, ΠMS(k2) depends explicitly on the fake parameter µ, whereas ΠOS(k2) does not. What does this all mean? First, in the MS scheme, the propagator ∆MS(k2) will no longer have a pole atk2=−m2. The pole will be somewhere else. However, by deﬁnition , the actual physical mass mphof the particle is determined by the location of this pole: k2=−m2 ph. Thus, the lagrangian parameter mis no longer the same as mph. Furthermore, the residue of this pole is no longer one. Let us call the residueR. The LSZ formula must now be corrected by multiplying its right-hand side by a factor of R−1/2for each external particle (incoming or outgoing). This is because it is the ﬁeld R−1/2ϕ(x) that now has unit amplitude to create a one-particle state. Note also that, in the LSZ formula, each Klein-Gordon wave op erator should be −∂2+m2 ph, and not −∂2+m2; also, each external four-momentum should square to −m2 ph, and not −m2. A review of the derivation of the LSZ formula clearly shows that each of these mass parameters must be the actual particle mass, and not the parameter in the lagrangia n. Finally, in the LSZ formula, each external line will contrib ute a factor of Rwhen the associated Klein-Gordon wave operator hits the ext ernal prop- agator and cancels its momentum-space pole, leaving behind the residue R. Combined with the correction factor of R−1/2for each ﬁeld, we get a net factor ofR1/2for each external line when using the MS scheme. Internal lines each contribute a factor of ( −i)/(k2+m2), wheremis the lagrangian- parameter mass, and each vertex contributes a factor of iZgg, wheregis the lagrangian-parameter coupling. Let us now compute the relation between mandmph, and then compute R. We have ∆MS(k2)−1=k2+m2−ΠMS(k2), (27.6) 27: Other Renormalization Schemes 174 and, by deﬁnition, ∆MS(−m2 ph)−1= 0. (27.7) Settingk2=−m2 phin eq.(27.6), using eq.(27.7), and rearranging, we ﬁnd m2 ph=m2−ΠMS(−m2 ph). (27.8) Since ΠMS(k2) isO(α), we see that the diﬀerence between m2 phandm2is O(α). Therefore, on the right-hand side, we can replace m2 phwithm2, and only make an error of O(α2). Thus m2 ph=m2−ΠMS(−m2) +O(α2). (27.9) Working this out, we get m2 ph=m2−1 2α/bracketleftbigg 1 6m2−m2+/integraldisplay1 0dxD 0ln(D0/µ2)/bracketrightbigg +O(α2),(27.10) whereD0= [1−x(1−x)]m2. Doing the integrals yields m2 ph=m2/bracketleftig 1 +5 12α/parenleftig ln(µ2/m2) +c′/parenrightig +O(α2)/bracketrightig . (27.11) wherec′= (34−3π√ 3)/15 = 1.18. Now, physics should be independent of the fake parameter µ. However, the right-hand side of eq.(27.11) depends explicitly on µ. It must, be, then, thatmandαtake on diﬀerent numerical values as µis varied, in just the right way to leave physical quantities (like mph) unchanged. We can use this information to ﬁnd diﬀerential equations tha t tell us howmandαchange with µ. For example, take the logarithm of eq.(27.11) and divide by two to get lnmph= lnm+5 12α/parenleftig ln(µ/m) +1 2c′/parenrightig +O(α2). (27.12) Now diﬀerentiate with respect to ln µand require mphto remain ﬁxed: 0 =d dlnµlnmph =1 mdm dlnµ+5 12α+O(α2). (27.13) To get the second line, we had to assume that dα/dlnµ=O(α2), which we will verify shortly. Then, rearranging eq.(27.13) gives dm dlnµ=/parenleftig −5 12α+O(α2)/parenrightig m. (27.14) 27: Other Renormalization Schemes 175 The factor in large parentheses on the right is called the anomalous dimen- sionof the mass parameter, and it is often given the name γm(α). Turning now to the residue R, we have R−1=d dk2/bracketleftig ∆MS(k2)−1/bracketrightig/vextendsingle/vextendsingle/vextendsingle/vextendsingle/vextendsingle k2=−m2 ph. (27.15) Using eq.(27.6), we get R−1= 1−Π′ MS(−m2 ph) = 1−Π′ MS(−m2) +O(α2) = 1 +1 12α/parenleftig ln(µ2/m2) +c′′/parenrightig +O(α2), (27.16) wherec′′= (17−3π√ 3)/3 = 0.23. We can also use MS to deﬁne the vertex function. We take C=−α1ε+O(α2), (27.17) and so V3,MS(k1,k2,k3) =g/bracketleftbigg 1−1 2α/integraldisplay dF3ln(D/µ2) +O(α2)/bracketrightbigg (27.18) whereD=xyk2 1+yzk2 2+zxk2 3+m2. Let us now compute the ϕϕ→ϕϕscattering amplitude in our fancy new renormalization scheme. In the low-mass limit, repeati ng the steps that led to eq.(26.1), and including the LSZ correction fact or (R1/2)4, we get T=R2T0/bracketleftig 1−11 12α/parenleftig ln(s/µ2) +O(m0)/parenrightig +O(α2)/bracketrightig , (27.19) where T0=−g2(s−1+t−1+u−1) is the tree-level result. Now using Rfrom eq.(27.16), we ﬁnd T=T0/bracketleftig 1−α/parenleftig 11 12ln(s/µ2) +1 6ln(µ2/m2) +O(m0)/parenrightig +O(α2)/bracketrightig .(27.20) To get an observable amplitude-squared with an imperfect de tector, we must square eq.(27.20) and multiply it by the correction fac tor we derived in section 26, |T |2 obs=|T |2/bracketleftig 1 +1 3α/parenleftig ln(δ2s/m2) +O(m0)/parenrightig +O(α2)/bracketrightig , (27.21) whereδis the angular resolution of the detector. Combining this wi th eq.(27.20), we get |T |2 obs=|T0|2/bracketleftig 1−α/parenleftig 3 2ln(s/µ2) +1 3ln(1/δ2) +O(m0)/parenrightig +O(α2)/bracketrightig . (27.22) 27: Other Renormalization Schemes 176 All factors of ln m2have disappeared! Finally, we have obtained an expres- sion that has a well-deﬁned m→0 limit. Of course,µis still a fake parameter, and so |T |2 obscannot depend on it. It must be, then, that the explicit dependence on µin eq.(27.22) is canceled by the implicit µdependence of α. We can use this information to ﬁgure out how αmust vary with µ. Noting that |T0|2=O(g4) =O(α2), we have ln|T |2 obs=C1+ 2lnα+ 3α/parenleftig lnµ+C2/parenrightig +O(α2), (27.23) whereC1andC2are independent of µandα(but depend on the Mandel- stam variables). Diﬀerentiating with respect to ln µthen gives 0 =d dlnµln|T |2 obs =2 αdα dlnµ+ 3α+O(α2), (27.24) or, after rearranging, dα dlnµ=−3 2α2+O(α3). (27.25) The right-hand side of this equation is called the beta function . Returning to eq.(27.22), we are free to choose any convenien t value ofµ that we might like. To avoid introducing unnecessary large l ogs, we should chooseµ2∼s. To compare the results at diﬀerent values of s, we need to solve eq.(27.25). Keeping only the leading term in the beta function, the solut ion is α(µ2) =α(µ1) 1 +3 2α(µ1)ln(µ2/µ1). (27.26) Thus, asµincreases,α(µ) decreases. A theory with this property is said to beasymptotically free . In this case, the tree-level approximation (in the MS scheme with µ2∼s) becomes better and better at higher and higher energies. Of course, the opposite is true as well: as µdecreases,α(µ) increases. As we go to lower and lower energies, the theory becomes more a nd more strongly coupled. If the particle mass is nonzero, this process stops at µ∼m. This is because the minimum value of sis 4m2, and so the factor of ln( s/µ2) becomes an unwanted large log for µ≪m. We should therefore not use 27: Other Renormalization Schemes 177 values ofµbelowm. Perturbation theory is still good at these low energies ifα(m)≪1. If the particle mass is zero, α(µ) continues to increase at lower and lower energies, and eventually perturbation theory breaks down. This is a signal that the low-energy physics may be quite diﬀerent from what w e expect on the basis of a perturbative analysis. In the case of ϕ3theory, we know what the correct low-energy physics is: the perturbative ground state is unstable against tunne ling through the potential barrier, and there is no true ground state. Asympt otic freedom is, in this case, a signal of this impending disaster. Much more interesting is asymptotic freedom in a theory that doeshave a true ground state, such as quantum chromodynamics. In this example, the particle excitations are colorless hadrons, rather tha n the quarks and gluons we would expect from examining the lagrangian. If the sign of the beta function is positive, then the theory i sinfrared free. The coupling increases as µincreases, and, at suﬃciently high en- ergy, perturbation theory breaks down. On the other hand, th e coupling decreases as we go to lower energies. Once again, though, we s hould stop this process at µ∼mif the particles have nonzero mass. Quantum elec- trodynamics with massive electrons (but, of course, massle ss photons) is in this category. Still more complicated behaviors are possible if the beta fu nction has a zero at a nonzero value of α. We brieﬂy consider this case in the next section. Reference Notes Minimal subtraction is treated in more detail in Brown ,Collins , andRa- mond I . Problems 27.1) Suppose that we have a theory with β(α) =b1α2+O(α3), (27.27) γm(α) =c1α+O(α2). (27.28) Neglecting the higher-order terms, show that m(µ2) =/bracketleftbiggα(µ2) α(µ1)/bracketrightbiggc1/b1 m(µ1). (27.29) 28: The Renormalization Group 178 28The Renormalization Group Prerequisite: 27 In section 27 we introduced the MS renormalization scheme, and used the fact that physical observables must be independent of the fa ke parameter µto ﬁgure out how the lagrangian parameters mandgmust change with µ. In this section we re-derive these results from a much more f ormal (but calculationally simpler) point of view, and see how they ext end to all or- ders of perturbation theory. Equations that tells us how the lagrangian parameters (and other objects that are not directly measura ble, like cor- relation functions) vary with µare collectively called the equations of the renormalization group . Let us recall the lagrangian of our theory, and write it in two diﬀerent ways. Ind= 6−εdimensions, we have L=−1 2Zϕ∂µϕ∂µϕ−1 2Zmm2ϕ2+1 6Zgg˜µε/2ϕ3+Yϕ (28.1) and L=−1 2∂µϕ0∂µϕ0−1 2m2 0ϕ2 0+1 6g0ϕ3 0+Y0ϕ0. (28.2) The ﬁelds and parameters in eq.(28.1) are the renormalized ﬁelds and pa- rameters. (And in particular, they are renormalized using t heMS scheme, withµ2= 4πe−γ˜µ2.) The ﬁelds and parameters in eq.(28.2) are the bare ﬁelds and parameters. Comparing eqs.(28.1) and (28.2) give s us the rela- tionships between them: ϕ0(x) =Z1/2 ϕϕ(x), (28.3) m0=Z−1/2 ϕZ1/2 mm, (28.4) g0=Z−3/2 ϕZgg˜µε/2, (28.5) Y0=Z−1/2 ϕY . (28.6) Recall that, after using dimensional regularization, the i nﬁnities coming from loop integrals take the form of inverse powers of ε= 6−d. In the MS renormalization scheme, we choose the Z’s to cancel oﬀ these powers of 1/ε, and nothing more. Therefore the Z’s can be written as Zϕ= 1 +∞/summationdisplay n=1an(α) εn, (28.7) Zm= 1 +∞/summationdisplay n=1bn(α) εn, (28.8) Zg= 1 +∞/summationdisplay n=1cn(α) εn, (28.9) 28: The Renormalization Group 179 whereα=g2/(4π)3. Computing Π( k2) andV3(k1,k2,k3) in perturbation theory in the MS scheme gives us Taylor series in αforan(α),bn(α), and cn(α). So far we have found a1(α) =−1 6α+O(α2), (28.10) b1(α) =−α+O(α2), (28.11) c1(α) =−α+O(α2), (28.12) and thatan(α),bn(α), andcn(α) are all at least O(α2) forn≥2. Next we turn to the trick that we will employ to compute the bet a function for α, the anomalous dimension of m, and other useful things. This is the trick: bare ﬁelds and parameters must be independent of µ. Why is this so? Recall that we introduced µwhen we found that we had to regularize the theory to avoid inﬁnities in the loop integ rals of Feynman diagrams. We argued at the time (and ever since) that physica l quantities had to be independent of µ. Thusµis not really a parameter of the theory, but just a crutch that we had to introduce at an intermediate s tage of the calculation. In principle, the theory is completely speciﬁ ed by the values of the bare parameters, and, if we were smart enough, we would be able to compute the exact scattering amplitudes in terms of them, without ever introducing µ. The point is this: since the exact scattering amplitudes ar e independent of µ, the bare parameters must be as well. Let us start with g0. It is convenient to deﬁne α0≡g2 0/(4π)3=Z2 gZ−3 ϕ˜µεα, (28.13) and also G(α,ε)≡ln(Z2 gZ−3 ϕ). (28.14) From the general structure of eqs.(28.7) and (28.9), we have G(α,ε) =∞/summationdisplay n=1Gn(α) εn, (28.15) where, in particular, G1(α) = 2c1(α)−3a1(α) =−3 2α+O(α2). (28.16) The logarithm of eq.(28.13) can now be written as lnα0=G(α,ε) + lnα+εln ˜µ. (28.17) 28: The Renormalization Group 180 Next, diﬀerentiate eq.(28.17) with respect to ln µ, and require α0to be independent of it: 0 =d dlnµlnα0 =∂G(α,ε) ∂αdα dlnµ+1 αdα dlnµ+ε. (28.18) Now regroup the terms, multiply by α, and use eq.(28.15) to get 0 =/parenleftbigg 1 +αG′ 1(α) ε+αG′ 2(α) ε2+.../parenrightbiggdα dlnµ+εα. (28.19) Next we use some physical reasoning: dα/dlnµis the rate at which α must change to compensate for a small change in ln µ. If compensation is possible at all, this rate should be ﬁnite in the ε→0 limit. Therefore, in a renormalizable theory, we should have dα dlnµ=−εα+β(α). (28.20) The ﬁrst term, −εα, is ﬁxed by matching the O(ε) terms in eq.(28.19). The second term, the beta function β(α), is similarly determined by matching theO(ε0) terms; the result is β(α) =α2G′ 1(α). (28.21) Terms that are higher-order in 1 /εmust also cancel, and this determines all the other G′ n(α)’s in terms of G′ 1(α). Thus, for example, cancellation of theO(ε−1) terms ﬁxes G′ 2(α) =αG′ 1(α)2. These relations among the G′ n(α)’s can of course be checked order by order in perturbation th eory. From eq.(28.21) and eq.(28.16), we ﬁnd that the beta functio n is β(α) =−3 2α2+O(α3). (28.22) Hearteningly, this is the same result we found in section 27 b y requiring the observed scattering cross section |T |2 obsto be independent of µ. However, simply as a matter of practical calculation, it is much easie r to compute G1(α) than it is to compute |T |2 obs. Next consider the invariance of m0. We begin by deﬁning M(α,ε)≡ln(Z1/2 mZ−1/2 ϕ) =∞/summationdisplay n=1Mn(α) εn. (28.23) 28: The Renormalization Group 181 From eqs.(28.10) and (28.12) we have M1(α) =1 2b1(α)−1 2a1(α) =−5 12α+O(α2). (28.24) Then, from eq.(28.4), we have lnm0=M(α,ε) + lnm. (28.25) Take the derivative with respect to ln µand require m0to be unchanged: 0 =d dlnµlnm0 =∂M(α,ε) ∂αdα dlnµ+1 mdm dlnµ. =∂M(α,ε) ∂α/parenleftig −εα+β(α)/parenrightig +1 mdm dlnµ. (28.26) Rearranging, we ﬁnd 1 mdm dlnµ=/parenleftig εα−β(α)/parenrightig∞/summationdisplay n=1M′ n(α) εn =αM′ 1(α) +... , (28.27) where the ellipses stand for terms with powers of 1 /ε. In a renormalizable theory,dm/d lnµshould be ﬁnite in the ε→0 limit, and so these terms must actually all be zero. Therefore, the anomalous dimensi on of the mass, deﬁned via γm(α)≡1 mdm dlnµ, (28.28) is given by γm(α) =αM′ 1(α) =−5 12α+O(α2). (28.29) Comfortingly, this is just what we found in section 27. Let us now consider the propagator in the MS renormalization scheme, ˜∆(k2) =i/integraldisplay d6xeikx∝an}b∇acketle{t0|Tϕ(x)ϕ(0)|0∝an}b∇acket∇i}ht. (28.30) The bare propagator, ˜∆0(k2) =i/integraldisplay d6xeikx∝an}b∇acketle{t0|Tϕ0(x)ϕ0(0)|0∝an}b∇acket∇i}ht, (28.31) 28: The Renormalization Group 182 should be (by the now-familiar argument) independent of µ. The bare and renormalized propagators are related by ˜∆0(k2) =Zϕ˜∆(k2). (28.32) Taking the logarithm and diﬀerentiating with respect to ln µ, we get 0 =d dlnµln˜∆0(k2) =dlnZϕ dlnµ+d dlnµln˜∆(k2) =dlnZϕ dlnµ+1 ˜∆(k2)/parenleftbigg∂ ∂lnµ+dα dlnµ∂ ∂α+dm dlnµ∂ ∂m/parenrightbigg ˜∆(k2).(28.33) We can write lnZϕ=a1(α) ε+a2(α)−1 2a2 1(α) ε2+... . (28.34) Then we have dlnZϕ dlnµ=∂lnZϕ ∂αdα dlnµ =/parenleftbigga′ 1(α) ε+.../parenrightbigg/parenleftig −εα+β(α)/parenrightig =−αa′ 1(α) +... , (28.35) where the ellipses in the last line stand for terms with power s of 1/ε. Since ˜∆(k2) should vary smoothly with µin theε→0 limit, these must all be zero. We then deﬁne the anomalous dimension of the ﬁeld γϕ(α)≡1 2dlnZϕ dlnµ. (28.36) From eq.(28.35) we ﬁnd γϕ(α) =−1 2αa′ 1(α) = +1 12α+O(α2). (28.37) Eq.(28.33) can now be written as /parenleftbigg∂ ∂lnµ+β(α)∂ ∂α+γm(α)m∂ ∂m+ 2γϕ(α)/parenrightbigg ˜∆(k2) = 0 (28.38) 28: The Renormalization Group 183 in theε→0 limit. This is the Callan-Symanzik equation for the propagator. The Callan-Symanzik equation is most interesting in the mas sless limit, and for a theory with a zero of the beta function at a nonzero va lue ofα. So, let us suppose that β(α∗) = 0 for some α∗∝ne}ationslash= 0. Then, for α=α∗and m= 0, the Callan-Symanzik equation becomes /parenleftbigg∂ ∂lnµ+ 2γϕ(α∗)/parenrightbigg ˜∆(k2) = 0. (28.39) The solution is ˜∆(k2) =C(α∗) k2/parenleftigg µ2 k2/parenrightigg−γϕ(α∗) , (28.40) whereC(α∗) is an integration constant. (We used the fact that ˜∆(k2) has mass dimension −2 to get the k2dependence in addition to the µ dependence.) Thus the naive scaling law ˜∆(k2)∼k−2is changed to ˜∆(k2)∼k−2[1−γϕ(α∗)]. This has applications in the theory of critical phe- nomena, which is beyond the scope of this book. Reference Notes The formal development of the renormalization group is expl ored in more detail in Brown ,Collins , andRamond I . Problems 28.1) Consider ϕ4theory, L=−1 2Zϕ∂µϕ∂µϕ−1 2Zmm2ϕ2−1 24Zλλ˜µεϕ4, (28.41) ind= 4−εdimensions. Compute the beta function to O(λ2), the anomalous dimension of mtoO(λ), and the anomalous dimension of ϕtoO(λ). 28.2) Repeat problem 28.1 for the theory of problem 9.3. 28.3) Consider the lagrangian density L=−1 2Zϕ∂µϕ∂µϕ−1 2Zmm2ϕ2+Yϕ −1 2Zχ∂µχ∂µχ−1 2ZMM2χ2 +1 6Zgg˜µε/2ϕ3+1 2Zhh˜µε/2ϕχ2(28.42) ind= 6−εdimensions, where ϕandχare real scalar ﬁelds, and Y is adjusted to make ∝an}b∇acketle{t0|ϕ(x)|0∝an}b∇acket∇i}ht= 0. (Why is no such term needed for χ?) 28: The Renormalization Group 184 a) Compute the one-loop contributions to each of the Z’s in the MS renormalization scheme. b) The bare couplings are related to the renormalized ones vi a g0=Z−3/2 ϕZgg˜µε/2, (28.43) h0=Z−1 ϕZ−1/2 χZhh˜µε/2. (28.44) Deﬁne G(g,h,ε) =/summationtext∞ n=1Gn(g,h)ε−n≡ln(Z−3/2 ϕZg), (28.45) H(g,h,ε) =/summationtext∞ n=1Hn(g,h)ε−n≡ln(Z−1 ϕZ−1/2 χZh).(28.46) By requiring g0andh0to be independent of µ, and by assuming that dg/dµ anddh/dµ are ﬁnite as ε→0, show that µdg dµ=−1 2εg+1 2g/parenleftbigg g∂G1 ∂g+h∂G1 ∂h/parenrightbigg , (28.47) µdh dµ=−1 2εh+1 2h/parenleftbigg g∂H1 ∂g+h∂H1 ∂h/parenrightbigg . (28.48) c) Use your results from part (a) to compute the beta function s βg(g,h)≡limε→0µdg/dµ andβh(g,h)≡limε→0µdh/dµ . You should ﬁnd terms of order g3,gh2, andh3inβg, and terms of order g2h,gh2, andh3inβh. d) Without loss of generality, we can choose gto be positive; hcan then be positive or negative, and the diﬀerence is physicall y signiﬁ- cant. (You should understand why this is true.) For what nume rical range(s) of h/gareβgandβh/hboth negative? Why is this an inter- esting question? 29: Eﬀective Field Theory 185 29Effective Field Theory Prerequisite: 28 So far we have been discussing only renormalizable theories . In this section, we investigate what meaning can be assigned to nonrenormali zable theories, following an approach pioneered by Ken Wilson. We will begin by analyzing a renormalizable theory from a new point of view. Consider, as an example, ϕ4theory in four spacetime dimensions: L=−1 2Zϕ∂µϕ∂µϕ−1 2Zmm2 phϕ2−1 24Zλλphϕ4. (29.1) (This example is actually problematic, because this theory istrivial , a tech- nical term that we will exlain later. For now we proceed with a perturbative analysis.) We take the renormalizing Zfactors to be deﬁned in an on-shell scheme, and have emphasized this by writing the particle mas s asmphand the coupling constant as λph. We deﬁne λphas the value of the exact 1PI four-point vertex with zero external four-momenta: λph≡V4(0,0,0,0). (29.2) The path integral is given by Z(J) =/integraldisplay DϕeiS+i/integraltext Jϕ, (29.3) whereS=/integraltextd4xLand/integraltextJϕis short for/integraltextd4xJϕ. Our ﬁrst step in analyzing this theory will be to perform the W ick rotation (applied to loop integrals in section 14) directly on the action. We deﬁne a euclidean time τ≡it. Then we have Z(J) =/integraldisplay Dϕe−SE−/integraltext Jϕ, (29.4) whereSE=/integraltextd4xLE,d4x=d3xdτ, LE=1 2Zϕ∂µϕ∂µϕ+1 2Zmm2 phϕ2+1 24Zλλphϕ4, (29.5) and ∂µϕ∂µϕ= (∂ϕ/∂τ )2+ (∇ϕ)2. (29.6) Note that each term in SEis always positive (or zero) for any ﬁeld conﬁgu- rationϕ(x). This is the advantage of working in euclidean space: eq.(2 9.4), theeuclidean path integral , is strongly damped (rather than rapidly oscil- lating) at large values of the ﬁeld and/or its derivatives, a nd this makes its convergence properties more obvious. 29: Eﬀective Field Theory 186 Next, we Fourier transform to (euclidean) momentum space vi a ϕ(x) =/integraldisplayd4k (2π)4eikx/tildewideϕ(k). (29.7) The euclidean action becomes SE=1 2/integraldisplayd4k (2π)4/tildewideϕ(−k)/parenleftig Zϕk2+Zmm2 ph/parenrightig /tildewideϕ(k) +1 24Zλλph/integraldisplayd4k1 (2π)4...d4k4 (2π)4(2π)4δ4(k1+k2+k3+k4) ×/tildewideϕ(k1)/tildewideϕ(k2)/tildewideϕ(k3)/tildewideϕ(k4). (29.8) Note thatk2=k2+k2 τ≥0. We now introduce an ultraviolet cutoﬀ Λ. It should be much larger than the particle mass mph, or any other energy scale of practical interest. Then we perform the path integral over all /tildewideϕ(k) with |k|>Λ. We also take /tildewideJ(k) = 0 for |k|>Λ. Then we ﬁnd Z(J) =/integraldisplay Dϕ|k|<Λe−Seﬀ(ϕ;Λ)−/integraltext Jϕ, (29.9) where e−Seﬀ(ϕ;Λ)=/integraldisplay Dϕ|k|>Λe−SE(ϕ). (29.10) Seﬀ(ϕ;Λ) is called the Wilsonian eﬀective action . We can write the corre- sponding lagrangian density as Leﬀ(ϕ;Λ) =1 2Z(Λ)∂µϕ∂µϕ+1 2m2(Λ)ϕ2+1 24λ(Λ)ϕ4 +/summationdisplay d≥6/summationdisplay icd,i(Λ)Od,i, (29.11) where the Fourier components of ϕ(x) are now cut oﬀ at |k|>Λ: ϕ(x) =/integraldisplayΛ 0d4k (2π)4eikx/tildewideϕ(k). (29.12) The operators Od,iin eq.(29.11) consist of all terms that have mass dimen- siond≥6 and that are even under ϕ↔ −ϕ;iis an index that distinguishes operators of the same dimension that are inequivalent after integrations by parts of any derivatives that act on the ﬁelds. (The operator s must be even underϕ↔ −ϕin order to respect the ϕ↔ −ϕsymmetry of the original lagrangian.) The coeﬃcients Z(Λ),m2(Λ),λ(Λ), andcd,i(Λ) in eq.(29.11) are all ﬁnitefunctions of Λ. This is established by the following argumen t. We can 29: Eﬀective Field Theory 187 .. . Figure 29.1: A one-loop 1PI diagram with 2 nexternal lines. Each external line represents a ﬁeld with |k|<Λ. The internal (dashed) line represents a ﬁeld with |k|>Λ. diﬀerentiate eq.(29.9) with respect to J(x) to compute correlation functions of the renormalized ﬁeld ϕ(x), and correlation functions of renormalized ﬁelds are ﬁnite. Using eq.(29.9), we can compute these corre lation functions as a series of Feynman diagrams, with Feynman rules based on Leﬀ. These rules include an ultraviolet cutoﬀ Λ on the loop momenta, sin ce the ﬁelds with higher momenta have already been integrated out . Thus all of the loop integrals in these diagrams are ﬁnite. Therefore the ot her parameters that enter the diagrams— Z(Λ),m2(Λ),λ(Λ), andcd,i(Λ)—must be ﬁnite as well, in order to end up with ﬁnite correlation functions. To compute these parameters, we can think of eq.(29.8) as the action for two kinds of ﬁelds, those with |k|<Λ and those with |k|>Λ. Then we draw all 1PI diagrams with external lines for |k|<Λ ﬁelds only. For λph≪1, the dominant contribution to cd,i(Λ) for an operator Od,iwith 2n ﬁelds andd−2nderivatives is then given by a one-loop diagram with 2 n external lines (representing |k|<Λ ﬁelds),nvertices, and a |k|>Λ ﬁeld circulating in the loop; see ﬁg.(29.1). The simplest case to consider is O2n,1≡ϕ2n. With 2nexternal lines, there are (2 n)! ways of assigning the external momenta to the lines, but 2n×n×2 of these give the same diagram: 2nfor exchanging the two external lines that meet at any one vertex; nfor rotations of the diagram; and 2 for reﬂection of the diagram. Since there are no derivatives on the external ﬁelds, we can set all of the external momenta to zero; then all (2n)!/(2n2n) diagrams have the same value. With a euclidean action, each i nternal line contributes a factor of 1 /(k2+m2 ph), and each vertex contributes a factor of−Zλλph=−λph+O(λ2 ph). The vertex factor associated with the term 29: Eﬀective Field Theory 188 c2n,1(Λ)ϕ2ninLeﬀis−(2n)!c2n,1(Λ). Thus we have −(2n)!c2n,1(Λ) =(−λph)n(2n)! 2n2n/integraldisplay∞ Λd4k (2π)4/parenleftigg 1 k2+m2 ph/parenrightiggn +O(λn+1 ph). (29.13) For 2n≥6, the integral converges, and we ﬁnd c2n,1(Λ) = −(−λph/2)n 32π2n(n−2)1 Λ2n−4+O(λn+1 ph). (29.14) We have taken Λ ≫mph, and dropped terms down by powers of mph/Λ. For 2n= 4, we have to include the tree-level vertex; in this case, we have −λ(Λ) = −Zλλph+3 2(−λph)2/integraldisplay∞ Λd4k (2π)4/parenleftigg 1 k2+m2 ph/parenrightigg2 +O(λ3 ph). (29.15) This integral diverges. To evaluate it, we note that the one- loop contribu- tion to the exact four-point vertex is given by the same diagram, but with ﬁelds of allmomenta circulating in the loop. Thus we have −V4(0,0,0,0) =−Zλλph+3 2(−λph)2/integraldisplay∞ 0d4k (2π)4/parenleftigg 1 k2+m2 ph/parenrightigg2 +O(λ3 ph). (29.16) Then, using V4(0,0,0,0) =λphand subtracting eq.(29.15) from eq.(29.16), we get −λph+λ(Λ) =3 2(−λph)2/integraldisplayΛ 0d4k (2π)4/parenleftigg 1 k2+m2 ph/parenrightigg2 +O(λ3 ph).(29.17) Evaluating the (now ﬁnite!) integral and rearranging, we ha ve λ(Λ) =λph+3 16π2λ2 ph/bracketleftig ln(Λ/mph)−1 2/bracketrightig +O(λ3 ph). (29.18) Note that this result has the problem of a large log ; the second term is smaller than the ﬁrst only if λphln(Λ/mph)≪1. To cure this problem, we must change the renormalization scheme. We will take up this issue shortly, but ﬁrst let us examine the case of two external lines while co ntinuing to use the on-shell scheme. 29: Eﬀective Field Theory 189 For the case of two external lines, the one-loop diagram has j ust one vertex, and by momentum conservation, the loop integral is c ompletely indepedent of the external momentum. This implies that the o ne-loop contribution to Z(Λ) vanishes, and so we have Z(Λ) = 1 +O(λ2 ph). (29.19) The one-loop diagram does, however, give a nonzero contribu tion tom2(Λ); after including the tree-level term, we ﬁnd −m2(Λ) = −Zmm2 ph+1 2(−λph)/integraldisplay∞ Λd4k (2π)41 k2+m2 ph+O(λ2 ph).(29.20) This integral diverges. To evaluate it, recall that the one- loop contribution to the exact particle mass-squared is given by the same diagram, but with ﬁelds of allmomenta circulating in the loop. Thus we have −m2 ph=−Zmm2 ph+1 2(−λph)/integraldisplay∞ 0d4k (2π)41 k2+m2 ph+O(λ2 ph).(29.21) Then, subtracting eq.(29.20) from eq.(29.21), we get −m2 ph+m2(Λ) =1 2(−λph)/integraldisplayΛ 0d4k (2π)41 k2+m2 ph+O(λ2 ph). (29.22) Evaluating the (now ﬁnite!) integral and rearranging, we ha ve m2(Λ) =m2 ph−λph 16π2/bracketleftig Λ2−m2 phln(Λ2/m2 ph)/bracketrightig +O(λ2 ph). (29.23) We see that we now have an even worse situation than we did with the large log in λ(Λ): the correction term is quadratically divergent . As already noted, to ﬁx these problems we must change the reno rmal- ization scheme. In the context of an eﬀective action with a sp eciﬁc value of the cutoﬀ Λ 0, there is a simple way to do so: we simply treat this eﬀective action as the fundamental starting point, with Z(Λ0),m2(Λ0),λ(Λ0), and cd,i(Λ0) as input parameters. We then see what physics emerges at ene rgy scales well below Λ 0. We can set Z(Λ0) = 1, with the understanding that the ﬁeld no longer has the LSZ normalization (and that we will have to correct the LSZ formula to account for this). We will also ass ume that the parameters λ(Λ0),m2(Λ0), andcd,i(Λ0) are all small when measured in units of the cutoﬀ: λ(Λ0)≪1, (29.24) |m2(Λ0)| ≪Λ2 0, (29.25) cd,i(Λ0)≪Λ−(d−4) 0. (29.26) 29: Eﬀective Field Theory 190 The proposal to treat the eﬀective action as the fundamental starting point may not seem very appealing. For one thing, we now have a n inﬁnite number of parameters to specify, rather than two! Also, we no w have an explicit cutoﬀ in place, rather than trying to have a theory t hat works at all energy scales. On the other hand, it may well be that quantum ﬁeld theory does not work at arbitrarily high energies. For example, quantum ﬂuc tuations in spacetime itself should become important above the Planck scale , which is given by the inverse square root of Newton’s constant, and ha s a numerical value of ∼1019GeV (compared to, say, the proton mass, which is ∼1GeV). So, let us leave the cutoﬀ Λ 0in place for now. We will then make a two- pronged analysis. First, we will see what happens at much low er energies. Then, we will see what happens if we try to take the limit Λ 0→ ∞. We begin by examining lower energies. To make things more tra ctable, we will set cd,i(Λ0) = 0 ; (29.27) later we will examine the eﬀects of a more general choice. A nice way to see what happens at lower energies is to integrat e out some more high-energy degrees of freedom. Let us, then, perform t he functional integral over Fourier modes /tildewideϕ(k) with Λ<|k|<Λ0; we have e−Seﬀ(ϕ;Λ)=/integraldisplay DϕΛ<|k|<Λ0e−Seﬀ(ϕ;Λ0). (29.28) We can do this calculation in perturbation theory, mimickin g the procedure that we used earlier. We ﬁnd m2(Λ) =m2(Λ0) +1 2λ(Λ0)/integraldisplayΛ0 Λd4k (2π)41 k2+m2(Λ0)+... , (29.29) λ(Λ) =λ(Λ0)−3 2λ2(Λ0)/integraldisplayΛ0 Λd4k (2π)4/parenleftbigg1 k2+m2(Λ0)/parenrightbigg2 +... , (29.30) c2n,1(Λ) = −(−1)n 2n2nλn(Λ0)/integraldisplayΛ0 Λd4k (2π)4/parenleftbigg1 k2+m2(Λ0)/parenrightbiggn +... , (29.31) where the ellipses stand for higher-order corrections. For Λ not too much less than Λ 0(and, in particular, for |m2(Λ0)| ≪Λ2), we ﬁnd m2(Λ) =m2(Λ0) +1 16π2λ(Λ0)/parenleftig Λ2 0−Λ2/parenrightig +... , (29.32) λ(Λ) =λ(Λ0)−3 16π2λ2(Λ0)lnΛ0 Λ+... , (29.33) c2n,1(Λ) = −(−1)n 32π22nn(n−2)λn(Λ0)/parenleftigg 1 Λ2n−4−1 Λ2n−4 0/parenrightigg +... . (29.34) 29: Eﬀective Field Theory 191 Figure 29.2: A one-loop contribution to the ϕ4vertex for ﬁelds with |k|<Λ. The internal (dashed) line represents a ﬁeld with |k|>Λ. We see from this that the corrections to m2(Λ), which is the only coeﬃcient with positive mass dimension, are dominated by contributio ns from the highend of the integral. On the other hand, the corrections to cd,i(Λ), coeﬃcients with negative mass dimension, are dominated by c ontributions from the lowend of the integral. And the corrections to λ(Λ), which is dimensionless, come equally from all portions of the range o f integration. For thecd,i(Λ), this means that their starting values at Λ 0were not very important, as long as eq.(29.26) is obeyed. Nonzero sta rting values would contribute another term of order 1 /Λ2n−4 0toc2n,1(Λ), but all such terms are less important than the one of order 1 /Λ2n−4that comes from doing the integral down to |k|= Λ. Similarly, nonzero values of cd,i(Λ0) would make subdominant contri- butions toλ(Λ). As an example, consider the contribution of the diagram in ﬁg.(29.2). Ignoring numerical factors, the vertex facto r isc6,1(Λ0), and the loop integral is the same as the one that enters into m2(Λ); it yields a factor of Λ2 0−Λ2∼Λ2 0. Thus the contribution of this diagram to λ(Λ) is of orderc6,1(Λ0)Λ2 0. This is a pure number that, according to eq.(29.26), is small. This contribution is missing the logarithmic enha ncement factor ln(Λ0/Λ) that we see in eq.(29.33). On the other hand, for m2(Λ), there are inﬁnitely many contributions of order Λ2 0whencd,i(Λ0)∝ne}ationslash= 0. These must add up to give m2(Λ) a value that is much smaller. Indeed, we want to continue the process down to lower and lower values of Λ, with m2(Λ) dropping until it becomes of order m2 ph at Λ∼mph. For this to happen, there must be very precise cancellation s among all the terms of order Λ2 0that contribute to m2(Λ). In some sense, it is more “natural” to have m2 ph∼λ(Λ0)Λ2 0, rather than to arrange for these very precise cancellations. This philosophical issue is ca lled the ﬁne-tuning problem , and it generically arises in theories with spin-zero ﬁelds . In theories with higher-spin ﬁelds only, the action typical ly has more 29: Eﬀective Field Theory 192 symmetry when these ﬁelds are massless, and this typically p revents diver- gences that are worse than logarithmic. These theories are s aid to be tech- nically natural , while theories with spin-zero ﬁelds (with physical masses well below the cutoﬀ) generally are not. (The only exception s are theories where supersymmetry relates spin-zero and spin-one-half ﬁelds; the spin- zero ﬁelds then inherent the technical naturalness of their spin-one-half partners.) For now, in ϕ4theory, we will simply accept the necessity of ﬁne-tuning in order to have mph≪Λ. Returning to eqs.(29.32–29.34), we can recast them as diﬀer ential equa- tions that tell us how these parameters change with the vaule of the cutoﬀ Λ. In particular, let us do this for λ(Λ). We take the derivative of eq.(29.33) with respect to Λ, multiply by Λ, and then set Λ 0= Λ to get d dln Λλ(Λ) =3 16π2λ2(Λ) +... . (29.35) Notice that the right-hand side of eq.(29.33) is apparently the same as the beta function β(λ)≡dλ/dlnµthat we calculated in problem 28.1, where it represented the rate of change in the MS parameter λthat was need to compensate for a change in the MS renormalization scale µ. Eq.(29.33) gives us a new physical interpretation of the beta function: it is the rate of change in the coeﬃcient of the ϕ4term in the eﬀective action as we vary the ultraviolet cutoﬀ in that action. Actually, though, there is a technical detail: it is really Z(Λ)−2λ(Λ) that is most closely analogous to the MS parameter λ. This is because, if we rescaleϕso that it has a canonical kinetic term of1 2∂µϕ∂µϕ, then the coeﬃcient of the ϕ4term isZ(Λ)−2λ(Λ). SinceZ(Λ) = 1 +O(λ2(Λ)), this has no eﬀect at the one-loop level, but it does matter at two lo ops. We can account for the eﬀect of this wave function renormalization (in all the couplings) by writing, instead of eq.(29.11), Leﬀ(ϕ;Λ) =1 2Z(Λ)∂µϕ∂µϕ+1 2Z(Λ)m2(Λ)ϕ2+1 24Z2(Λ)λ(Λ)ϕ4 +/summationdisplay d≥6/summationdisplay iZnd,i/2(Λ)cd,i(Λ)Od,i, (29.36) wherend,iis the number of ﬁelds in the operator Od,i. Now the beta function for λis universal up through two loops; see problem 29.1. At three and higher loops, diﬀerences with the MS beta function can arise, due to the diﬀerent underlying deﬁnitions of the coupling λin the cutoﬀ scheme and the MS scheme. We now have the overall picture of Wilson’s approach to quant um ﬁeld theory. First, deﬁne a quantum ﬁeld theory via an action with an explicit 29: Eﬀective Field Theory 193 momentum cutoﬀ in place.1Then, lower the cutoﬀ by integrating out higher-momentum degrees of freedom. As a result, the coeﬃci ents in the eﬀective action will change. If the ﬁeld theory is weakly cou pled—which in practice means eqs.(29.24–29.26) are obeyed—then the coeﬃ cients of the operators with negative mass dimension will start to take on the values we would have computed for them in perturbation theory, rega rdless of their precise initial values. If we continuously rescale th e ﬁelds to have canonical kinetic terms, then the dimensionless coupling c onstant(s) will change according to their beta functions. The ﬁnal results, at an energy scaleEwell below the initial cutoﬀ Λ 0, are the same as we would predict via renormalized perturbation theory, up to small correcti ons by powers of E/Λ0. The advantage of the Wilson scheme is that it gives a nonpertu rbative deﬁnition of the theory which is applicable even if the theor y isnotweakly coupled. With a spacetime lattice providing the cutoﬀ, othe r techniques (typically requiring large-scale computer calculations) can be brought to bear on strongly-coupled theories. The Wilson scheme also allows us to give physical meaning to n onrenor- malizable theories. Given an action for a nonrenormalizabl e theory, we can regard it as an eﬀective action. We should then impose a momen tum cutoﬀ Λ0, where Λ 0can be deﬁned by saying that the coeﬃcient of every operator Oiwith mass dimension Di>4 is given by ci/ΛDi−4 0withci≤1. Then we can use this theory for physics at energies below Λ 0. At energies Efar below Λ 0, the eﬀective theory will look like a renormalizable one, up to corrections by powers of E/Λ0. (This renormalizable theory might simply be a free-ﬁeld theory with nointeractions, or no theory at all if there are no particles with physical masses well below Λ 0.) We now turn to the ﬁnal issue: can we remove the cutoﬀ complete ly? Returning to the example of ϕ4theory, let us suppose that we are somehow able to compute the exact beta function. Then we can i ntegrate the renormalization-group equation dλ/dln Λ =β(λ) from Λ = mphto Λ = Λ 0to get/integraldisplayλ(Λ0) λ(mph)dλ β(λ)= lnΛ0 mph. (29.37) We would like to take the limit Λ 0→ ∞. Obviously, the right-hand side of eq.(29.37) becomes inﬁnite in this limit, and so the left-ha nd side must as well. However, it may not. Recall that, for small λ,β(λ) is positive, and it 1This can be done in various ways: for example, we could replac e continuous spacetime with a discrete lattice of points with lattice spacing a; then there is an eﬀective largest momentum of order 1 /a. 29: Eﬀective Field Theory 194 increases faster than λ. If this is true for all λ, then the left-hand side of eq.(29.37) will approach a ﬁxed, ﬁnite value as we take the up per limit of integration to inﬁnity. This yields a maximum possible valu e for the initial cutoﬀ, given by lnΛmax mph≡/integraldisplay∞ λ(mph)dλ β(λ). (29.38) If we approximate the exact beta function with its leading te rm, 3λ2/16π2, and use the leading term in eq.(29.18) to get λ(mph) =λph, then we ﬁnd Λmax≃mphe16π2/3λph. (29.39) The existence of a maximum possible value for the cutoﬀ means that we cannot take the limit as the cutoﬀ goes to inﬁnity; we mustuse an eﬀective action with a cutoﬀ as our starting point. If we insist on taki ng the cutoﬀ to inﬁnity, then the only possible value of λphisλph= 0. Thus, ϕ4theory istrivial in the limit of inﬁnite cutoﬀ: there are no interactions. (Th ere is much evidence for this, but as yet no rigorous proof. The same is true of quantum electrodynamics, as was ﬁrst conjectured by Landau ; in this case, Λmaxis known as the location of the Landau pole .) However, the cutoﬀ canbe removed if the beta function grows no faster thanλat largeλ; then the left-hand side of eq.(29.37) would diverge as we take the upper limit of integration to inﬁnity. Or, β(λ) could drop to zero (and then become negative) at some ﬁnite value λ∗. Then, ifλph<λ∗, the left-hand side of eq.(29.37) would diverge as the upper limi t of integration approaches λ∗. In this case, the eﬀective coupling at higher and higher energies would remain ﬁxed at λ∗, andλ=λ∗is called an ultravioldet ﬁxed point of the renormalization group. If the beta function is negative for λ=λ(mph), the theory is said to beasymptotically free , andλ(Λ)decreases as the cutoﬀ is increased. In this case, there is no barrier to taking the limit Λ → ∞. In four space- time dimensions, the only asymptotically free theories are nonabelian gauge theories; see section 69. Reference Notes Eﬀective ﬁeld theory is discussed in Georgi ,Peskin & Schroeder , andWein- berg I. An introduction to lattice theory can be found in Smit. Problems 29.1) Consider a theory with a single dimensionless couplin ggwhose beta function takes the form β(g) =b1g2+b2g3+.... Now consider a new deﬁnition of the coupling of the form ˜ g=g+c2g2+.... 29: Eﬀective Field Theory 195 a) Show that β(˜g) =b1˜g2+b2˜g3+.... b) Generalize this result to the case of multiple dimensionl ess cou- plings. 29.2) Consider ϕ3theory in six euclidean spacetime dimensions, with a cutoﬀ Λ 0and lagrangian L=1 2Z(Λ0)∂µϕ∂µϕ+1 6Z3/2(Λ0)g(Λ0)ϕ3. (29.40) We assume that we have ﬁne-tuned to keep m2(Λ)≪Λ2, and so we neglect the mass term. a) Show that Z(Λ) =Z(Λ0)/parenleftigg 1−1 2g2(Λ0)d dk2/bracketleftigg/integraldisplayΛ0 Λd6ℓ (2π)61 (k+ℓ)2ℓ2/bracketrightigg k2=0+.../parenrightigg , g(Λ) =Z3/2(Λ0) Z3/2(Λ)g(Λ0)/parenleftigg 1 +g2(Λ0)/integraldisplayΛ0 Λd6ℓ (2π)61 (ℓ2)3+.../parenrightigg . Hint: note that the tree-level propagator is ˜∆(k) = [Z(Λ0)k2]−1. b) Use your results to compute the beta function β(g(Λ))≡d dln Λg(Λ), (29.41) and compare with the result in section 27. 30: Spontaneous Symmetry Breaking 196 30Spontaneous Symmetry Breaking Prerequisite: 21 Considerϕ4theory, where ϕis a real scalar ﬁeld with lagrangian L=−1 2∂µϕ∂µϕ−1 2m2ϕ2−1 24λϕ4. (30.1) As we discussed in section 23, this theory has a Z 2symmetry: Lis invari- ant underϕ(x)→ −ϕ(x), and we can deﬁne a unitary operator Zthat implements this: Z−1ϕ(x)Z=−ϕ(x). (30.2) We also have Z2= 1, and so Z−1=Z. Since unitarity implies Z−1=Z†, this makes Zhermitian as well as unitary. Now suppose that the parameter m2is, in spite of its name, negative rather than positive. We can write Lin the form L=−1 2∂µϕ∂µϕ−V(ϕ), (30.3) where the potential is V(ϕ) =1 2m2ϕ2+1 24λϕ4 =1 24λ(ϕ2−v2)2−1 24λv4. (30.4) In the second line, we have deﬁned v≡+(6|m2|/λ)1/2. (30.5) We can (and will) drop the last, constant, term in eq.(30.4). From eq.(30.4) it is clear that there are two classical ﬁeld c onﬁgurations that minimize the energy: ϕ(x) = +vandϕ(x) =−v. This is in contrast to the usual case of positive m2, for which the minimum-energy classical ﬁeld conﬁguration is ϕ(x) = 0. We can expect that the quantum theory will follow suit. For m2<0, there will be two ground states, |0+∝an}b∇acket∇i}htand|0−∝an}b∇acket∇i}ht, with the property that ∝an}b∇acketle{t0+|ϕ(x)|0+∝an}b∇acket∇i}ht= +v, ∝an}b∇acketle{t0−|ϕ(x)|0−∝an}b∇acket∇i}ht=−v, (30.6) up to quantum corrections from loop diagrams that we will tre at in detail in section 30. These two ground states are exchanged by the op eratorZ, Z|0+∝an}b∇acket∇i}ht=|0−∝an}b∇acket∇i}ht, (30.7) and they are orthogonal: ∝an}b∇acketle{t0+|0−∝an}b∇acket∇i}ht= 0. 30: Spontaneous Symmetry Breaking 197 This last claim requires some comment. Consider a similar pr oblem in quantum mechanics, H=1 2p2+1 24λ(x2−v2)2. (30.8) There are two approximate ground states in this case, speciﬁ ed by the approximate wave functions ψ±(x) =∝an}b∇acketle{tx|0±∝an}b∇acket∇i}ht ∼ exp[−ω(x∓v)2/2], (30.9) whereω= (λv2/3)1/2is the frequency of small oscillations about the mini- mum. However, the true ground state is a symmetric linear com bination of these. The antisymmetric linear combination has a slightly higher energy, due to the eﬀects of quantum tunneling. We can regard a ﬁeld theory as an inﬁnite set of oscillators, o ne for each point in space, each with a hamiltonian like eq.(30.8), and c oupled together by the ( ∇ϕ)2term in the ﬁeld-theory hamiltonian. There is a tunneling amplitude for each oscillator, but to turn the ﬁeld-theoret ic state |0+∝an}b∇acket∇i}htinto |0−∝an}b∇acket∇i}ht,allthe oscillators have to tunnel, and so the tunneling amplitu de gets raised to the power of the number of oscillators, that is, to t he power of inﬁnity (more precisely, to a power that scales like the volu me of space). Therefore, in the limit of inﬁnite volume, ∝an}b∇acketle{t0+|0−∝an}b∇acket∇i}htvanishes. Thus we can pick either |0+∝an}b∇acket∇i}htor|0−∝an}b∇acket∇i}htto use as the ground state. Let us choose |0+∝an}b∇acket∇i}ht. Then we can deﬁne a shifted ﬁeld, ρ(x) =ϕ(x)−v, (30.10) which obeys ∝an}b∇acketle{t0+|ρ(x)|0+∝an}b∇acket∇i}ht= 0. (We must still worry about loop corrections, which we will do at the end of this section.) The potential bec omes V(ϕ) =1 24λ[(ρ+v)2−v2]2 =1 6λv2ρ2+1 6λvρ3+1 24λρ4, (30.11) and so the lagrangian is now L=−1 2∂µρ∂µρ−1 6λv2ρ2−1 6λvρ3−1 24λρ4. (30.12) We see that the coeﬃcient of the ρ2term is1 6λv2=|m2|. This coeﬃcient should be identiﬁed as1 2m2 ρ, wheremρis the mass of the corresponding ρ particle. Also, we see that the shifted ﬁeld now has a cubic as well as a quartic interaction. Eq.(30.12) speciﬁes a perfectly sensible, renormalizable quantum ﬁeld theory, but it no longer has an obvious Z 2symmetry. We say that the Z 2 symmetry is spontaneously broken . 30: Spontaneous Symmetry Breaking 198 This leads to a question about renormalization. If we includ e renormal- izingZfactors in the original lagrangian, we get L=−1 2Zϕ∂µϕ∂µϕ−1 2Zmm2ϕ2−1 24Zλλϕ4. (30.13) For positive m2, these three Zfactors are suﬃcient to absorb inﬁnities for d≤4, where the mass dimension of λis positive or zero. On the other hand, looking at the lagrangian for negative m2after the shift, eq.(30.12), we would seem to need an extra Zfactor for the ρ3term. Also, once we have aρ3term, we would expect to need to add a ρterm to cancel tadpoles. So, the question is, are the original three Zfactors suﬃcient to absorb all the divergences in the Feynman diagrams derived from eq.(30 .13)? The answer is yes. To see why, consider the quantum action (in troduced in section 21) Γ(ϕ) =1 2/integraldisplayd4k (2π)4/tildewideϕ(−k)/parenleftig k2+m2−Π(k2)/parenrightig /tildewideϕ(k) +∞/summationdisplay n=31 n!/integraldisplayd4k1 (2π)4...d4kn (2π)4(2π)4δ4(k1+...+kn) ×Vn(k1,...,kn)/tildewideϕ(k1).../tildewideϕ(kn), (30.14) computed with m2>0. The ingredients of Γ( ϕ)—the self-energy Π( k2) and the exact 1PI vertices Vn—are all made ﬁnite and well-deﬁned (in, say, the MS renormalization scheme) by adjusting the three Zfactors in eq.(30.13). Furthermore, for m2>0, the quantum action inherits the Z 2 symmetry of the classical action. To see this directly, we no te that Vn must zero for odd n, simply because there is no way to draw a 1PI diagram with an odd number of external lines using only a four-point v ertex. Thus Γ(ϕ) also has the Z 2symmetry. This is a simple example of a more general result that we proved in problem 21.2: the quantum action inh erits any linear symmetry of the classical action, provided that it is also a symmetry of the integration measure Dϕ. (Linear means that the transformed ﬁelds are linear functions of the original ones.) The integration measure is almost always invariant; when it is not, the symmetry is said to be anomalous . We will meet an anomalous symmetry in section 75. Once we have computed the quantum action for m2>0, we can go ahead and consider the case of m2<0. Recall from section 21 that the quantum equation of motion in the presence of a source is δΓ/δϕ(x) = −J(x), and that the solution of this equation is also the vacuum ex pectation value ofϕ(x). Now set J(x) = 0, and look for a translationally invariant (that is, constant) solution ϕ(x) =v. If there is more than one such solution, we want the one(s) with the lowest energy. This is e quivalent to 30: Spontaneous Symmetry Breaking 199 minimizing the quantum potential U(ϕ), where Γ(ϕ) =/integraldisplay d4x/bracketleftig − U(ϕ)−1 2Z(ϕ)∂µϕ∂µϕ+.../bracketrightig , (30.15) and the ellipses stand for terms with more derivatives. In a w eakly coupled theory, we can expect the loop-corrected potential U(ϕ) to be qualitatively similar to the classical potential V(ϕ). Therefore, for m2<0, we expect that there are two minima of U(ϕ) with equal energy, located at ϕ(x) =±v, wherev=∝an}b∇acketle{t0|ϕ(x)|0∝an}b∇acket∇i}htis the exact vacuum expectation value of the ﬁeld. Thus we have a description of spontaneous symmetry breaking in the quantum theory based on the quantum action, and the quantum a ction is made ﬁnite by adjusting only the three Zfactors that appear in the original, symmetric form of the lagrangian. In the next section, we will see how this works in explicit cal culations. 31: Broken Symmetry and Loop Corrections 200 31Broken Symmetry and Loop Corrections Prerequisite: 30 Considerϕ4theory, where ϕis a real scalar ﬁeld with lagrangian L=−1 2Zϕ∂µϕ∂µϕ−1 2Zmm2ϕ2−1 24Zλλϕ4. (31.1) Ind= 4 spacetime dimensions, the coupling λis dimensionless. We begin by considering the case m2>0, where the Z 2symmetry of L underϕ→ −ϕis manifest. We wish to compute the three renormalizing Zfactors. We work in d= 4−εdimensions, and take λ→λ˜µε(where ˜µ has dimensions of mass) so that λremains dimensionless. The propagator correction Π( k2) is given by the diagrams of ﬁg.(31.1), which yield iΠ(k2) =1 2(−iλ˜µε)1 i˜∆(0)−i(Ak2+Bm2), (31.2) whereA=Zϕ−1 andB=Zm−1, and ˜∆(0) =/integraldisplayddℓ (2π)d1 ℓ2+m2. (31.3) Using the usual bag of tricks from section 14, we ﬁnd ˜µε˜∆(0) =−i (4π)2/bracketleftbigg2 ε+ 1 + ln(µ2/m2)/bracketrightbigg m2, (31.4) whereµ2= 4πe−γ˜µ2. Thus Π(k2) =λ 2(4π)2/bracketleftbigg2 ε+ 1 + ln(µ2/m2)/bracketrightbigg m2−Ak2−Bm2. (31.5) From eq.(31.5) we see that we must have A=O(λ2), (31.6) B=λ 16π2/parenleftbigg1 ε+κB/parenrightbigg +O(λ2), (31.7) whereκBis a ﬁnite constant (that may depend on µ). In the MS renor- malization scheme, we take κB= 0, but we will leave κBarbitrary for now. Next we turn to the vertex correction, given by the diagram of ﬁg.(31.2), plus two others with k2↔k3andk2↔k4; all momenta are treated as incoming. We have iV4(k1,k2,k3,k4) =−iZλλ+1 2(−iλ)2/parenleftig 1 i/parenrightig2/bracketleftig iF(−s) +iF(−t) +iF(−u)/bracketrightig +O(λ3). (31.8) 31: Broken Symmetry and Loop Corrections 201 k kl k k Figure 31.1: O(λ) corrections to Π( k2). k k4l 1l + k + k2 1k k23 Figure 31.2: The O(λ2) correction to V4(k1,k2,k3,k4). Two other dia- grams, obtained from this one via k2↔k3andk2↔k4, also contribute. Here we have deﬁned s=−(k1+k2)2,t=−(k1+k3)2,u=−(k1+k4)2, and iF(k2)≡˜µε/integraldisplayddℓ (2π)d1 ((ℓ+k)2+m2)(ℓ2+m2) =i 16π2/bracketleftbigg2 ε+/integraldisplay1 0dxln(µ2/D)/bracketrightbigg , (31.9) whereD=x(1−x)k2+m2. SettingZλ= 1 +Cin eq.(31.8), we see that we need C=3λ 16π2/parenleftbigg1 ε+κC/parenrightbigg +O(λ2), (31.10) whereκCis a ﬁnite constant. We may as well pause to compute the beta function, β(λ) =dλ/dlnµ, where the derivative is taken with the bare coupling λ0held ﬁxed, and the ﬁnite parts of the counterterms set to zero, in accord wit h the MS prescription. We have λ0=ZλZ−2 ϕλ˜µε, (31.11) with ln/parenleftig ZλZ−2 ϕ/parenrightig =3λ 16π21 ε+O(λ2). (31.12) LetL1(λ) be the coeﬃcient of 1 /εin eq.(31.12). Our analysis in section 28 shows that the beta function is then given by β(λ) =λ2L′ 1(λ). Thus we ﬁnd β(λ) =3λ2 16π2+O(λ3). (31.13) 31: Broken Symmetry and Loop Corrections 202 k kl Figure 31.3: The O(λ) correction to the vacuum expectation value of the ρﬁeld. The beta function is positive, which means that the theory be comes more and more strongly coupled at higher and higher energies. Now we consider the more interesting case of m2<0, which results in the spontaneous breakdown of the Z 2symmetry. Following the procedure of section 30, we set ϕ(x) =ρ(x) +v, where v= (6|m2|/λ)1/2minimizes the potential (without Zfactors). Then the lagrangian becomes (with Zfactors) L=−1 2Zϕ∂µρ∂µρ−1 2(3 4Zλ−1 4Zm)m2 ρρ2 +1 2(Zm−Zλ)(3/λ˜µε)1/2m3 ρρ −1 6Zλ(3λ˜µε)1/2mρρ3−1 24Zλλ˜µερ4, (31.14) wherem2 ρ= 2|m2|. Now we can compute various one-loop corrections. We begin with the vacuum expectation value of ρ. TheO(λ) correction is given by the diagrams of ﬁg.(31.3). The three-point verte x factor is −iZλg3, whereg3can be read oﬀ of eq.(31.14): g3= (3λ˜µε)1/2mρ. (31.15) The one-point vertex factor is iY, whereYcan also be read oﬀ of eq.(31.14): Y=1 2(Zm−Zλ)(3/λ˜µε)1/2m3 ρ. (31.16) Following the discussion of section 9, we then ﬁnd that ∝an}b∇acketle{t0|ρ(x)|0∝an}b∇acket∇i}ht=/parenleftig iY+1 2(−iZλg3)1 i˜∆(0)/parenrightig/integraldisplay d4y1 i∆(x−y), (31.17) plus higher-order corrections. Using eqs.(31.15) and (31. 16), and eq.(31.4) withm2→m2 ρ, the factor in large parentheses in eq.(31.17) becomes i 2(3/λ)1/2m3 ρ/parenleftbigg Zm−Zλ+λ 16π2/bracketleftbigg2 ε+ 1 + ln(µ2/m2 ρ)/bracketrightbigg +O(λ2)/parenrightbigg .(31.18) UsingZm= 1 +BandZλ= 1 +C, withBandCfrom eqs.(31.7) and (31.10), the factor in large parentheses in eq.(31.18) beco mes λ 16π2/bracketleftig κB−κC+ 1 + ln(µ2/m2 ρ)/bracketrightig . (31.19) 31: Broken Symmetry and Loop Corrections 203 All the 1/ε’s have canceled. The remaining ﬁnite vacuum expectation va lue forρ(x) can now be removed by choosing κB−κC=−1−ln(µ2/m2 ρ). (31.20) This will also cancel all diagrams with one-loop tadpoles. Next we consider the ρpropagator. The diagrams contributing to the O(λ) correction are shown in ﬁg.(31.4). The counterterm insert ion is−iX, where, again reading oﬀ of eq.(31.14), X=Ak2+ (3 4C−1 4B)m2 ρ. (31.21) Putting together the results of eq.(31.2) for the ﬁrst diagr am (withm2→ m2 ρ), eq.(31.9) for the second (ditto), and eq.(31.21) for the t hird, we get Π(k2) =−1 2(λ˜µε)1 i˜∆(0) +1 2g2 3F(k2)−X+O(λ2) =λ 32π2m2 ρ/bracketleftbigg2 ε+ 1 + ln(µ2/m2 ρ)/bracketrightbigg +λ 32π2m2 ρ/bracketleftbigg2 ε+/integraldisplay1 0dxln(µ2/D)/bracketrightbigg −Ak2−(3 4C−1 4B)m2 ρ+O(λ2). (31.22) Again using eqs.(31.7) and (31.10) for BandC, we see that all the 1 /ε’s cancel, and we’re left with Π(k2) =λ 32π2m2 ρ/bracketleftbigg 1 + ln(µ2/m2 ρ) +/integraldisplay1 0dxln(µ2/D) +1 2(9κC−κB)/bracketrightbigg +O(λ2). (31.23) We can now choose to work in an OS scheme, where we require Π( −m2 ρ) = 0 and Π′(−m2 ρ) = 0. We see that, to this order in λ, Π(k2) is independent of k2. Thus, we automatically have Π′(−m2 ρ) = 0, and we can choose 9 κC−κB to ﬁx Π( −m2 ρ) = 0. Together with eq.(31.20), this completely determines κBandκCto this order in λ. Next we consider the one-loop correction to the three-point vertex, given by the diagrams of ﬁg.(31.5). We wish to show that the inﬁniti es are canceled by the value of Zλ= 1+Cthat we have already determined. The ﬁrst diagram in ﬁg.(31.5) is ﬁnite, and so for our purposes we can ignore it. The remaining three, plus the original vertex, sum up to g ive iV3(k1,k2,k3)div=−iZλg3+1 2(−iλ)(−ig3)/parenleftig 1 i/parenrightig2 ×/bracketleftig iF(k2 1) +iF(k2 2) +iF(k2 3)/bracketrightig +O(λ5/2), (31.24) 31: Broken Symmetry and Loop Corrections 204 k k ll + k k k kkl Figure 31.4: O(λ) corrections to the ρpropagator. k3 k2 k1k2 k1 k3 k2 k3k1 k1 k2 k3 Figure 31.5: O(λ) corrections to the vertex for three ρﬁelds. where the subscript div means that we are keeping only the div ergent part. Using eq.(31.9), we have V3(k1,k2,k3)div=−g3/parenleftbigg 1 +C−3λ 16π21 ε+O(λ2)/parenrightbigg . (31.25) From eq.(31.10), we see that the divergent terms do indeed ca ncel to this order inλ. Finally, we have the correction to the four-point vertex. In this case, the divergent diagrams are just those of ﬁg.(31.1), and so th e calculation of the divergent part of V4is exactly the same as it is when m2>0 (but withmρin place of m). Since we have already done that calculation (it was how we determined Cin the ﬁrst place), we need not repeat it. We have thus seen how we can compute the divergent parts of the coun- terterms in the simpler case of m2>0, where the Z 2symmetry is unbroken, and that these counterterms will also serve to cancel the div ergences in the more complicated case of m2<0, where the Z 2symmetry is spontaneously broken. This a general rule for renormalizable theories wit h spontaneous symmetry breaking, regardless of the nature of the symmetry group. Reference Notes Another example of renormalization of a spontaneously brok en theory is worked out in Peskin & Schroeder . 32: Spontaneous Breaking of Continuous Symmetries 205 32Spontaneous Breaking of Continuous Symmetries Prerequisite: 22, 30 Consider the theory (introduced in section 22) of a complex s calar ﬁeldϕ with L=−∂µϕ†∂µϕ−m2ϕ†ϕ−1 4λ(ϕ†ϕ)2. (32.1) This lagrangian is obviously invariant under the U(1) trans formation ϕ(x)→e−iαϕ(x), (32.2) whereαis a real number. Now suppose that m2is negative. The minimum of the potential of eq.(32.1) is achieved for ϕ(x) =1√ 2ve−iθ, (32.3) where v= (4|m2|/λ)1/2, (32.4) and the phase θis arbitrary; the factor of root-two in eq.(32.3) is con- ventional. Thus we have a continuous family of minima of the p otential, parameterized by θ. Under the U(1) transformation of eq.(32.2), θchanges toθ+α; thus the diﬀerent minimum-energy ﬁeld conﬁgurations are a ll related to each other by the symmetry. In the quantum theory, we therefore expect to ﬁnd a continuou s family of ground states, labeled by θ, with the property that ∝an}b∇acketle{tθ|ϕ(x)|θ∝an}b∇acket∇i}ht=1√ 2ve−iθ. (32.5) Also, according to the discussion in section 30, we expect ∝an}b∇acketle{tθ′|θ∝an}b∇acket∇i}ht= 0 for θ′∝ne}ationslash=θ. Returning to classical language, there is a ﬂat direction in ﬁeld space that we can move along without changing the energy. The physi cal con- sequence of this is the existence of a massless particle call ed aGoldstone boson. Let us see how this works in more detail. We ﬁrst choose the pha se θ= 0, and then write ϕ(x) =1√ 2[v+a(x) +ib(x)], (32.6) whereaandbare real scalar ﬁelds. Substituting eq.(32.6) into eq.(32. 1), we ﬁnd L=−1 2∂µa∂µa−1 2∂µb∂µb − |m2|a2−1 2λ1/2|m|a(a2+b2)−1 16λ(a2+b2)2.(32.7) 32: Spontaneous Breaking of Continuous Symmetries 206 We see from this that the aﬁeld has a mass given by1 2m2 a=|m2|. Theb ﬁeld, on the other hand, is massless, and we identify it as the Goldstone boson. A diﬀerent parameterization brings out the role of the massl ess ﬁeld more clearly. We write ϕ(x) =1√ 2(v+ρ(x))e−iχ(x)/v, (32.8) whereρandχare real scalar ﬁelds. Substituting eq.(32.8) into eq.(32. 1), we get L=−1 2∂µρ∂µρ−1 2/parenleftig 1 +ρ v/parenrightig2∂µχ∂µχ − |m2|ρ2−1 2λ1/2|m|ρ3−1 16λρ4. (32.9) We see from this that the ρﬁeld has a mass given by1 2m2 ρ=|m2|, and that theχﬁeld is massless. These are the same particle masses we found using the parameterization of eq.(32.6). This is not an accident: the particle masses and scattering amplitudes are independent of ﬁeld re deﬁnitions. Note that the χﬁeld does not appear in the potential at all. Thus it parameterizes the ﬂat direction. In terms of the ρandχﬁelds, the U(1) transformation takes the simple form χ(x)→χ(x) +α. Does the masslessness of the χﬁeld survive loop corrections? It does. To see this, we note that if the χﬁeld remains massless, its exact propagator ˜∆χ(k2) should have a pole at k2= 0; equivalently, the self-energy Π χ(k2), related to the propagator by ˜∆χ(k2) = 1/[k2−Πχ(k2)], should satisfy Π(0) = 0. We can evaluate Π χ(0) by summing all 1PI diagrams with two external χlines, each with four-momentum k= 0. We note from eq.(32.9) that the derivatives acting on the χﬁelds imply that the vertex factors for the ρχχ andρρχχ vertices are each proportional to k1·k2, wherek1andk2are the momenta of the two χlines that meet at that vertex. Since the external lines have zero momentum, the attached vertices vanish; hen ce, Πχ(0) = 0, and theχparticle remains massless. The same conclusion can be reached by considering the quantu m action Γ(ϕ), which includes all loop corrections. According to our dis cussion in section 29, the quantum action has the same symmetries as the classical action. Therefore, in the case at hand, Γ(ϕ) = Γ(e−iαϕ). (32.10) Spontaneous symmetry breaking occurs if the minimum of Γ( ϕ) is at a constant, nonzero value of ϕ. Because of eq.(32.10), the phase of this 32: Spontaneous Breaking of Continuous Symmetries 207 constant is arbitrary. Therefore, there must be a ﬂat direct ion in ﬁeld space, corresponding to the phase of ϕ(x). The physical consequence of this ﬂat direction is a massless particle, the Goldstone bos on. All of this has a straightforward extension to the nonabelia n case. Con- sider L=−1 2∂µϕi∂µϕi−1 2m2ϕiϕi−1 16λ(ϕiϕi)2, (32.11) where a repeated index is summed. This lagrangian is invaria nt under the inﬁnitesimal SO( N) transformation δϕi=−iθa(Ta)ijϕj, (32.12) where runs from 1 to1 2N(N−1),θais a set of1 2N(N−1) real, inﬁnitesi- mal parameters, and each antisymmetric generator matrix Tahas a single nonzero entry −iabove the main diagonal, and a corresponding + ibelow the main diagonal. Now let us take m2<0 in eq.(32.11). The minimum of the potential is achieved for ϕi(x) =vi, wherev2=vivi= 4|m2|/λ, and the direction in which the N-component vector /vector vpoints is arbitrary. In the quantum theory, we interpret vias the vacuum expectation value (VEV for short) of the quantum ﬁeld ϕi(x). We can choose our coordinate system so that vi=vδiN; that is, the VEV lies entirely in the last component. Now consider making an inﬁnitesimal SO( N) transformation. This changes the VEV; we have vi→vi−iθa(Ta)ijvj =vδiN−iθa(Ta)iNv. (32.13) For some choices of θa, the second term on the right-hand side of eq.(32.13) vanishes. This happens if the corresponding Tahas no nonzero entry in the last column. There are N−1Ta’s with a nonzero entry in the last column: those with the −iin the ﬁrst row and last column, in the second row and last column, etc, down to the N−1throw and last column. These Ta’s are said to be broken generators. A generator is broken if ( Ta)ijvj∝ne}ationslash= 0, and unbroken if ( Ta)ijvj= 0. An inﬁnitesimal SO( N) transformation that involves a broken genera- tor changes the VEV of the ﬁeld, but not the energy. Thus, each broken generator corresponds to a ﬂat direction in ﬁeld space. Each ﬂat direction implies the existence of a corresponding massless particle . This is Gold- stone’s theorem : there is one massless Goldstone boson for each broken generator. The unbroken generators, on the other hand, do not change the VEV of the ﬁeld. Therefore, after rewriting the lagrangian in te rms of shifted 32: Spontaneous Breaking of Continuous Symmetries 208 ﬁelds (each with zero VEV), there should still be a manifest s ymmetry cor- responding to the set of unbroken generators. In the present case, the num- ber of unbroken generators is1 2N(N−1)−(N−1) =1 2(N−1)(N−2). This is the number of generators of SO( N−1). Therefore, we expect SO( N−1) to be an obvious symmetry of the lagrangian after it is writte n in terms of shifted ﬁelds. Let us see how this works in the present case. We can rewrite eq .(32.11) as L=−1 2∂µϕi∂µϕi−V(ϕ), (32.14) with V(ϕ) =1 16λ(ϕiϕi−v2)2, (32.15) wherev= (4|m2|/λ)1/2, and the repeated index iis implicitly summed from 1 toN. Now letϕN(x) =v+ρ(x), and plug this into eq.(32.14). With the repeated index inow implicitly summed from 1 to N−1, we have L=−1 2∂µϕi∂µϕi−1 2∂µρ∂µρ−V(ρ,ϕ), (32.16) where V(ρ,ϕ) =1 16λ[(v+ρ)2+ϕiϕi−v2]2 =1 16λ(2vρ+ρ2+ϕiϕi)2 =1 4λv2ρ2+1 4λvρ(ρ2+ϕiϕi) +1 16λ(ρ2+ϕiϕi)2.(32.17) There is indeed a manifest SO( N−1) symmetry in eqs.(32.16) and (32.17). Also, theN−1ϕiﬁelds are massless: they are the expected N−1 Goldstone bosons. Reference Notes Further discussion of Goldstone’s theorem can be found in Georgi ,Peskin & Schroeder , andWeinberg II . Problems 32.1) Consider the Noether current jµfor the U(1) symmetry of eq.(32.1), and the corresponding charge Q. a) Show that e−iαQϕe+iαQ=e+iαϕ. b) Use eq.(32.5) to show that e−iαQ|θ∝an}b∇acket∇i}ht=|θ+α∝an}b∇acket∇i}ht. c) Show that Q|0∝an}b∇acket∇i}ht ∝ne}ationslash= 0; that is, the charge does not annihilate the θ= 0 vacuum. Contrast this with the case of an unbroken symmetr y. 32: Spontaneous Breaking of Continuous Symmetries 209 32.2) In problem 24.3, we showed that [ ϕi,Qa] = (Ta)ijϕj, whereQais the Noether charge in the SO( N) symmetric theory. Use this result to show thatQa|0∝an}b∇acket∇i}ht ∝ne}ationslash= 0 if and only if Qais broken. 32.3) We deﬁne the decay constant fof the Goldstone boson via ∝an}b∇acketle{tk|jµ(x)|0∝an}b∇acket∇i}ht=ifkµe−ikx, (32.18) where |k∝an}b∇acket∇i}htis the state of a single Goldstone boson with four-momentum k, normalized in the usual way, |0∝an}b∇acket∇i}htis theθ= 0 vacuum, and jµ(x) is the Noether current. a) Compute fat tree level. (That is, express jµin terms of the ρ andχﬁelds, and then use free ﬁeld theory to compute the matrix element.) A nonvanishing value of findicates that the corresponding current is spontaneously broken. b) Discuss how your result would be modiﬁed by higher-order c orrec- tions. Part II Spin One Half 33: Representations of the Lorentz Group 211 33Representations of the Lorentz Group Prerequisite: 2 In section 2, we saw that we could deﬁne a unitary operator U(Λ) that implemented a Lorentz transformation on a scalar ﬁeld ϕ(x) via U(Λ)−1ϕ(x)U(Λ) =ϕ(Λ−1x). (33.1) As shown in section 2, this implies that the derivative of the ﬁeld transforms as U(Λ)−1∂µϕ(x)U(Λ) = Λµρ¯∂ρϕ(Λ−1x), (33.2) where the bar on he derivative means that it is with respect to the argument ¯x= Λ−1x. Eq.(33.2) suggests that we could deﬁne a vector ﬁeld Aµ(x) that would transform as U(Λ)−1Aρ(x)U(Λ) = ΛµρAρ(Λ−1x), (33.3) or atensor ﬁeld Bµν(x) that would transform as U(Λ)−1Bµν(x)U(Λ) = ΛµρΛνσBρσ(Λ−1x). (33.4) Note that if Bµνis either symmetric, Bµν(x) =Bνµ(x), or antisymmetric, Bµν(x) =−Bνµ(x), then the symmetry is preserved by the Lorentz trans- formation. Also, if we take the trace to get T(x)≡gµνBµν(x), then, using gµνΛµρΛνσ=gρσ, we ﬁnd that T(x) transforms like a scalar ﬁeld, U(Λ)−1T(x)U(Λ) =T(Λ−1x). (33.5) Thus, given a tensor ﬁeld Bµν(x) with no particular symmetry, we can write Bµν(x) =Aµν(x) +Sµν(x) +1 4gµνT(x), (33.6) whereAµνis antisymmetric ( Aµν=−Aνµ) andSµνis symmetric ( Sµν= Sνµ) and traceless ( gµνSµν= 0). The key point is that the ﬁelds Aµν,Sµν, andTdo not mix with each other under Lorentz transformations. Is it possible to further break apart these ﬁelds into still s maller sets that do not mix under Lorentz transformations? How do we make this decomposition into irreducible representations of the Lorentz group for a ﬁeld carrying nvector indices? Are there any other kinds of indices we could consistently assign to a ﬁeld? If so, how do these behav e under a Lorentz transformation? The answers to these questions are to be found in the theory of group representations . Let us see how this works for the Lorentz group (in four spacetime dimensions). 33: Representations of the Lorentz Group 212 Consider a ﬁeld (not necessarily hermitian) that carries a g eneric Lorentz index,ϕA(x). Under a Lorentz transformation, we have U(Λ)−1ϕA(x)U(Λ) =LAB(Λ)ϕB(Λ−1x), (33.7) whereLAB(Λ) is a matrix that depends on Λ. These ﬁnite-dimensional matrices must obey the group composition rule LAB(Λ′)LBC(Λ) =LAC(Λ′Λ). (33.8) We say that the matrices LAB(Λ) form a representation of the Lorentz group. For an inﬁnitesimal transformation Λµν=δµν+δωµν, we can write U(1+δω) =I+i 2δωµνMµν, (33.9) where the operators Mµνare the generators of the Lorentz group. As shown in section 2, the generators obey the commutation rela tions [Mµν,Mρσ] =i/parenleftig gµρMνσ−(µ↔ν)/parenrightig −(ρ↔σ), (33.10) which specify the Lie algebra of the Lorentz group. We can identify the components of the angular momentum opera tor/vectorJas Ji≡1 2εijkMjkand the components of the boost operator /vectorKasKi≡Mi0. We then ﬁnd from eq.(33.10) that [Ji,Jj] = +iεijkJk, (33.11) [Ji,Kj] = +iεijkKk, (33.12) [Ki,Kj] =−iεijkJk. (33.13) For an inﬁnitesimal transformation, we also have LAB(1+δω) =δAB+i 2δωµν(Sµν)AB, (33.14) Eq.(33.7) then becomes [ϕA(x),Mµν] =LµνϕA(x) + (Sµν)ABϕB(x), (33.15) where Lµν≡1 i(xµ∂ν−xν∂µ). Both the diﬀerential operators Lµνand the representation matrices ( Sµν)ABmust separately obey the same commuta- tion relations as the generators themselves; see problems 2 .8 and 2.9. Our problem now is to ﬁnd all possible sets of ﬁnite-dimensio nal matri- ces that obey eq.(33.10), or equivalently eqs.(33.11–33.1 3). Although the operatorsMµνmust be hermitian, the matrices ( Sµν)ABneed not be. 33: Representations of the Lorentz Group 213 If we restrict our attention to eq.(33.11) alone, we know (fr om stan- dard results in the quantum mechanics of angular momentum) t hat we can ﬁnd three (2 j+1)×(2j+1) hermitian matrices J1,J2, and J3that obey eq.(33.11), and that the eigenvalues of (say) J3are−j,−j+1,...,+j, wherejhas the possible values 0 ,1 2,1,.... We further know that these ma- trices constitute all of the inequivalent, irreducible rep resentations of the Lie algebra of SO(3), the rotation group in three dimensions .Inequivalent means not related by a unitary transformation; irreducible means cannot be made block-diagonal by a unitary transformation. (The stan dard deriva- tion assumes that the matrices are hermitian, but allowing n onhermitian matrices does not enlarge the set of solutions.) Also, when jis a half inte- ger, a rotation by 2 πresults in an overall minus sign; these representations of the Lie algebra of SO(3) are therefore actually not represenations of the group SO(3), since a 2 πrotation should be equivalent to no rotation. As we saw in section 24, the Lie algebra of SO(3) is the same as the Lie algebra of SU(2); the half-integer representations of this Lie alge bradoqualify as representations of the group SU(2). We would like to extend these conclusions to encompass the fu ll set of eqs.(33.11–33.13). In order to do so, it is helpful to deﬁne s ome nonher- mitian operators whose physical signiﬁcance is obscure, bu t which simplify the commutation relations. These are Ni≡1 2(Ji−iKi), (33.16) N† i≡1 2(Ji+iKi). (33.17) In terms of NiandN† i, eqs.(33.11–33.13) become [Ni,Nj] =iεijkNk, (33.18) [N† i,N† j] =iεijkN† k, (33.19) [Ni,N† j] = 0. (33.20) We see that we have two diﬀerent SU(2) Lie algebras that are ex changed by hermitian conjugation. As we just discussed, a represent ation of the SU(2) Lie algebra is speciﬁed by an integer or half integer; w e therefore conclude that a representation of the Lie algebra of the Lore ntz group in four spacetime dimensions is speciﬁed by twointegers or half-integers n andn′. We will label these representations as (2 n+1,2n′+1); the number of components of a representation is then (2 n+1)(2n′+1). Diﬀerent compo- nents within a representation can also be labeled by their an gular mo- mentum representations. To do this, we ﬁrst note that, from e qs.(33.16) and (33.17), we have Ji=Ni+N† i. Thus, deducing the allowed values 33: Representations of the Lorentz Group 214 ofjgivennandn′becomes a standard problem in the addition of an- gular momenta. The general result is that the allowed values ofjare |n−n′|,|n−n′|+1,...,n +n′, and each of these values appears exactly once. The four simplest and most often encountered representatio ns are (1,1), (2,1), (1,2), and (2,2). These are given special names: (1,1) =Scalar orsinglet (2,1) =Left-handed spinor (1,2) =Right-handed spinor (2,2) =Vector (33.21) It may seem a little surprising that (2 ,2) is to be identiﬁed as the vector representation. To see that this must be the case, we ﬁrst not e that the vector representation is irreducible: all the components o f a four-vector mix with each other under a general Lorentz transformation. Secondly, the vector representation has four components. The only candid ate irreducible representations are (4 ,1), (1,4), and (2,2). The ﬁrst two of these contain angular momenta j=3 2only, whereas (2 ,2) contains j= 0 andj= 1. This is just right for a four-vector, whose time component is a scalar under spatial rotations, and whose space components are a three-v ector. In order to gain a better understanding of what it means for (2 ,2) to be the vector representation, we must ﬁrst investigate the s pinor represen- tations (1,2) and (2,1), which contain angular momenta j=1 2only. Reference Notes An extended treatment of representations of the Lorentz gro up in four dimensions can be found in Weinberg I . Problems 33.1) Express Aµν(x),Sµν(x), andT(x) in terms of Bµν(x). 33.2) Verify that eqs.(33.18–33.20) follow from eqs.(33.1 1–33.13). 34: Left- and Right-Handed Spinor Fields 215 34Left- and Right-Handed Spinor Fields Prerequisite: 3, 33 Consider a left-handed spinor ﬁeld ψa(x), also known as a left-handed Weyl ﬁeld, which is in the (2 ,1) representation of the Lie algebra of the Lorentz group. Here the index ais aleft-handed spinor index that takes on two possible values. Under a Lorentz transformation, we have U(Λ)−1ψa(x)U(Λ) =Lab(Λ)ψb(Λ−1x), (34.1) whereLab(Λ) is a matrix in the (2 ,1) representation. These matrices satisfy the group composition rule Lab(Λ′)Lbc(Λ) =Lac(Λ′Λ). (34.2) For an inﬁnitesimal transformation Λµν=δµν+δωµν, we can write Lab(1+δω) =δab+i 2δωµν(Sµν L)ab, (34.3) where (Sµν L)ab=−(Sνµ L)abis a set of 2 ×2 matrices that obey the same commutation relations as the generators Mµν, namely [Sµν L,Sρσ L] =i/parenleftig gµρSνσ L−(µ↔ν)/parenrightig −(ρ↔σ). (34.4) Using U(1+δω) =I+i 2δωµνMµν, (34.5) eq.(34.1) becomes [ψa(x),Mµν] =Lµνψa(x) + (Sµν L)abψb(x), (34.6) where Lµν=1 i(xµ∂ν−xν∂µ). The Lµνterm in eq.(34.6) would also be present for a scalar ﬁeld, and is not the focus of our current i nterest; we will suppress it by evaluating the ﬁelds at the spacetime ori gin,xµ= 0. Recalling that Mij=εijkJk, whereJkis the angular momentum operator, we have εijk[ψa(0),Jk] = (Sij L)abψb(0). (34.7) Recall that the (2 ,1) representation of the Lorentz group includes an- gular momentum j=1 2only. For a spin-one-half operator, the standard convention is that the matrix on the right-hand side of eq.(3 4.7) is1 2εijkσk, where we have suppressed the row index aand the column index b, and whereσkis a Pauli matrix: σ1=/parenleftbigg0 1 1 0/parenrightbigg , σ 2=/parenleftbigg0−i i0/parenrightbigg , σ 3=/parenleftbigg1 0 0−1/parenrightbigg . (34.8) 34: Left- and Right-Handed Spinor Fields 216 We therefore conclude that (Sij L)ab=1 2εijkσk, (34.9) Thus, for example, setting i=1 andj=2 yields ( S12 L)ab=1 2ε12kσk=1 2σ3, and so (S12 L)11= +1 2, (S12 L)22=−1 2, and (S12 L)12= (S12 L)21= 0. Once we have the (2 ,1) representation matrices for the angular momen- tum operator Ji, we can easily get them for the boost operator Kk=Mk0. This is because Jk=Nk+N† kandKk=i(Nk−N† k), and, acting on a ﬁeld in the (2 ,1) representation, N† kis zero. Therefore, the representation matrices for Kkare simplyitimes those for Jk, and so (Sk0 L)ab=1 2iσk. (34.10) Now consider taking the hermitian conjugate of the left-han ded spinor ﬁeldψa(x). Recall that hermitian conjugation swaps the two SU(2) Lie algebras that comprise the Lie algebra of the Lorentz group. Therefore, the hermitian conjugate of a ﬁeld in the (2 ,1) representation should be a ﬁeld in the (1 ,2) representation; such a ﬁeld is called a right-handed spinor ﬁeldor aright-handed Weyl ﬁeld . We will distinguish the indices of the (1,2) representation from those of the (2 ,1) representation by putting dots over them. Thus, we write [ψa(x)]†=ψ† ˙a(x). (34.11) Under a Lorentz transformation, we have U(Λ)−1ψ† ˙a(x)U(Λ) =R˙a˙b(Λ)ψ† ˙b(Λ−1x), (34.12) whereR˙a˙b(Λ) is a matrix in the (1 ,2) representation. These matrices satisfy the group composition rule R˙a˙b(Λ′)R˙b˙c(Λ) =R˙a˙c(Λ′Λ). (34.13) For an inﬁnitesimal transformation Λµν=δµν+δωµν, we can write R˙a˙b(1+δω) =δ˙a˙b+i 2δωµν(Sµν R)˙a˙b, (34.14) where (Sµν R)˙a˙b=−(Sνµ R)˙a˙bis a set of 2 ×2 matrices that obey the same commutation relations as the generators Mµν. We then have [ψ† ˙a(0),Mµν] = (Sµν R)˙a˙bψ† ˙b(0). (34.15) Taking the hermitian conjugate of this equation, we get [Mµν,ψa(0)] = [(Sµν R)˙a˙b]∗ψb(0). (34.16) 34: Left- and Right-Handed Spinor Fields 217 Comparing this with eq.(34.6), we see that (Sµν R)˙a˙b=−[(Sµν L)ab]∗. (34.17) In the previous section, we examined the Lorentz-transform ation prop- erties of a ﬁeld carrying two vector indices. To help us get be tter acquainted with the properties of spinor indices, let us now do the same f or a ﬁeld that carries two (2 ,1) indices. Call this ﬁeld Cab(x). Under a Lorentz transfor- mation, we have U(Λ)−1Cab(x)U(Λ) =Lac(Λ)Lbd(Λ)Ccd(Λ−1x). (34.18) The question we wish to address is whether or not the four comp onents of Cabcan be grouped into smaller sets that do not mix with each othe r under Lorentz transformations. To answer this question, recall from quantum mechanics that two spin- one-half particles can be in a state of total spin zero, or tot al spin one. Furthermore, the single spin-zero state is the unique antisymmetric com- bination of the two spin-one-half states, and the three spin -one states are the three symmetric combinations of the two spin-one-half states. We can write this schematically as 2 ⊗2 = 1 A⊕3S, where we label the representa- tion of SU(2) by the number of its components, and the subscri pts S and A indicate whether that representation appears in the symme tric or anti- symmetric combination of the two 2’s. For the Lorentz group, the relevant relation is (2 ,1)⊗(2,1) = (1,1)A⊕(3,1)S. This implies that we should be able to write Cab(x) =εabD(x) +Gab(x), (34.19) whereD(x) is a scalar ﬁeld, εab=−εbais an antisymmetric set of constants, andGab(x) =Gba(x). The symbol εabis uniquely determined by its sym- metry properties up to an overall constant; we will choose ε21=−ε12= +1. SinceD(x) is a Lorentz scalar, eq.(34.19) is consistent with eq.(34. 18) only if Lac(Λ)Lbd(Λ)εcd=εab. (34.20) This means that εabis aninvariant symbol of the Lorentz group: it does not change under a Lorentz transformation that acts on all of its indices. In this way, εabis analogous to the metric gµν, which is also an invariant symbol, since ΛµρΛνσgρσ=gµν. (34.21) We usegµνand its inverse gµνto raise and lower vector indices, and we can use εaband and its inverse εabto raise and lower left-handed spinor indices. Here we deﬁne εabvia ε12=ε21= +1, ε21=ε12=−1. (34.22) 34: Left- and Right-Handed Spinor Fields 218 With this deﬁnition, we have εabεbc=δac, εabεbc=δac. (34.23) We can then deﬁne ψa(x)≡εabψb(x). (34.24) We also have (suppressing the spacetime argument of the ﬁeld ) ψa=εabψb=εabεbcψc=δacψc, (34.25) as we would expect. However, the antisymmetry of εabmeans that we must be careful with minus signs; for example, eq.(34.24) ca n be written in various ways, such as ψa=εabψb=−εbaψb=−ψbεba=ψbεab. (34.26) We must also be careful about signs when we contract indices, since ψaχa=εabψbχa=−εbaψbχa=−ψbχb. (34.27) In section 35, we will (mercifully) develop an index-free no tation that au- tomatically keeps track of these essential (but annoying) m inus signs. An exactly analogous discussion applies to the second SU(2) factor; from the group-theoretic relation (1 ,2)⊗(1,2) = (1,1)A⊕(1,3)S, we can deduce the existence of an invariant symbol ε˙a˙b=−ε˙b˙a. We will normalize ε˙a˙baccording to eq.(34.22). Then eqs.(34.23–34.27) hold if allthe undotted indices are replaced by dotted indices. Now consider a ﬁeld carrying one undotted and one dotted inde x,Aa˙a(x). Such a ﬁeld is in the (2 ,2) representation, and in section 33 we concluded that the (2,2) representation was the vector representation. We would m ore naturally write a ﬁeld in the vector representation as Aµ(x). There must, then, be a dictionary that gives us the components of Aa˙a(x) in terms of the components of Aµ(x); we can write this as Aa˙a(x) =σµ a˙aAµ(x), (34.28) whereσµ a˙ais another invariant symbol. That such a symbol must exist ca n be deduced from the group-theoretic relation (2,1)⊗(1,2)⊗(2,2) = (1,1)⊕... . (34.29) As we will see in section 35, it turns out to be consistent with our already established conventions for Sµν LandSµν Rto choose σµ a˙a= (I,/vector σ). (34.30) 34: Left- and Right-Handed Spinor Fields 219 Thus, for example, σ3 1˙1= +1,σ3 2˙2=−1,σ3 1˙2=σ3 2˙1= 0. In general, whenever the product of a set of representations includes the singlet, there is a corresponding invariant symbol. For example, we can deduce the existence of gµν=gνµfrom (2,2)⊗(2,2) = (1,1)S⊕(1,3)A⊕(3,1)A⊕(3,3)S. (34.31) Another invariant symbol, the Levi-Civita symbol , follows from (2,2)⊗(2,2)⊗(2,2)⊗(2,2) = (1,1)A⊕... , (34.32) where the subscript A denotes the completely antisymmetric part. The Levi-Civita symbol is εµνρσ, which is antisymmetric on exchange of any pair of its indices, and is normalized via ε0123= +1. To see that εµνρσis invariant, we note that ΛµαΛνβΛργΛσδεαβγδis antisymmetric on exchange of any two of its uncontracted indices, and therefore must be proportional toεµνρσ. The constant of proportionality works out to be detΛ, which is +1 for a proper Lorentz transformation. We are ﬁnally ready to answer a question we posed at the beginn ing of section 33. There we considered a ﬁeld Bµν(x) carrying two vector indices, and we decomposed it as Bµν(x) =Aµν(x) +Sµν(x) +1 4gµνT(x), (34.33) whereAµνis antisymmetric ( Aµν=−Aνµ) andSµνis symmetric ( Sµν= Sνµ) and traceless ( gµνSµν= 0). We asked whether further decomposition into still smaller irreducible representations was possib le. The answer to this question can be found in eq.(34.31). Obviously, T(x) corresponds to (1,1), andSµν(x) to (3,3).1But, according to eq.(34.31), the antisymmet- ric ﬁeldAµν(x) should correspond to (3 ,1)⊕(1,3). A ﬁeld in the (3 ,1) representation carries a symmetric pair of left-handed (un dotted) spinor indices; its hermitian conjugate is a ﬁeld in the (1 ,3) representation that carries a symmetric pair of right-handed (dotted) spinor in dices. We should, then, be able to ﬁnd a mapping, analogous to eq.(34.28), that givesAµν(x) in terms of a ﬁeld Gab(x) and its hermitian conjugate G† ˙a˙b(x). This mapping is provided by the generator matrices Sµν LandSµν R. We ﬁrst note that the Pauli matrices are traceless, and so eqs.( 34.9) and (34.10) imply that ( Sµν L)aa= 0. Using eq.(34.24), we can rewrite this asεab(Sµν L)ab= 0. Since εabis antisymmetric, ( Sµν L)abmust be symmetric on exchange of its two spinor indices. An identical argument shows that 1Note that a symmetric traceless tensor has three independen t diagonal components, and six independent oﬀ-diagonal components, for a total of n ine, the number of compo- nents of the (3 ,3) representation. 34: Left- and Right-Handed Spinor Fields 220 (Sµν R)˙a˙bmust be symmetric on exchange of its two spinor indices. Furt her- more, according to eqs.(34.9) and (34.10), we have (S10 L)ab=−i(S23 L)ab. (34.34) This can be written covariantly with the Levi-Civita symbol as (Sµν L)ab=−i 2εµνρσ(SLρσ)ab. (34.35) Similarly, (Sµν R)˙a˙b= +i 2εµνρσ(SRρσ)˙a˙b. (34.36) Eq.(34.36) follows from taking the complex conjugate of eq. (34.35) and using eq.(34.17). Now, given a ﬁeld Gab(x) in the (3,1) representation, we can map it into a self-dual antisymmetric tensor Gµν(x) via Gµν(x)≡(Sµν L)abGab(x). (34.37) Byself-dual , we mean that Gµν(x) obeys Gµν(x) =−i 2εµνρσGρσ(x). (34.38) Taking the hermitian conjugate of eq.(34.37), and using eq. (34.17), we get G†µν(x) =−(Sµν R)˙a˙bG† ˙a˙b(x), (34.39) which is anti-self-dual , G†µν(x) = +i 2εµνρσG† ρσ(x). (34.40) Given a hermitian antisymmetric tensor ﬁeld Aµν(x), we can extract its self-dual and anti-self-dual parts via Gµν(x) =1 2Aµν(x)−i 4εµνρσAρσ(x), (34.41) G†µν(x) =1 2Aµν(x) +i 4εµνρσAρσ(x). (34.42) Then we have Aµν(x) =Gµν(x) +G†µν(x). (34.43) The ﬁeldGµν(x) is in the (3 ,1) representation, and the ﬁeld G†µν(x) is in the (1,3) representation; these do not mix under Lorentz transform ations. Problems 34.1) Verify that eq.(34.6) follows from eq.(34.1). 34: Left- and Right-Handed Spinor Fields 221 34.2) Verify that eqs.(34.9) and (34.10) obey eq.(34.4). 34.3) Show that the Levi-Civita symbol obeys εµνρσεαβγσ=−δµαδνβδργ−δµβδνγδρα−δµγδναδρβ +δµβδναδργ+δµαδνγδρβ+δµγδνβδρα,(34.44) εµνρσεαβρσ=−2(δµαδνβ−δµβδνα), (34.45) εµνρσεανρσ=−6δµα. (34.46) 34.4) Consider a ﬁeld Ca...c˙a...˙c(x), withNundotted indices and Mdotted indices, that is furthermore symmetric on exchange of any pa ir of un- dotted indices, and also symmetric on exchange of any pair of dotted indices. Show that this ﬁeld corresponds to a single irreduc ible rep- resentation (2 n+1,2n′+1) of the Lorentz group, and identify nand n′. 35: Manipulating Spinor Indices 222 35Manipulating Spinor Indices Prerequisite: 34 In section 34 we introduced the invariant symbols εab,εab,ε˙a˙b, andε˙a˙b, where ε12=ε˙1˙2=ε21=ε˙2˙1= +1, ε21=ε˙2˙1=ε12=ε˙1˙2=−1.(35.1) We use the εsymbols to raise and lower spinor indices, contracting the second index on the ε. (If we contract the ﬁrst index instead, then there is an extra minus sign). Another invariant symbol is σµ a˙a= (I,/vector σ), (35.2) whereIis the 2 ×2 identity matrix, and σ1=/parenleftbigg0 1 1 0/parenrightbigg , σ 2=/parenleftbigg0−i i0/parenrightbigg , σ 3=/parenleftbigg1 0 0−1/parenrightbigg (35.3) are the Pauli matrices. Now let us consider some combinations of invariant symbols w ith some indices contracted, such as gµνσµ a˙aσν b˙b. This object must also be invariant. Then, since it carries two undotted and two dotted spinor ind ices, it must be proportional to εabε˙a˙b. Using eqs.(35.1) and (35.2), we can laboriously check this; it turns out to be correct.1The proportionality constant works out to be minus two: σµ a˙aσµb˙b=−2εabε˙a˙b. (35.4) Similarly,εabε˙a˙bσµ a˙aσν b˙bmust be proportional to gµν, and the proportionality constant is again minus two: εabε˙a˙bσµ a˙aσν b˙b=−2gµν. (35.5) Next, let’s see what we can learn about the generator matrice s (Sµν L)ab and (Sµν R)˙a˙bfrom the fact that εab,ε˙a˙b, andσµ a˙aare all invariant symbols. Begin with εab=L(Λ)acL(Λ)bdεcd, (35.6) which expresses the Lorentz invariance of εab. For an inﬁnitesimal trans- formation Λµν=δµν+δωµν, we have Lab(1+δω) =δab+i 2δωµν(Sµν L)ab, (35.7) 1If it did not turn out to be correct, then eq.(35.2) would not b e a viable choice of numerical values for this symbol. 35: Manipulating Spinor Indices 223 and eq.(35.6) becomes εab=εab+i 2δωµν/bracketleftig (Sµν L)acεcb+ (Sµν L)bdεad/bracketrightig +O(δω2) =εab+i 2δωµν/bracketleftig −(Sµν L)ab+ (Sµν L)ba/bracketrightig +O(δω2). (35.8) Since eq.(35.8) holds for any choice of δωµν, it must be that the factor in square brackets vanishes. Thus we conclude that ( Sµν L)ab= (Sµν L)ba, which we had already deduced in section 34 by a diﬀerent method. Sim ilarly, starting from the Lorentz invariance of ε˙a˙b, we can show that ( Sµν R)˙a˙b= (Sµν R)˙b˙a. Next, start from σρ a˙a= ΛρτL(Λ)abR(Λ)˙a˙bστ b˙b, (35.9) which expresses the Lorentz invariance of σρ a˙a. For an inﬁnitesimal trans- formation, we have Λρτ=δρτ+i 2δωµν(Sµν V)ρτ, (35.10) Lab(1+δω) =δab+i 2δωµν(Sµν L)ab, (35.11) R˙a˙b(1+δω) =δ˙a˙b+i 2δωµν(Sµν R)˙a˙b, (35.12) where (Sµν V)ρτ≡1 i(gµρδντ−gνρδµτ). (35.13) Substituting eqs.(35.10–35.13) into eq.(35.9) and isolat ing the coeﬃcient ofδωµνyields (gµρδντ−gνρδµτ)στ a˙a+i(Sµν L)abσρ b˙a+i(Sµν R)˙a˙bσρ a˙b= 0. (35.14) Now multiply by σρc˙cto get σµ c˙cσν a˙a−σν c˙cσµ a˙a+i(Sµν L)abσρ b˙aσρc˙c+i(Sµν R)˙a˙bσρ a˙bσρc˙c= 0. (35.15) Next use eq.(35.4) in each of the last two terms to get σµ c˙cσν a˙a−σν c˙cσµ a˙a+ 2i(Sµν L)acε˙a˙c+ 2i(Sµν R)˙a˙cεac= 0. (35.16) If we multiply eq.(35.16) by ε˙a˙c, and remember that ε˙a˙c(Sµν R)˙a˙c= 0 and thatε˙a˙cε˙a˙c=−2, we get a formula for ( Sµν L)ac, namely (Sµν L)ac=i 4ε˙a˙c(σµ a˙aσν c˙c−σν a˙aσµ c˙c). (35.17) 35: Manipulating Spinor Indices 224 Similarly, if we multiply eq.(35.16) by εac, we get (Sµν R)˙a˙c=i 4εac(σµ a˙aσν c˙c−σν a˙aσµ c˙c). (35.18) These formulae can be made to look a little nicer if we deﬁne ¯σµ˙aa≡εabε˙a˙bσµ b˙b. (35.19) Numerically, it turns out that ¯σµ˙aa= (I,−/vector σ). (35.20) Using ¯σµ, we can write eqs.(35.17) and (35.18) as (Sµν L)ab= +i 4(σµ¯σν−σν¯σµ)ab, (35.21) (Sµν R)˙a˙b=−i 4(¯σµσν−¯σνσµ)˙a˙b. (35.22) In eq.(35.22), we have suppressed a contracted pair of undot ted indices arranged ascc, and in eq.(35.21), we have suppressed a contracted pair of dotted indices arranged as ˙c˙c. We will adopt this as a general convention: a missing pair of c ontracted, undotted indices is understood to be written ascc, and a missing pair of contracted, dotted indices is understood to be written as ˙c˙c. Thus, ifχand ψare two left-handed Weyl ﬁelds, we have χψ=χaψaandχ†ψ†=χ† ˙aψ†˙a. (35.23) We expect Weyl ﬁelds to describe spin-one-half particles, a nd (by the spin- statistics theorem) these particles must be fermions . Therefore the corre- spoding ﬁelds must anticommute , rather than commute. That is, we should have χa(x)ψb(y) =−ψb(y)χa(x). (35.24) Thus we can rewrite eq.(35.23) as χψ=χaψa=−ψaχa=ψaχa=ψχ. (35.25) The second equality follows from anticommutation of the ﬁel ds, and the third from switching aatoaa(which introduces an extra minus sign). Eq.(35.25) tells us that χψ=ψχ, which is a nice feature of this notation. Furthermore, if we take the hermitian conjugate of χψ, we get (χψ)†= (χaψa)†= (ψa)†(χa)†=ψ† ˙aχ†˙a=ψ†χ†. (35.26) That (χψ)†=ψ†χ†is just what we would expect if we ignored the indices completely. Of course, by analogy with eq.(35.25), we also h aveψ†χ†= χ†ψ†. 35: Manipulating Spinor Indices 225 In order to tell whether a spinor ﬁeld is left-handed or right -handed when its spinor index is suppressed, we will adopt the conven tion that a right-handed ﬁeld is always written as the hermitian conjug ate of a left- handed ﬁeld. Thus, a right-handed ﬁeld is always written wit h a dagger, and a left-handed ﬁeld is always written without a dagger. Let’s try computing the hermitian conjugate of something a l ittle more complicated: ψ†¯σµχ=ψ† ˙a¯σµ˙acχc. (35.27) This behaves like a vector ﬁeld under Lorentz transformatio ns, U(Λ)−1[ψ†¯σµχ]U(Λ) = Λµν[ψ†¯σνχ]. (35.28) (To avoid clutter, we suppressed the spacetime argument of t he ﬁelds; as usual, it is xon the left-hand side and Λ−1xon the right.) The hermitian conjugate of eq.(35.27) is [ψ†¯σµχ]†= [ψ† ˙a¯σµ˙acχc]† =χ† ˙c(¯σµa˙c)∗ψa =χ† ˙c¯σµ˙caψa =χ†¯σµψ. (35.29) In the third line, we used the hermiticity of the matrices ¯ σµ= (I,−/vector σ). We will get considerably more practice with this notation in the follow- ing sections. Problems 35.1) Verify that eq.(35.20) follows from eqs.(35.2) and (3 5.19). Hint: write everything in “matrix multiplication” order, and not e that, nu- merically,εab=−εab=iσ2. Then make use of the properties of the Pauli matrices. 35.2) Verify that eq.(35.21) is consistent with eqs.(34.9) and (34.10). 35.3) Verify that eq.(35.22) is consistent with eq.(34.17) . 35.4) Verify eq.(35.5). 36: Lagrangians for Spinor Fields 226 36Lagrangians for Spinor Fields Prerequisite: 22, 35 Suppose we have a left-handed spinor ﬁeld ψa. We would like to ﬁnd a suitable lagrangian for it. This lagrangian must be Lorentz invariant, and it must be hermitian. We would also like it to be quadratic in ψand its hermitian conjugate ψ† ˙a, because this will lead to a linear equation of motion, with plane-wave solutions. We want plane-wave solu tions because these describe free particles, the starting point for a theo ry of interacting particles. Let us begin with terms with no derivatives. The only possibi lity is ψψ=ψaψa=εabψbψa, plus its hermitian conjugate. Because of anticom- mutation of the ﬁelds ( ψbψa=−ψaψb), this expression does not vanish (as it would if the ﬁelds commuted), and so we can use it as a term in L. Next we need a term with derivatives. The obvious choice is ∂µψ∂µψ, plus its hermitian conjugate. This, however, yields a hamil tonian that is unbounded below, which is unacceptable. To get a bounded ham iltonian, the kinetic term must involve both ψandψ†. A candidate is iψ†¯σµ∂µψ. This not hermitian, but (iψ†¯σµ∂µψ)†= (iψ† ˙a¯σµ˙ac∂µψc)† =−i∂µψ† ˙c(¯σµa˙c)∗ψa =−i∂µψ† ˙c¯σµ˙caψa =iψ† ˙c¯σµ˙ca∂µψa−i∂µ(ψ† ˙c¯σµ˙caψa). =iψ†¯σµ∂µψ−i∂µ(ψ†¯σµψ). (36.1) In the third line, we used the hermiticity of the matrices ¯ σµ= (I,−/vector σ). In the fourth line, we used −(∂A)B=A∂B−∂(AB). In the last line, the second term is a total divergence, and vanishes (with suitab le boundary conditions on the ﬁelds at inﬁnity) when we integrate it over d4xto get the actionS. Thusiψ†¯σµ∂µψhas the hermiticity properties necessary for a term in L. Our complete lagrangian for ψis then L=iψ†¯σµ∂µψ−1 2mψψ−1 2m∗ψ†ψ†, (36.2) wheremis a complex parameter with dimensions of mass. The phase of m is actually irrelevant: if m=|m|eiα, we can set ψ=e−iα/2˜ψin eq.(36.2); then we get a lagrangian for ˜ψthat is identical to eq.(36.2), but with m replaced by |m|. So we can, without loss of generality, take mto be real 36: Lagrangians for Spinor Fields 227 and positive in the ﬁrst place, and that is what we will do, set tingm∗=m in eq.(36.2). The equation of motion for ψis then 0 =−δS δψ†=−i¯σµ∂µψ+mψ†, (36.3) Restoring the spinor indices, this reads 0 =−i¯σµ˙ac∂µψc+mψ†˙a. (36.4) Taking the hermitian conjugate (or, equivalently, computi ng−δS/δψ ), we get 0 = +i(¯σµa˙c)∗∂µψ† ˙c+mψa = +i¯σµ˙ca∂µψ† ˙c+mψa =−iσµ a˙c∂µψ†˙c+mψa. (36.5) In the second line, we used the hermiticity of the matrices ¯ σµ= (I,−/vector σ). In the third, we lowered the undotted index, and switched˙c˙cto˙c˙c, which gives an extra minus sign. Eqs.