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Problem_Book_Quantum_Field_Theory

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A textbook of problems and solutions (Springer, 2006) by Voja Radovanovic of the University of Belgrade, a downloaded book rather than Phil's own work. Part I states the problems and Part II solves them. Chapters cover Lorentz and Poincare symmetries, the Klein-Gordon and Dirac equations, gamma matrices, classical fields and Noether's theorem, Green functions, canonical quantization of scalar, Dirac and electromagnetic fields, tree-level processes, and renormalization and regularization.

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Problem Book Quantum Field Theory V oja Radovanovic Problem Book Quantum Field Theory ABC V oja Radovanovic Faculty of Physics University of BelgradeStudentski trg 12-16 11000 Belgrade Yugoslavia Library of Congress Control Number: 2005934040 ISBN-10 3-540-29062-1 Springer Berlin Heidelberg New York ISBN-13 978-3-540-29062-9 Springer Berlin Heidelberg New York This work is subject to copyright. All rights are reserved, whether the whole or part of the material is concerned, speciﬁcally the rights of translation, reprinting, reuse of illustrations, recitation, broadcasting,reproduction on microﬁlm or in any other way, and storage in data banks. Duplication of this publication or parts thereof is permitted only under the provisions of the German Copyright Law of September 9, 1965, in its current version, and permission for use must always be obtained from Springer. Violations areliable for prosecution under the German Copyright Law. Springer is a part of Springer Science+Business Media springeronline.com c/circlecopyrtSpringer-Verlag Berlin Heidelberg 2006 Printed in The Netherlands The use of general descriptive names, registered names, trademarks, etc. in this publication does not imply, even in the absence of a speciﬁc statement, that such names are exempt from the relevant protective laws and regulations and therefore free for general use. Typesetting: by the author and TechBooks using a Springer L ATEX macro package Cover design: design & production GmbH, Heidelberg Printed on acid-free paper SPIN: 11544920 56/TechBooks 543210 To my daughter Natalija Preface This Problem Book is based on the exercises and lectures which I have given to undergraduate and graduate students of the Faculty of Physics, University of Belgrade over many years. Nowadays, there are a lot of excellent QuantumField Theory textbooks. Unfortunately, there is a shortage of Problem Books in this ﬁeld, one of the exceptions being the Problem Book of Cheng and Li [7]. The overlap between this Problem Book and [7] is very small, since the lattermostly deals with gauge ﬁeld theory and particle physics. Textbooks usually contain problems without solutions. As in other areas of physics doing more problems in full details improves both understanding and eﬃciency. So, I feelthat the absence of such a book in Quantum Field Theory is a gap in the literature. This was my main motivation for writing this Problem Book. To students: You cannot start to do problems without previous study- ing your lecture notes and textbooks. Try to solve problems without using solutions; they should help you to check your results. The level of this Prob- lem Book corresponds to the textbooks of Mandl and Show [15]; Greiner and Reinhardt [11] and Peskin and Schroeder [16]. Each Chapter begins with a short introduction aimed to deﬁne notation. The ﬁrst Chapter is devoted tothe Lorentz and Poincar´ e symmetries. Chapters 2, 3 and 4 deal with the rela- tivistic quantum mechanics with a special emphasis on the Dirac equation. In Chapter 5 we present problems related to the Euler-Lagrange equations andthe Noether theorem. The following Chapters concern the canonical quanti- zation of scalar, Dirac and electromagnetic ﬁelds. In Chapter 10 we consider tree level processes, while the last Chapter deals with renormalization andregularization. There are many colleagues whom I would like to thank for their support and help. Professors Milutin Blagojevi´ c and Maja Buri´ cg a v em a n yu s e f u l ideas concerning problems and solutions. I am grateful to the Assistants at the Faculty of Physics, University of Belgrade: Marija Dimitrijevi´ c, Duˇsko Latas and Antun Balaˇ z who checked many of the solutions. Duˇ sko Latas also drew all the ﬁgures in the Problem Book. I would like to mention the contribution of the students: Branislav Cvetkovi´ c, Bojan Nikoli´ c, Mihailo Vanevi´ c, Marko VIII Preface Vojinovi´ c, Aleksandra Stojakovi´ c, Boris Grbi´ c, Igor Salom, Irena Kneˇ zevi´c, Zoran Ristivojevi´ c and Vladimir Juriˇ ci´c. Branislav Cvetkovi´ c, Maja Buri´ c, Milutin Blagojevi´ c and Dejan Stojkovi´ c have corrected my English translation of the Problem Book. I thank them all, but it goes without saying that allthe errors that have crept in are my own. I would be grateful for any readers’ comments. Belgrade, August 2005 Voja Radovanovi´ c Contents Part I Problems 1 Lorentz and Poincar´ e symmetries .......................... 3 2 The Klein–Gordon equation ............................... 9 3T h e γ–matrices ............................................ 1 3 4 The Dirac equation ........................................ 1 7 5 Classical ﬁeld theory and symmetries ..................... 2 5 6 Green functions ........................................... 3 1 7 Canonical quantization of the scalar ﬁeld .................. 3 5 8 Canonical quantization of the Dirac ﬁeld .................. 4 3 9 Canonical quantization of the electromagnetic ﬁeld ........ 4 9 10 Processes in the lowest order of perturbation theory ....... 5 5 11 Renormalization and regularization ........................ 6 1 Part II Solutions 1 Lorentz and Poincar´ e symmetries .......................... 6 7 2 The Klein–Gordon equation ............................... 7 7 3T h e γ–matrices ............................................ 8 5 X Contents 4 The Dirac equation ........................................ 9 3 5 Classical ﬁelds and symmetries ............................1 2 1 6 Green functions ...........................................1 3 1 7 Canonical quantization of the scalar ﬁeld .................1 4 1 8 Canonical quantization of the Dirac ﬁeld ..................1 6 1 9 Canonical quantization of the electromagnetic ﬁeld ........1 7 9 10 Processes in the lowest order of the perturbation theory . . . 191 11 Renormalization and regularization ........................2 1 1 References .....................................................2 3 9 Index ..........................................................2 4 1 Part I Problems 1 Lorentz and Poincar´ e symmetries •Minkowski space, M4is a real 4-dimensional vector space with metric tensor deﬁned by gµν= 10 0 0 0−10 0 00 −10 00 0 −1 . (1.A) Vectors can be written in the form x=xµeµ,where xµarethe contravariant components of the vector xin the basis e0= 1 0 00 ,e 1= 0 1 00 ,e 2= 0 0 10 ,e 3= 0 0 01 . The square of the length of a vector in M 4isx2=gµνxµxν. The square of the line element between two neighboring points xµandxµ+dxµtakes the form ds2=gµνdxµdxν=c2dt2−dx2. (1.B) The space M4is also a manifold; xµare global (inertial) coordinates. The covariant components of a vector are deﬁned by xµ=gµνxν. •Lorentz transformations, x/primeµ=Λµ νxν, (1.C) leave the square of the length of a vector invariant, i.e. x/prime2=x2. The matrix Λ is a constant matrix1;xµandx/primeµare the coordinates of the same event in two diﬀerent inertial frames. In Problem 1.1 we shall show that from the previous deﬁnition it follows that the matrix Λmust satisfy the condition ΛTgΛ=g. The transformation law of the covariant components is given by x/prime µ=(Λ−1)ν µxν=Λν µxν. (1.D) 1The ﬁrst index in Λµ νis the row index, the second index the column index. 4P r o b l e m s •Letu=uµeµbe an arbitrary vector in tangent space2, where uµare its contravariant components. A dual space can be associated to the vector space in the following way. The dual basis, θµis determined by θµ(eν)=δµ ν.T h e vectors in the dual space, ω=ωµθµare called dual vectors or one–forms. The components of the dual vector transform like (1.D). The scalar (inner) product of vectors uandvis given by u·v=gµνuµvν=uµvµ. At e n s o ro fr a n k (m,n) in Minkowski spacetime is T=Tµ1...µmν1...νn(x)eµ1⊗...⊗eµm⊗θν1⊗...⊗θνn. The components of this tensor transform in the following way T/primeµ1...µmn1...νn(x/prime)=Λµ1ρ1...Λµmρm(Λ−1)σ1 ν1...(Λ−1)σn νnTρ1...ρmσ1...σn(x), under Lorentz transformations. A contravariant vector is tensor of rank (1 ,0), while the rank of a covariant vector (one-form) is (0 ,1). The metric tensor is a symmetric tensor of rank (0 ,2). •Poincar´ e transformations,3(Λ,a) consist of Lorentz transformations and translations, i.e. (Λ,a)x=Λx+a. (1.E) These are the most general transformations of Minkowski space which do not change the interval between any two vectors, i.e. (y/prime−x/prime)2=(y−x)2. •In a certain representation the elements of the Poincar´ e group near the identity are U(ω,/epsilon1)=e−i 2Mµνωµν+iPµ/epsilon1µ, (1.F) where ωµνandMµνare parameters and generators of the Lorentz subgroup respectively, while /epsilon1µandPµare the parameters and generators of the trans- lation subgroup. The Poincar´ e algebra is given in Problem 1.11. •The Levi-Civita tensor, /epsilon1µνρσis a totaly antisymmetric tensor. We will use the convention that /epsilon10123=+ 1 . 2The tangent space is a vector space of tangent vectors associated to each point of spacetime. 3Poincar´ e transformations are very often called inhomogeneous Lorentz transfor- mations. Chapter 1. Lorentz and Poincare symmetries 5 1.1.Show that Lorentz transformations satisfy the condition ΛTgΛ=g. Also, prove that they form a group. 1.2.Given an inﬁnitesimal Lorentz transformation Λµ ν=δµ ν+ωµ ν, show that the inﬁnitesimal parameters ωµνare antisymmetric. 1.3.Prove the following relation /epsilon1αβγδAα µAβ νAγ λAδ σ=/epsilon1µνλσdetA, where Aαµare matrix elements of the matrix A. 1.4.Show that the Kronecker δsymbol and Levi-Civita /epsilon1symbol are form invariant under Lorentz transformations. 1.5.Prove that /epsilon1µνρσ/epsilon1αβγδ=−/vextendsingle/vextendsingle/vextendsingle/vextendsingle/vextendsingle/vextendsingle/vextendsingleδ µαδµβδµγδµδ δναδνβδνγδνδ δραδρβδργδρδ δσαδσβδσγδσδ/vextendsingle/vextendsingle/vextendsingle/vextendsingle/vextendsingle/vextendsingle/vextendsingle, and calculate the following contractions /epsilon1 µνρσ/epsilon1µβγδ,/epsilon1µνρσ/epsilon1µνγδ,/epsilon1µνρσ/epsilon1µνρδ, /epsilon1µνρσ/epsilon1µνρσ. 1.6.Let us introduce the notations σµ=(I,σ); ¯σµ=(I,−σ), where Iis a unit matrix, while σare Pauli matrices4and deﬁne the matrix X=xµσµ. (a) Show that the transformation X→X/prime=SXS†, where S∈SL(2,C)5, describes the Lorentz transformation xµ→Λµ νxν. This is a homomorphism between proper orthochronous Lorentz transfor- mations6and the SL(2 ,C) group. (b) Show that xµ=1 2tr(¯σµX). 1.7.Prove that Λµν=1 2tr(¯σµSσνS†),andΛ(S)=Λ(−S).The last relation shows that the map is not unique. 4The Pauli matrices are σ1=/parenleftbigg 01 10/parenrightbigg ,σ2=/parenleftbigg 0−i i0/parenrightbigg and σ3=/parenleftbigg 10 0−1/parenrightbigg . 5SL(2,C) matrices are 2 ×2 complex matrices of unit determinant. 6The proper orthochronous Lorentz transformations satisfy the conditions: Λ0 0≥ 1,detΛ=1 . 6P r o b l e m s 1.8.Find the matrix elements of generators of the Lorentz group Mµνin its natural (deﬁning) representation (1.C). 1.9.Prove that the commutation relations of the Lorentz algebra [Mµν,Mρσ]=i (gµσMνρ+gνρMµσ−gµρMνσ−gνσMµρ) lead to [Mi,Mj]=i/epsilon1ijlMl,[Ni,Nj]=−i/epsilon1ijlNl,[Mi,Nj]=i/epsilon1ijlNl, where Mi=1 2/epsilon1ijkMjkandNk=Mk0.Further, one can introduce the following linear combinations Ai=1 2(Mi+iNi)a n d Bi=1 2(Mi−iNi). Prove that [Ai,Aj]=i/epsilon1ijlAl,[Bi,Bj]=i/epsilon1ijlBl,[Ai,Bj]=0. This is a well known result which gives a connection between the Lorentz algebra and ”two” SU(2) algebras. Irreducible representations of the Lorentz group are classiﬁed by two quantum numbers ( j1,j2) which come from above two SU(2) groups. 1.10. The Poincar´ e transformation ( Λ,a) is deﬁned by: x/primeµ=Λµ νxν+aµ. Determine the multiplication rule i.e. the product ( Λ1,a1)(Λ2,a2), as well as the unit and inverse element in the group. 1.11. (a) Verify the multiplication rule U−1(Λ,0)U(1,/epsilon1)U(Λ,0) =U(1,Λ−1/epsilon1), in the Poincar´ e group. In addition, show that from the previous relation follows: U−1(Λ,0)PµU(Λ,0) = ( Λ−1)ν µPν. Calculate the commutator [ Mµν,Pρ]. (b) Show that U−1(Λ,0)U(Λ/prime,0)U(Λ,0) =U(Λ−1Λ/primeΛ,0), and ﬁnd the commutator [ Mµν,Mρσ]. (c) Finally show that the generators of translations commute between them- selves, i.e. [ Pµ,Pν]=0 . 1.12. Consider the representation in which the vectors xof Minkowski space are (x,1)T, while the element of the Poincar´ e group, ( Λ,a)a r e5 ×5 matrices given by/parenleftbigg Λa 01/parenrightbigg . Check that the generators in this representation satisfy the commutation re- lations from the previous problem. Chapter 1. Lorentz and Poincare symmetries 7 1.13. Find the generators of the Poincar´ e group in the representation of a clas- sical scalar ﬁeld7. Prove that they satisfy the commutation relations obtained in Problem 1.11. 1.14.The Pauli–Lubanski vector is deﬁned by Wµ=1 2/epsilon1µνλσMνλPσ. (a) Show that WµPµ=0a n d[ Wµ,Pν]=0. (b) Show that W2=−1 2MµνMµνP2+MµσMνσPµPν. (c) Prove that the operators W2andP2commute with the generators of the Poincar´ e group. These operators are Casimir operators . They are used to classify the irreducible representations of the Poincar´ e group. 1.15. Show that W2|p=0,m,s,σ /angbracketright=−m2s(s+1 )|p=0,m,s,σ /angbracketright, where |p=0,m,s,σ /angbracketrightis a state vector for a particle of mass m, momentum p, spin swhile σis the z–component of the spin. The mass and spin classify the irreducible representations of the Poincar´ e group. 1.16. Verify the following relations (a) [Mµν,Wσ]=i (gνσWµ−gµσWν), (b) [Wµ,Wν]=−i/epsilon1µνσρWσPρ. 1.17. Calculate the commutators (a) [Wµ,M2], (b) [Mµν,WµWν], (c) [M2,Pµ], (d) [/epsilon1µνρσMµνMρσ,Mαβ]. 1.18. The standard momentum for a massive particle is ( m,0,0,0), while for a massless particle it is ( k,0,0,k). Show that the little group in the ﬁrst case is SU(2), while in the second case it is E(2) group8. 1.19. Show that conformal transformations consisting of dilations: xµ→x/primeµ=e−ρxµ, special conformal transformations (SCT): xµ→x/primeµ=xµ+cµx2 1+2c·x+c2x2, and usual Poincar´ e transformations form a group. Find the commutation re- lations in this group. 7Scalar ﬁeld transforms as φ/prime(Λx+a)=φ(x) 8E(2) is the group of rotations and translations in a plane. 2 The Klein–Gordon equation •The Klein–Gordon equation, (/unionsq /intersectionsq+m2)φ(x)=0, (2.A) is an equation for a free relativistic particle with zero spin. The transformation law of a scalar ﬁeld φ(x) under Lorentz transformations is given by φ/prime(Λx)= φ(x). •The equation for the spinless particle in an electromagnetic ﬁeld, Aµis ob- tained by changing ∂µ→∂µ+iqAµin equation (2.A), where qis the charge of the particle. 2.1.Solve the Klein–Gordon equation. 2.2.Ifφis a solution of the Klein–Gordon equation calculate the quantity Q=iq/integraldisplay d3x/parenleftbigg φ∗∂φ ∂t−φ∂φ∗ ∂t/parenrightbigg . 2.3.The Hamiltonian for a free real scalar ﬁeld is H=1 2/integraldisplay d3x[(∂0φ)2+(∇φ)2+m2φ2]. Calculate the Hamiltonian Hfor a general solution of the Klein–Gordon equa- tion. 2.4.The momentum for a real scalar ﬁeld is given by P=−/integraldisplay d3x∂0φ∇φ. Calculate the momentum Pfor a general solution of the Klein–Gordon equa- tion. 10 Problems 2.5.Show that the current1 jµ=−i 2(φ∂µφ∗−φ∗∂µφ) satisﬁes the continuity equation, ∂µjµ=0. 2.6.Show that the continuity equation ∂µjµ= 0 is satisﬁed for the current jµ=−i 2(φ∂µφ∗−φ∗∂µφ)−qAµφ∗φ, where φis a solution of Klein–Gordon equation in external electromagnetic potential Aµ. 2.7.A scalar particle in the s–state is moving in the potential qA0=/braceleftbigg −V, r < a 0,r > a, where Vis a positive constant. Find the dispersion relation, i.e. the relation between energy and momentum, for discrete particle states. Which condition has to be satisﬁed so that there is only one bound state in the case V<2m? 2.8.Find the energy spectrum and the eigenfunctions for a scalar particle in a constant magnetic ﬁeld, B=Bez. 2.9.Calculate the reﬂection and the transmission coeﬃcients of a Klein– Gordon particle with energy E, at the potential A0=/braceleftbigg0,z < 0 U0,z > 0, where U0is a positive constant. 2.10. A particle of charge qand mass mis incident on a potential barrier A0=/braceleftbigg0,z < 0,z>a U0,0<z<a, where U0is a positive constant. Find the transmission coeﬃcient. 2.11. A scalar particle of mass mand charge −emoves in the Coulomb ﬁeld of a nucleus. Find the energy spectrum of the bounded states for this systemif the charge of the nucleus is Ze. 2.12. Using the two-component wave function/parenleftbigg θ χ/parenrightbigg , where θ= 1 2(φ+i m∂φ ∂t) andχ=1 2(φ−i m∂φ ∂t), instead of φrewrite the Klein–Gordon equation in the Schr¨odinger form. 1Actually this is current density. Chapter 2. The Klein–Gordon equation 11 2.13. Find the eigenvalues of the Hamiltonian from the previous problem. Find the nonrelativistic limit of this Hamiltonian. 2.14. Determine the velocity operator v=i [H,x],where His the Hamiltonian obtained in Problem 2.12. Solve the eigenvalue problem for v. 2.15. In the space of two–component wave functions the scalar product is deﬁned by /angbracketleftψ1|ψ2/angbracketright=1 2/integraldisplay d3xψ† 1σ3ψ2. (a) Show that the Hamiltonian Hobtained in Problem 2.12 is Hermitian. (b) Find expectation values of the Hamiltonian /angbracketleftH/angbracketright, and the velocity /angbracketleftv/angbracketrightin the state/parenleftbigg 1 0/parenrightbigg e−ip·x. 3 Theγ–matrices •In Minkowski space M 4,t h eγ–matrices satisfy the anticommutation relations1 {γµ,γν}=2gµν. (3.A) •Inthe Dirac representation γ–matrices take the form γ0=/parenleftbigg I0 0−I/parenrightbigg ,γ=/parenleftbigg 0σ −σ0/parenrightbigg . (3.B) Other representations of the γ–matrices can be obtained by similarity trans- formation γ/prime µ=SγµS−1.The transformation matrix Sneed to be uni- tary if the transformed matrices are to satisfy the Hermicity condition: (γ/primeµ)†=γ/prime0γ/primeµγ/prime0.The Weyl representation of the γ–matrices is given by γ0=/parenleftbigg 0I I0/parenrightbigg ,γ=/parenleftbigg 0σ −σ0/parenrightbigg , (3.C) while in the Majorana representation we have γ0=/parenleftbigg 0σ2 σ20/parenrightbigg ,γ1=/parenleftbigg iσ30 0iσ3/parenrightbigg , γ2=/parenleftbigg 0−σ2 σ20/parenrightbigg ,γ3=/parenleftbigg −iσ10 0−iσ1/parenrightbigg .(3.D) •The matrix γ5is deﬁned by γ5=iγ0γ1γ2γ3,while γ5=−iγ0γ1γ2γ3.In the Dirac representation, γ5has the form γ5=/parenleftbigg 0I I0/parenrightbigg . 1The same type of relations hold in M d,w h e r e dis the dimension of spacetime. 14 Problems •σµνmatrices are deﬁned by σµν=i 2[γµ,γν]. (3.E) •Slash is deﬁned as /a=aµγµ. (3.F) •Sometimes we use the notation: β=γ0,α=γ0γ. The anticommutation relations (3.A) become {αi,αj}=2δij,{αi,β}=0. 3.1.Prove: (a)γ† µ=γ0γµγ0, (b)σ† µν=γ0σµνγ0. 3.2.Show that: (a)γ† 5=γ5=γ5=γ−1 5, (b)γ5=−i 4!/epsilon1µνρσγµγνγργσ, (c) (γ5)2=1, (d) (γ5γµ)†=γ0γ5γµγ0. 3.3.Show that: (a){γ5,γµ}=0, (b) [γ5,σµν]=0. 3.4.Prove / a2=a2. 3.5.Derive the following identities with contractions of the γ–matrices: (a)γµγµ=4, (b)γµγνγµ=−2γν, (c)γµγαγβγµ=4gαβ, (d)γµγαγβγγγµ=−2γγγβγα, (e)σµνσµν=1 2, (f)γµγ5γµγ5=−4, (g)σαβγµσαβ=0, (h)σαβσµνσαβ=−4σµν, (i)σαβγ5γµσαβ=0, (j)σαβγ5σαβ=1 2γ5. Chapter 3. The γ–matrices 15 3.6.Prove the following identities with traces of γ–matrices: (a) trγµ=0, (b) tr( γµγν)=4gµν, (c) tr( γµγνγργσ)=4 ( gµνgρσ−gµρgνσ+gµσgνρ), (d) trγ5=0, (e) tr( γ5γµγν)=0 , (f) tr( γ5γµγνγργσ)=−4i/epsilon1µνρσ, (g) tr(/ a1···/a2n+1)=0 , (h) tr(/ a1···/a2n) = tr(/ a2n···/a1), (i) tr( γ5γµ)=0 , 3.7.Calculate tr(/ a1/a2···/a6). 3.8.Calculate tr[(/ p−m)γµ(1−γ5)(/q+m)γν]. 3.9.Calculate γµ(1−γ5)(/p−m)γµ. 3.10. Verify the identity exp(γ5/a) = cos/radicalbig aµaµ+1√aµaµγ5/asin/radicalbig aµaµ, where a2>0. 3.11. Show that the set Γa={I, γµ,γ5,γµγ5,σµν}, is made of linearly independent 4 ×4 matrices. Also, show that the product of any two of them is again one of the matrices Γa,u pt o ±1,±i. 3.12. Show that any matrix A∈C44can be written in terms of Γa= {I, γµ,γ5,γµγ5,σµν},i.e.A=/summationtext acaΓawhere ca=1 4tr(AΓa). 3.13. Expand the following products of γ–matrices in terms of Γa: (a)γµγνγρ, (b)γ5γµγν, (c)σµνγργ5. 3.14. Expand the anticommutator {γµ,σνρ}in terms of Γ–matrices. 3.15. Calculate tr( γµγνγργσγaγβγ5). 3.16. Verify the relation γ5σµν=i 2/epsilon1µνρσσρσ. 3.17. Show that the commutator [ σµν,σρσ] can be rewritten in terms of σµν. Find the coeﬃcients in this expansion. 16 Problems 3.18. Show that if a matrix commutes with all gamma matrices γµ, then it is proportional to the unit matrix. 3.19. LetU=e x p ( βα·n),where βandαare Dirac matrices; nis a unit vector. Verify the following relation: α/prime≡UαU†=α−(I−U2)(α·n)n. 3.20. Show that the set of matrices (3.C) is a representation of γ–matrices. Find the unitary matrix which transforms this representation into the Dirac one. Calculate σµν,a n d γ5in this representation. 3.21. Find Dirac matrices in two dimensional spacetime. Deﬁne γ5and cal- culate tr(γ5γµγν). Simplify the product γ5γµ. 4 The Dirac equation •The Dirac equation , (iγµ∂µ−m)ψ(x)=0 , (4.A) is an equation of the free relativistic particle with spin 1 /2. The general solu- tion of this equation is given by ψ(x)=1 (2π)3 22/summationdisplay r=1/integraldisplay d3p/radicalbiggm Ep/parenleftbig ur(p)cr(p)e−ip·x+vr(p)d† r(p)eip·x/parenrightbig ,(4.B) where ur(p)a n d vr(p) are the basic bispinors which satisfy equations (/p−m)ur(p)=0 , (/p+m)vr(p)=0 .(4.C) We use the normalization ¯ur(p)us(p)=−¯vr(p)vs(p)=δrs, ¯ur(p)vs(p)=¯vr(p)us(p)=0.(4.D) The coeﬃcients cr(p)a n d dr(p) in (4.B) being given determined by boundary conditions. Equation (4.A) can be rewritten in the form i∂ψ ∂t=HDψ, where HD=α·p+βmis the so-called Dirac Hamiltonian. •Under the Lorentz transformation, x/primeµ=Λµνxν, Dirac spinor, ψ(x) trans- forms as ψ/prime(x/prime)=S(Λ)ψ(x)=e−i 4σµνωµνψ(x). (4.E) S(Λ) is the Lorentz transformation matrix in spinor representation, and it satisﬁes the equations: S−1(Λ)=γ0S†(Λ)γ0, 18 Problems S−1(Λ)γµS(Λ)=Λµ νγν. •The equation for an electron with charge −ein an electromagnetic ﬁeld Aµis given by [iγµ(∂µ−ieAµ)−m]ψ(x)=0 . (4.F) •Under parity, Dirac spinors transform as ψ(t,x)→ψ/prime(t,−x)=γ0ψ(t,x). (4.G) •Time reversal is an antiunitary operation: ψ(t,x)→ψ/prime(−t,x)=Tψ∗(t,x). (4.H) The matrix T, satisﬁes TγµT−1=γµ∗=γT µ. (4.I) The solution of the above condition is T=iγ1γ3, in the Dirac representation ofγ–matrices. It is easy to see that T†=T−1=T=−T∗. •Under charge conjugation, spinors ψ(x) transform as follows ψ(x)→ψc(x)=C¯ψT. (4.J) The matrix Csatisﬁes the relations: CγµC−1=−γT µ,C−1=CT=C†=−C. (4.K) In the Dirac representation, the matrix Cis given by C=iγ2γ0. 4.1.Find which of the operators given below commute with the Dirac Hamil- tonian: (a)p=−i∇, (b)L=r×p, (c)L2, (d)S=1 2Σ,where Σ=i 2γ×γ, (e)J=L+S, (f)J2, (g)Σ·p |p|, (h)Σ·n,where nis a unit vector. 4.2.Solve the Dirac equation for a free particle, i.e. derived (4.B). 4.3.Find the energy of the states us(p)e−ip·xandvs(p)eip·xfor the Dirac particle. Chapter 4. The Dirac equation 19 4.4.Using the solution of Problem 4.2 show that 2/summationdisplay r=1ur(p)¯ur(p)=/p+m 2m≡Λ+(p), −2/summationdisplay r=1vr(p)¯vr(p)=−/p−m 2m≡Λ−(p). The quantities Λ+(p)a n d Λ−(p) are energy projection operators. 4.5.Show that Λ2 ±=Λ±,andΛ+Λ−=0.How do these projectors act on the basic spinors ur(p)a n d vr(p)? Derive these results with and without using explicit expressions for spinors. 4.6.The spin operator in the rest frame for a Dirac particle is deﬁned by S=1 2Σ.Prove that: (a)Σ=γ5γ0γ, (b) [Si,Sj]=i/epsilon1ijkSk, (c)S2=−3 4. 4.7.Prove that:Σ·p |p|ur(p)=(−1)r+1ur(p), Σ·p |p|vr(p)=(−1)rvr(p). Are spinors ur(p)a n d vr(p) eigenstates of the operator Σ·n,where nis a unit vector? Check the same property for the spinors in the rest frame. 4.8.Find the boost operator for the transition from the rest frame to the frame moving with velocity valong the z–axis, in the spinor representation. Is this operator unitary? 4.9.Solve the previous problem upon transformation to the system rotated around the z–axis for an angle θ. Is this operator a unitary one? 4.10. The Pauli–Lubanski vector is deﬁned by Wµ=1 2/epsilon1µνρσMνρPσ,where Mνρ=1 2σνρ+i (xν∂ρ−xρ∂ν) is angular momentum, while Pµis linear mo- mentum. Show that W2ψ(x)=−1 2(1 +1 2)m2ψ(x), where ψ(x) is a solution of the Dirac equation. 20 Problems 4.11. The covariant operator which projects the spin operator onto an arbi- trary normalized four-vector sµ(s2=−1) is given by Wµsµ, where s·p=0, i.e. the vector polarization sµis orthogonal to the momentum vector. Show thatWµsµ m=1 2mγ5/s/p. Find this operator in the rest frame. 4.12. In addition to the spinor basis, one often uses the helicity basis. The helicity basis is obtained by taking n=p/|p|in the rest frame. Find the equations for the spin in this case. 4.13. Find the form of the equations for the spin, deﬁned in Problem 4.12 in the ultrarelativistic limit. 4.14. Show that the operator γ5/scommutes with the operator / p, and that the eigenvalues of this operator are ±1.Find the eigen-projectors of the operator γ5/s. Prove that these projectors commute with projectors onto positive and negative energy states, Λ±(p). 4.15. Consider a Dirac’s particle moving along the z–axis with momentum p. The nonrelativistic spin wave function is given by ϕ=1/radicalbig |a|2+|b|2/parenleftbigg a b/parenrightbigg . Calculate the expectation value of the spin projection onto a unit vector n, i.e./angbracketleftΣ·n/angbracketright.Find the nonrelativistic limit. 4.16. Find the Dirac spinor for an electron moving along the z−axis with momentum p. The electron is polarized along the direction n=(θ,φ=π 2). Calculate the expectation value of the projection spin on the polarization vector in that state. 4.17. Is the operator γ5a constant of motion for the free Dirac particle? Find the eigenvalues and projectors for this operator. 4.18. Let us introduce ψL=1 2(1−γ5)ψ, ψR=1 2(1 +γ5)ψ, where ψis a Dirac spinor. Derive the equations of motion for these ﬁelds. Show that they are decoupled in the case of a massless spinor. The ﬁelds ψL ψRare known as Weyl ﬁelds. Chapter 4. The Dirac equation 21 4.19. Let us consider the system of the following two–component equations: iσµ∂ψR(x) ∂xµ=mψL(x), i¯σµ∂ψL(x) ∂xµ=mψR(x), where σµ=(I,σ); ¯σµ=(I,−σ). (a) Is it possible to rewrite this system of equations as a Dirac equation? If this is possible, ﬁnd a unitary matrix which relates the new set of γ–matrices with the Dirac ones. (b) Prove that the system of equations given above is relativistically covariant. Find 2 ×2 matrices SRandSL, which satisfy ψ/prime R,L(x/prime)=SR,LψR,L(x), where ψ/prime R,Lis a wave function obtained from ψR,L(x) by a boost along the x–axis. 4.20. Prove that the operator K=β(Σ·L+1),whereΣ=−i 2α×αis the spin operator and Lis orbital momentum, commutes with the Dirac Hamiltonian. 4.21. Prove the Gordon identities: 2m¯u(p1)γµu(p2)=¯u(p1)[(p1+p2)µ+iσµν(p1−p2)ν]u(p2), 2m¯v(p1)γµv(p2)=−¯v(p1)[(p1+p2)µ+iσµν(p1−p2)ν]v(p2). Do not use any particular representation of Dirac spinors. 4.22. Prove the following identity: ¯u(p/prime)σµν(p+p/prime)νu(p)=i ¯u(p/prime)(p/prime−p)µu(p). 4.23. The current Jµis given by Jµ=¯u(p2)/p1γµ/p2u(p1), where u(p)a n d ¯u(p) are Dirac spinors. Show that Jµcan be written in the following form: Jµ=¯u(p2)[F1(m,q2)γµ+F2(m,q2)σµνqν]u(p1), where q=p2−p1.Determine the functions F1andF2. 4.24. Rewrite the expression ¯u(p)1 2(1−γ5)u(p) as a function of the normalization factor N=u†(p)u(p). 4.25. Consider the current Jµ=¯u(p2)pρqλσµργλu(p1), where u(p1)a n d u(p2) are Dirac spinors; p=p1+p2andq=p2−p1.Show thatJµhas the following form: Jµ=¯u(p2)(F1γµ+F2qµ+F3σµρqρ)u(p1), and determine the functions Fi=Fi(q2,m), (i=1,2,3). 22 Problems 4.26. Prove that if ψ(x) is a solution of the Dirac equation, that it is also a solution of the Klein-Gordon equation. 4.27. Determine the probability density ρ=¯ψγ0ψand the current density j=¯ψγψ, for an electron with momentum pand in an arbitrary spin state. 4.28. Find the time dependence of the position operator rH(t)=eiHtre−iHt for a free Dirac particle. 4.29. The state of the free electron at time t= 0 is given by ψ(t=0,x)=δ(3)(x) 1 00 0 . Findψ(t>0,x). 4.30. Determine the time evolution of the wave packet ψ(t=0,x)=1 (πd2)3 4exp/parenleftbigg −x2 2d2/parenrightbigg 1 00 0 , for the Dirac equation. 4.31. An electron with momentum p=pe zand positive helicity meets a potential barrier −eA0=/braceleftbigg0,z < 0 V, z > 0. Calculate the coeﬃcients of reﬂection and transmission. 4.32. Find the coeﬃcients of reﬂection and transmission for an electron mov- ing in a potential barrier: −eA0=/braceleftbigg0,z < 0,z>a V,0<z<a. The energy of the electron is E, while its helicity is 1 /2. Also, ﬁnd the energy of particle for which the transmission coeﬃcient is equal to one. 4.33. Let an electron move in a potential hole 2 awide and Vdeep. Consider only bound states of the electron. (a) Find the dispersion relations. (b) Determine the relation between Vandaif there are Nbound states. Take V<2m. If there is only one bound state present in the spectrum, is it odd or even? Chapter 4. The Dirac equation 23 (c) Give a rough description of the dispersion relations for V>2m. 4.34. Determine the energy spectrum of an electron in a constant magnetic ﬁeldB=Bez. 4.35. Show that if ψ(x) is a solution of the Dirac equation in an electromag- netic ﬁeld, then it satisﬁes the ”generalize” Klein-Gordon equation: [(∂µ−ieAµ)(∂µ−ieAµ)−e 2σµνFµν+m2]ψ(x)=0 , where Fµν=∂µAν−∂νAµis the ﬁeld strength tensor. 4.36. Find the nonrelativistic approximation of the Dirac Hamiltonian H= α·(p+eA)−eA0+mβ, including terms of orderv2 c2. 4.37. IfVµ(x)=¯ψ(x)γµψ(x) is a vector ﬁeld, show that Vµis a real quantity. Find the transformation properties of this quantity under proper orthochro-nous Lorentz transformations, charge conjugation C, parity Pand time re- versal T. 4.38. Investigate the transformation properties of the quantity A µ(x)= ¯ψ(x)γµγ5ψ(x),under proper orthochronous Lorentz transformations and the discrete transformations C,PandT. 4.39. Prove that the quantity ¯ψ(x)γµ∂µψ(x) is a Lorentz scalar. Find its transformation rules under the discrete transformations. 4.40. Using the Dirac equation, show that C¯uT(p,s)=v(p,s),where Cis charge conjugation. Also, prove the above relation in a concrete representa- tion. 4.41. The matrix Cis deﬁned by CγµC−1=−γT µ. Prove that if matrices C/primeandC/prime/primesatisfy the above relation, then C/prime=kC/prime/prime, where kis a constant. 4.42. If ψ(x)=Np /parenleftbigg 1 0/parenrightbigg σ3p Ep+m/parenleftbigg 1 0/parenrightbigg e−iEt+ipz, is the wave function in frame Sof the relativistic particle whose spin is 1 /2, ﬁnd: (a) the wave function ψc(x)=C¯ψT(x) of the antiparticle, (b) the wave function of this particle for an observer moving with momentum p=pez, 24 Problems (c) the wave functions which are obtained after space and time inversion, (d) the wave function in a frame which is obtained from Sby a rotation about thex–axis through θ. 4.43. Find the matrices CandPin the Weyl representation of the γ–matrices. 4.44. Prove that the helicity of the Dirac particle changes sign under space inversion, but not under time reversal. 4.45. The Dirac Hamiltonian is H=α·p+βm.Determine the parameter θfrom the condition that the new Hamiltonian H/prime=UHU†,where U= eβα·pθ(p)has even form, i.e. H/prime∼β. (Foldy–Wouthuysen transformation). 4.46. Show that the spin operator Σ=i 2γ×γand the angular momentum L=r×p, in Foldy-Wouthuysen representation, have the following form: ΣFW=m EpΣ+p(p·Σ) 2Ep(m+Ep)+iβ(α×p) 2Ep, LFW=L−p(p·Σ) 2Ep(m+Ep)+p2Σ 2Ep(m+Ep)−iβ(α×p) 2Ep. 4.47. Find the Foldy-Wouthuysen transform of the position operator xand the momentum operator p.Calculate the commutator [ xFW,pFW]. 5 Classical ﬁeld theory and symmetries •Iff(x) is a function and F[f(x)] a functional, the functional derivative ,δF[f(x)] δf(y) is deﬁned by the relation δF=/integraldisplay dyδF[f(x)] δf(y)δf(y), (5.A) where δFis a variation of the functional. •The action is given by S=/integraldisplay d4xL(φr,∂µφr), (5.B) where Lis the Lagrangian density, which is a function of the ﬁelds φr(x),r= 1,...,n and their ﬁrst derivatives. The Euler–Lagrange equations of motion are ∂µ/parenleftbigg∂L ∂(∂µφr)/parenrightbigg −∂L ∂φr=0. (5.C) •The canonical momentum conjugate to the ﬁeld variable φris πr(x)=∂L ∂˙φr. (5.D) The canonical Hamiltonian is H=/integraldisplay d3xH=/integraldisplay d3x(˙φrπr−L). (5.E) •Noether theorem : If the action is invariant with respect to the continous in- ﬁnitesimal transformations: xµ→x/prime µ=xµ+δxµ, φr(x)→φ/prime r(x/prime)=φr(x)+δφr(x), 26 Problems then the divergence of the current jµ=∂L ∂(∂µφr)δφr(x)−Tµνδxν, (5.F) is equal to zero, i.e. ∂µjµ= 0. The quantity Tµν=∂L ∂(∂µφr)∂νφr−Lgµν, (5.G) is the energy–momentum tensor . The Noether charges Qa=/integraltext d3xja 0(x)a r e constants of motion under suitable asymptotic conditions. The index ais related to a symmetry group. 5.1.Let (a)Fµ=∂µφ, (b)S=/integraltext d4x/bracketleftbig1 2(∂µφ)2−V(φ)/bracketrightbig , be functionals. Calculate the functional derivativesδFµ δφin the ﬁrst case, and δ2S δφ(x)δφ(y)in the second case. 5.2.Find the Euler–Lagrange equations for the following Lagrangian densi- ties: (a)L=−(∂µAν)(∂νAµ)+1 2m2AµAµ+λ 2(∂µAµ)2, (b)L=−1 4FµνFµν+1 2m2AµAµ,where Fµν=∂µAν−∂νAµ, (c)L=1 2(∂µφ)(∂µφ)−1 2m2φ2−1 4λφ4, (d)L=(∂µφ−ieAµφ)(∂µφ∗+ieAµφ∗)−m2φ∗φ−1 4FµνFµν, (e)L=¯ψ(iγµ∂µ−m)ψ+1 2(∂µφ)2−1 2m2φ2+1 4λφ4−ig¯ψγ5ψφ . 5.3.The action of a free scalar ﬁeld in two dimensional spacetime is S=/integraldisplay∞ −∞dt/integraldisplayL 0dx/parenleftbigg1 2∂µφ∂µφ−m2 2φ2/parenrightbigg . The spatial coordinate xvaries in the region 0 <x<L . Find the equation of motion and discuss the importance of the boundary term. 5.4.Prove that the equations of motion remain unchanged if the divergence of an arbitrary ﬁeld function is added to the Lagrangian density. 5.5.Show that the Lagrangian density of a real scalar ﬁeld can be taken as L=−1 2φ(/unionsq /intersectionsq+m2)φ. Chapter 5. Classical ﬁeld theory and symmetries 27 5.6.Show that the Lagrangian density of a free spinor ﬁeld can be taken in the form L=i 2(¯ψ/∂ψ−(∂µ¯ψ)γµψ)−m¯ψψ. 5.7.The Lagrangian density for a massive vector ﬁeld Aµis given by L=−1 4FµνFµν+1 2m2AµAµ. Prove that the equation ∂µAµ= 0 is a consequence of the equations of motion. 5.8.Prove that the Lagrangian density of a massless vector ﬁeld is invariant under the gauge transformation: Aµ→Aµ+∂µΛ(x),where Λ=Λ(x)i sa n arbitrary function. Is the relation ∂µAµ= 0 a consequence of the equations of motion? 5.9.The Einstein–Hilbert gravitation action is S=κ/integraldisplay d4x√−gR , where gµνis the metric of four-dimensional curved spacetime; Ris scalar curvature and κis a constant. In the weak-ﬁeld approximation the metric is small perturbation around the ﬂat metric g(0) µν, i.e. gµν(x)=g(0) µν+hµν(x). The perturbation hµν(x) is a symmetric second rank tensor ﬁeld. The Einstein– Hilbert action in this approximation becomes an action in ﬂat spacetime (any- one familiar with general relativity can easily prove this): S=/integraldisplay d4x/parenleftbigg1 2∂σhµν∂σhµν−∂σhµν∂νhµσ+∂σhµσ∂µh−1 2∂µh∂µh/parenrightbigg , where h=hµ µ.Derive the equations of motion for hµν. These are the linearized Einstein equations. Show that the linearized theory is invariant under thegauge symmetry: h µν→hµν+∂µΛν+∂νΛµ, where Λµ(x) is any four-vector ﬁeld. 5.10. Find the canonical Hamiltonian for free scalar and spinor ﬁelds. 5.11. Show that the Lagrangian density L=1 2[(∂φ1)2+(∂φ2)2]−m2 2(φ2 1+φ2 2)−λ 4(φ2 1+φ2 2)2, is invariant under the transformation φ1→φ/prime 1=φ1cosθ−φ2sinθ, φ2→φ/prime 2=φ1sinθ+φ2cosθ. Find the corresponding Noether current and charge. 28 Problems 5.12. Consider the Lagrangian density L=(∂µφ†)(∂µφ)−m2φ†φ, where/parenleftbigg φ1 φ2/parenrightbigg is an SU(2) doublet. Show that the Lagrangian density has SU(2) symmetry. Find the related Noether currents and charges. 5.13. The Lagrangian density is given by L=¯ψ(iγµ∂µ−m)ψ, where ψ=/parenleftbigg ψ1 ψ2/parenrightbigg is a doublet of SU(2) group. Show that Lhas SU(2) sym- metry. Find Noether currents and charges. Derive the equations of motion for spinor ﬁelds ψi, where i=1,2. 5.14. Prove that the following Lagrangian densities are invariant under phase transformations (a)L=¯ψ(iγµ∂µ−m)ψ, (b)L=(∂µφ†)(∂µφ)−m2φ†φ. Find the Noether currents. 5.15. The Lagrangian density of a real three-component scalar ﬁeld is given by L=1 2∂µφT∂µφ−m2 2φTφ, where φ= φ1 φ2 φ3 . Find the equations of motions for the scalar ﬁelds φi. Prove that the Lagrangian density is SO(3) invariant and ﬁnd the Noether currents. 5.16. Investigate the invariance property of the Dirac Lagrangian density un- der chiral transformations ψ(x)→ψ/prime(x)=eiαγ5ψ(x), where αis a constant. Find the Noether current and its four-divergence. 5.17. The Lagrangian density of a σ-model is given by L=1 2[(∂µσ)(∂µσ)+(∂µπ)·(∂µπ)] + i¯Ψ/∂Ψ +g¯Ψ(σ+iτ·πγ5)Ψ−m2 2(σ2+π2)+λ 4(σ2+π2)2, Chapter 5. Classical ﬁeld theory and symmetries 29 where σis a scalar ﬁeld, πis a tree-component scalar ﬁeld, Ψa doublet of spinor ﬁelds, while τare Pauli matrices. Prove that the Lagrangian density Lhas the symmetry: σ(x)→σ(x), π(x)→π(x)−α×π(x), Ψ(x)→Ψ(x)+iα·τ 2Ψ(x), where αis an inﬁnitesimal constant vector. Find the corresponding conserved current. 