Problem Book in Quantum Field Theory

Problem Book Quantum Field Theory

Voja Radovanovic

Problem Book Quantum Field Theory

ABC

Voja Radovanovic Faculty of Physics University of Belgrade Studentski 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 are liable for prosecution under the German Copyright Law. Springer is a part of Springer Science+Business Media springeronline.com c Springer-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 LATEX macro package Cover design: design & production GmbH, Heidelberg Printed on acid-free paper

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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 Quantum Field 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 latter mostly 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 feel that 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 studying your lecture notes and textbooks. Try to solve problems without using solutions; they should help you to check your results. The level of this Problem 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 to the Lorentz and Poincar´e symmetries. Chapters 2, 3 and 4 deal with the relativistic quantum mechanics with a special emphasis on the Dirac equation. In Chapter 5 we present problems related to the Euler-Lagrange equations and the Noether theorem. The following Chapters concern the canonical quantization of scalar, Dirac and electromagnetic fields. In Chapter 10 we consider tree level processes, while the last Chapter deals with renormalization and regularization. There are many colleagues whom I would like to thank for their support and help. Professors Milutin Blagojevi´c and Maja Buri´c gave many useful 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 Balaˇz who checked many of the solutions. Duˇsko 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 Kneˇzevi´c, Zoran Ristivojevi´c and Vladimir Juriˇci´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 all the 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

3

The γ–matrices . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 13

4

The Dirac equation . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 17

5

Classical field theory and symmetries . . . . . . . . . . . . . . . . . . . . . 25

6

Green functions . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 31

7

Canonical quantization of the scalar field . . . . . . . . . . . . . . . . . . 35

8

Canonical quantization of the Dirac field . . . . . . . . . . . . . . . . . . 43

9

Canonical quantization of the electromagnetic field . . . . . . . . 49

10 Processes in the lowest order of perturbation theory . . . . . . . 55 11 Renormalization and regularization . . . . . . . . . . . . . . . . . . . . . . . . 61

Part II Solutions 1

Lorentz and Poincar´ e symmetries . . . . . . . . . . . . . . . . . . . . . . . . . . 67

2

The Klein–Gordon equation . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 77

3

The γ–matrices . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 85

X

Contents

4

The Dirac equation . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 93

5

Classical fields and symmetries . . . . . . . . . . . . . . . . . . . . . . . . . . . . 121

6

Green functions . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 131

7

Canonical quantization of the scalar field . . . . . . . . . . . . . . . . . 141

8

Canonical quantization of the Dirac field . . . . . . . . . . . . . . . . . . 161

9

Canonical quantization of the electromagnetic field . . . . . . . . 179

10 Processes in the lowest order of the perturbation theory . . . 191 11 Renormalization and regularization . . . . . . . . . . . . . . . . . . . . . . . . 211 References . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 239 Index . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 241

Part I

Problems

1 Lorentz and Poincar´ e symmetries

• Minkowski space, M4 is a real 4-dimensional vector space with metric tensor defined by   1 0 0 0 0   0 −1 0 gµν =  (1.A)  . 0 0 −1 0 0 0 0 −1 Vectors can be written in the form x = xµ eµ , where xµ are the contravariant components of the vector x in the basis         1 0 0 0 0 1 0 0 e0 =   , e1 =   , e2 =   , e3 =   . 0 0 1 0 0 0 0 1 The square of the length of a vector in M4 is x2 = gµν xµ xν . The square of the line element between two neighboring points xµ and xµ + dxµ takes the form (1.B) ds2 = gµν dxµ dxν = c2 dt2 − dx2 . The space M4 is also a manifold; xµ are global (inertial) coordinates. The covariant components of a vector are defined by xµ = gµν xν . • Lorentz transformations, (1.C) x
µ = Λµν xν , leave the square of the length of a vector invariant, i.e. x
2 = x2 . The matrix Λ is a constant matrix1 ; xµ and x
µ are 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 Λ must satisfy the condition ΛT gΛ = g. The transformation law of the covariant components is given by x
µ = (Λ−1 )νµ xν = Λµν xν . 1

(1.D)

The first index in Λµν is the row index, the second index the column index.

4

Problems

• Let u = uµ eµ be an arbitrary vector in tangent space2 , where uµ are its contravariant components. A dual space can be associated to the vector space in the following way. The dual basis, θ µ is determined by θ µ (eν ) = δνµ . The vectors in the dual space, ω = ωµ θ µ are called dual vectors or one–forms. The components of the dual vector transform like (1.D). The scalar (inner) product of vectors u and v is given by u · v = gµν uµ v ν = uµ vµ . A tensor of rank (m, n) in Minkowski spacetime is T = T µ1 ...µm ν1 ...νn (x)eµ1 ⊗ . . . ⊗ eµm ⊗ θ ν1 ⊗ . . . ⊗ θ νn . The components of this tensor transform in the following way σn

T
µ1 ...µm n1 ...νn (x
) = Λµ1 ρ1 . . . Λµm ρm (Λ−1 )σ1 ν1 . . . (Λ−1 )νn T ρ1 ...ρm σ1 ...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
− x
)2 = (y − x)2 . • In a certain representation the elements of the Poincar´e group near the identity are µν µ i (1.F) U (ω, ) = e− 2 Mµν ω +iPµ  , where ω µν and Mµν are parameters and generators of the Lorentz subgroup respectively, while µ and Pµ are the parameters and generators of the translation subgroup. The Poincar´e algebra is given in Problem 1.11. • The Levi-Civita tensor, µνρσ is a totaly antisymmetric tensor. We will use the convention that 0123 = +1.

2

3

The tangent space is a vector space of tangent vectors associated to each point of spacetime. Poincar´e transformations are very often called inhomogeneous Lorentz transformations.

Chapter 1. Lorentz and Poincare symmetries

5

1.1. Show that Lorentz transformations satisfy the condition ΛT gΛ = g. Also, prove that they form a group. 1.2. Given an infinitesimal Lorentz transformation Λµ ν = δ µ ν + ω µ ν , show that the infinitesimal parameters ωµν are antisymmetric. 1.3. Prove the following relation αβγδ Aα µ Aβ ν Aγ λ Aδ σ = µνλσ detA , where Aα µ are matrix elements of the matrix A. 1.4. Show that the Kronecker δ symbol and Levi-Civita  symbol are form invariant under Lorentz transformations. 1.5. Prove that µνρσ αβγδ

 µ δ α  ν δ = −  ρα δ α  σ δ α

δµ β δν β δρβ δσ β

δµ γ δν γ δργ δσ γ

 δµ δ  ν  δ δ  , δρδ  σ  δ δ

and calculate the following contractions µνρσ µβγδ , µνρσ µνγδ , µνρσ µνρδ , µνρσ µνρσ . ¯ µ = (I, −σ), where I is a 1.6. Let us introduce the notations σ µ = (I, σ); σ 4 unit matrix, while σ are Pauli matrices and define the matrix X = xµ σ µ . (a) Show that the transformation X → X
= SXS † , where S ∈ SL(2, C)5 , describes the Lorentz transformation xµ → Λµν xν . This is a homomorphism between proper orthochronous Lorentz transformations6 and the SL(2, C) group. σ µ X). (b) Show that xµ = 12 tr(¯ σ µ Sσν S † ), and Λ(S) = Λ(−S). The last relation 1.7. Prove that Λµ ν = 12 tr(¯ shows that the map is not unique. 4

The Pauli matrices are



σ1 = 5 6

0 1

1 0



 ,

σ2 =

0 i

−i 0



 and

σ3 =

1 0

0 −1

.

SL(2, C) matrices are 2 × 2 complex matrices of unit determinant. The proper orthochronous Lorentz transformations satisfy the conditions: Λ00 ≥ 1, detΛ = 1.

6

Problems

1.8. Find the matrix elements of generators of the Lorentz group Mµν in its natural (defining) representation (1.C). 1.9. Prove that the commutation relations of the Lorentz algebra [Mµν , Mρσ ] = i(gµσ Mνρ + gνρ Mµσ − gµρ Mνσ − gνσ Mµρ ) lead to [Mi , Mj ] = iijl Ml ,

[Ni , Nj ] = −iijl Nl ,

[Mi , Nj ] = iijl Nl ,

where Mi = 12 ijk Mjk and Nk = Mk0 . Further, one can introduce the following linear combinations Ai = 12 (Mi + iNi ) and Bi = 12 (Mi − iNi ). Prove that [Ai , Aj ] = iijl Al ,

[Bi , Bj ] = iijl Bl ,

[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
µ = Λµ ν xν + 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, )U (Λ, 0) = U (1, Λ−1 ) , in the Poincar´e group. In addition, show that from the previous relation follows: U −1 (Λ, 0)Pµ U (Λ, 0) = (Λ−1 )ν µ Pν . Calculate the commutator [Mµν , Pρ ]. (b) Show that U −1 (Λ, 0)U (Λ
, 0)U (Λ, 0) = U (Λ−1 Λ
Λ, 0) , and find the commutator [Mµν , Mρσ ]. (c) Finally show that the generators of translations commute between themselves, i.e. [Pµ , Pν ] = 0. 1.12. Consider the representation in which the vectors x of Minkowski space are (x, 1)T , while the element of the Poincar´e group, (Λ, a) are 5 × 5 matrices given by  Λ a . 0 1 Check that the generators in this representation satisfy the commutation relations 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 classical scalar field7 . Prove that they satisfy the commutation relations obtained in Problem 1.11. 1.14. The Pauli–Lubanski vector is defined by Wµ = 12 µνλσ M νλ P σ . (a) Show that Wµ P µ = 0 and [Wµ , Pν ] = 0. (b) Show that W 2 = − 12 Mµν M µν P 2 + Mµσ M νσ P µ Pν . (c) Prove that the operators W 2 and P 2 commute 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 W 2 |p = 0, m, s, σ = −m2 s(s + 1)|p = 0, m, s, σ , where |p = 0, m, s, σ is a state vector for a particle of mass m, momentum p, spin s while σ is 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(gνσ Wµ − gµσ Wν ) , (b) [Wµ , Wν ] = −iµνσρ W σ P ρ . 1.17. Calculate the commutators (a) [Wµ , M 2 ] , (b) [Mµν , W µ W ν ] , (c) [M 2 , Pµ ] , (d) [µνρσ Mµν Mρσ , 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
µ = e−ρ xµ , special conformal transformations (SCT): xµ → x
µ =

xµ + cµ x2 , 1 + 2c · x + c2 x2

and usual Poincar´e transformations form a group. Find the commutation relations in this group. 7 8

Scalar field transforms as φ
(Λx + a) = φ(x) E(2) is the group of rotations and translations in a plane.

2 The Klein–Gordon equation

• The Klein–Gordon equation, (  + 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
(Λx) = φ(x). • The equation for the spinless particle in an electromagnetic field, Aµ is obtained by changing ∂µ → ∂µ + iqAµ in equation (2.A), where q is the charge of the particle.

2.1. Solve the Klein–Gordon equation. 2.2. If φ is a solution of the Klein–Gordon equation calculate the quantity 

∂φ∗ 3 ∗ ∂φ Q = iq d x φ −φ . ∂t ∂t 2.3. The Hamiltonian for a free real scalar field is

1 d3 x[(∂0 φ)2 + (∇φ)2 + m2 φ2 ] . H= 2 Calculate the Hamiltonian H for a general solution of the Klein–Gordon equation. 2.4. The momentum for a real scalar field is given by

P = − d3 x∂0 φ∇φ . Calculate the momentum P for a general solution of the Klein–Gordon equation.

10

Problems

2.5. Show that the current1 i jµ = − (φ∂µ φ∗ − φ∗ ∂µ φ) 2 satisfies the continuity equation, ∂ µ jµ = 0. 2.6. Show that the continuity equation ∂µ j µ = 0 is satisfied for the current i jµ = − (φ∂µ φ∗ − φ∗ ∂µ φ) − qAµ φ∗ φ , 2 where φ is 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  −V, r < a 0 , qA = 0, r>a where V is 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  0, z 0 where U0 is a positive constant. 2.10. A particle of charge q and mass m is incident on a potential barrier  0, z < 0, z > a , A0 = U0 , 0 < z < a where U0 is a positive constant. Find the transmission coefficient. 2.11. A scalar particle of mass m and charge −e moves in the Coulomb field of a nucleus. Find the energy spectrum of the bounded states for this system if the charge of the nucleus is Ze.  θ 2.12. Using the two-component wave function , where θ = 12 (φ + mi ∂φ ∂t ) χ and χ = 12 (φ − mi ∂φ ∂t ), instead of φ rewrite the Klein–Gordon equation in the Schr¨ odinger form. 1

Actually 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 H is 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

1 ψ1 |ψ2  = d3 xψ1† σ3 ψ2 . 2 (a) Show that the Hamiltonian H obtained in Problem 2.12 is Hermitian. (b) Find expectation  values of the Hamiltonian H, and the velocity v in 1 the state e−ip·x . 0

3 The γ–matrices

• In Minkowski space M4 , the γ–matrices satisfy the anticommutation relations 1 {γ µ , γ ν } = 2g µν . • In the Dirac representation γ–matrices take the form   I 0 0 σ , γ= . γ0 = 0 −I −σ 0

(3.A)

(3.B)

Other representations of the γ–matrices can be obtained by similarity transformation γµ
= Sγµ S −1 . The transformation matrix S need to be unitary if the transformed matrices are to satisfy the Hermicity condition: (γ
µ )† = γ
0 γ
µ γ
0 . The Weyl representation of the γ–matrices is given by   0 I 0 σ , γ= , (3.C) γ0 = I 0 −σ 0 while in the Majorana representation we have   0 σ2 iσ3 0 γ0 = , , γ1 = σ2 0 0 iσ3   0 −σ2 −iσ1 0 . , γ3 = γ2 = σ2 0 0 −iσ1

(3.D)

• The matrix γ 5 is defined by γ 5 = iγ 0 γ 1 γ 2 γ 3 , while γ5 = −iγ0 γ1 γ2 γ3 . In the Dirac representation, γ5 has the form  0 I . γ5 = I 0

1

The same type of relations hold in Md , where d is the dimension of spacetime.

14

Problems

• σµν matrices are defined by σµν = • Slash is defined as

i [γµ , γν ] . 2

(3.E)

/a = aµ γµ .

(3.F)

• Sometimes we use the notation: β = γ , α = γ γ. The anticommutation relations (3.A) become 0

0

{αi , αj } = 2δ ij , {αi , β} = 0 .

3.1. Prove: (a) 㵆 = γ 0 γµ γ 0 , † = γ 0 σµν γ 0 . (b) σµν 3.2. Show that: (a) γ5† = γ5 = γ 5 = γ5−1 , (b) γ5 = − 4!i µνρσ γ µ γ ν γ ρ γ σ , (c) (γ5 )2 = 1 , (d) (γ5 γµ )† = γ 0 γ5 γµ γ 0 . 3.3. Show that: (a) {γ5 , γ µ } = 0 , (b) [γ5 , σ µν ] = 0 . 3.4. Prove a /2 = a2 . 3.5. Derive the following identities with contractions of the γ–matrices: (a) γµ γ µ = 4 , (b) γµ γ ν γ µ = −2γ ν , (c) γµ γ α γ β γ µ = 4g αβ , (d) γµ γ α γ β γ γ γ µ = −2γ γ γ β γ α , (e) σ µν σµν = 12 , (f) γµ γ5 γ µ γ 5 = −4 , (g) σαβ γµ σ αβ = 0 , (h) σαβ σ µν σ αβ = −4σ µν , (i) σ αβ γ 5 γ µ σαβ = 0 , (j) σ αβ γ 5 σαβ = 12γ 5 .

Chapter 3. The γ–matrices

15

3.6. Prove the following identities with traces of γ–matrices: (a) trγµ = 0 , (b) tr(γµ γν ) = 4gµν , (c) tr(γµ γν γρ γσ ) = 4(gµν gρσ − gµρ gνσ + gµσ gνρ ) , (d) trγ5 = 0 , (e) tr(γ5 γµ γν ) = 0 , (f) tr(γ5 γµ γν γρ γσ ) = −4iµνρσ , a2n+1 ) = 0 , (g) tr(/ a1 · · · / a2n ) = tr(/ a2n · · · /a1 ) , (h) tr(/ a1 · · · / (i) tr(γ5 γµ ) = 0 , a2 · · · / a6 ). 3.7. Calculate tr(/a1 / 3.8. Calculate tr[(/ p − m)γµ (1 − γ5 )(/ q + m)γν ]. 3.9. Calculate γµ (1 − γ5 )(/ p − m)γ µ . 3.10. Verify the identity exp(γ5 / a) = cos



aµ aµ + √

1 γ5 /a sin aµ aµ , µ aµ a

where a2 > 0 . 3.11. Show that the set Γ a = {I, γ µ , γ 5 , γ µ γ 5 , σ µν } , is made of linearly independent 4 × 4 matrices. Also, show that the product of any two of them is again one of the matrices Γ a , up to ±1, ±i. 3.12. Show that any matrix A ∈ C 44 can be written in terms of Γ a = µ 5 µ 5 µν {I, γ , γ , γ γ , σ }, i.e. A = a ca Γ a where ca = 14 tr(AΓa ). 3.13. Expand the following products of γ–matrices in terms of Γ a : (a) γµ γν γρ , (b) γ5 γµ γν , (c) σµν γρ γ5 . 3.14. Expand the anticommutator {γ µ , σ νρ } in terms of Γ –matrices. 3.15. Calculate tr(γµ γν γρ γσ γa γβ γ5 ). 3.16. Verify the relation γ5 σ µν = 2i µνρσ σρσ . 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 γ µ , then it is proportional to the unit matrix. 3.19. Let U = exp(βα · n), where β and α are Dirac matrices; n is a unit vector. Verify the following relation: α
≡ U αU † = α − (I − U 2 )(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 , and γ5 in this representation. 3.21. Find Dirac matrices in two dimensional spacetime. Define γ5 and calculate tr(γ 5 γ µ γ ν ) . Simplify the product γ 5 γ µ .

4 The Dirac equation

• The Dirac equation,

(iγ µ ∂µ − m)ψ(x) = 0 ,

(4.A)

is an equation of the free relativistic particle with spin 1/2. The general solution of this equation is given by ψ(x) =

2



1 3

(2π) 2

r=1

  m  ur (p)cr (p)e−ip·x + vr (p)d†r (p)eip·x , (4.B) d p Ep 3

where ur (p) and 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 u ¯r (p)us (p) = −¯ vr (p)vs (p) = δrs , u ¯r (p)vs (p) = v¯r (p)us (p) = 0.

(4.D)

The coefficients cr (p) and dr (p) in (4.B) being given determined by boundary conditions. Equation (4.A) can be rewritten in the form i

∂ψ = HD ψ , ∂t

where HD = α · p + βm is the so-called Dirac Hamiltonian. • Under the Lorentz transformation, x
µ = Λµ ν xν , Dirac spinor, ψ(x) transforms as i µν (4.E) ψ
(x
) = S(Λ)ψ(x) = e− 4 σ ωµν ψ(x) . S(Λ) is the Lorentz transformation matrix in spinor representation, and it satisfies the equations: S −1 (Λ) = γ0 S † (Λ)γ0 ,

18

Problems

S −1 (Λ)γ µ S(Λ) = Λµ ν γ ν . • The equation for an electron with charge −e in an electromagnetic field Aµ is given by (4.F) [iγ µ (∂µ − ieAµ ) − m] ψ(x) = 0 . • Under parity, Dirac spinors transform as ψ(t, x) → ψ
(t, −x) = γ0 ψ(t, x) .

(4.G)

• Time reversal is an antiunitary operation: ψ(t, x) → ψ
(−t, x) = T ψ ∗ (t, x) . The matrix T , satisfies

T γµ T −1 = γ µ∗ = γµT .

(4.H) (4.I)

The solution of the above condition is T = iγ 1 γ 3 , in the Dirac representation of γ–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 ψ¯T .

(4.J)

The matrix C satisfies the relations: Cγµ C −1 = −γµT ,

C −1 = C T = C † = −C .

(4.K)

In the Dirac representation, the matrix C is given by C = iγ 2 γ 0 .

4.1. Find which of the operators given below commute with the Dirac Hamiltonian: (a) p = −i∇ , (b) L = r × p , (c) L2 , (d) S = 12 Σ , where Σ = 2i γ × γ , (e) J = L + S , (f) J2 , p , (g) Σ · |p| (h) Σ · n, where n is 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·x and vs (p)eip·x for the Dirac particle.

Chapter 4. The Dirac equation

19

4.4. Using the solution of Problem 4.2 show that 2  r=1

−

2  r=1

ur (p)¯ ur (p) =

/p + m ≡ Λ+ (p) , 2m

vr (p)¯ vr (p) = −

/p − m ≡ Λ− (p) . 2m

The quantities Λ+ (p) and Λ− (p) are energy projection operators. 4.5. Show that Λ2± = Λ± , and Λ+ Λ− = 0. How do these projectors act on the basic spinors ur (p) and 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 = 12 Σ. Prove that: (a) Σ = γ5 γ0 γ , (b) [S i , S j ] = iijk S k , (c) S 2 = − 34 . 4.7. Prove that:

Σ·p ur (p) = (−1)r+1 ur (p) , |p| Σ·p vr (p) = (−1)r vr (p) . |p|

Are spinors ur (p) and vr (p) eigenstates of the operator Σ · n, where n is 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 v along 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 = 12 µνρσ M νρ P σ , where M νρ = 12 σ νρ + i(xν ∂ ρ − xρ ∂ ν ) is angular momentum, while P µ is linear momentum. Show that 1 1 W 2 ψ(x) = − (1 + )m2 ψ(x) , 2 2 where ψ(x) is a solution of the Dirac equation.