(36.5) and (36.4) can be combined to read /parenleftiggmδac−iσµ a˙c∂µ −i¯σµ˙ac∂µmδ˙a˙c/parenrightigg/parenleftiggψc ψ†˙c/parenrightigg = 0. (36.6) We can write this more compactly by introducing the 4 ×4gamma matrices γµ≡/parenleftigg0σµ a˙c ¯σµ˙ac0/parenrightigg . (36.7) Using the sigma-matrix relations, (σµ¯σν+σν¯σµ)ac=−2gµνδac, (¯σµσν+ ¯σνσµ)˙a˙c=−2gµνδ˙a˙c, (36.8) which are most easily derived from the numerical formulae σµ a˙a= (I,/vector σ) and ¯σµ˙aa= (I,−/vector σ), we see that the gamma matrices obey {γµ,γν}=−2gµν, (36.9) where {A,B} ≡AB+BAdenotes the anticommutator, and there is an understood 4 ×4 identity matrix on the right-hand side. We also introduce a four-component Majorana ﬁeld Ψ≡/parenleftiggψc ψ†˙c/parenrightigg . (36.10) 36: Lagrangians for Spinor Fields 228 Then eq.(36.6) becomes (−iγµ∂µ+m)Ψ = 0. (36.11) This is the Dirac equation . We ﬁrst encountered it in section 1, where the gamma matrices were given diﬀerent names ( β=γ0andαk=γ0γk). Also, in section 1 we were trying (and failing) to interpret Ψ as a wa ve function, rather than as a quantum ﬁeld. Now consider a theory of two left-handed spinor ﬁelds with an SO(2) symmetry, L=iψ† i¯σµ∂µψi−1 2mψiψi−1 2mψ† iψ† i, (36.12) where the spinor indices are suppressed and i= 1,2 is implicitly summed. As in the analogous case of two scalar ﬁelds discussed in sect ions 22 and 23, this lagrangian is invariant under the SO(2) transforma tion /parenleftiggψ1 ψ2/parenrightigg →/parenleftiggcosαsinα −sinαcosα/parenrightigg/parenleftiggψ1 ψ2/parenrightigg . (36.13) We can write the lagrangian so that the SO(2) symmetry appear s as a U(1) symmetry instead; let χ=1√ 2(ψ1+iψ2), (36.14) ξ=1√ 2(ψ1−iψ2). (36.15) In terms of these ﬁelds, we have L=iχ†¯σµ∂µχ+iξ†¯σµ∂µξ−mχξ−mξ†χ†. (36.16) Eq.(36.16) is invariant under the U(1) version of eq.(36.13 ), χ→e−iαχ, ξ→e+iαξ. (36.17) Next, let us derive the equations of motion that we get from eq .(36.16), following the same procedure that ultimately led to eq.(36. 6). The result is /parenleftiggmδac−iσµ a˙c∂µ −i¯σµ˙ac∂µmδ˙a˙c/parenrightigg/parenleftiggχc ξ†˙c/parenrightigg = 0. (36.18) We can now deﬁne a four-component Dirac ﬁeld Ψ≡/parenleftiggχc ξ†˙c/parenrightigg , (36.19) 36: Lagrangians for Spinor Fields 229 which obeys the Dirac equation , eq.(36.11). (We have annoyingly used the same symbol Ψ to denote both a Majorana ﬁeld and a Dirac ﬁeld; t hese are diﬀerent objects, and so we must always announce which is mea nt when we write Ψ.) We can also write the lagrangian, eq.(36.16), in terms of the Dirac ﬁeld Ψ, eq.(36.19). First we take the hermitian conjugate of Ψ to g et Ψ†= (χ† ˙a, ξa). (36.20) Introduce the matrix β≡/parenleftigg0δ˙a˙c δac0/parenrightigg . (36.21) Numerically, β=γ0. However, the spinor index structure of βandγ0is diﬀerent, and so we will distinguish them. Given β, we deﬁne Ψ≡Ψ†β= (ξa, χ† ˙a). (36.22) Then we have ΨΨ =ξaχa+χ† ˙aξ†˙a. (36.23) Also, Ψγµ∂µΨ =ξaσµ a˙c∂µξ†˙c+χ† ˙a¯σµ˙ac∂µχc. (36.24) UsingA∂B =−(∂A)B+∂(AB), the ﬁrst term on the right-hand side of eq.(36.24) can be rewritten as ξaσµ a˙c∂µξ†˙c=−(∂µξa)σµ a˙cξ†˙c+∂µ(ξaσµ a˙cξ†˙c). (36.25) Then the ﬁrst term on the right-hand side of eq.(36.25) can be rewritten as −(∂µξa)σµ a˙cξ†˙c= +ξ†˙cσµ a˙c∂µξa= +ξ† ˙c¯σµ˙ca∂µξa. (36.26) Here we used anticommutation of the ﬁelds to get the ﬁrst equa lity, and switched˙c˙cto˙c˙candaatoaa(thus generating two minus signs) to get the second. Combining eqs.(36.24–36.26), we get Ψγµ∂µΨ =χ†¯σµ∂µχ+ξ†¯σµ∂µξ+∂µ(ξσµξ†). (36.27) Therefore, up to an irrelevant total divergence, we have L=iΨγµ∂µΨ−mΨΨ. (36.28) This form of the lagrangian is invariant under the U(1) trans formation Ψ→e−iαΨ, Ψ→e+iαΨ, (36.29) 36: Lagrangians for Spinor Fields 230 which, given eq.(36.19), is the same as eq.(36.17). The Noet her current associated with this symmetry is jµ=ΨγµΨ =χ†¯σµχ−ξ†¯σµξ. (36.30) In quantum electrodynamics, the electromagnetic current i seΨγµΨ, where eis the charge of the electron. As in the case of a complex scalar ﬁeld with a U(1) symmetry, th ere is an additional discrete symmetry, called charge conjugation , that enlarges SO(2) to O(2). Charge conjugation simply exchanges χandξ. We can deﬁne a unitary charge conjugation operator Cthat implements this, C−1χa(x)C=ξa(x), C−1ξa(x)C=χa(x), (36.31) where, for the sake of precision, we have restored the spinor index and spacetime argument. We then have C−1L(x)C=L(x). To express eq.(36.31) in terms of the Dirac ﬁeld, eq.(36.19) , we ﬁrst introduce the charge conjugation matrix C ≡/parenleftiggεac0 0ε˙a˙c/parenrightigg . (36.32) Next we notice that, if we take the transpose of Ψ, eq.(36.22), we get ΨT=/parenleftigg ξa χ† ˙a/parenrightigg . (36.33) Then, if we multiply by C, we get a ﬁeld that we will call ΨC, thecharge conjugate of Ψ, ΨC≡ CΨT=/parenleftigg ξa χ†˙a/parenrightigg . (36.34) We see that ΨCis the same as the original ﬁeld Ψ, eq.(36.19), except that the roles of χandξhave been switched. We therefore have C−1Ψ(x)C= ΨC(x) (36.35) for a Dirac ﬁeld. The charge conjugation matrix has a number of useful propert ies. As a numerical matrix, it obeys CT=C†=C−1=−C, (36.36) 36: Lagrangians for Spinor Fields 231 and we can also write it as C=/parenleftigg−εac0 0−ε˙a˙c/parenrightigg . (36.37) A result that we will need later is C−1γµC=/parenleftiggεab0 0ε˙a˙b/parenrightigg/parenleftigg0σµ b˙c ¯σµ˙bc0/parenrightigg/parenleftiggεce0 0ε˙c˙e/parenrightigg =/parenleftigg0εabσµ b˙cε˙c˙e ε˙a˙b¯σµ˙bcεce 0/parenrightigg =/parenleftigg0−¯σµa˙e −σµ ˙ae 0/parenrightigg . (36.38) The minus signs in the last line come from raising or lowering an index by contracting with the ﬁrst (rather than the second) index of a nεsymbol. Comparing with γµ=/parenleftigg0σµ e˙a ¯σµ˙ea0/parenrightigg , (36.39) we see that C−1γµC=−(γµ)T. (36.40) Now let us return to the Majorana ﬁeld, eq.(36.10). It is obvi ous that a Majorana ﬁeld is its own charge conjugate, that is, ΨC= Ψ. This condition is analogous to the condition ϕ†=ϕthat is satisﬁed by a real scalar ﬁeld. A Dirac ﬁeld, with its U(1) symmetry, is analogous to a comple x scalar ﬁeld, while a Majorana ﬁeld, which has no U(1) symmetry, is an alogous to a real scalar ﬁeld. We can write our original lagrangian for a single left-hande d spinor ﬁeld, eq.(36.2), in terms of a Majorana ﬁeld, eq.(36.10), by retra cing eqs.(36.20– 36.28) with χ→ψandξ→ψ. The result is L=i 2Ψγµ∂µΨ−1 2mΨΨ. (36.41) However, we cannot yet derive the equation of motion from eq. (36.41) because it does not yet incorporate the Majorana condition ΨC= Ψ. To remedy this, we use eq.(36.36) to write the Majorana conditi on Ψ = CΨT asΨ = ΨTC. Then we can replace Ψ in eq.(36.41) by ΨTCto get L=i 2ΨTCγµ∂µΨ−1 2mΨTCΨ. (36.42) The equation of motion that follows from this lagrangian is o nce again the Dirac equation. 36: Lagrangians for Spinor Fields 232 We can also recover the Weyl components of a Dirac or Majorana ﬁeld by means of a suitable projection matrix. Deﬁne γ5≡/parenleftigg−δac0 0 +δ˙a˙c/parenrightigg , (36.43) where the subscript 5 is simply part of the traditional name o f this matrix, rather than the value of some index. Then we can deﬁne left and right projection matrices PL≡1 2(1−γ5) =/parenleftiggδac0 0 0/parenrightigg , PR≡1 2(1 +γ5) =/parenleftigg0 0 0δ˙a˙c/parenrightigg . (36.44) Thus we have, for a Dirac ﬁeld, PLΨ =/parenleftiggχc 0/parenrightigg , PRΨ =/parenleftigg 0 ξ†˙c/parenrightigg . (36.45) The matrix γ5can also be expressed as γ5=iγ0γ1γ2γ3 =−i 24εµνρσγµγνγργσ, (36.46) whereε0123=−1. Finally, let us consider the behavior of a Dirac or Majorana ﬁ eld under a Lorentz transformation. Recall that left- and right-hand ed spinor ﬁelds transform according to U(Λ)−1ψa(x)U(Λ) =L(Λ)acψc(Λ−1x), (36.47) U(Λ)−1ψ† ˙a(x)U(Λ) =R(Λ)˙a˙cψ† ˙c(Λ−1x), (36.48) where, for an inﬁnitesimal transformation Λµν=δµν+δωµν, L(1+δω)ac=δac+i 2δωµν(Sµν L)ac, (36.49) R(1+δω)˙a˙c=δ˙a˙c+i 2δωµν(Sµν R)˙a˙c, (36.50) 36: Lagrangians for Spinor Fields 233 and where (Sµν L)ac= +i 4(σµ¯σν−σν¯σµ)ac, (36.51) (Sµν R)˙a˙c=−i 4(¯σµσν−¯σνσµ)˙a˙c. (36.52) From these formulae, and the deﬁnition of γµ, eq.(36.7), we can see that i 4[γµ,γν] =/parenleftigg+(Sµν L)ac0 0 −(Sµν R)˙a˙c/parenrightigg ≡Sµν. (36.53) Then, for either a Dirac or Majorana ﬁeld Ψ, we can write U(Λ)−1Ψ(x)U(Λ) =D(Λ)Ψ(Λ−1x), (36.54) where, for an inﬁnitesimal transformation, the 4 ×4 matrixD(Λ) is D(1+δω) = 1 +i 2δωµνSµν, (36.55) withSµνgiven by eq.(36.53). The minus sign in front of Sµν Rin eq.(36.53) is compensated by the switch from a˙c˙ccontraction in eq.(36.50) to a ˙c˙c contraction in eq.(36.54). Problems 36.1) Using the results of problem 2.9, show that, for a rotat ion by an angle θabout thezaxis, we have D(Λ) = exp( −iθS12), (36.56) and that, for a boost by rapidity ηin thezdirection, we have D(Λ) = exp(+ iηS30). (36.57) 36.2) Verify that eq.(36.46) is consistent with eq.(36.43) . 36.3) a) Prove the Fierz identities (χ† 1¯σµχ2)(χ† 3¯σµχ4) =−2(χ† 1χ† 3)(χ2χ4), (36.58) (χ† 1¯σµχ2)(χ† 3¯σµχ4) = (χ† 1¯σµχ4)(χ† 3¯σµχ2). (36.59) b) Deﬁne the Dirac ﬁelds Ψi≡/parenleftiggχi ξ† i/parenrightigg ,ΨC i≡/parenleftiggξi χ† i/parenrightigg . (36.60) 36: Lagrangians for Spinor Fields 234 Use eqs.(36.58) and (36.59) to prove the Dirac form of the Fie rz identities, (Ψ1γµPLΨ2)(Ψ3γµPLΨ4) =−2(Ψ1PRΨC 3)(ΨC 4PLΨ2),(36.61) (Ψ1γµPLΨ2)(Ψ3γµPLΨ4) = (Ψ1γµPLΨ4)(Ψ3γµPLΨ2).(36.62) c) By writing both sides out in terms of Weyl ﬁelds, show that Ψ1γµPRΨ2=−ΨC 2γµPLΨC 1, (36.63) Ψ1PLΨ2= +ΨC 2PLΨC 1, (36.64) Ψ1PRΨ2= +ΨC 2PRΨC 1. (36.65) Combining eqs.(36.63–36.65) with eqs.(36.61–36.62) yiel ds more use- ful forms of the Fierz identities. 36.4) Consider a ﬁeld ϕA(x) in an unspeciﬁed representation of the Lorentz group, indexed by A, that obeys U(Λ)−1ϕA(x)U(Λ) =LAB(Λ)ϕB(Λ−1x). (36.66) For an inﬁnitesimal transformation, LAB(1+δω) =δAB+i 2δωµν(Sµν)AB. (36.67) a) Following the procedure of section 22, show that the energ y-momentum tensor is Tµν=gµνL −∂L ∂(∂µϕA)∂νϕA. (36.68) b) Show that the Noether current corresponding to a Lorentz t rans- formation is Mµνρ=xνTµρ−xρTµν+Bµνρ, (36.69) where Bµνρ≡ −i∂L ∂(∂µϕA)(Sνρ)ABϕB. (36.70) c) Use the conservation laws ∂µTµν= 0 and∂µMµνρ= 0 to show that Tνρ−Tρν+∂µBµνρ= 0. (36.71) d) Deﬁne the improved energy-momentum tensor orBelinfante tensor Θµν≡Tµν+1 2∂ρ(Bρµν−Bµρν−Bνρµ). (36.72) 36: Lagrangians for Spinor Fields 235 Show that Θµνis symmetric: Θµν= Θνµ. Also show that Θµνis conserved, ∂µΘµν= 0, and that/integraltextd3xΘ0ν=/integraltextd3xT0ν=Pν, where Pνis the energy-momentum four-vector. (In general relativit y, it is the Belinfante tensor that couples to gravity.) e) Show that the improved tensor Ξµνρ≡xνΘµρ−xρΘµν(36.73) obeys∂µΞµνρ= 0, and that/integraltextd3xΞ0νρ=/integraltextd3xM0νρ=Mνρ, where Mνρare the Lorentz generators. f) Compute Θµνfor a left-handed Weyl ﬁeld with Lgiven by eq.(36.2), and for a Dirac ﬁeld with Lgiven by eq.(36.28). 36.5)Symmetries of fermion ﬁelds. (Prerequisite: 24.) Consider a theory withNmassless Weyl ﬁelds ψj, L=iψ† jσµ∂µψj, (36.74) where the repeated index jis summed. This lagrangian is clearly invariant under the U( N) transformation, ψj→Ujkψk, (36.75) whereUis a unitary matrix. State the invariance group for the following cases: a)NWeyl ﬁelds with a common mass m, L=iψ† jσµ∂µψj−1 2m(ψjψj+ψ† jψ† j). (36.76) b)Nmassless Majorana ﬁelds, L=i 2ΨT jCγµ∂µΨj. (36.77) c)NMajorana ﬁelds with a common mass m, L=i 2ΨT jCγµ∂µΨj−1 2mΨT jCΨj. (36.78) d)Nmassless Dirac ﬁelds, L=iΨjγµ∂µΨj. (36.79) e)NDirac ﬁelds with a common mass m, L=iΨjγµ∂µΨj−mΨjΨj. (36.80) 37: Canonical Quantization of Spinor Fields I 236 37Canonical Quantization of Spinor Fields I Prerequisite: 36 Consider a left-handed Weyl ﬁeld ψwith lagrangian L=iψ†¯σµ∂µψ−1 2m(ψψ+ψ†ψ†). (37.1) The canonically conjugate momentum to the ﬁeld ψa(x) is then1 πa(x)≡∂L ∂(∂0ψa(x)) =iψ† ˙a(x)¯σ0˙aa. (37.2) The hamiltonian is H=πa∂0ψa− L =iψ† ˙a¯σ0˙aaψa− L =−iψ†¯σi∂iψ+1 2m(ψψ+ψ†ψ†). (37.3) The appropriate canonical anticommutation relations are {ψa(x,t),ψc(y,t)}= 0, (37.4) {ψa(x,t),πc(y,t)}=iδacδ3(x−y). (37.5) Substituting in eq.(37.2) for πc, we get {ψa(x,t),ψ† ˙c(y,t)}¯σ0˙cc=δacδ3(x−y). (37.6) Then, using ¯ σ0=σ0=I, we have {ψa(x,t),ψ† ˙c(y,t)}=σ0 a˙cδ3(x−y), (37.7) or, equivalently, {ψa(x,t),ψ†˙c(y,t)}= ¯σ0˙caδ3(x−y). (37.8) We can also translate this into four-component notation for either a Dirac or a Majorana ﬁeld. A Dirac ﬁeld is deﬁned in terms of two left- handed Weyl ﬁelds χandξvia Ψ≡/parenleftiggχc ξ†˙c/parenrightigg . (37.9) 1Here we gloss over a subtlety about diﬀerentiating with resp ect to an anticommuting object; we will take up this topic in section 44, and for now si mply assume that eq.(37.2) is correct. 37: Canonical Quantization of Spinor Fields I 237 We also deﬁne Ψ≡Ψ†β= (ξa, χ† ˙a), (37.10) where β≡/parenleftigg0δ˙a˙c δac0/parenrightigg . (37.11) The lagrangian is L=iχ†¯σµ∂µχ+iξ†¯σµ∂µξ−m(χξ+ξ†χ†) =iΨγµ∂µΨ−mΨΨ. (37.12) The ﬁeldsχandξeach obey the canonical anticommutation relations of eq.(37.5). This translates into {Ψα(x,t),Ψβ(y,t)}= 0, (37.13) {Ψα(x,t),Ψβ(y,t)}= (γ0)αβδ3(x−y), (37.14) whereαandβare four-component spinor indices, and γµ≡/parenleftigg0σµ a˙c ¯σµ˙ac0/parenrightigg . (37.15) Eqs.(37.13) and (37.14) can also be derived directly from th e four-component form of the lagrangian, eq.(37.12), by noting that the canon ically conjugate momentum to the ﬁeld Ψ is ∂L/∂(∂0Ψ) =iΨγ0, and that ( γ0)2= 1. A Majorana ﬁeld is deﬁned in terms of a single left-handed Wey l ﬁeld ψvia Ψ≡/parenleftiggψc ψ†˙c/parenrightigg . (37.16) We also deﬁne Ψ≡Ψ†β= (ψa, ψ† ˙a). (37.17) A Majorana ﬁeld obeys the Majorana condition Ψ = ΨTC, (37.18) where C ≡/parenleftigg−εac0 0−ε˙a˙c/parenrightigg (37.19) is the charge conjugation matrix. The lagrangian is L=iψ†¯σµ∂µψ−1 2m(ψψ+ψ†ψ†) =i 2Ψγµ∂µΨ−1 2mΨΨ =i 2ΨTCγµ∂µΨ−1 2mΨTCΨ. (37.20) 37: Canonical Quantization of Spinor Fields I 238 The ﬁeldψobeys the canonical anticommutation relations of eq.(37.5 ). This translates into {Ψα(x,t),Ψβ(y,t)}= (Cγ0)αβδ3(x−y), (37.21) {Ψα(x,t),Ψβ(y,t)}= (γ0)αβδ3(x−y), (37.22) whereαandβare four-component spinor indices. To derive eqs.(37.21) a nd (37.22) directly from the four-component form of the lagran gian, eq.(37.20), requires new formalism for the quantization of constrained systems . This is because the canonically conjugate momentum to the ﬁeld Ψ is ∂L/∂(∂0Ψ) = i 2ΨTCγ0, and this is linearly related to Ψ itself; this relation cons titutes a constraint that must be solved before imposition of the anti commutation relations. In this case, solving the constraint simply retu rns us to the Weyl formalism with which we began. The equation of motion that follows from either eq.(37.12) o r eq.(37.20) is the Dirac equation, (−i/∂+m)Ψ = 0. (37.23) Here we have introduced the Feynman slash : given any four-vector aµ, we deﬁne /a≡aµγµ. (37.24) To solve the Dirac equation, we ﬁrst note that if we act on it wi th i/∂+m, we get 0 = (i/∂+m)(−i/∂+m)Ψ = (/∂/∂+m2)Ψ = (−∂2+m2)Ψ. (37.25) Here we have used /a/a=aµaνγµγν =aµaν/parenleftig 1 2{γµ,γν}+1 2[γµ,γν]/parenrightig =aµaν/parenleftig −gµν+1 2[γµ,γν]/parenrightig =−aµaνgµν+ 0 =−a2. (37.26) From eq.(37.25), we see that Ψ obeys the Klein-Gordon equati on. There- fore, the Dirac equation has plane-wave solutions. Let us co nsider a speciﬁc solution of the form Ψ(x) =u(p)eipx+v(p)e−ipx. (37.27) 37: Canonical Quantization of Spinor Fields I 239 wherep0=ω≡(p2+m2)1/2, andu(p) andv(p) are four-component constant spinors. Plugging eq.(37.27) into the eq.(37.23) , we get (/p+m)u(p)eipx+ (−/p+m)v(p)e−ipx= 0. (37.28) Thus we require (/p+m)u(p) = 0, (−/p+m)v(p) = 0. (37.29) Each of these equations has two linearly independent soluti ons that we will callu±(p) andv±(p); their detailed properties will be worked out in the next section. The general solution of the Dirac equation can then be written as Ψ(x) =/summationdisplay s=±/integraldisplay /tildewiderdp/bracketleftig bs(p)us(p)eipx+d† s(p)vs(p)e−ipx/bracketrightig , (37.30) where the integration measure is as usual /tildewiderdp≡d3p (2π)32ω. (37.31) Problems 37.1) Verify that eqs.(37.13) and (37.14) follow from eqs.( 37.4) and (37.5). 38: Spinor Technology 240 38Spinor Technology Prerequisite: 37 The four-component spinors us(p) andvs(p) obey the equations (/p+m)us(p) = 0, (−/p+m)vs(p) = 0. (38.1) Each of these equations has two solutions, which we label via s= + and s=−. Form∝ne}ationslash= 0, we can go to the rest frame, p=0. We will then distinguish the two solutions by the eigenvalue of the spin m atrix Sz=i 4[γ1,γ2] =i 2γ1γ2=/parenleftigg1 2σ30 01 2σ3/parenrightigg . (38.2) Speciﬁcally, we will require Szu±(0) =±1 2u±(0), Szv±(0) =∓1 2v±(0). (38.3) The reason for the opposite sign for the vspinor is that this choice results in [Jz,b† ±(0)] =±1 2b† ±(0), [Jz,d† ±(0)] =±1 2d† ±(0), (38.4) whereJzis thezcomponent of the angular momentum operator. Eq.(38.4) implies that b† +(0) andd† +(0) each creates a particle with spin up along the zaxis. We will verify eq.(38.4) in problem 39.2. Forp=0, we have /p=−mγ0, where γ0=/parenleftigg0I I0/parenrightigg . (38.5) Eqs.(38.1) and (38.3) are then easy to solve. Choosing (for l ater conve- nience) a speciﬁc normalization and phase for each of u±(0) andv±(0), we get u+(0) =√m 1 0 1 0 , u−(0) =√m 0 1 0 1 , v+(0) =√m 0 1 0 −1 , v−(0) =√m −1 0 1 0 . (38.6) 38: Spinor Technology 241 For later use we also compute the barred spinors us(p)≡u† s(p)β , vs(p)≡v† s(p)β , (38.7) where β=/parenleftigg0I I0/parenrightigg (38.8) satisﬁes βT=β†=β−1=β. (38.9) We get u+(0) =√m(1,0,1,0), u−(0) =√m(0,1,0,1), v+(0) =√m(0,−1,0,1), v−(0) =√m(1,0,−1,0). (38.10) We can now ﬁnd the spinors corresponding to an arbitrary thre e-momentum pby applying to us(0) andvs(0) the matrix D(Λ) that corresponds to an appropriate boost. This is given by D(Λ) = exp(iηˆp·K), (38.11) where ˆpis a unit vector in the pdirection,Kj=i 4[γj,γ0] =i 2γjγ0is the boost matrix, and η≡sinh−1(|p|/m) is the rapidity (see problem 2.9). Thus we have us(p) = exp(iηˆp·K)us(0), vs(p) = exp(iηˆp·K)vs(0). (38.12) We also have us(p) =us(0)exp(−iηˆp·K), vs(p) =vs(0)exp(−iηˆp·K). (38.13) This follows from Kj=Kj, where for any general combination of gamma matrices, A≡βA†β. (38.14) In particular, it turns out that γµ=γµ, Sµν=Sµν, iγ5=iγ5, γµγ5=γµγ5, iγ5Sµν=iγ5Sµν. (38.15) 38: Spinor Technology 242 The barred spinors satisfy the equations us(p)(/p+m) = 0, vs(p)(−/p+m) = 0. (38.16) It is not very hard to work out us(p) andvs(p) from eq.(38.12), but it is even easier to use various tricks that will sidestep any need for the explicit formulae. Consider, for example, us′(p)us(p); from eqs.(38.12) and (38.13), we see that us′(p)us(p) =us′(0)us(0), and this is easy to compute from eqs.(38.6) and (38.10). We ﬁnd us′(p)us(p) = +2mδs′s, vs′(p)vs(p) =−2mδs′s, us′(p)vs(p) = 0, vs′(p)us(p) = 0. (38.17) Also useful are the Gordon identities , 2mus′(p′)γµus(p) =us′(p′)/bracketleftig (p′+p)µ−2iSµν(p′−p)ν/bracketrightig us(p), −2mvs′(p′)γµvs(p) =vs′(p′)/bracketleftig (p′+p)µ−2iSµν(p′−p)ν/bracketrightig vs(p).(38.18) To derive them, start with γµ/p=1 2{γµ,/p}+1 2[γµ,/p] =−pµ−2iSµνpν, (38.19) /p′γµ=1 2{γµ,/p′} −1 2[γµ,/p′] =−p′µ+ 2iSµνp′ ν. (38.20) Add eqs.(38.19) and (38.20), sandwich them between u′anduspinors (or v′andvspinors), and use eqs.(38.1) and (38.16). An important spec ial case isp′=p; then, using eq.(38.17), we ﬁnd us′(p)γµus(p) = 2pµδs′s, vs′(p)γµvs(p) = 2pµδs′s. (38.21) With a little more eﬀort, we can also show us′(p)γ0vs(−p) = 0, vs′(p)γ0us(−p) = 0. (38.22) We will need eqs.(38.21) and (38.22) in the next section. Consider now the spin sums/summationtext s=±us(p)us(p) and/summationtext s=±vs(p)vs(p), each of which is a 4 ×4 matrix. The sum over eigenstates of Szshould remove any memory of the spin-quantization axis, and so the r esult should 38: Spinor Technology 243 be expressible in terms of the four-vector pµand various gamma matrices, with all vector indices contracted. In the rest frame, / p=−mγ0, and it is easy to check that/summationtext s=±us(0)us(0) =mγ0+mand/summationtext s=±vs(0)vs(0) = mγ0−m. We therefore conclude that /summationdisplay s=±us(p)us(p) =−/p+m, /summationdisplay s=±vs(p)vs(p) =−/p−m. (38.23) We will make extensive use of eq.(38.23) when we calculate sc attering cross sections for spin-one-half particles. From eq.(38.23), we can get u+(p)u+(p), etc, by applying appropriate spin projection matrices. In the rest frame, we have 1 2(1 + 2sSz)us′(0) =δss′us′(0), 1 2(1−2sSz)vs′(0) =δss′vs′(0). (38.24) In order to boost these projection matrices to a more general frame, we ﬁrst recall that γ5≡iγ0γ1γ2γ3=/parenleftigg−I0 0I/parenrightigg . (38.25) This allows us to write Sz=i 2γ1γ2asSz=−1 2γ5γ3γ0. In the rest frame, we can write γ0as−/p/m, andγ3as /z, wherezµ= (0,ˆz); thus we have Sz=1 2mγ5/z/p. (38.26) Now we can boost Szto any other frame simply by replacing / zand /pwith their values in that frame. (Note that, in any frame, zµsatisﬁesz2= 1 and z·p= 0.) Boosting eq.(38.24) then yields 1 2(1−sγ5/z)us′(p) =δss′us′(p), 1 2(1−sγ5/z)vs′(p) =δss′vs′(p), (38.27) where we have used eq.(38.1) to eliminate / p. Combining eqs.(38.23) and (38.27) we get us(p)us(p) =1 2(1−sγ5/z)(−/p+m), vs(p)vs(p) =1 2(1−sγ5/z)(−/p−m). (38.28) It is interesting to consider the extreme relativistic limi t of this formula. Let us take the three-momentum to be in the zdirection, so that it is parallel to the spin-quantization axis. The component of th e spin in the 38: Spinor Technology 244 direction of the three-momentum is called the helicity . A fermion with helicity +1/2 is said to be right-handed , and a fermion with helicity −1/2 is said to be left-handed . For rapidity η, we have 1 mpµ= (coshη,0,0,sinhη), zµ= (sinhη,0,0,coshη). (38.29) The ﬁrst equation is simply the deﬁnition of η, and the second follows from z2= 1 andp·z= 0 (along with the knowledge that a boost of a four-vector in thezdirection does not change its xandycomponents). In the limit of largeη, we see that zµ=1 mpµ+O(e−η). (38.30) Hence, in eq.(38.28), we can replace / zwith /p/m, and then use the matrix relation (/p/m)(−/p±m) =∓(−/p±m), which holds for p2=−m2. For consistency, we should then also drop the mrelative to /p, since it is down by a factor of O(e−η). We get us(p)us(p)→1 2(1 +sγ5)(−/p), vs(p)vs(p)→1 2(1−sγ5)(−/p). (38.31) The spinor corresponding to a right-handed fermion (helici ty +1/2) is u+(p) for ab-type particle and v−(p) for ad-type particle. According to eq.(38.31), either of these is projected by1 2(1 +γ5) = diag(0,0,1,1) onto the lower two components only. In terms of the Dirac ﬁeld Ψ(x), this is the part that corresponds to the right-handed Weyl ﬁeld. S imilarly, left- handed fermions are projected (in the extreme relativistic limit) onto the upper two spinor components only, corresponding to the left -handed Weyl ﬁeld. The case of a massless particle follows from the extreme rela tivistic limit of a massive particle. In particular, eqs.(38.1), (38.16), (38.17), (38.21), (38.22), and (38.23) are all valid with m= 0, and eq.(38.31) becomes exact. Finally, for our discussion of parity, time reversal, and ch arge conjuga- tion in section 40, we will need a number of relationships amo ng theuand vspinors. First, note that βus(0) = +us(0) andβvs(0) =−vs(0). Also, βKj=−Kjβ. We then have us(−p) = +βus(p), vs(−p) =−βvs(p). (38.32) Next, we need the charge conjugation matrix C= 0−1 0 0 +1 0 0 0 0 0 0 +1 0 0 −1 0 . (38.33) 38: Spinor Technology 245 which obeys CT=C†=C−1=−C, (38.34) βC=−Cβ , (38.35) C−1γµC=−(γµ)T. (38.36) Using eqs.(38.6), (38.10), and (38.33), we can show that Cus(0)T=vs(0) andCvs(0)T=us(0). Also, eq.(38.36) implies C−1KjC=−(Kj)T. From this we can conclude that Cus(p)T=vs(p), Cvs(p)T=us(p). (38.37) Taking the complex conjugate of eq.(38.37), and using uT∗=u†=βu, we get u∗ s(p) =Cβvs(p), v∗ s(p) =Cβus(p). (38.38) Next, note that γ5us(0) = +sv−s(0) andγ5vs(0) =−su−s(0), and that γ5Kj=Kjγ5. Therefore γ5us(p) = +sv−s(p), γ5vs(p) =−su−s(p). (38.39) Combining eqs.(38.32), (38.38), and (38.39) results in u∗ −s(−p) =−sCγ5us(p), v∗ −s(−p) =−sCγ5vs(p). (38.40) We will need eq.(38.32) in our discussion of parity, eq.(38. 37) in our dis- cussion of charge conjugation, and eq.(38.40) in our discus sion of time reversal. Problems 38.1) Use eq.(38.12) to compute us(p) andvs(p) explicity. Hint: show that the matrix 2 iˆ p·Khas eigenvalues ±1, and that, for any matrix Awith eigenvalues ±1,ecA= (coshc) + (sinhc)A, wherecis an arbitrary complex number. 38.2) Verify eq.(38.15). 38.3) Verify eq.(38.22). 38.4) Derive the Gordon identities us′(p′)/bracketleftig (p′+p)µ−2iSµν(p′−p)ν/bracketrightig γ5us(p) = 0, vs′(p′)/bracketleftig (p′+p)µ−2iSµν(p′−p)ν/bracketrightig γ5vs(p) = 0.(38.41) 39: Canonical Quantization of Spinor Fields II 246 39Canonical Quantization of Spinor Fields II Prerequisite: 38 A Dirac ﬁeld Ψ with lagrangian L=iΨ/∂Ψ−mΨΨ (39.1) obeys the canonical anticommutation relations {Ψα(x,t),Ψβ(y,t)}= 0, (39.2) {Ψα(x,t),Ψβ(y,t)}= (γ0)αβδ3(x−y), (39.3) and has the Dirac equation (−i/∂+m)Ψ = 0 (39.4) as its equation of motion. The general solution is Ψ(x) =/summationdisplay s=±/integraldisplay /tildewiderdp/bracketleftig bs(p)us(p)eipx+d† s(p)vs(p)e−ipx/bracketrightig , (39.5) wherebs(p) andd† s(p) are operators; the properties of the four-component spinorsus(p) andvs(p) were belabored in the previous section. Let us express bs(p) andd† s(p) in terms of Ψ( x) and Ψ(x). We begin with /integraldisplay d3xe−ipxΨ(x) =/summationdisplay s′=±/bracketleftig 1 2ωbs′(p)us′(p) +1 2ωe2iωtd† s′(−p)vs′(−p)/bracketrightig . (39.6) Next, multiply on the left by us(p)γ0, and useus(p)γ0us′(p) = 2ωδss′and us(p)γ0vs′(−p) = 0 from section 38. The result is bs(p) =/integraldisplay d3xe−ipxus(p)γ0Ψ(x). (39.7) Note thatbs(p) is time independent. To getb† s(p), take the hermitian conjugate of eq.(39.7), using /bracketleftig us(p)γ0Ψ(x)/bracketrightig†=us(p)γ0Ψ(x) =Ψ(x)γ0us(p) =Ψ(x)γ0us(p), (39.8) where, for any general combination of gamma matrices A, A≡βA†β. (39.9) 39: Canonical Quantization of Spinor Fields II 247 Thus we ﬁnd b† s(p) =/integraldisplay d3xeipxΨ(x)γ0us(p). (39.10) To extract d† s(p) from Ψ(x), we start with /integraldisplay d3xeipxΨ(x) =/summationdisplay s′=±/bracketleftig 1 2ωe−2iωtbs′(−p)us′(−p) +1 2ωd† s′(p)vs′(p)/bracketrightig . (39.11) Next, multiply on the left by vs(p)γ0, and usevs(p)γ0vs′(p) = 2ωδss′and vs(p)γ0us′(−p) = 0 from section 38. The result is d† s(p) =/integraldisplay d3xeipxvs(p)γ0Ψ(x). (39.12) To getds(p), take the hermitian conjugate of eq.(39.12), which yields ds(p) =/integraldisplay d3xe−ipxΨ(x)γ0vs(p). (39.13) Next, let us work out the anticommutation relations of the banddop- erators (and their hermitian conjugates). From eq.(39.2), it is immediately clear that {bs(p),bs′(p′)}= 0, {ds(p),ds′(p′)}= 0, {bs(p),d† s′(p′)}= 0, (39.14) because these involve only the anticommutator of Ψ with itse lf, and this vanishes. Of course, hermitian conjugation also yields {b† s(p),b† s′(p′)}= 0, {d† s(p),d† s′(p′)}= 0, {b† s(p),ds′(p′)}= 0. (39.15) Now consider {bs(p),b† s′(p′)}=/integraldisplay d3xd3ye−ipx+ip′yus(p)γ0{Ψ(x),Ψ(y)}γ0us′(p′) =/integraldisplay d3xe−i(p−p′)xus(p)γ0γ0γ0us′(p′) = (2π)3δ3(p−p′)us(p)γ0us′(p) = (2π)3δ3(p−p′)2ωδss′. (39.16) In the ﬁrst line, we are free to set x0=y0becausebs(p) andb† s′(p′) are actually time independent. In the third, we used ( γ0)2= 1, and in the fourth,us(p)γ0us′(p) = 2ωδss′. 39: Canonical Quantization of Spinor Fields II 248 Similarly, {d† s(p),ds′(p′)}=/integraldisplay d3xd3yeipx−ip′yvs(p)γ0{Ψ(x),Ψ(y)}γ0vs′(p′) =/integraldisplay d3xei(p−p′)xvs(p)γ0γ0γ0vs′(p′) = (2π)3δ3(p−p′)vs(p)γ0vs′(p) = (2π)3δ3(p−p′)2ωδss′. (39.17) And ﬁnally, {bs(p),ds′(p′)}=/integraldisplay d3xd3ye−ipx−ip′yus(p)γ0{Ψ(x),Ψ(y)}γ0vs′(p′) =/integraldisplay d3xe−i(p+p′)xus(p)γ0γ0γ0vs′(p′) = (2π)3δ3(p+p′)us(p)γ0vs′(−p) = 0. (39.18) According to the discussion in section 3, eqs.(39.14–39.18 ) are exactly what we need to describe the creation and annihilation of fermion s. In this case, we have two diﬀerent kinds: b-type andd-type, each with two possible spin states,s= + ands=−. Next, let us evaluate the hamiltonian H=/integraldisplay d3xΨ(−iγi∂i+m)Ψ (39.19) in terms of the banddoperators. We have (−iγi∂i+m)Ψ =/summationdisplay s=±/integraldisplay /tildewiderdp/parenleftig −iγi∂i+m/parenrightig/parenleftig bs(p)us(p)eipx +d† s(p)vs(p)e−ipx/parenrightig =/summationdisplay s=±/integraldisplay /tildewiderdp/bracketleftig bs(p)(+γipi+m)us(p)eipx +d† s(p)(−γipi+m)vs(p)e−ipx/bracketrightig =/summationdisplay s=±/integraldisplay /tildewiderdp/bracketleftig bs(p)(γ0ω)us(p)eipx +d† s(p)(−γ0ω)vs(p)e−ipx/bracketrightig . (39.20) Therefore H=/summationdisplay s,s′/integraldisplay /tildewiderdp/tildewiderdp′d3x/parenleftig b† s′(p′)us′(p′)e−ip′x+ds′(p′)vs′(p′)eip′x/parenrightig 39: Canonical Quantization of Spinor Fields II 249 ×ω/parenleftig bs(p)γ0us(p)eipx−d† s(p)γ0vs(p)e−ipx/parenrightig =/summationdisplay s,s′/integraldisplay /tildewiderdp/tildewiderdp′d3xω/bracketleftig b† s′(p′)bs(p)us′(p′)γ0us(p)e−i(p′−p)x −b† s′(p′)d† s(p)us′(p′)γ0vs(p)e−i(p′+p)x +ds′(p′)bs(p)vs′(p′)γ0us(p)e+i(p′+p)x −ds′(p′)d† s(p)vs′(p′)γ0vs(p)e+i(p′−p)x/bracketrightig =/summationdisplay s,s′/integraldisplay /tildewiderdp1 2/bracketleftig b† s′(p)bs(p)us′(p)γ0us(p) −b† s′(−p)d† s(p)us′(−p)γ0vs(p)e+2iωt +ds′(−p)bs(p)vs′(−p)γ0us(p)e−2iωt −ds′(p)d† s(p)vs′(p)γ0vs(p)/bracketrightig =/summationdisplay s/integraldisplay /tildewiderdp ω/bracketleftig b† s(p)bs(p)−ds(p)d† s(p)/bracketrightig . (39.21) Using eq.(39.17), we can rewrite this as H=/summationdisplay s=±/integraldisplay /tildewiderdp ω/bracketleftig b† s(p)bs(p) +d† s(p)ds(p)/bracketrightig −4E0V , (39.22) where E0=1 2(2π)−3/integraltextd3k ωis the zero-point energy per unit volume that we found for a real scalar ﬁeld in section 3, and V= (2π)3δ3(0) =/integraltextd3x is the volume of space. That the zero-point energy is negativ e rather than positive is characteristic of fermions; that it is larger in magnitude by a factor of four is due to the four types of particles that are as sociated with a Dirac ﬁeld. We can cancel oﬀ this constant energy by includin g a constant term Ω 0=−4E0in the original lagrangian density; from here on, we will assume that this has been done. The ground state of the hamiltonian (39.22) is the vacuum state |0∝an}b∇acket∇i}ht that is annihilated by every bs(p) andds(p), bs(p)|0∝an}b∇acket∇i}ht=ds(p)|0∝an}b∇acket∇i}ht= 0. (39.23) Then, we can interpret the b† s(p) operator as creating a b-type particle with momentum p, energyω= (p2+m2)1/2, and spinSz=1 2s, and the d† s(p) operator as creating a d-type particle with the same properties. The b-type andd-type particles are distinguished by the value of the charge Q=/integraltextd3xj0, wherejµ=ΨγµΨ is the Noether current associated with the invariance of Lunder the U(1) transformation Ψ →e−iαΨ,Ψ→e+iαΨ. 39: Canonical Quantization of Spinor Fields II 250 Following the same procedure that we used for the hamiltonia n, we can show that Q=/integraldisplay d3xΨγ0Ψ =/summationdisplay s=±/integraldisplay /tildewiderdp/bracketleftig b† s(p)bs(p) +ds(p)d† s(p)/bracketrightig =/summationdisplay s=±/integraldisplay /tildewiderdp/bracketleftig b† s(p)bs(p)−d† s(p)ds(p)/bracketrightig + constant .(39.24) Thus the conserved charge Qcounts the total number of b-type particles minus the total number of d-type particles. (We are free to shift the overall value ofQto remove the constant term, and so we shall.) In quantum electrodynamics, we will identify the b-type particles as electrons and the d-type particles as positrons. Now consider a Majorana ﬁeld Ψ with lagrangian L=i 2ΨTC/∂Ψ−1 2mΨTCΨ. (39.25) The equation of motion for Ψ is once again the Dirac equation, and so the general solution is once again given by eq.(39.5). However, Ψ must also obey the Majorana condition Ψ = CΨT. Starting from the barred form of eq.(39.5), Ψ(x) =/summationdisplay s=±/integraldisplay /tildewiderdp/bracketleftig b† s(p)us(p)e−ipx+ds(p)vs(p)eipx/bracketrightig , (39.26) we have CΨT(x) =/summationdisplay s=±/integraldisplay /tildewiderdp/bracketleftig b† s(p)CuT s(p)e−ipx+ds(p)CvT s(p)eipx/bracketrightig .(39.27) From section 38, we have Cus(p)T=vs(p), Cvs(p)T=us(p), (39.28) and so CΨT(x) =/summationdisplay s=±/integraldisplay /tildewiderdp/bracketleftig b† s(p)vs(p)e−ipx+ds(p)us(p)eipx/bracketrightig . (39.29) Comparing eqs.(39.5) and (39.29), we see that we will have Ψ = CΨTif ds(p) =bs(p). (39.30) 39: Canonical Quantization of Spinor Fields II 251 Thus a free Majorana ﬁeld can be written as Ψ(x) =/summationdisplay s=±/integraldisplay /tildewiderdp/bracketleftig bs(p)us(p)eipx+b† s(p)vs(p)e−ipx/bracketrightig . (39.31) The anticommutation relations for a Majorana ﬁeld, {Ψα(x,t),Ψβ(y,t)}= (Cγ0)αβδ3(x−y), (39.32) {Ψα(x,t),Ψβ(y,t)}= (γ0)αβδ3(x−y), (39.33) can be used to show that {bs(p),bs′(p′)}= 0, {bs(p),b† s′(p′)}= (2π)3δ3(p−p′)2ωδss′, (39.34) as we would expect. The hamiltonian for the Majorana ﬁeld Ψ is H=1 2/integraldisplay d3xΨTC(−iγi∂i+m)Ψ =1 2/integraldisplay d3xΨ(−iγi∂i+m)Ψ, (39.35) and we can work through the same manipulations that led to eq. (39.21); the only diﬀerences are an extra overall factor of one-half, and ds(p) =bs(p). Thus we get H=1 2/summationdisplay s=±/integraldisplay /tildewiderdp ω/bracketleftig b† s(p)bs(p)−bs(p)b† s(p)/bracketrightig . (39.36) Note that this would reduce to a constant if we tried to use com mutators rather than anticommutators in eq.(39.34), a reﬂection of t he spin-statistics theorem. Using eq.(39.34) as it is, we ﬁnd H=/summationdisplay s=±/integraldisplay /tildewiderdpωb† s(p)bs(p)−2E0V. (39.37) Again, we can (and will) cancel oﬀ the zero-point energy by in cluding a term Ω 0=−2E0in the original lagrangian density. The Majorana lagrangian has no U(1) symmetry. Thus there is n o associated charge, and only one kind of particle (with two po ssible spin states). Problems 39: Canonical Quantization of Spinor Fields II 252 39.1) Verify eq.(39.24). 39.2) Use [Ψ( x),Mµν] =−i(xµ∂ν−xν∂µ)Ψ(x) +SµνΨ(x), plus whatever spinor identities you need, to show that Jzb† s(pˆ z)|0∝an}b∇acket∇i}ht=1 2sb† s(pˆ z)|0∝an}b∇acket∇i}ht, Jzd† s(pˆ z)|0∝an}b∇acket∇i}ht=1 2sd† s(pˆ z)|0∝an}b∇acket∇i}ht, (39.38) where p=pˆ zis the three-momentum, and ˆ zis a unit vector in the z direction. 39.3) Show that U(Λ)−1b† s(p)U(Λ) =b† s(Λ−1p), U(Λ)−1d† s(p)U(Λ) =d† s(Λ−1p), (39.39) and hence that U(Λ)|p,s,q∝an}b∇acket∇i}ht=|Λp,s,q∝an}b∇acket∇i}ht, (39.40) where |p,s,+∝an}b∇acket∇i}ht ≡b† s(p)|0∝an}b∇acket∇i}ht, |p,s,−∝an}b∇acket∇i}ht ≡d† s(p)|0∝an}b∇acket∇i}ht (39.41) are single-particle states. 39.4)The spin-statistics theorem for spin-one-half particles. We will follow the proof for spin-zero particles in section 4. We start with bs(p) andb† s(p) as the fundamental objects; we take them to have either commutation ( −) or anticommutation (+) relations of the form [bs(p),bs′(p′)]∓= 0, [b† s(p),b† s′(p′)]∓= 0, [bs(p),b† s′(p′)]∓= (2π)32ωδ3(p−p′)δss′. (39.42) Deﬁne Ψ+(x)≡/summationdisplay s=±/integraldisplay /tildewiderdpbs(p)us(p)eipx, Ψ−(x)≡/summationdisplay s=±/integraldisplay /tildewiderdpb† s(p)vs(p)e−ipx. (39.43) a) Show that U(Λ)−1Ψ±(x)U(Λ) =D(Λ)Ψ±(Λ−1x). 39: Canonical Quantization of Spinor Fields II 253 b) Show that [Ψ+(x)]†= [Ψ−(x)]TCβ. Thus a hermitian interaction term in the lagrangian must involve both Ψ+(x) and Ψ−(x). c) Show that [Ψ+ α(x),Ψ− β(y)]∓∝ne}ationslash= 0 for (x−y)2>0. d) Show that [Ψ+ α(x),Ψ− β(y)]∓=−[Ψ+ β(y),Ψ− α(x)]∓for (x−y)2>0. e) Consider Ψ( x)≡Ψ+(x)+λΨ−(x), whereλis an arbitrary complex number, and evaluate both [Ψ α(x),Ψβ(y)]∓and [Ψα(x),Ψβ(y)]∓for (x−y)2>0. Show these can both vanish if and only if |λ|= 1 and we use anticommutators. 40: Parity, Time Reversal, and Charge Conjugation 254 40Parity, Time Reversal, and Charge Conjugation Prerequisite: 23, 39 Recall that, under a Lorentz transformation Λ implemented b y the unitary operatorU(Λ), a Dirac (or Majorana) ﬁeld transforms as U(Λ)−1Ψ(x)U(Λ) =D(Λ)Ψ(Λ−1x). (40.1) For an inﬁnitesimal transformation Λµν=δµν+δωµν, the matrix D(Λ) is given by D(1+δω) =I+i 2δωµνSµν, (40.2) where the Lorentz generator matrices are Sµν=i 4[γµ,γν]. (40.3) In this section, we will consider the two Lorentz transforma tions that cannot be reached via a sequence of inﬁnitesimal transformations a way from the identity: parity and time reversal. We begin with parity. Deﬁne the parity transformation Pµν= (P−1)µν= +1 −1 −1 −1 (40.4) and the corresponding unitary operator P≡U(P). (40.5) Now we have P−1Ψ(x)P=D(P)Ψ(Px). (40.6) The question we wish to answer is, what is the matrix D(P)? First of all, if we make a second parity transformation, we ge t P−2Ψ(x)P2=D(P)2Ψ(x), (40.7) and it is tempting to conclude that we should have D(P)2= 1, so that we return to the original ﬁeld. This is correct for scalar ﬁelds , since they are themselves observable. With fermions, however, it takes an even number of ﬁelds to construct an observable. Therefore we need only r equire the weaker condition D(P)2=±1. 40: Parity, Time Reversal, and Charge Conjugation 255 We will also require the particle creation and annihilation operators to transform in a simple way. Because P−1PP=−P, (40.8) P−1JP= +J, (40.9) where Pis the total three-momentum operator and Jis the total angu- lar momentum operator, a parity transformation should reve rse the three- momentum while leaving the spin direction unchanged. We the refore re- quire P−1b† s(p)P=ηb† s(−p), P−1d† s(p)P=ηd† s(−p), (40.10) whereηis a possible phase factor that (by the previous argument abo ut observables) should satisfy η2=±1. We could in principle assign diﬀerent phase factors to the banddoperators, but we choose them to be the same so that the parity transformation is compatible with the Major ana condition ds(p) =bs(p). Writing the mode expansion of the free ﬁeld Ψ(x) =/summationdisplay s=±/integraldisplay /tildewiderdp/bracketleftig bs(p)us(p)eipx+d† s(p)vs(p)e−ipx/bracketrightig , (40.11) the parity transformation reads P−1Ψ(x)P =/summationdisplay s=±/integraldisplay /tildewiderdp/bracketleftig/parenleftig P−1bs(p)P/parenrightig us(p)eipx+/parenleftig P−1d† s(p)P/parenrightig vs(p)e−ipx/bracketrightig =/summationdisplay s=±/integraldisplay /tildewiderdp/bracketleftig η∗bs(−p)us(p)eipx+ηd† s(−p)vs(p)e−ipx/bracketrightig =/summationdisplay s=±/integraldisplay /tildewiderdp/bracketleftig η∗bs(p)us(−p)eipPx+ηd† s(p)vs(−p)e−ipPx/bracketrightig .(40.12) In the last line, we have changed the integration variable fr ompto−p. We now use a result from section 38, namely that us(−p) = +βus(p), vs(−p) =−βvs(p), (40.13) where β=/parenleftigg0I I0/parenrightigg . (40.14) 40: Parity, Time Reversal, and Charge Conjugation 256 Then, if we choose η=−i, eq.(40.12) becomes P−1Ψ(x)P=/summationdisplay s=±/integraldisplay /tildewiderdp/bracketleftig ibs(p)βus(p)eipPx+id† s(p)βvs(p)e−ipPx/bracketrightig =iβΨ(Px). (40.15) Thus we see that D(P) =iβ. (We could also have chosen η=i, resulting inD(P) =−iβ; either choice is acceptable.) The factor of ihas an interesting physical consequence. Consider a state of an electron and positron with zero center-of-mass m omentum, |φ∝an}b∇acket∇i}ht=/integraldisplay /tildewiderdpφ(p)b† s(p)d† s′(−p)|0∝an}b∇acket∇i}ht; (40.16) hereφ(p) is the momentum-space wave function. Let us assume that the vacuum is parity invariant: P|0∝an}b∇acket∇i}ht=P−1|0∝an}b∇acket∇i}ht=|0∝an}b∇acket∇i}ht. Let us also assume that the wave function has deﬁnite parity: φ(−p) = (−)ℓφ(p). Then, applying the inverse parity operator on |φ∝an}b∇acket∇i}ht, we get P−1|φ∝an}b∇acket∇i}ht=/integraldisplay /tildewiderdpφ(p)/parenleftig P−1b† s(p)P/parenrightig (P−1d† s′(−p)P/parenrightig P−1|0∝an}b∇acket∇i}ht. = (−i)2/integraldisplay /tildewiderdpφ(p)b† s(−p)d† s′(p)|0∝an}b∇acket∇i}ht = (−i)2/integraldisplay /tildewiderdpφ(−p)b† s(p)d† s′(−p)|0∝an}b∇acket∇i}ht =−(−)ℓ|φ∝an}b∇acket∇i}ht. (40.17) Thus, the parity of this state is opposite to that of its wave f unction; an electron-positron pair has an intrinsic parity of−1. This also applies to a pair of Majorana fermions. This inﬂuences the selection rul es for fermion pair annihilation in theories that conserve parity. (A pair of electrons also has negative intrinsic parity, but this is less interesting because the electrons are prevented from annihilating by charge conservation.) Let us see what eq.(40.15) implies for the two Weyl ﬁelds that comprise the Dirac ﬁeld. Recalling that Ψ =/parenleftiggχa ξ†˙a/parenrightigg , (40.18) we see from eqs.(40.14) and (40.15) that P−1χa(x)P=iξ†˙a(Px), P−1ξ†˙a(x)P=iχa(Px). (40.19) 40: Parity, Time Reversal, and Charge Conjugation 257 Thus a parity transformation exchanges a left-handed ﬁeld f or a right- handed one. If we take the hermitian conjugate of eq.(40.19), then raise the index on one side while lowering it on the other (and remember that thi s introduces a relative minus sign!), we get P−1χ†˙a(x)P=iξa(Px), P−1ξa(x)P=iχ†˙a(Px). (40.20) Comparing eqs.(40.19) and (40.20), we see that they are comp atible with the Majorana condition χa(x) =ξa(x). Next we take up time reversal. Deﬁne the time-reversal trans formation Tµν= (T−1)µν= −1 +1 +1 +1 (40.21) and the corresponding operator T≡U(T). (40.22) Now we have T−1Ψ(x)T=D(T)Ψ(Tx). (40.23) The question we wish to answer is, what is the matrix D(T)? As with parity, we can conclude that D(T)2=±1, and we will require the particle creation and annihilation operators to transf orm in a simple way. Because T−1PT=−P, (40.24) T−1JT=−J, (40.25) where Pis the total three-momentum operator and Jis the total angu- lar momentum operator, a time-reversal transformation sho uld reverse the direction of both the three-momentum and the spin. We theref ore require T−1b† s(p)T=ζsb† −s(−p), T−1d† s(p)T=ζsd† −s(−p). (40.26) This time we allow for possible s-dependence of the phase factor. Also, we recall from section 23 that Tmust be an antiunitary operator, so that 40: Parity, Time Reversal, and Charge Conjugation 258 T−1iT=−i. Then we have T−1Ψ(x)T =/summationdisplay s=±/integraldisplay /tildewiderdp/bracketleftig/parenleftig T−1bs(p)T/parenrightig u∗ s(p)e−ipx+/parenleftig T−1d† s(p)T/parenrightig v∗ s(p)eipx/bracketrightig =/summationdisplay s=±/integraldisplay /tildewiderdp/bracketleftig ζ∗ sb−s(−p)u∗ s(p)e−ipx+ζsd† −s(−p)v∗ s(p)eipx/bracketrightig =/summationdisplay s=±/integraldisplay /tildewiderdp/bracketleftig ζ∗ −sbs(p)u∗ −s(−p)eipTx+ζ−sd† s(p)v∗ −s(−p)e−ipTx/bracketrightig . (40.27) In the last line, we have changed the integration variable fr ompto−p, and the summation variable from sto−s. We now use a result from section 38, namely that u∗ −s(−p) =−sCγ5us(p), v∗ −s(−p) =−sCγ5vs(p). (40.28) Then, if we choose ζs=s, eq.(40.27) becomes T−1Ψ(x)T=Cγ5Ψ(Tx). (40.29) Thus we see that D(T) =Cγ5. (We could also have chosen ζs=−s, resulting in D(T) =−Cγ5; either choice is acceptable.) As with parity, we can consider the eﬀect of time reversal on t he Weyl ﬁelds. Using eqs.(40.18), (40.29), C=/parenleftigg−εab0 0−ε˙a˙b/parenrightigg , (40.30) and γ5=/parenleftigg−δac0 0 +δ˙a˙c/parenrightigg , (40.31) we see that T−1χa(x)T= +χa(Tx), T−1ξ†˙a(x)T=−ξ† ˙a(Tx). (40.32) Thus left-handed Weyl ﬁelds transform into left-handed Wey l ﬁelds (and right-handed into right-handed) under time reversal. 40: Parity, Time Reversal, and Charge Conjugation 259 If we take the hermitian conjugate of eq.(40.32), then raise the index on one side while lowering it on the other (and remember that thi s introduces a relative minus sign!), we get T−1χ†˙a(x)T=−χ† ˙a(Tx), T−1ξa(x)T= +ξa(Tx). (40.33) Comparing eqs.(40.32) and (40.33), we see that they are comp atible with the Majorana condition χa(x) =ξa(x). It is interesting and important to evaluate the transformat ion properties of fermion bilinears of the form ΨAΨ, whereAis some combination of gamma matrices. We will consider A’s that satisfy A=A, whereA≡ βA†β; in this case, ΨAΨ is hermitian. Let us begin with parity transformations. From Ψ = Ψ†βand eq.(40.15) we get P−1Ψ(x)P=−iΨ(Px)β , (40.34) Combining eqs.(40.15) and (40.34) we ﬁnd P−1/parenleftig ΨAΨ/parenrightig P=Ψ/parenleftig βAβ/parenrightig Ψ, (40.35) where we have suppressed the spacetime arguments (which tra nsform in the obvious way). For various particular choices of Awe have β1β= +1, βiγ5β=−iγ5, βγ0β= +γ0, βγiβ=−γi, βγ0γ5β=−γ0γ5, βγiγ5β= +γiγ5. (40.36) Therefore, the corresponding hermitian bilinears transfo rm as P−1/parenleftig ΨΨ/parenrightig P= +ΨΨ, P−1/parenleftig Ψiγ5Ψ/parenrightig P=−Ψiγ5Ψ, P−1/parenleftig ΨγµΨ/parenrightig P= +PµνΨγνΨ, P−1/parenleftig Ψγµγ5Ψ/parenrightig P=−PµνΨγνγ5Ψ, (40.37) Thus we see that ΨΨ and ΨγµΨ are even under a parity transformation, while Ψiγ5Ψ and Ψγµγ5Ψ are odd. We say that ΨΨ is a scalar, ΨγµΨ 40: Parity, Time Reversal, and Charge Conjugation 260 is a vector or polar vector ,Ψiγ5Ψ is a pseudoscalar, and Ψγµγ5Ψ is a pseudovector oraxial vector . Turning to time reversal, from eq.(40.29) we get T−1Ψ(x)T=Ψ(Tx)γ5C−1. (40.38) Combining eqs.(40.29) and (40.38), along with T−1AT=A∗, we ﬁnd T−1/parenleftig ΨAΨ/parenrightig T=Ψ/parenleftig γ5C−1A∗Cγ5/parenrightig Ψ, (40.39) where we have suppressed the spacetime arguments (which tra nsform in the obvious way). Recall that C−1γµC=−(γµ)Tand that C−1γ5C=γ5. Also, γ0andγ5are real, hermitian, and square to one, while γiis antihermitian. Finally,γ5anticommutes with γµ. Using all of this info, we ﬁnd γ5C−11∗Cγ5= +1, γ5C−1(iγ5)∗Cγ5=−iγ5, γ5C−1(γ0)∗Cγ5= +γ0, γ5C−1(γi)∗Cγ5=−γi, γ5C−1(γ0γ5)∗Cγ5= +γ0γ5, γ5C−1(γiγ5)∗Cγ5=−γiγ5. (40.40) Therefore, T−1/parenleftig ΨΨ/parenrightig T= +ΨΨ, T−1/parenleftig Ψiγ5Ψ/parenrightig T=−Ψiγ5Ψ, T−1/parenleftig ΨγµΨ/parenrightig T=−TµνΨγνΨ, T−1/parenleftig Ψγµγ5Ψ/parenrightig T=−TµνΨγνγ5Ψ. (40.41) Thus we see that ΨΨ is even under time reversal, while Ψiγ5Ψ,ΨγµΨ, and Ψγµγ5Ψ are odd. For completeness we will also consider the transformation p roperties of bilinears under charge conjugation. Recall that C−1Ψ(x)C=CΨT(x), C−1Ψ(x)C= ΨT(x)C. (40.42) The bilinear ΨAΨ therefore transforms as C−1/parenleftig ΨAΨ/parenrightig C= ΨTCACΨT. (40.43) 40: Parity, Time Reversal, and Charge Conjugation 261 Since all indices are contracted, we can rewrite the right-h and side as its transpose, with an extra minus sign for exchanging the order of the two fermion ﬁelds. We get C−1/parenleftig ΨAΨ/parenrightig C=−ΨCTATCTΨ. (40.44) Recalling that CT=C−1=−C, we have C−1/parenleftig ΨAΨ/parenrightig C=Ψ/parenleftig C−1ATC/parenrightig Ψ. (40.45) Once again we can go through the list: C−11TC= +1, C−1(iγ5)TC= +iγ5, C−1(γµ)TC=−γµ, C−1(γµγ5)TC= +γµγ5. (40.46) Therefore, C−1/parenleftig ΨΨ/parenrightig C= +ΨΨ, C−1/parenleftig Ψiγ5Ψ/parenrightig C= +Ψiγ5Ψ, C−1/parenleftig ΨγµΨ/parenrightig C=−ΨγµΨ, C−1/parenleftig Ψγµγ5Ψ/parenrightig C= +Ψγµγ5Ψ. (40.47) Thus we see that ΨΨ,Ψiγ5Ψ, and Ψγµγ5Ψ are even under charge conju- gation, while ΨγµΨ is odd. For a Majorana ﬁeld, we have C−1ΨC= Ψ andC−1ΨC=Ψ; this impliesC−1(ΨAΨ)C=ΨAΨ for any combination of gamma matrices A. Since eq.(40.47) tells that C−1(ΨγµΨ)C=−ΨγµΨ for either a Dirac or Majorana ﬁeld, it must be that ΨγµΨ = 0 for a Majorana ﬁeld. Let us consider the combined eﬀects of the three transformat ions (C, P, andT) on the bilinears. From eqs.(40.37), (40.41), and (40.47), we have (CPT)−1/parenleftig ΨΨ/parenrightig CPT = +ΨΨ, (CPT)−1/parenleftig Ψiγ5Ψ/parenrightig CPT = +Ψiγ5Ψ, (CPT)−1/parenleftig ΨγµΨ/parenrightig CPT =−ΨγµΨ, (CPT)−1/parenleftig Ψγµγ5Ψ/parenrightig CPT =−Ψγµγ5Ψ, (40.48) where we have used PµνTνρ=−δµρ. We see that ΨΨ and Ψiγ5Ψ are both even under CPT, while ΨγµΨ and Ψγµγ5Ψ are both odd. These are (it 40: Parity, Time Reversal, and Charge Conjugation 262 turns out) examples of a more general rule: a fermion bilinea r withnvector indices (and no uncontracted spinor indices) is even (odd) u nderCPT ifn is even (odd). This also applies if we allow derivatives acti ng on the ﬁelds, since each component of ∂µis odd under the combination PTand even underC. For scalar and vector ﬁelds, it is always possible to choose t he phase factors in the C,P, andTtransformations so that, overall, they obey the same rule: a hermitian combination of ﬁelds and derivatives is even or odd depending on the total number of uncontracted vector indice s. Putting this together with our result for fermion bilinears, we see that a ny hermitian combination of any set of ﬁelds (scalar, vector, Dirac, Majo rana) and their derivatives that is a Lorentz scalar (and so carries no indic es) is even under CPT. Since the lagrangian must be formed out of such combination s, we haveL(x)→ L(−x) underCPT, and so the action S=/integraltextd4xLis invariant. This is the CPT theorem. Reference Notes A detailed treatment of CPT for ﬁelds of any spin is given in Weinberg I . Problems 40.1) Find the transformation properties of ΨSµνΨ and ΨiSµνγ5Ψ under P,T, andC. Verify that they are both even under CPT, as claimed. Do either or both vanish if Ψ is a Majorana ﬁeld? 41: LSZ Reduction for Spin-One-Half Particles 263 41LSZ Reduction for Spin-One-Half Particles Prerequisite: 5, 39 Let us now consider how to construct appropriate initial and ﬁnal states for scattering experiments. We will ﬁrst consider the case o f a Dirac ﬁeld Ψ, and assume that its interactions respect the U(1) symmetr y that gives rise to the conserved current jµ=ΨγµΨ and its associated charge Q. In the free theory, we can create a state of one particle by act ing on the vacuum state with a creation operator: |p,s,+∝an}b∇acket∇i}ht=b† s(p)|0∝an}b∇acket∇i}ht, (41.1) |p,s,−∝an}b∇acket∇i}ht=d† s(p)|0∝an}b∇acket∇i}ht, (41.2) where the label ±on the ket indicates the value of the U(1) charge Q, and b† s(p) =/integraldisplay d3xeipxΨ(x)γ0us(p), (41.3) d† s(p) =/integraldisplay d3xeipxvs(p)γ0Ψ(x). (41.4) Recall that b† s(p) andd† s(p) are time independent in the free theory. The states |p,s,±∝an}b∇acket∇i}hthave the Lorentz-invariant normalization ∝an}b∇acketle{tp,s,q|p′,s′,q′∝an}b∇acket∇i}ht= (2π)32ωδ3(p−p′)δss′δqq′, (41.5) whereω= (p2+m2)1/2. Let us consider an operator that (in the free theory) creates a particle with deﬁnite spin and charge, localized in momentum space ne arp1, and localized in position space near the origin: b† 1≡/integraldisplay d3pf1(p)b† s1(p), (41.6) where f1(p)∝exp[−(p−p1)2/4σ2] (41.7) is an appropriate wave packet, and σis its width in momentum space. If we time evolve (in the Schr¨ odinger picture) the state creat ed by this time- independent operator, then the wave packet will propagate ( and spread out). The particle will thus be localized far from the origin ast→ ±∞ . If we consider instead an initial state of the form |i∝an}b∇acket∇i}ht=b† 1b† 2|0∝an}b∇acket∇i}ht, where p1∝ne}ationslash=p2, then we have two particles that are widely separated in the fa r past. Let us guess that this still works in the interacting theory. One compli- cation is that b† s(p) will no longer be time independent, and so b† 1, eq.(41.6), 41: LSZ Reduction for Spin-One-Half Particles 264 becomes time dependent as well. Our guess for a suitable init ial state for a scattering experiment is then |i∝an}b∇acket∇i}ht= lim t→−∞b† 1(t)b† 2(t)|0∝an}b∇acket∇i}ht. (41.8) By appropriately normalizing the wave packets, we can make ∝an}b∇acketle{ti|i∝an}b∇acket∇i}ht= 1, and we will assume that this is the case. Similarly, we can consid er a ﬁnal state |f∝an}b∇acket∇i}ht= lim t→+∞b† 1′(t)b† 2′(t)|0∝an}b∇acket∇i}ht, (41.9) where p′ 1∝ne}ationslash=p′ 2, and∝an}b∇acketle{tf|f∝an}b∇acket∇i}ht= 1. This describes two widely separated par- ticles in the far future. (We could also consider acting with more creation operators, if we are interested in the production of some ext ra particles in the collision of two, or using d†operators instead of b†operators for some or all of the initial and ﬁnal particles.) Now the scattering amplitude is simply given by ∝an}b∇acketle{tf|i∝an}b∇acket∇i}ht. We need to ﬁnd a more useful expression for ∝an}b∇acketle{tf|i∝an}b∇acket∇i}ht. To this end, let us note that b† 1(−∞)−b† 1(+∞) =−/integraldisplay+∞ −∞dt∂0b† 1(t) =−/integraldisplay d3pf1(p)/integraldisplay d4x∂0/parenleftig eipxΨ(x)γ0us1(p)/parenrightig . =−/integraldisplay d3pf1(p)/integraldisplay d4xΨ(x)/parenleftig γ0←∂0−iγ0p0/parenrightig us1(p)eipx =−/integraldisplay d3pf1(p)/integraldisplay d4xΨ(x)/parenleftig γ0←∂0−iγipi−im/parenrightig us1(p)eipx =−/integraldisplay d3pf1(p)/integraldisplay d4xΨ(x)/parenleftig γ0← ∂0−γi→ ∂i−im/parenrightig us1(p)eipx =−/integraldisplay d3pf1(p)/integraldisplay d4xΨ(x)/parenleftig γ0← ∂0+γi← ∂i−im/parenrightig us1(p)eipx =i/integraldisplay d3pf1(p)/integraldisplay d4xΨ(x)(+i← /∂+m)us1(p)eipx. (41.10) The ﬁrst equality is just the fundamental theorem of calculu s. To get the second, we substituted the deﬁnition of b† 1(t), and combined the d3xfrom this deﬁnition with the dtto getd4x. The third comes from straightforward evaluation of the time derivatives. The fourth uses (/ p+m)us(p) = 0. The ﬁfth writes ipias∂iacting oneipx. The sixth uses integration by parts to move the∂ionto the ﬁeld Ψ(x); here the wave packet is needed to avoid a surface term. The seventh simply identiﬁes γ0∂0+γi∂ias /∂. In free-ﬁeld theory, the right-hand side of eq.(41.10) is ze ro, since Ψ( x) obeys the Dirac equation, which, after barring it, reads Ψ(x)(+i← /∂+m) = 0. (41.11) 41: LSZ Reduction for Spin-One-Half Particles 265 In an interacting theory, however, the right-hand side of eq .(41.10) will not be zero. We will also need the hermitian conjugate of eq.(41.10), whi ch (after some slight rearranging) reads b1(+∞)−b1(−∞) =i/integraldisplay d3pf1(p)/integraldisplay d4xe−ipxus1(p)(−i/∂+m)Ψ(x),(41.12) and the analogous formulae for the doperators, d† 1(−∞)−d† 1(+∞) =−i/integraldisplay d3pf1(p)/integraldisplay d4xeipxvs1(p)(−i/∂+m)Ψ(x),(41.13) d1(+∞)−d1(−∞) =−i/integraldisplay d3pf1(p)/integraldisplay d4xΨ(x)(+i← /∂+m)vs1(p)e−ipx.(41.14) Let us now return to the scattering amplitude we were conside ring, ∝an}b∇acketle{tf|i∝an}b∇acket∇i}ht=∝an}b∇acketle{t0|b2′(+∞)b1′(+∞)b† 1(−∞)b† 2(−∞)|0∝an}b∇acket∇i}ht. (41.15) Note that the operators are in time order. Thus, if we feel lik e it, we can put in a time-ordering symbol without changing anything: ∝an}b∇acketle{tf|i∝an}b∇acket∇i}ht=∝an}b∇acketle{t0|Tb2′(+∞)b1′(+∞)b† 1(−∞)b† 2(−∞)|0∝an}b∇acket∇i}ht. (41.16) The symbol T means the product of operators to its right is to b e ordered, not as written, but with operators at later times to the left o f those at earlier times. However, there is an extra minus sign if this rearrangement involves an odd number of exchanges of these anticommuting o perators . Now let us use eqs.(41.10) and (41.12) in eq.(41.16). The tim e-ordering symbol automatically moves all bi′(−∞)’s to the right, where they anni- hilate |0∝an}b∇acket∇i}ht. Similarly, all b† i(+∞)’s move to the left, where they annihilate ∝an}b∇acketle{t0|. The wave packets no longer play a key role, and we can take the σ→0 limit in eq.(41.7), so that f1(p) =δ3(p−p1). The initial and ﬁnal states now have a delta-function normalization, the multiparticl e generalization of eq.(41.5). We are left with the Lehmann-Symanzik-Zimmermann reduction formula for spin-one-half particles, ∝an}b∇acketle{tf|i∝an}b∇acket∇i}ht=i4/integraldisplay d4x1d4x2d4x1′d4x2′ ×e−ip′ 1x′ 1[us1′(p1′)(−i/∂1′+m)]α1′ 41: LSZ Reduction for Spin-One-Half Particles 266 ×e−ip′ 2x′ 2[us2′(p2′)(−i/∂2′+m)]α2′ × ∝an}b∇acketle{t0|T Ψα2′(x2′)Ψα1′(x1′)Ψα1(x1)Ψα2(x2)|0∝an}b∇acket∇i}ht ×[(+i← /∂1+m)us1(p1)]α1eip1x1 ×[(+i← /∂2+m)us2(p2)]α2eip2x2. (41.17) The generalization of the LSZ formula to other processes sho uld be clear; insert a time-ordering symbol, and make the following repla cements: b† s(p)in→+i/integraldisplay d4xΨ(x)(+i← /∂+m)us(p)e+ipx, (41.18) bs(p)out→+i/integraldisplay d4xe−ipxus(p)(−i/∂+m)Ψ(x), (41.19) d† s(p)in→ −i/integraldisplay d4xe+ipxvs(p)(−i/∂+m)Ψ(x), (41.20) ds(p)out→ −i/integraldisplay d4xΨ(x)(+i← /∂+m)vs(p)e−ipx, (41.21) where we have used the subscripts “in” and “out” to denote t→ −∞ and t→+∞, respectively. All of this holds for a Majorana ﬁeld as well. In that case, ds(p) =bs(p), and we can use either eq.(41.18) oreq.(41.20) for the incoming particles, andeither eq.(41.19) oreq.(41.21) for the outgoing particles, whichever is more convenient. The Majorana condition Ψ = ΨTCguarantees that the results will be equivalent. As in the case of a scalar ﬁeld, we cheated a little in our deriv ation of the LSZ formula, because we assumed that the creation oper ators of freeﬁeld theory would work comparably in the interacting theory. After performing an analysis that is entirely analogous to what we did for the scalar in section 5, we come to the same conclusion: the LSZ fo rmula holds provided the ﬁeld is properly normalized. For a Dirac ﬁeld, w e must require ∝an}b∇acketle{t0|Ψ(x)|0∝an}b∇acket∇i}ht= 0, (41.22) ∝an}b∇acketle{tp,s,+|Ψ(x)|0∝an}b∇acket∇i}ht= 0, (41.23) ∝an}b∇acketle{tp,s,−|Ψ(x)|0∝an}b∇acket∇i}ht=vs(p)e−ipx, (41.24) ∝an}b∇acketle{tp,s,+|Ψ(x)|0∝an}b∇acket∇i}ht=us(p)e−ipx, (41.25) ∝an}b∇acketle{tp,s,−|Ψ(x)|0∝an}b∇acket∇i}ht= 0, (41.26) where ∝an}b∇acketle{t0|0∝an}b∇acket∇i}ht= 1, and the one-particle states are normalized according to eq.(41.5). 41: LSZ Reduction for Spin-One-Half Particles 267 The zeros on the right-hand sides of eqs.(41.23) and (41.26) are required by charge conservation. To see this, start with [Ψ( x),Q] = +Ψ(x), take the matrix elements indicated, and use Q|0∝an}b∇acket∇i}ht= 0 andQ|p,s,±∝an}b∇acket∇i}ht=±|p,s,±∝an}b∇acket∇i}ht. The zero on the right-hand side of eq.(41.22) is required by L orentz invariance. To see this, start with [Ψ(0) ,Mµν] =SµνΨ(0), and take the expectation value in the vacuum state |0∝an}b∇acket∇i}ht. If|0∝an}b∇acket∇i}htis Lorentz invariant (as we will assume), then it is annihilated by the Lorentz generato rsMµν, which means that we must have Sµν∝an}b∇acketle{t0|Ψ(0)|0∝an}b∇acket∇i}ht= 0; this is possible for all µand νonly if ∝an}b∇acketle{t0|Ψ(0)|0∝an}b∇acket∇i}ht= 0, which (by translation invariance) is possible only if∝an}b∇acketle{t0|Ψ(x)|0∝an}b∇acket∇i}ht= 0. The right-hand sides of eqs.(41.24) and (41.25) are similar ly ﬁxed by Lorentz invariance: only the overall scale might be diﬀeren t in an interact- ing theory. However, the LSZ formula is correctly normalize d if and only if eqs.(41.24) and (41.25) hold as written. We will enforce thi s by rescaling (or, one might say, renormalizing ) Ψ(x) by an overall constant. This is just a change of the name of the operator of interest, and does not aﬀect the physics. However, the rescaled Ψ( x) will obey eqs.(41.24) and (41.25). (These two equations are related by charge conjugation, and so actually constitute only one condition on Ψ.) For a Majorana ﬁeld, there is no conserved charge, and we have ∝an}b∇acketle{t0|Ψ(x)|0∝an}b∇acket∇i}ht= 0, (41.27) ∝an}b∇acketle{tp,s|Ψ(x)|0∝an}b∇acket∇i}ht=vs(p)e−ipx, (41.28) ∝an}b∇acketle{tp,s|Ψ(x)|0∝an}b∇acket∇i}ht=us(p)e−ipx, (41.29) instead of eqs.(41.22–41.26). The renormalization of Ψ necessitates including appropria teZfactors in the lagrangian. Consider, for example, L=iZΨ/∂Ψ−ZmmΨΨ−1 4Zgg(ΨΨ)2, (41.30) where Ψ is a Dirac ﬁeld, and gis a coupling constant. We choose the three constantsZ,Zm, andZgso that the following three conditions are satisﬁed: mis the mass of a single particle; gis ﬁxed by some appropriate scattering cross section; and eq.(41.24) and is obeyed. [Eq.(41.25) th en follows by charge conjugation.] Next, we must develop the tools needed to compute the correla tion functions ∝an}b∇acketle{t0|TΨα1′(x1′)...Ψα1(x1)...|0∝an}b∇acket∇i}htin an interacting quantum ﬁeld theory. Problems 41.1) Assuming that eq.(39.40) holds for the exact single-p article states, verify eqs.(41.23) and (41.26), up to overall scale. 42: The Free Fermion Propagator 268 42The Free Fermion Propagator Prerequisite: 39 Consider a free Dirac ﬁeld Ψ(x) =/summationdisplay s=±/integraldisplay /tildewiderdp/bracketleftig bs(p)us(p)eipx+d† s(p)vs(p)e−ipx/bracketrightig , (42.1) Ψ(y) =/summationdisplay s′=±/integraldisplay /tildewiderdp′/bracketleftig b† s′(p′)us′(p′)e−ip′y+ds′(p′)vs′(p′)eip′y/bracketrightig ,(42.2) where bs(p)|0∝an}b∇acket∇i}ht=ds(p)|0∝an}b∇acket∇i}ht= 0, (42.3) and {bs(p),b† s′(p′)}= (2π)3δ3(p−p′)2ωδss′, (42.4) {ds(p),d† s′(p′)}= (2π)3δ3(p−p′)2ωδss′, (42.5) and all the other possible anticommutators between banddoperators (and their hermitian conjugates) vanish. We wish to compute the Feynman propagator S(x−y)αβ≡i∝an}b∇acketle{t0|TΨα(x)Ψβ(y)|0∝an}b∇acket∇i}ht, (42.6) where T denotes the time-ordered product, TΨα(x)Ψβ(y)≡θ(x0−y0)Ψα(x)Ψβ(y)−θ(y0−x0)Ψβ(y)Ψα(x),(42.7) andθ(t) is the unit step function. Note the minus sign in the second t erm; this is needed because Ψ α(x)Ψβ(y) =−Ψβ(y)Ψα(x) whenx0∝ne}ationslash=y0. We can now compute ∝an}b∇acketle{t0|Ψα(x)Ψβ(y)|0∝an}b∇acket∇i}htand∝an}b∇acketle{t0|Ψβ(y)Ψα(x)|0∝an}b∇acket∇i}htby in- serting eqs.(42.1) and (42.2), and then using eqs.(42.3–42 .5). We get ∝an}b∇acketle{t0|Ψα(x)Ψβ(y)|0∝an}b∇acket∇i}ht =/summationdisplay s,s′/integraldisplay /tildewiderdp/tildewiderdp′eipxe−ip′yus(p)αus′(p′)β∝an}b∇acketle{t0|bs(p)b† s′(p′)|0∝an}b∇acket∇i}ht =/summationdisplay s,s′/integraldisplay /tildewiderdp/tildewiderdp′eipxe−ip′yus(p)αus′(p′)β(2π)3δ3(p−p′)2ωδss′ =/summationdisplay s/integraldisplay /tildewiderdpeip(x−y)us(p)αus(p)β =/integraldisplay /tildewiderdpeip(x−y)(−/p+m)αβ. (42.8) 42: The Free Fermion Propagator 269 To get the last line, we used a result from section 38. Similar ly, ∝an}b∇acketle{t0|Ψβ(y)Ψα(x)|0∝an}b∇acket∇i}ht =/summationdisplay s,s′/integraldisplay /tildewiderdp/tildewiderdp′e−ipxeip′yvs(p)αvs′(p′)β∝an}b∇acketle{t0|ds′(p′)d† s(p)|0∝an}b∇acket∇i}ht =/summationdisplay s,s′/integraldisplay /tildewiderdp/tildewiderdp′e−ipxeip′yvs(p)αvs′(p′)β(2π)3δ3(p−p′)2ωδss′ =/summationdisplay s/integraldisplay /tildewiderdpe−ip(x−y)vs(p)αvs(p)β =/integraldisplay /tildewiderdpe−ip(x−y)(−/p−m)αβ. (42.9) We can combine eqs.(42.8) and (42.9) into a compact formula f or the time- ordered product by means of the identity /integraldisplayd4p (2π)4eip(x−y)f(p) p2+m2−iǫ=iθ(x0−y0)/integraldisplay /tildewiderdpeip(x−y)f(p) +iθ(y0−x0)/integraldisplay /tildewiderdpe−ip(x−y)f(−p),(42.10) wheref(p) is a polynomial in p; the derivation of eq.(42.10) was sketched in section 8. We get ∝an}b∇acketle{t0|TΨα(x)Ψβ(y)|0∝an}b∇acket∇i}ht=1 i/integraldisplayd4p (2π)4eip(x−y)(−/p+m)αβ p2+m2−iǫ, (42.11) and so S(x−y)αβ=/integraldisplayd4p (2π)4eip(x−y)(−/p+m)αβ p2+m2−iǫ. (42.12) Note thatS(x−y) is a Green’s function for the Dirac wave operator: (−i/∂x+m)αβS(x−y)βγ=/integraldisplayd4p (2π)4eip(x−y)(/p+m)αβ(−/p+m)βγ p2+m2−iǫ =/integraldisplayd4p (2π)4eip(x−y)(p2+m2)δαγ p2+m2−iǫ =δ4(x−y)δαγ. (42.13) Similarly, S(x−y)αβ(+i← /∂y+m)βγ=/integraldisplayd4p (2π)4eip(x−y)(−/p+m)αβ(/p+m)βγ p2+m2−iǫ =/integraldisplayd4p (2π)4eip(x−y)(p2+m2)δαγ p2+m2−iǫ =δ4(x−y)δαγ. (42.14) 42: The Free Fermion Propagator 270 We can also consider ∝an}b∇acketle{t0|TΨα(x)Ψβ(y)|0∝an}b∇acket∇i}htand∝an}b∇acketle{t0|TΨα(x)Ψβ(y)|0∝an}b∇acket∇i}ht, but it is easy to see that now there is no way to pair up a bwith ab†or adwith a d†, and so ∝an}b∇acketle{t0|TΨα(x)Ψβ(y)|0∝an}b∇acket∇i}ht= 0, (42.15) ∝an}b∇acketle{t0|TΨα(x)Ψβ(y)|0∝an}b∇acket∇i}ht= 0. (42.16) Next, consider a Majorana ﬁeld Ψ(x) =/summationdisplay s=±/integraldisplay /tildewiderdp/bracketleftig bs(p)us(p)eipx+b† s(p)vs(p)e−ipx/bracketrightig , (42.17) Ψ(y) =/summationdisplay s′=±/integraldisplay /tildewiderdp′/bracketleftig b† s′(p′)us′(p′)e−ip′y+bs′(p′)vs′(p′)eip′y/bracketrightig .(42.18) It is easy to see that ∝an}b∇acketle{t0|TΨα(x)Ψβ(y)|0∝an}b∇acket∇i}htis the same as it is in the Dirac case; the only diﬀerence in the calculation is that we would h avebandb†in place ofdandd†in the second line of eq.(42.9), and this does not change the ﬁnal result. Thus, i∝an}b∇acketle{t0|TΨα(x)Ψβ(y)|0∝an}b∇acket∇i}ht=S(x−y)αβ, (42.19) whereS(x−y) is given by eq.(42.12). However, eqs.(42.15) and (42.16) no longer hold for a Majora na ﬁeld. Instead, the Majorana condition Ψ = ΨTC, which can be rewritten as ΨT=ΨC−1, implies i∝an}b∇acketle{t0|TΨα(x)Ψβ(y)|0∝an}b∇acket∇i}ht=i∝an}b∇acketle{t0|TΨα(x)Ψγ(y)|0∝an}b∇acket∇i}ht(C−1)γβ = [S(x−y)C−1]αβ. (42.20) Similarly, using CT=C−1, we can write the Majorana condition as ΨT= C−1Ψ, and so i∝an}b∇acketle{t0|TΨα(x)Ψβ(y)|0∝an}b∇acket∇i}ht=i(C−1)αγ∝an}b∇acketle{t0|TΨγ(x)Ψβ(y)|0∝an}b∇acket∇i}ht = [C−1S(x−y)]αβ. (42.21) Of course, C−1=−C, but it will prove more convenient to leave eqs.(42.20) and (42.21) as they are. We can also consider the vacuum expectation value of a time-o rdered product of more than two ﬁelds. In the Dirac case, we must have an equal number of Ψ’s and Ψ’s to get a nonzero result; and then, the Ψ’s and Ψ’s must pair up to form propagators. There is an extra minus sign if the 42: The Free Fermion Propagator 271 ordering of the ﬁelds in their pairs is an odd permutation of t he original ordering. For example, i2∝an}b∇acketle{t0|TΨα(x)Ψβ(y)Ψγ(z)Ψδ(w)|0∝an}b∇acket∇i}ht= +S(x−y)αβS(z−w)γδ −S(x−w)αδS(z−y)γβ.(42.22) In the Majorana case, we may as well let all the ﬁelds be Ψ’s (si nce we can always replace a Ψ with ΨTC). Then we must pair them up in all possible ways. There is an extra minus sign if the ordering of the ﬁelds in their pairs is an odd permutation of the original ordering. For example, i2∝an}b∇acketle{t0|TΨα(x)Ψβ(y)Ψγ(z)Ψδ(w)|0∝an}b∇acket∇i}ht= + [S(x−y)C−1]αβ[S(z−w)C−1]γδ −[S(x−z)C−1]αγ[S(y−w)C−1]βδ + [S(x−w)C−1]αδ[S(y−z)C−1]βγ. (42.23) Note that the ordering within a pair does not matter, since [S(x−y)C−1]αβ=−[S(y−x)C−1]βα. (42.24) This follows from anticommutation of the ﬁelds and eq.(42.2 0). Problems 42.1) Prove eq.(42.24) directly, using properties of the Cmatrix. 43: The Path Integral for Fermion Fields 272 43The Path Integral for Fermion Fields Prerequisite: 9, 42 We would like to write down a path integral formula for the vac uum- expectation value of a time-ordered product of free Dirac or Majorana ﬁelds. Recall that for a real scalar ﬁeld with L0=−1 2∂µϕ∂µϕ−1 2m2ϕ2 =−1 2ϕ(−∂2+m2)ϕ−1 2∂µ(ϕ∂µϕ), (43.1) we have ∝an}b∇acketle{t0|Tϕ(x1)...|0∝an}b∇acket∇i}ht=1 iδ δJ(x1)...Z 0(J)/vextendsingle/vextendsingle/vextendsingle J=0, (43.2) where Z0(J) =/integraldisplay Dϕexp/bracketleftbigg i/integraldisplay d4x(L0+Jϕ)/bracketrightbigg . (43.3) In this formula, we use the epsilon trick (see section 6) of re placingm2with m2−iǫto construct the vacuum as the initial and ﬁnal state. Then we get Z0(J) = exp/bracketleftbiggi 2/integraldisplay d4xd4yJ(x)∆(x−y)J(y)/bracketrightbigg , (43.4) where the Feynman propagator ∆(x−y) =/integraldisplayd4k (2π)4eik(x−y) k2+m2−iǫ(43.5) is the inverse of the Klein-Gordon wave operator: (−∂2 x+m2)∆(x−y) =δ4(x−y). (43.6) For a complex scalar ﬁeld with L0=−∂µϕ†∂µϕ−m2ϕ†ϕ =−ϕ†(−∂2+m2)ϕ−∂µ(ϕ†∂µϕ), (43.7) we have instead ∝an}b∇acketle{t0|Tϕ(x1)...ϕ†(y1)...|0∝an}b∇acket∇i}ht=1 iδ δJ†(x1)...1 iδ δJ(y1)...Z 0(J†,J)/vextendsingle/vextendsingle/vextendsingle J=J†=0, (43.8) where Z0(J†,J) =/integraldisplay Dϕ†Dϕexp/bracketleftbigg i/integraldisplay d4x(L0+J†ϕ+ϕ†J)/bracketrightbigg = exp/bracketleftbigg i/integraldisplay d4xd4yJ†(x)∆(x−y)J(y)/bracketrightbigg . (43.9) 43: The Path Integral for Fermion Fields 273 We treatJandJ†as independent variables when evaluating eq.(43.8). In the case of a fermion ﬁeld, we should have something simila r, except that we need to account for the extra minus signs from anticom mutation. For this to work out, a functional derivative with respect to an anticom- muting variable must itself be treated as anticommuting. Th us if we deﬁne an anticommuting source η(x) for a Dirac ﬁeld, we can write δ δη(x)/integraldisplay d4y/bracketleftig η(y)Ψ(y) +Ψ(y)η(y)/bracketrightig =−Ψ(x), (43.10) δ δη(x)/integraldisplay d4y/bracketleftig η(y)Ψ(y) +Ψ(y)η(y)/bracketrightig = +Ψ(x). (43.11) The minus sign in eq.(43.10) arises because the δ/δηmust pass through Ψ before reaching η. Thus, consider a free Dirac ﬁeld with L0=iΨ/∂Ψ−mΨΨ =−Ψ(−i/∂+m)Ψ. (43.12) A natural guess for the appropriate path-integral formula, based on analogy with eq.(43.9), is ∝an}b∇acketle{t0|TΨα1(x1)...Ψβ1(y1)...|0∝an}b∇acket∇i}ht =1 iδ δηα1(x1)... iδ δηβ1(y1)... Z 0(η,η)/vextendsingle/vextendsingle/vextendsingle η=η=0,(43.13) where Z0(η,η) =/integraldisplay DΨDΨ exp/bracketleftbigg i/integraldisplay d4x(L0+ηΨ +Ψη)/bracketrightbigg = exp/bracketleftbigg i/integraldisplay d4xd4yη(x)S(x−y)η(y)/bracketrightbigg , (43.14) and the Feynman propagator S(x−y) =/integraldisplayd4p (2π)4(−/p+m)eip(x−y) p2+m2−iǫ(43.15) is the inverse of the Dirac wave operator: (−i/∂x+m)S(x−y) =δ4(x−y). (43.16) Note that each δ/δηin eq.(43.13) comes with a factor of irather than the usual 1/i; this reﬂects the extra minus sign of eq.(43.10). We treat ηandη 43: The Path Integral for Fermion Fields 274 as independent variables when evaluating eq.(43.13). It is straightforward to check (by working out a few examples) that eqs.(43.13–43. 16) do indeed reproduce the result of section 42 for the vacuum expectatio n value of a time-ordered product of Dirac ﬁelds. This is really all we need to know. Recall that, for a complex s calar ﬁeld with interactions speciﬁed by L1(ϕ†,ϕ), we have Z(J†,J)∝exp/bracketleftbigg i/integraldisplay d4xL1/parenleftbigg1 iδ δJ(x),1 iδ δJ†(x)/parenrightbigg/bracketrightbigg Z0(J†,J),(43.17) where the overall normalization is ﬁxed by Z(0,0) = 1. Thus, for a Dirac ﬁeld with interactions speciﬁed by L1(Ψ,Ψ), we have Z(η,η)∝exp/bracketleftbigg i/integraldisplay d4xL1/parenleftbigg iδ δη(x),1 iδ δη(x)/parenrightbigg/bracketrightbigg Z0(η,η), (43.18) where again the overall normalization is ﬁxed by Z(0,0) = 1. Vacuum expectation values of time-ordered products of Dirac ﬁelds in an interact- ing theory will now be given by eq.(43.13), but with Z0(η,η) replaced by Z(η,η). Then, just as for a scalar ﬁeld, this will lead to a Feynman- diagram expansion for Z(η,η). There are two extra complications: we must keep track of the spinor indices, and we must keep track of the extr a minus signs from anticommutation. Both tasks are straightforward; we w ill take them up in section 45. Next, let us consider a Majorana ﬁeld with L0=i 2ΨTC/∂Ψ−1 2mΨTCΨ =−1 2ΨTC(−i/∂+m)Ψ. (43.19) A natural guess for the appropriate path-integral formula, based on analogy with eq.(43.2), is ∝an}b∇acketle{t0|TΨα1(x1)...|0∝an}b∇acket∇i}ht=1 iδ δηα1(x1)... Z 0(η)/vextendsingle/vextendsingle/vextendsingle η=0, (43.20) where Z0(η) =/integraldisplay DΨ exp/bracketleftbigg i/integraldisplay d4x(L0+ηTΨ)/bracketrightbigg = exp/bracketleftbigg −i 2/integraldisplay d4xd4yηT(x)S(x−y)C−1η(y)/bracketrightbigg .(43.21) The Feynman propagator S(x−y)C−1is the inverse of the Majorana wave operator C(−i/∂+m): C(−i/∂x+m)S(x−y)C−1=δ4(x−y). (43.22) 43: The Path Integral for Fermion Fields 275 The extra minus sign in eq.(43.21), as compared with eq.(43. 14), arises because all functional derivatives in eq.(43.20) are accom panied by 1 /i, rather than half by 1 /iand half by i, as in eq.(43.13). It is now straight- forward to check (by working out a few examples) that eqs.(43 .20–43.22) do indeed reproduce the result of section 42 for the vacuum ex pectation value of a time-ordered product of Majorana ﬁelds. 44: Formal Development of Fermionic Path Integrals 276 44Formal Development of Fermionic Path Integrals Prerequisite: 43 In 43, we formally deﬁned the fermionic path integral for a fr ee Dirac ﬁeld Ψ via Z0(η,η) =/integraldisplay DΨDΨ exp/bracketleftbigg i/integraldisplay d4xΨ(i/∂−m)Ψ +ηΨ +Ψη/bracketrightbigg = exp/bracketleftbigg i/integraldisplay d4xd4yη(x)S(x−y)η(y)/bracketrightbigg , (44.1) where the Feynman propagator S(x−y) is the inverse of the Dirac wave operator: (−i/∂x+m)S(x−y) =δ4(x−y). (44.2) We would like to ﬁnd a mathematical framework that allows us t o derive this formula, rather than postulating it by analogy. Consider a set of anticommuting numbers orGrassmann variables ψi that obey {ψi,ψj}= 0, (44.3) wherei= 1,...,n . Let us begin with the very simplest case of n= 1, and thus a single anticommuting number ψthat obeys ψ2= 0. We can deﬁne a functionf(ψ) of such an object via a Taylor expansion; because ψ2= 0, this expansion ends with the second term: f(ψ) =a+ψb. (44.4) The reason for writing the coeﬃcient bto the right of the variable ψwill become clear in a moment. Next we would like to deﬁne the derivative of f(ψ) with respect to ψ. Before we can do so, we must decide if f(ψ) itself is to be commuting or anticommuting; generally we will be interested in functi ons that are themselves commuting. In this case, ain eq.(44.4) should be treated as an ordinary commuting number, but bshould be treated as an anticommuting number: {b,b}={b,ψ}= 0. In this case, f(ψ) =a+ψb=a−bψ. Now we can deﬁne two kinds of derivatives. The left derivative off(ψ) with respect to ψis given by the coeﬃcient of ψwhenf(ψ) is written with theψalways on the far left: ∂ψf(ψ) = +b. (44.5) 44: Formal Development of Fermionic Path Integrals 277 Similarly, the right derivative off(ψ) with respect to ψis given by the coeﬃcient of ψwhenf(ψ) is written with the ψalways on the far right: f(ψ)←∂ψ=−b. (44.6) Generally, when we write a derivative with respect to a Grass mann vari- able, we mean the left derivative. However, in section 37, wh en we wrote the canonical momentum for a fermionic ﬁeld ψasπ=∂L/∂(∂0ψ), we actually meant the right derivative. (This is a standard, th ough rarely stated, convention.) Correspondingly, we wrote the hamilt onian density as H=π∂0ψ− L, with∂0ψto the right of π. Finally, we would like to deﬁne a deﬁnite integral, analogou s to inte- grating a real variable xfrom minus to plus inﬁnity. The key features of such an integral over x(when it converges) are linearity, /integraldisplay+∞ −∞dxcf(x) =c/integraldisplay+∞ −∞dxf(x), (44.7) and invariance under shifts of the dependent variable xby a constant: /integraldisplay+∞ −∞dxf(x+a) =/integraldisplay+∞ −∞dxf(x). (44.8) Up to an overall numerical factor that is the same for every f(ψ), the only possible nontrivial deﬁnition of/integraltextdψf(ψ) that is both linear and shift invariant is /integraldisplay dψf(ψ) =b. (44.9) Now let us generalize this to n>1. We have f(ψ) =a+ψibi+1 2ψi1ψi2ci1i2+...+1 n!ψi1...ψindi1...in, (44.10) where the indices are implicitly summed. Here we have writte n the coef- ﬁcients to the right of the variables to facilitate left-diﬀ erentiation. These coeﬃcients are completely antisymmetric on exchange of any two indices. The left derivative of f(ψ) with respect to ψjis ∂ ∂ψjf(ψ) =bj+ψicji+...+1 (n−1)!ψi2...ψindji2...in. (44.11) Next we would like to ﬁnd a linear, shift-invariant deﬁnitio n of the integral of f(ψ). Note that the antisymmetry of the coeﬃcients implies that di1...in=dεi1...in. (44.12) wheredis a just a number (ordinary if fis commuting and nis even, Grass- mann iffis commuting and nis odd, etc.), and εi1...inis the completely 44: Formal Development of Fermionic Path Integrals 278 antisymmetric Levi-Civita symbol with ε1...n= +1. This number dis a candidate (in fact, up to an overall numerical factor, the on ly candidate!) for the integral of f(ψ):/integraldisplay dnψf(ψ) =d. (44.13) Although eq.(44.13) really tells us everything we need to kn ow about/integraltextdnψ, we can, if we like, write dnψ=dψn...dψ 1(note the backwards ordering), and treat the individual diﬀerentials as anticommuting: {dψi,dψj}= 0, {dψi,ψj}= 0. Then we take/integraltextdψi= 0 and/integraltextdψiψj=δijas our basic formulae, and use them to derive eq.(44.13). Let us work out some consequences of eq.(44.13). Consider wh at hap- pens if we make a linear change of variable, ψi=Jijψ′ j, (44.14) whereJjiis a matrix of commuting numbers (and therefore can be writte n on either the left or right of ψ′ j). We now have f(ψ) =a+...+1 n!(Ji1j1ψ′ j1)...(Jinjnψ′ jn)εi1...ind. (44.15) Next we use εi1...inJi1j1...Jinjn= (detJ)εj1...jn, (44.16) which holds for any n×nmatrixJ, to get f(ψ) =a+...+1 n!ψ′ i1...ψ′ inεi1...in(detJ)d. (44.17) If we now integrate f(ψ) overdnψ′, eq.(44.13) tells us that the result is (detJ)d. Thus, /integraldisplay dnψf(ψ) = (detJ)−1/integraldisplay dnψ′f(ψ). (44.18) Recall that, for integrals over commuting real numbers xiwithxi=Jijx′ j, we have instead /integraldisplay dnxf(x) = (detJ)+1/integraldisplay dnx′f(x). (44.19) Note the opposite sign on the power of the determinant. Now consider a quadratic form ψTMψ=ψiMijψj, whereMis an anti- symmetric matrix of commuting numbers (possibly complex). Let’s evalu- ate the gaussian integral/integraltextdnψexp(1 2ψTMψ). For example, for n= 2, we have M=/parenleftigg0 +m −m0/parenrightigg , (44.20) 44: Formal Development of Fermionic Path Integrals 279 andψTMψ= 2mψ1ψ2. Thus exp(1 2ψTMψ) = 1 +mψ1ψ2, and so /integraldisplay dnψexp(1 2ψTMψ) =m. (44.21) For largern, we use the fact that a complex antisymmetric matrix can be brought to a block-diagonal form via UTMU= 0 +m1 −m10 ... , (44.22) whereUis a unitary matrix, and each mIis real and positive. (If nis odd there is a ﬁnal row and column of all zeroes; from here on, we as sumenis even.) We can now let ψi=Uijψ′ j; then, we have /integraldisplay dnψexp(1 2ψTMψ) = (detU)−1n/2/productdisplay I=1/integraldisplay d2ψIexp(1 2ψTMIψ),(44.23) whereMIrepresents one of the 2 ×2 blocks in eq.(44.22). Each of these two-dimensional integrals can be evaluated using eq.(44.2 1), and so /integraldisplay dnψexp(1 2ψTMψ) = (detU)−1n/2/productdisplay I=1mI. (44.24) Taking the determinant of eq.(44.22), we get (detU)2(detM) =n/2/productdisplay I=1m2 I. (44.25) We can therefore rewrite the right-hand side of eq.(44.24) a s /integraldisplay dnψexp(1 2ψTMψ) = (detM)1/2. (44.26) In this form, there is a sign ambiguity associated with the sq uare root; it is resolved by eq.(44.24). However, the overall sign (more g enerally, any overall numerical factor) will never be of concern to us, so w e can use eq.(44.26) without worrying about the correct branch of the square root. It is instructive to compare eq.(44.26) with the correspond ing gaussian integral for commuting real numbers, /integraldisplay dnxexp(−1 2xTMx) = (2π)n/2(detM)−1/2. (44.27) 44: Formal Development of Fermionic Path Integrals 280 HereMis a complex symmetric matrix. Again, note the opposite sign on the power of the determinant. Now let us introduce the notion of complex Grassmann variables via χ≡1√ 2(ψ1+iψ2), ¯χ≡1√ 2(ψ1−iψ2). (44.28) We can invert this to get /parenleftiggψ1 ψ2/parenrightigg =1√ 2/parenleftigg1 1 i−i/parenrightigg/parenleftigg¯χ χ/parenrightigg . (44.29) The determinant of this transformation matrix is −i, and so d2ψ=dψ2dψ1= (−i)−1dχd¯χ. (44.30) Also,ψ1ψ2=−i¯χχ. Thus we have /integraldisplay dχd¯χ¯χχ= (−i)(−i)−1/integraldisplay dψ2dψ1ψ1ψ2= 1. (44.31) Thus, if we have a function f(χ,¯χ) =a+χb+ ¯χc+ ¯χχd, (44.32) its integral is/integraldisplay dχd¯χf(χ,¯χ) =d. (44.33) In particular,/integraldisplay dχd¯χexp(m¯χχ) =m. (44.34) Let us now consider ncomplex Grassmann variables χiand their com- plex conjugates, ¯ χi. We deﬁne dnχdn¯χ≡dχnd¯χn...dχ 1d¯χ1. (44.35) Then under a change of variable, χi=Jijχ′ jand ¯χi=Kij¯χ′ j, we have dnχdn¯χ= (detJ)−1(detK)−1dnχ′dn¯χ′. (44.36) Note that we need not require Kij=J∗ ij, because, as far as the integral is concerned, it is does not matter whether or not ¯ χiis the complex conjugate ofχi. We now have enough information to evaluate/integraltextdnχdn¯χexp(χ†Mχ), whereMis a general complex matrix. We make the change of variable 44: Formal Development of Fermionic Path Integrals 281 χ=Uχ′andχ†=χ′†V, whereUandVare unitary matrices with the property that VMU is diagonal with positive real entries mi. Then we get /integraldisplay dnχdn¯χexp(χ†Mχ) = (detU)−1(detV)−1n/productdisplay i=1/integraldisplay dχid¯χiexp(mi¯χiχi) = (detU)−1(detV)−1n/productdisplay i=1mi = detM . (44.37) This can be compared to the analogous integral for commuting complex variableszi= (xi+iyi)/√ 2 and ¯z= (xi−iyi)/√ 2, withdnzdn¯z=dnxdny, namely/integraldisplay dnzdn¯zexp(−z†Mz) = (2π)n(detM)−1. (44.38) We can now generalize eqs.(44.26) and (44.37) by shifting th e integra- tion variables, and using shift invariance of the integrals . Thus, by making the replacement ψ→ψ−M−1ηin eq.(44.26), we get /integraldisplay dnψexp(1 2ψTMψ+ηTψ) = (detM)1/2exp(1 2ηTM−1η). (44.39) (In verifying this, remember that Mand its inverse are both antisym- metric.) Similarly, by making the replacements χ→χ−M−1ηand χ†→χ†−η†M−1in eq.(44.37), we get /integraldisplay dnχdn¯χexp(χ†Mχ+η†χ+χ†η) = (detM)exp(−η†M−1η).(44.40) We can now see that eq.(44.1) is simply a particular case of eq .(44.40), with the index on the complex Grassmann variable generalize d to include both the ordinary spin index αand the continuous spacetime argument x of the ﬁeld Ψ α(x). Similarly, eq.(43.21) for the path integral for a free Majorana ﬁeld is simply a particular case of eq.(44.39). In b oth cases, the determinant factors are constants (that is, independent of the ﬁelds and sources) that we simply absorb into the overall normalizati on of the path integral. We will meet determinants that cannot be so neatly absorbed in sections 53 and 71. 45: The Feynman Rules for Dirac Fields 282 45The Feynman Rules for Dirac Fields Prerequisite: 10, 12, 41, 43 In this section we will derive the Feynman rules for Yukawa theory , a theory with a Dirac ﬁeld Ψ (with mass m) and a real scalar ﬁeld ϕ(with mass M), interacting via L1=gϕΨΨ, (45.1) wheregis a coupling constant. In this section, we will be concerned with tree-level processes only, and so we omit renormalizing Zfactors. In four spacetime dimensions, ϕhas mass dimension [ ϕ] = 1 and Ψ has mass dimension [Ψ] =3 2; thus the coupling constant gis dimensionless: [g] = 0. As discussed in section 12, this is generally the most in teresting situation. Note that L1is invariant under the U(1) transformation Ψ →e−iαΨ, as is the free Dirac lagrangian. Thus, the corresponding Noe ther current ΨγµΨ is still conserved, and the associated charge Q(which counts the number ofb-type particles minus the number of d-type particles) is constant in time. We can think of Qas electric charge, and identify the b-type particle as the electron e−, and thed-type particle as the positron e+. The scalar particle is electrically neutral (and could, for example, b e thought of as the Higgs boson; see section 88). We now use the general result of sections 9 and 43 to write Z(η,η,J )∝exp/bracketleftbigg ig/integraldisplay d4x/parenleftbigg1 iδ δJ(x)/parenrightbigg/parenleftbigg iδ δηα(x)/parenrightbigg/parenleftbigg1 iδ δηα(x)/parenrightbigg/bracketrightbigg Z0(η,η,J ), (45.2) where Z0(η,η,J ) = exp/bracketleftbigg i/integraldisplay d4xd4yη(x)S(x−y)η(y)/bracketrightbigg ×exp/bracketleftbiggi 2/integraldisplay d4xd4yJ(x)∆(x−y)J(y)/bracketrightbigg ,(45.3) and S(x−y) =/integraldisplayd4p (2π)4(−/p+m)eip(x−y) p2+m2−iǫ, (45.4) ∆(x−y) =/integraldisplayd4k (2π)4eik(x−y) k2+M2−iǫ(45.5) are the appropriate Feynman propagators for the correspond ing free ﬁelds. We impose the normalization Z(0,0,0) = 1, and write Z(η,η,J ) = exp[iW(η,η,J )]. (45.6) 45: The Feynman Rules for Dirac Fields 283 (a) (c) (d)(b) Figure 45.1: Tree contributions to iW(η,η,J ) with four or fewer sources. TheniW(η,η,J ) can be expressed as a series of connected Feynman dia- grams with sources. We use a dashed line to stand for the scalar propagator1 i∆(x−y), and a solid line to stand for the fermion propagator1 iS(x−y). The only allowed vertex joins two solid lines and one dashed line; the associated vertex factor is ig. The blob at the end of a dashed line stands for the ϕ sourcei/integraltextd4xJ(x), and the blob at the end of a solid line for either the Ψ sourcei/integraltextd4xη(x), or the Ψ sourcei/integraltextd4xη(x). To tell which is which, we adopt the “arrow rule” of problem 9.3: the blob stands for i/integraltextd4xη(x) if the arrow on the attached line points away from the blob, and the blob stands fori/integraltextd4xη(x) if the arrow on the attached line points towards the blob. Because L1involves one Ψ and one Ψ, we also have the rule that, at each vertex, one arrow must point towards the vertex, and one away . The ﬁrst few tree diagrams that contribute to iW(η,η,J ) are shown in ﬁg.(45.1). We omit tadpole diagrams; as in ϕ3theory, these can be cancelled by shifting theϕﬁeld, or, equivalently, adding a term linear in ϕtoL. The LSZ formula is valid only after all tadpole diagrams have been ca ncelled in this way. The spin indices on the fermionic sources and propagators ar e all con- tracted in the obvious way. For example, the complete expres sion corre- sponding to ﬁg.(45.1)(b) is Fig.(45.1)(b) = i3/parenleftig 1 i/parenrightig3(ig)/integraldisplay d4xd4yd4zd4w ×/bracketleftig η(x)S(x−y)S(y−z)η(z)/bracketrightig ×∆(y−w)J(w). (45.7) Our main purpose in this section is to compute the tree-level amplitudes 45: The Feynman Rules for Dirac Fields 284 2 w ww1 x z1z2y 2 w ww1 x z zy 2 1 Figure 45.2: Diagrams corresponding to eq.(45.8). for various two-body elastic scattering processes, such as e−ϕ→e−ϕand e+e−→ϕϕ; for these, we will need to evaluate the tree-level contribu tions to connected correlation functions of the form ∝an}b∇acketle{t0|TΨΨϕϕ|0∝an}b∇acket∇i}htC. Other pro- cesses of interest include e−e−→e−e−ande+e−→e+e−; for these, we will need to evaluate the tree-level contributions to conne cted correlation functions of the form ∝an}b∇acketle{t0|TΨΨΨΨ|0∝an}b∇acket∇i}htC. For∝an}b∇acketle{t0|TΨΨϕϕ|0∝an}b∇acket∇i}htC, the relevant tree-level contribution to iW(η,η,J ) is given by ﬁg.(45.1)(c). We have ∝an}b∇acketle{t0|TΨα(x)Ψβ(y)ϕ(z1)ϕ(z2)|0∝an}b∇acket∇i}htC =1 iδ δηα(x)iδ δηβ(y)1 iδ δJ(z1)1 iδ δJ(z2)iW(η,η,J )/vextendsingle/vextendsingle/vextendsingle η=η=J=0 =/parenleftig 1 i/parenrightig5(ig)2/integraldisplay d4w1d4w2 ×[S(x−w1)S(w1−w2)S(w2−y)]αβ ×∆(z1−w1)∆(z2−w2) +/parenleftig z1↔z2/parenrightig +O(g4). (45.8) The corresponding diagrams, with sources removed, are show n in ﬁg.(45.2). For∝an}b∇acketle{t0|TΨΨΨΨ|0∝an}b∇acket∇i}htC, the relevant tree-level contribution to iW(η,η,J ) is given by ﬁg.(45.1)(d), which has a symmetry factor S= 2. We have ∝an}b∇acketle{t0|TΨα1(x1)Ψβ1(y1)Ψα2(x2)Ψβ2(y2)|0∝an}b∇acket∇i}htC =1 iδ δηα1(x1)iδ δηβ1(y1)1 iδ δηα2(x2)iδ δηβ2(y2)iW(η,η,J )/vextendsingle/vextendsingle/vextendsingle η=η=J=0. (45.9) The twoηderivatives can act on the two η’s in the diagram in two diﬀerent ways; ditto for the two ηderivatives. This results in four diﬀerent terms, but two of them are algebraic duplicates of the other two; thi s duplication cancels the symmetry factor (which is a general result for tr ee diagrams). 45: The Feynman Rules for Dirac Fields 285 x y y xww2 2 21 1 1 x y y xww2 21 1 12 Figure 45.3: Diagrams corresponding to eq.(45.10). We get ∝an}b∇acketle{t0|TΨα1(x1)Ψβ1(y1)Ψα2(x2)Ψβ2(y2)|0∝an}b∇acket∇i}htC =/parenleftig 1 i/parenrightig5(ig)2/integraldisplay d4w1d4w2 ×[S(x1−w1)S(w1−y1)]α1β1 ×∆(w1−w2) ×[S(x2−w2)S(w2−y2)]α2β2 −/parenleftig (y1,β1)↔(y2,β2)/parenrightig +O(g4). (45.10) The corresponding diagrams, with sources removed, are show n in ﬁg.(45.3). Note that we now have a relative minus sign between the two diagrams, due to the anticommutation of the derivatives with respect t oη. In general, the overall sign for a diagram can be determined b y the following procedure. First, draw each diagram with all the f ermion lines horizontal, with their arrows pointing from left to right, a nd with the left endpoints labeled in the same ﬁxed order (from top to bottom) . Next, in each diagram, note the ordering (from top to bottom) of the la bels on the right endpoints of the fermion lines. If this ordering is an e ven permutation of an arbitrarily chosen ﬁxed ordering, then the sign of that diagram is positive, and if it is an odd permutation, the sign is negativ e. (This rule arises because endpoints with arrows pointing away from the vertex come from derivatives with respect to ηthat anticommute. Of course, we could equally well put the right endpoints in a ﬁxed order, and get t he sign from the permutation of the left endpoints, which come from deriv atives with respect toηthat anticommute.) Also, in loop diagrams, a closed fermion loop yields an extra minus sign; we will discuss this rule in s ection 51. Let us now consider a particular scattering process: e−ϕ→e−ϕ. The scattering amplitude is ∝an}b∇acketle{tf|i∝an}b∇acket∇i}ht=∝an}b∇acketle{t0|Ta(k′)outbs′(p′)outb† s(p)ina†(k)in|0∝an}b∇acket∇i}ht. (45.11) Next we make the replacements b† s(p)in→i/integraldisplay d4yΨ(y)(+i← /∂+m)us(p)e+ipy, (45.12) 45: The Feynman Rules for Dirac Fields 286 p k k kp+kp p k kp p Figure 45.4: Diagrams for e−ϕ→e−ϕ, corresponding to eq.(45.16). bs′(p′)out→i/integraldisplay d4xe−ip′xus′(p′)(−i/∂+m)Ψ(x), (45.13) a†(k)in→i/integraldisplay d4z1e+ikz1(−∂2+m2)ϕ(z1), (45.14) a(k′)out→i/integraldisplay d4z2e−ik′z2(−∂2+m2)ϕ(z2) (45.15) in eq.(45.11), and then use eq.(45.8). The wave operators (e ither Klein- Gordon or Dirac) act on the external propagators, and conver t them to delta functions. After using eqs.(45.4) and (45.5) for the i nternal propa- gators, all dependence on the various spacetime coordinate s is in the form of plane-wave factors, as in section 10. Integrating over th e internal co- ordinates then generates delta functions that conserve fou r-momentum at each vertex. The only new feature arises from the spinor fact orsus(p) and us′(p′). We ﬁnd that us(p) is associated with the external fermion line whose arrow points towards the vertex, and that us′(p′) is associated with the external fermion line whose arrow points away from the vertex. We can therefore draw the momentum-space diagrams of ﬁg.(45.4). S ince there is only one fermion line in each diagram, the relative sign is po sitive. The tree-levele−ϕ→e−ϕscattering amplitude is then given by iTe−ϕ→e−ϕ=1 i(ig)2us′(p′)/bracketleftigg −/p−/k+m −s+m2+−/p+ /k′+m −u+m2/bracketrightigg us(p),(45.16) wheres=−(p+k)2andu=−(p−k′)2. (We can safely ignore the iǫ’s in the propagators, because their denominators cannot vanish for any physically allowed values of sandu.) Next consider the process e+ϕ→e+ϕ. We now have ∝an}b∇acketle{tf|i∝an}b∇acket∇i}ht=∝an}b∇acketle{t0|Ta(k′)outds′(p′)outd† s(p)ina†(k)in|0∝an}b∇acket∇i}ht. (45.17) The relevant replacements are d† s(p)in→ −i/integraldisplay d4xe+ipxvs(p)(−i/∂+m)Ψ(x), (45.18) 45: The Feynman Rules for Dirac Fields 287 p k k kp+kp p k kp p Figure 45.5: Diagrams for e+ϕ→e+ϕ, corresponding to eq.(45.22). ds′(p′)out→ −i/integraldisplay d4yΨ(y)(+i← /∂+m)vs′(p′)e−ipy,(45.19) a†(k)in→i/integraldisplay d4z1e+ikz1(−∂2+m2)ϕ(z1), (45.20) a(k′)out→i/integraldisplay d4z2e−ikz2(−∂2+m2)ϕ(z2). (45.21) We substitute these into eq.(45.17), and then use eq.(45.8) . This ulti- mately leads to the momentum-space Feynman diagrams of ﬁg.( 45.5). Note that we must now label the external fermion lines with minus their four- momenta; this is characteristic of d-type particles. (The same phenomenon occurs for a complex scalar ﬁeld; see problem 10.2.) Regardi ng the spinor factors, we ﬁnd that −vs(p) is associated with the external fermion line whose arrow points away from the vertex, and −vs′(p′) with the external fermion line whose arrow points towards the vertex. The minus signs at- tached to each vandvcan be consistently dropped, however, as they only aﬀect the overall sign of the amplitude (and not the relative signs among contributing diagrams). The tree-level expression for the e+ϕ→e+ϕam- plitude is then iTe+ϕ→e+ϕ=1 i(ig)2vs(p)/bracketleftigg /p−/k′+m −u+m2+/p+ /k+m −s+m2/bracketrightigg vs′(p′),(45.22) where again s=−(p+k)2andu=−(p−k′)2. After working out a few more of these (you might try your hand a t some of them before reading ahead), we can abstract the follo wing set of Feynman rules. 