5.18. In general, the canonical energy–momentum tensor is not symmetric under the permutation of indices. The energy–momentum tensor is not unique: a new equivalent energy–momentum tensor can be deﬁned by adding a four-divergence ˜T µν=Tµν+∂ρχρµν, where χρµν=−χµρν. The two energy–momentum tensors are equivalent since they lead to the same conserved charges, i.e. both satisfy the continuity equa- tion. If we take that the tensor χµνρis given by1 χµνρ=1 2/parenleftbigg −∂L ∂(∂µφr)(Iρν)rs+∂L ∂(∂ρφr)(Iµν)rs+∂L ∂(∂νφr)(Iµρ)rs/parenrightbigg then ˜Tµνis symmetric2. The quantities ( Iρν)rsin the previous formula are deﬁned by the transformation law of ﬁelds under Lorentz transformations: δφr≡φ/prime r(x/prime)−φr(x)=1 2ωµν(Iµν)rsφs(x). (a) Find the energy–momentum and angular momentum tensors for scalar, Dirac and electromagnetic ﬁelds employing the Noether theorem. (b) Applying the previously described procedure, ﬁnd the symmetrized (or Be- linfante) energy–momentum tensors for the Dirac and the electromagnetic ﬁeld. 5.19. Under dilatation the coordinates are transformed as x→x/prime=e−ρx. The corresponding transformation rule for a scalar ﬁeld is given by φ(x)→φ/prime(x/prime)=eρφ(x), 1Belinfante, Physica 6, 887 (1939) 2Symmetric energy–momentum tensors are not only simpler to work with but give the correct coupling to gravity. 30 Problems where ρis a constant parameter. Determine the inﬁnitesimal form variation3 of the scalar ﬁeld φ. Does the action for the scalar ﬁeld possess dilatation invariance? Find the Noether current. 5.20. Prove that the action for the massless Dirac ﬁeld is invariant under the dilatations: x→x/prime=e−ρx, ψ (x)→ψ/prime(x/prime)=e3ρ/2ψ(x). Calculate the Noether current and charge. 3A form variation is deﬁned by δ0φr(x)=φ/prime r(x)−φr(x); total variation is δφr(x)= φ/prime r(x/prime)−φr(x). 6 Green functions •The Green function (or propagator) of the Klein-Gordon equation, ∆(x−y) satisﬁes the equation (/unionsq /intersectionsqx+m2)∆(x−y)=−δ(4)(x−y). (6.A) To deﬁne the Green function entirely, one also needs to ﬁx the boundary condition. •The Green function (or propagator) S(x−y) of the Dirac equation is deﬁned by (iγµ∂x µ−m)S(x−y)=δ(4)(x−y), (6.B) naturally, again with the appropriate boundary conditions ﬁxed. •The retarded (advanced) Green function is deﬁned to be nonvanishing for positive (negative) values of time x0−y0. The boundary conditions for the Feynman propagator are causal, i.e. positive (negative) energy solutions prop- agate forward (backward) in time. The Dyson propagator is anticausal. 6.1.Using Fourier transform determine the Green functions for the Klein– Gordon equation. Discus how one goes around singularities. 6.2.If∆Fis the Feynman propagator, and ∆Ris the retarded propagator of the Klein–Gordon equation, prove that the diﬀerence between them, ∆F−∆R is a solution of the homogeneous Klein–Gordon equation. 6.3.Show that /integraldisplay d4kδ(k2−m2)θ(k0)f(k)=/integraldisplayd3k 2ωkf(k), where ωk=√ k2+m2. 32 Problems 6.4.Prove the following properties: ∆R(−x)=∆A(x), ∆F(−x)=∆F(x). ∆Aand∆Rare the advanced and retarded Green functions; ∆Fis the Feyn- man propagator. 6.5.If the Green function ¯∆(x) of the Klein–Gordon equation is deﬁned as1 ¯∆(x)=P/integraldisplayd4k (2π)4e−ik·x k2−m2, prove the relations: ¯∆(x)=1 2(∆R(x)+∆A(x)), ¯∆(−x)=¯∆(x). P denotes the principal value. 6.6.Write ∆(x)=−1 (2π)4/contintegraldisplay Cd4ke−ik·x k2−m2, and ∆±(x)=−1 (2π)4/contintegraldisplay C±d4ke−ik·x k2−m2 in terms of integrals over three momentum, k. The integration contours are given in Fig. 6.1. Fig. 6.1. The integration contours CandC±. In addition, prove that ∆(x)=∆+(x)+∆−(x). 1¯∆(x) is also called the principal-part propagator. Chapter 6. Green functions 33 6.7.Show that ∂∆(x) ∂xi/vextendsingle/vextendsingle/vextendsingle/vextendsingle x0=0=0, ∂∆(x) ∂x0/vextendsingle/vextendsingle/vextendsingle/vextendsingle x0=0=−δ(3)(x). 6.8.Prove that ∆(x) is a solution of the homogeneous Klein–Gordon equation. 6.9.Prove the following relation: ∆F(x)|m=0=−1 4πδ(x2)+i 4π2P1 x2, where ∆Fis the Feynman propagator of the Klein–Gordon equation. 6.10. Prove that ∆R,A|m2=0=−1 2πθ(±t)δ(x2). 6.11. If the source ρis given by ρ(y)=gδ(3)(y), show that φR=g 4πexp(−m|x|) |x|, where φR(x)=−/integraltext d4y∆R(x−y)ρ(y). 6.12. Show that the Green function of the Dirac equation, S(x) has the fol- lowing form S(x) = (i/∂+m)∆(x), where ∆(x) is the Green function of the Klein–Gordon equation with corre- sponding boundary conditions. 6.13. Starting from deﬁnition (6.B), determine the retarded, advanced, Feyn- man and Dyson propagators of the Dirac equation. Also, prove that the diﬀer- ence between any two of them is a solution of the homogenous Dirac equation. 6.14. If the source is given by j(y)=gδ(y0)eiq·y(1,0,0,0)T, where gis a constant while qis a constant vector, calculate ψ(x)=/integraldisplay d4ySF(x−y)j(y). SFis the Feynman propagator of the Dirac ﬁeld. 6.15. Calculate the Green function in momentum space for a massive vector ﬁeld, described by the Lagrangian density L=−1 4FµνFµν+1 2m2AµAµ. Fµν=∂µAν−∂νAµis the ﬁeld strength. 34 Problems 6.16. Calculate the Green function of a massless vector ﬁeld for which the Lagrangian density is given by L=−1 4FµνFµν+1 2λ(∂A)2. The second term is known as the gauge ﬁxing term; λis a constant. 7 Canonical quantization of the scalar ﬁeld •The operators of a complex free scalar ﬁeld are given by φ(x)=1 (2π)3 2/integraldisplayd3k√2ωk(a(k)e−ik·x+b†(k)eik·x), (7.A) φ†(x)=1 (2π)3 2/integraldisplayd3k√2ωk(b(k)e−ik·x+a†(k)eik·x), (7.B) where a(k)a n d b(k)a r eannihilation operators ;a†(k)a n d b†(k)creation op- erators anda(k)=b(k) is valid for a real scalar ﬁeld. Real scalar ﬁelds are associated to neutral particles, while complex ﬁelds describe charged particles. •The ﬁelds canonically conjugate to φandφ†are π=∂L ∂˙φ=˙φ†,π†=∂L ∂˙φ†=˙φ. Equal–time commutation relations take the following form: [φ(x,t),π(y,t)] = [φ†(x,t),π†(y,t)] = iδ(3)(x−y), [φ(x,t),φ(y,t)] = [φ(x,t),φ†(y,t)] = [π(x,t),π(y,t)] = 0 , (7.C) [π(x,t),π†(y,t)] = [φ(x,t),π†(y,t)] = 0 . From (7.C) we obtain: [a(k),a†(q)] = [b(k),b†(q)] =δ(3)(k−q), [a(k),a(q)] = [a†(k),a†(q)] = [a(k),b†(q)] = [a†(k),b†(q)] = 0 ,(7.D) [b(k),b(q)] = [b†(k),b†(q)] = [a(k),b(q)] = [a†(k),b(q)] = 0 . •The vacuum |0/angbracketrightis deﬁned by a(k)|0/angbracketright=0,b(k)|0/angbracketright=0,for all k.A state a†(k)|0/angbracketrightdescribes scalar particle with momentum k,b†(k)|0/angbracketrightan antiparticle with momentum k. Many–particle states are obtained by acting repeatedly with creation operators on the vacuum state. 36 Problems •In normal ordering, denoted by : :, the creation operators stand to the left of all the annihilation operators. For example: :a1a2a† 3a4a†5:=a† 3a†5a1a2a4. •The Hamiltonian, linear momentum and angular momentum of a scalar ﬁeld are H=1 2/integraldisplay d3x[(∂0φ)2+(∇φ)2+m2φ2], P=−/integraldisplay d3x∂0φ∇φ, Mµν=/integraldisplay d3x(xµT0ν−xνT0µ). •The Feynman propagator of a complex ﬁeld is deﬁned by i∆F(x−y)=/angbracketleft0|T(φ(x)φ†(y))|0/angbracketright. (7.E) Time ordering is deﬁned by T/parenleftbig φ(x)φ†(y)/parenrightbig =θ(x0−y0)φ(x)φ†(y)+θ(y0−x0)φ†(y)φ(x). •The transformation rules for a scalar ﬁeld under Poincar´ e transformations are given in Problem 7.20. Problems 7.21, 7.22 and 7.23 present the transforma-tions of a scalar ﬁeld under discrete transformations. 7.1.Starting from the canonical commutators [φ(x,t),˙φ(y,t)] = iδ(3)(x−y), [φ(x,t),φ(y,t)] = [ ˙φ(x,t),˙φ(y,t)] = 0 , derive the following commutation relations for creation and annihilation op- erators: [a(k),a†(q)] =δ(3)(k−q), [a(k),a(q)] = [a†(k),a†(q)] = 0 . 7.2.Att= 0, a real scalar ﬁeld and its time derivative are given by φ(t=0,x)=0,˙φ(t=0,x)=c, where cis a constant. Find the scalar ﬁeld φ(t,x) at an arbitrary moment t>0. Chapter 7. Canonical quantization of the scalar ﬁeld 37 7.3.Calculate the energy : H:, momentum : P: and charge : Q: of a complex scalar ﬁeld. Compare these results to the results obtained in Problems 2.2, 2.3 and 2.4. 7.4.Prove that the modes uk=1/radicalbig 2(2π)3ωke−iωkt+ik·x, are orthonormal with respect to the scalar product /angbracketleftf|g/angbracketright=−i/integraldisplay d3x[f(x)∂0g∗(x)−g∗(x)∂0f(x)]. 7.5.Show that the vacuum expectation value of the scalar ﬁeld Hamiltonian is given by /angbracketleft0|H|0/angbracketright=−1 4πm4δ(3)(0)Γ(−2). As one can see, this expression is the product of two divergent terms. Note that normal ordering gets rid of this c–number divergent term. 7.6.Calculate the following commutators: (Assume that the scalar ﬁeld is a real one except for case (d)) (a) [Pµ,φ(x)], (b) [Pµ,F(φ(x),π(x))],where Fis an arbitrary polynomial function of ﬁelds and momenta, (c) [H,a†(k)a(q)], (d) [Q,Pµ], (e) [N,H], where N=/integraltext d3ka†(k)a(k) is the particle number operator, (f)/integraltext d3x[H,φ(x)]e−ip·x. 7.7.Prove that eiQφ(x)e−iQ=e−iqφ(x). 7.8.The angular momentum of a scalar ﬁeld Mµν, is obtained in Problem 5.18. Instead of the classical ﬁeld, use the corresponding operator. Prove the following relations: (a) [Mµν,φ(x)] =−i(xµ∂ν−xν∂µ)φ(x), (b) [Mµν,Pλ]=i (gλνPµ−gλµPν), (c) [Mµν,Mρσ]=i (gµσMνρ+gνρMµσ−gµρMνσ−gνσMµρ). 7.9.Prove that φk(x)=/angbracketleftk|φ(x)|0/angbracketrightsatisﬁes the Klein–Gordon equation. 7.10. Calculate the charges Qa=/integraltext d3xja 0(x), where ja 0are zero components of the Noether currents for the symmetries deﬁned in Problems 5.12 and 5.15. (a) Prove that in both cases the charges satisfy the commutation relations of the SU(2) algebra. 38 Problems (b) Calculate [Qa,φi],[Qa,φ† i],(i=1,2), for the symmetry deﬁned in Problem 5.12 and [Qk,φi],(i=1,2,3), for the symmetry deﬁned in 5.15. 7.11. In Problem 5.19, it is shown that the action of a free massless scalar ﬁeld is invariant under dilatations. (a) Calculate the conserved charge D=/integraltext d3xj0. (b) Prove that relations ρ[D,φ(x)] = iδ0φ(x)a n d ρ[D,π(x)] = iδ0π(x) hold. (c) Calculate the commutator [ D,F(φ,π)], where Fis an arbitrary analytic function. (d) Prove that [ D,Pµ]=iPµ. 7.12. If, instead of the ﬁeld φ(x), we deﬁne the smeared ﬁeld φf(x,t)=/integraldisplay d3yφ(t,y)f(x−y), where fis given by f(x)=1 (a2π)3/2e−x2/a2, calculate the vacuum expectation value /angbracketleft0|φf(t,x)φf(t,x)|0/angbracketright.Find the result in the limit of vanishing mass. 7.13. The creation and annihilation operators of the free bosonic string αµ m (0<m∈Z), and αµ m(0>m∈Z), satisfy the commutation relations [αµ m,αν n]=−mδm+n,0gµν. Show that the operators Lm=−1 2/summationtextαµ m−nαnµsatisfy [Lm,Ln]=(m−n)Lm+n. The operators Lmform the classical Virasora algebra. Upon normal ordering of the L/prime ms one can obtain the full algebra (with central charge): [Lm,Ln]=(m−n)Lm+n+D−2 12(m3−m)δm+n,0. Dis number of scalar ﬁelds. 7.14. Calculate the vacuum expectation value /angbracketleft0|{φ(x),φ(y)}|0/angbracketright, where {,}is the anticommutator. Assume that the scalar ﬁeld is massless. Prove that the obtained expression satisﬁes the Klein–Gordon equation. Chapter 7. Canonical quantization of the scalar ﬁeld 39 7.15. Calculate /angbracketleft0|φ(x1)φ(x2)φ(x3)φ(x4)|0/angbracketright for a free scalar ﬁeld. 7.16. Find /angbracketleft0|φ(x)φ(y)|0/angbracketright in two dimensions, for a massless scalar ﬁeld. 7.17. Prove the relation (/unionsq /intersectionsqx+m2)/angbracketleft0|T(φ(x)φ(y))|0/angbracketright=−iδ(4)(x−y). 7.18. The Lagrangian density of a spinless Schr¨ odinger ﬁeld ψ, is given by L=iψ†∂ψ ∂t−1 2m∇ψ†·∇ψ−V(r)ψ†ψ. (a) Find the equations of motion. (b) Express the free ﬁelds ψandψ†in terms of creation and annihilation operators and ﬁnd commutation relations between them. (c) Calculate the Green function G(x0,x,y0,y)=−i/angbracketleft0|ψ(x0,x)ψ†(y0,y)|0/angbracketrightθ(x0−y0) and prove that it satisﬁes the equation /parenleftbigg i∂ ∂t+1 2m/triangle/parenrightbigg G(t,x,0,0) =δ(t)δ(3)(x). (d) Calculate the Green function for one-dimensional particle in the potential V=/braceleftbigg0,x > 0 ∞,x < 0. (e) Show that the free Schr¨ odinger equation is invariant under Galilean trans- formations, which contain: - spatial translations ψ/prime(t,r+/epsilon1)=ψ(t,r), - time translations ψ/prime(t+δ,r)=ψ(t,r), - spatial rotations ψ/prime(t,r+θ×r)=ψ(t,r), - ”boost” ψ/prime(t,r−vt)=e−imv·r+imv2t/2ψ(t,r). Without the phase factor in the last transformation rule the Schr¨ odinger equation will not be invariant, unless m= 0. Consequently this represen- tation of the Galilean group is projective. (f) Find the conserved quantities associated with these transformations and commutations relations between them, i.e. the Galilean algebra. 40 Problems 7.19. Let f(x)=/integraldisplayd3p 2ωp˜f(p)e−ip·x, be a classical function which satisﬁes the Klein–Gordon equation. Introduce the operators a=C/integraldisplayd3p/radicalbig2ωp˜f∗(p)a(p), a†=C/integraldisplayd3p/radicalbig2ωp˜f(p)a†(p), where a(p)a n d a†(p) are annihilation and creation operators for scalar ﬁeld, andCis a constant given by C=1/radicalBig/integraltextd3p 2ωp|˜f(p)|2. A coherent state is deﬁned by |z/angbracketright=e−|z|2/2eza†|0/angbracketright, where zis a complex number. (a) Calculate the following commutators: [a(p),a†],[a(p),a]. (b) Prove the relation [a(p),(a†)n]=Cn˜f(p)/radicalbig2ωp(a†)n−1. (c) Show that the coherent state is an eigenstate of the operator a(p). (d) Calculate the standard deviation of a scalar ﬁeld in the coherent state /radicalbig /angbracketleftz|:φ2(x):|z/angbracketright−(/angbracketleftz|φ(x)|z/angbracketright)2. (e) Find the expectation value of the Hamiltonian in the coherent state, /angbracketleftz|H|z/angbracketright. 7.20. Under the Poincar´ e transformation, x→x/prime=Λx+a, the real scalar ﬁeld transforms as follows: U(Λ,a)φ(x)U−1(Λ,a)=φ(Λx+a), where U(Λ,a) is a representation of the Poincar´ e group in space of the ﬁelds. Chapter 7. Canonical quantization of the scalar ﬁeld 41 (a) Prove the following transformation rules for creation and annihilation op- erators: U(Λ,a)a(k)U−1(Λ,a)=/radicalbiggωk/prime ωkexp(−iΛµ νkνaµ)a(Λk), U(Λ,a)a†(k)U−1(Λ,a)=/radicalbiggωk/prime ωkexp(iΛµ νkνaµ)a†(Λk). (b) Prove that the transformation rule of the n–particle state |k1,...,k n/angbracketrightis given by U(Λ,a)|k1,k2,...,k n/angbracketright=/radicalbiggωk/prime 1···ωk/primen ωk1···ωkneiaµΛµ ν(kν 1+...+kν n)|Λk1,...,Λk n/angbracketright. (c) Prove that the momentum operator, Pµof a scalar ﬁeld is a vector under Lorentz transformations: U(Λ,0)PµU−1(Λ,0) =ΛνµPν. (d) Prove that the commutator [ φ(x),φ(y)] is invariant with respect to Lorentz transformations. 7.21. The parity operator of a scalar ﬁeld is given by P=e x p/bracketleftbigg −iπ 2/integraldisplay d3k/parenleftbig a†(k)a(k)−ηpa†(k)a(−k)/parenrightbig/bracketrightbigg , where ηp=±1 is the intrinsic parity of the ﬁeld. (a) Prove that Pcommutes with the Hamiltonian. (b) Prove the relation PM ijP−1=Mij, where Mijis the angular momentum for scalar ﬁeld. 7.22. Under time reversal, the scalar ﬁeld is transformed according to τφ(x)τ−1=ηφ(−t,x), where τis an antiunitary operator, while ηis a phase. (a) Prove the relations: τa(k)τ−1=ηa(−k), τa†(k)τ−1=η∗a†(−k). (b) Derive the transformation rules for the Hamiltonian and momentum under the time reversal. 7.23. Charge conjugation for the charged scalar ﬁeld is deﬁned by Cφ(x)C−1=ηcφ†(x), where ηcis a phase factor. Prove that CQC−1=−Q, where Qis the charge operator. 8 Canonical quantization of the Dirac ﬁeld •The operators of a Dirac ﬁeld are: ψ(x)=1 (2π)3 22/summationdisplay r=1/integraldisplay d3p/radicalbiggm Ep/parenleftbig ur(p)cr(p)e−ip·x+vr(p)d† r(p)eip·x/parenrightbig ,(8.A) ¯ψ(x)=1 (2π)3 22/summationdisplay r=1/integraldisplay d3p/radicalbiggm Ep/parenleftbig ¯ur(p)c† r(p)eip·x+¯vr(p)dr(p)e−ip·x/parenrightbig .(8.B) The operators c† r(p)a n d d† r(p)a r ecreation operators , while cr(p),dr(p)a r e annihilation operators . •From the Dirac Lagrangian density, L=¯ψ(iγµ∂µ−m)ψ, one obtains the expressions for the conjugate momenta: πψ=∂L ∂˙ψ=iψ†,π¯ψ=∂L ∂˙¯ψ=0. Particles of spin 1 /2 obey Fermi-Dirac statistics. We impose the canonical equal-time anticommutation relations: {ψa(t,x),ψ† b(t,y)}=δabδ(3)(x−y), (8.C) {ψa(t,x),ψb(t,y)}={ψ† a(t,x),ψ† b(t,y)}=0. (8.D) From this we obtain the corresponding anticommutation relations between creation and annihilation operators: {cr(p),c† s(q)}={dr(p),d† s(q)}=δrsδ(3)(p−q). (8.E) All other anticommutators are zero. 44 Problems •T h eF o c ks p a c e of states is obtained as usual, by acting with creation operators on the vacuum |0/angbracketright.The states c†(p,r)|0/angbracketright,andd†(p,r)|0/angbracketrightare the electron and positron one–particle states, respectively with deﬁned momentum and polarization. •Normal ordering is deﬁned as in the case scalar ﬁeld but now the anticommu- tation relations (8.E) have to be taken into account, e.g. :c(q)c†(p): =−c†(p)c(q), :c(q)c(k)c†(p): =c†(p)c(q)c(k). •The Hamiltonian, momentum and angular moment of the Dirac ﬁeld are: H=/integraldisplay d3x¯ψ[−iγ∇+m]ψ, P=−i/integraldisplay d3xψ†∇ψ, Mµν=/integraldisplay d3xψ†(i(xµ∂ν−xν∂µ)+1 2σµν)ψ. •The Feynman propagator is given by iSF(x−y)=/angbracketleft0|T/parenleftbig ψ(x)¯ψ(y)/parenrightbig |0/angbracketright. (8.F) Time ordering is deﬁned by T/parenleftbig ψ(x)¯ψ(y)/parenrightbig =θ(x0−y0)ψ(x)¯ψ(y)−θ(y0−x0)¯ψ(y)ψ(x). •Under the Lorentz transformation ,x/prime=Λxthe operator ψ(x) transforms according to: U(Λ)ψ(x)U−1(Λ)=S−1(Λ)ψ(Λx). (8.G) HereU(Λ) is a unitary operator in spinor representation which generates the Lorentz transformation. •Parity ,t/prime=t,x/prime=−xchanges the Dirac ﬁeld as follows Pψ(t,x)P−1=γ0ψ(t,−x), (8.H) where Pis the appropriate unitary operator. •Time reversal ,t/prime=−t,x/prime=xis represented by an antiunitary operator. The transformation law is given by τψ(t,x)τ−1=Tψ(−t,x). (8.I) Properties of the matrix T, are given in Chapter 4. One should not forget that time reversal includes complex conjugation: τ(c...)τ−1=c∗τ...τ−1. Chapter 8. Canonical quantization of the Dirac ﬁeld 45 •The operator Cgenerates charge conjugation in the space of spinors: Cψa(x)C−1=(CγT 0)abψ† b(x). (8.J) Properties of the matrix Care given in Chapter 4. The charge conjugation transforms a particle into an antiparticle and vice–versa. •In this chapter we will very often use the identities: [AB,C ]=A[B,C]+[A,C]B, [AB,C ]=A{B,C}−{A,C}B. (8.K) 8.1.Starting from the anticommutation relations (8.E) show that: iS(x−y)={ψ(x),¯ψ(y)}= i(iγµ∂µ+m)∆(x−y) {ψ(x),ψ(y)}=0, where the function ∆(x−y) is to be determined. Prove that for x0=y0the function i S(x−y) becomes γ0δ(3)(x−y), i.e. the equal-time anticommutation relations for the Dirac ﬁeld is obtained. 8.2.Express the following quantities in terms of creation and annihilation operators: (a) charge Q=−e/integraltext d3x:ψ+ψ:, (b) energy H=/integraltext d3x[:¯ψ(−iγi∂i+m)ψ:], (c) momentum P=−i/integraltext d3x:ψ†∇ψ:. 8.3.(a) Show that i[ H, ψ(x)] =∂ ∂tψ(x).Comment on this result. (b) If the Dirac ﬁeld is quantized according to the Bose-Einstein rather than Fermi-Dirac statistics, what would be the energy of the ﬁeld? 8.4.Calculate [ H, c† r(p)cr(p)]. 8.5.Starting from the transformation law for the classical Dirac ﬁeld under Lorentz transformations show that the generators of these transformationsare given by M µν=i (xµ∂ν−xν∂µ)+1 2σµν. 8.6.The angular momentum of the Dirac ﬁeld is Mµν=/integraldisplay d3xψ†(x)/bracketleftbigg i(xµ∂ν−xν∂µ)+1 2σµν/bracketrightbigg ψ(x). 46 Problems (a) Prove that [Mµν,ψ(x)] =−i(xµ∂ν−xν∂µ)ψ(x)−1 2σµνψ(x), and comment on this result. (b) Also, prove [Mµν,Pρ]=i (gνρPµ−gµρPν), where Pµis the four-vector of momentum. 8.7.Show that the helicity of the Dirac ﬁeld is given by Sp=1 2/summationdisplay r/integraldisplay d3p(−1)r+1[c† r(p)cr(p)+d† r(p)dr(p)]. 8.8.Let|p1,r1;p2,r2/angbracketright=c† r1(p1)c† r2(p2)|0/angbracketrightbe a two-particle state. Find the energy, charge and helicity of this state. Here r1,2are helicities of one-particle states. 8.9.Prove that the charges found in Problem 5.13 satisfy the commutation relation: [Qa,Qb]=i/epsilon1abcQc. 8.10. Find conserved charges for the symmetry in Problem 5.17 and calculate the commutators: (a) [Qa,Qb], (b) [Qb,πa(x)],[Qb,ψi(x)],[Qb,¯ψi(x)]. 8.11. In Problem 5.20 we showed that the action for a massless Dirac ﬁeld is invariant under dilatations. Find the conserved charge D=/integraltext d3xj0for this symmetry and show that the relation [D,Pµ]=iPµ, is satisﬁed. 8.12. Let the Lagrangian density be given by L=i¯ψγµ∂µψ−gx2¯ψψ , where gis a constant. (a) Derive the expression for the energy–momentum tensor Tµν. Find its di- vergence, ∂µTµν. Comment on this result. (b) Calculate the commutator [ P0(t),Pi(t)]. (c) Find the four divergence of the angular momentum operator Mµαβ. 8.13. Consider the current commutator [ Jµ(x),Jν(y)] where Jµ=¯ψγµψ. Chapter 8. Canonical quantization of the Dirac ﬁeld 47 (a) Prove that the commutator given above is Lorentz covariant. (b) Show that the commutator is equal to zero for space–like interval, i.e. for (x−y)2<0. 8.14. Calculate /angbracketleft0|¯ψ(x1)ψ(x2)ψ(x3)¯ψ(x4)|0/angbracketright.The result should be expressed in terms of vacuum expectation value of two ﬁelds. 8.15. Prove that : ¯ψγµψ:=1 2[¯ψ,γµψ]. 8.16. Prove that /angbracketleft0|T(¯ψ(x)Γψ(y))|0/angbracketrightis equal to zero for Γ={γ5,γ5γµ}, while for Γ=γµγνone gets the result −4imgµν∆F(y−x). 8.17. The Dirac spinor in terms of two Weyl spinors ϕandχis of the form ψ=/parenleftbigg ϕ −iσ2χ∗/parenrightbigg . (a) Show that the Majorana spinor equals ψM=/parenleftbigg χ −iσ2χ∗/parenrightbigg . (b) Prove the identities: ¯ψMφM=¯φMψM, ¯ψMγµφM=−¯φMγµψM, ¯ψMγ5φM=¯φMγ5ψM, ¯ψMγµγ5φM=¯φMγµγ5ψM, ¯ψMσµνφM=−¯φMσµνψM. (c) Express the Majorana ﬁeld operator, ψM=1√ 2(ψ+ψc) using creation and annihilation operators of a Dirac ﬁeld. Introduce creation and annihilation operators for Majorana spinors and ﬁnd corresponding anticomutation re- lations. (d) Rewrite the QED Lagrangian density using Majorana spinors. 8.18. Find the transformation laws of the quantities Vµ(x)=¯ψ(x)γµψ(x)a n d Aµ(x)=¯ψ(x)γ5∂µψ(x) under Lorentz and discrete transformations. 8.19. Show that the Lagrangian density L=i¯ψ(x)γµ∂µψ(x)+m¯ψ(x)ψ(x), is invariant under the Lorentz and discrete transformations. 8.20. Show that the quantity Tµν(x)=¯ψ(x)σµνψ(x) transforms as a tensor under Lorentz transformations. Find its transformation rules under discrete symmetries. 9 Canonical quantization of the electromagnetic ﬁeld •The Lagrangian density of the electromagnetic ﬁeld in the presence of an exterior current jµis L=−1 4FµνFµν−jµAµ. From this expression we derive the equations of motion to be: ∂µFµν=jν⇒(δν µ/unionsq/intersectionsq−∂µ∂ν)Aµ=jν. (9.A) It is easy to see that the ﬁeld strength Fµνsatisﬁes the identity: ∂µFνρ+∂νFρµ+∂ρFµν=0. (9.B) Equations (9.A-B) are the Maxwell equations ; (9.B) is the so–called, Bianchi identity and is a kinematical condition. •Electrodynamics is invariant under the gauge transformation Aµ→Aµ+∂µΛ(x), where Λ(x) is an arbitrary function. The gauge symmetry can be ﬁxed by imposing a ”gauge condition”. The following choices are often convenient: Lorentz gauge ∂µAµ=0, Coulomb gauge ∇·A=0, Time gauge A0=0, Axial gauge A3=0. •The general solution of the vacuum Maxwell equations ( jµ= 0) takes the form: Aµ(x)=3/summationdisplay λ=01 (2π)3 2/integraldisplayd3k√2ωk/parenleftBig aλ(k)/epsilon1µ λ(k)e−ik·x+a† λ(k)/epsilon1µ λ(k)eik·x/parenrightBig ,(9.C) where ωk=|k|,/epsilon1µ λ(k) are polarization vectors. The transverse polarization vectors which satisfy /epsilon1(k)·k= 0 we denote by /epsilon1µ 1(k)a n d /epsilon1µ 2(k). The scalar 50 Problems polarization vector is /epsilon1µ 0=nµ, where nµis a unit time–like vector. We can choose nµ=( 1,0,0,0). The longitudinal polarization vector, /epsilon1µ 3(k) is given by /epsilon1µ 3(k)=kµ−(n·k)nµ (n·k). Due to gauge symmetry only two polarizations are independent. The polar- ization vectors satisfy the orthonormality relations: gµν/epsilon1µ λ(k)/epsilon1ν λ/prime(k)=−δλλ/prime. In (9.C) we assumed the polarization vectors to be real valued. •The polarization vectors satisfy the following completeness relations : 3/summationdisplay λ=0gλλ/epsilon1µ λ(k)/epsilon1ν λ(k)=gµν. (9.D) From (9.D) follows that the sum over transverse photons is 2/summationdisplay λ=1/epsilon1i λ(k)/epsilon1j λ(k)=−gij−kikj (k·n)2+kinj+kjni k·n. (9.E) •In the Lorentz gauge the equal-time commutation relations are: [Aµ(t,x),πν(t,y)] = igµνδ(3)(x−y), [Aµ(t,x),Aν(t,y)] = 0 , (9.F) [πµ(t,x),πν(t,y)] = 0 . where πν=−˙Aν. Creation and annihilation operators of the photon ﬁeld satisfy the following commutation relations: [aλ(k),a† λ/prime(q)] =−gλλ/primeδ(3)(k−q), [aλ(k),aλ/prime(q)] = 0 , (9.G) [a† λ(k),a† λ/prime(q)] = 0 . The physical states, |Φ/angbracketrightsatisfy the operator condition ∂µA(+) µ|Φ/angbracketright=0. This is the Gupta–Bleuler method of quantization. •In the Coulomb gauge we have A(x)=2/summationdisplay λ=11 (2π)3 2/integraldisplayd3k√2ωk/parenleftBig aλ(k)/epsilon1λ(k)e−ik·x+a† λ(k)/epsilon1λ(k)eik·x/parenrightBig ,(9.H) Chapter 9. Canonical quantization of the electromagnetic ﬁeld 51 while A0=0 .The equal-time commutation relations are: [Ai(t,x),πj(t,y)] =−iδ(3) ⊥ij(x−y), [Ai(t,x),Aj(t,y)] = 0 , (9.I) [πi(t,x),πj(t,y)] = 0 , where π=Eandδ(3) ⊥ij(x−y) is the transversal delta function given by δ(3) ⊥ij(x−y)=1 (2π)3/integraldisplay d3keik·(x−y)/parenleftbigg δij−kikj k2/parenrightbigg . Creation and annihilation operators obey [aλ(k),a† λ/prime(q)] =δλλ/primeδ(3)(k−q), [aλ(k),aλ/prime(q)] = 0 , (9.J) [a† λ(k),a† λ/prime(q)] = 0 . •The Feynman propagator for the electromagnetic ﬁeld is given by iDµν F(x−y)=/angbracketleft0|T(Aµ(x)Aν(y))|0/angbracketright. (9.K) 9.1.Starting from the commutation relations (9.G) prove that [Aµ(t,x),˙Aν(t,y)] =−igµνδ(3)(x−y). 9.2.Find the commutator iDµν(x−y)=[Aµ(x),Aν(y)], in the Lorentz gauge. 9.3.Calculate the commutators between components of the electric and the magnetic ﬁelds: [Ei(x),Ej(y)], [Bi(x),Bj(y)], [Ei(x),Bj(y)]. Also calculate the previous commutators for equal times, x0=y0. 9.4.Prove that [ Pµ,Aν]=−i∂µAν. 52 Problems 9.5.Determine the helicity of photons described by polarization vectors /epsilon1µ +(kez)=2−1/2(0,1,i,0)Tand/epsilon1µ −(kez)=2−1/2(0,1,−i,0)T. 9.6.A photon linearly polarized along the x–axis is moving along the z– direction with momentum k. Determine the polarization of the photon for observer S/primemoving in the x–direction with velocity v. 9.7.The arbitrary state not containing transversal photons has the form |Φ/angbracketright=/summationdisplay nCn|Φn/angbracketright, where Cnare constants and |Φn/angbracketright=/integraldisplay d3k1...d3knf(k1,...,kn)n/productdisplay i=1(a† 0(ki)−a† 3(ki))|0/angbracketright, where f(k1,...,kn) are arbitrary functions. The state |Φ0/angbracketrightis a vacuum. (a) Prove that /angbracketleftΦn|Φn/angbracketright=δn,0. (b) Show that /angbracketleftΦ|Aµ(x)|Φ/angbracketrightis a pure gauge. 9.8.Let Pµν=gµν−kµ¯kν+kν¯kµ k·¯k, and Pµν ⊥=kµ¯kν+kν¯kµ k·¯k, where ¯kµ=(k0,−k). Calculate: PµνPνσ,Pµν ⊥P⊥ νσ,Pµν+Pµν ⊥,gµνPµν,gµνP⊥ µν,PµνPνσ ⊥,ifk2=0 . 9.9.The angular momentum of the photon ﬁeld is deﬁned by Jl=1 2/epsilon1lijMij, where Mijwas found in Problem 5.18. (a) Express Jin terms of the potentials in the Coulomb gauge. (b) Express the spin part of the angular momentum in terms of aλ(k),a† λ(k) and diagonalize it. (c) Show that the states a† ±(q)|0/angbracketright=1√ 2(a† 1(q)±ia† 2(q))|0/angbracketright, are the eigenstates of the helicity operator with the eigenvalues ±1. (d) Calculate the commutator [ Jl,Am(y,t)]. 9.10. Calculate: (a)/angbracketleft0|{Ei(x),Bj(y)}|0/angbracketright, (b)/angbracketleft0|{Bi(x),Bj(y)}|0/angbracketright, Chapter 9. Canonical quantization of the electromagnetic ﬁeld 53 (c)/angbracketleft0|{Ei(x),Ej(y)}|0/angbracketright. 9.11. Consider the quantization of the electromagnetic ﬁeld in space between two parallel square plates located at z=0a n d z=a. The plates are squares with size of length L. They are perfect conductors. (a) Find the general solution for the electromagnetic potential inside this ca- pacitor. (b) Quantize the electromagnetic ﬁeld using canonical quantization. (c) Find the Hamiltonian Hand show that the vacuum energy is E=1 2L2/integraldisplayd2k (2π)2/bracketleftBigg 2∞/summationdisplay n=1/radicalbigg k2 1+k2 2+/parenleftBignπ a/parenrightBig2 +/radicalBig k2 1+k2 2/bracketrightBigg .(9.1) (d) Deﬁne the quantity /epsilon1=E−E0 L2, which is the diﬀerence between the vacuum energies per unit area in the presence and in the absent of plates. This quantity is divergent and can be regularized introducing the function f(k)=/braceleftbigg 1,k < Λ 0,k > Λ, into the integral; Λis a cutoﬀ parameter. Calculate /epsilon1and show that there is an attractive force between the plates. This is the Casimir eﬀect . (e) The energy per unit area, E/L2can be regularized in a diﬀerent way. Calculate integral I=/integraldisplay d2k1 (k2+m2)α, for Re α>0, and then analitically continue this integral to Re α≤0. Show that E/L2=−π2 6a3∞/summationdisplay n=1n3. Regularize the sum in the previous expression using the Rieman ζ–function ζ(s)=∞/summationdisplay n=1n−s. Calculate the energy and the force per unit area. 10 Processes in the lowest order of perturbation theory •The Wick’s theorem states T(A BC...YZ )= :{A BC...YZ + ”all contractions” }:. (10.A) In the case of fermions we have to take care about anticommutation relations, i.e. every time when we interchange neighboring fermionic operators a minussign appears. •TheS–matrix is given by S=∞/summationdisplay n=0(−i)n n!/integraldisplay .../integraldisplay d4x1...d4xnT(HI(x1)···H I(xn)), (10.B) where HIis the Hamiltonian density of interaction in the interaction pictures. •S–matrix elements have the general form Sﬁ=( 2π)4δ(4)(pf−pi)iM/productdisplay b1√ 2VE/productdisplay f/radicalbiggm VE, (10.C) where piandpfare the initial and the ﬁnal momenta, respectively; i Mis the Feynman amplitude for the process, which will be determined using Feynman diagrams. The delta function in (10.C) is a consequence of the conservation of energy and momentum in the process. Normalization factors also appear in the expression (10.C) and they are diﬀerent for bosonic and fermionic particles. In this Chapter we will use so–called box normalization. •The diﬀerential cross section for the scattering of two particles into Nﬁnal particles is dσ=|Sﬁ|2 T1 |Jin|N/productdisplay i=1Vd3pi (2π)3, (10.D) where Jinis the ﬂux of initial particles: |Jin|=vrel V. 56 Problems The relative velocity vrelis given by vrel=|p1| E1, in the laboratory frame of reference (particle 2 is at rest), while in the center– of–mass frame we have vrel=|p1|E1+E2 E1E2, p1is the momentum of particle 1, and E1,2are energies of particles. In ex- pression (10.D), Vd3p/(2π)3is the volume element of phase space. •Feynman rules for QED : ◦Vertex: =ieγµ ◦Photon and lepton propagators: iDFµν= =−igµν k2+i/epsilon1, iSF(p)= =i /p−m+i/epsilon1. ◦External lines: =u(p,s) initial a) leptons (i.e. electron):=¯u(p,s) ﬁnal =v(p,s) initial b) antileptons (i.e. positron):=¯v(p,s) ﬁnal =εµ(k,λ) initial c) photons:=ε∗ µ(k,λ) ﬁnal ◦Spinor factor are written from the left to the right along each of the fermionic lines. The order of writing is important, because it is a ques- tion of matrix multiplication of the corresponding factors. ◦For all loops with momentum k, we must integrate over the momentum:/integraltext d4k/(2π)4. This corresponds to the addition of quantum mechanical amplitudes. ◦For fermion loops we have to take the trace and multiply it by the factor −1. Chapter 10. Processes in the lowest order of perturbation theory 57 ◦If two diagrams diﬀer for an odd number of fermionic interchanges, then they must diﬀer by a relative minus sign. 10.1. For the process A(E1,p1)+B ( E2,p2)→C(E/prime 1,p/prime 1)+D ( E/prime 2,p/prime 2) prove that the diﬀerential cross section in the center of mass frame is given by/parenleftbiggdσ dΩ/parenrightbigg cm=1 4π2(E1+E2)2|p/prime 1| |p1|mAmBmCmD|M|2, where i Mis the Feynman amplitude. Assume that all particles in the process are fermions. 10.2. Consider the following integral: I=/integraldisplayd3p 2Epd3q 2Eqδ(3)(p+q−P)δ(Ep+Eq−P0), where E2 p=p2+m2andE2 q=q2+m/prime2. Show that the integral Iis Lorentz invariant. Calculate it in the frame where P=0. 10.3. If iM=¯u(p,r)γµ(1−γ5)u(q,s)/epsilon1µ(k,λ), calculate the sum 2/summationdisplay λ=12/summationdisplay r,s=1|M|2. 10.4. Using the Wick theorem evaluate: (a)/angbracketleft0|T(φ4(x)φ4(y))|0/angbracketright, (b)T(:φ4(x)::φ4(y): ), (c)/angbracketleft0|T(¯ψ(x)ψ(x)¯ψ(y)ψ(y))|0/angbracketright. 10.5. Inφ4theory the interaction Lagrangian density is Lint=−λ 4!φ4.U s - ing the Wick theorem determine the symmetry factor S, for the following diagrams: (a)x2 x1 58 Problems (b) x2 x1 (c)x2 x1 Also, check the results using the formula [6]: S=g/productdisplay n=2,3,..2β(n!)αn, where gis the number of possible permutations of vertices which leave un- changed the diagram with ﬁxed external lines, αnis the number of vertex pairs connected by nidentical lines, and βis the number of lines connecting a vertex with itself. 10.6. Inφ3theory calculate 1 2/parenleftbigg−iλ 3!/parenrightbigg2/integraldisplay d4y1d4y2/angbracketleft0|T(φ(x1)φ(x2)φ3(y1)φ3(y2))|0/angbracketright. 10.7. For the QED processes : (a)µ−µ+→e−e+, (b)e−µ+→e−µ+, write the expressions for amplitudes using Feynman rules. Calculate/angbracketleftbig |M|2/angbracketrightbig averaging over all initial polarization states and summing over the ﬁnal polar- ization states of particles. Calculate the diﬀerential cross sections in center–of–mass system in an ultrarelativistic limit. 10.8. Show that the Feynman amplitude for the Compton scattering is a gauge invariant quantity. 10.9. Find the diﬀerential cross section for the scattering of an electron in the external electromagnetic ﬁeld ( a,g,k are constants) (a)A µ(x)=(ae−k2x2,0,0,0), (b)Aµ(x)=( 0 ,0,0,g re−r/a). 10.10. Calculate the cross section per unit volume for the creation of electron– positron pairs by the electromagnetic potential Aµ=( 0,0,ae−iωt,0), where ωandaare constants. Chapter 10. Processes in the lowest order of perturbation theory 59 10.11. Find the diﬀerential cross section for the scattering of an electron in the external potential Aµ=( 0,0,0,ae−k2x2), for a theory which is the same as QED except the fact that the vertex i eγµis replaced by i eγµ(1−γ5). 10.12. Find the diﬀerential cross section for the scattering of a positron in the external potential Aµ=(g r,0,0,0), where gis a constant. The S–matrix element is given by Sﬁ=ie/integraldisplay d4x¯ψf(x)∂µψi(x)Aµ(x). 10.13. Calculate the cross section for the scattering of an electron with posi- tive helicity in the electromagnetic potential Aµ=(aδ(3)(x),0,0,0), where ais a constant. 10.14. Calculate the diﬀerential cross section for scattering of e−and a muon µ+ e−µ+→e−µ+, in the center–of–mass system. Assume that initial particles have negative he- licity, while the spin states of ﬁnal particles are arbitrary. 10.15. Consider the theory of interaction of a spinor and scalar ﬁeld: L=1 2(∂φ)2−M2 2φ2+¯ψ(iγµ∂µ−m)ψ−g¯ψγ5ψφ . Calculate the cross section for the scattering of two fermions in the lowest order. 10.16. Write the expressions for the Feynman amplitudes for diagrams given in the ﬁgure. (a) (b) (c) (d) (e) 60 Problems (f) (g) (h) (i) 11 Renormalization and regularization •Table of D-dimensional integrals in Minkowski spacetime: /integraldisplay dDk1 (k2+2p·k−m2+i/epsilon1)n=i(−1)nπD 2 Γ(n)(m2+p2)n−D 2Γ(n−D 2),(11.A) /integraldisplay dDkkµ (k2+2p·k−m2+i/epsilon1)n=−i(−1)nπD 2 Γ(n)(m2+p2)n−D 2pµΓ(n−D 2),(11.B) /integraldisplay dDkkµkν (k2+2p·k−m2+i/epsilon1)n=i(−1)nπD 2 Γ(n)(m2+p2)n−D 2/bracketleftbigg pµpνΓ(n−D 2) −1 2gµν(p2+m2)Γ(n−D 2−1)/bracketrightbigg , (11.C) /integraldisplay dDkkµkνkρ (k2+2p·k−m2+i/epsilon1)n=−i(−1)nπD 2 Γ(n)(m2+p2)n−D 2/bracketleftbigg pµpνpρΓ(n−D 2) −1 2(gµνpρ+gµρpν+gνρpµ)(p2+m2)Γ(n−D 2−1)/bracketrightbigg , (11.D) /integraldisplay dDkkµkνkρkσ (k2+2p·k−m2+i/epsilon1)n=i(−1)nπD/2 Γ(n)(m2+p2)n−D 2/bracketleftbigg pµpνpρpσΓ(n−D 2) −1 2(gµνpρpσ+gµρpνpσ+gµσpνpρ+gνρpµpσ+gνσpρpµ+gρσpµpν) ×(p2+m2)Γ(n−D 2−1) +1 4(gµνgρσ+gµρgνσ+gµσgρν)(p2+m2)2Γ(n−D 2−2)/bracketrightbigg . (11.E) 62 Problems •The gamma–function obeys Γ(−n+/epsilon1)=(−1)n n!/parenleftbigg1 /epsilon1+ψ(n+1 )+ o(/epsilon1)/parenrightbigg , (11.F) where n∈Nand ψ(n+1 )=1+1 2+...+1 n−γ. Theγ=0,5772 is the Euler–Mascheroni constant. •The general expression for Feynman parametrization is given in Problem 11.1. The most frequently used parameterizations are: 1 AB=/integraldisplay1 0dx1 [xA+( 1−x)B]2, (11.G) 1 ABC=2/integraldisplay1 0dx/integraldisplay1−x 0dz1 [A+(B−A)x+(C−A)z]3. (11.H) •Cutkosky rule for computing discontinuity of any Feynman diagram contains the following steps: 1. Cut through the diagram in all possible ways such that the cut propagators can be put on–shell. 2. For each cut, make the replacement 1 p2−m2→(−2iπ)δ(4)(p2−m2)θ(p0). 3. Sum the contributions of all possible cuts. 11.1. Prove the following formula (the Feynman parametrization) 1 A1...A n=(n−1)!/integraldisplay1 0.../integraldisplay1 0dx1...dxnδ(x1+...+xn−1) (x1A1+...+xnAn)n. 11.2. Show that expression (11.A) holds. 11.3. Prove the formula (11.F). 11.4. Regularize the integral I=/integraldisplay d4k1 k21 (k+p)2−m2, using Pauli–Villars regularization. Chapter 11. Renormalization and regularization 63 11.5. Compute Iαβµνρσ =/integraldisplay dDkkαkβkµkνkρkσ (k2)n. Also, ﬁnd the divergent part of the previous integral for n= 5. Apply the dimensional regularization. 11.6. Consider the interacting theory of two scalar ﬁelds φandχ: L=1 2(∂φ)2−1 2m2φ2+1 2(∂χ)2−1 2M2χ2−gφ2χ. (a) Find the self–energy of the χparticle, −iΠ(p2). (b) Calculate the decay rate of the χparticle into two φparticles. (c) Prove that ImΠ(M2)=−MΓ. 11.7. Consider the theory L=1 2(∂µφ)2−m2 2φ2−g 3!φ3−λ 4!