20

Problems

4.11. The covariant operator which projects the spin operator onto an arbitrary normalized four-vector sµ (s2 = −1) is given by Wµ sµ , where s · p = 0, i.e. the vector polarization sµ is orthogonal to the momentum vector. Show that Wµ sµ 1 = γ5 /s/p . m 2m 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 /s commutes with the operator /p, and that the eigenvalues of this operator are ±1. Find the eigen-projectors of the operator s. Prove that these projectors commute with projectors onto positive and γ5 / 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  1 a ϕ= . |a|2 + |b|2 b Calculate the expectation value of the spin projection onto a unit vector n, i.e. Σ · n. 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 = π2 ). Calculate the expectation value of the projection spin on the polarization vector in that state. 4.17. Is the operator γ5 a constant of motion for the free Dirac particle? Find the eigenvalues and projectors for this operator. 4.18. Let us introduce

1 (1 − γ5 )ψ , 2 1 ψR = (1 + γ5 )ψ , 2 where ψ is 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 ψR are known as Weyl fields. ψL =

Chapter 4. The Dirac equation

21

4.19. Let us consider the system of the following two–component equations: ∂ψR (x) = mψL (x) , ∂xµ ∂ψL (x) = mψR (x) , i¯ σµ ∂xµ µ µ where σ = (I, σ); σ ¯ = (I, −σ). iσ µ

(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.
(x
) = SR,L ψR,L (x), Find 2 × 2 matrices SR and SL , which satisfy ψR,L
is a wave function obtained from ψR,L (x) by a boost along the where ψR,L x–axis. 4.20. Prove that the operator K = β(Σ·L+1), where Σ = − 2i α×α is the spin operator and L is orbital momentum, commutes with the Dirac Hamiltonian. 4.21. Prove the Gordon identities: 2m¯ u(p1 )γµ u(p2 ) = u ¯(p1 )[(p1 + p2 )µ + iσµν (p1 − p2 )ν ]u(p2 ) , 2m¯ v (p1 )γµ v(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
)σµν (p + p
)ν u(p) = i¯ u(p
)(p
− p)µ u(p) . 4.23. The current Jµ is given by Jµ = u ¯(p2 )/ p1 γµ /p2 u(p1 ), where u(p) and u ¯(p) are Dirac spinors. Show that Jµ can be written in the following form: Jµ = u ¯(p2 )[F1 (m, q 2 )γµ + F2 (m, q 2 )σµν q ν ]u(p1 ) , where q = p2 − p1 . Determine the functions F1 and F2 . 4.24. Rewrite the expression 1 u ¯(p) (1 − γ5 )u(p) 2 as a function of the normalization factor N = u† (p)u(p). 4.25. Consider the current Jµ = u ¯(p2 )pρ q λ σµρ γλ u(p1 ) , where u(p1 ) and u(p2 ) are Dirac spinors; p = p1 + p2 and q = p2 − p1 . Show that Jµ has the following form: Jµ = u ¯(p2 )(F1 γµ + F2 qµ + F3 σµρ q ρ )u(p1 ) , and determine the functions Fi = Fi (q 2 , 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. ¯ 0 ψ and the current density 4.27. Determine the probability density ρ = ψγ ¯ j = ψγψ, for an electron with momentum p and in an arbitrary spin state. 4.28. Find the time dependence of the position operator rH (t) = eiHt re−iHt for a free Dirac particle. 4.29. The state of the free electron at time t = 0 is given by   1 0 (3) ψ(t = 0, x) = δ (x)   . 0 0 Find ψ(t > 0, x). 4.30. Determine the time evolution of the wave packet    1 2 x 1 0 − 2   , ψ(t = 0, x) = 3 exp 2 0 2d 4 (πd ) 0 for the Dirac equation. 4.31. An electron with momentum p = pez and positive helicity meets a potential barrier  0, z < 0 0 −eA = . V, z > 0 Calculate the coefficients of reflection and transmission. 4.32. Find the coefficients of reflection and transmission for an electron moving in a potential barrier:  0, z < 0, z > a . −eA0 = 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 2a wide and V deep. Consider only bound states of the electron. (a) Find the dispersion relations. (b) Determine the relation between V and a if there are N bound 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 field B = Bez . 4.35. Show that if ψ(x) is a solution of the Dirac equation in an electromagnetic field, then it satisfies the ”generalize” Klein-Gordon equation: e [(∂µ − ieAµ )(∂ µ − ieAµ ) − σµν F µν + m2 ]ψ(x) = 0 , 2 where F µν = ∂ µ Aν − ∂ ν Aµ is the field strength tensor. 4.36. Find the nonrelativistic approximation of the Dirac Hamiltonian H = 2 α · (p + eA) − eA0 + mβ, including terms of order vc2 . ¯ 4.37. If Vµ (x) = ψ(x)γ µ ψ(x) is a vector field, show that Vµ is a real quantity. Find the transformation properties of this quantity under proper orthochronous Lorentz transformations, charge conjugation C, parity P and time reversal T . 4.38. Investigate the transformation properties of the quantity Aµ (x) = µ ¯ γ5 ψ(x), under proper orthochronous Lorentz transformations and the ψ(x)γ discrete transformations C, P and T . µ ¯ 4.39. Prove that the quantity ψ(x)γ µ ∂ ψ(x) is a Lorentz scalar. Find its transformation rules under the discrete transformations.

4.40. Using the Dirac equation, show that C u ¯T (p, s) = v(p, s), where C is charge conjugation. Also, prove the above relation in a concrete representation. 4.41. The matrix C is defined by Cγµ C −1 = −γµT . Prove that if matrices C
and C

satisfy the above relation, then C
= kC

, where k is a constant. 4.42. If

  1  0   e−iEt+ipz , ψ(x) = Np    σp 1 3 Ep +m 0 

is the wave function in frame S of the relativistic particle whose spin is 1/2, find: (a) the wave function ψc (x) = C ψ¯T (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 S by a rotation about the x–axis through θ. 4.43. Find the matrices C and P in 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 θ from the condition that the new Hamiltonian H
= U HU † , where U = eβα·pθ(p) has even form, i.e. H
∼ β. (Foldy–Wouthuysen transformation). 4.46. Show that the spin operator Σ = 2i γ × γ and the angular momentum L = r × p, in Foldy-Wouthuysen representation, have the following form: ΣFW = LFW = L −

iβ(α × p) m p(p · Σ) + Σ+ , Ep 2Ep (m + Ep ) 2Ep

p2 Σ iβ(α × p) p(p · Σ) + − . 2Ep (m + Ep ) 2Ep (m + Ep ) 2Ep

4.47. Find the Foldy-Wouthuysen transform of the position operator x and the momentum operator p. Calculate the commutator [xFW , pFW ].

5 Classical field theory and symmetries

• If f (x) is a function and F [f (x)] a functional, the functional derivative, is defined by the relation

δF [f (x)] δf (y) , δF = dy δf (y) where δF is a variation of the functional. • The action is given by

S = d4 xL(φr , ∂µ φr ),

δF [f (x)] δf (y)

(5.A)

(5.B)

where L is 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  ∂L ∂L =0. (5.C) ∂µ − ∂(∂µ φr ) ∂φr • The canonical momentum conjugate to the field variable φr is πr (x) =

∂L . ∂ φ˙ r

The canonical Hamiltonian is



H = d3 xH = d3 x(φ˙ r πr − L) .

(5.D)

(5.E)

• Noether theorem: If the action is invariant with respect to the continous infinitesimal transformations: xµ → x
µ = xµ + δxµ , φr (x) → φ
r (x
) = φr (x) + δφr (x) ,

26

Problems

then the divergence of the current jµ =

∂L δφr (x) − T µν δxν , ∂(∂µ φr )

(5.F)

is equal to zero, i.e. ∂µ j µ = 0. The quantity Tµν =

∂L ∂ν φr − Lgµν , ∂(∂ µ φr )

(5.G)

 is the energy–momentum tensor . The Noether charges Qa = d3 xj0a (x) are constants of motion under suitable asymptotic conditions. The index a is related to a symmetry group.

5.1. Let (a) Fµ = ∂µ φ ,  (b) S = d4 x 12 (∂µ φ)2 − V (φ) , be functionals. Calculate the functional derivatives 2

δ S δφ(x)δφ(y)

δFµ δφ

in the first case, and

in the second case.

5.2. Find the Euler–Lagrange equations for the following Lagrangian densities: (a) L = −(∂µ Aν )(∂ν Aµ ) + 12 m2 Aµ Aµ + λ2 (∂µ Aµ )2 , (b) L = − 41 Fµν F µν + 12 m2 Aµ Aµ , where Fµν = ∂µ Aν − ∂ν Aµ , (c) L = 12 (∂µ φ)(∂ µ φ) − 12 m2 φ2 − 14 λφ4 , (d) L = (∂µ φ − ieAµ φ)(∂ µ φ∗ + ieAµ φ∗ ) − m2 φ∗ φ − 14 Fµν F µν , ¯ 5 ψφ . ¯ µ ∂ µ − m)ψ + 1 (∂µ φ)2 − 1 m2 φ2 + 1 λφ4 − ig ψγ (e) L = ψ(iγ 2 2 4 5.3. The action of a free scalar field in two dimensional spacetime is 

∞ L 1 m2 2 µ ∂µ φ∂ φ − φ S= dt dx . 2 2 −∞ 0 The spatial coordinate x varies 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  + m2 )φ. L = − 12 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 ¯∂ ψ − (∂µ ψ)γ ¯ ¯ µ ψ) − mψψ. the form L = 2i (ψ/ 5.7. The Lagrangian density for a massive vector field Aµ is given by 1 1 L = − Fµν F µν + m2 Aµ Aµ . 4 2 Prove that the equation ∂µ Aµ = 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 + ∂µ Λ(x), where Λ = Λ(x) is an arbitrary function. Is the relation ∂µ Aµ = 0 a consequence of the equations of motion? 5.9. The Einstein–Hilbert gravitation action is

√ S = κ d4 x −gR , where gµν is the metric of four-dimensional curved spacetime; R is scalar curvature and κ is a constant. In the weak-field approximation the metric is (0) small perturbation around the flat metric gµν , i.e. (0) gµν (x) = gµν + 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 (anyone familiar with general relativity can easily prove this): 

1 1 ∂σ hµν ∂ σ hµν − ∂σ hµν ∂ ν hµσ + ∂σ hµσ ∂µ h − ∂µ h∂ µ h , S = d4 x 2 2 where h = hµµ . Derive the equations of motion for hµν . These are the linearized Einstein equations. Show that the linearized theory is invariant under the gauge symmetry: hµν → hµν + ∂µ Λν + ∂ν Λµ , 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 m2 2 λ [(∂φ1 )2 + (∂φ2 )2 ] − (φ1 + φ22 ) − (φ21 + φ22 )2 , 2 2 4

is invariant under the transformation φ1 → φ
1 = φ1 cos θ − φ2 sin θ , φ2 → φ
2 = φ1 sin θ + φ2 cos θ . Find the corresponding Noether current and charge.

28

Problems

5.12. Consider the Lagrangian density L = (∂µ φ† )(∂ µ φ) − m2 φ† φ , 

φ1 is an SU(2) doublet. Show that the Lagrangian density has φ2 SU(2) symmetry. Find the related Noether currents and charges. where

5.13. The Lagrangian density is given by ¯ µ ∂µ − m)ψ , L = ψ(iγ 

ψ1 is a doublet of SU(2) group. Show that L has SU(2) symψ2 metry. Find Noether currents and charges. Derive the equations of motion for spinor fields ψi , where i = 1, 2. where ψ =

5.14. Prove that the following Lagrangian densities are invariant under phase transformations ¯ µ ∂ µ − m)ψ , (a) L = ψ(iγ (b) L = (∂µ φ† )(∂ µ φ) − m2 φ† φ . Find the Noether currents. 5.15. The Lagrangian density of a real three-component scalar field is given by 1 m2 T L = ∂µ φT ∂ µ φ − φ φ, 2 2   φ1 where φ =  φ2 . Find the equations of motions for the scalar fields φi . φ3 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 under chiral transformations ψ(x) → ψ
(x) = eiαγ5 ψ(x) , where α is a constant. Find the Noether current and its four-divergence. 5.17. The Lagrangian density of a σ-model is given by 1 [(∂µ σ)(∂ µ σ) + (∂µ π) · (∂ µ π)] + iΨ¯ /∂ Ψ 2 m2 2 λ + g Ψ¯ (σ + iτ · πγ5 )Ψ − (σ + π 2 ) + (σ 2 + π 2 )2 , 2 4

L=

Chapter 5. Classical field theory and symmetries

29

where σ is a scalar field, π is a tree-component scalar field, Ψ a doublet of spinor fields, while τ are Pauli matrices. Prove that the Lagrangian density L has the symmetry: σ(x) → σ(x), π(x) → π(x) − α × π(x), α·τ Ψ (x) , Ψ (x) → Ψ (x) + i 2 where α is 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 fourdivergence T˜µν = Tµν + ∂ ρ χρµν , where χρµν = −χµρν . The two energy–momentum tensors are equivalent since they lead to the same conserved charges, i.e. both satisfy the continuity equation. If we take that the tensor χµνρ is given by1  1 ∂L ∂L ∂L (I (I (I χµνρ = ) + ) + ) − ρν rs µν rs µρ rs 2 ∂(∂ µ φr ) ∂(∂ ρ φr ) ∂(∂ ν φr ) then T˜µν is symmetric2 . The quantities (Iρν )rs in the previous formula are defined by the transformation law of fields under Lorentz transformations: δφr ≡ φ
r (x
) − φr (x) =

1 µν ω (Iµν )rs φs (x) . 2

(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 Belinfante) energy–momentum tensors for the Dirac and the electromagnetic field. 5.19. Under dilatation the coordinates are transformed as x → x
= e−ρ x . The corresponding transformation rule for a scalar field is given by φ(x) → φ
(x
) = eρ φ(x) , 1 2

Belinfante, Physica 6, 887 (1939) Symmetric energy–momentum tensors are not only simpler to work with but give the correct coupling to gravity.

30

Problems

where ρ is 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
= e−ρ x,

ψ(x) → ψ
(x
) = e3ρ/2 ψ(x) .

Calculate the Noether current and charge.

3

A form variation is defined by δ0 φr (x) = φ
r (x)−φr (x); total variation is δφr (x) = φ
r (x
) − φr (x).

6 Green functions

• The Green function (or propagator) of the Klein-Gordon equation, ∆(x − y) satisfies the equation ( x + 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 (6.B) (iγ µ ∂µx − m)S(x − y) = δ (4) (x − y) , 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 propagate 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 ∆F is the Feynman propagator, and ∆R is 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



d4 kδ(k 2 − m2 )θ(k0 )f (k) =

where ωk =

√

k2 + m 2 .

d3 k f (k) , 2ωk

32

Problems

6.4. Prove the following properties: ∆R (−x) = ∆A (x) , ∆F (−x) = ∆F (x) . ∆A and ∆R are the advanced and retarded Green functions; ∆F is the Feynman propagator. ¯ 6.5. If the Green function ∆(x) of the Klein–Gordon equation is defined as1

¯ ∆(x) =P

d4 k e−ik·x , (2π)4 k 2 − m2

prove the relations: 1 ¯ ∆(x) = (∆R (x) + ∆A (x)) , 2 ¯ ¯ ∆(−x) = ∆(x) . P denotes the principal value. 6.6. Write ∆(x) = −

1 (2π)4



and ∆± (x) = −

1 (2π)4

d4 k C

e−ik·x , k 2 − m2

 d4 k C±

e−ik·x k 2 − m2

in terms of integrals over three momentum, k. The integration contours are given in Fig. 6.1.

Fig. 6.1. The integration contours C and C± .

In addition, prove that ∆(x) = ∆+ (x) + ∆− (x).

1

¯ ∆(x) is also called the principal-part propagator.

Chapter 6. Green functions

6.7. Show that

33

 ∂∆(x)  =0, ∂xi x0 =0  ∂∆(x)  = −δ (3) (x) . ∂x0 x0 =0

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 i 1 δ(x2 ) + 2 P 2 , 4π 4π x

where ∆F is the Feynman propagator of the Klein–Gordon equation. 6.10. Prove that ∆R,A |m2 =0 = −

1 θ(±t)δ(x2 ) . 2π

6.11. If the source ρ is given by ρ(y) = gδ (3) (y), show that φR =

g exp(−m|x|) , 4π |x|

 where φR (x) = − d4 y∆R (x − y)ρ(y).

6.12. Show that the Green function of the Dirac equation, S(x) has the following form S(x) = (i/ ∂ + m)∆(x) , where ∆(x) is the Green function of the Klein–Gordon equation with corresponding boundary conditions. 6.13. Starting from definition (6.B), determine the retarded, advanced, Feynman and Dyson propagators of the Dirac equation. Also, prove that the difference 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 g is a constant while q is a constant vector, calculate

ψ(x) = d4 ySF (x − y)j(y) . SF is 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 1 1 L = − Fµν F µν + m2 Aµ Aµ . 4 2 Fµν = ∂µ Aν − ∂ν Aµ is the field strength.

34

Problems

6.16. Calculate the Green function of a massless vector field for which the Lagrangian density is given by 1 1 L = − Fµν F µν + λ(∂A)2 . 4 2 The second term is known as the gauge fixing term; λ is a constant.

7 Canonical quantization of the scalar field

• The operators of a complex free scalar field are given by

1 d3 k √ φ(x) = (a(k)e−ik·x + b† (k)eik·x ) , 3 2ωk (2π) 2

1 d3 k √ φ† (x) = (b(k)e−ik·x + a† (k)eik·x ) , 3 2ωk (2π) 2

(7.A)

(7.B)

where a(k) and b(k) are annihilation operators; a†(k) and b† (k) creation operators and a(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 φ and φ† 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 is defined by a(k) |0 = 0, b(k) |0 = 0, for all k. A state a† (k) |0 describes scalar particle with momentum k, b† (k) |0 an 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: : a1 a2 a†3 a4 a†5 := a†3 a†5 a1 a2 a4 . • The Hamiltonian, linear momentum and angular momentum of a scalar field are

1 d3 x[(∂0 φ)2 + (∇φ)2 + m2 φ2 ] , H= 2

P = − d3 x∂0 φ∇φ ,

M µν =

d3 x(xµ T 0ν − xν T 0µ ) .

• The Feynman propagator of a complex field is defined by i∆F (x − y) = 0| T (φ(x)φ† (y)) |0 .

(7.E)

Time ordering is defined by   T φ(x)φ† (y) = 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 transformations 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 operators: [a(k), a† (q)] = δ (3) (k − q) , [a(k), a(q)] = [a† (k), a† (q)] = 0 . 7.2. At t = 0, a real scalar field and its time derivative are given by φ(t = 0, x) = 0,

˙ = 0, x) = c , φ(t

where c is 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 2(2π)3 ωk

e−iωk t+ik·x ,

are orthonormal with respect to the scalar product

f |g = −i d3 x[f (x)∂0 g ∗ (x) − g ∗ (x)∂0 f (x)] . 7.5. Show that the vacuum expectation value of the scalar field Hamiltonian is given by 1 0| H |0 = − πm4 δ (3) (0)Γ (−2) . 4 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) [P µ , φ(x)] , (b) [P µ , F (φ(x), π(x))], where F is an arbitrary polynomial function of fields and momenta, (c) [H, a† (k)a(q)] , (d) [Q, P µ ] ,  (e) [N, = d3 ka† (k)a(k) is the particle number operator,  3H], where N−ip·x . (f) d x[H, φ(x)]e 7.7. Prove that eiQ φ(x)e−iQ = e−iq φ(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(gλν Pµ − gλµ Pν ) , (c) [Mµν , Mρσ ] = i(gµσ Mνρ + gνρ Mµσ − gµρ Mνσ − gνσ Mµρ ) . 7.9. Prove that φk (x) = k|φ(x)|0 satisfies the Klein–Gordon equation.  7.10. Calculate the charges Qa = d3 xj0a (x), where j0a are 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 = d3 xj 0 . (b) Prove that relations ρ[D, φ(x)] = iδ0 φ(x) and ρ[D, π(x)] = iδ0 π(x) hold. (c) Calculate the commutator [D, F (φ, π)], where F is an arbitrary analytic function. (d) Prove that [D, P µ ] = iP µ . 7.12. If, instead of the field φ(x), we define the smeared field

φf (x, t) = d3 yφ(t, y)f (x − y) , where f is given by f (x) =

1

e−x

2

3/2

(a2 π)

/a2

,

calculate the vacuum expectation value 0| φf (t, x)φf (t, x) |0 . Find the result in the limit of vanishing mass. µ 7.13. The creation and annihilation operators of the free bosonic string αm µ (0 < m ∈ Z), and αm (0 > m ∈ Z), satisfy the commutation relations µ , αnν ] = −mδm+n,0 g µν . [αm

µ αm−n αnµ satisfy Show that the operators Lm = − 12

[Lm , Ln ] = (m − n)Lm+n . The operators Lm form the classical Virasora algebra. Upon normal ordering of the Lm
s one can obtain the full algebra (with central charge): [Lm , Ln ] = (m − n)Lm+n +

D−2 3 (m − m)δm+n,0 . 12

D is number of scalar fields. 7.14. Calculate the vacuum expectation value 0| {φ(x), φ(y)} |0 , 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 0| φ(x1 )φ(x2 )φ(x3 )φ(x4 ) |0 for a free scalar field. 7.16. Find 0| φ(x)φ(y) |0 in two dimensions, for a massless scalar field. 7.17. Prove the relation ( x + m2 ) 0| T (φ(x)φ(y)) |0 = −iδ (4) (x − y) . 7.18. The Lagrangian density of a spinless Schr¨ odinger field ψ, is given by L = iψ †

1 ∂ψ − ∇ψ † · ∇ψ − V (r)ψ † ψ . ∂t 2m

(a) Find the equations of motion. (b) Express the free fields ψ and ψ † in terms of creation and annihilation operators and find commutation relations between them. (c) Calculate the Green function G(x0 , x, y0 , y) = −i 0| ψ(x0 , x)ψ † (y0 , y) |0 θ(x0 − y0 ) and prove that it satisfies the equation  1 ∂ G(t, x, 0, 0) = δ(t)δ (3) (x) . i + ∂t 2m (d) Calculate the Green function for one-dimensional particle in the potential  0, x>0 V = . ∞, x < 0 (e) Show that the free Schr¨ odinger equation is invariant under Galilean transformations, which contain: - spatial translations ψ
(t, r + ) = ψ(t, r) , - time translations ψ
(t + δ, r) = ψ(t, r) , - spatial rotations ψ
(t, r + θ × r) = ψ(t, r) , 2 - ”boost” ψ
(t, r − vt) = e−imv·r+imv t/2 ψ(t, r) . Without the phase factor in the last transformation rule the Schr¨ odinger equation will not be invariant, unless m = 0. Consequently this representation 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) =

d3 p ˜ f (p)e−ip·x , 2ωp

be a classical function which satisfies the Klein–Gordon equation. Introduce the operators

d3 p ˜∗ a=C f (p)a(p) , 2ωp

d3 p ˜ a† = C f (p)a† (p) , 2ωp where a(p) and a† (p) are annihilation and creation operators for scalar field, and C is a constant given by C = 

1 d3 p ˜ 2 2ωp |f (p)|

.