1. For each incoming electron , draw a solid line with an arrow pointed towards the vertex, and label it with the electron’s four-momentum, pi. 2. For each outgoing electron , draw a solid line with an arrow pointed awayfrom the vertex, and label it with the electron’s four-momen tum, p′ i. 45: The Feynman Rules for Dirac Fields 288 3. For each incoming positron , draw a solid line with an arrow pointed away from the vertex, and label it with minus the positron’s four- momentum, −pi. 4. For each outgoing positron , draw a solid line with an arrow pointed towards the vertex, and label it with minus the positron’s four-momentum, −p′ i. 5. For each incoming scalar , draw a dashed line with an arrow pointed towards the vertex, and label it with the scalar’s four-momentum, ki. 6. For each outgoing scalar , draw a dashed line with an arrow pointed away from the vertex, and label it with the scalar’s four-momentu m, k′ i. 7. The only allowed vertex joins two solid lines, one with an a rrow point- ing towards it and one with an arrow pointing away from it, and one dashed line (whose arrow can point in either direction). Usi ng this vertex, join up all the external lines, including extra inte rnal lines as needed. In this way, draw all possible diagrams that are topologically inequivalent . 8. Assign each internal line its own four-momentum. Think of the four- momenta as ﬂowing along the arrows, and conserve four-momen tum at each vertex. For a tree diagram, this ﬁxes the momenta on al l the internal lines. 9. The value of a diagram consists of the following factors: for each incoming or outgoing scalar, 1; for each incoming electron, usi(pi); for each outgoing electron, us′ i(p′ i); for each incoming positron, vsi(pi); for each outgoing positron, vs′ i(p′ i); for each vertex, ig; for each internal scalar, −i/(k2+M2−iǫ); for each internal fermion, −i(−/p+m)/(p2+m2−iǫ). 10. Spinor indices are contracted by starting at one end of a f ermion line: speciﬁcally, the end that has the arrow pointing away from th e vertex. The factor associated with the external line is either uorv. Go along the complete fermion line, following the arrows backwards, and write down (in order from left to right) the factors associated wit h the 45: The Feynman Rules for Dirac Fields 289 1 p2k2k1 1p p1k1 p2k2k1 1p p1k1 p2kk1p p1k2 2 Figure 45.6: Diagrams for e+e−→ϕϕ, corresponding to eq.(45.23). vertices and propagators that you encounter. The last facto r is either auorv. Repeat this procedure for the other fermion lines, if any. 11. The overall sign of a tree diagram is determined by drawin g all con- tributing diagrams in a standard form: all fermion lines hor izontal, with their arrows pointing from left to right, and with the le ft end- points labeled in the same ﬁxed order (from top to bottom); if the ordering of the labels on the right endpoints of the fermion l ines in a given diagram is an even (odd) permutation of an arbitrarily chosen ﬁxed ordering, then the sign of that diagram is positive (neg ative). 12. The value of iT(at tree level) is given by a sum over the values of the contributing diagrams. There are additional rules for counterterms and loops; in pa rticular, each closed fermion loop contributes an extra minus sign. We will postpone discussion of loop corrections to section 51. Let us apply these rules to e+e−→ϕϕ. Let the initial electron and positron have four-momenta p1andp2, respectively, and the two ﬁnal scalars have four-momenta k′ 1andk′ 2. The relevant diagrams are shown in ﬁg.(45.6); there is only one fermion line, and so the relat ive sign is positive. The result is iTe+e−→ϕϕ=1 i(ig)2vs2(p2)/bracketleftigg −/p1+ /k′ 1+m −t+m2+−/p1+ /k′ 2+m −u+m2/bracketrightigg us1(p1), (45.23) wheret=−(p1−k′ 1)2andu=−(p1−k′ 2)2. Next, consider e−e−→e−e−. Let the initial electrons have four- momentap1andp2, and the ﬁnal electrons have four-momenta p′ 1and p′ 2. The relevant diagrams are shown in ﬁg.(45.7), and accordin g to rule #11 the relative sign is negative. Thus the result is iTe−e−→e−e−=1 i(ig)2/bracketleftbigg(u′ 1u1)(u′ 2u2) −t+M2−(u′ 2u1)(u′ 1u2) −u+M2/bracketrightbigg , (45.24) 45: The Feynman Rules for Dirac Fields 290 2p1 p1 p1p1 p1 p p2 2 pp2 p p11 2p Figure 45.7: Diagrams for e−e−→e−e−, corresponding to eq.(45.24). p1 p1p1 p2 p2p1 p1 p2p1 p2p1p2+ Figure 45.8: Diagrams for e+e−→e+e−, corresponding to eq.(45.25). whereu1is short for us1(p1), etc., and t=−(p1−p′ 1)2,u=−(p1−p′ 2)2. One more: e+e−→e+e−. Let the initial electron and positron have four-momenta p1andp2, respectively, and the ﬁnal electron and positron have four-momenta p′ 1andp′ 2, respectively. The relevant diagrams are shown in ﬁg.(45.8). If we redraw them in the the standard form of rule #11, as shown in ﬁg.(45.9), we see that the relative sign is ne gative. Thus the result is iTe+e−→e+e−=1 i(ig)2/bracketleftbigg(u′ 1u1)(v2v′ 2) −t+M2−(v2u1)(u′ 1v′ 2) −u+M2/bracketrightbigg , (45.25) p2 p1 p1p1 p2 p2 p2 p1p1 p1p2p1 Figure 45.9: Same as ﬁg.(45.8), but with the diagrams redraw n in the standard form given in rule #11. 45: The Feynman Rules for Dirac Fields 291 wheres=−(p1+p2)2andt=−(p1−p′ 1)2. Problems 45.1) a) Determine how ϕ(x) must transform under parity, time reversal, and charge conjugation in order for these to all be symmetrie s of the theory. (Prerequisite: 40) b) Same question, but with the interaction given by L1=igϕΨγ5Ψ instead of eq.(45.1). 45.2) Use the Feynman rules to write down (at tree level) iTfor the pro- cessese+e+→e+e+andϕϕ→e+e−. 46: Spin Sums 292 46Spin Sums Prerequisite: 45 In the last section, we calculated various tree-level scatt ering amplitudes in Yukawa theory. For example, for e−ϕ→e−ϕwe found T=g2us′(p′)/bracketleftigg −/p−/k+m −s+m2+−/p+ /k′+m −u+m2/bracketrightigg us(p), (46.1) wheres=−(p+k)2andu=−(p−k′)2. In order to compute the cor- responding cross section, we must evaluate |T |2=T T∗. We begin by simplifying eq.(46.1) a little; we use (/ p+m)us(p) = 0 to replace the −/pin each numerator with m. We then abbreviate eq.(46.1) as T=u′Au, (46.2) where A≡g2/bracketleftigg −/k+ 2m m2−s+/k′+ 2m m2−u/bracketrightigg . (46.3) Then we have T∗=T=u′Au=uAu′, (46.4) where in general A≡βA†β, and, for the particular Aof eq.(46.3), A=A. Thus we have |T |2= (u′Au)(uAu′) =/summationdisplay αβγδu′ αAαβuβuγAγδu′ δ =/summationdisplay αβγδu′ δu′ αAαβuβuγAγδ = Tr/bracketleftig (u′u′)A(uu)A/bracketrightig . (46.5) Next, we use a result from section 38: us(p)us(p) =1 2(1−sγ5/z)(−/p+m), (46.6) wheres=±tells us whether the spin is up or down along the spin quan- tization axis z. We then have |T |2=1 4Tr/bracketleftig (1−s′γ5/z′)(−/p′+m)A(1−sγ5/z)(−/p+m)A/bracketrightig . (46.7) We now simply need to take traces of products of gamma matrice s; we will work out the technology for this in the next section. 46: Spin Sums 293 However, in practice, we are often not interested in (or are u nable to easily measure or prepare) the spin states of the scattering particles. Thus, if we know that an electron with momentum p′landed in our detector, but know nothing about its spin, we should sum|T |2over the two possible spin states of this outgoing electron. Similarly, if the spin sta te of the initial electron is not specially prepared for each scattering even t, then we should average |T |2over the two possible spin states of this initial electron. T hen we can use/summationdisplay s=±us(p)us(p) =−/p+m (46.8) in place of eq.(46.6). Let us, then, take |T |2, sum over all ﬁnal spins, and average over all initial spins, and call the result ∝an}b∇acketle{t|T |2∝an}b∇acket∇i}ht. In the present case, we have ∝an}b∇acketle{t|T |2∝an}b∇acket∇i}ht ≡1 2/summationdisplay s,s′|T |2 =1 2Tr/bracketleftig (−/p′+m)A(−/p+m)A/bracketrightig , (46.9) which is much less cumbersome than eq.(46.7). Next let’s try something a little harder, namely e+e−→e+e−. We found in section 45 that T=g2/bracketleftbigg(u′ 1u1)(v2v′ 2) M2−t−(v2u1)(u′ 1v′ 2) M2−s/bracketrightbigg . (46.10) We then have T=g2/bracketleftigg (u1u′ 1)(v′ 2v2) M2−t−(u1v2)(v′ 2u′ 1) M2−s/bracketrightigg . (46.11) When we multiply TbyT, we will get four terms. We want to arrange the factors in each of them so that every uand everyvstands just to the left of the corresponding uandv. In this way, we get |T |2= +g4 (M2−t)2Tr/bracketleftig u1u1u′ 1u′ 1/bracketrightig Tr/bracketleftig v′ 2v′ 2v2v2/bracketrightig +g4 (M2−s)2Tr/bracketleftig u1u1v2v2/bracketrightig Tr/bracketleftig v′ 2v′ 2u′ 1u′ 1/bracketrightig −g4 (M2−t)(M2−s)Tr/bracketleftig u1u1v2v2v′ 2v′ 2u′ 1u′ 1/bracketrightig −g4 (M2−s)(M2−t)Tr/bracketleftig u1u1u′ 1u′ 1v′ 2v′ 2v2v2/bracketrightig .(46.12) 46: Spin Sums 294 Then we average over initial spins and sum over ﬁnal spins, an d use eq.(46.8) and/summationdisplay s=±vs(p)vs(p) =−/p−m. (46.13) We then must evaluate traces of products of up to four gamma ma trices. 47: Gamma Matrix Technology 295 47Gamma Matrix Technology Prerequisite: 36 In this section, we will learn some tricks for handling gamma matrices. We need the following information as a starting point: {γµ,γν}=−2gµν, (47.1) γ2 5= 1, (47.2) {γµ,γ5}= 0, (47.3) Tr 1 = 4. (47.4) Now consider the trace of the product of ngamma matrices. We have Tr[γµ1...γµn] = Tr[γ2 5γµ1γ2 5...γ2 5γµn] = Tr[(γ5γµ1γ5)...(γ5γµnγ5)] = Tr[( −γ2 5γµ1)...(−γ2 5γµn)] = (−1)nTr[γµ1...γµn]. (47.5) We used eq.(47.2) to get the ﬁrst equality, the cyclic proper ty of the trace for the second, eq.(47.3) for the third, and eq.(47.2) again for the fourth. Ifnis odd, eq.(47.5) tells us that this trace is equal to minus it self, and must therefore be zero: Tr[odd # of γµ’s ] = 0. (47.6) Similarly, Tr[γ5(odd # ofγµ’s )] = 0. (47.7) Next, consider Tr[ γµγν]. We have Tr[γµγν] = Tr[γνγµ] =1 2Tr[γµγν+γνγµ] =−gµνTr 1 =−4gµν. (47.8) The ﬁrst equality follows from the cyclic property of the tra ce, the second averages the left- and right-hand sides of the ﬁrst, the thir d uses eq.(47.1), and the fourth uses eq.(47.4). A slightly nicer way of expressing eq.(47.8) is to introduce two arbitrary four-vectors aµandbµ, and write Tr[/a/b] =−4(ab), (47.9) 47: Gamma Matrix Technology 296 where /a=aµγµ, /b=bµγµ, and (ab) =aµbµ. Next consider Tr[/ a/b/c/d]. We evaluate this by moving / ato the right, using eq.(47.1), which is now more usefully written as /a/b=−/b/a−2(ab). (47.10) Using this repeatedly, we have Tr[/a/b/c/d] =−Tr[/b/a/c/d]−2(ab)Tr[/c/d] = +Tr[/b/c/a/d] + 2(ac)Tr[/b/d]−2(ab)Tr[/c/d] =−Tr[/b/c/d/a]−2(ad)Tr[/b/c] + 2(ac)Tr[/b/d]−2(ab)Tr[/c/d].(47.11) Now we note that the ﬁrst term on the right-hand side of the las t line is, by the cyclic property of the trace, equal to minus the left-h and side. We can then move this term to the left-hand side to get 2Tr[/a/b/c/d] =−2(ad)Tr[/b/c] + 2(ac)Tr[/b/d]−2(ab)Tr[/c/d]. (47.12) Finally, we evaluate each Tr[/ a/b] with eq.(47.9), and divide by two: Tr[/a/b/c/d] = 4/bracketleftig (ad)(bc)−(ac)(bd) + (ab)(cd)/bracketrightig . (47.13) This is our ﬁnal result for this trace. Clearly, we can use the same technique to evaluate the trace o f the product of any even number of gamma matrices. Next, let’s consider traces that involve γ5’s andγµ’s. Since {γ5,γµ}= 0, we can always bring all the γ5’s together by moving them through the γµ’s (generating minus signs as we go). Then, since γ2 5= 1, we end up with either oneγ5or none. So we need only consider Tr[ γ5γµ1...γµn]. And, according to eq.(47.7), we need only be concerned with even n. Recall that an explicit formula for γ5is γ5=iγ0γ1γ2γ3. (47.14) Eq.(47.13) then implies Trγ5= 0. (47.15) Similarly, we can show that Tr[γ5γµγν] = 0. (47.16) Finally, consider Tr[ γ5γµγνγργσ]. The only way to get a nonzero result is to have the four vector indices take on four diﬀerent values. If we consider the 47: Gamma Matrix Technology 297 special case Tr[ γ5γ3γ2γ1γ0], plug in eq.(47.14), and then use ( γi)2=−1 and (γ0)2= 1, we get i(−1)3Tr 1 = −4i, or equivalently Tr[γ5γµγνγργσ] =−4iεµνρσ, (47.17) whereε0123=ε3210= +1. Another category of gamma matrix combinations that we will e ventually encounter is γµ/a...γµ. The simplest of these is γµγµ=gµνγµγν =1 2gµν{γµ,γν} =−gµνgµν =−d. (47.18) To get the second equality, we used the fact that gµνis symmetric, and so only the symmetric part of γµγνcontributes. In the last line, dis the number of spacetime dimensions. Of course, our entire spino r formalism has been built around d= 4, but we will need formal results for d= 4−ε when we dimensionally regulate loop diagrams involving fer mions. We move on to evaluate γµ/aγµ=γµ(−γµ/a−2aµ) =−γµγµ/a−2/a = (d−2)/a. (47.19) We continue with γµ/a/bγµ= 4(ab)−(d−4)/a/b (47.20) and γµ/a/b/cγµ= 2/c/b/a+ (d−4)/a/b/c; (47.21) the derivations are left as an exercise. Problems 47.1) Verify eq.(47.16). 47.2) Verify eqs.(47.20) and (47.21). 47.3) Show that the most general 4 ×4 matrix can be written as a linear combination (with complex coeﬃcients) of 1, γµ,Sµν,γµγ5, andγ5, where 1 is the identity matrix and Sµν=i 4[γµ,γν]. Hint: ifAandB are two diﬀerent members of this set, prove linear independe nce by showing that Tr A†B= 0 vanishes. Then count. 48: Spin-Averaged Cross Sections 298 48Spin-Averaged Cross Sections Prerequisite: 46, 47 In section 46, we computed |T |2for (among other processes) e+e−→e+e−. We take the incoming and outgoing electrons to have momenta p1andp′ 1, respectively, and the incoming and outgoing positrons to ha ve momenta p2 andp′ 2, respectively. We have p2 i=p′2 i=−m2, wheremis the electron (and positron) mass. The Mandelstam variables are s=−(p1+p2)2=−(p′ 1+p′ 2)2, t=−(p1−p′ 1)2=−(p2−p′ 2)2, u=−(p1−p′ 2)2=−(p2−p′ 1)2, (48.1) and they obey s+t+u= 4m2. Our result was |T |2=g4/bracketleftbiggΦss (M2−s)2−Φst+ Φts (M2−s)(M2−t)+Φtt (M2−t)2/bracketrightbigg ,(48.2) whereMis the scalar mass, and Φss= Tr/bracketleftig u1u1v2v2/bracketrightig Tr/bracketleftig v′ 2v′ 2u′ 1u′ 1/bracketrightig , Φtt= Tr/bracketleftig u1u1u′ 1u′ 1/bracketrightig Tr/bracketleftig v′ 2v′ 2v2v2/bracketrightig , Φst= Tr/bracketleftig u1u1u′ 1u′ 1v′ 2v′ 2v2v2/bracketrightig , Φts= Tr/bracketleftig u1u1v2v2v′ 2v′ 2u′ 1u′ 1/bracketrightig . (48.3) Next, we average over the two initial spins and sum over the tw o ﬁnal spins to get ∝an}b∇acketle{t|T |2∝an}b∇acket∇i}ht=1 4/summationdisplay s1,s2,s′ 1,s′ 2|T |2. (48.4) Then we use /summationdisplay s=±us(p)us(p) =−/p+m, /summationdisplay s=±vs(p)vs(p) =−/p−m, (48.5) to get ∝an}b∇acketle{tΦss∝an}b∇acket∇i}ht=1 4Tr/bracketleftig (−/p1+m)(−/p2−m)/bracketrightig Tr/bracketleftig (−/p′ 2−m)(−/p′ 1+m)/bracketrightig ,(48.6) ∝an}b∇acketle{tΦtt∝an}b∇acket∇i}ht=1 4Tr/bracketleftig (−/p1+m)(−/p′ 1+m)/bracketrightig Tr/bracketleftig (−/p′ 2−m)(−/p2−m)/bracketrightig ,(48.7) ∝an}b∇acketle{tΦst∝an}b∇acket∇i}ht=1 4Tr/bracketleftig (−/p1+m)(−/p′ 1+m)(−/p′ 2−m)(−/p2−m)/bracketrightig , (48.8) ∝an}b∇acketle{tΦts∝an}b∇acket∇i}ht=1 4Tr/bracketleftig (−/p1+m)(−/p2−m)(−/p′ 2−m)(−/p′ 1+m)/bracketrightig . (48.9) 48: Spin-Averaged Cross Sections 299 It is now merely tedious to evaluate these traces with the tec hnology of section 47. For example, Tr/bracketleftig (−/p1+m)(−/p2−m)/bracketrightig = Tr[/p1/p2]−m2Tr 1 =−4(p1p2)−4m2, (48.10) It is convenient to write four-vector products in terms of th e Mandelstam variables. We have p1p2=p′ 1p′ 2=−1 2(s−2m2), p1p′ 1=p2p′ 2= +1 2(t−2m2), p1p′ 2=p′ 1p2= +1 2(u−2m2), (48.11) and so Tr/bracketleftig (−/p1+m)(−/p2−m)/bracketrightig = 2s−8m2. (48.12) Thus, we can easily work out eqs.(48.6) and (48.7): ∝an}b∇acketle{tΦss∝an}b∇acket∇i}ht= (s−4m2)2, (48.13) ∝an}b∇acketle{tΦtt∝an}b∇acket∇i}ht= (t−4m2)2. (48.14) Obviously, if we start with ∝an}b∇acketle{tΦss∝an}b∇acket∇i}htand make the swap s↔t, we get ∝an}b∇acketle{tΦtt∝an}b∇acket∇i}ht. We could have anticipated this from eqs.(48.6) and (48.7): i f we start with the right-hand side of eq.(48.6) and make the swap p2↔ −p′ 1, we get the right-hand side of eq.(48.7). But from eq.(48.11), we se e that this momentum swap is equivalent to s↔t. Let’s move on to ∝an}b∇acketle{tΦst∝an}b∇acket∇i}htand∝an}b∇acketle{tΦts∝an}b∇acket∇i}ht. These two are also related by p2↔ −p′ 1, and so we only need to compute one of them. We have ∝an}b∇acketle{tΦst∝an}b∇acket∇i}ht=1 4Tr[/p1/p′ 1/p′ 2/p2] +1 4m2Tr[/p1/p′ 1−/p1/p′ 2−/p1/p2−/p′ 1/p′ 2−/p′ 1/p2+ /p′ 2/p2] +1 4m4Tr 1 = (p1p′ 1)(p2p′ 2)−(p1p′ 2)(p2p′ 1) + (p1p2)(p′ 1p′ 2) −m2[p1p′ 1−p1p′ 2−p1p2−p′ 1p′ 2−p′ 1p2+p2p′ 2] +m4 =−1 2st+ 2m2u. (48.15) To get the last line, we used eq.(48.11), and then simpliﬁed i t as much as possible via s+t+u= 4m2. Since our result is symmetric on s↔t, we have∝an}b∇acketle{tΦts∝an}b∇acket∇i}ht=∝an}b∇acketle{tΦst∝an}b∇acket∇i}ht. Putting all of this together, we get ∝an}b∇acketle{t|T |2∝an}b∇acket∇i}ht=g4/bracketleftigg (s−4m2)2 (M2−s)2+st−4m2u (M2−s)(M2−t)+(t−4m2)2 (M2−t)2/bracketrightigg .(48.16) 48: Spin-Averaged Cross Sections 300 This can then be converted to a diﬀerential cross section (in any frame) via the formulae of section 11. Let’s do one more: e−ϕ→e−ϕ. We take the incoming and outgoing electrons to have momenta pandp′, respectively, and the incoming and outgoing scalars to have momenta kandk′, respectively. We then have p2=p′2=−m2andk2=k′2=−M2. The Mandelstam variables are s=−(p+k)2=−(p′+k′)2, t=−(p−p′)2=−(k−k′)2, u=−(p−k′)2=−(k−p′)2, (48.17) and they obey s+t+u= 2m2+ 2M2. Our result in section 46 was ∝an}b∇acketle{t|T |2∝an}b∇acket∇i}ht=1 2Tr/bracketleftig A(−/p+m)A(−/p′+m)/bracketrightig , (48.18) where A=g2/bracketleftigg −/k+ 2m m2−s+/k′+ 2m m2−u/bracketrightigg . (48.19) Thus we have ∝an}b∇acketle{t|T |2∝an}b∇acket∇i}ht=g4/bracketleftbigg∝an}b∇acketle{tΦss∝an}b∇acket∇i}ht (m2−s)2+∝an}b∇acketle{tΦsu∝an}b∇acket∇i}ht+∝an}b∇acketle{tΦus∝an}b∇acket∇i}ht (m2−s)(m2−u)+∝an}b∇acketle{tΦuu∝an}b∇acket∇i}ht (m2−u)2/bracketrightbigg ,(48.20) where now ∝an}b∇acketle{tΦss∝an}b∇acket∇i}ht=1 2Tr/bracketleftig (−/p′+m)(−/k+2m)(−/p+m)(−/k+2m)/bracketrightig ,(48.21) ∝an}b∇acketle{tΦuu∝an}b∇acket∇i}ht=1 2Tr/bracketleftig (−/p′+m)(+/k′+2m)(−/p+m)(+/k′+2m)/bracketrightig ,(48.22) ∝an}b∇acketle{tΦsu∝an}b∇acket∇i}ht=1 2Tr/bracketleftig (−/p′+m)(−/k+2m)(−/p+m)(+/k′+2m)/bracketrightig ,(48.23) ∝an}b∇acketle{tΦus∝an}b∇acket∇i}ht=1 2Tr/bracketleftig (−/p′+m)(+/k′+2m)(−/p+m)(−/k+2m)/bracketrightig .(48.24) We can evaluate these in terms of the Mandelstam variables by using our trace technology, along with pk=p′k′=−1 2(s−m2−M2), pp′= +1 2(t−2m2), kk′= +1 2(t−2M2), pk′=p′k= +1 2(u−m2−M2). (48.25) Examining eqs.(48.21) and (48.22), we see that ∝an}b∇acketle{tΦss∝an}b∇acket∇i}htand∝an}b∇acketle{tΦuu∝an}b∇acket∇i}htare trans- formed into each other by k↔ −k′. Examining eqs.(48.23) and (48.24), we 48: Spin-Averaged Cross Sections 301 see that ∝an}b∇acketle{tΦsu∝an}b∇acket∇i}htand∝an}b∇acketle{tΦus∝an}b∇acket∇i}htare also transformed into each other by k↔ −k′. From eq.(48.25), we see that this is equivalent to s↔u. Thus we need only compute ∝an}b∇acketle{tΦss∝an}b∇acket∇i}htand∝an}b∇acketle{tΦsu∝an}b∇acket∇i}ht, and then take s↔uto get ∝an}b∇acketle{tΦuu∝an}b∇acket∇i}htand∝an}b∇acketle{tΦus∝an}b∇acket∇i}ht. This is, again, merely tedious, and the results are ∝an}b∇acketle{tΦss∝an}b∇acket∇i}ht=−su+m2(9s+u) + 7m4−8m2M2+M4,(48.26) ∝an}b∇acketle{tΦuu∝an}b∇acket∇i}ht=−su+m2(9u+s) + 7m4−8m2M2+M4,(48.27) ∝an}b∇acketle{tΦsu∝an}b∇acket∇i}ht= +su+ 3m2(s+u) + 9m4−8m2M2−M4,(48.28) ∝an}b∇acketle{tΦus∝an}b∇acket∇i}ht= +su+ 3m2(s+u) + 9m4−8m2M2−M4.(48.29) Problems 48.1) The tedium of these calculations is greatly alleviate d by making use of a symbolic manipulation program like Mathematica or Maple. One approach is brute force: compute 4 ×4 matrices like / pin the CM frame, and take their products and traces. If you are familia r with a symbolic-manipulation program, write one that does this. S ee if you can verify eqs.(48.26–48.29). 48.2) Compute ∝an}b∇acketle{t|T |2∝an}b∇acket∇i}htfore+e−→ϕϕ. You should ﬁnd that your result is the same as that for e−ϕ→e−ϕ, but with s↔t, and an extra overall minus sign. This relationship is known as crossing symmetry . There is an overall minus sign for each fermion that is moved f rom the initial to the ﬁnal state. 48.3) Compute ∝an}b∇acketle{t|T |2∝an}b∇acket∇i}htfore−e−→e−e−. You should ﬁnd that your result is the same as that for e+e−→e+e−, but withs↔u. This is another example of crossing symmetry. 48.4) Suppose that M > 2m, so that the scalar can decay to an electron- positron pair. a) Compute the decay rate, summed over ﬁnal spins. b) Compute |T |2for decay into an electron with spin s1and a positron with spins2. Take the fermion three-momenta to be along the zaxis, and let the x-axis be the spin-quantization axis. You should ﬁnd that|T |2= 0 ifs1=−s2, or ifM= 2m(so that the outgoing three-momentum of each fermion is zero). Discuss this in lig ht of conservation of angular momentum and of parity. (Prerequis ite: 40.) c) Compute |T |2for decay into an electron with helicity s1and a positron with helicity s2. (See section 38 for the deﬁnition of helicity.) 48: Spin-Averaged Cross Sections 302 You should ﬁnd that the decay rate is zero if s1=−s2. Discuss this in light of conservation of angular momentum and of parity. d) Now consider changing the interaction to L1=igϕΨγ5Ψ, and com- pute the spin-summed decay rate. Explain (in light of conser vation of angular momentum and of parity) why the decay rate is larger t han it was without the iγ5in the interaction. e) Repeat parts (b) and (c) for the new form of the interaction , and explain any diﬀerences in the results. 48.5) The charged pion π−is represented by a complex scalar ﬁeld ϕ, the muonµ−by a Dirac ﬁeld M, and the muon neutrino νµby a spin- projected Dirac ﬁeld PLN, wherePL=1 2(1−γ5). The charged pion can decay to a muon and a muon antineutrino via the interactio n L1= 2c1GFfπ∂µϕMγµPLN+ h.c., (48.30) wherec1is the cosine of the Cabibbo angle ,GFis theFermi constant , andfπis the pion decay constant . a) Compute the charged pion decay rate Γ. b) The charged pion mass is mπ= 139.6MeV, the muon mass is mµ= 105.7MeV, and the muon neutrino is massless. The Fermi constant is measured in muon decay to be GF= 1.166×10−5GeV−2, and the cosine of the Cabibbo angle is measured in nuclear bet a decays to bec1= 0.974. The measured value of the charged pion lifetime is 2.603×10−8s. Determine the value of fπin MeV. Your result is too large by 0.8%, due to neglect of electromagnetic loop correc tions. 49: The Feynman Rules for Majorana Fields 303 49The Feynman Rules for Majorana Fields Prerequisite: 45 In this section we will deduce the Feynman rules for Yukawa th eory, but with a Majorana ﬁeld instead of a Dirac ﬁeld. We can think of th e particles associated with the Majorana ﬁeld as massive neutrinos. We have L1=1 2gϕΨΨ =1 2gϕΨTCΨ, (49.1) where Ψ be a Majorana ﬁeld (with mass m),ϕis a real scalar ﬁeld (with massM), andgis a coupling constant. In this section, we will be concerned with tree-level processes only, and so we omit renormalizin gZfactors. From section 41, we have the LSZ rules appropriate for a Major ana ﬁeld, b† s(p)in→ −i/integraldisplay d4xe+ipxvs(p)(−i/∂+m)Ψ(x) (49.2) = +i/integraldisplay d4xΨT(x)C(+i← /∂+m)us(p)e+ipx, (49.3) bs′(p′)out→+i/integraldisplay d4xe−ip′xus′(p′)(−i/∂+m)Ψ(x), (49.4) =−i/integraldisplay d4xe−ip′xΨT(x)C(+i← /∂+m)vs′(p′)e−ip′x.(49.5) Eq.(49.3) follows from eq.(49.2) by taking the transpose of the right-hand side, and using vs′(p′)T=−Cus′(p′) and ( −i/∂+m)T=C(+i/∂+m)C−1; similarly, eq.(49.5) follows from eq.(49.4). Which form we use depends on convenience, and is best chosen on a diagram-by-diagram bas is, as we will see shortly. Eqs.(49.2–49.5) lead us to compute correlation functions c ontaining Ψ’s, but not Ψ’s. In position space, this leads to Feynman rules where the fermion propagator is1 iS(x−y)C−1, and theϕΨΨ vertex is igC; the factor of1 2inL1is canceled by a symmetry factor of 2! that arises from having two identical Ψ ﬁelds in L1. In a particular diagram, as we move along a fermion line, the C−1in the propagator will cancel against the Cin the vertex, leaving over a ﬁnal C−1at one end. This C−1can be canceled by aCfrom eq.(49.3) (for an incoming particle) or eq.(49.5) (for an outgoing particle). On the other hand, for the other end of the same lin e, we should use either eq.(49.2) (for an incoming particle) or eq.(49.4 ) (for an outgoing 49: The Feynman Rules for Majorana Fields 304 particle) to avoid introducing an extra Catthatend. In this way, we can avoid ever having explicit factors of Cin our Feynman rules.1 Using this approach, the Feynman rules for this theory are as follows. 1. The total number of incoming and outgoing neutrinos is alw ays even; call this number 2 n. Drawnsolid lines. Connect them with internal dashed lines, using a vertex that joins one dashed and two sol id lines. Also, attach an external dashed line for each incoming or out going scalar. In this way, draw all possible diagrams that are topologically inequivalent . 2. Draw arrows on each segment of each solid line; keep the arr ow di- rection continuous along each line. 3. Label each external dashed line with the momentum of an inc oming or outgoing scalar. If the particle is incoming, draw an arro w on the dashed line that points towards the vertex; If the particle is outgoing, draw an arrow on the dashed line that points away from the vertex. 4. Label each external solid line with the momentum of an inco ming or outgoing neutrino, but include a minus sign with the momentu m if (a) the particle is incoming and the arrow points away from the vertex, or (b) the particle is outgoing and the arrow points towards the vertex. Do this labeling of external lines in all possible inequivalent ways. Two diagrams are considered equivalent if they can be transformed into each other by reversing all the arrows on one or more ferm ion lines, and correspondingly changing the signs of the extern al momenta on each arrow-reversed line. 5. Assign each internal line its own four-momentum. Think of the four- momenta as ﬂowing along the arrows, and conserve four-momen tum at each vertex. For a tree diagram, this ﬁxes the momenta on al l the internal lines. 6. The value of a diagram consists of the following factors: for each incoming or outgoing scalar, 1; for each incoming neutrino labeled with + pi,usi(pi); for each incoming neutrino labeled with −pi,vsi(pi); for each outgoing neutrino labeled with + p′ i,us′ i(p′ i); for each outgoing neutrino labeled with −p′ i,vs′ i(p′ i); 1This is not always possible if the Majorana ﬁelds interact wi th Dirac ﬁelds, and we use the usual rules for the Dirac ﬁelds. 49: The Feynman Rules for Majorana Fields 305 p2p1kp1 p2k Figure 49.1: Two equivalent diagrams for ϕ→νν. for each vertex, ig; for each internal scalar, −i/(k2+M2−iǫ); for each internal fermion, −i(−/p+m)/(p2+m2−iǫ). 7. Spinor indices are contracted by starting at one end of a fe rmion line: speciﬁcally, the end that has the arrow pointing away f rom the vertex. The factor associated with the external line is eith eruorv. Go along the complete fermion line, following the arrows bac kwards, and writing down (in order from left to right) the factors ass ociated with the vertices and propagators that you encounter. The la st factor is either auorv. Repeat this procedure for the other fermion lines, if any. 8. The overall sign of a tree diagram is determined by drawing all con- tributing diagrams in a standard form: all fermion lines hor izontal, with their arrows pointing from left to right, and with the le ft end- points labeled in the same ﬁxed order (from top to bottom); if the ordering of the labels on the right endpoints of the fermion l ines in a given diagram is an even (odd) permutation of an arbitrarily chosen ﬁxed ordering, then the sign of that diagram is positive (neg ative). To compare two diagrams, it may be necessary to use the arrow-re versing equivalence relation of rule #4; there is then an extra minus sign for each arrow-reversed line. 9. The value of iTis given by a sum over the values of all these diagrams. There are additional rules for counterterms and loops, but w e will postpone those to section 51. Let’s look at the simplest process, ϕ→νν. There are two possible diagrams for this, shown in ﬁg.(49.1). However, according t o rule #4, these two diagrams are equivalent, and we should keep only on e of them. The ﬁrst diagram yields iT1=igv′ 2u′ 1and the second iT2=igv′ 1u′ 2. Rule #8 then implies that we should have T1=−T2. To check this, we note that (after dropping primes to simplify the notation) v1u2= [v1u2]T 49: The Feynman Rules for Majorana Fields 306 2p1 p1p1 p2 p2p1 p2p2 p1p2 p2 p1p1 p1p2p1 p1p Figure 49.2: Diagrams for νν→νν, corresponding to eq.(49.7). =uT 2vT 1 =v2C−1C−1u1 =−v2u1, (49.6) as required. In general, for processes with a total of just two incoming an d outgo- ing neutrinos, such as νϕ→νϕorνν→ϕϕ, these rules give (up to an irrelevant overall sign) the same result for iTas we would get for the cor- responding process in the Dirac case, e−ϕ→e−ϕore+e−→ϕϕ. (Note, however, that in the Dirac case, we have L1=gϕΨΨ, as compared with L1=1 2gϕΨΨ in the Majorana case.) The diﬀerences between Dirac and Majorana fermions become m ore pronounced for νν→νν. Now there are three inequivalent contributing diagrams, shown in ﬁg.(49.2). The corresponding amplitude can be written as iT=1 i(ig)2/bracketleftbigg(u′ 1u1)(u′ 2u2) −t+M2−(u′ 2u1)(u′ 1u2) −u+M2+(v2u1)(u′ 1v′ 2) −s+M2/bracketrightbigg ,(49.7) wheres=−(p1+p2)2,t=−(p1−p′ 1)2andu=−(p1−p′ 2)2. After arbitrarily assigning the ﬁrst diagram a plus sign, the minus sign of the s econd diagram follows from rule #8. To get the sign of the third diagram, we c ompare it with the ﬁrst. To do so, we reverse the arrow direction on the l ower line of the ﬁrst diagram (which yields an extra minus sign), and th en redraw it in standard form. Comparing this modiﬁed ﬁrst diagram wit h the third diagram (and invoking rule #8) reveals a relative minus sign . Since the modiﬁed ﬁrst diagram has a minus sign from the arrow reversal , we conclude that the third diagram has an overall plus sign. After taking the absolute square of eq.(49.7), we can use rel ations like eq.(49.6) on a term-by-term basis to put everything into a fo rm that allows the spin sums to be performed in the standard way. In fact, we h ave already done all the necessary work in the Dirac case. The s-s,s-t, andt-tterms in ∝an}b∇acketle{t|T |2∝an}b∇acket∇i}htforνν→ννare the same as those for e+e−→e+e−, while the t-t, 49: The Feynman Rules for Majorana Fields 307 t-u, andu-uterms are the same as those for the crossing-related process e−e−→e−e−. Finally, the s-uterms can be obtained from the s-tterms viat↔u, or equivalently from the t-uterms viat↔s. Thus the result is ∝an}b∇acketle{t|T |2∝an}b∇acket∇i}ht=g4/bracketleftigg (s−4m2)2 (M2−s)2+st−4m2u (M2−s)(M2−t) +(t−4m2)2 (M2−t)2+tu−4m2s (M2−t)(M2−u) +(u−4m2)2 (M2−u)2+us−4m2t (M2−u)(M2−s)/bracketrightigg , (49.8) which is neatly symmetric on permutations of s,t, andu. Problems 49.1) Let Ψ be a Dirac ﬁeld (representing the electron and pos itron),X be a Majorana ﬁeld (represeting the photino , the hypothetical super- symmetric partner of the photon, with mass m˜γ), andELandERbe two diﬀerent complex scalar ﬁelds (representing the two selectrons , the hypothetical supersymmetric partners of the left-hand ed electron and the right-handed electron, with masses ML, andMR; note that the subscripts L and R are just part of their names, and do not s ig- nify anything about their Lorentz transformation properti es). They interact via L1=√ 2eE† LXPLΨ +√ 2eE† RXPRΨ + h.c., (49.9) whereα=e2/4π≃1/137 is the ﬁne-structure constant, and PL,R= 1 2(1∓γ5). a) Write down the hermitian conjugate term explicitly. b) Find the tree-level scattering amplitude for e+e−→˜γ˜γ. Hint: there are four contributing diagrams, two each in the tanduchannels, with exchange of either ELorER. c) Compute the spin-averaged diﬀerential cross section for this process in the case that me(the electron mass) can be neglected, and |t|,|u| ≪ ML=MR. Express it as a function of sand the center-of-mass scattering angle θ. 50: Massless Particles and Spinor Helicity 308 50Massless Particles and Spinor Helicity Prerequisite: 48 Scattering amplitudes often simplify greatly if the partic les are massless (or can be approximated as massless because the Mandelstam v ariables all have magnitudes much larger than the particle masses square d). In this section we will explore this phenomenon for spin-one-half ( and spin-zero) particles. We will begin developing the technology of spinor helicity , which will prove to be of indispensible utility in Part III. Recall from section 38 that the uspinors for a massless spin-one-half particle obey us(p)us(p) =1 2(1 +sγ5)(−/p), (50.1) wheres=±speciﬁes the helicity , the component of the particle’s spin measured along the axis speciﬁed by its three-momentum; in t his notation the helicity is1 2s. Thevspinors obey a similar relation, vs(p)vs(p) =1 2(1−sγ5)(−/p). (50.2) In fact, in the massless case, with the phase conventions of s ection 38, we havevs(p) =u−s(p). Thus we can conﬁne our discussion to u-type spinors only, since we need merely change the sign of sto accomodate v-type spinors. Consider a uspinor for a particle of negative helicity. We have u−(p)u−(p) =1 2(1−γ5)(−/p). (50.3) Let us deﬁne pa˙a≡pµσµ a˙a. (50.4) Then we also have p˙aa=εacε˙a˙cpc˙c=pµ¯σµ˙aa. (50.5) Then, using γµ=/parenleftigg0σµ ¯σµ0/parenrightigg ,1 2(1−γ5) =/parenleftigg1 0 0 0/parenrightigg (50.6) in eq.(50.3), we ﬁnd u−(p)u−(p) =/parenleftigg0−pa˙a 0 0/parenrightigg . (50.7) On the other hand, we know that the lower two components of u−(p) vanish, and so we can write u−(p) =/parenleftiggφa 0/parenrightigg . (50.8) 50: Massless Particles and Spinor Helicity 309 Hereφais a two-component numerical spinor; it is not an anticommut ing object. Such a commuting spinor is sometimes called a twistor . An explicit numerical formula for it (veriﬁed in problem 50.2) is φa=√ 2ω/parenleftigg−sin(1 2θ)e−iφ + cos(1 2θ)/parenrightigg , (50.9) whereθandφare the polar and azimuthal angles that specify the directio n of the three-momentum p, andω=|p|. Barring eq.(50.8) yields u−(p) = (0, φ∗ ˙a), (50.10) whereφ∗ ˙a= (φa)∗. Now, combining eqs.(50.8) and (50.10), we get u−(p)u−(p) =/parenleftigg0φaφ∗ ˙a 0 0/parenrightigg . (50.11) Comparing with eq.(50.7), we see that pa˙a=−φaφ∗ ˙a. (50.12) This expresses the four-momentum of the particle neatly in t erms of the twistor that describes its spin state. The essence of the spi nor helicity method is to treat φaas the fundamental object, and to express the parti- cle’s four-momentum in terms of it, via eq.(50.12). Given eq.(50.8), and the phase conventions of section 38, th e positive- helicity spinor is u+(p) =/parenleftigg0 φ∗˙a/parenrightigg , (50.13) whereφ∗˙a=ε˙a˙cφ∗ ˙c. Barring eq.(50.13) yields u+(p) = (φa,0 ). (50.14) Computation of u+(p)u+(p) via eqs.(50.13) and (50.14), followed by com- parison with eq.(50.1) with s= +, then reproduces eq.(50.12), but with the indices raised. In fact, the decomposition of pa˙ainto the direct product of a twistor and its complex conjugate is unique (up to an overall phase fo r the twistor). To see this, use σµ= (I,/vector σ) to write pa˙a=/parenleftigg−p0+p3p1−ip2 p1+ip2−p0−p3/parenrightigg . (50.15) The determinant of this matrix is −(p0)2+p2, and this vanishes because the particle is (by assumption) massless. Thus pa˙ahas a zero eigenvalue. 50: Massless Particles and Spinor Helicity 310 Therefore, it can be written as a projection onto the eigenve ctor correspond- ing to the nonzero eigenvalue. That is what eq.(50.12) repre sents, with the nonzero eigenvalue absorbed into the normalization of the e igenvectorφa. Let us now introduce some useful notation. Let pandkbe two four- momenta, and φaandκathe corresponding twistors. We deﬁne the twistor product [pk]≡φaκa. (50.16) Becauseφaκa=εacφcκa, and the twistors commute, we have [kp] =−[pk]. (50.17) From eqs.(50.8) and (50.14), we can see that u+(p)u−(k) = [pk]. (50.18) Similarly, let us deﬁne ∝an}b∇acketle{tpk∝an}b∇acket∇i}ht ≡φ∗ ˙aκ∗˙a. (50.19) Comparing with eq.(50.16) we see that ∝an}b∇acketle{tpk∝an}b∇acket∇i}ht= [kp]∗, (50.20) which implies that this product is also antisymmetric, ∝an}b∇acketle{tkp∝an}b∇acket∇i}ht=−∝an}b∇acketle{tpk∝an}b∇acket∇i}ht. (50.21) Also, from eqs.(50.10) and (50.13), we have u−(p)u+(k) =∝an}b∇acketle{tpk∝an}b∇acket∇i}ht. (50.22) Note that the other two possible spinor products vanish: u+(p)u+(k) =u−(p)u−(k) = 0. (50.23) The twistor products ∝an}b∇acketle{tpk∝an}b∇acket∇i}htand [pk] satisfy another important relation, ∝an}b∇acketle{tpk∝an}b∇acket∇i}ht[kp] = (φ∗ ˙aκ∗˙a)(κaφa) = (φ∗ ˙aφa)(κaκ∗˙a) =p˙aaka˙a =−2pµkµ, (50.24) where the last line follows from ¯ σµ˙aaσν a˙a=−2gµν. 50: Massless Particles and Spinor Helicity 311 Let us apply this notation to the tree-level scattering ampl itude for e−ϕ→e−ϕin Yukawa theory, which we ﬁrst computed in Section 44, and which reads Ts′s=g2us′(p′)/bracketleftig˜S(p+k) +˜S(p−k′)/bracketrightig us(p). (50.25) For a massless fermion, ˜S(p) =−/p/p2. If the scalar is also massless, then (p+k)2= 2p·kand (p−k′)2=−2p·k′. Also, we can remove the / p’s in the propagator numerators in eq.(50.25), because / pus(p) = 0. Thus we have Ts′s=g2us′(p′)/bracketleftigg −/k 2p·k+−/k′ 2p·k′/bracketrightigg us(p). (50.26) Now consider the case s′=s= +. From eqs.(50.13), (50.14), and −/k=/parenleftigg0κaκ∗ ˙a κ∗˙aκa0/parenrightigg , (50.27) we get u+(p′)(−/k)u+(p) =φ′aκaκ∗ ˙aφ∗˙a = [p′k]∝an}b∇acketle{tkp∝an}b∇acket∇i}ht. (50.28) Similarly, for s′=s=−, we ﬁnd u−(p′)(−/k)u−(p) =φ′∗ ˙aκ∗˙aκaφa =∝an}b∇acketle{tp′k∝an}b∇acket∇i}ht[kp], (50.29) while fors′∝ne}ationslash=s, the amplitude vanishes: u−(p′)(−/k)u+(p) =u+(p′)(−/k)u−(p) = 0. (50.30) Then, using eq.(50.24) on the denominators in eq.(50.26), w e ﬁnd T++=−g2/parenleftbigg[p′k] [pk]+[p′k′] [pk′]/parenrightbigg , T−−=−g2/parenleftbigg∝an}b∇acketle{tp′k∝an}b∇acket∇i}ht ∝an}b∇acketle{tpk∝an}b∇acket∇i}ht+∝an}b∇acketle{tp′k′∝an}b∇acket∇i}ht ∝an}b∇acketle{tpk′∝an}b∇acket∇i}ht/parenrightbigg , (50.31) while T+−=T−+= 0. (50.32) Thus we have rather simple expressions for the ﬁxed-helicit y scattering amplitudes in terms of twistor products. 50: Massless Particles and Spinor Helicity 312 Reference Notes Spinor-helicity methods are discussed by Siegel. Problems 50.1) Consider a bra-ket notation for twistors, |p] =u−(p) =v+(p), |p∝an}b∇acket∇i}ht=u+(p) =v−(p), [p|=u+(p) =v−(p), ∝an}b∇acketle{tp|=u−(p) =v+(p). (50.33) We then have ∝an}b∇acketle{tk| |p∝an}b∇acket∇i}ht=∝an}b∇acketle{tkp∝an}b∇acket∇i}ht, [k| |p] = [kp], ∝an}b∇acketle{tk| |p] = 0, [k| |p∝an}b∇acket∇i}ht= 0. (50.34) a) Show that −/p=|p∝an}b∇acket∇i}ht[p|+|p]∝an}b∇acketle{tp|, (50.35) wherepis any massless four-momentum. b) Use this notation to rederive eqs. (50.28–50.30). 50.2) a) Use eqs.(50.9) and (50.15) to verify eq.(50.12). b) Let the three-momentum pbe in the + ˆ zdirection. Use eq.(38.12) to compute u±(p) explicitly in the massless limit (corresponding to the limitη→ ∞, where sinh η=|p|/m). Verify that, when θ= 0, your results agree with eqs.(50.8), (50.9), and (50.13). 50.3) Prove the Schouten identity , ∝an}b∇acketle{tpq∝an}b∇acket∇i}ht∝an}b∇acketle{trs∝an}b∇acket∇i}ht+∝an}b∇acketle{tpr∝an}b∇acket∇i}ht∝an}b∇acketle{tsq∝an}b∇acket∇i}ht+∝an}b∇acketle{tps∝an}b∇acket∇i}ht∝an}b∇acketle{tqr∝an}b∇acket∇i}ht= 0. (50.36) Hint: note that the left-hand side is completely antisymmet ric in the three labels q,r, ands, and that each corresponding twistor has only two components. 50: Massless Particles and Spinor Helicity 313 50.4) Show that ∝an}b∇acketle{tpq∝an}b∇acket∇i}ht[qr]∝an}b∇acketle{trs∝an}b∇acket∇i}ht[sp] = Tr1 2(1−γ5)/p/q/r/s, (50.37) and evaluate the right-hand side. 50.5) a) Prove the useful identities ∝an}b∇acketle{tp|γµ|k] = [k|γµ|p∝an}b∇acket∇i}ht, (50.38) ∝an}b∇acketle{tp|γµ|k]∗=∝an}b∇acketle{tk|γµ|p], (50.39) ∝an}b∇acketle{tp|γµ|p] = 2pµ, (50.40) ∝an}b∇acketle{tp|γµ|k∝an}b∇acket∇i}ht= 0, (50.41) [p|γµ|k] = 0. (50.42) b) Extend the last two identies of part (a): show that the prod uct of an odd number of gamma matrices sandwiched between either ∝an}b∇acketle{tp| and|k∝an}b∇acket∇i}htor [p|and|k] vanishes. Also show that the product of an even number of gamma matrices between either ∝an}b∇acketle{tp|and|k] or [p|and|k∝an}b∇acket∇i}ht vanishes. c) Prove the Fierz identities, −1 2∝an}b∇acketle{tp|γµ|q]γµ=|q]∝an}b∇acketle{tp|+|p∝an}b∇acket∇i}ht[q|, (50.43) −1 2[p|γµ|q∝an}b∇acket∇i}htγµ=|q∝an}b∇acket∇i}ht[p|+|p]∝an}b∇acketle{tq|. (50.44) Now take the matrix element of eq.(50.44) between ∝an}b∇acketle{tr|and|s] to get another useful form of the Fierz identity, [p|γµ|q∝an}b∇acket∇i}ht∝an}b∇acketle{tr|γµ|s] = 2[ps]∝an}b∇acketle{tqr∝an}b∇acket∇i}ht. (50.45) 51: Loop Corrections in Yukawa Theory 314 51Loop Corrections in Yukawa Theory Prerequisite: 19, 40, 48 In this section we will compute the one-loop corrections in Y ukawa theory with a Dirac ﬁeld. The basic concepts are all the same as for a s calar ﬁeld, and so we will mainly be concerned with the extra technicalit ies arising from spin indices and anticommutation. First let us note that the general discussion of sections 18 a nd 29 leads us to expect that we will need to add to the lagrangian all poss ible terms whose coeﬃcients have positive or zero mass dimension, and t hat respect the symmetries of the original lagrangian. These include Lo rentz symmetry, the U(1) phase symmetry of the Dirac ﬁeld, and the discrete sy mmetries of parity, time reversal, and charge conjugation. The mass dimensions of the ﬁelds (in four spacetime dimensio ns) are [ϕ] = 1 and [Ψ] =3 2. Thus any power of ϕup toϕ4is allowed. But there are no additional required terms involving Ψ: the only candi dates contain eitherγ5(e.g.,iΨγ5Ψ) and are forbidden by parity, or C(e.g, ΨTCΨ) and are forbidden by the U(1) symmetry. Nevertheless, having to deal with the addition of three new t erms (ϕ, ϕ3,ϕ4) is annoying enough to prompt us to look for a simpler example . Consider, then, a modiﬁed form of the Yukawa interaction, LYuk=igϕΨγ5Ψ. (51.1) This interaction will conserve parity if and only if ϕis a pseudoscalar: P−1ϕ(x,t)P=−ϕ(−x,t). (51.2) Then,ϕandϕ3are odd under parity, and so we will notneed to add them toL. The one term we will need to add is ϕ4. Therefore, the theory we will consider is L=L0+L1, (51.3) L0=iΨ/∂Ψ−mΨΨ−1 2∂µϕ∂µϕ−1 2M2ϕ2, (51.4) L1=iZggϕΨγ5Ψ−1 24Zλλϕ4+Lct, (51.5) Lct=i(ZΨ−1)Ψ/∂Ψ−(Zm−1)mΨΨ −1 2(Zϕ−1)∂µϕ∂µϕ−1 2(ZM−1)M2ϕ2(51.6) whereλis a new coupling constant. We will use an on-shell renormali za- tion scheme. The lagrangian parameter mis then the actual mass of the electron. We will deﬁne the couplings gandλas the values of appropri- ate vertex functions when the external four-momenta vanish . Finally, the 51: Loop Corrections in Yukawa Theory 315 ﬁelds are normalized according to the requirements of the LS Z formula. In practice, this means that the scalar and fermion propagator s must have appropriate poles with unit residue. We will assume that M <2m, so that the scalar is stable against decay into an electron-positron pair. The exact scalar propagato r (in momentum space) can be then written in Lehmann-K¨ all´ en form as ˜∆(k2) =1 k2+M2−iǫ+/integraldisplay∞ M2 thdsρ(s) k2+s−iǫ, (51.7) where the spectral density ρ(s) is real and nonnegative. The threshold mass Mthis either 2m(corresponding to the contribution of an electron-positro n pair) or 3M(corresponding to the contribution of three scalars; by par ity, there is no contribution from two scalars), whichever is les s. We can also write ˜∆(k2)−1=k2+M2−iǫ−Π(k2), (51.8) whereiΠ(k2) is given by the sum of one-particle irreducible (1PI for sho rt; see section 14) diagrams with two external scalar lines, and the external propagators removed. The fact that ˜∆(k2) has a pole at k2=−M2with residue one implies that Π( −M2) = 0 and Π′(−M2) = 0; this ﬁxes the coeﬃcients ZϕandZM. All of this is mimicked for the Dirac ﬁeld. When parity is cons erved, the exact propagator (in momentum space) can be written in Lehma nn-K¨ all´ en form as ˜S(/p) =−/p+m p2+m2−iǫ+/integraldisplay∞ m2 thds−/pρ1(s) +√sρ2(s) p2+s−iǫ, (51.9) where the spectral densities ρ1(s) andρ2(s) are both real, and ρ1(s) is non- negative and greater than ρ2(s). The threshold mass mthism+M(corre- sponding to the contribution of a fermion and a scalar), whic h, by assump- tion, is less than 3 m(corresponding to the contribution of three fermions; by Lorentz invariance, there is no contribution from two fer mions). Sincep2=−/p/p, we can rewrite eq.(51.9) as ˜S(/p) =1 /p+m−iǫ+/integraldisplay∞ m2 thds−/pρ1(s) +√sρ2(s) (−/p+√s−iǫ)(/p+√s−iǫ),(51.10) with the understanding that 1 /(...) refers to the matrix inverse. However, since /pis the only matrix involved, we can think of ˜S(/p) as an analytic function of the single variable / p. With this idea in mind, we see that ˜S(/p) has an isolated pole at / p=−mwith residue one. This residue corresponds to the ﬁeld normalization that is needed for the validity of t he LSZ formula. 51: Loop Corrections in Yukawa Theory 316 lk k kkl kk k+l Figure 51.1: The one-loop and counterterm corrections to th e scalar prop- agator in Yukawa theory. We can also write the exact fermion propagator in the form ˜S(/p)−1= /p+m−iǫ−Σ(/p), (51.11) whereiΣ(/p) is given by the sum of 1PI diagrams with two external fermion lines, and the external propagators removed. The fact that ˜S(/p) has a pole at /p=−mwith residue one implies that Σ( −m) = 0 and Σ′(−m) = 0; this ﬁxes the coeﬃcients ZΨandZm. We proceed to the diagrams. The Yukawa vertex carries a facto r of i(iZgg)γ5=−Zggγ5. SinceZg= 1 +O(g2), we can set Zg= 1 in the one-loop diagrams. Consider ﬁrst Π( k2), which receives the one-loop (and counterterm) corrections shown in ﬁg.(51.1). The ﬁrst diagram has a close d fermion loop. As we will see in problem 51.1 (and section 53), anticommutat ion of the fermion ﬁelds results in an extra factor of minus one for each closed fermion loop. The spin indices on the propagators and vertices are co ntracted in the usual way, following the arrows backwards. Since the loo p closes on itself, we end up with a trace over the spin indices. Thus we ha ve iΠΨ loop(k2) = (−1)(−g)2/parenleftig 1 i/parenrightig2/integraldisplayd4ℓ (2π)4Tr/bracketleftig˜S(/ℓ+/k)γ5˜S(/ℓ)γ5/bracketrightig ,(51.12) where ˜S(/p) =−/p+m p2+m2−iǫ(51.13) is the free fermion propagator in momentum space. We now proceed to evaluate eq.(51.12). We have Tr[(−/ℓ−/k+m)γ5(−/ℓ+m)γ5] = Tr[( −/ℓ−/k+m)(+/ℓ+m)] = 4[(ℓ+k)ℓ+m2] ≡4N . (51.14) 51: Loop Corrections in Yukawa Theory 317 The ﬁrst equality follows from γ2 5= 1 andγ5/pγ5=−/p. Next we combine the denominators with Feynman’s formula. Su ppress- ing theiǫ’s, we have 1 (ℓ+k)2+m21 ℓ2+m2=/integraldisplay1 0dx1 (q2+D)2, (51.15) whereq=ℓ+xkandD=x(1−x)k2+m2. We then change the integration variable in eq.(51.12) from ℓtoq; the result is iΠΨ loop(k2) = 4g2/integraldisplay1 0dx/integraldisplayd4q (2π)4N (q2+D)2, (51.16) where nowN= (q+(1−x)k)(q−xk)+m2. The integral diverges, and so we analytically continue it to d= 4−εspacetime dimensions. (Here we ignore a subtlety with the deﬁnition of γ5inddimensions, and assume that γ2 5= 1 andγ5/pγ5=−/pcontinue to hold.) We also make the replacement g→ g˜µε/2, where ˜µhas dimensions of mass, so that gremains dimensionless. Expanding out the numerator, we have N=q2−x(1−x)k2+m2+ (1−2x)kq. (51.17) The term linear in qintegrates to zero. For the rest, we use the general result of section 14 to get ˜µε/integraldisplayddq (2π)d1 (q2+D)2=i 16π2/bracketleftbigg2 ε−ln(D/µ2)/bracketrightbigg , (51.18) ˜µε/integraldisplayddq (2π)dq2 (q2+D)2=i 16π2/bracketleftbigg2 ε+1 2−ln(D/µ2)/bracketrightbigg (−2D),(51.19) whereµ2= 4πe−γ˜µ2, and we have dropped terms of order ε. Plugging eqs.(51.18) and (51.19) into eq.(51.16) yields ΠΨloop(k2) =−g2 4π2/bracketleftigg 1 ε(k2+ 2m2) +1 6k2+m2 −/integraldisplay1 0dx/parenleftig 3x(1−x)k2+m2/parenrightig ln(D/µ2)/bracketrightigg .(51.20) We see that the divergent term has (as expected) a form that pe rmits cancellation by the counterterms. We evaluated the second diagram of ﬁg.(51.1) in section 31, w ith the result Πϕloop(k2) =λ (4π)2/bracketleftbigg1 ε+1 2−1 2ln(M2/µ2)/bracketrightbigg M2. (51.21) 51: Loop Corrections in Yukawa Theory 318 p ppl p p+l Figure 51.2: The one-loop and counterterm corrections to th e fermion prop- agator in Yukawa theory. The third diagram gives the contribution of the counterterm s, Πct(k2) =−(Zϕ−1)k2−(ZM−1)M2. (51.22) Adding up eqs.(51.20–51.22), we see that ﬁniteness of Π( k2) requires Zϕ= 1−g2 4π2/parenleftbigg1 ε+ ﬁnite/parenrightbigg , (51.23) ZM= 1 +/parenleftigg λ 16π2−g2 2π2m2 M2/parenrightigg/parenleftbigg1 ε+ ﬁnite/parenrightbigg , (51.24) plus higher-order (in gand/orλ) corrections. Note that, although there is anO(λ) correction to ZM, there is not an O(λ) correction to Zϕ. We can impose Π( −M2) = 0 by writing Π(k2) =g2 4π2/bracketleftbigg/integraldisplay1 0dx/parenleftig 3x(1−x)k2+m2/parenrightig ln(D/D 0) +κϕ(k2+M2)/bracketrightbigg , (51.25) whereD0=−x(1−x)M2+m2, andκϕis a constant to be determined. We ﬁxκϕby imposing Π′(−M2) = 0, which yields κϕ=/integraldisplay1 0dxx(1−x)[3x(1−x)M2−m2]/D0. (51.26) Note that, in this on-shell renormalization scheme, there i s noO(λ) correc- tion to Π(k2). Next we turn to the Ψ propagator, which receives the one-loop (and counterterm) corrections shown in ﬁg.(51.2). The spin indi ces are con- tracted in the usual way, following the arrows backwards. We have iΣ1loop(/p) = (−g)2/parenleftig 1 i/parenrightig2/integraldisplayd4ℓ (2π)4/bracketleftig γ5˜S(/p+ /ℓ)γ5/bracketrightig˜∆(ℓ2), (51.27) where ˜S(/p) is given by eq.(51.13), and ˜∆(ℓ2) =1 ℓ2+M2−iǫ(51.28) 51: Loop Corrections in Yukawa Theory 319 is the free scalar propagator in momentum space. We evaluate eq.(51.27) with the usual bag of tricks. The resu lt is iΣ1loop(/p) =−g2/integraldisplay1 0dx/integraldisplayd4q (2π)4N (q2+D)2, (51.29) whereq=ℓ+xpand N= /q+ (1−x)/p+m, (51.30) D=x(1−x)p2+xm2+ (1−x)M2. (51.31) The integral diverges, and so we analytically continue it to d= 4−ε spacetime dimensions, make the replacement g→g˜µε/2, and take the limit asε→0. The term linear in qin eq.(51.30) integrates to zero. Using eq.(51.18), we get Σ1loop(/p) =−g2 16π2/bracketleftigg 1 ε(/p+ 2m)−/integraldisplay1 0dx/parenleftig (1−x)/p+m/parenrightig ln(D/µ2)/bracketrightigg .(51.32) We see that the divergent term has (as expected) a form that pe rmits cancellation by the counterterms, which give Σct(/p) =−(ZΨ−1)/p−(Zm−1)m. (51.33) Adding up eqs.(51.32) and (51.33), we see that ﬁniteness of Σ (/p) requires ZΨ= 1−g2 16π2/parenleftbigg1 ε+ ﬁnite/parenrightbigg , (51.34) Zm= 1−g2 8π2/parenleftbigg1 ε+ ﬁnite/parenrightbigg , (51.35) plus higher-order corrections. We can impose Σ( −m) = 0 by writing Σ(/p) =g2 16π2/bracketleftbigg/integraldisplay1 0dx/parenleftig (1−x)/p+m/parenrightig ln(D/D 0) +κΨ(/p+m)/bracketrightbigg ,(51.36) whereD0isDevaluated at p2=−m2, andκΨis a constant to be deter- mined. We ﬁx κΨby imposing Σ′(−m) = 0. In diﬀerentiating with respect to /p, we take the p2inD, eq.(51.31), to be −/p2; we ﬁnd κΨ=−2/integraldisplay1 0dxx2(1−x)m2/D0. (51.37) 51: Loop Corrections in Yukawa Theory 320 +ll p+lp p kp Figure 51.3: The one-loop correction to the scalar-fermion -fermion vertex in Yukawa theory. Next we turn to the correction to the Yukawa vertex. We deﬁne t he vertex function iVY(p′,p) as the sum of one-particle irreducible diagrams with one incoming fermion with momentum p, one outgoing fermion with momentum p′, and one incoming scalar with momentum k=p′−p. The original vertex −Zggγ5is the ﬁrst term in this sum, and the diagram of ﬁg.(51.3) is the second. Thus we have iVY(p′,p) =−Zggγ5+iVY,1loop(p′,p) +O(g5), (51.38) where iVY,1loop(p′,p) = (−g)3/parenleftig 1 i/parenrightig3/integraldisplayddℓ (2π)d/bracketleftig γ5˜S(/p′+/ℓ)γ5˜S(/p+/ℓ)γ5/bracketrightig˜∆(ℓ2). (51.39) The numerator can be written as N= (/p′+ /ℓ+m)(−/p−/ℓ+m)γ5, (51.40) and the denominators combined in the usual way. We then get iVY,1loop(p′,p) =−ig3/integraldisplay dF3/integraldisplayd4q (2π)4N (q2+D)3, (51.41) where the integral over Feynman parameters was deﬁned in sec tion 16, and now q=ℓ+x1p+x2p′, (51.42) N= [/q−x1/p+ (1−x2)/p′+m][−/q−(1−x1)/p+x2/p′+m]γ5,(51.43) D=x1(1−x1)p2+x2(1−x2)p′2−2x1x2p·p′ + (x1+x2)m2+x3M2. (51.44) Using /q/q=−q2, we can write Nas N=q2γ5+/tildewideN+ (linear in q), (51.45) 51: Loop Corrections in Yukawa Theory 321 l k kk k 12 3 4 Figure 51.4: One of six diagrams with a closed fermion loop an d four external scalar lines; the other ﬁve are obtained by permuti ng the external momenta in all possible inequivalent ways. where /tildewideN= [−x1/p+ (1−x2)/p′+m][−(1−x1)/p+x2/p′+m]γ5. (51.46) The terms linear in qin eq.(51.45) integrate to zero, and only the ﬁrst term is divergent. Performing the usual manipulations, we ﬁnd iVY,1 loop(p′,p) =g3 8π2/bracketleftigg/parenleftbigg1 ε−1 4−1 2/integraldisplay dF3ln(D/µ2)/parenrightbigg γ5+1 4/integraldisplay dF3/tildewideN D/bracketrightigg . (51.47) From eq.(51.38), we see that ﬁniteness of VY(p′,p) requires Zg= 1 +g2 8π2/parenleftbigg1 ε+ ﬁnite/parenrightbigg , (51.48) plus higher-order corrections. To ﬁx the ﬁnite part of Zg, we need a condition to impose on VY(p′,p). We will mimic what we did in ϕ3theory in section 16, and require VY(0,0) to have the tree-level value igγ5. We leave the details to problem 51.2. Next we turn to the corrections to the ϕ4vertexiV4(k1,k2,k3,k4); the tree-level contribution is −iZλλ. There are diagrams with a closed fermion loop, as shown in ﬁg.(51.4), plus one-loop diagrams with ϕparticles only that we evaluated in section 31. We have iV4,Ψ loop = (−1)(−g)4/parenleftig 1 i/parenrightig4/integraldisplayd4ℓ (2π)4Tr/bracketleftig˜S(/ℓ)γ5˜S(/ℓ−/k1)γ5 ×˜S(/ℓ+/k2+/k3)γ5˜S(/ℓ+/k2)γ5/bracketrightig + 5 permutations of ( k2,k3,k4). (51.49) Again we can employ the standard methods; there are no unfami liar as- pects. This being the case, let us concentrate on obtaining t he divergent 51: Loop Corrections in Yukawa Theory 322 part; this will give us enough information to calculate the o ne-loop contri- butions to the beta functions for gandλ. To obtain the divergent part of eq.(51.49), it is suﬃcient to setki= 0. The term in the numerator that contributes to the divergen t part is Tr (/ℓγ5)4= 4(ℓ2)2, and the denominator is ( ℓ2+m2)4. Then we ﬁnd, after including identical contributions from the other ﬁve permu tations of the external momenta, V4,Ψ loop=−3g4 π2/parenleftbigg1 ε+ ﬁnite/parenrightbigg . (51.50) From section 31, we have V4,ϕloop=3λ 16π2/parenleftbigg1 ε+ ﬁnite/parenrightbigg . (51.51) Then, using V4=−Zλλ+V4,Ψloop+V4,ϕloop+... , (51.52) we see that ﬁniteness of V4requires Zλ= 1 +/parenleftigg 3λ 16π2−3g4 π2λ/parenrightigg/parenleftbigg1 ε+ ﬁnite/parenrightbigg , (51.53) plus higher-order corrections. Reference Notes A detailed derivation of the Lehmann-K¨ all´ en form of the fe rmion propaga- tor can be found in Itzykson & Zuber . Problems 51.1) Derive the fermion-loop correction to the scalar prop agator by work- ing through eq.(45.2), and show that it has an extra minus sig n rel- ative to the case of a scalar loop. 51.2) Finish the computation of VY(p′,p), imposing the condition VY(0,0) =igγ5. (51.54) 51.3) Consider making ϕa scalar rather than a pseudoscalar, so that the Yukawa interaction is LYuk=gϕΨΨ. In this case, renormalizability requires us to add a term Lϕ3=1 6Zκκϕ3, as well as term linear in ϕto cancel tadpoles. Find the one-loop contributions to the ren ormalizing Zfactors for this theory in the MS scheme. 52: Beta Functions in Yukawa Theory 323 52Beta Functions in Yukawa Theory Prerequisite: 28, 51 In this section we will compute the beta functions for the Yuk awa coupling gand theϕ4couplingλin Yukawa theory, using the methods of section 28. The relations between the bare and renormalized couplings a re g0=Z−1/2 ϕZ−1 ΨZg˜µε/2g, (52.1) λ0=Z−2 ϕZλ˜µελ. (52.2) Let us deﬁne ln/parenleftig Z−1/2 ϕZ−1 ΨZg/parenrightig =∞/summationdisplay n=1Gn(g,λ) εn, (52.3) ln/parenleftig Z−2 ϕZλ/parenrightig =∞/summationdisplay n=1Ln(g,λ) εn. (52.4) From our results in section 51, we have G1(g,λ) =5g2 16π2+... , (52.5) L1(g,λ) =3λ 16π2+g2 2π2−3g4 π2λ+... , (52.6) where the ellipses stand for higher-order (in g2and/orλ) corrections. Taking the logarithm of eqs.(52.1) and (52.2), and using eqs .(52.3) and (52.4), we get lng0=∞/summationdisplay n=1Gn(g,λ) εn+ lng+1 2εln ˜µ, (52.7) lnλ0=∞/summationdisplay n=1Ln(g,λ) εn+ lnλ+εln ˜µ. (52.8) We now use the fact that g0andλ0must be independent of µ. We diﬀeren- tiate eqs.(52.7) and (52.8) with respect to ln µ; the left-hand sides vanish, and we multiply the right-hand sides by gandλ, respectively. The result is 0 =∞/summationdisplay n=1/parenleftbigg g∂Gn ∂gdg dlnµ+g∂Gn ∂λdλ dlnµ/parenrightbigg1 εn+dg dlnµ+1 2εg, (52.9) 0 =∞/summationdisplay n=1/parenleftbigg λ∂Ln ∂gdg dlnµ+λ∂Ln ∂λdλ dlnµ/parenrightbigg1 εn+dλ dlnµ+ελ. (52.10) 52: Beta Functions in Yukawa Theory 324 In a renormalizable theory, dg/dlnµanddλ/dlnµmust be ﬁnite in the ε→0 limit. Thus we can write dg dlnµ=−1 2εg+βg(g,λ), (52.11) dλ dlnµ=−ελ+βλ(g,λ). (52.12) Substituting these into eqs.(52.9) and (52.10), and matchi ng powers of ε, we ﬁnd βg(g,λ) =g/parenleftbigg 1 2g∂ ∂g+λ∂ ∂λ/parenrightbigg G1, (52.13) βλ(g,λ) =λ/parenleftbigg 1 2g∂ ∂g+λ∂ ∂λ/parenrightbigg L1. (52.14) The coeﬃcients of all higher powers of 1 /εmust also vanish, but this gives us no more information about the beta functions. Using eqs.(52.5) and (52.6) in eqs.(52.13) and (52.14), we g et βg(g,λ) =5g3 16π2+... , (52.15) βλ(g,λ) =1 16π2/parenleftig 3λ2+ 8λg2−48g4/parenrightig +... . (52.16) The higher-order corrections have extra factors of g2and/orλ. Problems 52.1) Compute the one-loop contributions to the anomalous d imensions of m,M, Ψ, andϕ. 52.2) Consider the theory of problem 51.3. Compute the one-l oop contri- butions to the beta functions for g,λ, andκ, and to the anomalous dimensions of m,M, Ψ, andϕ. 52.3) Consider the beta functions of eqs.(52.15) and (52.16 ). a) Letρ≡λ/g2, and compute dρ/dlnµ. Express your answer in terms ofgandρ. Explain why it is better to work with gandρ rather than gandλ. Hint: the answer is mathematical, not physical. b) Show that there are two ﬁxed points ,ρ∗ +andρ∗ −, wheredρ/dlnµ= 0, and ﬁnd their values. c) Suppose that, for some particular value of the renormaliz ation scale µ, we haveρ= 0 andg≪ ≪1. What happens to ρat much higher 52: Beta Functions in Yukawa Theory 325 values ofµ(but still low enough to keep g≪1)? At much lower values ofµ? d) Same question, but with an initial value of ρ= 5. e) Same question, but with an initial value of ρ=−5. f) Find the trajectory in the ( ρ,g) plane that is followed for each of the three starting points as µis varied up and down. Hint: you should ﬁnd that the trajectories take the form g=g0/vextendsingle/vextendsingle/vextendsingle/vextendsingle/vextendsingleρ−ρ∗ + ρ−ρ∗−/vextendsingle/vextendsingle/vextendsingle/vextendsingle/vextendsingleν for some particular exponent ν. Put arrows on the trajectories that point in the direction of increasing µ. g) Explain why ρ∗ −is called an ultraviolet stable ﬁxed point, and why ρ∗ +is called an infrared stable ﬁxed point. 53: Functional Determinants 326 53Functional Determinants Prerequisite: 44, 45 In the section we will explore the meaning of the functional determinants that arise when doing gaussian path integrals, either boson ic or fermionic. We will be interested in situations where the path integral o ver one partic- ular ﬁeld is gaussian, but generates a functional determina nt that depends on some other ﬁeld. We will see how to relate this functional d eterminant to a certain inﬁnite set of Feynman diagrams. We will need the technology we develop here to compute the path integral for nonabelian g auge theory in section 70. We begin by considering a theory of a complex scalar ﬁeld χwith L=−∂µχ†∂µχ−m2χ†χ+gϕχ†χ, (53.1) whereϕis a real scalar background ﬁeld . That is,ϕ(x) is treated as a ﬁxed function of spacetime. Next we deﬁne the path integral Z(ϕ) =/integraldisplay Dχ†Dχei/integraltext d4xL, (53.2) where we use the ǫtrick of section 6 to impose vacuum boundary conditions, and the normalization Z(0) = 1 is ﬁxed by hand. Recall from section 44 that if we have ncomplex variables zi, then we can evaluate gaussian integrals by the general formula /integraldisplay dnzdn¯zexp (−i¯ziMijzj)∝(detM)−1. (53.3) In the case of the functional integral in eq.(53.2), the inde xion the inte- gration variable is replaced by the continuous spacetime la belx, and the “matrix”Mbecomes M(x,y) = [−∂2 x+m2−gϕ(x)]δ4(x−y). (53.4) In order to apply eq.(53.3), we have to understand what it mea ns to com- pute the determinant of this expression. To this end, let us ﬁrst note that we can write M=M0/tildewiderM, which is shorthand for M(x,z) =/integraldisplay d4yM0(x,y)/tildewiderM(y,z), (53.5) where M0(x,y) = (−∂2 x+m2)δ4(x−y), (53.6) /tildewiderM(y,z) =δ4(y−z)−g∆(y−z)ϕ(z). (53.7) 53: Functional Determinants 327 Here ∆(y−z) is the Feynman propagator, which obeys (−∂2 y+m2)∆(y−z) =δ4(y−z). (53.8) After various integrations by parts, it is easy to see that eq s.(53.5–53.7) reproduce eq.(53.4). Now we can use the general matrix relation det AB= detAdetBto conclude that detM= detM0det/tildewiderM . (53.9) The advantage of this decomposition is that M0is independent of the background ﬁeld ϕ, and so the resulting factor of (det M0)−1inZ(ϕ) can simply be absorbed into the overall normalization. Further more, we have /tildewiderM=I−G, where I(x,y) =δ4(x−y) (53.10) is the identity matrix, and G(x,y) =g∆(x−y)ϕ(y). (53.11) Thus, forϕ(x) = 0, we have/tildewiderM=Iand so det/tildewiderM= 1. Then, using eq.(53.3) and the normalization condition Z(0) = 1, we see that for nonzero ϕ(x) we must have simply Z(ϕ) = (det/tildewiderM)−1. (53.12) Next, we need the general matrix relation det A= exp Tr ln A, which is most easily proved by working in a basis where Ais in Jordan form (that is, all entries below the main diagonal are zero). Thus we can write det/tildewiderM= exp Tr ln/tildewiderM = exp Tr ln( I−G) = exp Tr/bracketleftigg −∞/summationdisplay n=11 nGn/bracketrightigg . (53.13) Combining eqs.(53.12) and (53.13) we get Z(ϕ) = exp∞/summationdisplay n=11 nTrGn, (53.14) where TrGn=gn/integraldisplay d4x1...d4xn∆(x1−x2)ϕ(x2)...∆(xn−x1)ϕ(x1).(53.15) 53: Functional Determinants 328 ... Figure 53.1: All connected diagrams with ϕ(x) treated as an external ﬁeld. Each of the ndots represents a factor of igϕ(x), and each solid line is a χ or Ψ propagator. This is our ﬁnal result for Z(ϕ). To better understand what it means, we will rederive it in a di ﬀerent way. Consider treating the gϕχ†χterm in Las an interaction. This leads to a vertex that connects two χpropagators; the associated vertex factor isigϕ(x). According to the general analysis of section 9, we have Z(ϕ) = expiΓ(ϕ), whereiΓ(ϕ) is given by a sum of connected diagrams. (We have called the exponent Γ rather than Wbecause it is naturally interpreted as a quantum action for ϕafterχhas been integrated out.) The only connected diagrams we can draw with these Feynman rules are those of ﬁg. (53.1), withninsertions of the vertex, where n≥1. The diagram with nvertices has ann-fold cyclic symmetry, leading to a symmetry factor of S=n. The factor of iassociated with each vertex is canceled by the factor of 1 /i associated with each propagator. Thus the value of the n-vertex diagram is 1 ngn/integraldisplay d4x1...d4xn∆(x1−x2)ϕ(x2)...∆(xn−x1)ϕ(x1). (53.16) Summing up these diagrams, and using eq.(53.15), we ﬁnd iΓ(ϕ) =∞/summationdisplay n=11 nTrGn. (53.17) This neatly reproduces eq.(53.14). Thus we see that a functi onal determi- nant can be represented as an inﬁnite sum of Feynman diagrams . Next we consider a theory of a Dirac fermion Ψ with L=iΨ/∂Ψ−mΨΨ +gϕΨΨ, (53.18) whereϕis again a real scalar background ﬁeld. We deﬁne the path inte gral Z(ϕ) =/integraldisplay DΨDΨei/integraltext d4xL, (53.19) 53: Functional Determinants 329 where we again use the ǫtrick to impose vacuum boundary conditions, and the normalization Z(0) = 1 is ﬁxed by hand. Recall from section 44 that if we have ncomplex Grassmann variables ψi, then we can evaluate gaussian integrals by the general form ula /integraldisplay dn¯ψdnψexp/parenleftbig−i¯ψiMijψj/parenrightbig∝detM . (53.20) In the case of the functional integral in eq.(53.19), the ind exion the integration variable is replaced by the continuous spaceti me labelxplus the spinor index α, and the “matrix” Mbecomes Mαβ(x,y) = [−i/∂x+m−gϕ(x)]αβδ4(x−y). (53.21) In order to apply eq.(53.20), we have to understand what it me ans to compute the determinant of this expression. To this end, let us ﬁrst note that we can write M=M0/tildewiderM, which is shorthand for Mαγ(x,z) =/integraldisplay d4yM0αβ(x,y)/tildewiderMβγ(y,z), (53.22) where M0αβ(x,y) = (−i/∂x+m)αβδ4(x−y), (53.23) /tildewiderMβγ(y,z) =δβγδ4(y−z)−gSβγ(y−z)ϕ(z). (53.24) HereSβγ(y−z) is the Feynman propagator, which obeys (−i/∂y+m)αβSβγ(y−z) =δαγδ4(y−z). (53.25) After various integrations by parts, it is easy to see that eq s.(53.22–53.24) reproduce eq.(53.21). Now we can use eq.(53.9). The advantage of this decompositio n is that M0is independent of the background ﬁeld ϕ, and so the resulting factor of detM0inZ(ϕ) can simply be absorbed into the overall normalization. Furthermore, we have/tildewiderM=I−G, where Iαβ(x,y) =δαβδ4(x−y) (53.26) is the identity matrix, and Gαβ(x,y) =gSαβ(x−y)ϕ(y). (53.27) Thus, forϕ(x) = 0, we have/tildewiderM=Iand so det/tildewiderM= 1. Then, using eq.(53.20) and the normalization condition Z(0) = 1, we see that for nonzeroϕ(x) we must have simply Z(ϕ) = det/tildewiderM . (53.28) 53: Functional Determinants 330 Next, we use eqs.(53.13) and (53.28) to get Z(ϕ) = exp −∞/summationdisplay n=11 nTrGn, (53.29) where now TrGn=gn/integraldisplay d4x1...d4xntrS(x1−x2)ϕ(x2)...S(xn−x1)ϕ(x1),(53.30) and “tr” denotes a trace over spinor indices. This is our ﬁnal result for Z(ϕ). To better understand what it means, we will rederive it in a di ﬀerent way. Consider treating the gϕΨΨ term in Las an interaction. This leads to a vertex that connects two Ψ propagators; the associated v ertex factor isigϕ(x). According to the general analysis of section 9, we have Z(ϕ) = expiΓ(ϕ), whereiΓ(ϕ) is given by a sum of connected diagrams. (We have called the exponent Γ rather than Wbecause it is naturally interpreted as a quantum action for ϕafter Ψ has been integrated out.) The only connected diagrams we can draw with these Feynman rules are t hose of ﬁg.(53.1), with ninsertions of the vertex, where n≥1. The diagram with nvertices has an n-fold cyclic symmetry, leading to a symmetry factor of S=n. The factor of iassociated with each vertex is canceled by the factor of 1/iassociated with each propagator. The closed fermion loop im plies a trace over the spinor indices. Thus the value of the n-vertex diagram is 1 ngn/integraldisplay d4x1...d4xntrS(x1−x2)ϕ(x2)...S(xn−x1)ϕ(x1).(53.31) Summing up these diagrams, we ﬁnd that we are missing the over all minus sign in eq.(53.29). The appropriate conclusion is that we mu st associate an extra minus sign with each closed fermion loop. Part TIT Spin One 54: Maxwell’s Equations 332 54Maxwell’s Equations Prerequisite: 3 The most common (and important) spin-one particle is the pho ton. Emis- sion and absorption of photons by matter is an important phen omenon in many areas of physics, and so that is the context in which most physicists ﬁrst encounter a serious treatment of photons. We will use a b rief review of this subject (in this section and the next) as our entry poi nt into the theory of quantum electrodynamics. Let us begin with classical electrodynamics. Maxwell’s equ ations are ∇·E=ρ, (54.1) ∇ ×B−˙E=J, (54.2) ∇ ×E+˙B= 0, (54.3) ∇·B= 0, (54.4) whereEis the electric ﬁeld, Bis the magnetic ﬁeld, ρis the charge density, andJis the current density. We have written Maxwell’s equations in Heaviside-Lorentz units , and also set c= 1. In these units, the magnitude of the force between two charges of magnitude QisQ2/4πr2. Maxwell’s equations must be supplemented by formulae that g ive us the dynamics of the charges and currents (such as the Lorentz force law for point particles). For now, however, we will treat the charge s and currents as speciﬁed sources, and focus on the dynamics of the electroma gnetic ﬁelds. The last two of Maxwell’s equations, the ones with no sources on the right-hand side, can be solved by writing the EandBﬁelds in terms of a scalar potential ϕand a vector potential A, E=−∇ϕ−˙A, (54.5) B=∇ ×A. (54.6) The potentials uniquely determine the ﬁelds, but the ﬁelds d o not uniquely determine the potentials. Given a particular ϕandAthat result in a particular EandB, we will get the same EandBfrom any other potentials ϕ′andA′that are related by ϕ′=ϕ+˙Γ, (54.7) A′=A− ∇Γ, (54.8) where Γ is an arbitrary function of spacetime. A change of pot entials that does not change the ﬁelds is called a gauge transformation . The EandB ﬁelds are gauge invariant . 54: Maxwell’s Equations 333 All this becomes more compact and elegant in a relativistic n otation. We deﬁne the four-vector potential or gauge ﬁeld Aµ≡(ϕ,A). (54.9) We also deﬁne the ﬁeld strength Fµν≡∂µAν−∂νAµ. (54.10) Obviously,Fµνis antisymmetric: Fµν=−Fνµ. Comparing eqs.(54.5) and (54.6) with eqs.(54.9) and (54.10), we see that F0i=Ei, (54.11) Fij=εijkBk. (54.12) The ﬁrst two of Maxwell’s equations can now be written as ∂νFµν=Jµ, (54.13) where Jµ≡(ρ,J) (54.14) is the charge-current density four-vector. If we take the four-divergence of eq.(54.13), we get ∂µ∂νFµν=∂µJµ. The left-hand side of this equation vanishes, because ∂µ∂νis symmetric on exchange of µandν, whileFµνis antisymmetric. We conclude that we must have ∂µJµ= 0, (54.15) or equivalently ˙ρ+∇·J= 0 ; (54.16) that is, the electromagnetic current must be conserved. The last two of Maxwell’s equations can be written as εµνρσ∂ρFµν= 0, (54.17) whereεµνρσis the completely antisymmetric Levi-Civita tensor; see se ction 34. Plugging in eq.(54.10), we see that eq.(54.17) is automa tically satisﬁed, since the antisymmetric combination of two derivatives van ishes. Eqs.(54.7) and (54.8) can be combined into A′µ=Aµ−∂µΓ. (54.18) SettingF′µν=∂µA′ν−∂νA′µand using eq.(54.18), we get F′µν=Fµν−(∂µ∂ν−∂ν∂µ)Γ. (54.19) 54: Maxwell’s Equations 334 The last term vanishes because derivatives commute; thus th e ﬁeld strength is gauge invariant, F′µν=Fµν. (54.20) Next we will ﬁnd an action that results in Maxwell’s equation s as the equations of motion. We will treat the current as an external source. The action we seek should be Lorentz invariant, gauge invariant , parity and time-reversal invariant, and no more than second order in de rivatives. The only candidate is S=/integraltextd4xL, where L=−1 4FµνFµν+JµAµ. (54.21) The ﬁrst term is obviously gauge invariant, because Fµνis. After a gauge transformation, eq.(54.18), the second term becomes JµA′ µ, and the diﬀer- ence is Jµ(A′ µ−Aµ) =−Jµ∂µΓ =−(∂µJµ)Γ−∂µ(JµΓ). (54.22) The ﬁrst term in eq.(54.22) vanishes because the current is c onserved. The second term is a total divergence, and its integral over d4xvanishes (assum- ing suitable boundary conditions at inﬁnity). Thus the acti on speciﬁed by eq.(54.21) is gauge invariant. SettingFµν=∂µAν−∂νAµand multiplying out the terms, eq.(54.21) becomes L=−1 2∂µAν∂µAν+1 2∂µAν∂νAµ+JµAµ (54.23) = +1 2Aµ(gµν∂2−∂µ∂ν)Aν+JµAµ−∂µKµ, (54.24) whereKµ=1 2Aν(∂µAν−∂νAµ). The last term is a total divergence, and can be dropped. From eq.(54.24), we can see that varying Aµwhile requiring Sto be unchanged yields the equation of motion (gµν∂2−∂µ∂ν)Aν+Jµ= 0. (54.25) Noting that ∂νFµν=∂ν(∂µAν−∂νAµ) = (∂µ∂ν−gµν∂2)Aν, we see that eq.(54.25) is equivalent to eq.(54.13), and hence to Maxwel l’s equations. 55: Electrodynamics in Coulomb Gauge 335 55Electrodynamics in Coulomb Gauge Prerequisite: 54 Next we would like to construct the hamiltonian, and quantiz e the electro- magnetic ﬁeld. There is an immediate diﬃculty, caused by the gauge invarian ce: we have too many degrees of freedom. This problem manifests its elf in several ways. For example, the lagrangian L=−1 4FµνFµν+JµAµ (55.1) =−1 2∂µAν∂µAν+1 2∂µAν∂νAµ+JµAµ (55.2) does not contain the time derivative of A0. Thus, this ﬁeld has no canoni- cally conjugate momentum and no dynamics. To deal with this problem, we must eliminate the gauge freedo m. We do this by choosing a gauge . We choose a gauge by imposing a gauge condition . This is a condition that we require Aµ(x) to satisfy. The idea is that there should be only one Aµ(x) that results in a given Fµν(x) and that also satisﬁes the gauge condition. One possible class of gauge conditions is nµAµ(x) = 0, where nµis a constant four-vector. If nis spacelike ( n2>0), then we have chosen axial gauge; ifnis lightlike, ( n2= 0), it is lightcone gauge ; and ifnis timelike, (n2<0), it is temporal gauge . Another gauge is Lorenz gauge , where the condition is ∂µAµ= 0. We will meet a family of closely related gauges in section 62. In this section, we will work in Coulomb gauge , also known as radiation gauge ortransverse gauge . The condition for Coulomb gauge is ∇·A(x) = 0. (55.3) We can impose eq.(55.3) by acting on Ai(x) with a projection operator, Ai(x)→/parenleftbigg δij−∇i∇j ∇2/parenrightbigg Aj(x). (55.4) We construct the right-hand side of eq.(55.4) by Fourier-tr ansforming Ai(x) to/tildewideAi(k), multiplying/tildewideAi(k) by the matrix δij−kikj/k2, and then Fourier- transforming back to position space. From now on, whenever w e writeAi, we will implicitly mean the right-hand side of eq.(55.4). Now let us write out the lagrangian in terms of the scalar and v ector potentials, ϕ=A0andAi, withAiobeying the Coulomb gauge condition. Starting from eq.(55.2), we get L=1 2˙Ai˙Ai−1 2∇jAi∇jAi+JiAi 55: Electrodynamics in Coulomb Gauge 336 +1 2∇iAj∇jAi+˙Ai∇iϕ +1 2∇iϕ∇iϕ−ρϕ. (55.5) In the second line of eq.(55.5), the ∇iin each term can be integrated by parts; in the ﬁrst term, we will then get a factor of ∇j(∇iAi), and in the second term, we will get a factor of ∇i˙Ai. Both of these vanish by virtue of the gauge condition ∇iAi= 0, and so both of these terms can simply be dropped. If we now vary ϕ(and require S=/integraltextd4xLto be stationary), we ﬁnd thatϕobeys Poisson’s equation, −∇2ϕ=ρ. (55.6) The solution is ϕ(x,t) =/integraldisplay d3yρ(y,t) 4π|x−y|. (55.7) This solution is unique if we impose the boundary conditions thatϕandρ both vanish at spatial inﬁnity. Eq.(55.7) tells us that ϕ(x,t) is given entirely in terms of the charge density at the same time, and so has no dynamics of its own. It i s therefore legitimate to plug eq.(55.7) back into the lagrangian. Afte r an integration by parts to turn ∇iϕ∇iϕinto−ϕ∇2ϕ=ϕρ, the result is L=1 2˙Ai˙Ai−1 2∇jAi∇jAi+JiAi+Lcoul, (55.8) where Lcoul=−1 2/integraldisplay d3yρ(x,t)ρ(y,t) 4π|x−y|. (55.9) We can now vary Ai; keeping proper track of the implicit projection opera- tor in eq.(55.4), we ﬁnd that Aiobeys the massless Klein-Gordon equation with the projected current as a source, −∂2Ai(x) =/parenleftbigg δij−∇i∇j ∇2/parenrightbigg Jj(x). (55.10) For a free ﬁeld ( Ji= 0), the general solution is A(x) =/summationdisplay λ=±/integraldisplay /tildewiderdk/bracketleftig ε∗ λ(k)aλ(k)eikx+ελ(k)a† λ(k)e−ikx/bracketrightig , (55.11) wherek0=ω=|k|,/tildewiderdk=d3k/(2π)32ω, and ε+(k) and ε−(k) are po- larization vectors. In order to satisfy the Coulomb gauge co ndition, the polarization vectors must be orthogonal to the wave vector k. We will 55: Electrodynamics in Coulomb Gauge 337 choose them to correspond to right- and left-handed circula r polarizations; fork= (0,0,k), we then have ε+(k) =1√ 2(1,−i,0), ε−(k) =1√ 2(1,+i,0). (55.12) More generally, the two polarization vectors along with the unit vector in thekdirection form an orthonormal and complete set, k·ελ(k) = 0, (55.13) ελ′(k)·ε∗ λ(k) =δλ′λ, (55.14) /summationdisplay λ=±ε∗ iλ(k)εjλ(k) =δij−kikj k2. (55.15) The coeﬃcients aλ(k) anda† λ(k) will become operators after quantization, which is why we have used the dagger symbol for conjugation. In complete analogy with the procedure used for a scalar ﬁeld in section 3, we can invert eq.(55.11) and its time derivative to get aλ(k) = +iελ(k)·/integraldisplay d3xe−ikx↔∂0A(x), (55.16) a† λ(k) =−iε∗ λ(k)·/integraldisplay d3xe+ikx↔ ∂0A(x), (55.17) wheref↔ ∂µg=f(∂µg)−(∂µf)g. Now we can proceed to the hamiltonian formalism. First, we co mpute the canonically conjugate momentum to Ai, Πi=∂L ∂˙Ai=˙Ai. (55.18) Note that ∇iAi= 0 implies ∇iΠi= 0. The hamiltonian density is then H= Πi˙Ai− L =1 2ΠiΠi+1 2∇jAi∇jAi−JiAi+Hcoul, (55.19) where Hcoul=−Lcoul. To quantize the ﬁeld, we impose the canonical commutation re lations. Keeping proper track of the implicit projection operator in eq.(55.4), we have [Ai(x,t),Πj(y,t)] =i/parenleftbigg δij−∇i∇j ∇2/parenrightbigg δ3(x−y) =i/integraldisplayd3k (2π)3eik·(x−y)/parenleftbigg δij−kikj k2/parenrightbigg .(55.20) 55: Electrodynamics in Coulomb Gauge 338 The commutation relations of the aλ(k) anda† λ(k) operators follow from eq.(55.20) and [ Ai,Aj] = [Πi,Πj] = 0 (at equal times). The result is [aλ(k),aλ′(k′)] = 0, (55.21) [a† λ(k),a† λ′(k′)] = 0, (55.22) [aλ(k),a† λ′(k′)] = (2π)32ωδ3(k′−k)δλλ′. (55.23) We interpret a† λ(k) andaλ(k) as creation and annihilation operators for photons of deﬁnite helicity, with helicity +1 correspondin g to right-circular polarization and helicity −1 to left-circular polarization. It is now straightfoward to write the hamiltonian explicitl y in terms of these operators. We ﬁnd H=/summationdisplay λ=±/integraldisplay /tildewiderdkωa† λ(k)aλ(k) + 2E0V−/integraldisplay d3xJ(x)·A(x) +Hcoul,(55.24) where E0=1 2(2π)−3/integraltextd3k ωis the zero-point energy per unit volume that we found for a real scalar ﬁeld in section 3, Vis the volume of space, the Coulomb hamiltonian is Hcoul=1 2/integraldisplay d3xd3yρ(x,t)ρ(y,t) 4π|x−y|, (55.25) and we use eq.(55.11) to express Ai(x) in terms of aλ(k) anda† λ(k) at any one particular time (say, t= 0). This is suﬃcient, because Hitself is time independent. This form of the hamiltonian of electrodynamics is often use d as the starting point for calculations of atomic transition rates , with the charges and currents treated via the nonrelativistic Schr¨ odinger equation. The Coulomb interaction appears explicitly, and the J·Aterm allows for the creation and annihilation of photons of deﬁnite polarizati on. Reference Notes A more rigorous treatment of quantization in Coulomb gauge c an be found inWeinberg I . Problems 55.1) Use eqs.(55.13–55.20) and [ Ai,Aj] = [Πi,Πj] = 0 (at equal times) to verify eqs.(55.21–55.23). 55.2) Use eqs.(55.11), (55.14), (55.19), and (55.21–55.23 ) to verify eq.(55.24). 56: LSZ Reduction for Photons 339 56LSZ Reduction for Photons Prerequisite: 5, 55 In section 55, we found that the creation and annihilation op erators for free photons could be written as a† λ(k) =−iε∗ λ(k)·/integraldisplay d3xe+ikx↔∂0A(x), (56.1) aλ(k) = +iελ(k)·/integraldisplay d3xe−ikx↔∂0A(x), (56.2) where ελ(k) is a polarization vector. From here, we can follow the analy sis of section 5 line by line to deduce the LSZ reduction formula f or photons. The result is that the creation operator for each incoming ph oton should be replaced by a† λ(k)in→iεµ∗ λ(k)/integraldisplay d4xe+ikx(−∂2)Aµ(x), (56.3) and the destruction operator for each outgoing photon shoul d be replaced by aλ(k)out→iεµ λ(k)/integraldisplay d4xe−ikx(−∂2)Aµ(x), (56.4) and then we should take the vacuum expectation value of the ti me-ordered product. Note that, in writing eqs.(56.3) and (56.4), we hav e made them look nicer by introducing ε0 λ(k)≡0, and then using four-vector dot prod- ucts rather than three-vector dot products. The LSZ formula is valid provided the ﬁeld is normalized acco rding to the free-ﬁeld formulae ∝an}b∇acketle{t0|Ai(x)|0∝an}b∇acket∇i}ht= 0, (56.5) ∝an}b∇acketle{tk,λ|Ai(x)|0∝an}b∇acket∇i}ht=εi λ(k)eikx, (56.6) where |k,λ∝an}b∇acket∇i}htis a single photon state, normalized according to ∝an}b∇acketle{tk′,λ′|k,λ∝an}b∇acket∇i}ht= (2π)32ωδ3(k′−k)δλλ′. (56.7) The zero on the right-hand side of eq.(56.5) is required by ro tation invari- ance, and only the overall scale of the right-hand side of eq. (56.6) might be diﬀerent in an interacting theory. The renormalization of Ainecessitates including appropriate Zfactors in the lagrangian, L=−1 4Z3FµνFµν+Z1JµAµ. (56.8) 56: LSZ Reduction for Photons 340 HereZ3andZ1are the traditional names; we will meet Z2in section 62. We must choose Z3so that eq.(56.6) holds. We will ﬁx Z1by requiring the corresponding vertex function to take on a certain value for a particular set of external momenta. Next we must compute the correlation functions ∝an}b∇acketle{t0|TAi(x)...|0∝an}b∇acket∇i}ht. As usual, we begin by working with free ﬁeld theory. The analysi s is again almost identical to the case of a scalar ﬁeld; see problem 8.4 . We ﬁnd that, in free ﬁeld theory, ∝an}b∇acketle{t0|TAi(x)Aj(y)|0∝an}b∇acket∇i}ht=1 i∆ij(x−y), (56.9) where the propagator is ∆ij(x−y) =/integraldisplayd4k (2π)4eik(x−y) k2−iǫ/summationtext λ=±εi∗ λ(k)εj λ(k). (56.10) As with a free scalar ﬁeld, correlations of an odd number of ﬁe lds van- ish, and correlations of an even number of ﬁelds are given in t erms of the propagator by Wick’s theorem; see section 8. We would now like to evaluate the path integral for the free el ectromag- netic ﬁeld Z0(J)≡ ∝an}b∇acketle{t0|0∝an}b∇acket∇i}htJ=/integraldisplay DAei/integraltext d4x[−1 4FµνFµν+JµAµ]. (56.11) Here we treat the current Jµ(x) as an external source. We will evaluate Z0(J) in Coulomb gauge. This means that we will integrate over only those ﬁeld conﬁgurations that satisfy ∇·A= 0. We begin by integrating over A0. Because the action is quadratic in Aµ, this is equivalent to solving the variational equation for A0, and then substituting the solution back into the lagrangian. The res ult is that we have the Coulomb term in the action, Scoul=−1 2/integraldisplay d4xd4yδ(x0−y0)J0(x)J0(y) 4π|x−y|. (56.12) Since this term does not depend on the vector potential, we si mply get a factor of exp( iScoul) in front of the remaining path integral over Ai. We wll perform this integral formally (as we did for fermion ﬁelds i n section 43) by requiring it to yield the correct results for the correlat ion functions of Aiwhen we take functional derivatives with respect to Ji. In this way we ﬁnd that Z0(J) = exp/bracketleftbigg iScoul+i 2/integraldisplay d4xd4yJi(x)∆ij(x−y)Jj(y)/bracketrightbigg . (56.13) 56: LSZ Reduction for Photons 341 We can make Z0(J) look prettier by writing it as Z0(J) = exp/bracketleftbiggi 2/integraldisplay d4xd4yJµ(x)∆µν(x−y)Jν(y)/bracketrightbigg , (56.14) where we have deﬁned ∆µν(x−y)≡/integraldisplayd4k (2π)4eik(x−y)˜∆µν(k), (56.15) ˜∆µν(k)≡ −1 k2δµ0δν0+1 k2−iǫ/summationtext λ=±εµ∗ λ(k)εν λ(k).(56.16) The ﬁrst term on the right-hand side of eq.(56.16) reproduce s the Coulomb term in eq.(56.13) by virtue of the facts that /integraldisplay+∞ −∞dk0 2πe−ik0(x0−y0)=δ(x0−y0), (56.17) /integraldisplayd3k (2π)3eik·(x−y) k2=1 4π|x−y|. (56.18) The second term on the right-hand side of eq.(56.16) reprodu ces the second term in eq.(56.13) by virtue of the fact that ε0 λ(k) = 0. Next we will simplify eq.(56.16). We begin by introducing a u nit vector in the time direction, ˆtµ= (1,0). (56.19) Next we need a unit vector in the kdirection, which we will call ˆ zµ. We ﬁrst note that ˆt·k=−k0, and so we can write (0,k) =kµ+ (ˆt·k)ˆtµ. (56.20) The square of this four-vector is k2=k2+ (ˆt·k)2, (56.21) where we have used ˆt2=−1. Thus the unit vector that we want is ˆzµ=kµ+ (ˆt·k)ˆtµ [k2+ (ˆt·k)2]1/2. (56.22) Now we recall from section 55 that /summationdisplay λ=±εi∗ λ(k)εj λ(k) =δij−kikj k2. (56.23) 56: LSZ Reduction for Photons 342 This can be extended to i→µandj→νby writing /summationdisplay λ=±εµ∗ λ(k)εν λ(k) =gµν+ˆtµˆtν−ˆzµˆzν. (56.24) It is not hard to check that the right-hand side of eq.(56.24) vanishes if µ= 0 orν= 0, and agrees with eq.(56.23) for µ=iandν=j. Putting all this together, we can now write eq.(56.16) as ˜∆µν(k) =−ˆtµˆtν k2+ (ˆt·k)2+gµν+ˆtµˆtν−ˆzµˆzν k2−iǫ. (56.25) The next step is to consider the terms in this expression that contain factors ofkµorkν; from eq.(56.22), we see that these will arise from the ˆzµˆzνterm. In eq.(56.15), a factor of kµcan be written as a derivative with respect toxµacting oneik(x−y). This derivative can then be integrated by parts in eq.(56.14) to give a factor of ∂µJµ(x). But∂µJµ(x) vanishes, because the current must be conserved. Similarly, a factor o fkνcan be turned into ∂νJν(y), and also leads to a vanishing contribution. Therefore, we can ignore any terms in ˜∆µν(k)that contain factors of kµorkν. From eq.(56.22), we see that this means we can make the substi tution ˆzµ→(ˆt·k)ˆtµ [k2+ (ˆt·k)2]1/2. (56.26) Then eq.(56.25) becomes ˜∆µν(k) =1 k2−iǫ/bracketleftigg gµν+/parenleftigg −k2 k2+ (ˆt·k)2+ 1−(ˆt·k)2 k2+ (ˆt·k)2/parenrightigg ˆtµˆtν/bracketrightigg , (56.27) where the three coeﬃcients of ˆtµˆtνcome from the Coulomb term, the ˆtµˆtν term in the polarization sum, and the ˆ zµˆzνterm, respectively. A bit of algebra now reveals that the net coeﬃcient of ˆtµˆtνvanishes, leaving us with the elegant expression ˜∆µν(k) =gµν k2−iǫ. (56.28) Written in this way, the photon propagator is said to be in Feynman gauge . (It would still be in Coulomb gauge if we had retained the kµandkνterms that we previously dropped.) In the next section, we will rederive eq.(56.28) from a more e xplicit path-integral point of view. Problems 56.1) Use eqs.(55.11) and (55.21–55.23) to verify eqs.(56. 9–56.10). 57: The Path Integral for Photons 343 57The Path Integral for Photons Prerequisite: 8, 56 In this section, in order to get a better understanding of the photon path integral, we will evaluate it directly, using the methods of section 8. We begin with Z0(J) =/integraldisplay DAeiS0, (57.1) S0=/integraldisplay d4x/bracketleftig −1 4FµνFµν+JµAµ/bracketrightig . (57.2) Following section 8, we Fourier-transform to momentum spac e, where we ﬁnd S0=1 2/integraldisplayd4k (2π)4/bracketleftig −/tildewideAµ(k)/parenleftig k2gµν−kµkν/parenrightig /tildewideAν(−k) +/tildewideJµ(k)/tildewideAµ(−k) +/tildewideJµ(−k)/tildewideAµ(k)/bracketrightig . (57.3) The next step is to shift the integration variable/tildewideAso as to “complete the square”. This involves inverting the 4 ×4 matrixk2gµν−kµkν. However, this matrix has a zero eigenvalue, and cannot be inverted. To see this, let us write k2gµν−kµkν=k2Pµν(k), (57.4) where we have deﬁned Pµν(k)≡gµν−kµkν/k2. (57.5) This is a projection matrix because, as is easily checked, Pµν(k)Pνλ(k) =Pµλ(k). (57.6) Thus the only allowed eigenvalues of Pare zero and one. There is at least one zero eigenvalue, because Pµν(k)kν= 0. (57.7) On the other hand, the sum of the eigenvalues is given by the tr ace gµνPµν(k) = 3. (57.8) Thus the remaining three eigenvalues must all be one. 57: The Path Integral for Photons 344 Now let us imagine carrying out the path integral of eq.(57.1 ), with S0given by eq.(57.3). Let us decompose the ﬁeld/tildewideAµ(k) into components aligned along a set of linearly independent four-vectors, o ne of which is kµ. (It will not matter whether or not this basis set is orthonorm al.) Because the term quadratic in/tildewideAµinvolves the matrix k2Pµν(k), andPµν(k)kν= 0, the component of/tildewideAµ(k) that lies along kµdoes not contribute to this quadratic term. Furthermore, it does not contribute to the l inear term either, because ∂µJµ(x) = 0 implies kµ/tildewideJµ(k) = 0. Thus this component does not appear in the path integal at all! It then makes no sen se to integrate over it. We therefore deﬁne/integraltextDAto mean integration over only those components that are spanned by the remaining three bas is vectors, and therefore satisfy kµ/tildewideAµ(k) = 0. This is equivalent to imposing Lorenz gauge,∂µAµ(x) = 0. The matrix Pµν(k) is simply the matrix that projects a four-vector into the subspace orthogonal to kµ. Within the subspace, Pµν(k) is equiva- lent to the identity matrix. Therefore, within the subspace , the inverse ofk2Pµν(k) is (1/k2)Pµν(k). Employing the ǫtrick to pick out vacuum boundary conditions replaces k2withk2−iǫ. We can now continue following the procedure of section 8, wit h the result that Z0(J) = exp/bracketleftigg i 2/integraldisplayd4k (2π)4/tildewideJµ(k)Pµν(k) k2−iǫ/tildewideJν(−k)/bracketrightigg = exp/bracketleftbiggi 2/integraldisplay d4xd4yJµ(x)∆µν(x−y)Jν(y)/bracketrightbigg , (57.9) where ∆µν(x−y) =/integraldisplayd4k (2π)4eik(x−y)Pµν(k) k2−iǫ(57.10) is the photon propagator in Lorenz gauge (also known as Landau gauge ). Of course, because the current is conserved, the kµkνterm inPµν(k) does not contribute, and so the result is equivalent to that of Fey nman gauge, wherePµν(k) is replaced by gµν. 58: Spinor Electrodynamics 345 58Spinor Electrodynamics Prerequisite: 45, 57 In the section, we will study spinor electrodynamics : the theory of photons interacting with the electrons and positrons of a Dirac ﬁeld . (We will use the term quantum electrodynamics to denote any theory of photons, irrespective of the kinds of particles with which they inter act.) We construct spinor electrodynamics by taking the electrom agnetic cur- rentjµ(x) to be proportional to the Noether current corresponding to the U(1) symmetry of a Dirac ﬁeld; see section 36. Speciﬁcally, jµ(x) =eΨ(x)γµΨ(x). (58.1) Heree=−0.302822 is the charge of the electron in Heaviside-Lorentz units, with ¯ h=c= 1. (We will rely on context to distinguish this e from the base of natural logarithms.) In these units, the ﬁne -structure constant is α=e2/4π= 1/137.036. With the normalization of eq.(58.1), Q=/integraltextd3xj0(x) is the electric charge operator. Of course, when we specify a number in quantum ﬁeld theory, we must always have a renormalization scheme in mind; e=−0.302822 corresponds to a speciﬁc version of on-shell renormalization that we wil l explore in sections 62 and 63. The value of eis diﬀerent in other renormalization schemes, such as MS, as we will see in section 66. The complete lagrangian of our theory is thus L=−1 4FµνFµν+iΨ/∂Ψ−mΨΨ +eΨγµΨAµ. (58.2) In this section, we will be concerned with tree-level proces ses only, and so we omit renormalizing Zfactors. We have a problem, though. A Noether current is conserved onl y when the ﬁelds obey the equations of motion, or, equivalently, on ly at points in ﬁeld space where the action is stationary. On the other han d, in our development of photon path integrals in sections 56 and 57, w e assumed that the current was always conserved. This issue is resolved by enlarging the deﬁnition of a gauge t ransfor- mation to include a transformation on the Dirac ﬁeld as well a s the elec- tromagnetic ﬁeld. Speciﬁcally, we deﬁne a gauge transforma tion to consist of Aµ(x)→Aµ(x)−∂µΓ(x), (58.3) Ψ(x)→exp[−ieΓ(x)]Ψ(x), (58.4) Ψ(x)→exp[+ieΓ(x)]Ψ(x). (58.5) 58: Spinor Electrodynamics 346 It is not hard to check that L(x) isinvariant under this transformation, whether or not the ﬁelds obey their equations of motion. To pe rform this check most easily, we ﬁrst rewrite Las L=−1 4FµνFµν+iΨ /DΨ−mΨΨ, (58.6) where we have deﬁned the gauge covariant derivative (or just covariant derivative for short) Dµ≡∂µ−ieAµ. (58.7) In the last section, we found that Fµνis invariant under eq.