φ4. Find the expression for the self–energy and the mass shift δm. 11.8. The Lagrangian density is given by L=1 2(∂µσ)2+1 2(∂µπ)2−m2 2σ2−λvσ3−λvσπ2−λ 4(σ2+π2)2, where σandπare scalar ﬁelds, and v2=m2 2λis constant. Classically, πﬁeld is massless. Show that it also remains massless when the one–loop corrections are included. 11.9. Find the divergent part of the diagram Prove that this diagram cancels with the diagram of the reverse orientation inside the fermion loop. 11.10. The polarization of vacuum in QED has form −iΠµν(q)=−i(qµqν−q2gµν)Π(q2). Prove the following expression: ImΠ(q2)=−e2 12π/parenleftbigg 1+2m2 q2/parenrightbigg/radicalBigg 1−4m2 q2θ/parenleftbigg 1−4m2 q2/parenrightbigg . 64 Problems 11.11. In scalar electrodynamics two diagrams give contribution to the po- larization of vacuum. Using dimensional regularization derive the following expression for the divergent part of the vacuum polarization: ie2 24π21 /epsilon1(pµpν−p2gµν). 11.12. The Lagrangian density for the pseudoscalar Yukawa theory is given by L=1 2(∂φ)2−m2 2φ2+¯ψ(iγµ∂µ−M)ψ−ig¯ψγ5ψφ−λ 4!φ4. (a) Find the superﬁcial degree of divergence for this theory and the corre- sponding divergent amplitudes. Write the bare Lagrangian density as a sum of the initial Lagrangian density and counterterms. Write out the Feynman rules in the renormalized theory. (b) Find the self–energy of the spinor ﬁeld at one–loop and determine the corresponding counterterms. (c) Find the self–energy of the scalar ﬁeld at one–loop and determine the corresponding counterterms. (d) Calculate the one–loop vertex correction φ¯ψψandδg. (e) Calculate the one–loop vertex correction φ4andδλ. 11.13. Consider massless two-dimensional QED, the so–called Schwinger mo- del. (a) Calculate the vacuum polarization at one–loop. (b) Find the full photon propagator and read oﬀ the mass of the photon. 11.14. Consider φ3theory in six–dimensional spacetime, with the Lagrangian density given by L=1 2(∂φ)2−m2 2φ2−g 3!φ3−hφ . (a) Determine the superﬁcial divergent amplitudes. Write the renormalized Lagrangian density and derive the Feynman rules. (b) Calculate the tadpole one–loop diagram and explain why the contribution of the tadpole diagrams can be ignored. (c) Calculate the propagator correction at one–loop order and determine δZ andδm. Use the minimal subtraction (MS) scheme. (d) Calculate the vertex correction and ﬁnd δg. (e) Derive the relations m0=m0(m,g,/epsilon1 )a n d g0=g0(m,g,/epsilon1 ). Part II Solutions 1 Lorentz and Poincar´ e symmetries 1.1The square of the length of a four-vector, xisx2=gµνxµxν. By substi- tuting x/primeµ=Λµ ρxρinto the condition x/prime2=x2one obtains: gµνΛµ ρΛν σxρxσ=gρσxρxσ. (1.1) Since (1.1) is valid for any vector x∈M4,w eg e t ΛµρgµνΛνσ=gρσ.The previous condition can be rewritten in the following form (ΛT)ρµgµνΛν σ=gρσ⇒ΛTgΛ=g, (1.2) and we have obtained the requested expression. Now, we shall show that the Lorentz transformations form a group. If Λ1andΛ2are Lorentz transformations then their product, Λ1Λ2is Lorentz transformation because it satisﬁes the condition (1.2): (Λ1Λ2)Tg(Λ1Λ2)=ΛT 2(ΛT 1gΛ1)Λ2=ΛT 2gΛ2=g. Thus, we have shown the closure axiom. Multiplication of matrices is generally an associative operation, so this property is valid for Lorentz matrices Λ. Identity matrix I satisﬁes the condition (1.2) and it is the unit element of thegroup. Taking determinant of the expression (1.2) we obtain det Λ=±1. Since detΛ/negationslash= 0 the inverse element Λ −1exists for every Lorentz matrix. From (1.2) we see that the inverse element is given by Λ−1=g−1ΛTg. In the component notation the previous relation takes the following form: (Λ−1)µ ν=gµρΛσ ρgσν=Λνµ. 1.2By substituting inﬁnitesimal form of the Lorentz transformation into the formula (1.2), one gets: (δµ ρ+ωµ ρ)gµν(δν σ+ων σ)+o(ω2)=gρσ, 68 Solutions gρσ+ωµ ρgµνδν σ+ων σgµνδµ ρ+o(ω2)=gρσ. from which follows that ωρσ+ωσρ=0⇒ωρσ=−ωσρ. Since the parameters of the Lorentz group ωµνare antisymmetric only six of them are independent, so the Lorentz group is six–parameters group. Moreover the Lorentz group is a Lie group. 1.3Given relation is in agreement with deﬁnitions of the /epsilon1symbol and determinant. 1.4From (1.2) follows that δσ ρ=δν µΛµρΛνσ, so we conclude that δ/primeσ ρ=δσ ρ. In the same way we have /epsilon1/prime µνρσ=ΛµαΛνβΛργΛσδ/epsilon1αβγδ= det( Λ−1)/epsilon1µνρσ=/epsilon1µνρσ, since det Λ−1= 1 for the proper orthochronous Lorentz transformations. Thus, Levi-Civita symbol is deﬁned independently of the inertial frame. Note that the components /epsilon1µνρσare obtained by applying the antisymmetric tensor /epsilon1on basis vectors e0,...,e3: /epsilon1(eµ,eν,eρ,eσ)=/epsilon1µνρσ. The/epsilon1tensor can be written in the form /epsilon1=θ0∧θ1∧θ2∧θ3, where θµare basic one-forms. 1.5The results are given below /epsilon1µνρσ/epsilon1µβγδ=−δν βδρ γδσ δ+δν γδρ βδσ δ+δν βδρ δδσ γ−δν γδρ δδσ β−δν δδρ βδσ γ+δν δδρ γδσ β, /epsilon1µνρσ/epsilon1µνγδ=−2(δρ γδσ δ−δρ δδσ γ), /epsilon1µνρσ/epsilon1µνρδ=−6δσ δ, /epsilon1µνρσ/epsilon1µνρσ=−24. 1.6 (a) The matrix Xis X=/parenleftbigg x0−x3−x1+ix2 −x1−ix2x0+x3/parenrightbigg , so det X=(x0)2−(x)2=x2.It is not diﬃcult to see that from the transformation law, X/prime=SXS†,follows that detX/prime= detSdetXdetS†= detX, which means that x/prime2=x2. Chapter 1. Lorentz and Poincar´ e symmetries 69 (b) Multiplying the expression X=xµσµby ¯σνand taking trace we obtain the requested relation. The matrices σµsatisfy the following orthogonality relation tr[¯ σµσν]=2gµν. 1.7The result follows from x/primeµ=1 2tr(¯σµX/prime)=1 2xνtr(¯σµSσνS†)=Λµ νxν. 1.8An arbitrary Lorentz transformation, which is connected with the unit element, can be written in the form U(ω) = exp/parenleftbig −i 2Mµνωµν/parenrightbig ,where Mµν are generators. There are three (independent) rotations and three (also in- dependent) boosts. Rotation around z−axis for angle θ3is represented by matrix Λ(θ3)= 10 0 0 0c o s θ3sinθ30 0−sinθ3cosθ30 00 0 1 ≈I+ 00 0 0 00 θ30 0−θ300 00 0 1 . From the previous expression we conclude that ω1 2=−ω12=θ3. The gener- ator of this transformation is M12=idΛ(θ3) dω12/vextendsingle/vextendsingle/vextendsingle/vextendsingle ω12=0=−idΛ(θ3) dθ3/vextendsingle/vextendsingle/vextendsingle/vextendsingle θ3=0=i 0 000 00 −10 0 100 0 000 .(1.3) In the same way we obtain the other two generators: M13=i 000 0 000 −1 000 0 010 0 ,M 23=i 000 0 000 0 000 −1 001 0 . (1.4) In this case the relation between the parameters ωijand the angles of rotations θiaround xi−axis is θi=−1 2/epsilon1ijkωjk. The matrix of the boost along x−axis is Λ(ϕ1)= chϕ1−shϕ100 −shϕ1shϕ100 00 1 000 0 1 ≈I+ 0−ϕ 100 −ϕ100 0 00 0 000 0 0 , where ω 0 1=−ϕ1=−arc th v1. The corresponding generator is M01=idΛ(ϕ1) dω01/vextendsingle/vextendsingle/vextendsingle/vextendsingle ϕ1=0=idΛ(ϕ1) dϕ1/vextendsingle/vextendsingle/vextendsingle/vextendsingle ϕ1=0=−i 0100 10000000 0000 .(1.5) 70 Solutions The other two generators are M03=−i 0001 0000 0000 1000 ,M 02=−i 0010 0000 1000 0000 . (1.6) The boost parameters (rapidity) are ωoi=−ϕi=−arc th( vi), where viis the velocity of the inertial frame moving along the xi−axis. 1.10 The multiplication rule is (Λ1,a1)(Λ2,a2)=(Λ1Λ2,Λ1a2+a1). Unit element is ( I,0), while the inverse is ( Λ,a)−1=(Λ−1,−Λ−1a). 1.11 (a) Since this relation is valid in the deﬁning representation then it is also valid in any arbitrary representation. By using this relation one gets: U−1(Λ,0)(1 + i /epsilon1µPµ)U(Λ,0) = 1 + i( Λ−1)µ ν/epsilon1νPµ. (1.7) From the expression (1.7) we obtain U−1(Λ,0)PµU(Λ,0) = ( Λ−1)ν µPν. (1.8) The formula (1.8) is transformation law of the momentum Pµunder Lorentz transformations; the momentum is a four-vector. By substitut- ing U(ω,0) = exp/parenleftbigg −i 2Mµνωµν/parenrightbigg =1−i 2Mµνωµν+o(ω2) into (1.8) we get (1 +i 2Mρσωρσ)Pµ(1−i 2Mρσωρσ)=(δα µ−ωα µ)Pα, (1.9) and then iωρσ(MρσPµ−PµMρσ)=−ωρσ(gµσPρ−gµρPσ). (1.10) We had to antisymmetrize the right hand side of Equation (1.10) in order to eliminate antisymmetric parameters ωρσ. Finally, we obtain [Mρσ,Pµ]=i (gµσPρ−gµρPσ). (1.11) (b) If we take an inﬁnitesimal transformation Λ/prime=I+ω/primethen (Λ−1Λ/primeΛ)µ ν=δµ ν+(Λ−1)µ ρΛσ νω/primeρ σ, (1.12) Chapter 1. Lorentz and Poincar´ e symmetries 71 so that U−1(Λ,0)(1−i 2ω/primeρσMρσ)U(Λ,0) = 1 −i 2Mµν(Λ−1)µρΛσνω/prime ρσ.(1.13) From the last expression follows U−1(Λ,0)MρσU(Λ,0) = ( Λ−1)µ ρ(Λ−1)ν σMµν. (1.14) The last equation is the transformation law of the second rank tensor. For an inﬁnitesimal Lorentz transformation Λµ ν=δµ ν+ωµ νfrom Equation (1.14) follows i 2ωµν[Mµν,Mρσ]=1 2ωµν(gσµMρν−gρνMµσ−gσνMρµ+gρµMνσ), or [Mµν,Mρσ]=i (gσµMνρ+gρνMµσ−gρµMνσ−gσνMµρ). (1.15) (c) It is easy to prove that [Pµ,Pν]=0. (1.16) The relations (1.11), (1.15) and (1.16) are the commutation relations of the Poincar´ e algebra. 1.12 In the given representation the generator of the rotation around z–axis is M12=i 0 000 0 00 −100 0 100 00 000 0 0 000 0 . The time translation generator has the form T 0=−i 00001 0000000000 00000 00000 . The other generators have similar structure and they can be computed easily. The relations (1.11), (1.15) and (1.16) are fulﬁlled. 1.13 Under the Poincar´ e transformation x /prime=Λx+a≈x+δx , a classical scalar ﬁeld transforms as follows φ/prime(x+δx)=φ(x). 72 Solutions From the last relation we have φ/prime(x)=φ(x−δx)=φ(x)−δxµ∂µφ. (1.17) Form variation of a scalar ﬁeld is given by δ0φ=φ/prime(x)−φ(x)=−δxµ∂µφ. (1.18) For the Lorentz transformation δxµ=ωµ νxν,and therefore δ0φ=−ωµνxν∂µφ=−1 2ωµν(xν∂µ−xµ∂ν)φ. (1.19) On the other hand δ0φ=−i 2ωµνMµνφ. (1.20) By comparing two previous results we get that Lorentz’s generators are Mµν=i (xµ∂ν−xν∂µ). (1.21) For translations δxµ=/epsilon1µand δ0φ=−/epsilon1µ∂µφ=i/epsilon1µPµφ. (1.22) Hence Pµ=i∂µ. (1.23) Since [xµ∂ν,xρ∂σ]=gνρxµ∂σ−gσµxρ∂ν, (1.24) and [xµ∂ν,∂ρ]=−gρµ∂ν (1.25) we get the commutation relations of the Poincar´ e algebra: [Pµ,Pν]=0 [Mρσ,Pµ]=i (gµσPρ−gµρPσ) [Mµν,Mρσ]=i (gσµMνρ+gρνMµσ−gρµMνσ−gσνMµρ). 1.14 (a)WµPµ=1 2/epsilon1µνρσMνρPσPµ=0,since product of an antisymmetric with a symmetric tensor equals zero. Using the same argument, we obtain [Wµ,Pν]=0 . (b) Using the result of Problem 1.11 we obtain W2=1 4/epsilon1µνρσ/epsilon1µαβγMνρPσMαβPγ =1 4/epsilon1µνρσ/epsilon1µαβγMνρ/parenleftbig MαβPσ−iδσ βPα+iδσ αPβ/parenrightbig Pγ =1 4/epsilon1µνρσ/epsilon1µαβγMνρMαβPσPγ. (1.26) Chapter 1. Lorentz and Poincar´ e symmetries 73 The contraction of two /epsilon1symbols in the last line of (1.26) has been calcu- lated in 1.5 so that: W2=−1 4(δα νδβ ρδγ σ+δβ νδγ ρδα σ+δγ νδα ρδβ σ−δβ νδα ρδγ σ−δα νδγ ρδβ σ−δγ νδβ ρδα σ) ×MνρMαβPσPγ =−1 4/parenleftbig 2MνρMνρP2−MνρMνσPσPρ+MνρMσνPσPρ+ +MνρMρσPσPν−MνρMσρPσPν) =−1 2MνρMνρP2+MνρMνσPσPρ. (1.27) (c) Using the previous result we have [W2,Mρσ]=−1 2[MµνMµνP2,Mρσ]+[MµαMναPµPν,Mρσ].(1.28) The ﬁrst commutator in (1.28) we denote by A, while the second one by B. Using (1.15) we obtain that A= 0; this result is obvious since the P2 andMµνMµνare Lorentz scalars. The commutator Bis B=MµαMνα(Pµ[Pν,Mρσ]+[Pµ,Mρσ]Pν)+ +Mµα[Mνα,Mρσ]PµPν+[Mµα,Mρσ]MναPµPν.(1.29) Using the commutation relations (1.11) and (1.15) we get B= 0. There- fore, we have [W2,Mρσ]=0. 1.15 By using the result of Problem 1.14 (b) and Pµ|pµ,s,σ/angbracketright=pµ|pµ,s,σ/angbracketright we get W2|p=0,m,s,σ /angbracketright=−m2/parenleftbigg1 2MµνMµν−M0iM0i/parenrightbigg |p=0,m,s,σ /angbracketright =−1 2MijMijm2|p=0,m,s,σ /angbracketright =−m2/parenleftbig (M12)2+(M13)2+(M23)2/parenrightbig |p=0,m,s,σ /angbracketright =−m2J2|p=0,m,s,σ /angbracketright =−m2s(s+1 )|p=0,m,s,σ /angbracketright, because Ji=1 2/epsilon1ijkMjkare the components of the angular momentum tensor. 1.16 (a) Under Lorentz transformations Wµtransforms according to: U−1(Λ)WσU(Λ)=ΛσαWα. (1.30) From Equation (1.30) we have 74 Solutions i 2[Mµν,Wσ]ωµν=ωµνgσµWν=1 2(gσµWν−gσνWµ)ωµν. From the previous expression we easily obtain the requested result. (b) Using the result of the previous part we have [Wµ,Wν]=1 2/epsilon1µαβγ[MαβPγ,Wν] =1 2/epsilon1µαβγ/parenleftbig Mαβ[Pγ,Wν]+[Mαβ,Wν]Pγ/parenrightbig =i/epsilon1µανγWαPγ. 1.17 (a) Applying the result of Problem 1.16 (a) we get [Wµ,M2]=−2i(WαMαµ+MαµWα). (b) [Mµν,WµWν] = 0. Take care that δµ µ=4 . (c) Using the formula (1.11) we obtain [ M2,Pµ] = 2i( PαMαµ+MαµPα). This result and the result in the ﬁrst part of this Problem are similar, sinceWµandPµare both four-vectors. (d) [/epsilon1µνρσMµνMρσ,Mαβ]=0. 1.18 In the case of massive particles, m2>0 since the Lorentz transfor- mations, Λµν=δµ ν+ωµνleave pµinvariant (i.e. Λµνpν=pµ) the following relation is satisﬁed:  0ω01ω02 ω03 ω010−ω12−ω13 ω02ω12 0−ω23 ω03ω13ω23 0  m 0 00 = 0 0 00 . From here follows ω 01=ω02=ω03=0,ωij/negationslash=0. The corresponding generators are M12,M13andM23and they are gener- ators of the spatial rotations. Therefore, for massive particles little group is SO(3). The little group for the quantum mechanical Lorentz group, i.e. SL(2,C) group, is SO(3) = SU(2) . For massless particles we have  0ω01ω02 ω03 ω010−ω12−ω13 ω02ω12 0−ω23 ω03ω13ω23 0  k 0 0 k = 0 0 0 0 , which gives ω03=0,ω01=ω13,ω02=ω23while the parameter ω12is arbitrary. It corresponds to the rotation around z–axis. The generator of this transformation is M12.From the conditions derived above follows that there Chapter 1. Lorentz and Poincar´ e symmetries 75 are two independent generators M01+M13and−(M02+M23). Note that W1=(M02+M23)k,W2=−(M01+M13)kas well as W0=−M12k. Then, using Problem 1.16 (b) we obtain [W1,W2]=0,[W0/k,W 1]=−iW2,[W0/k,W 2]=iW1. These commutation relations deﬁne E(2) algebra. Thus, for massless particles little group is euclidian group E(2) in two dimensions. 1.19 It is easy to prove that Lorentz transformations, dilatations and SCT form a group. It is the conformal group, C(1,3). An arbitrary element of this group is U(ω,/epsilon1,ρ,c )=ei(Pµ/epsilon1µ−1 2Mµνωµν+ρD+cµKµ), where Dis generator of dilatation, and Kµare four generators for SCT . Conformal group has 15 parameters. The commutation relations of the algebracan be evaluated from multiplication rules of the group. Let ( Λ,a,ρ,c ) denote group element. If we start from (Λ −1,0,0,0)(I,0,0,c)(Λ,0,0,0) = ( I,0,0,Λ−1c) for inﬁnitesimal SCT we obtain U−1(Λ)KρU(Λ)=(Λ−1)µ ρKµ. For inﬁnitesimal Lorentz transformations we get: [Mµν,Kρ]=i (gνρKµ−gµρKν). (1.31) From U−1(Λ,0,0,0)U(I,0,ρ,0)U(Λ,0,0,0) =U(I,0,ρ,0),follows [Mµν,D]=0. (1.32) Starting from (I,0,ρ,0)−1(I,0,0,c)(I,0,ρ,0)xµ=(I,0,ρ,0)−1(I,0,0,c)e−ρxµ =(I,0,ρ,0)−1e−ρxµ+cµe−2ρx2 1+2 ( c·x)e−ρ+c2e−2ρx2 =xµ+cµe−ρx2 1+2 ( c·x)e−ρ+c2e−2ρx2 =(I,0,0,e−ρc)xµ, we obtain e−iρD(1 +iKµcµ)eiρD=1+i Kµe−ρcµ, for inﬁnitesimal SCT. From the last expression follows e−iρDKµeiρD=e−ρKµ. 76 Solutions This is the transformation law of SCT generators under dilatation. For inﬁn- itesimal dilatations we get: [D,Kµ]=−iKµ. (1.33) Similar procedure gives us the following commutators: [Pµ,D]=−iPµ, (1.34) [D,D]=0, (1.35) [Kµ,Kν]=0, (1.36) [Pµ,Kν] = 2i( gµνD+Mµν). (1.37) Equations (1.31)–(1.37) together with (1.11), (1.15) and (1.16) are commuta- tion relations of the conformal algebra. 2 The Klein–Gordon equation 2.1A particular solution of the Klein–Gordon equation (/unionsq /intersectionsq+m2)φ(x)=0 , (2.1) is plane wave, e−ik·x=e−iEt+ik·x, (2.2) where Eandkare energy and momentum respectively. We see that from i∂ ∂te−ik·x=Ee−ik·x, and −i∇e−ik·x=ke−ik·x. By inserting the solution (2.2) into (2.1) we obtain k2=m2i.e.E= ±√ k2+m2=±ωk. Therefore, the plane wave (2.2) is a solution of the Klein– Gordon equation if the previous relation is satisﬁed. For momentum kthere are two independent solutions e−iωkt+ik·xand e+iωkt+ik·x.The general solution of (2.1) is φ(x)=1 (2π)3/2/integraldisplayd3k√2ωk/parenleftBig a(k)e−i(ωkt−k·x)+b†(−k)ei(ωkt+k·x)/parenrightBig ,(2.3) where a(k)a n d b†(k) are complex coeﬃcients. In the second term in (2.3) we make the following change k→−k. After that the formula (2.3) becomes φ(x)=1 (2π)3/2/integraldisplayd3k√2ωk/parenleftbig a(k)e−ik·x+b†(k)eik·x/parenrightbig , (2.4) where kµ=(ωk,k).Ifφ(x) is a real ﬁeld then a(k)=b(k). 2.2Using (2.4) we get 78 Solutions Q=iq/integraldisplay d3x/parenleftbigg φ∗∂φ ∂t−φ∂φ∗ ∂t/parenrightbigg =iq 2(2π)3/integraldisplayd3xd3kd3k/prime √ωkωk/prime/bracketleftbig/parenleftbig a†(k)eik·x+b(k)e−ik·x/parenrightbig ×/parenleftBig −iωk/primea(k/prime)e−ik/prime·x+iωk/primeb†(k/prime)eik/prime·x/parenrightBig −/parenleftbig a(k)e−ik·x+b†(k)eik·x/parenrightbig ×/parenleftBig iωk/primea†(k/prime)eik/prime·x−iωk/primeb(k/prime)e−ik/prime·x/parenrightBig/bracketrightBig . (2.5) By integrating over xin (2.5), we obtain Q=−q 2/integraldisplay d3kd3k/prime/radicalbiggωk/prime ωk/parenleftBig −a†(k)a(k/prime)ei(ωk−ωk/prime)tδ(3)(k−k/prime) +a†(k)b†(k/prime)ei(ωk+ωk/prime)tδ(3)(k+k/prime)−b(k)a(k/prime)e−i(ωk+ωk/prime)tδ(3)(k+k/prime) +b†(k)b(k/prime)e−i(ωk−ωk/prime)tδ(3)(k−k/prime)+c.c./parenrightBig . (2.6) where c .c.denotes complex conjugation. If in expression (2.6) we integrate over the momentum k/primewe obtain Q=q 2/integraldisplay d3k/bracketleftbig a†(k)a(k)+a(k)a†(k)−b†(k)b(k)−b(k)b†(k)/bracketrightbig .(2.7) In the result (2.7) we do not take care about ordering of a(k),a†(k)a n d b(k),b†(k) since they are complex numbers. This will be diﬀerent in the Chap- ter 7 where a(k)a n d b†(k) are going to be operators. 2.3If we ﬁrst integrate over xwe get H=−1 4/integraldisplayd3kd3k/prime √ωkωk/prime/parenleftBig a(k)a(k/prime)(ωkωk/prime+k·k/prime−m2)e−i(ωk+ωk/prime)tδ(3)(k+k/prime) +a†(k)a†(k/prime)(ωkωk/prime+k·k/prime−m2)ei(ωk+ωk/prime)tδ(3)(k+k/prime) −a(k)a†(k/prime)(ωkωk/prime+k·k/prime+m2)e−i(ωk−ωk/prime)tδ(3)(k−k/prime) −a†(k)a(k/prime)(ωkωk/prime+k·k/prime+m2)ei(ωk−ωk/prime)tδ(3)(k−k/prime)/parenrightBig . (2.8) Performing integration over momentum k/prime, and using the relation k2+m2= ω2 k, we obtain H=1 2/integraldisplay d3kωk/parenleftbig a†(k)a(k)+a(k)a†(k)/parenrightbig . (2.9) 2.4Solution of this problem is very similar to the solutions of the previous two. The result is P=/integraldisplay d3kka†(k)a(k). 2.5The four-divergence of the current jµis Chapter 2. The Klein–Gordon equation 79 ∂µjµ=−i 2(∂µφ∂µφ∗+φ/unionsq /intersectionsqφ∗−∂µφ∂µφ∗−φ∗/unionsq /intersectionsqφ). Using the equations of motion we obtain the requested result ∂µjµ=0 . 2.6It is easy to see that ∂µjµ=−i 2(∂µφ∂µφ∗+φ/unionsq /intersectionsqφ∗−∂µφ∂µφ∗−φ∗/unionsq /intersectionsqφ)− −q(φAµ∂µφ∗+φφ∗∂µAµ+φ∗Aµ∂µφ). (2.10) The equations of motion are /bracketleftbig /unionsq/intersectionsq−iq(∂µAµ+2Aµ∂µ−iqAµAµ)+m2/bracketrightbig φ∗(x)=0 , (2.11) /bracketleftbig /unionsq /intersectionsq+iq(∂µAµ+2Aµ∂µ+iqAµAµ)+m2/bracketrightbig φ(x)=0 . (2.12) If we multiply Equation (2.11) by φand Equation (2.12) by φ∗and then subtract obtained equations we get φ/unionsq /intersectionsqφ∗−φ∗/unionsq /intersectionsqφ−2iq(φφ∗∂µAµ+Aµφ∗∂µφ+Aµφ∂µφ∗)=0 . Combining the previous expression and (2.10), one easily obtains ∂µjµ=0. 2.7The equation of motion for a scalar particle in a electromagnetic ﬁeld is /bracketleftbig (∂µ+iqAµ)(∂µ+iqAµ)+m2/bracketrightbig φ(x)=0 . (2.13) In the region r<a Equation (2.13) becomes /bracketleftbigg/parenleftbigg∂ ∂t−iV/parenrightbigg/parenleftbigg∂ ∂t−iV/parenrightbigg −∆+m2/bracketrightbigg φ(x)=0 . (2.14) For stationary states φ(x)=e−iEtF(r) one gets /bracketleftbig −(E+V)2−∆+m2/bracketrightbig F(r)=0 . (2.15) If we assume that a solution of the previous equation is given by F=f(r) rQ(θ,ϕ), then from (2.15) we get the following two equations: d2f dr2+/bracketleftbig (E+V)2−m2/bracketrightbig f=l(l+1 ) r2f, (2.16) 1 sinθ∂ ∂θ/parenleftbigg sinθ∂Q ∂θ/parenrightbigg +1 sin2θ∂2Q ∂ϕ2=−l(l+1 )Q. (2.17) 80 Solutions The particular solutions of (2.17) are spherical harmonics, Ylm. In the case l= 0, the corresponding spherical harmonic Y00is a constant. The solution of (2.16) is f=Asin(qr)+Bcos(qr), (2.18) where q2=[ (E+V)2−m2]>0. (2.19) Constant Bhas to be zero since function f(r)/rshould not be singular in the r→0 limit. In the region r>a (A0= 0) the solution is given by f=Ce−kr+Dekr, (2.20) where k2=m2−E2.But, the constant Dhas to be zero since the wave function has to be ﬁnite in the large rlimit. Therefore, the wave function is φ<=Asinqr r,r < a (2.21) φ>=Ce−kr r,r > a . (2.22) Atr=awe should apply the continuity conditions: φ<(a)=φ>(a)a n d φ/prime <(a)=φ/prime >(a) for the wave function and its ﬁrst derivative. These boundary conditions give: Asin(qa)−Ce−ka=0, (2.23) Aqcos(qa)+Cke−ka=0. (2.24) The homogenous system (2.23–2.24) has non-trivial solutions if and only if its determinant is equal to zero. Finally, we obtain the condition tan(qa) q=−1 k. (2.25) The dispersion relation (2.25) will be analyzed graphically in the case V<2m. Solid line in Fig. 2.1 is function tan( qa)/qwhile dashed line is f(q)=−1 k=−1/radicalBig 2V/radicalbig q2+m2−V2−q2. There is only one bound state (in case V<2m) if the condition π 2a</radicalbig V(V+2m)≤3π 2a. is satisﬁed. 2.8The wave equation is Chapter 2. The Klein–Gordon equation 81 Fig. 2.1. Graphical solution of the dispersion relation (2.25) for V<2m /bracketleftBigg ∂2 ∂t2−/parenleftbigg∂ ∂x+iqBy/parenrightbigg2 −∂2 ∂y2−∂2 ∂z2+m2/bracketrightBigg φ(x)=0 . (2.26) It is easy to see that the operators ˆ px=−i∂ ∂xand ˆpz=−i∂ ∂zcommute with the Hamiltonian, so we can assume that the solution of (2.26) has the following form φ=e−i(Et−kxx−kzz)ϕ(y). (2.27) From (2.26) and (2.27) we get /parenleftbiggd2 dy2−(kx+qBy)2+E2−k2 z−m2/parenrightbigg ϕ(y)=0 . (2.28) Introducing the new variable ξ=kx+qBy, Equation (2.28) takes the same form as the Schr¨ odinger equation for the oscillator /parenleftbiggd2 dξ2−1 (qB)2ξ2+E2−k2 z−m2 (qB)2/parenrightbigg ˜ϕ(ξ)=0 . Then the energy levels are En=/radicalbig m2+k2z+( 2n+1 )qB , n =0,1,2,... . Eigenfunctions are φn(x)=(qπB)−1/41√ 2nn!e−iEnt+ikxx+ikzze−(kx+qBy)2/2qBHn(kx+qBy√qB), (2.29) where Hnare the Hermite polynomials. 2.9In the region z>0 the equation of motion is /bracketleftbigg /unionsq/intersectionsq−q2U2 0+2 iqU0∂ ∂t+m2/bracketrightbigg φII(x)=0 . (2.30) 82 Solutions Substituting φII=Ce−iEt+ikzin (2.30), we get k=±K=±/radicalbig (E−qU0)2−m2, (2.31) or E=±/radicalbig k2+m2+qU0. (2.32) Forz<0 the particle is free and the solution is φI=Ae−iEt+ipz+Be−iEt−ipz, (2.33) where p=√ E2−m2.The ﬁrst term in (2.33) is the incident wave, the second one is the reﬂected wave. At z= 0 we have to apply the continuity conditions: φI(0) = φII(0),φ/prime I(0) = φ/prime II(0). They give A=1 2/parenleftbigg 1+k p/parenrightbigg C, B =1 2/parenleftbigg 1−k p/parenrightbigg C. (2.34) We will separately discuss three diﬀerent possibilities: Case 1: E>m +qU0. For this value of energy the sign in the expressions (2.31) and (2.32) is plus. The formula for the current has been given in Problem 2.5. The reﬂection coeﬃcient is R=−(jr)z (jin)z=|B|2 |A|2=/vextendsingle/vextendsingle/vextendsingle/vextendsinglep−K p+K/vextendsingle/vextendsingle/vextendsingle/vextendsingle2 , while the transmission coeﬃcient is T=1−R. Case 2: E<−m+qU0. In this case the momentum is negative, k=−K. The reﬂection coeﬃcient is diﬀerent comparing to the previous case: R=/vextendsingle/vextendsingle/vextendsingle/vextendsinglep+K p−K/vextendsingle/vextendsingle/vextendsingle/vextendsingle2 ,T=1−R. As we immediately see the reﬂection coeﬃcient is larger than 1: the potential is strong enough to create particle–antiparticle pairs. The antiparticles aremoving to the right producing a negative charge current and therefore we obtain negative transmission coeﬃcient. This is Klein paradox Case 3: |E−qU 0|<m. We leave to the reader to show that in this case R=1,T=0. 2.10 Forz<0a n d z>0 a wave function satisﬁes the free Klein–Gordon equation, while in the region 0 <z<a the equation is /bracketleftbigg /unionsq/intersectionsq−q2U2 0+2 iqU0∂ ∂t+m2/bracketrightbigg φII(x)=0 . The solution is given by: Chapter 2. The Klein–Gordon equation 83 φI=Ae−iEt+ipz+Be−iEt−ipz, φII=Ce−iEt+ikz+De−iEt−ikz φIII=Fe−iEt+ipz, (2.35) where k=/radicalbig (E−qU0)2−m2andp=√ E2−m2. From the continuity con- ditions follows: A+B=C+D, A−B=k p(C−D), Ceika+De−ika=Feipa, Ceika−De−ika=p kFeipa. (2.36) Thus, one gets: T=/vextendsingle/vextendsingle/vextendsingleF A/vextendsingle/vextendsingle/vextendsingle2 =16 |2+p k+k p+( 2−p k−k p)e2ika|2. If (E−qU0)2−m2<0 the momentum kbecomes imaginary, i.e. k=iκ=i/radicalbig m2−(E−qU0)2. 2.11 The Klein–Gordon equation for a particle in the Coulomb potential is /bracketleftBigg/parenleftbigg∂ ∂t−ieZe r/parenrightbigg2 −∆+m2/bracketrightBigg φ(x)=0 . (2.37) By substituting φ=e−iEtR(r)Y(θ,ϕ) in (2.37) and using (2.17) we obtain: −1 2m1 rd2 dr2(rR)+l(l+1 )−Z2e4 2mr2R−Ze2E mrR=E2−m2 2mR. This equation has the same form as the Schr¨ odinger equation for hydrogen atom. By comparing these equations we get En,l=m1/radicalbigg 1+Z2e4(n−l−1 2)+/radicalBig (l+1 2)2−Z2e4. In the nonrelativistic limit the result is En−m=−mZ2e4 2n2−Z3e6m n3/parenleftbigg1 2l+1−3 8n/parenrightbigg . 2.12 The Klein–Gordon equation in the Schr¨ odinger form is i∂ ∂t/parenleftbigg θ χ/parenrightbigg =H/parenleftbigg θ χ/parenrightbigg , (2.38) 84 Solutions where the Hamiltonian is given by H=/bracketleftbigg −∆ 2m/parenleftbigg 11 −1−1/parenrightbigg +m/parenleftbigg 10 0−1/parenrightbigg/bracketrightbigg . 2.13 The eigenequation, Hφ=Eφin the momentum representation takes the following form /parenleftBigg p2 2m+mp2 2m −p2 2m−p2 2m−m/parenrightBigg/parenleftbigg θ0 χ0/parenrightbigg =E/parenleftbigg θ0 χ0/parenrightbigg . (2.39) The eigenvalues of the Hamiltonian are evaluated easily and they are E= ±ωp=±/radicalbig p2+m2. In order to ﬁnd nonrelativistic limit we suppose that the solution has the following form/parenleftbigg θ χ/parenrightbigg =/parenleftbigg θ0 χ0/parenrightbigg e−i(m+T)t, (2.40) where Tis the kinetic energy of the particle. From (2.38) we get /parenleftbigg−/triangle 2m+m−/triangle 2m /triangle 2m/triangle 2m−m/parenrightbigg/parenleftbigg θ0 χ0/parenrightbigg =(m+T)/parenleftbigg θ0 χ0/parenrightbigg , (2.41) i.e. /parenleftbigg −/triangle 2m+m/parenrightbigg θ0−/triangle 2mχ0=(m+T)θ0, /triangle 2mθ0+/parenleftbigg/triangle 2m−m/parenrightbigg χ0=(T+m)χ0. (2.42) From the second equation in (2.42) we obtain χ0≈/triangle 4m2θ0, (2.43) in nonrelativistic limit. Using this the ﬁrst equation in (2.42) becomes Tθ0=/parenleftbigg −/triangle 2m−/triangle2 8m3/parenrightbigg θ0. (2.44) Also, from (2.43) we see that χ0/lessmuchθ0andχis so called small component. From the expression (2.44) follows that ﬁrst relativistic correction of nonrel- ativistic Hamiltonian is −∇4/8m3. 2.14 Velocity operator is v=∂H ∂p=p m/parenleftbigg 11 −1−1/parenrightbigg . The eigenvalue of the velocity operator is zero. 2.15 Show that/angbracketleftbig ψ†|Hψ/angbracketrightbig =/angbracketleftbig Hψ†|ψ/angbracketrightbig . The average value is /angbracketleftv/angbracketright=p m. 3 Theγ–matrices 3.1 (a) In the Dirac representation of γ–matrices we have (γ0)†=/parenleftbigg I0 0−I/parenrightbigg† =/parenleftbigg I0 0−I/parenrightbigg =γ0γ0γ0=γ0, (γi)†=/parenleftbigg 0σi −σi0/parenrightbigg† =−/parenleftbigg 0σi −σi0/parenrightbigg =−γ0γ0γi=γ0γiγ0, where we used the facts that ( γ0)2=1 ,γ0andγianticommute, and the Pauli matrices are hermitian. This relation is true in any representation ofγ–matrices which is obtained by a unitary transformation from the Dirac representation. (b) Using the previous result we ﬁnd σ † µν=−i 2(γµγν−γνγµ)† =−i 2(γ† νγ† µ−γ† µγ† ν) =−i 2γ0(γνγµ−γµγν)γ0 =γ0σµνγ0. 3.2 (a) Taking the adjoint of γ5we obtain γ† 5=iγ† 3γ† 2γ† 1γ† 0 =iγ0γ3γ0γ0γ2γ0γ0γ1γ0γ0γ0γ0 =iγ0γ3γ2γ1 =−iγ0γ1γ2γ3=γ5. 86 Solutions The property γ−1 5=γ5c a nb ep r o v e db yu s i n g γ−1 0=γ0andγ−1 i= −γi=γi. Both of these relations follow from anticommutation relations {γµ,γν}=2δν µ. (b) Using the deﬁnition of the /epsilon1symbol we ﬁnd −i 4!/epsilon1µνρσγµγνγργσ=i 4!(γ0γ1γ2γ3−γ0γ1γ3γ2+...+γ3γ2γ1γ0) =iγ0γ1γ2γ3=γ5. (c) This is a consequence of (a) result. (d) In a similar manner, we have: (γ5γµ)†=γ† µγ† 5=γ0γµγ0γ5=γ0γ5γµγ0. 3.3 (a) For µ=0w eh a v e {γ5,γ0}=γ5γ0+γ0γ5 =−iγ0γ1γ2γ3γ0−iγ0γ0γ1γ2γ3 =iγ1γ2γ3−iγ1γ2γ3=0, (3.1) and similarly for other three cases. (b) By a straightforward calculation one gets: [σµν,γ5]=i 2[γµγν−γνγµ,γ5] =i 2(γµ{γν,γ5}−{γµ,γ5}γν−γν{γµ,γ5}+{γµ,γ5}γν) =0 since{γµ,γ5}=0. 3.4/a/a=aµaνγµγν=1 2aµaν(γµγν+γνγµ)=gµνaµaν=a2 3.5 (a) From the relation {γµ,γµ}=2γµγµ=2δµ µ= 8 it follows that γµγµ=4 . (b)γµγνγµ=( 2gµν−γνγµ)γµ=2γν−4γν=−2γν. (c)γµγαγβγµ=( 2gµα−γαγµ)γβγµ=2γβγα+2γαγβ=4δβ α,where we used the second part of this problem and (3.A). (d) By commuting γµandγαand making use of the previous result, one gets: γµγαγβγγγµ=( 2δα µ−γαγµ)γβγγγµ =2γβγγγα−4γαgβγ =−2(2gβγ−γβγγ)γα =−2γγγβγα. Chapter 3. The γ–matrices 87 (e) By using the deﬁnition σµν–matrices, one obtains: σµνσµν=−1 4(γµγνγµγν−γµγνγνγµ−γνγµγµγν+γνγµγνγµ). By using parts (a) and (b) of this problem, one gets σµνσµν= 12. (f) Use Problem 3.3 and parts (a) and (b) of this problem. (g) By direct calculation, one ﬁnds σαβγµσαβ=−1 4(γαγβγµγαγβ−γαγβγµγβγα −γβγαγµγαγβ+γβγαγµγβγα) =−1 4(4δβ µγβ−4γµ−4γµ+4gµβγβ)=0 . (h) σαβσµνσαβ=−i 8(γαγβγµγνγαγβ−γαγβγµγνγβγα −γαγβγνγµγαγβ+γαγβγνγµγβγα−γβγαγµγνγαγβ +γβγαγµγνγβγα+γβγαγνγµγαγβ−γβγαγνγµγβγα) =−i 8(−8γνγµ−16gµν+8γµγν +16gµν−16gµν−8γνγµ+1 6gµν+8γµγν) =−2i(γµγν−γνγµ)=−4σµν. (i) Use part (g) of this problem. (j)σµνγ5σµν=i 2(γµγν−γνγµ)γ5σµν=γ5σµνσµν=1 2γ5. 3.6 (a) By using the trace property tr( A1A2...A n)=t r ( A2A3...A nA1), Prob- lem 3.3(a), and ( γ5)2= 1, it follows that tr(γµ)=t r ( γµγ5γ5) =−tr(γ5γµγ5) =−tr((γ5)2γµ) =−tr(γµ). From the previous expression we get tr( γµ)=0 . (b) Taking trace of the relation {γµ,γν}=2gµν,we easy obtain the requested result. (c) By applying the basic anticommutation relation (3.A), one gets: tr(γµγνγργσ) = tr [(2 gµν−γνγµ)γργσ] =2gµνtr(γργσ)−tr[γν(2gµρ−γργµ)γσ] =2gµνtr(γργσ)−2gµρtr(γνγσ)+2gµσtr(γνγρ) −tr(γνγργσγµ). 88 Solutions From the previous part of this problem and relation tr( γµγνγργσ)= tr(γνγργσγµ),one easily obtains the requested result. (d) trγ5=t r (γ5γ0γ0)=−tr(γ0γ5γ0), where we used Problem 3.3 (a). Further, from the trace property and ( γ0)2= 1 it follows that: trγ5=−tr(γ0γ0γ5)=−trγ5, which implies tr γ5=0 . (e) Since γαγα=4 ,w eh a v e tr(γ5γµγν)=1 4tr(γ5γαγαγµγν) =1 4tr(γαγµγνγ5γα) =−1 4tr(γ5γαγµγνγα) =−gµνtr(γ5)=0 . In the previous calculation we used the trace property and Problem 3.5 (c). (f) The quantity tr( γ5γµγνγργσ) is an antisymmetric tensor with respect to the indexes ( µ, ν, ρ, σ ). Thus, it must be proportional to the Levi-Civita tensor. The constant of proportionality can be determined by substituting µ=0,ν=1,ρ=2a n d σ=3. (g) From ( γ5)2=1 ,{γ5,γµ}= 0 and the trace property follows: tr(/a1.../a2n+1)=t r ( γ5γ5/a1···/a2n+1) =(−1)2n+1tr(γ5/a1···/a2n+1γ5) =−tr(γ5γ5/a1···/a2n+1) =−tr(/a1.../a2n+1). Hence, tr(/ a1.../a2n+1)=0 . (h) tr(/ a1···/a2n)=t r ( C/a1C−1C···C−1C/a2nC−1),where the matrix Csat- isﬁes the relation CγµC−1=−γT µ.Thus, tr(/a1···/a2n)=(−1)2ntr(/aT 1···/aT 2n) = tr(/ a2n···/a1). (i) tr( γ5γµ)=−itr(γ0γ1γ2γ3γµ)=0,since it is the trace of odd number of γ–matrices. 3.7 tr(/a1/a2···/a6)= 4{(a1·a2)[(a3·a4)(a5·a6)−(a3·a5)(a4·a6)+(a3·a6)(a4·a5)] −(a1·a3)[(a2·a4)(a5·a6)−(a2·a5)(a4·a6)+(a2·a6)(a4·a5)] +(a1·a4)[(a2·a3)(a5·a6)−(a2·a5)(a3·a6)+(a2·a6)(a3·a5)] −(a1·a5)[(a2·a3)(a4·a6)−(a2·a4)(a3·a6)+(a2·a6)(a3·a4)] +(a1·a6)[(a2·a3)(a4·a5)−(a2·a4)(a3·a5)+(a2·a5)(a3·a4)]}. Chapter 3. The γ–matrices 89 3.84/bracketleftbig pµqν−(p·q)gµν+pνqµ+i/epsilon1αµβνpαqβ−m2gµν/bracketrightbig . 3.9−2/p−2γ5/p−4m−4mγ5. 3.10 Expanding the exponential function in series, we ﬁnd eγ5/a=1+( γ5/a)+1 2(γ5/a)2+1 3!(γ5/a)3+···. (3.2) By substituting ( γ5/a)2=−a2,(γ5/a)3=−a2(γ5/a),...into expression (3.2), we get eγ5/a=( 1−a2 2!+a4 4!+···)+(γ5/a)(1−a2 3!+a4 5!−···) =c o s (√ a2)+1√ a2sin(√ a2)γ5/a, where a2=aµaµ. 3.11 The fact that the product of any two Γ–matrices is again a Γmatrix (modulo ±1,±i) can be proved directly. For example, γ5σ01=−iσ23. Now, we shall prove that Γ–matrices are linearly independent. Multiplying the relation/summationtext acaΓa=0b y Γb=(Γb)−1,we obtain cbΓbΓb+/summationdisplay a/negationslash=bcaΓaΓb=0, where the b–term is separated. Using the ordering lemma, the last expression becomes cbI+/summationdisplay d,Γd/negationslash=IcdηΓd=0, (3.3) where η∈{ ±1,±i}.After taking trace of (3.3) and using the fact that tr(Γa)=/braceleftbigg 0,Γa/negationslash=I 4,Γa=I, one obtains cb=0(∀b). This means that Γ–matrices are linearly independent one. 3.12 Multiplying the equation A=/summationtext acaΓabyΓbfrom the right and sepa- rating the b–term in the sum, we have AΓb=cbΓbΓb+/summationdisplay a/negationslash=bcaΓaΓb=cbI+/summationdisplay d,Γd/negationslash=IcdηΓd. Taking the trace of previous relation we obtain the requesting relation. 3.13 The coeﬃcients can be calculated by using the formula obtained in the previous problem. 90 Solutions (a) From the traces (which were actually calculated in Problem 3.6): tr(γµγνγρ)=0 , tr(γµγνγργσ)=4 ( gµνgρσ−gµρgνσ+gµσgνρ), tr(γµγνγργσγ5)=−4i/epsilon1µνρσ, tr(γµγνγργ5)=t r ( γµγνγρσαβ)=0 , follows γµγνγρ=(gµνgρσ−gµρgσν+gµσgρν)γσ+i/epsilon1σµνργ5γσ. (b)γ5γµγν=gµνγ5+1 2/epsilon1αβ µνσαβ, (c)σµνγργ5=/epsilon1αµνργα−igνργ5γµ+igµργ5γν. 3.14 From Problem 3.13 (a), it follows that {γµ,σνρ}=−2/epsilon1αµνργ5γα. 3.15 By applying the result of Problem 3.13 (a) the trace can be transformed as follows tr(γµγνγργσγαγβγ5)=(gµνgρδ−gµρgνδ+gµδgρν)tr(γδγσγαγβγ5) +i/epsilon1δµνρtr(γδγσγαγβ). Using 3.6 (c), (f), we get tr(γµγνγργσγαγβγ5) = 4i( −gµν/epsilon1ρσαβ+gµρ/epsilon1νσαβ −gρν/epsilon1µσαβ+gαβ/epsilon1σµνρ−gσβ/epsilon1αµνρ+gσα/epsilon1βµνρ). 3.16 Use the solution of Problem 3.13 (b). 3.17 Applying the formulae [A,BC ]=[A,B]C+B[A,C], and [AB,C ]=A{B,C}−{A,C}B, as well as the anticommutation relations (3.A), we obtain [γµγν,γργσ]=γµ{γν,γρ}γσ−{γµ,γρ}γνγσ +γργµ{γν,γσ}−γρ{γµ,γσ}γν =2gνργµγσ+2gνσγργµ−2gµσγργν−2gµργνγσ. From the above result we obtain: [σµν,σρσ] = 2i( gνρσµσ+gµσσνρ−gµρσνσ−gνσσµρ). The matrices1 2σµνare generators of the Lorentz group in the spinor repre- sentation. 3.18 LetMbe a matrix which commutes with all γ–matrices. Using the Problem 3.11, we can write ( Γb/negationslash=I) Chapter 3. The γ–matrices 91 M=cbΓb+/summationdisplay a/negationslash=bcaΓa. (3.4) On the other hand, we know that there is always a matrix Γdwhich anticom- mute with Γb/negationslash= I. Multiplying the expression (3.4) by matrix Γdfrom the left, and by Γdfrom the right, we get: ΓdMΓd=−cbΓb+/summationdisplay a/negationslash=bηcaΓa. (3.5) The matrix Mcommutes with γµ, and therefore with Γd,s ow eg e t M=−cbΓb+/summationdisplay a/negationslash=bηcaΓa. (3.6) If we now multiply equations (3.4) and (3.6) by Γband take trace of the re- sulting expressions, we get cb= 0. So, each of the coeﬃcients in the expansion (3.4) is equal to zero except the unit matrix coeﬃcient. 3.19 By applying the Baker–Hausdorﬀ formula eBAe−B=A+[B,A]+1 2![B,[B,A]] +··· we get UαU†=α+2βn−2(n·α)n−8 3!βn+16 4!(α·n)n+··· =α+∞/summationdisplay k=1(−1)k22k (2k)!(α·n)n+∞/summationdisplay k=0(−1)k22k+1 (2k+ 1)!βn,(3.7) since [βα·n,αi]=nj(β{αj,αi}−{β,αi}αj)=2βni, [βα·n,[βα·n,αi]] =−4(α·n)ni, [βα·n,[βα·n,[βα·n,αi]]] =−8βni, [βα·n,[βα·n,[βα·n,[βα·n,αi]]]] = 16( α·n)ni,etc. On the other hand, we have the following identities ( βα·n)2=−1,(βα·n)3= −(βαn),(βα·n)4=1,...so that α+(U2−I)(α·n)n=α+2βn−2(α·n)n−8 3!βn+··· =α+∞/summationdisplay k=1(−1)k22k (2k)!(α·n)n+∞/summationdisplay k=0(−1)k22k+1 (2k+ 1)!βn. (3.8) It is clear that the results (3.7) and (3.8) are equal. 92 Solutions 3.20 It is straightforward to show that the γ–matrices satisfy the relation {γµ,γν}=2gµν.The connection with Dirac representation γDirac µis given by γµS=SγDirac µ. (3.9) This statement is known as the fundamental (Pauli) theorem. If we substitute S=/parenleftbigg ab cd/parenrightbigg ,where a,b,c,d are 2×2 matrices, into (3.9) we ﬁnd /parenleftbigg cd ab/parenrightbigg =/parenleftbigg a−b c−d/parenrightbigg ,/parenleftbigg −σic−σid σiaσib/parenrightbigg =/parenleftbigg bσi−aσi dσi−cσi/parenrightbigg .(3.10) The solution of (3.10) is a=−b=c=d= I. A particular solution for Sis given by S=1√ 2/parenleftbigg I−I II/parenrightbigg . The matrices σµνare σoi=−i/parenleftbigg −σi0 0σi/parenrightbigg ,σ ij=/epsilon1ijk/parenleftbigg σk0 0σk/parenrightbigg , (3.11) while γ5=iγ0γ1γ2γ3=/parenleftbigg −I0 0I/parenrightbigg . (3.12) 3.21 Matrices γ0=σ1=/parenleftbigg 01 10/parenrightbigg and γ1=−iσ2=/parenleftbigg 0−1 10/parenrightbigg have the following properties: (γ0)2=1,(γ1)2=−1,γ0γ1=−γ1γ0, hence, they satisfy the Cliﬀord algebra (3.A). The matrix γ5is deﬁned by γ5=γ0γ1=/parenleftbigg 10 0−1/parenrightbigg . tr(γ5γµγν) is an antisymmetric tensor and it should be proportional to /epsilon1µν: tr(γ5γµγν)=C/epsilon1µν. By ﬁxing µ=0,ν= 1 we obtain1C= 2. One can easily show that γ5γµ=/epsilon1µνγν. 1Our sign convention is /epsilon101=+ 1 . 4 The Dirac equation 4.1In terms of αandβmatrices, the Dirac Hamiltonian has the form HD=α·p+βm, so that: (a) [HD,p]=0 , (b) [HD,Li]=/epsilon1ijk[α·p+βm,xjpk]=/epsilon1ijkαl[pl,xj]pk=−i/epsilon1ijkαjpk=i (p× α)i, (c) [HD,L2]=−i/epsilon1ijkαj(Lipk+pkLi)/negationslash=0 , (d) [HD,Si]=−i 4[HD,/epsilon1ijkαjαk]=i/epsilon1ijkpkαj=−i(p×α)i, (e) By applying (b) and (d) we get that this commutator vanishes.(f) [H D,J2]=0 , (g) From (d) we have [ HD,Σ·ˆp]=−i 2|p|/epsilon1ijkpjαkpi=0 , (h) If vectors nandpare collinear, then the commutator vanishes. In the opposite case it is not zero. 4.2The plane wave ψ=/parenleftbigg ϕ χ/parenrightbigg e−ip·x, (4.1) is a particular solution of the Dirac equation, (iγµ∂µ−m)ψ(x)=0 . (4.2) By substituting (4.1) in (4.2) (in the Dirac representation of γ–matrices) we obtain /parenleftbigg E−m−σ·p σ·p−E−m/parenrightbigg/parenleftbigg ϕ χ/parenrightbigg =0, (4.3) where Eandpare the energy and momentum of the particle, respectively. Nontrivial solutions of the homogeneous system (4.3) exist if and only if its determinant vanishes. This gives the following relation between energy andmomentum: E=±/radicalbig p2+m2=±Ep,which tells us that there are solutions of positive and negative energy as we expected. 