A coherent state is defined by |z = e−|z|

2

/2 za†

e

|0 ,

where z is a complex number. (a) Calculate the following commutators: [a(p), a† ],

[a(p), a] .

(b) Prove the relation nf˜(p) † n−1 (a ) . [a(p), (a† )n ] = C 2ωp (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 z| : φ2 (x) : |z − (z| φ(x) |z)2 . (e) Find the expectation value of the Hamiltonian in the coherent state, z| H |z . 7.20. Under the Poincar´e transformation, x → x
= 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 operators:  ωk
−1 exp(−iΛµ ν k ν aµ )a(Λk) , U (Λ, a)a(k)U (Λ, a) = ωk  ωk
† −1 U (Λ, a)a (k)U (Λ, a) = exp(iΛµ ν k ν aµ )a† (Λk) . ωk (b) Prove that the transformation rule of the n–particle state |k1 , . . . , kn  is given by  ωk1
· · · ωkn
iaµ Λµ ν (kν +...+kν ) 1 n |Λk , . . . , Λk  . e U (Λ, a) |k1 , k2 , . . . , kn  = 1 n ωk 1 · · · ωk n (c) Prove that the momentum operator, P µ of a scalar field is a vector under Lorentz transformations: U (Λ, 0)P µ U −1 (Λ, 0) = Λν µ P ν . (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  

 †  π 3 † P = exp −i d k a (k)a(k) − ηp a (k)a(−k) , 2 where ηp = ±1 is the intrinsic parity of the field. (a) Prove that P commutes with the Hamiltonian. (b) Prove the relation P Mij P −1 = Mij , where Mij is the angular momentum for scalar field. 7.22. Under time reversal, the scalar field is transformed according to τ φ(x)τ −1 = ηφ(−t, x) , where τ is an antiunitary operator, while η is a phase. (a) Prove the relations:

τ a(k)τ −1 = ηa(−k) , τ a† (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 = ηc φ† (x) , where ηc is a phase factor. Prove that CQC −1 = −Q , where Q is the charge operator.

8 Canonical quantization of the Dirac field

• The operators of a Dirac field are: ψ(x) =

¯ ψ(x) =

2



1 3

(2π) 2

r=1 2



1 (2π)

3 2

r=1

  m  ur (p)cr (p)e−ip·x + vr (p)d†r (p)eip·x , (8.A) d p Ep 3

  m  u ¯r (p)c†r (p)eip·x + v¯r (p)dr (p)e−ip·x . (8.B) d3 p Ep

The operators c†r (p) and d†r (p) are creation operators, while cr (p), dr (p) are annihilation operators. • From the Dirac Lagrangian density, ¯ µ ∂µ − m)ψ , L = ψ(iγ one obtains the expressions for the conjugate momenta: πψ =

∂L ∂L = iψ † , πψ¯ = =0. ˙ ∂ψ ∂ ψ˙¯

Particles of spin 1/2 obey Fermi-Dirac statistics. We impose the canonical equal-time anticommutation relations: {ψa (t, x), ψb† (t, y)} = δab δ (3) (x − y) ,

(8.C)

{ψa (t, x), ψb (t, y)} = {ψa† (t, x), ψb† (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)} = δrs δ (3) (p − q) . All other anticommutators are zero.

(8.E)

44

Problems

• The Fock space of states is obtained as usual, by acting with creation operators on the vacuum |0 . The states c† (p, r) |0 , and d† (p, r) |0 are 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 anticommutation 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:

¯ + m]ψ , H= d3 xψ[−iγ∇

P= −i

Mµν =

d3 xψ † ∇ψ ,

1 d3 xψ † (i(xµ ∂ν − xν ∂µ ) + σµν )ψ . 2

• The Feynman propagator is given by   ¯ |0 . iSF (x − y) = 0| T ψ(x)ψ(y)

(8.F)

Time ordering is defined by   ¯ ¯ ¯ − θ(y0 − x0 )ψ(y)ψ(x) . T ψ(x)ψ(y) = θ(x0 − y0 )ψ(x)ψ(y) • Under the Lorentz transformation, x
= Λx the operator ψ(x) transforms according to: (8.G) U (Λ)ψ(x)U −1 (Λ) = S −1 (Λ)ψ(Λx) . Here U (Λ) is a unitary operator in spinor representation which generates the Lorentz transformation. • Parity, t
= t, x
= −x changes the Dirac field as follows P ψ(t, x)P −1 = γ0 ψ(t, −x) ,

(8.H)

where P is the appropriate unitary operator. • Time reversal, t
= −t, x
= x is represented by an antiunitary operator. The transformation law is given by τ ψ(t, x)τ −1 = T ψ(−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 C generates charge conjugation in the space of spinors: Cψa (x)C −1 = (Cγ0T )ab ψb† (x) .

(8.J)

Properties of the matrix C are 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 = y0 the function iS(x − y) becomes γ0 δ (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:  3 + (a) charge Q = −e :,  3d x:ψ ψ ¯ (b) energy H = d x[: ψ(−iγ i ∂i + m)ψ :] , (c) momentum P = −i d3 x : ψ † ∇ψ : . ∂ ψ(x). Comment on this result. 8.3. (a) Show that i[H, ψ(x)] = ∂t (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 transformations are given by 1 Mµν = i(xµ ∂ν − xν ∂µ ) + σµν . 2 8.6. The angular momentum of the Dirac field is  

1 Mµν = d3 xψ † (x) i(xµ ∂ν − xν ∂µ ) + σµν ψ(x) . 2

46

Problems

(a) Prove that 1 [Mµν , ψ(x)] = −i(xµ ∂ν − xν ∂µ )ψ(x) − σµν ψ(x) , 2 and comment on this result. (b) Also, prove [Mµν , Pρ ] = i(gνρ Pµ − gµρ Pν ) , where Pµ is the four-vector of momentum. 8.7. Show that the helicity of the Dirac field is given by

1 Sp = d3 p(−1)r+1 [c†r (p)cr (p) + d†r (p)dr (p)] . 2 r 8.8. Let |p1 , r1 ; p2 , r2  = c†r1 (p1 )c†r2 (p2 ) |0 be a two-particle state. Find the energy, charge and helicity of this state. Here r1,2 are helicities of one-particle states. 8.9. Prove that the charges found in Problem 5.13 satisfy the commutation relation: [Qa , Qb ] = iabc Qc . 8.10. Find conserved charges for the symmetry in Problem 5.17 and calculate the commutators: (a) [Qa , Qb ] , (b) [Qb , π a (x)], [Qb , ψi (x)], [Qb , ψ¯i (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 = d3 xj 0 for this symmetry and show that the relation [D, P µ ] = iP µ , is satisfied. 8.12. Let the Lagrangian density be given by ¯ µ ∂µ ψ − gx2 ψψ ¯ , L = iψγ where g is a constant. (a) Derive the expression for the energy–momentum tensor Tµν . Find its divergence, ∂µ T µν . Comment on this result. (b) Calculate the commutator [P 0 (t), P i (t)]. (c) Find the four divergence of the angular momentum operator M µαβ . ¯ µ ψ. 8.13. Consider the current commutator [Jµ (x), Jν (y)] where Jµ = ψγ

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. ¯ 1 )ψ(x2 )ψ(x3 )ψ(x ¯ 4 ) |0 . The result should be expressed 8.14. Calculate 0| ψ(x in terms of vacuum expectation value of two fields. ¯ µ ψ := 1 [ψ, ¯ γ µ ψ]. 8.15. Prove that : ψγ 2 ¯ 8.16. Prove that 0| T (ψ(x)Γ ψ(y)) |0 is equal to zero for Γ = {γ5 , γ5 γµ }, while for Γ = γµ γν one gets the result −4imgµν ∆F (y − x). 8.17. The Dirac spinor in terms of two Weyl spinors ϕ and χ is of the form  ϕ ψ= . −iσ2 χ∗ (a) Show that the Majorana spinor equals  χ ψM = . −iσ2 χ∗ (b) Prove the identities: ψ¯M φM = φ¯M ψM , ψ¯M γ µ φM = −φ¯M γ µ ψM , ψ¯M γ5 φM = φ¯M γ5 ψM , ψ¯M γ µ γ5 φM = φ¯M γ µ γ5 ψM , ψ¯M σµν φM = −φ¯M σµν ψM . (c) Express the Majorana field operator, ψM = √12 (ψ + ψc ) using creation and annihilation operators of a Dirac field. Introduce creation and annihilation operators for Majorana spinors and find corresponding anticomutation relations. (d) Rewrite the QED Lagrangian density using Majorana spinors. ¯ 8.18. Find the transformation laws of the quantities Vµ (x) = ψ(x)γ µ ψ(x) and ¯ Aµ (x) = ψ(x)γ5 ∂µ ψ(x) under Lorentz and discrete transformations. 8.19. Show that the Lagrangian density µ ¯ ¯ L = iψ(x)γ ∂µ ψ(x) + mψ(x)ψ(x) ,

is invariant under the Lorentz and discrete transformations. ¯ 8.20. Show that the quantity Tµν (x) = ψ(x)σ µν ψ(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 jµ is 1 L = − Fµν F µν − j µ Aµ . 4 From this expression we derive the equations of motion to be:  − ∂µ ∂ ν )Aµ = j ν . ∂µ F µν = j ν ⇒ (δµν 

(9.A)

It is easy to see that the field strength Fµν satisfies the identity: ∂µ Fνρ + ∂ν Fρµ + ∂ρ Fµν = 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 + ∂ µ Λ(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 Coulomb Time Axial

gauge gauge gauge gauge

∂µ Aµ = 0 , ∇·A=0 , A0 = 0 , A3 = 0 .

• The general solution of the vacuum Maxwell equations (j µ = 0) takes the form:

3   1 d3 k  µ † µ −ik·x ik·x √ a , (9.C) Aµ (x) = (k) (k)e + a (k) (k)e λ 3 λ λ λ 2ωk (2π) 2 λ=0 where ωk = |k|,µλ (k) are polarization vectors. The transverse polarization vectors which satisfy (k) · k = 0 we denote by µ1 (k) and µ2 (k). The scalar

50

Problems

polarization vector is µ0 = nµ , where nµ is a unit time–like vector. We can choose nµ = (1, 0, 0, 0). The longitudinal polarization vector, µ3 (k) is given by k µ − (n · k)nµ . µ3 (k) = (n · k) Due to gauge symmetry only two polarizations are independent. The polarization vectors satisfy the orthonormality relations: gµν µλ (k)νλ
(k) = −δλλ
. In (9.C) we assumed the polarization vectors to be real valued. • The polarization vectors satisfy the following completeness relations: 3 

gλλ µλ (k)νλ (k) = g µν .

(9.D)

λ=0

From (9.D) follows that the sum over transverse photons is 2 

iλ (k)jλ (k) = −g ij −

λ=1

ki kj k i nj + k j ni . + (k · n)2 k·n

(9.E)

• In the Lorentz gauge the equal-time commutation relations are: [Aµ (t, x), π ν (t, y)] = ig µν δ (3) (x − y) , [Aµ (t, x), Aν (t, y)] = 0 , µ

(9.F)

ν

[π (t, x), π (t, y)] = 0 . where π ν = −A˙ ν . Creation and annihilation operators of the photon field satisfy the following commutation relations: [aλ (k), a†λ
(q)] = −gλλ
δ (3) (k − q) , [aλ (k), aλ
(q)] = 0 , [a†λ (k), a†λ
(q)]

(9.G)

=0.

The physical states, |Φ satisfy the operator condition ∂ µ A(+) µ |Φ = 0. This is the Gupta–Bleuler method of quantization. • In the Coulomb gauge we have A(x) =

2  λ=1



1 (2π)

3 2

 d3 k  √ aλ (k)λ (k)e−ik·x + a†λ (k)λ (k)eik·x , (9.H) 2ωk

Chapter 9. Canonical quantization of the electromagnetic field

51

while A0 = 0. The equal-time commutation relations are: (3)

[Ai (t, x), π j (t, y)] = −iδ⊥ij (x − y) , [Ai (t, x), Aj (t, y)] = 0 ,

(9.I)

[π i (t, x), π j (t, y)] = 0 , (3)

where π = E and δ⊥ij (x − y) is the transversal delta function given by (3) δ⊥ij (x

1 − y) = (2π)3



3

d ke

ik·(x−y)

ki kj δij − 2 k

.

Creation and annihilation operators obey [aλ (k), a†λ
(q)] = δλλ
δ (3) (k − q) , [aλ (k), aλ
(q)] = 0 ,

(9.J)

[a†λ (k), a†λ
(q)] = 0 . • The Feynman propagator for the electromagnetic field is given by iDFµν (x − y) = 0| T (Aµ (x)Aν (y)) |0 .

(9.K)

9.1. Starting from the commutation relations (9.G) prove that [Aµ (t, x), A˙ ν (t, y)] = −ig µν δ (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: [E i (x), E j (y)] , [B i (x), B j (y)] , [E i (x), B j (y)] . Also calculate the previous commutators for equal times, x0 = y 0 . 9.4. Prove that [P µ , Aν ] = −i∂ µ Aν .

52

Problems

9.5. Determine the helicity of photons described by polarization vectors µ+ (kez ) = 2−1/2 (0, 1, i, 0)T and µ− (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
moving in the x–direction with velocity v. 9.7. The arbitrary state not containing transversal photons has the form  |Φ = Cn |Φn  , n

where Cn are constants and

n  |Φn  = d3 k1 . . . d3 kn f (k1 , . . . , kn ) (a†0 (ki ) − a†3 (ki )) |0 , i=1

where f (k1 , . . . , kn ) are arbitrary functions. The state |Φ0  is a vacuum. (a) Prove that Φn |Φn  = δn,0 . (b) Show that Φ| Aµ (x) |Φ is a pure gauge. 9.8. Let P µν = g µν − and P⊥µν =

k µ k¯ν + k ν k¯µ , k · k¯

k µ k¯ν + k ν k¯µ , k · k¯

where k¯µ = (k 0 , −k). ⊥ ⊥ , P µν + P⊥µν , g µν Pµν , g µν Pµν , P µ ν P⊥νσ , if k 2 = 0. Calculate: P µν Pνσ , P⊥µν Pνσ 9.9. The angular momentum of the photon field is defined by J l = 12 lij M ij , where M ij was found in Problem 5.18. (a) Express J in 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 1 a†± (q) |0 = √ (a†1 (q) ± ia†2 (q)) |0 , 2 are the eigenstates of the helicity operator with the eigenvalues ±1. (d) Calculate the commutator [J l , Am (y, t)]. 9.10. Calculate: (a) 0| {E i (x), B j (y)} |0 , (b) 0| {B i (x), B j (y)} |0 ,

Chapter 9. Canonical quantization of the electromagnetic field

53

(c) 0| {E i (x), E j (y)} |0 . 9.11. Consider the quantization of the electromagnetic field in space between two parallel square plates located at z = 0 and 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 capacitor. (b) Quantize the electromagnetic field using canonical quantization. (c) Find the Hamiltonian H and show that the vacuum energy is  ∞  

 nπ 2   1 2 d2 k 2 2 2 2 E= L k1 + k2 + + k1 + k2 . 2 (9.1) 2 (2π)2 a n=1 (d) Define the quantity

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  1, k < Λ f (k) = , 0, k > Λ =

into the integral; Λ is a cutoff parameter. Calculate  and show that there is an attractive force between the plates. This is the Casimir effect. (e) The energy per unit area, E/L2 can be regularized in a different way. Calculate integral

1 , I = d2 k 2 (k + m2 )α for Re α > 0, and then analitically continue this integral to Re α ≤ 0. Show that ∞ π2  3 E/L2 = − 3 n . 6a n=1 Regularize the sum in the previous expression using the Rieman ζ–function ζ(s) =

∞ 

n−s .

n=1

Calculate the energy and the force per unit area.

10 Processes in the lowest order of perturbation theory

• The Wick’s theorem states T (ABC . . . Y Z) =: {ABC . . . Y Z + ”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 minus sign appears. • The S–matrix is given by



∞  (−i)n . . . d4 x1 . . . d4 xn T (HI (x1 ) · · · HI (xn )) , (10.B) S= n! n=0 where HI is the Hamiltonian density of interaction in the interaction pictures. • S–matrix elements have the general form    1 m 4 (4) √ , (10.C) Sfi = (2π) δ (pf − pi )iM VE 2V E b

f

where pi and pf are the initial and the final momenta, respectively; iM is 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 N final particles is N |Sfi |2 1  V d3 pi , (10.D) dσ = T |Jin | i=1 (2π)3 where Jin is the flux of initial particles: |Jin | =

vrel . V

56

Problems

The relative velocity vrel is given by |p1 | , E1

vrel =

in the laboratory frame of reference (particle 2 is at rest), while in the center– of–mass frame we have E1 + E 2 vrel = |p1 | , E1 E2 p1 is the momentum of particle 1, and E1,2 are energies of particles. In expression (10.D), V d3 p/(2π)3 is the volume element of phase space. • Feynman rules for QED: ◦ Vertex:

= ieγ µ

◦ Photon and lepton propagators: =−

iDF µν = iSF (p) =

=

igµν , k 2 + i

i . /p − m + i

◦ 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 question of matrix multiplication of the corresponding factors. ◦ For with momentum k, we must integrate over the momentum:  4all loops d k/(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(E1
, p
1 ) + D(E2
, p
2 ) prove that the differential cross section in the center of mass frame is given by  dσ |p
1 | 1 mA mB mC mD |M|2 , = dΩ cm 4π 2 (E1 + E2 )2 |p1 | where iM is the Feynman amplitude. Assume that all particles in the process are fermions. 10.2. Consider the following integral:

3 d p d3 q (3) I= δ (p + q − P)δ(Ep + Eq − P 0 ) , 2Ep 2Eq where Ep2 = p2 + m2 and Eq2 = q2 + m
2 . Show that the integral I is Lorentz invariant. Calculate it in the frame where P = 0. 10.3. If iM = u ¯(p, r)γµ (1 − γ5 )u(q, s)µ (k, λ) , calculate the sum

2  2 

|M|2 .

λ=1 r,s=1

10.4. Using the Wick theorem evaluate: (a) 0| T (φ4 (x)φ4 (y)) |0 , (b) T (: φ4 (x) : : φ4 (y) :) , ¯ ¯ (c) 0| T (ψ(x)ψ(x) ψ(y)ψ(y)) |0 . λ 4 10.5. In φ4 theory the interaction Lagrangian density is Lint = − 4! φ . Using the Wick theorem determine the symmetry factor S, for the following diagrams:

(a)

x1

x2

58

Problems

(b)

(c)

x1

x2

x1

x2

Also, check the results using the formula [6]:  2β (n!)αn , S=g n=2,3,..

where g is the number of possible permutations of vertices which leave unchanged the diagram with fixed external lines, αn is the number of vertex pairs connected by n identical lines, and β is the number of lines connecting a vertex with itself. 10.6. In φ3 theory calculate 1 2



−iλ 3!

2

d4 y1 d4 y2 0| T (φ(x1 )φ(x2 )φ3 (y1 )φ3 (y2 )) |0 .