(58.3), and so theFFterm in Lis obviously invariant as well. It is also obvious that themΨΨ term in Lis invariant under eqs.(58.4) and (58.5). This leaves the Ψ /DΨ term. This term will also be invariant if, under the gauge transformation, the covariant derivative of Ψ transforms a s DµΨ(x)→exp[−ieΓ(x)]DµΨ(x). (58.8) To see if this is true, we note that DµΨ→/parenleftig ∂µ−ie[Aµ−∂µΓ]/parenrightig/parenleftig exp[−ieΓ]Ψ/parenrightig = exp[ −ieΓ]/parenleftig ∂µΨ−ie(∂µΓ)Ψ−ie[Aµ−∂µΓ]Ψ/parenrightig = exp[ −ieΓ]/parenleftig ∂µ−ieAµ/parenrightig Ψ = exp[ −ieΓ]DµΨ. (58.9) So eq.(58.8) holds, and Ψ /DΨ is gauge invariant. We can also write the transformation rule for Dµa little more abstractly as Dµ→e−ieΓDµe+ieΓ, (58.10) where the ordinary derivative in Dµis deﬁned to act on anything to its right, including any ﬁelds that are left unwritten in eq.(58 .10). Thus we have DµΨ→/parenleftig e−ieΓDµe+ieΓ/parenrightig/parenleftig e−ieΓΨ/parenrightig =e−ieΓDµΨ, (58.11) which is, of course, the same as eq.(58.9). We can also expres s the ﬁeld strength in terms of the covariant derivative by noting that [Dµ,Dν]Ψ(x) =−ieFµν(x)Ψ(x). (58.12) 58: Spinor Electrodynamics 347 We can write this more abstractly as Fµν=ie[Dµ,Dν], (58.13) where, again, the ordinary derivative in each covariant der ivative acts on anything to its right. From eqs.(58.10) and (58.13), we see t hat, under a gauge transformation, Fµν→ie/bracketleftig e−ieΓDµe+ieΓ,e−ieΓDνe+ieΓ/bracketrightig =e−ieΓ/parenleftigie[Dµ,Dν]/parenrightig e+ieΓ =e−ieΓFµνe+ieΓ =Fµν. (58.14) In the last line, we are able to cancel the e±ieΓfactors against each other because no derivatives act on them. Eq.(58.14) shows us that (as we already knew)Fµνis gauge invariant. It is interesting to note that the gauge transformation on th e fermion ﬁelds, eqs.(58.4–58.5), is a generalization of the U(1) tra nsformation Ψ→e−iαΨ, (58.15) Ψ→e+iαΨ, (58.16) that is a symmetry of the free Dirac lagrangian. The diﬀerenc e is that, in the gauge transformation, the phase factor is allowed to b e a function of spacetime, rather than a constant that is the same everywh ere. Thus, the gauge transformation is also called a localU(1) transformation, while eqs.(58.15–58.16) correspond to a global U(1) transformation. We say that, in a gauge theory, the global U(1) symmetry is promoted to a lo cal U(1) symmetry, or that we have gauged the U(1) symmetry. In section 57, we argued that the path integral over Aµshould be re- stricted to those components of/tildewideAµ(k) that are orthogonal to kµ, because the component parallel to kµdid not appear in the integrand. Now we must make a slightly more subtle argument. We argue that the path i ntegral over the parallel component is redundant, because the fermionic path integral over Ψ and Ψ already includes all possible values of Γ( x). Therefore, as in section 57, we should not integrate over the parallel compon ent. (We will make a more precise and careful version of this argument when we discuss the quantization of nonabelian gauge theories in section 71 .) 58: Spinor Electrodynamics 348 By the standard procedure, this leads us to the following for m of the path integral for spinor electrodynamics: Z(η,η,J )∝exp/bracketleftigg ie/integraldisplay d4x/parenleftbigg1 iδ δJµ(x)/parenrightbigg/parenleftbigg iδ δηα(x)/parenrightbigg (γµ)αβ/parenleftigg 1 iδ δηβ(x)/parenrightigg/bracketrightigg ×Z0(η,η,J ), (58.17) where Z0(η,η,J ) = exp/bracketleftbigg i/integraldisplay d4xd4yη(x)S(x−y)η(y)/bracketrightbigg ×exp/bracketleftbiggi 2/integraldisplay d4xd4yJµ(x)∆µν(x−y)Jν(y)/bracketrightbigg ,(58.18) and S(x−y) =/integraldisplayd4p (2π)4(−/p+m) p2+m2−iǫeip(x−y), (58.19) ∆µν(x−y) =/integraldisplayd4k (2π)4gµν k2−iǫeik(x−y)(58.20) are the appropriate Feynman propagators for the correspond ing free ﬁelds, with the photon propagator in Feynman gauge. We impose the no rmaliza- tionZ(0,0,0) = 1, and write Z(η,η,J ) = exp[iW(η,η,J )]. (58.21) TheniW(η,η,J ) can be expressed as a series of connected Feynman dia- grams with sources. The rules for internal and external Dirac fermions were work ed out in the context of Yukawa theory in section 45, and they follow he re with no change. For external photons, the LSZ analysis of section 56 implies that each external photon line carries a factor of the polarizati on vectorεµ(k). Putting everything together, we get the following set of Fey nman rules for tree-level processes in spinor electrodynamics. 1. For each incoming electron , draw a solid line with an arrow pointed towards the vertex, and label it with the electron’s four-momentum, pi. 2. For each outgoing electron , draw a solid line with an arrow pointed awayfrom the vertex, and label it with the electron’s four-momen tum, p′ i. 58: Spinor Electrodynamics 349 3. For each incoming positron , draw a solid line with an arrow pointed away from the vertex, and label it with minus the positron’s four- momentum, −pi. 4. For each outgoing positron , draw a solid line with an arrow pointed towards the vertex, and label it with minus the positron’s four-momentum, −p′ i. 5. For each incoming photon , draw a wavy line with an arrow pointed towards the vertex, and label it with the photon’s four-momentum, ki. (Wavy lines for photons is a standard convention.) 6. For each outgoing photon , draw a wavy line with an arrow pointed away from the vertex, and label it with the photon’s four-momentu m, k′ i. 7. The only allowed vertex joins two solid lines, one with an a rrow point- ing towards it and one with an arrow pointing away from it, and one wavy line (whose arrow can point in either direction). Using this vertex, join up all the external lines, including extra inte rnal lines as needed. In this way, draw all possible diagrams that are topologically inequivalent . 8. Assign each internal line its own four-momentum. Think of the four- momenta as ﬂowing along the arrows, and conserve four-momen tum at each vertex. For a tree diagram, this ﬁxes the momenta on al l the internal lines. 9. The value of a diagram consists of the following factors: for each incoming photon, εµ∗ λi(ki); for each outgoing photon, εµ λ′i(k′ i); for each incoming electron, usi(pi); for each outgoing electron, us′ i(p′ i); for each incoming positron, vsi(pi); for each outgoing positron, vs′ i(p′ i); for each vertex, ieγµ; for each internal photon, −igµν/(k2−iǫ); for each internal fermion, −i(−/p+m)/(p2+m2−iǫ). 10. Spinor indices are contracted by starting at one end of a f ermion line: speciﬁcally, the end that has the arrow pointing away f rom the vertex. The factor associated with the external line is eith eruorv. 58: Spinor Electrodynamics 350 Go along the complete fermion line, following the arrows bac kwards, and write down (in order from left to right) the factors assoc iated with the vertices and propagators that you encounter. The last fa ctor is either auorv. Repeat this procedure for the other fermion lines, if any. The vector index on each vertex is contracted with the ve ctor index on either the photon propagator (if the attached photo n line is internal) or the photon polarization vector (if the attac hed photon line is external). 11. The overall sign of a tree diagram is determined by drawin g all con- tributing diagrams in a standard form: all fermion lines hor izontal, with their arrows pointing from left to right, and with the le ft end- points labeled in the same ﬁxed order (from top to bottom); if the ordering of the labels on the right endpoints of the fermion l ines in a given diagram is an even (odd) permutation of an arbitrarily chosen ﬁxed ordering, then the sign of that diagram is positive (neg ative). 12. The value of iT(at tree level) is given by a sum over the values of all the contributing diagrams. In the next section, we will do a sample calculation. Problems 58.1) Compute P−1Aµ(x,t)P,T−1Aµ(x,t)T, andC−1Aµ(x,t)C, assuming thatP,T, andCare symmetries of the lagrangian. (Prerequisite: 40.) 58.2)Furry’s theorem. Show that any scattering amplitude with no exter- nal fermions, and an odd number of external photons, is zero. 59: Scattering in Spinor Electrodynamics 351 59Scattering in Spinor Electrodynamics Prerequisite: 48, 58 In the last section, we wrote down the Feynman rules for spino r electro- dynamics. In this section, we will compute the scattering am plitude (and its spin-averaged square) at tree level for the process of el ectron-positron annihilation into a pair of photons, e+e−→γγ. The contributing diagrams are shown in ﬁg.(59.1), and the as sociated expression for the scattering amplitude is T=e2εµ 1′εν 2′v2/bracketleftbigg γν/parenleftbigg−/p1+ /k′ 1+m −t+m2/parenrightbigg γµ+γµ/parenleftbigg−/p1+ /k′ 2+m −u+m2/parenrightbigg γν/bracketrightbigg u1, (59.1) whereεµ 1′is shorthand for εµ λ′1(k′ 1),v2is shorthand for vs2(p2), and so on. The Mandelstam variables are s=−(p1+p2)2=−(k′ 1+k′ 2)2, t=−(p1−k′ 1)2=−(p2−k′ 2)2, u=−(p1−k′ 2)2=−(p2−k′ 1)2, (59.2) and they obey s+t+u= 2m2. Following the procedure of section 46, we write eq.(59.1) as T=εµ 1′εν 2′v2Aµνu1, (59.3) where Aµν≡e2/bracketleftbigg γν/parenleftbigg−/p1+ /k′ 1+m −t+m2/parenrightbigg γµ+γµ/parenleftbigg−/p1+ /k′ 2+m −u+m2/parenrightbigg γν/bracketrightbigg .(59.4) We also have T∗=T=ερ∗ 1′εσ∗ 2′u1Aρσv2. (59.5) Using /a/b...=.../b/a, we see from eq.(59.4) that Aρσ=Aσρ. (59.6) Thus we have |T |2=εµ 1′εν 2′ερ∗ 1′εσ∗ 2′(v2Aµνu1)(u1Aσρv2). (59.7) Next, we will average over the initial electron and positron spins, using the technology of section 46; the result is 1 4/summationdisplay s1,s2|T |2=1 4εµ 1′εν 2′ερ∗ 1′εσ∗ 2′Tr/bracketleftig Aµν(−/p1+m)Aσρ(−/p2−m)/bracketrightig . (59.8) 59: Scattering in Spinor Electrodynamics 352 21k k21p p2p1k11p p2k2 1kp1k Figure 59.1: Diagrams for e+e−→γγ, corresponding to eq.(59.1). We would also like to sum over the ﬁnal photon polarizations. From eq.(59.8), we see that we must evaluate /summationdisplay λ=±εµ λ(k)ερ∗ λ(k). (59.9) We did this polarization sum in Coulomb gauge in section 56, w ith the result that/summationdisplay λ=±εµ λ(k)ερ∗ λ(k) =gµρ+ˆtµˆtρ−ˆzµˆzρ, (59.10) where ˆtµis a unit vector in the time direction, and ˆ zµis a unit vector in thekdirection that can be expressed as ˆzµ=kµ+ (ˆt·k)ˆtµ [k2+ (ˆt·k)2]1/2. (59.11) It is tempting to drop the kµandkρterms in eq.(59.10), on the grounds that the photons couple to a conserved current, and so these t erms should not contribute. (We indeed used this argument to drop the ana logous terms in the photon propagator.) This also follows from the n otion that the scattering amplitude should be invariant under a gauge t ransformation, as represented by a transformation of the external polariza tion vectors of the form εµ λ(k)→εµ λ(k)−i˜Γ(k)kµ. (59.12) Thus, if we write a scattering amplitude Tfor a process that includes a particular outgoing photon with four-momentum kµas T=εµ λ(k)Mµ, (59.13) or a particular incoming photon with four-momentum kµas T=εµ∗ λ(k)Mµ, (59.14) 59: Scattering in Spinor Electrodynamics 353 then in either case we should have kµMµ= 0. (59.15) Eq.(59.15) is in fact valid; we will give a proof of it, based o n the Ward identity for the electromagnetic current, in section 67. Fo r now, we will take eq.(59.15) as given, and so drop the kµandkρterms in eq.(59.10). This leaves us with /summationdisplay λ=±εµ λ(k)ερ∗ λ(k)→gµρ+ˆtµˆtρ−(ˆt·k)2 k2+ (ˆt·k)2ˆtµˆtρ. (59.16) But, for an external photon, k2= 0. Thus the second and third terms in eq.(59.16) cancel, leaving us with the beautifully simple s ubstitution rule /summationdisplay λ=±εµ λ(k)ερ∗ λ(k)→gµρ. (59.17) Using eq.(59.17), we can sum |T |2over the polarizations of the outgoing photons, in addition to averaging over the spins of the incom ing fermions; the result is ∝an}b∇acketle{t|T |2∝an}b∇acket∇i}ht ≡1 4/summationdisplay λ′ 1,λ′ 2/summationdisplay s1,s2|T |2 =1 4Tr/bracketleftig Aµν(−/p1+m)Aνµ(−/p2−m)/bracketrightig =e4/bracketleftbigg∝an}b∇acketle{tΦtt∝an}b∇acket∇i}ht (m2−t)2+∝an}b∇acketle{tΦtu∝an}b∇acket∇i}ht+∝an}b∇acketle{tΦut∝an}b∇acket∇i}ht (m2−t)(m2−u)+∝an}b∇acketle{tΦuu∝an}b∇acket∇i}ht (m2−u)2/bracketrightbigg ,(59.18) where ∝an}b∇acketle{tΦtt∝an}b∇acket∇i}ht=1 4Tr/bracketleftig γν(−/p1+/k′ 1+m)γµ(−/p1+m)γµ(−/p1+/k′ 1+m)γν(−/p2−m)/bracketrightig , ∝an}b∇acketle{tΦuu∝an}b∇acket∇i}ht=1 4Tr/bracketleftig γµ(−/p1+/k′ 2+m)γν(−/p1+m)γν(−/p1+/k′ 2+m)γµ(−/p2−m)/bracketrightig , ∝an}b∇acketle{tΦtu∝an}b∇acket∇i}ht=1 4Tr/bracketleftig γν(−/p1+/k′ 1+m)γµ(−/p1+m)γν(−/p1+/k′ 2+m)γµ(−/p2−m)/bracketrightig , ∝an}b∇acketle{tΦut∝an}b∇acket∇i}ht=1 4Tr/bracketleftig γµ(−/p1+/k′ 2+m)γν(−/p1+m)γµ(−/p1+/k′ 1+m)γν(−/p2−m)/bracketrightig . (59.19) Examinging ∝an}b∇acketle{tΦtt∝an}b∇acket∇i}htand∝an}b∇acketle{tΦuu∝an}b∇acket∇i}ht, we see that they are transformed into each other byk′ 1↔k′ 2, which is equivalent to t↔u. The same is true of ∝an}b∇acketle{tΦtu∝an}b∇acket∇i}ht and∝an}b∇acketle{tΦut∝an}b∇acket∇i}ht. Thus we need only compute ∝an}b∇acketle{tΦtt∝an}b∇acket∇i}htand∝an}b∇acketle{tΦtu∝an}b∇acket∇i}ht, and then take t↔uto get ∝an}b∇acketle{tΦuu∝an}b∇acket∇i}htand∝an}b∇acketle{tΦut∝an}b∇acket∇i}ht. 59: Scattering in Spinor Electrodynamics 354 Now we can apply the gamma-matrix technology of section 47. I n particular, we will need the d= 4 relations γµγµ=−4, γµ/aγµ= 2/a, γµ/a/bγµ= 4(ab), γµ/a/b/cγµ= 2/c/b/a, (59.20) in addition to the trace formulae. We also need p1p2=−1 2(s−2m2), k′ 1k′ 2=−1 2s, p1k′ 1=p2k′ 2= +1 2(t−m2), p1k′ 2=p2k′ 1= +1 2(u−m2). (59.21) which follow from eq.(59.2) plus the mass-shell conditions p2 1=p2 2=−m2 andk′2 1=k′2 2= 0. After a lengthy and tedious calculation, we ﬁnd ∝an}b∇acketle{tΦtt∝an}b∇acket∇i}ht= 2[tu−m2(3t+u)−m4], (59.22) ∝an}b∇acketle{tΦtu∝an}b∇acket∇i}ht= 2m2(s−4m2), (59.23) which then implies ∝an}b∇acketle{tΦuu∝an}b∇acket∇i}ht= 2[tu−m2(3u+t)−m4], (59.24) ∝an}b∇acketle{tΦut∝an}b∇acket∇i}ht= 2m2(s−4m2). (59.25) This completes our calculation. Other tree-level scattering processes in spinor electrody namics pose no new calculational diﬃculties, and are left to the problems. In the high-energy limit, where the electron can be treated a s massless, we can reduce our labor with the method of spinor helicity , which was introduced in section 50. We take this up in the next section. Problems 59.1) Compute ∝an}b∇acketle{t|T |2∝an}b∇acket∇i}htforCompton scattering ,e−γ→e−γ. You should ﬁnd that your result is the same as that for e+e−→γγ, but with s↔t, and an extra overall minus sign. This is an example of crossing symmetry ; there is an overall minus sign for each fermion that is moved from the initial to the ﬁnal state. 59: Scattering in Spinor Electrodynamics 355 59.2) Compute ∝an}b∇acketle{t|T |2∝an}b∇acket∇i}htforBhabha scattering ,e+e−→e+e−. 59.3) Compute ∝an}b∇acketle{t|T |2∝an}b∇acket∇i}htforMøller scattering ,e−e−→e−e−. You should ﬁnd that your result is the same as that for e+e−→e+e−, but with s↔u. This is another example of crossing symmetry. 60: Spinor Helicity for Spinor Electrodynamics 356 60Spinor Helicity for Spinor Electrodynamics Prerequisite: 50, 59 In section 50, we introduced a special notation for uandvspinors of deﬁnite helicity for massless electrons and positrons. This notation greatly simpli- ﬁes calculations in the high-energy limit ( s,|t|, and |u|all much greater thanm2). We deﬁne the twistors |p]≡u−(p) =v+(p), |p∝an}b∇acket∇i}ht ≡u+(p) =v−(p), [p| ≡u+(p) =v−(p), ∝an}b∇acketle{tp| ≡u−(p) =v+(p). (60.1) We then have [k| |p] = [kp], ∝an}b∇acketle{tk| |p∝an}b∇acket∇i}ht=∝an}b∇acketle{tkp∝an}b∇acket∇i}ht, [k| |p∝an}b∇acket∇i}ht= 0, ∝an}b∇acketle{tk| |p] = 0, (60.2) where the twistor products [kp] and∝an}b∇acketle{tkp∝an}b∇acket∇i}htare antisymmetric, [kp] =−[pk], ∝an}b∇acketle{tkp∝an}b∇acket∇i}ht=−∝an}b∇acketle{tpk∝an}b∇acket∇i}ht, (60.3) and related by complex conjugation, ∝an}b∇acketle{tpk∝an}b∇acket∇i}ht∗= [kp]. They can be expressed explicitly in terms of the components of the massless four-m omentakand p. However, more useful are the relations ∝an}b∇acketle{tkp∝an}b∇acket∇i}ht[pk] = Tr1 2(1−γ5)/k/p =−2k·p =−(k+p)2(60.4) and ∝an}b∇acketle{tpq∝an}b∇acket∇i}ht[qr]∝an}b∇acketle{trs∝an}b∇acket∇i}ht[sp] = Tr1 2(1−γ5)/p/q/r/s = 2[(p·q)(r·s)−(p·r)(q·s) + (p·s)(q·r) +iεµνρσpµqνrρsσ]. (60.5) 60: Spinor Helicity for Spinor Electrodynamics 357 Finally, for any massless four-momentum pwe can write −/p=|p∝an}b∇acket∇i}ht[p|+|p]∝an}b∇acketle{tp|. (60.6) We will quote other results from section 50 as we need them. To apply this formalism to spinor electrodynamics, we need t o write photon polarization vectors in terms of twistors. The formu lae we need are εµ +(k) =−∝an}b∇acketle{tq|γµ|k]√ 2∝an}b∇acketle{tqk∝an}b∇acket∇i}ht, (60.7) εµ −(k) =−[q|γµ|k∝an}b∇acket∇i}ht√ 2[qk], (60.8) whereqis an arbitrary massless reference momentum . We will verify eqs.(60.7) and (60.8) for a speciﬁc choice of k, and then rely on the Lorentz transformation properties of twistors t o conclude that the result must hold in any frame (and therefore for any massl ess four- momentum k). We will choose kµ= (ω,ωˆ z) =ω(1,0,0,1). Then, the most general form ofεµ +(k) is εµ +(k) =eiφ1√ 2(0,1,−i,0) +Ckµ. (60.9) Hereeiφis an arbitrary phase factor, and Cis an arbitrary complex num- ber; the freedom to add a multiple of kcomes from the underlying gauge invariance. To verify that eq.(60.7) reproduces eq.(60.9), we need the e xplicit form of the twistors |k] and|k∝an}b∇acket∇i}htwhen the three-momentum is in the zdirection. Using results in section 50 we ﬁnd |k] =√ 2ω 0 1 0 0 , |k∝an}b∇acket∇i}ht=√ 2ω 0 0 1 0 . (60.10) For any value of q, the twistor ∝an}b∇acketle{tq|takes the form ∝an}b∇acketle{tq|= (0,0, α, β), (60.11) whereαandβare complex numbers. Plugging eqs.(60.10) and (60.11) into eq.(60.7), and using γµ=/parenleftigg0σµ ¯σµ0/parenrightigg (60.12) 60: Spinor Helicity for Spinor Electrodynamics 358 along with σµ= (I,/vector σ) and ¯σµ= (I,−/vector σ), we ﬁnd that we reproduce eq.(60.9) with eiφ= 1 andC=−β/(√ 2αω). There is now no need to check eq.(60.8), because εµ −(k) =−[εµ +(k)]∗, as can be seen by using ∝an}b∇acketle{tqk∝an}b∇acket∇i}ht∗= −[qk] along with another result from section 50, ∝an}b∇acketle{tq|γµ|k]∗=∝an}b∇acketle{tk|γµ|q]. In spinor electodynamics, the vector index on a photon polar ization vector is always contracted with the vector index on a gamma m atrix. We can get a convenient formula for / ε±(k) by using the Fierz identities −1 2γµ∝an}b∇acketle{tq|γµ|k] =|k]∝an}b∇acketle{tq|+|q∝an}b∇acket∇i}ht[k|, (60.13) −1 2γµ[q|γµ|k∝an}b∇acket∇i}ht=|k∝an}b∇acket∇i}ht[q|+|q]∝an}b∇acketle{tk|. (60.14) We then have /ε+(k;q) =√ 2 ∝an}b∇acketle{tqk∝an}b∇acket∇i}ht/parenleftig |k]∝an}b∇acketle{tq|+|q∝an}b∇acket∇i}ht[k|/parenrightig , (60.15) /ε−(k;q) =√ 2 [qk]/parenleftig |k∝an}b∇acket∇i}ht[q|+|q]∝an}b∇acketle{tk|/parenrightig , (60.16) where we have added the reference momentum as an explicit arg ument on the left-hand sides. Now we have all the tools we need for doing calculations. Howe ver, we can simplify things even further by making maximal use of c rossing symmetry. Note from eq.(60.1) that u−(which is the factor associated with an in- coming electron) and v+(an outgoing positron) are both represented by the twistor |p], whileu+(an outgoing electron) and v−(an incoming positron) are both represented by [ p|. Thus the square-bracket twistors correspond to outgoing fermions with positive helicity, and incoming f ermions with negative helicity. Similarly, the angle-bracket twistors correspond to out- going fermions with negative helicity, and incoming fermio ns with positive helicity. Let us adopt a convention in which all particles are assigned four- momenta that are treated as outgoing. A particle that has an a ssigned four- momentum pthen has physical four-momentum ǫpp, whereǫp= sign(p0) = +1 if the particle is physically outgoing, and ǫp= sign(p0) =−1 if the particle is physically incoming. Since the physical three-momentum of an incoming particle i s opposite to its assigned three-momentum, a particle with negative he licity relative to its physical three-momentum has positive helicity relat ive to its assigned three-momentum. From now on, we will refer to the helicity of a particle relative to its assigned momentum. Thus a particle that we sa y has “pos- itive helicity” actually has negative physical helicity if it is incoming, and positive physical helicity if it is outgoing. 60: Spinor Helicity for Spinor Electrodynamics 359 4p1 p2p1 p2p4 p4 p3pp1 3p3 pp1 Figure 60.1: Diagrams for fermion-fermion scattering, wit h all momenta treated as outgoing. With this convention, the square-bracket twistors |p] and [p|represent positive-helicity fermions, and the angle-bracket twisto rs|p∝an}b∇acket∇i}htand∝an}b∇acketle{tp|repre- sent negative-helicity fermions. When ǫp= sign(p0) =−1, we analytically continue the twistors by replacing each ω1/2in eq.(60.10) with i|ω|1/2. Then all of our formulae for twistors and polarizations hold with out change, with the exception of the rule for complex conjugation of a twisto r product, which becomes ∝an}b∇acketle{tpk∝an}b∇acket∇i}ht∗=ǫpǫk[kp]. (60.17) Now we are ready to calculate some amplitudes. Consider ﬁrst the process of fermion-fermion scattering. The contributing t ree-level diagrams are shown in ﬁg.(60.1). The ﬁrst thing to notice is that a diagram is zero if two extern al fermion lines that meet at a vertex have the same helicity. This is bec ause (as shown in section 50) we get zero if we sandwich the product of an odd n umber of gamma matrices between two twistors of the same helicity. In particular, we have ∝an}b∇acketle{tp|γµ|k∝an}b∇acket∇i}ht= 0 and [p|γµ|k] = 0. Thus, we will get a nonzero result for the tree-level amplitude only if two of the helicities ar e positive, and two are negative. This means that, of the 24= 16 possible combinations of helicities, only six give a nonzero tree-level amplitude :T++−−,T+−+−, T+−−+,T−−++,T−+−+, andT−++−, where the notation is Ts1s2s3s4. Fur- thermore, the last three of these are related to the ﬁrst thre e by complex conjugation, so we only have three amplitudes to compute. Let us begin with T+−−+. Only the ﬁrst diagram of ﬁg.(60.1) con- tributes, because the second has two postive-helicity line s meeting at a vertex. To evaluate the ﬁrst diagram, we note that the two ver tices con- tribute a factor of ( ie)2=−e2, and the internal photon line contributes a factor ofigµν/s13, where we have deﬁned the Mandelstam variable sij≡ −(pi+pj)2. (60.18) Following the charge arrows backwards on each fermion line, and dividing 60: Spinor Helicity for Spinor Electrodynamics 360 byito get T(rather than iT), we ﬁnd T+−−+=−e2∝an}b∇acketle{t3|γµ|1][4|γµ|2∝an}b∇acket∇i}ht/s13 = +2e2[14]∝an}b∇acketle{t23∝an}b∇acket∇i}ht/s13, (60.19) where ∝an}b∇acketle{t3|is short for ∝an}b∇acketle{tp3|, etc, and we have used yet another form of the Fierz identity to get the second line. The computation of T+−+−is exactly analogous, except that now it is only the second diagram of ﬁg.(60.1) that contributes. Acco rding to the Feynman rules, this diagram comes with a relative minus sign , and so we have T+−+−=−2e2[13]∝an}b∇acketle{t24∝an}b∇acket∇i}ht/s14. (60.20) Finally, we turn to T++−−. Now both diagrams contribute, and we have T++−−=−e2/parenleftbigg∝an}b∇acketle{t3|γµ|1]∝an}b∇acketle{t4|γµ|2] s13−∝an}b∇acketle{t4|γµ|1]∝an}b∇acketle{t3|γµ|2] s14/parenrightbigg =−2e2[12]∝an}b∇acketle{t34∝an}b∇acket∇i}ht/parenleftbigg1 s13+1 s14/parenrightbigg = +2e2[12]∝an}b∇acketle{t34∝an}b∇acket∇i}ht/parenleftbiggs12 s13s14/parenrightbigg , (60.21) where we used the Mandelstam relation s12+s13+s14= 0 to get the last line. To get the cross section for a particular set of helicities, w e must take the absolute squares of the amplitudes. These follow from eqs.( 60.4), (60.17), and (60.18) : |∝an}b∇acketle{t12∝an}b∇acket∇i}ht|2=|[12]|2=ǫ1ǫ2s12=|s12|. (60.22) We can then compute the spin-averaged cross section by summi ng the ab- solute squares of eqs.(60.19–60.21), multiplying by two to account for the processes in which all helicities are opposite (and which ha ve amplitudes that are related by complex conjugation), and then dividing by four to aver- age over the initial helicities. Making use of s34=s12and its permutations, we ﬁnd ∝an}b∇acketle{t|T |2∝an}b∇acket∇i}ht= 2e4/parenleftigg s2 14 s2 13+s2 13 s2 14+s4 12 s2 13s2 14/parenrightigg = 2e4/parenleftigg s4 12+s4 13+s4 14 s2 13s2 14/parenrightigg . (60.23) For the processes of e−e−→e−e−ande+e+→e+e+, we haves12=s, s13=t, ands14=u; fore+e−→e+e−, we haves13=s,s14=t, and s12=u. 60: Spinor Helicity for Spinor Electrodynamics 361 4k3 k4p2 p2 kp1p1 p1p1k4 3k3 k Figure 60.2: Diagrams for fermion-photon scattering, with all momenta treated as outgoing. Now we turn to processes with two external fermions and two ex ternal photons, as shown in ﬁg.(60.2). The ﬁrst thing to notice is th at a diagram is zero if the two external fermion lines have the same helicity . This is because the corresponding twistors sandwich an odd number of gamma m atrices: one from each vertex, and one from the massless fermion propa gator ˜S(p) = −/p/p2. Thus we need only compute T+−λ3λ4sinceT−+λ3λ4is related by complex conjugation. Next we use eqs.(60.15–60.16) and (60.2–60.3) to get /ε−(k;p)|p] = 0, (60.24) [p|/ε−(k;p) = 0. (60.25) /ε+(k;p)|p∝an}b∇acket∇i}ht= 0, (60.26) ∝an}b∇acketle{tp|/ε+(k;p) = 0, (60.27) Thus we can get some amplitudes to vanish with appropriate ch oices of the reference momenta in the photon polarizations. So, let us consider T+−λ3λ4=−e2∝an}b∇acketle{t2|/ελ4(k4;q4)(/p1+ /k3)/ελ3(k3;q3)|1]/s13 −e2∝an}b∇acketle{t2|/ελ3(k3;q3)(/p1+ /k4)/ελ4(k4;q4)|1]/s14.(60.28) If we takeλ3=λ4=−, then we can get both terms in eq.(60.28) to vanish by choosing q3=q4=p1, and using eq.(60.24). If we take λ3=λ4= +, then we can get both terms in eq.(60.28) to vanish by choosing q3=q4=p2, and using eq.(60.27). Thus, we need only compute T+−−+andT+−+−. For T+−+−, we can get the second term in eq.(60.28) to vanish by choosing q3=p2, and using eq.(60.27). Then we have T+−+−=−e2∝an}b∇acketle{t2|/ε−(k4;q4)(/p1+ /k3)/ε+(k3;p2)|1]/s13 60: Spinor Helicity for Spinor Electrodynamics 362 =−e2√ 2 [q44]∝an}b∇acketle{t24∝an}b∇acket∇i}ht[q4|(/p1+ /k3)|2∝an}b∇acket∇i}ht[31]√ 2 ∝an}b∇acketle{t23∝an}b∇acket∇i}ht1 s13.(60.29) Next we note that [ p|/p= 0, and so it is useful to choose either q4=p1 orq4=k3. There is no obvious advantage in one choice over the other, and they must give equivalent results, so let us take q4=k3. Then, using eq.(60.6) for / k3, we get T+−+−= 2e2∝an}b∇acketle{t24∝an}b∇acket∇i}ht[31]∝an}b∇acketle{t12∝an}b∇acket∇i}ht[31] [34]∝an}b∇acketle{t23∝an}b∇acket∇i}hts13(60.30) Now we use [31] ∝an}b∇acketle{t12∝an}b∇acket∇i}ht=−[34]∝an}b∇acketle{t42∝an}b∇acket∇i}htin the numerator (see problem 60.2), and sets13=∝an}b∇acketle{t13∝an}b∇acket∇i}ht[31] in the denominator. Canceling common factors and using antisymmetry of the twistor product then yields T+−+−= 2e2∝an}b∇acketle{t24∝an}b∇acket∇i}ht2 ∝an}b∇acketle{t13∝an}b∇acket∇i}ht∝an}b∇acketle{t23∝an}b∇acket∇i}ht. (60.31) We can now get T+−−+simply by exchanging the labels 3 and 4, T+−−+= 2e2∝an}b∇acketle{t23∝an}b∇acket∇i}ht2 ∝an}b∇acketle{t14∝an}b∇acket∇i}ht∝an}b∇acketle{t24∝an}b∇acket∇i}ht. (60.32) We can compute the spin-averaged cross section by summing th e abso- lute squares of eqs.(60.31) and (60.32), multiplying by two to account for the processes in which all helicities are opposite (and whic h have ampli- tudes that are related by complex conjugation), and then div iding by four to average over the initial helicities. The result is ∝an}b∇acketle{t|T |2∝an}b∇acket∇i}ht= 2e4/parenleftbigg/vextendsingle/vextendsingle/vextendsingle/vextendsingles13 s14/vextendsingle/vextendsingle/vextendsingle/vextendsingle+/vextendsingle/vextendsingle/vextendsingle/vextendsingles14 s13/vextendsingle/vextendsingle/vextendsingle/vextendsingle/parenrightbigg . (60.33) For the processes of e−γ→e−γande+γ→e+γ, we haves13=s,s12=t, ands14=u; fore+e−→γγandγγ→e+e−we haves12=s,s13=t, and s14=u. Problems 60.1) a) Show that p·ε+(k;q) =∝an}b∇acketle{tqp∝an}b∇acket∇i}ht[pk]√ 2∝an}b∇acketle{tqk∝an}b∇acket∇i}ht, (60.34) p·ε−(k;q) =[qp]∝an}b∇acketle{tpk∝an}b∇acket∇i}ht√ 2[qk]. (60.35) 60: Spinor Helicity for Spinor Electrodynamics 363 Use this result to show that k·ε±(k;q) = 0, (60.36) which is required by gauge invariance, and also that q·ε±(k;q) = 0. (60.37) b) Show that ε+(k;q)·ε+(k′;q′) =∝an}b∇acketle{tqq′∝an}b∇acket∇i}ht[kk′] ∝an}b∇acketle{tqk∝an}b∇acket∇i}ht∝an}b∇acketle{tq′k′∝an}b∇acket∇i}ht, (60.38) ε−(k;q)·ε−(k′;q′) =[qq′]∝an}b∇acketle{tkk′∝an}b∇acket∇i}ht [qk][q′k′], (60.39) ε+(k;q)·ε−(k′;q′) =∝an}b∇acketle{tqk′∝an}b∇acket∇i}ht[kq′] ∝an}b∇acketle{tqk∝an}b∇acket∇i}ht[q′k′]. (60.40) Note that the right-hand sides of eqs.(60.38) and (60.39) va nish if q′=q, and that the right-hand side of eq.(60.40) vanishes if q=k′ orq′=k. 60.2) a) For a process with nexternal particles, and all momenta treated as outgoing, show that n/summationdisplay j=1∝an}b∇acketle{tij∝an}b∇acket∇i}ht[jk] = 0 andn/summationdisplay j=1[ij]∝an}b∇acketle{tjk∝an}b∇acket∇i}ht= 0. (60.41) Hint: make use of eq.(60.6). b) Forn= 4, show that [31] ∝an}b∇acketle{t12∝an}b∇acket∇i}ht=−[34]∝an}b∇acketle{t42∝an}b∇acket∇i}ht. 60.3) Use various identities to show that eq.(60.31) can als o be written as T+−+−=−2e2[13]2 [14][24]. (60.42) 60.4) a) Show explicitly that you would get the same result as eq.(60.31) if you setq4=p1in eq.(60.29). b) Show explicitly that you would get the same result as eq.(6 0.31) if you setq4=p2in eq.(60.29). 61: Scalar Electrodynamics 364 61Scalar Electrodynamics Prerequisite: 58 In this section, we will consider how charged spin-zero part icles interact with photons. We begin with the lagrangian for a complex scal ar ﬁeld with a quartic interaction, L=−∂µϕ†∂µϕ−m2ϕ†ϕ−1 4λ(ϕ†ϕ)2. (61.1) This lagrangian is obviously invariant under the global U(1 ) symmetry ϕ(x)→e−iαϕ(x), ϕ†(x)→e+iαϕ†(x). (61.2) We would like to promote this global symmetry to a local symme try, ϕ(x)→exp[−ieΓ(x)]ϕ(x), (61.3) ϕ†(x)→exp[+ieΓ(x)]ϕ†(x). (61.4) To do so, we must replace each ordinary derivative in eq.(61. 1) with a covariant derivative Dµ≡∂µ−ieAµ, (61.5) whereAµtransforms as Aµ(x)→Aµ(x)−∂µΓ(x), (61.6) which implies that Dµtransforms as Dµ→exp[−ieΓ(x)]Dµexp[+ieΓ(x)]. (61.7) Our complete lagrangian for scalar electrodynamics is then L=−(Dµϕ)†Dµϕ−m2ϕ†ϕ−1 4λ(ϕ†ϕ)2−1 4FµνFµν. (61.8) We have added the usual gauge-invariant kinetic term for the gauge ﬁeld. The quartic interaction term has a dimensionless coeﬃcient , and so is nec- essary for renormalizability. For now, we omit the renormal izingZfactors. Of course, eq.(61.8) is invariant under a global U(1) transf ormation as well as a local U(1) transformation: we simply set Γ( x) to a constant. Then we can ﬁnd the conserved Noether current corresponding to th is symmetry, following the procedure of section 22. In the case of spinor e lectrodynamics, this current is same as it is for a free Dirac ﬁeld, jµ=ΨγµΨ. In the case of a complex scalar ﬁeld, we ﬁnd jµ=−i[ϕ†Dµϕ−(Dµϕ)†ϕ]. (61.9) 61: Scalar Electrodynamics 365 k k Figure 61.1: The three vertices of scalar electrodynamics; the corresponding vertex factors are ie(k+k′)µ,−2ie2gµν, and−iλ. With a factor of e, this current should be identiﬁed as the electromag- netic current. Because the covariant derivative appears in eq.(61.9), the electromagnetic current depends explicitly on the gauge ﬁe ld. We had not previously contemplated this possibility, but in scalar el ectrodynamics it arises naturally, and is essential for gauge invariance. It also poses no special problem in the quantum theory. We wil l make the same assumption that we did for spinor electrodynamics: namely, that the correct procedure is to omit integration over the compon ent of ˜Aµ(k) that is parallel to kµ, on the grounds that this integration is redundant. This leads to the same Feynman rules for internal and externa l photons as in section 58. The Feyman rules for internal and external s calars are the same as those of problem 10.2. We will call the spin-zero p article with electric charge + eascalar electron orselectron (recall that our convention is thateis negative), and the spin-zero particle with electric char ge−e ascalar positron orspositron . Scalar lines (traditionally drawn as dashed in scalar electrodynamics) carry a charge arrow whose direc tion must be preserved when lines are joined by vertices. To determine the kinds of vertices we have, we ﬁrst write out t he inter- action terms in the lagrangian of eq.(61.8): L1=ieAµ[(∂µϕ†)ϕ−ϕ†∂µϕ]−e2AµAµϕ†ϕ−1 4λ(ϕ†ϕ)2. (61.10) This leads to the vertices shown in ﬁg.(61.1). The vertex fac tors associated with the last two terms are −2ie2gµνand−iλ. To get the vertex factor for the ﬁrst term, we note that if |k∝an}b∇acket∇i}htis an incoming selectron state, then ∝an}b∇acketle{t0|ϕ(x)|k∝an}b∇acket∇i}ht=eikxand∝an}b∇acketle{t0|ϕ†(x)|k∝an}b∇acket∇i}ht= 0; and if ∝an}b∇acketle{tk′|is an outgoing selectron state, then ∝an}b∇acketle{tk′|ϕ†(x)|0∝an}b∇acket∇i}ht=e−ik′xand∝an}b∇acketle{t0|ϕ(x)|k∝an}b∇acket∇i}ht= 0. Therefore, in free ﬁeld theory, ∝an}b∇acketle{tk′|(∂µϕ†)ϕ|k∝an}b∇acket∇i}ht=−ik′ µe−i(k′−k)x, (61.11) ∝an}b∇acketle{tk′|ϕ†∂µϕ|k∝an}b∇acket∇i}ht= +ikµe−i(k′−k)x. (61.12) 61: Scalar Electrodynamics 366 This implies that the vertex factor for the ﬁrst term in eq.(6 1.10) is given byi(ie)[(−ik′ µ)−(ikµ)] =ie(k+k′)µ. Putting everything together, we get the following set of Fey nman rules for tree-level processes in scalar electrodynamics. 1. For each incoming selectron , draw a dashed line with an arrow pointed towards the vertex, and label it with the selectron’s four-momentum , ki. 2. For each outgoing selectron , draw a dashed line with an arrow pointed awayfrom the vertex, and label it with the selectron’s four-mome ntum, k′ i. 3. For each incoming spositron , draw a dashed line with an arrow pointed away from the vertex, and label it with minus the spositron’s four- momentum, −ki. 4. For each outgoing spositron , draw a dashed line with an arrow pointed towards the vertex, and label it with minus the spositron’s four- momentum, −k′ i. 5. For each incoming photon , draw a wavy line with an arrow pointed towards the vertex, and label it with the photon’s four-momentum, ki. 6. For each outgoing photon , draw a wavy line with an arrow pointed away from the vertex, and label it with the photon’s four-momentu m, k′ i. 7. There are three allowed vertices, shown in ﬁg.(61.1). Usi ng these vertices, join up all the external lines, including extra in ternal lines as needed. In this way, draw all possible diagrams that are topologically inequivalent . 8. Assign each internal line its own four-momentum. Think of the four- momenta as ﬂowing along the arrows, and conserve four-momen tum at each vertex. For a tree diagram, this ﬁxes the momenta on al l the internal lines. 9. The value of a diagram consists of the following factors: for each incoming photon, εµ∗ λi(ki); for each outgoing photon, εµ λi(ki); for each incoming or outgoing selectron or spositron, 1; for each scalar-scalar-photon vertex, ie(k+k′)µ; 61: Scalar Electrodynamics 367 2 2k2k1k 1k 1k 2kk k1 21k 1kk k k1 k21k2 k2 Figure 61.2: Diagrams for /tildewidee+/tildewidee−→γγ. for each scalar-scalar-photon-photon vertex, −2ie2gµν; for each four-scalar vertex, −iλ; for each internal photon, −igµν/(k2−iǫ); for each internal scalar, −i/(k2+m2−iǫ). 10. The vector index on each vertex is contracted with the vec tor index on either the photon propagator (if the attached photon line is inter- nal) or the photon polarization vector (if the attached phot on line is external). 11. The value of iT(at tree level) is given by a sum over the values of all the contributing diagrams. Let us compute the scattering amplitude for a particular pro cess,/tildewidee+/tildewidee−→ γγ, where/tildewidee−denotes a selectron. We have the diagrams of ﬁg.(61.2). The amplitude is iT= (ie)21 i(2k1−k′ 1)µεµ 1′(k1−k′ 1−k2)νεν 2′ m2−t+ (1′↔2′) −2ie2gµνεµ 1′εν 2′, (61.13) wheret=−(k1−k′ 1)2andu=−(k1−k′ 2)2. This expression can be simpliﬁed by noting that k1−k′ 1−k2=k′ 2−2k2, and thatk′ i·ε′ i= 0. Then we have T=−e2/bracketleftbigg4(k1·ε1′)(k2·ε2′) m2−t+4(k1·ε2′)(k2·ε1′) m2−u+ 2(ε1′·ε2′)/bracketrightbigg .(61.14) To get the polarization-summed cross section, we take the ab solute square of eq.(61.14), and use the substitution rule /summationdisplay λ=±εµ λ(k)ερ∗ λ(k)→gµρ. (61.15) This is a straightforward calculation, which we leave to the problems. 61: Scalar Electrodynamics 368 Problems 61.1) Compute ∝an}b∇acketle{t|T |2∝an}b∇acket∇i}htfor/tildewidee+/tildewidee−→γγ, and express your answer in terms of the Mandelstam variables. 61.2) Compute ∝an}b∇acketle{t|T |2∝an}b∇acket∇i}htfor the process /tildewidee−γ→/tildewidee−γ. You should ﬁnd that your result is the same as that for /tildewidee+/tildewidee−→γγ, but with s↔t, an example of crossing symmetry. 62: Loop Corrections in Spinor Electrodynamics 369 62Loop Corrections in Spinor Electrodynamics Prerequisite: 51, 59 In this section we will compute the one-loop corrections in s pinor electro- dynamics. First let us note that the general discussion of sections 18 a nd 29 leads us to expect that we will need to add to the free lagrangian L0=iΨ/∂Ψ−mΨΨ−1 4FµνFµν (62.1) all possible terms whose coeﬃcients have positive or zero ma ss dimension, and that respect the symmetries of the original lagrangian. These include Lorentz symmetry, the U(1) gauge symmetry, and the discrete symmetries of parity, time reversal, and charge conjugation. The mass dimensions of the ﬁelds (in four spacetime dimensio ns) are [Aµ] = 1 and [Ψ] =3 2. Gauge invariance requires that Aµappear only in the form of a covariant derivative Dµ. (Recall that the ﬁeld strength Fµν can be expressed as the commutator of two covariant derivati ves.) Thus, the only possible term we could add to L0that does not involve the Ψ ﬁeld, and that has mass dimension four or less, is εµνρσFµνFρσ. This term, however, is odd under parity and time reversal. Similarly, t here are no terms meeting all the requirements that involve Ψ: the only candid ates contain eitherγ5(e.g.,iΨγ5Ψ) and are forbidden by parity, or C(e.g, ΨTCΨ) and are forbidden by the U(1) symmetry. Therefore, the theory we will consider is speciﬁed by L=L0+L1, where L0is given by eq.(62.1), and L1=Z1eΨ /AΨ +Lct, (62.2) Lct=i(Z2−1)Ψ/∂Ψ−(Zm−1)mΨΨ−1 4(Z3−1)FµνFµν.(62.3) We will use an on-shell renormalization scheme. We can write the exact photon propagator (in momentum space) as a geometric series of the form ˜∆µν(k) =˜∆µν(k) +˜∆µρ(k)Πρσ(k)˜∆σν(k) +... , (62.4) whereiΠµν(k) is given by a sum of one-particle irreducible (1PI for short ; see section 14) diagrams with two external photon lines (and the external propagators removed), and ˜∆µν(k) is the free photon propagator, ˜∆µν(k) =1 k2−iǫ/parenleftbigg gµν−(1−ξ)kµkν k2/parenrightbigg . (62.5) 62: Loop Corrections in Spinor Electrodynamics 370 Here we have used the freedom to add kµorkνterms to put the propa- gator into generalized Feynman gauge orRξgauge. (The name Rξgauge has historically been used only in the context of spontaneou s symmetry breaking—see section 85—but we will use it here as well. Rstands for renormalizable andξstands forξ.) Settingξ= 1 gives Feynman gauge, and setting ξ= 0 gives Lorenz gauge (also known as Landau gauge). Observable squared amplitudes should not depend on the valu e ofξ. This suggests that Πµν(k) should be transverse , kµΠµν(k) =kνΠµν(k) = 0, (62.6) so that the ξdependent term in ˜∆µν(k) vanishes when an internal photon line is attached to Πµν(k). Eq.(62.6) is in fact valid; we will give a proof of it, based on the Ward identity for the electromagnetic curre nt, in problem 68.1. For now, we will take eq.(62.6) as given. This implies t hat we can write Πµν(k) = Π(k2)/parenleftig k2gµν−kµkν/parenrightig (62.7) =k2Π(k2)Pµν(k), (62.8) where Π(k2) is a scalar function, and Pµν(k) =gµν−kµkν/k2is the pro- jection matrix introduced in section 57. Note that we can also write ˜∆µν(k) =1 k2−iǫ/parenleftbigg Pµν(k) +ξkµkν k2/parenrightbigg . (62.9) Then, using eqs.(62.8) and (62.9) in eq.(62.4), and summing the geometric series, we ﬁnd ˜∆µν(k) =Pµν(k) k2[1−Π(k2)]−iǫ+ξkµkν/k2 k2−iǫ. (62.10) Theξdependent term should be physically irrelevant (and can be s et to zero by the gauge choice ξ= 0, corresponding to Lorenz gauge). The remaining term has a pole at k2= 0 with residue Pµν(k)/[1−Π(0)]. In our on-shell renormalization scheme, we should have Π(0) = 0. (62.11) This corresponds to the ﬁeld normalization that is needed fo r validity of the LSZ formula. 62: Loop Corrections in Spinor Electrodynamics 371 lk k k+l Figure 62.1: The one-loop and counterterm corrections to th e photon prop- agator in spinor electrodynamics. Let us now turn to the calculation of Πµν(k). The one-loop and coun- terterm contributions are shown in ﬁg.(62.1). We have iΠµν(k) = (−1)(iZ1e)2/parenleftig 1 i/parenrightig2/integraldisplayd4ℓ (2π)4Tr/bracketleftig˜S(/ℓ+/k)γµ˜S(/ℓ)γν/bracketrightig −i(Z3−1)(k2gµν−kµkν) +O(e4), (62.12) where the factor of minus one is for the closed fermion loop, a nd˜S(/p) = (−/p+m)/(p2+m2−iǫ) is the free fermion propagator in momentum space. Anticipating that Z1= 1 +O(e2), we setZ1= 1 in the ﬁrst term. We can write Tr/bracketleftig˜S(/ℓ+/k)γµ˜S(/ℓ)γν/bracketrightig =/integraldisplay1 0dx4Nµν (q2+D)2, (62.13) where we have combined denominators in the usual way: q=ℓ+xkand D=x(1−x)k2+m2−iǫ. (62.14) The numerator is 4Nµν= Tr/bracketleftig (−/ℓ−/k+m)γµ(−/ℓ+m)γν/bracketrightig (62.15) Completing the trace, we get Nµν= (ℓ+k)µℓν+ℓµ(ℓ+k)ν−[ℓ(ℓ+k) +m2]gµν. (62.16) Settingℓ=q−xkand and dropping terms linear in q(because they inte- grate to zero), we ﬁnd Nµν→2qµqν−2x(1−x)kµkν−[q2−x(1−x)k2+m2]gµν.(62.17) The integrals diverge, and so we analytically continue to d= 4−εdimen- sions, and replace ewithe˜µε/2(so thateremains dimensionless for any d). 62: Loop Corrections in Spinor Electrodynamics 372 Next we recall a result from problem 14.3, /integraldisplay ddqqµqνf(q2) =1 dgµν/integraldisplay ddqq2f(q2). (62.18) This allows the replacement Nµν→ −2x(1−x)kµkν+/bracketleftig/parenleftig 2 d−1/parenrightig q2+x(1−x)k2−m2/bracketrightig gµν.(62.19) Using the results of section 14, along with a little manipula tion of gamma functions, we can show that /parenleftig 2 d−1/parenrightig/integraldisplayddq (2π)dq2 (q2+D)2=D/integraldisplayddq (2π)d1 (q2+D)2. (62.20) Thus we can make the replacement (2 d−1)q2→Din eq.(62.19), and we ﬁnd Nµν→2x(1−x)(k2gµν−kµkν). (62.21) This guarantees that the one-loop contribution to Πµν(k) is transverse (as we expected) in any number of spacetime dimensions. Now we evaluate the integral over q, using ˜µε/integraldisplayddq (2π)d1 (q2+D)2=i 16π2Γ(ε 2)/parenleftig 4π˜µ2/D/parenrightigε/2 =i 8π2/bracketleftbigg1 ε−1 2ln(D/µ2)/bracketrightbigg , (62.22) whereµ2= 4πe−γ˜µ2, and we have dropped terms of order εin the last line. Combining eqs.(62.7), (62.12), (62.13), (62.21), and (62. 22), we get Π(k2) =−e2 π2/integraldisplay1 0dxx(1−x)/bracketleftbigg1 ε−1 2ln(D/µ2)/bracketrightbigg −(Z3−1)+O(e4).(62.23) Imposing Π(0) = 0 ﬁxes Z3= 1−e2 6π2/bracketleftbigg1 ε−ln(m/µ)/bracketrightbigg +O(e4) (62.24) and Π(k2) =e2 2π2/integraldisplay1 0dxx(1−x)ln(D/m2) +O(e4). (62.25) Next we turn to the fermion propagator. The exact propagator can be written in Lehmann-K¨ all´ en form as ˜S(/p) =1 /p+m−iǫ+/integraldisplay∞ m2 thdsρΨ(s) /p+√s−iǫ. (62.26) 62: Loop Corrections in Spinor Electrodynamics 373 We see that the ﬁrst term has a pole at / p=−mwith residue one. This residue corresponds to the ﬁeld normalization that is neede d for the validity of the LSZ formula. There is a problem, however: in quantum electrodynamics, th e thresh- old massmthism, corresponding to the contribution of a fermion and a zero-energy photon. Thus the second term has a branch point a t /p=−m. The pole in the ﬁrst term is therefore not isolated, and its re sidue is ill deﬁned. This is a reﬂection of an underlying infrared divergence, as sociated with the massless photon. To deal with it, we must impose an infrar ed cutoﬀ that moves the branch point away from the pole. The most direc t method is to change the denominator of the photon propagator from k2tok2+m2 γ, wheremγis a ﬁctitious photon mass. Ultimately, as in section 26, we m ust deal with this issue by computing cross-sections that take i nto account detector ineﬃciencies. In quantum electrodynamics, we mus t specify the lowest photon energy ωminthat can be detected. Only after computing cross sections with extra undectable photons, and then summing ov er them, is it safe to take the limit mγ→0. It turns out that it is not also necessary to abandon the on-shell shell renormalization scheme (as we were forced to do in massless ϕ3theory in section 27), as long as the electron is massive. An alternative is to use dimensional regularization for the infrared di- vergences as well as the ultraviolet ones. As discussed in se ction 25, there are no soft-particle infrared divergences for d >4 (and no colinear diver- gences at all in quantum electrodynamics with massive charg ed particles). In practice, infrared-divergent integrals are ﬁnite away f rom even-integer dimensions, just like ultraviolet-divergent integrals. T hus we simply keep d= 4−εall the way through to the very end, taking the ε→0 limit only after summing over cross sections with extra undetecta ble photons, all computed in 4 −εdimensions. This method is calculationally the simplest, but requires careful bookkeeping to segregate the infrared and ultraviolet singularities. For that reason, we will not pursue it furthe r. We can write the exact fermion propagator in the form ˜S(/p)−1= /p+m−iǫ−Σ(/p), (62.27) whereiΣ(/p) is given by the sum of 1PI diagrams with two external fermion lines (and the external propagators removed). The fact that ˜S(/p) has a pole at /p=−mwith residue one implies that Σ( −m) = 0 and Σ′(−m) = 0; this ﬁxes the coeﬃcients Z2andZm. As we will see, we must have an infrared cutoﬀ in place in order to have a ﬁnite value for Σ′(−m). Let us now turn to the calculation of Σ(/ p). The one-loop and counter- 62: Loop Corrections in Spinor Electrodynamics 374 p p p p+ll p Figure 62.2: The one-loop and counterterm corrections to th e fermion prop- agator in spinor electrodynamics. term contributions are shown in ﬁg.(62.2). We have iΣ(/p) = (iZ1e)2/parenleftig 1 i/parenrightig2/integraldisplayd4ℓ (2π)4/bracketleftig γν˜S(/p+ /ℓ)γµ/bracketrightig˜∆µν(ℓ) −i(Z2−1)/p−i(Zm−1)m+O(e4). (62.28) It is simplest to work in Feynman gauge, where we take ˜∆µν(ℓ) =gµν ℓ2+m2γ−iǫ; (62.29) here we have included the ﬁctitious photon mass mγas an infrared cutoﬀ. We now apply the usual bag of tricks to get iΣ(/p) =e2˜µε/integraldisplay1 0dx/integraldisplayddq (2π)dN (q2+D)2 −i(Z2−1)/p−i(Zm−1)m+O(e4), (62.30) whereq=ℓ+xpand D=x(1−x)p2+xm2+ (1−x)m2 γ, (62.31) N=γµ(−/p−/ℓ+m)γµ =−(d−2)(/p+ /ℓ)−dm =−(d−2)[/q+ (1−x)/p]−dm, (62.32) where we have used (from section 47) γµγµ=−dandγµ/pγµ= (d−2)/p. The term linear in qintegrates to zero, and then, using eq.(62.22), we get Σ(/p) =−e2 8π2/integraldisplay1 0dx/parenleftig (2−ε)(1−x)/p+ (4−ε)m/parenrightig/bracketleftbigg1 ε−1 2ln(D/µ2)/bracketrightbigg −(Z2−1)/p−(Zm−1)m+O(e4). (62.33) 62: Loop Corrections in Spinor Electrodynamics 375 p +lp p+lpl Figure 62.3: The one-loop correction to the photon-fermion -fermion vertex in spinor electrodynamics. We see that ﬁniteness of Σ(/ p) requires Z2= 1−e2 8π2/parenleftbigg1 ε+ ﬁnite/parenrightbigg +O(e4), (62.34) Zm= 1−e2 2π2/parenleftbigg1 ε+ ﬁnite/parenrightbigg +O(e4). (62.35) We can impose Σ( −m) = 0 by writing Σ(/p) =e2 8π2/bracketleftbigg/integraldisplay1 0dx/parenleftig (1−x)/p+ 2m/parenrightig ln(D/D 0) +κ2(/p+m)/bracketrightbigg +O(e4), (62.36) whereD0isDevaluated at p2=−m2, D0=x2m2+ (1−x)m2 γ, (62.37) andκ2is a constant to be determined. We ﬁx κ2by imposing Σ′(−m) = 0. In diﬀerentiating with respect to / p, we take the p2inD, eq.(62.31), to be −/p2; we ﬁnd κ2=−2/integraldisplay1 0dxx(1−x2)m2/D0 =−2ln(m/mγ) + 1, (62.38) where we have dropped terms that go to zero with the infrared c utoﬀmγ. Next we turn to the loop correction to the vertex. We deﬁne the vertex functioniVµ(p′,p) as the sum of one-particle irreducible diagrams with one incoming fermion with momentum p, one outgoing fermion with momentum p′, and one incoming photon with momentum k=p′−p. The original vertex iZ1eγµis the ﬁrst term in this sum, and the diagram of ﬁg.(62.3) is th e second. Thus we have iVµ(p′,p) =iZ1eγµ+iVµ 1 loop(p′,p) +O(e5), (62.39) 62: Loop Corrections in Spinor Electrodynamics 376 where iVµ 1loop(p′,p) = (ie)3/parenleftig 1 i/parenrightig3/integraldisplayd4ℓ (2π)4/bracketleftig γρ˜S(/p′+/ℓ)γµ˜S(/p+/ℓ)γν/bracketrightig˜∆νρ(ℓ). (62.40) We again use eq.(62.29) for the photon propagator, and combi ne denomi- nators in the usual way. We then get iVµ 1 loop(p′,p) =e3/integraldisplay dF3/integraldisplayd4q (2π)4Nµ (q2+D)3, (62.41) where the integral over Feynman parameters is /integraldisplay dF3≡2/integraldisplay1 0dx1dx2dx3δ(x1+x2+x3−1), (62.42) and q=ℓ+x1p+x2p′, (62.43) D=x1(1−x1)p2+x2(1−x2)p′2−2x1x2p·p′ + (x1+x2)m2+x3m2 γ, (62.44) Nµ=γν(−/p′−/ℓ+m)γµ(−/p−/ℓ+m)γν =γν[−/q+x1/p−(1−x2)/p′+m]γµ[−/q−(1−x1)/p+x2/p′+m]γν =γν/qγµ/qγν+/tildewideNµ+ (linear in q), (62.45) where /tildewideNµ=γν[x1/p−(1−x2)/p′+m]γµ[−(1−x1)/p+x2/p′+m]γν.(62.46) The terms linear in qin eq.(62.45) integrate to zero, and only the ﬁrst term is divergent. After continuing to ddimensions, we can use eq.(62.18) to make the replacement γν/qγµ/qγν→1 dq2γνγργµγργν. (62.47) Then we use γργµγρ= (d−2)γµtwice to get γν/qγµ/qγν→(d−2)2 dq2γµ. (62.48) Performing the usual manipulations, we ﬁnd Vµ 1loop(p′,p) =e3 8π2/bracketleftigg/parenleftbigg1 ε−1−1 2/integraldisplay dF3ln(D/µ2)/parenrightbigg γµ+1 4/integraldisplay dF3/tildewideNµ D/bracketrightigg . (62.49) 62: Loop Corrections in Spinor Electrodynamics 377 From eq.(62.39), we see that ﬁniteness of Vµ(p′,p) requires Z1= 1−e2 8π2/parenleftbigg1 ε+ ﬁnite/parenrightbigg +O(e4). (62.50) To completely ﬁx Vµ(p′,p), we need a suitable condition to impose on it. We take this up in the next section. Problems 62.1) Show that adding a gauge ﬁxing term −1 2ξ−1(∂µAµ)2toLresults in eq.(62.9) as the photon propagator. Explain why ξ= 0 corresponds to Lorenz gauge, ∂µAµ= 0. 62.2) Find the coeﬃcients of e2/εinZ1,2,3,minRξgauge. In particular, show thatZ1=Z2= 1 +O(e4) in Lorenz gauge. 62.3) Consider the six one-loop diagrams with four external photons (and no external fermions). Show that, even though each diagram i s log- arithmically divergent, their sum is ﬁnite. Use gauge invar iance to explain why this must be the case. 63: The Vertex Function in Spinor Electrodynamics 378 63The Vertex Function in Spinor Electrodynamics Prerequisite: 62 In the last section, we computed the one-loop contribution t o the vertex function Vµ(p′,p) in spinor electrodynamics, where pis the four-momentum of an incoming electron (or outgoing positron), and p′is the four-momentum of an outgoing electron (or incoming positron). We left open the issue of the renormalization condition we wish to impose on Vµ(p′,p). For the theories we have studied previously, we have usually made the mathematically convenient (but physically obscure) choic e to deﬁne the coupling constant as the value of the vertex function when al l external four- momenta are set to zero. However, in the case of spinor electr odynamics, the masslessness of the photon gives us the opportunity to do something more physically meaningful: we can deﬁne the coupling const ant as the value of the vertex function when all three particles are on s hell:p2=p′2= −m2, andq2= 0, where q≡p′−pis the photon four-momentum. Because the photon is massless, these three on-shell conditions are compatible with momentum conservation. To be more precise, let us sandwich Vµ(p′,p) between the spinor factors that are appropriate for an incoming electron with momentum pand an outgoing electron with momentum p′, impose the on-shell conditions, and deﬁne the electron charge evia us′(p′)Vµ(p′,p)us(p)/vextendsingle/vextendsingle/vextendsingle/vextendsinglep2=p′2=−m2 (p′−p)2=0=eus′(p′)γµus(p)/vextendsingle/vextendsingle/vextendsingle/vextendsinglep2=p′2=−m2 (p′−p)2=0.(63.1) This deﬁnition is in accord with the usual one provided by Cou lomb’s law. To see why, consider the process of electron-electron s cattering. Ac- cording to the general discussion in section 19, we compute t he exact ampli- tude for this process by using tree diagrams with exact inter nal propagators and vertices, as shown in ﬁg.(63.1). In the last section, we r enormalized the photon propagator so that it approaches its tree-level v alue˜∆µν(q) whenq2→0. And we have just chosen to renormalize the electron-photo n vertex function by requiring it to approach its tree-level v alueeγµwhen q2→0, and when sandwiched between external spinors for on-shel l incom- ing and outgoing electrons. Therefore, as q2→0, the ﬁrst two diagrams in ﬁg.(63.1) approach the tree-level scattering amplitude , with the electron charge equal to e. Furthermore, the third diagram does not have a pole at q2= 0, and so can be neglected in this limit. Physically, q2→0 means that the electron’s momentum changes very little during the scattering. Measuring a slight deﬂection in the trajectory of one charge d particle (due 63: The Vertex Function in Spinor Electrodynamics 379 p1 p2p1 p2p1 p2p2 p1p1 p1 p2 p2p1p1 p1p2 Figure 63.1: Diagrams for the exact electron-electron scat tering amplitude. The vertices and photon propagator are exact; external line s stand for the usualuanduspinor factors, times the unit residue of the pole at p2=−m2. to the presence of another) is how we measure the coeﬃcient in Coulomb’s law. Thus, eq.(63.1) corresponds to this traditional deﬁni tion of the charge of the electron. We can simplify eq.(63.1) by noting that the on-shell condit ions actually enforcep′=p. So we can rewrite eq.(63.1) as us(p)Vµ(p,p)us(p) =eus(p)γµus(p) = 2epµ, (63.2) wherep2=−m2is implicit. We have taken s′=s, because otherwise the right-hand side vanishes (and hence does not specify a value fore). Now we can use eq.(63.2) to completely determine Vµ(p′,p). Using the freedom to choose the ﬁnite part of Z1, we write Vµ(p′,p) =eγµ−e3 16π2/integraldisplay dF3/bracketleftigg/parenleftig ln(D/D 0)+2κ1/parenrightig γµ−Nµ 2D/bracketrightigg +O(e5),(63.3) where D=x1(1−x1)p2+x2(1−x2)p′2−2x1x2p·p′ + (x1+x2)m2+x3m2 γ, (63.4) D0isDevaluated at p′=pandp2=−m2, D0= (x1+x2)2m2+x3m2 γ = (1−x3)2m2+x3m2 γ, (63.5) and Nµ=γν[x1/p−(1−x2)/p′+m]γµ[−(1−x1)/p+x2/p′m]γν; (63.6) 63: The Vertex Function in Spinor Electrodynamics 380 Nµwas called/tildewideNµin section 62, but we have dropped the tilde for notational convenience. We ﬁx the constant κ1in eq.(63.3) by imposing eq.(63.2). This yields 4κ1pµ=/integraldisplay dF3us(p)Nµ 0us(p) 2D0, (63.7) whereNµ 0isNµwithp′=pandp2=p′2=−m2. So now we must evaluate uNµ 0u. To do so, we ﬁrst write Nµ=γν(/a1+m)γµ(/a2+m)γν, (63.8) where a1=x1p−(1−x2)p′, a2=x2p′−(1−x1)p. (63.9) Now we use the gamma matrix contraction identities to get Nµ= 2/a2γµ/a1+ 4m(a1+a2)µ+ 2m2γµ. (63.10) Here we have set d= 4, because we have already removed the divergence and taken the limit ε→0. Setting p′=p, and using / pu=−muand u/p=−mu, along with uγµu= 2pµanduu= 2m, and recalling that x1+x2+x3= 1, we ﬁnd uNµ 0u= 4(1−4x3+x2 3)m2pµ. (63.11) Using eqs.(63.5), (63.7), and (63.11), we get κ1=1 2/integraldisplay dF31−4x3+x2 3 (1−x3)2+x3m2γ/m2 =/integraldisplay1 0dx3(1−x3)1−4x3+x2 3 (1−x3)2+x3m2γ/m2 =−2ln(m/mγ) +5 2(63.12) in the limit of mγ→0. We see that an infrared regulator is necessary for the vertex function as well as the fermion propagator. Now that we have Vµ(p′,p), we can extract some physics from it. Con- sider again the process of electron-electron scattering, s hown in ﬁg.(63.1). In order to compute the contributions of these diagrams, we m ust evalu- ateus′(p′)Vµ(p′,p)us(p) withp2=−p′2=−m2, but withq2= (p′−p)2 arbitrary. 63: The Vertex Function in Spinor Electrodynamics 381 To evaluate u′Nµu, we start with eq.(63.10), and use the anticommu- tation relations of the gamma matrices to move all the / p’s inNµto the far right (where we can use / pu=−mu) and all the / p′’s to the far left (where we can use u′/p′=−mu′). This results in Nµ→[4(1−x1−x2+x1x2)p·p′+ 2(2x1−x2 1+2x2−x2 2)m2]γµ + 4m(x2 1−x2+x1x2)pµ+ 4m(x2 2−x1+x1x2)p′µ.(63.13) Next, replace p·p′with−1 2q2−m2, group the pµandp′µterms intop′+p andp′−pcombinations, and make use of x1+x2+x3= 1 to simplify some coeﬃcients. The result is Nµ→2[(1−2x3−x2 3)m2−(x3+x1x2)q2]γµ −2m(x3−x2 3)(p′+p)µ −2m[(x1+x2 1)−(x2+x2 2)](p′−p)µ. (63.14) In the denominator, set p2=p′2=−m2andp·p′=−1 2q2−m2to get D→x1x2q2+ (1−x3)2m2+x3m2 γ. (63.15) Note that the right-hand side of eq.(63.15) is symmetric und erx1↔x2. Thus the last line of eq.(63.14) will vanish when we integrat eu′Nµu/D over the Feynman parameters. Finally, we use the Gordon iden tity from section 38, u′(p′+p)µu=u′[2mγµ+ 2iSµνqν]u, (63.16) whereSµν=i 4[γµ,γν], to get Nµ→2[(1−4x3+x2 3)m2−(x3+x1x2)q2]γµ −4im(x3−x2 3)Sµνqν. (63.17) So now we have us′(p′)Vµ(p′,p)us(p) =eu′/bracketleftig F1(q2)γµ−imF2(q2)Sµνqν/bracketrightig u, (63.18) where we have deﬁned the form factors F1(q2) = 1−e2 16π2/integraldisplay dF3/bracketleftigg ln/parenleftigg 1 +x1x2q2/m2 (1−x3)2/parenrightigg +1−4x3+x2 3 (1−x3)2+x3m2γ/m2 +(x3+x1x2)q2/m2−(1−4x3+x2 3) x1x2q2/m2+ (1−x3)2+x3m2γ/m2/bracketrightigg +O(e4),(63.19) F2(q2) =e2 8π2/integraldisplay dF3x3−x2 3 x1x2q2/m2+ (1−x3)2+O(e4). (63.20) 63: The Vertex Function in Spinor Electrodynamics 382 We have set mγ= 0 in eq.(63.20), and in the logarithm term in eq.(63.19), because these terms do not suﬀer from infrared divergences. We can simplify F2(q2) by using the delta function in dF3to do the integral over x2(which replaces x2with 1 −x3−x1), making the change of variablex1= (1−x3)y, and performing the integral over x3from zero to one; the result is F2(q2) =e2 8π2/integraldisplay1 0dy 1−y(1−y)q2/m2+O(e4). (63.21) This last integral can also be done in closed form, but we will be mostly interested in its value at q2= 0, corresponding to an on-shell photon: F2(0) =α 2π+O(α2), (63.22) whereα=e2/4π= 1/137.036 is the ﬁne-structure constant. We will explore the physical consequences of eq.(63.22) in the next section . Problems 63.1) The most general possible form of u′Vµ(p′,p)uis a linear combination ofγµ,pµ, andp′µsandwiched between u′andu, with coeﬃcients that depend onq2. (The only other possibility is to include terms with γ5, butγ5does not appear in the tree-level propagators or vertex, and so it cannot be generated in any Feynman diagram; this is a conse quence of parity conservation.) Thus we can write us′(p′)Vµ(p′,p)us(p) =eu′[A(q2)γµ+B(q2)(p′+p)µ +C(q2)(p′−p)µ]u. (63.23) a) Use gauge invariance to show that qµu′Vµ(p′,p)u= 0, and deter- mine the consequences for A,B, andC. b) Express F1andF2in terms of A,B, andC. 64: The Magnetic Moment of the Electron 383 64The Magnetic Moment of the Electron Prerequisite: 63 In the last section, we computed the one-loop contribution t o the vertex function Vµ(p′,p) in spinor electrodynamics, where pis the four-momentum of an incoming electron, and p′is the four-momentum of an outgoing elec- tron. We found us′(p′)Vµ(p′,p)us(p) =eu′/bracketleftig F1(q2)γµ−imF2(q2)Sµνqν/bracketrightig u, (64.1) whereq=p′−pis the four-momentum of the photon (treated as incoming), and with complicated expressions for the form factors F1(q2) andF2(q2). For our purposes in this section, all we will need to know is th at F1(0) = 1 exactly , F2(0) =α 2π+O(α2). (64.2) Eq.(64.1) follows from a quantum action of the form Γ =/integraldisplay d4x/bracketleftig eF1(0)Ψ /AΨ +e 2mF2(0)FµνΨSµνΨ +.../bracketrightig , (64.3) where the ellipses stand for terms with more derivatives. Th e displayed terms yield the vertex factor of eq.(64.1) with q2= 0. To see this, recall that an incoming photon translates into a factor of Aµ∼ε∗ µeiqx, and there- fore ofFµν∼i(qµε∗ ν−qνε∗ µ)eiqx; the two terms in Fµνcancel the extra factor of one half in the second term in eq.(64.3). Now we will see what eq.(64.3) predicts for the magnetic moment of the electron. We deﬁne the magnetic moment by the following p rocedure. We take the photon ﬁeld Aµto be a classical ﬁeld that corresponds to a constant magnetic ﬁeld in the zdirection:A0= 0 and A= (0,Bx,0). This yieldsF12=−F21=B, with all other components of Fµνvanishing. Then we deﬁne a normalized state of an electron at rest, with spin u p along the zaxis: |e∝an}b∇acket∇i}ht ≡/integraldisplay /tildewiderdpf(p)b† +(p)|0∝an}b∇acket∇i}ht, (64.4) where the wave packet is rotationally invariant (so that the re is no orbital angular momentum) and sharply peaked at p= 0, something like f(p)∼exp(−a2p2/2) (64.5) witha≪1/m. We normalize the wave packet by/integraltext/tildewiderdp|f(p)|2= 1; then we have∝an}b∇acketle{te|e∝an}b∇acket∇i}ht= 1. 64: The Magnetic Moment of the Electron 384 Now we deﬁne the interaction hamiltonian as what we get from t he two displayed terms in eq.(64.3), using our speciﬁed ﬁeld Aµ, and with the form-factor values of eq.(64.2): H1≡ −eB/integraldisplay d3xΨ/bracketleftig xγ2+α 2πmS12/bracketrightig Ψ. (64.6) Then the electron’s magnetic moment µis speciﬁed by µB≡ −∝an}b∇acketle{te|H1|e∝an}b∇acket∇i}ht. (64.7) In quantum mechanics in general, if we identify H1as the piece of the hamiltonian that is linear in the external magnetic ﬁeld, th en eq.(64.7) deﬁnes the magnetic moment of a normalized quantum state wit h deﬁnite angular momentum in the Bdirection. Now we turn to the computation. We need to evaluate ∝an}b∇acketle{te|Ψα(x)Ψβ(x)|e∝an}b∇acket∇i}ht. Using the usual plane-wave expansions, we have ∝an}b∇acketle{t0|b+(p′)Ψα(x)Ψβ(x)b† +(p)|0∝an}b∇acket∇i}ht=u+(p′)αu+(p)βei(p−p′)x. (64.8) Thus we get ∝an}b∇acketle{te|H1|e∝an}b∇acket∇i}ht=−eB/integraldisplay /tildewiderdp/tildewiderdp′d3xei(p−p′)x ×f∗(p′)u+(p′)/bracketleftig xγ2+α 2πmS12/bracketrightig u+(p)f(p).(64.9) We can write the factor of xas−i∂p1acting onei(p−p′)x, and integrate by parts to put this derivative onto u+(p)f(p); the wave packets kill any surface terms. Then we can complete the integral over d3xto get a factor of (2π)3δ3(p′−p), and do the integral over/tildewiderdp′. The result is ∝an}b∇acketle{te|H1|e∝an}b∇acket∇i}ht=−eB/integraldisplay/tildewiderdp 2ωf∗(p)u+(p)/bracketleftig iγ2∂p1+α 2πmS12/bracketrightig u+(p)f(p). (64.10) Suppose the ∂p1acts onf(p). Sincef(p) is rotationally invariant, the result is odd in p1. We then use u+(p)γiu+(p) = 2pito conclude that this term is odd in both p1andp2, and hence integrates to zero. The remaining contribution from the ﬁrst term has the ∂p1acting on u+(p). Recall from section 38 that us(p) = exp(iηˆp·K)us(0), (64.11) whereKj=Sj0=i 2γjγ0is the boost matrix, ˆpis a unit vector in the p direction, and η= sinh−1(|p|/m) is the rapidity. Since the wave packet is 64: The Magnetic Moment of the Electron 385 sharply peaked at p= 0, we can expand eq.(64.11) to linear order in p, take the derivative with respect to p1, and then set p= 0; the result is ∂p1u+(p)/vextendsingle/vextendsingle/vextendsingle p=0=imK1u+(0) =−1 2mγ1γ0u+(0) =−1 2mγ1u+(0), (64.12) where we used γ0us(0) =us(0) to get the last line. Then we have u+(p)iγ2∂p1u+(p)/vextendsingle/vextendsingle/vextendsingle p=0=u+(0)−i 2mγ2γ1u+(0) =1mu+(0)S12u+(0) (64.13) Plugging this into eq.(64.10) yields ∝an}b∇acketle{te|H1|e∝an}b∇acket∇i}ht=−eB/integraldisplay/tildewiderdp 2ω|f(p)|2/parenleftig 1 +α 2π/parenrightig1mu+(0)S12u+(0) =−eB 2m2/parenleftig 1 +α 2π/parenrightig u+(0)S12u+(0). (64.14) Next we use S12u±(0) =±1 2u±(0) andu±(0)u±(0) = 2mto get ∝an}b∇acketle{te|H1|e∝an}b∇acket∇i}ht=−eB 2m/parenleftig 1 +α 2π/parenrightig . (64.15) Comparing with eq.(64.7), we see that the magnetic moment of the electron is µ=g1 2e 2m, (64.16) wheree/2mis the Bohr magneton , the extra factor of one-half is for the electron’s spin (a classical spinning ball of charge would h ave a magnetic moment equal to the Bohr magneton times its angular momemtum ), and gis the Land´ e g factor , given by g= 2/parenleftig 1 +α 2π+O(α2)/parenrightig . (64.17) Sincegcan be measured to high precision, calculations of µprovide a stringent test of spinor electrodynamics. Corrections up t hrough the α4 term have been computed; the result is currently in good agre ement with experiment. Problems 64.1) Let the wave packet be f(p)∼exp(−a2p2/2)Yℓm(ˆ p), whereYℓm(ˆ p) is a spherical harmonic. Find the contribution of the orbita l angular momentum to the magnetic moment. 65: Loop Corrections in Scalar Electrodynamics 386 65Loop Corrections in Scalar Electrodynamics Prerequisite: 61, 62 In this section we will compute the one-loop corrections in s calar electro- dynamics. We will concentrate on the divergent parts of the d iagrams, enabling us to compute the renormalizing Zfactors in the MS scheme, and hence the beta functions. This gives us the most important qu alitative in- formation about the theory: whether it becomes strongly cou pled at high or low energies. Our lagrangian for scalar electrodynamics in section 61 alr eady includes all possible terms whose coeﬃcients have positive or zero ma ss dimension, and that respect Lorentz symmetry, the U(1) gauge symmetry, parity, time reversal, and charge conjugation. Therefore, the theory we will consider is L=L0+L1, (65.1) L0=−∂µϕ†∂µϕ−m2ϕ†ϕ−1 4FµνFµν, (65.2) L1=iZ1e[ϕ†∂µϕ−(∂µϕ†)ϕ]Aµ−Z4e2ϕ†ϕAµAµ −1 4Zλλ(ϕ†ϕ)2+Lct, (65.3) Lct=−(Z2−1)∂µϕ†∂µϕ−(Zm−1)m2ϕ†ϕ−1 4(Z3−1)FµνFµν.(65.4) We will use the MS renormalization scheme to ﬁx the values of the Z’s. We begin with the photon self-energy, Πµν(k). The one-loop and coun- terterm contributions are shown in ﬁg.(65.1). We have iΠµν(k) = (iZ1e)2/parenleftig 1 i/parenrightig2/integraldisplayd4ℓ (2π)4(2ℓ+k)µ(2ℓ+k)ν ((ℓ+k)2+m2)(ℓ2+m2) + (−2iZ4)e2gµν/integraldisplayd4ℓ (2π)41 ℓ2+m2 −i(Z3−1)(k2gµν−kµkµ) +... , (65.5) where the ellipses stand for higher-order (in e2and/orλ) terms. We can setZi= 1 +O(e2,λ) in the ﬁrst two terms. It will prove convenient to combine these ﬁrst two terms into iΠµν(k) =e2/integraldisplayd4ℓ (2π)4Nµν ((ℓ+k)2+m2)(ℓ2+m2) −i(Z3−1)(k2gµν−kµkµ) +... , (65.6) 65: Loop Corrections in Scalar Electrodynamics 387 where Nµν= (2ℓ+k)µ(2ℓ+k)ν−2[(ℓ+k)2+m2]gµν. (65.7) Then we continue to ddimensions, replace ewithe˜µε/2, and combine the denominators with Feynman’s formula; the result is iΠµν(k) =e2˜µε/integraldisplay1 0dx/integraldisplayddq (2π)dNµν (q2+D)2 −i(Z3−1)(k2gµν−kµkµ) +... , (65.8) whereq=ℓ+xkandD=x(1−x)k2+m2. The numerator is Nµν= (2q+ (1−2x)k)µ(2q+ (1−2x)k)ν−2[(q+ (1−x)k)2+m2]gµν = 4qµqν+ (1−2x)2kµkν−2[q2+ (1−x)2k2+m2]gµν + (linear in q) →4d−1gµνq2+ (1−2x)2kµkν−2[q2+ (1−x)2k2+m2]gµν,(65.9) where we used the symmetric-integration identity from prob lem 14.3 to get the last line. We can rearrange eq.(65.9) into Nµν= 2/parenleftig 2 d−1/parenrightig gµνq2+ (1−2x)2kµkν−2[(1−x)2k2+m2]gµν.(65.10) Now we recall from section 62 that, when q2is integrated against ( q2+D)−2, we can make the replacement (2 d−1)q2→D; thus we have Nµν→2Dgµν+ (1−2x)2kµkν−2[(1−x)2k2+m2]gµν = (1−2x)2kµkν−2(1−2x)(1−x)k2gµν. (65.11) Next we note that if we make the change of variable x=y+1 2, then we haveD= (1−1 4y2)k2+m2, andyis integrated from −1 2to +1 2. Therefore, any term in Nµνthat is even in ywill integrate to zero. We then get Nµν= 4y2kµkν−2(2y2−y)k2gµν → −4y2(k2gµν−kµkν). (65.12) Thus we see that Πµν(k) is transverse, as expected. Performing the integral over qin eq.(65.8), and focusing on the diver- gent part, we get ˜µε/integraldisplayddq (2π)d1 (q2+D)2=i 8π21 ε+O(ε0). (65.13) 65: Loop Corrections in Scalar Electrodynamics 388 l k k k k k k k+l l Figure 65.1: The one-loop and counterterm corrections to th e photon prop- agator in scalar electrodynamics. l k k k kl k kl k+lk k Figure 65.2: The one-loop and counterterm corrections to th e scalar prop- agator in scalar electrodynamics. Then performing the integral over yyields /integraldisplay1/2 −1/2dyNµν=−1 3(k2gµν−kµkν). (65.14) Combining eqs.(65.8), (65.13), and (65.14), we get Πµν(k) = Π(k2)(k2gµν−kµkν), (65.15) where Π(k2) =−e2 24π21 ε+ ﬁnite −(Z3−1) +... . (65.16) Thus we ﬁnd, in the MS scheme, Z3= 1−e2 24π21 ε+... . (65.17) Now we turn to the one-loop corrections to the scalar propaga tor, shown in ﬁg.(65.2). It will prove very convenient to work in Lorenz gauge, where 65: Loop Corrections in Scalar Electrodynamics 389 the photon propagator is ˜∆µν(ℓ) =Pµν(ℓ) ℓ2−iǫ, (65.18) withPµν(ℓ) =gµν−ℓµℓν/ℓ2. The diagrams in ﬁg.(65.2) then yield iΠϕ(k2) = (iZ1e)2/parenleftig 1 i/parenrightig2/integraldisplayd4ℓ (2π)4Pµν(ℓ)(ℓ+ 2k)µ(ℓ+ 2k)ν ℓ2((ℓ+k)2+m2) + (−2iZ4e2gµν)/parenleftig 1 i/parenrightig/integraldisplayd4ℓ (2π)4Pµν(ℓ) ℓ2+m2γ + (−iλ)/parenleftig 1 i/parenrightig/integraldisplayd4ℓ (2π)41 ℓ2+m2 −i(Z2−1)k2−i(Zm−1)m2+... . (65.19) In the second line, mγis a ﬁctitious photon mass; it appears here as an infrared regulator. We can set Zi= 1 +O(e2,λ) in the ﬁrst three lines. Continuing to d dimensions, making the replacements e→e˜µε/2andλ→λ˜µε, and using the relations ℓµPµν(ℓ) =ℓνPµν(ℓ) = 0 andgµνPµν(ℓ) =d−1, we get iΠϕ(k2) = 4e2˜µε/integraldisplayddℓ (2π)dPµν(ℓ)kµkν ℓ2((ℓ+k)2+m2) −2(d−1)e2˜µε/integraldisplayddℓ (2π)d1 ℓ2+m2γ −λ˜µε/integraldisplayd4ℓ (2π)41 ℓ2+m2 −i(Z2−1)k2−i(Zm−1)m2+... . (65.20) We evaluate the second and third lines via ˜µε/integraldisplayddℓ (2π)d1 ℓ2+m2=−i 8π21 εm2+O(ε0). (65.21) Then, taking the limit m2→0 (withεﬁxed) in eq.(65.21) shows that the second line of eq.(65.20) vanishes when the infrared regula tor is removed. To evaluate the ﬁrst line of eq.(65.20), we multiply the nume rator and denominator by ℓ2and use Feynman’s formula to get /integraldisplayddℓ (2π)dℓ2Pµν(ℓ)kµkν ℓ2ℓ2((ℓ+k)2+m2)=/integraldisplay dF3/integraldisplayddq (2π)dN (q2+D)3,(65.22) 65: Loop Corrections in Scalar Electrodynamics 390 ll k l+kl l+k l k l l k l+kk Figure 65.3: The one-loop corrections to the three-point ve rtex in scalar electrodynamics. whereq=ℓ+x3k,D=x3(1−x3)k2+x3m2, and N=ℓ2k2−(ℓ·k)2 = (q−x3k)2k2−(q·k−x3k2)2 =q2k2−(q·k)2+ (linear in q) →q2k2−d−1q2k2. (65.23) Now we use ˜µε/integraldisplayddq (2π)dq2 (q2+D)3=i 8π21 ε+O(ε0). (65.24) Combining eqs.(65.20–65.24), and requiring Π ϕ(k2) to be ﬁnite, we ﬁnd Z2= 1 +3e2 8π21 ε+... , (65.25) Zm= 1 +λ 8π21 ε+... (65.26) in the MS scheme. Now we turn to the one-loop corrections to the three-point (s calar– scalar–photon) vertex, shown in ﬁg.(65.3). In order to simp lify the calcu- lation of the divergent terms as much as possible, we have cho sen a special set of external momenta. (If we wanted the complete vertex fu nction, in- cluding the ﬁnite terms, we would need to use a general set of e xternal momenta.) We take the incoming scalar to have zero four-mome ntum, and 65: Loop Corrections in Scalar Electrodynamics 391 the photon (treated as incoming) to have four-momentum k; then, by mo- mentum conservation, the outgoing scalar also has four-mom entumk. We take the internal photon to have four-momentum ℓ. Now comes the magic of Lorenz gauge: in the second and third di agrams of ﬁg.(65.3), the vertex factor for the leftmost vertex is ieℓµ, and this is zero when contracted with the Pµν(ℓ) of the internal photon propagator. Thus the second and third diagrams vanish. Alas, we will have to do some more work to evaluate the ﬁrst and fourth diagrams. We have iVµ 3(k,0) =ieZ1kµ + (iZ1e)(−2iZ4e2gµν)/parenleftig 1 i/parenrightig2/integraldisplayd4ℓ (2π)4Pνρ(ℓ)(ℓ+ 2k)ρ ℓ2((ℓ+k)2+m2) + (−iZλλ)(iZ1e)/parenleftig 1 i/parenrightig2/integraldisplayd4ℓ (2π)4(2ℓ+k)µ (ℓ2+m2)((ℓ+k)2+m2) +... . (65.27) We can set Zi= 1 +O(e2,λ) in the second and third lines. We then do the usual manipulations; the integral in the third line becomes /integraldisplay1 0dx/integraldisplayddq (2π)d2qµ+ (1−2x)kµ (q2+D)2, (65.28) whereq=ℓ+xkandD=x(1−x)k2+m2. The term linear in qvanishes upon integration over q, and the term linear in kvanishes upon integration overx. Thus the third line of eq.(65.27) evaluates to zero. To evaluate the second line of eq.(65.27), we note that, sinc ePνρ(ℓ)ℓρ= 0, it already has an overall factor of k. We can then treat kas inﬁnitesi- mal, and set k= 0 in the denominator. We can then use the symmetric- integration identity to make the replacement ℓνℓρ/ℓ2→d−1gνρin the nu- merator. Putting all of this together, and using eq.(65.13) , we ﬁnd Vµ 3(k,0)/e=Z1kµ−3e2 8π21 εkµ+O(ε0) +... , (65.29) and so Z1= 1 +3e2 8π21 ε+... , (65.30) in the MS scheme. Next up is the four-point, scalar–scalar–photon–photon ve rtex. Because the tree-level vertex factor, −2iZ4e2gµν, does not depend on the external four-momenta, we can simply set them all to zero. Then, whene ver an 65: Loop Corrections in Scalar Electrodynamics 392 ll l l Figure 65.4: The nonvanishing one-loop corrections to the s calar–scalar– photon–photon vertex in scalar electrodynamics (in Lorenz gauge with van- ishing external momenta). internal photon line attaches to an external scalar with a th ree-point vertex, the diagram is zero, for the same reason that the second and th ird diagrams of ﬁg.(65.3) were zero. This kills a lot of diagrams; the surv ivors are shown in ﬁg.(65.4). We have iVµν 4(0,0,0) =−2iZ4e2gµν + (−2iZ4e2)2/parenleftig 1 i/parenrightig2/integraldisplayd4ℓ (2π)4gµρPρσ(ℓ)gσν ℓ2(ℓ2+m2)+ (µ↔ν) + (iZ1e)2(−iZλλ)/parenleftig 1 i/parenrightig3/integraldisplayd4ℓ (2π)4(2ℓ)µ(2ℓ)ν (ℓ2+m2)3+ (µ↔ν) + (−iZλλ)(−2iZ4e2gµν)/parenleftig 1 i/parenrightig2/integraldisplayd4ℓ (2π)41 (ℓ2+m2)2 +... . (65.31) The notation +( µ↔ν) in the second and third lines means that we must add the same expression with these indices swapped; this is b ecause the original and swapped versions of each diagram are topologic ally distinct, and contribute separately to the vertex function. As usual, we set Zi= 1 +O(e2,λ) in the second through fourth lines. After using the symmetric-integration identity, along wit h eqs.(65.13) and (65.24), we can see that the divergent parts of the third and f ourth lines cancel each other. The ﬁrst line is easily evaluated with sym metric inte- gration and eq.(65.13). Then we have Vµν 4(0,0,0)/e2=−2Z4gµν+3e2 4π21 εgµν+O(ε0) +... , (65.32) and so Z4= 1 +3e2 8π21 ε+... (65.33) 65: Loop Corrections in Scalar Electrodynamics 393 2 4l 1 3 1 234l 2 31 4l2 31 4l 1 32 4l Figure 65.5: The nonvanishing one-loop corrections to the f our-scalar ver- tex in scalar electrodynamics (in Lorenz gauge with vanishi ng external mo- menta). in the MS scheme. Finally, we have the one-loop corrections to the four-scala r vertex. Once again, because the tree-level vertex factor, −iZλλ, does not depend on the external four-momenta, we can set them all to zero. Then, whe never an internal photon line attaches to an external scalar with a th ree-point vertex, the diagram is zero. The remaining diagrams are shown in ﬁg.( 65.5). Even though we have set the external momenta to zero, we still have to keep track of which particle is which, in order to count the diagrams correctly; thus the external lines are labelled 1 through 4. Lines 1 and 2 have arrows pointing towards their vertices, and 3 and 4 have arro ws pointing away from their vertices. The symmetry factor for each of the ﬁrst three diagrams is S= 2; for each of the last two, it is S= 1. The diﬀerence arises because the last two diagrams have the charge arrows pointin g in opposite directions on the two internal propagators, and so these pro pagators cannot be exchanged. It is clear that the ﬁrst two diagrams will yield identical co ntributions to the vertex function (when the external momenta are all zer o). Similarly, except for symmetry factors, the contributions of the last t hree diagrams are also identical. Thus we have iV4ϕ(0,0,0) =−iZλλ +/parenleftig 1 2+1 2/parenrightig (−2iZ4e2)2/parenleftig 1 i/parenrightig2/integraldisplayd4ℓ (2π)4gµνPνρ(ℓ)gρσPρµ(ℓ) (ℓ2+m2γ)2 65: Loop Corrections in Scalar Electrodynamics 394 +/parenleftig 1 2+ 1 + 1/parenrightig (−iZλλ)2/parenleftig 1 i/parenrightig2/integraldisplayd4ℓ (2π)41 (ℓ2+m2)2 +... . (65.34) Using the familiar techniques, we ﬁnd V4ϕ(0,0,0) =−iZλλ+3e4 2π21 ε+5λ2 16π21 ε+O(ε0) +... , (65.35) and so Zλ= 1 +/parenleftigg 3e4 2π2λ+5λ 16π2/parenrightigg 1 ε+... (65.36) in the MS scheme. Problems 65.1) What conditions should be imposed on Vµ 3(p′,p) and Vµν 4(k,p′,p) in the OS scheme? (Here kis the incoming four-momentum of the photon at the µvertex, and p′andpare the four-momenta of the outgoing and incoming scalars, respectively.) 65.2) Consider a gauge transformaton Aµ→Aµ−∂µΓ. Show that there is a transformation of ϕthat leaves the lagrangian of eqs.(65.1–65.4) invariant if and only if Z4=Z2 1/Z2. 66: Beta Functions in Quantum Electrodynamics 395 66Beta Functions in Quantum Electrodynamics Prerequisite: 52, 62 In this section we will compute the beta function for the elec tromagnetic couplingein spinor electrodynamics and scalar electrodynamics. We w ill also compute the beta function for the ϕ4couplingλin scalar electrody- namics. In spinor electrodynamics, the relation between the bare an d renormal- ized couplings is e0=Z−1/2 3Z−1 2Z1˜µε/2e. (66.1) It is convenient to recast this formula in terms of the ﬁne-st ructure constant α=e2/4πand its bare counterpart α0=e2 0/4π, α0=Z−1 3Z−2 2Z2 1˜µεα. (66.2) From section 62, we have Z1= 1−α 2π1 ε+O(α2), (66.3) Z2= 1−α 2π1 ε+O(α2), (66.4) Z3= 1−2α 3π1 ε+O(α2), (66.5) in the MS scheme. Let us write ln/parenleftig Z−1 3Z−2 2Z2 1/parenrightig =∞/summationdisplay n=1En(α) εn. (66.6) Then we have lnα0=∞/summationdisplay n=1En(α) εn+ lnα+εln ˜µ. (66.7) From eqs.(66.3–66.5), we get E1(α) =2α 3π+O(α2). (66.8) Then, the general analysis of section 28 yields β(α) =α2E′ 1(α), (66.9) where the prime denotes diﬀerentiation with respect to α. Thus we ﬁnd β(α) =2α2 3π+O(α3) (66.10) 66: Beta Functions in Quantum Electrodynamics 396 in spinor electrodynamics, We can, if we like, restate this i n terms ofeas β(e) =e3 12π2+O(e5). (66.11) To go from eq.(66.10) to eq.(66.11), we use α=e2/4πand ˙α=e˙e/2π, where the dot denotes d/dlnµ. The most important feature of either eq.(66.10) or eq.(66.1 1) is that the beta function is positive: the electromagnetic couplin g in spinor elec- trodynamics gets stronger at high energies, and weaker at lo w energies. It is easy to generalize eqs.(66.10) and (66.11) to the case o fNDirac ﬁelds with electric charges Qie. There is now a factor of Z2ifor each ﬁeld, and ofZ1ifor each interaction. These are found by replacing αin eqs.(66.3) and (66.4) with Q2 iα. Then we ﬁnd Z1i/Z2i= 1 +O(α2), so that this ratio is universal, at least through O(α). In fact, as we will see in section 67, Z1i/Z2iis always exactly equal to one, and so it always cancels in eq. (66.6). As forZ3, now each Dirac ﬁeld contributes separately to the fermion l oop in the photon self-energy, and so we should replace αin eq.(66.5) with/summationtext iQ2 iα. Thus we ﬁnd that the generalization of eq.(66.11) is β(e) =/summationtextN i=1Q2 i 12π2e3+O(e5). (66.12) Now we turn to scalar electrodynamics. (Prerequisite: 65.) The rela- tions between the bare and renormalized couplings are e0=Z−1/2 3Z−1 ϕZ1˜µε/2e. (66.13) e2 0=Z−2 3Z−2 ϕZ4˜µεe2. (66.14) λ0=Z−2 ϕZλ˜µελ. (66.15) We have two diﬀerent relations between eande0, coming from the two types of vertices. We can guess (and will demonstrate in sect ion 67) that these two renormalizations must work out to give the same ans wer. Indeed, from section 65, we have Z1= 1 +3e2 8π21 ε+... , (66.16) Z2= 1 +3e2 8π21 ε+... , (66.17) Z3= 1−e2 24π21 ε+... , (66.18) 66: Beta Functions in Quantum Electrodynamics 397 Z4= 1 +3e2 8π21 ε+... , (66.19) Zλ= 1 +/parenleftigg 3e4 2π2λ+5λ 16π2/parenrightigg 1 ε+... , (66.20) in Lorenz gauge in the MS scheme; the ellipses stand for higher powers of e2and/orλ. We see that Z1=Z2=Z4, at least through O(e2,λ). The correct guess is that this is true exactly. Thus eqs.(66.13) and (66.14) both collapse to e0=Z−1/2 3e, just as in spinor electrodynamics. Thus we can write ln/parenleftig Z−1/2 3/parenrightig =∞/summationdisplay n=1En(e,λ) εn, (66.21) ln/parenleftig Z−2 2Zλ/parenrightig =∞/summationdisplay n=1Ln(e,λ) εn. (66.22) Then we have lne0=∞/summationdisplay n=1En(e,λ) εn+ lne+1 2eln ˜µ, (66.23) lnλ0=∞/summationdisplay n=1Ln(e,λ) εn+ lnλ+eln ˜µ. (66.24) Using eqs.(66.17), (66.18) and (66.20), we have E1(e,λ) =e2 24π2+... , (66.25) L1(e,λ) =1 16π2/parenleftig 5λ+ 24e4/λ−12e2/parenrightig +... , (66.26) Now applying the general analysis of section 52 yields βe(e,λ) =e3 48π2+... , (66.27) βλ(e,λ) =1 16π2/parenleftig 5λ2−6λe2+ 24e4/parenrightig +... . (66.28) Both right-hand sides are strictly positive, and so both eandλbecome large at high energies, and small at low energies. Generalizing eq.(66.25) to the case of several complex scal ar ﬁelds with chargesQieworks in the same way as it does in spinor electrodynamics. 66: Beta Functions in Quantum Electrodynamics 398 For a theory with both Dirac ﬁelds and complex scalar ﬁelds, t he one-loop contributions to Z3simply add, and so the beta function for eis βe(e,λ) =1 12π2/parenleftig/summationtext ΨQ2 Ψ+1 4/summationtext ϕQ2 ϕ/parenrightig e3+... . (66.29) Problems 66.1) Compute the one-loop contributions to the anomalous d imensions of m, Ψ, andAµin spinor electrodynamics in Feynman gauge. 66.2) Compute the one-loop contributions to the anomalous d imensions of m,ϕ, andAµin scalar electrodynamics in Lorenz gauge. 66.3) Use the results of problem 62.2 to compute the anomalou s dimension ofmand the beta function for ein spinor electrodynamics in Rξ gauge. You should ﬁnd that the results are independent of ξ. 66.4)The value of α(MW).The solution of eq.(66.12) is 1 α(MW)=1 α(µ)−2 3π/summationdisplay iQ2 iln(MW/µ), (66.30) where the sum is over all quarks and leptons (each color of qua rk counts separately), and we have chosen the W±boson mass MWas a reference scale. We can deﬁne a diﬀerent renormalization s cheme, modiﬁed decoupling subtraction orDS, where we imagine integrating out a ﬁeld when µis below its mass. In this scheme, eq.(66.30) becomes 1 α(MW)=1 α(µ)−2 3π/summationdisplay iQ2 iln[MW/min(mi,µ)], (66.31) where the sum is now over all quarks and leptons with mass less than MW. Forµ<me, theDS scheme coincides with the OS scheme, and we have1 α(MW)=1 α−2 3π/summationdisplay iQ2 iln(MW/mi), (66.32) whereα= 1/137.036 is the ﬁne-structure constant in the OS scheme. Usingmu=md=ms∼300MeV for the light quark masses (because quarks should be replaced by hadrons at lower energies), and other quark and lepton masses from sections 83 and 88, compute α(MW). 67: Ward Identities in Quantum Electrodynamics I 399 67Ward Identities in Quantum Electrodynamics I Prerequisite: 22, 59 In section 59, we assumed that scattering amplitudes would b e gauge in- variant, in the sense that they would be unchanged if we repla ced any photon polarization vector εµwithεµ+ckµ, wherekµis the photon’s four-momentum and cis an arbitrary constant. Thus, if we write a scat- tering amplitude Tfor a process that includes an external photon with four-momentum kµas T=εµMµ, (67.1) then we should have kµMµ= 0. (67.2) In this section, we will use the Ward identity for the electro magnetic current to prove eq.(67.2). We begin by recalling the LSZ formula for scalar ﬁelds, ∝an}b∇acketle{tf|i∝an}b∇acket∇i}ht=i/integraldisplay d4x1e−ik1x1(−∂2 1+m2)...∝an}b∇acketle{t0|Tϕ(x1)...|0∝an}b∇acket∇i}ht. (67.3) We have treated all external particles as outgoing; an incom ing particle has k0 i<0. We can rewrite eq.(67.3) as ∝an}b∇acketle{tf|i∝an}b∇acket∇i}ht= lim k2 i→−m2(k2 1+m2)...∝an}b∇acketle{t0|T ˜ϕ(k1)...|0∝an}b∇acket∇i}ht. (67.4) Here ˜ϕ(k) =i/integraltextd4xe−ikxϕ(x) is the ﬁeld in momentum space (with an extra factor of i), and we do not ﬁx k2=−m2. We know that the right-hand side of eq.(67.4) must include an overall energy-momentum delta function, so let us write ∝an}b∇acketle{t0|T ˜ϕ(k1)...|0∝an}b∇acket∇i}ht= (2π)4δ4(/summationtext iki)F(k2 i,ki·kj), (67.5) where F(k2 i,ki·kj) is a function of the Lorentz scalars k2 iandki·kj. Then, since ∝an}b∇acketle{tf|i∝an}b∇acket∇i}ht=i(2π)4δ4(/summationtext iki)T, (67.6) eq.(67.4) tells us that Fshould have a multivariable pole as each k2 iap- proaches −m2, and that iTis the residue of this pole. That is, near k2 i=−m2,Ftakes the form F(k2 i,ki·kj) =iT (k2 1+m2)...(k2n+m2)+ nonsingular . (67.7) 67: Ward Identities in Quantum Electrodynamics I 400 The key point is this: contributions to Fthat do nothave this multivariable pole do notcontribute to T. We have framed this discussion in terms of scalar ﬁelds in ord er to keep the notation as simple as possible, but the general point hol ds for ﬁelds of any spin. In section 22, we analyzed how various classical ﬁeld equati ons apply to quantum correlation functions. For example, we derived t heSchwinger- Dyson equations ∝an}b∇acketle{t0|TδS δφa(x)φa1(x1)...φan(xn)|0∝an}b∇acket∇i}ht =in/summationdisplay j=1∝an}b∇acketle{t0|Tφa1(x1)...δaajδ4(x−xj)...φan(xn)|0∝an}b∇acket∇i}ht.(67.8) Here we have used φa(x) to denote any kind of ﬁeld, not necessarily a scalar ﬁeld, carrying any kind of index or indices. The class ical equation of motion for the ﬁeld φa(x) isδS/δφa(x) = 0. Thus, eq.(67.8) tells us that the classical equation of motion holds for a ﬁeld inside a quantum correlation function, as long as its spacetime argument and indices do not match up exactly with those of any other ﬁeld in the correlati on function. These matches, which constitute the right-hand side of eq.( 67.8), are called contact terms . Suppose we have a correlation function that, for whatever re ason, in- cludes a contact term with a factor of, say, δ4(x1−x2). After Fourier- transforming to momentum space, this contact term is a funct ion ofk1+k2, but is independent of k1−k2; hence it cannot take the form of the singular term in eq.(67.7). Therefore, contact terms in a correlation function Fdo not contribute to the scattering amplitude T. Now let us consider a scattering process in quantum electrod yamics that involves an external photon with four-momentum k. In Lorenz gauge (the simplest for this analysis), the LSZ formula reads ∝an}b∇acketle{tf|i∝an}b∇acket∇i}ht=iεµ/integraldisplay d4xe−ikx(−∂2)...∝an}b∇acketle{t0|TAµ(x)...|0∝an}b∇acket∇i}ht, (67.9) and the classical equation of motion for Aµis −Z3∂2Aµ=∂L ∂Aµ. (67.10) In spinor electrodynamics, the right-hand side of eq.(67.1 0) isZ1jµ, where jµis the electromagnetic current. (This is also true for scala r electrodynam- ics ifZ4=Z2 1/Z2; we saw in problem 65.2 that this condition is necessary 67: Ward Identities in Quantum Electrodynamics I 401 for gauge invariance.) We therefore have ∝an}b∇acketle{tf|i∝an}b∇acket∇i}ht=iZ−1 3Z1εµ/integraldisplay d4xe−ikx.../bracketleftig ∝an}b∇acketle{t0|Tjµ(x)...|0∝an}b∇acket∇i}ht+ contact terms/bracketrightig . (67.11) The contact terms arise because, as we saw in eq.(67.8), the c lassical equa- tions of motion hold inside quantum correlation functions o nly up to contact terms. However, the contact terms cannot generate singular ities in thek2’s of the other particles, and so they do not contribute to the le ft-hand side. (Remember that, for each of the other particles, there is sti ll an appropriate wave operator, such as the Klein-Gordon wave operator for a s calar, acting on the correlation function. These wave operators kill any t erm that does not have an appropriate singularity.) Now let us try replacing εµin eq.(67.11) with kµ. We are attempting to prove that the result is zero, and we are almost there. We ca n write the factor ofikµas−∂µacting on the e−ikx, and then we can integrate by parts to get∂µacting on the correlation function. (Strictly speaking, we need a wave packet for the external photon to kill surface terms.) T hen we have ∂µ∝an}b∇acketle{t0|Tjµ(x)...|0∝an}b∇acket∇i}hton the right-hand side. Now we use another result from section 22, namely that a Noether current for an exact symmet ry obeys ∂µjµ= 0 classically, and ∂µ∝an}b∇acketle{t0|Tjµ(x)...|0∝an}b∇acket∇i}ht= contact terms (67.12) quantum mechanically; this is the Ward (orWard-Takahashi )identity . But once again, the contact terms do not have the right singul arities to contribute to ∝an}b∇acketle{tf|i∝an}b∇acket∇i}ht. Thus we conclude that ∝an}b∇acketle{tf|i∝an}b∇acket∇i}htvanishes if we replace an external photon’s polarization vector εµwith its four-momentum kµ, quo erat demonstratum. Reference Notes Diagrammatic proofs of the Ward identity in spinor electrod ynamics can be found in Peskin & Schroeder andZee. Problems 67.1) Show explicitly that the tree-level /tildewidee+/tildewidee−→γγscattering amplitude in scalar electrodynamics, T=−e2/bracketleftbigg4(k1·ε1′)(k2·ε2′) m2−t+4(k1·ε2′)(k2·ε1′) m2−u+ 2(ε1′·ε2′)/bracketrightbigg , vanishes ifεµ 1′is replaced with k′µ 1. 67: Ward Identities in Quantum Electrodynamics I 402 67.2) Show explicitly that the tree-level e+e−→γγscattering amplitude in spinor electrodynamics, T=e2v2/bracketleftbigg /ε2′/parenleftbigg−/p1+ /k′ 1+m m2−t/parenrightbigg /ε1′+ /ε1′/parenleftbigg−/p1+ /k′ 2+m m2−u/parenrightbigg /ε2′/bracketrightbigg u1, vanishes ifεµ 1′is replaced with k′µ 1. 68: Ward Identities in Quantum Electrodynamics II 403 68Ward Identities in Quantum Electrodynamics II Prerequisite: 63, 67 In this section, we will show that Z1=Z2in spinor electrodynamics, and thatZ1=Z2=Z4in scalar electrodyanmics (in the OS and MS renormal- ization schemes). Let us specialize to the case of spinor electrodynamics with a single Dirac ﬁeld, and consider the correlation function Cµ αβ(k,p′,p)≡iZ1/integraldisplay d4xd4yd4zeikx−ip′y+ipz∝an}b∇acketle{t0|Tjµ(x)Ψα(y)Ψβ(z)|0∝an}b∇acket∇i}ht, (68.1) wherejµ=eΨγµΨ is the electromagnetic current. As we saw in section 67, including Z1jµ(x) inside a correlation function adds a vertex for an external photon; the factor of Z1provides the necessary renormalization of this vertex. The explicit fermion ﬁelds on the right-hand si de of eq.(68.1) combine with the fermion ﬁelds in the current to generate pro pagators. Thus we have Cµ αβ(k,p′,p) = (2π)4δ4(k+p−p′)/bracketleftig 1 i˜S(p′)iVµ(p′,p)1 i˜S(p)/bracketrightig αβ,(68.2) where ˜S(p) is the exact fermion propagator, and Vµ(p′,p) is the exact 1PI photon–fermion–fermion vertex function. Now let us consider kµCµ αβ(k,p′,p). Using eq.(68.1), we can write the factor ofikµon the right-hand side as ∂µacting oneikx, and then integrate by parts to get −∂µacting onjµ(x). (Strictly speaking, we need a wave packet for the external photon to kill surface terms.) Thus w e have kµCµ αβ(k,p′,p) =−/integraldisplay d4xd4yd4zeikx−ip′y+ipz∂µ∝an}b∇acketle{t0|Tjµ(x)Ψα(y)Ψβ(z)|0∝an}b∇acket∇i}ht. (68.3) Now we use the Ward identity from section 22, which in general reads −∂µ∝an}b∇acketle{t0|TJµ(x)φa1(x1)...φan(xn)|0∝an}b∇acket∇i}ht =in/summationdisplay j=1∝an}b∇acketle{t0|Tφa1(x1)...δφaj(x)δ4(x−xj)...φan(xn)|0∝an}b∇acket∇i}ht.(68.4) Hereδφa(x) is the change in a ﬁeld φa(x) under an inﬁnitesimal transfor- mation that leaves the action invariant, and Jµ=∂L ∂(∂µφa)δφa (68.5) 68: Ward Identities in Quantum Electrodynamics II 404 is the corresponding Noether current. In the case of spinor e lectrodynamics, δΨ(x) =−ieΨ(x), δΨ(x) = +ieΨ(x), (68.6) where we have dropped an inﬁnitesimal parameter on the right -hand sides, but included a factor of the electron charge e. Using∂L/∂(∂µΨ) =iZ2Ψγµ and∂L/∂(∂µΨ) = 0 we ﬁnd that, with these conventions, the Noether current isJµ=Z2eΨγµΨ =Z2jµ. Thus the Ward identity becomes −Z2∂µ∝an}b∇acketle{t0|Tjµ(x)Ψα(y)Ψβ(z)|0∝an}b∇acket∇i}ht= +eδ4(x−y)∝an}b∇acketle{t0|TΨα(y)Ψβ(z)|0∝an}b∇acket∇i}ht −eδ4(x−z)∝an}b∇acketle{t0|TΨα(y)Ψβ(z)|0∝an}b∇acket∇i}ht.