94 Solutions For the positive energy solution, E=Ep, the system (4.3) has the following form: (Ep−m)ϕ−(σ·p)χ=0, (σ·p)ϕ−(Ep+m)χ=0. (4.4) These relations imply: χ=σ·p Ep+mϕ, (4.5) or u(Ep,p)=/parenleftbigg ϕ χ/parenrightbigg =/parenleftbiggϕ σ·p Ep+mϕ/parenrightbigg , (4.6) where ϕis arbitrary. For the negative energy solution, E=−Ep, the system (4.3) is solved by u(−Ep,p)=/parenleftbigg ϕ χ/parenrightbigg =/parenleftbigg −σ·p Ep+mχ χ/parenrightbigg . (4.7) If we introduce the notation v(p)=u(−Ep,−p)a n d u(p)=u(Ep,p), linearly independent solutions of Equation (4.2), for ﬁxed p, are given as u(p)e−ip·x,v(p)eip·x, where pµ=(Ep,p). Note the change of sign in the negative energy solu- tion. The energy and momentum of the solution u(p)e−ip·xareEpandp, respectively, while for v(p)eip·x,they are −Epand−p. In order to ﬁnd the additional degrees of freedom, let us recall that the helicity operator1 2Σ·ˆp, where ˆp=p/|p|, commutes with the Dirac Hamiltonian [see Problem 4.1 (g)]. From the eigenequation σ·ˆpϕ=±ϕ, (and a similar equation for χ) we obtain ϕ1=1/radicalbig 2(1 + ˆ p3)/parenleftbigg ˆp3+1 ˆp1+iˆp2/parenrightbigg ,ϕ 2=1/radicalbig 2(1 + ˆ p3)/parenleftbigg −ˆp1+iˆp2 ˆp3+1/parenrightbigg ,(4.8) (and similarly for χr,r=1,2). If we take p=pez, the basis vectors become /parenleftbigg 1 0/parenrightbigg ,/parenleftbigg 0 1/parenrightbigg . (4.9) Then, the basis bispinors are Chapter 4. The Dirac equation 95 u1(p)=Np /parenleftbigg 1 0/parenrightbigg σ·p Ep+m/parenleftbigg 1 0/parenrightbigg ,u2(p)=Np /parenleftbigg 0 1/parenrightbigg σ·p Ep+m/parenleftbigg 0 1/parenrightbigg , v1(p)=Np σ·p Ep+m/parenleftbigg 0 1/parenrightbigg /parenleftbigg 0 1/parenrightbigg ,v2(p)=Np σ·p Ep+m/parenleftbigg 1 0/parenrightbigg /parenleftbigg 1 0/parenrightbigg ,(4.10) where Np=/radicalBig Ep+m 2mis the normalization factor. Do not forget that p=pez i.e.p·σ=pσ3. In this case, the bispinors (4.10) form the helicity basis. For arbitrary momentum pwe have to use (4.8) instead of (4.9), if we want to construct the helicity basis. Although, in that case vectors in (4.10) are also a base, but not the helicity one. Spinors uandvare normalized according to (4.D). General solution of (4.2) is given by ψ=1 (2π)3/22/summationdisplay r=1/integraldisplay d3p/radicalbiggm Ep/parenleftBig ur(p)cr(p)e−ip·x+vr(p)d† r(p)eip·x/parenrightBig .(4.11) The Dirac spinor (bispinor) ψcontains two SL(2 ,C) spinors, as is easily seen in the chiral (Weyl) representation. The Dirac spinor is transformed according to the (1 /2,0)⊕(0,1/2) reducible representation of the quantum Lorentz group (i.e. SL(2 ,C) group, which is universally covering group for the Lorentz group). 4.3The states us(p),vs(p) are eigenstates of the energy operator, i∂ ∂twith eigenvalues Epand−Ep, respectively. 4.4By using the expressions for the Dirac spinors found in Problem 4.2, we obtain /summationtext rur(p)¯ur(p)= Ep+m 2m/parenleftBigg ϕ1ϕ† 1+ϕ2ϕ† 2 −(ϕ1ϕ† 1+ϕ2ϕ† 2)σ·p Ep+m σ·p Ep+m(ϕ1ϕ† 1+ϕ2ϕ† 2)−σ·p Ep+m(ϕ1ϕ† 1+ϕ2ϕ† 2)σ·p Ep+m/parenrightBigg , where ϕr(r={1,2}) are given by (4.8). They satisfy the completeness relation ϕ1ϕ† 1+ϕ2ϕ† 2= I. Using also ( p·σ)2=p2=E2 p−m2,we get 2/summationdisplay r=1ur(p)¯ur(p)=1 2m/parenleftbigg Ep+m−σ·p σ·p−Ep+m/parenrightbigg =/p+m 2m. The second identity can be shown in a similar manner. 4.5Using the expressions for the projectors given in Problem 4.4, we see that 96 Solutions Λ2 +=1 4m2(/p2+2m/p+m2)=Λ+, w h e r ew eh a v eu s e d/ p2=p2=m2. Similarly, we obtain Λ2 −=Λ−. Orthogo- nality of the projectors follows from the identity (/p+m)(/p−m)=p2−m2=0. At this stage we apply the Dirac equation in momentum space (4.C). Namely, Λ+ur(p)=1 2m(/p+m)ur(p)=1 2m(m+m)ur(p)=ur(p), Λ−ur(p)=1 2m(/p−m)ur(p)=1 2m(m−m)ur(p)=0 . Similarly, one can prove the identities Λ−vr(p)=0,Λ+vr(p)=vr(p). 4.6 (a) We can directly prove this property. For example, the x–component of the vector Σis Σ1=i 2(γ2γ3−γ3γ2)=iγ2γ3. On the other hand, γ5γ0γ1=iγ1γ2γ3γ1=iγ2γ3. The corresponding iden- tities for the yandz–components can be proven in a similar way. (b) By applying the deﬁnition of Σ,w eh a v e [Σi,Σj]=−1 4/epsilon1ilm/epsilon1jpq[γlγm,γpγq] =−1 4/epsilon1ilm/epsilon1jpq/parenleftbig [γlγm,γp]γq+γp[γlγm,γq]/parenrightbig .(4.12) Next step is to expand the commutators in terms of the anticommutators: [Σi,Σj]=−1 4/epsilon1ilm/epsilon1jpq/parenleftbig γl{γm,γp}γq−{γl,γp}γmγq +γpγl{γm,γq}−γp{γl,γq}γm/parenrightbig . (4.13) Then, using anticommutation relations (3.A) we get [Σi,Σj]=−1 2/epsilon1ilm/epsilon1jpq/parenleftbig gmpγlγq−glpγmγq+gmqγpγl−glqγpγm/parenrightbig . (4.14) The ﬁrst term in (4.14) has the form /epsilon1ilm/epsilon1jpqgmpγlγq=(δijδlq−δiqδlj)γlγq=−3δij−γjγi. Others terms in (4.14) can be transformed in the same way. Finally, [Σi,Σj]=γjγi−γiγj. Chapter 4. The Dirac equation 97 On the other hand, 2i/epsilon1ijkΣk=−/epsilon1ijk/epsilon1klmγlγm=γjγi−γiγj, so that [Σi,Σj]=2 i/epsilon1ijkΣk. We conclude that operators1 2Σare the generators of SU(2) subgroup of the Lorentz group1 (c)S2=−1 4Σ2=−1 4(γ5γ0γ)2=1 4γ·γ=−3 4. 4.7Use the expressions σ·ˆpϕr=(−1)r+1ϕrandσ·ˆpχr=(−1)rχrfrom Problem 4.2. For example: Σ·p |p|ur(p)=Σ·p |p|N/parenleftbiggϕr σ·p Ep+mϕr/parenrightbigg =N/parenleftbigg σ·ˆp0 0σ·ˆp/parenrightbigg/parenleftbiggϕr σ·p Ep+mϕr/parenrightbigg =N/parenleftbiggσ·ˆpϕr (σ·p)(σ·ˆp) Ep+mϕr/parenrightbigg =(−1)r+1N/parenleftbiggϕr σ·p Ep+mϕr/parenrightbigg =(−1)r+1ur(p), where Nis the normalization factor. It is easy to see that the spinors ur(p) andvr(p) are not eigenspinors of the operator Σ·n, unless vectors nandp are parallel. 4.8The transformation operator from the rest frame to the frame moving along the z–axis with velocity v,i sS(Λ(vez)) = e−i 2ω03σ03. By using the relation ω03=−ϕ=−arctan( v), we obtain S(Λ)=c o s h/parenleftBigϕ 2/parenrightBig I−sinh/parenleftBigϕ 2/parenrightBig/parenleftbigg 0σ3 σ3o/parenrightbigg =/radicalbigg Ep+m 2m/parenleftbiggI −pσ3 Ep+m −pσ3 Ep+mI/parenrightbigg . For arbitrary boost, σ3pshould be replaced by σ·p. The operator S(Λ)i sn o t unitary one. Since the Lorentz group is noncompact, it does not have ﬁnite dimensional irreducible unitary representations. 4.9In this case we have S=/parenleftbiggcos/parenleftbigθ 2/parenrightbig +is i n/parenleftbigθ 2/parenrightbig σ30 0c o s/parenleftbigθ 2/parenrightbig +is i n/parenleftbigθ 2/parenrightbig σ3/parenrightbigg . 1Recall that Σk=1 2/epsilon1kijσij. 98 Solutions This operator is unitary because SO(3) is a compact subgroup of the Lorentz group. 4.10 The Pauli–Lubanski vector is Wµ=1 2/epsilon1µνρσ(ixν∂ρ−ixρ∂ν+1 2σνρ)i∂σ=i 4/epsilon1µνρσσνρ∂σ, (4.15) since the product of a symmetric and an antisymmetric tensors vanishes. Then W2ψ(x)=−1 16/epsilon1µνρσ/epsilon1µαβγσνρσαβ∂σ∂γψ(x) =1 16/parenleftBig δν αδρ βδσ γ−δν αδσ βδρ γ+δρ αδσ βδν γ− −δρ αδν βδσ γ+δσ αδν βδρ γ−δσ αδρ βδν γ/parenrightBig σνρσαβ∂σ∂γψ(x) =1 16/parenleftbig 2σαβσαβ/unionsq/intersectionsq−4σαγσαρ∂ρ∂γ/parenrightbig ψ =3 4/unionsq /intersectionsqψ =−3 4m2ψ, where we used identity σµσσµν=2γσγν+δν σ and the results of Problems 1.5 and 3.5. 4.11 It is easy to see (Problem 3.16 and the condition s·p= 0) that Wµsµ m=1 4m/epsilon1µνρσσνρPσsµ=1 2mγ5σµσsµ∂σ =i 2mγ5(γµγσ−gµσ)(∓ipσ)sµ=±1 2mγ5/s/p=1 2γ5/s. The previous equation holds on space of plane wave solutions; upper (lower) sing is related to positive (negative) energy solutions. In the rest frame, the vector sµbecomes (0 ,n), so /s=−n·γ, and we can use/p m=p0γ0 m=γ0,so thatW·s m=±1 2γ5γ0n·γ=±1 2Σ·n. where Problem 4.6 has been used. 4.12 Positive energy solutions satisfy γ5/su(p,±s)=±u(p,±s). (4.16) If we choose that polarization vector sµin the rest frame equals (0 ,n=p |p|), according to the formulation of this problem, then in the frame in which Chapter 4. The Dirac equation 99 electron has momentum p, the polarization vector is obtained by applying a Lorentz boost: sµ=/parenleftBiggEp mpj m pi mδij+pipj m(Ep+m)/parenrightBigg/parenleftbigg 0 nj/parenrightbigg =/parenleftbigg p·n m n+(n·p)p m(Ep+m)/parenrightbigg . Forn=p/|p|we get sµ=(|p| m,Ep mn).Using that, we ﬁnd γ5/su(p,±s)=1 mγ5/s/pu(p,±s) =1 mγ5/parenleftbigg|p| mγ0−Ep mγ·n/parenrightbigg (Epγ0−p·γ)u(p,±s). If we insert ( p·γ)2=−p2in the previous formula we obtain: γ5/su(p,±s)=γ5γ0γ·p |p|u(p,±s)=Σ·p |p|u(p,±s). (4.17) From the expressions (4.16) and (4.17) we get Σ·p |p|u(p,±s)=±u(p,±s). The similar procedure can be done for negative energy solutions. Starting from γ5/sv(p,±s)=±v(p,±s), one gets Σ·p |p|v(p,±s)=∓v(p,±s). 4.13 In the ultrarelativistic limit, m/lessmuchEp, the vector sµis given by sµ≈/parenleftbiggEp m,p m/parenrightbigg ≈pµ m. Then we have γ5/su(p,±s)≈γ5/p mu(p,±s)=γ5u(p,±s), (4.18) where we used the Dirac equation / pu(p,±s)=mu(p,±s).From (4.18) we conclude that the helicity operator Σ·p/|p|is equal to the chirality operator γ5. The eigenequation becomes γ5u(p,±s)=±u(p,±s). 100 Solutions Forvspinors the situation is similar. So, for the particles of high energy (i.e. neglected mass) helicity and chirality are approximatively equal, while for massless particles these two quantities exactly are equal. 4.14 The commutator between γ5/sand /pis [γ5/s,/p]=γ5/s/p−/pγ5/s =γ5(/s/p+/p/s) =γ5sµpν{γµ,γν} =2s·pγ5=0. From ( γ5/s)2=−s2= 1 it follows that eigenvalues of the operator γ5/sare±1. Then the eigen projectors are Σ(±s)=1±γ5/s 2. 4.15 The average value of Σ·nin state ψ(x)=/radicalbigg Ep+m 2m/parenleftbiggϕ σ·p Ep+mϕ/parenrightbigg e−ip·x, (4.19) is /angbracketleftΣ·n/angbracketright=/integraltext d3xψ†(x)Σ·nψ(x)/integraltext d3xψ†(x)ψ(x) =Ep+m 2Ep/parenleftbigg ϕ†σ·nϕ+ϕ†(σ·p)(σ·n)(σ·p)ϕ (Ep+m)2/parenrightbigg .(4.20) Since (σ·A)(σ·B)=A·B+i (A×B)·σ (4.21) it follows that (σ·p)(σ·n)(σ·p)=|p|2(n3σ3−n2σ2−n1σ1). (4.22) By substituting (4.22) into (4.20) we get: /angbracketleftΣ·n/angbracketright=1 |a|2+|b|2 ×/bracketleftbiggEp+m 2Ep/parenleftbig n3|a|2+(n1+in2)b∗a+(n1−in2)a∗b−n3|b|2/parenrightbig +Ep−m 2Ep/parenleftbig n3|a|2+(−n1+in2)a∗b−(n1+in2)b∗a−n3|b|2/parenrightbig/bracketrightbigg . In the nonrelativistic limit we obtain /angbracketleftΣ·n/angbracketright=ϕ†σ·nϕ=n3|a|2+(n1+in2)b∗a+(n1−in2)a∗b−n3|b|2 |a|2+|b|2. Chapter 4. The Dirac equation 101 4.16 In the rest frame a spinor takes the following form/parenleftbigg ϕ 0/parenrightbigg e−imt,where ϕsatisﬁes 1 2Σ·n/parenleftbigg ϕ 0/parenrightbigg =1 2/parenleftbigg ϕ 0/parenrightbigg . The last condition becomes /parenleftbigg cosθ−isinθ isinθ−cosθ/parenrightbigg/parenleftbigg a b/parenrightbigg =/parenleftbigg a b/parenrightbigg , (4.23) where we put ϕ=/parenleftbigg a b/parenrightbigg .From the last expression we obtain ϕ=/parenleftbiggcosθ 2 isinθ 2/parenrightbigg . (4.24) In the rest frame the Dirac spinor takes the form ψ0= cosθ 2 isinθ 2 0 0 e−imt. (4.25) Applying the boost along z−axis, we obtain ψ(x)=S(−pez)ψ0, (4.26) where Sis given in Problem 4.8. Note a minus sign appearing in S(−pez)! After a simple calculation, we obtain ψ(x)=/radicalbigg Ep+m 2m cosθ 2 isinθ 2 p·σ Ep+m/parenleftbiggcosθ 2 isinθ 2/parenrightbigg e−ip·x. (4.27) The mean value of the operator1 2γ5/sis /angbracketleftbigg1 2γ5/s/angbracketrightbigg =1 2/integraltext d3xψ†γ5/sψ/integraltext d3xψ†ψ, (4.28) where the vector sµis obtained from (0 ,n) by the Lorentz boost along the z–axis. The components of vector sµare (see Problem 4.12) s0=n·p m,s=n+(n·p)p m(Ep+m). In our case we have 102 Solutions sµ=/parenleftbiggp mcosθ,0,sinθ,Ep mcosθ/parenrightbigg . Thus, in the Dirac representation of γ–matrices, γ5/sis given by γ5/s=/parenleftbigg s·σ−s0I s0I−s·σ/parenrightbigg , (4.29) and ﬁnally γ5/s= Ep mcosθ−isinθ−p mcosθ 0 isinθ−Ep mcosθ 0 −p mcosθ p mcosθ 0 −Ep mcosθisinθ 0p mcosθ−isinθEp mcosθ . (4.30) By substituting (4.30) and (4.27) in the formula (4.28), we obtain: /angbracketleftbigg1 2γ5/s/angbracketrightbigg =1 2, as we expected, because ψ(x) is the eigenstate of the operator1 2γ5/s, with eigenvalue1 2. 4.17 The Dirac Hamiltonian can be rewritten in terms of γ–matrices so that [HD,γ5]=[γ0γ·p+γ0m,γ5]=2mγ0γ5. Thus, the operator γ5is a constant of motion in the case of massless Dirac particle. Its eigenvalues and eigen projectors are ±1,Σ±=1 2(1±γ5), respec- tively. The operator γ5is known as the chirality operator. 4.18 By multiplying the Dirac equation from the left by γ5, we obtain (i/ ∂+ m)γ5ψ= 0. By adding and subtracting the previous equations and the Dirac equation, we get i/∂ψL−mψR=0, i/∂ψR−mψL=0. 4.19 (a) The system of equations can be rewritten as the Dirac equation. The Dirac spinor takes form ψ=/parenleftbigg ψL ψR/parenrightbigg , while γµ=/parenleftbigg 0σµ ¯σµ0/parenrightbigg , areγ–matrices (see Problem 3.20). Chapter 4. The Dirac equation 103 (b) In order to be covariant, these equations have to have the following form iσµ∂/prime µψ/prime R(x/prime)=mψ/prime L(x/prime), (4.31) i¯σµ∂/prime µψ/prime L(x/prime)=mψ/prime R(x/prime), (4.32) in the primed frame ( x/prime=Λx). If we assume that the new spinors take the form ψ/prime L(x/prime)=SLψL(x)a n d ψ/prime R(x/prime)=SRψR(x),where SLandSRare nonsingular 2 ×2 matrices, Equations (4.31) and (4.32) become iσµSRΛµν∂νψR(x)=mSLψL(x), (4.33) i¯σµSLΛµν∂νψL(x)=mSRψR(x). (4.34) By multiplying Equation (4.33) by S−1 Lfrom left, and (4.34) by S−1 Ralso from left we obtain iS−1 LσµSRΛµν∂νψR(x)=mψL(x), (4.35) iS−1 R¯σµSLΛµν∂νψL(x)=mψR(x). (4.36) The system of equations is covariant if the conditions S−1 R¯σµSL=Λµ ν¯σν, S−1 LσµSR=Λµ νσν hold. The solution for matrices SLandSRis given as SL=e x p/parenleftbigg1 2ϕiσi+i 2θkσk/parenrightbigg ≈1+1 2ϕiσi+i 2θkσk, (4.37) SR=e x p/parenleftbigg −1 2ϕiσi+i 2θkσk/parenrightbigg ≈1−1 2ϕiσi+i 2θkσk. (4.38) The parameters θiandϕiwere deﬁned in Problem 1.8. Boost along the x–axis is deﬁned by : SL=c o s h/parenleftBigϕ1 2/parenrightBig +σ1sinh/parenleftBigϕ1 2/parenrightBig (4.39) SR=c o s h/parenleftBigϕ1 2/parenrightBig −σ1sinh/parenleftBigϕ1 2/parenrightBig . (4.40) Note that ψLandψRtransform in the same way under rotations, but dif- ferently under boosts. The left ψL, and right ψRspinors transform under (1 2,0) and (0 ,1 2) irreducible representation of the Lorentz group respec- tively. 104 Solutions 4.20 First note that [HD,K]=[α·p,β(Σ·L)] + [α·p,β]+m[β,β(Σ·L)]. (4.41) The ﬁrst term in the expression (4.41) is [α·p,β(Σ·L)] =β[α·p,Σ·L]+[α·p,β]Σ·L =−i 2/epsilon1mnp/epsilon1mjlβ/parenleftbig pi{αi,αn}αpxjpl−piαn{αi,αp}xjpl+ +αnαpαi[pi,xj]pl−2αiαnαppixjpl/parenrightbig . Using the relations {αi,αj}=2δijand [xi,pj]=iδij, we obtain [α·p,β(Σ·L)] =−i 2β/parenleftbig 4αlpnxnpl−4αjplxjpl− −iαjαlαjpl+3 iαipi−2αiαjαlpixjpl+2αiαlαjpixjpl/parenrightbig =iβ/parenleftbig 2αiplxipl−2iα·p−αjαiαlpixjpl−αiαlαjpixjpl/parenrightbig , where we used αiαjαi=−αj. By substituting pixj=xjpi−iδijinto the last line of previous formula, we have [α·p,β(Σ·L)] = 2 β(α·p). (4.42) The second term in (4.41) is −2β(α·p), while the third term vanishes. Thus, [HD,K]=0. 4.21 From (3.E) we have i¯u(p1)σµν(p1−p2)νu(p2)=1 2¯u(p1)(γνγµ−γµγν)(p1−p2)νu(p2) =1 2¯u(p1)[−γµ(/p1−/p2)+( /p1−/p2)γµ]u(p2) =1 2¯u(p1)[−γµ(/p1−m)+(m−/p2)γµ]u(p2). By using γµ/p1=2pµ 1−/p1γµand /p2γµ=2pµ 2−γµ/p2we obtain i¯u(p1)σµν(p1−p2)νu(p2)=2m¯u(p1)γµu(p2)−(p1+p2)µ¯u(p1)u(p2), where we used that u(p) and ¯ u(p) satisfy the Dirac equation. The last expres- sion is the requested identity. The second identity can be proven similarly. 4.23 It is easy to see that γαγµγβ=2gαµγβ−2gαβγµ+2gµβγα−γβγµγα. (4.43) From (4.43) we have Chapter 4. The Dirac equation 105 ¯u(p2)/p1γµ/p2u(p1)=¯u(p2)[2m(p1+p2)µ−(2p1·p2+m2)γµ]u(p1),(4.44) where we used the Dirac equation (4.C). The ﬁrst term in (4.44) can be transformed by using the Gordon identity (Problem 4.21) ¯u(p2)/p1γµ/p2u(p1)=¯u(p2)[−2p1·p2+3m2]γµu(p1)−2mi¯u(p2)σµνqνu(p1) =¯u(p2)/braceleftbig (q2+m2)γµ−2imσµνqν/bracerightbig u(p1). (4.45) From the last expression we can make the following identiﬁcations: F1= q2+m2andF2=−2im. 4.24 By using u(p)=/pu(p)/mand {γµ,γ5}=0, we have ¯u(p)γ5u(p)=1 m¯u(p)γ5/pu(p)=−1 m¯u(p)/pγ5u(p). By applying the Dirac equation (3.C) we obtain ¯u(p)γ5u(p)=−¯u(p)γ5u(p). Thus ¯ u(p)γ5u(p) = 0. By using the Gordon identity (for µ= 0) it ﬁnally follows that1 2¯u(p)(1−γ5)u(p)=m 2EpN. 4.25 F1=−iq2,F2=−2im, F 3=−2m. 4.26 By applying the operator (i/ ∂+m) to the Dirac equation we obtain (i/∂+m)(i/∂−m)ψ=−(/unionsq /intersectionsq+m2)ψ=0. 4.27 The probability density is ρ(x)=ψ†(x)ψ(x). By using the expression for the wave function from Problem 4.2, we easily get ρ=Ep m. The current density is j=¯ψγψ=p m¯ψψ,where the Gordon identity (for µ=i) has been applied. Finally j=p m. 4.28 The position operator in the Heisenberg picture satisﬁes the following equation drH dt=−i[rH,H]=αH. In order to integrate the last equation we have to ﬁnd the Dirac matrices in the Heisenberg picture αH=eiHtαe−iHt=∞/summationdisplay n=0(it)n n![H,[H,... [H,α]...]]. Since 106 Solutions [H,α]=2 (p−αH), (4.46) [H,[H,α]] =−22(p−αH)H, (4.47) [H,[H,[H,α]]] = 23(p−αH)H2,etc. (4.48) we get αH=α+(αH−p)/parenleftbigg −2it+(2it)2 2!H−(2it)3 3!H2+.../parenrightbigg =p H+/parenleftBig α−p H/parenrightBig e−2itH. (4.49) Then, equation drH dt=p H+/parenleftBig α−p H/parenrightBig e−2itH(4.50) implies rH=r+p Ht−i/parenleftBig α−p H/parenrightBig1 2H+i/parenleftBig α−p H/parenrightBig1 2He−2iHt. The integration constant is determined using the condition rH(t=0 )= r. As we see ”the motion of particle” is a superposition of classical uniform and rapid oscillatory motions. 4.29 We should calculate the coeﬃcients cr(p)a n d d∗ r(p) in the expansion ψ(0,x)=1 (2π)3/2/summationdisplay r/integraldisplay d3p/radicalbiggm Ep(cr(p)ur(p)eip·x+d∗ r(p)vr(p)e−ip·x). (4.51) If we multiply this expression by u† s(q)e−iq·xfrom left and integrate over x, we get cs(q)=1 (2π)3/2/radicalbiggm Eq/integraldisplay d3xu† s(q)ψ(0,x)e−iq·x, where we applied the relations u† r(p)us(p)=v† r(p)vs(p)=Ep mδrs,v† r(−p)us(p)=u† r(−p)vs(p)=0 . (4.52) These relations can be obtained from (4.D) by using the Gordon identity. Similarly for dcoeﬃcients we get d∗ s(q)=1 (2π)3/2/radicalbiggm Eq/integraldisplay d3xv† s(q)ψ(0,x)eiq·x. Carrying out the integrations, we ﬁnd Chapter 4. The Dirac equation 107 c1(p)=1 (2π)3/2/radicalBigg Ep+m 2Ep, c2(p)=0, d∗ 1(p)=1 (2π)3/21/radicalbig 2Ep(Ep+m)(px+ipy), d∗ 2(p)=1 (2π)3/21/radicalbig 2Ep(Ep+m)pz. (4.53) The wave function at time t>0i s ψ(x)=1 (2π)3/2/summationdisplay r/integraldisplay d3p/radicalbiggm Ep(cr(p)ur(p)re−ip·x+d∗ r(p)vr(p)eip·x),(4.54) where the coeﬃcients cr(p)a n d d∗ r(p) are given in (4.53). 4.30 In this case the coeﬃcients cr(p)a n d d∗ r(p) in expansion (4.51) are: c1(p)=/parenleftbiggd2 π/parenrightbigg3/4/radicalBigg Ep+m 2Epe−d2p2/2, c2(p)=0 , d∗ 1(p)=/parenleftbiggd2 π/parenrightbigg3/41/radicalbig 2Ep(Ep+m)e−d2p2/2(px+ipy), d∗ 2(p)=/parenleftbiggd2 π/parenrightbigg3/41/radicalbig 2Ep(Ep+m)pze−d2p2/2. 4.31 The equation for spin 1 /2 particle in the electromagnetic ﬁeld has the following form [iγµ(∂µ−ieAµ)−m]ψ=0. (4.55) If we assume that a wave function for z>0 has the form ψ=/parenleftbigg ϕ χ/parenrightbigg e−iEt+iqz, (4.56) then (4.55) becomes /parenleftbigg E−m−V −σ3q σ3q −E−m+V/parenrightbigg/parenleftbigg ϕ χ/parenrightbigg =0. (4.57) The system of equations (4.57) has a nontrivial solution if and only if E=V±/radicalbig q2+m2. (4.58) The wave function2is 2From the boundary conditions it follows that there is no spin ﬂip. 108 Solutions ψI=a 1 0 pσ3 (E+m)/parenleftbigg 1 0/parenrightbigg e−iEt+ipz +b 1 0 −pσ3 (E+m)/parenleftbigg 1 0/parenrightbigg e−iEt−ipz,z < 0, (4.59) ψII=d 1 0 qσ3 (E+m−V)/parenleftbigg 1 0/parenrightbigg e−iEt+iqz,z > 0, where p=√ E2−m2.The terms proportional to the coeﬃcient a, bandd in (4.59) are the initial ψin, reﬂected ψrand transmitted wave ψt. Since the Dirac equation is the ﬁrst order equation, the continuity condition is satisﬁed for the wave function only. The condition ψI(0) = ψII(0) gives a+b=d, (4.60) a−b=rd , (4.61) where r=E+m E+m−Vq p. Now, we will consider three cases: 1.If|E−V|≤m, the momentum qis imaginary, q=iκso that the wave function exponentially decreases in the region z>0, as is the case in nonrela- tivistic quantum mechanics. The transmitted, reﬂected and incident currents are: jr=¯ψtrγ3ψtrez=0, (4.62) jr=¯ψrγ3ψrez=−2p E+m|b|2ez, (4.63) jin=¯ψinγ3ψinez=2p E+m|a|2ez. (4.64) Sincejtr= 0 the transmission coeﬃcient is zero. The reﬂection coeﬃcient is R=−jr jin=/vextendsingle/vextendsingle/vextendsingle/vextendsinglep(E+m−V)−iκ(E+m) p(E+m−V)+iκ(E+m)/vextendsingle/vextendsingle/vextendsingle/vextendsingle2 =1. (4.65) 2.IfV< E −m, the momentum qis real. The currents are: jtr=2q E+m−V|d|2ez, (4.66) jr=−2p E+m|b|2ez, (4.67) jin=2p E+m|a|2ez. (4.68) Chapter 4. The Dirac equation 109 The transmission coeﬃcient is T=jtr jin=r/vextendsingle/vextendsingle/vextendsingle/vextendsingled a/vextendsingle/vextendsingle/vextendsingle/vextendsingle2 =4r (1 +r)2, (4.69) while the reﬂection coeﬃcient is R=−jr jin=/parenleftbigg1−r 1+r/parenrightbigg2 . (4.70) 3.IfE+m<V , the momentum qis real, which implies that the wave function in region z>0 becomes oscillating. This is caused by the fact that there are two parts of electron spectrum separated by a gap, whose width is equal to 2m. The expressions for the coeﬃcients of reﬂection and transmission are the same as in the second case. But in this case, the coeﬃcient of reﬂection is greater then 1, while T<0. The described eﬀect is known as the Klein paradox .T h e explanation of this paradox is given in Problem 2.9. 4.32 The solution of the Dirac equation is ψI= 1 0 pσ3 (E+m)/parenleftbigg 1 0/parenrightbigg eipz +B 1 0 −pσ3 (E+m)/parenleftbigg 1 0/parenrightbigg e−ipz,z < 0, ψII=C 1 0 qσ3 (E+m−V)/parenleftbigg 1 0/parenrightbigg eiqz +D 1 0 −qσ3 (E+m−V)/parenleftbigg 1 0/parenrightbigg e−iqz,0<z<a, ψIII=F 1 0 pσ3 (E+m)/parenleftbigg 1 0/parenrightbigg eipz,z > a , where p=√ E2−m2andq=/radicalbig (E−V)2−m2.From the boundary con- ditions ψI(0) = ψII(0) and ψII(a)=ψIII(a), we obtain the transmission coeﬃcient T=|F|2=1 6|r|2 |(1 +r)2e−iqa−(1−r)2eiqa|2, where r=q pE+m E+m−V. It is easy to show that the transmission coeﬃcient is equal to one if E=V 2. 110 Solutions 4.33 (a) The wave function is ψI= B B/prime −iκσ3 (E+m)/parenleftbigg B B/prime/parenrightbigg eκz,z < −a, ψII= C C/prime qσ3 (E+m+V)/parenleftbigg C C/prime/parenrightbigg eiqz(4.71) + D D/prime −qσ3 (E+m+V)/parenleftbigg D D/prime/parenrightbigg e−iqz,−a<z<a , ψIII= F F/prime iκσ3 (E+m)/parenleftbigg F F/prime/parenrightbigg e−κz,z > a , where κ=√ m2−E2andq=/radicalbig (E+V)2−m2.Since there is no spin ﬂip, we can take B/prime=C/prime=D/prime=F/prime= 0. From the boundary conditions ψI(−a)=ψII(−a)a n d ψII(a)=ψIII(a), it follows that e−κaB=e−iqaC+eiqaD e−κaF=eiqaC+e−iqaD −ire−κaB=e−iqaC−eiqaD ire−κaF=eiqaC−e−iqaD, where r=κ qE+m+V E+m. By combining previous equations we obtain e−κa(B−F)=2 is i n ( qa)(D−C) ire−κa(B−F)=2c o s ( qa)(D−C) e−κa(B+F)=2c o s ( qa)(D+C) re−κa(B+F)=2s i n ( qa)(D+C). Further, we will distinguish two classes of solutions: odd and even. If B=F andC=D, the third and the fourth equations give the following dispersion relation: tan(qa)=κ qE+m+V E+m. These solutions satisfy the following property: ψ/prime(z)=γ0ψ(−z)=ψ(z); thus they are even. On the other hand, if B=−FandC=−D,t h e dispersion relation is Chapter 4. The Dirac equation 111 cot(qa)=−κ qE+m+V E+m. This class of solutions satisfy ψ/prime(z)=γ0ψ(−z)=−ψ(z), and therefore they are odd. (b) The dispersion relations are transcendental equations and they cannot be solved analytically. We can analyze them graphically. For even solutions, the dispersion relation has the form qtan(qa)=f(q), (4.72) where f(q)=/radicalBig 2V/radicalbig q2+m2−q2−V2m+/radicalbig q2+m2 m+/radicalbig q2+m2−V, and its graphical solution is given in Fig. 4.1. Fig. 4.1. Graphical solution of Equation (4.72) for even states ( V<2m) In the case of odd solutions, the dispersion relation qcot(qa)=−f(q) (4.73) is shown in Fig. 4.2. From these ﬁgures we see that the spectrum of electron bound states will contain Nstates if the condition (N−1)π 2a≤/radicalbig V(V+2m)<Nπ 2a is satisﬁed. It is easy to see that if N= 1 then this solution is even. (c) Graphical solutions for odd and even part of spectrum are given in Fig. 4.3 and Fig. 4.4. 112 Solutions Fig. 4.2. Graphical solution of Equation (4.73) for odd states ( V<2m) Fig. 4.3. Graphical solution for odd states ( V>2m) Fig. 4.4. Graphical solution for even states ( V>2m) Chapter 4. The Dirac equation 113 4.34 The Dirac equation in this case has following form /bracketleftbigg iγ0∂ ∂t+iγ1/parenleftbigg∂ ∂x−ieBy/parenrightbigg +iγ2∂ ∂y+iγ3∂ ∂z−m/bracketrightbigg ψ=0. (4.74) A particular solution of (4.74) is ψ=e−iEt+ipxx+ipzz/parenleftbigg ϕ(y) χ(y)/parenrightbigg . (4.75) By substituting (4.75) in (4.74) we obtain /parenleftbiggE−m (eBy−px)σ1−pzσ3+iσ2d dy (px−eBy)σ1+pzσ3−iσ2d dy−E−m/parenrightbigg/parenleftbigg ϕ χ/parenrightbigg =0. (4.76) From the second equation in (4.76), follows χ(y)=1 E+m/parenleftbigg pxσ1+pzσ3−eByσ 1−iσ2d dy/parenrightbigg ϕ(y), (4.77) and plugging it into the ﬁrst equation of (4.76), we get /parenleftbiggd2 dy2−(px−eBy)2+E2−m2−p2 z−eBσ3/parenrightbigg ϕ=0, (4.78) where we used the following identity σiσj=δij+i/epsilon1ijkσk. By introducing new variable ξ=px−eBy, Equation (4.78) becomes the Schr¨odinger equation for a linear oscillator (parameters M, ω and/epsilon1), where M2ω2=1 (eB)2,2M/epsilon1=E2−m2−p2 z∓eB (eB)2. We assumed that the spinor ϕis an eigenstate of σ3/2, i.e. 1 2σ3ϕ=±1 2ϕ. The energy eigenvalues are En,pz=/radicalbig m2+p2z±eB+( 2n+1 )eB , (4.79) where n=0,1,2,... 4.35 Acting by (i/ ∂+e/A+m) on (i/ ∂+e/A−m)ψ(x)=0,we get [/unionsq/intersectionsq−ieγµγν∂µAν−2ieAµ∂µ−e2A2+m2]ψ=0. 114 Solutions On the other hand, one can show that −e 2σµνFµν=ie(∂µAµ−γµγν∂µAν). The requested result can be obtained by combining these expressions. 4.36 By substituting ψ=/parenleftbigg ϕ χ/parenrightbigg e−imt in the Dirac equation (i/∂+e/A−m)ψ(x)=0, we obtain the following equations: /parenleftbigg i∂ ∂t+eA0/parenrightbigg ϕ=cσ·(p+eA)χ, /parenleftbigg i∂ ∂t+2mc2+eA0/parenrightbigg χ=cσ·(p+eA)ϕ. In the case A= 0, the second equation yields: χ=1 2mc/parenleftbigg σ·pϕ−i 2mc2σ·p∂ϕ ∂t−eA0 2mc2σ·pϕ/parenrightbigg . Combining this relation with the ﬁrst equation, we obtain i∂ϕ ∂t=H/primeϕ, where H/prime=/bracketleftbiggp2 2m−eA0−p4 8m3c2+e 4m2c2(2iE·p−∆A0) −e 4m2c2(iE·p+σ·(E×p))/bracketrightBig . The operator H/primeis not the Hamiltonian, since it is not hermitian. This is related to the fact that ϕ†ϕis not the probability density. Actually, the prob- ability density should be taken in the following form: ρ=¯ψψ=ϕ†ϕ−χ†χ =ϕ†(1 +p2 4m2c2)ϕ+o/parenleftbiggv2 c2/parenrightbigg . We introduce the new wave function ϕs=/parenleftbigg 1+p2 8m2c2/parenrightbigg ϕ. Chapter 4. The Dirac equation 115 Then, the new Hamiltonian is given by H=/parenleftbigg 1+p2 8m2c2/parenrightbigg H/prime/parenleftbigg 1−p2 8m2c2/parenrightbigg . After that, we obtain H=p2 2m−eA0−p4 8m3c2−e 8m2c2∆A0+e 4m2c2σ·(E×p). In the case A/negationslash= 0, the Hamiltonian is H=(p+eA)2 2m−eA0+e 2mcσ·B−p4 8m3c2 −e 8m2c2∆A0+e 4m2c2σ·(E×(p+eA)). 4.37 First, we are going to show that Vµ(x) is a real quantity: V∗ µ=V† µ=(¯ψγµψ)† =ψ†γ† µ(ψ†γ0)† =ψ†γ0γµγ0γ0ψ =¯ψγµψ =Vµ. (4.80) Under proper orthochronous Lorentz transformations, Vµis transformed in the following way: V/prime µ(x/prime)=¯ψ/prime(x/prime)γµψ/prime(x/prime)=ψ†(x)γ0S−1γµSψ(x), where we used the fact that γ0S−1=S†γ0.Using S−1γµS=Λν µγν, we obtain V/prime µ(x/prime)=Λν µVν(x).So, the quantity Vµis a Lorentz four-vector. Under parity we have Vµ(t,x)→V/prime µ(t,−x)=¯ψ(t,x)γ0γµγ0ψ(t,x). This implies V/prime 0(t,x)=V0(t,−x),V/prime i(t,x)=−Vi(t,−x). As we know, under charge conjugation the spinors transform according to: ψ(x)→ψc(x)=C¯ψT, ¯ψ=ψ†γ0→(C¯ψT)†γ0 =(C(γ0)Tψ∗)†γ0 =ψT((γ0)TC(γ0)†)T =−ψT(Cγ0γ0)T =ψTC. (4.81) 116 Solutions Then, we can ﬁnd the transformation law for Vµ: Vµ→−ψTCγµC−1¯ψT=(¯ψγµψ)T=Vµ. The following formulae CγµC−1=−γT µ,C=−C−1have been used (Prove the last one).For time reversal we have ψ(x)→ψ /prime(−t,x)=Tψ∗(t,x),where matrix T satisﬁes TγµT−1=γµ∗=γT µandT†=T−1=T=−T∗.I ti se a s yt os e e that ¯ψ(x)→¯ψ/prime(−t,x)=ψT(t,x)Tγ0. Then Vµ(t,x)→ψTTγ0γµTψ∗ =ψTTγ0T−1TγµT−1ψ∗ =ψT(γ0)T(γµ)Tψ∗ =(ψ†γµγ0ψ)T =ψ†γµγ0ψ. (4.82) Therefore, V/prime 0(−t,x)=V0(t,x),V/prime i(−t,x)=−Vi(t,x). 4.38 The quantity Aµtransforms under Lorentz transformations in the fol- lowing way: A/primeµ(x/prime)=Λµ ν¯ψ(x)γνS−1γ5Sψ(x) = detΛΛµ ν¯ψ(x)γνγ5ψ(x) = det ΛΛµ νAν(x), w h e r ew eu s e d S−1γ5S=−i 4!/epsilon1µνρσS−1γµSS−1γνSS−1γρSS−1γσS =−i 4!/epsilon1µνρσΛµ αΛν βΛσ γΛρ δγαγβγγγδ =−i 4!/epsilon1αβγδdetΛγαγβγγγδ = detΛγ5. The charge conjugation changes the sign of Aµ. The parity changes the sign of the time component, but does not change the sign of spatial components. The eﬀect of time reversal is exactly opposite. 4.39 The quantity ¯ψγµ∂µψtransforms as a scalar under Lorentz transfor- mations. The parity does not change it. The action of the charge conjugation yields ( ∂µ¯ψ)γµψ, while the time reversal produces −(∂µ¯ψ)γµψ. 4.40 By transposing the Dirac equation, Chapter 4. The Dirac equation 117 ¯u(p,s)(/p−m)=0 , and using C−1γµC=−(γµ)T, one gets the requested result. 4.41 Let us assume that there are two diﬀerent matrices C/primeandC/prime/prime, which both satisfy the relation CγµC−1=−(γµ)T.Then from C/prime/primeγµC/prime/prime−1= C/primeγµC/prime−1follows that [ C/prime−1C/prime/prime,γµ] = 0, whereupon (see Problem 3.18) the requested relation follows. 4.42 We directly obtain: (a) ψc(x)=Np −p E+m/parenleftbigg 0 1/parenrightbigg /parenleftbigg 0 1/parenrightbigg e−iEt−ipz. (b) ψ/prime(x/prime)= 1 0 00 e −imt/prime. (c) ψp(t,x)=Np /parenleftbigg 1 0/parenrightbigg −p Ep+m/parenleftbigg 1 0/parenrightbigg e−i(Et+pz). Momentum is inverted under parity. Time reversal transforms the wave function into ψt(t,x)=−iNp /parenleftbigg 0 1/parenrightbigg p Ep+m/parenleftbigg 0 1/parenrightbigg ei(−Et−pz), and we see that spin and the direction of the momentum are inverted. (d) The wave function for S/primeobserver is ψ/prime(x/prime)=Np/parenleftbiggϕ p Ep+mϕ/parenrightbigg ei(Et−p/primez/prime) where ϕ=/parenleftbiggcos/parenleftbigθ 2/parenrightbig isin/parenleftbigθ 2/parenrightbig/parenrightbigg . 4.43 P=γ0=/parenleftbigg 0I I0/parenrightbigg ,C=iγ2γ0=i/parenleftbigg σ20 0−σ2/parenrightbigg . 4.44 Multiplying the equation 118 Solutions Σ·p |p|ur(p)=(−1)r+1ur(p), (4.83) byγ0from left, we obtain Σ·(−p) |p|ur(−p)=(−1)rur(−p), (4.84) sinceγ0ur(p)=ur(−p). From (4.84) we see that the helicity is inverted. Under the time reversal, the wave function of the Dirac particle (4.6) becomes ψt(t,x)=iγ1γ3ψ∗ r(−t,x) =−N/parenleftbiggσ2ϕ∗ r σ2(σ∗·p) Ep+mϕ∗ r/parenrightbigg ei(−Ept−p·x) =−N/parenleftbiggσ2ϕ∗ r −(σ·p)σ2 Ep+mϕ∗ r/parenrightbigg ei(−Ept−p·x), (4.85) w h e r ew eu s e d σ2σ∗=−σσ2in the second step. From the last expression, we conclude that the momentum changes its direction, i.e. p→−p. Prove that σ2ϕ∗ 1=iϕ2andσ2ϕ∗ 2=−iϕ1. Now, we consider the case r= 1 (the other caser= 2 is similar). From (4.85) it follows that ψt(t,x)=−iN/parenleftbiggϕ2 −p·σ Ep+mϕ2/parenrightbigg ei(−Ept−p·x). (4.86) By applyingΣ·(−p) |p|on (4.86), we see that the helicity is unchanged. The same result can be obtained by complex conjugation and multiplication of Equation (4.83) from left by i γ1γ3. You can prove the same for vspinors. 4.45 The transformed Hamiltonian is H/prime=α·p/parenleftbigg cos(2pθ)−m psin(2pθ)/parenrightbigg +mβ/parenleftBig cos(2pθ)+p msin(2pθ)/parenrightBig , where p=|p|. In order to have even form of the Hamiltonian, the coeﬃcient multiplying α·phas to be zero. This is satisﬁed if tan(2 pθ)=p/m . 4.47 First prove that: U=c o s ( pθ)+βα·p psin(pθ)=/radicalBigg Ep+m 2Ep+βα·p/radicalbig 2Ep(Ep+m), hence xFW=/parenleftBigg/radicalBigg Ep+m 2Ep+βα·p/radicalbig 2Ep(Ep+m)/parenrightBigg x/parenleftBigg/radicalBigg Ep+m 2Ep−βα·p/radicalbig 2Ep(Ep+m)/parenrightBigg . From the well known identity [ x,f(p)] = i∇f(p) we get two auxiliary results: Chapter 4. The Dirac equation 119 x/radicalBigg Ep+m 2Ep=−i 2/radicalBigg Ep 2(Ep+m)m E3pp+/radicalBigg Ep+m 2Epx, xβα·p/radicalbig 2Ep(Ep+m)=iβα/radicalbig 2Ep(Ep+m)−iβ(α·p)(2Ep+m) 2√ 2(Ep(Ep+m))3/2p Ep +βα·p/radicalbig 2Ep(Ep+m)x. Using these formulae we get xFW=x−ip 2Ep(Ep+m)+ip(βα·p) 2E2p(Ep+m)−iβα 2Ep+iα(α·p) 2Ep(Ep+m). The last expression can be rewritten in the form xFW=x+ip(βα·p) 2E2p(Ep+m)−iβα 2Ep−Σ×p 2Ep(Ep+m). The Foldy–Wouthuysen transformation does not change the momentum, so that [xk FW,plFW]=iδkl. 5 Classical ﬁelds and symmetries 5.1We apply the deﬁnition of functional derivative (5.A). (a) From δFµ=∂µδφ=/integraldisplay d4y(∂µδφ)yδ(4)(y−x)=−/integraldisplay d4y∂y µδ(4)(y−x)δφ(y), we have δFµ[φ(x)] δφ(y)=−∂y µδ(4)(y−x), (b) The ﬁrst functional derivative of the action with respect to φis δS δφ(x)=−/unionsq/intersectionsqφ−∂V ∂φ. Then δ/parenleftbiggδS δφ(x)/parenrightbigg =−/unionsq/intersectionsqδφ(x)−∂2V ∂φ2(x)δφ(x) =/integraldisplay d4y/bracketleftBig −/unionsq/intersectionsqyδ(4)(x−y)− −∂2V ∂φ(x)∂φ(y)δ(4)(x−y)/bracketrightbigg δφ(y). Hence, δ2S δφ(x)δφ(y)=−/unionsq/intersectionsqyδ(4)(y−x)−∂2V ∂φ(x)∂φ(y)δ(4)(x−y). 5.2In this problem we use the Euler–Lagrange equations of motion (5.B). (a) First note that∂L ∂Aρ=m2Aρand∂L ∂(∂σAρ)=−2∂ρAσ+λgρσ(∂µAµ) so that the equations of motion are given by (λ−2)∂σ∂ρAσ−m2Aρ=0. 122 Solutions (b) The derivative of the Lagrangian density with respect to ∂σAρis ∂L ∂(∂σAρ)=−1 2Fµν∂Fµν ∂(∂σAρ)=−1 2Fµν(δσ µδρ ν−δσ νδρ µ)=−Fσρ. In the last step we used the fact that Fρσis an antisymmetric tensor, i.e. Fρσ=−Fσρ. The Euler-Lagrange equations of motion are ∂σFσρ+m2Aρ=0. By using the deﬁnition of ﬁeld strength Fρσ, the Euler-Lagrange equations become/parenleftbig δρ σ/unionsq/intersectionsq−∂σ∂ρ+m2δρ σ/parenrightbig Aσ=0. (c) (/unionsq /intersectionsq+m2)φ=−λφ3. (d) The equations of motion are: −/unionsq/intersectionsqAρ+∂σ∂ρAσ=−ie[φ(∂ρφ∗+ieAρφ∗)−φ∗(∂ρφ−ieAρφ)], /unionsq /intersectionsqφ∗+2 ieAρ∂ρφ∗+ieφ∗∂ρAρ−e2A2φ∗+m2φ∗=0, /unionsq /intersectionsqφ−2ieAρ∂ρφ−ieφ∂ρAρ−e2A2φ+m2φ=0. (e) The equations are: (iγµ∂µ−m)ψ=igγ5ψφ , ¯ψ(iγµ←−∂µ+m)=−ig¯ψγ5φ, /unionsq /intersectionsqφ+m2φ=λφ3−ig¯ψγ5ψ. 5.3The variation of the action is δS=/integraldisplay∞ −∞dt/integraldisplayL 0dx/parenleftbig ∂µφ∂µ(δφ)−m2φδφ/parenrightbig =/integraldisplay∞ −∞dt/integraldisplayL 0dx[∂µ(∂µφδφ)−(/unionsq /intersectionsq+m2)φδφ] =/integraldisplayL 0dx∂0φδφ/vextendsingle/vextendsingle/vextendsinglet=∞ t=−∞−/integraldisplay∞ −∞dt∂φ ∂xδφ/vextendsingle/vextendsingle/vextendsinglex=L x=0 −/integraldisplay∞ −∞dt/integraldisplayL 0dx(/unionsq /intersectionsqφ+m2φ)δφ , where we integrated by parts. As the ﬁrst term vanishes, from Hamiltonian principe one obtains the equation of motion (/unionsq /intersectionsq+m2)φ=0, and the boundary conditions: δφ(t,x=0 )= δφ(t,x=L)=0 ,(Dirichlet boundary conditions) Chapter 5. Classical ﬁelds and symmetries. 123 or φ/prime(t,x=0 )= φ/prime(t,x=L)=0 ,(Neumann boundary conditions) , where prime denote the partial derivative with respect to x. Here, we see that beside the equation of motion we get the boundary conditions in order to elim- inate the surface term. Let us mention that the mixed boundary conditions can be imposed. 5.4In order to show that the change L→L +∂µFµ(φr) does not change the equations of motion, we have to prove that δ/integraldisplay Ωd4x∂µFµ(φr)=0 . Applying the Gauss theorem we get δ/integraldisplay Ωd4x∂µFµ(φr)=/contintegraldisplay ∂ΩdΣµδFµ=/contintegraldisplay ∂ΩdΣµ∂Fµ ∂φrδφr=0, since the variation of ﬁelds on the boundary is equal to zero. 5.5Add to the Lagrangian density the term −1 2∂µ(φ∂µφ). Note that it does not have the form as in Problem 5.4, because here the function Fµdepends on the ﬁeld derivatives. However, δ/integraldisplay Ωd4x∂µ(φ∂µφ)=/contintegraldisplay ∂ΩdΣµδ(φ∂µφ)=/contintegraldisplay ∂ΩdΣµ(δφ∂µφ+φδ∂µφ). The ﬁrst term is zero since δφ|∂Ω= 0 . If we take that the boundary is at inﬁnity ( r→∞), the second term is also zero because the ﬁelds tend to zero at inﬁnity. 5.6Use the similar reasoning as in the previous problem. 5.7The equation of motion for the vector ﬁeld was derived in Problem 5.2 (b). Acting by ∂ρon this equation we obtain m2∂ρAρ=0.Since m/negationslash=0 ,w e conclude that ∂ρAρ=0. 5.8The ﬁeld strength tensor, Fµνis invariant under the gauge transforma- tions. From this, it follows that the Lagrangian is also invariant. The condition∂ µAµ= 0 does not follow from the equations of motion, but by using gauge symmetry we can transform the potential so that it satisﬁes this condition. This condition is called the Lorentz gauge. 5.9Firstly, show that ∂L ∂(∂αhρσ)=∂αhρσ−∂σhρα−∂ρhσα+1 2gρα∂σh +1 2gσα∂ρh+gρσ∂µhµα−gρσ∂αh. 124 Solutions The equations of motion are /unionsq /intersectionsqhρσ−∂α∂σhρα−∂α∂ρhσα+∂ρ∂σh +gρσ∂µ∂νhµν−gρσ/unionsq /intersectionsqh=0. In order to prove gauge invariance of the action show that the Lagrangian density is changed up to four-divergence term. 5.11 This transformation is an internal one, so it is enough to prove the invariance of the Lagrangian density. The transformation law for the kineticterm is 1 2[(∂φ1)2+(∂φ2)2]→1 2[(∂φ/prime 1)2+(∂φ/prime 2)2] =1 2[(∂φ1cosθ−∂φ2sinθ)2+(∂φ1sinθ+∂φ2cosθ)2] =1 2[(∂φ1)2+(∂φ2)2]. Similarly, we can prove that the other two terms are invariant. The inﬁnites- imal variations of the ﬁelds φiareδφ1=−θφ2andδφ2=θφ1, so that jµ=∂L ∂(∂µφi)δφi=θ(φ1∂µφ2−φ2∂µφ1). The parameter θcan be dropped out since it is a constant. The charge corre- sponding to the SO(2) symmetry is Q=/integraltext d3x(φ1˙φ2−φ2˙φ1). 5.12 Under the SU(2) transformations, the ﬁelds are transformed accord- ing to φ/prime=ei 2τaθaφ,where τa(a=1,2,3) are the Pauli matrices. For an inﬁnitesimal transformation we obtain δφi=i 2τa ijθaφj,δ φ∗i=−i 2φ∗ jτa jiθa. The Noether current is determined by jµ=∂L ∂(∂µφi)δφi+δφ∗ i∂L ∂(∂µφ∗ i) =i 2θa/parenleftbig ∂µφ∗ iτa ijφj−φ∗ iτa ij∂µφj/parenrightbig . From the previous relation ( θaare constant independent parameters) it follows that the conserved currents are: ja µ=−i 2/parenleftbig ∂µφ∗ iτa ijφj−φ∗ iτa ij∂µφj/parenrightbig . The charges are Qa=−i 2/integraldisplay d3x(∂0φ∗ iτa ijφj−φ∗ iτa ij∂0φj). Chapter 5. Classical ﬁelds and symmetries. 