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 |M|2 averaging over all initial polarization states and summing over the final polarization 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) Aµ (x) = (ae−k x , 0, 0, 0) , (b) Aµ (x) = (0, 0, 0, gr e−r/a ) . 2

2

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 ω and a are 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−k

2

x2

),

for a theory which is the same as QED except the fact that the vertex ieγµ is replaced by ieγµ (1 − γ5 ). 10.12. Find the differential cross section for the scattering of a positron in the external potential g Aµ = ( , 0, 0, 0) , r where g is a constant. The S–matrix element is given by

Sfi = ie d4 xψ¯f (x)∂µ ψi (x)Aµ (x) . 10.13. Calculate the cross section for the scattering of an electron with positive helicity in the electromagnetic potential Aµ = (aδ (3) (x), 0, 0, 0) , where a is 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 helicity, while the spin states of final particles are arbitrary. 10.15. Consider the theory of interaction of a spinor and scalar field: L=

1 M2 2 ¯ ¯ 5 ψφ . (∂φ)2 − φ + ψ(iγµ ∂ µ − m)ψ − g ψγ 2 2

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. (b)

(a)

(c)

(d)

(e)

60

Problems

(f)

(h)

(g)

(i)

11 Renormalization and regularization

• Table of D-dimensional integrals in Minkowski spacetime:

D 1 i(−1)n π 2 D ), = dD k 2 D Γ (n − n− 2 2 (k + 2p · k − m2 + i)n 2 Γ (n)(m + p ) 2

D

dD k

kµ −i(−1)n π 2 D µ ) , (11.B) = D p Γ (n − 2 2 n n− 2 2 (k + 2p · k − m + i) 2 Γ (n)(m + p ) 2 D

kµ kν i(−1)n π 2 d k 2 = D (k + 2p · k − m2 + i)n Γ (n)(m2 + p2 )n− 2  1 µν 2 D 2 − g (p + m )Γ (n − − 1) , 2 2 D



 D pµ pν Γ (n − ) 2 (11.C)

 D pµ pν pρ Γ (n − ) 2  1 D − (g µν pρ + g µρ pν + g νρ pµ )(p2 + m2 )Γ (n − − 1) , (11.D) 2 2 D

dD k



(11.A)

kµ kν kρ −i(−1)n π 2 = D 2 2 n (k + 2p · k − m + i) Γ (n)(m2 + p2 )n− 2

kµ kν kρ kσ i(−1)n π D/2 d k 2 = D 2 n (k + 2p · k − m + i) Γ (n)(m2 + p2 )n− 2 D

 D pµ pν pρ pσ Γ (n − ) 2

1 − (g µν pρ pσ + g µρ pν pσ + g µσ pν pρ + g νρ pµ pσ + g νσ pρ pµ + g ρσ pµ pν ) 2 D − 1) × (p2 + m2 )Γ (n − 2  1 D + (gµν gρσ + gµρ gνσ + gµσ gρν )(p2 + m2 )2 Γ (n − − 2) . (11.E) 4 2

62

Problems

• The gamma–function obeys Γ (−n + ) =

(−1)n n!



1 + ψ(n + 1) + o() , 

(11.F)

where n ∈ N and

1 1 + ... + − γ . 2 n The γ = 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 1 1 = dx , (11.G) AB [xA + (1 − x)B]2 0

1

1−x 1 1 =2 dx dz . (11.H) ABC [A + (B − A)x + (C − A)z]3 0 0 ψ(n + 1) = 1 +

• 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 p2

1 → (−2iπ)δ (4) (p2 − m2 )θ(p0 ) . − m2

3. Sum the contributions of all possible cuts.

11.1. Prove the following formula (the Feynman parametrization)

1

1 1 δ(x1 + . . . + xn − 1) = (n − 1)! ... dx1 . . . dxn . A1 . . . An (x1 A1 + . . . + xn An )n 0 0 11.2. Show that expression (11.A) holds. 11.3. Prove the formula (11.F). 11.4. Regularize the integral

1 1 I = d4 k 2 , k (k + p)2 − m2 using Pauli–Villars regularization.

Chapter 11. Renormalization and regularization

11.5. Compute

63



kα kβ kµ kν kρ kσ . (k 2 )n Also, find the divergent part of the previous integral for n = 5. Apply the dimensional regularization. Iαβµνρσ =

dD k

11.6. Consider the interacting theory of two scalar fields φ and χ: L=

1 1 1 1 (∂φ)2 − m2 φ2 + (∂χ)2 − M 2 χ2 − gφ2 χ . 2 2 2 2

(a) Find the self–energy of the χ particle, −iΠ(p2 ). (b) Calculate the decay rate of the χ particle into two φ particles. (c) Prove that Im Π(M 2 ) = −M Γ. 11.7. Consider the theory m2 2 g λ 1 (∂µ φ)2 − φ − φ 3 − φ4 . 2 2 3! 4! Find the expression for the self–energy and the mass shift δm. L=

11.8. The Lagrangian density is given by L=

1 1 m2 2 λ (∂µ σ)2 + (∂µ π)2 − σ − λvσ 3 − λvσπ 2 − (σ 2 + π 2 )2 , 2 2 2 4 2

where σ and π are scalar fields, and v 2 = m 2λ is constant. Classically, π field 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(qµ qν − q 2 gµν )Π(q 2 ) . Prove the following expression:  e2 2m2 2 Im Π(q ) = − 1+ 2 12π q

 4m2 4m2 1− 2 θ 1− 2 . q q

64

Problems

11.11. In scalar electrodynamics two diagrams give contribution to the polarization of vacuum. Using dimensional regularization derive the following expression for the divergent part of the vacuum polarization: ie2 1 (pµ pν − p2 gµν ) . 24π 2  11.12. The Lagrangian density for the pseudoscalar Yukawa theory is given by m2 2 ¯ 1 ¯ 5 ψφ − λ φ4 . φ + ψ(iγµ ∂ µ − M )ψ − ig ψγ L = (∂φ)2 − 2 2 4! (a) Find the superficial degree of divergence for this theory and the corresponding 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. ¯ and δg. (d) Calculate the one–loop vertex correction φψψ (e) Calculate the one–loop vertex correction φ4 and δλ. 11.13. Consider massless two-dimensional QED, the so–called Schwinger model. (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 φ3 theory in six–dimensional spacetime, with the Lagrangian density given by m2 2 g 1 φ − φ3 − hφ . L = (∂φ)2 − 2 2 3! (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 and δm. Use the minimal subtraction (MS) scheme. (d) Calculate the vertex correction and find δg. (e) Derive the relations m0 = m0 (m, g, ) and g0 = g0 (m, g, ).

Part II

Solutions

1 Lorentz and Poincar´ e symmetries

1.1 The square of the length of a four-vector, x is x2 = gµν xµ xν . By substituting x
µ = Λµρ xρ into the condition x
2 = x2 one obtains: gµν Λµ ρ Λν σ xρ xσ = gρσ xρ xσ .

(1.1)

Since (1.1) is valid for any vector x ∈ M4 , we get Λµ ρ gµν Λν σ = gρσ . The previous condition can be rewritten in the following form µ

(ΛT )ρ gµν Λν σ = gρσ ⇒ ΛT gΛ = g ,

(1.2)

and we have obtained the requested expression. Now, we shall show that the Lorentz transformations form a group. If Λ1 and Λ2 are Lorentz transformations then their product, Λ1 Λ2 is Lorentz transformation because it satisfies the condition (1.2): (Λ1 Λ2 )T g(Λ1 Λ2 ) = ΛT2 (ΛT1 gΛ1 )Λ2 = ΛT2 gΛ2 = 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 the group. Taking determinant of the expression (1.2) we obtain detΛ = ±1. Since detΛ = 0 the inverse element Λ−1 exists for every Lorentz matrix. From (1.2) we see that the inverse element is given by Λ−1 = g −1 ΛT g. In the component notation the previous relation takes the following form: (Λ−1 )µ ν = g µρ Λσ ρ gσν = Λν µ . 1.2 By substituting infinitesimal form of the Lorentz transformation into the formula (1.2), one gets: (δρµ + ω µ ρ )gµν (δσν + ω ν σ ) + o(ω 2 ) = gρσ ,

68

Solutions

gρσ + ω µ ρ gµν δσν + ω ν σ gµν δρµ + o(ω 2 ) = gρσ . from which follows that ωρσ + ωσρ = 0 ⇒ ωρσ = −ωσρ . Since the parameters of the Lorentz group ωµν are antisymmetric only six of them are independent, so the Lorentz group is six–parameters group. Moreover the Lorentz group is a Lie group. 1.3 Given relation is in agreement with definitions of the  symbol and determinant. 1.4 From (1.2) follows that δρσ = δµν Λµ ρ Λν σ , so we conclude that δρ
σ = δρσ . In the same way we have 
µνρσ = Λµ α Λν β Λρ γ Λσ δ αβγδ = det(Λ−1 )µνρσ = µνρσ , since detΛ−1 = 1 for the proper orthochronous Lorentz transformations. Thus, Levi-Civita symbol is defined independently of the inertial frame. Note that the components µνρσ are obtained by applying the antisymmetric tensor  on basis vectors e0 , . . . , e3 : (eµ , eν , eρ , eσ ) = µνρσ . The  tensor can be written in the form  = θ0 ∧ θ1 ∧ θ2 ∧ θ3 , where θ µ are basic one-forms. 1.5 The results are given below µνρσ µβγδ = −δβν δγρ δδσ + δγν δβρ δδσ + δβν δδρ δγσ − δγν δδρ δβσ − δδν δβρ δγσ + δδν δγρ δβσ , µνρσ µνγδ = −2(δγρ δδσ − δδρ δγσ ) , µνρσ µνρδ = −6δδσ , µνρσ µνρσ = −24 . 1.6 (a) The matrix X is  X=

x0 − x3 −x1 − ix2

−x1 + ix2 x0 + x3

,

so detX = (x0 )2 − (x)2 = x2 . It is not difficult to see that from the † transformation law, X
= SXS , follows that †

detX
= detSdetXdetS = detX , which means that x
2 = x2 .

Chapter 1. Lorentz and Poincar´e symmetries

69

(b) Multiplying the expression X = xµ σ µ by σ ¯ ν and taking trace we obtain µ the requested relation. The matrices σ satisfy the following orthogonality relation tr[¯ σ µ σ ν ] = 2g µν . 1.7 The result follows from x
µ =

† 1 1 tr(¯ σ µ X
) = xν tr(¯ σ µ Sσν S ) = Λµν xν . 2 2

1.8 An arbitrary Lorentz transformation, which is connectedwith the unit element, can be written in the form U (ω) = exp − 2i Mµν ω µν , where Mµν are generators. There are three (independent) rotations and three (also independent) boosts. Rotation around z−axis for angle θ3 is represented by matrix     1 0 0 0 0 0 0 0 sin θ3 0  0 θ3 0   0 cos θ3 0 Λ(θ3 ) =   ≈I +  . 0 − sin θ3 cos θ3 0 0 −θ3 0 0 0 0 0 1 0 0 0 1 From the previous expression we conclude that ω 12 = −ω12 = θ3 . The generator of this transformation is   0 0 0 0     dΛ(θ3 )  dΛ(θ3 )   0 0 −1 0  =−i = i M12 = i (1.3) . 0 1 0 0 dω 12 ω12 =0 dθ3 θ3 =0 0 0 0 0 In the same way we obtain the other two generators:    0 0 0 0 0  0 0 0 −1  0 M13 = i   , M23 = i  0 0 0 0 0 0 1 0 0 0

 0 0 0 0  . 0 −1 1 0

0 0 0 0

(1.4)

In this case the relation between the parameters ωij and the angles of rotations θi around xi −axis is θi = − 21 ijk ωjk . The matrix of the boost along x−axis is     chϕ1 −shϕ1 0 0 0 −ϕ1 0 0 0 0 0  −shϕ1 shϕ1 0 0   −ϕ1 Λ(ϕ1 ) =  ≈I +  , 0 0 1 0 0 0 0 0 0 0 0 1 0 0 0 0 where ω 01 = −ϕ1 = −arc th v1 . The corresponding generator is

M01

 dΛ(ϕ1 )  =i dω 01 

ϕ1 =0

 dΛ(ϕ1 )  =i dϕ1 



ϕ1 =0

0 1 = −i  0 0

1 0 0 0

0 0 0 0

 0 0  . 0 0

(1.5)

70

Solutions

The other two generators are 

M03

0 0 = −i  0 1

0 0 0 0

0 0 0 0

 1 0 , 0 0



M02

0 0 = −i  1 0

0 0 0 0

1 0 0 0

 0 0 . 0 0

(1.6)

The boost parameters (rapidity) are ωoi = −ϕi = −arc th (vi ), where vi is the velocity of the inertial frame moving along the xi −axis. 1.10 The multiplication rule is (Λ1 , a1 )(Λ2 , a2 ) = (Λ1 Λ2 , Λ1 a2 + a1 ) . Unit element is (I, 0), while the inverse is (Λ, a)−1 = (Λ−1 , −Λ−1 a) . 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 + iµ Pµ )U (Λ, 0) = 1 + i(Λ−1 )µν ν Pµ .

(1.7)

From the expression (1.7) we obtain U −1 (Λ, 0)Pµ U (Λ, 0) = (Λ−1 )νµ Pν .

(1.8)

The formula (1.8) is transformation law of the momentum Pµ under Lorentz transformations; the momentum is a four-vector. By substituting  i i U (ω, 0) = exp − Mµν ω µν = 1 − Mµν ω µν + o(ω 2 ) 2 2 into (1.8) we get i i (1 + Mρσ ω ρσ )Pµ (1 − Mρσ ω ρσ ) = (δµα − ω α µ )Pα , 2 2

(1.9)

iω ρσ (Mρσ Pµ − Pµ Mρσ ) = −ω ρσ (gµσ Pρ − gµρ Pσ ) .

(1.10)

and then

We had to antisymmetrize the right hand side of Equation (1.10) in order to eliminate antisymmetric parameters ω ρσ . Finally, we obtain [Mρσ , Pµ ] = i(gµσ Pρ − gµρ Pσ ) .

(1.11)

(b) If we take an infinitesimal transformation Λ
= I + ω
then (Λ−1 Λ
Λ)µ ν = δνµ + (Λ−1 )µ ρ Λσ ν ω
ρ σ ,

(1.12)

Chapter 1. Lorentz and Poincar´e symmetries

71

so that i i
. (1.13) U −1 (Λ, 0)(1 − ω
ρσ Mρσ )U (Λ, 0) = 1 − Mµν (Λ−1 )µρ Λσν ωρσ 2 2 From the last expression follows U −1 (Λ, 0)Mρσ U (Λ, 0) = (Λ−1 )µ ρ (Λ−1 )ν σ Mµν .

(1.14)

The last equation is the transformation law of the second rank tensor. For an infinitesimal Lorentz transformation Λµν = δνµ + ω µν from Equation (1.14) follows i µν 1 ω [Mµν , Mρσ ] = ω µν (gσµ Mρν − gρν Mµσ − gσν Mρµ + gρµ Mνσ ) , 2 2 or [Mµν , Mρσ ] = i(gσµ Mνρ + gρν Mµσ − gρµ Mνσ − gσν Mµρ ) .

(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 is  0 0 0 0  0 0 −1 0  M12 = i  0 1 0 0  0 0 0 0 0 0 0 0 The time translation generator has the  0 0 0 0  T0 = −i  0 0  0 0 0 0

of the rotation around z–axis  0 0  0 .  0 0

form 0 0 0 0 0

0 0 0 0 0

 1 0  0 .  0 0

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
= Λx + a ≈ x + δx , a classical scalar field transforms as follows φ
(x + δx) = φ(x) .

72

Solutions

From the last relation we have φ
(x) = φ(x − δx) = φ(x) − δxµ ∂µ φ .

(1.17)

Form variation of a scalar field is given by δ0 φ = φ
(x) − φ(x) = −δxµ ∂µ φ .

(1.18)

For the Lorentz transformation δxµ = ω µν xν , and therefore 1 δ0 φ = −ω µν xν ∂µ φ = − ω µν (xν ∂µ − xµ ∂ν )φ . 2

(1.19)

On the other hand

i δ0 φ = − ω µν Mµν φ . (1.20) 2 By comparing two previous results we get that Lorentz’s generators are Mµν = i(xµ ∂ν − xν ∂µ ) .

(1.21)

For translations δxµ = µ and δ0 φ = −µ ∂µ φ = iµ Pµ φ .

(1.22)

Pµ = i∂µ .

(1.23)

[xµ ∂ν , xρ ∂σ ] = gνρ xµ ∂σ − gσµ xρ ∂ν ,

(1.24)

[xµ ∂ν , ∂ρ ] = −gρµ ∂ν

(1.25)

Hence Since and we get the commutation relations of the Poincar´e algebra: [Pµ , Pν ] = 0 [Mρσ , Pµ ] = i(gµσ Pρ − gµρ Pσ ) [Mµν , Mρσ ] = i(gσµ Mνρ + gρν Mµσ − gρµ Mνσ − gσν Mµρ ) . 1.14 (a) Wµ P µ = 12 µνρσ M νρ P σ P µ = 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 1 µνρσ µαβγ M νρ P σ Mαβ Pγ 4   1 = µνρσ µαβγ M νρ Mαβ P σ − iδβσ Pα + iδασ Pβ Pγ 4 1 = µνρσ µαβγ M νρ Mαβ P σ Pγ . 4

W2 =

(1.26)

Chapter 1. Lorentz and Poincar´e symmetries

73

The contraction of two  symbols in the last line of (1.26) has been calculated in 1.5 so that: 1 W 2 = − (δνα δρβ δσγ + δνβ δργ δσα + δνγ δρα δσβ − δνβ δρα δσγ − δνα δργ δσβ − δνγ δρβ δσα ) 4 × M νρ Mαβ P σ Pγ 1 = − 2M νρ Mνρ P 2 − M νρ Mνσ P σ Pρ + M νρ Mσν P σ Pρ + 4 + M νρ Mρσ P σ Pν − M νρ Mσρ P σ Pν ) 1 (1.27) = − M νρ Mνρ P 2 + M νρ Mνσ P σ Pρ . 2 (c) Using the previous result we have 1 [W 2 , Mρσ ] = − [M µν Mµν P 2 , Mρσ ] + [Mµα M να P µ Pν , Mρσ ] . 2

(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 P 2 and Mµν M µν are Lorentz scalars. The commutator B is B = Mµα M να (P µ [Pν , Mρσ ] + [P µ , Mρσ ]Pν ) + +Mµα [M να , Mρσ ]P µ Pν + [Mµα , Mρσ ]M να P µ Pν .

(1.29)

Using the commutation relations (1.11) and (1.15) we get B = 0. Therefore, we have [W 2 , Mρσ ] = 0 . 1.15 By using the result of Problem 1.14 (b) and P µ |pµ , s, σ = pµ |pµ , s, σ we get  1 µν M Mµν − M0i M 0i |p = 0, m, s, σ W 2 |p = 0, m, s, σ = −m2 2 1 = − Mij M ij m2 |p = 0, m, s, σ 2   = −m2 (M12 )2 + (M13 )2 + (M23 )2 |p = 0, m, s, σ = −m2 J2 |p = 0, m, s, σ = −m2 s(s + 1) |p = 0, m, s, σ , because Ji = 12 ijk Mjk are the components of the angular momentum tensor. 1.16 (a) Under Lorentz transformations Wµ transforms according to: U −1 (Λ)Wσ U (Λ) = Λσ α Wα . From Equation (1.30) we have

(1.30)

74

Solutions

i 1 [Mµν , Wσ ]ω µν = ω µν gσµ Wν = (gσµ Wν − gσν Wµ )ω µν . 2 2 From the previous expression we easily obtain the requested result. (b) Using the result of the previous part we have 1 µαβγ [M αβ P γ , Wν ] 2   1 = µαβγ M αβ [P γ , Wν ] + [M αβ , Wν ]P γ 2 = iµανγ W α P γ .

[Wµ , Wν ] =

1.17 (a) Applying the result of Problem 1.16 (a) we get [Wµ , M 2 ] = −2i(W α Mαµ + Mαµ W α ) . (b) [Mµν , W µ W ν ] = 0. Take care that δµµ = 4. (c) Using the formula (1.11) we obtain [M 2 , Pµ ] = 2i(P α Mαµ + Mαµ P α ) . This result and the result in the first part of this Problem are similar, since Wµ and Pµ are both four-vectors. (d) [µνρσ Mµν Mρσ , Mαβ ] = 0. 1.18 In the case of massive particles, m2 > 0 since the Lorentz transformations, Λµ ν = δνµ + ω µ ν leave pµ invariant (i.e. Λµ ν pν = pµ ) the following relation is satisfied:      m ω03 0 ω01 ω02 0 0 −ω12 −ω13   0   0   ω01   =   .  0 ω02 ω12 0 −ω23 0 ω03 ω13 ω23 0 0 0 From here follows ω01 = ω02 = ω03 = 0, ωij = 0 . The corresponding generators are M 12 , M 13 and M 23 and they are generators 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      k ω03 0 ω01 ω02 0 0 −ω12 −ω13   0   0   ω01   =   ,  0 ω02 ω12 0 −ω23 0 ω03 ω13 ω23 0 k 0 which gives ω03 = 0, ω01 = ω13 , ω02 = ω23 while the parameter ω12 is 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 M 01 + M 13 and −(M 02 + M 23 ). Note that W1 = (M 02 + M 23 )k , W2 = −(M 01 + M 13 )k as well as W0 = −M 12 k. Then, using Problem 1.16 (b) we obtain [W1 , W2 ] = 0,

[W0 /k, W1 ] = −iW2 ,

[W0 /k, W2 ] = 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 µ µν µ 1 U (ω, , ρ, c) = ei(Pµ  − 2 Mµν ω +ρD+cµ K ) , where D is generator of dilatation, and K µ are four generators for SCT . Conformal group has 15 parameters. The commutation relations of the algebra can 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−1 c) for infinitesimal SCT we obtain U −1 (Λ)Kρ U (Λ) = (Λ−1 )µ ρ Kµ . For infinitesimal Lorentz transformations we get: [Mµν , Kρ ] = i(gνρ Kµ − gµρ Kν ).