(68.7) Recall that ∝an}b∇acketle{t0|TΨα(y)Ψβ(z)|0∝an}b∇acket∇i}ht=1 i/integraldisplayd4q (2π)4eiq(y−z)˜S(q)αβ. (68.8) Using eqs.(68.7) and (68.8) in eq.(68.3), and carrying out t he coordinate integrals, we get kµCµ αβ(k,p′,p) =−iZ−1 2Z1(2π)4δ4(k+p−p′)/bracketleftig e˜S(p)−e˜S(p′)/bracketrightig αβ.(68.9) On the other hand, from eq.(68.2) we have kµCµ αβ(k,p′,p) =−i(2π)4δ4(k+p−p′)/bracketleftig ˜S(p′)kµVµ(p′,p)˜S(p)/bracketrightig αβ.(68.10) Comparing eqs.(68.9) and (68.10) shows that (p′−p)µ˜S(p′)Vµ(p′,p)˜S(p) =Z−1 2Z1e/bracketleftig˜S(p)−˜S(p′)/bracketrightig , (68.11) where we have dropped the spin indices. We can simplify eq.(6 8.11) by multiplying on the left by ˜S(p′)−1, and on the right by ˜S(p)−1, to get (p′−p)µVµ(p′,p) =Z−1 2Z1e/bracketleftig˜S(p′)−1−˜S(p)−1/bracketrightig . (68.12) Thus we ﬁnd a relation between the exact photon–fermion–fer mion vertex function Vµ(p′,p) and the exact fermion propagator ˜S(p). Since both ˜S(p) andVµ(p′,p) are ﬁnite, eq.(68.12) implies that Z1/Z2 must be ﬁnite as well. In the MS scheme (where all corrections to Zi= 1 are divergent), this immediately implies that Z1=Z2. (68.13) 68: Ward Identities in Quantum Electrodynamics II 405 In the OS scheme, we recall that near the on-shell point p2=p′2=−m2 and (p′−p)2= 0 we have ˜S(p) = /p+mandVµ(p′,p) =eγµ. Plugging these expressions into eq.(68.12) then yields Z1=Z2for the OS scheme. To better understand this result, we note that when Z1=Z2, we can combine the fermion kinetic term iZ2Ψ/∂Ψ and the interaction term Z1eΨ /AΨ intoiZ2Ψ /DΨ, whereDµ=∂µ−ieAµis the covariant derivative. Recall that it is Dµthat has a simple gauge transformation, and so we might expect the lagrangian, written in terms of renormaliz ed ﬁelds, to include∂µandAµonly in the combination Dµ. It is still necessary to go through the analysis that led to eq.(68.12), however, becau se quantization requires ﬁxing a gauge, and this renders suspect any naive ar guments based on gauge invariance. Still, in this case, those arguments yi eld the correct result. We can make a similar analysis in scalar electrodynamics. We leave the details to the problems. Reference Notes BRST symmetry (see section 74) can be used to derive the Ward identities; seeRamond I . Problems 68.1) Consider the current correlation function ∝an}b∇acketle{t0|Tjµ(x)jν(y)|0∝an}b∇acket∇i}htin spinor electrodynamics. a) Show that its Fourier transform is proportional to Πµν(k) + Πµρ(k)˜∆ρσ(k)Πσν(k) +... . (68.14) b) Use this to prove that Πµν(k) is transverse: kµΠµν(k) = 0. 68.2) Verify that eq.(68.12) holds at the one-loop level in a ny renormaliza- tion scheme with Z1=Z2. 68.3)Scalar electrodynamics. (Prerequisite: 65.) a) Consider the Fourier transform of ∝an}b∇acketle{t0|TJµ(x)ϕ(y)ϕ†(z)|0∝an}b∇acket∇i}ht, where Jµ=−ieZ2[ϕ†∂µϕ−(∂µϕ†)ϕ]−2Z1e2Aµϕ†ϕ (68.15) is the Noether current. You may assume that Z4=Z2 1/Z2(which is necessary for gauge invariance). Show that (p′−p)µVµ 3(p′,p) =Z−1 2Z1e/bracketleftig˜∆(p′)−1−˜∆(p)−1/bracketrightig , (68.16) 68: Ward Identities in Quantum Electrodynamics II 406 whereVµ 3(p′,p) is the exact scalar-scalar-photon vertex function, and ˜∆(p) is the exact scalar propagator. b) Use this result to show that Z1=Z2in both the MS and OS renormalization schemes. c) Consider the Fourier transform of ∝an}b∇acketle{t0|TJµ(x)Aν(w)ϕ(y)ϕ†(z)|0∝an}b∇acket∇i}ht. Show that kµVµν 4(k,p′,p) =Z−1 1Z4e/bracketleftig Vν 3(p′−k,p)−Vν 3(p′,p+k)/bracketrightig ,(68.17) where Vµν 4(k,p′,p) is the exact scalar-scalar-photon-photon vertex function, with kthe incoming momentum of the photon at the µ vertex. 69: Nonabelian Gauge Theory 407 69Nonabelian Gauge Theory Prerequisite: 24, 58 Consider a lagrangian with Nscalar or spinor ﬁelds φi(x) that is invariant under a continuous SU( N) or SO(N) symmetry, φi(x)→Uijφj(x), (69.1) whereUijis anN×Nspecial unitary matrix in the case of SU( N), or an N×Nspecial orthogonal matrix in the case of SO( N). (Special means that the determinant of Uis one.) Eq.(69.1) is called a global symmetry transformation, because the matrix Udoes not depend on the spacetime labelx. In section 58, we saw that quantum electrodynamics could be u nder- stood as having a localU(1) symmetry, φ(x)→U(x)φ(x), (69.2) whereU(x) = exp[ −ieΓ(x)] can be thought of as a 1 ×1 unitary matrix thatdoesdepend on the spacetime label x. Eq.(69.2) can be a symmetry of the lagrangian only if we include a U(1) gauge ﬁeld Aµ(x), and promote ordinary derivatives ∂µofφ(x) to covariant derivatives Dµ=∂µ−ieAµ. Under the transformation of eq.(69.2), we have Dµ→U(x)DµU†(x). (69.3) With this transformation rule, a scalar kinetic term like −(Dµϕ)†Dµϕ, or a fermion kinetic term like iΨ /DΨ, is invariant, as are mass terms like m2ϕ†ϕ andmΨΨ. We call eq.(69.3) a gauge transformation , and say that the lagrangian is gauge invariant . Eq.(69.3) implies that the gauge ﬁeld transforms as Aµ(x)→U(x)Aµ(x)U†(x) +ieU(x)∂µU†(x). (69.4) If we useU(x) = exp[ −ieΓ(x)], then eq.(69.4) simpliﬁes to Aµ(x)→Aµ(x)−∂µΓ(x), (69.5) which is what we originally had in section 54. We can now easily generalize this construction of U(1) gauge theory to SU(N) or SO(N). (We will consider other possibilities later.) To be concrete, let us consider SU( N). Recall from section 24 that we can write an inﬁnitesimal SU( N) transformation as Ujk(x) =δjk−igθa(x)(Ta)jk+O(θ2), (69.6) 69: Nonabelian Gauge Theory 408 where we have inserted a coupling constant gfor later convenience. The indicesjandkrun from 1 to N, the index aruns from 1 to N2−1 (and is implicitly summed), and the generator matrices Taare hermitian and traceless. (These properties of Tafollow immediately from the special uni- tarity ofU.) The generator matrices obey commutation relations of the form [Ta,Tb] =ifabcTc, (69.7) where the real numerical factors fabcare called the structure coeﬃcients of the group. If fabcdoes not vanish, the group is nonabelian . We can choose the generator matrices so that they obey the nor maliza- tion condition Tr(TaTb) =1 2δab; (69.8) then eqs.(69.7) and (69.8) can be used to show that fabcis completely antisymmetric. For SU(2), we have Ta=1 2σa, whereσais a Pauli matrix, andfabc=εabc, whereεabcis the completely antisymmetric Levi-Civita symbol. Now we deﬁne an SU( N) gauge ﬁeld Aµ(x) as a traceless hermitian N×Nmatrix of ﬁelds with the gauge transformation property Aµ(x)→U(x)Aµ(x)U†(x) +igU(x)∂µU†(x). (69.9) Note that this is identical to eq.(69.4), except that now U(x) is a spe- cial unitary matrix (rather than a phase factor), and Aµ(x) is a traceless hermitian matrix (rather than a real number). (Also, the ele ctromagnetic couplingehas been replaced by g.) We can write U(x) in terms of the generator matrices as U(x) = exp[ −igΓa(x)Ta], (69.10) where the real parameters Γa(x) are no longer inﬁnitesimal. The covariant derivative is Dµ=∂µ−igAµ(x), (69.11) where there is an understood N×Nidentity matrix multiplying ∂µ. Acting on the set of Nﬁeldsφi(x) that transform according to eq.(69.2), the covariant derivative can be written more explicitly as (Dµφ)j(x) =∂µφj(x)−igAµ(x)jkφk(x), (69.12) with an understood sum over k. The covariant derivative transforms ac- cording to eq.(69.3). Replacing all ordinary derivatives i nLwith covariant 69: Nonabelian Gauge Theory 409 derivatives renders Lgauge invariant (assuming, of course, that Loriginally had a global SU( N) symmetry). We still need a kinetic term for Aµ(x). Let us deﬁne the ﬁeld strength Fµν(x)≡ig[Dµ,Dν] (69.13) =∂µAν−∂νAµ−ig[Aµ,Aν]. (69.14) BecauseAµis a matrix, the ﬁnal term in eq.(69.14) does not vanish, as it does in U(1) gauge theory. Eqs.(69.3) and (69.13) imply th at, under a gauge transformation, Fµν(x)→U(x)Fµν(x)U†(x). (69.15) Therefore, Lkin=−1 2Tr(FµνFµν) (69.16) is gauge invariant, and can serve as a kinetic term for the SU( N) gauge ﬁeld. (Note, however, that the ﬁeld strength itself is notgauge invariant, in contrast to the situation in U(1) gauge theory.) Since we have taken Aµ(x) to be hermitian and traceless, we can expand it in terms of the generator matrices: Aµ(x) =Aa µ(x)Ta. (69.17) Then we can use eq.(69.8) to invert eq.(69.17): Aa µ(x) = 2TrAµ(x)Ta. (69.18) Similarly, we have Fµν(x) =Fa µνTa, (69.19) Fa µν(x) = 2TrFµνTa. (69.20) Using eq.(69.19) in eq.(69.14), we get Fc µνTc= (∂µAc ν−∂νAc µ)Tc−igAa µAb ν[Ta,Tb] = (∂µAc ν−∂νAc µ+gfabcAa µAb ν)Tc. (69.21) Then using eqs.(69.20) and (69.8) yields Fc µν=∂µAc ν−∂νAc µ+gfabcAa µAb ν. (69.22) Also, using eqs.(69.19) and (69.8) in eq.(69.16), we get Lkin=−1 4FcµνFc µν. (69.23) 69: Nonabelian Gauge Theory 410 From eq.(69.22), we see that Lkinincludes interactions among the gauge ﬁelds. A theory of this type, with nonzero fabc, is called nonabelian gauge theory orYang–Mills theory . Everything we have just said about SU( N) also goes through for SO( N), withunitary replaced by orthogonal , and traceless replaced by antisym- metric . There is also another class of compact nonabelian groups ca lled Sp(2N), and ﬁve exceptional compact groups: G(2), F(4), E(6), E(7 ), and E(8). Compact means that Tr( TaTb) is a positive deﬁnite matrix. Non- abelian gauge theory must be based on a compact group, becaus e otherwise some of the terms in Lkinwould have the wrong sign, leading to a hamil- tonian that is unbounded below. As a speciﬁc example, let us consider quantum chromodynamics , or QCD, which is based on the gauge group SU(3). There are severa l Dirac ﬁelds corresponding to quarks . Each quark comes in three colors ; these are the values of the SU(3) index. (These colors have nothing to do with ordinary color.) There are also six ﬂavors : up, down, strange, charm, bottom (or beauty) and top (or truth). Thus we consider the Di rac ﬁeld ΨiI(x), whereiis the color index and Iis the ﬂavor index. The lagrangian is L=iΨiI/DijΨjI−mIΨIΨI−1 2Tr(FµνFµν), (69.24) where all indices are summed. The diﬀerent quark ﬂavors have diﬀerent masses, ranging from a few MeV for the up and down quarks to 178 GeV for the top quark. (The quarks also have electric charges: +2 3|e|for theu, c, andtquarks, and −1 3|e|for thed,s, andbquarks. For now, however, we omit the appropriate coupling to the electromagnetic ﬁeld. ) The covariant derivative in eq.(69.24) is (Dµ)ij=δij∂µ−igAa µTa ij. (69.25) The indexaonAa µruns from 1 to 8, and the corresponding massless spin- one particles are the eight gluons . In a nonabelian gauge theory in general, we can consider scal ar or spinor ﬁelds in diﬀerent representations of the group. A representation of a com- pact nonabelian group is a set of ﬁnite-dimensional hermiti an matrices Ta R (the R is part of the name, not an index) that obey that same com muta- tion relations as the original generator matrices Ta. Given such a set of D(R)×D(R) matrices (where D(R) is the dimension of the representa- tion), and a ﬁeld φ(x) withD(R) components, we can write its covariant derivative as Dµ=∂µ−igAa µTa R, with an understood D(R)×D(R) identity matrix multiplying ∂µ. Under a gauge transformation, φ(x)→UR(x)φ(x), whereUR(x) is given by eq.(69.10) with Tareplaced by Ta R. The theory will be gauge invariant provided that the transformation ru le forAa µis in- 69: Nonabelian Gauge Theory 411 depedent of the representation used in eq.(69.9); we show in problem 69.1 that it is. We will not need to know a lot of representation theory, but we collect some useful facts in the next section. Problems 69.1) Show that eq.(69.9) implies a transformation rule for Aa µthat is in- dependent of the representation used in eq.(69.9). Hint: co nsider an inﬁnitesimal transformation. 69.2) Show that [ TaTa,Tb] = 0. 70: Group Representations 412 70Group Representations Prerequisite: 69 Given the structure coeﬃcients fabcof a compact nonabelian group, a rep- resentation of that group is speciﬁed by a set of D(R)×D(R) traceless hermitian matrices Ta R(the R is part of the name, not an index) that obey that same commutation relations as the original generators matricesTa, namely [Ta R,Tb R] =ifabcTc R. (70.1) The number D(R) is the dimension of the representation. The original Ta’s correspond to the fundamental ordeﬁning representation. Consider taking the complex conjugate of the commutation re lations, eq.(70.1). Since the structure coeﬃcients are real, we see t hat the matrices −(Ta R)∗also obey these commutation relations. If −(Ta R)∗=Ta R, or if we can ﬁnd a unitary transformation Ta R→U−1Ta RUthat makes −(Ta R)∗=Ta Rfor everya, then the representation R is real. If such a unitary transformation does not exist, but we can ﬁnd a unitary matrix V∝ne}ationslash=Isuch that −(Ta R)∗= V−1Ta RVfor everya, then the representation R is pseudoreal . If such a unitary matrix also does not exist, then the representation R iscomplex . In this case, the complex conjugate representation R is speciﬁed by Ta R=−(Ta R)∗. (70.2) One way to prove that a representation is complex is to show th at at least one generator matrix Ta R(or a real linear combination of them) has eigenvalues that do not come in plus-minus pairs. This is the case for the fundamental representation of SU( N) withN≥3. For SU(2), the fundamental representation is pseudoreal, because −(1 2σa)∗∝ne}ationslash=1 2σa, but−(1 2σa)∗=V−1(1 2σa)VwithV=σ2. For SO(N), the fundamental representation is real, because the generator matrices are antisymmetric, and every antisymmetric hermitian matrix is equal to minus i ts complex conjugate. An important representation for any compact nonabelian gro up is the adjoint representation A. This is given by (Ta A)bc=−ifabc. (70.3) Becausefabcis real and completely antisymmetric, Ta Ais manifestly her- mitian, and also satisﬁes eq.(70.2); thus the adjoint repre sentation is real. The dimension of the adjoint representation D(A) is equal to the number of generators of the group; this number is also called the dim ension of the group. 70: Group Representations 413 To see that the Ta A’s satisfy the commutation relations, we use the Jacobi identity fabdfdce+fbcdfdae+fcadfdbe= 0, (70.4) which holds for the structure coeﬃcients of any group. To pro ve the Jacobi identity, we note that TrTe/parenleftig [[Ta,Tb],Tc] + [[Tb,Tc],Ta] + [[Tc,Ta],Tb]/parenrightig = 0, (70.5) where the Ta’s are the original generator matrices. That the left-hand side of eq.(70.5) vanishes can be seen by writing out all the c ommutators as matrix products, and noting that they cancel in pairs. Emp loying the commutation relations twice in each term, followed by Tr(TaTb) =1 2δab, (70.6) ultimately yields eq.(70.4). Then, using the antisymmetry of the structure coeﬃcients, inserting some judicious factors of i, and moving the last term of eq.(70.4) to the right-hand side, we can rewrite it as (−ifabd)(−ifcde)−(−ifcbd)(−ifade) =ifacd(−ifdbe). (70.7) Now we use eq.(70.3) in eq.(70.7) to get (Ta A)bd(Tc A)de−(Tc A)bd(Ta A)de=ifacd(Td A)be, (70.8) or equivalently [ Ta A,Tc A] =ifacdTd A. Thus the Ta A’s satisfy the appropriate commutation relations. Two related numbers usefully characterize a representatio n: the index T(R) and the quadratic Casimir C(R). The index is deﬁned via Tr(Ta RTb R) =T(R)δab. (70.9) Next we recall from problem 69.2 that the matrix Ta RTa Rcommutes with every generator, and so must be a number times the identity ma trix; this number is the quadratic Casimir C(R). It is easy to show that T(R)D(A) =C(R)D(R). (70.10) With the standard normalization conventions for the genera tors, we have T(N) =1 2for the fundamental representation of SU( N) andT(N) = 2 for the fundamental representation of SO( N). We show in problem 70.2 that T(A) =Nfor the adjoint representation of SU( N), and in problem 70.3 thatT(A) = 2N−4 for the adjoint representation of SO( N). A representation R is reducible if there is a unitary transformation Ta R→ U−1Ta RUthat puts all the nonzero entries into the same diagonal bloc ks for 70: Group Representations 414 eacha; otherwise it is irreducible . Consider a reducible representation R whose generators can be put into (for example) two blocks, wi th the blocks forming the generators of the irreducible representations R1and R 2. Then R is the direct sum representation R = R 1⊕R2, and we have D(R1⊕R2) =D(R1) +D(R2), (70.11) T(R1⊕R2) =T(R1) +T(R2). (70.12) Suppose we have a ﬁeld ϕiI(x) that carries two group indices, one for the representation R 1and one for the representation R 2, denoted by iand Irespectively. This ﬁeld is in the direct product representation R 1⊗R2. The corresponding generator matrix is (Ta R1⊗R2)iI,jJ= (Ta R1)ijδIJ+δij(Ta R2)IJ, (70.13) whereiandItogether constitute the row index, and jandJtogether constitute the column index. We then have D(R1⊗R2) =D(R1)D(R2), (70.14) T(R1⊗R2) =T(R1)D(R2) +D(R1)T(R2). (70.15) To get eq.(70.15), we need to use the fact that the generator m atrices are traceless, (Ta R)ii= 0. At this point it is helpful to introduce a slightly more reﬁne d notation for the indices of a complex representation. Consider a ﬁeld ϕin the complex representation R. We will adopt the convention that such a ﬁe ld carries a “down” index: ϕi, wherei= 1,2,...,D (R). Hermitian conjugation changes the representation from R to R, and we will adopt the convention that this also raises the index on the ﬁeld, (ϕi)†=ϕ†i. (70.16) Thus a down index corresponds to the representation R, and an up index to R. Indices can be contracted only if one is up and one is down. G enerator matrices for R are then written with the ﬁrst index down and th e second index up: ( Ta R)ij. An inﬁnitesimal group transformation of ϕitakes the form ϕi→(1−iθaTa R)ijϕj =ϕi−iθa(Ta R)ijϕj. (70.17) The generator matrices for R are then given by (Ta R)ij=−(Ta R)ji, (70.18) 70: Group Representations 415 where we have used the hermiticity to trade complex conjugat ion for trans- position of the indices. An inﬁnitesimal group transformat ion ofϕ†itakes the form ϕ†i→(1−iθaTa R)ijϕ†j =ϕ†i−iθa(Ta R)ijϕ†j =ϕ†i+iθa(Ta R)jiϕ†j, (70.19) where we used eq.(70.18) to get the last line. Note that eqs.( 70.17) and (70.19) together imply that ϕ†iϕiis invariant, as expected. Consider the Kronecker delta symbol with one index down and o ne up: δij. Under a group transformation, we have δij→(1 +iθaTa R)ik(1 +iθaTa R)jlδkl = (1 +iθaTa R)ikδkl(1−iθaTa R)lj =δij+O(θ2). (70.20) Eq.(70.20) shows that δijis aninvariant symbol of the group. This exis- tence of this invariant symbol, which carries one index for R and one for R, tells us that the product of the representations R and R must contain thesinglet representation 1, speciﬁed by Ta 1= 0. We therefore can write R⊗R = 1 ⊕... . (70.21) The generator matrix ( Ta R)ij, which carries one index for R, one for R, and one for the adjoint representation A, is also an invarian t symbol. To see this, we make a simultaneous inﬁnitesimal group transforma tion on each of these indices, (Tb R)ij→(1−iθaTa R)ik(1−iθaTa R)jl(1−iθaTa A)bc(Tc R)kl = (Tb R)ij−iθa[(Ta R)ik(Tb R)kj+ (Ta R)jl(Tb R)il+ (Ta A)bc(Tc R)ij] +O(θ2). (70.22) The factor in square brackets should vanish if (as we claim) t he generator matrix is an invariant symbol. Using eqs.(70.3) and (70.18) , we have [...] = (Ta R)ik(Tb R)kj−(Ta R)lj(Tb R)il−ifabc(Tc R)ij = (Ta RTb R)ij−(Tb RTa R)ij−ifabc(Tc R)ij = 0, (70.23) 70: Group Representations 416 where the last line follows from the commutation relations. The fact that (Ta R)ijis an invariant symbol implies that R⊗R⊗A = 1 ⊕... . (70.24) If we now multiply both sides of eq.(70.24) by A, and use A ⊗A = 1 ⊕... [which follows from eq.(70.21) and the reality of A], we ﬁnd R ⊗R = A ⊕... . Combining this with eq.(70.21), we have R⊗R = 1 ⊕A⊕... . (70.25) That is, the product of a representation with its complex con jugate is al- ways reducible into a sum that includes (at least) the single t and adjoint representations. For the fundamental representation N of SU( N), we have N⊗N = 1 ⊕A, (70.26) with no other representations on the right-hand side. To see this, recall that D(1) = 1,D(N) =D(N) =N, and, as shown in section 24, D(A) =N2−1. From eq.(70.14), we see that there is no room for anything els e on the right-hand side of eq.(70.26). Consider now a real representation R. From eq.(70.25), with R = R, we have R⊗R = 1 ⊕A⊕... . (70.27) The singlet on the right-hand side implies the existence of a n invariant symbol with two R indices; this symbol is the Kronecker delta δij. It is invariant because δij→(1−iθaTa R)ik(1−iθaTa R)jlδkl =δij−iθa[(Ta R)ij+ (Ta R)ji] +O(θ2). (70.28) The term in square brackets vanishes by hermiticity and eq.( 70.18). The fact thatδij=δjiimplies that the singlet on the right-hand side of eq.(70.28 ) appears in the symmetric part of this product of two identica l representa- tions. The fundamental representation N of SO( N) is real, and we have N⊗N = 1 S⊕AA⊕SS. (70.29) The subscripts tell whether the representation appears in t he symmetric or antisymmetric part of the product. The representation S cor responds to a ﬁeld with a symmetric traceless pair of fundamental indice s:ϕij=ϕji, 70: Group Representations 417 ϕii= 0, where the repeated index is summed. We have D(1) = 1,D(N) = N, and, as shown in section 24, D(A) =1 2N(N−1). Also, a traceless symmetric tensor has D(S) =1 2N(N+1)−1 independent components; thus eq.(70.14) is fulﬁlled. Consider now a pseudoreal representation R. Since R is equiv alent to its complex conjugate, up to a change of basis, eq.(70.27) still holds. However, we cannot identify δijas the corresponding invariant symbol, because then eq.(70.28) shows that R would have to be real, rather than pse udoreal. From the perspective of the direct product, the only alterna tive is to have the singlet appear in the antisymmetric part of the product, rather than the symmetric part. The corresponding invariant symbol mus t then be antisymmetric on exchange of its two R indices. An example (the only one that will be of interest to us) is the f unda- mental representation of SU(2). For SU( N) in general, another invariant symbol is the Levi-Civita tensor εi1...iN, which carries Nfundamental in- dices and is completely antisymmetric. It is invariant beca use, under an SU(N) transformation, εi1...iN→Ui1j1...UiNjNεj1...jN = (detU)εi1...iN. (70.30) Since detU= 1 for SU( N), we see that the Levi-Civita symbol is invariant. We can similarly consider εi1...iN, which carries Ncompletely antisymmetric antifundamental indices. For SU(2), the Levi-Civita symbo l isεij=−εji; this is the two-index invariant symbol that corresponds to t he singlet in the product 2⊗2 = 1 A⊕3S, (70.31) where 3 is the adjoint representation. We can use εijandεijto raise and lower SU(2) indices. This is another way to see that there is no distinction between the fundament al represen- tation 2 and its complex conjugate 2. That is, if we have a ﬁeld ϕiin the representation 2, we can get a ﬁeld in the representation 2 by raising the index:ϕi=εijϕj. The structure constants fabcare another invariant symbol. This follows from (Ta A)bc=−ifabc, since we have seen that generator matrices (in any representation) are invariant. Alternatively, given the g enerator matrices in a representation R, we can write T(R)fabc=−iTr(Ta R[Tb R,Tc R]). (70.32) Since the right-hand side is invariant, the left-hand side m ust be as well. 70: Group Representations 418 If we use an anticommutator in place of the commutator in eq.( 70.32), we get another invariant symbol, A(R)dabc≡1 2Tr(Ta R{Tb R,Tc R}), (70.33) whereA(R) is the anomaly coeﬃcient of the representation. The cyclic property of the trace implies that A(R)dabcis symmetric on exchange of any pair of indices. Using eq.(70.18), we can see that A(R) =−A(R). (70.34) Thus, if R is real or pseudoreal, A(R) = 0. We also have A(R1⊕R2) =A(R1) +A(R2), (70.35) A(R1⊗R2) =A(R1)D(R2) +D(R1)A(R2). (70.36) We normalize the anomaly coeﬃcient so that it equals one for t he smallest complex representation. In particular, for SU( N) withN≥3, the smallest complex representation is the fundamental, and A(N) = 1. For SU(2), all representations are real or pseudoreal, and A(R) = 0 for all of them. Reference Notes More group and representation theory can be found in Ramond II . Problems 70.1) Verify eq.(70.10). 70.2) a) Use eqs.(70.12) and (70.26) to compute T(A) for SU( N). b) For SU(2), the adjoint representation is speciﬁed by ( Ta A)bc= −iεabc. Use this to compute T(A) explicitly for SU(2). Does your result agree with part (a)? c) Consider the SU(2) subgroup of SU(N) that acts on the ﬁrst two components of the fundamental representation of SU( N). Under this SU(2) subgroup, the N of SU( N) transforms as 2 ⊕(N−2)1’s. Us- ing eq.(70.26), ﬁgure out how the adjoint representation of SU(N) transforms under this SU(2) subgroup. d) Use your results from parts (b) and (c) to compute T(A) for SU(N). Does your result agree with part (a)? 70: Group Representations 419 70.3) a) Consider the SO(3) subgroup of SO( N) that acts on the ﬁrst three components of the fundamental representation of SO( N). Under this SO(3) subgroup, the N of SO( N) transforms as 3 ⊕(N−3)1’s. Us- ing eq.(70.29), ﬁgure out how the adjoint representation of SO(N) transforms under this SO(3) subgroup. b) Use your results from part (a) and from problem 70.2 to comp ute T(A) for SO( N). 70.4) a) For SU( N), we have N⊗N =AA⊕ SS, (70.37) where Acorresponds to a ﬁeld with two antisymmetric fundamental SU(N) indices,ϕij=−ϕji, and Scorresponds to a ﬁeld with two symmetric fundamental SU( N) indices,ϕij= +ϕji. Compute D(A) andD(S). b) By considering an SU(2) subgroup of SU( N), compute T(A) and T(S). c) For SU(3), show that A=3. d) By considering an SU(3) subgroup of SU( N), compute A(A) and A(S). 70.5) Consider a ﬁeld ϕiin the representation R 1and a ﬁeld χIin the representation R 2. Their product ϕiχIis then in the direct product representation R 1⊗R2, with generator matrices given by eq.(70.13). a) Prove the distribution rule for the covariant derivative , [Dµ(ϕχ)]iI= (Dµϕ)iχI+ϕi(Dµχ)I. (70.38) b) Consider a ﬁeld ϕiin the complex representation R. Show that ∂µ(ϕ†iϕi) = (Dµϕ†)iϕi+ϕ†i(Dµϕ)i. (70.39) Explain why this is a special case of eq.(70.38). 70.6) The ﬁeld strength in Yang-Mills theory is in the adjoin t representa- tion, and so its covariant derivative is (DρFµν)a=∂ρFa µν−igAc ρ(Tc A)abFb µν. (70.40) Prove the Bianchi identity , (DµFνρ)a+ (DνFρµ)a+ (DρFµν)a= 0. (70.41) 71: The Path Integral for Nonabelian Gauge Theory 420 71The Path Integral for Nonabelian Gauge Theory Prerequisite: 53, 69 We wish to evaluate the path integral for nonabelian gauge th eory (also known as Yang–Mills theory), Z(J)∝/integraldisplay DAeiSYM(A,J), (71.1) SYM(A,J) =/integraldisplay d4x/bracketleftig −1 4FaµνFa µν+JaµAa µ/bracketrightig . (71.2) In section 57, we evaluated the path integral for U(1) gauge t heory by arguing that, in momentum space, the component of the U(1) ga uge ﬁeld parallel to the four-momentum kµdid not appear in the action, and hence should not be integrated over. This argument relied on the fo rm of the U(1) gauge transformation, Aµ(x)→Aµ(x)−∂µΓ(x). (71.3) In the nonabelian case, however, the gauge transformation i s nonlinear, Aµ(x)→U(x)Aµ(x)U†(x) +igU(x)∂µU†(x), (71.4) whereAµ(x) =Aa µ(x)Ta. For an inﬁnitesimal transformation, U(x) =I−igθ(x) +O(θ2) =I−igθa(x)Ta+O(θ2), (71.5) we have Aµ(x)→Aµ(x) +ig[Aµ(x),θ(x)]−∂µθ(x), (71.6) or equivalently Aa µ(x)→Aa µ(x)−gfabcAb µ(x)θc(x)−∂µθa(x) =Aa µ(x)−[δac∂µ+gfabcAb µ(x)]θc(x) =Aa µ(x)−[δac∂µ−igAb µ(−ifbac)]θc(x) =Aa µ(x)−[δac∂µ−igAb µ(Tb A)ac]θc(x) =Aa µ(x)−Dac µθc(x), (71.7) whereDac µis the covariant derviative in the adjoint representation. We see the similarity with the abelian case, eq.(71.3). Howeve r, the fact that 71: The Path Integral for Nonabelian Gauge Theory 421 it isDµthat appears in eq.(71.7), rather than ∂µ, means that we cannot account for gauge redundancy in the path integral by simply e xcluding the components of Aa µthat are parallel to kµ. We will have to do something more clever. Consider an ordinary integral of the form Z∝/integraldisplay dxdyeiS(x), (71.8) where both xandyare integrated from minus to plus inﬁnity. Because y does not appear in S(x), the integral over yis redundant. We can then deﬁneZby simply dropping the integral over y, Z≡/integraldisplay dxeiS(x). (71.9) This is how we dealt with gauge redundancy in the abelian case . We could get the same answer by inserting a delta function, ra ther than by dropping the yintegral: Z=/integraldisplay dxdyδ (y)eiS(x). (71.10) Furthermore, the argument of the delta function can be shift ed by an ar- bitrary function of x, without changing the result: Z=/integraldisplay dxdyδ (y−f(x))eiS(x). (71.11) Suppose we are not given f(x) explicitly, but rather are told that y=f(x) is the unique solution, for ﬁxed x, ofG(x,y) = 0. Then we can write δ(G(x,y)) =δ(y−f(x)) |∂G/∂y |, (71.12) where we have used a standard rule for delta functions. We can drop the absoute-value signs if we assume that ∂G/∂y is positive when evaluated at y=f(x). Then we have Z=/integraldisplay dxdy∂G ∂yδ(G)eiS. (71.13) Now let us generalize this result to an integral over dnxdny. We will need nfunctionsGi(x,y) to ﬁx all ncomponents of y. The generalization of eq.(71.13) is Z=/integraldisplay dnxdnydet/parenleftigg ∂Gi ∂yj/parenrightigg /producttext iδ(Gi)eiS. (71.14) 71: The Path Integral for Nonabelian Gauge Theory 422 Now we are ready to translate these results to path integrals over non- abelian gauge ﬁelds. The role of the redundant integration v ariableyis played by the set of all gauge transformations θa(x). The role of the inte- gration variables xandytogether is played by the gauge ﬁeld Aa µ(x). The role ofGis played by a gauge-ﬁxing function . We will use the gauge-ﬁxing function appropriate for Rξgauge, which is Ga(x)≡∂µAa µ(x)−ωa(x), (71.15) whereωa(x) is a ﬁxed, arbitrarily chosen function of x. (We will see how the parameter ξenters later.) In eq.(71.15), the spacetime argument xand the indexaplay the role of the index iin eq.(71.14). Our path integral becomes Z(J)∝/integraldisplay DAdet/parenleftbiggδG δθ/parenrightbigg/producttext x,aδ(G)eiSYM, (71.16) whereSYMis given by eq.(71.2). Now we have to evaluate the functional derivative δGa(x)/δθb(y), and then its functional determinant. From eqs.(71.7) and (71.1 5), we ﬁnd that, under an inﬁnitesimal gauge transformation, Ga(x)→Ga(x)−∂µDab µθb(x). (71.17) Thus we have δGa(x) δθb(y)=−∂µDab µδ4(x−y), (71.18) where the derivatives are with respect to x. Now we need to compute the functional determinant of eq.(71. 18). Luckily, we learned how to do this in section 53. A functional determi- nant can be written as a path integral over complex Grassmann variables. So let us introduce the complex Grassmann ﬁeld ca(x), and its hermitian conjugate ¯ca(x). (We use a bar rather than a dagger to keep the notation a little less cluttered.) These ﬁelds are called Faddeev-Popov ghosts . Then we can write detδGa(x) δθb(y)∝/integraldisplay DcD¯ceiSgh, (71.19) where the ghost action is Sgh=/integraltextd4xLgh, and the ghost lagrangian is Lgh= ¯ca∂µDab µcb =−∂µ¯caDab µcb =−∂µ¯ca∂µca+ig∂µ¯caAc µ(Tc A)abcb =−∂µ¯ca∂µca+gfabcAc µ∂µ¯cacb. (71.20) 71: The Path Integral for Nonabelian Gauge Theory 423 We dropped a total divergence in the second line. We see that ca(x) has the standard kinetic term for a complex scalar ﬁeld. (We need the factor of iin front ofSghin eq.(71.19) for this to work out; this factor aﬀects only the overall phase of Z(J), and so we can choose it at will.) The ghost ﬁeld is also a Grassmann ﬁeld, and so a closed loop of ghost lines in a Feynman diagram carries an extra factor of minus one. We see from eq.( 71.20) that the ghost ﬁeld interacts with the gauge ﬁeld, and so we will ha ve such loops. Since the particles associated with the ghost ﬁeld do not in f act exist (and would violate the spin-statistics theorem if they did) , it must be that the amplitude to produce them in any scattering process is ze ro. This is indeed the case, as we will discuss in section 74. We note that in abelian gauge theory, where fabc= 0, there is no interaction term for the ghost ﬁeld. In that case, it is simpl y an extra free ﬁeld, and we can absorb its path integral into the overall nor malization. We have one ﬁnal trick to perform. Our gauge-ﬁxing function, Ga(x) contains an arbitrary function ωa(x). The path integral Z(J) is, however, independent of ωa(x). So, we can multiply Z(J) by an arbitrary functional ofω, and then perform a path integral over ω; the result can change only the overall normalization of Z(J). In particular, let us multiply Z(J) by exp/bracketleftbigg −i 2ξ/integraldisplay d4xωaωa/bracketrightbigg . (71.21) Because of the delta-functional in eq.(71.16), it is easy to integrate over ω. The ﬁnal result for Z(J) is Z(J)∝/integraldisplay DAD¯cDcexp/parenleftig iSYM+iSgh+iSgf/parenrightig , (71.22) whereSYMis given by eq.(71.2), Sghis given by the integral over d4xof eq.(71.20), and Sgf(gf stands for gauge ﬁxing ) is given by the integral over d4xof Lgf=−1 2ξ−1∂µAa µ∂νAa ν. (71.23) In the next section, we will derive the Feynman rules that fol low from this path integral. 72: The Feynman Rules for Nonabelian Gauge Theory 424 72The Feynman Rules for Nonabelian Gauge Theory Prerequisite: 71 Let us begin by considering nonabelian gauge theory without any scalar or spinor ﬁelds. The lagrangian is LYM=−1 4FeµνFe µν =−1 4(∂µAeν−∂νAeµ+gfabeAaµAbν)(∂µAe ν−∂νAe µ+gfcdeAc µAd ν) =−1 2∂µAeν∂µAe ν+1 2∂µAeν∂νAe µ −gfabeAaµAbν∂µAe ν−1 4g2fabefcdeAaµAbνAc µAd ν. (72.1) To this we should add the gauge-ﬁxing term for Rξgauge, Lgf=−1 2ξ−1∂µAe µ∂νAe ν. (72.2) Adding eqs.(72.1) and (72.2), and doing some integrations- by-parts in the quadratic terms, we ﬁnd LYM+Lgf=1 2Aeµ(gµν∂2−∂µ∂ν)Aeν+1 2ξ−1Aeµ∂µ∂νAeν −gfabcAaµAbν∂µAc ν−1 4g2fabefcdeAaµAbνAc µAd ν.(72.3) The ﬁrst line of eq.(72.3) yields the gluon propagator in Rξgauge, ˜∆ab µν(k) =δab k2−iǫ/parenleftbigg gµν−kµkν k2+ξkµkν k2/parenrightbigg . (72.4) The second line of eq.(72.3) yields three- and four-gluon ve rtices, shown in ﬁg.(72.1). The three-gluon vertex factor is iVabc µνρ(p,q,r) =i(−gfabc)(−irµgνρ) + [5 permutations of ( a,µ,p),(b,ν,q),(c,ρ,r)] =gfabc[(q−r)µgνρ+ (r−p)νgρµ+ (p−q)ρgµν].(72.5) The four-gluon vertex factor is iVabcd µνρσ=−ig2fabefcdegµρgνσ + [5 permutations of ( b,ν),(c,ρ),(d,σ) ] =−ig2[fabefcde(gµρgνσ−gµσgνρ) +facefdbe(gµσgρν−gµνgρσ) +fadefbce(gµνgσρ−gµρgσν)]. (72.6) 72: The Feynman Rules for Nonabelian Gauge Theory 425 ba bµ νp qr aµ cρ σ d c ρ ν Figure 72.1: The three-gluon and four-gluon vertices in non abelian gauge theory. These vertex factors are quite a bit more complicated that th e ones we are used to, and they lead to rather involved formulae for sca ttering cross sections. For example, the tree-level gg→ggcross section (where gis a gluon), averaged over initial spins and colors and summed ov er ﬁnal spins and colors, has 12,996 terms! Of course, many are identical a nd the ﬁnal result can be expressed much more simply, but this is no help t o us at the initial stages of computation. For this reason, we postpone any attempt at tree-level calculations until section 81, where we will m ake use of some techniques (color ordering and Gervais-Neveu gauge) that, combined with spinor-helicity methods, greatly reduce the necessary lab or. For loop calculations, we need to include the ghosts. The gho st la- grangian is Lgh=−∂µ¯cbDbc µcc =−∂µ¯cc∂µcc+ig∂µ¯cbAa µ(Ta A)bccc =−∂µ¯cc∂µcc+gfabcAa µ∂µ¯cbcc. (72.7) The ghost propagator is ˜∆ab(k2) =δab k2−iǫ. (72.8) Because the ghosts are complex scalars, their propagators c arry a charge arrow. The ghost-ghost-gluon vertex shown in ﬁg.(72.2); th e associated vertex factor is iVabc µ(q,r) =i(gfabc)(−iqµ) =gfabcqµ. (72.9) 72: The Feynman Rules for Nonabelian Gauge Theory 426 r aµc b q Figure 72.2: The ghost-ghost-gluon vertex in nonabelian ga uge theory. i aµj Figure 72.3: The quark-quark-gluon vertex in nonabelian ga uge theory. If we include a quark coupled to the gluons, we have the quark l a- grangian Lq=iΨi/DijΨj−mΨiΨi =iΨi/∂Ψi−mΨiΨi+gAa µΨiγµTa ijΨj. (72.10) The quark propagator is ˜Sij(p) =(−/p+m)δij p2+m2−iǫ. (72.11) The quark-quark-gluon vertex shown in ﬁg.(72.3); the assoc iated vertex factor is iVµa ij=igγµTa ij. (72.12) If the quark is in a representation R other than the fundament al, thenTa ij becomes (Ta R)ij. Problems 72.1) Consider a complex scalar ﬁeld ϕiin a representation R of the gauge group. Find the vertices that involve this ﬁeld, and the asso ciated vertex factors. 73: The Beta Function in Nonabelian Gauge Theory 427 73The Beta Function in Nonabelian Gauge Theory Prerequisite: 70, 72 In this section, we will do enough loop calculations to compu te the beta function for the Yang–Mills coupling g. We can write the complete lagrangian, including Zfactors, as L=1 2Z3Aaµ(gµν∂2−∂µ∂ν)Aaν+1 2ξ−1Aaµ∂µ∂νAaν −Z3ggfabcAaµAbν∂µAc ν−1 4Z4gg2fabefcdeAaµAbνAc µAd ν −Z2′∂µ¯Ca∂µCa+Z1′gfabcAc µ∂µ¯CaCb +iZ2Ψi/∂Ψi−ZmmΨiΨi+Z1gAa µΨiγµTa ijΨj. (73.1) Note that the gauge-ﬁxing term in the ﬁrst line does not need a Zfactor; we saw in section 62 that the ξ-dependent term in the propagator is not renormalized. We see that gappears in several places in L, and gauge invariance leads us to expect that it will renormalize in the same way in each pl ace. If we rewrite Lin terms of bare ﬁelds and parameters, and compare with eq.(73.1), we ﬁnd that g2 0=Z2 1 Z2 2Z3g2˜µε=Z2 1′ Z2 2′Z3g2˜µε=Z2 3g Z3 3g2˜µε=Z4g Z2 3g2˜µε, (73.2) whered= 4−εis the number of spacetime dimensions. To prove eq.(73.2), we have to derive the nonabelian analogs of the Ward identiti es, known as Slavnov-Taylor identities . For now, we simply assume that eq.(73.2) holds; we will return to this issue in section 74. The simplest computation to perform is the renormalization of the quark-quark-gluon vertex. This is partly because much of th e calculation is the same as it is in spinor electrodynamics, and so we can ma ke use of our results in section 62. We then must compute Z1,Z2, andZ3. We will work in Feynman gauge, and use the MS renormalization scheme. We begin with Z2. TheO(g2) corrections to the fermion propagator are shown in ﬁg.(73.1). These diagrams are the same as in spin or elec- trodynamics, except for the factors related to the color ind ices. The loop diagram has a color-factor of ( TaTa)ij=C(R)δij. (Here we have allowed the quark to be in an arbitrary representation R; for notatio nal simplicity, we will continue to omit the label R on the generator matrices .) In section 62, we found that, in spinor electrodynamics, the divergent part of this diagram contributes −(e2/8π2ε)/pto the electron self-energy Σ(/ p). Thus 73: The Beta Function in Nonabelian Gauge Theory 428 nl p p p p p+la b i i j j k Figure 73.1: The one-loop and counterterm corrections to th e quark prop- agator in quantum chromodynamics. bl l l ijb aµν ν aµi j lllν ρ c Figure 73.2: The one-loop corrections to the quark–quark–g luon vertex in in quantum chromodynamics. in Yang–Mills gauge theory, the divergent part of this diagr am contributes −(g2/8π2ε)C(R)δij/pto the quark self-energy Σ ij(/p). This divergent term must be cancelled by the counterterm contribution of −(Z2−1)δij/p. There- fore, in Yang–Mills theory, with a quark in the representati on R, using Feynman gauge and the MS renormalization scheme, we have Z2= 1−C(R)g2 8π21 ε+O(g4). (73.3) Moving on to the quark-quark-gluon vertex, we the contribut ing one- loop diagrams are shown in ﬁg.(73.2). The ﬁrst diagram is aga in the same as it is in spinor electrodynamics, except for the color fact or of (TbTaTb)ij. We can simplify this via TbTaTb=Tb/parenleftig TbTa+ifabcTc/parenrightig =C(R)Ta+1 2ifabc[Tb,Tc] =C(R)Ta+1 2(ifabc)(ifbcd)Td =C(R)Ta−1 2(Ta A)bc(Td A)cbTd =/bracketleftig C(R)−1 2T(A)/bracketrightig Ta. (73.4) In the second line, we used the complete antisymmetry of fabcto replace TbTcwith1 2[Tb,Tc]. To get the last line, we used Tr( Ta ATd A) =T(A)δad. In 73: The Beta Function in Nonabelian Gauge Theory 429 section 62, we found that, in spinor electrodynamics, the di vergent part of this diagram contributes ( e2/8π2ε)ieγµto the vertex function iVµ(p′,p). Thus in Yang–Mills theory, the divergent part of this diagra m contributes /bracketleftig C(R)−1 2T(A)/bracketrightigg2 8π2εigTa ijγµ(73.5) to the quark-quark-gluon vertex function iVaµ ij(p′,p). This divergent term, along with any divergent term from the second diagram of ﬁg.( 73.2), must be cancelled by the tree-level vertex iZ1gγµTa ij. Now we must evaluate the second diagram of ﬁg.(73.2). The div ergent part is independent of the external momenta, and so we can set them to zero. Then we get a contribution to iVaµ ij(0,0) of (ig)2gfabc(TcTb)ij/parenleftig 1 i/parenrightig3/integraldisplayd4ℓ (2π)4γρ(−/ℓ+m)γν ℓ2ℓ2(ℓ2+m2) ×[(ℓ−(−ℓ))µgνρ+ (−ℓ−0)νgρµ+ (0−ℓ)ρgµν].(73.6) We can simplify the color factor with the manipulations of eq .(73.4), fabcTcTb=1 2fabc[Tc,Tb] =1 2ifabcfcbdTd =−1 2iT(A)Ta. (73.7) The numerator in eq.(73.6) is Nµ=γρ(−γσℓσ+m)γν(2ℓµgνρ−ℓνgρµ−ℓρgµν). (73.8) We can drop the terms linear in ℓ, and make the replacement ℓσℓµ→ d−1ℓ2gσµ. Thus we have Nµ→ −d−1ℓ2(γργσγν)(2gσµgνρ−gσνgρµ−gσρgµν) → −d−1ℓ2(2γνγµγν−γµγνγν−γργργµ) → −d−1ℓ2(2(d−2) +d+d)γµ. (73.9) Because we are only keeping track of the divergent term, we ar e free to set d= 4, which yields Nµ→ −3ℓ2γµ. (73.10) Using eqs.(73.7) and (73.10) in eq.(73.6), we get 3 2T(A)g3Ta ijγµ/integraldisplayd4ℓ (2π)41 ℓ2(ℓ2+m2). (73.11) 73: The Beta Function in Nonabelian Gauge Theory 430 lkµa kb νcρ kb νkµal l k+l σd kµa kb νkµa kb νkµa kb ν k+lc l d k+l Figure 73.3: The one-loop and counterterm corrections to th e gluon prop- agator in quantum chromodynamics. After continuing to ddimensions, the integral becomes i/8π2ε+O(ε0). Combining eqs.(73.5) and (73.6), we ﬁnd that the divergent p art of the quark-quark-gluon vertex function is Vaµ ij(0,0)div=/parenleftigg Z1+/bracketleftig C(R)−1 2T(A)/bracketrightigg2 8π2ε+3 2T(A)g2 8π2ε/parenrightigg gTa ijγµ. (73.12) Requiring Vaµ ij(0,0) to be ﬁnite yields Z1= 1−/bracketleftig C(R) +T(A)/bracketrightigg2 8π21 ε+O(g4) (73.13) in Feynman gauge and the MS renormalization scheme. Note that we have found that Z1does not equal Z2. In electrodynamics, we argued that gauge invariance requires all derivatives in the lagrangian to be covariant derivatives, and that both pieces of Dµ=∂µ−ieAµshould therefore be renormalized by the same factor; this then impl ies thatZ1 must equal Z2. In Yang–Mills theory, however, this argument fails. This failure is due to the introduction of the ordinary derivativ e in the gauge- ﬁxing function for Rξgauge: once we have added LgfandLghtoLYM, we ﬁnd that both ordinary and covariant derivatives appear. (T his is especially obvious for Lgh.) Therefore, to be certain of what gauge invariance does and does not imply, we must derive the appropriate Slavnov-Tayl or identities, a subject we will take up in section 74. Next we turn to the calculation of Z3. TheO(g2) corrections to the gluon propagator are shown in ﬁg.(73.3). The ﬁrst diagram is proportional to/integraltextd4ℓ/ℓ2; as we saw in section 65, this integral vanishes after dimens ional regularization. 73: The Beta Function in Nonabelian Gauge Theory 431 The second diagram yields a contribution to iΠµνab(k) of 1 2g2facdfbcd/parenleftig 1 i/parenrightig2/integraldisplayd4ℓ (2π)4Nµν ℓ2(ℓ+k)2, (73.14) where the one-half is a symmetry factor, and Nµν= [(k+ℓ)−(−ℓ))µgρσ+ (−ℓ−(−k))ρgσµ+ ((−k)−(k+ℓ))σgµρ] ×[(−k−ℓ)−ℓ)νgρσ+ (ℓ−k)ρδσν+ (k−(−k−ℓ))σδνρ] =−[(2ℓ+k)µgρσ−(ℓ−k)ρgσµ−(ℓ+2k)σgµρ] ×[(2ℓ+k)νgρσ−(ℓ−k)ρδσν−(ℓ+2k)σδνρ] (73.15) The color factor can be simpliﬁed via facdfbcd=T(A)δab. We combine de- nominators with Feynman’s formula, and continue to d= 4−εdimensions; we now have −1 2g2T(A)δab˜µε/integraldisplay1 0dx/integraldisplayddq (2π)dNµν (q2+D)2, (73.16) whereD=x(1−x)k2andq=ℓ+xk. The numerator is Nµν=−[(2q+(1−2x)k)µgρσ−(q−(1+x)k)ρgσµ−(q+(2−x)k)σgµρ] ×[(2q+(1−2x)k)νgρσ−(q−(1+x)k)ρδσν−(q+(2−x)k)σδνρ]. (73.17) Terms linear in qwill integrate to zero, and so we have Nµν→ − 2q2gµν−(4d−6)qµqν −[(1+x)2+ (2−x)2]k2gµν −[d(1−2x)2+ 2(1−2x)(1+x) −2(2−x)(1+x)−2(2−x)(1−2x)]kµkν.(73.18) Since we are only interested in the divergent part, we can go a head and set d= 4 in the numerator. We can also make the replacement qµqν→1 4q2gµν. Then we ﬁnd Nµν→ −9 2q2gµν−(5−2x+2x2)k2gµν+ (2+10x−10x2)kµkν.(73.19) We saw in section 62 that, when integrated against ( q2+D)−2,q2can be replaced with (2 d−1)−1D; in our case this is −2x(1−x)k2. This yields Nµν→ −(5−11x+11x2)k2gµν+ (2+10x−10x2)kµkν. (73.20) 73: The Beta Function in Nonabelian Gauge Theory 432 We now use ˜µε/integraldisplayddq (2π)d1 (q2+D)2=i 8π2ε+O(ε0) (73.21) in eq.(73.16) to get −ig2 16π2T(A)δab1 ε/integraldisplay1 0dxNµν+O(ε0). (73.22) Performing the integral over xyields −ig2 16π2T(A)δab1 ε/parenleftig −19 6k2gµν+11 3kµkν/parenrightig (73.23) as the divergent contribution of the second diagram to iΠµνab(k). Next we have the third diagram of ﬁg.(73.3), which makes a con tribu- tion toiΠµνab(k) of (−1)g2facdfbdc/parenleftig 1 i/parenrightig2/integraldisplayd4ℓ (2π)4(ℓ+k)µℓν ℓ2(ℓ+k)2, (73.24) where the factor of minus one is from the closed ghost loop. Th e color factor isfacdfbdc=−T(A)δab. After combining denominators, the numerator becomes (ℓ+k)µℓν= (q+ (1−x)k)µ(q−xk)ν →1 4q2gµν−x(1−x)kµkν → −1 2x(1−x)k2gµν−x(1−x)kµkν. (73.25) We then use eq.(73.21) in eq.(73.24); performing the integr al overxyields −ig2 8π2T(A)δab1 ε/parenleftig −1 12k2gµν−1 6kµkν/parenrightig (73.26) as the divergent contribution of the third diagram to iΠµνab(k). Finally, we have the fourth diagram. This is the same as it is i n spinor electrodynamics, except for the color factor of Tr( TaTb) =T(R)δab. If there is more than one ﬂavor of quark, each contributes separ ately, leading to a factor of the number of ﬂavors nF. Then, using our results in section 62, we ﬁnd −ig2 6π2nFT(R)δab1 ε/parenleftig k2gµν−kµkν/parenrightig (73.27) as the divergent contribution of the fourth diagram to iΠµνab(k). 73: The Beta Function in Nonabelian Gauge Theory 433 Adding up eqs.(73.23), (73.26), and (73.27), as well as the c ounterterm contriubtion, we ﬁnd that the gluon self-energy is transver se, Πµνab(k) = Π(k2)(k2gµν−kµkν)δab, (73.28) and that Π(k2)div=−(Z3−1) +/bracketleftig 5 3T(A)−4 3nFT(R)/bracketrightigg2 8π21 ε+O(g4).(73.29) Thus we ﬁnd Z3= 1 +/bracketleftig 5 3T(A)−4 3nFT(R)/bracketrightigg2 8π21 ε+O(g4) (73.30) in Feynman gauge and the MS renormalization scheme. Let us collect our results: Z1= 1−/bracketleftig C(R) +T(A)/bracketrightigg2 8π21 ε+O(g4), (73.31) Z2= 1−C(R)g2 8π21 ε+O(g4), (73.32) Z3= 1 +/bracketleftig 5 3T(A)−4 3nFT(R)/bracketrightigg2 8π21 ε+O(g4), (73.33) in Feynman gauge and the MS renormalization scheme. We deﬁne α≡g2 4π. (73.34) Then we have α0=Z2 1 Z2 2Z3α˜µε. (73.35) Let us write ln/parenleftig Z−1 3Z−2 2Z2 1/parenrightig =∞/summationdisplay n=1Gn(α) εn. (73.36) Then we have lnα0=∞/summationdisplay n=1Gn(α) εn+ lnα+εln ˜µ. (73.37) From eqs.(73.31–73.33), we get G1(α) =−/bracketleftig 11 3T(A)−4 3nFT(R)/bracketrightigα 2π+O(α2). (73.38) Then, the general analysis of section 28 yields β(α) =α2G′ 1(α), (73.39) 73: The Beta Function in Nonabelian Gauge Theory 434 where the prime denotes diﬀerentiation with respect to α. Thus we ﬁnd β(α) =−/bracketleftig 11 3T(A)−4 3nFT(R)/bracketrightigα2 2π+O(α3). (73.40) in nonabelian gauge theory with nFDirac fermions in the representation R of the gauge group. We can, if we like, restate eq.(73.40) in terms of gas β(g) =−/bracketleftig 11 3T(A)−4 3nFT(R)/bracketrightigg3 16π2+O(g5). (73.41) To go from eq.(73.40) to eq.(73.41), we use α=g2/4πand ˙α=g˙g/2π, where the dot denotes d/dlnµ. In quantum chromodynamics, the gauge group is SU(3), and the quarks are in the fundamental representation. Thus T(A) = 3 and T(R) =1 2, and the factor in square brackets in eq.(73.41) is 11 −2 3nF. So fornF≤16, the beta function is negative: the gauge coupling in quantum chr omodynamics gets weaker at high energies, and stronger at low energies. This has dramatic physical consequences. Perturbation the ory cannot serve as a reliable guide to the low-energy physics. And inde ed, in nature we do not see isolated quarks or gluons. (Quarks, in particular , have fractional electric charges and would be easy to discover.) The appropr iate conclusion is that color is conﬁned : all ﬁnite-energy states are invariant under a global SU(3) transformation. This has not yet been rigorously prov en, but it is the only hypothesis that is consistent with all of the availa ble theoretical and experimental information. Problems 73.1) Compute the beta function for gin Yang–Mills theory with a complex scalar ﬁeld in the representation R of the gauge group. Hint: all the real work has been done already in this section, problem 72.1 , and section 66. 73.2) Write down the beta function for the gauge coupling in Y ang–Mills theory with several Dirac fermions in the representations R i, and several complex scalars in the representations R′ j. 73.3) Compute the one-loop contributions to the anomalous d imensions of m, Ψ, andAµ. 74: BRST Symmetry 435 74BRST Symmetry Prerequisite: 70, 71 In this section we will rederive the gauge-ﬁxed path integra l for nonabelian gauge theory from a diﬀerent point of view. We will discover t hat the complete gauge-ﬁxed lagrangian, L ≡ L YM+Lgf+Lgh, still has a residual form of the gauge symmetry, known as Becchi-Rouet-Stora-Tyutin symme- try, orBRST symmetry for short. BRST symmetry can be used to derive the Slavnov-Taylor identities that, among other useful thi ngs, show that the coupling constant is renormalized by the same factor at e ach of its ap- pearances in L. Also, we can use BRST symmetry to show that gluons whose polarizations are not both spacelike and transverse ( perpendicular to the four-momentum) decouple from physical scattering am plitudes (as do particles that are created by the ghost ﬁeld). Consider a nonabelian gauge theory with a gauge ﬁeld Aa µ(x), and a scalar or spinor ﬁeld φi(x) in the representation R. Then, under an in- ﬁnitesimal gauge transformation parameterized by θa(x), we have δAa µ(x) =−Dabθb(x), (74.1) δφi(x) =−igθa(x)(Ta R)ijφj(x). (74.2) We now introduce a scalar Grassmann ﬁeld ca(x) in the adjoint represen- tation; this ﬁeld will turn out to be the ghost ﬁeld that we int roduced in section 71. We deﬁne an inﬁnitesimal BRST transformation vi a δBAa µ(x)≡Dab µcb(x) (74.3) =∂µca(x)−gfabcAc µ(x)cb(x), (74.4) δBφi(x)≡igca(x)(Ta R)ijφj(x). (74.5) This is simply an inﬁnitesimal gauge transformation, with t he ghost ﬁeld ca(x) in place of the inﬁnitesimal parameter −θa(x). Therefore, any combi- nation of ﬁelds that is gauge invariant is also BRST invarian t. In particular, the Yang–Mills lagrangian LYM(including the appropriate lagrangian for the scalar or spinor ﬁeld φi) is BRST invariant, δBLYM= 0. (74.6) We now place a further restriction on the BRST transformatio n: we require a BRST variation of a BRST variation to be zero. This r equirement will determine the BRST transformation of the ghost ﬁeld. Co nsider δB(δBφi) =ig(δBca)(Ta R)ijφj−igca(Ta R)ijδBφj. (74.7) 74: BRST Symmetry 436 There is a minus sign in front of the second term because δBacts as an anticommuting object, and it generates a minus sign when it p asses through another anticommuting object, in this case ca. Using eq.(74.5), we have δB(δBφi) =ig(δBca)(Ta R)ijφj−g2cacb(Ta RTb R)ikφk. (74.8) We now use cbca=−cacbin the second term to replace Ta RTb Rwith its anti- symmetric part,1 2[Ta R,Tb R] =i 2fabcTc R. Then, after relabeling some dummy indices, we have δB(δBφi) =ig(δBcc+1 2gfabccacb)(Tc R)ijφj. (74.9) The right-hand side of eq.(74.9) will vanish for all φj(x) if and only if δBcc(x) =−1 2gfabcca(x)cb(x). (74.10) We therefore adopt eq.(74.10) as the BRST variation of the gh ost ﬁeld. Let us now check to see that the BRST variation of the BRST vari ation of the gauge ﬁeld also vanishes. From eq.(74.4) we have δB(δBAa µ) = (δab∂µ−gfabcAc µ)(δBcb)−gfabc(δBAc µ)cb =Dab µ(δBcb)−gfabc(Dcd µcd)cb =Dab µ(δBcb)−gfabc(∂µcc)cb+g2fabcfcdeAe µcdcb.(74.11) We now use the antisymmetry of fabcin the second term to replace ( ∂µcc)cb with its antisymmetric part, (∂µc[c)cb]≡1 2(∂µcc)cb−1 2(∂µcb)cc =1 2(∂µcc)cb+1 2cc(∂µcb) =1 2∂µ(cccb). (74.12) Similarly, we use the antisymmetry of cdcbin the third term to replace fabcfcdewith its antisymmetric part, 1 2(fabcfcde−fadcfcbe) =−1 2[(Tb A)ac(Td A)ce−(Td A)ac(Tb A)ce] =−1 2ifbdh(Th A)ae =−1 2fbdhfhae, (74.13) which is just the Jacobi identity. Now we have δB(δBAa µ) =Dab µ(δBcb)−1 2gfabc(∂µcccb)−1 2g2fbdhfhaeAe µcdcb =Dah µ(δBch)−(δah∂µ−gfaheAe µ)1 2gfbchcccb =Dah µ(δBch+1 2gfbchcbcc). (74.14) 74: BRST Symmetry 437 We see that this vanishes if the BRST variation of the ghost ﬁe ld is given by eq.(74.10). Now we introduce the antighost ﬁeld ¯ca(x). We take its BRST trans- formation to be δB¯ca(x) =Ba(x), (74.15) whereBa(x) is a commuting (as opposed to Grassmann) scalar ﬁeld, the Lautrup-Nakanishi auxiliary ﬁeld . BecauseBa(x) is itself a BRST variation, we have δBBa(x) = 0. (74.16) Note that eq.(74.15) is in apparent contradiction with eq.( 74.10). How- ever, there is actually no need to identify ¯ ca(x) as the hermitian conjugate ofca(x). The role of these ﬁelds (in producing the functional deter minant that must accompany the gauge-ﬁxing delta functional) is fu lﬁlled as long asca(x) and ¯ca(x) are treated as independent when we integrate them; whether or not they are hermitian conjugates of each other is irrelevant. We identiﬁed them as hermitian conjugates in section 71 only for the sake of familiarity in deriving the associated Feynman rules. No w, however, we must abandon this notion. In fact, it will be most convenient to treatca(x) and ¯ca(x) as two real Grassmann ﬁelds. Now that we have introduced a collection of new ﬁelds— ca(x), ¯ca(x), andBa(x)—what are we to do with them? Consider adding to LYMa new term that is the BRST variation of some object O, L=LYM+δBO. (74.17) Clearly Lis BRST invariant, because LYMis, and because δB(δBO) = 0. We will see that adding δBOcorresponds to ﬁxing a gauge; which gauge we get depends on O. We will choose O(x) = ¯ca(x)/bracketleftig 1 2ξBa(x)−Ga(x)/bracketrightig , (74.18) whereGa(x) is a gauge-ﬁxing function, and ξis a parameter. If we further choose Ga(x) =∂µAa µ(x), (74.19) then we end up with Rξgauge. Let us see how this works. We have δBO= (δB¯ca)/bracketleftig 1 2ξBa−∂µAa µ/bracketrightig −¯ca/bracketleftig 1 2ξ(δBBa)−∂µ(δBAa µ)/bracketrightig .(74.20) There is a minus sign in front of the second set of terms becaus eδBacts as an anticommuting object, and so it generates a minus sign w hen it 74: BRST Symmetry 438 passes through another anticommuting object, in this case ¯ ca. Now using eqs.(74.3), (74.15), and (74.16), we get δBO=1 2ξBaBa−Ba∂µAa µ+ ¯ca∂µDab µcb. (74.21) We see that the last term is the ghost lagrangian Lghthat we found in section 71. If we like, we can integrate the ordinary derivat ive by parts, so that it acts on the antighost ﬁeld, δBO →1 2ξBaBa−Ba∂µAa µ−∂µ¯caDab µcb. (74.22) Examining the ﬁrst two terms in eq.(74.22), we see that no der ivatives act on the auxiliary ﬁeld Ba(x). Furthermore, it appears only quadratically and linearly in δBO. We can, therefore, perform the path integral over it; the result is equivalent to solving the classical equation o f motion ∂(δBO) ∂Ba(x)=ξBa(x)−∂µAa µ(x) = 0, (74.23) and substituting the result back into δBO. This yields δBO → −1 2ξ−1∂µAa µ∂νAa ν−∂µ¯caDab µcb. (74.24) We see that the ﬁrst term is the gauge-ﬁxing lagrangian Lgfthat we found in section 71. We now take note of all the symmetries of the action S=/integraltextd4xL, where L=LYM+δBO. With our choice of O, they are: (1) Lorentz invariance; (2) the discrete symmetries of parity, time reversal, and charg e conjugation; (3) global gauge invariance (that is, invariance under a gauge t ransformation with a spacetime-independent parameter θa); (4) BRST invariance; (5) ghost number conservation; and (6) antighost translation i nvariance. Global gauge invariance simply requires every term in Lto have all the group indices contracted in a group-invariant manner. G host number conservation corresponds to assigning ghost number +1 to ca,−1 to ¯ca, and zero to all other ﬁelds, and requiring every term in Lto have ghost number zero. Antighost translation invariance corresponds to ¯ ca(x)→¯ca(x) +χ, whereχis a Grassmann constant. This leaves Linvariant because, in the form of eq.(74.22), Lcontains only a derivative of ¯ cb(x). We now claim that Lalready includes all terms consistent with these symmetries that have coeﬃcients with positive or zero mass d imension. This means that we will not encounter any divergences in pert urbation theory that cannot be absorbed by including a Zfactor for each term inL. Furthermore, loop corrections should respect the symmetr ies, and BRST symmetry requires that grenormalize in the same way at each of 74: BRST Symmetry 439 its appearances. (Filling in the mathematical details of th ese claims is a lengthy project that we will not undertake.) We can regard a BRST transformation as inﬁnitesimal, and hen ce con- struct the associated Noether current via the standard form ula jµ B(x) =/summationdisplay I∂L ∂(∂µΦI(x))δBΦI(x), (74.25) where ΦI(x) stands for all the ﬁelds, including the matter (scalar and/ or spinor), gauge, ghost, antighost, and auxiliary ﬁelds. We c an then deﬁne the BRST charge QB=/integraldisplay d3xj0 B(x). (74.26) If we think of ca(x) and ¯ca(x) as independent hermitian ﬁelds, then QBis hermitian. The BRST charge generates a BRST transformation , i[QB,Aa µ(x)] =Dab µcb(x), (74.27) i{QB,ca(x)}=−1 2gfabccb(x)cc(x), (74.28) i{QB,¯ca(x)}=Ba(x), (74.29) i[QB,Ba(x)] = 0, (74.30) i[QB,φi(x)]±=igca(x)(Ta R)ijφj(x). (74.31) where {A,B}=AB+BAis the anticommutator, and [ ,]±is the commu- tator ifφiis a scalar ﬁeld, and the anticommutator if φiis a spinor ﬁeld. Also, since the BRST transformation of a BRST transformatio n is zero,QB must be nilpotent , Q2 B= 0. (74.32) Eq.(74.32) has far-reaching consequences. In order for it t o be satisﬁed, many states must be annihilated by QB; such states are said to be in the kernel ofQB. A state |ψ∝an}b∇acket∇i}htwhich is annihilated by QBmay take the form ofQBacting on some other state; such states are said to be in the image ofQB. There may be some states in the kernel of QBthat are not in the image; such states are said to be in the cohomology ofQB. Two states in the cohomology of QBare identiﬁed if their diﬀerence is in the image; that is, ifQB|ψ∝an}b∇acket∇i}ht= 0 but |ψ∝an}b∇acket∇i}ht ∝ne}ationslash=QB|χ∝an}b∇acket∇i}htfor any state |χ∝an}b∇acket∇i}ht, and if |ψ′∝an}b∇acket∇i}ht=|ψ∝an}b∇acket∇i}ht+QB|ζ∝an}b∇acket∇i}ht for some state |ζ∝an}b∇acket∇i}ht, then we identify |ψ∝an}b∇acket∇i}htand|ψ′∝an}b∇acket∇i}htas a single element of the cohomology of QB. Note any state in the image of QBhas zero norm, since if |ψ∝an}b∇acket∇i}ht=QB|χ∝an}b∇acket∇i}ht, then∝an}b∇acketle{tψ|ψ∝an}b∇acket∇i}ht=∝an}b∇acketle{tψ|QB|χ∝an}b∇acket∇i}ht= 0. (Here we have used the hermiticity of QBto conclude that QB|ψ∝an}b∇acket∇i}ht= 0 implies ∝an}b∇acketle{tψ|QB= 0.) 74: BRST Symmetry 440 Now consider starting at some initial time with a normalized state|ψ∝an}b∇acket∇i}ht in the cohomology: ∝an}b∇acketle{tψ|ψ∝an}b∇acket∇i}ht= 1,QB|ψ∝an}b∇acket∇i}ht= 0,|ψ∝an}b∇acket∇i}ht ∝ne}ationslash=QB|χ∝an}b∇acket∇i}ht. (This last equation is actually redundant, because if |ψ∝an}b∇acket∇i}ht=QB|χ∝an}b∇acket∇i}htfor some state |χ∝an}b∇acket∇i}ht, then |ψ∝an}b∇acket∇i}ht has zero norm.) Since Lis BRST invariant, the hamiltonian that we derive from it must commute with the BRST charge: [ H,Q B] = 0. Thus, an initial state |ψ∝an}b∇acket∇i}htthat is annihilated by QBmust still be annihilated by it at later times, since QBe−iHt|ψ∝an}b∇acket∇i}ht=e−iHtQB|ψ∝an}b∇acket∇i}ht= 0. Also, since unitary time evolution does not change the norm of a state, the time-evolv ed state must still be in the cohomology. We now claim that the physical states of the theory correspond to the cohomology of QB. We have already shown that if we start with a state in the cohomology, it remains in the cohomology under time ev olution. Consider, then, an initial state of widely separated wave pa ckets of incoming particles. According to our discussion in section 5, we can t reat these states as being created by the appropriate Fourier modes of the ﬁeld s, and ignore interactions. We will suppress the group index (because it p lays no essential role when interactions can be neglected) and write the mode e xpansions Aµ(x) =/summationdisplay λ=>,<, +,−/integraldisplay /tildewiderdk/bracketleftig εµ∗ λ(k)aλ(k)eikx+εµ λ(k)a† λ(k)e−ikx/bracketrightig ,(74.33) c(x) =/integraldisplay /tildewiderdk/bracketleftig c(k)eikx+c†(k)e−ikx/bracketrightig , (74.34) ¯c(x) =/integraldisplay /tildewiderdk/bracketleftig b(k)eikx+b†(k)e−ikx/bracketrightig , (74.35) φ(x) =/integraldisplay /tildewiderdk/bracketleftig aφ(k)eikx+a† φ(k)e−ikx/bracketrightig , (74.36) Here, for maximum simplicity, we have taken φ(x) to be a real scalar ﬁeld. (This is possible if R is a real representation.) In eq.(74.3 3), we have included four polarization vectors that span four-dimensi onal spacetime. Forkµ= (ω,k) =ω(1,0,0,1), we choose these four polarization vectors to be εµ >(k) =1√ 2(1,0,0,1), εµ <(k) =1√ 2(1,0,0,−1), εµ +(k) =1√ 2(0,1,−i,0), εµ −(k) =1√ 2(0,1,+i,0). (74.37) The ﬁrst two of these, >and<, are lightlike vectors; εµ >(k) is parallel to kµ, andεµ <(k) is spatially opposite. The latter two, + and −, are spacelike 74: BRST Symmetry 441 and transverse: they correspond to physical photon polariz ations of deﬁnite helicity. We setg= 0, plug eqs.(74.33–74.36) into eqs.(74.27–74.31), and us e eq.(74.23) to eliminate the auxiliary ﬁeld. Matching coeﬃc ients ofe−ikx, we ﬁnd [QB,a† λ(k)] =√ 2ωδλ>c†(k), (74.38) {QB,c†(k)}= 0, (74.39) {QB,b†(k)}=ξ−1√ 2ωa† <(k), (74.40) [QB,a† φ(k)] = 0. (74.41) Consider a normalized state |ψ∝an}b∇acket∇i}htin the cohomology: ∝an}b∇acketle{tψ|ψ∝an}b∇acket∇i}ht= 1,QB|ψ∝an}b∇acket∇i}ht= 0. Eq.(74.38) tells us that if we add a photon with the unphysica l polarization >by acting on |ψ∝an}b∇acket∇i}htwitha† >(k), then this state is not annihilated by QB; hence the state a† >(k)|ψ∝an}b∇acket∇i}htis not in the cohomology. Eq.(74.40) tells us that the statea† <(k)|ψ∝an}b∇acket∇i}htis proportional to QBb†(k)|ψ∝an}b∇acket∇i}ht; hence the state a† <(k)|ψ∝an}b∇acket∇i}ht is also not in the cohomology. On the other hand, the states a† +(k)|ψ∝an}b∇acket∇i}htand a† −(k)|ψ∝an}b∇acket∇i}htare annihilated by QB, but they cannot be written as QBacting on some other state; hence these states arein the cohomology. Also, by similar reasoning, the state with one extra φparticle,a† φ(k)|ψ∝an}b∇acket∇i}ht, is in the cohomology. Eq.(74.38) tells us that if we add a ghost particle by acting o n|ψ∝an}b∇acket∇i}ht withc†(k), then this state is proportional to QBa† >(k)|ψ∝an}b∇acket∇i}ht; hence the state c†(k)|ψ∝an}b∇acket∇i}htis not in the cohomology. Eq.(74.40) tells us that if we add an antighost particle by acting on |ψ∝an}b∇acket∇i}htwithb†(k), then this state is not anni- hilated byQB; hence the state b†(k)|ψ∝an}b∇acket∇i}htis also not in the cohomology. We conclude that the only particle creation operators that d o not take a state out of the cohomology are a† φ(k),a† +(k), anda† −(k). Of course, it is precisely these operators that create the expected physi cal particles. Finally, we note that the vacuum |0∝an}b∇acket∇i}htmust be in the cohomology, because it is the unique state with zero energy and positive norm. Thus we can conclude that we can build an initial state of wide ly sep- arated particles that is in the cohomology only if we do not in clude any ghost or antighost particles, or photons with polarization s other than + and−. Since a state in the cohomology must evolve to another state in the cohomology, no ghosts, antighosts, or unphysically polari zed photons can be produced in the scattering process. Reference Notes A detailed treatment of BRST symmetry can be found in Weinberg II . 74: BRST Symmetry 442 Problems 74.1) The creation operator for a photon of positive helicit y can be written as a† +(k) =−iεµ∗ +(k)/integraldisplay d3xe+ikx↔∂0Aµ(x). (74.42) Consider the state a† +(k)|ψ∝an}b∇acket∇i}ht, where |ψ∝an}b∇acket∇i}htis in the BRST cohomology. Deﬁne a gauge-transformed polarization vector ˜εµ +(k) =εµ +(k) +ckµ, (74.43) wherecis a constant, and a corresponding creation operator ˜ a† +(k). Show that ˜a† +(k)|ψ∝an}b∇acket∇i}ht=a† +(k)|ψ∝an}b∇acket∇i}ht+QB|χ∝an}b∇acket∇i}ht, (74.44) which implies that ˜ a† +(k)|ψ∝an}b∇acket∇i}htanda† +(k)|ψ∝an}b∇acket∇i}htrepresent the same element of the cohomology, and hence are physically equivalent. Fin d the state |χ∝an}b∇acket∇i}ht. 75: Chiral Gauge Theories and Anomalies 443 75Chiral Gauge Theories and Anomalies Prerequisite: 70, 72 So far, we have only discussed gauge theories with Dirac ferm ion ﬁelds. Recall that a Dirac ﬁeld Ψ can be written in terms of two left-h anded Weyl ﬁeldsχandξas Ψ =/parenleftiggχ ξ†/parenrightigg . (75.1) If Ψ is in a representation R of the gauge group, then χandξ†must be as well. Equivalently, χmust be in the representation R, and ξmust be in the complex conjugate representation R. (For an abelian theory, this means that if Ψ has charge + Q, thenχhas charge + Qandξhas charge −Q.) Thus a Dirac ﬁeld in a representation R is equivalent to two le ft-handed Weyl ﬁelds, one in R and one in R. If the representation R is real, then we can have a Majorana ﬁe ld Ψ =/parenleftiggψ ψ†/parenrightigg (75.2) instead of a Dirac ﬁeld; the left-handed Weyl ﬁeld ψand and its hermitian conjugateψ†are both in the representation R. Thus a Majorana ﬁeld in a real representation R is equivalent to a single left-handed Weyl ﬁeld in R. Now suppose that we have a single left-handed Weyl ﬁeld ψin acomplex representation R. Such a gauge theory is automatically pari ty violating (because the right-handed hermitian conjugate of the left- handed Weyl ﬁeld is in an inequivalent represetnation of the gauge group), an d is said to be chiral. The lagrangian is L=iψ†¯σµDµψ−1 4FaµνFa µν, (75.3) whereDµ=∂µ−igAa µTa R. SinceTa Ris a hermitian matrix (even when R is a complex representation), iψ†¯σµDµψis hermitian (up to a total diver- gence, as usual). We cannot include a mass term for ψ, though, because ψψtransforms as R ⊗R, and R ⊗R does not contain a singlet if R is complex. Thus, ψψis not gauge invariant. But without a mass term, this lagrangian would appear to possess all the required propert ies: Lorentz invariance, gauge invariance, and no terms with coeﬃcients with negative mass dimension. However, it turns out that most chiral gauge theories do not e xist as quantum ﬁeld theories; they are anomalous . The problem can ultimately be traced back to the functional measure for the fermion ﬁeld ; it turns out 75: Chiral Gauge Theories and Anomalies 444 that this measure is, in general, not gauge invariant. We wil l explore this surprising fact in section 77. For now we will content ourselves with analyzing Feynman dia grams. We will ﬁnd an insuperable problem with gauge invariance at t he one-loop level that aﬄicts most chiral gauge theories. We will work with the simplest possible example: a U(1) theor y with a single Weyl ﬁeld ψwith charge +1. The lagrangian is L=iψ†¯σµ(∂µ−igAµ)ψ−1 4FµνFµν. (75.4) We can use the following trick to write this theory in terms of a Dirac ﬁeld Ψ. We note that PLΨ =/parenleftiggψ 0/parenrightigg , (75.5) wherePL=1 2(1−γ5) is the left-handed projection matrix, does not involve the right-handed components of Ψ. Then we can write eq.(75.4 ) as L=iΨγµ(∂µ−igAµ)PLΨ−1 4FµνFµν, (75.6) and treat Ψ as a Dirac ﬁeld when we derive the Feynman rules. To better understand the physical consequences of eq.(75.5 ), consider the case of a free ﬁeld. The mode expansion is PLΨ(x) =/summationdisplay s=±/integraldisplay /tildewiderdp/bracketleftig bs(p)PLus(p)eipx+d† s(p)PLvs(p)e−ipx/bracketrightig .(75.7) For a massless ﬁeld, we learned in section 38 that PLu+(p) = 0 and PLv−(p) = 0. Thus we can write eq.(75.7) as PLΨ(x) =/integraldisplay /tildewiderdp/bracketleftig b−(p)u−(p)eipx+d† +(p)v+(p)e−ipx/bracketrightig . (75.8) Eq.(75.8) shows us that there are only two kinds of particles associated with this ﬁeld (as opposed to four with a Dirac ﬁeld): b† −(p) creates a particle with charge +1 and helicity −1/2, andd† +(p) creates a particle with charge −1 and helicity +1 /2. In this theory, charge and spin are correlated. We can easily read the Feynman rules oﬀ of eq.(75.6). In parti cular, the fermion propagator in momentum space is −PL/p/p2, and the fermion– fermion–photon vertex is igγµPL. When we go to evaluate loop diagrams, we need a method of regul at- ing the divergent integrals. However, our usual choice, dim ensional reg- ularization, is problematic, due to the close connection be tweenγ5and four-dimensional spacetime. In particular, in four dimens ions we have Tr[γ5γµγνγργσ] =−4iεµνρσ, (75.9) 75: Chiral Gauge Theories and Anomalies 445 lk k k+l Figure 75.1: The one-loop and counterterm corrections to th e photon prop- agator. whereε0123= +1. It is not obvious what should be done with this formula inddimensions. One possibility is to take d>4 and deﬁne γ5≡iγ0γ1γ2γ3. Then eq.(75.9) holds, but with each of the four vector indice s restricted to span 0, 1, 2, 3. We also have {γµ,γ5}= 0 forµ= 0,1,2,3, but [γµ,γ5] = 0 forµ>3. This approach is workable, but cumbersome in practice. It is therefore tempting to abandon dimensional regulariza tion in favor of, say, Pauli–Villars regularization, which involves the replacement PL/p p2→PL/parenleftbigg−/p p2−−/p+ Λ p2+ Λ2/parenrightbigg . (75.10) Pauli–Villars regularization is equivalent to adding an ex tra fermion ﬁeld with mass Λ, and a propagator with the wrong sign (correspond ing to changing the signs of the kinetic and mass terms in the lagran gian). But, a Dirac ﬁeld with a chiral coupling to the gauge ﬁeld cannot ha ve a mass, since the mass term would not be gauge invariant. So, in a chir al gauge theory, Pauli–Villars regularization violates gauge inva riance, and hence is unacceptable. Given the diﬃculty with regulating chiral gauge theories (w hich is a hint that they may not make sense), we will sidestep the issue for now, and see what we can deduce about loop diagrams without a regulato r in place. Consider the correction to the photon propagator, shown in ﬁ g.(75.1). We have iΠµν(k) = (−1)(ig)2/parenleftig 1 i/parenrightig2/integraldisplayd4ℓ (2π)4Nµν (ℓ+k)2ℓ2 −i(Z3−1)(k2gµν−kµkν) +O(g4), (75.11) where the numerator is Nµν= Tr[PL(/ℓ+/k)γµPLPL/ℓγνPL]. (75.12) We haveP2 L=PLandPLγµ=γµPR(and hence PLγµγν=γµγνPL), and so all thePL’s in eq.(75.12) can be collapsed into just one; this is gener ically 75: Chiral Gauge Theories and Anomalies 446 true along any fermion line. Thus we have Nµν= Tr[(/ℓ+/k)γµ/ℓγνPL]. (75.13) The term in eq.(75.13) with PL→1 2simply yields half the result that we get in spinor electrodynamics with a Dirac ﬁeld. The term in eq.(75.13) with PL→ −1 2γ5, on the other hand, yields a vanishing contribution to Πµν(k). To see this, ﬁrst note that Nµν→ −1 2Tr[(/ℓ+/k)γµ/ℓγνγ5] = 2iεαµβν(ℓ+k)αℓβ = 2iεαµβνkαℓβ. (75.14) Thus we have Πµν(k) =1 2Πµν(k)Dirac−2g2εαµβνkα/integraldisplayd4ℓ (2π)4ℓβ (ℓ+k)2ℓ2 −(Z3−1)(k2gµν−kµkν) +O(g4). (75.15) The integral is logarithmically divergent. But, it carries a single vector indexβ, and the only vector it depends on is k. Therefore, any Lorentz- invariant regularization must yield a result that is propor tional tokβ. This then vanishes when contracted with εαµβνkα. We therefore conclude that, at the one-loop level, the contribution to Πµν(k) of a single charged Weyl ﬁeld is half that of a Dirac ﬁeld. This is physically reasonab le, since a Dirac ﬁeld is equivalent to two charged Weyl ﬁelds. Nothing interesting happens in the one-loop corrections to the fermion propagator, or the fermion–fermion–photon vertex. There i s simply an extra factor of PLalong the fermion line, which can be moved to the far right. Except for this factor, the results exactly duplicat e those of spinor electrodynamics. All of this implies that a single Weyl ﬁeld makes half the cont ribution of a Dirac ﬁeld to the leading term in the beta function for the ga uge coupling. Next we turn to diagrams with three external photons, and no e xternal fermions, shown in ﬁg.(75.2). In spinor electrodynamics, t he fact that the vector potential is odd under charge conjugation implies th at the sum of these diagrams must vanish; see problem 58.2. For the presen t case of a single Weyl ﬁeld, there is no charge-conjugation symmetry, and so we must evaluate these diagrams. The second diagram in ﬁg.(75.2) is the same as the ﬁrst, with p↔q andµ↔ν. Thus we have iVµνρ(p,q,r) = (−1)(ig)3/parenleftig 1 i/parenrightig3/integraldisplayd4ℓ (2π)4Nµνρ (ℓ−p)2ℓ2(ℓ+q)2 75: Chiral Gauge Theories and Anomalies 447 pµ ν qρ rpµ ν qρ r l ql l l+q l p l+p Figure 75.2: One-loop contributions to the three-photon ve rtex. + (p,µ↔q,ν) +O(g5), (75.16) where Nµνρ= Tr[( −/ℓ+/p)γµ(−/ℓ)γν(−/ℓ−/q)γρPL]. (75.17) The term in eq.(75.17) with PL→1 2simply yields half the result that we get in spinor electrodynamics with a Dirac ﬁeld, which gives a vanishing contribution to Vµνρ(p,q,r). Hence, we can make the replacement PL→ −1 2γ5in eq.(75.17). Then, after cancelling some minus signs, we h ave Nµνρ→1 2Tr[(/ℓ−/p)γµ/ℓγν(/ℓ+/q)γργ5]. (75.18) We would now like to verify that Vµνρ(p,q,r) is gauge invariant. We should have pµVµνρ(p,q,r) = 0, (75.19) qνVµνρ(p,q,r) = 0, (75.20) rρVµνρ(p,q,r) = 0. (75.21) Let us ﬁrst check the last of these. From eq.(75.16) we ﬁnd rρVµνρ(p,q,r) =ig3/integraldisplayd4ℓ (2π)4rρNµνρ (ℓ−p)2ℓ2(ℓ+q)2 + (p,µ↔q,ν) +O(g5), (75.22) where rρNµνρ=1 2Tr[(/ℓ−/p)γµ/ℓγν(/ℓ+/q)rργργ5]. (75.23) It will be convenient to use the cyclic property of the trace t o rewrite eq.(75.23) as rρNµνρ=1 2Tr[/ℓγν(/ℓ+/q)rργρ(/ℓ−/p)γµγ5]. (75.24) 75: Chiral Gauge Theories and Anomalies 448 To simplify eq.(75.24), we write rργρ= /r=−(/q+/p) =−(/ℓ+/q) + (/ℓ−/p). Then (/ℓ+/q)rργρ(/ℓ−/p) = (/ℓ+/q)[−(/ℓ+/q) + (/ℓ−/p)](/ℓ−/p) = (ℓ+q)2(/ℓ−/p)−(ℓ−p)2(/ℓ+/q). (75.25) Now we have rρNµνρ=1 2(ℓ+q)2Tr[/ℓγν(/ℓ−/p)γµγ5]−1 2(ℓ−p)2Tr[/ℓγν(/ℓ+/q)γµγ5] =−2iεανβµ/bracketleftig (ℓ+q)2ℓα(ℓ−p)β−(ℓ−p)2ℓα(ℓ+q)β/bracketrightig = +2iεανβµ/bracketleftig (ℓ+q)2ℓαpβ+ (ℓ−p)2ℓαqβ/bracketrightig (75.26) Putting eq.(75.26) into eq.(75.22), we get rρVµνρ(p,q,r) =−2g3εανβµ/integraldisplayd4ℓ (2π)4/bracketleftbiggℓαpβ ℓ2(ℓ−p)2+ℓαqβ ℓ2(ℓ+q)2/bracketrightbigg + (p,µ↔q,ν) +O(g5), (75.27) Consider the ﬁrst term in the integrand. Because the only fou r-vector that it depends on is p, any Lorentz-invariant regularization of its integral must yield a result proportional to pαpβ. Similarly, any Lorentz-invariant regularization of the integral of the second term must yield a result pro- portional to qαqβ. Bothpαpβandqαqβvanish when contracted with εµανβ. Therefore, we have shown that rρVµνρ(p,q,r) = 0, (75.28) as required by gauge invariance. It might seem that now we are done: we can invoke symmetry amon g the external lines to conclude that we must also have pµVµνρ(p,q,r) = 0 andqνVµνρ(p,q,r) = 0. However, eq.(75.16) is not manifestly symmetric on the exchanges ( p,µ↔r,ρ) and (q,ν↔r,ρ). So it still behooves us to compute either pµVµνρ(p,q,r) orqνVµνρ(p,q,r). From eq.(75.16) we ﬁnd pµVµνρ(p,q,r) =ig3/integraldisplayd4ℓ (2π)4pµNµνρ (ℓ−p)2ℓ2(ℓ+q)2 + (p,µ↔q,ν) +O(g5), (75.29) where pµNµνρ=1 2Tr[(/ℓ−/p)pµγµ/ℓγν(/ℓ+/q)γργ5]. (75.30) 75: Chiral Gauge Theories and Anomalies 449 To simplify eq.(75.30), we write pµγµ= /p=−(/ℓ−/p) + /ℓ. Then (/ℓ−/p)pµγµ/ℓ= (/ℓ−/p)[−(/ℓ−/p) + /ℓ]/ℓ = (ℓ−p)2/ℓ−ℓ2(/ℓ−/p). (75.31) Now we have pµNµνρ=1 2(ℓ−p)2Tr[/ℓγν(/ℓ+/q)γργ5]−1 2ℓ2Tr[(/ℓ−/p)γν(/ℓ+/q)γργ5] =−2iεανβρ/bracketleftig (ℓ−p)2ℓαqβ−ℓ2(ℓ−p)α(ℓ+q)β/bracketrightig =−2iεανβρ/bracketleftig (ℓ−p)2ℓαqβ−ℓ2(ℓ−p)α(ℓ−p+p+q)β/bracketrightig =−2iεανβρ/bracketleftig (ℓ−p)2ℓαqβ−ℓ2(ℓ−p)α(p+q)β/bracketrightig . (75.32) Putting eq.(75.32) into eq.(75.29), we get pµVµνρ(p,q,r) = 2g3εανβρ/integraldisplayd4ℓ (2π)4/bracketleftbiggℓαqβ ℓ2(ℓ+q)2−(ℓ−p)α(p+q)β (ℓ−p)2(ℓ+q)2/bracketrightbigg + (p,µ↔q,ν) +O(g5), (75.33) The ﬁrst term on the right-hand side of eq.(75.33) must vanis h, because any Lorentz-invariant regularization of the integral must yield a result pro- portional to qαqβ, and this vanishes when contracted with εανβρ. As for the second term, we can shift the loop momentum from ℓtoℓ+p, which results in (ℓ−p)α(p+q)β (ℓ−p)2(ℓ+q)2→ℓα(p+q)β ℓ2(ℓ+p+q)2. (75.34) We can now use Lorentz invariance to argue that the integral o f the right- hand side of eq.(75.34) must yield something proportional t o (p+q)α(p+q)β; this vanishes when contracted with with εανβρ. Thus, we have shown that pµVµνρ(p,q,r) = 0, as required by gauge invariance, provided that the shift of the loop momentum did not change the value of the inte gral. This would of course be true if the integral was convergent. Inste ad, however, the integral is linearly divergent, and so we must be more car eful. Consider a one-dimensional example of a linearly divergent integral: let I(a)≡/integraldisplay+∞ −∞dxf(x+a), (75.35) wheref(±∞) =c±, withc+andc−two ﬁnite constants. If the integral con- verged, then I(a) would be independent of a. In the present case, however, 75: Chiral Gauge Theories and Anomalies 450 we can Taylor expand f(x+a) in powers of a, and note that f(±∞) =c± implies that every derivative of f(x) vanishes at x=±∞. Thus we have I(a) =/integraldisplay+∞ −∞dx/bracketleftig f(x) +af′(x) +1 2a2f′′(x) +.../bracketrightig =I(0) +a(c+−c−). (75.36) We see that I(a) isnotindependent of a. Furthermore, even if we cannot assign a deﬁnite value to I(0) (because the integral is divergent), we can assign a deﬁnite value to the diﬀerence I(a)−I(0) =a(c+−c−). (75.37) Now let us return to eqs.(75.33) and (75.34). Deﬁne fα(ℓ)≡ℓα ℓ2(ℓ+p+q)2. (75.38) Using Lorentz invariance, we can argue that /integraldisplayd4ℓ (2π)4fα(ℓ) =A(p+q)α, (75.39) whereAis a scalar that will depend on the regularization scheme. No w consider fα(ℓ−p) =fα(ℓ)−pβ∂ ∂ℓβfα(ℓ) +... . (75.40) The integral of the ﬁrst term on the right-hand side of eq.(75 .40) is given by eq.(75.39). The integrals of the remaining terms can be co nverted to surface integrals at inﬁnity. Only the second term in eq.(75 .40) falls oﬀ slowly enough to contribute. To determine the value of its in tegral, we make a Wick rotation to euclidean space, which yields a facto r ofias usual; then we have /integraldisplayd4ℓ (2π)4∂ ∂ℓβfα(ℓ) =ilim ℓ→∞/integraldisplaydSβ (2π)4fα(ℓ), (75.41) wheredSβ=ℓ2ℓβdΩ is a surface-area element, and dΩ is the diﬀerential solid angle in four dimensions. We thus ﬁnd /integraldisplayd4ℓ (2π)4∂ ∂ℓβfα(ℓ) =ilim ℓ→∞/integraldisplaydΩ (2π)4ℓβℓα (ℓ+p+q)2 =iΩ4 (2π)41 4gαβ =i 32π2gαβ, (75.42) 75: Chiral Gauge Theories and Anomalies 451 where we used Ω 4= 2π2. Combining eqs.(75.38–75.42), we ﬁnd /integraldisplayd4ℓ (2π)4(ℓ−p)α (ℓ−p)2(ℓ+q)2=A(p+q)α−i 32π2pα. (75.43) Using this in eq.(75.33), we ﬁnd pµVµνρ(p,q,r) =ig3 16π2εανβρpα(p+q)β+ (p,µ↔q,ν) +O(g5) =ig3 8π2εανβρpαqβ+O(g5). (75.44) An exactly analogous calculation results in qνVµνρ(p,q,r) =ig3 8π2εαρβµqαpβ+O(g5). (75.45) Eqs.(75.44) and (75.45) show that the three-photon vertex i snotgauge invariant. Since rρVµνρ(p,q,r) = 0, eqs.(75.44) and (75.45) also show that the three-photon vertex does not exhibit the expected symme try among the external lines. This is a puzzle, because the only asymmetric aspects of the d iagrams in ﬁg.(75.2) are the momentum labels on the internal lines. The resolution of the puzzle lies in the fact that the integral in eq.(75.16) is linearly divergent, and so shifting the loop momentum changes its value. To accou nt for this, let us write Vµνρ(p,q,r) withℓreplaced with ℓ+a, whereais an arbitrary linear combination of pandq. We deﬁne Vµνρ(p,q,r;a)≡1 2ig3/integraldisplayd4ℓ (2π)4Tr[(/ℓ+a−/p)γµ(/ℓ+a)γν(/ℓ+a+/q)γργ5] (ℓ+a−p)2(ℓ+a)2(ℓ+a+q)2 + (p,µ↔q,ν) +O(g5). (75.46) Our previous expression, eq.(75.16), corresponds to a= 0. The integral in eq.(75.46) is linearly divergent, and so we can express th e diﬀerence between Vµνρ(p,q,r;a) and Vµνρ(p,q,r;0) as a surface integral. Let us write Vµνρ(p,q,r;a) =1 2ig3Iαβγ(a)Tr[γαγµγβγνγγγργ5] + (p,µ↔q,ν) +O(g5), (75.47) where Iαβγ(a)≡/integraldisplayd4ℓ (2π)4(ℓ+a−p)α(ℓ+a)β(ℓ+a−q)γ (ℓ+a−p)2(ℓ+a)2(ℓ+a+q)2. (75.48) 75: Chiral Gauge Theories and Anomalies 452 Then we have Iαβγ(a)−Iαβγ(0) =aδ/integraldisplayd4ℓ (2π)4∂ ∂ℓδ/bracketleftbigg(ℓ−p)αℓβ(ℓ−q)γ (ℓ−p)2ℓ2(ℓ+q)2/bracketrightbigg =iaδlim ℓ→∞/integraldisplaydΩ (2π)4ℓδ(ℓ−p)αℓβ(ℓ−q)γ (ℓ−p)2(ℓ+q)2 =iaδΩ4 (2π)41 24/parenleftig gδαgβγ+gδβgγα+gδγgαβ/parenrightig =i 192π2/parenleftig aαgβγ+aβgγα+aγgαβ/parenrightig . (75.49) Using this in eq.(75.47), we get contractions of the form gαβγαγµγβ= 2γµ. The three terms in eq.(75.49) all end up contributing equall y, and after using eq.(75.9) to compute the trace, we ﬁnd Vµνρ(p,q,r;a)−Vµνρ(p,q,r;0) = −ig3 16π2εµνρβaβ + (p,µ↔q,ν) +O(g5).(75.50) Since the Levi-Civita symbol is antisymmetric on µ↔ν, only the part of athat is antisymmetric on p↔qcontributes to Vµνρ(p,q,r;a). Therefore we will set a=c(p−q), wherecis a numerical constant. Then we have Vµνρ(p,q,r;a)−Vµνρ(p,q,r;0) =−ig3 8π2cεµνρβ(p−q)β+O(g5).(75.51) Using this, along with eqs.(75.28), (75.44), and (75.45), a nd making some simplifying rearrangements of the indices and momenta on th e right-hand sides (using p+q+r= 0), we ﬁnd pµVµνρ(p,q,r;a) =−ig3 8π2(1−c)ενραβqαrβ+O(g5),(75.52) qνVµνρ(p,q,r;a) =−ig3 8π2(1−c)ερµαβrαpβ+O(g5),(75.53) rρVµνρ(p,q,r;a) =−ig3 8π2(2c)εµναβpαqβ+O(g5). (75.54) We see that choosing c= 1 removes the anomalous right-hand side from eqs.(75.52) and (75.53), but it then necessarily appears in eq.(75.54). Chos- ingc=1 3restores symmetry among the external lines, but now all thre e right-hand sides are anomalous. (This is what results from d imensional reg- ularization of this theory with γ5=iγ0γ1γ2γ3.) We have therefore failed to construct a gauge-invariant U(1) theory with a single charg ed Weyl ﬁeld. 75: Chiral Gauge Theories and Anomalies 453 Consider now a U(1) gauge theory with several left-handed We yl ﬁelds ψi, with charges Qi, so that the covariant derivative of ψiis (∂µ−igQiAµ)ψi. Then each of these ﬁelds circulates in the loop in ﬁg.(75.2), and each vertex has an extra factor of Qi. The right-hand sides of eqs.(75.52–75.54) are now multiplied by/summationtext iQ3 i. And if/summationtext iQ3 ihappens to be zero, then gauge invariance is restored! The simplest possibility is to have theψ’s come in pairs with equal and opposite charges. (In this case, they ca n be assembled into Dirac ﬁelds.) But there are other possibilities as well : for example, one ﬁeld with charge +2 and eight with charge −1. Such a gauge theory is still chiral, but it is anomaly free . (It could be that further obstacles to gauge invariance arise with more external photons and/or mo re loops, but this turns out not to be the case. We will discuss this in secti on 77.) All of this has a straightforward generalization to nonabel ian gauge theories. Suppose we have a single Weyl ﬁeld in a (possibly re ducible) representation R of the gauge group. Then we must attach an ex tra factor of Tr(Ta RTb RTc R) to the ﬁrst diagram in ﬁg.(75.2), and a factor of Tr( Ta RTc RTb R) to the second; here the group indicies a,b,c go along with the momenta p,q,r, respectively. Repeating our analysis shows that the diagr ams with PL→1 2come with an extra factor of1 2Tr([Ta R,Tb R]Tc R) =i 2T(R)fabcTc R; these contribute to the renormalization of the tree-level t hree-gluon vertex. Diagrams with PL→ −1 2γ5come with an extra factor of 1 2Tr({Ta R,Tb R}Tc R) =A(R)dabc. (75.55) Heredabcis a completely symmetric tensor that is independent of the r ep- resentation, and A(R) is the anomaly coeﬃcient of R, introduced in sec- tion 70. In order for this theory to exist, we must have A(R) = 0. As shown in section 70, A(R) =−A(R); thus a theory whose left-handed Weyl ﬁelds come in R ⊕R pairs is automatically anomaly free (as is one whose Weyl ﬁelds are all in real representations). Otherwise, we m ust arrange the cancellation by hand. For SU(2) and SO( N), all representations have A(R) = 0. For SU( N) withN > 3, the fundamental representation has A(N) = 1, and most complex SU( N) representations R have A(R)∝ne}ationslash= 0. So the cancellation is nontrivial. We mention in passing two other kinds of anomalies: if we coup le our theory to gravity, we can draw a triangle diagram with two gra vitons and one gauge boson. This diagram violates general coordinate i nvariance (the gauge symmetry of gravity). If the gauge boson is from a nonab elian group, the diagram is accompanied by a factor of Tr Ta R= 0, and so there is no anomaly. If the gauge boson is from a U(1) group, the diagram i s accom- panied by a factor of/summationtext iQi, and this must vanish to cancel the anomaly. There is also a global anomaly that aﬄicts theories with an odd number of Weyl fermions in a pseudoreal representation, such as the fundamental 75: Chiral Gauge Theories and Anomalies 454 representation of SU(2). The global anomaly cannot be seen i n perturba- tion theory; we will discuss it brieﬂy in section 77. Reference Notes Discussions of anomalies emphasizing diﬀerent aspects can be found in Georgi ,Peskin & Schroeder , andWeinberg I . Problems 75.1) Consider a theory with a nonabelian gauge symmetry, an d also a U(1) gauge symmetry. The theory contains left-handed Weyl ﬁ elds in the representations (R i,Qi), where R iis the representation of the nonabelian group, and Qiis the U(1) charge. Find the conditions for this theory to be anomaly free. 76: Anomalies in Global Symmetries 455 76Anomalies in Global Symmetries Prerequisite: 75 In this section we will study anomalies in global symmetries that can arise in gauge theories that are free of anomalies in the localsymmetries (and are therefore consistent quantum ﬁeld theories). A phenome nological ap- plication will be discussed in section 90. The simplest example is electrodynamics with a massless Dir ac ﬁeld Ψ with charge Q= +1. The lagrangian is L=iΨ /DΨ−1 4FaµνFa µν, (76.1) where /D=γµDµandDµ=∂µ−igAµ. (We call the coupling constant g rather than ebecause we are using this theory as a formal example rather than a physical model.) We can write Ψ in terms of two left-han ded Weyl ﬁeldsχandξvia Ψ =/parenleftiggχ ξ†/parenrightigg , (76.2) whereχhas charge Q= +1 andξhas charge Q=−1. In terms of χand ξ, the lagrangian is L=iχ†¯σµ(∂µ−igAµ)χ+iξ†¯σµ(∂µ+igAµ)ξ−1 4FaµνFa µν. (76.3) The lagrangian is invariant under a U(1) gauge transformati on Ψ(x)→e−igΓ(x)Ψ(x), (76.4) Ψ(x)→e+igΓ(x)Ψ(x), (76.5) Aµ(x)→Aµ(x)−∂µΓ(x). (76.6) In terms of the Weyl ﬁelds, eqs.(76.4) and (76.5) become χ(x)→e−igΓ(x)χ(x), (76.7) ξ(x)→e+igΓ(x)ξ(x). (76.8) Because the fermion ﬁeld is massless, the lagrangian is also invariant under a global symmetry in which χandξtransform with the same phase, χ(x)→e+iαχ(x), (76.9) ξ(x)→e+iαξ(x). (76.10) 76: Anomalies in Global Symmetries 456 In terms of Ψ, this is Ψ(x)→e−iαγ5Ψ(x), (76.11) Ψ(x)→Ψ(x)e−iαγ5. (76.12) This is called axial U(1) symmetry , because the associated Noether current jµ A(x)≡Ψ(x)γµγ5Ψ(x) (76.13) is an axial vector (that is, its spatial part is odd under pari ty). Noether’s theorem leads us to expect that this current is conserved: ∂µjµ A= 0. How- ever, in this section we will show that the axial current actu ally has an anomalous divergence , ∂µjµ A=−g2 16π2εµνρσFµνFρσ. (76.14) We will see in section 77 that eq.(76.14) is exact; there are n o higher-order corrections. We will demonstrate eq.(76.14) by making use of our results i n section 75. Consider the matrix element ∝an}b∇acketle{tp,q|jρ A(z)|0∝an}b∇acket∇i}ht, where ∝an}b∇acketle{tp,q|is a state of two outgoing photons with four-momenta pandq, and polarization vectors εµ andε′ ν, respectively. (We omit the helicity label, which will play no essential role.) Using the LSZ formula for photons (see section 67), we have ∝an}b∇acketle{tp,q|jρ A(z)|0∝an}b∇acket∇i}ht= (ig)2εµε′ ν/integraldisplay d4xd4ye−i(px+qy)∝an}b∇acketle{t0|Tjµ(x)jν(y)jρ A(z)|0∝an}b∇acket∇i}ht, (76.15) where jµ(x)≡Ψ(x)γµΨ(x) (76.16) is the Noether current corresponding to the U(1) gauge symme try. Since bothjµ(x) andjµ A(x) are Noether currents, we expect the Ward identities ∂ ∂xµ∝an}b∇acketle{t0|Tjµ(x)jν(y)jρ A(z)|0∝an}b∇acket∇i}ht= 0, (76.17) ∂ ∂yν∝an}b∇acketle{t0|Tjµ(x)jν(y)jρ A(z)|0∝an}b∇acket∇i}ht= 0, (76.18) ∂ ∂zρ∝an}b∇acketle{t0|Tjµ(x)jν(y)jρ A(z)|0∝an}b∇acket∇i}ht= 0, (76.19) to be satisﬁed. Note that there are no contact terms in eqs.(7 6.17–76.19), because both jµ(x) andjµ A(x) are invariant under both U(1) transforma- tions. If we use eq.(76.19) in eq.(76.15), we see that we expe ct ∂ ∂zρ∝an}b∇acketle{tp,q|jρ A(z)|0∝an}b∇acket∇i}ht= 0. (76.20) 76: Anomalies in Global Symmetries 457 However, our experience in section 75 leads us to proceed mor e cautiously. Let us deﬁne Cµνρ(p,q,r) via (2π)4δ4(p+q+r)Cµνρ(p,q,r) ≡/integraldisplay d4xd4yd4ze−i(px+qy+rz)∝an}b∇acketle{t0|Tjµ(x)jµ(y)jρ A(z)|0∝an}b∇acket∇i}ht.(76.21) Then we can rewrite eq.(76.15) as ∝an}b∇acketle{tp,q|jρ A(z)|0∝an}b∇acket∇i}ht=−g2εµε′ νCµνρ(p,q,r)eirz/vextendsingle/vextendsingle/vextendsingle r=−p−q. (76.22) Taking the divergence of the current yields ∝an}b∇acketle{tp,q|∂ρjρ A(z)|0∝an}b∇acket∇i}ht=−ig2εµε′ νrρCµνρ(p,q,r)eirz/vextendsingle/vextendsingle/vextendsingle r=−p−q. (76.23) The expected Ward identities become pµCµνρ(p,q,r) = 0, (76.24) qνCµνρ(p,q,r) = 0, (76.25) rρCµνρ(p,q,r) = 0. (76.26) To check eqs.(76.24–76.26), we compute Cµνρ(p,q,r) with Feynman dia- grams. At the one-loop level, the contributing diagrams are exactly those we computed in section 75, except that the three vertex facto rs are now γµ,γν, andγργ5, instead of igγµPL,igγνPL, andigγρPL. But, as we saw, the threePL’s can be combined into just one at the last vertex, and then this one can be replaced by −1 2γ5. Thus, the vertex function iVµνρ(p,q,r) of section 75 is related to Cµνρ(p,q,r) by iVµνρ(p,q,r) =−1 2(ig)3Cµνρ(p,q,r) +O(g5). (76.27) In section 75, we saw that we could choose a regularization sc heme that preserved eqs.(76.24) and (76.25), but not also (76.26). Fo r the theory of this section, we deﬁnitely want to preserve eqs.(76.24) and (76.25), because these imply conservation of the current coupled to the gauge ﬁeld, which is necessary for gauge invariance. On the other hand, we are les s enamored of eq.(76.26), because it implies conservation of the current for a mere global symmetry. Using eq.(76.27) and our results from section 75, we ﬁnd that preserving eqs.(76.24) and (76.25) results in rρCµνρ(p,q,r) =−i 2π2εµναβpαqβ+O(g2) (76.28) 76: Anomalies in Global Symmetries 458 in place of eq.(76.26). Using this in eq.(76.23), we ﬁnd ∝an}b∇acketle{tp,q|∂ρjρ A(z)|0∝an}b∇acket∇i}ht=−g2 2π2εµναβpαqβεµε′ νe−i(p+q)z+O(g4). (76.29) Now we come to the point. The right-hand side of eq.(76.29) is exactly what we get in free-ﬁeld theory for the matrix element of the r ight-hand side of eq.(76.14). We conclude that eq.