125 5.13 The currents and charges are ja µ=1 2¯ψiγµτa ijψj,Qa=1 2/integraldisplay d3xψ† iτa ijψj. The equations of motion are (i γµ∂µ−m)ψi=0a n d ¯ψi(iγµ←−∂µ+m)=0.The current conversation law, ∂µjµa=0c a nb ep r o v e de a s i l y : 2∂µjµa=(∂µ¯ψi)γµτa ijψj+¯ψiγµτa ij∂µψj=im¯ψiτa ijψj+¯ψiτa ij(−imψj)=0 , where we used the equations of motion. The Noether theorem is valid on–shell. 5.14 (a) The phase invariance is the U(1) symmetry, where ψ→ψ/prime=eiθψand ¯ψ→¯ψ/prime=e−iθ¯ψ.The Noether current is jµ=¯ψγµψ,while the charge is given by Q=−e/integraltext d3xψ†ψ.Note that the current does not have additional indices since U(1) is a one–parameter group. (b)jµ=i (φ∗∂µφ−φ∂µφ∗),Q =iq/integraltext d3x(φ∗∂0φ−φ∂0φ∗). 5.15 The equations of motion are ( /unionsq /intersectionsq+m2)φi= 0. The expression φTφis invariant under SO(3) transformations, hence the Lagrangian density has thesame symmetry. The generators of SO(3) group are J 1= 00 0 00 −i 0i 0 ,J2= 00 i 00 0 −i00 ,J3= 0−i0 i00 000 .(5.1) Note that we can write (Jk)ij=−i/epsilon1kij. Under SO(3) transformations, the inﬁnitesimal variations of the ﬁelds are δφi=i (Jk)ijθkφj=/epsilon1kijθkφjand the Noether current is jµ=∂L ∂(∂µφi)δφi =/epsilon1kijφj∂µφiθk =−θ·(φ×∂µφ). The parameters of rotations θk, are arbitrary and therefore the currents jµ k=−/epsilon1kijφj∂µφi are also conserved. 5.16 First, derive the following formula eiαγ5=c o s α+iγ5sinα. The transfor- mation law for the Dirac Lagrangian density under the chiral transformationis given by 126 Solutions L→ψ†e−iαγ5γ0(iγµ∂µ−m)eiαγ5ψ = (cos2α+s i n2α)¯ψiγµ∂µψ−m¯ψ(cosα+iγ5sinα)2ψ =¯ψiγµ∂µψ−m¯ψ(cos 2α+iγ5sin 2α)ψ. From the previous expression we can conclude that the Lagrangian density is invariant only for massless fermions. The Noether current is jµ=¯ψγµγ5ψ. Prove that ∂µjµis proportional to the mass mof the ﬁeld. 5.17 The current is given by jµ=∂L ∂(∂µσ)δσ+∂L ∂(∂µπa)δπa+∂L ∂(∂µΨi)δΨi+δ¯Ψi∂L ∂(∂µ¯Ψi) =−/epsilon1abcαb∂µπaπc−1 2¯Ψiγµαaτa ijΨj. The ﬁnal result has the form jµ=π×∂µπ+1 2¯ΨγµτΨ. 5.18 (a) For translations, we have δxµ=/epsilon1µ,while the total variations of the ﬁelds equal zero. The Noether current is Tµν=∂L ∂(∂µφr)∂φr ∂xν−Lgµν. (5.2) The index νin (5.2) comes from the group of translations. For a real scalar ﬁeld, from (5.2) we obtain Tµν=∂µφ∂νφ−1 2/bracketleftbig (∂φ)2−m2φ2/bracketrightbig gµν. (5.3) The conserved charges are the Hamiltonian (for ν= 0), H=/integraldisplay d3xT00=1 2/integraldisplay d3x/bracketleftbig (∂0φ)2+(∇φ)2+m2φ2/bracketrightbig ,(5.4) and the momentum (for ν=i) Pi=/integraldisplay d3xT0i=/integraldisplay d3x∂0φ∂iφ. (5.5) For the Dirac ﬁeld the energy–momentum tensor is given by Tµν=i¯ψγµ∂νψ−Lgµν. The Hamiltonian and momentum are given by Chapter 5. Classical ﬁelds and symmetries. 127 H=/integraldisplay d3x¯ψ[−iγ∇+m]ψ, (5.6) P=−i/integraldisplay d3xψ†∇ψ. (5.7) For electromagnetic ﬁeld the energy–momentum tensor is Tµν=∂L ∂(∂µAρ)∂Aρ ∂xν−Lgµν from which we obtain Tµν=−Fµρ∂νAρ+1 4F2gµν. (5.8) For the Lorentz transformations δxν=ωνρxρand δφ=0,δψ=−i 4σνρωνρψ, δA µ=ων µAν, The Noether currents for scalar, spinor and electromagnetic ﬁeld are jµ=[xνTµρ−xρTµν]ωνρ, jµ=[1 2¯ψγµσνρψ+xνTµρ−xρTµν]ωνρ, (5.9) jµ=[FµρAν−FµνAρ+(xνTµρ−xρTµν)]ωνρ. Dropping the parameters of the Lorentz transformations ωνρ, the con- served currents have the form Mµνρ, and they are given by the expression in square brackets in (5.9). The angular-momentum is Mνρ=/integraltext d3xM0νρ. (b) As we see, the energy–momentum tensors for Dirac and electromagnetic ﬁelds are not symmetric. To ﬁnd the symmetrized energy–momentum ten- sors we employ the procedure given in the problem. For the Dirac ﬁeld we have χρµν=1 4(−¯ψγµσρνψ+¯ψγρσµνψ+¯ψγνσµρψ) =i 8¯ψ(4gµνγρ−4gρνγµ+γµγνγρ−γργνγµ)ψ. Using (4.43) we ﬁnd ∂ρχρµν=−i 4∂ν¯ψγµψ−i 4∂µ¯ψγνψ−3i 4¯ψγµ∂νψ +i 4¯ψγν∂µψ+gµνi 2(∂ν¯ψγνψ+¯ψ/∂ψ). The symmetrized energy–momentum tensor for Dirac ﬁeld is 128 Solutions ˜Tµν=i 4(¯ψγν∂µψ+¯ψγµ∂νψ−∂µ¯ψγνψ−∂ν¯ψγµψ)− −gµν(i 2∂ν¯ψγνψ−i 2¯ψ/∂ψ−m¯ψψ). Similarly we determine the symmetrized energy–momentum tensor for the electromagnetic ﬁeld. From transformation rule of the electromagnetic po- tential with respect to Lorentz transformations δAα=ωαβAβ≡1 2ωµν(Iµν)αβAβ, follows that (Iµν)αβ=gµαgνβ−gµβgνα. Then χρµν=FµρAνand the new energy–momentum tensor is ˜Tµν=−FµρFν ρ+1 4F2gµν. (5.10) If we introduce the electric and magnetic ﬁelds: F0i=−Ei,Fij= −/epsilon1ijkBk,then the components of energy–momentum tensor are: ˜T00=−F0iF0 i+1 4(2F0iF0i+FijFij) =E2+1 4(−2E2+2B2) =1 2(E2+B2), ˜T0i=−F0jFi j =/epsilon1ijkEjBk =(E×B)i, (5.11) ˜Tij=−EiEj+/epsilon1ikl/epsilon1jknBlBn+1 2(E2−B2)δij =−/parenleftbig EiEj+BiBj−δijT00/parenrightbig . From the expression (5.11) we conclude that ˜T00˜T0i,−˜Tijare the energy density of electromagnetic ﬁeld, the Poynting vector, and the components of the Maxwell stress tensor. 5.19 The variation of form is deﬁned by δ0φ(x)=φ/prime(x)−φ(x). From δ0φ=δφ−∂µφδxµ, where δφ=φ/prime(x/prime)−φ(x) is the total variation of a ﬁeld, it follows that the inﬁnitesimal form variation of φis δ0φ=ρ(φ(x)+xµ∂µφ). (5.12) Chapter 5. Classical ﬁelds and symmetries. 129 The induced change of the action is S/prime−S=1 2/integraldisplay d4x/prime/bracketleftbig (∂/primeφ/prime)2−m2φ/prime2(x/prime)/bracketrightbig −1 2/integraldisplay d4x/bracketleftbig (∂φ)2−m2φ2(x)/bracketrightbig . (5.13) The transformed volume of integration is given by d4x/prime=|J|d4x= det(e−ρI)d4x=e−4ρd4x. (5.14) The ﬁeld derivative is changed according to the following rule: ∂µφ(x)→∂φ/prime ∂x/primeµ=∂xν ∂x/primeµ∂ ∂xν(eρφ)=e2ρ∂µφ. (5.15) Thus, the change of the action is S/prime−S=1 2/integraldisplay d4xe−4ρ/bracketleftbig e4ρ(∂φ)2−m2e2ρφ2(x)/bracketrightbig −1 2/integraldisplay d4x/bracketleftbig (∂φ)2−m2φ2(x)/bracketrightbig =1 2m2(1−e−2ρ)/integraldisplay d4xφ2(x). For an inﬁnitesimal dilatation ( ρ/lessmuch1), the variation of the action is δS=m2ρ/integraldisplay d4xφ2(x). (5.16) From (5.16) it is clear that the theory of massless scalar ﬁeld is invariant under dilatations.The conserved current is j µ=−φ∂µφ−xν∂µφ∂νφ+Lxµ. (5.17) By calculating ∂µjµone obtains that ∂µjµis proportional to the mass m. 5.20 From d4x/prime=e−4ρd4x≈(1−4ρ)d4x, (5.18) and ¯ψ/prime(x/prime)γµ∂/prime µψ/prime(x/prime)=e4ρ¯ψγµ∂µψ≈(1 + 4 ρ)¯ψγµ∂µψ, (5.19) it follows that this transformation leaves the action unchanged. The Noether current is jµ=−3 2i¯ψγµψ−ixν¯ψγµ∂νψ+xµL. 6 Green functions 6.1The Green function of the Klein-Gordon equation satisﬁes the equation (/unionsq /intersectionsqx+m2)∆(x−y)=−δ(4)(x−y). (6.1) Fourier transformations of the Green function and the δ-function in (6.1) gives (/unionsq /intersectionsqx+m2)1 (2π)4/integraldisplay d4k˜∆(k)e−ik·(x−y)=−1 (2π)4/integraldisplay d4ke−ik·(x−y).(6.2) From (6.2) follows ˜∆(k)=1 k2−m2=1 k2 0−k2−m2. Then, the Green function is deﬁned by ∆(x−y)=/integraldisplayd4k (2π)41 k2 0−k2−m2e−ik·(x−y). (6.3) The integral (6.3) is divergent, since the integrand has the poles in k0=±ωk. We shall modify the contour of integration to make the integral (6.3) conver- gent. It is clear that we have to give the physical reasons for this modiﬁcation of integral. The poles can be evaded in four diﬀerent ways. The ﬁrst one isfrom the upper side (Fig. 6.1). The exponential term in (6.3) for large energy k 0behaves as e(x0−y0)Imk0, therefore the contour for x0>y0has to be closed from the lower side (Im k0<0), while in the case x0<y0we will close the integration contour on the upper side. By applying the Cauchy theorem we get ∆(x−y)=−1 (2π)4/integraldisplay d3keik·(x−y)2πi(Res ωk+R e s −ωk)θ(x0−y0).(6.4) From (6.4) follows 132 Solution ∆R=−i (2π)3/integraldisplayd3k 2ωkeik·(x−y)(e−iωk(x0−y0)−eiωk(x0−y0))θ(x0−y0).(6.5) ∆R(x−y)i sthe retarded Green function . The solution of the inhomogeneous equation ( /unionsq /intersectionsq+m2)φ=Jis φ(x)=−/integraldisplay d4y∆(x−y)J(y)+φ0, (6.6) where φ0is a solution of homogeneous equation. From the expressions (6.5) and (6.6) (because of θ−function), we conclude that we integrate over y0from −∞tox0. The value of the ﬁeld φat time x0is determined by the source Jat earlier times. For this reason this function is called the retarded Green function. Fig. 6.1. The integration contour for the retarded boundary conditions Fig. 6.2. The integration contour for the advanced boundary conditions By evading poles as in Fig. 6.2 we get the so-called advanced Green function ∆A=i (2π)3/integraldisplayd3k 2ωkeik·(x−y)(e−iωk(x0−y0)−eiωk(x0−y0))θ(y0−x0).(6.7) The advanced Green function contributes nontrivially to the ﬁeld φ(x)f o r y0>x0. If we evade poles as in Fig. 6.3, we get the Feynman propagator : Chapter 6. Green functions 133 Fig. 6.3. The integration contour which deﬁned the Feynman propagator Fig. 6.4. The integration contour for the Dyson Green function ∆F=i (2π)3/integraldisplay d3keik·(x−y)/bracketleftbig Res−ωkθ(y0−x0)−Resωkθ(x0−y0)/bracketrightbig =−i (2π)3/integraldisplayd3k 2ωkeik·(x−y)/bracketleftBig e−iωk(x0−y0)θ(x0−y0) (6.8) +eiωk(x0−y0)θ(y0−x0)/bracketrightBig . We can conclude that positive (negative) energy solutions propagate forward (backward) in spacetime. This is what we need in the relativistic quantumphysics in contrast to the classical theory (for example in classical electrody- namics), where all physically relevant information is contained in the retarded Green function. Dyson Green function is obtained by evading poles as in Fig. 6.4. This Green function can be evaluated in a way similar to the previous three cases. It is recommended to do this calculation as an exercise. 6.2From (6.5) and (6.8) it follows that (we take y=0 ) ∆ F(x)−∆R(x)=−i (2π)3/integraldisplayd3k 2ωkei(ωkt+k·x), (6.9) sinceθ(t)+θ(−t)=1.By applying ( /unionsq /intersectionsq+m2) on (6.9) we get (/unionsq /intersectionsq+m2)[∆F(x)−∆R(x)] = 0 . 6.3 I=/integraldisplay d4kδ(k2−m2)θ(k0)f(k) 134 Solution =/integraldisplay d4kδ(k2 0−ω2 k)θ(k0)f(k) =/integraldisplay d3kdk01 2ωk[δ(k0−ωk)+δ(k0+ωk)]θ(k0)f(k) =/integraldisplayd3k 2ωkf(k)/vextendsingle/vextendsingle/vextendsingle/vextendsingle k0=ωk. From this calculation it is clear that the expression d3k/(2ωk)i saL o r e n t z invariant measure. 6.5Let us take x0<0. The integral over the contour in Fig. 6.5 vanishes since there are no poles within the contour of integration. So, we get /integraldisplay−ωk−ρ −R+/integraldisplay C− ρ+/integraldisplayωk−ρ −ωk+ρ+/integraldisplay C+ ρ+/integraldisplayR ωk+ρ+/integraldisplay CR=0. (6.10) Fig. 6.5. The integration contour that deﬁned the principal-part propagator The integral along the half–circle, CRtends to zero for large R, which can be seen if we take that limit in the integrand. If in the integral/integraltext C+ ρwe take k0=ωk+ρeiϕ, it becomes /integraldisplay C+ ρ=/integraldisplay0 πie−ix0(ωk+ρeiϕ) 1 ρeiϕ+2ωkdϕ. (6.11) By taking ρ→0 in (6.11) we get /integraldisplay C+ ρ=−iπ 2ωke−iωkx0. (6.12) In the same way we can show that /integraldisplay C− ρ=iπ 2ωkeiωkx0. (6.13) From (6.10), (6.12) and (6.13) we get (for x0<0) Chapter 6. Green functions 135 ¯∆(x)=iπ (2π)4/integraldisplayd3k 2ωkeik·x/bracketleftbig e−iωkx0−eiωkx0/bracketrightbig θ(−x0). (6.14) The case x0>0 is analogous to the previous one. The result is ¯∆(x)=−iπ (2π)4/integraldisplayd3k 2ωkeik·x/bracketleftbig e−iωkx0−eiωkx0/bracketrightbig θ(x0). (6.15) By comparing equations (6.14) and (6.15) with the expressions for ∆Rand ∆Awe obtain ¯∆(x)=1 2(∆R(x)+∆A(x)). 6.6 ∆(x)=−i (2π)3/integraldisplayd3k 2ωkeik·x(e−iωkt−eiωkt), (6.16) ∆±(x)=∓i (2π)3/integraldisplayd3k 2ωkei(k·x∓ωkt). (6.17) 6.7By using the expression for ∆obtained in Problem 6.6 we get ∂i∆(x)=−i (2π)3/integraldisplayd3k 2ωkikieik·x(e−iωkt−eiωkt)=0 , (6.18) since the integrand is an odd function of k. The second identity can be proven easily. 6.8By applying the operator ( /unionsq /intersectionsq+m2) to the expression (6.16) we get (/unionsq /intersectionsq+m2)∆(x)=−i (2π)3/integraldisplayd3k 2ωk(−ω2 k+k2+m2)/bracketleftBig ei(−ωkt+k·x)−ei(ωkt+k·x)/bracketrightBig , from which follows that ( /unionsq /intersectionsq+m2)∆(x) = 0, as k2=m2. 6.9Form= 0 from (6.8) it follows that ∆F|m=0=−i (2π)3/integraldisplayd3k 2keik·x/bracketleftBig e−ikx0θ(x0)+eikx0θ(−x0)/bracketrightBig =−i 2(2π)2/integraldisplay∞ 0/integraldisplayπ 0ksinθdkdθ ×/bracketleftBig eik(−t+rcosθ)θ(t)+eik(t+rcosθ)θ(−t)/bracketrightBig , (6.19) where in the second line we integrated over the polar angle ϕ. Integration over θgives ∆F(x)|m=0=−1 2(2π)2r/integraldisplay∞ 0dk/bracketleftBig (e−ik(t−r)−e−ik(t+r))θ(t) +(eik(t+r)−eik(t−r))θ(−t)/bracketrightBig . (6.20) 136 Solution Now, we shall consider separately two cases: t>0a n d t<0. In the ﬁrst one,t>0 the second term in the integrand of (6.20) is zero. The ﬁrst part of the integrand has bad behavior for large k. We regularize it by making substitution t→t−i/epsilon1, where /epsilon1→0+.In this way we ensure convergence of this integral. Then from (6.20) it follows that ∆F|m=0=i 2(2π)2r/parenleftbigg1 t−r−i/epsilon1−1 t+r−i/epsilon1/parenrightbigg (6.21) =i (2π)21 t2−r2−i/epsilon1=i (2π)21 x2−i/epsilon1. (6.22) By applying the formula 1 z±i/epsilon1=P1 z∓iπδ(z), (6.23) in expression (6.22) we get ∆F(x)|m=0=−1 4πδ(x2)+i 4π2P1 x2. (6.24) For the case t<0 one also obtains the expression (6.24); this is left as an exercise. 6.10 We shall start from (6.5) and use spherical coordinates. Integration over angles θandϕleads to ∆R(x)=−1 2(2π)2r/integraldisplay∞ 0dk/bracketleftBig e−ik(t−r)−eik(t+r)−e−ik(t+r)+eik(t−r)/bracketrightBig θ(t). (6.25) The change of variable k/prime=−kin the third and the fourth integral in expres- sion (6.25) gives ∆R(x)=−1 2(2π)2r/integraldisplay∞ −∞dk(e−ik(t−r)−eik(t+r))θ(t). (6.26) Note the change of the lower integration limit in the expression (6.26). From (6.26) follows ∆R|m=0(x)=−1 4πr[δ(t−r)−δ(t+r)]θ(t). (6.27) The second term in (6.27) has a ”wrong” sign but it is irrelevant as this term vanishes ( t>0a n d r>0). By changing this minus into a plus in (6.27) we ﬁnally obtain: ∆R|m=0(x)=−1 2πδ(t2−r2)θ(t)=−1 2πδ(x2)θ(t). (6.28) The case of advanced Green function is left for an exercise. Chapter 6. Green functions 137 6.11 In the Problem 6.1, we modiﬁed the the contour of integration according to the boundary conditions, while the poles were not moved. Sometimes it is useful to do the opposite, i.e. to move the poles and to integrate over the real k0–axis. For the retarded Green function this can be done by changing k2−m2→k2−m2+iηk0 in the propagator denominator, where ηis a small positive number. Therefore, ∆R(x−y)=/integraldisplayd4k (2π)4e−ik·(x−y) k2−m2+iηk0. (6.29) Now the poles of the integrand in (6.29) are k0=±ωk−iη/2.From (6.6) and (6.29) we have φR(x)=−g (2π)4/integraldisplay d4ke−ik·x k2−m2+iηk0/integraldisplay dy0eik0y0/integraldisplay d3yδ(3)(y)e−ik·y. (6.30) First in (6.30) we shall integrate over y0, then over yand ﬁnally over k0; this gives φR(x)=g (2π)3/integraldisplay d3keik·x k2+m2. (6.31) In order to compute this three-dimensional momentum integral we introduce spherical coordinates; also we take x=rez. The angular integrations give (in one integral use the change k/prime=−k) φR(x)=−g (2π)2ir/integraldisplay∞ −∞kdk k2+m2e−ikr. (6.32) Fig. 6.6. The integral in (6.32) has the poles at k0=±im. The integration contour is given in Fig. 6.6. By applying the Cauchy theorem in (6.32) we obtain: φR(x)=g 4πre−mr, (6.33) which is the requested result. 138 Solution 6.12 Apply i/ ∂−monS(x). 6.13 The Fourier transformation of the equation (i/ ∂−m)S(x−y)=δ(4)(x−y) leads to (i/∂−m)1 (2π)4/integraldisplay d4p˜S(p)e−ip·(x−y)=1 (2π)4/integraldisplay d4pe−ip·(x−y).(6.34) From (6.34) follows ˜S(p)=/p+m p2−m2. Therefore, the Green function is given by S(x−y)=/integraldisplayd4p (2π)4/p+m p2 0−p2−m2e−ip·(x−y). (6.35) The poles of the integrand in (6.35) are p0=±Ep=±/radicalbig p2+m2.The propagator is SF(x−y)=1 (2π)4/integraldisplay d3peip·(x−y)/integraldisplay CFdp0p0γ0+piγi+m p2 0−E2pe−ip0(x0−y0), (6.36) where the integration contour CFis deﬁned in Problem 6.1. Applying the Cauchy theorem we get SF(x−y)=−i (2π)3/integraldisplayd3p 2Epeip·(x−y) /bracketleftBig (Epγ0+piγi+m)e−iEp(x0−y0)θ(x0−y0)+ +(−Epγ0+piγi+m)eiEp(x0−y0)θ(y0−x0)/bracketrightBig =−i (2π)3/integraldisplayd3p 2Ep/bracketleftBig (/p+m)e−ip·(x−y)θ(x0−y0)− −(/p−m)eip·(x−y)θ(y0−x0)/bracketrightBig . (6.37) The advanced Green function can be found in the same way. The result is SA(x−y)=i (2π)3/integraldisplayd3p 2Epeip·(x−y)/bracketleftBig (Epγ0+piγi+m)e−iEp(x0−y0)− −(−Epγ0+piγi+m)eiEp(x0−y0)/bracketrightBig θ(y0−x0). (6.38) For simplicity we take y= 0 in (6.37) and (6.38). We have SF−SA=−i (2π)3/integraldisplayd3p 2Epei(p·x−Epx0)(Epγ0+piγi+m)(θ(x0)+θ(−x0)) =−i (2π)3/integraldisplayd3p 2Epei(p·x−Epx0)(Epγ0+piγi+m). (6.39) Chapter 6. Green functions 139 Thus, SF−SA=−i (2π)3/integraldisplayd3p 2Ep(Epγ0+piγi+m)e−ip·x. (6.40) By applying i/ ∂−mon (6.40) we get (i/ ∂−m)(SF−SA)=0,since (/p+m)(/p−m)=p2−m2=0. 6.14 The integration along the curve CFis equivalent to the integration along the real p0–axis if we make the replacement p2−m2→p2−m2+i/epsilon1,where /epsilon1 is a small positive number in the propagator denominator. The simple poles arep0=±Ep∓i/epsilon1.S ow eg e t ψ(x)=g (2π)4/integraldisplay d4y/integraldisplay dp0/integraldisplay d3p/p+m p2−m2+i/epsilon1e−ip·(x−y)δ(y0)eiq·y 1 0 00 . After the integration over the variables y 0andywe get ψ(x)=g 2π/integraldisplay dp0d3p/p+m p2−m2+i/epsilon1e−i(p0x0−p·x)δ(3)(p−q) 1 0 00 .(6.41) Integration over the momentum pis simple and it gives ψ(x)=g 2πeiq·x/integraldisplay∞ −∞dp0p0γ0−q·γ+m p2 0−q2−m2+i/epsilon1e−ip0x0 1 0 0 0 . (6.42) Employing the Cauchy theorem we ﬁnd that ψ(x)=−ig 2Eqeiq·x/bracketleftbig (−Eqγ0−q·γ+m)eiEqx0θ(−x0) +(Eqγ0−q·γ+m)e−iEqx0θ(x0)/bracketrightbig 1 0 0 0 , (6.43) which ﬁnally gives: ψ(x)=−ig 2Eqeiq·x × eiEqx0 −Eq+m 0 q3 q1+iq2 θ(−x0)+e−iEqx0 Eq+m 0 q3 q1+iq2 θ(x0) .(6.44) 140 Solution 6.15 The equation for the free massive vector ﬁeld Aµis given by (gρσ/unionsq/intersectionsq−∂ρ∂σ+m2gρσ)Aσ=0. (6.45) The Green function (it is in fact the inverse kinetic operator) is deﬁned by (gρσ/unionsq/intersectionsq−∂ρ∂σ+m2gρσ)xGσν(x−y)=δ(4)(x−y)δρ ν. (6.46) If we introduce Gσν=1 (2π)4/integraldisplay d4ke−ik·(x−y)˜Gσν(k), in (6.46), we get (−k2gρσ+kρkσ+m2gρσ)˜Gσν=δρ ν. (6.47) We shall assume that the solution of (6.47) has the form ˜Gρσ=Ak2gρσ+ Bkρkσ,where AandBare scalars, i.e. they depend on k2andm2. Inserting the solution into (6.47), after comparing of the appropriate coeﬃcients, we get A=1 −k4+k2m2,B =−1 m2(m2−k2). The ﬁnal result takes the following form ˜Gµν=1 k2−m2/parenleftbigg −gµν+kµkν m2/parenrightbigg . (6.48) 6.16 Use the same procedure as in the previous problem. The result is ˜Gµν=−gµν k2+1+λ λk4kµkν. 7 Canonical quantization of the scalar ﬁeld 7.1Starting from the expressions for scalar ﬁeld φand its canonical momen- tumπ=˙φ, φ=/integraldisplayd3k/radicalbig 2(2π)3ωk/bracketleftbig a(k)e−ik·x+a†(k)eik·x/bracketrightbig , ˙φ=i/integraldisplayd3k/radicalbig (2π)32ωkωk/bracketleftbig −a(k)e−ik·x+a†(k)eik·x/bracketrightbig , we have /integraldisplay d3xφ(x)e−ik/prime·x=(2π)3/2 √2ωk/prime/bracketleftbig a(k/prime)e−iωk/primet+a†(−k/prime)eiωk/primet/bracketrightbig , (7.1) /integraldisplay d3x˙φ(x)e−ik/prime·x= i(2π)3/2/radicalbiggωk/prime 2/bracketleftbig a†(−k/prime)eiωk/primet−a(k/prime)e−iωk/primet/bracketrightbig .(7.2) From (7.1) and (7.2) it follows that a(k)=1 (2π)3/21√2ωk/integraldisplay d3xeik·x/bracketleftBig ωkφ(x)+i˙φ(x)/bracketrightBig , (7.3) a†(k)=1 (2π)3/21√2ωk/integraldisplay d3xe−ik·x/bracketleftBig ωkφ(x)−i˙φ(x)/bracketrightBig . (7.4) By using the expressions (7.3) and (7.4), we ﬁnd: [a(k),a†(k/prime)] =i 2(2π)31√ωkωk/prime/integraldisplay d3xd3yei(k·x−k/prime·y)/parenleftBig −ωk[φ(x),˙φ(y)]+ +ωk/prime[˙φ(x),φ(y)]/parenrightBig =1 2(2π)31√ωkωk/prime/integraldisplay d3xei(ωk−ωk/prime)t+i(k/prime−k)·x(ωk+ωk/prime) =δ(3)(k−k/prime). (7.5) 142 Solutions In the previous formula, we used the equal–time commutation relations for real scalar ﬁeld (7.C) i.e. we took1x0=y0. We can do this because the creation and annihilation operators are time independent. This can be proved directly: da(k) dt=1 (2π)3/21√2ωk/integraldisplay d3xeik·x/bracketleftbig iω2 kφ+i∇2φ−im2φ/bracketrightbig . After two partial integrations in the second term we get da(k) dt=i (2π)3/21√2ωk/integraldisplay d3xeik·x/bracketleftbig ω2 k−k2−m2/bracketrightbig φ. The dispersion relation, ω2 k=m2+k2gives d a(k)/dt=0.It is clear that a†(k) is also time independent. Similarly, we can prove that: [a(k),a(k/prime)] = [a†(k),a†(k/prime)] = 0 . 7.2In this problem, φ(x) is a classical ﬁeld, so that a(k)a n d a†(k)a r et h e coeﬃcients rather then operators. We can calculate them from the expressions (7.3) and (7.4) inserting φ(t=0,x)=0a n d ˙φ(t=0,x)=c: a(k)=1 (2π)3/21√2ωk/integraldisplay d3xe−ik·xic =ic√ 2m(2π)3/2δ(3)(k). Then, the scalar ﬁeld is φ(t,x)=c msin(mt). Generally, if we know a ﬁeld and its normal derivative on some space–like surface σ, then the ﬁeld at an arbitrary point is given by φ(y)=/integraldisplay σ[φ(x)∂x µ∆(x−y)−∆(y−x)∂µφ(x)]dΣµ. Solve this problem using the previous theorem. 7.3The results are: :H:=/integraldisplay d3kωk/bracketleftbig a†(k)a(k)+b†(k)b(k)/bracketrightbig , (7.6) :Q:=q/integraldisplay d3k/bracketleftbig a†(k)a(k)−b†(k)b(k)/bracketrightbig , (7.7) :P:=/integraldisplay d3kk/bracketleftbig a†(k)a(k)+b†(k)b(k)/bracketrightbig . (7.8) 1This will be done in the forthcoming problems, too. Chapter 7. Canonical quantization of the scalar ﬁeld 143 7.4(up,uk)=δ(3)(k−p), (up,u∗ k)=0. 7.5From (2.9), we have /angbracketleft0|H|0/angbracketright=1 2/integraldisplay d3kωk/angbracketleft0|(a†(k)a(k)+a(k)a†(k))|0/angbracketright =1 2/integraldisplay d3kωk/angbracketleft0|a(k)a†(k)|0/angbracketright =1 2/integraldisplay d3kωk(δ(3)(0)−/angbracketleft0|a†(k)a(k)|0/angbracketright) =1 2δ(3)(0)/integraldisplay d3k/radicalbig k2+m2 =2πδ(3)(0)/integraldisplay∞ 0dkk2/radicalbig k2+m2. By change of variable k=m√ t, the last integral becomes Euler’s beta function /angbracketleft0|H|0/angbracketright=πm4δ(3)(0)B(3 2,−2) =−πm4 4δ(3)(0)Γ(−2). 7.6Use the formulae from Problem 7.3 and the commutation relations (7.D). (a) Direct calculation yields [Pµ,φ]=1 (2π)3/2/integraldisplayd3kd3k/prime √2ωk/primekµ/bracketleftBig a†(k)a(k),a(k/prime)e−ik/prime·x+a†(k/prime)eik/prime·x/bracketrightBig =1 (2π)3/2/integraldisplayd3k√2ωkkµ/parenleftbig −a(k)e−ik·x+a†(k)eik·x/parenrightbig =−i∂µφ. (7.9) The same result can be obtained if we start from the transformation law of the ﬁeld φunder translations (see Problem 7.20): φ(x+/epsilon1)=ei/epsilon1·Pφ(x)e−i/epsilon1·P=φ(x)+i/epsilon1µ[Pµ,φ(x)] +o(/epsilon12). (7.10) On the other hand, we have φ(x+/epsilon1)=φ(x)+/epsilon1µ∂µφ+o(/epsilon12). (7.11) From (7.11) and (7.10) the result (7.9) comes. (b) First, we calculate the commutator [ Pµ,φn(x)]: [Pµ,φn(x)] =n/summationdisplay k=1φk−1[Pµ,φ]φn−k =n/summationdisplay k=1φk−1(−i∂µφ)φn−k =−i∂µφn. 144 Solutions In the same way one can prove that [Pµ,πn(x)] =−i∂µπn. As a consequence, [Pµ,φn(x)πm(x)] =−i∂µ(φn(x)πm(x)). An arbitrary analytical function F(φ,π) can be expanded in series as F(φ,π)=/summationdisplay nmCnmφnπm. Then [Pµ,F(φ,π)] =−i∂µF. (c) [H,a†(k)a(q)] = (ωk−ωq)a†(k)a(q). (d) [Q,Pµ]=0 . (e) [H,N]=0 . (f)/integraltext d3x[H,φ(x)]e−ip·x=( 2π)3/2/radicalBig ωp 2/parenleftbig −a(p)e−iωpt+a†(−p)eiωpt/parenrightbig 7.7From the Baker–Hausdorﬀ relation follows eiQφe−iQ=φ+i [Q,φ]+i2 2![Q,[Q,φ]] +... . (7.12) The ﬁrst commutator in the previous expansion is given by [Q,φ]=iq/integraldisplay d3y[φ†(y)π†(y)−φ(y)π(y),φ(x)] =−q/integraldisplay d3yδ(3)(x−y)φ(y)=−qφ(x). Then [Q,[Q,φ]] = (−q)2φ,[Q,[Q,[Q,φ]]] = (−q)3φ ,... (7.13) Finally, eiQφe−iQ=/parenleftbigg 1−iq+(−iq)2 2+.../parenrightbigg φ=e−iqφ. (7.14) 7.8The angular momentum of a scalar ﬁeld has the form Mµν=/integraldisplay d3x(xµT0ν−xνT0µ). (a) By inserting the previous formula in the commutator, we have [Mµν,φ(x)] =/integraldisplay d3y[yµ(˙φ∂νφ−g0νL)−yν(˙φ∂µφ−g0µL),φ(x)].(7.15) The following equal–time commutators can be easily evaluated: Chapter 7. Canonical quantization of the scalar ﬁeld 145 [L(y),φ(x)] =−iδ(3)(x−y)π(y), [π(y)∂µφ(y),φ(x)] =−i∂µφδ(3)(x−y)−iδµ0π(y)δ(3)(x−y). By substituting these expressions in (7.15) and performing integration, we get [Mµν,φ(x)] = i( xν∂µ−xµ∂ν)φ(x). (7.16) The same result can be obtained if we start from the transformation law for the ﬁeld φ(x) under Lorentz transformations, ei 2ωµνMµνφ(x)e−i 2ωµνMµν=φ(Λ−1(ω)x). (b) We ﬁrst calculate the commutator [ Mµν,P0]: [Mµν,P0]=/integraldisplay d3x[xµT0ν−xνT0µ,P0] =/integraldisplay d3x(xµ[T0ν,P0]−xν[T0µ,P0]) =i/integraldisplay d3x(xµ∂0T0ν−xν∂0T0µ) =i/integraldisplay d3x/parenleftbig −xµ∂iTi ν+xν∂iTi µ/parenrightbig =i/integraldisplay d3x/parenleftbig gµiTi ν−giνTi µ/parenrightbig =i/integraldisplay d3x/parenleftbig Tµν−gµ0T0 ν−Tνµ+g0νT0 µ/parenrightbig =−i(gµ0Pν−gν0Pµ). (7.17) In (7.17), we used the results of Problem 7.6 (b), the continuity equation ∂µTµν= 0 and integrated by parts. In the case λ=iwe can use of a partial integration. The result is [ Mµν,Pi]=−i(giµPν−giνPµ). Thus, [Mµν,Pλ]=i (gλνPµ−gλµPν). (7.18) (c) Let us calculate ﬁrstly the commutator [ Mij,Mkl]. 146 Solutions [Mij,Mkl]=/integraldisplay d3xd3y/bracketleftBig xi˙φ(x)∂jφ(x)−xj˙φ(x)∂iφ(x), yk˙φ(y)∂lφ(y)−yl˙φ(y)∂kφ(y)/bracketrightBig =/integraldisplay d3xd3y/parenleftBig xiyk[˙φ(x)∂jφ(x),˙φ(y)∂lφ(y)]− −xiyl[˙φ(x)∂jφ(x),˙φ(y)∂kφ(y)] −xjyk[˙φ(x)∂iφ(x),˙φ(y)∂lφ(y)] +xjyl[˙φ(x)∂iφ(x),˙φ(y)∂kφ(y)]/parenrightBig . (7.19) Applying the equal–time commutation relations, we obtain2 [Mij,Mkl]=i/integraldisplay d3xd3y/bracketleftBig xiyk/parenleftBig ˙φ(x)∂lφ(y)∂x j −˙φ(y)∂jφ(x)∂y l/parenrightBig δ(3)(x−y) −xiyl/parenleftBig ˙φ(x)∂kφ(y)∂x j−˙φ(y)∂jφ(x)∂y k/parenrightBig δ(3)(x−y) −xjyk/parenleftBig ˙φ(x)∂lφ(y)∂x i−˙φ(y)∂iφ(x)∂y l/parenrightBig δ(3)(x−y) +xjyl/parenleftBig ˙φ(x)∂kφ(y)∂x i−˙φ(y)∂iφ(x)∂y k/parenrightBig δ(3)(x−y)/bracketrightBig . If we use the relation ∂x mδ(3)(x−y)=−∂y mδ(3)(x−y) we obtain [Mij,Mkl]=−i/integraldisplay d3xd3y /bracketleftbigg xiyk/parenleftBig ˙φ(x)∂lφ(y)∂y jδ(3)(x−y)−˙φ(y)∂jφ(x)∂x lδ(3)(x−y)/parenrightBig −xiyl/parenleftBig ˙φ(x)∂kφ(y)∂y jδ(3)(x−y)−˙φ(y)∂jφ(x)∂x kδ(3)(x−y)/parenrightBig −xjyk/parenleftBig ˙φ(x)∂lφ(y)∂y iδ(3)(x−y)−˙φ(y)∂iφ(x)∂x lδ(3)(x−y)/parenrightBig +xjyl/parenleftBig ˙φ(x)∂kφ(y)∂y iδ(3)(x−y)−˙φ(y)∂iφ(x)∂x kδ(3)(x−y)/parenrightBig/bracketrightbigg . By performing partial integrations in the last expression, we obtain 2We have used the following notation: ∂x m=∂ ∂xm;∂m x=∂ ∂xm. Chapter 7. Canonical quantization of the scalar ﬁeld 147 [Mij,Mkl]=−i/integraldisplay d3x/bracketleftbigg gjk(xl˙φ(x)∂iφ(x)−xi˙φ(x)∂lφ(x)) +gil(xk˙φ(x)∂jφ(x)−xj˙φ(x)∂kφ(x)) +gik(xj˙φ(x)∂lφ(x)−xl˙φ(x)∂jφ(x)) +gjl(xi˙φ(x)∂kφ(x)−xk˙φ(x)∂iφ(x))/bracketrightbigg =i (gjkMil+gliMjk−gikMjl−gjlMik). (7.20) The next two commutators [ Mij,M0k], [M0j,M0k] can be evaluated in the same way. Do this explicitly, please. 7.10 (a) The commutator is given by [Qa,Qb]=−1 4/integraldisplay d3xd3yτa ijτb mn ×/bracketleftBig ˙φ† i(x)φj(x)−φ† i(x)˙φj(x),˙φ† m(y)φn(y)−φ† m(y)˙φn(y)/bracketrightBig . Recall that as the charges are time-independent we can work with the equal–time commutators and we have [Qa,Qb]=−i 4/integraldisplay d3x/parenleftBig ˙φ†[τa,τb]φ−φ†[τa,τb]˙φ/parenrightBig . By using [ τa,τb]=2 i/epsilon1abcτc,w eg e t [Qa,Qb]=i/epsilon1abcQc. The second case is similar to the previous one: [Qi,Qj]=/epsilon1imn/epsilon1jpq/integraldisplay d3x/integraldisplay d3y[φm(x)˙φn(x),φp(y)˙φq(y)] =i/integraldisplay d3x(−/epsilon1imn/epsilon1jnqφm˙φq+/epsilon1imn/epsilon1jpmφp˙φn) =i/integraldisplay d3x(δijφm˙φm−φj˙φi−δijφm˙φm+φi˙φj) =i/integraldisplay d3x(φi˙φj−φj˙φi) =i/epsilon1ijk/epsilon1kmn/integraldisplay d3xφm˙φn =i/epsilon1ijkQk. As in the ﬁrst part of this problem, we used the equal–time commutation relations and the formula for appropriate product of two three–dimensional /epsilon1symbols. 148 Solutions (b) The commutator between the charges Qaand the ﬁeld φmcan be found similarly: [Qa,φm(x)] =−i 2/integraldisplay d3yτa ij[˙φ† i(y)φj(y)−φ† i(y)˙φj(y),φm(x)] =−i 2τa ij/integraldisplay d3y[˙φ† i(y),φm(x)]φj(y) =−1 2τa ij/integraldisplay d3yδ(3)(x−y)δimφj(y) =−1 2τa mjφj(x). In the same way, we ﬁnd: [Qa,φ† m(x)] =1 2τa imφ† i. The previous two results can be rewritten in the form [θaQa,φm(x)] = iδ0φm(x), [θaQa,φ† m(x)] = iδ0φ† m(x). In the case of SO(3) symmetry, the calculation is the same as above. The result is [Qk,φm(x)] = i/epsilon1kmjφj(x). 7.11 The dilatation current is jµ=−φ∂µφ−xν∂µφ∂νφ+Lxµ. (a) The dilatation generator is D=−/integraldisplay d3x/parenleftbigg φ˙φ+xi˙φ∂iφ+1 2x0(˙φ2−∂iφ∂iφ)/parenrightbigg . (b) The commutator between the generator Dand the ﬁeld φ(x) is given by [D,φ(y)] =−/integraldisplay d3x[φ(x)π(x)+xiπ(x)∂iφ(x) +1 2x0π2(x)−1 2x0∂iφ(x)∂iφ(x),φ(y)] =−/integraldisplay d3x/parenleftbig φ(x)[π(x),φ(y)] +x0π(x)[π(x),φ(y)] +xi[π(x),φ(y)]∂iφ(x)/parenrightbig . By using the commutation relations (7.C), we have Chapter 7. Canonical quantization of the scalar ﬁeld 149 ρ[D,φ(y)] = iρ(φ(y)+y0π(y)+yi∂iφ) =iρ(φ(y)+yµ∂µφ(y)) = iδ0φ. In the same way, we obtain: ρ[D,π(x)] = iρ(2π+xµ∂µπ)=iδ0π. (c) By applying the previous result, we easily get ρ[D,φ2]=ρ([D,φ]φ+φ[D,φ]) = i((δ0φ)φ+φδ0φ)=iδ0(φ2), and generally ρ[D,φa]=iδ0(φa). Similarly, one can show that ρ[D,πa]=iδ0(πa). An arbitrary analytic function can be expanded in the following form F(φ,π)=/summationdisplay abcabφaπb, so that ρ[D,F]=ρ/summationdisplay a,bcab[D,φaπb] =ρ/summationdisplay a,bcab/parenleftbig [D,φa]πb+φb[D,πb]/parenrightbig =i/summationdisplay a,bcab/parenleftbig δ0(φa)πb+φaδ0(πb)/parenrightbig =iδ0 /summationdisplay a,bcabφaπb  =iδ0F. (d) We ﬁrst consider the case µ=i: [D,Pi]=/integraldisplay d3x[D,π∂iφ] =/integraldisplay d3x/parenleftbig π[D,∂iφ]+[D,π]∂iφ/parenrightbig . By using part (b) of this problem, we obtain 150 Solutions [D,Pi]=i/integraldisplay d3x/bracketleftbig (2π+x0∂0π+xj∂jπ)∂iφ +π(2∂iφ+x0∂iπ+xj∂i∂jφ)/bracketrightbig . (7.21) The second term in this expression is transformed in the following way /integraldisplay d3xx0∂k∂kφ∂iφ=−/integraldisplay d3xx0∂kφ∂k∂iφ=−1 2/integraldisplay d3x∂i(x0∂kφ∂kφ), where we used the Klein-Gordon equation, ∂0π=−∂i∂iφand then per- formed a partial integration. Thus, we conclude that the second term canbe dropped as a surface term. The expression/integraltext d 3xπx0∂iπis also a surface term. Similarly, one can show that /integraldisplay d3xxj∂jπ∂iφ=−3/integraldisplay d3xπ∂iφ−/integraldisplay d3xxjπ∂j∂iφ. Inserting these results in the formula (7.21) we obtain [D,Pi]=iPi. The commutator [ D,P0]=iP0can be calculated in the same way. 7.12 In the expression for the vacuum expectation value, express the ﬁelds φfin terms of the creation and annihilations operators. From four terms, only one, which is proportional to /angbracketleft0|a(k)a†(k/prime)|0/angbracketright=δ(3)(k−k/prime), is nonzero. Then, we have /angbracketleft0|φf(t,x)φf(t,x)|0/angbracketright=1 (a2π)31 (2π)3/integraldisplayd3k 2ωk/parenleftbigg/integraldisplay d3ye−(x−y)2/a2+ik·(x−y)/parenrightbigg2 . Calculating the Poisson integral in this formula, we obtain /angbracketleft0|φf(t)φf(t)|0/angbracketright=1 2(2π)3/integraldisplayd3k ωke−k2a2/2 =1 (2π)2/integraldisplay∞ 0k2dk√ k2+m2e−k2a2/2. By the change of variable k2=t, the last integral becomes /angbracketleft0|φf(t)φf(t)|0/angbracketright=1 8π2/integraldisplay∞ 0√ tdt√ t+m2e−ta2/2 =m2 16π2em2a2/4/bracketleftbigg K1(m2a2 4)−K0(m2a2 4)/bracketrightbigg ,(7.22) where Kν(x) are modiﬁed Bessel functions of the third kind (MacDonald functions). Using the asymptotic expansions: Chapter 7. Canonical quantization of the scalar ﬁeld 151 K1(x)=1 x, K0(x)=−(log(x/2) + 0 ,5772) forx/lessmuch1, we obtain in the limit m→0 /angbracketleft0|φf(t)φf(t)|0/angbracketright=1 4π2a2. 7.13 Express the operators LmandLnin terms of αµ mand use the commu- tation relations. 7.14 After a very simple calculation, we ﬁnd that /angbracketleft0|{φ(x),φ(y)}|0/angbracketright=i 2(2π)21 |x−y|lim/epsilon1→0/integraldisplay∞ 0dke−/epsilon1k/parenleftBig eik(y0−x0−|x−y|) −eik(y0−x0+|x−y|)+eik(x0−y0−|x−y|) −eik(x0−y0+|x−y|)/parenrightBig . (7.23) The integrals in the previous expression are regularized by introducing /epsilon1as a regularization parameter. At the end we have to take the limit /epsilon1→0. The result is /angbracketleft0|{φ(x),φ(y)}|0/angbracketright=−1 2π21 (x−y)2. 7.15 The vacuum expectation value /angbracketleftφ(x)φ(y)/angbracketrightis given by /angbracketleftφ(x)φ(y)/angbracketright=/angbracketleftbig φ+(x)φ−(y)/angbracketrightbig =/integraldisplayd3k (2π)3/2√2ωkd3q (2π)3/2/radicalbig2ωqei(q·y−k·x)δ(3)(k−q) =1 (2π)3/integraldisplayd3k 2ωke−ik·(x−y), where we split the ﬁeld φinto positive and negative energy parts, φ=φ++φ−. If we do the same in the vacuum expectation value of four scalar ﬁelds, we see that only two terms remain: /angbracketleftφ(x1)φ(x2)φ(x3)φ(x4)/angbracketright=/angbracketleftbig φ+(x1)φ+(x2)φ−(x3)φ−(x4)/angbracketrightbig +/angbracketleftbig φ+(x1)φ−(x2)φ+(x3)φ−(x4)/angbracketrightbig .(7.24) The ﬁrst term in the last expression is /angbracketleftbig φ+(x1)φ+(x2)φ−(x3)φ−(x4)/angbracketrightbig =4/productdisplay i=1/integraldisplayd3qi (2π)3/2√2ωi/angbracketleftBig a1a2a† 3a†4/angbracketrightBig ×ei(−q1·x1−q2·x2+q3·x3+q4·x4), 152 Solutions where ai=a(qi). Using the relation /angbracketleftBig a1a2a† 3a†4/angbracketrightBig =/angbracketleftBig a1(δ(3)(q2−q3)+a† 3a2)a† 4/angbracketrightBig =δ(3)(q2−q3)δ(3)(q1−q4)+/angbracketleftBig a1a† 3(δ(3)(q2−q4)−a† 4a2)/angbracketrightBig =δ(3)(q2−q3)δ(3)(q1−q4)+δ(3)(q1−q3)δ(3)(q2−q4), we obtain /angbracketleftbig φ+(x1)φ+(x2)φ−(x3)φ−(x4)/angbracketrightbig =1 (2π)6/integraldisplayd3q1 2ω1d3q2 2ω2e−iq2·(x2−x3)−iq1·(x1−x4) +1 (2π)6/integraldisplayd3q1 2ω1d3q2 2ω2e−iq2·(x2−x4)−iq1·(x1−x3) =/angbracketleftφ(x2)φ(x3)/angbracketright/angbracketleftφ(x1)φ(x4)/angbracketright +/angbracketleftφ(x1)φ(x3)/angbracketright/angbracketleftφ(x2)φ(x4)/angbracketright. The following result can be derived in the same way: /angbracketleftbig φ+(x1)φ−(x2)φ+(x3)φ−(x4)/angbracketrightbig =/angbracketleftφ(x1)φ(x2)/angbracketright/angbracketleftφ(x3)φ(x4)/angbracketright. By adding two last expressions, we get /angbracketleftφ(x1)φ(x2)φ(x3)φ(x4)/angbracketright=/angbracketleftφ(x1)φ(x3)/angbracketright/angbracketleftφ(x2)φ(x4)/angbracketright +/angbracketleftφ(x1)φ(x4)/angbracketright/angbracketleftφ(x2)φ(x3)/angbracketright+ +/angbracketleftφ(x1)φ(x2)/angbracketright/angbracketleftφ(x3)φ(x4)/angbracketright. This result is a special case of Wick’ s theorem. 7.16 Scalar ﬁeld in two dimensional spacetime can be represented as φ(x)=/integraldisplay∞ −∞dk/radicalbig (2π)2ωk/bracketleftBig a(k)e−ikµxµ+a†(k)eikµxµ/bracketrightBig , so that /angbracketleftφ(x)φ(y)/angbracketright=1 4π/integraldisplay∞ −∞dk |k|ei|k|(y0−x0)−ik(y−x). (7.25) If we introduce the notation y0−x0=τ,y−x=r, the previous integral becomes /angbracketleftφ(x)φ(y)/angbracketright=1 4π/integraldisplay∞ 0dk k/parenleftBig eik(τ−r)+eik(τ+r)/parenrightBig . (7.26) Denoting the integral in (7.26) by Iand introducing the regularization para- meter /epsilon1, we get: ∂I ∂τ=i 4πlim/epsilon1→0/integraldisplay∞ 0dke−/epsilon1k/parenleftBig eik(τ−r)+eik(τ+r)/parenrightBig =−1 2πτ τ2−r2. (7.27) Chapter 7. Canonical quantization of the scalar ﬁeld 153 From (7.27), it follows that /angbracketleftφ(x)φ(y)/angbracketright=−1 4πlogτ2−r2 µ2=−1 4πlog(x−y)2 µ2, where µis an integration constant which has the dimension of length. 7.17 By taking partial derivative of the expression /angbracketleft0|T(φ(x)φ(y))|0/angbracketrightwith respect to x0, we get: ∂x0/angbracketleft0|T(φ(x)φ(y))|0/angbracketright=δ(x0−y0)/angbracketleft0|[φ(x),φ(y)]|0/angbracketright+ +θ(x0−y0)/angbracketleft0|∂x0φ(x)φ(y)|0/angbracketright+θ(y0−x0)/angbracketleft0|φ(y)∂x0φ(x)|0/angbracketright. The ﬁrst term is equal to zero as a consequence of the equal–time commutation relation. By taking second order partial derivative with respect to x0, we get: ∂2 x0/angbracketleft0|T(φ(x)φ(y))|0/angbracketright=δ(x0−y0)[π(x),φ(y)] +θ(x0−y0)/angbracketleft0|∂2 x0φ(x)φ(y)|0/angbracketright+ +θ(y0−x0)/angbracketleft0|φ(y)∂2 x0φ(x)|0/angbracketright. In the ﬁrst term, we use the equal–time commutation relation, and ﬁnally get the result ∂2 x0/angbracketleft0|T(φ(x)φ(y))|0/angbracketright=−iδ(4)(x−y)+ +θ(x0−y0)/angbracketleft0|∂2 x0φ(x)φ(y)|0/angbracketright+ +θ(y0−x0)/angbracketleft0|φ(y)∂2 x0φ(x)|0/angbracketright, which implies (/unionsq /intersectionsqx+m2)/angbracketleft0|T(φ(x)φ(y))|0/angbracketright=−iδ(4)(x−y)+ +θ(x0−y0)/angbracketleft0|(/unionsq /intersectionsqx+m2)φ(x)φ(y)|0/angbracketright+ +θ(y0−x0)/angbracketleft0|φ(y)(/unionsq /intersectionsqx+m2)φ(x)|0/angbracketright. The last two terms vanish since the ﬁeld φsatisﬁes the Klein–Gordon equation. Therefore, (/unionsq /intersectionsqx+m2)/angbracketleft0|T(φ(x)φ(y))|0/angbracketright=−iδ(4)(x−y). (7.28) 7.18 (a) Applying the variational principle to the given action leads to the equa- tions: i∂ψ ∂t=/parenleftbigg −1 2m∆+V(r)/parenrightbigg ψ −i∂ψ† ∂t=/parenleftbigg −1 2m∆+V(r)/parenrightbigg ψ†. The ﬁrst of these equations is the Schr¨ odinger equation, the second one is its conjugation equation. 