(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−ρ xµ e−ρ xµ + cµ e−2ρ x2 = (I, 0, ρ, 0)−1 1 + 2(c · x)e−ρ + c2 e−2ρ x2 µ x + cµ e−ρ x2 = 1 + 2(c · x)e−ρ + c2 e−2ρ x2 = (I, 0, 0, e−ρ c)xµ , we obtain

e−iρD (1 + iK µ cµ )eiρD = 1 + iK µ e−ρ cµ ,

for infinitesimal SCT. From the last expression follows e−iρD K µ eiρD = e−ρ K µ .

76

Solutions

This is the transformation law of SCT generators under dilatation. For infinitesimal dilatations we get: [D, K µ ] = −iK µ .

(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(gµν D + Mµν ).

(1.37)

Equations (1.31)–(1.37) together with (1.11), (1.15) and (1.16) are commutation relations of the conformal algebra.

2 The Klein–Gordon equation

2.1 A particular solution of the Klein–Gordon equation

is plane wave,

(  + m2 )φ(x) = 0 ,

(2.1)

e−ik·x = e−iEt+ik·x ,

(2.2)

where E and k are energy and momentum respectively. We see that from i and

∂ −ik·x e = Ee−ik·x , ∂t

−i∇e−ik·x = ke−ik·x .

By√ inserting the solution (2.2) into (2.1) we obtain k 2 = m2 i.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 k there are two independent solutions e−iωk t+ik·x and +iωk t+ik·x . The general solution of (2.1) is e

 1 d3 k  −i(ωk t−k·x) † i(ωk t+k·x) √ φ(x) = + b (−k)e a(k)e , (2.3) (2π)3/2 2ωk where a(k) and 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

 1 d3 k  √ a(k)e−ik·x + b† (k)eik·x , (2.4) φ(x) = 3/2 (2π) 2ωk where k µ = (ωk , k). If φ(x) is a real field then a(k) = b(k). 2.2 Using (2.4) we get

78

Solutions



∂φ∗ −φ Q = iq d x φ ∂t ∂t

3 3 3
 d xd kd k  † q a (k)eik·x + b(k)e−ik·x =i √ 2(2π)3 ωk ωk
   

× −iωk
a(k
)e−ik ·x + iωk
b† (k
)eik ·x − a(k)e−ik·x + b† (k)eik·x  !

× iωk
a† (k
)eik ·x − iωk
b(k
)e−ik ·x . (2.5)

3

∗ ∂φ

By integrating over x in (2.5), we obtain 

ωk
 † q 3 3
−a (k)a(k
)ei(ωk −ωk
)t δ (3) (k − k
) Q=− d kd k 2 ωk + a† (k)b† (k
)ei(ωk +ωk
)t δ (3) (k + k
) − b(k)a(k
)e−i(ωk +ωk
)t δ (3) (k + k
)  + b† (k)b(k
)e−i(ωk −ωk
)t δ (3) (k − k
) + c.c. . (2.6) where c.c. denotes complex conjugation. If in expression (2.6) we integrate over the momentum k
we obtain

  q Q= d3 k a† (k)a(k) + a(k)a† (k) − b† (k)b(k) − b(k)b† (k) . (2.7) 2 In the result (2.7) we do not take care about ordering of a(k), a† (k) and b(k), b† (k) since they are complex numbers. This will be different in the Chapter 7 where a(k) and b† (k) are going to be operators. 2.3 If we first integrate over x we get

3 3
 1 d kd k H=− a(k)a(k
)(ωk ωk
+ k · k
− m2 )e−i(ωk +ωk
)t δ (3) (k + k
) √ 4 ωk ωk
+ a† (k)a† (k
)(ωk ωk
+ k · k
− m2 )ei(ωk +ωk
)t δ (3) (k + k
) − a(k)a† (k
)(ωk ωk
+ k · k
+ m2 )e−i(ωk −ωk
)t δ (3) (k − k
)  − a† (k)a(k
)(ωk ωk
+ k · k
+ m2 )ei(ωk −ωk
)t δ (3) (k − k
) .

(2.8)

Performing integration over momentum k
, and using the relation k2 + m2 = ωk2 , we obtain

  1 (2.9) H= d3 kωk a† (k)a(k) + a(k)a† (k) . 2 2.4 Solution of this problem is very similar to the solutions of the previous two. The result is

P = d3 kka† (k)a(k) . 2.5 The four-divergence of the current j µ is

Chapter 2. The Klein–Gordon equation

79

i ∂µ j µ = − (∂µ φ∂ µ φ∗ + φ φ∗ − ∂µ φ∂ µ φ∗ − φ∗  φ) . 2 Using the equations of motion we obtain the requested result ∂µ j µ = 0. 2.6 It is easy to see that i ∂µ j µ = − (∂µ φ∂ µ φ∗ + φ φ∗ − ∂µ φ∂ µ φ∗ − φ∗  φ) − 2 − q(φAµ ∂µ φ∗ + φφ∗ ∂µ Aµ + φ∗ Aµ ∂µ φ) . The equations of motion are    − iq(∂µ Aµ + 2Aµ ∂µ − iqAµ Aµ ) + m2 φ∗ (x) = 0 ,     + iq(∂µ Aµ + 2Aµ ∂µ + iqAµ Aµ ) + m2 φ(x) = 0 . 

(2.10)

(2.11) (2.12)

If we multiply Equation (2.11) by φ and Equation (2.12) by φ∗ and then subtract obtained equations we get φ φ∗ − φ∗  φ − 2iq(φφ∗ ∂µ Aµ + Aµ φ∗ ∂µ φ + Aµ φ∂µ φ∗ ) = 0 . Combining the previous expression and (2.10), one easily obtains ∂µ j µ = 0 . 2.7 The equation of motion for a scalar particle in a electromagnetic field is   (∂µ + iqAµ )(∂ µ + iqAµ ) + m2 φ(x) = 0 . (2.13) In the region r < a Equation (2.13) becomes    ∂ ∂ 2 − iV − iV − ∆ + m φ(x) = 0 . ∂t ∂t For stationary states φ(x) = e−iEt F (r) one gets   −(E + V )2 − ∆ + m2 F (r) = 0 .

(2.14)

(2.15)

If we assume that a solution of the previous equation is given by F =

f (r) Q(θ, ϕ) , r

then from (2.15) we get the following two equations:  d2 f  l(l + 1) + (E + V )2 − m2 f = f , 2 dr r2  1 ∂ ∂Q 1 ∂2Q = −l(l + 1)Q . sin θ + sin θ ∂θ ∂θ sin2 θ ∂ϕ2

(2.16) (2.17)

80

Solutions

The particular solutions of (2.17) are spherical harmonics, Ylm . In the case l = 0, the corresponding spherical harmonic Y00 is a constant. The solution of (2.16) is f = A sin(qr) + B cos(qr) , (2.18) where q 2 = [(E + V )2 − m2 ] > 0 .

(2.19)

Constant B has to be zero since function f (r)/r should 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 k 2 = m2 − E 2 . But, the constant D has to be zero since the wave function has to be finite in the large r limit. Therefore, the wave function is φ< = A

sin qr , r

ra. (2.22) r At r = a we should apply the continuity conditions: φ< (a) = φ> (a) and φ
< (a) = φ
> (a) for the wave function and its first derivative. These boundary conditions give: (2.23) A sin(qa) − Ce−ka = 0 , φ> = C

Aq cos(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) 1 =− . q 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)/q while dashed line is f (q) = −

1 = − k

2V



1

.

q 2 + m2 − V 2 − q 2

There is only one bound state (in case V < 2m) if the condition π 3π < V (V + 2m) ≤ . 2a 2a is satisfied. 2.8 The wave equation is

Chapter 2. The Klein–Gordon equation

81

Fig. 2.1. Graphical solution of the dispersion relation (2.25) for V < 2m



∂2 − ∂t2



∂ + iqBy ∂x

2

 ∂2 ∂2 2 − 2 − 2 + m φ(x) = 0 . ∂y ∂z

(2.26)

∂ ∂ and pˆz = −i ∂z commute with It is easy to see that the operators pˆx = −i ∂x the Hamiltonian, so we can assume that the solution of (2.26) has the following form (2.27) φ = e−i(Et−kx x−kz z) ϕ(y) .

From (2.26) and (2.27) we get  2 d 2 2 2 2 − (kx + qBy) + E − kz − m ϕ(y) = 0 . dy 2

(2.28)

Introducing the new variable ξ = kx + qBy, Equation (2.28) takes the same form as the Schr¨ odinger equation for the oscillator  2 d 1 E 2 − kz2 − m2 2 − ξ + ϕ(ξ) ˜ =0. dξ 2 (qB)2 (qB)2 Then the energy levels are En = m2 + kz2 + (2n + 1)qB ,

n = 0, 1, 2, . . . .

Eigenfunctions are 2 1 kx + qBy φn (x) = (qπB)−1/4 √ e−iEn t+ikx x+ikz z e−(kx +qBy) /2qB Hn ( √ ), qB 2n n! (2.29) where Hn are the Hermite polynomials.

2.9 In the region z > 0 the equation of motion is   ∂  − q 2 U02 + 2iqU0 + m2 φII (x) = 0 .  ∂t

(2.30)

82

Solutions

Substituting φII = Ce−iEt+ikz in (2.30), we get k = ±K = ± (E − qU0 )2 − m2 ,

(2.31)

E = ± k 2 + m2 + qU0 .

or

(2.32)

For z < 0 the particle is free and the solution is φI = Ae−iEt+ipz + Be−iEt−ipz ,

(2.33) √ where p = E 2 − 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
I (0) = φ
II (0) . They give 1 A= 2



k 1+ p

C,

1 B= 2



k 1− p

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    p − K 2 |B|2 −(jr )z  ,  = = R= (jin )z |A|2 p+K 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:    p + K 2  , T =1−R . R =  p−K As we immediately see the reflection coefficient is larger than 1: the potential is strong enough to create particle–antiparticle pairs. The antiparticles are moving to the right producing a negative charge current and therefore we obtain negative transmission coefficient. This is Klein paradox Case 3: |E − qU0 | < m. We leave to the reader to show that in this case R = 1, T = 0 . 2.10 For z < 0 and z > 0 a wave function satisfies the free Klein–Gordon equation, while in the region 0 < z < a the equation is   ∂ 2 2 2  − q U0 + 2iqU0 + m φII (x) = 0 .  ∂t 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 = F e−iEt+ipz , (2.35) √ where k = (E − qU0 )2 − m2 and p = E 2 − m2 . From the continuity conditions follows: A+B = C +D , k A − B = (C − D) , p ika −ika Ce + De = F eipa , p Ceika − De−ika = F eipa . k

(2.36)

Thus, one gets:  F 2   T =  = A |2 +

p k

+

k p

16 + (2 −

p k

− kp )e2ika |2

.

If (E − qU0 )2 − m2 < 0 the momentum k becomes imaginary, i.e. k = iκ = i m2 − (E − qU0 )2 . 2.11 The Klein–Gordon equation for a particle in the Coulomb potential is   2 ∂ Ze 2 − ie − ∆ + m φ(x) = 0 . (2.37) ∂t r By substituting φ = e−iEt R(r)Y (θ, ϕ) in (2.37) and using (2.17) we obtain: −

E 2 − m2 l(l + 1) − Z 2 e4 Ze2 E 1 1 d2 R= R. (rR) + R− 2 2 2m r dr 2mr mr 2m

This equation has the same form as the Schr¨odinger equation for hydrogen atom. By comparing these equations we get 1 En,l = m  .  1 1 2 2 4 2 4 1 + Z e (n − l − 2 ) + (l + 2 ) − Z e In the nonrelativistic limit the result is En − m = −

m mZ 2 e4 − Z 3 e6 3 2n2 n



1 3 − 2l + 1 8n



2.12 The Klein–Gordon equation in the Schr¨ odinger form is   ∂ θ θ i =H , χ ∂t χ

.

(2.38)

84

Solutions

where the Hamiltonian is given by    ∆ 1 1 1 H= − +m 0 2m −1 −1

0 −1

 .

2.13 The eigenequation, Hφ = Eφ in the momentum representation takes the following form # " 2  p p2 + m θ0 θ0 2m 2m = E . (2.39) p2 p2 χ χ 0 0 − 2m − 2m − m The eigenvalues of the Hamiltonian are evaluated easily and they are E = ±ωp = ± p2 + m2 . In order to find nonrelativistic limit we suppose that the solution has the following form   θ θ0 = (2.40) e−i(m+T )t , χ χ0 where T is the kinetic energy of the particle. From (2.38) we get      − 2m + m − 2m θ0 θ0 = (m + T ) ,   χ χ − m 0 0 2m 2m i.e.



(2.41)

+ m θ0 − χ0 = (m + T )θ0 , − 2m 2m  θ0 + − m χ0 = (T + m)χ0 . 2m 2m

(2.42)

From the second equation in (2.42) we obtain χ0 ≈

θ0 , 4m2

(2.43)

in nonrelativistic limit. Using this the first equation in (2.42) becomes  2 − T θ0 = − (2.44) θ0 . 2m 8m3 Also, from (2.43) we see that χ0  θ0 and χ is so called small component. From the expression (2.44) follows that first relativistic correction of nonrelativistic Hamiltonian is −∇4 /8m3 . 2.14 Velocity operator is ∂H p v= = ∂p m



1 −1

1 −1

.

The eigenvalue of the velocity operator is zero.     2.15 Show that ψ † |Hψ = Hψ † |ψ . The average value is v =

p m.

3 The γ–matrices

3.1 (a) In the Dirac representation of γ–matrices we have (γ 0 )† = (γ i )† =



0 −σi



I 0 0 −I σi 0

†

 = 

† =−

I 0 0 −I

0 −σi

σi 0

= γ0γ0γ0 = γ0 , = −γ 0 γ 0 γ i = γ 0 γ i γ 0 ,

where we used the facts that (γ 0 )2 = 1, γ 0 and γ i anticommute, and the Pauli matrices are hermitian. This relation is true in any representation of γ–matrices which is obtained by a unitary transformation from the Dirac representation. (b) Using the previous result we find i † = − (γµ γν − γν γµ )† σµν 2 i = − (γν† γµ† − 㵆 γν† ) 2 i = − γ0 (γν γµ − γµ γν )γ0 2 = γ0 σµν γ0 . 3.2 (a) Taking the adjoint of γ5 we obtain γ5† = iγ3† γ2† γ1† γ0† = iγ0 γ3 γ0 γ0 γ2 γ0 γ0 γ1 γ0 γ0 γ0 γ0 = iγ0 γ3 γ2 γ1 = −iγ0 γ1 γ2 γ3 = γ5 .

86

Solutions

The property γ5−1 = γ5 can be proved by using γ0−1 = γ0 and γi−1 = −γi = γ i . Both of these relations follow from anticommutation relations {γµ , γ ν } = 2δµν . (b) Using the definition of the  symbol we find −

i i µνρσ γ µ γ ν γ ρ γ σ = (γ 0 γ 1 γ 2 γ 3 − γ 0 γ 1 γ 3 γ 2 + . . . + γ 3 γ 2 γ 1 γ 0 ) 4! 4! = iγ 0 γ 1 γ 2 γ 3 = γ5 .

(c) This is a consequence of (a) result. (d) In a similar manner, we have: (γ5 γµ )† = 㵆 γ5† = γ0 γµ γ 0 γ5 = γ 0 γ5 γµ γ 0 . 3.3 (a) For µ = 0 we have {γ5 , γ 0 } = γ5 γ 0 + γ 0 γ5 = −iγ0 γ1 γ2 γ3 γ0 − iγ0 γ0 γ1 γ2 γ3 = iγ1 γ2 γ3 − iγ1 γ2 γ3 = 0 ,

(3.1)

and similarly for other three cases. (b) By a straightforward calculation one gets: i [γµ γν − γν γµ , γ5 ] 2 i = (γµ {γν , γ5 } − {γµ , γ5 }γν − γν {γµ , γ5 } + {γµ , γ5 }γν ) 2 =0

[σµν , γ5 ] =

since {γµ , γ5 } = 0. 3.4 / a/ a = aµ aν γµ γν = 12 aµ aν (γµ γν + γν γµ ) = g µν aµ aν = a2 3.5 (a) From the relation {γµ , γ µ } = 2γµ γ µ = 2δµµ = 8 it follows that γµ γ µ = 4. (b) γµ γν γ µ = (2gµν − γν γµ )γ µ = 2γν − 4γν = −2γν . (c) γµ γα γ β γ µ = (2gµα − γα γµ )γ β γ µ = 2γ β γα + 2γα γ β = 4δαβ , where we used the second part of this problem and (3.A). (d) By commuting γµ and γ α and making use of the previous result, one gets: γµ γ α γ β γ γ γ µ = (2δµα − γ α γµ )γ β γ γ γ µ = 2γ β γ γ γ α − 4γ α g βγ = −2(2g βγ − γ β γ γ )γ α = −2γ γ γ β γ α .

Chapter 3. The γ–matrices

87

(e) By using the definition σµν –matrices, one obtains: 1 σµν σ µν = − (γ µ γ ν γµ γν − γ µ γ ν γν γµ − γ ν γ µ γµ γν + γ ν γ µ γν γµ ) . 4 By using parts (a) and (b) of this problem, one gets σµν σ µν = 12. (f) Use Problem 3.3 and parts (a) and (b) of this problem. (g) By direct calculation, one finds 1 σαβ γµ σ αβ = − (γ α γ β γµ γα γβ − γ α γ β γµ γβ γα 4 −γ β γ α γµ γα γβ + γ β γ α γµ γβ γα ) 1 = − (4δµβ γβ − 4γµ − 4γµ + 4gµβ γ β ) = 0 . 4 (h) i σ αβ σ µν σαβ = − (γ α γ β γ µ γ ν γα γβ − γ α γ β γ µ γ ν γβ γα 8 −γ α γ β γ ν γ µ γα γβ + γ α γ β γ ν γ µ γβ γα − γ β γ α γ µ γ ν γα γβ +γ β γ α γ µ γ ν γβ γα + γ β γ α γ ν γ µ γα γβ − γ β γ α γ ν γ µ γβ γα ) i = − (−8γ ν γ µ − 16g µν + 8γ µ γ ν 8 +16g µν − 16g µν − 8γ ν γ µ + 16g µν + 8γ µ γ ν ) = −2i(γ µ γ ν − γ ν γ µ ) = −4σ µν . (i) Use part (g) of this problem. (j) σµν γ5 σ µν = 2i (γµ γν − γν γµ )γ5 σ µν = γ5 σµν σ µν = 12γ5 . 3.6 (a) By using the trace property tr(A1 A2 . . . An ) = tr(A2 A3 . . . An A1 ), Problem 3.3(a), and (γ5 )2 = 1, it follows that tr(γµ ) = tr(γµ γ5 γ5 ) = −tr(γ5 γµ γ5 ) = −tr((γ5 )2 γµ ) = −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(γµ γν γρ γσ ) = tr [(2gµν − γν γµ )γρ γσ ] = 2gµν tr(γρ γσ ) − tr[γν (2gµρ − γρ γµ )γσ ] = 2gµν tr(γρ γσ ) − 2gµρ tr(γν γσ ) + 2gµσ tr(γν γρ ) − tr(γν γρ γσ γµ ) .

88

Solutions

From the previous part of this problem and relation tr(γµ γν γρ γσ ) = tr(γν γρ γσ γµ ), one easily obtains the requested result. (d) trγ5 = tr(γ5 γ0 γ0 ) = −tr(γ0 γ5 γ0 ), where we used Problem 3.3 (a). Further, from the trace property and (γ0 )2 = 1 it follows that: trγ5 = −tr(γ0 γ0 γ5 ) = −trγ5 , which implies trγ5 = 0. (e) Since γα γ α = 4, we have 1 tr(γ5 γ α γα γµ γν ) 4 1 = tr(γα γµ γν γ5 γ α ) 4 1 = − tr(γ5 γα γµ γν γ α ) 4 = −gµν tr(γ5 ) = 0 .

tr(γ5 γµ γν ) =

In the previous calculation we used the trace property and Problem 3.5 (c). (f) The quantity tr(γ5 γµ γν γρ γσ ) 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 = 2 and σ = 3. (g) From (γ5 )2 = 1, {γ5 , γµ } = 0 and the trace property follows: tr(/ a1 . . . / a2n+1 ) = tr(γ5 γ5 /a1 · · · /a2n+1 ) = (−1)2n+1 tr(γ5 /a1 · · · /a2n+1 γ5 ) = −tr(γ5 γ5 /a1 · · · /a2n+1 ) = −tr(/ a1 .../a2n+1 ) . a2n+1 ) = 0 . Hence, tr(/a1 . . . / a2n ) = tr(C/ a1 C −1 C · · · C −1 C/a2n C −1 ) , where the matrix C sat(h) tr(/ a1 · · · / isfies the relation Cγµ C −1 = −γµT . Thus, tr(/ a1 · · · / a2n ) = (−1)2n tr(/ aT1 · · · /aT2n ) = tr(/ a2n · · · /a1 ) . (i) tr(γ5 γµ ) = −itr(γ0 γ1 γ2 γ3 γµ ) = 0, since it is the trace of odd number of γ–matrices. 3.7 a2 · · · / a6 ) = tr(/ a1 / 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

  3.8 4 pµ qν − (p · q)gµν + pν qµ + iαµβν pα q β − m2 gµν . 3.9 −2/p − 2γ5 / p − 4m − 4mγ5 . 3.10 Expanding the exponential function in series, we find 1 1 eγ5 /a = 1 + (γ5 /a) + (γ5 /a)2 + (γ5 /a)3 + · · · . 2 3!