(76.14) is correct, up to possible higher-order corrections. In the next section, we will see that eq.(76.14) is exact. Problems 76.1) Verify that the right-hand side of eq.(76.29) is exact ly what we get in free-ﬁeld theory for the matrix element of the right-hand side of eq.(76.14). 77: Anomalies and the Path Integral for Fermions 459 77Anomalies and the Path Integral for Fermions Prerequisite: 76 In the last section, we saw that in a U(1) gauge theory with a ma ssless Dirac ﬁeld Ψ with charge Q= +1, the axial vector current jµ A=Ψγµγ5Ψ, (77.1) which should (according to Noether’s theorem) be conserved , actually has an anomalous divergence, ∂µjµ A=−g2 16π2εµνρσFµνFρσ. (77.2) In this section, we will derive eq.(77.2) directly from the p ath integral, using the Fujikawa method . We will see that eq.(77.2) is exact; there are no higher-order corrections. We can also consider a nonabelian gauge theory with a massles s Dirac ﬁeld Ψ in a (possibly reducible) representation R of the gaug e group. In this case, the triangle diagrams that we analyzed in the last section carry an extra factor of Tr( Ta RTb R) =T(R)δab, and we have ∂µjµ A=−g2 16π2T(R)εµνρσ∂[µAa ν]∂[ρAa σ]+O(g3), (77.3) where∂[µAa ν]≡∂µAa ν−∂νAa µ. We expect the right-hand side of eq.(77.3) to be gauge invariant (since this theory is free of anomalies in the currents coupled to the gauge ﬁelds); this suggests that we should hav e ∂µjµ A=−g2 16π2T(R)εµνρσFa µνFa ρσ, (77.4) whereFa µν=∂µAa ν−∂νAa µ+gfabcAb µAc νis the nonabelian ﬁeld strength. We will see that eq.(77.4) is correct, and that there are no hi gher-order corrections. We can write eq.(77.4) more compactly by using the matrix-va lued gauge ﬁeld Aµ≡Ta RAa µ (77.5) and ﬁeld strength Fµν=∂µAν−∂νAµ−ig[Aµ,Aν]. (77.6) Then eq.(77.4) can be written as ∂µjµ A=−g2 16π2εµνρσTrFµνFρσ. (77.7) 77: Anomalies and the Path Integral for Fermions 460 We now turn to the derivation of eqs.(77.2) and (77.7). We beg in with the path integral over the Dirac ﬁeld, with the gauge ﬁeld tre ated as a ﬁxed background, to be integrated later. We have Z(A)≡/integraldisplay DΨDΨeiS(A), (77.8) where S(A)≡/integraldisplay d4xΨi/DΨ (77.9) is the Dirac action, i/D=iγµDµis the Dirac wave operator, and Dµ=∂µ−igAµ (77.10) is the covariant derivative. Here Aµis either the U(1) gauge ﬁeld, or the matrix-valued nonabelian gauge ﬁeld of eq.(77.5), dependi ng on the theory under consideration. Our notation allows us to treat both ca ses simultane- ously. We can formally evaluate eq.(77.8) as a functional determin ant, Z(A) = det(i/D). (77.11) However, this expression is not useful without some form of r egularization. We will take up this issue shortly. Now consider an axial U(1) transformation of the Dirac ﬁeld, but with a spacetime dependent parameter α(x): Ψ(x)→e−iα(x)γ5Ψ(x), (77.12) Ψ(x)→Ψ(x)e−iα(x)γ5. (77.13) We can think of eqs.(77.12) and (77.13) as a change of integra tion variable in eq.(77.8); then Z(A) should be independent of α(x). The corresponding change in the action is S(A)→S(A) +/integraldisplay d4xjµ A(x)∂µα(x). (77.14) We can integrate by parts to write this as S(A)→S(A)−/integraldisplay d4xα(x)∂µjµ A(x). (77.15) If we assume that the measure DΨDΨ is invariant under the axial U(1) transformation, then we have Z(A)→/integraldisplay DΨDΨeiS(A)e−i/integraltext d4xα(x)∂µjµ A(x). (77.16) 77: Anomalies and the Path Integral for Fermions 461 This must be equal to the original expression for Z(A), eq.(77.8). This implies that ∂µjµ A(x) = 0 holds inside quantum correlation functions, up to contact terms, as discussed in section 22. However, the assumption that the measure DΨDΨ is invariant under the axial U(1) transformation must be examined more closely . The change of variable in eqs.(77.12) and (77.13) is implemented by the functional matrix J(x,y) =δ4(x−y)e−iα(x)γ5. (77.17) Because the path integral is over fermionic variables (rath er than bosonic), we get a jacobian factor of (det J)−1(rather than det J) for each of the transformations in eqs.(77.12) and (77.13), so that we have DΨDΨ→(detJ)−2DΨDΨ. (77.18) Using log det J= Tr logJ, we can write (detJ)−2= exp/bracketleftbigg 2i/integraldisplay d4xα(x)Trδ4(x−x)γ5/bracketrightbigg , (77.19) where the explicit trace is over spin and group indices. Like eq.(77.11), this expression is not useful without some form of regulariz ation. We could try to replace the delta-function with a gaussian; t his is equiv- alent to δ4(x−y)→e∂2 x/M2δ4(x−y), (77.20) whereMis a regulator mass that we would take to inﬁnity at the end of t he calculation. However, the appearance of the ordinary deriv ative∂, rather than the covariant derivative D, implies that eq.(77.20) is not properly gauge invariant. So, another possibility is δ4(x−y)→eD2 x/M2δ4(x−y). (77.21) However, eq.(77.21) presents us with a more subtle problem. Our regular- ization scheme for eq.(77.19) should be compatible with our regularization scheme for eq.(77.11). It is not obvious whether or not eq.(7 7.21) meets this criterion, because D2has no simple relation to i/D. To resolve this issue, we use δ4(x−y)→e(i/ Dx)2/M2δ4(x−y) (77.22) to regulate the delta function in eq.(77.19). To evaluate eq.(77.22), we write the delta function on the ri ght-hand side of eq.(77.22) as a Fourier integral, δ4(x−y)→/integraldisplayd4k (2π)4e(i/ Dx)2/M2eik(x−y). (77.23) 77: Anomalies and the Path Integral for Fermions 462 Then we use f(∂)eikx=eikxf(∂+ik); eq.(77.23) becomes δ4(x−y)→/integraldisplayd4k (2π)4eik(x−y)e(i/ D−/ k)2/M2, (77.24) where a derivative acting on the far right now yields zero. We have (i/D−/k)2= /k2−i{/k,/D} −/D2 =−k2−i{γµ,γν}kµDν−γµγνDµDν. (77.25) Next we use γµγν=1 2({γµ,γν}+ [γµ,γν]) =−gµν−2iSµνto get (i/D−/k)2=−k2+ 2ik·D+D2+ 2iSµνDµDν. (77.26) In the last term, we can use the antisymmetry of Sµνto replaceDµDνwith 1 2[Dµ,Dν] =−1 2igFµν, which yields (i/D−/k)2=−k2+ 2ik·D+D2+gSµνFµν. (77.27) We use eq.(77.27) in eq.(77.24), and then rescale kbyM; the result is δ4(x−y)→M4/integraldisplayd4k (2π)4eiMk(x−y)e−k2e2ik·D/M+D2/M2+gSµνFµν/M2. (77.28) Thus we have Trδ4(x−x)γ5→M4/integraldisplayd4k (2π)4e−k2Tre2ik·D/M+D2/M2+gSµνFµν/M2γ5. (77.29) We can now expand the exponential in inverse powers of M; only terms up toM−4will survive the M→ ∞ limit. Furthermore, the trace over spin indices will vanish unless there are four or more gamma matri ces multiply- ingγ5. Together, these considerations imply that the only term th at can make a nonzero contribution is1 2(gSµνFµν)2/M4. Thus we ﬁnd Trδ4(x−x)γ5→1 2g2/integraldisplayd4k (2π)4e−k2(TrFµνFρσ)(TrSµνSρσγ5),(77.30) where the ﬁrst trace is over group indices (in the nonabelian case), and the second trace is over spin indices. The spin trace is TrSµνSρσγ5= Tr(i 2γµγν)(i 2γργσ)γ5 =−1 4Trγµγνγργσγ5 =iεµνρσ. (77.31) 77: Anomalies and the Path Integral for Fermions 463 To evaluate the integral over kin eq.(77.30), we analytically continue to euclidean spacetime; this results in an overall factor of i, as usual. Then each of the four gaussian integrals gives a factor of π1/2. So we ﬁnd Trδ4(x−x)γ5→ −g2 32π2εµνρσTrFµνFρσ. (77.32) Using this in eq.(77.19), we get (detJ)−2= exp/bracketleftigg −ig2 16π2/integraldisplay d4xα(x)εµνρσTrFµν(x)Fρσ(x)/bracketrightigg .(77.33) Including the transformation of the measure, eq.(77.18), i n the transfor- mation of the path integral, eq.(77.16), then yields Z(A)→/integraldisplay DΨDΨeiS(A)e−i/integraltext d4xα(x)[(g2/16π2)εµνρσTrFµν(x)Fρσ(x)+∂µjµ A(x)] (77.34) in place of eq.(77.16). This must be equal to the original exp ression for Z(A), eq.(77.8). This implies that eq.(77.7) holds inside quan tum correla- tion functions, up to possible contact terms. Note that this derivation of eq.(77.7) did not rely on an expa nsion in powers ofg, and so eq.(77.7) is exact; there are no higher-order correc tions. This result is known as the Adler-Bardeen theorem . It can also be (and originally was) established by a careful study of Feynman di agrams. The Fujikawa method can be used to ﬁnd the anomaly in the chira l gauge theories that we studied in section 75, but the analysi s is more in- volved. Here we will quote only the ﬁnal result. Consider a left-handed Weyl ﬁeld in a (possibly reducible) r epresen- tation R of the gauge group. We deﬁne the chiral gauge current jaµ≡ ΨTa RγµPLΨ. Its covariant divergence (which should be zero, accordin g to Noether’s theorem) is given by Dab µjbµ=g2 24π2εµνρσ∂µTr/bracketleftig Ta R(Aν∂ρAσ−1 2igAνAρAσ)/bracketrightig . (77.35) Note that the right-hand side of eq.(77.35) is notgauge invariant. The anomaly spoils gauge invariance in chiral gauge theories, u nless this right- hand side happens to vanish for group-theoretic reasons. We show in prob- lem 77.1 that this occurs if and only if A(R) = 0, where A(R) is the anomaly coeﬃcient of the representation R. For comparison, note that eq.(77.7) can be written as ∂µjµ A=−g2 4π2εµνρσ∂µTr/bracketleftig Aν∂ρAσ−2 3igAνAρAσ/bracketrightig . (77.36) 77: Anomalies and the Path Integral for Fermions 464 The relative value of the overall numerical prefactor in eqs .(77.35) and (77.36) is easy to understand: there is a minus one-half in eq .(77.35) from PL→ −1 2γ5, and a one-third from regularizing to preserve symmetry amo ng the three external lines in the triangle diagram. (The relat ive coeﬃcients of the second terms have no comparably simple explanation.) Finally, a related but more subtle problem, known as a global anomaly , arises for theories with an odd number of Weyl ﬁelds in a pseud oreal rep- resentation, such as the fundamental representation of SU( 2). In this case, every gauge ﬁeld conﬁguration Aµcan be smoothly deformed into another gauge ﬁeld conﬁguration A′ µthat has the same action, but has Z(A′) =−Z(A). Thus, when we integrate over A, the contribution from A′cancels the contribution from A, and the result is zero. Since its path integral is trivial, this theory does not exist. Problems 77.1) Show that the right-hand side of eq.(77.35) vanishes i f and only if A(R) = 0. 77.2) Show that the right-hand side of eq.(77.36) equals the right-hand side of eq.(77.7). 78: Background Field Gauge 465 78Background Field Gauge Prerequisite: 73 In the section, we will introduce a clever choice of gauge, background ﬁeld gauge, that greatly simpliﬁes the calculation of the beta functio n for Yang– Mills theory, especially at the one-loop level. We begin with the lagrangian for Yang–Mills theory, LYM=−1 4FaµνFa µν, (78.1) where the ﬁeld strength is Fa µν=∂µAa ν−∂νAa µ+gfabcAb µAc ν. (78.2) To evaluate the path integral, we must choose a gauge. As we sa w in section 71, one large class of gauges corresponds to choosing a gauge -ﬁxing function Ga(x), and adding Lgf+LghtoLYM, where Lgf=−1 2ξ−1GaGa, (78.3) Lgh= ¯ca∂Ga ∂AbµDbc µcc. (78.4) HereDbc µ=δbc∂µ−ig(Ta A)bcAa µ=δbc∂µ+gfbacAa µis the covariant derivative in the adjoint representation, and cand ¯care the ghost and antighost ﬁelds. The notation ∂Ga/∂Ab µmeans that any derivatives that act on Ab µinGa now act to the right in eq.(78.4). We getRξgauge by choosing Ga=∂µAa µ. To get background ﬁeld gauge , we ﬁrst introduce a ﬁxed, classical background ﬁeld ¯Aa µ(x), and the corresponding background covariant derivative, ¯Dµ≡∂µ−igTa A¯Aa µ. (78.5) Then we choose Ga= (¯Dµ)ab(A−¯A)b µ. (78.6) The ghost lagrangian becomes Lgh= ¯ca¯DµabDbc µcc, or, after an integration by parts, Lgh=−(¯Dµ¯c)a(Dµc)a, (78.7) where ( ¯Dµ¯c)a=¯Dµab¯cband (Dµc)a=Dac µcc. Under an inﬁnitesimal gauge transformation, the change in t he ﬁelds is δGAa µ(x) =−Dac µθc(x), (78.8) δGcb(x) =−igθa(x)(Ta A)bccc(x), (78.9) δG¯Aa µ(x) = 0. (78.10) 78: Background Field Gauge 466 The antighost ¯ ctransforms in the same way as c(since the adjoint repre- sentation is real). The background ﬁeld ¯Ais ﬁxed, and so does not change under a gauge transformation. Of course, this means that LgfandLghare notgauge invariant; their role is to ﬁx the gauge. We can, however, deﬁne a background ﬁeld gauge transformation , under which only the background ﬁeld transforms, δBG¯Aa µ(x) =−¯Dac µθc(x), (78.11) δBGAa µ(x) = 0, (78.12) δBGcb(x) = 0. (78.13) Obviously, LYMis invariant under this transformation (since it does not involve the background ﬁeld at all), but LgfandLghare not. However, LgfandLghareinvariant under the combined transformation δG+BG. For Lgh, as given by eq.(78.7), this follows immediately from the fa ct that Dµand¯Dµhave the same transformation property under the combined transformation, and that using covariant derivatives with all group indices contracted always yields a gauge-invariant expression. To use this argument on Lgf, as given by eqs.(78.3) and (78.6), we need to show that ( A−¯A)a µtransforms under the combined transformation in the same way as does an ordinary ﬁeld in the adjoint represent ation, such as the ghost ﬁeld in eq.(78.9). To show this, we write δG+BG(A−¯A)b µ=−(D−¯D)ba µθa = +ig(A−¯A)c µ(Tc A)baθa =−igθa(Ta A)bc(A−¯A)c µ. (78.14) We used the complete antisymmetry of ( Tc A)ba=−ifcbato get the last line. We see that ( A−¯A)a µtransforms like an ordinary ﬁeld in the adjoint rep- resentation, and so any expression that involves only covar iant derivatives (either ¯DµorDµ) acting on this ﬁeld, with all group indices contracted, is invariant under the combined transformation. Therefore, the complete lagrangian, L=LYM+Lgf+Lgh, is invariant under the combined transformation . Now consider constructing the quantum action Γ( A,c,¯c;¯A). Recall from section 21 that the quantum action can be expressed as the sum of all 1PI diagrams, with the external propagators replaced by the corresponding ﬁelds. In a gauge theory, the quantum action is in general notgauge invariant, because we had to ﬁx a gauge in order to carry out th e path integral. The quantum action thus depends on the choice of ga uge, and 78: Background Field Gauge 467 hence (in the case of background ﬁeld gauge) on the backgroun d ﬁeld ¯A. This is why we have written ¯Aas an argument of Γ, but separated by a semicolon to indicate its special role. An important property of the quantum action is that it inheri ts all lin- ear symmetries of the classical action; see problem 21.2. In the present case, these symmetries include the combined gauge transfor mationδG+BG. Therefore, the quantum action is also invariant under the combined tran s- formation . The quantum action takes its simplest form if we set the exte r- nal ﬁeldAequal to the background ﬁeld ¯A. Then, Γ( ¯A,c,¯c;¯A) is invariant under a gauge transformation of the form δG+BG¯Aa µ(x) =−¯Dac µθc(x), (78.15) δG+BGcb(x) =−igθa(x)(Ta A)bccc(x). (78.16) This is now simply an ordinary gauge transformation, with ¯Aas the gauge ﬁeld. The quantum action can be expressed as the classical action, plus loop corrections. For A=¯A, we have Γ(¯A,c,¯c;¯A) =/integraldisplay d4x/bracketleftig −1 4¯Faµν¯Fa µν−(¯Dµ¯c)a(¯Dµc)a/bracketrightig +... , (78.17) where the ellipses stand for the loop corrections. Note that Lgfhas dis- appeared [because we set A=¯Ain eq.(78.6)], and Lghhas the form of a kinetic term for a complex scalar ﬁeld in the adjoint represe ntation. This term is therefore manifestly gauge invariant, as is the ¯F¯Fterm. The gauge invariance of the quantum action has an important c onse- quence for the loop corrections. In background ﬁeld gauge, t he renormal- izingZfactors must respect the gauge invariance of the quantum act ion. Therefore, using the notation of section 73, we must have Z1=Z2, (78.18) Z1′=Z2′, (78.19) Z3=Z3g=Z4g. (78.20) Thus the relation between the bare and renormalized gauge co uplings be- comes g2 0=Z−1 3g2˜µε. (78.21) This relation now involves only Z3. We can therefore compute the beta function from Z3alone. This is the major advantage of background ﬁeld gauge. 78: Background Field Gauge 468 To compute the loop corrections, we need to evaluate 1PI diag rams in background-ﬁeld gauge with the external propagators remov ed and replaced with external ﬁelds; the external gauge ﬁeld should be set eq ual to the background ﬁeld. The easiest way to do this is to set A=¯A+A (78.22) at the beginning, and to write the path integral in terms of A. Then the A ﬁeld appears only on internal lines, and the ¯Aﬁeld only on external lines. The gauge-ﬁxing term now reads Lgf=−1 2ξ−1(¯DµAµ)a(¯DνAν)a, (78.23) and the ghost term is given by eq.(78.7). The Feynman rules that follow from LYM+Lgf+Lghare closely related to those we found in Rξgauge in section 72. The ghost and gluon prop- agators are the same, and vertices involving all internal lines are also the same. But if one or more gluon lines are external , then there are additional contributions to the vertices from LgfandLgh. We leave the details to problem 78.1. Further simpliﬁcations arise at the one-loop level. Using e q.(78.22) in eq.(78.2), we ﬁnd Fa µν=∂µ¯Aa ν−∂ν¯Aa µ+gfabc¯Ab µ¯Ac ν +∂µAa ν−∂νAa µ+gfabc(¯Ab µAc ν+Ab µ¯Ac ν) +gfabcAb µAc ν =¯Fa µν+ (¯DµAν)a−(¯DνAµ)a+gfabcAb µAc ν. (78.24) We then have LYM=−1 4¯Faµν¯Fa µν−1 2(¯DµAν)a(¯DµAν)a+1 2(¯DµAν)a(¯DνAµ)a −1 2gfabc¯FaµνAb µAc ν+... , (78.25) where the ellipses stand for terms that are linear, cubic, or quartic in A. Vertices arising from terms linear in Acannot appear in a 1PI diagram, and the cubic and quartic vertices do not appear in the one-loop c ontribution to the ¯Apropagator. The last term on the ﬁrst line of eq.(78.25) can be usefully ma nipulated with some dummy-index relabelings and integrations by part s; we have (¯DµAν)a(¯DνAµ)a= (¯DνAµ)c(¯DµAν)c =−Ab µ(¯Dν¯Dµ)bcAc ν =−Ab µ(¯Dµ¯Dν−[¯Dµ,¯Dν])bcAc ν =−Ab µ(¯Dµ¯Dν)bcAc ν−ig(Ta A)bc¯FaµνAb µAc ν = +( ¯DµAµ)c(¯DνAν)c−gfabc¯FaµνAb µAc ν.(78.26) 78: Background Field Gauge 469 Figure 78.1: The one-loop contributions to the ¯Apropagator in background ﬁeld gauge; the dashed lines can be either ghosts or internal Agauge ﬁelds. The dot denotes the ¯FAAvertex. Now the ﬁrst term on the right-hand side of eq.(78.26) has the same form as the gauge-ﬁxing term. If we choose ξ= 1, these two terms will cancel. Settingξ= 1, and including a renormalizing factor of Z3, the terms of interest in the complete lagrangian become L=−1 4Z3¯Faµν¯Fa µν−1 2Z3(¯DµAν)a(¯DµAν)a−(¯Dµ¯c)a(¯Dµc)a −Z3gfabc¯FaµνAb µAc ν. (78.27) In the ghost term, we have replaced Dµwith¯Dµ; the vertex corresponding to the dropped Aterm does not appear in the one-loop contribution to the¯Apropagator. Also, we can rescale Ato absorbZ3in all terms except the ﬁrst; since Anever appears on an external line, its normalization is irrelevant, and always cancels among propagators and verti ces. (The same is true of the ghost ﬁeld.) The one-loop diagrams that contribute to the ¯Apropagator are shown in ﬁg.(78.1). The dashed lines in the ﬁrst two diagrams represe nt either the A ﬁeld or the ghost ﬁelds. In either case, the second diagram va nishes, because it is proportional to/integraltextd4ℓ/ℓ2, which is zero after dimensional regularization. Note that the ghost term in eq.(78.27) has the form of a kineti c term for a complex scalar ﬁeld in the adjoint representation. In p roblem 73.1, we found the contribution of a complex scalar ﬁeld in a repres entation R CS to Π(k2) is ΠCS(k2) =−g2 24π2T(RCS)1 ε+ ﬁnite. (78.28) The ghost contribution is minus this, with R CS→A. (The minus sign is from the closed ghost loop.) Thus we have Πgh(k2) = +g2 24π2T(A)1 ε+ ﬁnite +O(g4). (78.29) For reference we recall that the counterterm contribution i s Πct(k2) =−(Z3−1). (78.30) 78: Background Field Gauge 470 Next we consider the diagrams with Aﬁelds in the loop. If the ¯FAA interaction term was absent, the calculation would again be a familiar one; the¯DA¯DAterm in eq.(78.27) has the form of a kinetic term for a real scalar ﬁeld that carries an extra index ν. That this index is a Lorentz vector index is immaterial for the diagrammatic calculatio n; the index is simply summed around the loop, yielding an extra factor of d= 4. There is also an extra factor of one-half (relative to the case of a c omplex scalar) because Ais real rather than complex. (Equivalently, the diagram has a symmetry factor of S= 2 from exchange of the top and bottom internal propagators when they do not carry charge arrows.) We thus ha ve Π¯DA¯DA(k2) =−g2 12π2T(A)1 ε+ ﬁnite +O(g4). (78.31) If we now include the ¯FAAinteraction, we can think of ¯Fa µνas a constant external ﬁeld. We can then draw the third diagram of ﬁg.(78.1 ), where and each dot denotes a vertex factor of −2igfabc¯Fa µν. This vacuum diagram has a symmetry factor of S= 2×2: one factor of two for exchanging the top and bottom propagators, and one for exchanging the left and r ight sources. Its contribution to the quantum action is iΓ¯FAA/VT=1 4(−2igfacd¯Fa µν)(−2igfbeg¯Fb ρσ)/parenleftig 1 i/parenrightig2˜µε/integraldisplayddℓ (2π)dgµρδce ℓ2gνσδdg ℓ2 =g2T(A)¯Faµν¯Fa µν/parenleftbiggi 8π2ε+ ﬁnite/parenrightbigg , (78.32) whereVTis the volume of spacetime. Comparing this with the tree-lev el lagrangian −1 4Z3¯F¯F, and recalling eq.(78.30), we see that eq.(78.32) is equivalent to a contribution to Π( k2) of Π¯FAA(k2) = +g2 2π2T(A)1 ε+ ﬁnite. (78.33) There is also a one-loop diagram with one ¯DA¯DAvertex and one ¯FAA vertex; however, contracting the vector indices on the Aﬁelds around the loop leads to a factor of ¯Fµνgµν= 0. Similarly, a one-loop diagram with a single ¯FAAvertex vanishes. We could also couple the gauge ﬁeld to a Dirac fermion in the re p- resentation R DF, and a complex scalar in the representation R CS. The corresponding contributions to Π( k2) were computed in section 73, and are given by eq.(78.28) and ΠDF(k2) =−g2 6π2T(RDF)1 ε+ ﬁnite, (78.34) 78: Background Field Gauge 471 Adding up eqs.(78.28), (78.29), (78.30), (78.31), (78.33) , and (78.34), we ﬁnd that ﬁniteness of Π( k2) requires Z3= 1+g2 24π2/bracketleftig/parenleftig +1−2+12/parenrightig T(A)−4T(RDF)−T(RCS)/bracketrightig1 ε+O(g4) (78.35) in the MS renormalization scheme. The analysis of section 28 now results in a beta function of β(g) =−g3 48π2/bracketleftig 11T(A)−4T(RDF)−T(RCS)/bracketrightig +O(g5). (78.36) A Majorana fermion or a Weyl fermion makes half the contribut ion of a Dirac fermion in the same representation; a real scalar ﬁeld makes half the contribution of a complex scalar ﬁeld. (Majorana fermions a nd real scalars must be in real representations of the gauge group.) In quantum chromodynamics, the gauge group is SU(3), and the re are nF= 6 ﬂavors of quarks (which are Dirac fermions) in the fundame ntal representation. We therefore have T(A) = 3,T(RDF) =1 2nF, andT(RCS) = 0; therefore β(g) =−g3 16π2/parenleftig 11−2 3nF/parenrightig +O(g5). (78.37) We see that the beta function is negative for nF≤16, and so QCD is asymptotically free. Problems 78.1) Compute the tree-level vertex factors in background ﬁ eld gauge for all vertices that connect one or more external gluons with two or more internal lines (ghost or gluon). 78.2) Our one-loop corrections can be interpreted as functi onal determi- nants. Deﬁne R,(a,b)≡¯D2+gTa R¯Fa µνSµν (a,b), (78.38) where ¯Dµ=∂µ−ig(Ta R)¯Aa µis the background-covariant derivative in the representation R, implicitly multiplied by the indenti ty matrix for the (a,b) representation of the Lorentz group, and Sµν (a,b)are the Lorentz generators for that representation; in particular , Sµν (1,1)= 0, (78.39) Sµν (2,1)⊕(1,2)=i 4[γµ,γν], (78.40) (Sµν (2,2))αβ=−i(δµαδνβ−δναδµβ). (78.41) 78: Background Field Gauge 472 Show that the one-loop contribution to the terms in the quant um action that do not depend on the ghost ﬁelds is given by expiΓ1−loop(¯A,0,0;¯A)∝(det A,(1,1))+1 ×(det A,(2,2))−1/2 ×(det RDF,(2,1)⊕(1,2))+1/2 ×(det RCS,(1,1))−1. (78.42) Verify that this expression agrees with the diagrammatic an alysis in this section. 79: Gervais–Neveu Gauge 473 79Gervais–Neveu Gauge Prerequisite: 78 In section 78, we used background ﬁeld gauge to set up the comp utation of a quantum action that is gauge invariant. Given this quant um action, we can use it to compute scattering amplitudes via the corres ponding tree diagrams, as discussed in section 19. Since the ghost ﬁelds i n the quantum action do not contribute to tree diagrams, we can simply drop all the ghost terms in the quantum action. Because the quantum action computed in background ﬁeld gaug e is itself gauge invariant, it requires further gauge ﬁxing to s pecify the gluon propagator and vertices. We can choose whatever gauge is mos t convenient; for example, Rξgauge. In principle, this gauge ﬁxing involves introducing newghost ﬁelds, but, once again, these do not contribute to tree diagrams, and so we can ignore them. If we start with the tree-level approximation to the quantum action, then, inRξgauge, we simply get the gluon propagator and vertices of section 72. As we noted there, the complexity of the three- an d four-gluon vertices in Rξgauge leads to long, involved computations of even simple processes like gluon-gluon scattering. In this section, we will introduce another gauge, Gervais–Neveu gauge , that simpliﬁes these tree-level computations. We begin by specializing to the gauge group SU( N), and working with the matrix-valued ﬁeld Aµ=Aa µTa. For later convenience, we will normal- ize the generators via TrTaTb=δab. (79.1) With this choice, their commutation relations become [Ta,Tb] =i√ 2fabcTc. (79.2) The tree-level action is speciﬁed by the Yang–Mills lagrang ian, LYM=−1 4TrFµνFµν, (79.3) where the matrix-valued ﬁeld strength is Fµν=∂µAν−∂νAµ−ig√ 2[Aµ,Aν]. (79.4) Let us introduce the matrix-valued complex tensor Hµν≡∂µAν−ig√ 2AµAν. (79.5) ThenFµνis the antisymmetric part of Hµν, Fµν=Hµν−Hνµ. (79.6) 79: Gervais–Neveu Gauge 474 The Yang–Mills lagrangian can now be written as LYM=−1 2Tr/parenleftig HµνHµν−HµνHνµ/parenrightig . (79.7) To ﬁx the gauge, we choose a matrix-valued gauge-ﬁxing funct ionG(x), and add Lgf=−1 2TrGG (79.8) toLYM. Here we have set the gauge parameter ξto one, and ignored the ghost lagrangian (since, as we have already discussed, it do es not aﬀect tree diagrams). The choice of Gthat yields Gervais–Neveu gauge is G=Hµµ. (79.9) At ﬁrst glance, this choice seems untenable, because we see f rom eq.(79.5) that thisG(and hence Lgf) is not hermitian. However, because the role of Lgfis merely to ﬁx the gauge, it is acceptable for Lgfto be nonhermitian. Combining eqs.(79.7) and (79.8), we get a total, gauge-ﬁxed lagrangian L=−1 2Tr/parenleftig HµνHµν−HµνHνµ+HµµHνν/parenrightig . (79.10) Consider the terms in Lwith two derivatives. After some integrations by parts, those from the third term in eq.(79.10) cancel those f rom the second, leading to L2∂=−1 2Tr∂µAν∂µAν, (79.11) just as inRξgauge with ξ= 1. Now consider the terms with no derivatives. Once again, those from the third term in eq.(79.10) cancel th ose from the second (after using the cyclic property of the trace), leadi ng to L0∂= +1 4g2TrAµAνAµAν. (79.12) Finally, we have the terms with one derivative, L1∂= +ig√ 2Tr/parenleftig ∂µAνAµAν−∂µAνAνAµ+∂µAµAνAν/parenrightig . (79.13) Each derivative acts only on the ﬁeld to its immediate right. If we integrate by parts in the last term in eq.(79.13), we generate two terms ; one of these cancels the ﬁrst term in eq.(79.13), and the other duplicate s the second. Thus we have L1∂=−i√ 2gTr∂µAνAνAµ. (79.14) Combining eqs.(79.11), (79.12), and (79.14), we ﬁnd L= Tr/parenleftig −1 2∂µAν∂µAν−i√ 2g∂µAνAνAµ+1 4g2AµAνAµAν/parenrightig .(79.15) 79: Gervais–Neveu Gauge 475 Because this lagrangian has a rather simple structure in ter ms of the matrix- valued ﬁeld Aµ, it is helpful to stick with this notation, rather than tryin g to reexpress Lin terms of Aa µ= Tr(TaAµ). In the next section, we explore the Feynman rules for a matrix-valued ﬁeld in a simpliﬁed con text. Reference Notes Gervais–Neveu gauge and some interesting variations are di scussed in Siegel. 80: The Feynman Rules for N×NMatrix Fields 476 80The Feynman Rules for N×NMatrix Fields Prerequisite: 10 In section 79, we found that the lagrangian for SU( N) Yang–Mills theory in Gervais–Neveu gauge is L= Tr/parenleftig −1 2∂µAν∂µAν−i√ 2g∂µAνAνAµ+1 4g2AµAνAµAν/parenrightig ,(80.1) whereAµ(x) is a traceless hermitian N×Nmatrix. In this section, we will work out the Feynman rules for a simpliﬁed model of a scalar ﬁe ld that keeps the essence of the matrix structure. LetB(x) be a hermitian N×Nmatrix that is nottraceless. Let Tabe a complete set of N2hermitianN×Nmatrices normalized according to TrTaTb=δab. (80.2) We will take one of these matrices, TN2, to be proportional to the identity matrix; then eq.(80.2) requires the rest of the Ta’s to be traceless. We can expandB(x) in theTa’s, with coeﬃcient ﬁelds Ba(x), B(x) =Ba(x)Ta, (80.3) Ba(x) = TrTaB(x), (80.4) where the repeated index in eq.(80.3) is implicitly summed o vera= 1 to N2. Consider a lagrangian for B(x) of the form L= Tr/parenleftig −1 2∂µB∂µB+1 3gB3−1 4λB4/parenrightig . (80.5) Using eqs.(80.2) and (80.3), we ﬁnd an expression for Lin terms of the coeﬃcient ﬁelds, L=−1 2∂µBa∂µBa+1 3gTr(TaTbTc)BaBbBc −1 4λTr(TaTbTcTd)BaBbBcBd. (80.6) It is easy to read oﬀ the Feynman rules from this form of L. The propagator for the coeﬃcient ﬁeld Bais ˜∆ab(k2) =δab k2−iǫ. (80.7) There is a three-point vertex with vertex factor 2 igTr(TaTbTc), and a four- point vertex with vertex factor −6iλTr(TaTbTcTd). This clearly leads to messy and complicated formulae for scattering amplitudes. 80: The Feynman Rules for N×NMatrix Fields 477 ki l j Figure 80.1: The double-line notation for the propagator of a hermitian matrix ﬁeld. Figure 80.2: 3- and 4-point vertices in the double-line nota tion. Instead, let us work with Lin the form of eq.(80.5). Writing the matrix indices explicitly, with one up and one down (and employing t he rule that two indices can be contracted only if one is up and one is down) , we have B(x)ij=Ba(x)(Ta)ij. This implies that the propagator for Bijis ˜∆ijkl(k2) =(Ta)ij(Ta)kl k2−iǫ. (80.8) Since theTamatrices form a complete set, there is a completeness relati on of the form ( Ta)ij(Ta)kl∝δilδkj. To get the constant of proportionality, set j=kandl=ito turn the left-hand side into ( Ta)ik(Ta)ki= Tr(TaTa), and the right-hand side into δiiδkk=N2. From eq.(80.2) we have Tr( TaTa) = δaa=N2. So the constant of proportionality is one, and (Ta)ij(Ta)kl=δilδkj. (80.9) We can represent the Bpropagator with a double-line notation, as shown in ﬁg.(80.1). The arrow on each line points from an up in dex to a down index. Since the interactions are simple matrix produ cts, with an up index from one ﬁeld contracted with a down index from an adj acent ﬁeld, the vertices follow the pattern shown in ﬁg.(80.2). Si nce ann-point vertex of this type has only an n-fold cyclic symmetry (rather than an n!- fold permutation symmetry), the vertex factor is itimes the coeﬃcient of Tr(Bn) inLtimesn(rather than n!). Thus, for the lagrangian of eq.(80.5), the 3- and 4-point vertex factors are igand−iλ. Now consider a scattering process. Particles correspondin g to the coef- ﬁcient ﬁelds labeled by the indices a1anda2(and with four-momenta k1 80: The Feynman Rules for N×NMatrix Fields 478 4 1 2 3411 22 334 Figure 80.3: Tree diagrams with four external lines. Five mo re diagrams of each of these three types, with the external labels 2, 3, an d 4 permuted, also contribute. andk2) scatter into particles corresponding to the coeﬃcient ﬁel ds labeled by the indices a3anda4(and with four-momenta k3andk4). We wish to compute the scattering amplitude for this process, at tree l evel. There are 18 contributing Feynman diagrams. Three are shown in ﬁg.(80.3); the remaining 15 are obtained by making noncylic permutations of the labels 1 ,2,3,4 (equivalent to making unrestricted permutations of 2,3,4). For simplicity, we will treat all external momenta as out going; thenk0 1andk0 2are negative, and k1+k2+k3+k4= 0. Each external line carries a factor of Tai, with its matrix indices contracted by following the arrows backward through the diagrams. Omitting the iǫ’s in the prop- agators (which are not relevant for tree diagrams), the resu lting tree-level amplitude is iT= Tr(Ta1Ta2Ta3Ta4)/parenleftigg (ig)2(−i) (k1+k2)2+(ig)2(−i) (k1+k4)2−iλ/parenrightigg +/parenleftig (234)→(342),(423),(243),(432),(324)/parenrightig . (80.10) More generally, we can see that the value of any tree-level di agram with nexternal lines is proportional to Tr( Tai1...Tain). If the diagram is drawn in planar fashion (that is, with no crossed lines), then the o rdering of the aiindices in the trace is determined by the cyclic ordering of t he labels on the external lines (which we take to be couterclockwise). Then, each internal line contributes a factor of −i/k2, each 3-point vertex a factor of ig, and each four-point vertex a factor of −iλ. These are the color-ordered Feynman rules for this theory. Return now to iTas given by eq.(80.10). Suppose that we wish to square this amplitude, and sum over all possible particle ty pes for each incoming or outgoing particle. We then have to evaluate expr essions like Tr(Ta1Ta2Ta3Ta4)[Tr(Ta1Ta2Ta4Ta3)]∗, (80.11) 80: The Feynman Rules for N×NMatrix Fields 479 341 2 3 412 Figure 80.4: Evaluation of Tr( Ta1Ta2Ta3Ta4)[Tr(Ta1Ta2Ta4Ta3)]∗, with all repeated indices summed. Each of the two closed single-line loops yields a factor ofδii=N. with all repeated indices summed. Using the hermiticity of t heTamatrices, we have [Tr(Ta1...Tan)]∗= Tr(Tan...Ta1), (80.12) It is then easiest to evaluate eq.(80.11) diagrammatically , as shown in ﬁg.(80.4). Each closed single-line loop yields a factor of δii=N. The result is that the absolute square of any particular trace yi elds a factor of N4, and the product of any trace times the complex conjugate of a ny other diﬀerent trace yields a factor of N2. The coeﬃcient of both Tr( Ta1Ta2Ta3Ta4) and Tr(Ta1Ta4Ta3Ta2) in eq.(80.10) is A3≡g2 (k1+k2)2+g2 (k1+k4)2−λ. (80.13) Similarly, the coeﬃcient of both Tr( Ta1Ta3Ta4Ta2) and Tr(Ta1Ta2Ta4Ta3) is A4≡g2 (k1+k3)2+g2 (k1+k2)2−λ, (80.14) and of both Tr( Ta1Ta4Ta2Ta3) and Tr(Ta1Ta3Ta2Ta4) is A2≡g2 (k1+k4)2+g2 (k1+k3)2−λ. (80.15) Thus we have /summationdisplay a1,a2,a3,a4|T |2= (2N4+ 2N2)/summationdisplay j|Aj|2+ 4N2/summationdisplay j/ne}ationslash=kA∗ jAk = (2N4−2N2)/summationdisplay j|Aj|2+ 4N2(/summationdisplay jA∗ j)(/summationdisplay kAk),(80.16) wherejandkare summed over 2 ,3,4. 80: The Feynman Rules for N×NMatrix Fields 480 i l j ki l k1 Nj Figure 80.5: The propagator for a traceless hermitian ﬁeld. Now suppose we wish to impose the condition that the matrix ﬁe ldB istraceless : TrB= 0. This means that we eliminate the component ﬁeld witha=N2, corresponding to the matrix TN2=N−1/2I. We must also eliminateTN2from the sum in eq.(80.9), leading to (Ta)ij(Ta)kl=δilδkj−1 Nδijδkl. (80.17) This can all be done diagrammatically by replacing the propa gator in ﬁg.(80.1) with the one in ﬁg.(80.5). (The kinematic factor, −i/k2, is un- changed.) Fig.(80.5) must now be used as the internal propag ator in the diagrams of ﬁg.(80.3). Also, when we multiply one diagram by the complex conjugate of another in the computation of/summationtext a1...an|T |2, we must use the propagator of ﬁg.(80.5) to connect the external line of one d iagram with the matching external line of the complex conjugate diagram . Although these computations are still straightforward, they can bec ome considerably more involved. Problems 80.1) Show that the color-ordered Feynman rules, and the rul es for compo- nent ﬁelds given after eq.(80.6), agree in the case N= 1. 80.2) Verify the results quoted after eq.(80.12). 80.3) Compute/summationtext a1,a2,a3,a4|Tr(Ta1Ta2Ta3Ta4)|2for the case of traceless Ta’s. 80.4)The large-N limit. Letλ=cg2, wherecis a number of order one. Now consider evaluating the path integral, without sources, as a function ofgandN, Z(g,N) =eiW(g,N)=/integraldisplay DBei/integraltext ddxL, (80.18) whereW(g,N) is normalized by W(0,N) = 0. As usual, Wcan be expressed as a sum of connected vacuum diagrams, which we dra w in the double-line notation. Consider a diagram with V3three-point vertices,V4four-point vertices, Epropagators or edges, andFclosed single-line loops or faces. 80: The Feynman Rules for N×NMatrix Fields 481 a) Find the dependence on gandNof a diagram speciﬁed by the values ofV3,V4,E, andF. b) Express Efor a vacuum diagram in terms of V3andV4. c) Recall, derive, or look up the formula for the Euler character χof the two-dimensional surface of a polyhedron in terms of the v alues of V≡V3+V4,E, andF. The Euler character is related to the genus Gof the surface by χ= 2−2G;Gcounts the number of handles, so that a sphere has genus zero, a torus has genus one, etc. d) Consider the limit g→0 andN→ ∞ with the ’t Hooft coupling ¯λ=g2Nheld ﬁxed (and not necessarily small). Show that W(¯λ,N) has atopological expansion of the form W(¯λ,N) =∞/summationdisplay G=0N2−2GWG(¯λ), (80.19) whereWG(¯λ) is given by a sum over diagrams that form polyhedra with genus G. In particular, the leading term, W0(¯λ), is given by a sum over diagrams with spherical topology, also known as planar diagrams . 81: Scattering in Quantum Chromodynamics 482 81Scattering in Quantum Chromodynamics Prerequisite: 60, 79, 80 In section 79, we found that the lagrangian for SU( N) Yang–Mills theory in Gervais–Neveu gauge is L= Tr/parenleftig −1 2∂µAν∂µAν−i√ 2g∂µAνAνAµ+1 4g2AµAνAµAν/parenrightig ,(81.1) whereAµ(x) is a traceless hermitian N×Nmatrix. For quantum chromo- dynamics,N= 3, but we will leave Nunspeciﬁed in our calculations. In section 80, we worked out the color-ordered Feynman rules for a scalar ma- trix ﬁeld; the same technology applies here as well. In parti cular, we draw each tree diagram in planar fashion (that is, with no crossed lines). Then the cyclic, counterclockwise ordering i1...inof the external lines ﬁxes the color factor as Tr( Tai1...Tain), where the generator matrices are normal- ized via Tr( TaTb) =δab. The tree-level n-gluon scattering amplitude is then written as T=gn−2/summationdisplay noncyclic permsTr(Ta1...Tan)A(1,...,n ), (81.2) where we have pulled out the coupling constant dependence, a ndA(1,...,n ) is apartial amplitude that we compute with the color-ordered Feynman rules. The partial amplitudes are cyclically symmetric, A(2,...,n, 1) =A(1,2,...,n ). (81.3) The sum in eq.(81.2) is over all noncyclic permutations of 1 ...n, which is equivalent to a sum over allpermutations of 2 ...n. From the ﬁrst term in eq.(81.1), we see that the gluon propaga tor is simply ˜∆µν(k) =gµν k2−iǫ. (81.4) Here we have left out the matrix indices since we have already accounted for them with the color factor in eq.(81.2). The second and th ird terms in eq.(81.1) yield three- and four-gluon vertices. The thre e-gluon vertex factor (again without the matrix indices) is iVµνρ(p,q,r) =i(−i√ 2g)(−ipρgµν) + [2 cyclic permutations of ( µ,p),(ν,q),(ρ,r)] =−i√ 2g(pρgµν+qµgνρ+rνgρµ), (81.5) 81: Scattering in Quantum Chromodynamics 483 where the four-momenta p,q, andrare all taken to be outgoing. The four-gluon vertex factor is simply iVµνρσ=ig2gµρgνσ. (81.6) However, in the context of the color-ordered rules, it is sim pler to designate the outgoing four-momentum on each external line as ki, and contract the vector index with the corresponding polarization vector εi. (For now we suppress the helicity label λ=±.) In this notation, the vertex factors become iV123=−i√ 2g/bracketleftig (ε1ε2)(k1ε3) + (ε2ε3)(k2ε1) + (ε3ε1)(k3ε2)/bracketrightig ,(81.7) iV1234= +ig2(ε1ε3)(ε2ε4), (81.8) where the external lines are numbered sequentially, counte rclockwise around the vertex. (Of course, if an attached line is internal, the c orresponding polarization vector is simply a placeholder for an internal propagator.) The color-ordered three-point vertex, eq.(81.7), is antis ymmetric on the reversal 123 ↔321, while the four-point vertex, eq.(81.8), is symmetric o n the reversal 1234 ↔4321. This implies the reﬂection identity , A(n,..., 2,1) = (−1)nA(1,2,...,n ), (81.9) which will be useful later. It is clear from eqs.(81.7) and (81.8) that every term in any t ree-level scattering amplitude is proportional to products of polari zation vectors with each other, or with external momenta. (Actually, this f ollows directly from Lorentz invariance, and the fact that the scattering am plitude is linear in each polarization.) We get one momentum factor from each t hree-point vertex. Since every tree diagram with nexternal lines has no more than n−2 vertices, there are no more than n−2 momenta to contract with the npolarizations. Therefore, every term in every tree-level a mplitude must include at least one product of two polarization vectors. Th en, if the prod- uct of every possible pair of polarization vectors vanishes , the tree-level amplitude for that process is zero. We will now show that this is indeed the case if all, or all but o ne, of the external gluons have the same helicity. (Here we are usin g the seman- tic convention of section 60: the helicity of an external glu on is speciﬁed relative to the outgoing four-momentum kithat labels the corresponding external line. If that gluon is actually incoming—as indica ted by a negative value ofk0 i—then its physical helicity is opposite to its labeled helic ity.) 81: Scattering in Quantum Chromodynamics 484 To proceed, we recall from section 60 some formulae for produ cts of polarization vectors in the spinor-helicity formalism, ε+(k;q)·ε+(k′;q′) =∝an}b∇acketle{tqq′∝an}b∇acket∇i}ht[kk′] ∝an}b∇acketle{tqk∝an}b∇acket∇i}ht∝an}b∇acketle{tq′k′∝an}b∇acket∇i}ht, (81.10) ε−(k;q)·ε−(k′;q′) =[qq′]∝an}b∇acketle{tkk′∝an}b∇acket∇i}ht [qk][q′k′], (81.11) ε+(k;q)·ε−(k′;q′) =∝an}b∇acketle{tqk′∝an}b∇acket∇i}ht[kq′] ∝an}b∇acketle{tqk∝an}b∇acket∇i}ht[q′k′]. (81.12) The ﬁrst argument of each εis the momentum of the corresponding line; the second argument is an arbitrary reference momentum. Rec all that the twistor producs ∝an}b∇acketle{tqk∝an}b∇acket∇i}htand [qk] are antisymmetric, and hence ∝an}b∇acketle{tqq∝an}b∇acket∇i}ht= [qq] = 0. Using this fact in eq.(81.10), we see that choosing the sam e reference momentum qfor all positive-helicity polarizations results in a vanis hing product for any pair of them. Furthermore, if we choose this qequal to the momentum k′of a negative-helicity gluon, eq.(81.12) tells us that the product of its polarization with that of any positive-helic ity gluon also vanishes. Thus, if all, or all but one, of the external gluons have positive helicity, all possible polarization products are zero, and hence the tree-level scattering amplitude is also zero. Thus we have shown that A(1±,2+,...,n+) = 0, (81.13) where the superscripts are the helicities. Of course, the sa me is true if all, or all but one, of the helicities are negative, A(1±,2−,...,n−) = 0. (81.14) Now we turn to the calculation of some nonzero tree-level par tial am- plitudes, beginning with A(1−,2−,3+,4+). The contributing color-ordered Feynman diagrams are shown in ﬁg.(81.1). We choose the refer ence mo- menta to be q1=q2=k3andq3=q4=k2. Then all polarization products vanish, with the exception of ε1·ε4=ε−(k1,q1)·ε+(k4,q4) =ε−(k1,k3)·ε+(k4,k2) =∝an}b∇acketle{t21∝an}b∇acket∇i}ht[43] ∝an}b∇acketle{t24∝an}b∇acket∇i}ht[31]. (81.15) With this choice of the reference momenta, the third diagram in ﬁg.(81.1) obviously vanishes, because it has a factor of ε1·ε3= 0 (and also, for good 81: Scattering in Quantum Chromodynamics 485 51 2 341 2 21 334 4 5 Figure 81.1: Diagrams for the partial amplitude A(1,2,3,4). measure,ε2·ε4= 0). Now consider the 235 vertex in the second diagram; we have V235∝(ε2ε3)(k2ε5) + (ε3ε5)(k3ε2) + (ε5ε2)(k5ε3), (81.16) whereε5is a placeholder for an internal propagator. The ﬁrst term in eq.(81.16) vanishes because ε2·ε3= 0. The second term vanishes because k3=q2andq2·ε2= 0. Finally, the third term vanishes because k5= −k2−k3=−q3−k3, andq3·ε3=k3·ε3= 0. Hence the 235 vertex vanishes, and therefore so does the second diagram. That leaves only the ﬁrst diagram. We then have ig2A(1−,2−,3+,4+) = (iV125)(iV345′)/vextendsingle/vextendsingle/vextendsingleεµ 5εν 5→igµν/s12, (81.17) where 5′means the momentum is −k5rather than k5, and s12≡ −(k1+k2)2=∝an}b∇acketle{t12∝an}b∇acket∇i}ht[21]. (81.18) We have iV125=−i√ 2g/bracketleftig (ε1ε2)(k1ε5) + (ε2ε5)(k2ε1) + (ε5ε1)(k5ε2)/bracketrightig ,(81.19) but the ﬁrst term vanishes because ε1·ε2= 0. Similarly, the ﬁrst term of iV345′=−i√ 2g/bracketleftig (ε3ε4)(k3ε5) + (ε4ε5)(k4ε3) + (ε5ε3)(−k5ε4)/bracketrightig (81.20) also vanishes. When we take the product of these two vertices , and replace the internal polarizations with the propagator, as indicat ed in eq.(81.17), only the product of the third term of eq.(81.19) with the seco nd term of eq.(81.20) is nonzero; all other terms include a vanishin g product of polarizations. We get ig2A(1−,2−,3+,4+) = (−i√ 2g)2(i/s12)(ε1ε4)(k5ε2)(k4ε3).(81.21) 81: Scattering in Quantum Chromodynamics 486 Sincek5=−k1−k2, andk2·ε2= 0, we have k5·ε2=−k1·ε2. We evaluate k1·ε2andk4·ε3via the general formulae p·ε+(k;q) =∝an}b∇acketle{tqp∝an}b∇acket∇i}ht[pk]√ 2∝an}b∇acketle{tqk∝an}b∇acket∇i}ht, (81.22) p·ε−(k;q) =[qp]∝an}b∇acketle{tpk∝an}b∇acket∇i}ht√ 2 [qk]. (81.23) Settingq2=k3andq3=k2, we get k1·ε2=[31]∝an}b∇acketle{t12∝an}b∇acket∇i}ht√ 2 [32], (81.24) k4·ε3=∝an}b∇acketle{t24∝an}b∇acket∇i}ht[43]√ 2∝an}b∇acketle{t23∝an}b∇acket∇i}ht. (81.25) Using eqs.(81.15), (81.18), (81.24), and (81.25) in eq.(81 .21), and using antisymmetry of the twistor products to cancel common facto rs, we get A(1−,2−,3+,4+) =∝an}b∇acketle{t21∝an}b∇acket∇i}ht[43]2 [21][32] ∝an}b∇acketle{t23∝an}b∇acket∇i}ht. (81.26) We can make our result look nicer by multiplying the numerato r and denominator by ∝an}b∇acketle{t34∝an}b∇acket∇i}ht. In the numerator, we use ∝an}b∇acketle{t34∝an}b∇acket∇i}ht[43] =s34=s12=∝an}b∇acketle{t12∝an}b∇acket∇i}ht[21], (81.27) and cancel the [21] with the one in the denominator. Now multi ply the nu- merator and denominator by ∝an}b∇acketle{t41∝an}b∇acket∇i}ht, and use the momentum-conservation identity (see problem 60.2) to replace ∝an}b∇acketle{t41∝an}b∇acket∇i}ht[43] in the numerator with −∝an}b∇acketle{t21∝an}b∇acket∇i}ht[23], and cancel the [23] with the [32] in the denominator (wh ich yields a minus sign). Finally, multiply the numerator and de nominator by ∝an}b∇acketle{t12∝an}b∇acket∇i}htto get A(1−,2−,3+,4+) =∝an}b∇acketle{t12∝an}b∇acket∇i}ht4 ∝an}b∇acketle{t12∝an}b∇acket∇i}ht∝an}b∇acketle{t23∝an}b∇acket∇i}ht∝an}b∇acketle{t34∝an}b∇acket∇i}ht∝an}b∇acketle{t41∝an}b∇acket∇i}ht. (81.28) This is our ﬁnal result for A(1−,2−,3+,4+). Now, using cyclic symmetry, we can get any partial amplitude where the two negative helicities are adjacent; for example, A(1+,2−,3−,4+) =∝an}b∇acketle{t23∝an}b∇acket∇i}ht4 ∝an}b∇acketle{t12∝an}b∇acket∇i}ht∝an}b∇acketle{t23∝an}b∇acket∇i}ht∝an}b∇acketle{t34∝an}b∇acket∇i}ht∝an}b∇acketle{t41∝an}b∇acket∇i}ht. (81.29) We must still calculate one partial amplitude where the nega tive helicities are not adjacent, such as A(1−,2+,3−,4+). Once we have it, we can use cyclic symmetry to get all the remaining partial amplitudes . 81: Scattering in Quantum Chromodynamics 487 Before turning to this calculation, let us consider the prob lem of squar- ing the total amplitude and summing over colors. Because the generator matrices are traceless, we should (as we discussed in sectio n 80) use the completeness relation (Ta)ij(Ta)kl=δilδkj−1 Nδijδkl. (81.30) However, recall that the Yang–Mills ﬁeld strength is Fµν=∂µAν−∂νAµ−ig√ 2[Aµ,Aν]. (81.31) If we allow a generator matrix proportional to the identity, which corre- sponds to a gauge group of U( N) rather than SU( N), then this extra U(1) generator commutes with every other generator. Thus the U(1 ) ﬁeld does not appear in the commutator term in eq.(81.31). Since it is t his commuta- tor term that is responsible for the interaction of the gluon s, the U(1) ﬁeld is a free ﬁeld. Therefore, any scattering amplitude involvi ng the associated particle (which we will call the ﬁctitious photon ) must be zero. Thus, if we write a scattering amplitude in the form of eq.(81.2), and re place one of theTa’s with the identity matrix, the result must be zero. Thisdecoupling of the ﬁctitious photon allows us to use the much simpler completeness relation (Ta)ij(Ta)kl=δilδkj(81.32) in place of eq.(81.30). There is no need to subtract the U(1) g enerator from the sum over the generators, as we did in eq.(81.30), bec ause the terms involving it vanish anyway. The decoupling of the ﬁctitious photon is useful in another w ay. Let us apply it to the case of n= 4, and set Ta4∝Iin eq.(81.2). Then we have 0 = Tr(Ta1Ta2Ta3)/bracketleftig A(1,2,3,4) +A(1,2,4,3) +A(1,4,2,3)/bracketrightig + Tr(Ta1Ta3Ta2)/bracketleftig A(1,3,2,4) +A(1,3,4,2) +A(1,4,3,2)/bracketrightig .(81.33) The contents of each square bracket must vanish. Requiring t his of the ﬁrst term yields A(1,2,3,4) =−A(1,2,4,3)−A(1,4,2,3). (81.34) Assigning some helicities, this reads A(1−,2+,3−,4+) =−A(1−,2+,4+,3−)−A(1−,4+,2+,3−).(81.35) Note that we have now expressed a partial amplitude with nona djacent negative helicities in terms of partial amplitudes with adj acent negative 81: Scattering in Quantum Chromodynamics 488 helicities, which we have already calculated. Thus we have A(1−,2+,3−,4+) =−/bracketleftigg ∝an}b∇acketle{t31∝an}b∇acket∇i}ht4 ∝an}b∇acketle{t31∝an}b∇acket∇i}ht∝an}b∇acketle{t12∝an}b∇acket∇i}ht∝an}b∇acketle{t24∝an}b∇acket∇i}ht∝an}b∇acketle{t43∝an}b∇acket∇i}ht+∝an}b∇acketle{t31∝an}b∇acket∇i}ht4 ∝an}b∇acketle{t31∝an}b∇acket∇i}ht∝an}b∇acketle{t14∝an}b∇acket∇i}ht∝an}b∇acketle{t42∝an}b∇acket∇i}ht∝an}b∇acketle{t23∝an}b∇acket∇i}ht/bracketrightigg =−∝an}b∇acketle{t13∝an}b∇acket∇i}ht3 ∝an}b∇acketle{t24∝an}b∇acket∇i}ht/bracketleftbigg1 ∝an}b∇acketle{t12∝an}b∇acket∇i}ht∝an}b∇acketle{t34∝an}b∇acket∇i}ht+1 ∝an}b∇acketle{t14∝an}b∇acket∇i}ht∝an}b∇acketle{t23∝an}b∇acket∇i}ht/bracketrightbigg =−∝an}b∇acketle{t13∝an}b∇acket∇i}ht3 ∝an}b∇acketle{t24∝an}b∇acket∇i}ht/bracketleftbigg∝an}b∇acketle{t14∝an}b∇acket∇i}ht∝an}b∇acketle{t23∝an}b∇acket∇i}ht+∝an}b∇acketle{t12∝an}b∇acket∇i}ht∝an}b∇acketle{t34∝an}b∇acket∇i}ht ∝an}b∇acketle{t12∝an}b∇acket∇i}ht∝an}b∇acketle{t34∝an}b∇acket∇i}ht∝an}b∇acketle{t14∝an}b∇acket∇i}ht∝an}b∇acketle{t23∝an}b∇acket∇i}ht/bracketrightbigg =−∝an}b∇acketle{t13∝an}b∇acket∇i}ht3 ∝an}b∇acketle{t24∝an}b∇acket∇i}ht/bracketleftbigg− ∝an}b∇acketle{t13∝an}b∇acket∇i}ht∝an}b∇acketle{t42∝an}b∇acket∇i}ht ∝an}b∇acketle{t12∝an}b∇acket∇i}ht∝an}b∇acketle{t34∝an}b∇acket∇i}ht∝an}b∇acketle{t14∝an}b∇acket∇i}ht∝an}b∇acketle{t23∝an}b∇acket∇i}ht/bracketrightbigg , (81.36) where the last line follows from the Schouten identity (see p roblem 50.3). A ﬁnal clean-up yields A(1−,2+,3−,4+) =∝an}b∇acketle{t13∝an}b∇acket∇i}ht4 ∝an}b∇acketle{t12∝an}b∇acket∇i}ht∝an}b∇acketle{t23∝an}b∇acket∇i}ht∝an}b∇acketle{t34∝an}b∇acket∇i}ht∝an}b∇acketle{t41∝an}b∇acket∇i}ht. (81.37) Now that we have all the partial amplitudes, we can compute th e color- summed |T |2. There are only three partial amplitudes that are not relate d by either cyclic permutations, eq.(81.3), or reﬂections, e q.(81.9); we can take these to be A3≡A(1,2,3,4), (81.38) A4≡A(1,3,4,2), (81.39) A2≡A(1,4,2,3), (81.40) where the subscript on the left-hand side is the third argume nt on the right-hand side. (Switching the second and fourth argument s is equivalent to a reﬂection and a cyclic permutation, and so leaves the par tial amplitude unchanged.) This mimics the notation we used at the end of sec tion 80, and we can apply our result from there to the color sum, /summationdisplay colors|T |2= 2N2(N2−1)g4/summationdisplay j|Aj|2+ 4N2g4(/summationdisplay jA∗ j)(/summationdisplay kAk),(81.41) wherejandkare summed over 2 ,3,4. In the present case, however, eq.(81.34) is equivalent to/summationtext jAj= 0, so the second term in eq.(81.41) vanishes. Our result for the color-summed squared amplitud e is then /summationdisplay colors|T |2= 2N2(N2−1)g4/parenleftig |A2|2+|A3|2+|A4|2/parenrightig . (81.42) 81: Scattering in Quantum Chromodynamics 489 41 2 34 51 25 3 Figure 81.2: Color-ordered diagrams for ¯ qqggscattering. 31 21 2 344 Figure 81.3: The double-line version of ﬁg.(81.2); the asso ciated color factor is (Ta3Ta4)i2i1. For the case where 1 and 2 are the incoming gluons, and 3 and 4 ar e the outgoing gluons, we can write this in terms of the usual Ma ndelstam variabless=s12=s34,t=s13=s24, andu=s14=s23by recalling that |∝an}b∇acketle{t12∝an}b∇acket∇i}ht|2=|[12]|2=|s12|, etc. Let us take the case where gluons 1 and 2 have negative helicity, and 3 and 4 have positive helicity. I n this case, we see from eqs.(81.28) and (81.37) that the numerator in every nonvanishing partial amplitude is ∝an}b∇acketle{t12∝an}b∇acket∇i}ht4. Then we get /summationdisplay colors|T |2 1−2−3+4+= 2N2(N2−1)g4s4/parenleftbigg1 s2t2+1 t2u2+1 u2s2/parenrightbigg .(81.43) We can also sum over helicities. There are six patterns of two positive and two negative helicities; −−++ and ++ −−yield a factor of s4,−+−+ and + −+−yieldt4, and−++−and + −−+ yieldu4. The helicity sum is therefore /summationdisplay colors helicities|T |2= 4N2(N2−1)g4(s4+t4+u4)/parenleftbigg1 s2t2+1 t2u2+1 u2s2/parenrightbigg .(81.44) Of course, we really want to average (rather than sum) over the initial colors and helicities; to do so we must divide eq.(81.44) by 4 (N2−1)2. Next we turn to scattering of quarks and gluons. We consider a single type of massless quark: a Dirac ﬁeld in the N representation o f SU(N). 81: Scattering in Quantum Chromodynamics 490 The lagrangian for this ﬁeld is L=iΨ /DΨ, where the covariant derivative isDµ=∂µ−(ig/√ 2)Aµ. Thus the color-ordered vertex factor is iVµ= (ig/√ 2)γµ. (81.45) To get the color factor, we use the double-line notation, wit h a single line for the quark. As an example, consider the process of ¯ qq→gg(and its crossing-related cousins). The contributing color-order ed tree diagrams are shown in ﬁg.(81.2). The corresponding double-line diagram s are shown in ﬁg.(81.3); the quark is represented by a single line, with an arrow direction that matches its charge arrow. To get the color factor, we sta rt with line 2, and follow the arrows backwards; the result is ( Ta3Ta4)i2i1. The complete amplitude can then be written as T=g2/bracketleftig (Ta3Ta4)i2i1A(1¯q,2q,3,4) + (Ta4Ta3)i2i1A(1¯q,2q,4,3)/bracketrightig ,(81.46) whereA(1¯q,2q,3,4) is the appropriate partial amplitude. The subscripts q and ¯qindicate the labels that correspond to an outgoing quark and outgoing antiquark, respectively. From our results for spinor electrodynamics in section 60, w e know that a nonzero amplitude requires opposite helicities on the two ends of any fermion line. Consider, then, the case of T−+λ3λ4. The partial amplitude corresponding to the diagrams of ﬁg.(81.2) is ig2A(1− ¯q,2+ q,3,4) = (ig/√ 2)2(1/i)[2|/ε3(−/p5/p2 5)/ε4|1∝an}b∇acket∇i}ht + (ig/√ 2)[2|/ε5|1∝an}b∇acket∇i}htiV345/vextendsingle/vextendsingle/vextendsingleεµ 5εν 5→igµν/s12.(81.47) Suppose both gluons have positive helicity. Then using /ε+(k;q) =√ 2 ∝an}b∇acketle{tqk∝an}b∇acket∇i}ht/parenleftig |k]∝an}b∇acketle{tq|+|q∝an}b∇acket∇i}ht[k|/parenrightig , (81.48) we can get both lines of eq.(81.47) to vanish by choosing q3=q4=p1. Similarly, if both gluons have negative helicity, then usin g /ε−(k;q) =√ 2 [qk]/parenleftig |k∝an}b∇acket∇i}ht[q|+|q]∝an}b∇acketle{tk|/parenrightig , (81.49) we can get both lines of eq.(81.47) to vanish by choosing q3=q4=p2. So the gluons must have opposite helicities to get a nonzero amp litude. Consider, then, the case of λ3= + andλ4=−. We can get V345to vanish by choosing q3=k4andq4=k3. The partial amplitude is then given by just the ﬁrst line of eq.(81.47), A(1− ¯q,2+ q,3+,4−) =1 2[2|/ε3+(/p1+/k4)/ε4−|1∝an}b∇acket∇i}ht/(−s14). (81.50) 81: Scattering in Quantum Chromodynamics 491 Withq3=k4andq4=k3, we have /ε3+=√ 2 ∝an}b∇acketle{t43∝an}b∇acket∇i}ht/parenleftig |4∝an}b∇acket∇i}ht[3|+|3]∝an}b∇acketle{t4|/parenrightig , (81.51) /ε4−=√ 2 [34]/parenleftig |4∝an}b∇acket∇i}ht[3|+|3]∝an}b∇acketle{t4|/parenrightig . (81.52) Using the identity /p=− |p∝an}b∇acket∇i}ht[p| − |p]∝an}b∇acketle{tp|, (81.53) eq.(81.50) becomes A(1− ¯q,2+ q,3+,4−) =[23]∝an}b∇acketle{t41∝an}b∇acket∇i}ht[13]∝an}b∇acketle{t41∝an}b∇acket∇i}ht ∝an}b∇acketle{t43∝an}b∇acket∇i}ht[34]s14. (81.54) In the numerator, we use [13] ∝an}b∇acketle{t41∝an}b∇acket∇i}ht=−[23]∝an}b∇acketle{t42∝an}b∇acket∇i}ht. In the denominator, we sets14=s23=∝an}b∇acketle{t23∝an}b∇acket∇i}ht[32]. Then we multiply the numerator and denomina- tor by ∝an}b∇acketle{t12∝an}b∇acket∇i}ht, and use [23] ∝an}b∇acketle{t12∝an}b∇acket∇i}ht=−[43]∝an}b∇acketle{t14∝an}b∇acket∇i}htin the numerator. Finally we multiply both by ∝an}b∇acketle{t14∝an}b∇acket∇i}ht, and rearrange to get A(1− ¯q,2+ q,3+,4−) =∝an}b∇acketle{t14∝an}b∇acket∇i}ht3∝an}b∇acketle{t24∝an}b∇acket∇i}ht ∝an}b∇acketle{t12∝an}b∇acket∇i}ht∝an}b∇acketle{t23∝an}b∇acket∇i}ht∝an}b∇acketle{t34∝an}b∇acket∇i}ht∝an}b∇acketle{t41∝an}b∇acket∇i}ht. (81.55) An analogous calculation yields A(1− ¯q,2+ q,3−,4+) =∝an}b∇acketle{t13∝an}b∇acket∇i}ht3∝an}b∇acketle{t23∝an}b∇acket∇i}ht ∝an}b∇acketle{t12∝an}b∇acket∇i}ht∝an}b∇acketle{t23∝an}b∇acket∇i}ht∝an}b∇acketle{t34∝an}b∇acket∇i}ht∝an}b∇acketle{t41∝an}b∇acket∇i}ht. (81.56) The remaining nonzero amplitudes are related by complex con jugation. Now that we have all the partial amplitudes, we can compute th e color- summed |T |2. To do so, we multiply eq.(81.46) by its complex conjugate, and use hermiticity of the generator matrices to get /summationdisplay colors|T |2=g4/bracketleftig Tr(TaTbTbTa)/parenleftig |A3|2+|A4|2/parenrightig + Tr(TaTbTaTb)/parenleftig A∗ 3A4+A∗ 4A3/parenrightig/bracketrightig , (81.57) whereA3≡A(1¯q,2q,3,4) andA4≡A(1¯q,2q,4,3). The traces are easily evaluated with the double-line technique of section 80; bec ause the ﬁcti- tious photon couples to the quark, we must use eq.(81.30) to p roject it out. The traces in eq.(81.57) are also easily evaluated with the group- theoretic methods of section 70, with the normalization tha t the index of the fundamental representation is one: T(N) = 1. Either way, the results are Tr(TaTbTbTa) = +(N2−1)2/N , (81.58) Tr(TaTbTaTb) =−(N2−1)/N . (81.59) 81: Scattering in Quantum Chromodynamics 492 The sum over the four possible helicity patterns ( −++−,−+−+, +−−+, +−+−) is left as an exercise. Now that we have calculated these scattering amplitudes for quarks and gluons, an important questions arises: why did we bother to do it? Quarks and gluons are conﬁned inside colorless bound states , thehadrons , and so apparently cannot appear as incoming and outgoing par ticles in a scattering event. To answer this question, suppose we collide two hadrons with a center- of-mass energy E=√slarge enough so that the QCD coupling gis small when renormalized in the MS scheme with µ=E. (In the real world, we haveα≡g2/4π= 0.12 forµ=MZ= 91GeV.) Then we can think of each hadron as being made up of a loose collection of quarks an d gluons, and these parts of a hadron, or partons , can be treated as independent participants in scattering processes. In order to extract q uantitative results for hadron scattering (a project beyond the scope of this boo k), we need to know how each hadron’s energy and momentum is shared among its partons. This is described by parton distribution functions . At present, these cannot be calculated from ﬁrst principles, but they ha ve to satisfy a variety of consistency conditions that canbe derived from perturbation theory, and that relate their values at diﬀerent energies. T hese conditions are well satisﬁed by current experimental data. Reference Notes More detail on how hadron scattering experiments can be comp ared with parton scaterring amplitudes can be found in Peskin & Schroeder ,Muta, Quigg , andSterman . Problems 81.1) Compute the four-gluon partial amplitude A(1−,2+,3−,4+) directly from the Feynman diagrams, and verify eq.(81.37). 81.2) Compute the ¯ qqggpartial amplitude A(1− ¯q,2+ q,3−,4+) withq4=p1 andq3=k4. Show that, with this choice of the reference momenta, the ﬁrst line of eq.(81.47) vanishes. Evaluate the second li ne, and veriﬁy eq.(81.55). 81.3) Compute A(1− ¯q,2+ q,3−,4+), and verify eq.(81.56). 81.4) a) Verify eqs.(81.58) and (81.59) using the double-li ne notation of section 80. b) Compute Tr( Ta RTb RTb RTa R) and Tr(Ta RTb RTa RTb R) in terms of the index T(R) and dimension D(R) of the representation R, and the index 81: Scattering in Quantum Chromodynamics 493 T(A) and dimension D(A) of the adjoint representation. Verify that your results reproduce eqs.(81.58) and (81.59). 81.5) Compute the sum over helicities of eq.(81.57). Expres s your answer in terms of s,t, andufor the process ¯ qq→gg. 81.6) Consider the partial amplitude A(1− ¯q,2+ q,3−,4+,5+). Show that, with the choiceq3=k2andq4=q5=k1, there are just two contributing diagrams. Evaluate them. After some manipulations, you sho uld be able to put your result in the form A(1− ¯q,2+ q,3−,4+,5+) =∝an}b∇acketle{t13∝an}b∇acket∇i}ht3∝an}b∇acketle{t23∝an}b∇acket∇i}ht ∝an}b∇acketle{t12∝an}b∇acket∇i}ht∝an}b∇acketle{t23∝an}b∇acket∇i}ht∝an}b∇acketle{t34∝an}b∇acket∇i}ht∝an}b∇acketle{t45∝an}b∇acket∇i}ht∝an}b∇acketle{t5 1∝an}b∇acket∇i}ht. (81.60) 82: Wilson Loops, Lattice Theory, and Conﬁnement 494 82Wilson Loops, Lattice Theory, and Confinement Prerequisite: 29, 73 In this section, we will contruct a gauge-invariant operato r, theWilson loop , whose vacuum expectation value (VEV for short) can diagnose whether or not a gauge theory exhibits conﬁnement . A theory is conﬁning if all ﬁnite-energy states are invariant under a global gauge tran sformation. U(1) gauge theory—quantum electrodynamics—is notconﬁning, because there are ﬁnite-energy states (such as the state of a single electr on) that have nonzero electric charge, and hence change by a phase under a g lobal gauge transformation. Conﬁnement is a nonperturbative phenomenon; it cannot be seen at any ﬁnite order in the kind of weak-coupling perturbation theor y that we have been doing. (This is why we had no trouble calculating quark a nd gluon scattering amplitudes.) In this section, we will introduce lattice gauge the- ory, in which spacetime is replaced by a discrete set of points; t he inverse lattice spacing 1 /athen acts as an ultraviolet cutoﬀ (see section 29). This cutoﬀ theory can be analyzed at strong coupling, and, as we will see, in this regime the VEV of the Wilson loop is indicative of conﬁne ment. The outstanding question is whether this phenomenon persists a s we simul- taneously lower the coupling and increase the ultraviolet c utoﬀ (with the relationship between the two governed by the beta function) , or whether we encounter a phase transition , signalled by a sudden change in the behavior of the Wilson loop VEV. We take the gauge group to be SU( N). Consider two spacetime points xµandxµ+εµ, whereεµis inﬁnitesimal. Deﬁne the Wilson link W(x+ε,x)≡exp[igεµAµ(x)], (82.1) whereAµ(x) is anN×Nmatrix-valued traceless hermitian gauge ﬁeld. Sinceεis inﬁnitesimal, we also have W(x+ε,x) =I+igεµAµ(x) +O(ε2). (82.2) Let us determine the behavior of the Wilson link under a gauge trans- formation. Using the gauge transformation of Aµ(x) from section 69, we ﬁnd W(x+ε,x)→1 +igεµU(x)Aµ(x)U†(x)−εµU(x)∂µU†(x),(82.3) whereU(x) is a spacetime-dependent special unitary matrix. Since UU†= 1, we have −U∂µU†= +(∂µU)U†; thus we can rewrite eq.(82.3) as W(x+ε,x)→/parenleftig (1 +εµ∂µ)U(x)/parenrightig U†(x) +igεµU(x)Aµ(x)U†(x).(82.4) 82: Wilson Loops, Lattice Theory, and Conﬁnement 495 In the ﬁrst term, we can use (1 + εµ∂µ)U(x) =U(x+ε) +O(ε2). In the second term, which already contains an explicit factor of εµ, we can replace U(x) withU(x+ε) at the cost of an O(ε2) error. Then we get W(x+ε,x)→U(x+ε)/parenleftig 1 +igεµAµ(x)/parenrightig U†(x), (82.5) which is equivalent to W(x+ε,x)→U(x+ε)W(x+ε,x)U†(x). (82.6) Note also that eq.(82.1) implies W†(x+ε,x) =W(x−ε,x). We can shiftxtox+εat the cost of an O(ε2) error, and so W†(x+ε,x) =W(x,x+ε), (82.7) which is consistent with eq.(82.6). Now consider mutiplying together a string of Wilson links, s peciﬁed by a starting point xandnsequential inﬁnitesimal displacement vectors εj. The ordered set of ε’s deﬁnes a pathPthrough spacetime that starts at x and ends at y=x+ε1+...+εn. The Wilson line for this path is WP(y,x)≡W(y,y−εn)...W(x+ε1+ε2,x+ε1)W(x+ε1,x).(82.8) Using eq.(82.6) and the unitarity of U(x), we see that, under a gauge transformation, the Wilson line transforms as WP(y,x)→U(y)WP(y,x)U†(x). (82.9) Also, since hermitian conjugation reverses the order of the product in eq.(82.8), using eq.(82.7) yields W† P(y,x) =W−P(x,y), (82.10) where −Pdenotes the reverse of the path P. Now consider a path that returns to its starting point, formi ng a closed, oriented curve Cin spacetime. The Wilson loop is the trace of the Wilson line for this path, WC≡TrWC(x,x). (82.11) Using eq.(82.9), we see that the Wilson loop is gauge invariant , WC→WC. (82.12) Also, eq.(82.10) implies W† C=W−C, (82.13) 82: Wilson Loops, Lattice Theory, and Conﬁnement 496 where −Cdenotes the curve Ctraversed in the opposite direction. To gain some intuition, we will calculate ∝an}b∇acketle{t0|WC|0∝an}b∇acket∇i}htfor U(1) gauge theory, without charged ﬁelds. This is simply a free-ﬁeld theory, an d the calculation can be done exactly. In order to avoid dealing with iǫissues, it is convenient to make a Wick rotation to euclidean spacetime (see section 29). The actio n is then S=/integraldisplay d4x1 4FµνFµν, (82.14) whereFµν=∂µAν−∂νAµ. The VEV of the Wilson loop is now given by the path integral ∝an}b∇acketle{t0|WC|0∝an}b∇acket∇i}ht=/integraldisplay DAeig/contintegraltext CdxµAµe−S. (82.15) If we formally identify g/contintegraltext Cdxµas a current Jµ(x), we can apply our results for the path integral from section 57. After including a fact or ofifrom the Wick rotation, we get ∝an}b∇acketle{t0|WC|0∝an}b∇acket∇i}ht= exp/bracketleftbigg −1 2g2/contintegraldisplay Cdxµ/contintegraldisplay Cdyν∆µν(x−y)/bracketrightbigg , (82.16) where ∆µν(x−y) is the photon propagator in euclidean spacetime. In Feyn- man gauge, we have ∆µν(x−y) =δµν/integraldisplayd4k (2π)4eik·(x−y) k2 =δµν4π (2π)4/integraldisplay∞ 0k3dk k2/integraldisplayπ 0dθsin2θeik|x−y|cosθ =δµν4π (2π)4/integraldisplay∞ 0k3dk k2πJ1(k|x−y|) k|x−y| =δµν 4π2(x−y)2/integraldisplay∞ 0duJ1(u) =δµν 4π2(x−y)2, (82.17) whereJ1(u) is a Bessel function. Since ∆ µν(x−y) depends only on x−y, the double line integral in eq.(82.16) will yield a factor of the perimeter P of the curve C. There is also an ultraviolet divergence as xapproaches y; we will cut this oﬀ at a length scale a. The result is then ∝an}b∇acketle{t0|WC|0∝an}b∇acket∇i}ht= exp[ −(˜cg2/a)P], (82.18) 82: Wilson Loops, Lattice Theory, and Conﬁnement 497 where ˜cis a numerical constant that depends on the shape of Cand the de- tails of the cutoﬀ procedure. This behavior of the Wilson loo p in euclidean spacetime—exponential decay with the length of the perimet er—is called theperimeter law . It is indicative of unconﬁned charges. We can gain more insight into the meaning of ∝an}b∇acketle{t0|WC|0∝an}b∇acket∇i}htby takingC to be a rectangle, with length Tin the time direction and Rin a space direction, where a≪R≪T. (Of course, in euclidean spacetime, the choice of the time direction is an arbitrary convention.) Th e reason for this particular shape is that the current g/contintegraltext Cdxµcorresponds to a point charge moving along the curve C. When the particle is moving backwards in time, the associated minus sign is equivalent to a change i n the sign of its charge. So when we compute ∝an}b∇acketle{t0|WC|0∝an}b∇acket∇i}ht, we are doing the path integral in the presence of a pair of point charges with opposite sign, separated by a distanceR, that exists for a time T. On general principles (see section 6), this path integral is proportional to exp( −EpairT), whereEpairis the ground-state energy of the quantum electrodynamic ﬁeld in t he presence of the charged particle pair. We now turn to the calculation. If both xandyare on the same side of the rectangle, we ﬁnd /integraldisplayL 0/integraldisplayL 0dxdy (x−y)2= 2L/a−2ln(L/a) +O(1), (82.19) whereLis the length of the side (either RorT), and theO(1) term is a numerical constant that depends on the details of the short- distance cutoﬀ. Ifxandyare on perpendicular sides, the double line integral is zero , because then dx·dy= 0. Ifxis on one short side and yon the other, the integral evaluates to R2/T2, and this we can neglect. Finally, if xis on one long side and yis on the other, we have /integraldisplayT 0/integraldisplayT 0dxdy (x−y)2+R2=πT/R −2ln(T/R)−2 +O(R2/T2).(82.20) Adding up all these contributions, we ﬁnd in the limit of larg eTthat /contintegraldisplay C/contintegraldisplay Cdx·dy (x−y)2=/parenleftig 4/a−2π/R/parenrightig T+O(lnT). (82.21) Combining this with eqs.(82.16) and (82.17), and setting α=g2/4π, we ﬁnd ∝an}b∇acketle{t0|WC|0∝an}b∇acket∇i}ht= exp/bracketleftbigg −/parenleftbigg2α πa−α R/parenrightbigg T/bracketrightbigg . (82.22) Comparing this with the general expectation ∝an}b∇acketle{t0|WC|0∝an}b∇acket∇i}ht ∝exp(−EpairT), we ﬁnd a cutoﬀ dependent contribution to Epairthat represents a divergent 82: Wilson Loops, Lattice Theory, and Conﬁnement 498 2 x 1 Figure 82.1: The minimal Wilson loop on a hypercubic lattice goes around an elementary plaquette; this one lies in the 1-2 plane. self-energy for each point particle, plus the Coulomb poten tial energy for the pair,V(R) =−α/R. In the nonabelian case, where there are interactions among t he gluons, we must expand everything in powers of g. Then we ﬁnd ∝an}b∇acketle{t0|WC|0∝an}b∇acket∇i}ht= Tr/bracketleftbigg 1−1 2g2TaTa/contintegraldisplay Cdxµ/contintegraldisplay Cdyν∆µν(x−y) +O(g4)/bracketrightbigg .(82.23) SinceTaTaequals the quadratic Casimir C(N) for the fundamental repre- sentation (times an identity matrix), we see that to leading order ing2we simply reproduce the results of the abelian case, but with g2→g2C(N). We can also consider a Wilson loop in a diﬀerent representati on by setting Aµ(x) =Aa µ(x)Ta R. Then, at leading order, we get a factor of C(R) instead ofC(N). Perturbative corrections can be computed via standard Feynman diagrams with gluon lines that, in position space, have at le ast one end on the curveC. Next we turn to a strong coupling analysis. We begin by constr ucting alattice action for nonabelian gauge theory. Consider a hypercubic lattice of points in four-dimensional euclidean spacetime, with a lattice spacing a between nearest-neighbor points. The smallest Wilson loop we can make on this lattice goes around an elementary square or plaquette , as shown in ﬁg.(82.1). Let ε1andε2be vectors of length ain the 1 and 2 directions, and letxbe the point at the center of the plaquette. Using the center o f each link as the argument of the gauge ﬁeld, and using the lowe r-left corner as the starting point, we have (multiplying the Wilson links from right to left along the path) Wplaq= Tre−igaA2(x−ε1/2)e−igaA1(x+ε2/2)e+igaA2(x+ε1/2)e+igaA1(x−ε2/2). (82.24) If we now treat the gauge ﬁeld as smooth and expand in a, we get Wplaq= Tre−igaA2(x)+iga2∂1A2(x)/2+...e−igaA1(x)−iga2∂2A1(x)/2+... ×e+igaA2(x)+iga2∂1A2(x)/2+...e+igaA1(x)−iga2∂2A1(x)/2+....(82.25) 82: Wilson Loops, Lattice Theory, and Conﬁnement 499 Next we use eAeB=eA+B+[A,B]/2+...to combine the two exponential factors on the ﬁrst line of eq.(82.25), and also the two exponential f actors on the second line. Then we use this formula once again to combine th e two results. We get Wplaq= Tre+iga2(∂1A2−∂2A1−ig[A1,A2])+..., (82.26) where all ﬁelds are evaluated at x. If we now take Wplaq+W−plaqand expand the exponentials, we ﬁnd Wplaq+W−plaq= 2N−g2a4TrF2 12+... , (82.27) whereF12=∂1A2−∂2A1−ig[A1,A2] is the Yang–Mills ﬁeld strength. From eq.(82.27), we conclude that an appropriate action for Yang –Mills theory on a euclidean spacetime lattice is S=−1 2g2/summationdisplay plaqWplaq, (82.28) where the sum includes both orientations of all plaquettes. EachWplaqis expressed as the trace of the product of four special unitary N×Nmatrices, one for each oriented link in the plaquette. If Uis the matrix associated with one orientation of a particular link, then U†is the matrix associated with the opposite orientation of that link. The path integra l for this lattice gauge theory is Z=/integraldisplay DU e−S, (82.29) where DU=/productdisplay linksdUlink, (82.30) anddUis theHaar measure for a special unitary matrix. The Haar measure is invariant under the transformation U→VU, whereVis a constant spe- cial unitary matrix, and is normalized via/integraltextdU= 1; this ﬁxes it uniquely. ForN≥3, it obeys /integraldisplay dU Uij= 0, (82.31) /integraldisplay dU UijUkl= 0, (82.32) /integraldisplay dU UijU∗ kl=1 Nδikδjl, (82.33) which is all we will need to know. Now consider a Wilson loop, expressed as the trace of the prod uct of theU’s associated with the oriented links that form a closed curv eC. For simplicity, we take this curve to lie in a plane. We have ∝an}b∇acketle{t0|WC|0∝an}b∇acket∇i}ht=Z−1/integraldisplay DU WCe−S. (82.34) 82: Wilson Loops, Lattice Theory, and Conﬁnement 500 We will evaluate eq.(82.34) in the strong coupling limit by expanding e−S in powers of 1 /g2. At zeroth order, e−S→1; then eq.(82.31) tells us that the integral over every link in Cvanishes. To get a nonzero result, we need to have a corresponding U†from the expansion of e−S. This can only come from a plaquette containing that link. But then the integral over the other links of this plaquette will vanish, unless there is a compen satingU†for each of them. We conclude that a nonzero result for ∝an}b∇acketle{t0|WC|0∝an}b∇acket∇i}htrequires us to ﬁll the interior of Cwith plaquettes from the expansion of e−S. Since each plaquette is accompanied by a factor of 1 /g2, we have ∝an}b∇acketle{t0|WC|0∝an}b∇acket∇i}ht ∼(1/g2)A/a2, (82.35) whereAis the areaof the surface bounded by C, andA/a2is the number of plaquettes in this surface. Eq.(82.35) yields the area law for a Wilson loop, ∝an}b∇acketle{t0|WC|0∝an}b∇acket∇i}ht ∝e−τA, (82.36) where τ=c(g)/a2(82.37) is the string tension . In the strong coupling limit, c(g) = ln(g2) +O(1). The area law for the Wilson loop implies conﬁnement. To see wh y, let us again consider a rectangular loop with area A=RT. Comparing eq.(82.36) with the general expectation ∝an}b∇acketle{t0|WC|0∝an}b∇acket∇i}ht ∝exp(−EpairT), we see thatEpair=V(R) =τR. This corresponds to a linear potential between nonabelian point charges in the fundamental representatio n. It takes an inﬁnite amount of energy to separate these charges by an inﬁn ite distance; the charges are therefore conﬁned . The coeﬃcient τofRinV(R) is called the string tension because a linear potential is what we get f rom two points joined by a string with a ﬁxed energy per unit length; the ener gy per unit length of a string is its tension. The string tension τis a physical quantity that should remain ﬁxed as we remove the cutoﬀ by lowering a. Thus lowering arequires us to lowerg. The outstanding question is whether c(g) reaches zero at a ﬁnite, nonzero value of g. If so, at this point there is a phase transition to an unconﬁned phase with zero string tension. This has been prov en to be the case for abelian gauge theory (which also exhibits an area la w at strong coupling, by the identical argument). In nonabelian gauge t heory, on the other hand, analytic and numerical evidence strongly sugge sts thatc(g) remains nonzero for all nonzero values of g. At smallaand smallg, the behavior of gas a function of ais governed by the beta function, β(g) =−adg/da . (The minus sign arises because the ultraviolet cutoﬀ is a−1.) Requiring τto be independent of ayields c′(g) =−1 2β(g)c(g). (82.38) 82: Wilson Loops, Lattice Theory, and Conﬁnement 501 At smallg, we have β(g) =−b1g3+O(g5), (82.39) whereb1= 11N/48π2for SU(N) gauge theory without quarks. Solving for c(g) yields c(g) =Cexp(1/b1g2), (82.40) whereCis an integration constant, which is nonzero if there is no ph ase transition. In this case, at small g, the string tension has the form τ=Cexp(1/b1g2)/a2. (82.41) We then want to take the continuum limit ofa→0 andg→0 withτheld ﬁxed. Note that eq.(82.41) shows that the string tension, at weak c oupling, is not analytic in g, and so cannot be computed via the Taylor expansion ingthat is provided by conventional weak-coupling perturbati on theory. Instead, the path integrals of eqs.(82.29) and (82.34) can b e performed on a ﬁnite-size lattice via numerical integration. The limiti ng factor in such a calculation is computer resources. Reference Notes An introduction to lattice theory is given in Smit. Problems 82.1) LetCbe a circle of radius R. Evaluate the constant ˜ cin eq.(82.18), whereP= 2πRis the circumference of the circle. Replace 1 /(x−y)2 with zero when |x−y|<a. Assumea≪R. 83: Chiral Symmetry Breaking 502 83Chiral Symmetry Breaking Prerequisite: 76, 82 In the previous section, we discussed conﬁnement in Yang–Mi lls theory without quarks. In the real world, there are six diﬀerent ﬂavors of quark; see Table 1. Each ﬂavor has a diﬀerent mass, and is represente d by a Dirac ﬁeld in the fundamental or 3 representation of the color grou p SU(3). Such a Dirac ﬁeld is equivalent to two left-handed Weyl ﬁelds, one in the funda- mental representation, and one in the antifundamental or 3 representation. The lightest quarks are the up and down quarks, with masses of a few MeV. These masses are small, in the following sense. The gaug e coupling gof QCD becomes large at low energies. If we truncate the beta f unction after some number of terms (in practice, four or fewer), and i ntegrate it, we ﬁnd that gbecomes inﬁnite at some ﬁnite, nonzero value of the MS parameterµ; this value is called Λ QCD. Measurements of the strength of the gauge coupling at high energies imply Λ QCD∼0.2GeV. The up and down quark masses are much less than Λ QCD. We can therefore begin with the approximation that the up and down quarks are massless. T he mass of the strange quark is also somewhat less than Λ QCD. It is sometimes useful (though clearly less justiﬁed) to treat the strange quark as massless as well. If we are interested in hadron physics at energies below ∼1GeV, we can ignore the charm, bottom, and top quarks entirely; we will al so ignore the strange quark for now. Let us, then, consider QCD with nF= 2 ﬂavors of massless quarks. We then have left-handed Weyl ﬁelds χαi, whereα= 1,2,3 is a color index for the 3 representation, and i= 1,2 is a ﬂavor index, and left-handed Weyl ﬁelds ξα¯ı, whereα= 1,2,3 is a color index for the 3 representation, and ¯ ı= 1,2 is a ﬂavor index; we distinguish this ﬂavor index from the one for the χ’s by putting a bar over it, and we write it as a superscript for later notational convenience. We suppres s the undotted spinor index carried by both χandξ. The lagrangian is L=iχ†αi¯σµ(Dµ)αβχβi+iξ† ¯ıα¯σµ(¯Dµ)αβξβ¯ı−1 4FaµνFa µν, (83.1) whereDµ=∂µ−igTa 3Aa µand¯Dµ=∂µ−igTa 3Aa µ, with (Ta 3)αβ=−(Ta 3)βα. In addition to the SU(3) color gauge symmetry, this lagrangi an has a global U(2)×U(2) ﬂavor symmetry: Lis invariant under χαi→Lijχαj, (83.2) ξα¯ı→(R∗)¯ı¯ξα¯, (83.3) whereLandR∗are independent 2 ×2 constant unitary matrices. (The complex conjugation of Ris a notational convention that turns out to be 83: Chiral Symmetry Breaking 503 name symbol mass Q (GeV) upu 0.0017 +2 /3 down d 0.0039 −1/3 strange s 0.076 −1/3 charm c 1.3 +2/3 bottom b 4.3 −1/3 topt 178 +2/3 Table 1: The six ﬂavors of quark. Each ﬂavor is represented by a Dirac ﬁeld in the 3 representation of the color group SU(3). Qis the electric charge in units of the proton charge. Masses are approximate , and are MS parameters. For the u,d, andsquarks, the MS scaleµis taken to be 2GeV. For thec,b, andtquarks,µis taken to be equal to the corresponding mass; e.g., the bottom quark mass is 4.3GeV when µ= 4.3GeV. convenient.) In terms of the Dirac ﬁeld Ψαi=/parenleftiggχαi ξ† α¯ı/parenrightigg , (83.4) eqs.(83.2) and (83.3) read PLΨαi→LijPLΨαi, (83.5) PRΨα¯ı→R¯ı¯PRΨα¯, (83.6) wherePL,R=1 2(1∓γ5). Thus the global ﬂavor symmetry is often called U(2) L×U(2) R. A symmetry that treats the left- and right-handed parts of a Dirac ﬁeld diﬀerently is said to be chiral. However, there is an anomaly in the axial U(1) symmetry corre sponding toL=R∗=eiαI(which is equivalent to Ψ →e−iαγ5Ψ for the Dirac ﬁeld). Thus the nonanomalous global ﬂavor symmetry is SU(2) L×SU(2) R×U(1) V, where V stands for vector. The U(1) Vtransformation corresponds to L=R=e−iαI, or equivalently Ψ →e−iαΨ. The corresponding con- served charge is quark number , the number of quarks minus the number of antiquarks; this is one third of the baryon number , the number of baryons minus the number of antibaryons. ( Baryons are color-singlet bound states of three quarks; the proton and neuton are baryons. Mesons are color- singlet bound states of a quark and an antiquark; pions are me sons.) Thus, U(1) Vresults in classiﬁcation of hadrons by their baryon num- ber. How is the SU(2) L×SU(2) Rsymmetry realized in nature? The vector 83: Chiral Symmetry Breaking 504 subgroup SU(2) V, obtained by setting R=Lin eq.(83.3), is known as iso- topic spin orisospin symmetry. Hadrons clearly come in representations of SU(2) V: the lightest spin-one-half hadrons (the proton, mass 0 .938GeV, and the neutron, mass 0.940GeV) form a doublet or 2 represent ation, while the lightest spin-zero hadrons (the π0, mass 0.135GeV, and the π±, mass 0.140GeV) form a triplet or 3 representation. Isospin is not an exact sym- metry; it is violated by the small mass diﬀerence between the up and down quarks, and by electromagnetism. Thus we see small diﬀerenc es in the masses of the hadrons assigned to a particular isotopic mult iplet. The role of the axial part of the SU(2) L×SU(2) Rsymmetry, obtained by settingR=L†in eq.(83.3), is harder to identify. The hadrons do not appear to be classiﬁed into multiplets by a second SU(2) symm etry group. In particular, there is no evidence for a classiﬁcation that distinguishes the left- and right-handed components of spin-one-half hadron s like the proton and neutron. Reconciliation of these observations with the SU(2) L×SU(2) Rsymmetry of the underlying lagrangian is only possible if the axial ge nerators are spontaneously broken . The three pions (which have spin zero, odd parity, and are by far the lightest hadrons) are then identiﬁed as the corresponding Goldstone bosons. They are not exactly massless (and hence a re sometimes called pseudogoldstone bosons ) because the SU(2) L×SU(2) Rsymmetry is, as we just discussed, not exact. To spontaneously break the axial part of the SU(2) L×SU(2) R, some op- erator that transforms nontrivially under it must acquire a nonzero vacuum expectation value, or VEV for short. To avoid spontaneous br eakdown of Lorentz invariance, this operator must be a Lorentz scalar, and to avoid spontaneous breakdown of the SU(3) gauge symmetry, it must b e a color singlet. Since we have no fundamental scalar ﬁelds that coul d acquire a nonzero VEV, we must turn to composite ﬁelds instead. The simplest can- didate isχa αiξα¯ a=Ψα¯PLΨαi, whereais an undotted spinor index. (The product of two ﬁelds is generically singular, and a renormal ization scheme must be speciﬁed to deﬁne it.) We assume that ∝an}b∇acketle{t0|χa αiξα¯ a|0∝an}b∇acket∇i}ht=−v3δi¯, (83.7) wherevis a parameter with dimensions of mass. Its numerical value depends on the renormalization scheme; for MS withµ= 2GeV,v≃ 0.23GeV. To see that this fermion condensate does the job of breaking the axial generators of SU(2) L×SU(2) Rwhile preserving the vector generators, we note that, under the transformation of eqs.(83.2) and (83.3 ), ∝an}b∇acketle{t0|χa αiξα¯ a|0∝an}b∇acket∇i}ht →Lik(R∗)¯¯n∝an}b∇acketle{t0|χa αkξα¯n a|0∝an}b∇acket∇i}ht 83: Chiral Symmetry Breaking 505 → −v3(LR†)i¯, (83.8) where we used eq.(83.7) to get the second line. If we take R=L, cor- responding to an SU(2) Vtransformation, the right-hand side of eq.(83.8) is unchanged from its value in eq.(83.7). This signiﬁes that SU(2) V[and also U(1) V] is unbroken. However, for a more general transformation wi th R∝ne}ationslash=L, the right-hand side of eq.(83.8) does not match that of eq.( 83.7), signifying the spontaneous breakdown of the axial generato rs. Eq.(83.7) is nonperturbative :∝an}b∇acketle{t0|χa αiξα¯ a|0∝an}b∇acket∇i}htvanishes at tree level. Pertur- bative corrections then also vanish, because of the chiral ﬂ avor symmetry of the lagrangian. Thus the value of vis not accessible in perturbation the- ory. On general grounds, we expect v∼ΛQCD, since Λ QCDis the only mass scale in the theory when the quarks are massless. Similarly, ΛQCDsets the scale for the masses of all the hadrons that are not pseudogol dstone bosons, including the proton and neutron. We can construct a low-energy eﬀective lagrangian for the th ree pseu- dogoldstone bosons (to be identiﬁed as the pions) in the foll owing way. We allow the orientation in ﬂavor space of the VEV of χa αiξα¯ ato vary slowly as a function of spacetime. That is, in place of eq.(83.7), we wr ite ∝an}b∇acketle{t0|χa αi(x)ξα¯ a(x)|0∝an}b∇acket∇i}ht=−v3Ui¯(x), (83.9) whereU(x) is a spacetime dependent unitary matrix. We can write it as U(x) = exp[2iπa(x)Ta/fπ], (83.10) whereTa=1 2σawitha= 1,2,3 are the generator matrices of SU(2), πa(x) are three real scalar ﬁelds to be identiﬁed with the pions, a ndfπis a parameter with dimensions of mass, the pion decay constant . We do not include a fourth generator matrix proportional to the ident ity, since the corresponding ﬁeld would be the Goldstone boson for the U(1) Asymmetry that is eliminated by the anomaly. Equivalently, we require detU(x) = 1. We will think of U(x) as an eﬀective, low energy ﬁeld. Its lagrangian should be the most general one that is consistent with the und erlying SU(2) L×SU(2) Rsymmetry.1Under a general SU(2) L×SU(2) Rtrans- formation, we have U(x)→LU(x)R†, (83.11) whereLandRare independent special unitary matrices. We can organize the terms in the eﬀective lagrangian for U(x) (also known as the chiral lagrangian ) by the number of derivatives they contain. Because U†U= 1, 1U(1) Vacts trivially on U(x), and so we need not be concerned with it. 83: Chiral Symmetry Breaking 506 there are no terms with no derivatives. There is one term with two (all others being equivalent after integrations by parts), L=−1 4f2 πTr∂µU†∂µU . (83.12) If we substitute in eq.(83.10) for U, and expand in inverse powers of fπ, we ﬁnd L=−1 2∂µπa∂µπa+1 6f−2 π(πaπa∂µπb∂µπb−πaπb∂µπb∂µπa) +... .(83.13) Thus the pion ﬁelds are conventionally normalized, and they have interac- tions that are dictated by the general form of eq.(83.12). Th ese interactions lead to Feynman vertices that contain factors of momenta pdivided byfπ. Therefore, we can think of p/fπas an expansion parameter. Of course, we should also add to Lall possible inequivalent terms with four or more derivatives, with coeﬃcients that include inverse powers o ffπ. These will lead to more vertices, but their eﬀects will be suppressed by additional powers ofp/fπ. Comparison with experiment then yields fπ= 92.4MeV. (In practice, the value of fπis more readily determined from the decay rate of the pion via the weak interaction; see section 90 and probl em 48.5.) This value for fπmay seem low; it is, for example, less than the mass of the “almost masless” pions. However, it turns out that tre e and loop diagrams contribute roughly equally to any particular proc ess if each extra derivative in Lis accompanied by a factor of (4 πfπ)−1rather that f−1 π, and each loop momentum is cut oﬀ at 4 πfπ. Thus it is 4 πfπ∼1GeV that sets the scale of the interactions, rather than fπ∼100MeV. Now let us consider the eﬀect of including the small masses fo r the up and down quarks. The most general mass term we can add to the la grangian is Lmass=−ξα¯M¯iχαi+ h.c. =−M¯iχαiξα¯+ h.c. =−TrMχαξα+ h.c., (83.14) whereMis a complex 2 ×2 matrix. By making an SU(2) L×SU(2) R transformation, we can bring Mto the form M=/parenleftiggmu0 0md/parenrightigg e−iθ/2, (83.15) wheremuandmdare real and positive. We cannot remove the overall phaseθ, however, without making a forbidden U(1) Atransformation. A 83: Chiral Symmetry Breaking 507 nonzero value of θhas physical consequences, as we will discuss in section 94. For now, we note that experimental observations ﬁx |θ|<10−9, and so we will set θ= 0 on this phenomenological basis. Next, we replace χαξαin eq.(83.14) with its spacetime dependent VEV, eq.(83.7). The result is a term in the chiral lagrangian that incorporates the leading eﬀect of the quark masses, Lmass=v3Tr(MU+M†U†). (83.16) Here we continue to distinguish MandM†, even though, with θ= 0, they are the same matrix. If we think of Mas transforming as M→RML†, whileUtransforms as U→LUR†, then TrMUis formally invariant. We then require all terms in the chiral lagrangian to exhibit th is formal invari- ance. If we expand Lmassin inverse powers of fπ, and useM†=M, we ﬁnd Lmass=−4(v3/f2 π)Tr(MTaTb)πaπb+... =−2(v3/f2 π)Tr(M{Ta,Tb})πaπb+... =−(v3/f2 π)(TrM)πaπa+... . (83.17) We used the SU(2) relation {Ta,Tb}=1 2δabto get the last line. From eq.(83.17), we see that all three pions have the same mass, gi ven by the Gell-Mann–Oakes–Renner relation , m2 π= 2(mu+md)v3/f2 π. (83.18) On the right-hand side, the quark masses and v3depend on the renormaliza- tion scheme, but their product does not. In the real world, el ectromagnetic interactions raise the mass of the π±slightly above that of the π0. This framework is easily expanded to include the strange qua rk. The three pions ( π+,π−, mass 0.140GeV;π0, mass 0.135GeV), the four kaons (K+,K−, mass 0.494GeV; K0,K0, mass 0.498GeV), and the eta ( η, mass 0.548GeV) are identiﬁed as the eight expected Goldstone boson s. We can assemble them into the hermitian matrix Π≡πaTa/fπ=1 2fπ π0+1√ 3η√ 2π+√ 2K+ √ 2π−−π0+1√ 3η√ 2K0 √ 2K−√ 2K0−2√ 3η . (83.19) The second line of eq.(83.17) still applies, but now the Ta’s are the genera- tors of SU(3), and Mincludes a third diagonal entry for the strange quark mass. We leave the details to the problems. 83: Chiral Symmetry Breaking 508 Next we turn to the coupling of the pions to the nucleons (the p roton and neutron). We deﬁne a Dirac ﬁeld Ni, whereN1=p(the proton) andN2=n(the neutron). We assume that, under an SU(2) L×SU(2) R transformation, PLNi→LijPLNj, (83.20) PRN¯ı→R¯ı¯PRN¯. (83.21) The standard Dirac kinetic term iN/∂Nis then SU(2) L×SU(2) Rinvariant, but the standard mass term mNNNis not. (Here mNis the value of the nucleon mass in the limit of zero up and down quark masses.) Ho wever, we can construct an invariant mass term by including appropr iate factors ofUandU†, Lmass=−mNN(U†PL+UPR)N . (83.22) There is one other parity, time-reversal, and SU(2) L×SU(2) Rinvariant term with one derivative. Including this term, we have L=iN/∂N−mNN(U†PL+UPR)N −1 2(gA−1)iNγµ(U∂µU†PL+U†∂µUPR)N , (83.23) wheregA= 1.27 is the axial vector coupling . Its value is determined from the decay rate of the neutron via the weak interaction; see se ction 90. The form of the lagrangian in eq.(83.23) is somewhat awkward . It can be simpliﬁed by ﬁrst deﬁning u(x)≡exp[iπa(x)Ta/fπ], (83.24) so thatU(x)≡u2(x). Then we deﬁne a new nucleon ﬁeld N ≡(u†PL+uPR)N . (83.25) (This is a ﬁeld redeﬁnition in the sense of problem 11.5.) Equivalently, using the unitarity of u, we have N= (uPL+u†PR)N. (83.26) Using eq.(83.26) in eq.