154 Solutions (b) A particular solution of the free Schr¨ odinger equation is a plane wave e−iEkt+ik·r, where Ek=k2/2mso that the general solution is ψ(t,r)=/integraldisplayd3k (2π)3/2a(k)e−iEkt+ik·r. (7.29) The negative energy solutions are not present in previous expression since Ek>0 in nonrelativistic quantum mechanics. The ﬁeld ψ†is ψ†(t,r)=/integraldisplayd3k (2π)3/2a†(k)eiEkt−ik·r. (7.30) In the quantum theory these classical ﬁelds are replaced by operators in the Hilbert space. The ﬁeld conjugate to ψis π=∂L ∂˙ψ=iψ†. The equal–time commutation relations are [ψ(t,x),ψ†(t,y)] =δ(3)(x−y), [ψ(t,x),ψ(t,y)] = [ψ†(t,x),ψ†(t,y)] = 0 . (7.31) From the relations (7.29) and (7.30) follows a(k)=1 (2π)3/2eiEkt/integraldisplay d3xψ(t,x)e−ik·x a†(k)=1 (2π)3/2e−iEkt/integraldisplay d3xψ†(t,x)eik·x. From (7.31) and previous relations one easily gets the commutation rela- tions: [a(k),a†(p)] =δ(3)(p−k), (7.32) [a(k),a(p)] = [a†(k),a†(p)] = 0 . (7.33) (c) Substituting (7.29) and (7.30) into the expression for the Green function one obtains G(x0,x,y0,y)=−i/angbracketleft0|ψ(x0,x)ψ†(y0,y)|0/angbracketrightθ(x0−y0) =−i (2π)3/integraldisplay d3kd3pe−i(Ekx0−k·x−Epy0+p·y) ×/angbracketleft0|a(k)a†(p)|0/angbracketrightθ(x0−y0) =−i (2π)3/integraldisplay d3kd3pe−i(Ekx0−k·x−Epy0+p·y) ×δ(3)(p−k)θ(x0−y0) =−i (2π)3/integraldisplay d3ke−ik2 2m(x0−y0)+ik·(x−y)θ(x0−y0) =−i/parenleftbiggm 2πi(x0−y0)/parenrightbigg3/2 eim(x−y)2 2(x0−y0)θ(x0−y0). Chapter 7. Canonical quantization of the scalar ﬁeld 155 (d) The eigenfunctions are uk=/radicalbigg 2 πsin(kx), hence the (nonrelativistic) ﬁeld operators are ψ=/radicalbigg 2 π/integraldisplay∞ 0dka(k)e−ik2 2mtsin(kx), (7.34) ψ†=/radicalbigg 2 π/integraldisplay∞ 0dka†(k)eik2 2mtsin(kx). (7.35) We shall leave to the reader to prove that G(x0,x,y0,y)=−i/parenleftbiggm 2πi(x0−y0)/parenrightbigg1/2/bracketleftbigg eim(x−y)2 2(x0−y0)−eim(x+y)2 2(x0−y0)/bracketrightbigg θ(x0−y0). (7.36) Generally, if the eigenfunctions of the Hamiltonian are un(x) the Green function is G(x0,x,y0,y)=−i/summationdisplay ne−iEn(x0−y0)un(x)u∗ n(y)θ(x0−y0).(7.37) (e) The invariance of the Schr¨ odinger equation can be proven directly. We leave that to reader. (f) In order to ﬁnd the conserved charges we should calculate only time com- ponents of the conserved currents. For the spatial translations the timecomponent of the current is j 0=−∂L ∂(∂0ψ)∂iψ/epsilon1i =−iψ†∂iψ/epsilon1i=−iψ†∇ψ·/epsilon1. (7.38) The conserved charge is the linear momentum P=−/integraldisplay d3xψ†(i∇)ψ. (7.39) The Hamiltonian H=/integraldisplay d3xψ†(−1 2m)∆ψ (7.40) is generator of time translations. The angular momentum J=−i/integraldisplay d3xψ†(x×∇)ψ (7.41) is generator of rotations. Under Galilean boosts we have δxi=−vit, δψ = −imv·xψso that 156 Solutions j0=v·j0=mv·xψ†ψ+ivtψ†∇ψ. (7.42) Consequently, the boost generator is G=/integraldisplay d3xψ†(mx+it∇)ψ. (7.43) The commutation relations can be found using the commutation relations (7.31). Let us start with [ Pi,Gj]: [Pi,Gj]=i/integraldisplay d3xd3y[−ψ†(y)∂y iψ(y),ψ†(x)(mxj+it∂j)ψ(x)] =−im/integraldisplay d3xd3y/parenleftbig ψ†(y)[∂iψ(y),ψ†(x)xjψ(x)] +[ψ†(y),ψ†(x)xjψ(x)]∂iψ(y)/parenrightbig =−im/integraldisplay d3x(−∂iψ†xjψ(x)−xjψ†∂iψ) =−iMδij, (7.44) where M=m/integraltext d3xψ†ψis the mass operator. It appears since the rep- resentation is projective. We have two possibilities either to enlarge the Galilean algebra with this operator or to add a superselection rule which forbids superposition of particles of diﬀerent masses.In the similar manner the other commutation relations can be obtained: [G i,Gj]=[H,P]=[H,J]=0 [Ji,Jj]=i/epsilon1ijkJk [Ji,Gj]=i/epsilon1ijkGk [Ji,Pj]=i/epsilon1ijkPk [H,G i]=−iPi. The Galilean algebra can also be derived from the Poincar´ e algebra [23]. 7.19(a) By using the ﬁrst commutation relation in (7.D), we get [a(p),a †]=C/integraldisplayd3q/radicalbig2ωq[a(p),a†(q)]˜f(q) =C/integraldisplayd3q/radicalbig2ωq˜f(q)δ(3)(p−q) =C1/radicalbig2ωp˜f(p). (7.45) The second commutator can be evaluated in the same way. The result is [a†(p),a]=−C1/radicalbig2ωp˜f∗(p). (7.46) Chapter 7. Canonical quantization of the scalar ﬁeld 157 (b) Using (7.45), we have a(p)(a†)n=C1/radicalbig2ωp˜f(p)(a†)n−1+a†a(p)(a†)n−1. (7.47) By repeating this procedure ntimes, we get a(p)(a†)n=C1/radicalbig2ωpn˜f(p)(a†)n−1+(a†)na(p). (7.48) Hence, [a(p),(a†)n]=Cn˜f(p)/radicalbig2ωp(a†)n−1. (7.49) (c) This calculation is straightforward: a(p)|z/angbracketright=e−|z|2/2a(p)∞/summationdisplay n=0zn(a†)n n!|0/angbracketright =e−|z|2/2∞/summationdisplay n=1C/radicalbig2ωpzn˜f(p) (n−1)!(a†)n−1|0/angbracketright =C/radicalbig2ωp˜f(p)z|z/angbracketright. (7.50) (d) By using the previous relation and the property /angbracketleftz|z/angbracketright=1 ,w eh a v e /angbracketleftz|φ|z/angbracketright=/integraldisplayd3p (2π)3/2/radicalbig2ωp/parenleftbig /angbracketleftz|a(p)|z/angbracketrighte−ip·x+/angbracketleftz|a†(p)|z/angbracketrighteip·x/parenrightbig =C/integraldisplayd3p (2π)3/22ωp/parenleftBig z˜f(p)e−ip·x+z∗˜f∗(p)eip·x/parenrightBig =C (2π)3/2(zf(x)+z∗f∗(x)). (7.51) In the same manner we have /angbracketleftz|:φ2:|z/angbracketright=/integraldisplayd3p (2π)3/2/radicalbig2ωpd3q (2π)3/2/radicalbig2ωq/parenleftBig /angbracketleftz|a(p)a(q)|z/angbracketrighte−i(p+q)·x +/angbracketleftz|a†(q)a(p)|z/angbracketrightei(q−p)·x +/angbracketleftz|a†(p)a(q)|z/angbracketrightei(p−q)·x+/angbracketleftz|a†(p)a†(q)|z/angbracketrightei(q+p)·x/parenrightBig =C2/integraldisplayd3p (2π)3/22ωpd3q (2π)3/22ωq/parenleftBig ˜f(p)˜f(q)z2e−i(p+q)·x +˜f(p)˜f∗(q)|z|2e−i(p−q)·x +˜f∗(p)˜f(q)|z|2ei(p−q)·x+˜f∗(p)˜f∗(q)(z∗)2ei(p+q)·x/parenrightBig =C2 (2π)3(zf(x)+z∗f∗(x))2. (7.52) 158 Solutions Hence, (∆φ)2=0. (7.53) (e) It is easy to see that /angbracketleftz|H|z/angbracketright=C2|z|2/integraldisplay d3p|˜f(p)|2. (7.54) 7.20 (a) By substituting the expression for φin the relation U(Λ,a)φ(x)U−1(Λ,a)=φ(Λx+a) we obtain /integraldisplayd3k (2π)3/2√2ωkU(Λ,a)/parenleftbig a(k)e−ik·x+a†(k)eik·x/parenrightbig U−1(Λ,a) =/integraldisplayd3k/prime (2π)3/2√2ωk/prime/parenleftBig a(k/prime)e−ik/prime·(Λx+a)+a†(k/prime)eik/prime·(Λx+a)/parenrightBig .(7.55) In the integral on the right hand side we make the changing of variables k/primeµΛν µ=kν. In Problem 6.3, we proved that d3k/(2ωk)i saL o r e n t z invariant measure, so that d3k/prime √2ωk/prime=/radicalbiggωk/prime 2d3k ωk. By performing the inverse Fourier transformation, we obtain the requested result. (b) It is easy to see that U(Λ,a)|k1,...,k n/angbracketright=U(Λ,a)a†(k1)U−1(Λ,a)U(Λ,a)··· ···U(Λ,a)a†(kn)U−1(Λ,a)|0/angbracketright =/radicalbiggωk/prime 1···ωk/prime n ωk1···ωkneiaµΛµ ν(kν 1+...+kν n)|Λk1,...,Λk n/angbracketright. (c) From the expressions (7.6) and (7.8) and the ﬁrst part of this problem, we have U(Λ)PµU−1(Λ)=/integraldisplay d3kkµU(Λ)a†(k)a(k)U−1(Λ) =/integraldisplay d3kkµωk/prime ωka†(Λk)a(Λk) =Λµ ν/integraldisplay d3k/primek/primeνa†(k/prime)a(k/prime) =Λµ νPν, where we made the change of variables kµ=Λµ νk/primeνin the integral. Chapter 7. Canonical quantization of the scalar ﬁeld 159 (d) First, you should prove the following formulae: U(Λ)[φ(x),φ(y)]U−1(Λ)=[φ(Λx),φ(Λx)], [φ(x),φ(y)] = i∆(x−y). From the integral expression for the function ∆(x−y) (Problem 6.6), it follows that ∆(Λx−Λy)=∆(x−y), i.e. it is a relativistic covariant quantity. 7.21 (a) In Problem 7.3, we obtained the Hamiltonian H=/integraldisplay d3kωka†(k)a(k). The Backer–Hausdorﬀ relation reads PHP−1=eAHe−A=H+[A,H]+1 2[A,[A,H]] +... (7.56) where A=−iπ 2/integraltext d3q/parenleftbig a†(q)a(q)−ηpa†(q)a(−q)/parenrightbig .The ﬁrst commutator in this expression is [A,H]=−iπ 2ηp/integraldisplay d3kωk/parenleftbig a†(k)a(−k)−a†(−k)a(k)/parenrightbig . By changing k→−kin the second term, we get [ A,H]=0.It is clear that the other commutators in (7.56) also vanish, hence [P,H]=0. (b) Starting from Problem 7.8, we obtain the requested result. 7.22 τPτ−1=−P,τ H τ−1=H 7.23 The ﬁrst step is to show that Cφ†C−1=η∗ cφ,CπC−1=ηcπ†and Cπ†C−1=ηcπ. 8 Canonical quantization of the Dirac ﬁeld 8.1If we use the anticommutation relation (8.E) the anticommutator i Sab(x− y)={ψa(x),¯ψb(y)},where a,b=1,...,4 are Dirac indices, becomes {ψa(x),¯ψb(y)}=/summationdisplay r,s1 (2π)3/integraldisplay d3pd3qm/radicalbig EpEqδrsδ(3)(p−q) ×/parenleftBig ua(p,r)¯ub(q,s)ei(q·y−p·x) +va(p,r)¯vb(q,s)e−i(q·y−p·x)/parenrightBig . Applying the solution of Problem 4.4 we have iSab=1 (2π)3/integraldisplayd3p 2Ep/bracketleftBig (/p+m)abe−ip·(x−y)+( /p−m)abeip·(x−y)/bracketrightBig .(8.1) The last expression can be easily transformed into the following form {ψa(x),¯ψb(y)}=( iγµ∂x µ+m)ab1 (2π)3/integraldisplayd3p 2Ep/bracketleftBig e−ip·(x−y)−eip·(x−y)/bracketrightBig .(8.2) From (8.2) we see that ∆(x−y) is given by ∆(x−y)=−i (2π)3/integraldisplayd3p 2Ep/bracketleftBig e−ip·(x−y)−eip·(x−y)/bracketrightBig . The function ∆(x−y) was deﬁned in Problem 6.6. In the special case x0=y0 we shall make change p→−pin the second term of expression (8.1) and obtain {ψa(x),¯ψb(y)}|x0=y0=(γ0)ab/integraldisplayd3p (2π)3eip·(x−y)=(γ0)abδ(3)(x−y).(8.3) 162 Solutions 8.2 (a) Substituting (8.A,B) in the expression for charge Qwe obtain Q=−e/integraldisplay d3x:ψ†ψ: =−e/summationdisplay r,s/integraldisplay d3pm Ep/bracketleftbig c† r(p)cs(p)u† r(p)us(p) +:dr(p)d† s(p):v† r(p)vs(p)+c† r(p)d† s(−p)u† r(p)vs(−p)e2iEpt +dr(p)cs(−p)v† r(p)us(−p)e−2iEpt/bracketrightbig . (8.4) From (4.52) and (8.4) we get Q=−e/summationdisplay r/integraldisplay d3p/parenleftbig c† r(p)cr(p)−d† r(p)dr(p)/parenrightbig . (8.5) (b) As ψsatisfy the Dirac equation, ( −iγi∂i+m)ψ=iγ0∂0ψthe Hamiltonian is H=i/integraldisplay d3x:ψ†∂0ψ: =/summationdisplay r,s1 (2π)3/integraldisplay d3xd3pd3q/radicalbiggm Ep/radicalbiggm Eq:/parenleftbig u† r(p)c† r(p)eip·x +v† r(p)dr(p)e−ip·x/parenrightbig Eq/parenleftbig us(q)cs(q)e−iq·x−vs(q)d† s(q)eiq·x/parenrightbig : =/summationdisplay r/integraldisplay d3pEp/parenleftbig c† r(p)cr(p)+d† r(p)dr(p)/parenrightbig . (8.6) (c) P=/summationdisplay r/integraldisplay d3pp/parenleftbig c† r(p)cr(p)+d† r(p)dr(p)/parenrightbig . (8.7) 8.3 (a) It is easy to see that [H,ψ]=/summationdisplay r,s1 (2π)3/2/integraldisplay d3pd3qEp/radicalbiggm Eq ×/bracketleftbig c† r(p)cr(p)+d† r(p)dr(p),cs(q)us(q)e−iq·x+d† s(q)vs(q)eiq·x/bracketrightbig =/summationdisplay r,s1 (2π)3/2/integraldisplay d3pd3qEp/radicalbiggm Eqδrsδ(3)(p−q) ×/parenleftbig −cr(p)us(q)e−iq·x+d† r(p)vs(q)eiq·x/parenrightbig =/summationdisplay r/integraldisplayd3p (2π)3/2/radicalbig mEp/parenleftbig −cr(p)ur(p)e−ip·x+d† r(p)vr(p)eip·x/parenrightbig =−i∂ψ ∂t, Chapter 8. Canonical quantization of the Dirac ﬁeld 163 where we have used: [c† r(p)cr(p),cs(q)] =−{c† r(p),cs(q)}cr(p) =−δrsδ(3)(p−q)cr(p), and the similar expression for d−operators. (b) If we had used commutation relations instead of anticommutation rela- tions in the quantization process we would have obtained: H=/summationdisplay r/integraldisplay d3pEp/parenleftBig c† r(p)cr(p)−d† r(p)dr(p)/parenrightBig . From here we conclude that the energy spectrum would have been un- bounded from below, which is physically unacceptable. 8.4 [H,c† r(p)cr(p)] =/summationdisplay s/integraldisplay d3qEq[c† s(q)cs(q)+d† s(q)ds(q),c† r(p)cr(p)] =/summationdisplay s/integraldisplay d3qEq/parenleftbig [c† s(q)cs(q),c† r(p)]cr(p) +c† r(p)[c† s(q)cs(q),cr(p)] =/summationdisplay s/integraldisplay d3qEq/parenleftbig c† s(q){cs(q),c† r(p)}cr(p) −{c† s(q),c† r(p)}cs(q)cr(p) +c† r(p)(c† s(q){cs(q),cr(p)}−{c† s(q),cr(p)}cs(q))/parenrightbig =Ep/parenleftbig c† r(p)cr(p)−c† r(p)cr(p)/parenrightbig =0 8.5The form variation of a spinor ﬁeld is δ0ψ=δψ−δxµ∂µψ= =−i 4ωµνσµνψ−ωµνxν∂µψ =1 2ωµν/parenleftbigg xµ∂ν−xν∂µ−i 2σµν/parenrightbigg ψ. On the other hand we have δ0ψ=−i 2ωµνMµνψ.Comparing these results we conclude that the generators are given by Mµν=i (xµ∂ν−xν∂µ)+1 2σµν. 8.6 (a) Applying the formula [ AB,C ]=A{B,C}−{A,C}Bwe obtain 164 Solutions [Mµν,ψa(x)] =/integraldisplay d3y/bracketleftbigg ψ† b(y)/parenleftbigg i(yµ∂ν−yν∂µ)+1 2σµν/parenrightbigg bcψc(y),ψa(x)/bracketrightbigg =−/integraldisplay d3y{ψ† b(y),ψa(x)}/parenleftbigg i(yµ∂ν−yν∂µ)+1 2σµν/parenrightbigg bcψc(y) =−[i(xµ∂ν−xν∂µ)+1 2σµν]acψc(x), where we have used anticommutation relations (8.C,D). This result is a consequence of Lorentz symmetry. (b) Substituting the expressions for angular momentum and momentum of the Dirac ﬁeld we get [Mµν,Pρ]=i/integraldisplay d3xd3y ×/bracketleftbigg ψ† a(x)/parenleftbigg i(xµ∂ν−xν∂µ)+1 2σµν/parenrightbigg abψb(x),ψ† c(y)∂ρψc(y)/bracketrightbigg . First we suppose that all indices are the spatial: µ=i,ν=j, ρ=k. Then, [Mij,Pk]=i/integraldisplay d3xd3y ×/parenleftbigg ψ† a(x)/braceleftbigg/parenleftbigg i(xi∂j−xj∂i)+1 2σij/parenrightbigg abψb(x),ψ† c(y)/bracerightbigg ∂kψc(y) −ψ† c(y){ψ† a(x),∂kψc(y)}/parenleftbigg i(xi∂j−xj∂i)+1 2σij/parenrightbigg abψb(x)/parenrightbigg =i/integraldisplay d3xd3y ×/parenleftbigg ψ† a(x)/parenleftbigg i(xi∂j−xj∂i)+1 2σij/parenrightbigg abδ(3)(x−y)∂kψb(y) −ψ† c(y)∂y kδ(3)(x−y)δac/parenleftbigg i(xi∂j−xj∂i)+1 2σij/parenrightbigg abψb(x)/parenrightbigg , where we used the equal-time anticommutation relations (8.C,D). The integration over yleads to [Mij,Pk]=i/integraldisplay d3x/parenleftBig igjkψ†∂iψ−igikψ†∂jψ/parenrightBig , or [Mij,Pk]=i (gjkPi−gikPj). Now we take µ=0,ν=i,andρ=k, i.e. we calculate the commutator [M0i,Pk]. In order to do it ﬁrst we are going to compute anticommutator {∂x0ψ(x),¯ψ(y)}|x0=y0. Chapter 8. Canonical quantization of the Dirac ﬁeld 165 Taking partial derivative of (8.1) with respect to x0and substituting x0= y0we get {∂x0ψa(x),¯ψb(y)}|x0=y0=i 2(2π)3/integraldisplay d3p/bracketleftBig (−Epγ0+p·γ−m)abeip·(x−y) +(Epγ0−p·γ−m)abe−ip·(x−y)/bracketrightBig =i (2π)3/integraldisplay d3p(p·γ−m)abeip·(x−y) =γab∇xδ(3)(x−y)−imδabδ(3)(x−y). Then [M0i,Pk]=i/integraldisplay d3xd3y ×/parenleftbigg ψ† a(x)/braceleftbigg/parenleftbigg i(x0∂i−xi∂0)+1 2σ0i/parenrightbigg abψb(x),ψ† c(y)/bracerightbigg ∂kψc(y) −ψ† c(y){ψ† a(x),∂kψc(y)}/parenleftbigg i(x0∂i−xi∂0)+1 2σ0i/parenrightbigg abψb(x)/parenrightbigg =i/integraldisplay d3xd3y/parenleftBig ix0ψ†(x)∂x iδ(3)(x−y)∂kψ(y) −ixiψ† a(x)(γγ0∇x−imγ0)acδ(3)(x−y)∂kψc(y) +ψ† a(x)1 2(σ0i)abδ(3)(x−y)∂kψb(y) −ix0ψ†(y)∂y kδ(3)(x−y)∂iψ(x) +ixiψ†(y)∂y kδ(3)(x−y)∂0ψ(x) −ψ† a(y)1 2(σ0i)ab∂y kδ(3)(x−y)ψb(x)/parenrightbigg =i/integraldisplay d3x/parenleftbig −ixiψ†γγ0∂k∇ψ−mxiψ†γ0∂kψ−ixi∂kψ†∂0ψ/parenrightbig =i/integraldisplay d3x/parenleftbig igikψ†∂0ψ+xi¯ψ(iγ0∂0+iγ∇−m)∂kψ/parenrightbig . The second term in the last line vanishes since ψsatisﬁes the Dirac equa- tion. Then we get [M0i,Pk]=igikP0. The remaining commutators [ M0i,P0] and [ Mij,P0] can be computed in the same way. 8.7The helicity operator is Sp=1 2/integraldisplay d3x:ψ†Σ·p |p|ψ:. (8.8) 166 Solutions Inserting expressions for ﬁelds ψandψ†in the previous formula and using the fact that ur(p)a n d vr(p) are eigenspinors of Σ·p/|p|with eigenvalues (−1)r+1and (−1)r,respectively (see Problem 4.7) we get Sp=1 2(2π)3/integraldisplay d3x2/summationdisplay r,s=1/integraldisplay d3pd3qm/radicalbig EpEq ×/bracketleftBig c† r(q)cs(p)(−1)s+1u† r(q)us(p)ei(q−p)·x +c† r(q)d† s(p)(−1)su† r(q)vs(p)ei(q+p)·x +dr(q)cs(p)(−1)s+1v† r(q)us(p)e−i(q+p)·x −d† s(p)dr(q)(−1)sv† r(q)vs(p)ei(p−q)·x/bracketrightBig . (8.9) Performing the xintegration and applying orthogonality relations (4.52) one gets that the second and the third term in the expression (8.9) vanish. Finally, integration over the momentum qgives Sp=1 22/summationdisplay r=1/integraldisplay d3p(−1)r+1/parenleftbig c† r(p)cr(p)+d† r(p)dr(p)/parenrightbig . (8.10) Let us emphasize that we have used the expansion of the ﬁelds with respect to helicity basis. 8.8The two-particle state given in the problem is eigenstate of the operators H, Q, andSp. Using the explicit form of the Hamiltonian from Problem 8.2 we have Hc† r1(p1)c† r2(p2)|0/angbracketright=/summationdisplay r/integraldisplay d3pEp/parenleftbig c† r(p)cr(p) +d† r(p)dr(p)/parenrightbig c† r1(p1)c† r2(p2)|0/angbracketright. (8.11) Let us calculate the ﬁrst term in the previous expression. Commuting cr(p) to the right we get c† r(p)cr(p)c† r1(p1)c† r2(p2)|0/angbracketright=δr1rδ(3)(p−p1)c† r(p)c† r2(p2)|0/angbracketright −c† r(p)c† r1(p1)cr(p)c† r2(p2)|0/angbracketright.(8.12) Repeating once more we get c† r(p)cr(p)c† r1(p1)c† r2(p2)|0/angbracketright=δr1rδ(3)(p−p1)c† r(p)c† r2(p2)|0/angbracketright −c† r(p)c† r1(p1)δrr2δ(3)(p−p2)|0/angbracketright.(8.13) It is easy to see that d† r(p)dr(p)c† r1(p1)c† r2(p2)|0/angbracketright=0. (8.14) Chapter 8. Canonical quantization of the Dirac ﬁeld 167 Inserting (8.13) and (8.14) in (8.11) and integrating over momentum pwe obtain Hc† r1(p1)c† r2(p2)|0/angbracketright=(Ep1+Ep2)c† r1(p1)c† r2(p2)|0/angbracketright. (8.15) Similar as before we have: Qc† r1(p1)c† r2(p2)|0/angbracketright=−2ec† r1(p1)c† r2(p2)|0/angbracketright, (8.16) for charge and Spc† r1(p1)c† r2(p2)|0/angbracketright =1 2/parenleftbig (−1)r1+1+(−1)r2+1/parenrightbig c† r1(p1)c† r2(p2)|0/angbracketright (8.17) for helicity. To summarize: energy, charge and helicity of the two–particle state |p1,r1;p2,r2/angbracketrightare Ep1+Ep2,−2e,1 2/parenleftbig (−1)r1+1+(−1)r2+1/parenrightbig , (8.18) respectively. 8.9The commutator is [Qa,Qb]=1 4/integraldisplay d3xd3yτa ijτb kl[ψ† i(x)ψj(x),ψ† k(y)ψl(y)] =1 4/integraldisplay d3xd3yτa ijτb kl(ψ† i(x)ψl(y)δjk−ψ† k(y)ψj(x)δil)δ(3)(x−y) =1 4/integraldisplay d3x(ψ† iτa ijτb jlψl−ψ† kτb klτa ljψj) =1 4/integraldisplay d3xψ†[τa,τb]ψ =i 2/epsilon1abc/integraldisplay d3xψ†τcψ=i/epsilon1abcQc. The generators Qasatisfy the commutation relations of SU(2) algebra as we expected. 8.10 The charges are Qb=/integraldisplay d3xjb 0=/integraldisplay d3x(/epsilon1abc˙πaπc+1 2Ψ† iτb ijΨj). (8.19) (a) The commutator is [Qb,Qe]=/integraldisplay d3xd3y/parenleftbig /epsilon1abc/epsilon1def[˙πa(x)πc(x),˙πd(y)πf(y)] +τb ij 2τe mn 2[Ψ† i(x)Ψj(x),Ψ† m(y)Ψn(y)]/parenrightbigg 168 Solutions =/integraldisplay d3xd3y/parenleftBig /epsilon1abc/epsilon1def(iδ(3)(x−y)δcd˙πa(x)πf(y) −iδ(3)(x−y)δaf˙πd(y)πc(x)) +τb ij 2τe mn 2δ(3)(x−y)(δjmΨ† i(x)Ψn(y)−δinΨ† m(y)Ψj(x))/parenrightBig =/integraldisplay d3x/parenleftbigg i(˙πeπb−˙πbπe)+i 2/epsilon1bedΨ†τdΨ/parenrightbigg =i/epsilon1bed/integraldisplay d3x/parenleftbigg /epsilon1adc˙πaπc+1 2Ψ†τdΨ/parenrightbigg =i/epsilon1bedQd. (b) The results are [Qb,πa(x)] =−i/epsilon1abcπc(x), [Qb,ψi(x)] =−τb in 2ψn(x), [Qb,¯ψi(x)] =−¯ψn(x)τb ni 2. 8.11 The conserved charge for dilatation is D=/integraldisplay d3xj0=−i/integraldisplay d3x/parenleftbigg3 2ψ†ψ+xjψ†∂jψ−x0¯ψγj∂jψ/parenrightbigg .(8.20) Let us ﬁnd the commutator between the operator Dand momentum Pi [D,Pi]=/integraldisplay d3xd3y/parenleftbigg [3 2ψ†(x)ψ(x)+xjψ†(x)∂jψ(x),ψ†(y)∂iψ(y)] −[x0¯ψ(x)γj∂jψ(x),ψ†(y)∂iψ(y)]/parenrightbig . We decompose the previous expression on three commutators. The ﬁrst one is [ψ†(x)ψ(x),ψ†(y)∂iψ(y)] = [ψ† a(x)ψa(x),ψ† b(y)]∂iψb(y) +ψ† b(y)[ψ† a(x)ψa(x),∂iψb(y)] =ψ† a(x){ψa(x),ψ† b(y)}∂iψb(y) −ψ† b(y){ψ† a(x),∂iψb(y)}ψa(x), where we have dropped the vanishing terms. The anticommutation relations (8.C–D) give the following result [ψ†(x)ψ(x),ψ†(y)∂iψ(y)] =ψ†(x)∂iψ(y)δ(3)(x−y) −ψ†(y)ψ(x)∂i yδ(3)(x−y). (8.21) The remaining commutators can be calculated in the same way. The result is: Chapter 8. Canonical quantization of the Dirac ﬁeld 169 [ψ†(x)∂jψ(x),ψ†(y)∂iψ(y)] =ψ†(x)∂iψ(y)∂x jδ(3)(x−y) −ψ†(y)∂jψ(x)∂i yδ(3)(x−y),(8.22) [¯ψ(x)γj∂jψ(x),ψ†(y)∂iψ(y)] =¯ψ(x)γj∂iψ(y)∂x jδ(3)(x−y) −¯ψ(y)γj∂jψ(x)∂i yδ(3)(x−y).(8.23) Inserting (8.21), (8.22) and (8.23) in (8.21) and applying ∂k xδ(3)(x−y)=−∂k yδ(3)(x−y), (8.24) we get [D,Pi]=−/integraldisplay d3xψ†∂iψ=iPi. (8.25) Similarly one can show that [D,P0]=iP0. (8.26) 8.12 (a) Using the expression (5.G) the energy–momentum tensor is Tαβ=i¯ψγα∂βψ−gαβ(i¯ψ/∂ψ−gx2¯ψψ). Taking derivative of the previous expression we get ∂αTαβ=2gxβ¯ψψ , where we have used the equations of motion: i/∂ψ−gx2ψ=0, i∂µ¯ψγµ+gx2¯ψ=0. The result ∂αTαβ/negationslash= 0 shows that there is no translation symmetry in the theory. As a consequence, the energy and momentum are not conserved inthis theory. (b) From the expression for the four-momentum (5.6) we have P 0(t)=/integraldisplay d3x(−i¯ψγj∂jψ+gx2¯ψψ), Pi(t)=i/integraldisplay d3xψ†∂iψ, so 170 Solutions [P0(t),Pi(t)] =/integraldisplay/integraldisplay d3xd3y ×/parenleftBig [¯ψ(t,x)γj∂jψ(t,x),ψ†(t,y)∂iψ(t,y)] +igx2[¯ψ(t,x)ψ(t,x),ψ†(t,y)∂iψ(t,y)]/parenrightBig =/integraldisplay/integraldisplay d3xd3y ×/parenleftBig (γ0γj)ab[ψ† a(t,x)∂jψb(t,x),ψ† c(t,y)∂iψc(t,y)] +igx2γ0 ab[ψ† a(t,x)ψb(t,x),ψ† c(t,y)∂iψc(t,y)]/parenrightBig . The commutators in the previous expression can be found in the same way as in the previous problem [P0(t),Pi(t)] =/integraldisplay d3x/parenleftbig −∂j¯ψγj∂iψ−¯ψγj∂j∂iψ +igx2(¯ψ∂iψ+(∂i¯ψ)ψ)/parenrightbig =/integraldisplay d3x/parenleftbig −∂j(¯ψγj∂iψ)+igx2∂i(¯ψψ)/parenrightbig =−2ig/integraldisplay d3xxi¯ψψ , where we dropped the surface terms. ( c )I ti se a s yt os h o wt h a t ∂µMµνρ= 0, which is a consequence of the Lorentz symmetry of the Lagrangian density. 8.13 (a) Under the Lorentz transformation the commutator [ Jµ(x),Jν(y)] trans- forms in the following way U(Λ)[Jµ(x),Jν(y)]U−1(Λ) =U(Λ)[¯ψa(x)γµ abψb(x),¯ψc(y)γν cdψd(y)]U−1(Λ) (8.27) =[U¯ψa(x)U−1γµ abUψb(x)U−1,U¯ψc(y)U−1γν cdUψd(y)U−1]. Taking the adjoint of (8.G) and multiplying by γ0we obtain U(Λ)¯ψ(x)U−1(Λ)=¯ψ(Λx)S(Λ). (8.28) By using (8.G), last expression and S−1γµS=Λµ νγνin (8.27) we get U(Λ)[Jµ(x),Jν(y)]U−1(Λ)=Λµ ρΛν σ[Jρ(Λx),Jσ(Λy)]. (8.29) From the last result we see that the commutator [ Jµ(x),Jν(y)] is a covari- ant quantity. Chapter 8. Canonical quantization of the Dirac ﬁeld 171 (b) Using the fact that the commutator is a Lorentz tensor we calculate it in the frame where x0=y0=t,x/negationslash=y.W eg e t [Jµ(t,x),Jν(t,y)] =(γ0γµ)ab(γ0γν)cd[ψ† a(t,x)ψb(t,x),ψ† c(t,y)ψd(t,y)] =(γ0γµ)ab(γ0γν)cd/parenleftbig ψ† a(t,x){ψb(t,x),ψ† c(t,y)}ψd(t,y) −ψ† c(t,y){ψ† a(t,x),ψd(t,y)}ψb(t,x)/parenrightbig . (8.30) Using the anticommutation relation (8.D) in (8.30) gives [Jµ(t,x),Jν(t,y)] =/parenleftbig¯ψ(t,x)γµγ0γνψ(t,y)−¯ψ(t,y)γνγ0γµψ(t,x)/parenrightbig δ(3)(x−y). Sincex/negationslash=ythenδ(3)(x−y) = 0 and the commutator is equal to zero in the special frame we have chosen. Because of the covariance it follows thatit is equal to zero for ( x−y) 2<0. Therefore, microcausality principle is valid. 8.14 First show that /angbracketleftbig ψa(x)¯ψb(y)/angbracketrightbig =1 (2π)3/integraldisplayd3p 2Ep(/p+m)abe−ip·(x−y), (8.31) /angbracketleftbig¯ψa(x)ψb(y)/angbracketrightbig =1 (2π)3/integraldisplayd3p 2Ep(/p−m)bae−ip·(x−y). (8.32) If in the expression/angbracketleftbig¯ψa(x1)ψb(x2)ψc(x3)¯ψd(x4)/angbracketrightbig ,we substitute the expan- sions (8.A–B), we obtain /angbracketleftbig¯ψa(x1)ψb(x2)ψc(x3)¯ψd(x4)/angbracketrightbig =/summationdisplay r1,...,r4m2 (2π)6/parenleftBigg4/productdisplay i=1/integraldisplayd3pi/radicalbig Epi/parenrightBigg ×/parenleftBig/angbracketleftBig d1c2d† 3c†4/angbracketrightBig ¯v1au2bv3c¯u4dei(−p1·x1−p2·x2+p3·x3+p4·x4) +/angbracketleftBig d1d† 2c3c†4/angbracketrightBig ¯v1av2bu3c¯u4dei(−p1·x1+p2·x2−p3·x3+p4·x4)/parenrightBig , where the vanishing terms are discarded. Also, we use the abbreviations: d1=dr1(p1),u1=ur1(p1),etc. Applying the expressions for projectors to positive and negative energy solu- tions from Problem 4.4 and using /angbracketleftBig d1c2d† 3c†4/angbracketrightBig =−δr1r3δr2r4δ(3)(p1−p3)δ(3)(p2−p4), 172 Solutions /angbracketleftBig d1d† 2c3c†4/angbracketrightBig =δr1r2δr3r4δ(3)(p1−p2)δ(3)(p3−p4) we have /angbracketleftbig¯ψa(x1)ψb(x2)ψc(x3)¯ψd(x4)/angbracketrightbig =−1 (2π)6/integraldisplayd3p1d3p2 4Ep1Ep2(/p1−m)ca(/p2+m)bde−ip1·(x1−x3)−ip2·(x2−x4) +1 (2π)6/integraldisplayd3p1d3p3 4Ep1Ep3(/p1−m)ba(/p3+m)cde−ip1·(x1−x2)−ip3·(x3−x4). By using (8.31) and (8.32) the last expression takes the form /angbracketleftbig¯ψa(x1)ψb(x2)ψc(x3)¯ψd(x4)/angbracketrightbig =−/angbracketleftbig¯ψa(x1)ψc(x3)/angbracketrightbig/angbracketleftbig ψb(x2)¯ψd(x4)/angbracketrightbig +/angbracketleftbig¯ψa(x1)ψb(x2)/angbracketrightbig/angbracketleftbig ψc(x3)¯ψd(x4)/angbracketrightbig . The previous formula is special case of the Wick theorem. 8.15 Substituting (8.A-B) in the commutator we obtain 1 2[¯ψ,γµψ]=1 2(2π)3/summationdisplay r,s/integraldisplay d3pd3qm/radicalbig EpEq[¯ur(p)γµus(q) ×(c† r(p)cs(q)−cs(q)c† r(p))ei(p−q)·x +¯ur(p)γµvs(q)(c† r(p)d† s(q)−d† s(q)c† r(p))ei(p+q)·x +¯vr(p)γµus(q)(dr(p)cs(q)−cs(q)dr(p))e−i(p+q)·x +¯vr(p)γµvs(q)(dr(p)d† s(q)−d† s(q)dr(p))ei(q−p)·x/bracketrightBig . (8.33) Using the anticommutation relations (8.E) we obtain 1 2[¯ψ,γµψ]=:¯ψγµψ:− −1 2(2π)3/integraldisplay d3ppµ Ep/summationdisplay r(¯ur(p)ur(p)+¯vr(p)vr(p)), where we have used the Gordon identities (Problem 4.21) in addition. The requested result follows after applying the orthogonality relations (4.D). 8.16 Let us ﬁrst prove that /angbracketleft0|T(¯ψa(x)ψb(y))|0/angbracketright=−iSFba(y−x). Using the deﬁnition of time ordering and the expressions (8.31) and (8.32) we obtain Chapter 8. Canonical quantization of the Dirac ﬁeld 173 /angbracketleft0|T(¯ψa(x)ψb(y))|0/angbracketright=1 (2π)3/integraldisplayd3p 2Ep/bracketleftBig (/p−m)baeip·(y−x)θ(x0−y0) −(/p+m)baeip·(x−y)θ(y0−x0)/bracketrightBig . (8.34) With a help of Problem 6.13 we see that right hand side of the expression (8.34) is −iSFba(y−x) and we have /angbracketleft0|T(¯ψ(x)Γψ(y))|0/angbracketright=Γab/angbracketleft0|T(¯ψa(x)ψb(y))|0/angbracketright =−iΓabSFba(y−x) =−it r[ΓSF(y−x)] =−i/integraldisplayd4p (2π)4e−ip·(y−x) p2−m2+i/epsilon1tr [(/p+m)Γ]. Using the identities from the Problems 3.6(b),(d),(e) and (i) we obtain tr [(/p+m)γ5] = tr [(/ p+m)γ5γµ]=0,tr [(/p+m)γµγν]=4mgµν. From here the requested result follows. 8.17 (a) In the Weyl representation for γ–matrices the charge conjugate spinor is ψc=C¯ψT =i/parenleftbigg σ20 0−σ2/parenrightbigg/parenleftbigg 01 10/parenrightbigg/parenleftbigg ϕ∗ −iσ2χ/parenrightbigg =/parenleftbigg χ −iσ2ϕ∗/parenrightbigg . The condition ψM=ψc Mgivesϕ=χ. (b) If ψM=/parenleftbigg χ −iσ2χ∗/parenrightbigg andφM=/parenleftbigg ϕ −iσ2ϕ∗/parenrightbigg , then ¯ψMφM=−iχ†σ2ϕ∗+iχTσ2ϕ =−iσ2abχ∗ aϕ∗b+iσ2abχaϕb =−iσ2baϕ∗ bχ∗a+iσ2baϕbχa =−iϕ†σ2χ∗+iϕTσ2χ=¯φMψM. In the last expression we used that ϕandχare Grassmann variables. The other identities can be proved in the same way. For the second one thefollowing identity is useful: σ 2σµσ2=¯σµT. 174 Solutions (c) The Majorana ﬁeld operator is ψM=1√ 2(ψ+ψc) =/integraldisplayd3p (2π)3/radicalbiggm Ep/summationdisplay r/parenleftbiggcr(p)+dr(p)√ 2ur(p)e−ip·x +c† r(p)+d† r(p)√ 2vr(p)eip·x/parenrightbigg . The annihilation and creation operators can easily be read oﬀ: bM(p,r)=cr(p)+dr(p)√ 2,b† M(p,r)=c† r(p)+d† r(p)√ 2. The anticommutation relations are derived from (8.E): {bM(p,r),b† M(q,s)}=δrsδ(3)(p−q), {bM(p,r),bM(q,s)}={b† M(p,r),b† M(q,s)}=0. (d) The Dirac spinor is ψD=ψ1+iψ2where ψ1,2are Majorana spinors. The Lagrangian density is L=i¯ψ1/∂ψ1+i¯ψ2/∂ψ2−m(¯ψ1ψ1+¯ψ2ψ2)+ie(¯ψ1/Aψ2−¯ψ2/Aψ1). 8.18 Under Lorentz transformations the operator Vµ(x)=¯ψ(x)γµψ(x) trans- forms in the following way: U(Λ)Vµ(x)U−1(Λ)=U(Λ)¯ψ(x)U−1(Λ)γµU(Λ)ψ(x)U−1(Λ) =¯ψ(Λx)S(Λ)γµS−1(Λ)ψ(Λx)=Λν µVν(Λx), (8.35) since SγµS−1=Λνµγν.The other operator Aµ(x)=¯ψ(x)γ5∂µψ(x) trans- forms as U(Λ)Aµ(x)U−1(Λ)=U(Λ)¯ψ(x)U−1(Λ)γ5∂µU(Λ)ψ(x)U−1(Λ) =¯ψ(Λx)γ5∂µψ(Λx), where we used well known relation Sγ5S−1=γ5(see Problem 4.38). Since ∂µ=Λρµ∂/prime ρwe have U(Λ)Aµ(x)U−1(Λ)=Λρ µAρ(Λx). (8.36) Under parity vector Vµtransforms as follows: Vµ(x)→PVµ(x)P−1=ψ†(t,−x)γµγ0ψ(t,−x) =/braceleftbigg V0(t,−x),forµ=0 −Vi(t,−x),forµ=i =Vµ(t,−x), Chapter 8. Canonical quantization of the Dirac ﬁeld 175 since P¯ψ(x)P−1=(Pψ(x)P−1)†γ0=(γ0ψ(t,−x))†γ0=ψ†(t,−x). In the similar way we get PAµ(x)P−1=−¯ψ(t,−x)γ5∂µψ(t,−x) =/braceleftbigg −¯ψ(t,−x)γ5∂/prime 0ψ(t,−x),forµ=0 ¯ψ(t,−x)γ5∂/prime iψ(t,−x),forµ=i =−Aµ(t,−x). From τψ(t,x)τ−1=Tψ(−t,x), where τis an antiunitary operator of time reversal follows τ¯ψ(t,x)τ−1=τψ†(t,x)τ−1γ∗ 0=ψ†(−t,x)T†γ∗ 0. From the previous expressions we get τVµ(t,x)τ−1=ψ†(−t,x)T†(γ0γµ)∗Tψ(−t,x). (8.37) With a help of TγµT−1=γµ∗we get τVµ(x)τ−1=¯ψ(−t,x)γµψ(−t,x)=Vµ(−t,x). (8.38) We would suggest to reader to prove the previous result by taking T=iγ1γ3. The identity (iγ1γ3)†γ∗ 0γ∗ µiγ1γ3=γ0γµ, (8.39) has to be shown. Under time reversal the operator Aµ(x) transforms as τAµ(x)τ−1=−¯ψ(−t,x)γ5∂/primeµψ(−t,x)=−Aµ(−t,x). (8.40) FromCψa(x)C−1=(CγT 0)abψ† b(x) follows C¯ψaC−1=−ψbC−1 ba,where Cis a unitary charge conjugation operator while Cis a matrix. It is easy to see CVµC−1=−ψcC−1 caγµ abCbd¯ψd =ψc(γµ)T cd¯ψd =ψc(γµ)dc¯ψd =−¯ψdγµ dcψc =−Vµ. The minus sign in the forth line of the previous calculation appears since the ﬁelds ψand¯ψanticommute. An inﬁnity constant is ignored. Compare this result with result of Problem 4.37. In the similar way result CAµC−1= ∂µ¯ψγ5ψis derived. 8.19 The Dirac Lagrangian density transforms as 176 Solutions U(Λ)...U−1(Λ), with respect to Lorentz transformations. Therefore, we have: U(Λ)L(x)U−1(Λ) =iU(Λ)¯ψ(x)U−1(Λ)γµ∂µU(Λ)ψ(x)U−1(Λ)−mU(Λ)¯ψ(x)ψ(x)U−1(Λ) =i¯ψ(Λx)Sγµ∂µS−1ψ(Λx)−m¯ψ(Λx)SS−1ψ(Λx) =i (Λ−1)µ ν¯ψ(Λx)γνΛρ µ∂/prime ρψ(Λx)−¯ψ(Λx)ψ(Λx) =i¯ψ(Λx)γµ∂/prime µψ(Λx)−m¯ψ(Λx)ψ(Λx) =L(Λx). Under the parity Ltransforms as follows PLP−1=iψ†(t,−x)γµ∂µγ0ψ(t,−x)− −m¯ψ(t,−x)ψ(t,−x). From γµγ0∂µ=γ0γ0∂/prime 0+γ0γi∂/prime i=γ0γµ∂/prime µ, we get PL(t,x)P−1=L(t,−x). The transformation rules under time reversal and charge conjugation in the previous problem were found using the general properties of matrices Tand C. Here, we use explicit expressions for them. Starting from τψ(t,x)τ−1=iγ1γ3ψ(−t,x), (8.41) we obtain τ¯ψ(t,x)τ−1=τψ†(t,x)τ−1γ∗ 0 =−iψ†(−t,x)(γ3)†(γ1)†(γ0)∗ =−i¯ψ(−t,x)γ3γ1. Further, τLτ−1=−i¯ψ(−t,x)γ3γ1(γµ)∗γ1γ3∂µψ(−t,x) −m¯ψ(−t,x)γ3γ1γ1γ3ψ(−t,x). Applying (γ0)∗=γ0,(γ1)∗=γ1,(γ2)∗=−γ2,(γ3)∗=γ3, the anticommutation relation among γ–matrices and introducing derivatives with respect to new coordinates t/prime=−t,x/prime=xinstead of the old ones gives Chapter 8. Canonical quantization of the Dirac ﬁeld 177 τLτ−1=i¯ψ(−t,x)γµ∂/prime µψ(−t,x)−m¯ψ(−t,x)ψ(−t,x) =L(−t,x). The transformation law for ﬁeld ψunder charge conjugation CψaC−1=i (γ2)abψ† b induces C¯ψaC−1=iψb(γ2γ0)ba. Then Lagrangian density transforms as CLC−1=−iψc(γ2γ0γµγ2)ca∂µψ† a+mψb(γ2γ0γ2)baψ† a. Since γ2γ0γµγ2∂µ=(−γ0∂0+γ1∂1−γ2∂2+γ3∂3)γ0, then the kinetic term becomes −iψc/bracketleftbig −γ0∂0+γ1∂1−γ2∂2+γ3∂3)/bracketrightbig cd¯ψd. In the Dirac representation of γ–matrices the following relations are satisﬁed: (γ0)T=γ0,(γ1)T=−γ1,(γ2)T=γ2,(γ3)T=−γ3, and the kinetic term is iψc(γµT)cd∂µ¯ψd=−i∂µ¯ψdγµ dcψc. As in the previous problem we anticommute the ﬁelds ¯ψandψ, and ignore the inﬁnity constant δ(3)(0). At the end we obtain CLC−1=−i∂µ¯ψγµψ−m¯ψψ , which is the starting Lagrangian density up to four divergence. 8.20 From S(Λ)σµνS−1(Λ)=Λρ µΛσ νσρσ, (8.42) follows U(Λ)TµνU−1(Λ)=Λρ µΛσ νTρσ(Λx), (8.43) and therefore Tµνis a second rank tensor. Under parity the transformation rule is: PT0i(t,x)P−1=−T0i(t,−x), PTij(t,x)P−1=Tij(t,−x). Charge conjugation act on a Tµνtensor according to CTµν(x)C−1=−Tµν(x). (8.44) 178 Solutions In order to conﬁrm the previous result you should to prove that C−1σµνC=−(σµν)T. (8.45) The identity TσµνT−1=−(σµν)∗, (8.46) can be derived easily. Consequently, τT0i(t,x)τ−1=T0i(−t,x), τTij(t,x)τ−1=−Tij(−t,x). 9 Canonical quantization of the electromagnetic ﬁeld 9.1The commutator is [Aµ(t,x),˙Aν(t,y)] =/summationdisplay λ,λ/primei (2π)3/integraldisplayd3kd3q 2√ωkωqωq/epsilon1µ λ(k)/epsilon1ν λ/prime(q) ×/parenleftBig [aλ(k),a† λ/prime(q)]ei(k·x−q·y) −[a† λ(k),aλ/prime(q)]e−i(k·x−q·y)/parenrightBig . Using the commutation relations (9.G) as well as orthogonality relations (9.D) we obtain [Aµ(t,x),˙Aν(t,y)] =−i 2(2π)3gµν/integraldisplay d3k/parenleftBig eik·(x−y)+eik·(y−x)/parenrightBig =−igµνδ(3)(x−y). 9.2Using the commutation relations (9.G) and the completeness relation (9.D) we get iDµν=[Aµ(x),Aν(y)] =−gµν1 (2π)3/integraldisplayd3k 2|k|/parenleftBig e−ik·(x−y)−eik·(x−y)/parenrightBig .(9.1) In order to calculate the integral (9.1) we shall use spherical coordinates (using notation x0−y0=t,|x−y|=r) iDµν(x−y)=−gµν1 2(2π)2/integraldisplay∞ 0kdk/integraldisplayπ 0dθsinθ ×/parenleftBig e−i(kt−krcosθ)−ei(kt−krcosθ)/parenrightBig =−gµν1 2(2π)21 ir/integraldisplay∞ 0dk/parenleftbig e−ikt(eikr−e−ikr)+eikt(e−ikr−eikr)/parenrightbig 180 Solutions =−gµν1 2(2π)21 ir/integraldisplay∞ −∞dk/parenleftbig e−ikt+ikr−e−ikt−ikr/parenrightbig =−gµν1 4πir(δ(t−r)−δ(t+r)) =igµν1 2π/epsilon1(t)δ(t2−r2), (9.2) where /epsilon1(t)=/braceleftBigg1,t > 0 −1,t < 0 0,t =0. The previous result in terms of xandycoordinates has the form iDµν(x−y)=−igµνD(x−y) =gµν i 4π|x−y|(δ(x0−y0−|x−y|)−δ(x0−y0+|x−y|)) =i 2πgµν/epsilon1(x0−y0)δ(4)((x−y)2). 9.3Both the electric and magnetic ﬁelds are gauge invariants. The simplest way to calculate the commutators is in the Lorentz gauge. The ﬁrst commu- tator is [Ei(x),Ej(y)] =∂i x∂j y[A0(x),A0(y)] +∂0 x∂0 y[Ai(x),Aj(y)],(9.3) where we used relation between the electric ﬁeld and the electromagnetic potential: E=−∇A0−∂A ∂t. Using Problem 9.2 we get [Ei(x),Ej(y)] = i( ∂i x∂j x−δij∂0 x∂0 x)D(x−y). The commutator between the components of the magnetic ﬁeld is: [Bi(x),Bj(y)] =/epsilon1ikl/epsilon1jmn∂x k∂y m[Al(x),An(y)] =i/epsilon1ikl/epsilon1jml∂x k∂y mD(x−y) =i (δijδkm−δimδkj)∂x k∂y mD(x−y) =i (−δij∆+∂x i∂x j)D(x−y). In the similar way one can get [Ei(x),Bj(y)] = i/epsilon1jki∂x 0∂x kD(x−y). Now, consider the equal–time commutators i.e. take that x0=y0. First show that ∂x0D(x−y)|x0=y0=−δ(3)(x−y), Chapter 9. Canonical quantization of the electromagnetic ﬁeld 181 ∂2 x0D(x−y)|x0=y0=0, ∂i xD(x−y)|x0=y0=0, ∂i x∂j xD(x−y)|x0=y0=0, ∂x i∂x 0D(x−y)|x0=y0=−∂x iδ(3)(x−y). The easiest way to prove the previous formulae is to start with the integral expression for D–function: D(x)=−i (2π)3/integraldisplayd3k 2|k|/parenleftBig e−ik·(x−y)−eik·(x−y)/parenrightBig . The results for the equal–time commutators are: [Ei(x),Ej(y)]|x0=y0=0, [Bi(x),Bj(y)]|x0=y0=0, [Ei(x),Bj(y)]|x0=y0=−i/epsilon1ijk∂x kδ(3)(x−y). 9.4We shall ﬁrst calculate the commutator between the Hamiltonian and the electromagnetic potential Aν(x): [H,Aν(x)] =−1 2/integraldisplay d3y[πµπµ+∇Aµ∇Aµ,Aν(x)] =−1 2/integraldisplay d3y(πµ(y)[πµ(y),Aν(x)] + [πµ(y),Aν(x)]πµ(y)) =−1 2/integraldisplay d3yδ(3)(x−y)/parenleftbig πµ(y)(−i)gν µ−igµνπµ(y)/parenrightbig =iπν(x) =−i∂0Aν. The commutator between three–momentum of electromagnetic ﬁeld and elec- tromagnetic potential can be calculated in the similar manner [Pi,Aν(x)] =−/integraldisplay d3y[˙Aρ(y)∂iAρ(y),Aν(x)] =−igρν/integraldisplay d3yδ(3)(x−y)∂iAρ(y) =−i∂iAν(x). 9.5The helicity of the state /epsilon1µ (±)(k) is determined under the rotation for angle θaboutk/|k|=ez–axis. Namely, 182 Solutions /epsilon1/prime ±=Λ(θ)/epsilon1± = 10 0 0 0c o s θsinθ0 0−sinθcosθ0 00 0 1  0 1/√ 2 ±i/√ 2 0  =e±iθ 0 1/√ 2 ±i/√ 2 0  =e±iθ/epsilon1±. From the last line we can read oﬀ that helicity is λ=±1.Polarization of these photons is circular. 9.6The four–momentum of the photon for observer S/primeis k/primeµ=Λµ νkν= γ−βγ00 −βγ γ 00 00 1 000 0 1  k 0 0 k = kγ −kβγ 0 k . Under the Lorentz transformation the polarization vector /epsilon1 µ(k) transforms as /epsilon1/primeµ(k/prime)=Λµ ν/epsilon1ν(k)−iα(k/prime)k/primeµ. The second term comes from the gauge transformation of the electromagnetic potential; α(κ/prime) is an arbitrary function of the momentum. This term can be easily obtained by substituting A/primeµ=/epsilon1/primeµ(k/prime)e−ik/prime·x/prime, and Λ(x/prime)=αe−ik/prime·x/prime in the gauge transformation rule ˜Aµ=A/primeµ+∂/primeµΛ(x/prime). If we choose the function α=iβ/kwe get /epsilon1/primeµ(k/prime)= 0 γ−1 0 β . Note that the vector /epsilon1/primeis orthogonal to the photon direction of motion k/prime/k/prime. This was a condition to determine the function α(k/prime). Thus, the polarization of photon is transversal for both observers. Chapter 9. Canonical quantization of the electromagnetic ﬁeld 183 9.7 (a) In the ﬁrst step use the commutation relations (9.G) to derive the expres- sion: [a3(k)−a0(k),a† 3(q)−a† 0(q)] = 0 . From the previous result it is not hard to show that /angbracketleftΦn|Φn/angbracketright=δn0. (b) There are only two terms in the expression /angbracketleftΦ|Aµ|Φ/angbracketrightwhich are not equal to zero: /angbracketleftΦ|Aµ|Φ/angbracketright=C∗ 0C1/angbracketleftΦ0|Aµ|Φ1/angbracketright+C0C∗ 1/angbracketleftΦ1|Aµ|Φ0/angbracketright. It is easy to see that /angbracketleftΦ0|Aµ|Φ1/angbracketright=−1 (2π)3/2/integraldisplayd3k/radicalbig 2|k|f(k)e−ik·x/parenleftBig /epsilon1µ (0)(k)+/epsilon1µ (3)(k)/parenrightBig . By applying the relation /epsilon1µ (0)(k)+/epsilon1µ (3)(k)=kµ |k|, we get /angbracketleftΦ|Aµ|Φ/angbracketright=∂µΛ, where Λis given by Λ=−i (2π)3/2/integraldisplayd3k/radicalbig 2|k||k|/parenleftbig C∗ 0C1f(k)e−ik·x−C0C∗ 1f∗(k)eik·x/parenrightbig . 9.8The quantities deﬁned in this problem are projectors on massless states with the helicities ±1 and 0. Let us ﬁrst calculate Pµν ⊥Pνσ⊥: Pµν ⊥Pνσ⊥=kµ¯kν+kν¯kµ k·¯kkν¯kσ+kσ¯kν k·¯k =kµ¯kσ+kσ¯kµ k·¯k =Pµ σ⊥, since¯k·¯k= 0. The other expressions can be evaluated in the same way. The results are: PµνPνσ=Pµ σ,Pµν+Pµν ⊥=gµν, gµνPµν=2,gµνP⊥ µν=2,P µνPνσ ⊥=0. 9.9 (a) The components of the angular momentum Mijwere calculated in Prob- lem 5.18 using the Nether technique. It follows that (in the Coulomb gauge) Jl=/epsilon1lij/integraldisplay d3x/parenleftBig ˙AjAi+xi˙Ak∂jAk/parenrightBig . 184 Solutions (b) The spin part of the angular momentum is Sl=/epsilon1lij/integraldisplay d3x˙AjAi. By substituting the explicit expression for the electromagnetic potential we get Sl=i 2/epsilon1lij/summationdisplay λ,λ/prime/integraldisplay d3k/parenleftBig −/epsilon1j λ(k)/epsilon1i λ/prime(−k)aλ(k)aλ/prime(−k)e−2iωkt− −/epsilon1j λ(k)/epsilon1i λ/prime(k):aλ(k)a† λ/prime(k):+/epsilon1j λ(k)/epsilon1i λ/prime(k)a† λ(k)aλ/prime(k)+ +/epsilon1j λ(k)/epsilon1i λ/prime(−k)a† λ(k)a† λ/prime(−k)e2iωkt/parenrightBig . The ﬁrst and the last term are symmetric under the change of indices iand j, so that the multiplication by the antisymmetric /epsilon1symbol give vanishing contribution. Then: S=i 2/summationdisplay λ,λ/prime/integraldisplay d3k(/epsilon1λ/prime(k)×/epsilon1λ(k))/parenleftBig a† λ(k)aλ/prime(k)−a† λ/prime(k)aλ(k)/parenrightBig . By using /epsilon11(k)×/epsilon12(k)=k/|k|we get S=i/integraldisplay d3kk |k|/parenleftBig a† 2(k)a1(k)−a† 1(k)a2(k)/parenrightBig . By using the operators a±(k) which were deﬁned in the problem, the spin Sbecomes diagonal S=/integraldisplay d3kk |k|/parenleftBig a† +(k)a+(k)−a† −(k)a−(k)/parenrightBig . From the previous result we conclude that the operator Λ=/integraldisplay d3k/parenleftBig a† +(k)a+(k)−a† −(k)a−(k)/parenrightBig , is the helicity. (c) By applying the commutation relations (9.J) we get [a† ±(k),a±(q)] =−δ(3)(k−q), from which we have Λa† ±(q)|0/angbracketright=[Λ,a† ±(q)]|0/angbracketright =±/integraldisplay d3kδ(3)(k−q)a† ±(k)|0/angbracketright =±a† ±(q)|0/angbracketright. Chapter 9. Canonical quantization of the electromagnetic ﬁeld 185 (d) The commutator between the angular momentum and the electromagnetic potential is: [Jl,Am(t,x)] =/epsilon1lij/integraldisplay d3y/bracketleftBig ˙Aj(t,y),Am(t,x)/bracketrightBig Ai(t,y)+ +yi[˙An(t,y),Am(t,x)]∂jAn(t,y) =−i/epsilon1lij/integraldisplay d3yδ(3) ⊥nm(x−y)/parenleftbig δnjAi(t,y)+yi∂jAn(t,y)/parenrightbig =−i/epsilon1lij1 (2π)3/integraldisplay d3y/integraldisplay d3keik·(x−y)/parenleftbigg δnm−knkm k2/parenrightbigg ×/parenleftbig δjnAi(t,y)+yi∂jAn(t,y)/parenrightbig . (9.4) The term which contains knkm/k2is equal to zero: /integraldisplay d3y/integraldisplay d3kknkm k2eik·(x−y)/parenleftbig Aiδnj+yi∂jAn/parenrightbig =/integraldisplay d3y/integraldisplay d3k/parenleftbig Aiδnj+yi∂jAn/parenrightbigkm k2(i∂ ∂yneik·(x−y)).