(3.2)

By substituting (γ5 / a)2 = −a2 , (γ5 /a)3 = −a2 (γ5 /a), . . . into expression (3.2), we get a4 a4 a2 a2 + + · · ·) + (γ5 /a)(1 − + − · · ·) eγ5 /a = (1 − 2! 4! 3! 5! √ √ 1 = cos( a2 ) + √ sin( a2 )γ5 /a , a2 where a2 = aµ aµ . 3.11 The fact that the product of any two Γ –matrices is again a Γ matrix (modulo ±1, ±i) can be proved directly. For example, γ5 σ01 = −iσ23 . Now, we shall

prove that Γ –matrices are linearly independent. Multiplying the relation a ca Γ a = 0 by Γb = (Γ b )−1 , we obtain cb Γ b Γb +



ca Γ a Γb = 0 ,

a=b

where the b–term is separated. Using the ordering lemma, the last expression becomes  cd ηΓ d = 0 , (3.3) cb I + d,Γ d =I

where η ∈ {±1, ±i}. After taking trace of (3.3) and using the fact that  0, Γ a = I tr(Γ a ) = , 4, Γ a = I one obtains cb = 0 (∀b). This means that Γ –matrices are linearly independent one.

3.12 Multiplying the equation A = a ca Γ a by Γb from the right and separating the b–term in the sum, we have   ca Γ a Γb = cb I + cd ηΓ d . AΓb = cb Γ b Γb + a=b

d,Γ d =I

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(γµ γν γρ ) = 0 , tr(γµ γν γρ γσ ) = 4(gµν gρσ − gµρ gνσ + gµσ gνρ ) , tr(γµ γν γρ γσ γ5 ) = −4iµνρσ , tr(γµ γν γρ γ5 ) = tr(γµ γν γρ σαβ ) = 0 , follows γµ γν γρ = (gµν gρσ − gµρ gσν + gµσ gρν )γ σ + iσµνρ γ5 γ σ . (b) γ5 γµ γν = gµν γ5 + 12 αβµν σαβ , (c) σµν γρ γ5 = αµνρ γ α − igνρ γ5 γµ + igµρ γ5 γν . 3.14 From Problem 3.13 (a), it follows that {γµ , σνρ } = −2αµνρ γ 5 γ α . 3.15 By applying the result of Problem 3.13 (a) the trace can be transformed as follows tr(γµ γν γρ γσ γα γβ γ5 ) = (gµν gρδ − gµρ gνδ + gµδ gρν )tr(γ δ γσ γα γβ γ5 ) + iδµνρ tr(γ δ γσ γα γβ ) . Using 3.6 (c), (f), we get tr(γµ γν γρ γσ γα γβ γ5 ) = 4i(−gµν ρσαβ + gµρ νσαβ − gρν µσαβ + gαβ σµνρ − gσβ αµνρ + gσα βµνρ ) . 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 [γµ γν , γρ γσ ] = γµ {γν , γρ }γσ − {γµ , γρ }γν γσ + γρ γµ {γν , γσ } − γρ {γµ , γσ }γν = 2gνρ γµ γσ + 2gνσ γρ γµ − 2gµσ γρ γν − 2gµρ γν γσ . From the above result we obtain: [σµν , σρσ ] = 2i(gνρ σµσ + gµσ σνρ − gµρ σνσ − gνσ σµρ ) . The matrices sentation.

1 2 σµν

are generators of the Lorentz group in the spinor repre-

3.18 Let M be a matrix which commutes with all γ–matrices. Using the Problem 3.11, we can write (Γ b = I)

Chapter 3. The γ–matrices

M = cb Γ b +



ca Γ a .

91

(3.4)

a=b

On the other hand, we know that there is always a matrix Γ d which anticommute with Γ b = I. Multiplying the expression (3.4) by matrix Γd from the left, and by Γ d from the right, we get:  Γd M Γ d = −cb Γ b + ηca Γ a . (3.5) a=b

The matrix M commutes with γµ , and therefore with Γ d , so we get  ηca Γ a . M = −cb Γ b +

(3.6)

a=b

If we now multiply equations (3.4) and (3.6) by Γb and take trace of the resulting 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 eB Ae−B = A + [B, A] +

1 [B, [B, A]] + · · · 2!

we get 16 8 βn + (α · n)n + · · · 3! 4! ∞ ∞   (−1)k 22k (−1)k 22k+1 = α+ (α · n)n + βn , (2k)! (2k + 1)!

U αU † = α + 2βn − 2(n · α)n −

k=1

(3.7)

k=0

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 + (U 2 − I)(α · n)n = α + 2βn − 2(α · n)n − = α+

∞  (−1)k 22k k=1

(2k)!

(α · n)n +

8 βn + · · · 3!

∞  (−1)k 22k+1 k=0

(2k + 1)!

It is clear that the results (3.7) and (3.8) are equal.

βn .

(3.8)

92

Solutions

3.20 It is straightforward to show that the γ–matrices satisfy the relation {γµ , γν } = 2gµν . The connection with Dirac representation γµDirac is given by γµ S = SγµDirac .

(3.9)

This  statement is known as the fundamental (Pauli) theorem. If we substitute a b S= , where a, b, c, d are 2 × 2 matrices, into (3.9) we find c d  i    bσ −aσ i c d a −b −σ i c −σ i d . (3.10) = = , σi b dσ i −cσ i σi a a b c −d The solution of (3.10) is a = −b = c = d = I. A particular solution for S is given by  1 I −I S=√ . 2 I I The matrices σµν are   k −σ i 0 σ 0 σoi = −i , σij = ijk , (3.11) 0 σi 0 σk 

while γ 5 = iγ 0 γ 1 γ 2 γ 3 = 

3.21 Matrices 0

1

γ =σ = 

and γ = −iσ = 1

2

−I 0

0 1 1 0 0 1

0 I

.

(3.12)



−1 0



have the following properties: (γ 0 )2 = 1, (γ 1 )2 = −1, γ 0 γ 1 = −γ 1 γ 0 , hence, they satisfy the Clifford algebra (3.A). The matrix γ 5 is defined by  1 0 γ5 = γ0γ1 = . 0 −1 tr(γ 5 γ µ γ ν ) is an antisymmetric tensor and it should be proportional to µν : tr(γ 5 γ µ γ ν ) = Cµν . By fixing µ = 0, ν = 1 we obtain1 C = 2. One can easily show that γ 5 γ µ = µν γν . 1

Our sign convention is 01 = +1 .

4 The Dirac equation

4.1 In terms of α and β matrices, the Dirac Hamiltonian has the form HD = α · p + βm, so that: (a) [HD , p ] = 0, (b) [HD , Li ] = ijk [α · p + βm, xj pk ] = ijk αl [pl , xj ]pk = −iijk αj pk = i(p × α)i , (c) [HD , L2 ] = −iijk αj (Li pk + pk Li ) = 0, (d) [HD , S i ] = − 4i [HD , ijk αj αk ] = iijk pk αj = −i(p × α)i , (e) By applying (b) and (d) we get that this commutator vanishes. (f) [HD , J2 ] = 0, i ˆ ] = − 2|p| ijk pj αk pi = 0, (g) From (d) we have [HD , Σ · p (h) If vectors n and p are collinear, then the commutator vanishes. In the opposite case it is not zero. 4.2 The plane wave

 ϕ ψ= e−ip·x , χ

(4.1)

is a particular solution of the Dirac equation, (iγ µ ∂µ − m)ψ(x) = 0 .

(4.2)

By substituting (4.1) in (4.2) (in the Dirac representation of γ–matrices) we obtain   E − m −σ · p ϕ =0, (4.3) σ · p −E − m χ where E and p are 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 and momentum: E = ± 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 . These relations imply: χ= or

σ·p ϕ, Ep + m

  ϕ ϕ u(Ep , p) = = , σ·p χ Ep +m ϕ

(4.4)

(4.5)

(4.6)

where ϕ is arbitrary. For the negative energy solution, E = −Ep , the system (4.3) is solved by   ϕ − Eσ·p χ +m p = u(−Ep , p) = . (4.7) χ χ If we introduce the notation v(p) = u(−Ep , −p) and 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 solution. The energy and momentum of the solution u(p)e−ip·x are Ep and p, respectively, while for v(p)eip·x , they are −Ep and −p. In order to find the ˆ, additional degrees of freedom, let us recall that the helicity operator 12 Σ · p ˆ where p = p/|p|, commutes with the Dirac Hamiltonian [see Problem 4.1 (g)]. From the eigenequation ˆ ϕ = ±ϕ , σ·p (and a similar equation for χ) we obtain   1 1 p2 pˆ3 + 1 −ˆ p1 + iˆ , ϕ2 = ϕ1 = , (4.8) p2 pˆ3 + 1 2(1 + pˆ3 ) pˆ1 + iˆ 2(1 + pˆ3 ) (and similarly for χr , r = 1, 2). If we take p = pez , the basis vectors become   1 0 , . (4.9) 0 1 Then, the basis bispinors are

Chapter 4. The Dirac equation

95

     1 0   0  1   , u2 (p) = Np  , u1 (p) = Np    σ·p  1 0  σ·p Ep +m Ep +m 0 1       (4.10) 0 1 σ·p σ·p  Ep +m 1   Ep +m 0   , v2 (p) = Np  ,   v1 (p) = Np      0 1 1 0  Ep +m where Np = is the normalization factor. Do not forget that p = pez 2m i.e. p · σ = pσ3 . In this case, the bispinors (4.10) form the helicity basis. For arbitrary momentum p we 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 u and v are normalized according to (4.D). General solution of (4.2) is given by 

ψ=

 2

  † m  1 3 ur (p)cr (p)e−ip·x + vr (p)dr (p)eip·x . (4.11) p d 3/2 Ep (2π) r=1

The Dirac spinor (bispinor) ψ contains 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). ∂ 4.3 The states us (p), vs (p) are eigenstates of the energy operator, i ∂t with eigenvalues Ep and −Ep , respectively.

4.4 By using the expressions for the Dirac spinors found in Problem 4.2, we obtain

ur (p) = r ur (p)¯ " # † † ϕ 1 ϕ1 + ϕ 2 ϕ2 −(ϕ1 ϕ†1 + ϕ2 ϕ†2 ) Eσ·p Ep +m +m p , † † † † σ·p σ·p σ·p 2m Ep +m (ϕ1 ϕ1 + ϕ2 ϕ2 ) − Ep +m (ϕ1 ϕ1 + ϕ2 ϕ2 ) Ep +m where ϕr (r = {1, 2}) are given by (4.8). They satisfy the completeness relation ϕ1 ϕ†1 + ϕ2 ϕ†2 = I. Using also (p · σ)2 = p2 = Ep2 − m2 , we get 2 

1 ur (p)¯ ur (p) = 2m r=1



Ep + m σ·p

−σ · p −Ep + m

=

/p + m . 2m

The second identity can be shown in a similar manner. 4.5 Using the expressions for the projectors given in Problem 4.4, we see that

96

Solutions

Λ2+ =

1 (/ p2 + 2m/p + m2 ) = Λ+ , 4m2

where we have used /p2 = p2 = m2 . Similarly, we obtain Λ2− = Λ− . Orthogonality 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 1 (/ p + m)ur (p) = (m + m)ur (p) = ur (p) , 2m 2m

1 1 (/ p − m)ur (p) = (m − m)ur (p) = 0 . 2m 2m Similarly, one can prove the identities Λ− vr (p) = 0, Λ+ vr (p) = vr (p). Λ− ur (p) =

4.6 (a) We can directly prove this property. For example, the x–component of the vector Σ is i Σ 1 = (γ 2 γ 3 − γ 3 γ 2 ) = iγ 2 γ 3 . 2 On the other hand, γ5 γ0 γ 1 = iγ1 γ2 γ3 γ 1 = iγ 2 γ 3 . The corresponding identities for the y and z–components can be proven in a similar way. (b) By applying the definition of Σ, we have 1 [Σ i , Σ j ] = − ilm jpq [γ l γ m , γ p γ q ] 4  1 ilm jpq  l m p q [γ γ , γ ]γ + γ p [γ l γ m , γ q ] . =−   4

(4.12)

Next step is to expand the commutators in terms of the anticommutators:  1 [Σ i , Σ j ] = − ilm jpq γ l {γ m , γ p }γ q − {γ l , γ p }γ m γ q 4  +γ p γ l {γ m , γ q } − γ p {γ l , γ q }γ m .

(4.13)

Then, using anticommutation relations (3.A) we get   1 [Σ i , Σ j ] = − ilm jpq g mp γ l γ q − g lp γ m γ q + g mq γ p γ l − g lq γ p γ m . 2 (4.14) The first term in (4.14) has the form ilm jpq g mp γ l γ q = (δ ij δ lq − δ iq δ lj )γ l γ q = −3δ ij − γ j γ i . Others terms in (4.14) can be transformed in the same way. Finally, [Σ i , Σ j ] = γ j γ i − γ i γ j .

Chapter 4. The Dirac equation

97

On the other hand, 2iijk Σ k = −ijk klm γ l γ m = γ j γ i − γ i γ j , so that [Σ i , Σ j ] = 2iijk Σ k . We conclude that operators 12 Σ are the generators of SU(2) subgroup of the Lorentz group1 (c) S 2 = − 14 Σ2 = − 14 (γ5 γ0 γ)2 = 14 γ · γ = − 34 . ˆ ϕr = (−1)r+1 ϕr and σ · p ˆ χr = (−1)r χr from 4.7 Use the expressions σ · p Problem 4.2. For example:  ϕr Σ·p Σ·p ur (p) = N σ·p |p| |p| Ep +m ϕr   ϕr ˆ σ·p 0 =N σ·p ˆ 0 σ·p Ep +m ϕr  ˆ ϕr σ·p = N (σ·p)(σ·ˆp) Ep +m ϕr  ϕr = (−1)r+1 N σ·p Ep +m ϕr = (−1)r+1 ur (p) , where N is the normalization factor. It is easy to see that the spinors ur (p) and vr (p) are not eigenspinors of the operator Σ · n, unless vectors n and p are parallel. 4.8 The transformation operator from the rest frame to the frame moving 03 i along the z–axis with velocity v, is S(Λ(vez )) = e− 2 ω03 σ . By using the relation ω03 = −ϕ = − arctan(v), we obtain ϕ  ϕ 0 σ3 I − sinh S(Λ) = cosh σ3 o 2 2   3 I − Epσ Ep + m p +m . = pσ3 − Ep +m I 2m For arbitrary boost, σ3 p should be replaced by σ · p. The operator S(Λ) is not unitary one. Since the Lorentz group is noncompact, it does not have finite dimensional irreducible unitary representations. 4.9 In this case we have      cos θ2 + i sin θ2 σ 3 S= 0 1

Recall that Σ k = 12 kij σ ij .

0   θ cos 2 + i sin θ2 σ 3

.

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 µνρσ 1 i  (ixν ∂ρ − ixρ ∂ν + σνρ )i∂σ = µνρσ σνρ ∂σ , 2 2 4

(4.15)

since the product of a symmetric and an antisymmetric tensors vanishes. Then 1 µνρσ  µαβγ σνρ σ αβ ∂σ ∂ γ ψ(x) 16 1  ν ρ σ δ δ δ − δαν δβσ δγρ + δαρ δβσ δγν − = 16 α β γ  − δαρ δβν δγσ + δασ δβν δγρ − δασ δβρ δγν σνρ σ αβ ∂σ ∂ γ ψ(x)

W 2 ψ(x) = −

 1  αβ 2σ σαβ   − 4σ αγ σαρ ∂ ρ ∂γ ψ 16 3 ψ =  4 3 = − m2 ψ , 4 =

where we used identity σµσ σ µν = 2γσ γ ν + δσν 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 Wµ sµ 1 1 = µνρσ σ νρ P σ sµ = γ5 σµσ sµ ∂ σ m 4m 2m i 1 1 γ5 (γµ γσ − gµσ )(∓ipσ )sµ = ± γ5 /s/p = γ5 /s . = 2m 2m 2 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 0 0 /p vector sµ becomes (0, n), so s/ = −n · γ, and we can use m = p mγ = γ 0 , so that W ·s 1 1 = ± γ5 γ0 n · γ = ± Σ · n . m 2 2 where Problem 4.6 has been used. 4.12 Positive energy solutions satisfy γ5 / su(p, ±s) = ±u(p, ±s) .

(4.16)

p If we choose that polarization vector sµ in the rest frame equals (0, n = |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: "E # pj p 0 m m sµ = j pi pi pj n δij + m(Ep +m) m  p·n m = . (n·p)p n + m(E p +m) For n = p/|p| we get sµ = ( |p| m,

Ep m n).

Using that, we find

1 γ5 / s/ pu(p, ±s) m  |p| Ep 1 γ0 − γ · n (Ep γ0 − p · γ)u(p, ±s) . = γ5 m m m

γ5 / su(p, ±s) =

If we insert (p · γ)2 = −p2 in the previous formula we obtain: γ5 / su(p, ±s) = γ5 γ0 γ ·

Σ·p p u(p, ±s) = u(p, ±s) . |p| |p|

(4.17)

From the expressions (4.16) and (4.17) we get Σ·p u(p, ±s) = ±u(p, ±s) . |p| The similar procedure can be done for negative energy solutions. Starting from γ5 / sv(p, ±s) = ±v(p, ±s) , one gets

Σ·p v(p, ±s) = ∓v(p, ±s) . |p|

4.13 In the ultrarelativistic limit, m  Ep , the vector sµ is given by  Ep p pµ µ , . s ≈ ≈ m m m Then we have γ5 / su(p, ±s) ≈ γ5

/p u(p, ±s) = γ5 u(p, ±s) , m

(4.18)

where we used the Dirac equation p /u(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 γ5 u(p, ±s) = ±u(p, ±s) .

100

Solutions

For v spinors 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 /s and p / is [γ5 /s, /p] = γ5 /s/p − /pγ5 /s = γ5 (/ s/p + p //s) µ ν = γ5 s p {γµ , γν } = 2s · pγ5 = 0 . s)2 = −s2 = 1 it follows that eigenvalues of the operator γ5 /s are ±1. From (γ5 / Then the eigen projectors are Σ(±s) =

1 ± γ5 /s . 2

4.15 The average value of Σ · n in state   ϕ Ep + m ψ(x) = e−ip·x , σ·p 2m Ep +m ϕ is

(4.19)



d3 xψ † (x)Σ · nψ(x)  d3 xψ † (x)ψ(x)  Ep + m ϕ† (σ · p)(σ · n)(σ · p)ϕ † = ϕ σ · nϕ + . 2Ep (Ep + m)2

Σ · n =

(4.20)

Since (σ · A)(σ · B) = A · B + i(A × B) · σ

(4.21)

(σ · p)(σ · n)(σ · p) = |p|2 (n3 σ3 − n2 σ2 − n1 σ1 ) .

(4.22)

it follows that

By substituting (4.22) into (4.20) we get: 1 + |b|2   Ep + m  n3 |a|2 + (n1 + in2 )b∗ a + (n1 − in2 )a∗ b − n3 |b|2 × 2Ep   Ep − m  2 ∗ ∗ 2 n3 |a| + (−n1 + in2 )a b − (n1 + in2 )b a − n3 |b| + . 2Ep

Σ · n =

|a|2

In the nonrelativistic limit we obtain Σ · n = 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

 ϕ 4.16 In the rest frame a spinor takes the following form e−imt , where 0 ϕ satisfies   1 1 ϕ ϕ Σ·n = . 0 2 2 0 The last condition becomes    cos θ −i sin θ a a = , i sin θ − cos θ b b  a where we put ϕ = . From the last expression we obtain b  cos θ2 . ϕ= i sin θ2 In the rest frame the Dirac spinor takes the form   cos θ2  i sin θ  −imt 2  . ψ0 =   0 e 0

(4.23)

(4.24)

(4.25)

Applying the boost along z−axis, we obtain ψ(x) = S(−pez )ψ0 ,

(4.26)

where S is given in Problem 4.8. Note a minus sign appearing in S(−pez )! After a simple calculation, we obtain   cos θ2  θ  −ip·x Ep + m  i sin   2 θ  e . (4.27) ψ(x) =  cos 2  2m p·σ Ep +m i sin θ2 The mean value of the operator 12 γ5 /s is  % $ 1 1 d3 xψ † γ5 /sψ  γ5 /s = , 2 2 d3 xψ † ψ

(4.28)

where the vector sµ is obtained from (0, n) by the Lorentz boost along the z–axis. The components of vector sµ are (see Problem 4.12) s0 = In our case we have

(n · p)p n·p , s=n+ . m m(Ep + m)

102

Solutions

 µ

s =

p Ep cos θ, 0, sin θ, cos θ m m

.