(83.23), along with the identities ∂µU= (∂µu)u+ u(∂µu), (∂µu†)u=−u†(∂µu), etc., we ultimately ﬁnd L=iN/∂N −mNNN+N/vN −gAN/aγ5N, (83.27) where we have deﬁned the hermitian vector ﬁelds vµ≡1 2i[u†(∂µu) +u(∂µu†)], (83.28) aµ≡1 2i[u†(∂µu)−u(∂µu†)]. (83.29) 83: Chiral Symmetry Breaking 509 If we now expand uandu†in inverse powers of fπ, we get L=iN/∂N −mNNN+ (gA/fπ)∂µπaNTaγµγ5N+... . (83.30) We can integrate by parts in the interaction term to put the de rivative on theNandNﬁelds. Then, if we consider a process where an oﬀ-shell pion is scattered by an on-shell nucleon, we can use the Dirac equa tion to replace the derivatives of NandNwith factors of mN. We then ﬁnd a coupling of the pion to an on-shell nucleon of the form LπNN=−igπNNπaNσaγ5N, (83.31) where we have set Ta=1 2σa, and identiﬁed the pion-nucleon coupling constant , gπNN=gAmN/fπ. (83.32) The value of gπNNcan be determined from measurements of the neutron- proton scattering cross section, assuming that it is domina ted by pion ex- change; the result is gπNN= 13.5. Eq.(83.32), known as the Goldberger– Treiman relation , is then satisﬁed to within about 5%. Reference Notes The chiral lagrangian is treated in Georgi ,Ramond II , and Weinberg II . Light quark masses are taken from MILC . Problems 83.1) Suppose that the color group is SO(3) rather than SU(3) , and that each quark ﬂavor is represented by a Dirac ﬁeld in the 3 repres entation of SO(3). a) WithnFﬂavors of massless quarks, what is the nonanomalous ﬂavor symmetry group? b) Assume the formation of a color-singlet, Lorentz scalar, fermion condensate. Assume that it preserves the largest possible u nbroken subgroup of the ﬂavor symmetry. What is this unbroken subgro up? c) For the case nF= 2, how many massless Goldstone bosons are there? d) Now suppose that the color group is SU(2) rather than SU(3) , and that each quark ﬂavor is represented by a Dirac ﬁeld in the 2 representation of SU(2). Repeat parts (a), (b), and (c) for t his case. Hint: at least one of the answers is diﬀerent! 83: Chiral Symmetry Breaking 510 83.2) Why is there a minus sign on the right-hand side of eq.(8 3.7)? 83.3) Verify that eq.(83.13) follows from eq.(83.12). 83.4) Use eqs.(83.12) and (83.16) to compute the tree-level contribution to the scattering amplitude for πaπb→πcπd. Work in the isospin limit, mu=md≡m. Express your answer in terms of the Mandelstam variables and the pion mass mπ. 83.5) Verify that eq.(83.27) follows from eqs.(83.26) and e q.(83.23). 83.6) Consider the case of three light quark ﬂavors, with mas sesmu,md, andms. a) Find the masses-squared of the eight pseudogoldstone bos ons. Take the limitmu,d≪ms, and drop terms that are of order m2 u,d/ms. b) Assume that m2 π±andm2 K±each receive an electromagnetic con- tribution; to zeroth order in the quark masses, this contrib ution is the same for both, but the comparatively large strange quark mas s results in an electromagnetic contribution to m2 K±that is roughly twice as large as the electromagnetic contribution ∆ m2 EMtom2 π±. Use the observed masses of the π±,π0,K±, andK0to compute muv3/f2 π, mdv3/f2 π,msv3/f2 π, and ∆m2 EM. c) Compute the quark mass ratios mu/mdandms/md. d) Use your results from part (b) to predict the ηmass. How good is your prediction? 83.7) Suppose that the U(1) Asymmetry is not anomalous, so that we must include a ninth Goldstone boson. We can write U(x) = exp[2iπa(x)Ta/fπ+iπ9(x)/f9]. (83.33) The ninth Goldstone boson is given its own decay constant f9, since there is no symmetry that forces it to be equal to fπ. We write the two-derivative terms in the lagrangian as L=−1 4f2 πTr∂µU†∂µU−1 4F2∂µ(detU†)∂µ(detU). (83.34) a) By requiring all nine Goldstone ﬁelds to have canonical ki netic terms, determine Fin terms of fπandf9. b) To simplify the analysis, let mu=md≡m≪ms. Find the masses of the nine pseudogoldstone bosons. Identify the three ligh test as the pions, and call their mass mπ. Show that another one of the nine has a mass less than or equal to√ 3mπ. (The nonexistence of such a 83: Chiral Symmetry Breaking 511 particle in nature is the U(1) problem ; the axial anomaly solves this problem.) 83.8) a) Write down all possible parity and time-reversal in variant terms with no derviatives that are bilinear in the nucleon ﬁeld Nand that have one factor of the quark mass matrix M. b) Reexpress your result in terms of the nucleon ﬁeld N. c) Use the observed neutron-proton mass diﬀerence, mn−mp= 1.293MeV, and the mu/mdratio you found in problem 83.6, to de- temine as much as you can about the coeﬃcients of the terms wro te down. (Ignore the mass diﬀerence due to electromagnetism.) 84: Spontaneous Breaking of Gauge Symmetries 512 84Spontaneous Breaking of Gauge Symmetries Prerequisite: 32, 70 Consider scalar electrodynamics, speciﬁed by the lagrangi an L=−(Dµϕ)†Dµϕ−V(ϕ)−1 4FµνFµν, (84.1) whereϕis a complex scalar ﬁeld, Dµ=∂µ−igAµ, and V(ϕ) =m2ϕ†ϕ+1 4λ(ϕ†ϕ)2. (84.2) (We call the gauge coupling constant grather than ebecause we are using this theory as a formal example rather than a physical model. ) So far we have always taken m2>0, but now let us consider m2<0. We analyzed this model in the absence of the gauge ﬁeld in section 32. Clas sically, the ﬁeld has a nonzero vacuum expectation value (VEV for short), given by ∝an}b∇acketle{t0|ϕ(x)|0∝an}b∇acket∇i}ht=1√ 2v, (84.3) where we have made a global U(1) transformation to set the pha se of the VEV to zero, and v= (4|m2|/λ)1/2. (84.4) We therefore write ϕ(x) =1√ 2(v+ρ(x))e−iχ(x)/v, (84.5) whereρ(x) andχ(x) are real scalar ﬁelds. The scalar potential depends only onρ, and is given by V(ϕ) =1 4λv2ρ2+1 4λvρ3+1 16λρ4. (84.6) Sinceχdoes not appear in the potential, it is massless; it is the Gol dstone boson for the spontaneously broken U(1) symmetry. The big diﬀerence in the gauge theory is that we can make a gaug e transformation that shifts the phase of ϕ(x) by an arbitrary spacetime function. We can use this gauge freedom to set χ(x) = 0; this choice is called unitary gauge . Using eq.(84.5) with χ(x) = 0 in eq.(84.1), we have −(Dµϕ)†Dµϕ=−1 2(∂µρ+ig(v+ρ)Aµ)(∂µρ−ig(v+ρ)Aµ) =−1 2∂µρ∂µρ−1 2g2(v+ρ)2AµAµ. (84.7) Expanding out the last term, we see that the gauge ﬁeld now has a mass M=gv. (84.8) 84: Spontaneous Breaking of Gauge Symmetries 513 This is the Higgs mechanism : the Goldstone boson disappears, and the gauge ﬁeld acquires a mass. Note that this leaves the countin g of particle spin states unchanged: a massless spin-one particle has two spin states, but a massive one has three. The Goldstone boson has become th e third orlongitudinal state of the now-massive gauge ﬁeld. A scalar ﬁeld whose VEV breaks a gauge symmetry is generically called a Higgs ﬁeld . This generalizes in a straightforward way to a nonabelian ga uge theory. Consider a complex scalar ﬁeld ϕin a representation R of the gauge group. The kinetic term for ϕis−(Dµϕ)†Dµϕ, where the covariant derivative is (Dµϕ)i=∂µϕi−igaAa µ(Ta R)ijϕj, and the indices iandjrun from 1 to d(R). We assume that ϕacquires a VEV ∝an}b∇acketle{t0|ϕi(x)|0∝an}b∇acket∇i}ht=1√ 2vi, (84.9) where the value of viis determined (up to a global gauge transformation) by minimizing the potential. If we replace ϕby its VEV in −(Dµϕ)†Dµϕ, we ﬁnd a mass term for the gauge ﬁelds, Lmass=−1 2(M2)abAaµAb µ, (84.10) where the mass-squared matrix is (M2)ab=1 2g2v∗ i{Ta R,Tb R}ijvj. (84.11) The anticommutator appears because AaµAb µis symmetric on a↔b, and so we replaced Ta RTb Rwith1 2{Ta R,Tb R}. If the ﬁeld ϕis real rather than complex (which is possible only if R is a real representation), then we remove the factor of root- two from the right-hand side of eq.(84.9), but this is compensated by an e xtra factor of one-half from the kinetic term for a real scalar ﬁeld; thus eq .(84.11) holds as written. If there is more than one gauge group, then the g2in eq.(84.11) is replaced by gagb, wheregais the coupling constant that goes along with the generator Ta, and all generators of all gauge groups are included in the mass-squared matrix. Recall from section 32 that a generator Tais spontaneously broken if (Ta R)ijvj∝ne}ationslash= 0. From eq.(84.11), we see that gauge ﬁelds corresponding to broken generators get a mass, while those corresponding t o unbroken generators do not. The unbroken generators (if any) form a ga uge group with massless gauge ﬁelds. The massive gauge ﬁelds (and all o ther ﬁelds) form representations of this unbroken group. Let us work out some simple examples. Consider the gauge group SU( N), with a complex scalar ﬁeld ϕin the fundamental representation. We can make a global SU( N) transforma- tion to bring the VEV entirely into the last component, and fu rthermore 84: Spontaneous Breaking of Gauge Symmetries 514 make it real. Any generator ( Ta)ijthat does not have a nonzero entry in the last column will remain unbroken. These generators form an unbro- ken SU(N−1) gauge group. There are three classes of broken generators : those with ( Ta)iN=1 2fori∝ne}ationslash=N(there areN−1 of these); those with (Ta)iN=−1 2ifori∝ne}ationslash=N(there are also N−1 of these), and ﬁnally the sin- gle generator TN2−1= [2N(N−1)]−1/2diag(1,...,1,−(N−1)). The gauge ﬁelds corresponding to the generators in the ﬁrst two classe s get a mass M=1 2gv; we can group them into a complex vector ﬁeld that transforms in the fundamental representation of the unbroken SU( N−1) subgroup. The gauge ﬁeld corresponding to TN2−1gets a mass M= [(N−1)/2N]1/2gv; it is a singlet of SU( N−1). Consider the gauge group SO( N), with a real scalar ﬁeld in the fun- damental representation. We can make a global SO( N) transformation to bring the VEV entirely into the last component. Any generato r (Ta)ijthat does not have a nonzero entry in the last column will remain un broken. These generators form an unbroken SO( N−1) subgroup. There are N−1 broken generators, those with ( Ta)iN=−ifori∝ne}ationslash=N. The corresponding gauge ﬁelds get a mass M=gv; they form a fundamental representation of the unbroken SO( N−1) subgroup. In the case N= 3, this subgroup is SO(2), which is equivalent to U(1). Consider the gauge group SU( N), with a real scalar ﬁeld Φain the adjoint representation. It will prove more convenient to wo rk with the matrix-valued ﬁeld Φ = ϕaTa; the covariant derivative of Φ is DµΦ = ∂µΦ−igAa µ[Ta,Φ], and the VEV of ϕis a traceless hermitian N×N matrixV. Thus the mass-squared matrix for the gauge ﬁelds is ( M2)ab= −1 2g2Tr{[Ta,V],[Tb,V]}. We can make a global SU( N) transformation to bringVinto diagonal form. Suppose the diagonal entries consist of N1v1’s, followed by N2v2’s, etc., where v1<v2<... , and/summationtext iNivi= 0. Then all generators whose nonzero entries lie entirely within the ithblock commute withV, and hence form an unbroken SU( Ni) subgroup. Furthermore, the linear combination of diagonal generators that is proporti onal toValso commutes with V, and forms a U(1) subgroup. Thus the unbroken gauge group is SU( N1)×SU(N2)×...×U(1). The gauge coupling constants for the diﬀerent groups are all the same, and equal to the origina l SU(N) gauge coupling constant. As a speciﬁc example, consider the case of SU(5), which has 24 gener- ators. Let the diagonal entries of Vbe given by ( −1 3,−1 3,−1 3,+1 2,+1 2)v. The unbroken subgroup is then SU(3) ×SU(2) ×U(1). The number of broken generators is 24 −8−3−1 = 12. The generator of the U(1) sub- group isT24= diag( −1 3c,−1 3c,−1 3c,+1 2c,+1 2c), wherec2= 3/5. Under the unbroken SU(3) ×SU(2) ×U(1) subgroup, the 5 representation of SU(5) 84: Spontaneous Breaking of Gauge Symmetries 515 transforms as 5→(3,1,−1 3)⊕(1,2,+1 2). (84.12) Here the last entry is the value of T24/c. The 5 of SU(5) then transforms as 5→(¯3,1,+1 3)⊕(1,2,−1 2). (84.13) To ﬁnd out how the adjoint or 24 representation of SU(5) trans forms under the SU(3) ×SU(2) ×U(1) subgroup, we use the SU(5) relation 5⊗5 = 24 ⊕1. (84.14) From eqs.(84.12) and (84.13), we have 5⊗5→[(3,1,−1 3)⊕(1,2,+1 2)]⊗[(¯3,1,+1 3)⊕(1,2,−1 2)].(84.15) If we expand this out, and compare with eq.(84.14), we see tha t 24→(8,1,0)⊕(1,3,0)⊕(1,1,0) ⊕(3,2,−5 6)⊕(¯3,2,+5 6). (84.16) The ﬁrst line on the right-hand side of eq.(84.16) is the adjo int repre- sentation of SU(3) ×SU(2) ×U(1); the corresponding gauge ﬁelds remain massless. The second line shows us that the gauge ﬁelds corre sponding to the twelve broken generators can be grouped into a complex ve ctor ﬁeld in the representation (3 ,2,−5 6). Since it is an irreducible representation of the unbroken subgroup, all twelve vectors ﬁelds must have th e same mass. This mass is most easily computed from ( M2)44=−g2Tr([T4,V][T4,V]), where we have deﬁned ( T4)ij=1 2(δi1δj4+δi4δj1); the result is M=5 6√ 2gv. Problems 84.1) Conside a theory with gauge group SU( N), with a real scalar ﬁeld Φ in the adjoint representation, and potential V(Φ) =1 2m2Tr Φ2+1 4λ1Tr Φ4+1 4λ2(Tr Φ2)2. (84.17) This is the most general potential consistent with SU( N) symmetry and a Z 2symmetry Φ ↔ −Φ, which we impose to keep things sim- ple. We assume m2<0. We can work in a basis in which Φ = vdiag(α1,...,αN), with the constraints/summationtext iαi= 0 and/summationtext iα2 i= 1. a) Extremize V(Φ) with respect to v. Solve forv, and plug your result back intoV(Φ). You should ﬁnd V(Φ) =−1 4(m2)2 λ1A(α) +λ2B(α), (84.18) 84: Spontaneous Breaking of Gauge Symmetries 516 whereA(α) andB(α) are functions of αi. b) Show that λ1A(α) +λ2B(α) must be everywhere positive in order for the potential to be bounded below. c) Show that the absolute mimimum of the potential (assuming that it is bounded below) occurs at the absolute minimum of λ1A(α) + λ2B(α). d) Show that, at any extremum of the potential, the αitake on at most three diﬀerent values, and that these three values sum t o zero. Hint: impose the constraints with Lagrange multipliers. e) Show that, for λ1>0 andλ2>0, at the absolute minimum of V(Φ) the unbroken symmetry group is SU( N+)×SU(N−)×U(1), whereN+=N−=1 2NifNis even, and N±=1 2(N±1) ifNis odd. 85: Spontaneously Broken Abelian Gauge Theory 517 85Spontaneously Broken Abelian Gauge Theory Prerequisite: 61, 84 Consider scalar electrodynamics, speciﬁed by the lagrangi an L=−(Dµϕ)†Dµϕ−V(ϕ)−1 4FµνFµν, (85.1) whereϕis a complex scalar ﬁeld and Dµ=∂µ−igAµ. We choose V(ϕ) =1 4λ(ϕ†ϕ−1 2v2)2, (85.2) which yields a nonzero VEV for ϕ. We therefore write ϕ(x) =1√ 2(v+ρ(x))e−iχ(x)/v, (85.3) whereρ(x) andχ(x) are real scalar ﬁelds. The scalar potential depends only onρ, and is given by V(ϕ) =1 4λv2ρ2+1 4λvρ3+1 16λρ4. (85.4) We can now make a gauge transformation to set χ(x) = 0. (85.5) This is unitary gauge . The kinetic term for ϕbecomes −(Dµϕ)†Dµϕ=−1 2∂µρ∂µρ−1 2g2(v+ρ)2AµAµ. (85.6) We see that the gauge ﬁeld has acquired a mass M=gv. (85.7) The terms in Lthat are quadratic in Aµare L0=−1 4FµνFµν−1 2M2AµAµ. (85.8) The equation of motion that follows from eq.(85.8) is [(−∂2+M2)gµν+∂µ∂ν]Aν= 0. (85.9) If we act with ∂µon this equation, we get M2∂νAν= 0. (85.10) 85: Spontaneously Broken Abelian Gauge Theory 518 If we now use eq.(85.10) in eq.(85.9), we ﬁnd that each compon ent ofAν obeys the Klein-Gordon equation, (−∂2+M2)Aν= 0. (85.11) The general solution of eqs.(85.9) and (85.10) is Aµ(x) =/summationdisplay λ=−,0,+/integraldisplay /tildewiderdk/bracketleftig εµ∗ λ(k)aλ(k)eikx+εµ λ(k)a† λ(k)e−ikx/bracketrightig ,(85.12) where the polarization vectors must satisfy kµεµ λ(k) = 0. In the rest frame, wherek= (M,0,0,0), we choose the polarization vectors to correspond to deﬁnite spin along the ˆ zaxis, ε+(0) =1√ 2(0,1,−i,0), ε−(0) =1√ 2(0,1,+i,0), ε0(0) = (0,0,0,1). (85.13) More generally, the three polarization vectors along with t he timelike unit vectorkµ/Mform an orthonormal and complete set, k·εµ λ(k) = 0, (85.14) ελ′(k)·ε∗ λ(k) =δλ′λ, (85.15) /summationdisplay λ=−,0,+εµ∗ λ(k)εν λ(k) =gµν+kµkν M2. (85.16) Since the lagrangian of eq.(85.8) has no manifest gauge inva riance, quantization is straightforward. The coeﬃcients a† λ(k) andaλ(k) become particle creation and annihilation operators in the usual w ay, and the prop- agator of the Aµﬁeld is given by i∝an}b∇acketle{t0|TAµ(x)Aν(y)|0∝an}b∇acket∇i}ht=/integraldisplayd4k (2π)4eik(x−y) k2+M2−iǫ/summationdisplay λεµ∗ λ(k)εν λ(k) =/integraldisplayd4k (2π)4eik(x−y) k2+M2−iǫ/parenleftbigg gµν+kµkν M2/parenrightbigg .(85.17) The interactions of the massive vector ﬁeld Aµwith the real scalar ﬁeld ρ can be read oﬀ of eq.(85.6). The self-interactions of the ρﬁeld can be read oﬀ of eq.(85.4). The resulting Feynman rules can be used for t ree-level calculations. Loop calculations are more subtle. We have imposed the gauge condi- tionχ(x) = 0, which corresponds to inserting a functional delta func tion 85: Spontaneously Broken Abelian Gauge Theory 519 /producttext xδ(χ(x)) into the path integral. In order to integrate over χ, we must make a change of integration variables from Re ϕand Imϕtoρandχ; this is simply a transformation from cartesian to polar coordina tes, analogous todxdy =rdrdφ . So we must include a factor analogous to rin the functional measure; this factor is /productdisplay x/parenleftig v+ρ(x)/parenrightig = det(v+ρ) ∝det(1 +v−1ρ) ∝/integraldisplay D¯cDce−im2 gh/integraltext d4x¯c(1+v−1ρ)c. (85.18) In the last line, we have written the functional determinant as an integral over ghost ﬁelds. We see that they have no kinetic term, and we have chosen the overall nomalization of their action so that thei r mass ismgh, wheremghis an arbitrary mass parameter. Thus the momentum-space propagator for the ghosts is simply ˜∆(k2) = 1/m2 gh. We also see that there is a ghost-ghost-scalar vertex, with vertex factor −im2 ghv−1, but there is no interaction between the ghosts and the vector ﬁeld. This seems like a fairly convenient gauge for loop calculati ons, but there is a complication. The fact that the ghost propagator is inde pendent of the momentum means that additional internal ghost propagators do not help the convergence of loop-momentum integrals. The same is tru e of vector- ﬁeld propagators; from eq.(85.17) we see that, in momentum s pace, the propagator scales like 1 /M2in the limit that all components of kbecome large. Thus, in unitary gauge, loop diagrams with arbitrari ly many external lines diverge. This makes it diﬃcult to establish renormali zability. A gauge that does not suﬀer from this problem is a generalizat ion of Rξgauge (and in fact this name has traditionally been applied o nly to this generalization). We begin by using a cartesian basis for ϕ, ϕ=1√ 2(v+h+ib), (85.19) wherehandbare real scalar ﬁelds. In terms of handb, the potential is V(ϕ) =1 4λv2h2+1 4λvh(h2+b2) +1 16λ(h2+b2)2, (85.20) and the covariant derivative of ϕis Dµϕ=1√ 2/bracketleftig (∂µh+gbAµ) +i(∂µb−g(v+h)Aµ)/bracketrightig . (85.21) Thus the kinetic term for ϕbecomes −(Dµϕ)†Dµϕ=−1 2(∂µh+gbAµ)2−1 2(∂µb−g(v+h)Aµ)2.(85.22) 85: Spontaneously Broken Abelian Gauge Theory 520 Expanding this out, and rearranging, we get −(Dµϕ)†Dµϕ=−1 2∂µh∂µh−1 2∂µb∂µb−1 2g2v2AµAµ+gvAµ∂µb +gAµ(h∂µb−b∂µh) −gvhAµAµ−1 2g2(h2+b2)AµAµ. (85.23) The ﬁrst line on the right-hand side of eq.(85.23) contains a ll the terms that are quadratic in the ﬁelds. The ﬁrst two are the kinetic t erms for the handbﬁelds. The third is the mass term for the vector ﬁeld. The four th is an annoying cross term between the vector ﬁeld and the deri vative ofb. In abelian gauge theory, in the absence of spontaneous symme try break- ing, we ﬁxRξgauge by adding to Lthe gauge-ﬁxing and ghost terms Lgf+Lgh=−1 2ξ−1G2−¯cδG δθc, (85.24) whereG=∂µAµ, andθ(x) parameterizes an inﬁnitesimal gauge transfor- mation, Aµ→Aµ−∂µθ, (85.25) ϕ→ϕ−igθϕ. (85.26) WithG=∂µAµ, we haveδG/δθ =−∂2. Thus the ghost ﬁelds have no interactions, and can be ignored. In the presence of spontaneous symmetry breaking, we choose instead G=∂µAµ−ξgvb, (85.27) which reduces to ∂µAµwhenv= 0. Multiplying out G2, we have Lgf=−1 2ξ−1∂µAµ∂νAν+gvb∂µAµ−1 2ξg2v2b2 =−1 2ξ−1∂µAν∂νAµ−gvAµ∂µb−1 2ξg2v2b2, (85.28) where we integrated by parts in the ﬁrst two terms to get the se cond line. Note that the second term on the second line of eq.(85.28) can cels the annoying last term on the ﬁrst line of eq.(85.23). Also, the l ast term on the second line of eq.(85.28) gives a mass ξ1/2Mto thebﬁeld. We must still evaluate Lgh. To do so, we ﬁrst translate eq.(85.26) into h→h+gθb, (85.29) b→b−gθ(v+h). (85.30) Then we haveδG δθ=−∂2+ξg2v(v+h). (85.31) 85: Spontaneously Broken Abelian Gauge Theory 521 From eq.(85.24) we see that the ghost lagrangian is Lgh=−¯c/bracketleftig −∂2+ξg2v(v+h)/bracketrightig c =−∂µ¯c∂µc−ξg2v2¯cc−ξg2vh¯cc. (85.32) We see from the second term that the ghost has acquired the sam e mass as thebﬁeld,ξ1/2M. Now let us examine the vector ﬁeld. Including Lgf, the terms in Lthat are quadratic in the vector ﬁeld can be written as L0=−1 2Aµ/bracketleftig gµν(−∂2+M2) + (1−ξ−1)∂µ∂ν/bracketrightig Aν. (85.33) In momentum space, this reads ˜L0=−1 2˜Aµ(−k)/bracketleftig (k2+M2)gµν+ (1−ξ−1)kµkν/bracketrightig˜Aν(k). (85.34) The kinematic matrix is /bracketleftig .../bracketrightig = (k2+M2)gµν+ (1−ξ−1)kµkν = (k2+M2)/parenleftig Pµν(k) +kµkν/k2/parenrightig + (1−ξ−1)kµkν = (k2+M2)Pµν(k) +ξ−1(k2+ξM2)kµkν/k2,(85.35) wherePµν(k) =gµν−kµkν/k2projects onto the transverse subspace; Pµν(k) andkµkν/k2are orthogonal projection matrices. Using this fact, it is easy to invert eq.(85.35) to get the propagator for the m assive vector ﬁeld inRξgauge, ˜∆µν(k) =Pµν(k) k2+M2−iǫ+ξkµkν/k2 k2+ξM2−iǫ. (85.36) We see that the transverse components of the vector ﬁeld prop agate with massM, while the longitudinal component propagates with the same mass as theband ghost ﬁelds, ξ1/2M. Eq.(85.36) simpliﬁes greatly if we choose ξ= 1; then we have ˜∆µν(k) =gµν k2+M2−iǫ(ξ= 1). (85.37) On the other hand, leaving ξas a free parameter allows us to check that all ξdependence cancels out of any physical scattering amplitud e. Since their masses depend on ξ, the ghosts, the bﬁeld, and the longitudinal component of the vector ﬁeld must all represent unphysical particles t hat do not appear in incoming or outgoing states. 85: Spontaneously Broken Abelian Gauge Theory 522 To summarize, in Rξgauge we have the physical hﬁeld with mass- squaredm2 h=1 2λv2and propagator 1 /(k2+m2 h), the unphysical bﬁeld with propagator 1 /(k2+ξM2), the ghost ﬁelds ¯ candcwith propagator 1/(k2+ξM2), and the vector ﬁeld with the propagator of eq.(85.36). For external vectors, the polarizations are still given by eq.( 85.13), and obey the sum rules of eq.(85.16). The mass parameter Mis given by M=gv. The interactions of these ﬁelds are governed by L1=−1 4λvh(h2+b2)−1 16λ(h2+b2)2 +gAµ(h∂µb−b∂µh)−gvhAµAµ−1 2g2(h2+b2)AµAµ −ξg2vh¯cc. (85.38) It is interesting to consider the limit ξ→ ∞. In this limit, the vector propagator in Rξgauge, eq.(85.36), turns into the massive vector propa- gator of eq.(85.17), ˜∆µν(k) =gµν+kµkν/M2 k2+M2−iǫ(ξ=∞). (85.39) Thebﬁeld becomes inﬁnitely heavy, and we can drop it. (Equivalen tly, its propagator goes to zero.) The ghost ﬁelds also become inﬁ nitely heavy, but we must be more careful with them because their interacti on term, the last line of eq.(85.38), also contains a factor of ξ. The vertex factor for this interaction is −iξg2v=−i(ξM2)v−1. Note that this is the same vertex factor that we found in unitary gauge for the ineraction betw een theρﬁeld and the ghost ﬁelds; see eq.(85.18) and take m2 gh=ξM2. Thus we cannot drop the ghost ﬁelds, but we can take their propagator to be 1 /m2 ghrather than 1/(k2+m2 gh), sincek2≪m2 gh=ξM2in the limit ξ→ ∞. This is the ghost propagator that we found in unitary gauge. We concl ude that Rξgauge in the limit ξ→ ∞ is equivalent to unitary gauge. Of course, in this limit, we reencounter the problems with divergent diag rams that led us to consider alternative gauge choices in the ﬁrst place. For practical loop calculations, Rξgauge with ξ= 1 is typically the most convenient. In the next section, we consider Rξgauge for nonabelian theories. 86: Spontaneously Broken Nonabelian Gauge Theory 523 86Spontaneously Broken Nonabelian Gauge Theory Prerequisite: 85 In the previous section, we worked out the lagrangian for a U( 1) gauge theory with spontaneous symmetry breaking in Rξgauge. In this section, we extend this analysis to a general nonabelian gauge theory . As in section 85, it will be convenient to work with real scala r ﬁelds. We therefore decompose any complex scalar ﬁelds into pairs o f real ones, and organize all the real scalar ﬁelds into a big list φi,i= 1,...,N . These real scalar ﬁelds form a (possibly reducible) representati on R of the gauge group. Let Tabe the gauge-group generator matrices that act on φ; they are linear combinations of the generators of the SO( N) group that rotates all components of φiinto each other. Because these SO( N) generators are hermitian and antisymmetric, so are the Ta’s. Thusi(Ta)ijis a real, antisymmetric matrix. The lagrangian for our theory can now be written as L=−1 2DµφDµφ−V(φ)−1 4FaµνFa µν, (86.1) where (Dµφ)i=∂µφi−igaAa µ(Ta)ijφj (86.2) is the covariant derivative, and the adjoint index aruns over all generators of all gauge groups. Because φiandAa µare real ﬁelds, and i(Ta)ijis a real matrix, (Dµφ)iis real. Now we suppose that the potential V(φ) is minimized when φhas a VEV ∝an}b∇acketle{t0|φi(x)|0∝an}b∇acket∇i}ht=vi. (86.3) A generator Tais unbroken if ( Ta)ijvj= 0, and broken if ( Ta)ijvj∝ne}ationslash= 0. Each broken generator results in a massless Goldstone boson . To see this, we note that the potential must be invariant under a glo bal gauge transformation, V((1−iθaTa)φ) =V(φ). (86.4) Expanding to linear order in the inﬁnitesimal parameter θ, we ﬁnd ∂V ∂φj(Ta)jkφk= 0. (86.5) We diﬀerentiate eq.(86.5) with respect to φkto get ∂2V ∂φi∂φj(Ta)jkφk+∂V ∂φj(Ta)jk= 0. (86.6) 86: Spontaneously Broken Nonabelian Gauge Theory 524 Now setφi=vi; then∂V/∂φivanishes, because φi=viminimizesV(φ). Also, we can identify (m2)ij=∂2V ∂φi∂φj/vextendsingle/vextendsingle/vextendsingle/vextendsingle/vextendsingleφi=vi(86.7) as the mass-squared matrix for the scalars (after spontaneo us symmetry breaking). Thus eq.(86.6) becomes (m2)ij(Tav)j= 0. (86.8) We see that if Tav∝ne}ationslash= 0, then Tavis an eigenvector of the mass-squared matrix with eigenvalue zero. So there is a zero eigenvalue fo r every linearly independent broken generator. Let us write φi(x) =vi+χi(x), (86.9) whereχiis a real scalar ﬁeld. The covariant derivative of φbecomes (Dµφ)i=∂µχi−igaAa µ(Ta)ij(v+χ)j (86.10) It is now convenient to deﬁne a set of real antisymmetric matr ices (τa)ij≡iga(Ta)ij, (86.11) and the real rectangular matrix Fai≡(τa)ijvj. (86.12) We can now write (Dµφ)i=∂µχi−Aa µ(Fa+τaχ)i. (86.13) The kinetic term for φbecomes −1 2DµφDµφ=−1 2∂µχi∂µχi−1 2(FaiFbi)AaµAb µ+FaiAa µ∂µχi +Aa µχi(τa)ij∂µχj−AaµAb µFai(τb)ijχj −1 2AaµAb µχi(τaτb)ijχj. (86.14) We see (from the second term on the right-hand side) that the m ass-squared matrix for the vector ﬁelds is (M2)ab=FaiFbi= (FFT)ab. (86.15) A theorem of linear algebra states that every real rectangul ar matrix can be written as Fai=Sab(Mbδbj)Rji, (86.16) 86: Spontaneously Broken Nonabelian Gauge Theory 525 whereSandRare orthogonal matrices, and the diagonal entries Maare real and nonnegative. From eq.(86.15) we see that these diag onal entries are the masses of the vector ﬁelds. The vector ﬁelds of deﬁnit e mass are then given by ˜Aa µ=SbaAb µ. Now we are ready to ﬁx Rξgauge. To do so, we add to Lthe gauge- ﬁxing and ghost terms Lgf+Lgh=−1 2ξ−1GaGa−¯caδGa δθbcb, (86.17) where we choose Ga=∂µAa µ−ξFaiχi. (86.18) Then we have Lgf=−1 2ξ−1∂µAa µ∂νAa ν+Faiχi∂µAa µ−1 2ξ(FaiFaj)χiχj =−1 2ξ−1∂µAa ν∂νAa µ−FaiAa µ∂µχi−1 2ξ(FaiFaj)χiχj.(86.19) We integrated by parts in the ﬁrst two terms to get the second l ine. Note that the second term on the second line of eq.(86.19) cancels the annoying last term on the ﬁrst line of eq.(86.14). Also, the last term o n the second line of eq.(86.19) makes a contribution to the mass-squared matrix for the χﬁelds, ξ(M2)ij=ξFaiFaj=ξ(FTF)ij. (86.20) Eq.(86.16) tells us that the eigenvalues of this matrix are ξ1/2Ma, where Maare the vector-boson masses. The mass-squared matrix ξM2should be added to the mass-squared matrix m2that we get from the potential, eq.(86.7). Note that eqs.(86.8) and (86.12) imply that ( m2)ijFaj= 0; eq.(86.20) then yields ( m2)ij(ξM2)jk= 0. Thus these two contributions to the mass-squared matrix of the scalar ﬁelds live in orthogon al subspaces. The scalar ﬁelds of deﬁnite mass are ˜ χi=Rijχj, where the block of Rin the m2subspace is chosen to diagonalize m2. Them2subspace consists of the physical, massive scalars, and the ξM2subspace consists of the unphysical Goldstone bosons; these are the ﬁelds that would be set to zer o in unitary gauge. We must still evaluate Lgh. To do so, we recall that θa(x) parameterizes an inﬁnitesimal gauge transformation, Aa µ→Aa µ−Dab µθb, (86.21) χi→ −θa(τa)ij(v+χ)j. (86.22) Thus we have δGa δθb=−∂µDab µ+ξFaj(τb)jk(v+χ)k 86: Spontaneously Broken Nonabelian Gauge Theory 526 =−∂µDab µ+ξFajFbj+ξFaj(τb)jkχk =−∂µDab µ+ξ(M2)ab+ξFaj(τb)jkχk, (86.23) and so the ghost lagrangian is Lgh=−∂µ¯caDab µcb−ξ(M2)ab¯cacb−ξFaj(τb)jkχk¯cacb. (86.24) The ghost ﬁelds of deﬁnite mass are ˜ ca=Sbacband˜¯ca=Sba¯cb. The complete gauge-ﬁxed lagrangian is now given by eqs.(86. 1), (86.14) (86.19), and (86.24). We can rewrite it in terms of the ﬁelds o f deﬁnite mass. This results in the replacements Fai→Maδai, (86.25) (τa)ij→Sab(RTτbR)ij, (86.26) fabc→SadSbeScgfdeg(86.27) throughout L. The Feynman rules then follow in the usual way. Problems 86.1) Letϕibe a complex scalar ﬁeld in a complex representation R of the gauge group. Under an inﬁnitesimal gauge transformation, w e have δϕi=−iθa(Ta R)ijϕj. Let us write ϕi=1√ 2(φi+iφi+d(R)), whereφiis a real scalar ﬁeld with the index irunning from 1 to 2 d(R). Then, under an inﬁnitesimal gauge transformation, we have δφi=−iθa(Ta)ijφj. a) Express Tain terms of the real and imaginary parts of Ta R. b) Show that the Tamatrices satisfy the appropriate commutation relations. 87: The Standard Model: Gauge and Higgs Sector 527 87The Standard Model: Gauge and Higgs Sector Prerequisite: 84 We now turn to the construction of the Standard Model of elementary particles, also called the Glashow–Weinberg–Salam model . This is the com- plete (except for gravity) quantum ﬁeld theory that appears to describe our world. It can be succinctly speciﬁed as a gauge theory with ga uge group SU(3) ×SU(2) ×U(1), with left-handed Weyl ﬁelds in three copies of the representation (1 ,2,−1 2)⊕(1,1,+1)⊕(3,2,+1 6)⊕(¯3,1,−2 3)⊕(¯3,1,+1 3), and a complex scalar ﬁeld in the representation (1 ,2,−1 2). Here the last entry of each triplet gives the value of the U(1) charge, known as hypercharge . The lagrangian includes all terms of mass dimension four or l ess that are allowed by the gauge symmetries and Lorentz invariance. We will construct the Standard Model over several sections. We begin with the electroweak part of the gauge group, SU(2) ×U(1), and the complex scalar ﬁeld ϕ, known as the Higgs ﬁeld , in the representation (2 ,−1 2). The Higgs ﬁeld acquires a nonzero VEV that spontaneously breaks SU(2)×U(1) to U(1); the unbroken U(1) is identiﬁed as electromagnetism . We begin with the covariant derivative of the Higgs ﬁeld ϕ, (Dµϕ)i=∂µϕi−i[g2Aa µTa+g1BµY]ijϕj, (87.1) whereTa=1 2σaandY=−1 2I;Yis the hypercharge generator. It will prove useful to write out g2Aa µTa+g1BµYin matrix form, g2Aa µTa+g1BµY=1 2/parenleftiggg2A3 µ−g1Bµg2(A1 µ−iA2 µ) g2(A1 µ+iA2 µ)−g2A3 µ−g1Bµ/parenrightigg . (87.2) Now suppose that ϕhas a potential V(ϕ) =1 4λ(ϕ†ϕ−1 2v2)2. (87.3) This potential gives ϕa nonzero VEV. We can make a global gauge trans- formation to bring this VEV entirely into the ﬁrst component , and further- more make it real, so that ∝an}b∇acketle{t0|ϕ(x)|0∝an}b∇acket∇i}ht=1√ 2/parenleftiggv 0/parenrightigg . (87.4) The kinetic term for ϕis−(Dµϕ)†Dµϕ. After replacing ϕby its VEV, we ﬁnd a mass term for the gauge ﬁelds, Lmass=−1 8v2(1,0)/parenleftiggg2A3 µ−g1Bµg2(A1 µ−iA2 µ) g2(A1 µ+iA2 µ)−g2A3 µ−g1Bµ/parenrightigg2/parenleftigg1 0/parenrightigg .(87.5) 87: The Standard Model: Gauge and Higgs Sector 528 To diagonalize this mass-squared matrix, we ﬁrst deﬁne the weak mixing angle θW≡tan−1(g1/g2), (87.6) and the ﬁelds W± µ≡1√ 2(A1 µ∓iA2 µ), (87.7) Zµ≡cWA3 µ−sWBµ, (87.8) Aµ≡sWA3 µ+cWBµ, (87.9) wheresW≡sinθW,cW≡cosθW. In terms of these ﬁelds, eq.(87.5) becomes Lmass=−1 8g2 2v2( 1,0 )/parenleftigg1cWZµ√ 2W+ µ √ 2W− µ.../parenrightigg 1 0  =−(g2v/2)2W+µW− µ−1 2(g2v/2cW)2ZµZµ =−M2 WW+µW− µ−1 2M2 ZZµZµ, (87.10) where we have identiﬁed MW=g2v/2, (87.11) MZ=MW/cosθW. (87.12) The observed masses of the W±andZ0particles are MW= 80.4GeV and MZ= 91.2GeV. Eq.(87.12) then implies cos θW= 0.882, or, as it is more usually expressed, sin2θW= 0.223.1 Note that the Aµﬁeld remains massless; this signiﬁes that there is an unbroken U(1) subgroup. We will identify this unbroken U(1) with the gauge group of electromagnetism. Before introducing leptons and quarks (which we do in sectio ns 87 and 88), let us work out the complete lagrangian for the gauge and Higgs ﬁelds, in unitary gauge. This is suﬃcient for tree-level calculati ons. The two complex components of the ϕﬁeld yield four real scalar ﬁelds; three of these become the longitudinal components of the W±andZ0. The 1Of course, this number is only meaningful once a renormaliza tion scheme has been speciﬁed. We are implicitly using an on-shell scheme in whic hθWisdeﬁned by the relation cos θW=MW/MZ, where MWandMZare the actual particle masses. The relation g1=g2tanθWis then subject to loop corrections that depend on the precis e deﬁnitions adopted for g1andg2. In the MS scheme, on the other hand, θWis deﬁned by eq.(87.6), and for µ=MZ, we have sin2θW= 0.231. 87: The Standard Model: Gauge and Higgs Sector 529 remaining scalar ﬁeld must be able to account for shifts in th e overall scale ofϕ. Thus we can write, in unitary gauge, ϕ(x) =1√ 2/parenleftiggv+H(x) 0/parenrightigg , (87.13) whereHis a real scalar ﬁeld; the corresponding particle is the Higgs boson . The potential now reads V(ϕ) =1 4λv2H2+1 4λvH3+1 16λH4. (87.14) We see that the mass of the Higgs boson is given by m2 H=1 2λv2. (As of this writing, the Higgs boson has not been observed; the lowe r limit on its mass ismH>115GeV.) The kinetic term for Hcomes from the kinetic term forϕ, and is the usual one for a real scalar ﬁeld, −1 2∂µH∂µH. Finally, recall that the mass term for the gauge ﬁelds, eq.(87.10), is proportional tov2. Hence it should be multiplied by a factor of (1 + v−1H)2. Now we have to work out the kinetic terms for the gauge ﬁelds. W e have L=−1 4FaµνFa µν−1 4BµνBµν, (87.15) where F1 µν=∂µA1 ν−∂νA1 µ+g2(A2 µA3 ν−A2 νA3 µ), (87.16) F2 µν=∂µA2 ν−∂νA2 µ+g2(A3 µA1 ν−A3 νA1 µ), (87.17) F3 µν=∂µA3 ν−∂νA3 µ+g2(A1 µA2 ν−A1 νA2 µ), (87.18) Bµν≡∂µBν−∂νBµ. (87.19) Next, form the combinations F1 µν±iF2 µν. Using eq.(87.7), we ﬁnd 1√ 2(F1 µν−iF2 µν) =DµW+ ν−DνW+ µ, (87.20) 1√ 2(F1 µν+iF2 µν) =D† µW− ν−D† νW− µ, (87.21) where we have deﬁned a covariant derivative that acts on W+ µ, Dµ≡∂µ−ig2A3 µ =∂µ−ig2(sWAµ+cWZµ). (87.22) If we identify Aµas the electromagnetic vector potential, and assign electr ic chargeQ= +1 (in units of the proton charge) to the W+, then we see from 87: The Standard Model: Gauge and Higgs Sector 530 eq.(87.22) that we must identify the electromagnetic coupl ing constant e as e=g2sinθW. (87.23) Here we are adopting the convention that eis positive. (In our treatment of quantum electrodynamics, we used the convention that eis negative, but that is less convenient in the present context.) We also have F3 µν=∂µA3 ν−∂νA3 µ−ig2(W+ µW− ν−W+ νW− µ) =sWFµν+cWZµν−ig2(W+ µW− ν−W+ νW− µ), (87.24) Bµν=cWFµν−sWZµν, (87.25) whereFµν=∂µAν−∂νAµis the usual electromagnetic ﬁeld strength, and Zµν≡∂µZν−∂νZµ (87.26) is the abelian ﬁeld strength associated with the Zµﬁeld. Now we can assemble all of this into the complete lagrangian f or the electroweak gauge ﬁelds and the Higgs boson in unitary gauge . We will expressg2in terms of eandθWviag2=e/sinθW, andλin terms of mH andvviaλ= 2m2 H/v2. We ultimately get L=−1 4FµνFµν−1 4ZµνZµν−D†µW−νDµW+ ν+D†µW−νDνW+ µ +ie(Fµν+ cotθWZµν)W+ µW− ν −1 2(e2/sin2θW)(W+µW− µW+νW− ν−W+µW+ µW−νW− ν) −(M2 WW+µW− µ+1 2M2 ZZµZµ)(1 +v−1H)2 −1 2∂µH∂µH−1 2m2 HH2−1 2m2 Hv−1H3−1 8m2 Hv−2H4,(87.27) where Dµ=∂µ−ie(Aµ+ cotθWZµ). (87.28) With theW+ µﬁeld assigned electric charge Q= +1, this lagrangian exhibits manifest electromagnetic gauge invariance. The full under lying SU(2) × U(1) gauge invariance is not manifest, however, because we h ave ﬁxed uni- tary gauge. Reference Notes Discussions of the Standard Model in Rξgauge can be found in Cheng & LiandRamond II . Problems 87: The Standard Model: Gauge and Higgs Sector 531 87.1) Find the generator Qof the unbroken U(1) subroup as a linear com- bination of the Ta’s andY. 87.2) a) Ignoring loop corrections, ﬁnd the numerical value s ofv,g1, and g2. Takee2/4π=α(MZ) = 1/127.9 and sin2θW= 0.231. b) The Fermi constant is deﬁned (at tree level) as GF≡e2 4√ 2sin2θWM2 W. (87.29) Find its numerical value in GeV−2. c) Express GFin terms of v. 87.3) In this problem we will work out the generator matrices introduced in section 86 for the case of the Standard Model. a) Write the Higgs ﬁeld as ϕ=1√ 2/parenleftiggφ1+iφ3 φ2+iφ4/parenrightigg . (87.30) whereφiis a real scalar ﬁeld. Express the SU(2) generators Taand the hypercharge generator Yas 4×4 matrices TaandYthat act on φi. Hint: see problem 86.1. b) Compute the matrix Fai, deﬁned in eq.(86.12). c) Compute the mass-squared matrix for the vector ﬁelds, ( M2)ab= FaiFbi, and ﬁnd its eigenvalues. 87.4) Work out the Feynman rules for the lagrangian of eq.(87 .27). 87.5) Assume that mH>2MZ, and compute (at tree level) the decay rate of the Higgs boson into W+W−andZ0Z0pairs. Express your answer in GeV for mH= 200GeV. 88: The Standard Model: Lepton Sector 532 88The Standard Model: Lepton Sector Prerequisite: 75, 87 Leptons are spin-one-half particles that are singlets of the color g roup. There are six diﬀerent ﬂavors of lepton; see Table 2. The six ﬂ avors are naturally grouped into three families orgenerations :eandνe,µandνµ, τandντ. Let us begin by describing a single lepton family, the electr on and its neutrino. We introduce left-handed Weyl ﬁelds ℓand ¯ein the representa- tions (2,−1 2) and (1,+1) of SU(2) ×U(1). Here the bar over the ein the ﬁeld ¯eispart of the name of the ﬁeld , and does not denote any sort of conjugation. The covariant derivatives of these ﬁelds are (Dµℓ)i=∂µℓi−ig2Aa µ(Ta)ijℓj−ig1(−1 2)Bµℓi, (88.1) Dµ¯e=∂µ¯e−ig1(+1)Bµ¯e, (88.2) and their kinetic terms are Lkin=iℓ†i¯σµ(Dµℓ)i+i¯e†¯σµDµ¯e. (88.3) The representation (2 ,−1 2)⊕(1,+1) for the left-handed Weyl ﬁelds is com- plex; hence the gauge theory is chiral, and therefore parity violating. We cannot write down a mass term involving ℓand/or ¯ebecause there is no gauge-group singlet contained in any of the products (2,−1 2)⊗(2,−1 2), (2,−1 2)⊗(1,+1), (1,+1)⊗(1,+1). (88.4) However, we are able to write down a Yukawa coupling of the for m LYuk=−yεijϕiℓj¯e+ h.c., (88.5) whereϕis the Higgs ﬁeld in the (2 ,−1 2) representation that we introduced in the last section, and yis the Yukawa coupling constant. A gauge-invariant Yukawa coupling is possible because there is a singlet on the right-hand side of (2,−1 2)⊗(2,−1 2)⊗(1,+1) = (1,0)⊕(3,0). (88.6) There are no other gauge-invariant terms involving ℓor ¯ethat have mass dimension four or less. Hence there are no other terms that we could add toLwhile preserving renormalizability. We add eqs.(88.3) and (88.5) to the lagrangian for ϕand the gauge ﬁelds that we worked out in the last section. In unitary gauge , we replace 88: The Standard Model: Lepton Sector 533 name symbol mass Q (MeV) electron e 0.511 −1 electron neutrino νe 0 0 muonµ 105.7 −1 muon neutrino νµ 0 0 tauτ 1777 −1 tau neutrino ντ 0 0 Table 2: The six ﬂavors of lepton. Qis the electric charge in units of the proton charge. Each charged ﬂavor is represented by a Dir ac ﬁeld, each neutral ﬂavor by a Majorana ﬁeld (or, equivalently, a left-h aned Weyl ﬁeld). Neutrino masses are exactly zero in the Standard Model. ϕ1with1√ 2(v+H), whereHis the real scalar ﬁeld representing the physical Higgs boson, and ϕ2with zero. The Yukawa term becomes LYuk=−1√ 2y(v+H)(ℓ2¯e+ h.c.). (88.7) It is now convenient to assign new names to the SU(2) componen ts ofℓ, ℓ=/parenleftiggν e/parenrightigg . (88.8) (We will rely on context to distinguish the ﬁeld efrom the electromagnetic coupling constant e.) Then eq.(88.7) becomes LYuk=−1√ 2y(v+H)(e¯e+ ¯e†e†) =−1√ 2y(v+H)EE (88.9) where we have deﬁned a Dirac ﬁeld for the electron, E ≡/parenleftigge ¯e†/parenrightigg . (88.10) We see that the electron has acquired a mass me=yv√ 2. (88.11) The neutrino has remained massless. 88: The Standard Model: Lepton Sector 534 We can describe the neutrino with a Majorana ﬁeld N ≡/parenleftiggν ν†/parenrightigg . (88.12) However, it is often more convenient to work with NL≡PLN=/parenleftiggν 0/parenrightigg , (88.13) wherePL=1 2(1−γ5). We can think of NLas a Dirac ﬁeld; for example, the neutrino kinetic term iν†¯σµ∂µνcan be written as iNL/∂NL. Now we return to eqs.(88.1) and (88.2), and express the covar iant derivatives in the terms of the W± µ,Zµ, andAµﬁelds. From our results in section 87, we have g2A1 µT1+g2A2 µT2=g2√ 2/parenleftigg0W+ µ W− µ0/parenrightigg (88.14) and g2A3 µT3+g1BµY=esW(sWAµ+cWZµ)T3+ecW(cWAµ−sWZµ)Y =e(Aµ+ cotθWZµ)T3+e(Aµ−tanθWZµ)Y =e(T3+Y)Aµ+e(cotθWT3−tanθWY)Zµ.(88.15) Since we identify Aµas the electromagnetic ﬁeld and eas the electromag- netic coupling constant (with the convention that eis positive), we identify Q=T3+Y (88.16) as the generator of electric charge. Then, since T3ν= +1 2ν , T3e=−1 2e, T3¯e= 0, (88.17) Yν=−1 2ν , Ye =−1 2e, Y ¯e= +¯e, (88.18) we see from eq.(88.16) that Qν= 0, Qe =−e, Q ¯e= +¯e. (88.19) This is just the set of electric charge assignments that we ex pect for the electron and the neutrino. Then (since the action of Qon the ﬁelds is more 88: The Standard Model: Lepton Sector 535 familiar than the action of Y) it is convenient to replace Yin eq.(88.15) withQ−T3. We ﬁnd g2A3 µT3+g1BµY=eQAµ+e[(cotθW+ tanθW)T3−tanθWQ]Zµ =eQAµ+esWcW(T3−s2 WQ)Zµ. (88.20) In terms of the four-component ﬁelds, we have (g2A3 µT3+g1BµY)E=/bracketleftig −eAµ+esWcW(−1 2PL+s2 W)Zµ/bracketrightig E,(88.21) (g2A3 µT3+g1BµY)NL=esWcW(+1 2)ZµNL. (88.22) Using eqs.(88.14) and (88.21–88.22) in eqs.(88.1–88.3), w e ﬁnd the coup- ings of the gauge ﬁelds to the leptons, Lint=1√ 2g2W+ µJ−µ+1√ 2g2W− µJ+µ+esWcWZµJµ Z+eAµJµ EM,(88.23) where we have deﬁned the currents J+µ≡ELγµNL, (88.24) J−µ≡NLγµEL, (88.25) Jµ Z≡Jµ 3−s2 WJµ EM, (88.26) Jµ 3≡1 2NLγµNL−1 2ELγµEL, (88.27) Jµ EM≡ −EγµE. (88.28) In many cases, we are interested in scattering amplitudes fo r leptons whose momenta are all well below the W±andZ0masses. In this case, we can integrate the W± µandZµﬁelds out of the path integral, as discussed in section 29. We get the leading term (in a double expansion i n powers of the gauge couplings and inverse powers of MWandMZ) by ignoring the kinetic energy and other interactions of the W± µandZµﬁelds, solving the equations of motion for them that follow from Lmass+Lint, where Lintis given by eq.(88.23) and Lmass=−M2 WW+µW− µ−1 2M2 ZZµZµ, (88.29) and ﬁnally substituting the solutions back into Lmass+Lint. This is equiv- alent to evaluating tree-level Feynman diagrams with a sing leW±orZ0 exchanged, with the propagator gµν/M2 W,Z. The result is Leﬀ=g2 2 2M2 WJ+µJ− µ+e2 2s2 Wc2 WM2 ZJµ ZJZµ =e2 2s2 WM2 W(J+µJ− µ+Jµ ZJZµ) = 2√ 2GF(J+µJ− µ+Jµ ZJZµ). (88.30) 88: The Standard Model: Lepton Sector 536 eνµp1 p1 p3p2νeµ Figure 88.1: Feynman diagram for muon decay. The wavy line is aW propagator. We usede=g2sinθWandMW=MZcosθWto get the second line, and we deﬁned the Fermi constant GF≡e2 4√ 2 sin2θWM2 W(88.31) in the third line. We can use Leﬀto compute the tree-level scattering amplitude for processes like νee−→νee−; we leave this to the problems. Having worked out the interactions of a single lepton genera tion, we now examine what happens when there is more than one of them. L et us consider the ﬁelds ℓiIand ¯eI, whereI= 1,2,3 is a generation index. The kinetic term for all these ﬁelds is Lkin=iℓ†i I¯σµ(Dµ)ijℓjI+i¯e† I¯σµDµ¯eI, (88.32) where the repeated generation index is summed. The most gene ral Yukawa term we can write down now reads LYuk=−εijϕiℓjIyIJ¯eJ+ h.c., (88.33) whereyIJis a complex 3 ×3 matrix, and the generation indices are summed. We can make unitary transformations in generation space on t he ﬁelds: ℓI→LIJℓJand ¯eI→¯EIJ¯eJ, whereLand¯Eare independent unitary matrices. The kinetic terms are unchanged, and the Yukawa ma trixyis replaced with LTy¯E. We can choose Land¯Eso thatLTy¯Eis diagonal with positive real entries yI. The charged leptons EIthen have masses meI= yIv/√ 2, and the neutrinos remain massless. In the currents, eqs.( 88.24– 88.28), we simply add a generation index Ito each ﬁeld, and sum over it. Let us work out the details for one process of particular impo rtance: muon decay, µ−→e−νeνµ. Let the four-component ﬁelds be Efor the electron, Mfor the muon, Nefor the electron neutrino, and Nmfor the muon neutrino. Only the charged currents J± µare relevant; the neutral 88: The Standard Model: Lepton Sector 537 currentJµ Zand the electromagnetic current Jµ EMdo not contribute. Ignoring theτterms in the charged currents, we have J+µ=ELγµNeL+MLγµNmL, (88.34) J−µ=NeLγµEL+NmLγµML. (88.35) The relevant term in the eﬀective interaction is Leﬀ= 2√ 2GF(ELγµNeL)(NmLγµML). (88.36) This can be simpliﬁed by means of a Fierz identity (see proble m 36.3), Leﬀ=−4√ 2GF(MCPLNe)(EPRNC m). (88.37) Assigning momenta as shown in ﬁg.(88.1), and using the usual Feynman rules for incoming and outgoing particles and antiparticle s, the scattering amplitude is T=−4√ 2GF(uT 1CPLv′ 2)(u′ 3PRCu′T 1) =−4√ 2GF(v1PLv′ 2)(u′ 3PRv′ 1). (88.38) Taking the complex conjugate, and using PL=PR, we ﬁnd T∗=−4√ 2GF(v′ 2PRv1)(v′ 1PLu′ 3). (88.39) Multiplying eqs.(88.38) and (88.39), summing over ﬁnal spi ns and averag- ing over the initial spin, we get ∝an}b∇acketle{t|T |2∝an}b∇acket∇i}ht=1 2(4√ 2)2G2 FTr[(−/p1−mµ)PL(−/p′ 2)PR] ×Tr[(−/p′ 3+me)PR(−/p′ 1)PL]. (88.40) The traces are easily evaluated, with the result ∝an}b∇acketle{t|T |2∝an}b∇acket∇i}ht= 64G2 F(p1p′ 2)(p′ 1p′ 3). (88.41) We get the decay rate Γ by multiplying ∝an}b∇acketle{t|T |2∝an}b∇acket∇i}htbydLIPS 3(p1) and integrat- ing overp′ 1,2,3. We worked out the result (in the limit me≪mµ) in problem 11.3, Γ =G2 Fm5 µ 192π3. (88.42) After including one-loop corrections from electromagneti sm, and account- ing for the nonzero electron mass, the measured muon decay ra te is used to determine the value of GF, with the result GF= 1.166×10−5GeV−2. 88: The Standard Model: Lepton Sector 538 Reference Notes Lepton phenomonology is covered in more detail in in Cheng & Li ,Georgi , Peskin & Schroeder ,Quigg , andRamond II . Problems 88.1) Verify the claim made immediately after eq.(88.6). 88.2) Show that a neutrino always has negative helicity, and that an an- tineutrino always has positive helicity. Hint: see section 75. 88.3) Show that the sum of eqs.(88.32) and (88.33), when rewr itten in terms of ﬁelds of deﬁnite mass, has a global symmetry U(1) ×U(1)×U(1). The corresponding charges are called electron number ,muon number , andtau number ; the sum of the charges is the lepton number . List the value of each charge for each Dirac ﬁeld EIandNLI. 88.4) Compute ∝an}b∇acketle{t|T |2∝an}b∇acket∇i}htfor muon decay using eq.(88.36), without making the Fierz transformation to eq.(88.37), and verify eq.(88.41) . 88.5) a) Write down the term in Leﬀthat is relevant for and νµe−→νµe−. Express your answer in the form Leﬀ=1√ 2GFNγµ(1−γ5)NEγµ(CV−CAγ5)E, (88.43) where Nis the muon neutrino ﬁeld, and determine the values of CV andCA. b) Repeat part (a) for νee−→νee−. c) Compute ∝an}b∇acketle{t|T |2∝an}b∇acket∇i}htas a function of the Mandelstam variables, the electron mass, and CVandCA. 88.6) Compute the rates for the decay processes W+→e+νe,Z0→e+e−, andZ0→νeνe. Neglect the electron mass. Express your results in GeV. 88.7)Anomalous dimension of the Fermi constant. The coeﬃcient of the eﬀective interaction for muon decay, eq.(88.36), is subjec t to renor- malization by quantum electrodynamic processes. In partic ular, we can compute its anomalous dimension γG, deﬁned via µd dµGF(µ) =γG(α)GF(µ), (88.44) whereα=e2/4πis the ﬁne-structure constant in the MS scheme with renormalization scale µ. 88: The Standard Model: Lepton Sector 539 a) Argue that it is GF(MW) that is given by eq.(88.31). b) Multiply eq.(88.36) by a renormalzing factor ZG, and deﬁne ln(ZG/Z2) =∞/summationdisplay n=1Gn(α) εn, (88.45) whereZ2is the renormalizing factor for a ﬁeld of unit charge in spino r electrodynamics. Show that γG(α) =αG′ 1(α). (88.46) c) IfγG(α) =c1α+O(α2) andβ(α) =b1α2+O(α3), show that GF(µ) =/bracketleftbiggα(µ) α(MW)/bracketrightbiggc1/b1 GF(MW) (88.47) forµ < M W. (Forµ > M W, we should not be using an eﬀective interaction.) d) Ifα(µ)ln(MW/µ)≪1, show that eq.(88.47) becomes GF(µ) =/bracketleftig 1−c1α(µ)ln(MW/µ)/bracketrightig GF(MW). (88.48) e) Use a Fierz identity to rewrite eq.(88.36) in charge retention form , Leﬀ= 2√ 2ZGGF(ELγµML)(NmLγµNeL). (88.49) f) Consider the process of muon decay with an extra photon con - necting the µandelines. Work in Lorenz gauge, and with the four- fermion vertex provided by eq.(88.49). Use your results fro m problem 62.2 to show that, in this gauge, there is no O(α) contribution to ZG in the MS scheme. g) Use your result from part (d), and your result for Z2in Lorenz gauge from problem 62.2, to show that c1= 0, and hence that GF(µ) =GF(MW) at the one-loop level. 89: The Standard Model: Quark Sector 540 89The Standard Model: Quark Sector Prerequisite: 88 Quarks are spin-one-half particles that are triplets of the color g roup. There are six diﬀerent ﬂavors of quark; see Table 1 in section 83. Th e six ﬂavors are naturally grouped into three families orgenerations :uandd,cands, tandb. Let us begin by describing a single quark family, the up and do wn quarks. We introduce left-handed Weyl ﬁelds q, ¯u, and ¯din the represen- tations (3,2,+1 6), (¯3,1,−2 3), and ( ¯3,1,+1 3) of SU(3) ×SU(2) ×U(1). Here the bar over the letter in the ﬁelds ¯ uand¯dispart of the name of the ﬁeld , and does not denote any sort of conjugation. The covariant derivatives of these ﬁelds are (Dµq)αi=∂µqαi−ig3Aa µ(Ta 3)αβqβi−ig2Aa µ(Ta 2)ijqβj −ig1(+1 6)Bµqαi,(89.1) (Dµ¯u)α=∂µ¯uα−ig3Aa µ(Ta¯3)αβ¯uβ−ig1(−2 3)Bµ¯uα, (89.2) (Dµ¯d)α=∂µ¯dα−ig3Aa µ(Ta 3)αβ¯dβ−ig1(+1 3)Bµ¯dα. (89.3) We rely on context to distinguish the SU(3) gauge ﬁelds from t he SU(2) gauge ﬁelds. The kinetic terms for q, ¯u, and ¯dare Lkin=iq†αi¯σµ(Dµq)αi+i¯u† α¯σµ(Dµ¯u)α+i¯d† α¯σµ(Dµ¯d)α. (89.4) The representation (3 ,2,+1 6)⊕(¯3,1,−2 3)⊕(¯3,1,+1 3) for the left-handed Weyl ﬁelds is complex; hence the gauge theory is chiral, and t herefore parity violating. We cannot write down a mass term involving q, ¯u, and/or ¯dbecause there is no gauge-group singlet contained in any of the produ cts of their representations. But we are able to write down Yukawa coupli ngs of the form LYuk=−y′εijϕiqαj¯dα−y′′ϕ†iqαi¯uα+ h.c., (89.5) whereϕis the Higgs ﬁeld in the (1 ,2,−1 2) representation that we introduced in section 87, and y′andy′′are the Yukawa coupling constants. These gauge-invariant Yukawa couplings are possible because the re are singlets on the right-hand sides of (1,2,−1 2)⊗(3,2,+1 6)⊗(¯3,1,+1 3) = (1,1,0)⊕... , (89.6) (1,2,+1 2)⊗(3,2,+1 6)⊗(¯3,1,−2 3) = (1,1,0)⊕... . (89.7) 89: The Standard Model: Quark Sector 541 There are no other gauge-invariant terms involving q, ¯u, or¯dthat have mass dimension four or less. Hence there are no other terms th at we could add to Lwhile preserving renormalizability. In unitary gauge, we replace ϕ1with1√ 2(v+H), whereHis the real scalar ﬁeld representing the physical Higgs boson, and ϕ2with zero. The Yukawa term becomes LYuk=−1√ 2y′(v+H)qα2¯dα−1√ 2y′′(v+H)qα1¯uα+ h.c.. (89.8) It is now convenient to assign new names to the SU(2) componen ts ofq, q=/parenleftiggu d/parenrightigg . (89.9) Then eq.(89.8) becomes LYuk=−1√ 2y′(v+H)(dα¯dα+¯d† αd†α)−1√ 2y′′(v+H)(uα¯uα+ ¯u† αu†α) =−1√ 2y′(v+H)DαDα−1√ 2y′′(v+H)UαUα, (89.10) where we have deﬁned Dirac ﬁelds for the down and up quarks, Dα≡/parenleftiggdα ¯d† α/parenrightigg ,Uα≡/parenleftigguα ¯u† α/parenrightigg . (89.11) We see from eq.(89.10) that the up and down quarks have acquir ed masses md=y′v√ 2, mu=y′′v√ 2. (89.12) Now we return to eqs.(89.1–89.3), and express the covariant derivatives in the terms of the W± µ,Zµ, andAµﬁelds. From our results in section 88, we have g2A1 µT1+g2A2 µT2=g2√ 2/parenleftigg0W+ µ W− µ0/parenrightigg , (89.13) g2A3 µT3+g1BµY=eQAµ+esWcW(T3−s2 WQ)Zµ,(89.14) where Q=T3+Y (89.15) is the generator of electric charge. Then, since T3u= +1 2u, T3d=−1 2d, T3¯u= 0, T3¯d= 0, (89.16) Yu= +1 6u, Yd = +1 6d, Y ¯u=−2 3¯u, Y ¯d= +1 3¯d, (89.17) we see from eq.(89.15) that 89: The Standard Model: Quark Sector 542 Qu= +2 3u Qd =−1 3d, Q ¯u=−2 3¯u, Q ¯d= +1 3¯d. (89.18) This is just the set of electric charge assignments that we ex pect for the up and down quarks. In terms of the four-component ﬁelds, we hav e (g2A3 µT3+g1BµY)U=/bracketleftig +2 3eAµ+esWcW(+1 2PL−2 3s2 W)Zµ/bracketrightig U,(89.19) (g2A3 µT3+g1BµY)D=/bracketleftig −1 3eAµ+esWcW(−1 2PL+1 3s2 W)Zµ/bracketrightig D,(89.20) Using eqs.(89.13) and (89.19–89.20) in eqs.(89.1–89.4), w e ﬁnd the coup- ings of the electroweak gauge ﬁelds to the quarks, Lint=1√ 2g2W+ µJ−µ+1√ 2g2W− µJ+µ+esWcWZµJµ Z+eAµJµ EM,(89.21) where we have deﬁned the currents J+µ≡DLγµUL, (89.22) J−µ≡ULγµDL, (89.23) Jµ Z≡Jµ 3−s2 WJµ EM, (89.24) Jµ 3≡1 2ULγµUL−1 2DLγµDL, (89.25) Jµ EM≡+2 3UγµU −1 3DγµD. (89.26) Having worked out the interactions of a single quark generat ion, we now examine what happens when there is more than one of them. L et us consider the ﬁelds qαiI, ¯uI, and ¯dI, whereI= 1,2,3 is a generation index. The kinetic term for all these ﬁelds is Lkin=iq†αiI¯σµ(Dµ)αiβjqβjI+i¯u† αI¯σµ(Dµ)αβ¯uβ I+i¯d† αI¯σµ(Dµ)αβ¯dβ I, (89.27) where the repeated generation index is summed. The most gene ral Yukawa term we can write down now reads LYuk=−εijϕiqαjIy′ IJ¯dα J−ϕ†iqαiIy′′ IJ¯uα J+ h.c., (89.28) wherey′ IJandy′′ IJare complex 3 ×3 matrices, and the generation indices are summed. In unitary gauge, this becomes LYuk=−1√ 2(v+H)dαIy′ IJ¯dα J−1√ 2(v+H)uαIy′′ IJ¯uα J+ h.c.. (89.29) We can make unitary transformations in generation space on t he ﬁelds: dI→DIJdJ,¯dI→¯DIJ¯dJ,uI→UIJuJ, and ¯uI→¯UIJ¯uJ, whereU,D,¯U and¯Dare independent unitary matrices. The kinetic terms are unc hanged 89: The Standard Model: Quark Sector 543 (except for the couplings to the W±, as we will discuss momentarily), and the Yukawa matrices y′andy′′are replaced with DTy′¯DandUTy′′¯U. We can choose D,¯D,U, and ¯Uso thatDTy′¯DandUTy′′¯Uare diagonal with positive real entries y′ Iandy′′ I. The down quarks DIthen have masses mdI=y′ Iv/√ 2, and the up quarks UIhave masses muI=y′′ Iv/√ 2. In the neutral currents Jµ 3andJµ EM, we simply add a generation index Ito each ﬁeld, and sum over it. The charged currents are more comp licated, however; they become J+µ=DLI(V†)IJγµULJ, (89.30) J−µ=ULIVIJγµDLK, (89.31) whereV≡U†Dis the Cabibbo–Kobayashi–Maskawa matrix (orCKM ma- trixfor short). Note that we did not have this complication in the lepton sector, because there we had only one Yukawa term. A 3×3 unitary matrix has 9 real parameters. However, we are still free to make the independent phase rotations DI→eiαIDIandUI→eiβIUI, as these leave the kinetic and mass terms invariant. These ph ase changes allow us to make the ﬁrst row and column of VIJreal, eliminating 5 of the 9 parameters. The remaining four can be chosen as θ1(theCabibbo angle ), θ2,θ3, andδ, where V= c1 +s1c3 +s1s3 −s1c2c1c2c3−s2s3eiδc1c2s3+s2c3eiδ −s1s2c1s2c3+c2s3eiδc1s2s3−c2c3eiδ , (89.32) andci= cosθiandsi= sinθi. The measured values of these angles are s1= 0.224,s2= 0.041,s3= 0.016, andδ= 40◦. Note that the charged currents now have some terms with a phase factor eiδ, and some without. Since the time-reversal operator Tis antiunitary ( T−1iT=−i), the charged currents do not transform in a simple way under time reversal . This implies that the charged current terms in Lintare is not time-reversal invariant; hence the electroweak interactions violate time-reversal symmetry. Since CPT is always a good symmetry, time-reversal violation is equiv alent to CPviolation;δis therefore sometimes called the CPviolating phase . At high energies, we can use our results to compute electrowe ak con- tributions to scattering amplitudes involving quarks. Thi s is because, at high energies, the SU(3) coupling g3is weak; we can, for example, re- liably compute the decay rates of the W±andZ0into quarks, because α3(MZ)≡g2 3(MZ)/4π= 0.12 is small enough to make QCD loop correc- tions a few-percent eﬀect. 89: The Standard Model: Quark Sector 544 To understand low-energy processes such as neutron decay, w e must ﬁrst write the currents in terms of hadron ﬁelds. We take this up in the next section. For now, we simply note that the terms in the charged currents that involve only up and down quarks are J+µ=c1DLγµUL, (89.33) J−µ=c1ULγµDL, (89.34) wherec1is the cosine of the Cabibbo angle. Reference Notes Electroweak interactions of quarks are discussed in more de tail in Cheng & Li,Georgi ,Peskin & Schroeder ,Quigg , andRamond II . Problems 89.1) Verify the claims made immediately after eqs.(89.4) a nd (89.7). 89.2) Compute the rates for the decay processes W+→u¯d,Z0→¯uu, andZ0→¯dd. Neglect the quark masses. Express your results in GeV. Combine your answers with those of problem 88.6, and sum over generations to get the total decay rates for the W±andZ0. You can neglect the masses of all quarks and leptons except the to p quark, and takeθ2=θ3= 0. 89.3) Show that the Standard Model is anomaly free. Hint: you must consider 3–3–3, 2–2–2, 3–3–1, 2–2–1, and 1–1–1 anomalies, w here the number denotes the gauge group of one of the external gauge ﬁe lds in the triangle diagram. Why do we not need to worry about the unlisted combinations? 89.4) Compute the leading term in the beta function for each o f the three gauge couplings of the Standard Model. 89.5) After integrating out the W±ﬁelds, we get an eﬀective interaction between the hadron and lepton currents that includes Leﬀ= 2√ 2ZCC(ELγµNeL)(ULγµDL), (89.35) where we have deﬁned C≡c1GFat a renormalization scale µ=MW, andZCis a renormalizing factor. This interaction contributes to neutron decay; see section 90. In this problem, following th e analysis of problem 88.7, we will compute the anomalous dimension γCofC due to one-loop photon and gluon exchange. 89: The Standard Model: Quark Sector 545 a) Use Fierz identities to show that eq.(89.35) can be rewrit ten as Leﬀ= 2√ 2ZCC(ELγµDL)(ULγµNeL), (89.36) and also as Leﬀ=−4√ 2ZCC(DCPLNe)(EPRUC). (89.37) b) Working in Lorenz gauge and using the results of problem 88 .7, show that gluon exchange does not make a one-loop contributi on to ZC. c) Show that only a photon connecting the eandulines makes a one-loop contribution to ZC. d) Note that EPRUC=e†u†, and compare this with EE=e†¯e†+ h.c.. Argue that the photon-exchange contribution to ZCis given by the one-loop contribution to Zmin spinor electrodynamics in Lorenz gauge, with the replacement ( −1)(+1)e2→(−1)(+2 3)e2. e) LetγC(α) =c1α+..., whereα=e2/4π, and ﬁndc1. (Thisc1 should not be confused with the cosine of the Cabibbo angle.) 90: Electroweak Interactions of Hadrons 546 90Electroweak Interactions of Hadrons Prerequisite: 83, 89 Now that we know how quarks couple to the electroweak gauge ﬁe lds, we can use this information to obtain amplitudes for various pr ocesses in- volving hadrons. We will focus on three of the most important : neutron decay,n→pe−¯ν; charged pion decay, π−→µ−¯νµ; and neutral pion decay, π0→γγ. Recall from section 83 the chiral lagrangian for pions and nu cleons, L=−1 4f2 πTr∂µU†∂µU+v3Tr(MU+M†U†) +iN/∂N−mNN(U†PL+UPR)N −1 2(gA−1)iNγµ(U∂µU†PL+U†∂µUPR)N , (90.1) whereU(x) = exp[2iπa(x)Ta/fπ],πais the pion ﬁeld, Nis the nucleon ﬁeld, fπis the pion decay constant, Mis the quark mass matrix, v3is the value of the quark condensate, mNis the nucleon mass, and gAis the axial vector coupling. The electroweak gauge group SU(2) ×U(1) is a subgroup of the SU(2) L×SU(2) R×U(1) Vﬂavor group that we have in the limit of zero quark mass. It will prove convenient to go through the formal procedure of gauging the full ﬂavor group, and only later identifying t he electroweak subgroup. We therefore deﬁne matrix-valued gauge ﬁelds lµ(x) andrµ(x) that transform as lµ→LlµL†+iL∂µL†, (90.2) rµ→RrµR†+iR∂µR†. (90.3) HereL(x) andR(x) are 2 ×2 unitary matrices that correspond to a general SU(2) L×SU(2) R×U(1) Vgauge transformation; we restrict the U(1) part of the transformation to the vector subgroup by requiring de tL= detR and Trlµ= Trrµ. The transformation rules for the pion and nucleon ﬁelds are U→LUR†, (90.4) NL→LNL, (90.5) NR→RNR, (90.6) whereNL≡PLNandNR≡PRNare the left- and right-handed parts of the nucleon ﬁeld. We can make the chiral lagrangian gauge invariant (except fo r terms involving the quark masses) by replacing ordinary derivati ves with appro- priate covariant derivatives. We determine the covariant d erivative of each 90: Electroweak Interactions of Hadrons 547 ﬁeld by requiring it to transform in the same way as the ﬁeld it self; for example,DµU→L(DµU)R†. We thus ﬁnd DµU=∂µU−ilµU+iUrµ, (90.7) DµU†=∂µU†+iU†lµ−irµU†, (90.8) DµNL= (∂µ−ilµ)NL, (90.9) DµNR= (∂µ−irµ)NR. (90.10) Making the substitution ∂→DinL, we learn how the pions and nucleons couple to these gauge ﬁelds. As in section 83, it is more convenient to work with the nucleo n ﬁeld N, deﬁned via N= (uPL+u†PR)N, (90.11) whereu2=U. Making this transformation, we ultimately ﬁnd L=−1 4f2 πTr(∂µU†∂µU−ilµU↔∂µU†−irµU†↔∂µU +lµlµ+rµrµ−2lµUrµU†) +v3Tr(MU+M†U†) +iN/∂N −mNNN +N(/v+1 2/˜ℓ+1 2/ ˜r)N −gAN(/a+1 2/˜ℓ−1 2/ ˜r)γ5N,(90.12) where vµ≡1 2i[u†(∂µu) +u(∂µu†)], (90.13) aµ≡1 2i[u†(∂µu)−u(∂µu†)], (90.14) ˜lµ≡u†lµu, (90.15) ˜rµ≡urµu†. (90.16) It is now convenient to set lµ=la µTa+bµ, (90.17) rµ=ra µTa+bµ. (90.18) We have normalized bµso that the corresponding charge is baryon number. The SU(2) gauge ﬁelds of the Standard Model can now be identiﬁ ed as g2Aa µ=la µ, (90.19) and the electromagnetic gauge ﬁeld as eAµ=l3 µ+r3 µ+1 2bµ. (90.20) 90: Electroweak Interactions of Hadrons 548 Eq.(90.20) follows from reconciling eqs.(90.9) and (90.10 ) with the re- quirement that the electromagnetic covariant derivatives of the proton ﬁeld p=N1and the neutron ﬁeld n=N2be given by ( ∂µ−ieAµ)pand∂µn. We can now ﬁnd the hadronic parts of the currents that couple t o the gauge ﬁelds by diﬀerentiating Lwith respect to them, and then setting them to zero. We ﬁnd Jaµ L= (∂L/∂la µ)/vextendsingle/vextendsingle/vextendsingle l=r=0 =1 4if2 πTrTaU↔ ∂µU†+1 2Nu†Taγµ(1−gAγ5)uN = +1 2fπ∂µπa−1 2εabcπb∂µπc+1 2NTaγµ(1−gAγ5)N+... , (90.21) Jaµ R= (∂L/∂ra µ)/vextendsingle/vextendsingle/vextendsingle l=r=0 =1 4if2 πTrTaU†↔ ∂µU+1 2NuTaγµ(1+gAγ5)u†N =−1 2fπ∂µπa−1 2εabcπb∂µπc+1 2NTaγµ(1+gAγ5)N+... , (90.22) Jµ B= (∂L/∂bµ)/vextendsingle/vextendsingle/vextendsingle l=r=0 =NγµN. (90.23) In the third lines of eqs.(90.21) and (90.22), we have expand ed in inverse powers offπ. We can now identify the currents that couple to the physical W± µ,Zµ, andAµﬁelds as J+µ=c1(J1µ L−iJ2µ L) =1√ 2c1(fπ∂µπ++iπ0↔∂µπ+) +1 2c1nγµ(1−gAγ5)p+... , (90.24) J−µ=c1(J1µ L+iJ2µ L) =1√ 2c1(fπ∂µπ−−iπ0↔ ∂µπ−) +1 2c1pγµ(1−gAγ5)n+... , (90.25) Jµ Z=Jµ 3−s2 WJµ EM, (90.26) Jµ 3=J3µ L =1 2(fπ∂µπ0+iπ+↔ ∂µπ−) +1 4pγµ(1−gAγ5)p−1 4nγµ(1−gAγ5)n+... , (90.27) Jµ EM=J3µ L+J3µ R+1 2Jµ B =iπ+↔ ∂µπ−+pγµp+... , (90.28) 90: Electroweak Interactions of Hadrons 549 wherec1is the cosine of the Cabibbo angle, and the interactions are s pec- iﬁed by Lint=1√ 2g2W+ µJ−µ+1√ 2g2W− µJ+µ+esWcWZµJµ Z+eAµJµ EM.(90.29) For low-energy processes involving W±orZ0exchange, we can use the eﬀective current-current interaction that we derived in se ction 88, Leﬀ= 2√ 2GF(J+µJ− µ+Jµ ZJZµ). (90.30) We should include both hadronic and leptonic contributions to the currents. Consider charged pion decay, π−→µ−νµ. The relevant terms in the charged currents (neutral currents do not contribute) are J−µ=1√ 2c1fπ∂µπ−, (90.31) J+µ=1 2Mγµ(1−γ5)Nm, (90.32) where Mis the muon ﬁeld and Nmis the muon neutrino ﬁeld. The relevant term in the eﬀective interaction is then Leﬀ=GFc1fπ∂µπ−Mγµ(1−γ5)Nm. (90.33) The corresponding decay amplitude is T=GFc1fπkµu1γµ(1−γ5)v2, (90.34) where the four-momenta of the pion, muon, and antineutrino a rek,p1, andp2. Eq.(90.34) can be simpliﬁed by using / k= /p1+ /p2along with u1/p1=−mµu1and /p2v2= 0; we get T=−GFc1fπmµu1(1−γ5)v2. (90.35) We see that Tis proportional to the muon mass; since mµ≫me, decay to µ−¯νµis preferred over decay to e−¯νe. Squaring Tand summing over ﬁnal spins, we ﬁnd ∝an}b∇acketle{t|T |2∝an}b∇acket∇i}ht= (GFc1fπmµ)2(−8p1·p2) = 4(GFc1fπmµ)2(m2 π−m2 µ). (90.36) We used −2p1·p2=p2 1+p2 2−(p1+p2)2=−m2 µ+ 0 +m2 πto get the second line. We now have Γ =1 2mπ/integraldisplay dLIPS 2(k)∝an}b∇acketle{t|T |2∝an}b∇acket∇i}ht =|p1| 8πm2π∝an}b∇acketle{t|T |2∝an}b∇acket∇i}ht =G2 Fc2 1f2 πm2 µmπ 4π/parenleftigg 1−m2 µ m2π/parenrightigg2 , (90.37) 90: Electroweak Interactions of Hadrons 550 where we used |p1|= (m2 π−m2 µ)/2mπto get the last line. Since we deter- mine the value of GFfrom the decay rate of the muon (see section 88), the charged pion decay rate allows us to ﬁx the value of c1fπ. The value of c1can be determined from the rate for the decay pro- cessπ−→π0e−¯νe, which we will calculate in problem 90.6. The relevant hadronic term in the charged current is J−µ=−1√ 2ic1π0↔ ∂µπ−, (90.38) which depends on c1but notfπ. Comparison with experiment then yields c1= 0.974. We note that the key feature of eq.(90.38) is that it invo lves spin-zero hadrons that are members of an isospin triplet; eq .(90.38) applies to any such hadrons, including nuclei. Thus c1can also be measured in superallowed Fermi decays , which take a nucleus from one spin-zero state to another spin-zero state with the same parity in the same isos pin multiplet. Having thus determined c1, the charged pion decay rate yields fπ= 92.4MeV. Next we consider neutron decay, n→pe−¯νe. The relevant terms in the charged currents (neutral currents do not contribute) are J−µ=1 2c1pγµ(1−gAγ5)n, (90.39) J+µ=1 2Eγµ(1−γ5)Ne, (90.40) where Eis the electron ﬁeld and Neis the electron neutrino ﬁeld. The relevant term in the eﬀective interaction is then Leﬀ=1√ 2GFc1pγµ(1−gAγ5)nEγµ(1−γ5)Ne. (90.41) Consider a neutron with four-momentum pn= (mn,0), and spin up along thezaxis; the decay amplitude is T=1√ 2GFc1[upγµ(1−gAγ5)un][ueγµ(1−γ5)v¯ν], (90.42) whereunun=1 2(1−γ5/z)(−/pn+mn). We take the absolute square of Tand sum over the ﬁnal spins. Since the maximum available kinetic energy is mn−mp−me= 0.782MeV ≪mp, the proton is nonrelativistic, and we can use the approximations pp·pe≃ −mpEeandpp·p¯ν≃ −mpE¯νin addition to the exact formulae pn·pe=−mnEeandpn·p¯ν=−mnE¯ν. After a tedious but straightforward calculation, we ﬁnd ∝an}b∇acketle{t|T |2∝an}b∇acket∇i}ht= 16G2 Fc2 1(1 + 3g2 A)mnmpEeE¯ν ×/bracketleftbigg 1 +ape·p¯ν EeE¯ν+Aˆ z·pe Ee+Bˆ z·p¯ν E¯ν/bracketrightbigg , (90.43) 90: Electroweak Interactions of Hadrons 551 k 1 k1 k k2 k2l k l klµ νlµν1 l+k 2l+k 21 k Figure 90.1: One-loop diagrams contributing to π0→γγ. The solid line is a proton. where the correlation coeﬃcients are given by a=1−g2 A 1 + 3g2 A, A=2gA(1−gA) 1 + 3g2 A, B =2gA(1 +gA) 1 + 3g2 A. (90.44) When we integrate over the ﬁnal momenta to get the total decay rate, the correlation terms vanish, and so the rate is proportional to G2 Fc2 1(1 + 3g2 A). Since we get the value of G2 Fc2 1from the rate for π−→π0e−¯νe. the neutron decay rate allows us to determine 1 + 3 g2 A. To get the sign of gA, we need a measurement of either AorB. (The antineutrino three-momentum can be determined from the electron and proton three-momenta.) The result is thatgA= +1.27. The measured values of the three correlation coeﬃcients are all consistent with eq.(90.44). Finally, we consider the decay of the neutral pion into two ph otons, π0→γγ. None of the terms in our chiral lagrangian, eq.(90.12), cou ple a singleπ0to two photons. Therefore, without adding more terms, this process does not occur at tree level. However, at the one-loo p level, we have the diagrams of ﬁg.(90.1); a proton circulates in the lo op. Let us evaluate these diagrams. In section 83, we found that the cou pling of the π0to the nucleons is given by Lπ0NN=−1 2(gA/fπ)∂µπ0(pγµγ5p−nγµγ5n). (90.45) This leads to a π0ppvertex factor of1 2(gA/fπ)kργργ5. The diagrams in ﬁg.(90.1) are then identical to the diagrams we evaluated in section 76, and so the one-loop decay amplitude is iT1−loop=1 2(gA/fπ)(ie)2ε1µε2νkρCµνρ(k1,k2,k), (90.46) where kρCµνρ(k1,k2,k) =−i 2π2εµναβk1αk2β. (90.47) Here we have chosen to renormalize so as to have k1µCµνρ(k1,k2,k) = 0 and k2νCµνρ(k1,k2,k) = 0; this is required by electromagnetic gauge invariance. 90: Electroweak Interactions of Hadrons 552 Combining eqs.(90.46) and (90.47), we get T1−loop=−gAe2 4π2fπεαµβνk1αε1µk2βε2ν. (90.48) This result is subject to higher-loop corrections. Note tha t diagrams with extra internal pion lines attached to the nucleon loop are no t suppressed by any small expansion parameter. Thus we cannot trust the over all coeﬃcient in eq.(90.48). Note that this amplitude would arise at tree level from an int eraction of the form Lπ0γγ∝π0εαµβνFµαFνβ. If we integrate out the nucleon ﬁelds to get an eﬀective lagrangian for the pions and photons alone , such a term should appear. There is a problem, however. The SU(2) L×SU(2) R×U(1) Vsymmetry of the eﬀective lagrangian implies that a pion ﬁeld that has n o derivatives acting on it must be accompanied by at least one factor of a qua rk mass. For example, we could have Lπ0γγ∝iTr(MU−M†U†)εαµβνFµαFνβ. The problem is that there are no quark-mass factors in eq.(90.48 ). So we have an apparent contradiction between our explicit one-loop resu lt, and a general argument based on symmetry. This contradiction is resolved by noting that the electroma gnetic gauge ﬁeld results in an anomaly in the axial current J3µ A≡J3µ L−J3µ R. In terms of the quark doublet Q=/parenleftiggU D/parenrightigg , (90.49) this current is J3µ A=QT3γµγ5Q =1 2Uγµγ5U −1 2Dγµγ5D, (90.50) where we have suppressed the color indices. Using our result s in sections 76 and 77, the anomalous divergence of this current is given b y ∂µJ3µ A=−e2 16π2Tr(T3Q2)εµνρσFµνFρσ, (90.51) where Q=/parenleftigg+2 30 0−1 3/parenrightigg (90.52) is the electric charge matrix acting on the quark ﬁelds, and t he trace in- cludes a factor of three for color; we thus have Tr(T3Q2) = 3/parenleftig 1 2(+2 3)2−1 2(−1 3)2/parenrightig = +1 2, (90.53) 90: Electroweak Interactions of Hadrons 553 and so ∂µJ3µ A=−e2 32π2εµνρσFµνFρσ. (90.54) This formula is exact in the limit of zero quark mass. Now using eqs.(90.21) and (90.22), we can write the axial cur rent in terms of the pion ﬁelds as J3µ A≡J3µ L−J3µ R =fπ∂µπ0+... . (90.55) (We do not include the nucleon contribution because we are co nsidering the eﬀective lagrangian for pions and photons after integra ting out the nucleons.) From eq.(90.55) we have ∂µJ3µ A=fπ∂2π0+...; Combining this with eq.(90.54), we get −∂2π0=e2 32π2fπεµνρσFµνFρσ+O(f−2 π). (90.56) This equation of motion would follow from an eﬀective lagran gian that included an interaction term of the form Lπ0γγ=e2 32π2fππ0εµνρσFµνFρσ. (90.57) This interaction leads to a π0→γγdecay amplitude of T=−e2 4π2fπεµνρσk1µε1νk2ρε2σ. (90.58) This amplitude receives no higher-order corrections in e2, but is subject to quark-mass corrections; these are suppressed by powers o fm2 π/(4πfπ)2. Comparing eq.(90.58) with eq.(90.48), we see the one-loop r esult (which receives unsuppressed corrections) is too large by a factor ofgA= 1.27. Squaring T, summing over ﬁnal spins, integrating over dLIPS 2(k), and multiplying by a symmetry factor of one half (because there a re two iden- tical particles in the ﬁnal state), we ultimately ﬁnd that th e decay rate is Γ =α2m3 π 64π3f2π. (90.59) This prediction is in agreement with the experimental resul t, which has an uncertainty of about 7%. Reference Notes Electoweak interactions of hadrons are treated in Georgi andRamond II . 90: Electroweak Interactions of Hadrons 554 Problems 90.1) Verify that the covariant derivatives in eqs.(90.7–9 0.10) transform appropriately. 90.2) Verify that substituting eq.(90.11) into eq.(90.1) y ields eq.(90.12). 90.3) Compute the rate for the decay process τ−→π−ντ. Look up the measured value and compare with your result. 90.4) a) Verify eq.(90.43). b) Compute the total neutron decay rate. Given the measured n eu- tron lifetime τ= 886s, and using GF= 1.166×10−5GeV−2and c1= 0.974, compute gA. Your answer is about 4% too high, because we neglected loop corrections, and the Coulomb interaction between the outgoing electron and proton. 90.5) Use your results from problems 88.7 and 89.5 to show tha t the neutron decay rate is enhanced by a factor of 1 +2 παln(MW/mp). How much of the 4% discrepancy is accounted for by this eﬀect? 90.6) Compute the rate for the decay process π−→π0e−¯νe. Note that, sincemπ+−mπ0= 4.594MeV ≪mπ0, the outgoing π0is nonrela- tivistic. Compare your calculated rate with the measured va lue of 0.397s−1to determine c1. Your answer is about 1% too low, due to neglect of loop corrections. 90.7) Verify eq.(90.59). Express Γ in eV. 91: Neutrino Masses 555 91Neutrino Masses Prerequisite: 89 Recall from sections 88 and 89 that a single generation of qua rks and leptons consists of left-handed Weyl ﬁelds qαi, ¯uα,¯dα,ℓi, and ¯ein the representa- tions (3,2,+1 6), (¯3,1,−2 3), (¯3,1,+1 3), (1,2,−1 2), and (1,1,+1) of the gauge group SU(3) ×SU(2) ×U(1). The Higgs ﬁeld is a complex scalar ϕiin the representation (1 ,2,−1 2). The Yukawa couplings among these ﬁelds that are allowed by the gauge symmetry are LYuk=−yεijϕiℓj¯e−y′εijϕiqαj¯dα−y′′ϕ†iqαi¯uα+ h.c.. (91.1) After the Higgs ﬁeld acquires its VEV, these three terms give masses to the electron, down quark, and up quark, respectively. The neutr ino remains massless. Thus, massless neutrinos are a prediction of the S tandard Model. However, there is now good experimental evidence that the th ree neutri- nos actually have small masses. The data implies that mass of the heaviest neutrino is in the range from 0 .04eV to 0.5eV. To account for this, we must extend the Standard Model. Let us continue to consider a single generation. We introduc e a new left-handed Weyl ﬁeld ¯ νin the representation (1 ,1,0); this ﬁeld does not couple to the gauge ﬁelds at all, and its kinetic term is simpl yi¯ν†¯σµ∂µ¯ν. (The bar over the νin the ﬁeld ¯ νispart of the name of the ﬁeld , and does not denote any sort of conjugation.) With this new ﬁeld, we can introduce a new Yukawa coupling of the form LνYuk=−˜yϕ†iℓi¯ν+ h.c.. (91.2) In unitary gauge, this becomes LνYuk=−1√ 2˜y(v+H)(ν¯ν+ ¯ν†ν†). (91.3) We see that the neutrino mass is ˜ m= ˜yv/√ 2. If this was the end of the story, we would have no understandin g of why the neutrino mass is so much less than the other ﬁrst-generat ion quark and lepton masses; we would simply have to take ˜ ymuch less than y,y′, and y′′. However, because ¯ νis in a real representation of the gauge group, we are allowed by the gauge symmetry to add a mass term of the form L¯νmass=−1 2M(¯ν¯ν+ ¯ν†¯ν†). (91.4) HereMis an arbitrary mass parameter. In particular, it could be qu ite large. 91: Neutrino Masses 556 Adding eqs.(91.3) and (91.4), we ﬁnd a mass matrix of the form Lν¯νmass=−1 2(ν¯ν)/parenleftigg0 ˜m ˜m M/parenrightigg/parenleftiggν ¯ν/parenrightigg + h.c.. (91.5) If we takeM≫˜m, then the eigenvalues of this mass matrix are Mand −˜m2/M. (The sign of the smaller eigenvalue can be absorbed into the phase of the corresponding eigenﬁeld.) Thus, if ˜ mis of the order of the electron mass, then ˜ m2/Mis less than 1eV if Mis greater than 103GeV. So ˜ycan be of the same order as the other Yukawa couplings, provid edM is large. This is called the seesaw mechanism for getting small neutrino masses. The eigenﬁeld corresponding to the smaller eigenva lue is mostly ν, and the eigenﬁeld corresponding to the larger eigenvalue is mostly ¯ν. Another way to get this result is to integrate out the heavy ¯ νﬁeld at the beginning of our analysis. We get the leading term (in an e xpansion in inverse powers of M) by ignoring the kinetic energy of the ¯ νﬁeld, solving the equation of motion for it that follows from L¯νmass+LνYuk, and ﬁnally substituting the solution back into L¯νmass+LνYuk. The result is LνYuk+mass =˜y2 2M/bracketleftig (ϕ†iℓi)(ϕ†jℓj) + h.c./bracketrightig =−1 2mν(νν+ν†ν†)(1 +H/v)2, (91.6) where mν≡ −˜m2 M=−˜y2v2 2M. (91.7) Again, we can absorb the minus sign in eq.(91.7) by making the ﬁeld re- deﬁnitionν→iν. The seesaw mechanism has a straightforward extension to thr ee gen- erations. Let us consider the ﬁelds ℓiI, ¯eI, and ¯νI, whereI= 1,2,3 is a generation index. The most general Yukawa and mass terms we c an write down now read LYuk+mass =−εijϕiℓjIyIJ¯eJ−ϕ†iℓiI˜yIJ¯νJ−1 2MIJ¯νI¯νJ+ h.c.,(91.8) whereyIJand ˜yIJare complex 3 ×3 matrices, MIJis a complex symmetric 3×3 matrix, and the generation indices are summed. In unitary g auge, this becomes LYuk+mass =−1√ 2(v+H)eIyIJ¯eJ−1√ 2(v+H)νI˜yIJ¯νJ−1 2MIJ¯νI¯νJ+ h.c.. (91.9) We can now integrate out the ¯ νIﬁelds; eq.(91.9) is then replaced with LYuk+mass =−1√ 2(v+H)eIyIJ¯eJ−1 2(mν)IJ(νIνJ+ν† Iν† J)(1+H/v)2,(91.10) 91: Neutrino Masses 557 where we have deﬁned the complex symmetric neutrino mass mat rix (mν)IJ≡ −1 2v2(˜yTM−1˜y)IJ. (91.11) We can make unitary transformations in generation space on t he ﬁelds: eI→EIJeJ, ¯eI→¯EIJ¯eJ, andνI→NIJνJ, whereE,¯E, andNare inde- pendent unitary matrices. The kinetic terms are unchanged ( except for the couplings to the W±, as we will discuss momentarily), and the matrices y andmνare replaced with ETy¯EandNTmνN. We can choose the unitary matricesE,¯E, andNso thatETy¯EandNTmνNare diagonal with posi- tive real entries yIandmνI. The neutrinos NIthen have masses mνI, and the charged leptons EIhave masses meI=yIv/√ 2. In the neutral currents Jµ 3andJµ EM, we simply add a generation index Ito each ﬁeld, and sum over it. The charged currents are more complicated, however ; they become J+µ=ELI(X†)IJγµNLJ, (91.12) J−µ=NLIXIJγµELK, (91.13) whereX≡N†Eis the analog in the lepton sector of the CKM matrix V in the quark sector. One diﬀerence, though, between XandVis that the phases of the Majorana NIﬁelds are ﬁxed by the requirement that the neutrino masses are real and positive. Thus we cannot change these phases to m ake the ﬁrst column ofXreal, as we did with V. Weareallowed to change the phases of the Dirac EIﬁelds, so we can make the ﬁrst row of Xreal. ThusXhas 9−3 = 6 parameters, two more than the CKM matrix V. The presence of Xin the charged currents leads to the phenomenon of neutrino oscillations . A neutrino that is produced by scattering an electron oﬀ a target will be a linear combination XIJνJof the neutrinos of deﬁnite mass. The diﬀerent mass eigenstates propagate at diﬀerent s peeds, and then (in a subsequent scattering) may become (if there is eno ugh energy) muons or taus rather than electrons. It is the observation of neutrino oscillations that leads us to believe that neutrinos do, in f act, have mass. Reference Notes Neutrino masses are discussed in detail in Ramond II . Problems 91.1) Show that introducing neutrino masses via the seesaw m echanism results in lepton number no longer being conserved. 92: Solitons and Monopoles 558 92Solitons and Monopoles Prerequisite: 84 Consider a real scalar ﬁeld ϕwith lagrangian L=−1 2∂µϕ∂µϕ−V(ϕ), (92.1) with V(ϕ) =1 8λ(ϕ2−v2)2. (92.2) As we discussed in section 30, this potential yields two grou nd states or vacua, corresponding to the classical ﬁeld conﬁgurations ϕ(x) = +vand ϕ(x) =−v. After shifting the ﬁeld by its VEV (either + vor−v), we ﬁnd that the particle mass is m=λ1/2v. Let us consider this theory in two spacetime dimensions (one space dimensionxand timet). In this case, ϕandvare dimensionless, and λhas dimensions of mass squared. In the quantum theory, the coupl ing is weak ifλ≪m2. The case of one space dimension is interesting for the follow ing reason. The boundary of one-dimensional space consists of two point s,x=−∞ andx= +∞. This topology of the spatial boundary is mirrored by the topology of the space of vacuum ﬁeld conﬁgurations, which al so consists of two points, ϕ(x) =−vandϕ(x) = +v. In each vacuum, bothspatial boundary points ( x=−∞andx= +∞) are mapped to the same ﬁeld value (either −vor +v). This is a trivial map . More interesting is the identity map , wherex=−∞is mapped to ϕ=−v, andx= +∞is mapped to ϕ= +v. This map does notcorrespond to a vacuum; the ﬁeld must smoothly interpolate between ϕ=−vatx=−∞andϕ= +vat x= +∞, and this requires energy. The interesting question is whet her it can be done at the cost of a ﬁnite amount of energy. To make these notions more precise, we will look for a minimum en- ergy, time-independent solution of the classical ﬁeld equa tions, with the boundary conditions lim x→±∞ϕ(x) =±v. (92.3) The total energy is given by E=/integraldisplay+∞ −∞dx/bracketleftig 1 2˙ϕ2+1 2ϕ′2+V(ϕ)/bracketrightig . (92.4) The solution of interest is time independent, so we can set ˙ ϕ= 0. We can also rewrite the remaining terms in Eas E=/integraldisplay+∞ −∞dx/bracketleftig 1 2/parenleftig ϕ′−√ 2V(ϕ)/parenrightig2+√ 2V(ϕ)ϕ′/bracketrightig 92: Solitons and Monopoles 559 =/integraldisplay+∞ −∞dx1 2/parenleftig ϕ′−√ 2V(ϕ)/parenrightig2+/integraldisplay+v −v√ 2V(ϕ)dϕ =/integraldisplay+∞ −∞dx1 2/parenleftig ϕ′−√ 2V(ϕ)/parenrightig2+2 3(m2/λ)m. (92.5) Since the ﬁrst term in eq.(92.5) is positive, the minimum pos sible energy is M≡2 3(m2/λ)m; this is much larger than the particle mass mif the theory is weakly coupled ( λ≪m2). Requiring the ﬁrst term in eq.(92.5) to vanish yieldsϕ′=√ 2V(ϕ), which is easily integrated to get ϕ(x) =vtanh/parenleftig 1 2m(x−x0)/parenrightig , (92.6) wherex0is a constant of integration. The energy density is localize d near x=x0, and goes to zero exponentially fast for |x−x0|>1/m. This solution is a soliton , a solution of the classical ﬁeld equations with an energy density that is localized in space, and that does no t dissipate or change its shape with time. In this case (and in all cases of in terest to us), its existence is related to the topology of the boundary of sp ace and the topology of the set of vacua, and the existence of a nontrivia l map from the boundary of space to the set of vacua. Given eq.(92.6), we can get other soliton solutions by makin g a Lorentz boost; these solutions take the form ϕ(x,t) =vtanh/parenleftig 1 2γm(x−x0−βt)/parenrightig , (92.7) whereγ= (1−β2)−1/2; their energy is E=γM= (p2+M2)1/2, where M=2 3(m2/λ)mis the energy of the soliton at rest, and p=γβM is the momentum of the soliton, found by integrating the moment um density T01=ϕ′∂0ϕ. We see that the soliton behaves very much like a particle. We m ay expect that, in the quantum theory, the soliton will corresp ond to a new species of particle with mass M, in addition to the elementary ﬁeld excita- tion with mass m. The soliton solution is still interesting if there is more th an one spatial dimension. In that case, eq.(92.6) describes a domain wall , a structure that is localized in one particular spatial direction, but exten ded in the others. The wall has a surface tension (energy per unit transverse ar ea) given by σ=2 3m3/λ. Having found a theory that has a soliton that is localized in onespatial direction, let us try to ﬁnd a theory that has a soliton that is localized in twospatial directions. In two space dimensions, the spatial bo undary has the topology of a circle, denoted by the symbol S1. There is no smooth 92: Solitons and Monopoles 560 nontrivial map from a circle to two points; continuity of the map requires the entire circle to be mapped into one of the two points. But t here do exist smooth nontrivial maps from one circle to another circ le, as we will discuss momentarily. So, we would like to ﬁnd a theory whose vacua have the topology of a circle. To this end, let us consider a complex scalar ﬁeld ϕ(x), with lagrangian L=−∂µϕ†∂µϕ−V(ϕ), (92.8) where V(ϕ) =1 4λ(ϕ†ϕ−v2)2. (92.9) The vacuum ﬁeld conﬁgurations are ϕ(x) =veiα, (92.10) whereαis an arbitrary angle. This angle speciﬁes a point on a circle , and so the space of vacua does indeed have the topology of S1. Let us write x=r(cosφ,sinφ); then the angle φspeciﬁes a point on the spatial circle at inﬁnity. We can specify a map from the sp atial circle to the vacuum circle by giving αas a function of φ. In order for ϕ(x) to be single valued, this function must obey α(φ+ 2π) =α(φ) + 2πn, where the integernis the winding number of the map: we wind around the vacuum circlentimes for every one time that we wind around the spatial circl e. (Ifnis negative, the vacuum winding is opposite in direction to t he spatial winding.) An example of a map with winding number nisU(φ) =einφ. Settingn= 0 then yields the trival map, n= 1 the identity map, and n=−1 the inverse of the identity map. Given a smooth map U(φ), its winding number can be written as n=i 2π/integraldisplay2π 0dφU∂φU†, (92.11) whereU†is the complex conjugate of U. To verify that eq.(92.11) agrees with our previous deﬁnition, we ﬁrst check that plugging in o ur example map indeed yields the correct value of the winding number. We then show that the right-hand side of eq.(92.11) is invariant under sm ooth deforma- tions ofU(φ); see problem 92.2. Thus any U(φ) that can be smoothly deformed to einφhas winding number n. Next, we want to look for a ﬁnite-energy solution of the class ical ﬁeld equations for the theory speciﬁed by eqs.(92.8) and (92.9), with the bound- ary condition limr→∞ϕ(r,φ) =vU(φ), (92.12) 92: Solitons and Monopoles 561 withU(φ) corresponding to a map with nonzero winding number. We therefore make the ansatz ϕ(r,φ) =vf(r)einφ, (92.13) wheref(r) is a real function that obeys f(∞) = 1. We must also have f(0) = 0 so that ∇ϕ(r,φ) is well deﬁned at r= 0. Alas, it is easy to see that there is no ﬁnite-energy solution of this form. The gradient of the ﬁeld is ∇ϕ=v/bracketleftig f′(r)ˆr+inr−1f(r)ˆφ/bracketrightig einφ, (92.14) and the gradient energy density is |∇ϕ|2=v2/bracketleftig f′(r)2+n2r−2f(r)2/bracketrightig . (92.15) At larger,f(r) must approach one; then the integral over the second term in eq.(92.15) diverges logarithmically, /integraldisplay d2x|∇ϕ|2∼2πn2v2/integraldisplay∞dr r. (92.16) So the energy is inﬁnite. This is, in fact, a very general resu lt, known as Derrick’s theorem : with scalar ﬁelds only, there are no ﬁnite-energy, time- independent solitons that are localized in more than one dim ension. The problem is that the gradient energy always diverges at large distances from the putative soliton’s core. To get solitons that are localized in more than one dimension , we must introduce gauge ﬁelds. Note that the lagrangian of eq.(92.8 ) has a global U(1) symmetry. Let us gauge this U(1) symmetry, so that the la grangian becomes L=−(Dµϕ)†Dµϕ−V(ϕ)−1 4FµνFµν, (92.17) where Dµϕ=∂µϕ−ieAµϕ, (92.18) andV(ϕ) still given by eq.(92.9). The gauge symmetry is therefore s ponta- neously broken, and the mass of the vector particle is mV=ev. The mass of the scalar particle is mS=λ1/2v. The gradient energy density of the scalar ﬁeld is now |/vectorDϕ|2=|(∇ −ieA)ϕ|2. (92.19) Thus we have the opportunity to choose Aso as to partially cancel the badly behaved second term in eq.(92.14). To see how to do this , recall that 92: Solitons and Monopoles 562 a gauge transformation in this theory takes the form ϕ→Uϕ, (92.20) Aµ→UAµU†+ieU∂µU†, (92.21) whereUis a 1×1 unitary matrix that is a function of spacetime. As r→ ∞, our ansatz for ϕ, eq.(92.13), corresponds to a gauge transformation of a vacuum,ϕ=v, byU=einφ. The corresponding transformation of Aµ= 0 is limr→∞A(r,φ) =ieU∇U† =n erˆφ. (92.22) Before making the gauge transformation, we have ϕ=vandAµ= 0, and soDµϕ= 0; by gauge invariance, this must be true after the transfor mation as well. Indeed, it is easy to check that, with Agiven by eq.(92.22), we have (∇ −ieA)veiφ= 0. Forn∝ne}ationslash= 0, the gauge transformation U=einφislarge: it cannot be smoothly deformed to U= 1. This implies that we cannot extend it from r=∞into the interior of space without meeting an obstruction , a point whereU(r,φ) is ill deﬁned. For example, the simplest attempt at such an extension,U(r,φ) =einφ, is ill deﬁned at r= 0. Near the obstruction, the ﬁeldsϕandAmust deviate from a gauge transformation of a vacuum. This deviation costs energy, and results in a soliton. Our ansatz for a soliton in the theory speciﬁed by eq.(92.17) is then ϕ(r,φ) =vf(r)U(φ), (92.23) A(r,φ) =iea(r)U(φ)∇U†(φ), (92.24) whereU(φ) =einφ, and we require f(∞) =a(∞) = 1 (so that the solution approaches a large gauge transformation of a vacuum as r→ ∞ ) and f(0) =a(0) = 0 (so that Aand∇ϕare well deﬁned at r= 0). Forn= 1, this soliton is a Nielsen–Olesen vortex . The nonzero vector potential results in a perpendicular mag netic ﬁeld B=∇ ×A =1 r/parenleftbigg∂ ∂r(rAφ)−∂ ∂φAr/parenrightbigg ˆz =n ea′(r) rˆz. (92.25) 92: Solitons and Monopoles 563 The corresponding magnetic ﬂux is Φ =/integraldisplay dS·B = limr→∞/integraldisplay dℓ·A =i elimr→∞a(r)/integraldisplay2π 0dφU∂φU† =2πn e. (92.26) Here the second line follows from Stokes’ theorem, the third from eq.(92.24), and the fourth from eq.(92.11). The energy of the soliton is E=/integraldisplay d2x/bracketleftig |(∇ −ieA)ϕ|2+V(ϕ) +1 2B2/bracketrightig . (92.27) Substituting in our ansatz, eqs.(92.23) and (92.24), we get E= 2πv2/integraldisplay∞ 0drr/bracketleftbigg f′2+n2 r2(a−1)2f2+1 4λv2(f2−1)2+n2 e2v2r2a′2/bracketrightbigg .(92.28) It is convenient to deﬁne a dimensionless radial coordinate ρ≡evr=mVr. Let us also deﬁne β2≡λ/e2=m2 S/m2 V. Then eq.(92.28) becomes E= 2πv2/integraldisplay∞ 0dρρ/bracketleftbigg f′2+n2 ρ2(a−1)2f2+1 4β2(f2−1)2+n2 ρ2a′2/bracketrightbigg ,(92.29) where a prime now denotes a derivative with respect to ρ. We can ﬁnd the equations obeyed by f(ρ) anda(ρ) either by substituting the ansatz into the equations of motion, or by applying the variational prin ciple directly to eq.(92.29). Either way, the result is f′′+f′ ρ−n2f ρ2(1−a)2+1 2β2(1−f2)f= 0, (92.30) a′′−a′ ρ+ (1−a)f2= 0, (92.31) with the boundary conditions a(0) =f(0) = 0 and a(∞) =f(∞) = 1. Eqs.(92.30) and (92.31) have no closed-form solution. Howe ver, for ρ≪1, we can show that a(ρ)∼ρ2andf(ρ)∼ρn; and forρ≫1, that 1−a(ρ)∼e−ρand 1 −f(ρ)∼e−cρ, wherec= min(β,2); see problem 92.4. Fornandβof order one, the integral in eq.