(9.5) Integrating by parts in (9.5) we get that it vanishes. Then from (9.4) follows [Jl,Am(t,x)] = i/epsilon1lmiAi+i (r×∇)lAm. 9.10 The electric ﬁeld is E=/integraldisplayd3k/radicalbig 2(2π)3ωk2/summationdisplay λ=1iωk/epsilon1λ(k)/parenleftBig aλ(k)e−ik·x−a† λ(k)eik·x/parenrightBig , while the magnetic ﬁeld is given by B=/integraldisplayd3k/radicalbig 2(2π)3ωk2/summationdisplay λ=1i(k×/epsilon1λ(k))/parenleftBig aλ(k)e−ik·x−a† λ(k)eik·x/parenrightBig . (a) The vacuum expectation value of the anticommutator between the electric and the magnetic ﬁeld is /angbracketleft0|{Em(x),Bn(y)}|0/angbracketright=/angbracketleft0|Em(x)Bn(y)|0/angbracketright+/angbracketleft0|Bn(y)Em(x)|0/angbracketright =/integraldisplayd3kd3q 2(2π)3√ωkωq2/summationdisplay λ=12/summationdisplay λ/prime=1ωk/epsilon1m λ(k)(q×/epsilon1λ/prime(q))n ×/parenleftBig /angbracketleft0|aλ(k)a† λ/prime(q)|0/angbracketrighte−ik·x+iq·y+/angbracketleft0|aλ/prime(q)a† λ(k)|0/angbracketrighteik·x−iq·y/parenrightBig =/integraldisplayd3k 2(2π)32/summationdisplay λ=1/epsilon1m λ(k)(k×/epsilon1λ(k))n/parenleftBig e−ik·(x−y)+eik·(x−y)/parenrightBig . (9.6) By using 186 Solutions 2/summationdisplay λ=1/epsilon1nijki/epsilon1j λ(k)/epsilon1m λ(k)=/epsilon1nimki, the formula (9.6) becomes /angbracketleft0|{Em(x),Bn(y)}|0/angbracketright=/integraldisplayd3k 2(2π)3/epsilon1njmkj/parenleftBig e−ik·(x−y)+eik·(x−y)/parenrightBig . The result can be rewritten in the following form: /angbracketleft0|{Em(x),Bn(y)}|0/angbracketright=/epsilon1njm ∂2 ∂x0∂xj/integraldisplayd3k 2(2π)3ωk ×/parenleftBig e−ik·(x−y)+eik·(x−y)/parenrightBig =−1 2π2/epsilon1njm ∂2 ∂xo∂xj1 (x−y)2. (9.7) The integral in the ﬁrst line was calculated in Problem 7.14. (b) As before, /angbracketleft0|{Bi(x),Bj(y)}|0/angbracketright=/integraldisplayd3k 2(2π)3ωk2/summationdisplay λ=1(k×/epsilon1λ(k))i(k×/epsilon1λ(k))j ×/parenleftBig e−ik·(x−y)+eik·(x−y)/parenrightBig . Since (k×/epsilon1λ(k))i(k×/epsilon1λ(k))j=2/summationdisplay λ=1/epsilon1imn/epsilon1jpqkmkp/epsilon1n λ(k)/epsilon1q λ(k) =/epsilon1imn/epsilon1jpnkmkp =(k2δij−kikj). we have /angbracketleft0|{Bi(x),Bj(y)}|0/angbracketright=/integraldisplayd3k 2(2π)3ωk(k2δij−kikj) ×/parenleftBig e−ik·(x−y)+eik·(x−y)/parenrightBig =−1 2π2/parenleftbigg∂2 ∂xi∂xj−/triangleδij/parenrightbigg1 (x−y)2. (c) This expectation value can be obtained in the same way as the previous ones. The result is /angbracketleft0|{Ei(x),Ej(y)}|0/angbracketright=−1 2π2/parenleftbigg −∂2 ∂(x0)2δij+∂2 ∂xi∂xj/parenrightbigg1 (x−y)2.(9.8) Chapter 9. Canonical quantization of the electromagnetic ﬁeld 187 9.11 (a) The vector potential Acan be decomposed into parallel and normal com- ponents: A=A⊥+A/bardbl. The normal component of the vector potential is along the z−axis, while A/bardblis parallel to the plates. In the Coulomb gauge ( A0=0,divA=0 ) the electric ﬁeld is E=−∂A ∂t. Since the plates are ideal conductors, the parallel component of the electric ﬁeld and the normal component of magnetic ﬁeld vanish on the plates, i.e. ∂A/bardbl ∂t/vextendsingle/vextendsingle/vextendsingle/vextendsingle z=0=∂A/bardbl ∂t/vextendsingle/vextendsingle/vextendsingle/vextendsingle z=a=0, (9.9) Bz|z=0=Bz|z=a=0. (9.10) The vector potential Asatisﬁes the equation /parenleftbigg∂2 ∂t2−∆/parenrightbigg A=0. If we assume that a particular solution of this equation has the following form A=F(t,x,y)(Z1(z)e1+Z2(z)e2+Z3(z)e3), (9.11) then we get: d2Zi dz2+k2 3Zi= 0 (9.12) and /parenleftbigg∂2 ∂t2−∂2 ∂x2−∂2 ∂y2+k2 3/parenrightbigg F=0. (9.13) The solution of the ﬁrst equation is Zi=aisin(k3z)+bicos(k3z). The boundary conditions (9.9–9.10) give b1=b2=0a n d k3=nπ/a (n= 0,1,2,...).A particular solution for the function FisF=e−iωt+ik1x+ik2y. Inserting it into (9.13) we obtain ω=±ωk,n=±/radicalbigg k2 1+k2 2+/parenleftBignπ a/parenrightBig2 . From the Coulomb gauge condition follows that a3=0a n d ia1k1+ia2k2−nπ ab3=0 188 Solutions for (n/negationslash= 0); obviously there are two independent states of polarization, unless n=0 .F o r n= 0 polarization vector is e3, and there is only one mode. Thus, a particular solution is A=F/parenleftbig /epsilon1/bardblsin(nπz/a )+b3cos(nπz/a )/parenrightbig , where /epsilon1/bardblbelongs to the xy–plane. Then, the general solution reads: A=∞/summationdisplay n=1/integraldisplayd2k 2π1/radicalbig2ωk,n2/summationdisplay λ=1[aλ(k1,k2,n)e−iωk,nt+ik1x+ik2y ×(sin(nπz/a )/epsilon1/bardbl(k,n,λ)+c o s ( nπz/a )e3)+ +a† λ(k1,k2,n)eiωk,nt−ik1x−ik2y ×(sin(nπz/a )/epsilon1∗ /bardbl(k,n,λ)+c o s ( nπz/a )ez)] + +/integraldisplayd2k 2π1√2ωk[a(k1,k2)e−iωkt+ik1x+ik2y+ +a†(k1,k2)eiωkt−ik1x−ik2y]e3, (9.14) where ωk=/radicalbig k2 1+k2 2. (b) The canonical commutation relations have the following form [aλ(k1,k2,n),a† λ/prime(k/prime 1,k/prime 2,m)] =δnmδλλ/primeδ(k1−k/prime 1)δ(k2−k/prime 2), [a(k1,k2),a†(k/prime 1,k/prime 2)] =δ(k1−k/prime 1)δ(k2−k/prime 2), while the other commutators vanish. The Hamiltonian is given by H=/integraldisplay d2k∞/summationdisplay n=11 2ωk,n2/summationdisplay λ=1[a† λ(k1,k2,n)aλ(k1,k2,n) +aλ(k1,k2,n)a† λ(k1,k2,n)] +1 2/integraldisplay d2kωk[a†(k1,k2)a(k1,k2)+a(k1,k2)a†(k1,k2)].(9.15) (c) The energy of the ground state |0/angbracketrightis /angbracketleft0|H|0/angbracketright=∞/summationdisplay n=12/summationdisplay λ=1/integraldisplay d2k1 2ωk,n/angbracketleft0|aλ(k1,k2,n)a† λ(k1,k2,n)|0/angbracketright +/integraldisplay d2k1 2ωk/angbracketleft0|a(k1,k2)a†(k1,k2)|0/angbracketright =∞/summationdisplay n=11 2/integraldisplay d2kωk,n2δ(2)(0) +1 2/integraldisplay d2kωkδ(2)(0). Since Chapter 9. Canonical quantization of the electromagnetic ﬁeld 189 δ(2)(0) =/integraldisplaydxdy (2π)2eik1x+ik2y/vextendsingle/vextendsingle/vextendsingle k/bardbl=0=L2 (2π)2 we have E=L2 2(2π)2/integraldisplay d2k/parenleftBigg 2∞/summationdisplay n=1/radicalbigg k2 1+k2 2+/parenleftBignπ a/parenrightBig2 +/radicalBig k2 1+k2 2/parenrightBigg .(9.16) (d) The vacuum energy of the same part of space in the absence of the plates is given by E0=1 2/integraldisplayL2d2k (2π)2/integraldisplayadk3 2π2/radicalBig k2 1+k2 2+k2 3 =/integraldisplayL2d2k (2π)2/integraldisplay∞ 0dn/radicalbigg k2 1+k2 2+/parenleftBignπ a/parenrightBig2 . Then /epsilon1is /epsilon1=1 2/integraldisplay∞ 0kdk 2π/bracketleftBigg k+2∞/summationdisplay n=1/radicalbigg k2+/parenleftBignπ a/parenrightBig2 −2/integraldisplay∞ 0dn/radicalbigg k2+/parenleftBignπ a/parenrightBig2/bracketrightBigg . (9.17) The last integral can be rewritten as follows /epsilon1=π2 8a3/integraldisplay∞ 0du/parenleftBigg √u+2∞/summationdisplay n=1/radicalbig u+n2−2/integraldisplay∞ 0dn/radicalbig u+n2/parenrightBigg ,(9.18) where a new variable u=a2k2/π2was introduced. After the regularization /epsilon1takes the form /epsilon1=π2 8a3/integraldisplay∞ 0du/parenleftBigg √uf(π√u a)+2∞/summationdisplay n=1/radicalbig u+n2f(π√ u+n2 a)− −2/integraldisplay∞ 0dn/radicalbig u+n2f(π√ u+n2 a)/parenrightBigg , (9.19) and becomes ﬁnite. If we deﬁne a new function F(n)=/integraldisplay∞ 0du/radicalbig u+n2f(π√ u+n2 a), /epsilon1becomes /epsilon1=π2 8a3/parenleftBigg F(0) + 2∞/summationdisplay n=1F(n)−2/integraldisplay∞ 0dnF(n)/parenrightBigg . (9.20) To calculate the previous expression we will use the Euler-Maclaurin for- mula: 190 Solutions ∞/summationdisplay n=1F(n)−/integraldisplay∞ 0dnF(n)+1 2F(0) =−1 2!B2F/prime(0)−1 4!B4F/prime/prime/prime(0) + ... . B2,B4,...are Bernouli numbers and they are deﬁned by y ey−1=∞/summationdisplay ν=0Bνyν ν!. Consequently, /epsilon1=π2 4a3/parenleftbigg −1 2!B2F/prime(0)−1 4!B4F/prime/prime/prime(0) + .../parenrightbigg . (9.21) It is easy to get F/prime(0) = 0 ,F/prime/prime/prime(0) =−4. Then the vacuum energy per unit surface is /epsilon1=−π2 720a3. From the expression for the energy we can derive the force: f=−∂/epsilon1 ∂a=−π2 240a4. Ifa=1µmandL=1cmthe force is 10−8N. The vacuum energy of the electromagnetic ﬁeld between the two conducting plates produces a weak attractive force between them. This eﬀect was measured in 1958. (e) The integral Ican be found in [9]: I=2π/integraldisplay∞ 0kdk (k2+m2)α =πΓ(α−1) Γ(α)1 (m2)α−1. (9.22) Then E L2=1 2/integraldisplayd2k (2π)2 lim µ→01 (k2+µ2)−1/2+2∞/summationdisplay n=11/radicalBig k2+/parenleftbignπ a/parenrightbig2  =−1 12π/parenleftBigg lim µ→0(µ2)3/2+2π3 a3∞/summationdisplay n=1n3/parenrightBigg =−π2 6a3∞/summationdisplay n=1n3. (9.23) From ζ(1−n)=(−1)1+nBn n, follows that ζ(−3) = 1 /120 since B4=−1/30. Finally, we get the same result as before E L2=−π2 720a3. 10 Processes in the lowest order of the perturbation theory 10.1 The transition probability is |Sﬁ|2=( 2π)8[δ(4)(p/prime 1+p/prime 2−p1−p2)]2mAmBmCmD V4E1E2E/prime 1E/prime 2|M|2. (10.1) The square of the four-dimensional delta function is [δ(4)(pf−pi)]2=δ(4)(pf−pi)δ(4)(0) =1 (2π)4δ(4)(pf−pi)/integraldisplay Vd3x/integraldisplayT 2 −T 2dt =TV (2π)4δ(4)(pf−pi), (10.2) where: pi=p1+p2andpf=p/prime 1+p/prime 2are initial and ﬁnal four-momentum respectively. The diﬀerential cross section (10.D) is dσ=|Sﬁ|2 T1 |Jin|V2d3p/prime 1d3p/prime2 (2π)6. (10.3) The current density ﬂux, in the center–of–mass frame is |Jin|=|¯ψγψ|=|p1|(E1+E2) VE1E2. (10.4) By substituting (10.1), (10.2 ) and (10.4) into (10.3) the following formula is obtained dσ=1 (2π)2δ(E/prime 1+E/prime 2−E1−E2)δ(3)(p/prime 1+p/prime 2−p1−p2)|M|2 ×mAmBmCmD (E1+E2)E/prime 1E/prime 2|p1|d3p/prime 1d3p/prime2. (10.5) By integrating over p/prime 2we get 192 Solutions dσ dΩ=1 (2π)2/integraldisplay δ(/radicalBig p/prime2 1+m2 C+/radicalBig p/prime2 1+m2 D−E1−E2)|M|2 ×mAmBmCmD (E1+E2)E/prime 1E/prime 2p/prime2 1dp/prime 1 p1, where the fact that we are doing calculations in the center–of–mass frame have been used. By applying formula /integraldisplay dxg(x)δ(f(x)) =g(x) |f/prime(x)|/vextendsingle/vextendsingle/vextendsingle/vextendsingle f(x)=0(10.6) the requested result is obtained. 10.2 Four–dimensional delta function and integration measure are Lorentz invariant quantities (Problem 6.3) so is the given integral. In the inertial frame in which P= 0 the integral becomes I=1 4/integraldisplayd3p/radicalbig p2+m2d3q/radicalbig q2+m/prime2δ(3)(p+q)δ(Ep+Eq−P0). (10.7) By integrating over qin (10.7) and introducing spherical coordinates we obtain I=π/integraldisplay∞ 0p2dp1/radicalbig p2+m2/radicalbig p2+m/prime2δ(/radicalbig p2+m2+/radicalbig p2+m/prime2−P0). By applying the formula (10.6) one gets I=π P0/radicalBigg (m2−m/prime2−P2 0)2 4P2 0−m/prime2. 10.3 The Feynman amplitude, i Mis a complex number so that (iM)∗=( iM)†=( ¯u(p,r)γµ(1−γ5)u(q,s))† /epsilon1µ∗(k,λ) =u†(q,s)(1−γ5)γ0γµγ0γ0u†(p,r)/epsilon1µ∗(k,λ) =¯u(q,s)(1 + γ5)γµu(p,r)/epsilon1µ∗(k,λ), where identities from Problems 3.1 and 3.3 are used. The average value of the squared amplitude is ( a,b,... are Dirac’s indices) 2/summationdisplay λ=12/summationdisplay r,s=1|M|2=2/summationdisplay λ=12/summationdisplay r,s=1¯ua(p,r)[γµ(1−γ5)]abub(q,s) ×¯uc(q,s)[(1 + γ5)γν]cdud(p,r)/epsilon1µ(k,λ)/epsilon1ν∗(k,λ) =/parenleftBigg2/summationdisplay r=1ud(p,r)¯ua(p,r)/parenrightBigg [γµ(1−γ5)]ab ×/parenleftBigg2/summationdisplay s=1ub(q,s)¯uc(q,s)/parenrightBigg [(1 +γ5)γν]cd2/summationdisplay λ=1/epsilon1µ∗(k,λ)/epsilon1ν(k,λ). Chapter 10. Processes in the lowest order of the perturbation theory 193 By applying expression for the projection operator into positive-energy solu- tions (Problem 4.4) we have 2/summationdisplay λ=12/summationdisplay r,s=1|M|2=/parenleftbigg/p+m 2m/parenrightbigg da[γµ(1−γ5)]ab ×/parenleftbigg/q+m 2m/parenrightbigg bc[(1 +γ5)γν]cd2/summationdisplay λ=1/epsilon1µ(k,λ)/epsilon1ν∗(k,λ) =1 4m22/summationdisplay λ=1/epsilon1µ(k,λ)/epsilon1ν∗(k,λ) ×tr [(/p+m)γµ(1−γ5)(/q+m)(1 + γ5)γν]. Using the facts that γ5anticommutes with γµmatrices and that ( γ5)2=1 , the last expression becomes 2/summationdisplay λ=12/summationdisplay r,s=1|M|2=1 2m2tr [(/p+m)γµ(1−γ5)/qγν]2/summationdisplay λ=1/epsilon1µ(k,λ)/epsilon1ν∗(k,λ). By applying the corresponding traces form Problem 3.6 one obtains 2/summationdisplay λ=12/summationdisplay r,s=1|M|2=2 m2/bracketleftbig pµqν+pνqµ−(p·q)gµν+i/epsilon1ανβµqαpβ/bracketrightbig ×2/summationdisplay λ=1/epsilon1µ(k,λ)/epsilon1ν∗(k,λ). (10.8) To sum over the photon polarizations is reduced to replacement 2/summationdisplay λ=1/epsilon1µ(k,λ)/epsilon1ν∗(k,λ)→−gµν(10.9) in the expression (10.8) because the other two terms in (9.E) do not give any contribution. The result is 4 p·q/m2. 10.4 In the ﬁrst part of the Problem we shall apply Wick’s theorem for bosons and in the second part we shall make use of the Wick’s theorem for fermions. (a) It is clear that all normal-ordered terms fall oﬀ, because their vacuum expectation value is equal to zero. Thus the only remaining terms are those with four contractions. If we contract one φ(x) with one φ(y)f o u r times we shall get ( /angbracketleft0|T(φ(x)φ(y))|0/angbracketright)4.This can be done in 4! = 24 ways. The next possibility is to make two contractions between ﬁelds φ(x)a n d φ(y).One ﬁeld φ(x) can be contracted in 4 ways with one of the φ(y)/primes. The next φ(x) can be contracted in three ways with one of the remaining 194 Solutions φ(y)/primes . The obtained result has to be multiplied by 6, because this is the number of ways in which two ﬁelds φ(x) can be chosen from the four possible. Thus, there are 4 ·3·6 = 72 possible contractions of this type. There are three mutual contractions between two ﬁelds φ(x), the similar is obtained for ﬁelds φ(y), so the corresponding coeﬃcient is 9. Thus, /angbracketleft0|T(φ4(x)φ4(y))|0/angbracketright) = 24( /angbracketleft0|T(φ(x)φ(y))|0/angbracketright)4 +7 2/angbracketleft0|T(φ(x)φ(x))|0/angbracketright/angbracketleft0|T(φ(y)φ(y))|0/angbracketright(/angbracketleft0|T(φ(x)φ(y))|0/angbracketright)2 +9 (/angbracketleft0|T(φ(x)φ(x))|0/angbracketright)2(/angbracketleft0|T(φ(y)φ(y))|0/angbracketright)2 = 24(i ∆F(x−y))4+ 72(i ∆F(x−x))i∆F(y−y)(i∆F(x−y))2 + 9(i∆F(x−x))2(i∆F(y−y))2. The last expression can be represented by the following diagram: 24· y x +72·x y +9·x y (b) Here, the equal-time contractions are forbidden. The result is T(:φ4(x): :φ4(y): )=1 6: φ3(x)φ3(y):i∆F(x−y) +7 2: φ2(x)φ2(y):( i∆F(x−y))2 +9 6: φ(x)φ(y):( i∆F(x−y))3 + 24(i ∆F(x−y))4. (10.10) (c) By applying Wick’s theorem for fermions one obtains /angbracketleft0|T(¯ψ(x)ψ(x)¯ψ(y)ψ(y))|0/angbracketright =iSF(x−x)iSF(y−y)−iSF(x−y)iSF(y−x). 10.5 (a) The given diagram is obtained from the expression −iλ 4!/integraldisplay d4y/angbracketleft0|T(φ(x1)φ(x2)φ4(y))|0/angbracketright, where φ(x1) is to be contracted with one φ(y) (there are four ways to do this) and φ(x2) with one of the remaining three φ(y)/primes. The symmetry factor is1 4!4·3=1 2. This result can be easily checked by using the formula given in the problem, where g=1,α=0i β=1 . Chapter 10. Processes in the lowest order of the perturbation theory 195 (b) This diagram is one of the terms in the 1 2!/parenleftbigg −iλ 4!/parenrightbigg2/integraldisplay d4y1d4y2/angbracketleft0|T(φ(x1)φ(x2)φ4(y1)φ4(y2))|0/angbracketright, where φ(x1) is contracted with one of the four φ(y1)/primes (there are four ways to do this); φ(x2) with one of the remaining φ(y1) ﬁelds (there are three ways to do this). It is necessary to make two more contractions between φ(y1)a n d φ(y2)w h i c hc a nb ed o n ei n4 ·3 = 12 ways. Thus we have: S−1=2 !1 2!/parenleftbigg1 4!/parenrightbigg2 4·3·4·3=1 4, so the symmetry factor is S= 4. The same result is obtained by plugging g=1,α2=1i β= 1 into the formula given in the problem. (c) In order to get this diagram it is necessary to make the following contrac- tions in this third-order expression: 1 3!/parenleftbigg −iλ 4!/parenrightbigg3/integraldisplay d4y1d4y2d4y3/angbracketleft0|T(φ(x1)φ(x2)φ4(y1)φ4(y2)φ4(y3))|0/angbracketright, (10.11) φ(x1) with one of the four φ(y1)/primes (four ways); φ(x2) with one of the remaining φ(y1) ﬁelds (three ways); two φ(y1) ﬁelds with four φ(y2) ﬁelds (4·2 = 8 ways); the remaining φ(y1) ﬁeld with φ(y3) ﬁelds (4 ways); three contractions between three φ(y2)/primes and three φ(y3) ﬁelds (3 ·2 = 6 ways). Finally, one has to divide the obtained expression by two, because of the symmetry y2↔y3. By combining all the factors we have: S−1=3 !1 3!/parenleftbigg1 4!/parenrightbigg3 4·3·4·2·4·3·2·1 2=1 12, (10.12) soS= 12. This result can be checked by applying the formula given in the problem: g=2,n=3,α3=1,β=0 . 10.6 The result is 1 2/parenleftbigg−iλ 3!/parenrightbigg2/integraldisplay d4y1d4y2/angbracketleft0|T(φ(x1)φ(x2)φ3(y1)φ3(y2))|0/angbracketright= =/integraldisplay d4y1d4y2(−iλ)2/bracketleftbigg1 2i∆F(x1−y1)i∆F(x2−y2)(i∆F(y1−y2))2 +1 12i∆F(x1−x2)(i∆F(y1−y2))3 +1 8i∆F(x1−x2)i∆F(y1−y1)i∆F(y2−y2)i∆F(y1−y2) +1 2i∆F(x1−y1)i∆F(x2−y1)i∆F(y1−y2)i∆F(y2−y2) +1 4i∆F(x1−y1)i∆F(x2−y2)i∆F(y1−y1)i∆F(y2−y2)/bracketrightbigg (10.13) 196 Solutions which can be represented by the following diagram: 1 2· x2x1 1y2y +1 12·x2 x1 1y2y+1 8·x2 x1 1y2y +1 2· x2 x1 1y2y+1 4· x2 x11y2y The coeﬃcient1 2in the ﬁrst term (10.13) can be obtained in the following way: contraction φ(x1) with φ(y1) can be done in three ways, as well as the contraction φ(x2) with φ(y2). Two contractions φ(y1) with φ(y2) can be done in two ways. The obtained result has to be multiplied by 2! which comes from the interchange y1-vertices with y2-vertices, because, for instance, we could contract φ(x1) with φ(y2) instead of φ(y1). Thus, the overall coeﬃcient is 1 23·3·2 3!·3!·2=1 2. (10.14) In the second and third term there is no additional multiplying by 2 which comes from the y1↔y2interchange! 10.7 (a) Diagram for this process is represented in Fig. 10.1. Fig. 10.1. The three-level Feynman diagram for the scattering µ−(p1)+µ+(p2)→ e−(q1)+e+(q2) The Feynman amplitude is given by the following expression iM=ie2 (p1+p2)2+i/epsilon1¯v(p2,s)γµu(p1,r)¯u(q1,r/prime)γµv(q2,s/prime), hence Chapter 10. Processes in the lowest order of the perturbation theory 197 /angbracketleftbig |M|2/angbracketrightbig =e4 41 (p1+p2)42/summationdisplay r,s=12/summationdisplay r/prime,s/prime=1¯va(p2,s)γµ abub(p1,r) ×¯uc(q1,r/prime)(γµ)cdvd(q2,s/prime)¯ue(p1,r)γν efvf(p2,s) ×¯vg(q2,s/prime)(γν)ghuh(q1,r/prime) =e4 4(p1+p2)4/summationdisplay s(vf(p2,s)¯va(p2,s))γµ ab ×/summationdisplay r(ub(p1,r)¯ue(p1,r)) (γν)ef ×/summationdisplay r/prime(uh(q1,r/prime)¯uc(q1,r/prime)) (γµ)cd ×/summationdisplay s/prime(vd(q2,s/prime)¯vg(q2,s/prime)) (γν)gh. By performing matrix multiplying in the preceding expression we obtain two traces (Problem 4.4) /angbracketleftbig |M|2/angbracketrightbig =e4 4(p1+p2)41 16m2em2µtr[(/q1+me)γµ(/q2−me)γν] ×tr[(/p2−mµ)γµ(/p1+mµ)γν]. By applying corresponding identities from Problem 3.6 we get /angbracketleftbig |M|2/angbracketrightbig =e4 4(p1+p2)41 m2em2µ/bracketleftbig q1µq2ν+q2µq1ν−(q1·q2)gµν−m2 egµν/bracketrightbig ×/bracketleftbig pµ 1pν 2+pµ 2pν 1−(p1·p2)gµν−m2 µgµν/bracketrightbig . After multiplying and reducing the preceding expression one obtains /angbracketleftbig |M|2/angbracketrightbig =e4 4(p1+p2)4m2em2µ[2(p2·q1)(p1·q2)+2 ( p2·q2)(p1·q1) +2m2 e(p1·p2)+2m2 µ(q1·q2)+4m2 em2µ/bracketrightbig . (10.15) In the center–of–mass frame the four-momenta are p1=(E,0,0,p), p2=(E,0,0,−p), q1=(E/prime,qsinθ,0,qcosθ), q2=(E/prime,−qsinθ,0,−qcosθ), where pandqare intensities of the corresponding three–momenta vectors. After simple scalar product computations in (10.15) one gets: /angbracketleftbig |M|2/angbracketrightbig =e4 32E4m2em2µ/bracketleftbig 2((EE/prime)2+m2 em2µ)(1 + cos2θ) +2 (E2m2 e+E/prime2m2 µ)(1−cos2θ)−m4 e−m4 µ/bracketrightbig .(10.16) 198 Solutions In the high energy limit ( p≈E) expression (10.16) becomes /angbracketleftbig |M|2/angbracketrightbig =e4 16m2em2µ(1 + cos2θ). (10.17) Using the previous expression and Problem 10.1 the diﬀerential cross sec- tion is dσ dΩ=e4 256π2E2(1 + cos2θ). (b) We shall discuss just the main results. From the diagram Fig. 10.2. The Feynman diagram for the scattering e−(p1)+µ+(q1)→e−(p2)+ µ+(q2) in the lowest order the amplitude is iM=¯u(p2,r2)(ieγµ)u(p1,r1)−igµν (p1−p2)2+i/epsilon1¯v(q1,s1)(ieγν)v(q2,s2). The squared Feynman amplitude module (averaged over spin states of the initial particles and summed over spin states of the ﬁnal particles) is: /angbracketleftbig |M|2/angbracketrightbig =e4 4(p1−p2)41 16m2em2µtr [(/p2+me)γµ(/p1+me)γν] ×tr [(/q1−mµ)γµ(/q2−mµ)γν] =e4 2(p1−p2)4m2em2µ[(p2·q1)(p1·q2)+(p1·q1)(p2·q2) −m2 µ(p1·p2)−m2 e(q1·q2)+2m2 em2µ/bracketrightbig . Finally in the center–of–mass frame (in the high energy limit) we have: /angbracketleftbig |M|2/angbracketrightbig =e4 8m2em2µ4 + (1 + cos θ)2 (1−cosθ)2. (10.18) The diﬀerential cross section in the center–of–mass frame is: dσ dΩ=e4 128π2E24 + (1 + cos θ)2 (1−cosθ)2. (10.19) Note that for θ≈0 diﬀerential cross section diverges. This is a consequence of the fact that for these angles the prevailing contribution in the expres-sion for i Mcomes from the virtual photon (this contribution is actually divergent because k 2=(p1−p2)2≈0). Chapter 10. Processes in the lowest order of the perturbation theory 199 10.8 The Compton scattering is the process e−γ→e−γ. In the lowest order contribution to this scattering is given by the following two diagrams: so that the Feynman amplitude is iM=¯u(p/prime,s/prime)(ieγµ)/epsilon1∗ µ(k/prime,λ/prime)i(p/+k/+m) (p+k)2−m2(ieγν)/epsilon1ν(k,λ)u(p,s)+ +¯u(p/prime,s/prime)(ieγν)/epsilon1ν(k,λ)i(p/−k//prime+m) (p−k/prime)2−m2(ieγµ)/epsilon1∗ µ(k/prime,λ/prime)u(p,s) =−ie2/epsilon1∗ µ(k/prime,λ/prime)/epsilon1ν(k,λ)¯u(p/prime,s/prime)/bracketleftbiggγµ(p/+k/+m)γν (p+k)2−m2+ +γν(p/−k//prime+m)γµ (p−k/prime)2−m2/bracketrightbigg u(p,s). (10.20) As we see the Feynman amplitude has the following form iM=iMµν/epsilon1∗ µ(k/prime,λ/prime)/epsilon1ν(k,λ). In order to prove the gauge invariance of Mit is enough to show that iMµνkν=iMµνk/prime µ=0. (10.21) First we prove that i Mµνkν= 0. In the second term in (10.20) we will use p−k/prime=p/prime−k. Hence iMµν=−ie2¯u(p/prime,s/prime)/bracketleftbiggγµ(p/+k/+m)γν (p+k)2−m2+γν(/p/prime−k/+m)γµ (p/prime−k)2−m2/bracketrightbigg u(p,s). (10.22) The numerators can be also simpliﬁed using: (p/+m)γνu(p)=(γµpµ+m)γνu(p)=( 2 gµν−γνγµ)pµu(p)+mγνu(p) =2pνu(p)−γν(p/−m)u(p)=2pνu(p), and similarly ¯u(p/prime)γν(/p/prime+m)=2p/primeν¯u(p/prime). (10.23) After performing these two simpliﬁcations i Mµνkνbecomes 200 Solutions iMµνkν=−ie2kν¯u(p/prime,s/prime)/bracketleftbiggγµ/kγν+2γµpν 2p·k+−γν/kγµ+2γµp/primeν −2p·k/prime/bracketrightbigg u(p,s) =−ie2¯u/bracketleftbiggγµk2+2γµp·k 2p·k+−k2γµ+2γµp/prime·k −2p·k/prime/bracketrightbigg u(p,s)=0 , where we used p2=m2andk2= 0. The second condition i Mµνk/prime µ= 0 can be proved in the same way. 10.9 The initial state, |i/angbracketright=c†(pi,r)|0/angbracketrightis the electron with momentum piand polarization r, while the ﬁnal state in the process is the electron with momen- tumpfand polarization s,i .e . |f/angbracketright=c†(pf,s)|0/angbracketright. The transition amplitude matrix element is: Sﬁ=ie/integraldisplay d4x/angbracketleftf|¯ψ(x)γµψ(x)|i/angbracketrightAµ(x), (10.24) where ψand¯ψare ﬁeld operators and Aµis a classical electromagnetic ﬁeld. (a) From (10.24) one obtains Sﬁ=iea/radicalbiggm EiV/radicalbiggm EfV/integraldisplay d4x¯u(pf,s)γ0u(pi,r)e−ipi·x+ipf·xe−k2x2. (10.25) Because of /integraldisplay d3xe−k2x2+i(pi−pf)·x=/parenleftBigπ k2/parenrightBig3/2 e−(pi−pf)2/4k2, we have Sﬁ=iea/radicalbiggm EiV/radicalbiggm EfV/parenleftBigπ k2/parenrightBig3/2 2πδ(Ei−Ef) ×e−(pi−pf)2 4k2¯u(pf,s)γ0u(pi,r). (10.26) Delta function which appears in the transition amplitude (10.26) indicates on the energy conservation law, which is satisﬁed because potential Aµ does not depend on time. As three–space is inhomogeneous (the potential depends on x), the three-momentum is not conserved. The average value of the squared transition amplitude is obtained from (10.26) /angbracketleftbig |Sﬁ|2/angbracketrightbig =1 2e2m2a2 V2EiEf2πTδ(Ei−Ef)/parenleftBigπ k2/parenrightBig3 ×e−(pi−pf)2 2k22/summationdisplay r,s=1|u(pf,s)γ0u(pi,r)|2. (10.27) Because of (¯u(pf,s)γ0u(pi,r))∗=¯u(pi,r)γ0u(pf,s), Chapter 10. Processes in the lowest order of the perturbation theory 201 we have: 2/summationdisplay r,s=1|¯u(pf,s)γ0u(pi,r)|2=2/summationdisplay r=1(ua(pf,s)¯ub(pf,s))γ0 bc ×2/summationdisplay r=1(uc(pi,r)¯ud(pi,r))γ0 da =1 4m2tr[(/pf+m)γ0(/pi+m)γ0] =1 m2(EiEf+pi·pf+m2).(10.28) By plugging (10.28) into (10.27) one obtains /angbracketleftbig |Sﬁ|2/angbracketrightbig =e2a2π V2EiEf/parenleftBigπ k2/parenrightBig3 Tδ(Ei−Ef) ×e−(pi−pf)2 2k2(EiEf+|pi||pf|cosθ+m2). (10.29) By substituting (10.29) into the expression for the diﬀerential cross section, dσ=|Sﬁ|2 TVEi |pi|Vd3pf (2π)3, one gets dσ=e2a2π 8k6/parenleftbig EiEf+|pi||pf|cosθ+m2/parenrightbig ×exp/parenleftbigg −|pi|21−cosθ k2/parenrightbigg δ(Ef−Ei)|pf| |pi|dEfdΩ. TheEf–integration gives dσ dΩ=e2a2π 8k6/parenleftbig E2 i+|p|2cosθ+m2/parenrightbig e−|p|21−cosθ k2. (b) This problem is analogous to the previous one, so we shall discuss only the main steps. The transition amplitude is: Sﬁ=−2iegm V√EiEf(2π)δ(Ef−Ei)2π q2+1 a2¯u(pf,s)γ3u(pi,r), where q=pf−pi.The next step is to calculate the squared amplitude: 2/summationdisplay r,s=1|¯u(pf,s)γ3u(pi,r)|2=1 4m2tr[(/pf+m)γ3(/pi+m)γ3] =1 m2(2p3 ip3f+pi·pf−m2) =1 m2(EiEf+|pi||pf|cosθ−m2). 202 Solutions The average value of the squared transition amplitude is: /angbracketleftbig |Sﬁ|2/angbracketrightbig =16π3e2g2T V2EiEf1 /parenleftbig q2+1 a2/parenrightbig2δ(Ef−Ei)(EiEf+|pi||pf|cosθ−m2). The diﬀerential cross section is: dσ dΩ=2e2g2 (E2−m2)(1 + cos θ) /parenleftbig1 a2+2 (E2−m2)(1−cosθ)/parenrightbig2. 10.10 The initial state is vacuum |0/angbracketright, while the ﬁnal state is |f/angbracketright=c†(p1,r)d†(p2,s)|0/angbracketright. The transition amplitude is Sﬁ=ie V/integraldisplay d4x/summationdisplay r/primes/prime/integraldisplay d3q1d3q2/radicalbiggm Eq1/radicalbiggm Eq2/angbracketleft0|d(p2,s)c(p1,r) ×(c†(q1,r/prime)d†(q2,s/prime)¯u(q1,r/prime)γµAµ(x)v(q2,s/prime)eiq1·x+iq2·x+...)|0/angbracketright, where we have dropped the vanishing terms. After reducing the last expression one obtains Sﬁ=iema V√E1E2/integraldisplay d4x¯u(p1,r)γ2v(p2,s)ei(p2+p1)·xe−iωt =ie(2π)4ma V√E1E2 ×¯u(p1,r)γ2v(p2,s)δ(3)(p1+p2)δ(E1+E2−ω). The average value of the squared transition amplitude is /angbracketleftbig |Sﬁ|2/angbracketrightbig =( 2π)4TVδ(3)(p1+p2)δ(E1+E2−ω) ×e2a2 4V2E1E2tr[(/p1+m)γ2(/p2−m)γ2] =( 2π)4Tδ(3)(p1+p2)δ(E1+E2−ω)e2a2 VE1E2 ×(E1E2+|p1||p2|−2|p1||p2|sin2θcos2φ+m2), since the four-momenta are: pµ 1=(E1,p1sinθcosφ,p1sinθsinφ,p1cosθ), pµ 2=(E2,−p2sinθcosφ,−p2sinθsinφ,−p2cosθ). The diﬀerential cross section is: dσ=/angbracketleftbig |Sﬁ|2/angbracketrightbig TVd3p1 (2π)3Vd3p2 (2π)3. Chapter 10. Processes in the lowest order of the perturbation theory 203 By integrating over p2andp1one obtains the scattering cross section (per unit volume) σ=e2a2 3πω(ω2+2m2)/radicalbigg ω2 4−m2. 10.11 The transition amplitude is Sﬁ=ieam V1√EiEf¯u(pf,s)γ3(1−γ5)u(pi,r)/integraldisplay d4xe−ipi·x+ipf·xe−k2x2. By integrating over tandxwe get Sﬁ=iea/radicalbiggm EiV/radicalbiggm EfV/parenleftBigπ k2/parenrightBig3/2 e−(pi−pf)2 4k2 ×2πδ(Ei−Ef)¯u(pf,s)γ3(1−γ5)u(pi,r). The average value of the squared transition amplitude is: /angbracketleftbig |Sﬁ|2/angbracketrightbig =e2a2m2 V2EiEf2πTδ(Ei−Ef)/parenleftBigπ k2/parenrightBig3 e−(pi−pf)2 2k2/angbracketleftbig |M|2/angbracketrightbig , where /angbracketleftbig |M|2/angbracketrightbig =1 22/summationdisplay r,s=1|¯u(pf,s)γ3(1−γ5)u(pi,r)|2 =1 21 4m2tr [(/pf+m)γ3(1−γ5)(/pi+m)(1 + γ5)γ3] =1 m2(2p3 fp3i+pi·pf). The diﬀerential cross section is: dσ dΩ=e2a2π 4k6/parenleftbig E2 i+|pi|2cosθ/parenrightbig e−1 k2|pi|2(1−cosθ). 10.12 We shall present the expression for the transition amplitude and ﬁnal result for the diﬀerential cross section only: Sﬁ=iem V√EiEf¯v(pi,s)v(pf,r)/integraldisplay d4x(iEf)g |x|e−i(pi−pf)·x, dσ dΩ=e2g2E2(E2+m2−p2cosθ) 2|p|4(1−cosθ)2. 10.13 The transition amplitude Sﬁis Sﬁ=iea/radicalbiggm VEi/radicalbiggm VEf¯u(pf,sf)γ0u(pi,si)/integraldisplay d4xδ(3)(x)e−i(pi−pf)·x =ieam V√EiEf(2π)δ(Ei−Ef)¯u(pf,sf)γ0u(pi,si), 204 Solutions where siisfare initial and ﬁnal electron polarizations. In order to calculate |Sﬁ|2it is necessary to compute squared spin-part of the amplitude. Since u(p,s)¯u(p,s)=1+γ5/s 2/p+m 2m, we have |¯u(pf,sf)γ0u(pi,si)|2=1 16m2tr [(1 + γ5/sf)(/pf+m)γ0(1 +γ5/si)(/pi+m)γ0] =1 16m2/parenleftbig tr[/pfγ0/piγ0]+m2tr[1] −tr[/sf/pfγ0/si/piγ0]+m2tr[/sfγ0/siγ0]/parenrightbig , (10.30) where we have kept only the nonvanishing traces. The components of momenta and polarization vectors are: pµ i=(Ei,0,0,|pi|), pµ f=(Ef,|pf|sinθcosφ,|pf|sinθsinφ,|pf|cosθ), sµ i=(|pi|/m,0,0,Ei/m), sµ f=(|pf|/m,(Ef/m)sinθcosφ,(Ef/m)sinθsinφ,(Ef/m)cosθ). The traces in the sum (10.30) are: tr[/sf/pfγ0/si/piγ0]=−4m2cosθ, trI = 4 , tr[/sfγ0/siγ0]=4/parenleftbiggk2 m2+E2 m2cosθ/parenrightbigg , tr[/pfγ0/piγ0]=4 ( E2+k2cosθ), where Ei=Ef=Ewhile k=|pi|=|pf|. By summing all the terms we get |¯u(pf,sf)γ0u(pi,si)|2=E2 m2cos2/parenleftbiggθ 2/parenrightbigg . (10.31) The diﬀerential cross section for the scattering is computed in the usual way. The result is: dσ dΩ=e2a2 4π2E2cos2(θ/2). 10.14 The amplitude for this process is (see Fig. 10.2) iM=ie2 k2¯u(p2,r)γµu2(p1)¯v2(q1)γµv(q2,s), where subscript 2 in uandvspinors indicates that these are negative helicity particles. The squared Feynman amplitude module is Chapter 10. Processes in the lowest order of the perturbation theory 205 /angbracketleftbig |M|2/angbracketrightbig =e4 64m2em2µk4tr[(/p2+me)γµ(/p1+me)(1−γ5/s1)γν] ×tr[(/q1−mµ)(1−γ5/s2)γµ(/q2−mµ)γν], where we have summed over polarization states of the ﬁnal particles in the process. Here s1ands2are polarization vectors of the initial electron and muon which are going to be evaluated later. By applying corresponding iden- tities from Problem 3.6 and corresponding expression for contractions of two /epsilon1symbols from Problem 1.5 we get /angbracketleftbig |M|2/angbracketrightbig =e4 2m2em2µk4[(p2·q1)(p1·q2)+(p2·q2)(p1·q1)− −m2 µ(p2·p1)−m2 e(q1·q2)+2m2 em2µ+ +memµ((s1·s2)(p2·q2)−(s1·s2)(p2·q1)− −(s1·s2)(p1·q2)+(s1·s2)(p1·q1)− −(s1·q2)(s2·p2)+(s1·q1)(s2·p2)+ +(s1·q2)(s2·p1)−(s1·q1)(s2·p1))]. (10.32) Since mµ≈200mewe will neglect the electron mass. In the center–of–mass frame four momenta are pµ 1=(E,0,0,p), qµ 1=(E/prime,0,0,−p), pµ 2=(E,psinθcosφ,psinθsinφ,pcosθ), qµ 2=(E/prime,−psinθcosφ,−psinθsinφ,−pcosθ). Polarization vectors s1ands2are sµ 1=(p me,0,0,E me), sµ 2=(p mµ,0,0,−E/prime mµ). After ﬁnding scalar products between four-vectors in (10.32) and reducing the obtained expression one gets /angbracketleftbig |M|2/angbracketrightbig =e4 32m2em2µp4sin4(θ 2)/bracketleftbigg (EE/prime+p2)2−2p2(m2 e+m2 µ)sin2/parenleftbiggθ 2/parenrightbigg +(EE/prime+p2cosθ)2+p2/parenleftbigg 4p2sin2/parenleftbiggθ 2/parenrightbigg +EE/primesin2θ/parenrightbigg/bracketrightbigg ,(10.33) hence the diﬀerential cross section is dσ dΩ=e4 128π2(E+E/prime)2p4sin4(θ/2)/bracketleftbigg (EE/prime+p2)2−2p2(m2 e+m2 µ)sin2/parenleftbiggθ 2/parenrightbigg +(EE/prime+p2cosθ)2+p2/parenleftbigg 4p2sin2/parenleftbiggθ 2/parenrightbigg +EE/primesin2θ/parenrightbigg/bracketrightbigg . (10.34) 206 Solutions 10.15 The interaction Hamiltonian is Hint=g/integraldisplay d3x¯ψγ5ψφ , where the ﬁeld operators are written in the interaction picture. In the lowest (”tree-level”) order of the perturbation theory the transition amplitude is: Sﬁ=1 2(−ig)2/angbracketleftp/primek/prime|/integraldisplay d4xd4yT{:(¯ψγ5ψφ)x:: (¯ψγ5ψφ)y:}|pk/angbracketright.(10.35) Because of ψ(x)|p,r/angbracketright=/radicalbiggm VEpu(p,r)e−ip·x, /angbracketleftp,r|¯ψ(x)=/radicalbiggm VEp¯u(p,r)eip·x, from the expression (10.35) we conclude that there are four ways to make contractions which correspond to the given process. In that way we obtain(note that there are two couples containing two identical terms) S ﬁ=−g2 m2 V2/radicalbig E1E2E/prime 1E/prime 2/integraldisplay d4xd4yi∆F(x−y) ×/bracketleftBig −¯u(k/prime,s/prime)γ5u(k,s)¯u(p/prime,r/prime)γ5u(p,r)ei(p/prime−p)·y+i(k/prime−k)·x +¯u(p/prime,r/prime)γ5u(k,s)¯u(k/prime,s/prime)γ5u(p,r)ei(k/prime−p)·y+i(p/prime−k)·x/bracketrightBig .(10.36) The minus sign in the ﬁrst term is a consequence of the Wick theorem for fermions. After integrating the last expression and having in mind that i∆F(x−y)=i (2π)4/integraldisplay d4qe−iq·(x−y) q2−M2+i/epsilon1, one obtains Sﬁ=i(2π)4g2m2 V2/radicalbig E1E2E/prime 1E/prime 2δ(4)(p/prime+k/prime−p−k) ×/bracketleftbigg1 (p/prime−p)2−M2+i/epsilon1¯u(k/prime,s/prime)γ5u(k,s)¯u(p/prime,r/prime)γ5u(p,r)− −1 (p/prime−k)2−M2+i/epsilon1¯u(p/prime,r/prime)γ5u(k,s)¯u(k/prime,s/prime)γ5u(p,r)/bracketrightbigg . Feynman diagrams for the scattering are represented in the ﬁgure. Chapter 10. Processes in the lowest order of the perturbation theory 207 The squared amplitude is /angbracketleftbig |Sﬁ|2/angbracketrightbig =g4(2π)4Tδ(4)(p/prime+k/prime−p−k) 4V3E1E2E/prime 1E/prime 2 ×/bracketleftbigg(k·k/prime)(p·p/prime)−(k·k/prime)m2−(p·p/prime)m2+m4 ((p/prime−p)2−M2)2+ +(p·k/prime)(k·p/prime)−(p·k/prime)m2−(k·p/prime)m2+m4 ((p/prime−k)2−M2)2 −1 21 (p/prime−p)2−M21 (p/prime−k)2−M2Re[(k ·k/prime)(p·p/prime) −(p/prime·k/prime)(k·p)+(p·k/prime)(k·p/prime) −(k·k/prime)m2−(p·p/prime)m2−(k·p/prime)m2 −(p·k/prime)m2+(k·p)m2+(k/prime·p/prime)m2+m4/bracketrightbig/bracketrightbig . The squared amplitude per unit time as viewed from the center–of–mass frame is: /angbracketleftbig |Sﬁ|2/angbracketrightbig T=g4(2π)4δ(4)(p/prime+k/prime−p−k) 4V3E2|p|4 ×/bracketleftbigg(1−cosθ)2 (2|p|2(cosθ−1)−M2)2+(1 + cos θ)2 (2|p|2(cosθ+1 )+ M2)2 −sin2θ (2|p|2(cosθ−1)−M2)(2|p|2(cosθ+1 )+ M2)/bracketrightbigg ,(10.37) where E1=E2=E/prime 1=E/prime 2=Eare the energies of the initial and ﬁnal particles. All four fermions carry the momenta of the identical intensity |p|. In the high energy limit from (10.37) one obtains /angbracketleftbig |Sﬁ|2/angbracketrightbig T=3g4(2π)4δ(4)(p/prime+k/prime−p−k) 16V3E2. (10.38) The total cross section for the scattering is σ=/integraldisplay/integraldisplay/angbracketleftbig |Sﬁ|2/angbracketrightbig TVE 2|p1|Vd3p/prime 1 (2π)3Vd3p/prime 2 (2π)3 =3g4 4π2δ(2E−2E/prime) 16EdE/prime 1dΩ/prime 1 2E1 =3g4 64πE2. 10.16 By direct application of the Feynman rules we obtain the expression for the corresponding amplitudes. In the following expressions we drop external lines. 208 Solutions (a) iM= =( ie)2/integraldisplayd4k (2π)4/parenleftbigg γν1 /p−/k−m+i/epsilon1γµgµν k2+i/epsilon1/parenrightbigg (b) iM= = i(ie)4/integraldisplay/integraldisplayd4k (2π)4d4q (2π)4/parenleftbigg γµ 1 /p−/k−m+i/epsilon1γσ ×1 /p−/k−/q−m+i/epsilon1γσ1 /p−/k−m+i/epsilon1 ×γµ1 k2+i/epsilon11 q2+i/epsilon1/parenrightbigg (c) iM= =−(ie)3i3/integraldisplayd4p (2π)4tr/bracketleftbigg γν 1 /p−/q−m+i/epsilon1γρ ×1 /p+/k−m+i/epsilon1γµ 1 /p−m+i/epsilon1/bracketrightbigg (d) iM= = i(ie)3/integraldisplayd4p (2π)4/parenleftbigg γν 1 /p+/k−/q−m+i/epsilon1 Chapter 10. Processes in the lowest order of the perturbation theory 209 ×γρ 1 /p−/q−m+i/epsilon1γν1 q2+i/epsilon1/parenrightbigg (e) iM= =( ie)7i6(−i)3/integraldisplay/integraldisplay/integraldisplayd4k1 (2π)4d4q (2π)4d4k (2π)4 ×/bracketleftbigg γν 1 /p1+/q−m+i/epsilon1γα 1 /q−m+i/epsilon1γµ ×gµρ (p−q)2+i/epsilon1gσν (p−q)2+i/epsilon1 ×tr/parenleftbigg1 /k−m+i/epsilon1γσ 1 /p−/q+/k−m+i/epsilon1γρ/parenrightbigg ×gαβ p2 1+i/epsilon1tr/parenleftbigg1 /p1+/k1−m+i/epsilon1γδ 1 /k1−m+i/epsilon1γβ/parenrightbigg/bracketrightbigg (f) −iΠµν(k)= =( ie)2/integraldisplayd4p (2π)4tr/bracketleftbigg1 /p−/k−m+i/epsilon1γν 1 /p−m+i/epsilon1γµ/bracketrightbigg (g) −iM= =(−i)Πµν(k)−igνρ k2+i/epsilon1(−i)Πρσ(k) (h) −iM= =−i4(−i)(ie)4/integraldisplayd4p (2π)4d4q (2π)4tr/bracketleftbigg1 /p−/k−m+i/epsilon1γσ ×1 /p+/q−/k−m+i/epsilon1γν 1 /p+/q−m+i/epsilon1γρ ×1 /p−m+i/epsilon1γµ/bracketrightbigggρσ q2+i/epsilon1 210 Solutions (i) iM= =−(ie)4/integraldisplayd4p (2π)4tr/bracketleftbigg1 /p−/k1−m+i/epsilon1γµ 1 /p−/k1−/k2−m+i/epsilon1 ×γσ 1 /p−/q1−m+i/epsilon1γρ 1 /p−m+i/epsilon1γν/bracketrightbigg 11 Renormalization and regularization 11.1 In order to prove the Feynman formula we shall use mathematical induction. For n=2w eh a v e I2=/integraldisplay1 0dx1/integraldisplay1 0dx2δ(x1+x2−1)1 [x1A1+x2A2]2 =/integraldisplay1 0dx11 [x1A1+( 1−x1)A2]2 =1 A1A2. (11.1) By taking n-th derivative of (11.1) we get the useful identity 1 ABn=/integraldisplay1 0dx/integraldisplay1 0dyδ(x+y−1)nyn−1 [xA+yB]n+1. (11.2) Now we shall assume that the Feynman formula is valid for n=kand show that it holds for n=k+1 1 A1...AkAk+1=/integraldisplay1 0dz1...dzkδ(z1+...+zk−1)(k−1)! [z1A1+...+zkAk]kAk+1 =/integraldisplay1 0dz1...dzkdyk!δ(z1+...+zk−1) ×yk−1 [yz1A1+...+yzkAk+( 1−y)Ak+1]k+1. (11.3) By using substitution x1=yz1,...,x k=yzk,xk+1=1−yand a well known property of the δ–function δ(ax)=1 |a|δ(x), we obtain 212 Solutions 1 A1...AkAk+1=/integraldisplay dx1...dxkdxk+1δ(x1+...+xk+xk+1−1) ×k! [x1A1+...+xk+1Ak+1]k+1, (11.4) which concludes the proof. 