Thus, in the Dirac representation of γ–matrices, γ5 /s is given by  s · σ −s0 I s= γ5 / , s0 I −s · σ

(4.29)

and finally  Ep   s= γ5 / 

cos θ i sin θ p m cos θ 0 m

−i sin θ E − mp cos θ 0 p m cos θ

 0  p −m cos θ   . i sin θ 

p −m cos θ 0 E − mp cos θ −i sin θ

Ep m

(4.30)

cos θ

By substituting (4.30) and (4.27) in the formula (4.28), we obtain: $ % 1 1 γ5 /s = , 2 2 as we expected, because ψ(x) is the eigenstate of the operator eigenvalue 12 .

1 s, 2 γ5 /

with

4.17 The Dirac Hamiltonian can be rewritten in terms of γ–matrices so that [HD , γ5 ] = [γ 0 γ · p + γ 0 m, γ5 ] = 2mγ 0 γ5 . Thus, the operator γ5 is a constant of motion in the case of massless Dirac particle. Its eigenvalues and eigen projectors are ±1, Σ± = 12 (1 ± γ5 ), respectively. The operator γ5 is known as the chirality operator. 4.18 By multiplying the Dirac equation from the left by γ5 , we obtain (i/ ∂+ m)γ5 ψ = 0. By adding and subtracting the previous equations and the Dirac equation, we get i/ ∂ ψL − mψ R = 0, i/ ∂ ψR − mψL = 0 . 4.19 (a) The system of equations can be rewritten as the Dirac equation. The Dirac spinor takes form  ψL ψ= , ψR 

while γµ =

are γ–matrices (see Problem 3.20).

0 σ ¯µ

σµ 0

,

Chapter 4. The Dirac equation

103

(b) In order to be covariant, these equations have to have the following form
iσ µ ∂µ
ψR (x
) = mψL
(x
) ,

(4.31)


i¯ σ µ ∂µ
ψL
(x
) = mψR (x
) ,

(4.32)

in the primed frame (x
= Λx). If we assume that the new spinors take
(x
) = SR ψR (x), where SL and SR are the form ψL
(x
) = SL ψL (x) and ψR nonsingular 2 × 2 matrices, Equations (4.31) and (4.32) become iσ µ SR Λµ ν ∂ν ψR (x) = mSL ψL (x) , i¯ σ µ SL Λµ ν ∂ν ψL (x) = mSR ψR (x) . By multiplying Equation (4.33) by from left we obtain

SL−1

from left, and (4.34) by

(4.33) (4.34) −1 SR

also

iSL−1 σ µ SR Λµ ν ∂ν ψR (x) = mψL (x) ,

(4.35)

−1 µ iSR σ ¯ SL Λµ ν ∂ν ψL (x) = mψR (x) .

(4.36)

The system of equations is covariant if the conditions −1 µ SR σ ¯ SL = Λ µ ν σ ¯ν ,

SL−1 σ µ SR = Λµ ν σ ν hold. The solution for matrices SL and SR is given as  1 i i 1 ϕi σ i + θ k σ k ≈ 1 + ϕi σ i + θ k σ k , SL = exp 2 2 2 2  1 i i 1 SR = exp − ϕi σ i + θk σ k ≈ 1 − ϕi σ i + θk σ k . 2 2 2 2

(4.37)

(4.38)

The parameters θi and ϕi were defined in Problem 1.8. Boost along the x–axis is defined by : ϕ  ϕ  1 1 + σ1 sinh (4.39) SL = cosh 2 2 ϕ  ϕ  1 1 − σ1 sinh . (4.40) SR = cosh 2 2 Note that ψL and ψR transform in the same way under rotations, but differently under boosts. The left ψL , and right ψR spinors transform under ( 12 , 0) and (0, 12 ) irreducible representation of the Lorentz group respectively.

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 = − mnp mjl β pi {αi , αn }αp xj pl − pi αn {αi , αp }xj pl + 2  + αn αp αi [pi , xj ]pl − 2αi αn αp pi xj pl . Using the relations {αi , αj } = 2δij and [xi , pj ] = iδij , we obtain i  [α · p, β(Σ · L)] = − β 4αl pn xn pl − 4αj pl xj pl − 2  − iαj αl αj pl + 3iαi pi − 2αi αj αl pi xj pl + 2αi αl αj pi xj pl   = iβ 2αi pl xi pl − 2iα · p − αj αi αl pi xj pl − αi αl αj pi xj pl , where we used αi αj αi = −αj . By substituting pi xj = xj pi − iδ ij into the last line of previous formula, we have [α · p, β(Σ · L)] = 2β(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 1 u ¯(p1 )(γ ν γ µ − γ µ γ ν )(p1 − p2 )ν u(p2 ) 2 1 ¯(p1 )[−γ µ (/ = u p1 − /p2 ) + (/ p1 − /p2 )γ µ ]u(p2 ) 2 1 ¯(p1 )[−γ µ (/ p1 − m) + (m − /p2 )γ µ ]u(p2 ) . = u 2

i¯ u(p1 )σ µν (p1 − p2 )ν u(p2 ) =

By using γ µ / p1 = 2pµ1 − / p1 γ µ and p /2 γ µ = 2pµ2 − γ µ /p2 we obtain u(p1 )γ µ u(p2 ) − (p1 + p2 )µ u ¯(p1 )u(p2 ) , i¯ u(p1 )σ µν (p1 − p2 )ν u(p2 ) = 2m¯ where we used that u(p) and u ¯(p) satisfy the Dirac equation. The last expression is the requested identity. The second identity can be proven similarly. 4.23 It is easy to see that γα γµ γβ = 2gαµ γβ − 2gαβ γµ + 2gµβ γα − γβ γµ γα . From (4.43) we have

(4.43)

Chapter 4. The Dirac equation

105

u ¯(p2 )/ p1 γµ / p2 u(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 )/ p1 γµ / p2 u(p1 ) = u ¯(p2 )[−2p1 · p2 + 3m2 ]γµ u(p1 ) − 2mi¯ u(p2 )σµν q ν u(p1 ) ' & 2 =u ¯(p2 ) (q + m2 )γµ − 2imσµν q ν u(p1 ). (4.45) From the last expression we can make the following identifications: F1 = q 2 + m2 and F2 = −2im. 4.24 By using u(p) = p /u(p)/m and {γµ , γ5 } = 0 , we have

1 1 u ¯(p)γ5 /pu(p) = − u ¯(p)/ pγ5 u(p) . m m By applying the Dirac equation (3.C) we obtain u ¯(p)γ5 u(p) =

u ¯(p)γ5 u(p) = −¯ u(p)γ5 u(p) . Thus u ¯(p)γ5 u(p) = 0. By using the Gordon identity (for µ = 0) it finally follows that 1 m u ¯(p)(1 − γ5 )u(p) = N . 2 2Ep 4.25 F1 = −iq 2 , F2 = −2im, F3 = −2m. 4.26 By applying the operator (i/ ∂ + m) to the Dirac equation we obtain (i/ ∂ + m)(i/ ∂ − m)ψ = −(  + m2 )ψ = 0 . 4.27 The probability density is ρ(x) = ψ † (x)ψ(x). By using the expression E for the wave function from Problem 4.2, we easily get ρ = mp . The current p ¯ ¯ density is j = ψγψ = m ψψ, where the Gordon identity (for µ = i) has been p . applied. Finally j = m 4.28 The position operator in the Heisenberg picture satisfies the following equation drH = −i[rH , H] = αH . dt In order to integrate the last equation we have to find the Dirac matrices in the Heisenberg picture αH = eiHt αe−iHt = Since

∞  (it)n [H, [H, . . . [H, α] . . .]] . n! n=0

106

Solutions

[H, α] = 2(p − αH) ,

(4.46)

[H, [H, α]] = −2 (p − αH)H ,

(4.47)

[H, [H, [H, α]]] = 2 (p − αH)H , etc.

(4.48)

2

3

2

we get 

(2it)3 2 (2it)2 αH = α + (αH − p) −2it + H− H + ... 2! 3!   p p −2itH = + α− e . H H Then, equation



p  p  −2itH drH = + α− e dt H H

(4.49)

(4.50)

implies rH = r +

  p 1 p  1 −2iHt p t−i α− +i α− e . H H 2H H 2H

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) and d∗r (p) in the expansion  

m 1 3 d ψ(0, x) = p (cr (p)ur (p)eip·x + d∗r (p)vr (p)e−ip·x ) . 3/2 E (2π) p r (4.51) If we multiply this expression by u†s (q)e−iq·x from left and integrate over x, we get

 m 1 cs (q) = d3 xu†s (q)ψ(0, x)e−iq·x , (2π)3/2 Eq where we applied the relations Ep δrs , vr† (−p)us (p) = u†r (−p)vs (p) = 0 . m (4.52) These relations can be obtained from (4.D) by using the Gordon identity. Similarly for d coefficients we get

 m 1 d∗s (q) = d3 xvs† (q)ψ(0, x)eiq·x . (2π)3/2 Eq u†r (p)us (p) = vr† (p)vs (p) =

Carrying out the integrations, we find

Chapter 4. The Dirac equation

c1 (p) =

1 (2π)3/2

107

Ep + m , 2Ep

c2 (p) = 0, 1 1 (px + ipy ) , (2π)3/2 2Ep (Ep + m) 1 1 pz . d∗2 (p) = (2π)3/2 2Ep (Ep + m) d∗1 (p) =

(4.53)

The wave function at time t > 0 is  

m 1 3 ψ(x) = (cr (p)ur (p)r e−ip·x +d∗r (p)vr (p)eip·x ) , (4.54) d p Ep (2π)3/2 r where the coefficients cr (p) and d∗r (p) are given in (4.53). 4.30 In this case the coefficients cr (p) and d∗r (p) in expansion (4.51) are:  c1 (p) =

d2 π

3/4

Ep + m −d2 p2 /2 e , 2Ep

c2 (p) = 0 ,  2 3/4 2 2 d 1 d∗1 (p) = e−d p /2 (px + ipy ) , π 2Ep (Ep + m)  2 3/4 2 2 d 1 d∗2 (p) = pz e−d p /2 . π 2Ep (Ep + m) 4.31 The equation for spin 1/2 particle in the electromagnetic field has the following form [iγ µ (∂µ − ieAµ ) − m]ψ = 0 . (4.55) If we assume that a wave function for z > 0 has the form  ϕ ψ= e−iEt+iqz , χ then (4.55) becomes  E−m−V σ3 q

−σ3 q −E − m + V

 ϕ =0. χ

(4.56)

(4.57)

The system of equations (4.57) has a nontrivial solution if and only if (4.58) E = V ± q 2 + m2 . The wave function2 is 2

From the boundary conditions it follows that there is no spin flip.

108

Solutions

 1  0   e−iEt+ipz ψI = a  1  pσ3 (E+m) 0   1  0   e−iEt−ipz , z < 0 , + b 1  −pσ3 (E+m) 0   1  0   e−iEt+iqz , z > 0 , ψII = d  1  qσ3 (E+m−V ) 0 

(4.59)

√ where p = E 2 − m2 . The terms proportional to the coefficient a, b and d in (4.59) are the initial ψin , reflected ψr and 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 , a − b = rd ,

(4.60) (4.61)

E+m q where r = E+m−V p . Now, we will consider three cases: 1. If |E − V | ≤ m, the momentum q is imaginary, q = iκ so that the wave function exponentially decreases in the region z > 0, as is the case in nonrelativistic quantum mechanics. The transmitted, reflected and incident currents are: (4.62) jr = ψ¯tr γ 3 ψtr ez = 0 ,

2p |b|2 ez , (4.63) E+m 2p jin = ψ¯in γ 3 ψin ez = |a|2 ez . (4.64) E+m Since jtr = 0 the transmission coefficient is zero. The reflection coefficient is    p(E + m − V ) − iκ(E + m) 2 −jr  =1.  R= = (4.65) jin p(E + m − V ) + iκ(E + m)  jr = ψ¯r γ 3 ψr ez = −

2. If V < E − m, the momentum q is real. The currents are: jtr =

2q |d|2 ez , E+m−V

2p |b|2 ez , E+m 2p |a|2 ez . jin = E+m

jr = −

(4.66) (4.67) (4.68)

Chapter 4. The Dirac equation

109

The transmission coefficient is

 2 d 4r jtr = r   = , T = jin a (1 + r)2

while the reflection coefficient is R=

−jr = jin



1−r 1+r

(4.69)

2 .

(4.70)

3. If E +m < V , the momentum q is 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. The explanation of this paradox is given in Problem 2.9. 4.32 The solution of the Dirac equation is   1  0   eipz ψI =  1  pσ3 (E+m) 0   1  0   e−ipz , z < 0 , + B 1  −pσ3 (E+m) 0   1  0   eiqz ψII = C  1  qσ3 (E+m−V ) 0   1  0   e−iqz , 0 < z < a , + D 1  −qσ3 (E+m−V ) 0   1  0   eipz , z > a , ψIII = F  1  pσ3 (E+m) 0 √ where p = E 2 − m2 and q = (E − V )2 − m2 . From the boundary conditions ψI (0) = ψII (0) and ψII (a) = ψIII (a), we obtain the transmission coefficient |r|2 T = |F |2 = 16 , |(1 + r)2 e−iqa − (1 − r)2 eiqa |2 E+m where r = pq E+m−V . It is easy to show that the transmission coefficient is V equal to one if E = 2 .

110

Solutions

4.33 (a) The wave function is   ψI =    ψII =    + 

B
B

−iκσ3 (E+m)

 B B


C C
 qσ3 (E+m+V )

D D


−qσ3 (E+m+V )

F
F

κz   e , z < −a,

 C C


iqz  e

(4.71)



 −iqz , −a < z < a, D e D


 −κz , z > a, F e F
√ where κ = m2 − E 2 and q = (E + V )2 − m2 . Since there is no spin flip, we can take B
= C
= D
= F
= 0. From the boundary conditions ψI (−a) = ψII (−a) and ψII (a) = ψIII (a), it follows that ψIII

 =

iκσ3 (E+m)

e−κa B = e−iqa C + eiqa D e−κa F = eiqa C + e−iqa D −ire−κa B = e−iqa C − eiqa D ire−κa F = eiqa C − e−iqa D , where r =

κ E+m+V q E+m

. By combining previous equations we obtain

e−κa (B − F ) = 2i sin(qa)(D − C) ire−κa (B − F ) = 2 cos(qa)(D − C) e−κa (B + F ) = 2 cos(qa)(D + C) re−κa (B + F ) = 2 sin(qa)(D + C) . Further, we will distinguish two classes of solutions: odd and even. If B = F and C = D, the third and the fourth equations give the following dispersion relation: κE+m+V . tan(qa) = q E+m These solutions satisfy the following property: ψ
(z) = γ0 ψ(−z) = ψ(z); thus they are even. On the other hand, if B = −F and C = −D, the dispersion relation is

Chapter 4. The Dirac equation

cot(qa) = −

111

κE+m+V . q E+m

This class of solutions satisfy ψ
(z) = γ0 ψ(−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 q tan(qa) = f (q) ,

(4.72)

where  f (q) =

2V



q2

+

m2

−

q2

m + q 2 + m2 , −V m + q 2 + m2 − V 2

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 q cot(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 N states if the condition (N − 1)π Nπ ≤ V (V + 2m) < 2a 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

4.34 The Dirac equation in this case has following form    ∂ ∂ ∂ ∂ − ieBy + iγ 2 + iγ 3 −m ψ =0 . iγ 0 + iγ 1 ∂t ∂x ∂y ∂z

113

(4.74)

A particular solution of (4.74) is ψ = e−iEt+ipx x+ipz z



ϕ(y) χ(y)

.

(4.75)

By substituting (4.75) in (4.74) we obtain  d  E−m (eBy − px )σ1 − pz σ3 + iσ2 dy ϕ =0. d χ (px − eBy)σ1 + pz σ3 − iσ2 dy −E − m (4.76) From the second equation in (4.76), follows  1 d χ(y) = px σ1 + pz σ3 − eByσ1 − iσ2 ϕ(y) , (4.77) E+m dy and plugging it into the first equation of (4.76), we get  2 d 2 2 2 2 − (p − eBy) + E − m − p − eBσ x 3 ϕ=0 , z dy 2

(4.78)

where we used the following identity σi σj = δij + iijk σk . By introducing new variable ξ = px − eBy, Equation (4.78) becomes the Schr¨ odinger equation for a linear oscillator (parameters M, ω and  ), where M 2 ω2 =

1 , (eB)2

2M  =

E 2 − m2 − p2z ∓ eB . (eB)2

We assumed that the spinor ϕ is an eigenstate of σ3 /2, i.e. 1 1 σ3 ϕ = ± ϕ . 2 2 The energy eigenvalues are En,pz =



m2 + p2z ± eB + (2n + 1)eB ,

where n = 0, 1, 2, . . . 4.35 Acting by (i/ ∂ + e/ A + m) on (i/ ∂ + e/A − m)ψ(x) = 0, we get [  − ieγ µ γ ν ∂µ Aν − 2ieAµ ∂µ − e2 A2 + m2 ]ψ = 0 .

(4.79)

114

Solutions

On the other hand, one can show that e − σµν F µν = ie(∂µ Aµ − γ µ γ ν ∂µ Aν ) . 2 The requested result can be obtained by combining these expressions. 4.36 By substituting

 ϕ ψ= e−imt χ

in the Dirac equation (i/ ∂ + e/A − m)ψ(x) = 0, we obtain the following equations:  ∂ i + eA0 ϕ = cσ · (p + eA)χ , ∂t  ∂ 2 i + 2mc + eA0 χ = cσ · (p + eA)ϕ . ∂t In the case A = 0, the second equation yields:  eA0 1 ∂ϕ i − χ= σ · p σ · pϕ . σ · pϕ − 2mc 2mc2 ∂t 2mc2 Combining this relation with the first equation, we obtain i

∂ϕ = H
ϕ , ∂t

where 

p2 p4 e − eA0 − + (2iE · p − ∆A0 ) 2m 8m3 c2 4m2 c2 ! e − (iE · p + σ · (E × p)) . 4m2 c2

H
=

The operator H
is not the Hamiltonian, since it is not hermitian. This is related to the fact that ϕ† ϕ is not the probability density. Actually, the probability density should be taken in the following form: ¯ = ϕ† ϕ − χ† χ ρ = ψψ  2 v p2 † )ϕ + o 2 . = ϕ (1 + 4m2 c2 c We introduce the new wave function  ϕs = 1 +

p2 8m2 c2

ϕ.

Chapter 4. The Dirac equation

115

Then, the new Hamiltonian is given by   p2 p2
H = 1+ H 1 − . 8m2 c2 8m2 c2 After that, we obtain H=

p2 p4 e e − eA0 − − ∆A0 + σ · (E × p) . 3 2 2 2 2m 8m c 8m c 4m2 c2

In the case A = 0, the Hamiltonian is (p + eA)2 p4 e − eA0 + σ·B− 2m 2mc 8m3 c2 e e − ∆A0 + σ · (E × (p + eA)) . 8m2 c2 4m2 c2

H=

4.37 First, we are going to show that Vµ (x) is a real quantity: ¯ µ ψ)† Vµ∗ = Vµ† = (ψγ = ψ † 㵆 (ψ † γ 0 )† = ψ † γ 0 γµ γ 0 γ 0 ψ ¯ µψ = ψγ = Vµ .

(4.80)

Under proper orthochronous Lorentz transformations, Vµ is transformed in the following way: Vµ
(x
) = ψ¯
(x
)γµ ψ
(x
) = ψ † (x)γ0 S −1 γµ Sψ(x) , where we used the fact that γ0 S −1 = S † γ0 . Using S −1 γµ S = Λµν γν , we obtain Vµ
(x
) = Λµν Vν (x). So, the quantity Vµ is a Lorentz four-vector. Under parity we have ¯ x)γ0 γµ γ0 ψ(t, x) . Vµ (t, x) → Vµ
(t, −x) = ψ(t, This implies V0
(t, x) = V0 (t, −x), Vi
(t, x) = −Vi (t, −x) . As we know, under charge conjugation the spinors transform according to: ψ(x) → ψc (x) = C ψ¯T , ψ¯ = ψ † γ0 → (C ψ¯T )† γ0 = (C(γ 0 )T ψ ∗ )† γ0 = ψ T ((γ 0 )T C(γ 0 )† )T = −ψ T (Cγ 0 γ 0 )T = ψT C .

(4.81)

116

Solutions

Then, we can find the transformation law for Vµ : ¯ µ ψ)T = Vµ . Vµ → −ψ T Cγµ C −1 ψ¯T = (ψγ The following formulae Cγµ C −1 = −γµT , C = −C −1 have been used (Prove the last one). For time reversal we have ψ(x) → ψ
(−t, x) = T ψ ∗ (t, x), where matrix T satisfies T γµ T −1 = γ µ∗ = γµT and T † = T −1 = T = −T ∗ . It is easy to see that ¯ ψ(x) → ψ¯
(−t, x) = ψ T (t, x)T γ0 . Then V µ (t, x) → ψ T T γ 0 γ µ T ψ ∗ = ψ T T γ 0 T −1 T γ µ T −1 ψ ∗ = ψ T (γ 0 )T (γ µ )T ψ ∗ = (ψ † γ µ γ 0 ψ)T = ψ† γ µ γ 0 ψ . Therefore,

(4.82)

V0
(−t, x) = V0 (t, x), Vi
(−t, x) = −Vi (t, x) .