(92.29) also results in a number of order one, and so we have E∼2πv2. Forβ >1, it is possible to prove a Bogomolny bound ,E >2πv2|n|. In this case, a soliton with 92: Solitons and Monopoles 564 winding number nis unstable against breaking up into |n|solitons, each with winding number one (or minus one, if nis negative). Once we have our soliton solution, we can translate and/or bo ost it; thus we expect the soliton to behave like a particle in two space di mensions. In three space dimensions, the soliton becomes a Nielsen-Olesen string (also called a gauge string ), a structure that is localized in two directions, but extended in the third. Such strings can bend, and even form cl osed loops. In certain uniﬁed theories (see section 97), gauge strings m ay have formed in the early universe; they are then called cosmic strings . Now let us try to ﬁnd a soliton that is localized in three spatial direc- tions. In three space dimensions, the spatial boundary has t he topology of a two-dimensional sphere S2. There are smooth nontrivial maps from S2to S2, as we will discuss momentarily, so let us look for a theory wh ose vacua have the topology of S2. Consider three real scalar ﬁelds ϕa,a= 1,2,3, with lagrangian L=−1 2∂µϕa∂µϕa−V(ϕ), (92.32) where V(ϕ) =1 8λ(ϕaϕa−v2)2. (92.33) The vacuum ﬁeld conﬁgurations are ϕa(x) =vˆϕa, (92.34) where ˆϕis an arbitrary unit vector. This unit vector speciﬁes a poin t on a two-sphere, and so the space of vacua does indeed have the top ology of S2. Let us write x=r(sinθcosφ,sinθsinφ,cosθ); then the polar and az- imuthal angles θandφspecify a point on the spatial two-sphere at inﬁnity. We can specify a map from the spatial two-sphere to the vacuum two- sphere by giving ˆ ϕas an (appropriately periodic) function of θandφ. We can deﬁne a winding number nthat counts the number of times the vacuum two-sphere covers the spatial two-sphere, with nnegative if the orientation is reversed. An example of a map with winding number ncan be con- structed by taking the polar angle of ˆ ϕto beθ, and the azimuthal angle to benφ. Settingn= 1 then yields the identity map, and n=−1 the inverse of the identity map. Given a smooth map ˆ ϕa(θ,φ), its winding number can be written as n=1 8π/integraldisplay d2θεabcεijˆϕa∂iˆϕb∂jˆϕc, (92.35) whered2θ=dθdφ,∂1=∂/∂θ,∂2=∂/∂φ, andε12=−ε21= +1. To verify that eq.(92.35) agrees with our previous deﬁnition, we ﬁrst check 92: Solitons and Monopoles 565 that plugging in our example map indeed yields the correct va lue of the winding number; see problem 92.5. We then show that the right -hand side of eq.(92.35) is invariant under smooth deformations o f ˆϕa(θ,φ); see problem 92.6. It is also worthwhile to note that the right-ha nd side of eq.(92.35) is invariant under a change of coordinates, beca use the jacobian ford2θis cancelled by the jacobian for ∂1∂2. This is of course closely related to the invariance under smooth deformations, since one way t o make such a deformation is via a coordinate change. Next, we want to look for a ﬁnite-energy solution of the class ical ﬁeld equations with nonzero winding number, but we already know t hat these will not exist unless we introduce gauge ﬁelds. We therefore takeϕato be in the adjoint representation of an SU(2) gauge group. The la grangian is now L=−1 2(Dµϕ)a(Dµϕ)a−V(ϕ)−1 4FaµνFa µν, (92.36) where (Dµϕ)a=∂µϕa+eεabcAb µϕc, (92.37) Fa µν=∂µAa ν−∂νAa µ+eεabcAb µAc ν, (92.38) andV(ϕ) is given by eq.(92.33). We have called the gauge coupling efor reasons that will become clear in a moment. The gauge symmetry is spontaneously broken to U(1). If we tak e the vacuum ﬁeld conﬁguration to be ϕa=vδa3, then the A3 µﬁeld remains massless; we will think of it as the electromagnetic ﬁeld. Th e complex vector ﬁelds W± µ= (A1 µ∓iA2 µ)/√ 2 get a mass mW=ev, and have electric charge ±e. (This is the reason for calling the gauge coupling e.) This theory, known as the Georgi-Glashow model , was once considered as an alternative to the Standard Model of electroweak interacti ons (but is now ruled out, because it does not have a Z0boson). When the vacuum ﬁeld conﬁguration is ϕa=vδa3, the electromagnetic ﬁeld strength is Fµν=∂µA3 ν−∂νA3 µ. We can write down a gauge-invariant expression that reduces to Fµνwhen we set ϕa=vδa3; this expression is Fµν= ˆϕaFa µν−e−1εabcˆϕa(Dµˆϕ)b(Dνˆϕ)c. (92.39) Here ˆϕa=ϕa/|ϕ|, where |ϕ|= (ϕaϕa)1/2. We can, in fact, use eq.(92.39) as the deﬁnition of the electromagnetic ﬁeld strength at any sp acetime point where |ϕ| ∝ne}ationslash= 0. (If |ϕ|= 0, the SU(2) symmetry is unbroken, and there is no gauge-invariant way to pick out a particular component of th e nonabelian ﬁeld strength Fa µν.) If we substitute in eqs.(92.37) and (92.38), and make repeated use of ˆ ϕaˆϕa= 1 and the identity εabcεade=δbdδce−δbeδcd, it is 92: Solitons and Monopoles 566 possible to rewrite eq.(92.39) as Fµν=∂µ(ˆϕaAa ν)−∂ν(ˆϕaAa µ)−e−1εabcˆϕa∂µˆϕb∂νˆϕc. (92.40) In particular, the magnetic ﬁeld is Bi=1 2εijkFjk =εijk∂j(ˆϕaAa k)−(2e)−1εijkεabcˆϕa∂jˆϕb∂kˆϕc. (92.41) Let us consider the magnetic ﬂux through a sphere at spatial i nﬁnity; this is given by Φ =/integraltextdS·B, wheredSk=r2sinθdθdφ ˆxk, and ˆx=x/r is a radially outward unit vector. The ﬁrst term in eq.(92.41 ) forBis ∇ ×(ˆϕaAa); since this is a curl, it has zero divergence, and therefore zero surface integral. From eq.(92.35), we see that the second te rm in eq.(92.41) results in Φ =−4πn e. (92.42) This ﬂux implies that any soliton with nonzero winding numbe r is amag- netic monopole with magnetic charge QM= Φ. (In Heaviside-Lorentz units, the Coulomb ﬁeld of an electric point charge QEisEi=QEˆxi/4πr2, and so the total electric ﬂux is QE. We adopt the same convention for magnetic charge.) If we add a ﬁeld in the fundamental representation of SU(2), t hen the component ﬁelds have electric charges ±1 2e. This is the smallest electric charge we can get, and all possible electric charges are inte ger multiples of it. Eq.(92.42) tells us that all possible magnetic charge s are integer multiples of 4 π/e. Thus the possible electric and magnetic charges obey theDirac charge quantization condition , which is QEQM= 2πk, (92.43) wherekis an integer. This condition can be derived from general con sid- erations of the quantum properties of monopoles. Now let us turn to the explicit construction of a soliton solu tion. This simplest case to consider is provided by the identity map (wh ich has winding numbern= 1); the soliton we will ﬁnd is the ’tHooft-Polyakov monopole . The boundary condition on the scalar ﬁeld is limr→∞ϕa(x) =vxa/r. (92.44) We can ﬁnd the appropriate boundary condition on the gauge ﬁe ld by requiring (Dµϕ)a= 0 in the limit of large r. This condition yields ∂i(xa/r) +eεabcAb ixc/r= 0. (92.45) 92: Solitons and Monopoles 567 We have∂i(xa/r) = (r2δai−xaxi)/r3. Next we multiply by rxjεjda, and use the identity εjdaεabc=δjbδdc−δjcδdbto get εdijxj+e(xdxjAj i−r2Ad i) = 0. (92.46) If we ingore the ﬁrst term in the parentheses, we ﬁnd Ad i=εdijxj/er2. But thenxdAd i= 0, and so the ﬁrst term in the parentheses vanishes. Thus we have found the needed asymptotic behavior of Aa i. Our ansatz is therefore ϕa(x) =vf(r)xa/r, (92.47) Aa i(x) =a(r)εaijxj/er2. (92.48) We require f(∞) =a(∞) = 1 (so that Aa iandϕahave the desired asymp- totic limits) and f(0) =a(0) = 0 (so that Aa iandϕaare well deﬁned at r= 0). The total energy of the soliton (which we will call M, because it is the mass of the monopole) is given by M=/integraldisplay d3x/bracketleftig 1 2Ba iBa i+1 2(Diϕ)a(Diϕ)a+V(ϕ)/bracketrightig . (92.49) The nonabelian magnetic ﬁeld is Ba i=1 2εijkFa jk =εijk∂jAa k+1 2eεijkεabcAb jAc k. (92.50) If we write eq.(92.48) as Aa i=εaijKj, then after some manipulation we ﬁnd that eq.(92.50) becomes Ba i=∂aKi−δai∂jKj+KaKi. Plugging in Ki=a(r)xi/er2then yields Ba i=−1 e/bracketleftigg a′ r/parenleftig δai−ˆxaˆxi/parenrightig +2a−a2 r2ˆxaˆxi/bracketrightigg . (92.51) The magnetic ﬁeld energy then becomes 1 2Ba iBa i=1 2e2r4/bracketleftig 2r2a′2+ (2a−a2)2/bracketrightig . (92.52) The covariant derivative of the scalar ﬁeld is (Diϕ)a=v/bracketleftbigg(1−a)f r/parenleftig δai−ˆxaˆxi/parenrightig +f′ˆxaˆxi/bracketrightbigg . (92.53) The scalar gradient energy density then becomes 1 2(Diϕ)a(Diϕ)a=v2 2r2/bracketleftig 2(1−a)2f2+r2f′2/bracketrightig . (92.54) 92: Solitons and Monopoles 568 The scalar potential energy density is V(ϕ) =1 8λv4(f2−1)2. (92.55) We can plug eqs.(92.52), (92.54), and (92.55) into eq.(92.4 9), and then use the variational principle to get the second-order diﬀer ential equations obeyed byf(r) anda(r). We can get a lower bound on Mby performing a trick analogous to the one we used in eq.(92.5). We write 1 2Ba iBa i+1 2(Diϕ)a(Diϕ)a=1 2[Ba i+ (Diϕ)a]2−Ba i(Diϕ)a.(92.56) We can apply the distribution rule for covariant derivative s (see problem 70.5) to rewrite the last term as Ba i(Diϕ)a=∂i(Ba iϕa)−(DiBi)aϕa. Then we note that the Bianchi identity (see problem 70.6) implies (DiBi)a= 0. ThusBa i(Diϕ)a=∂i(Ba iϕa), and this is a total divergence. Then Gauss’s theorem yields/integraldisplay d3x∂i(Ba iϕa) =/integraldisplay dSiBa iϕa, (92.57) where the integral is over the surface at spatial inﬁnity. On this surface, we haveϕa=vˆxa. Next we use eq.(92.39). At spatial inﬁnity, the covariant de rivatives ofϕvanish; thus we have Ba iϕa=vBi, whereBiis the magnetic ﬁeld of electromagnetism. We can now see that the right-hand side of eq.(92.57) evaluates to vΦ, where Φ = QM=−4πn/e is the magnetic charge of the monopole. In our case, n= 1 andQMis negative; thus the last term in eq.(92.56) integrates to v|QM|. (For the case of positive QM, we can swap the plus and minus signs in eq.(92.56) to get the same result.) Thus th e mass of the monopole, eq.(92.49), can be written as M=4π|n|v e+/integraldisplay d3x/bracketleftig 1 2[Ba i+ (signn)(Diϕ)a]2+V(ϕ)/bracketrightig , (92.58) Both terms in the integrand of eq.(92.58) are positive, and s o we have a Bogomolny bound on the mass of the monopole. For λ >0, a monopole with winding number nis unstable against breaking up into |n|monopoles, each with winding number one (or minus one, if nis negative). UsingmW=evandα=e2/4π, we can write the Bogomolny bound as M≥mW α|n|. (92.59) Sinceα≪1, the monopole is much heavier than the Wboson. Alas, the Georgi-Glashow model, which has monopole solutio ns, is not in accord with nature, while the Standard Model, which is in a ccord with 92: Solitons and Monopoles 569 nature, does not have monopole solutions. This is because, i n the Standard Model, electric charge is a linear combination of an SU(2) ge nerator and the U(1) hypercharge generator. Nothing prevents us from in troducing an SU(2) singlet ﬁeld with an arbitrarily small hypercharge . Such a ﬁeld would have an arbitrarily small electric charge (in units of e), and then the Dirac charge quantization condition would preclude the existence of magnetic monopoles. This disappointing situation is remedied in uniﬁed theorie s (see section 97), where the gauge group has a single nonabelian factor lik e SU(5). In uniﬁed theories, the monopole mass is of order mX/α, wheremXis the mass of a superheavy vector boson; typically mX∼1015GeV. Returning to the Georgi-Glashow model, we can saturate the B ogo- molny bound if we consider the formal limit of λ→0; thenV(ϕ) vanishes. (This limit is formal because we need a nonzero potential to ﬁ x the magni- tude ofϕat inﬁnity.) Then we saturate the bound if Ba i=−(signn)(Diϕ)a. In the case of the ’tHooft-Polyakov monopole, we have n= 1, andBa i and (Diϕ)aare given by eqs.(92.51) and (92.53). Matching the coeﬃ- cients ofδai−ˆxaˆxiand ˆxaˆxiyields a pair of ﬁrst-order diﬀerential equa- tions. These look nicer if we introduce the dimensionless ra dial coordinate ρ≡evr=mWr; then we ﬁnd a′= (1−a)f , (92.60) f′= (2a−a2)/ρ2, (92.61) where a prime now denotes a derivative with respect to ρ. These equations have a closed-form solution, a(ρ) = 1−ρ sinhρ, (92.62) f(ρ) = cothρ−1ρ. (92.63) This is the Bogomolny-Prasad-Sommerfeld (or BPS for short) solution. A soliton that saturates a Bogomolny bound is generically ca lled a BPS soliton . Reference Notes Discussions of solitons, and their relation to the theory of homotopy groups, can be found in Coleman andWeinberg II . Problems 92: Solitons and Monopoles 570 92.1)Derrick’s theorem says that, in a theory with scalar ﬁelds only, there are no solitons localized in more than one dimension. To prov e this, consider a theory in Dspace dimensions with a set of real scalar ﬁelds ϕi; any complex scalar ﬁelds are written as a pair of real ones. T he lagrangian is L=−1 2∂µϕi∂µϕi−V(ϕi), withV(ϕi)≥0. Suppose we have a soliton solution ϕi(x); its energy is E=T+U, where T=1 2/integraltextdDx(∇ϕi)2andU=/integraltextdDxV(ϕi). a) Now consider ϕi(x/α), whereαis a positive real number. Show that, for this ﬁeld conﬁguration, the energy is E(α) =αD−2T+αDU. b) Argue that we must have E′(1) = 0. c) Use this to prove the theorem. 92.2) The winding number nfor a map from S1→S1is given by eq.(92.11), whereU†U= 1. We will prove that nis invariant under an inﬁnites- imal deformation of U. Since any smooth deformation can be made by compounding inﬁnitesimal ones, this will prove that nis invariant under any smooth deformation. a) Consider an inﬁnitesimal deformation of U,U→U+δU. Show thatδU†=−U†2δU. b) Use this result to show that δ(U∂φU†) =−∂φ(U†δU). c) Use this to show that δn= 0. 92.3) Show that if Un(φ) andUk(φ) are maps from S1→S1with winding numbersnandk, thenUn(φ)Uk(φ) is a map with winding number n+k. Hint: consider smoothly deforming Un(φ) to equal one for 0≤φ≤π. How should Uk(φ) be deformed? 92.4) Verify the statements made about the solutions to eqs. (92.30) and (92.31) in the limit of large and small ρ. 92.5) Use eq.(92.35) to compute the winding number for the ma p speciﬁed by ˆϕ= (sinθcosnφ,sinθsinnφ,cosθ). 92.6) The winding number nfor a map from S2→S2is given by eq.(92.35), where ˆϕaˆϕa= 1. We will prove that nis invariant under an inﬁnites- imal deformation of ˆ ϕ. Since any smooth deformation can be made by compounding inﬁnitesimal ones, this will prove that nis invariant under any smooth deformation. a) Consider an inﬁnitesimal deformation of ˆ ϕ, ˆϕ→ˆϕ+δˆϕ. Show that ˆϕ·δˆϕ= 0 and that ˆ ϕ·∂iˆϕ= 0. b) Use these results to show that εabcδˆϕa∂iˆϕb∂jˆϕc= 0. c) Use this to show that δn= 0. 93: Instantons and Theta Vacua 571 93Instantons and Theta Vacua Prerequisite: 92 Consider SU(2) gauge theory, with gauge ﬁelds only. The clas sical ﬁeld con- ﬁguration corresponding to the ground state is Fa µν= 0. This implies that the vector potential Aa µis a gauge transformation of zero, Aµ=Aa µTa= igU∂µU†. Let us restrict our attention to gauge transformations that are time independent, U=U(x). This ﬁxes temporal gauge ,A0= 0. We will also impose the boundary condition that U(x) approaches a particular constant matrix as |x| → ∞ , independent of direction. This is equivalent to adding a spatial “point at inﬁnity” where Uhas a deﬁnite value; space then has the topology of a three-dimensional sphere S3. Can every U(x) be smoothly deformed into every other U(x)? If the answer is yes, then all these ﬁeld conﬁgurations are gauge equivalent, an d they correspond to a single quantum vacuum state. If the answ er isno, then there must be more than one quantum vacuum state. To see why, s uppose thatU(x) and ˜U(x) cannot be smoothly deformed into each other. The associated vector potentials, Aµ=igU∂µU†and˜Aµ=ig˜U∂µ˜U†, are both gauge transformations of zero, and so both Fµνand˜Fµνvanish. However, if we try to smoothly deform Aµinto˜Aµ, we must pass through vector poten- tials that are notgauge transformations of zero, and whose ﬁeld strengths therefore do notvanish. These nonzero ﬁeld strengths imply nonzero en- ergy: there is an energy barrier between the ﬁeld conﬁgurati onsAµand ˜Aµ. Therefore, they represent two diﬀerent minima of the hamil tonian in the space of classical ﬁeld conﬁgurations. Diﬀerent minima in the space of classical ﬁeld conﬁgurations correspond to diﬀerent vacuu m states in the quantum theory. It turns out that every U(x) cannotbe smoothly deformed into every otherU(x); the ﬁeld conﬁgurations speciﬁed by U(x) are classiﬁed by a winding number. To see this, we ﬁrst note that any 2 ×2 special unitary matrixUcan be written in the form U=a4+i/vector a·/vector σ, (93.1) wherea4and the three-vector /vector aare real, and /vector a2+a2 4= 1 ; (93.2) see problem 93.1. Thus aµ≡(/vector a,a4) speciﬁes a euclidean four-vector of unit length,aµaµ= 1, and hence a point on a three-sphere. We will call this the vacuum three-sphere. Since our boundary conditions giv e space the 93: Instantons and Theta Vacua 572 topology of a three-sphere, U(x) provides a map from the spatial three- sphere to the vacuum three-sphere. We can deﬁne a winding num bern that counts the number of times the vacuum three-sphere cove rs the spatial three-sphere, with nnegative if the orientation is reversed. It is convenient to specify the spatial three-sphere by a euc lidean four- vectorzµ≡(/vector z,z4) of unit length, zµzµ= 1. An explicit relation between zµandxcan be constructed by (for example) stereographic projecti on: we take ˆz=/vector z/|/vector z|= ˆx, and|/vector z|= 2r/(1+r2),z4= (1−r2)/(1+r2), where r=|x|. Then we can construct an example of a map from the spatial S3 to the vacuum S3with winding number nby taking the two polar angles ofaµto be equal to the two polar angles of zµ, and the azimuthal angle of aµto be equal to ntimes the azimuthal angle of zµ. (The polar angles run from 0 toπ, and the azimuthal angle from 0 to 2 π.) Given a smooth map U(x), its winding number can be written as n=−1 24π2/integraldisplay d3xεijkTr[(U∂iU†)(U∂jU†)(U∂kU†)]. (93.3) Here we have used the original xcoordinates, but we could also use the angles that specify zµ: the integral in eq.(93.3) is invariant under a change of coordinates, because the jacobian for d3xis cancelled by the jacobian for∂1∂2∂3. To verify that eq.(93.3) agrees with our previous deﬁnitio n, we ﬁrst check that plugging in our example map indeed yields the correct value of the winding number; see problem 93.5. We then show that the right- hand side of eq.(93.3) is invariant under smooth deformatio ns ofU(x); see problem 93.3. So, we have concluded that SU(2) gauge theory has an inﬁnite n umber of classical ﬁeld conﬁgurations of zero energy, distinguis hed by an integer n, and separated by energy barriers. This is analogous to a sca lar ﬁeld theory with a potential V(ϕ) =λv4[1−cos(2πϕ/v)]. (93.4) This potential has minima at ϕ=nv, wherenis an integer. Let |n∝an}b∇acket∇i}htbe the quantum state corresponding to the minimum at ϕ=nv. Generically, between two quantum states |n∝an}b∇acket∇i}htand|n′∝an}b∇acket∇i}htthat are separated by an energy barrier, there is a tunneling amplitude of the form ∝an}b∇acketle{tn′|H|n∝an}b∇acket∇i}ht ∼e−S, (93.5) whereHis the hamltonian, and Sis the euclidean action for a classical solution of the euclidean ﬁeld equations that mediates betw een the ﬁeld conﬁguration corresponding to natt=−∞, and the ﬁeld conﬁguration corresponding to n′att= +∞. In the scalar ﬁeld theory, this solution 93: Instantons and Theta Vacua 573 is independent of x. Thus,Sscales like the volume of space V, and so ∝an}b∇acketle{tn′|H|n∝an}b∇acket∇i}htvanishes in the inﬁnite volume limit. The minima of eq.(93.4 ) therefore remain exactly degenerate in the quantum theory. Things are diﬀerent in the SU(2) gauge theory. In this case, t here is a classical solution of the euclidean ﬁeld equations that me diates between states with winding numbers nandn′, and that has an action that stays ﬁxed and ﬁnite in the inﬁnite-volume limit. The value of this action is S=|n′−n|S1, whereS1= 8π2/g2, andgis the Yang-Mills coupling con- stant. For n′=n+ 1, this solution is the instanton . The instanton is localized in all four euclidean directions. For n′=n−1, the solution is theantiinstanton . For|n′−n|>1, the solution is a dilute gas of|n′−n| instantons (or antiinstantons, if n′−nis negative) distributed throughout euclidean spacetime. We will shortly construct the instanton and examine its prop erties, but ﬁrst we study the consequences of its existence. For SU(2) ga uge theory, eq.(93.5) reads ∝an}b∇acketle{tn′|H|n∝an}b∇acket∇i}ht ∼e−|n′−n|S1. (93.6) These matrix elements depend only on n′−n, and soHcan be diagonalized bytheta vacua of the form |θ∝an}b∇acket∇i}ht=+∞/summationdisplay n=−∞e−inθ|n∝an}b∇acket∇i}ht; (93.7) see problem 93.2. For weak coupling, S1≫1, and so we can neglect all matrix elements of Hexcept those with n′=n±1. Then we ﬁnd that the energy of a theta vacuum is proportional to −cosθ. (We are of course free to add a constant to Hso that the lowest lying state, the theta vacuum withθ= 0, has energy zero.) We have derived these results in the weak-coupling regime. H owever, we are discussing properties of low-energy states, and the g auge coupling becomes large at low energies. Therefore we must consider th e theory to be in the strong-coupling regime. How does this aﬀect our con clusions? The topological properties of the gauge ﬁelds are independe nt of the value of the coupling constant, so we still expect vacuum sta tes labeled by the winding number nto exist. We also expect that ∝an}b∇acketle{tn′|H|n∝an}b∇acket∇i}htwill depend only on |n′−n|. To see this, consider making a gauge transformation by Uk(x), whereUk(x) has winding number k. The product of two maps with winding numbers nandkis a map with winding number n+k; see problem 93.4. Thus, making a gauge transformation by Uk(x) converts a ﬁeld conﬁguration with winding number nto one with winding number n+k. In the quantum theory, the gauge transformation is impleme nted by 93: Instantons and Theta Vacua 574 a unitary operator Uk, and we should have Uk|n∝an}b∇acket∇i}ht=|n+k∝an}b∇acket∇i}ht. (93.8) On the other hand, the hamiltonian, which is built out of ﬁeld strengths, must be invariant under time-independent gauge transforma tions: UkHU† k=H . (93.9) Inserting factors of I=U† kUkon either side of Hin∝an}b∇acketle{tn′|H|n∝an}b∇acket∇i}ht, and using eqs.(93.8) and (93.9), we ﬁnd ∝an}b∇acketle{tn′|H|n∝an}b∇acket∇i}ht=∝an}b∇acketle{tn′+k|H|n+k∝an}b∇acket∇i}ht. (93.10) We conclude that ∝an}b∇acketle{tn′|H|n∝an}b∇acket∇i}htdepends only on n′−n. We can also note that winding number is reversed by parity, P|n∝an}b∇acket∇i}ht=|−n∝an}b∇acket∇i}ht, and that the Yang- Mills hamiltonian is parity invariant, PHP−1=H, to conclude similarly that∝an}b∇acketle{tn′|H|n∝an}b∇acket∇i}ht=∝an}b∇acketle{t−n′|H|−n∝an}b∇acket∇i}ht. Thus ∝an}b∇acketle{tn′|H|n∝an}b∇acket∇i}htdepends only on |n′−n|. The fact that ∝an}b∇acketle{tn′|H|n∝an}b∇acket∇i}htdepends only on |n′−n|tells us that the theta vacua are still eigenstates of H. Furthermore, their energies must be a periodic, even function of θ. Of course, the eigenvalues of Hshould scale with the volume of space V. Then, on dimensional grounds, we have H|θ∝an}b∇acket∇i}ht=VΛ4 QCDf(θ)|θ∝an}b∇acket∇i}ht, (93.11) where Λ QCDis the scale where the gauge coupling becomes strong. The functionf(θ) must obey f(θ+ 2π) =f(θ) andf(−θ) =f(θ). We expect the minimum of f(θ) to be atθ= 0. We turn now to the solutions of the euclidean ﬁeld equations. At eu- clidean time x4=−T, we setAµ(x) =igU−(x)∂µU† −(x), whereU−(x) has winding number n−. Similarly, at euclidean time x4= +T, we set Aµ(x) =igU+(x)∂µU† +(x), whereU+(x) has winding number n+. At |x|=R, for−T≤x4≤T, we set the boundary condition Aµ= 0. This is equivalent to Aµ=igU∂µU†with∂µU†= 0; we therefore set U(x) to a constant matrix at |x|=R. We want to take TandRto inﬁnity at the end of the calculation. We have now speciﬁed U(x,x4) on the cylindrical boundary of four- dimensional spacetime shown in ﬁg.(93.1). This boundary is topologically a three-sphere. The winding number of the map on this three-s phere is n+−n−. We see this by using eq.(93.3), and noting that the cylindri cal wall makes no contribution (because ∂µU†= 0 there), the upper cap contributes n+, and the lower cap contributes −n−; the sign is negative because the orientation of the cap as part of the boundary is reversed fro m its original orientation. 93: Instantons and Theta Vacua 575 R4 −T+T x Figure 93.1: The boundary in euclidean spacetime. We have a ﬁ eld con- ﬁguration with winding number n−on the cap at x4=−T, and one with winding number n+on the cap at x4= +T. On the cylindrical surface at |x|=R, the ﬁeld vanishes. Since we are interested in large RandT, and since the shape of the boundary should not matter in this limit, we instead conside r the boundary to be a three-sphere at ρ≡(xµxµ)1/2=∞. On this boundary, we have a mapU(ˆx), where ˆxµ=xµ/ρ; this map has winding number n≡n+−n−. Our ﬁrst task will be to construct a Bogomolny bound on the euclidean action S=1 2/integraldisplay d4xTr(FµνFµν) (93.12) of a ﬁeld that obeys the boundary condition limρ→∞Aµ(x) =igU(ˆx)∂µU†(ˆx), (93.13) whereU(ˆx) is a map with winding number n. The ﬁeld strength is given in terms of the vector potential by Fµν=∂µAν−∂νAµ−ig[Aµ,Aν]. (93.14) We begin by deﬁning the polar angles χandψ, and the azimuthal angle φ, via ˆxµ= (sinχsinψcosφ,sinχsinψsinφ,sinχcosψ,cosχ).(93.15) (We useψrather than θin order to avoid any possible confusion with the vaccum angle.) Next we write the winding number in terms of th ese angles, n=−1 24π2/integraldisplayπ 0dχ/integraldisplayπ 0dψ/integraldisplay2π 0dφεαβγTr[(U∂αU†)(U∂βU†)(U∂γU†)],(93.16) 93: Instantons and Theta Vacua 576 whereα,β,γ run overχ,ψ,φ ;∂φ=∂/∂φ, etc.; and εχψφ= +1. Now we write eq.(93.16) as a surface integral over a surface at in ﬁnity in four- dimensional euclidean space, n=1 24π2/integraldisplay dSµεµνστTr[(U∂νU†)(U∂σU†)(U∂τU†)], (93.17) where∂µ=∂/∂xµ, andε1234= +1. (This implies that ερχψφ=−1, as can be checked by computing the jacobian for this change of coord inates; that is why the overall minus sign disappeared.) Now we use eq.(93 .13) to write the winding number in terms of the vector potential, n=ig3 24π2/integraldisplay dSµεµνστTr(AνAσAτ). (93.18) Next, we will write this surface integral as a volume integra l. To do so, we ﬁrst deﬁne the Chern-Simons current , Jµ CS≡2εµνστTr(AνFστ+2 3igAνAσAτ). (93.19) This current is not gauge invariant, but the relative coeﬃci ent of its two terms has been chosen so that its divergence isgauge invariant, ∂µJµ CS=εµνστTr(FµνFστ) = 2Tr( ˜FµνFµν), (93.20) where ˜Fµν≡1 2εµνστFστ (93.21) is the dual ﬁeld strength . On the surface at inﬁnity, the vector potential is a gauge tra nsformation of zero, and so the ﬁeld strength Fµνvanishes there. Thus we can use eq.(93.19) to write eq.(93.18) as n=g2 32π2/integraldisplay dSµJµ CS. (93.22) Using Gauss’s theorem, this becomes n=g2 32π2/integraldisplay d4x∂µJµ CS. (93.23) Finally, we use eq.(93.20) to get n=g2 16π2/integraldisplay d4xTr(˜FµνFµν). (93.24) 93: Instantons and Theta Vacua 577 Thus we have expressed the winding number as a (four-dimensi onal) volume integral of a gauge-invariant expression. Now it is easy to construct a Bogomolny bound. We ﬁrst note tha t ˜Fµν˜Fµν=FµνFµν, and hence 1 2Tr(˜Fµν±Fµν)2= Tr(FµνFµν)±Tr(˜FµνFµν). (93.25) The left-hand side of eq.(93.25) is nonnegative, and so we ha ve /integraldisplay d4xTr(FµνFµν)≥/vextendsingle/vextendsingle/vextendsingle/vextendsingle/integraldisplay d4xTr(˜FµνFµν)/vextendsingle/vextendsingle/vextendsingle/vextendsingle. (93.26) The left-hand side of eq.(93.26) is twice the euclidean acti on, while the right-hand side is, according to eq.(93.24), 16 π2|n|/g2. Thus we have S≥8π2|n|/g2. (93.27) Eq.(93.27) gives us the minimum value of the euclidean actio n for a solution of the euclidean ﬁeld equations that mediates between a vacu um conﬁgu- ration with winding number n−atx4=−∞and a vacuum conﬁguration with winding number n+=n−+natx4= +∞. From eq.(93.25) we see that we can saturate the bound in eq.(9 3.27) if and only if ˜Fµν= (signn)Fµν. (93.28) We can ﬁnd an explicit solution of eq.(93.28) for a map with wi nding numbern= 1; this solution is the instanton. We take the map on the boundary to be the identity map, U(ˆx) =x4+i/vector x·/vector σ ρ =/parenleftigg cosχ+isinχcosψ i sinχsinψe−iφ isinχsinψe+iφcosχ−isinχcosψ/parenrightigg ,(93.29) which hasn= 1. We then make the ansatz Aµ(x) =igf(ρ)U(ˆx)∂µU†(ˆx), (93.30) wheref(∞) = 1 (so that the solution obeys the boundary condition) and f(0) = 0 (so that Aµis well deﬁned at ρ= 0). Using eq.(93.14), we ﬁnd that the ﬁeld strength is Fµν=ig/bracketleftig (∂µf)U∂νU†+f∂µU∂νU†+f2(U∂µU†)(U∂νU†) −(µ↔ν)/bracketrightig . (93.31) 93: Instantons and Theta Vacua 578 Using (∂µU†)U=−U†∂µUin the third term, eq.(93.31) can be simpliﬁed to Fµν=ig/bracketleftig (∂µf)U∂νU†+f(1−f)∂µU∂νU†−(µ↔ν)/bracketrightig . (93.32) In three-spherical coordinates, we have ∂= ˆρ∂ρ+ ˆχρ−1∂χ+ˆψ(ρsinχ)−1∂ψ+ˆφ(ρsinχsinψ)−1∂φ,(93.33) where∂φ=∂/∂φ, etc., and ˆ ρis the same radial unit vector as ˆ x. Note thatfis a function of ρonly, while Uis a function of the angles only. We therefore have Fρχ=igf′ρ−1U∂χU†, (93.34) Fψφ=igf(1−f)(ρ2sin2χsinψ)−1(∂ψU∂φU†−∂φU∂ψU†).(93.35) Next we note that that eq.(93.21) implies ˜Fρχ=−Fψφ(sinceερχψφ=−1), and since ˜Fµν=Fµνfor the instanton solution, we have Fρχ=−Fψφ. Thus the right-hand side of eq.(93.34) equals minus the righ t-hand side of eq.(93.35). We then use separation of variables to conclude that ρf′=cf(1−f), (93.36) U∂χU†=−c−1(sin2χsinψ)−1(∂ψU∂φU†−∂φU∂ψU†),(93.37) wherecis the separation constant. If we plug eq.(93.29) into eq.(9 3.37), we ﬁnd that it is satisﬁed if c= 2. The solution of eq.(93.36) is then f(ρ) =ρ2 ρ2+a2, (93.38) wherea, thesize of the instanton , is a constant of integration. The instan- ton solution is also parameterized by the location of its cen ter; we have used the spacetime origin, but translation invariance allo ws us to displace it. If we consider initial and ﬁnal states whose winding numbers diﬀer by more than one, we can construct a mediating solution by patch ing together instantons (or antiinstantons) whose centers are widely se parated on the scale set by their sizes. Each instanton (or antiinstanton) contributes S1= 8π2/g2to the action, and so the minimum total action is |n+−n−|S1. To better understand the role of the θparameter, let us consider the euclidean path integral, with the boundary condition that w e start with a state of winding number n−atx4=−∞, and end with a state with winding 93: Instantons and Theta Vacua 579 numbern+atx4= +∞. The only ﬁeld conﬁgurations that contribute are those with winding number n+−n−. We can therefore write Zn+←n−(J) =/integraldisplay DAn+−n−e−S+JA, (93.39) whereJAis short for/integraltextd4xTr(JµAµ), and the subscript on the ﬁeld diﬀer- ential means that we integrate only over ﬁelds with that wind ing number. We see that Zn+←n−(J) depends only on n+−n−, and not separately on n+andn−. This in accord with our previous conclusion that ∝an}b∇acketle{tn′|H|n∝an}b∇acket∇i}ht depends only on n′−n. Suppose now that we are interested in starting with a particu lar theta vacuum |θ∝an}b∇acket∇i}ht, and ending with a (possibly diﬀerent) theta vacuum |θ′∝an}b∇acket∇i}ht. Then, we see from eq.(93.7) that the corresponding path integral i s Zθ′←θ(J) =/summationdisplay n−,n+ei(n+θ′−in−θ)Zn+←n−(J). (93.40) Letn+=n−+n, so thatn+θ′−n−θ=n−(θ′−θ)+nθ′. SinceZn+←n−(J) depends only on n,n−appears in eq.(93.40) only through a factor of ein−(θ′−θ). Summing over n−then generates δ(θ′−θ), which implies that the value of θis time indpendent. (Of course, we already knew this, becaus e θlabels energy eigenstates.) We now have Zθ′←θ(J) =δ(θ′−θ)/summationdisplay neinθ/integraldisplay DAne−S+JA. (93.41) We can drop the delta function, and just deﬁne Zθ(J)≡/summationdisplay neinθ/integraldisplay DAne−S+JA. (93.42) Next, we combine the sum over nand the integral over Aninto an integral over allA. To account for the factor of einθ, we use eq.(93.24); we get Zθ(J) =/integraldisplay DAexp/integraldisplay d4xTr/bracketleftigg −1 2FµνFµν+ig2θ 16π2˜FµνFµν+JµAµ/bracketrightigg .(93.43) The vacuum angle θnow appears as the coeﬃcient of an extra term in the Yang-Mills lagrangian. We can write the path integral in Minkowski space by setting x4=it. The extra term contains one derivative with respect to x4, and thus picks up a factor of −i. Also,ε1234= +1 but ε1230=−1. Putting all this together, we get Zθ(J) =/integraldisplay DAexpi/integraldisplay d4xTr/bracketleftigg −1 2FµνFµν−g2θ 16π2˜FµνFµν+JµAµ/bracketrightigg (93.44) 93: Instantons and Theta Vacua 580 in Minkowski space. We see that the extra term is gauge invari ant, Lorentz invariant, hermitian, and has a dimensionless coeﬃcient. W e therefore could have included it when we ﬁrst considered Yang-Mills th eory. We did not do so because this term is a total divergence; see eq.(93. 20). We have always dropped total divergences from the lagrangian, beca use they do not aﬀect the equations of motion or the Feynman rules. In this ca se, however, the new term does change the quantum physics, as we have seen. We will explore this in more detail in the next section. So far, we have only discussed SU(2) gauge theory, without sc alar or fermion ﬁelds. What can we say more generally? Adding scalar ﬁelds has no eﬀect on our analysis. Changing fr om SU(2) to another simple nonabelian group also has no eﬀect; instan ton solutions always reside in an SU(2) subgroup. If the gauge group is U(1) , there are no instantons, and hence no vacuum angle. If the gauge group inc ludes more than one nonabelian factor, then there is an independent vac uum angle for each of these factors. On the other hand, adding fermions can signiﬁcantly change t he physics. We take this up in the next section. Reference Notes Instantons are treated in more detail in Coleman andWeinberg II . Problems 93.1) Show that any 2 ×2 special unitary matrix Ucan be written in the formU=a4+i/vector a·/vector σ, wherea4and the three-vector /vector aare real, and /vector a2+a2 4= 1. 93.2) Verify that a state of the form of eq.(93.7) is an eigens tate of any hamiltonian with matrix elements of the form ∝an}b∇acketle{tn′|H|n∝an}b∇acket∇i}ht=f(n′−n). 93.3) The winding number nfor a map from S3→S3is given by eq.(93.3), whereU†U= 1. We will prove that nis invariant under an inﬁnites- imal deformation of U. Since any smooth deformation can be made by compounding inﬁnitesimal ones, this will prove that nis invariant under any smooth deformation. a) Consider an inﬁnitesimal deformation of U,U→U+δU. Show thatδU†=−U†δUU†, and hence that δ(U∂kU†) =−U∂k(U†δU)U†. b) Show that δn=−3 24π2/integraldisplay d3xεijkTr[(U∂iU†)(U∂jU†)δ(U∂kU†)].(93.45) 93: Instantons and Theta Vacua 581 Plug in your result from part (a), and integrate ∂kby parts. Show that the resulting integrand vanishes. Hint: make repeated use of U∂iU†=−∂iUU†, and the antisymmetry of εijk. 93.4) Show that if Un(x) andUk(x) are maps from S3→S3with winding numbersnandk, thenUn(x)Uk(x) is a map with winding number n+k. Hint: consider smoothly deforming Un(x) to equal one for x3<0. How should Uk(x) be deformed? 93.5) Use eq.(93.16) to compute the winding number for the ma p given in eq.(93.29), and for a generalization where φis replaced by nφ. 93.6) Use eq.(93.16) to compute the winding number for Un, whereUis the map given in eq.(93.29). Hint: ﬁrst show that Ucan be written in the form U= exp[i/vector χ·/vector σ], where/vector χis a three-vector that you should specify. Is your result in accord with the theorem of problem 93.4? 94: Quarks and Theta Vacua 582 94Quarks and Theta Vacua Prerequisite: 77, 83, 93 Consider quantum chromodynamics with one ﬂavor of massless quark, rep- resented by a Dirac ﬁeld Ψ in the fundamental representation of the gauge group SU(3). The path integral is Z=/integraldisplay DADΨDΨ expi/integraldisplay d4x/bracketleftigg iΨ /DΨ−1 4FaµνFa µν−g2θ 32π2˜FaµνFa µν/bracketrightigg ; (94.1) for the sake of brevity, we have not written the source terms e xplicitly. In addition to the SU(3) gauge symmetry, there is a U(1) V×U(1) A global symmetry of the quark action. However, the U(1) Asymmetry is anomalous. As we saw in section 77, under a U(1) Atransformation Ψ→e−iαγ5Ψ, (94.2) Ψ→Ψe−iαγ5, (94.3) the integration measure picks up a phase factor, DΨDΨ→exp/bracketleftigg −i/integraldisplay d4xg2α 16π2˜FaµνFa µν/bracketrightigg DΨDΨ. (94.4) Using this in eq.(94.1), we see that the eﬀect of a U(1) Atransformation is to change the value of the theta angle fromθtoθ+2α. Since the value of θcan be changed by making a U(1) Atransformation (which is simply a change of the dummy integration variable in the path integral), we mus t conclude that Zdoes not depend on θ. Apparently (and surprisingly), adding a massless quark has turned the theta angle into a physically irrelevan t, unobservable parameter. How do we reconcile this with our analysis in the previous sec tion, where we concluded that instanton-mediated tunneling ampl itudes make the vacuum energy density depend on θ? The answer is that when we perform the integral over the quark ﬁeld in eq.(94.1), we get a functional determinant; the path integral becomes Z=/integraldisplay DAdet(i/D)eiSeinθ, (94.5) whereSis the Yang-Mills action and nis the winding number. We see from eq.(94.5) that Zwould be independent of θif gauge ﬁelds with nonzero winding number did not contribute. This will be the case if de t(i/D) van- ishes for gauge ﬁelds with n∝ne}ationslash= 0. We conclude that i/Dmust have a zero 94: Quarks and Theta Vacua 583 eigenvalue or zero mode whenever the gauge ﬁeld has nonzero winding num- ber. This renders Zindependent of θ. Now consider adding a mass term for the quark. If we write the D irac ﬁeld Ψ in terms of two left-handed Weyl ﬁelds χandξ, Ψ =/parenleftiggχ ξ†/parenrightigg , (94.6) the mass term reads Lmass=−mχξ−m∗ξ†χ†. (94.7) We have allowed mto be complex: m=|m|eiφ. In terms of Ψ, eq.(94.7) can be written as Lmass=−|m|Ψe−iφγ5Ψ. (94.8) A U(1) Atransformation changes φtoφ+ 2α. Sinceθsimultaneously changes to θ+ 2α, we see that φ−θ, or equivalently me−iθ, is unchanged. Thus, the path integral can (and does) depend on me−iθ, but not on mand θseparately. With more quark ﬁelds, the mass term is L=−Mijχiξj+h.c.; a U(1) A transformation changes the phase of every χiandξibyeiα, and so every matrix element of Mpicks up a factor of e2iα. Simultaneously, θchanges toθ+ 2Nα, whereNis the number of quark ﬁelds. Thus (det M)e−iθis invariant under a U(1) Atransformation. To understand the eﬀects of the theta angle on hadronic physi cs, we turn to the eﬀective lagranagian (for the case of two light ﬂa vors) that we developed in section 83, L=−1 4f2 πTr∂µU†∂µU+v3Tr(MU+M†U†) +iN/∂N−mNN(U†PL+UPR)N −1 2(gA−1)iNγµ(U∂µU†PL+U†∂µUPR)N −c1N(MP L+M†PR)N−c2N(U†M†U†PL+UMUP R)N −c3Tr(MU+M†U†)N(U†PL+UPR)N −c4Tr(MU−M†U†)N(U†PL−UPR)N , (94.9) whereUis a 2×2 special unitary matrix ﬁeld representing the pions, Nis the ﬁeld for the nucleon doublet, M=/parenleftiggmu0 0md/parenrightigg e−iθ/2(94.10) is the quark mass matrix, v3is the value of the quark condensate, and gA andciare numerical constants. (The terms in the last three lines w ere introduced in problem 83.8.) 94: Quarks and Theta Vacua 584 For the case of θ= 0, the potential V(U) =−v3TrMU+ h.c. (94.11) is minimized by U=I, and we can expand about this point in powers of the pion ﬁelds, as we did in section 83. However, for nonzero θ, the minimum ofV(U) occurs at U=U0, whereU0is diagonal (because Mis) and has unit determinant (because Uis required to have unit determinant). We can therefore write U0=/parenleftigge+iφ0 0e−iφ/parenrightigg . (94.12) We determine φby minimizing V(U0) =−2v3/bracketleftig mucos(φ−1 2θ) +mdcos(φ+1 2θ)/bracketrightig ; (94.13) the result is tanφ=mu−md mu+mdtan(1 2θ). (94.14) As we will see shortly, experimental results require the val ue of|θ|to be less than 10−9; therefore, we can work to ﬁrst order in an expansion in power s ofθ. Forθ≪1, eqs.(94.12) and (94.14) can be written in the elegant form MU0=M0−iθ˜mI+O(θ2), (94.15) where M0=/parenleftiggmu0 0md/parenrightigg (94.16) is the quark mass matrix with θset to zero, Iis the identity matrix, and ˜m=mumd mu+md(94.17) is the reduced mass of the up and down quarks. We can now expand in powers of the pion ﬁelds. Though it is not a t all obvious, it turns out that the most convenient way to deﬁne th e pion ﬁelds is by writing U(x) =u0u2(x)u0, (94.18) whereu2 0=U0andu(x) = exp[iπa(x)Ta/fπ]. We also deﬁne ˜U(x) =u2(x), and a new nucleon ﬁeld Nvia N= (u0uPL+u† 0u†PR)N. (94.19) 94: Quarks and Theta Vacua 585 Substituting eqs.(94.18) and (94.19) into eq.(94.9), and u singu0Mu0= MU0(which follows because u0andMare both diagonal, and hence com- mute), we ultimately get L=−1 4f2 πTr∂µ˜U†∂µ˜U+v3Tr[(MU0)˜U+ (MU0)†˜U†] +iN/∂N −mNNN+N/vN −gAN/aγ5N −1 2c+N[u(MU0)u+u†(MU0)†u†]N +1 2c−N[u(MU0)u−u†(MU0)†u†]γ5N −c3Tr[(MU0)˜U+ (MU0)†˜U†]NN +c4Tr[(MU0)˜U−(MU0)†˜U†]Nγ5N, (94.20) wherevµ=1 2i[u†(∂µu) +u(∂µu†)],aµ=1 2i[u†(∂µu)−u(∂µu†)], andc±= c1±c2. Eq.(94.20) is exactly what we found in section 83, except th at the quark mass matrix Mhas been replaced everywhere by MU0. We can now use eq.(94.15) to get the O(θ) contributions to L. Using the fact that Tr( ˜U−˜U†) vanishes in the case of two light ﬂavors, we ﬁnd Lθ=−iθ˜m/bracketleftig −1 2c+N(˜U−˜U†)N+1 2c−N(˜U+˜U†)γ5N +c4Tr(˜U+˜U†)Nγ5N/bracketrightig . (94.21) Expanding in powers of the pion ﬁelds yields Lθ=−iθ˜m(c−+4c4)Nγ5N −(θc+˜m/fπ)πaNσaN+... , (94.22) where we used Ta=1 2σa. We can eliminate the ﬁrst term with a ﬁeld redeﬁnition of the form N →e−iαγ5N. This generates some new terms in eq.(94.20), but all have at least two factors of quark mass es, and hence can be neglected. The second term in eq.(94.22) provides a pi on-nucleon coupling that violates both parity Pand time-reversal T(equivalently, CP). The value of c+can be ﬁxed by baryon mass diﬀerences. The c+ term in eq.(94.20) makes a contribution of c+(mu−md) to the proton- neutron mass diﬀerence, mp−mn=−1.3MeV. Using mu= 1.7MeV andmd= 3.9MeV yields c+= 0.6. However, there is a comparable electromagnetic contribution to the proton-neutron mass d iﬀerence. We get a better estimate from the masses of baryons with strange quarks, c+(ms−1 2mu−1 2md) =mΞ0−mΣ0= 122MeV; using ms= 76MeV yields c+= 1.7. (All these values assume the MS renormalization scheme with µ= 2GeV.) For comparison with the interaction of eq.(94.22), the domi nant (Pand CPconserving) pion-nucleon interaction comes from the last t erm in the 94: Quarks and Theta Vacua 586 π+π+γ n p nπ+π+γ n p n Figure 94.1: Diagrams contributing to the electric dipole m oment of the neutron that are enhanced by a chiral log. The CPviolating vertex is denoted with a cross. second line of eq.(94.20), and is LπNN= (gA/fπ)∂µπaNTaγµγ5N, (94.23) wheregA= 1.27. As we did in section 83, we can integrate by parts to put the derivative on the nucleon ﬁelds, and then (for on-shell n ucleons) use the Dirac equation to get LπNN=−i(gAmN/fπ)πaNσaγ5N. (94.24) The strongest limit on a CPviolating pion-nucleon coupling comes from measurements of the electric dipole moment of the neutron. T he Feynman diagrams of ﬁg.(94.1) contribute to an ampitude of the form T=−2iD(q2)ε∗ µ(q)us′(p′)Sµνqνiγ5us(p), (94.25) whereq=p′−p. In theq→0 limit, this corresponds to a term in the eﬀective lagrangian of L=D(0)FµνnSµνiγ5n, (94.26) wherenis the neutron ﬁeld; see section 64. If the factor of iγ5was absent, this would represent a contribution of D(0) to the magnetic dipole moment of the neutron. To account for the factor of iγ5, we use Sµνiγ5=−1 2εµνρσSρσ (94.27) to see that eq.(94.26) is equivalent to L=−D(0)˜FµνnSµνn, (94.28) where ˜Fµν=1 2εµνρσFρσis the dual ﬁeld strength. Since ˜B=−E, eq.(94.26) represents a contribution of D(0) to the electric dipole moment of the neutrondn. 94: Quarks and Theta Vacua 587 p q pl+q/2 l−q/2 l+(p+p )/2 Figure 94.2: Momentum ﬂow in the diagrams of ﬁg.(94.1). The salient feature of the diagrams shown in ﬁg.(94.1) is tha t the pho- ton line attaches to the pion line. These diagrams are enhanc ed by a chiral logln(Λ2/m2 π)∼4.2, where Λ ∼4πfπis the ultraviolet cutoﬀ in the eﬀec- tive theory. No other contributing diagrams have this enhan cement; it is an infrared eﬀect, due to the light pion. Of course, 4.2 is not an impressively large number, and so we cannot be certain that the remaining c ontribu- tions are not signiﬁcant. These contributions depend on coe ﬃcients in the eﬀective lagrangian that are not well determined by other ex perimental results. Using πaσa=/parenleftigg π0√ 2π+ √ 2π−−π0/parenrightigg , (94.29) we can write the charged-pion terms in eqs.(94.24) and (94.2 1) as LπNN=−i√ 2(gAmN/fπ)(π+pγ5n+π−nγ5p), (94.30) LθπNN=−√ 2(θc+˜m/fπ)(π+pn+π−np). (94.31) From these we read oﬀ the pion-nucleon vertex factors. We lab el the in- ternal momenta as shown in ﬁg.(94.2); for small q,ℓis a pion momentum that should be cut oﬀ at Λ ∼4πfπ. Because the terms of interest have a chiral log produced by an infrared divergence at small ℓ, we can treat ℓas much less than pandp′. Thus the internal proton is nearly on-shell, which justiﬁes the use of eq.(94.24) for the CPconserving interaction. The diagrams of ﬁg.(94.1) yield an amplitude of iT=/parenleftig 1 i/parenrightig3(ie)(√ 2gAmN/fπ)(−i√ 2θc+˜m/fπ)ε∗ µ/integraldisplayΛ 0d4ℓ (2π)4 ×(2ℓµ)u′[(−/ℓ−/ ¯p+mN)γ5+γ5(−/ℓ−/ ¯p+mN)]u ((ℓ+¯p)2+m2 N)((ℓ+1 2q)2+m2π)((ℓ−1 2q)2+m2π),(94.32) 94: Quarks and Theta Vacua 588 where ¯p=1 2(p′+p). Using {γµ,γ5}= 0, the spinor factors in the numerator simplify to u′[...]u= 2mNu′γ5u. (94.33) From the spinor properties established in section 38, it is e asy to check that u′γ5uvanishes when p′=p; thusu′γ5umust beO(q), and so we can set q= 0 everywhere else. Also, taking ℓ≪p, we can set ( ℓ+¯p)2+m2 N= 2p·ℓ in the denominator of eq.(94.32). We now have T= 4(eθgAc+˜mm2 N/f2 π)ε∗ µ/integraldisplayΛ 0d4ℓ (2π)4(2ℓµ)u′γ5u (2p·ℓ)(ℓ2+m2π)2. (94.34) Integrating over the direction of ℓresults in ℓµ p·ℓ→pµ p2=−pµ m2 N. (94.35) Next we use the Gordon identity (see problem 38.4) pµu′γ5u=u′Sµνqνiγ5u+O(q2), (94.36) which veriﬁes that u′γ5uis linear in q. We now have T=−4(eθgAc+˜m/f2 π)ε∗ µu′Sµνqνiγ5u/integraldisplayΛ 0d4ℓ (2π)41 (ℓ2+m2π)2.(94.37) In the limit mπ→0, the integral diverges at small ℓ, generating a chiral log. This infrared divergence can only arise from diagrams w ith two pion propagators, which is why it appears only if the photon is att ached to the pion. After a Wick rotation, the integral evaluates to ( i/16π2)ln(Λ2/m2 π). Comparing with eq.(94.25), we see that the electric dipole m oment of the neutron is dn=eθgAc+˜m 8π2f2π/bracketleftig ln(Λ2/m2 π) +O(1)/bracketrightig . (94.38) Putting in numbers ( gA= 1.27,c+= 1.7, ˜m= 1.2MeV), we ﬁnd dn= 3.2×10−16θecm. (94.39) The experimental upper limit is |dn|<6.3×10−26ecm, and so we must have|θ|<2×10−10. Such a small value for a fundamental parameter cries out for a n ex- planation; this is the strong CP problem . Several solutions have been pro- posed. (1) The up quark mass may actually be zero, since a mass less quark rendersθunobservable (and eﬀectively zero). This requires higher- order 94: Quarks and Theta Vacua 589 corrections in the quark masses to account for the masses of t he pseudo- goldstone bosons. (2) The fundamental lagrangian may be CPinvariant, and the observed CPviolation in weak interactions due to spontaneous breaking of CPsymmetry. (3) The theta parameter may be promoted to a ﬁeld, the axion, which would minimize its energy by rolling to θ= 0; see problem 94.2. All of these solutions have interesting physi cal consequences. Reference Notes Quarks and theta vacua are discussed in Coleman ,Ramond II , andWein- berg II . Problems 94.1) Carry out the ﬁeld redeﬁnition discussed after eq.(94 .22), and verify that all new terms generated in the lagrangian are suppresse d by at least two powers of quark masses. 94.2) Consider adding to the Standard Model a massless quark , represented by a pair of Weyl fermions χandξin the 3 and 3 representations of SU(3). Also add a complex scalar Φ in the singlet represent a- tion. Assume that these ﬁelds have a Yukawa interaction of th e form LYuk=yΦχξ+ h.c., whereyis the Yukawa coupling constant. As- sume that the scalar potential V(Φ) depends only on Φ†Φ. a) Show that the lagrangian is invariant under a Peccei-Quinn trans- formationχ→eiαχ,ξ→eiαξ, Φ→e−2iαΦ, all other ﬁelds un- changed. b) Show that this global U(1) PQsymmetry is anomalous, and that θ→θ+ 2αunder a U(1) PQtransformation. c) Suppose that V(Φ) has its minimum at |Φ|=f/√ 2, withf∝ne}ationslash= 0. Show that this gives a mass to the quark we introduced. d) Write Φ = 2−1/2(f+ρ)eia/f, whereρandaare ﬁelds. Argue that, in eq.(94.10), we should replace θwithθ+a/f, and add to eq.(94.9) a kinetic term −1 2∂µa∂µafor theaﬁeld. d) Show that the minimum of V(U), deﬁned in eq.(94.11), is at U=I anda=−fθ. Show that PandCPare conserved at this minimum. e) The particle corresponding to the aﬁeld is the axion; compute its mass, assuming f≫fπ. f) Note that if fis large, the extra quark becomes very heavy, and the axion becomes very light. Show that couplings of the axio n to the hadrons are all suppressed by a factor of 1 /f. 95: Supersymmetry 590 95Supersymmetry Prerequisite: 69 Supersymmetry is a continuous symmetry that mixes up bosonic and fer- mionic degrees of freedom. A supersymmetric theory (in four spacetime dimensions) has a set of supercharges QaA, whereais a left-handed spinor index, and Ais an internal index that runs from 1 to N, where the al- lowed values of Nare 1, 2, and 4. The supercharges can be obtained as integrals over d3xof the time component of a supercurrent . The supercur- rent is found via the Noether procedure, once we have identiﬁ ed the set of supersymmetry transformations that leaves the action inva riant. The supercharges QaAand their hermitian conjugates Q† ˙aA, together with the generators of the Poincare group PµandMµν, obey a supersym- metry algebra [QaA,Pµ] = 0, (95.1) [Q† ˙aA,Pµ] = 0, (95.2) [QaA,Mµν] = (Sµν L)acQcA, (95.3) [Q† ˙aA,Mµν] = (Sµν R)˙a˙cQ† ˙cA, (95.4) {QaA,QbB}=ZABεab, (95.5) {QaA,Q† ˙aB}=−2δABσµ a˙aPµ. (95.6) Eqs.(95.1) and (95.2) simply say that the supercharges are c onserved, and eqs.(95.3) and (95.4) simply say that their spinor indices a re indeed spinor indices. In eq.(95.5), ZAB=−ZBAmust commute with QaA,Pµ, andMµν, and so represents a central charge in the supersymmetry algebra. We will be concerned only with the case of N= 1 supersymmetry: the index A then takes on only one value (and so can be dropped), and ZAB= 0. N= 1 supersymmetric theories are most easily formulated in super- space, where we augment the usual spacetime coordinate xµwith an anti- commuting left-handed spinor coordinate θaand its right-handed complex conjugateθ∗ ˙a. We deﬁne superﬁelds Φ(x,θ,θ∗) that are functions of all these coordinates. The energy-momentum vector generates translations of the u sual space- time coordinate xµin the usual way, [Φ(x,θ,θ∗),Pµ] =−i∂µΦ(x,θ,θ∗). (95.7) By analogy, we would expect [Φ(x,θ,θ∗),Qa] =−iQaΦ(x,θ,θ∗), (95.8) 95: Supersymmetry 591 [Φ(x,θ,θ∗),Q† ˙a] =−iQ∗ ˙aΦ(x,θ,θ∗), (95.9) where QaandQ∗ ˙aare appropriate diﬀerential operators. To ﬁgure out what they should be, we ﬁrst introduce the anticommuting der ivatives∂a≡ ∂/∂θaand∂∗ ˙a≡∂/∂θ∗˙a, which obey ∂aθc=δacand∂∗ ˙aθ∗˙c=δ˙a˙c. Note, however, that complex conjugation should reverse the order of a product of Grassmann variables, in order to maintain consistency wi th hermitian conjugation. Then we have δac= (∂aθc)∗=θ∗˙c(∂a)∗=−(∂a)∗θ∗˙c, which implies (∂a)∗=−∂∗ ˙a. (95.10) Thus, our ﬁrst guess for the diﬀerential operators in eqs.(9 5.8) and (95.9) isQa=∂aandQ∗ ˙a=−∂∗ ˙a. However, this choice is inconsistent with {Qa,Q† ˙a}=−2σµ a˙aPµ. An alternative that avoids this pitfall is Qa= +∂a+iσµ a˙cθ∗˙c∂µ, (95.11) Q∗ ˙a=−∂∗ ˙a−iθcσµ c˙a∂µ. (95.12) These obey the anticommutation relations {Qa,Qb}={Q∗ ˙a,Q∗ ˙b}= 0, (95.13) {Qa,Q∗ ˙a}=−2iσµ a˙a∂µ. (95.14) It is straightforward to check that eqs.(95.5–95.12) are no w mutually com- patible. In particular, the Jacobi identity {[Φ,Q],Q†}+{[Φ,Q†],Q} −[Φ,{Q,Q†}] = 0 (95.15) is satisﬁed. Next we introduce the supercovariant derivatives Da= +∂a−iσµ a˙cθ∗˙c∂µ, (95.16) D∗ ˙a=−∂∗ ˙a+iθcσµ c˙a∂µ. (95.17) These obey {Da,Db}={D∗ ˙a,D∗ ˙b}= 0, (95.18) {Da,D∗ ˙a}= 2iσµ a˙a∂µ, (95.19) {Da,Qb}={Da,Q∗ ˙b}={D∗ ˙a,Qb}={D∗ ˙a,Q∗ ˙b}= 0. (95.20) Because of eq.(95.20), we could impose the condition DaΦ = 0 or D∗ ˙aΦ = 0 on a superﬁeld, and this condition would be preserved by the s upersymme- try transformations of eqs.(95.8) and (95.9). A superﬁeld t hat obeys D∗ ˙aΦ(x,θ,θ∗) = 0 (95.21) 95: Supersymmetry 592 is aleft-handed chiral superﬁeld . Its hermitian conjugate Φ†(x,θ,θ∗) obeys DaΦ†(x,θ,θ∗) = 0, (95.22) and is a right-handed chiral superﬁeld . We can solve eq.(95.21) by introducing yµ=xµ−iθcσµ c˙cθ∗˙c, (95.23) and noting that D∗ ˙aθa= 0 and D∗ ˙ayµ= 0. (95.24) (When verifying D∗ ˙ayµ= 0, remember that there is a minus sign from pulling the ∂∗ ˙athroughθc.) Thus, any superﬁeld Φ( y,θ) that is a function ofyandθonly is a left-handed chiral superﬁeld. We can expand Φ( y,θ) in powers of θ; becauseθis an anticommuting variable with a two-valued index, we have θaθbθc= 0, and so the expansion terminates after the quadratic term. We thus have Φ(y,θ) =A(y) +√ 2θψ(y) +θθF(y), (95.25) whereA(y) andF(y) are complex scalar ﬁelds, ψa(y) is a left-handed Weyl ﬁeld, and we have used our standard index-suppression conve ntions:θψ= θaψaandθθ=θaθa. The factor of root-two is conventional. We can now substitute in eq.(95.23), and continue to expand i n powers ofθandθ∗. Making use of the spinor identities θaθb= +1 2θθεab, θaθb=−1 2θθεab, (95.26) θ∗ ˙aθ∗ ˙b=−1 2θ∗θ∗ε˙a˙b, θ∗˙aθ∗˙b= +1 2θ∗θ∗ε˙a˙b, (95.27) whereθθ=θaθaandθ∗θ∗= (θθ)∗=θ∗ ˙aθ∗˙a, along with the Fierz identity (θσµθ∗)(θσνθ∗) =−1 2θθθ∗θ∗gµν, (95.28) we ﬁnd Φ(x,θ,θ∗) =A(x) +√ 2θψ(x) +θθF(x)−i(θσµθ∗)∂µA(x) −1√ 2iθθθ∗¯σµ∂µψ(x) +1 4θθθ∗θ∗∂2A(x). (95.29) Let us investigate the properties of a left-handed chiral su perﬁeld under a supersymmetry transformation, given by eqs.(95.8) and (9 5.9). It is easiest to use the yandθcoordinates, since Qaθb=δab,Qayµ= 0, Q∗ ˙aθb= 0, Q∗ ˙ayµ=−2iθcσµ c˙a. (95.30) 95: Supersymmetry 593 We thus have QaΦ(y,θ) =∂aΦ(y,θ) =√ 2ψa(y) + 2θaF(y), (95.31) Q∗ ˙aΦ(y,θ) =−2iθcσµ c˙a∂µΦ(y,θ) =−2iθcσµ c˙a∂µA(y) +i√ 2θθ∂µψc(y)σµ c˙a,(95.32) where∂µis with respect to y; we used eq.(95.26) to get the last line. We can now ﬁnd the supersymmetry transformations of the compon ent ﬁelds A,ψ, andFby matching powers of θon each side of eqs.(95.8) and (95.9). Remembering that the QandQ†operators anticommute with θandθ∗, we get [A,Qa] =−i√ 2ψa, [A,Q† ˙a] = 0, (95.33) {ψc,Qa}=−i√ 2εacF , {ψc,Q† ˙a}=−√ 2σµ c˙a∂µA, (95.34) [F,Qa] = 0, [F,Q† ˙a] =√ 2∂µψcσµ c˙a, (95.35) where all component ﬁelds have spacetime argument y. However, yis arbitrary, and so we are free to replace it with x. Eq.(95.35) is the most important: it tells us that the supers ymmetry transformation of the Fﬁeld is a total derivative. Therefore,/integraltextd4xF(x) is invariant under a supersymmetry transformation, and hen ce could be a term in the action of a supersymmetric theory. The product of two left-handed chiral superﬁelds is another left-handed chiral superﬁeld; this is obvious from eq.(95.25), and the f act that the θexpansion always terminates with the quadratic term. For tw o chiral superﬁelds Φ 1(y,θ) and Φ 2(y,θ), we have Φ1Φ2=A1A2+√ 2θ(A1ψ2+A2ψ1) +θθ(A1F2+A2F1−ψ1ψ2).(95.36) More generally, given a set of left-handed chiral superﬁeld s Φi, we can consider a function of them W(Φ); this function is itself a left-handed chiral superﬁeld. Its Fterm (the coeﬃcient of θθ) is W(Φ)/vextendsingle/vextendsingle/vextendsingle F=∂W(A) ∂AiFi−1 2∂2W(A) ∂Ai∂Ajψiψj, (95.37) where repeated indices are summed. The spacetime integral o f this term (like the spacetime integral of any Fterm) is invariant under supersymme- try, and hence could be a term in the action of a supersymmetri c theory. In this case, the function W(Φ) is called the superpotential . 95: Supersymmetry 594 We still need kinetic terms. To get them, we ﬁrst investigate the prop- erties of a vector superﬁeld . A vector superﬁeld V(x,θ,θ∗) is hermitian, [V(x,θ,θ∗)]†=V(x,θ,θ∗), (95.38) but is not subject to any other constraint. Its component exp ansion is V(x,θ,θ∗) =C(x) +θχ(x) +θ∗χ†(x) +θθM(x) +θ∗θ∗M†(x) +θσµθ∗vµ(x) +θθθ∗λ†(x) +θ∗θ∗θλ(x) +1 2θθθ∗θ∗D(x), (95.39) whereCandDare real scalar ﬁelds, Mis a complex scalar ﬁeld, χandλ are left-handed Weyl ﬁelds, and vµis a real vector ﬁeld. Following the analysis that led to eq.(95.35), we ﬁnd [D,Qa] =−σµ a˙c∂µλ†˙c,[D,Q† ˙a] = +∂µλcσµ c˙a. (95.40) We see that the supersymmetry transformation of the Dcomponent of a vector superﬁeld is a total derivative. Therefore,/integraltextd4xD(x) is invariant under a supersymmetry transformation, and hence could be a t erm in the action of a supersymmetric theory. Consider the product of a left-handed chiral superﬁeld Φ( x,θ,θ∗), as given by eq.(95.29), and its hermitian conjugate Φ†(x,θ,θ∗) =A(x) +√ 2θ∗ψ†(x) +θ∗θ∗F†(x) +i(θσµθ∗)∂µA†(x) +1√ 2iθ∗θ∗∂µψ†(x)¯σµθ+1 4θθθ∗θ∗∂2A†(x). (95.41) The product Φ†Φ is obviously hermitian, and so is a vector superﬁeld. After considerable use of eqs.(95.26–95.28), we ﬁnd that th eDterm (the coeﬃcient of θθθ∗θ∗) of this vector superﬁeld is Φ†Φ/vextendsingle/vextendsingle/vextendsingle D=−1 2∂µA†∂µA+1 4A∂2A†+1 4A†∂2A +1 2iψ†¯σµ∂µψ−1 2i∂µψ†¯σµψ +F†F . (95.42) The spacetime integral of this term (like the spacetime inte gral of any D term) is invariant under supersymmetry, and hence could be a term in the action of a supersymmetric theory. After some integrations by parts, and dropping total divergences, we ﬁnd Φ†Φ/vextendsingle/vextendsingle/vextendsingle D=−∂µA†∂µA+iψ†¯σµ∂µψ+F†F . (95.43) We see that we have standard kinetic terms for the complex sca lar ﬁeldA and the left-handed Weyl ﬁeld ψ. We also have a term with no derivatives 95: Supersymmetry 595 for the complex scalar ﬁeld F. TheFﬁeld is therefore called an auxiliary ﬁeld. If we consider a set of left-handed chiral superﬁelds Φ i, we get a hermi- tian, supersymmetric action if we take as the lagrangian L= Φ† iΦi/vextendsingle/vextendsingle/vextendsingle D+/parenleftig W(Φ)/vextendsingle/vextendsingle/vextendsingle F+ h.c./parenrightig , (95.44) where the index in the ﬁrst term is summed. Since Fiappears only quadrat- ically and without derivatives, we can easily perform the pa th integral over it. The result is equivalent to solving the classical equati on of motion for Fi, ∂L ∂Fi=F† i+∂W(A) ∂Ai= 0, (95.45) and substituting the solution back into the lagrangian; the result is L=−∂µA† i∂µAi+iψ† i¯σµ∂µψi −/vextendsingle/vextendsingle/vextendsingle/vextendsingle∂W(A) ∂Ai/vextendsingle/vextendsingle/vextendsingle/vextendsingle2 −1 2/bracketleftigg ∂2W(A) ∂Ai∂Ajψiψj+ h.c./bracketrightigg , (95.46) where the indices are summed in each term. As an example, let us consider a single left-handed chiral su perﬁeld, with superpotential W(A) =1 2mA2+1 6gA3. (95.47) This is the Wess-Zumino model . The scalar potential is V(A) =|∂W/∂A |2 =m2A†A+1 2gm(A†A2+A†2A) +1 4g2(A†A)2.(95.48) We see that the scalar has mass m. The last term in eq.(95.46) becomes Lmass+Yuk =−1 2mψψ−1 2gAψψ + h.c.. (95.49) We see that the fermion also has mass m, and a Yukawa interaction with the scalar. The Yukawa couping is related (by supersymmetry ) to the cubic and quartic self-interactions of the scalar. Next we would like to introduce gauge ﬁelds. Recall that the v ector superﬁeldV(x,θ,θ∗) has among its components a real vector ﬁeld vµ(x) that could be identiﬁed as an abelian gauge ﬁeld. (Later we wi ll add an adjoint index to the superﬁeld in order to get a nonabelian ga uge ﬁeld.) We need to generalize the notion of a gauge transformation to super- ﬁelds. We begin by noting that if Ξ is a left-handed chiral sup erﬁeld, then i(Ξ†−Ξ) is a vector superﬁeld. We then deﬁne a supergauge transformation V→V+i(Ξ†−Ξ). (95.50) 95: Supersymmetry 596 We will attempt to construct actions that are invariant unde r eq.(95.50). Following the pattern of eqs.(95.29) and (95.41), we write Ξ(x,θ,θ∗) =B(x) +θξ(x) +θθG(x) −i(θσµθ∗)∂µB(x) +... , (95.51) Ξ†(x,θ,θ∗) =B†(x) +θ∗ξ†(x) +θ∗θ∗G†(x) +i(θσµθ∗)∂µB†(x) +... . (95.52) If we setB=1 2(b+ia), whereaandbare real scalar ﬁelds, we ﬁnd i(Ξ†−Ξ) =a−iθξ+iθ∗ξ†−iθθG+iθ∗θ∗G†−(θσµθ∗)∂µb+... .(95.53) From eq.(95.39), we see that the supergauge transformation of eq.(95.50) results in C→C+a, χ→χ−iξ , M→M−iG, vµ→vµ−∂µb. (95.54) The last of these is the usual abelian gauge transformation. The ﬁrst three allow us to gauge away theC,χ, andMcomponents of a vector superﬁeld. That is, we can make a supergauge transformation with a=−C,ξ=−iχ andG=−iM; in this gauge, known as Wess-Zumino gauge , the vector superﬁeld becomes V= (θσµθ∗)vµ+θθθ∗λ†+θ∗θ∗θλ+1 2θθθ∗θ∗D. (95.55) Note that we still have the freedom to make the supergauge tra nsformation of eq.(95.50) with B(x) =1 2b(x), and that this still implements the ordinary abelian gauge transformation of eq.(95.54). Now consider a left-handed chiral superﬁeld Φ that has charg e +1 under a U(1) gauge group. We take the kinetic term for Φ to be Lkin= Φ†e−2gVΦ/vextendsingle/vextendsingle/vextendsingle D, (95.56) wheregis the gauge coupling. The vector superﬁeld Φ†e−2gVΦ is clearly invariant under the supergauge transformation Φ→e−2igΞΦ, (95.57) Φ†→Φ†e+2igΞ†, (95.58) V→V+i(Ξ†−Ξ). (95.59) 95: Supersymmetry 597 Let us evaluate eq.(95.56) in Wess-Zumino gauge, where we ha ve V2=−1 2θθθ∗θ∗vµvµ, (95.60) V3= 0. (95.61) The exponential factor in eq.(95.56) becomes e−2gV= 1−2g(θσµθ∗)vµ−2gθθθ∗λ†−2gθ∗θ∗θλ −θθθ∗θ∗(gD+g2vµvµ). (95.62) The relevant terms in Φ†Φ are Φ†Φ =A†A+√ 2θ∗ψ†A+√ 2θψA† + (θσµθ∗)(ψ†¯σµψ−iA†∂µA+iA∂µA†) +...+θθθ∗θ∗(Φ†Φ)D, (95.63) where we used 2( θ∗ψ†)(θψ) = (θσµθ∗)(ψ†¯σµψ) to get the ﬁrst term in the second line, and (Φ†Φ)Dis given by eq.(95.42). Combining eqs.(95.62) and (95.63), taking the Dterm, and performing the same integrations by parts that led to eq.(95.43), we ﬁnd Φ†e−2gVΦ/vextendsingle/vextendsingle/vextendsingle D=−(DµA)†DµA+iψ†¯σµDµψ+F†F +√ 2gψ†λ†A+√ 2gA†λψ−gA†DA, (95.64) whereDµ=∂µ−igvµis the usual gauge covariant derivative acting on ﬁeld of charge +1. We still need a kinetic term for the vector superﬁeld. To get i t, we ﬁrst introduce a superﬁeld that carries a left-handed spinor ind ex, Wa≡1 4D∗ ˙aD∗˙aDaV . (95.65) Since the two components of D∗anticommute, we have D∗ ˙aD∗ ˙bD∗ ˙c= 0. ThusWaobeys D∗ ˙aWa= 0, and is therefore a left-handed chiral super- ﬁeld. Furthermore, Wais invariant under the supergauge transformation of eq.(95.50). To see this, we ﬁrst note that D∗ ˙aD∗˙aDaannihilates Ξ†(be- causeDadoes). Then, we use eq.(95.19) to write D∗ ˙aD∗˙aDa=−(D∗ ˙aDa+ 2iσµ a˙a∂µ)D∗˙a. (95.66) Thus, D∗ ˙aD∗˙aDaalso annihilates Ξ (because D∗˙adoes). Therefore, Wais invariant under eq.(95.50). SinceWais a left-handed chiral superﬁeld, it has an expansion in the form of eq.(95.25). To ﬁnd the component ﬁelds of Wa, we setx=y+ iθσµθ∗in eq.(95.55), and expand in θandθ∗. The result is V= (θσµθ∗)vµ+θθθ∗λ†+θ∗θ∗θλ+1 2θθθ∗θ∗(D−i∂µvµ),(95.67) 95: Supersymmetry 598 where all component ﬁelds have spacetime argument y. From eq.(95.24), we see that D∗ ˙a=−∂∗ ˙awhen it acts on a function of y,θ, andθ∗. We also have Da= (Daθc)∂c+ (Daθ∗˙c)∂∗ ˙c+ (Dayµ)∂µ=∂a+ 0−2iσµ a˙aθ∗˙a∂µ,(95.68) where∂µis with respect to y. Using eq.(95.68), we ﬁnd DaV=θ∗θ∗/bracketleftig λa+θa(D−i∂·v)−i(σµ¯σνθ)a∂µvν+iθθ(σµ∂µλ†)a/bracketrightig +... . (95.69) When we act on DaVwithD∗ ˙aD∗˙a=∂∗ ˙a∂∗˙ato getWa, only the coeﬃcient ofθ∗θ∗survives. Since ∂∗ ˙a∂∗˙a(θ∗θ∗) = 4, we ﬁnd Wa=λa+θa(D−i∂·v)−i(σµ¯σνθ)a∂µvν+iθθ(σµ∂µλ†)a.(95.70) We can simplify eq.(95.70) by using the identity (σµ¯σν)ab=−gµνδac−2i(Sµν L)ab. (95.71) Remembering that Sµν Lis antisymmetric on µ↔ν, and deﬁning the ﬁeld strength Fµν≡∂µvν−∂νvµ, (95.72) we get Wa=λa+θaD−(Sµν L)acθcFµν+iθθσµ a˙a∂µλ†˙a. (95.73) We see that Wainvolves the vector ﬁeld vµonly through its gauge-invariant ﬁeld strength Fµν. SinceWais supergauge invariant, this is to be expected. Next, consider the Fterm ofWaWa. This term is Lorentz invariant, and its spacetime integral (like the spacetime integral of a nyFterm) is invariant under supersymmetry, and hence could be a term in t he action of a supersymmetric theory. Working out the components, we ﬁnd WaWa/vextendsingle/vextendsingle/vextendsingle F= 2iλaσµ a˙a∂µλ†˙a−1 2Tr(Sµν LSρσ L)FµνFρσ+D2, (95.74) where Tr(Sµν LSρσ L) = (Sµν L)ac(Sρσ L)ca. To get the spin matrices into this form, we used the fact that ( Sµν L)acis symmetric on a↔c. Now we use the identity Tr(Sµν LSρσ L) =1 2(gµρgνσ−gµσgνρ)−1 2iεµνρσ(95.75) to get WaWa/vextendsingle/vextendsingle/vextendsingle F= 2iλaσµ a˙a∂µλ†˙a−1 2FµνFµν−1 2i˜FµνFµν+D2, (95.76) 95: Supersymmetry 599 where ˜Fµν=1 2εµνρσFρσ. We can now identify the kinetic term for the vector superﬁeld as Lkin=1 4WaWa/vextendsingle/vextendsingle/vextendsingle F+ h.c. =iλ†¯σµ∂µλ−1 4FµνFµν+1 2D2. (95.77) We integrated by parts and dropped total divergences to get t he second line. We see that we have the standard kinetic terms for the ga uge ﬁeldvµ and the gaugino ﬁeldλ, whileDis an auxiliary ﬁeld. All of this generalizes in a straightforward way to the nonab elian case. We deﬁne a matrix-valued vector superﬁeld V=VaTa R, and a matrix- valued chiral superﬁeld Ξ = ΞaTa R, whereais an adjoint group index. A chiral superﬁeld Φ in the representation R still transforms according to eqs.(95.57) and (95.58), but for the vector ﬁeld we have e−2gV→e−2igΞ†e−2gVe+2igΞ; (95.78) this reduces to eq.(95.59) in the abelian case. The ﬁeld-strength superﬁeld is now Wa=−1 8gD∗ ˙aD∗˙ae+2gVDae−2gV. (95.79) Under a supergauge transformation, Wa→e−2igΞWae+2igΞ. (95.80) Eq.(95.73) still holds, but the derivative that acts on λ†is now the gauge co- variant derivative for the adjoint representation, and Fµνnow includes the usual nonabelian commutator term. These changes also apply to eq.(95.77), where we must also trace over the group indices and (for prope r normal- ization) divide by the index T(R). Reference Notes Introductions to supersymmetry can be found in Martin ,Siegel,Weinberg III, andWess & Bagger . Problems 95.1) Use eq.(95.6) to show that the hamiltonian is positive semideﬁnite, and that a state with zero energy must be annihilated by all th e supercharges. 95: Supersymmetry 600 95.2) Supersymmetry is spontaneously broken if the ground s tate|0∝an}b∇acket∇i}htis not annihilated by all the supercharges. a) Use the ﬁrst of eqs.(95.34) to show that supersymmetry is s pon- taneously broken if ∝an}b∇acketle{t0|F|0∝an}b∇acket∇i}ht ∝ne}ationslash= 0. b) Compute {λa,Qb}. Use the result to show that supersymmetry is spontaneously broken if ∝an}b∇acketle{t0|D|0∝an}b∇acket∇i}ht ∝ne}ationslash= 0. 95.3) Consider a supersymmetric theory with three chiral su perﬁeldsA,B, andC, and a superpotential W=mBC +κA(C2−v2), wheremand vare parameters with dimensions of mass, and κis a dimensionless coupling constant. This is the O’Raifeartaigh model . a) Show that one or more Fcomponents is nonzero at the minimum of the potential, and hence that supersymmetry is spontaneo usly bro- ken. b) Show that the potential is minimized along a line in ﬁeld sp ace, and ﬁnd the masses of the particles at an arbitrary point on th is line. You should ﬁnd that there is a massless Goldstone fermion or goldstino that is related by supersymmetry to the linear combination ofFﬁelds that gets a nonzero vacuum expectation value. 95.4)Supersymmetric Quantum Electrodynamics. Consider a supersym- metric U(1) gauge theory with chiral superﬁelds Φ and Φ with charges +1 and −1, respectively. Here the bar over the Φ in the ﬁeld Φ ispart of the name of the ﬁeld , and does not denote any sort of conjugation. We include a gauge invariant superpotential W=mΦΦ. a) Work out the lagrangian in terms of the component ﬁelds. b) Eliminate the auxilliary ﬁelds F,F, andD. 95.5) In a supersymmetric gauge theory with a U(1) factor, we can add a Fayet-Illiopoulos term LFI=eξDto the lagrangian, where Dis the auxilliary ﬁeld for the U(1) gauge ﬁeld, and ξis a parameter with dimensions of mass-squared. a) Explain why adding this term preserves supersymmetry. Ex plain why the corresponding gauge ﬁeld cannot be nonabelian. b) Add this term to the SQED lagrangian that you found in probl em 94.4, and eliminate the auxilliary ﬁelds. c) Minimize the resulting potential. Show that supersymmet ry is spontaneously broken if ξis in a certain range. 95.6)R symmetry. Given a supersymmetric gauge theory (abelian or non- abelian), consider a global U(1) transformation that chang es the 95: Supersymmetry 601 phase of the gaugino ﬁelds, λa→e−iαλa; we say that the gauginos haveR charge +1. a) IfRsymmetry is to be a good symmetry of the lagrangian, what relation must hold between the Rcharges of the scalar and fermion components of a chiral superﬁeld that couples to the gauge ﬁe lds? b) In additon, what conditions must be placed on the superpot ential? c) Identify the Rsymmetry, if any, of supersymmetric quantum elec- trodynamics. 96: The Minimal Supersymmetric Standard Model 602 96The Minimal Supersymmetric Standard Model Prerequisite: 89, 95 Having seen how to construct a general supersymmetric gauge theory in section 95, we can now write down a supersymmetric version of the Stan- dard Model. To do so, we introduce vector superﬁelds for the gauge group S U(3)× SU(2) ×U(1), which include the usual gauge bosons along with their f er- mionic partners, the gauginos ; chiral superﬁelds LiI,¯EI,QαiI,¯Uα I, and ¯Dα I that include three generations of quarks and leptons along w ith their scalar partners, the squarks andsleptons ; and chiral superﬁelds Hiand¯Hiin the representations (1 ,2,−1 2) and (1,2,+1 2), which include two copies of the usual Higgs ﬁeld along with their fermionic partners, the higgsinos . We give all these ﬁelds supersymmetric, gauge invariant kinet ic terms, and a superpotential W=−yIJεijHiLjI¯EJ−y′ IJHiQαjI¯Dα J−y′′ IJ¯HiQαiI¯Uα J−µεij¯HiHj,(96.1) whereµis a mass parameter. This superpotential generates the usua l Yukawa couplings, among others, and gives a positive mass-s quared to both Higgs ﬁelds. We forbid terms of the form ¯HiLiI(which are allowed by the gauge symmetry) by invoking a discrete symmetry, R parity . UnderR parity, all Standard Model ﬁelds (including both Higgs scal ars) are taken to be even, and all superpartner ﬁelds (gauginos, higgsinos, s quarks, sleptons) are taken to be odd. Supersymmetry is clearly not an exact symmetry of the real wo rld, and so if present must be spontaneously broken. Correct phen omenology requires supersymmetry breaking to be triggered by ﬁelds ot her than those listed above. There are many possibilities for the dynamics of the ﬁelds in this hidden sector , and an exploration of them is beyond our scope. However, we can parameterize the eﬀects of the hidden sector via aspurion ﬁeld. This is a constant chiral superﬁeld of the form S=m2 Sθθ, (96.2) wheremSis the supersymmetry breaking scale . We couple Sto the quark, lepton, and Higgs superﬁelds via Dterms of the form Lspur,D=m−2 MS†S/braceleftbigg C(H)H†H+C(¯H)¯H†¯H +/summationdisplay IJ/bracketleftig C(L) IJL† ILJ+C(¯E) IJ¯E† I¯EJ+C(Q) IJQ† IQJ +C(¯U) IJ¯U† I¯UJ+C(¯D) IJ¯D† I¯DJ/bracketrightig/bracerightbigg/vextendsingle/vextendsingle/vextendsingle/vextendsingle D.(96.3) 96: The Minimal Supersymmetric Standard Model 603 Here we have suppressed all but generation indices; each C(Φ) IJis a dimen- sionless hermitian matrix in generation space. The paramet ermMis the messenger scale . Eq.(96.3) gives masses of order m2 S/mMto the scalars. We also couple Sto the chiral gauge superﬁelds W(i) aviaFterms of the form Lspur,gauge=m−1 MS/summationdisplay i=1,2,3C(i)W(i)aW(i) a/vextendsingle/vextendsingle/vextendsingle/vextendsingle F+ h.c., (96.4) where the sum is over the three gauge-group factors; ais a spinor index. Eq.(96.4) gives masses of order m2 S/mMto the gauginos. Finally, we couple Sto the chiral superﬁelds via Fterms of the form Lspur,F=m−1 MS28/summationdisplay A=1CAWA/vextendsingle/vextendsingle/vextendsingle/vextendsingle F+ h.c., (96.5) whereWAis one of the 28 gauge-invariant terms in the superpotential . In particular, eq.(96.5) includes a mass term of the form ¯HH+ h.c.for the Higgs scalars. Eqs.(96.1) and (96.3–96.5) specify the Minimal Supersymmetric Stan- dard Model , or MSSM for short. Obviously, it has a complicated phe- nomenology that is beyond our scope to explore in detail. One point worth noting is that Rparity implies that the lightest superpartner, or LSP, is absolutely stable. Reference Notes Supersymmetric versions of the Standard Model are discusse d inMartin , Ramond II , andWeinberg III . Problems 96.1) Explain why we need two Higgs doublets. 96.2) a) Write the mass terms for the Higgs scalars as LHiggsmass =−m2 1H†H−m2 2¯H†¯H−m2 3εij(¯HiHj+ h.c.),(96.6) and compute the quartic terms in Hand¯Hthat arise from eliminating the auxilliary SU(2) and U(1) Dﬁelds. b) Find the conditions on the mass parameters in eq.(96.6) in order for the potential to be bounded below. c) Find the conditions on the mass parameters in eq.(96.6) in order to have spontaneous breaking of the SU(2) ×U(1) symmetry. 96: The Minimal Supersymmetric Standard Model 604 d) Show that there are ﬁve physical Higgs particles, two char ged and three neutral. e) Let tanβ≡¯v/vbe the ratio of the two Higgs VEVs. Show that tanβ+ cotβ= (m2 1+m2 2)/m2 3. (96.7) 97: Grand Uniﬁcation 605 97Grand Unification Prerequisite: 89 The Standard Model is based on the gauge group SU(3) ×SU(2) ×U(1), with left-handed Weyl ﬁelds in three copies of the represent ation (1,2,−1 2)⊕ (1,1,+1)⊕(3,2,+1 6)⊕(¯3,1,−2 3)⊕(¯3,1,+1 3), and a complex scalar ﬁeld in the representation (1 ,2,−1 2). The lagrangian includes all terms of mass dimension four or less that are allowed by the gauge symmetri es and Lorentz invariance. To complete the speciﬁcation of the Standard Model, we need t wenty real numbers: the three gauge couplings; the three diagonal entries of each of the three diagonalized Yukawa coupling matrices (for the up quarks, the down quarks, and the charged leptons); the four angles in the CKM mixing matrix for the quarks; the vacuum angles for the SU(3) and SU( 2) gauge groups; the scalar quartic coupling; and the scalar mass-sq uared. The goal of grand uniﬁcation is to construct a more compact model with fewer parameters by supposing that the Standard Model i s the result of the spontaneous breaking of a larger gauge symmetry. The s implest model along these lines is the Georgi–Glashow SU(5) model. Its start- ing point is the grand uniﬁed gauge group SU(5). We include a scalar ﬁeld Φ = ΦaTain the adjoint or 24 representation, and assume that the scalar potential for this ﬁeld results in a vacuum expect ation value (VEV) of the form ∝an}b∇acketle{t0|Φ|0∝an}b∇acket∇i}ht= diag( −1 3,−1 3,−1 3,+1 2,+1 2)V. As we saw in section 84, this VEV spontaneously breaks the gauge symmetr y down to SU(3) ×SU(2) ×U(1). The generator of the unbroken U(1) subgroup is T24=cdiag(−1 3,−1 3,−1 3,+1 2,+1 2), wherec2= 3/5. It will prove convenient to write T24=/radicalig 3 5Y , (97.1) and express the U(1) charge as the value of Yrather than the value of T24. We also note that the SU(5) breaking scale Vmust be considerably larger than the SU(2) ×U(1) breaking scale v∼250GeV in order to suppress the observable eﬀects of the extra gauge ﬁelds. Under the SU(3) ×SU(2) ×U(1) subgroup, the fundamental and anti- fundamental representations of SU(5) transform as 5→(3,1,−1 3)⊕(1,2,+1 2). (97.2) 5→(¯3,1,+1 3)⊕(1,2,−1 2). (97.3) Next we use eq.(97.2) to ﬁnd that the product 5 ⊗5 transforms as 5⊗5→(6,1,−2 3)S⊕(3,2,+1 6)S⊕(¯3,1,−2 3)S ⊕(¯3,1,−2 3)A⊕(3,2,+1 6)A⊕(1,3,+1)A, (97.4) 97: Grand Uniﬁcation 606 where the subscripts indicate the symmetric and antisymmet ric parts of the product. In terms of SU(5), we have 5⊗5 = 15 S⊕10A. (97.5) Comparing eqs.(97.4) and (97.5), we ﬁnd 10→(¯3,1,−2 3)⊕(3,2,+1 6)⊕(1,3,+1). (97.6) From eqs.(97.2) and (97.6), we see that one generation of quark and lepton ﬁelds ﬁts exactly into the representation 5⊕10ofSU(5). We therefore deﬁne a left-handed Weyl ﬁeld ψiin the 5 representation of SU(5), and a left-handed Weyl ﬁeld χij=−χjiin the 10 representation. The gauge covariant derivatives of these ﬁelds are (Dµψ)i=∂µψi−ig5Aa µ(Ta 5)ijψj =∂µψi+ig5Aa µ(Ta)jiψj, (97.7) (Dµχ)ij=∂µχij−ig5Aa µ(Ta 10)ijklχkl =∂µχij−ig5Aa µ[(Ta)ikχkj+ (Ta)jlχil], (97.8) whereg5is the SU(5) gauge coupling and Tais the generator matrix in the fundamental representation. The kinetic terms for these ﬁe lds are Lkin=iψ† i¯σµ(Dµψ)i+1 2iχ†ij¯σµ(Dµχ)ij, (97.9) where the implicit sum over iandjis unrestricted; this necessitates the prefactor of one-half in the second term to avoid double coun ting. The interaction terms with the gauge ﬁelds then work out to be Lint=−g5/bracketleftig ψ† i(AT µ)ij¯σµψj+χ†ji(Aµ)ik¯σµχkj/bracketrightig , (97.10) whereAµ=Aa µTais the matrix-valued gauge ﬁeld, and AT µis its transpose. Note that we have written the factors in matrix-multiplicat ion order (with a trace for the second term). We can identify the components of ψiandχijas ψi= ( ¯dr¯db¯dge−ν), (97.11) χij= 0 ¯ug−¯uburdr −¯ug0 ¯urubdb ¯ub−ur0ugdg −ur−ub−ug0 ¯e −dr−db−dg−¯e 0 , (97.12) 97: Grand Uniﬁcation 607 wherer,b, andgstand for the three colors (red, blue, and green). We can also write the gauge ﬁelds as AaTa= Grr−1 3cB G rbGrg 1√ 2X1 r1√ 2X2 r GbrGbb−1 3cB G bg 1√ 2X1 b1√ 2X2 b GgrGgbGgg−1 3cB1√ 2X1 g1√ 2X2 g 1√ 2X†r 11√ 2X†b 11√ 2X†g 11 2W3+1 2cB1√ 2W+ 1√ 2X†r 21√ 2X†b 21√ 2X†g 21√ 2W−−1 2W3+1 2cB , (97.13) where the Lorentz index has been omitted. Here Bis the hypercharge gauge ﬁeld, and W3and√ 2W±=W1±iW2are the SU(2) gauge ﬁelds. The gluon ﬁelds Gijare subject to the constraint Grr+Gbb+Ggg= 0. The Xi αﬁelds correspond to the broken generators of SU(5), and henc e become massive; here iis an SU(2) index and αis an SU(3) index. As we saw in section 84, the Xi αﬁelds transform as (3 ,2,−5 6) under SU(3) ×SU(2)×U(1), and their mass is MX=5 6√ 2g5V. If we substitute eqs.(97.11–97.13) into eq.(97.10), we ﬁnd the usual interactions of the SU(3) ×SU(2) ×U(1) gauge ﬁelds with the quarks and leptons, but with/radicalig 5 3g1=g2=g3=g5. (97.14) These relations among the gauge couplings hold in the MS renormalization scheme; later we will discuss a modiﬁed scheme that is more ap propriate at energies well below MX. We also ﬁnd the couplings of the Xﬁeld to quarks and leptons; these work out to be LX,int=−1√ 2g5/bracketleftig X†α 1µ(¯d† α¯σµe−¯e†¯σµdα+u†β¯σµ¯uγεαβγ) +X†α 2µ(−¯d† α¯σµν+ ¯e†¯σµuα+d†β¯σµ¯uγεαβγ)/bracketrightig + h.c. =−1√ 2g5X†α iµ(εij¯d† α¯σµℓj−εij¯e†¯σµqjα+q†βi¯σµ¯uγεαβγ) + h.c. ≡ −1√ 2g5X†α iµJiµ α+ h.c., (97.15) where the last line deﬁnes the current Jµthat couples to the Xµﬁeld. (To include more than one generation, we add a generation index Ito each quark and lepton ﬁeld, and sum over it.) The most interesting feature of eq.(97.15) is that the ﬁrst two terms in the current have bary on number B= +1 3and lepton number L= +1, while the third term has B=−2 3 andL= 0. Thus exchange of an Xboson can violate baryon and lepton 97: Grand Uniﬁcation 608 number conservation, leading to phenomena such as proton de cay. Proton decay has not been observed; the limit on the rate 1 /τforp→e+π0is τ >1033yr. A rough estimate of 1 /τfrom eq.(97.15) is g4 5m5 p/8πM4 X. Takingg5∼g2∼0.6, we ﬁnd that we must have MX>3×1015GeV. We still need a scalar ﬁeld in the representation (1 ,2,−1 2). The smallest complete representation of SU(5) that includes this piece i s the5. Call the corresponding ﬁeld Hi; we can identify its components as Hi= (φrφbφgϕ−−ϕ0). (97.16) The possible gauge-invariant Yukawa couplings with the ψiandχijﬁelds are LYuk=−yHiψjχij−1 8y′′εijklmH† iχjkχlm+ h.c.; (97.17) with three generations, yandy′′become matrices in generation space. We can write out LYukusing eqs.(97.11), (97.12), and (97.16); the result is LYuk=−yεijϕiℓj¯e−yεijϕiqαj¯dα−y′′ϕ†iqαi¯uα −yεαβγφα¯dβ¯uγ−yεijφαqαiℓj−y′′φ† α¯uα¯e+ h.c..(97.18) The terms on the ﬁrst line are those of the Standard Model, exc ept that the down-quark Yukawa coupling matrix (called y′in section 89) is the same as the charged-lepton Yukawa coupling matrix (called yin section 88). Since the quark and lepton masses are directly proporti onal to the Yukawa couplings, eq.(97.18) predicts mb=mτ, ms=mµ, md=me. (97.19) These relations hold in the MS renormalization scheme; later we will discuss a modiﬁed scheme that is more appropriate at energies well be lowMX. The terms on the second line of eq.(97.18) are the couplings of th e colored scalar ﬁeldφto the quarks and leptons; we see that these couplings, like t hose of theXµﬁeld, violate baryon and lepton number conservation. Since ﬁrst- generation Yukawa couplings are smaller than gauge couplin gs by a factor of 105, the limit on Mφfrom proton decay is roughly Mφ>1010GeV. To compute Mφ, we need the complete scalar potential. For simplic- ity, we assume a Z 2symmetry under Φ ↔ −Φ. Then the most general renormalizable potential is V(Φ,H) =−1 2m2 ΦTrΦ2+1 4λ1Tr Φ4+1 4λ2(Tr Φ2)2 +m2 HH†H+1 4κ1(H†H)2−1 2κ2H†Φ2H . (97.20) We take all the parameters ( m2 Φ,m2 H,λ1,λ2,κ1,κ2) to be positive. We found in problem 84.1 that in this case the ﬁrst line of eq.(97 .20) is mini- mized by Φ = diag( −1 3,−1 3,−1 3,+1 2,+1 2)V, withV2= 36m2 Φ/(7λ1+30λ2). 97: Grand Uniﬁcation 609 From the ﬁrst and third terms on the second line of eq.(97.20) , we ﬁnd that the masses-squared of the ϕ∼(1,2,−1 2) andφ∼(¯3,1,+1 3) scalar ﬁelds are m2 ϕ=m2 H−1 8κ2V2, (97.21) M2 φ=m2 H−1 18κ2V2. (97.22) We wantm2 ϕ∼ −(100GeV)2andM2 φ>+(1010GeV)2. This requires m2 Hto be equal to1 8κ2V2to at least sixteen signiﬁcant digits (but not exactly). There is no obvious reason for the parameters in eq .(97.20) to satisfy this odd relation; we would more naturally expect m2 ϕandM2 φto have the same order of magnitude (and furthermore to have κ2∼g2 5and henceMφ∼MX>1015GeV). This is the ﬁne tuning problem of grand uniﬁed theories. More generally, we can ask why the breaking scale of the grand uniﬁed group is so much larger than the breaking sca le of the electroweak subgroup; this is gauge hierarchy problem . Since theXµandφﬁelds are so heavy, we can integrate them out, generating eﬀective interactions among the quarks and lept ons.1For the Xµﬁeld, we get the leading term (in a double expansion in powers ofg5and inverse powers of MX) by ignoring the kinetic energy and other interactions of theXµﬁeld, solving the equations of motion for Xµthat follow from LX,mass+LX,int, where LX,intis given by eq.(97.15) and LX,mass=−M2 XX†α iµXiµ α, (97.23) and ﬁnally substituting the solutions back into LX,mass+LX,int. This is equivalent to evaluating tree-level Feynman diagrams with a singleXex- changed. The result is LX,eﬀ=1 2M2 XJ†α iµJiµ α. (97.24) Keeping only ﬁelds from the ﬁrst generation, we ﬁnd that the b aryon- and lepton-number violating terms in LX,eﬀare L|∆B|=1 X,eﬀ=g2 5 2M2 Xεijεαβγ(¯d† α¯σµℓi−¯e†¯σµqiα)¯u† β¯σµqjγ+ h.c. =−g2 5 M2 Xεijεαβγ/bracketleftig (ℓiqjγ)(¯d† α¯u† β) + (¯e†¯u† γ)(qiαqjβ)/bracketrightig + h.c.,(97.25) where the second line follows from a Fierz identity and a rela beling of the color indices in the second term. We can treat the φﬁeld similarly; 1We should also integrate out the heavy components of Φ, which transform as (8 ,1,0)⊕ (1,3,0)⊕(1,1,0) under SU(3) ×SU(2) ×U(1), but these do not couple directly to quarks and leptons. 97: Grand Uniﬁcation 610 see problem 97.2. We will compute the decay rate for p→e+π0from eq.(97.25) in problem 97.4. Once the heavy ﬁelds have been integrated out, we can apply th eMS renormalization scheme to the theory with the remaining lig ht ﬁelds. This is not the same as MS for the original theory, because now only light ﬁelds circulate in loops. (Loops of heavy ﬁelds contribute to corr ections to the eﬀective interactions.) With the usual ﬁelds of the Standar d Model, we ﬁnd from our results in sections 66 and 73 that the one-loop be ta functions for the three gauge couplings are given by µd dµgi=bi 16π2g3 i+O(g5 i), (97.26) with b3=−11 +4 3n, (97.27) b2=−22 3+4 3n+1 6, (97.28) b1= +20 9n+1 6, (97.29) wheren= 3 in the number of generations; the +1 6contributions to b2 andb1are from the ϕﬁeld. These formulae apply for µ < M X. (We are assuming that the heavy scalars do not have masses much le ss than MX.) Forµ≥MX, we must restore the heavy ﬁelds, and then eq.(97.14) applies. If we now neglect the higher-loop corrections, int egrate eq.(97.26) for each coupling, impose eq.(97.14) at µ=MX, and setg2=e/sinθWand g1=e/cosθW, we ﬁnd for µ<M Xthat 1 α3(µ)=1 α5(MX)+b3 2πln(MX/µ), (97.30) sin2θW(µ) α(µ)=1 α5(MX)+b2 2πln(MX/µ), (97.31) cos2θW(µ) α(µ)=5/3 α5(MX)+b1 2πln(MX/µ). (97.32) The quantities on the left-hand sides are measured at µ=MZto be α3(MZ) = 0.1187±0.0020. (97.33) 1/α(MZ) = 127.91±0.02, (97.34) sin2θW(MZ) = 0.23120 ±0.00015. (97.35) 97: Grand Uniﬁcation 611 We can use eqs.(97.30–97.32) to solve for 1 /α5(MX) and ln(MX/MZ) in terms of the known parameters α(MZ) andα3(MZ); the result is 1 α5(MX)=1 b1+b2−8 3b3/parenleftbigg−b3 α(MZ)+b1+b2 α3(MZ)/parenrightbigg , (97.36) ln(MX/MZ) =2π b1+b2−8 3b3/parenleftbigg1 α(MZ)−8/3 α3(MZ)/parenrightbigg . (97.37) Plugging in eqs.(97.27–97.29) and eqs.(97.33–97.34), we ﬁ nd 1/α5(MX) = 41.5 andMX= 7×1014GeV; two-loop corrections lower MXto 4×1014GeV. This value of MXis about an order of magnitude below the lower limit imposed by proton decay. We can also use eqs.(97.30–97.32) to express sin2θW(MZ) in terms of α(MZ) andα3(MZ); the result is sin2θW(MZ) =1 b1+b2−8 3b3/parenleftbigg b2−b3+ (b1−5 3b2)α(MZ) α3(MZ)/parenrightbigg . (97.38) This is a prediction of the SU(5) model that we can test. Plugg ing in eqs.(97.27–97.29) and eqs.(97.33–97.34), we ﬁnd sin2θW(MZ) = 0.207. Two-loop corrections raise this to 0 .210±0.001. Comparing with eq.(97.35), we see that the SU(5) prediction is too low by about 10%. The situation improves considerably if we consider the Mini mal Su- persymmetric Standard Model, discussed in section 96. In th is case the beta-function coeﬃcients become b3=−9 + 2n, (97.39) b2=−6 + 2n+ 1, (97.40) b1= +10 3n+ 1. (97.41) Now we ﬁnd sin2θW(MZ) = 0.231, with two-loop corrections raising it to 0.234; there are, however, numerous sources of uncertainty re lated to the masses of the supersymmetric particles. We also ﬁnd MX= 2×1016GeV; this result (which is not changed signiﬁcantly by two-loop c orrections) is high enough to avoid too-rapid proton decay. Next, let us consider the predicted equality of the down quar k and charged lepton masses, eq.(97.19). These relations are sub ject to renormal- ization; see problem 97.5. However, the one-loop correctio ns from gauge- boson exchange cancel in the predicted ratios me mµ=md ms,mµ mτ=ms mb. (97.42) 97: Grand Uniﬁcation 612 Alas, these predictions are not satisﬁed, in the ﬁrst case by an order of magnitude. Resolving this problem requires either a more co mplicated set of Higgs ﬁelds, and/or including higher-dimension, non renormalizable terms in the lagrangian that are suppressed by inverse power s of the some mass scale larger than MX, such as the Planck mass MP= 1.2×1019GeV. To get neutrino masses in the SU(5) model (see section 91), we must add left-handed Weyl ﬁelds ¯ νIthat are singlets of SU(5), and couple them to the neutrinos via LYuk=−˜yIJH† iψi I¯νJ. A more elegant scheme starts with SO(10) as the grand uniﬁed g auge group. SO(10) has a complex sixteen dimensional spinor representation that transforms as 1 ⊕¯5⊕10 under the SU(5) subgroup; thus, each gen- eration of fermions (including ¯ ν) ﬁts into a single Weyl ﬁeld in this 16 representation. A scalar ﬁeld in the 10 representation is ne eded for the Standard Model Higgs ﬁeld, and additional scalars in higher dimensional representations (such as the 45 dimesional adjoint represe ntation and the 16) are needed to break SO(10) down to SU(3) ×SU(2) ×U(1). A great variety of grand uniﬁed models can be constructed, wi th and without supersymmetry. Which, if any, are relevant to the na tural world is a question yet to be answered. Problems 97.1) Is the gauge symmetry of the SU(5) model anomalous? If i t is, modify the model to turn it into a consistent quantum ﬁeld theory. Pr ereq- uisite: section 75. Hint: see problem 70.4. 97.2) Compute the ∆ B=±1 terms in the eﬀective lagrangian that arise fromφexchange. 97.3) Let us write eq.(97.25) as Leﬀ=−ZC1C1O1−ZC2C2O2+ h.c., (97.43) where O1≡εijεαβγ(ℓiqjγ)(¯d† α¯u† β) and O2≡εijεαβγ(¯e†¯u† γ)(qiαqjβ), ZC1andZC2are renormalizing factors, and C1andC2are coeﬃcients that depend on the MS renormalization scale µ. Atµ=MX, we have C1(MX) =C2(MX) = 4πα5(MX)/M2 X. a) Working in Lorenz gauge, and using the results of problems 88.7 and 89.5, show that the one-loop contribution to ZC1from gauge- boson exchange is given by Zmin spinor electrodynamics in Lorenz gauge, with (−1)(+1)e2→/bracketleftig 0 +εα′β′γ(Ta 3)α′α(Ta 3)β′β/εαβγ/bracketrightig g2 3 97: Grand Uniﬁcation 613 +/bracketleftig εi′j′(Ta 2)i′i(Ta 2)j′j/εij+ 0/bracketrightig g2 2 +/bracketleftig (−1 2)(+1 6) + (+1 3)(−2 3)/bracketrightig g2 1. (97.44) Evaluate these coeﬃcients. b) Similarly, compute the one-loop contribution to ZC2from gauge- boson exchange. c) Compute the corresponding anomalous dimensions γ1andγ2ofC1 andC2. d) Compute the numerical values of C1(µ) andC2(µ) atµ= 2GeV. For simplicity, take the top quark mass equal to MZ, and all other quark masses less than 2GeV. Ignore electromagnetic renorm alization belowMZ. 97.4) Consider the proton decay mode with the most stringent experimental bound,p→e+π0. The terms in eq.(97.43) relevant for this mode are Leﬀ=C1εαβγ(euγ)(¯d† α¯u† β) + 2C2εαβγ(¯e†¯u† γ)(dαuβ) + h.c..(97.45) Note that, under the SU(2) L×SU(2) Rglobal symmetry of QCD that we discussed in section 83, the operator u(¯d†¯u†) transforms as the ﬁrst component of a (2 ,1) representation, while the operator ¯ u†(du) is related by parity, and transforms as the ﬁrst component of a (1,2) representation. At low energies, we can replace these opera tors (up to an overall constant factor) with hadron ﬁelds with the same p roperties under Lorentz and SU(2) L×SU(2) R×U(1) Vtransformations. a) Show that PL(uN)1andPR(u†N)1transform appropriately. Here u= exp[iπaTa/fπ], whereπais the triplet of pion ﬁelds, and Niis the Dirac ﬁeld for the proton-neutron doublet. b) Show that the low-energy version of eq.(97.45) is then Leﬀ=C1AECPL(uN)1+ 2C2AECPR(u†N)1+ h.c., (97.46) where ECis the charge conjugate of the Dirac ﬁeld for the electron (in other words, ECis the Dirac ﬁeld for the positron), and Ais a constant with dimensions of mass-cubed. Lattice calculati ons have yielded a value of A= 0.0090GeV3forµ= 2GeV. c) Write out the terms in eq.(97.46) that contain the proton ﬁ eld and either zero or one π0ﬁelds. d) Compute the amplitude for p→e+π0. Note that there are two contributing Feynman diagrams: one where eq.(97.46) suppl ies the 97: Grand Uniﬁcation 614 proton-positron-pion vertex, and one where the proton emit s a pion via the interaction in eq.(83.30), and then converts to a pos itron via the no-pion terms in eq.(97.46). Neglect the positron mass. Hint: your result should be proportional to 1 + gA. e) Compute the spin-averaged decay rate for p→e+π0. Use the values ofC1andC2forµ= 2GeV that you computed in problem 97.3. How does your answer compare with the naive estimate we made earlier? Hint: your result should be proportional to C2 1+ 4C2 2. 97.5) Consider the Yukawa couplings for the the down quark an d charged lepton of one generation, LYuk=−Zyyεijϕiℓj¯e−Zy′y′εijϕiqαj¯dα, (97.47) where we have included renormalizing factors. a) Consider one-loop contributions from gauge-boson excha nge to Zy′/Zy. 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