11.2 By introducing a new variable q=k+p, the integral Ibecomes I=/integraldisplay dDq1 (q2−m2−p2+i/epsilon1)n. (11.5) If we do a Wick rotation to the Euclidian space, q0=iq0 E,q=qE, the integral Ibecomes I=i/integraldisplay dDqE1 (−q2 E−m2−p2+i/epsilon1)n. (11.6) The contour of the integration along the real axis can be rotated to the imagi- nary axis without passing through the poles. Transition from Minkowski space to Euclidian space is so-called Wick rotation. The relation between the Cartesian and the spherical coordinates in the Ddimensional space is x1=rsinθD−2sinθD−3...sinθ1sinφ, x2=rsinθD−2sinθD−3...sinθ1cosφ, x3=rsinθD−2sinθD−3...sinθ2cosθ1, ... xD=rcosθD−2, where 0 <φ< 2π,0<θ1,...,θ D−2<π .The volume element, d VDis dVD=rD−1drdφD−2/productdisplay 1(sinθm)mdθm. Therefore I=i (−1)n2πD−2/productdisplay m=1/integraldisplayπ 0dθm(sinθm)m/integraldisplay∞ 0drrD−1 (r2+m2+p2)n.(11.7) If we use [9]/integraldisplayπ 0dθ(sinθ)m=√πΓ/parenleftbigm+1 2/parenrightbig Γ/parenleftbigm+2 2/parenrightbig, and/integraldisplay∞ 0dxxb (x2+M)a=Γ/parenleftbig1+b 2/parenrightbig Γ/parenleftbig a−1+b 2/parenrightbig 2Ma−1+b 2Γ(a), Chapter 11. Renormalization and regularization 213 we obtain I=i (−1)nπD 2Γ/parenleftbig n−D 2/parenrightbig Γ(n)1 (m2+p2)n−D 2. 11.3 As we know, the Gamma–function is deﬁned by Γ(z)=/integraldisplay∞ 0dte−ttz−1. (11.8) ¿From the property Γ(z)=Γ(z+1 )/zfollows that Γ(z)=Γ(z+n+1 )n/productdisplay k=01 z+k. (11.9) By using the deﬁnition of number e, the integral (11.8) becomes Γ(z) = lim n→∞/integraldisplayn 0dttz−1(1−t/n)n. By introducing a new variable, t/n=xthe last integral is Γ(z) = lim n→∞nz/integraldisplay1 0dxxz−1(1−x)n = lim n→∞nzB(n+1,z) = lim n→∞nzΓ(n+1 )Γ(z) Γ(n+z+1 ) = lim n→∞nz Γ(n+1 ) z(z+1 )...(z+n) =1 zlim n→∞nz 1 (1 +z)(1 +z 2)...(1 +z n), (11.10) where we used (11.9). Euler-Mascheroni constant, γis deﬁned by γ= lim n→∞/parenleftbigg 1+1 2+1 3+...+1 n−lnn/parenrightbigg . Then e−γz= lim n→∞nze−z(1+1 2+...+1 n). (11.11) From (11.10) and (11.11) follows Γ(z)=e−zγ1 z∞/productdisplay n=1ez/n 1+z n. By taking the logarithm of the previous formula we get 214 Solutions lnΓ(z)=−γz−lnz+∞/summationdisplay n=1/parenleftBigz n−ln(1 +z n)/parenrightBig . Hence ψ(z)=dl nΓ(z) dz=Γ/prime(z) Γ(z)=−γ−1 z+∞/summationdisplay k=1/parenleftbigg1 k−1 k+z/parenrightbigg . (11.12) Forz=nfrom the previous expression we get ψ(n)=−γ+1+1 2+1 3+...+1 n−1. (11.13) Expanding Γ(1 +/epsilon1) according the Taylor formula we obtain Γ(1 +/epsilon1)=Γ(1) + /epsilon1Γ/prime(1) + ... =1−γ/epsilon1+o(/epsilon1). (11.14) By using (11.9) and the previous expression we have Γ(−n+/epsilon1)=Γ(1 +/epsilon1) /epsilon1(/epsilon1−1)...(/epsilon1−n) =(−1)n(1−/epsilon1γ+o(/epsilon1)) n!/epsilon1(1−/epsilon1)(1−/epsilon1/2)...(1−/epsilon1/n) =(−1)n n!/parenleftbigg1 /epsilon1−γ/parenrightbigg/parenleftbigg 1+/epsilon1/parenleftbigg 1+1 2+...+1 n/parenrightbigg/parenrightbigg +o(/epsilon1) =(−1)n n!/parenleftbigg1 /epsilon1−γ+1+1 2+...+1 n+o(/epsilon1)/parenrightbigg =(−1)n n!/parenleftbigg1 /epsilon1+ψ(n+1 )+ o(/epsilon1)/parenrightbigg . (11.15) 11.4 By applying the Feynman parametrization (11.G), the integral becomes I=/integraldisplay1 0dx/integraldisplay d4k1 [(k+px)2−∆]2, where ∆=p2(x2−x)+m2x.By making change of variable l=k+pxand going to Euclidian space ( l0=il0 E,l=lE)w eg e t I=i/integraldisplay1 0dx/integraldisplay d4lE1 [l2 E+∆]2. In order to compute the integral we introduce spherical coordinates. The an- gular integration can be done immediately Chapter 11. Renormalization and regularization 215 I=i/integraldisplay1 0dx/integraldisplay2π 0dφ/integraldisplayπ 0dθ1sinθ1/integraldisplayπ 0dθ2sin2θ2/integraldisplay∞ 0dlEl3 E (l2 E+∆)2 =iπ2/integraldisplay1 0dx/integraldisplay∞ 0dl2 El2 E1 (l2 E+∆)2=iπ2/integraldisplay1 0dx/bracketleftbig ln(l2 E+∆)|∞ 0−1/bracketrightbig . The previous integral diverges logarithmically. Performing the Pauli–Villars regularization the propagator 1 /k2in the integral Ibecomes 1 k2→1 k2−1 k2−Λ2, where Λis a large parameter. A contribution of the second term in the previous expression to the integral is IΛ=iπ2/integraldisplay1 0dx/bracketleftbig ln(l2 E+∆Λ)|∞ 0−1/bracketrightbig , where we introduced ∆Λ=Λ2+p2(x2−x)+x(m2−Λ2). By subtracting these two results we get I−IΛ=iπ2/integraldisplay1 0dxln/parenleftbiggΛ2+p2(x2−x)+x(m2−Λ2) p2(x2−x)+m2x/parenrightbigg =iπ2/integraldisplay1 0dxln/parenleftbiggΛ2(1−x) p2(x2−x)+m2x/parenrightbigg . 11.5 The integrand is symmetric with respect to any two indices and therefore Iαβµνρσ is of the form Iαβµνρσ =C[gαβ(gµνgρσ+gµρgνσ+gµσgνρ) +gαµ(gβνgρσ+gβρgνσ+gβσgνρ) +gαν(gβµgρσ+gβρgµσ+gβσgµρ) +gαρ(gβµgνσ+gβνgµσ+gβσgνµ) +gασ(gβµgνρ+gβνgµρ+gβρgµν)], where Cis a constant. In order to determine Cwe will compute the contraction gαβgµνgρσIαβµνρσ .It is easy to get gαβgµνgρσIαβµνρσ =C(D3+6D2+8D). On the other hand gαβgµνgρσIαβµνρσ =/integraldisplaydDk (k2)n−3= lim µ→0/integraldisplaydDk (k2−µ2)n−3 = lim µ→0i(−1)n−3π2Γ(n−3−D 2) Γ(n−3)(µ2)3−n+D 2, 216 Solutions where µis a infrared parameter. Comparing these results we get C=1 D3+6D2+8Dlim µ→0i(−1)n−3π2Γ(n−3−D 2) Γ(n−3)(µ2)3−n+D 2. Specially, for n= 5 the divergent part of the integral Iαβµνρσ is Iαβµνρσ |div=iπ2 96/epsilon1[gαβ(gµνgρσ+gµρgνσ+gµσgνρ) +gαµ(gβνgρσ+gβρgνσ+gβσgνρ) +gαν(gβµgρσ+gβρgµσ+gβσgµρ) +gαρ(gβµgνσ+gβνgµσ+gβσgνµ) +gασ(gβµgνρ+gβνgµρ+gβρgµν). 11.6 InD–dimensional space the interaction term takes the form −gµ/epsilon1/2χφ2. (a) The self–energy of the χparticle is determined by the diagram p kk pp+ from which we read −iΠ(p2)=2g2µ/epsilon1/integraldisplaydDk (2π)D1 k2−m2+i 01 (k+p)2−m2+i 0.(11.16) By introducing the Feynman parametrization (11.G) and integrating over the momentum kwe get: −iΠ(p2)=ig2 8π2/parenleftbigg2 /epsilon1−γ−/integraldisplay1 0dxlnm2+p2x(x−1)−i0 4πµ2/parenrightbigg =ig2 8π2/bracketleftbigg2 /epsilon1−γ−lnm2 4πµ2 −/integraldisplay1 0dxln/parenleftbigg 1+p2 m2x(x−1)−i0/parenrightbigg/bracketrightbigg . (11.17) As we know from the complex analysis the logarithm function, w=l nz has a branch cut along the positive x–axis which starts at the branch point z= 0. This branch cut is necessary if we want that branches of logarithm function to be single valued and holomorphic functions. Let us ﬁnd the branch point for function ln[1 +p2 m2x(x−1)]. It is the smallest value of p2for which the argument of logarithm function vanishes: Chapter 11. Renormalization and regularization 217 1+p2 m2(x2−x)=0 , i.e. ∂p2 ∂x=m22x−1 (x2−x)2=0, from which we get x=1 2. The point p2=4m2, which is step energy for the decay χ→2φ, is the branch point. A branch cut starts at this point and goes along x–axis in the positive direction to the inﬁnity. Let us introduce the following notation I=g2 8π2/integraldisplay1 0dxln/parenleftbigg 1+p2 m2x(x−1)−iδ/parenrightbigg . We shall calculate ﬁrst this integral in the case p2>4m2.F o rX>0w e have log[−X−i0] = log |X|−iπ. The zeroes of 1 +p2 m2x(x−1) are x1,2=1±/radicalBig 1−4m2 p2 2. Forx1<x<x 2the expression Xis negative, otherwise it is positive. Then I=g2 8π2/bracketleftbigg/integraldisplayx1 0dxln/parenleftbigg 1+p2 m2x(x−1)/parenrightbigg +/integraldisplay1 x2dxln/parenleftbigg 1+p2 m2x(x−1)/parenrightbigg +/integraldisplayx2 x1dxln/parenleftbigg −1−p2 m2x(x−1)/parenrightbigg −iπ(x2−x1)/bracketrightbigg .(11.18) By doing partial integration we have I=g2 8π2/bracketleftbigg xln/parenleftbigg 1+p2 m2x(x−1)/parenrightbigg/vextendsingle/vextendsingle/vextendsinglex1 0−p2 m2/integraldisplayx1 0dxx(2x−1) 1+p2(x2−x)/m2 +xln/parenleftbigg 1+p2 m2x(x−1)/parenrightbigg/vextendsingle/vextendsingle/vextendsingle1 x2−p2 m2/integraldisplay1 x2dxx(2x−1) 1+p2(x2−x)/m2 +xln/parenleftbigg −1−p2 m2x(x−1)/parenrightbigg/vextendsingle/vextendsingle/vextendsinglex2 x1−p2 m2/integraldisplayx2 x1dxx(2x−1) 1+p2(x2−x)/m2 −iπ(x2−x1)]. (11.19) Combining the terms in the previous formula we get 218 Solutions I=g2 8π2/bracketleftbigg −iπ(x2−x1)−p2 m2/integraldisplay1 0dxx(2x−1) 1+p2(x2−x)/m2/bracketrightbigg .(11.20) The integral in the previous formula can be simpliﬁed by introducing the new variable t=2x−1. The result is (see [9]) I=−ig2 8π/radicalBigg 1−4m2 p2−g2 4π2 1+1 2/radicalBigg 1−4m2 p2ln1−/radicalBig 1−4m2 p2 1+/radicalBig 1−4m2 p2 . For 0 <p2<4m2we get [9] I=g2 4π2/bracketleftBigg −1+/radicalBigg 4m2 p2−1a rc si n/radicalbigg p2 4m2/bracketrightBigg . The ﬁnal result for the vacuum polarization, −iΠ(p2)i s −iΠ(p2)=ig2 8π2/parenleftbigg2 /epsilon1−γ−lnm2 4πµ2+2/parenrightbigg +π(p2), (11.21) where π(p2)=−ig2 4π2/radicalBigg 4m2 p2−1a rc si n/radicalbigg p2 4m2 for 0<p2<4m2and π(p2)=ig2 8π2 i/radicalBigg 1−4m2 p2+/radicalBigg 1−4m2 p2ln1−/radicalBig 1−4m2 p2 1+/radicalBig 1−4m2 p2  forp2>4m2. (b) In the lowest order of the perturbation theory the transition amplitude is given by Sfi=−ig/integraldisplay d4x/angbracketleftp1,p2|χ(x)φ(x)φ(x)|M,p=0/angbracketright =( 2π)4δ(4)(p−p1−p2)/radicalbigg 1 2VM/radicalbigg 1 2VE1/radicalbigg 1 2VE2(−2ig), where p1,2are the momenta of the decay products. Also we take that χ particle is in the rest. The decay rate is dΓ=|Sfi|2 TV2d3p1d3p2 (2π)6. By integrating over the momentum p2we get: Chapter 11. Renormalization and regularization 219 Γ=4g2 (2π)2/integraldisplay dEpE1 8ME2δ(M−2E)/integraldisplayπ 0dθ/integraldisplayπ 0dφ, and the space angle integration gives 2 π(not 4 π, because the ﬁnal particles are identical). The ﬁnal result is given by: Γ=g2 4πM2/radicalbigg M2 4−m2. (c) The imaginary part of Π(p2) can be read oﬀ the part (a): ImΠ(p2)=−g2 8π/radicalBigg 1−4m2 p2θ(p2−4m2). (11.22) This result also can be obtain using Cutkosky rule . The expression (11.16) can be rewritten in the following form −iΠ(p2)=2g2/integraldisplayd4k (2π)41 (−k)2−m2+i 01 (k+p)2−m2+i 0.(11.23) The discontinuity of the amplitude DiscΠ(p2)=Π(p2+i/epsilon1)−Π(p2−i/epsilon1), is obtained by making the substitution 1 p2−m2→(−2iπ)δ(4)(p2−m2)θ(p0), in the expression (11.23). Since Π(p2) is a Lorentz scalar we shall take thatpµ=(p0,p= 0) i.e. we shall calculate it in the rest frame of the particle χ. In this way we obtain DiscΠ(p2)=2 i g2(−2iπ)2/integraldisplayd4k (2π)4δ(4)(k2−m2) ×δ(4)((k+p)2−m2)θ(−k0)θ(k0+p0) =−g2i 8π2/integraldisplay d4k1 ω2 kδ(k0+ωk)δ(k0+p0−ωk) =−ig2 8π2/integraldisplay d3kδ(p0−2ωk) ω2 k. (11.24) By performing the integration over the momentum kwe get DiscΠ(p2)=−ig2 4π/radicalBigg 1−4m2 p2. Since 220 Solutions Im Π(p2)=1 2iDiscΠ(p2), we again obtain the result (11.22). From the expressions for ΓandΠ(M2) we immediately see that the relation which was given in problem is valid. This relation is a consequence of the optic theorem . 11.7 InD=4−/epsilon1dimensional spacetime the dimension of a scalar ﬁeld is D/2−1, while the dimensions of the coupling constants are the same as in four dimensions: [ λ]=0,[g] = 1. The dimension of the Lagrangian density must be [ L]=D,s oi ti sg i v e nb y L=1 2(∂µφ)2−m2 2φ2−gµ/epsilon1/2 3!φ3−λµ/epsilon1 4!φ4, where we introduced the parameter µwhich has the dimension of mass. The self–energy is determined by diagrams shown in Fig. 11.1. Fig. 11.1. The one-loop contribution to the self–energy of φﬁeld The contribution of the ﬁrst one is −iΣ1=−iλ 2µ/epsilon1/integraldisplaydDk (2π)Di k2−m2. By applying the formula (11.A) we get −iΣ1=−iλm2 32π2/parenleftbigg4πµ2 m2/parenrightbigg/epsilon1/2 Γ/parenleftBig −1+/epsilon1 2/parenrightBig , which, using (11.F), gives −iΣ1=iλm2 32π2/parenleftbigg 1+/epsilon1 2ln/parenleftbigg4πµ2 m2/parenrightbigg +o(/epsilon1)/parenrightbigg/parenleftbigg2 /epsilon1+1−γ+o(/epsilon1)/parenrightbigg =iλm2 32π2/parenleftbigg2 /epsilon1+1−γ+l n/parenleftbigg4πµ2 m2/parenrightbigg +o(/epsilon1)/parenrightbigg . The second integral is −iΣ2(p)=(−ig)2 2µ/epsilon1/integraldisplaydDk (2π)Di k2−m2i (k−p)2−m2. By using the Feynman parametrization formula (11.G) the last expression becomes Chapter 11. Renormalization and regularization 221 −iΣ2(p)=−(−ig)2 2µ/epsilon1/integraldisplay1 0dx/integraldisplaydDk (2π)D1 [k2−2k·px+p2x−m2]2. The integration over the momentum kgives −iΣ2(p)=i 2µ/epsilon1g21 (4π)2−/epsilon1/2Γ/parenleftBig/epsilon1 2/parenrightBig/integraldisplay1 0dx(m2−p2x+p2x2)−/epsilon1/2 =ig2(4πµ2)/epsilon1/2 2(4π)2/parenleftbigg2 /epsilon1−γ+o(/epsilon1)/parenrightbigg ×/bracketleftbigg 1−/epsilon1 2/integraldisplay1 0dx/parenleftbigg lnm2+l n ( 1+p2 m2x(x−1))/parenrightbigg/bracketrightbigg . Finally, the integration over the Feynman parameter xgives (for p2<4m2) −iΣ2(p)=ig2 32π2/bracketleftBigg 2 /epsilon1−γ+2+l n4πµ2 m2−2/radicalBigg 4m2 p2−1a rc si n/radicalbigg p2 4m2/bracketrightBigg . The self–energy of the particle is −iΣ(p)=−iΣ1(p)−iΣ2(p). The mass shift is δm2=Σ(m2)=Σ1(m2)+Σ2(m2). 11.8 The vertices in this theory are shown in Fig. 11.2. Fig. 11.2. Vertices in σ–model The self–energy of the πparticle is determined by the diagrams given in Fig. 11.3. The full line depict the πﬁeld, while the dashed line depict σ. The ﬁrst diagram is one of the terms in the second order of the perturbation theory 1 2(−iλv)22/integraldisplay dx1dx2/angbracketleft0|T(π(y1)π(y2)σ3(x1)σ(x2)π2(x2))|0/angbracketright,(11.25) 222 Solutions Fig. 11.3. The one-loop correction to the πpropagator so that −iΣ1(p2)=6 (−ivλ)2i −m2/integraldisplaydDk (2π)Di k2−m2. The symmetry factor of this diagram is 6, since one πﬁeld can be contracted toπﬁeld from ππσ-vertex in two ways, while σσcontraction in the vertex σσσcan be done in 3 ways. Other diagrams are: −iΣ2(p2)=λ/integraldisplaydDk (2π)D1 k2−m2, −iΣ3(p2)=−2v2λ2 m2/integraldisplaydDk (2π)D1 k2, −iΣ4(p2)=3λ/integraldisplaydDk (2π)D1 k2, −iΣ5(p2)=4λ2v2/integraldisplaydDk (2π)D1 k2−m21 (k+p)2. Note that only the last diagram depends on the momentum p. The renormal- ized mass is determined by m2 R=Σ(0).It is easy to see that −iΣ5(0) = 4 λ2v2/integraldisplaydDk (2π)D1 k2−m21 k2 =4λ2v2 m2/integraldisplaydDk (2π)D/parenleftbigg1 k2−m2−1 k2/parenrightbigg . By summing all diagrams we obtain Σ(0) = Σ1(0) + Σ2(0) + Σ3(0) + Σ4(0) + Σ5(0) = 0 , somR=0 . 11.9 The amplitude for the diagram Chapter 11. Renormalization and regularization 223 is iM=e3/integraldisplaydDk (2π)Dtr[γµ(/k−/p1+m)γν(/k+/p2+m)γρ(/k+m)] ((k−p1)2−m2)((k+p2)2−m2)(k2−m2).(11.26) By applying the Feynman parametrization (11.H) we get 1 ((k−p1)2−m2)((k+p2)2−m2)(k2−m2) =2/integraldisplay1 0dx/integraldisplay1−x 0dz1 [k2−m2+(p2 2+2k·p2)x+(p2 1−2k·p1)z]3 =2/integraldisplay1 0dx/integraldisplay1−x 0dz1 [(k+p2x−p1z)2−∆]3, where we introduce the notation ∆=(p2x−p1z)2−p2 2x−p2 1z+m2. The numerator of the integrand in (11.26) is tr[γµ(/k−/p1+m)γν(/k+/p2+m)γρ(/k+m)] =t r [γµ(/l+A/+m)γν(/l+B/+m)γρ(/l+C/+m)], (11.27) where l=k+p2x−p1z, A=p1z−p2x−p1, B=p1z−p2x+p2, C=p1z−p2x. Since the trace of the odd number of γ–matrices is zero, (11.27) becomes tr[γµ(/l+A/+m)γν(/l+B/+m)γρ(/l+C/+m)] =t r [γµ/lγν/lγρ/l]+t r [ γµ/lγν/lγρC/]+t r [ γµ/lγνB/γρ/l]+ +t r [γµ/lγνB/γρC/]+t r [ γµA/γν/lγρ/l]+t r [ γµA/γν/lγρC/]+ +t r [γµA/γνB/γρ/l]+t r [ γµA/γνB/γρC/]+m2tr[γµ/lγνγρ]+ +m2tr[γµA/γνγρ]+m2tr[γµγν/lγρ]+ +m2tr[γµγνB/γρ]+m2tr[γµγνγρ/l]+m2tr[C/γµγνγρ].(11.28) 224 Solutions To calculate the integral (11.26) we make substitution of variable k→l.Terms in (11.28) which contain odd number of momenta lafter integration vanish. The terms which are proportional to m2as well as the term proportional to tr[γµA/γνB/γρC/] are ﬁnite, and therefore we consider only the remaining terms. The ﬁrst of the divergent integrals is iM1=8e3/integraldisplay1 0dx/integraldisplay1−x 0dz/integraldisplaydDl (2π)D/bracketleftbigg2lν(lµCρ−gµρC·l+lρCµ) (l2−∆)3− −l2(gµνCρ−gµρCν+gνρCµ) (l2−∆)3/bracketrightbigg , since tr[γµ/lγν/lγρ/C]=2lνtr[γµ/lγρ/C]−l2tr[γµγνγρ/C]. By integrating over l(using (11.C)) we get iM1=4ie3 (4π)D/2Γ/parenleftBig/epsilon1 2/parenrightBig/integraldisplay1 0dx/integraldisplay1−x 0dz/bracketleftBig 1−/epsilon1 2ln∆+o(/epsilon12)/bracketrightBig ×(1−D 2)(gµνCρ−gµρCν+gνρCµ). The divergent part of this integral is iM1|div=−ie3 2π2/epsilon1/integraldisplay1 0dx/integraldisplay1−x 0dz(gµνCρ−gµρCν+gνρCµ). The other two integrals can be evaluated in the same way. The ﬁnal result is iM|div=−ie3 2π2/epsilon1/bracketleftbigg1 6(gµν(p1−p2)ρ+gµρ(p1−p2)ν+gρν(p1−p2)µ)+ +1 2(gµν(p1+p2)ρ+gµρ(p2−p1)ν−gρν(p1+p2)µ)]. The diagram where the orientation in the loop is opposite is shown in the following ﬁgure. The amplitude is the same as in (11.26) except that the trace in (11.26) should be replaced by tr[γρ(−/k−/p2+m)γν(/p1−/k+m)γµ(−/k+m)]. Chapter 11. Renormalization and regularization 225 By putting C−1Cin the previous expression, where matrix Cis the charge conjugation matrix (4.K), we get tr[CγρC−1C(−/k−/p2+m)C−1CγνC−1C ×(/p1−/k+m)C−1CγµC−1C(−/k+m)C−1]. By using (4.K) we have tr[γρ(−/k−/p2+m)γν(/p1−/k+m)γµ(−/k+m)] =(−)3tr[γρ(/k+m)γµ(/k−/p1+m)γν(/k+/p2+m)], from which the we get the requested result. The statement is valid for all diagrams of this type with the odd number of vertices and this is called the Furry theorem. 11.10 The vacuum polarization in QED is −iΠµν(q)=−e2/integraldisplayd4k (2π)4tr[(/k+m)γµ(/k+/q+m)γν] (k2−m2)((k+q)2−m2). (11.29) From the Ward identity we know that this expression has the following form −iΠµν(q)=−(qµqν−q2gµν)iΠ(q2). By multiplying the previous expression by gµνand using (11.29) we get iΠ(q2)=−1 3q2igµνΠµν =−4e2 3q2/integraldisplayd4k (2π)4−2k·(k+q)+4m2 (k2−m2)((k+q)2−m2). (11.30) Discontinuity in the expression Π(q2) can be calculated by applying the Cutkosky rule. Then DiscΠ(q2)=4ie2 3q21 (2π)4(−2πi)2/integraldisplay d4k(4m2−2k·(k+q))δ(4)(k2−m2) ×δ(4)((k+q)2−m2)θ(−k0)θ(k0+q0). (11.31) By using δ(x2−a2)=1 2|a|(δ(x−a)+δ(x+a)) and taking qµ=(q0,0)w eg e t DiscΠ(q2)=−16iπ2e2 3q21 (2π)4/integraldisplay d4k(4m2−2k·(k+q)) ×1 4ω2 kδ(k0+ωk)δ(k0+q0−ωk). (11.32) 226 Solutions Integration over k0gives DiscΠ(q2)=−4iπ2e2 3q21 (2π)4/integraldisplay d3k(2m2+2q0ωk)1 ω2 kδ(q0−2ωk).(11.33) Since d3k=|k|ωkdωksinθdφdθwe have DiscΠ(q2)=−ie2 3πq2/integraldisplay∞ mdωk2m2+2q0ωk ωk/radicalBig ω2 k−m2δ(q0−2ωk).(11.34) Integration over ωkgives DiscΠ(q2)=e2 6πi/parenleftbigg 1+2m2 q2/parenrightbigg/radicalBigg 1−4m2 q2θ(q2−4m2). (11.35) Finally ImΠ(q2+i/epsilon1)=1 2iDiscΠ(p2) =−e2 12π/parenleftbigg 1+2m2 q2/parenrightbigg/radicalBigg 1−4m2 q2θ(q2−4m2).(11.36) 11.11 Scalar electrodynamics has two vertices: =−ie(p+p/prime)µ =2 ie2gµν The Feynman rules are standard except that for every closed photon loop we have an extra factor 1 /2. The photon self–energy is determined by the diagrams: The ﬁrst one is −iΠ(1) µν=2 ie2gµν/integraldisplaydDk (2π)Di k2−m2. By applying (11.A) and (11.F) we obtain: −iΠ(1) µν=−ie2 4π2/epsilon1m2gµν+ﬁ n.part. (11.37) The second diagram is Chapter 11. Renormalization and regularization 227 −iΠ(2) µν=e2/integraldisplaydDk (2π)D(2k+p)µ(2k+p)ν (k2−m2)((k+p)2−m2). By using the Feynman parametrization in the previous integral we get −iΠ(2) µν=e2/integraldisplay1 0dx/integraldisplaydDk (2π)D4kµkν+2kµpν+2kνpµ+pµpν [k2+2xk·p+p2x−m2]2. Applying the formulae (11.A–C) it follows that : −iΠ(2) µν=ie2πD/2 (2π)D/integraldisplay1 0dx/bracketleftbigg Γ/parenleftBig/epsilon1 2/parenrightBig1 (m2+p2x2−p2x)/epsilon1/2(4x2−4x+1 )pµpν −2gµνΓ/parenleftbig/epsilon1 2−1/parenrightbig (m2+p2x2−p2x)/epsilon1/2−1/bracketrightBigg , which is equal to −iΠ(2) µν=ie2 16π2/parenleftbigg2 3/epsilon1(pµpν−p2gµν)+4m2 /epsilon1gµν/parenrightbigg +ﬁ n.part. (11.38) Adding the divergent parts of the expressions (11.37) and (11.38) we get the requested result. Note that the terms proportional to m2cancel. So, the ﬁnal result is gauge invariant, as expected. 11.12 (a) Let us introduce the following notation: Nf−the number of external fermionic lines Ns−the number of external scalar lines Pf−the number of internal fermionic lines Ps−the number of internal scalar lines V3−the number of ¯ψγ5ψφvertices V4−the number of φ4vertices L−the number of loops. Then the superﬁcial degree of divergence for a diagram is D=4L−2Ps−Pf. On the other hand, Lc a nb ee x p r e s s e da s L=Ps+Pf−(V−1), since it is a number of independent internal momenta. By combining the previous formulae with 2V3=Nf+2Pf, V3+4V4=Ns+2Ps, we get 228 Solutions Fig. 11.4. Superﬁcially divergent diagrams in the Yukawa theory D=4−Ns−3 2Nf. Superﬁcially divergent amplitudes are shown in Fig. 11.4. The ﬁrst diagram is the vacuum one and it can be ignored; the second andﬁfth are equal to zero. The bare Lagrangian density is L 0=1 2(∂φ0)2−m2 0 2φ2 0+¯ψ0(iγµ∂µ−M0)ψ0−ig0¯ψ0γ5ψ0φ0−λ0 4!φ4 0.(11.39) If we rescale the ﬁelds as φ0=/radicalbig Zφφ=/radicalbig 1+δZφφ, ψ0=/radicalbig Zψψ=/radicalbig 1+δZψψ, and introduce a new set of variables: Zφm2 0=m2+δm2 ZψM0=M+δM Zψ/radicalbig Zφg0=µ/epsilon1/2(g+δg) Z2 φλ0=µ/epsilon1(λ+δλ), the bare Lagrangian density becomes L0=1 2(1 +δZφ)(∂φ)2−m2+δm2 2φ2+ i(1 + δZψ)¯ψ/∂ψ −(M+δM)¯ψψ−i(g+δg)µ/epsilon1/2¯ψγ5ψφ−(λ+δλ)µ/epsilon1 4!φ4. The Feynman rules are given in the Fig. 11.5 (b) The one–loop fermionic propagator correction is represented in Fig. 11.6. The ﬁrst diagram is −iΣ2(p)=−g2µ/epsilon1/integraldisplaydDk (2π)D1 k2−m2+i 0γ5/p−/k+M (p−k)2−M2+i 0γ5. Chapter 11. Renormalization and regularization 229 Fig. 11.5. Feynman rules in renormalized Yukawa theory Fig. 11.6. The one–loop correction to fermionic propagator Since γ5/aγ5=−/aand (γ5)2=1w eh a v e −iΣ2(p)=−g2µ/epsilon1 (2π)D/integraldisplay dDk−/p+/k+M (k2−m2+ i0)(( p−k)2−M2+ i0) =−g2µ/epsilon1 (2π)D/integraldisplay dDk/integraldisplay1 0dx−/p+/k+M (k−px)2−∆+ i0)2 =−g2µ/epsilon1 (2π)DiπD/2Γ/parenleftBig/epsilon1 2/parenrightBig/integraldisplay1 0dx/p(x−1) +M ∆/epsilon1/2, (11.40) where ∆=M2x+m2(1−x)−p2x+p2x2.Since µ/epsilon1 2DπD/2=1 16π2(4πµ2)/epsilon1/2=1 16π2/parenleftBig 1+/epsilon1 2ln(4πµ2)+.../parenrightBig , we have −iΣ2(p)=−ig2 16π2/bracketleftbigg2 /epsilon1−γ+o(/epsilon1)/bracketrightbigg/integraldisplay1 0dx[M+(x−1)/p]/bracketleftbigg 1−/epsilon1 2ln∆ 4πµ2/bracketrightbigg =−ig2 8π2/epsilon1(M−1 2/p)+ﬁ n .part. (11.41) The full one–loop correction to the fermionic propagator is −iΣ(p)=−ig2 8π2/epsilon1(M−1 2/p)−iδM+iδZψ/p+ﬁ n.part. From the renormalization conditions: Σ(/p=M)=0 , dΣ d/p/vextendsingle/vextendsingle/vextendsingle/p=M=0, (11.42) 230 Solutions follows that δZψ=−g2 16π2/epsilon1+ﬁ n.part, δM=−g2M 8π2/epsilon1+ﬁ n.part. (11.43) (c) The one–loop correction to the scalar propagator is represented in Fig. 11.7. Fig. 11.7. The one-loop correction to the scalar propagator The ﬁrst diagram is −iΠ1(p2)=−i2g2µ/epsilon1 (2π)D/integraldisplay dDktr[γ5(/k+M)γ5(/p+/k+M)] (k2−M2+ i0)(( p+k)2−M2+ i0) =g2µ/epsilon1 (2π)D/integraldisplay dDk/integraldisplay1 0dxtr[(−/k+M)(/p+/k+M)] (k2+2k·px−M2+p2x)2 =g2µ/epsilon1 (2π)D/integraldisplay1 0dx/integraldisplay dDk4(−k·p−k2+M2) (k2+2k·px−M2+p2x)2, where we use the Feynman parametrization formula (11.G). Introducing a new variable l=k+pxwe further have −iΠ1(p2)=4g2µ/epsilon1/integraldisplay1 0dx/integraldisplaydDl (2π)D2M2−∆−l2 (l2−∆+ i0)2 =ig2 4π2/integraldisplay1 0dx/parenleftbigg 1−/epsilon1 2ln∆ 4πµ2/parenrightbigg ×/parenleftbigg (M2−p2(x2−x))(2 /epsilon1−γ+o(/epsilon1))+ +D 2(−2 /epsilon1−1+γ+o(/epsilon1))(M2+p2(x2−x))/parenrightbigg =ig2 2π2/epsilon1/parenleftbiggp2 2−M2/parenrightbigg +ﬁ n.part, where ∆=M2+p2(x2−x).The second diagram is −iΠ2=iλm2 16π2/epsilon1+ﬁ n.part. (11.44) Summing, we obtain Chapter 11. Renormalization and regularization 231 −iΠ(p2)=ig2 2π2/epsilon1/parenleftbiggp2 2−M2/parenrightbigg +iλm2 16π2/epsilon1+iδZφp2−iδm2+ﬁn.part.(11.45) Using the renormalization conditions: Π(p2=m2)=0 dΠ dp2/vextendsingle/vextendsingle/vextendsingle p2=m2=0, (11.46) we get δZφ=−g2 4π2/epsilon1+ﬁ n.part δm2=λm2 16π2/epsilon1−g2M2 2π2/epsilon1+ﬁ n.part. (11.47) (d) The amplitude of the diagram is iM3=( ig)3µ3/epsilon1/2/integraldisplaydDk (2π)Dγ5(/k+/q+M)γ5(/k+M)γ5 ((k+q)2−M2)(k2−M2)((k−p)2−m2) =−2ig3µ3/epsilon1/2 (2π)Dγ5/integraldisplay1 0dx/integraldisplay1−x 0dz/integraldisplay dDkM2−/q/k+M/q−k2 ((k+qx−pz)2−∆)3 =−2ig3µ3/epsilon1/2 (2π)Dγ5/integraldisplay1 0dx/integraldisplay1−x 0dz/integraldisplay dDlN (l2−∆)3, where ∆=x2q2+z2p2+( 1−z)M2−xq2+zm2−p2z−2xzq·p and N=M2−(l−xq+zp)2+M/q−/q(/l−x/q+z/p). In the previous formulae we introduced a variable l=k+xq−zp.A sw e are interested to ﬁnd only the divergent part of i M3,i ti su s e f u lt on o t e that only l2–term in the numerator of the integrand is divergent. So, by using (11.C) we get: 232 Solutions iM3=2 ig3µ3/epsilon1/2γ5/integraldisplaydDl (2π)D/integraldisplay1 0dx/integraldisplay1−x 0dzl2 (l2−∆)3+... =−g3µ/epsilon1/2(4−/epsilon1) 32π2γ5/parenleftbigg2 /epsilon1−γ+.../parenrightbigg/integraldisplay1 0dx ×/integraldisplay1−x 0dz/parenleftbigg 1−/epsilon1 2ln∆ 4πµ2/parenrightbigg . Finally iM3=−g3µ/epsilon1/2 8π2/epsilon1γ5+ﬁ n.part. (11.48) The vertex correction is so, from iV3=/parenleftbigg gγ5µ/epsilon1/2+δgγ5µ/epsilon1/2−g3µ/epsilon1/2 8π2/epsilon1+ﬁ n.part/parenrightbigg/vextendsingle/vextendsingle/vextendsingle q2=0=gγ5 follows δg=g3 8π2/epsilon1+ﬁ n.part. (e) Let us ﬁrst calculate the following diagram Since we have to ﬁnd the divergent part of this diagram we can put that the external momenta are equal to zero. Then, iM4(k1=k2=k3=k4=0 )= −g4µ2/epsilon1/integraldisplaydDp (2π)Dtr[γ5(/p+M)]4 (p2−M2)4. (11.49) Since γ5(/p+M)γ5(/p+M)=(−/p+M)(/p+M)=M2−p2 we have Chapter 11. Renormalization and regularization 233 iM4(k1=k2=k3=k4=0 )= −4g4µ2/epsilon1/integraldisplaydDp (2π)D1 (p2−M2)2 =−ig4µ/epsilon1 4π2/parenleftbigg2 /epsilon1−γ/parenrightbigg/parenleftbigg 1−/epsilon1 2lnM2 4πµ2/parenrightbigg =−ig4µ/epsilon1 2π2/epsilon1+ﬁ n.part. (11.50) The previous result should be multiplied by a factor 6 as there are six diagrams of this type. The complete four vertex is iV4=/parenleftbigg −iλµ/epsilon1−iδλµ/epsilon1−6ig4µ/epsilon1 2π2/epsilon1+3iλ2µ/epsilon1 16π2/epsilon1+ﬁ n.part/parenrightbigg/vextendsingle/vextendsingle/vextendsingle s=4m2,t=u=0 =−iλ, (11.51) and ﬁnally δλ=−3g4 π2/epsilon1+3λ2 16π2/epsilon1+ﬁ n.part. (11.52) 11.13 In this problem dimension of spacetime is D=2−/epsilon1. (a) The polarization of vacuum is given by: −iΠµν(p)=( ie)2(−i2)/integraldisplaydDq (2π)Dtr[(/q−/p)γν/qγµ] q2(q−p)2. (11.53) In D-dimensional space trace identities necessary to calculate the previous expression read: tr(γµγν)=f(D)gµν, tr(γµγνγργσ)=f(D)(gµνgρσ−gµρgνσ+gµσgρν), where f(D) is any analytical function which satisﬁes the condition f(2) = 2. Instead of f(D) we will write 2 as we did in the previous problems (of course, there f(D) = 4). The Feynman parametrization gives −iΠµν(p)=2e2 (2π)D/integraldisplay1 0dx/integraldisplay dDq ×2qµqν−q2gµν−pµqν−pνqµ+(p·q)gµν (q2−2p·qx+p2x)2.(11.54) By using (11.A–C) in (11.54) we obtain −iΠµν=−2ie2πD/2 (2π)D/integraldisplay1 0dx/bracketleftbigg 2/parenleftbiggx2pµpν (−p2x+p2x2)1+/epsilon1/2Γ(1 +/epsilon1 2) −1 2gµν (−p2x+p2x2)/epsilon1/2Γ(/epsilon1 2)/parenrightbigg 234 Solutions −gµν/parenleftbiggx2p2 (−p2x+p2x2)1+/epsilon1/2Γ(1 +/epsilon1 2) −2−/epsilon1 21 (−p2x+p2x2)/epsilon1/2Γ(/epsilon1 2)/parenrightbigg −2xpµpν (−p2x+p2x2)1+/epsilon1/2Γ(1 +/epsilon1 2) +gµνp2x (−p2x+p2x2)1+/epsilon1/2Γ(1 +/epsilon1 2)/bracketrightbigg . From the previous expression (for D→2 i.e./epsilon1→0) we obtain −iΠµν(p)=−i(pµpν−p2gµν)Π(p2) =−ie2 πp2(pµpν−p2gµν), (11.55) from which we see that the polarization of vacuum is a ﬁnite quantity. (b) The full photon propagator is obtained by summing the diagrams in the Figure iDµν(p)=−igµν p2+i 0+−igµρ p2+i 0[p2gρσ−pρpσ]iΠ(p2)−igσν p2+i 0+... =−i p2+i 0(gµν−pµpν p2)(1 + Π(p2)+Π2(p2)+...)−ipµpν p4 =−i(gµν−pµpν p2) p2(1−Π(p2) + i0), (11.56) were we discarded the i pµpν/p4-term in the last line since the propagator is coupled to a conserved current. Then the photon propagator is iDµν(p)=−i(gµν−pµpν p2) p2−e2 π. (11.57) Photon mass is e/√π. 11.14 The dimension of spacetime is D=6−/epsilon1. (a) The renormalized Lagrangian density is Lren=L+Lct, (11.58) where L=1 2(∂φ)2−m2 2φ2−gµ/epsilon1/2 3!φ3−hµ−/epsilon1/2φ, (11.59) Chapter 11. Renormalization and regularization 235 Lct=1 2δZ(∂φ)2−δm2 2φ2−µ/epsilon1/2δg 3!φ3−µ−/epsilon1/2δhφ . (11.60) By introducing new quantities Z=1+ δZ , (11.61) m2 0Z=m2+δm2, (11.62) g0Z3/2=(g+δg)µ/epsilon1/2, (11.63) h0Z1/2=(h+δh)µ−/epsilon1/2, (11.64) and rescaling the ﬁeld, φ0=√ Zφ, the renormalized Lagrangian density becomes Lren=1 2(∂φ0)2−m2 0 2φ2 0−g0 3!φ3 0−h0φ0. The quantities with index 0 are called bare. The Feynman rules are given in Figure 11.8. Fig. 11.8. Feynman rules in φ3theory Superﬁcially divergent amplitudes are: Fig. 11.9. Divergent amplitudes in φ3theory (b) The tadpole diagram in one–loop order is shown in the following ﬁhure . 236 Solutions The second term is −igµ/epsilon1/2/integraldisplaydDk (2π)Di k2−m2+i 0 =−igµ/epsilon1/2 (2π)DπD/2 (m2)−2+/epsilon1/2Γ/parenleftBig −2+/epsilon1 2/parenrightBig =−igm4µ−/epsilon1/2 128π3/parenleftbigg2 /epsilon1+l n/parenleftbigg4πµ2 m2/parenrightbigg +3 2−γ/parenrightbigg =−igm4µ−/epsilon1/2 64π3/epsilon1+ﬁ n.part, and it does not depend on momentum. Summing all diagrams we get iH=−ihµ−/epsilon1/2−igm4µ−/epsilon1/2 64π3/epsilon1−iδhµ−/epsilon1/2+ﬁ n.part. (11.65) Hence, δh=−gm4 64π3/epsilon1+ﬁ n.part. (11.66) Finite part in the previous expression can be chosen so that H=0a n d we can ignore all diagrams which contain tadpoles. (c) The full one–loop propagator is shown in Fig. 11.10. Fig. 11.10. The one–loop propagator in φ3theory The second diagram is −iΠ2=(ig)2µ/epsilon1 2/integraldisplaydDk (2π)Di2 (k2−m2+ i0)(( k−p)2−m2+ i0) =g2µ/epsilon1 2/integraldisplay1 0dx/integraldisplaydDk (2π)D1 (k2−2k·px+p2x−m2+ i0)2 =−ig2 128π3/parenleftbigg2 /epsilon1+1−γ+o(/epsilon1)/parenrightbigg ×/integraldisplay1 0dx(m2+p2x(x−1))/parenleftbigg 1−/epsilon1 2lnm2+p2x(x−1) 4πµ2/parenrightbigg =−ig2 64π3/epsilon1/parenleftbigg m2−p2 6/parenrightbigg +ﬁ n.part. (11.67) Chapter 11. Renormalization and regularization 237 Propagator correction is −iΠ(p2)=−ig2 64π3/epsilon1/parenleftbigg m2−p2 6/parenrightbigg +ip2δZ−iδm2+ﬁ n.part.(11.68) From the condition −iΠ(p2) = ﬁnite we get δZ=−g2 384π3/epsilon1+ﬁ n.part, (11.69) δm2=−m2g2 64π3/epsilon1+ﬁ n.part. (11.70) In MS scheme the ﬁnite parts in (11.69) and (11.70) are zero. (d) The vertex correction is given in the Fig 11.11. Fig. 11.11. Vertex correction in φ3theory The second diagram is iΓ=(−ig)3µ3/epsilon1/2/integraldisplaydDk (2π)Di3 (k2−m2)((k+p2)2−m2)((k−p1)2−m2). (11.71) By applying (11.H) and integrating over the momentum kwe get iΓ=−(−ig)3µ3/epsilon1/2πD/2 (2π)DΓ/parenleftBig/epsilon1 2/parenrightBig/integraldisplay1 0dx/integraldisplay1−x 0dz ×1 (m2−p2 2x−p2 1z+p2 2x2+p2 1z2)/epsilon1/2 =−ig3µ/epsilon1/2 26−/epsilon1π3−/epsilon1/2/parenleftbigg2 /epsilon1+.../parenrightbigg/integraldisplay1 0dx/integraldisplay1−x 0dz ×/parenleftbigg 1−/epsilon1 2lnm2−p2 2x−p2 1z+p2 2x2+p2 1z2 µ2/parenrightbigg .(11.72) From the last formula we ﬁnd that the divergent part of i Γis given by −ig3µ/epsilon1/2 64π3/epsilon1. (11.73) The full one–loop vertex in the renormalized theory is 238 Solutions iV3=−igµ/epsilon1/2−iδgµ/epsilon1/2+iΓ. In minimal subtraction scheme δgis δg=−g3 64π3/epsilon1. (11.74) (e) From (11.61), (11.69) and (11.70) follows Z=1−g2 384π3/epsilon1, (11.75) m2=m2 0/parenleftbigg 1−g2 384π3/epsilon1/parenrightbigg +m2g2 64π3/epsilon1 =m2 0+5m2 0g2 384π3/epsilon1, (11.76) in the one–loop order. Similarly, from (11.69) and (11.74) we have g0=(g+δg)µ/epsilon1/2 Z3/2(11.77) =gµ/epsilon1/2/parenleftbigg 1−g2 64π3/epsilon1+g2 256π3/epsilon1/parenrightbigg (11.78) =gµ/epsilon1/2/parenleftbigg 1−3g2 256π3/epsilon1/parenrightbigg . (11.79) The last expression is important for calculation of the βfunction. References 1. D. Bailin and A. Love, Introduction to Gauge Field Theory , Adam Hilger, Bris- tol, 1986 2. J. Bjorken and S. Drell, Relativistic Quantum Mechanics , McGraw-Hill, New York, 1964 3. J. Bjorken and S. Drell, Relativistic Quantum Fields , McGraw-Hill, New York, 1965 4. N. N. Bogoljubov and D.V. Shirkov, Introduction to the Theory of Quantized Fields , Wiley-Interscience, New York, 1980 5. M. Blagojevi´ c,Gravitation and Gauge Symmetries , IOP Publishing, Bristol, 2002 6. T.P. Cheng and L.F. Li, Gauge Theory of Elementary Particle Physics , Oxford University Press, New York, 1984 7. T.P. Cheng and L.F. Li, Gauge Theory of Elementary Particle Physics, Problems and Solutions , Oxford University Press, New York, 2000 8. M. Damnjanovi´ c,Hilbert spaces and group theory , Faculty of Physics, Beograd, 2000 (in Serbian) 9. I.S. Gradshteyn and I.M. Ryzhnik, T a b l eo fI n t e g r a l s ,S e r i e sa n dP r o d u c t s , (trans. and ed. by Alan Jeﬀrey), Academic Press, Orlando, Florida, 1980 10. W. Greiner and J, Reinhardt, Quantum Electrodinamics , Springer, Berlin, Hei- delberg, New York, 1996 11. W. Greiner and J, Reinhardt, Field Quantization , Springer, Berlin, Heidelberg, New York, 1996 12. F. Gross, Relativistic Quantum Mechanics and Field Theory , Wiley, New York, 1993 13. C. Itzykson and J.B. Zuber, Quantum Field Theory , McGraw-Hill, New York, 1980 14. M. Kaku, Quantum Field Theory: A Modern Introduction , Oxford University Press, New York, 1993 15. F. Mandl and G. Show, Quantum Field Theory , New York, 1999 16. M. E. Peskin and D. V. Schroeder, An Introduction to Quantum Field Theory , Addison Wesley, 1995 17. P. Ramond, Field Theory: A Modern Primer (second edition), Addison-Wesley, RedwoodCity, California, 1989 240 References 18. L. Rayder, Quantum Field Theory , Cambridge University Press, Cambridge, 1985 19. J. J. Sakurai, Advanced Quantum Mechanics , Addison-Wesley, Reading, 1967 20. S. S. Schweber, An Introduction to Relativistic Quantum Field Theory , Harpen and Row, New York, 1962 21. A. G. Sveshnikov and A. N. Tikhonov, The Theory of Functions of a Complex Variable , Mir Publisher, Moscow, 1978 22. G. Sterman, Introduction to Quantum Field Theory , Cambridge University Press, Cambridge, 1993 23. S. Weinberg, The Quantum Theory of Fields I and II , Cambridge University Press, New York, 1996 Index Action 25 Einstein–Hilbert 27 Advanced Green function Dirac equation 138 Klein–Gordon equation 132 Angular momentum tensor Dirac ﬁeld 44, 45, 164 electromagnetic ﬁeld 52, 183–185Klein–Gordon ﬁeld 36, 37, 144 Anticommutation relations Dirac ﬁeld 43 Baker–Hausdorﬀ formula 91, 144 Bianchi identity 49 Casimir eﬀect 53, 187–190 Casimir operator 7 Charge Dirac ﬁeld 45, 162Klein–Gordon ﬁeld 37, 142 Charge conjugation Dirac equation 18 bilinears 23–24, 115–118 Dirac ﬁeld 45 bilinears 47, 175–177 scalar ﬁeld 41, 159 Chiral transformations 28 Coherent states 40, 156–158Commutation relations electromagnetic ﬁeld 50 scalar ﬁeld 35 Conformal group 75 Conformal transformations 7 Continuity equation 10Cross section 55 Cutkosky rule 62, 225 Decay rate 218 Diﬀerential cross section 192 Dilatations Dirac ﬁeld 30, 46, 129, 168 scalar ﬁeld 29, 38, 129, 148–150 Dimensional regularization 63 Dirac equation 17 helicity 99, 118helicity basic 20, 95plane wave solutions 17, 18, 93–95 spinor basic 20 Dirac ﬁeld quantization 43 Dirac particle in a hole 22, 110–111 in a magnetic ﬁeld 23, 113 Dyson Green function Klein–Gordon equation 133 Electromagnetic ﬁeld quantization 49 Energy–momentum tensor 26, 126 symmetric or Belinfante tensor 29, 127 Euler–Lagrange equations 25, 121 Feynman parametrization 62, 211 Feynman propagator Dirac equation 138, 139 Dirac ﬁeld 44 242 Index Klein–Gordon equation 31, 33, 132, 136 Klein–Gordon ﬁeld 36, 153 Foldy–Wouthuysen transformation 24, 118–119 Functional derivative 25, 121 Furry theorem 225 Galilean algebra 39, 156 Gamma matrices 13 contraction identities 14, 86–87 Dirac representation 13Majorana representation 13trace identities 15, 87–89Weyl representation 13 Gamma–function 62, 213 γ 5–matrix 13, 86, 102 γ5/s–operator 98 gauge transformations 49 Gordon identity 21, 104 Grassmann variable 173Green function Dirac equation 31, 33 Klein-Gordon equation 31 massive vector ﬁeld 33, 140massless vector ﬁeld 34, 140Schr¨odinger equation 154 Gupta–Bleuler quantization 50 Hamiltonian Dirac ﬁeld 44, 45, 162Klein–Gordon ﬁeld 36, 37, 142 Helicity 94, 165, 181 Klein paradox Dirac particle 109scalar particle 82 Klein–Gordon equation 9 plane wave solutions 77 Klein–Gordon particle in a hole 10, 79 in a magnetic ﬁeld 10, 81 in the Coulomb potential 10, 83 Lagrangian density Dirac ﬁeld 43 massive vector ﬁeld 27 massless vector ﬁeld 49Schr¨odinger ﬁeld 39sigma model 28 Left/right spinors 102–103 Levi-Civita tensor 4, 5, 68 Little group 74Lorentz group 5, 67 generators in deﬁning repr. 69 Lorentz transformations Dirac equation 17 bilinears 23–24, 115–118 Dirac ﬁeld 44, 170 bilinears 47, 174–177 scalar ﬁeld 158–159 Majorana spinor 47, 173 Maxwell equations 49 Metric tensor 3 Minkowski space 3 Momentum Dirac ﬁeld 44, 45 Klein–Gordon ﬁeld 36, 37, 142 MS scheme 237 Noether theorem 26 Normal ordering Dirac ﬁeld 44, 47, 172 Klein–Gordon ﬁeld 36 Optic theorem 220Parity Dirac equation 18 bilinears 23–24, 115–118 Dirac ﬁeld 44 bilinears 47, 174–177 scalar ﬁeld 41, 159 Pauli matrices 5Pauli–Lubanski vector 7, 19, 72–74, 98 Pauli–Villars regularization 62, 215 Phase transformations 28, 125 φ 3theory in 4D 58 φ3theory in 6D 64, 234–238 Poincar´ e algebra 6, 71, 72 Poincar´ e group 4, 6 Poincar´ e transformations 4 scalar ﬁeld 40 Projection operators energy 19, 95–96spin 100 QED processes Index 243 scattering in an external electromag- netic ﬁeld 202 QED processes µ−µ+→e−e+58, 196–198 e−µ+→e−µ+58 e−µ+→e−µ+198 Compton scattering 58, 199scattering in an external electromag- netic ﬁeld 58, 200 Reﬂection and transmission coeﬃcients Dirac equation 22 Klein–Gordon equation 10 Reiman ζ–function 53 Retarded Green function Klein–Gordon equation 132, 137 S–matrix 55 Scalar electrodynamics 64, 226Scalar ﬁeld quantization 35 Scalar product 4Scattering of polarized particles 59, 203–205 Schr¨odinger equation 153 Schwinger model 64, 233 Σ–vector 96 σ µν–matrices 14, 85, 87SL(2,C) group 5 Superﬁcial degree of divergence 64, 227 Symmetry factor in φ4theory 57, 194–195 Tensor of rank ( m, n)4 Time reversal Dirac equation 18 bilinears 23–24, 115–118 Dirac ﬁeld 44 bilinears 47, 175–178 scalar ﬁeld 41, 159 Vacuum polarization 63, 225 Vector 3 contravariant components 3covariant components 4 dual vector or one–form 4 Vertex correction 231–232, 237Virasora algebra 38 Weyl ﬁelds 20 Wick rotation 212Wick theorem 55, 57, 152, 172, 193–196 Yukawa theory 64, 206, 227–233