4.38 The quantity Aµ transforms under Lorentz transformations in the following way: ν −1 ¯ A
µ (x
) = Λµν ψ(x)γ S γ5 Sψ(x) µ ¯ = detΛ Λ ν ψ(x)γ ν γ5 ψ(x) = detΛ Λµν Aν (x) ,

where we used i µνρσ S −1 γ µ SS −1 γ ν SS −1 γ ρ SS −1 γ σ S 4! i = − µνρσ Λµ α Λν β Λσ γ Λρ δ γ α γ β γ γ γ δ 4! i = − αβγδ detΛ γ α γ β γ γ γ δ 4! = detΛ γ5 .

S −1 γ5 S = −

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. ¯ µ ∂µ ψ transforms as a scalar under Lorentz transfor4.39 The quantity ψγ mations. The parity does not change it. The action of the charge conjugation ¯ µ ψ, while the time reversal produces −(∂µ ψ)γ ¯ µ ψ. yields (∂µ ψ)γ 4.40 By transposing the Dirac equation,

Chapter 4. The Dirac equation

117

u ¯(p, s)(/ p − m) = 0 , and using C −1 γ µ C = −(γ µ )T , one gets the requested result. 4.41 Let us assume that there are two different matrices C
and C

, which both satisfy the relation Cγ µ C −1 = −(γ µ )T . Then from C

γµ C

−1 = C
γµ C
−1 follows that [C
−1 C

, γµ ] = 0, whereupon (see Problem 3.18) the requested relation follows. 4.42 We directly obtain: (a)

  0   1 e−iEt−ipz .  0 1



p  − E+m

ψc (x) = Np   (b)

  1
0  ψ
(x
) =   e−imt . 0 0

(c)

  1  0    e−i(Et+pz) . ψp (t, x) = Np   1  p − Ep +m 0 

Momentum is inverted under parity. Time reversal transforms the wave function into    0  1   ei(−Et−pz) , ψt (t, x) = −iNp   p 0  Ep +m 1 and we see that spin and the direction of the momentum are inverted. (d) The wave function for S
observer is 

ϕ ei(Et−p z ) ψ
(x
) = Np p ϕ Ep +m

 4.43 P = γ0 =

  cos θ2   . ϕ= i sin θ2  2 I σ 0 2 0 , C = iγ γ = i . 0 0 −σ 2 

where

0 I

4.44 Multiplying the equation

118

Solutions

Σ·p ur (p) = (−1)r+1 ur (p) , |p|

(4.83)

by γ0 from left, we obtain Σ · (−p) ur (−p) = (−1)r ur (−p) , |p|

(4.84)

since γ0 ur (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) = iγ 1 γ 3 ψr∗ (−t, x)  σ 2 ϕ∗r 2 ∗ = −N σ (σ ·p) ∗ ei(−Ep t−p·x) Ep +m ϕr  σ 2 ϕ∗r 2 = −N ei(−Ep t−p·x) , ∗ − (σ·p)σ ϕ Ep +m r

(4.85)

where we used σ 2 σ ∗ = −σσ 2 in the second step. From the last expression, we conclude that the momentum changes its direction, i.e. p → −p. Prove that σ 2 ϕ∗1 = iϕ2 and σ 2 ϕ∗2 = −iϕ1 . Now, we consider the case r = 1 (the other case r = 2 is similar). From (4.85) it follows that  ϕ2 ψt (t, x) = −iN (4.86) ei(−Ep t−p·x) . p·σ − Ep +m ϕ2 on (4.86), we see that the helicity is unchanged. The same By applying Σ·(−p) |p| result can be obtained by complex conjugation and multiplication of Equation (4.83) from left by iγ 1 γ 3 . You can prove the same for v spinors. 4.45 The transformed Hamiltonian is    p m sin(2pθ) + mβ cos(2pθ) + sin(2pθ) , H
= α · p cos(2pθ) − p m where p = |p|. In order to have even form of the Hamiltonian, the coefficient multiplying α · p has to be zero. This is satisfied if tan(2pθ) = p/m . 4.47 First prove that: U = cos(pθ) + hence xFW =

"

βα · p sin(pθ) = p

Ep + m βα · p + 2Ep 2Ep (Ep + m)

Ep + m βα · p + , 2Ep 2Ep (Ep + m) # " x

Ep + m βα · p − 2Ep 2Ep (Ep + m)

# .

From the well known identity [x, f (p)] = i∇f (p) we get two auxiliary results:

Chapter 4. The Dirac equation

x

x

Ep + m i =− 2Ep 2

m Ep p+ 2(Ep + m) Ep3

119

Ep + m x, 2Ep

βα · p iβα iβ(α · p)(2Ep + m) p = − √ 2Ep (Ep + m) 2Ep (Ep + m) 2 2(Ep (Ep + m))3/2 Ep βα · p + x. 2Ep (Ep + m)

Using these formulae we get xFW = x − i

p(βα · p) βα p α(α · p) +i 2 −i . +i 2Ep (Ep + m) 2Ep (Ep + m) 2Ep 2Ep (Ep + m)

The last expression can be rewritten in the form xFW = x + i

βα p(βα · p) Σ×p −i . − 2Ep2 (Ep + m) 2Ep 2Ep (Ep + m)

The Foldy–Wouthuysen transformation does not change the momentum, so that [xkFW , plFW ] = iδ kl .

5 Classical fields and symmetries

5.1 We apply the definition of functional derivative (5.A). (a) From



δFµ = ∂µ δφ = we have

4

d y(∂µ δφ)y δ

(4)

(y − x) = −

d4 y∂µy δ (4) (y − x)δφ(y) ,

δFµ [φ(x)] = −∂µy δ (4) (y − x) , δφ(y)

(b) The first functional derivative of the action with respect to φ is ∂V δS = − φ − . δφ(x) ∂φ Then

 δ

δS δφ(x)



∂2V δφ(x) = − δφ(x) − ∂φ2 (x)

( = d4 y − y δ (4) (x − y)−  ∂2V δ (4) (x − y) δφ(y) . − ∂φ(x)∂φ(y)

Hence, δ2 S ∂2V = − y δ (4) (y − x) − δ (4) (x − y) . δφ(x)δφ(y) ∂φ(x)∂φ(y) 5.2 In this problem we use the Euler–Lagrange equations of motion (5.B). ∂L = m2 Aρ and ∂(∂∂L = −2∂ ρ Aσ + λg ρσ (∂µ Aµ ) so that (a) First note that ∂A ρ σ Aρ ) the equations of motion are given by

(λ − 2)∂σ ∂ ρ Aσ − m2 Aρ = 0 .

122

Solutions

(b) The derivative of the Lagrangian density with respect to ∂σ Aρ is 1 1 ∂Fµν ∂L = − F µν = − F µν (δµσ δνρ − δνσ δµρ ) = −F σρ . ∂(∂σ Aρ ) 2 ∂(∂σ Aρ ) 2 In the last step we used the fact that Fρσ is an antisymmetric tensor, i.e. Fρσ = −Fσρ . The Euler-Lagrange equations of motion are ∂σ F σρ + m2 Aρ = 0 . By using the definition of field strength F ρσ , the Euler-Lagrange equations become   ρ  − ∂σ ∂ ρ + m2 δσρ Aσ = 0 . δσ  (c) (  + m2 )φ = −λφ3 . (d) The equations of motion are: − Aρ + ∂σ ∂ ρ Aσ = −ie[φ(∂ ρ φ∗ + ieAρ φ∗ ) − φ∗ (∂ ρ φ − ieAρ φ)] , φ∗ + 2ieAρ ∂ρ φ∗ + ieφ∗ ∂ρ Aρ − e2 A2 φ∗ + m2 φ∗ = 0 ,  φ − 2ieAρ ∂ρ φ − ieφ∂ρ Aρ − e2 A2 φ + m2 φ = 0 .  (e) The equations are: (iγ µ ∂µ − m)ψ = igγ5 ψφ ,

− ¯ µ← ¯ 5φ , ∂µ + m) = −ig ψγ ψ(iγ

¯ 5ψ . φ + m2 φ = λφ3 − ig ψγ  5.3 The variation of the action is

∞ L   δS = dt dx ∂µ φ∂ µ (δφ) − m2 φδφ −∞ ∞

0





−∞ L

=

dx[∂µ (∂ µ φδφ) − (  + m2 )φδφ] 0

t=∞  dx∂0 φδφ −

0

−

L

dt

=



∞

−∞

dt

∂φ x=L δφ ∂x x=0

L

dx( φ + m2 φ)δφ ,

dt −∞

t=−∞

∞

0

where we integrated by parts. As the first term vanishes, from Hamiltonian principe one obtains the equation of motion (  + 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
(t, x = 0) = φ
(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 eliminate the surface term. Let us mention that the mixed boundary conditions can be imposed. 5.4 In order to show that the change L → L + ∂µ F µ (φr ) does not change the equations of motion, we have to prove that

δ d4 x∂µ F µ (φr ) = 0 . Ω

Applying the Gauss theorem we get

  δ d4 x∂µ F µ (φr ) = dΣ µ δFµ = Ω

∂Ω

∂Ω

dΣ µ

∂Fµ δφr = 0 , ∂φr

since the variation of fields on the boundary is equal to zero. 5.5 Add to the Lagrangian density the term − 21 ∂µ (φ∂ µ φ). Note that it does not have the form as in Problem 5.4, because here the function F µ depends on the field derivatives. However,

  δ d4 x∂µ (φ∂ µ φ) = dΣ µ δ(φ∂µ φ) = dΣ µ (δφ∂µ φ + φδ∂µ φ) . Ω

∂Ω

∂Ω

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. 5.6 Use the similar reasoning as in the previous problem. 5.7 The equation of motion for the vector field was derived in Problem 5.2 (b). Acting by ∂ρ on this equation we obtain m2 ∂ρ Aρ = 0 . Since m = 0, we conclude that ∂ρ Aρ = 0 . 5.8 The field strength tensor, Fµν is invariant under the gauge transformations. From this, it follows that the Lagrangian is also invariant. The condition ∂µ Aµ = 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. 5.9 Firstly, show that ∂L 1 = ∂ α hρσ − ∂ σ hρα − ∂ ρ hσα + g ρα ∂ σ h ∂(∂α hρσ ) 2 1 + g σα ∂ ρ h + g ρσ ∂µ hµα − g ρσ ∂ α h . 2

124

Solutions

The equations of motion are hρσ − ∂ α ∂σ hρα − ∂ α ∂ρ hσα + ∂ρ ∂σ h  + gρσ ∂µ ∂ν hµν − gρσ  h = 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 kinetic term is 1 1 [(∂φ1 )2 + (∂φ2 )2 ] → [(∂φ
1 )2 + (∂φ
2 )2 ] 2 2 1 = [(∂φ1 cos θ − ∂φ2 sin θ)2 + (∂φ1 sin θ + ∂φ2 cos θ)2 ] 2 1 = [(∂φ1 )2 + (∂φ2 )2 ] . 2 Similarly, we can prove that the other two terms are invariant. The infinitesimal variations of the fields φi are δφ1 = −θφ2 and δφ2 = θφ1 , so that jµ =

∂L δφi = θ(φ1 ∂µ φ2 − φ2 ∂µ φ1 ) . ∂(∂ µ φi )

The parameter θ can be dropped out since  it is a constant. The charge corresponding to the SO(2) symmetry is Q = d3 x(φ1 φ˙ 2 − φ2 φ˙ 1 ) . 5.12 Under the SU(2) transformations, the fields are transformed accordi a a ing to φ
= e 2 τ θ φ , where τ a (a = 1, 2, 3) are the Pauli matrices. For an infinitesimal transformation we obtain δφi =

i a a i a a τ θ φj , δφ∗i = − φ∗j τji θ . 2 ij 2

The Noether current is determined by ∂L ∂L δφi + δφ∗i ∂(∂ µ φi ) ∂(∂ µ φ∗i )  i  a a φj − φ∗i τij ∂µ φj . = θa ∂µ φ∗i τij 2

jµ =

From the previous relation (θa are constant independent parameters) it follows that the conserved currents are:  i a a jµa = − ∂µ φ∗i τij φj − φ∗i τij ∂µ φj . 2 The charges are i Q =− 2 a



a a φj − φ∗i τij ∂0 φj ) . d3 x(∂0 φ∗i τij

Chapter 5. Classical fields and symmetries.

125

5.13 The currents and charges are jµa =

1¯ a ψi γµ τij ψj , 2

Qa =

1 2



a d3 xψi† τij ψj .

← − The equations of motion are (iγ µ ∂µ − m)ψi = 0 and ψ¯i (iγ µ ∂µ + m) = 0. The current conversation law, ∂µ j µa = 0 can be proved easily: a a a a 2∂µ j µa = (∂µ ψ¯i )γ µ τij ψj + ψ¯i γ µ τij ∂µ ψj = imψ¯i τij ψj + ψ¯i τij (−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
= eiθ ψ and ¯ µ ψ, while the charge is ψ¯ → ψ¯
= e−iθ ψ¯ . The Noether current is jµ = ψγ given by Q = −e d3 xψ † ψ. Note that the current does not have additional indices since U(1) is a one–parameter  group. (b) jµ = i(φ∗ ∂µ φ − φ∂µ φ∗ ) , Q = iq d3 x(φ∗ ∂0 φ − φ∂0 φ∗ ) . 5.15 The equations of motion are (  + m2 )φi = 0. The expression φT φ is invariant under SO(3) transformations, hence the Lagrangian density has the same symmetry. The generators of SO(3) group are       0 0 0 0 0 i 0 −i 0 J 1 =  0 0 −i  , J 2 =  0 0 0  , J 3 =  i 0 0  . (5.1) 0 i 0 −i 0 0 0 0 0 Note that we can write (J k )ij = −ikij . Under SO(3) transformations, the infinitesimal variations of the fields are δφi = i(J k )ij θk φj = kij θk φj and the Noether current is ∂L δφi ∂(∂ µ φi ) = kij φj ∂µ φi θk = −θ · (φ × ∂µ φ) .

jµ =

The parameters of rotations θk , are arbitrary and therefore the currents jkµ = −kij φj ∂ µ φi are also conserved. 5.16 First, derive the following formula eiαγ5 = cos α+iγ5 sin α. The transformation law for the Dirac Lagrangian density under the chiral transformation is given by

126

Solutions

L → ψ † e−iαγ5 γ0 (iγµ ∂ µ − m)eiαγ5 ψ ¯ µ ∂ µ ψ − mψ(cos ¯ = (cos2 α + sin2 α)ψiγ α + iγ5 sin α)2 ψ ¯ µ ∂ µ ψ − mψ(cos ¯ = ψiγ 2α + iγ5 sin 2α)ψ . From the previous expression we can conclude that the Lagrangian density ¯ µ γ5 ψ. is invariant only for massless fermions. The Noether current is jµ = ψγ Prove that ∂µ j µ is proportional to the mass m of the field. 5.17 The current is given by ∂L ∂L ∂L ∂L δσ + δπ a + δΨi + δ Ψ¯i ∂(∂ µ σ) ∂(∂ µ π a ) ∂(∂ µ Ψi ) ∂(∂ µ Ψ¯i ) 1 a Ψj . = −abc αb ∂µ π a π c − Ψ¯i γµ αa τij 2

jµ =

The final result has the form 1 jµ = π × ∂µ π + Ψ¯ γµ τ Ψ . 2 5.18 (a) For translations, we have δxµ = µ , while the total variations of the fields equal zero. The Noether current is T µν =

∂L ∂φr − Lg µν . ∂(∂µ φr ) ∂xν

(5.2)

The index ν in (5.2) comes from the group of translations. For a real scalar field, from (5.2) we obtain Tµν = ∂µ φ∂ν φ −

 1 (∂φ)2 − m2 φ2 gµν . 2

The conserved charges are the Hamiltonian (for ν = 0),



  1 d3 x (∂0 φ)2 + (∇φ)2 + m2 φ2 , H = d3 xT 00 = 2 and the momentum (for ν = i)



i 3 0i P = d xT = d3 x∂0 φ∂ i φ . For the Dirac field the energy–momentum tensor is given by ¯ µ ∂ ν ψ − Lg µν . T µν = iψγ The Hamiltonian and momentum are given by

(5.3)

(5.4)

(5.5)

Chapter 5. Classical fields and symmetries.

127

¯ d3 xψ[−iγ∇ + m]ψ ,

P = −i d3 xψ † ∇ψ .

H=

(5.6) (5.7)

For electromagnetic field the energy–momentum tensor is T µν =

∂L ∂Aρ − Lg µν ∂(∂µ Aρ ) ∂xν

from which we obtain 1 T µν = −F µρ ∂ ν Aρ + F 2 g µν . 4

(5.8)

For the Lorentz transformations δxν = ω νρ xρ and i δφ = 0 , δψ = − σνρ ω νρ ψ, δAµ = ωµν Aν , 4 The Noether currents for scalar, spinor and electromagnetic field are jµ = [xν Tµρ − xρ Tµν ]ω νρ , 1¯ νρ , jµ = [ ψγ µ σνρ ψ + xν Tµρ − xρ Tµν ]ω 2 jµ = [Fµρ Aν − Fµν Aρ + (xν Tµρ − xρ Tµν )]ω νρ .

(5.9)

Dropping the parameters of the Lorentz transformations ω νρ , the conserved currents have the form Mµνρ , and they are given by theexpression in square brackets in (5.9). The angular-momentum is Mνρ = d3 xM0νρ . (b) As we see, the energy–momentum tensors for Dirac and electromagnetic fields are not symmetric. To find the symmetrized energy–momentum tensors we employ the procedure given in the problem. For the Dirac field we have 1 ¯ ¯ ρ σµν ψ + ψγ ¯ ν σµρ ψ) (−ψγµ σρν ψ + ψγ 4 i ¯ = ψ(4g µν γρ − 4gρν γµ + γµ γν γρ − γρ γν γµ )ψ . 8

χρµν =

Using (4.43) we find i ¯ i ¯ ν ψ − 3i ψγ ¯ µ ∂ν ψ ∂ν ψγµ ψ − ∂µ ψγ 4 4 4 i i ¯ ¯∂ ψ) . ¯ ν + ψγ ν ∂µ ψ + gµν (∂ν ψγ ψ + ψ/ 4 2

∂ ρ χρµν = −

The symmetrized energy–momentum tensor for Dirac field is

128

Solutions

i ¯ ¯ ¯ ¯ T˜µν = (ψγ ν ∂µ ψ + ψγµ ∂ν ψ − ∂µ ψγν ψ − ∂ν ψγµ ψ) − 4 i ¯ ν i ¯ ¯ . ∂ ψ − mψψ) ψ − ψ/ − gµν ( ∂ν ψγ 2 2 Similarly we determine the symmetrized energy–momentum tensor for the electromagnetic field. From transformation rule of the electromagnetic potential with respect to Lorentz transformations δAα = ωαβ Aβ ≡

1 µν ω (Iµν )αβ Aβ , 2

follows that (Iµν )αβ = gµα gνβ − gµβ gνα . Then χρµν = F µρ Aν and the new energy–momentum tensor is 1 T˜µν = −F µρ F νρ + F 2 g µν . 4

(5.10)

If we introduce the electric and magnetic fields: F 0i = −E i , Fij = −ijk B k , then the components of energy–momentum tensor are: 1 T˜00 = −F 0i F 0i + (2F0i F 0i + Fij F ij ) 4 1 = E2 + (−2E2 + 2B2 ) 4 1 = (E2 + B2 ) , 2 T˜0i = −F 0j F ij = ijk E j B k = (E × B)i , T˜ij

(5.11)

1 = −E i E j + ikl jkn B l B n + (E2 − B2 )δij 2  = − E i E j + B i B j − δij T00 .

From the expression (5.11) we conclude that T˜00 T˜0i , −T˜ij are 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 δ0 φ(x) = φ
(x) − φ(x). From δ0 φ = δφ − ∂µ φδxµ , where δφ = φ
(x
) − φ(x) is the total variation of a field, it follows that the infinitesimal form variation of φ is δ0 φ = ρ(φ(x) + xµ ∂µ φ) .

(5.12)

Chapter 5. Classical fields and symmetries.

129

The induced change of the action is



  1   1 d4 x
(∂
φ
)2 − m2 φ
2 (x
) − d4 x (∂φ)2 − m2 φ2 (x) . S
− S = 2 2 (5.13) The transformed volume of integration is given by d4 x
= |J|d4 x = det(e−ρ I)d4 x = e−4ρ d4 x .

(5.14)

The field derivative is changed according to the following rule: ∂µ φ(x) →

∂φ
∂xν ∂ = (eρ φ) = e2ρ ∂µ φ .
µ ∂x ∂x
µ ∂xν

(5.15)

Thus, the change of the action is

  1 d4 xe−4ρ e4ρ (∂φ)2 − m2 e2ρ φ2 (x) S
− S = 2

  1 − d4 x (∂φ)2 − m2 φ2 (x) 2

1 = m2 (1 − e−2ρ ) d4 xφ2 (x) . 2 For an infinitesimal dilatation (ρ  1), the variation of the action is

2 δS = m ρ d4 xφ2 (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∂ µ φ − xν ∂ µ φ∂ν φ + Lxµ .

(5.17)

By calculating ∂µ j µ one obtains that ∂µ j µ is proportional to the mass m. 5.20 From and

d4 x
= e−4ρ d4 x ≈ (1 − 4ρ)d4 x ,

(5.18)

¯ µ ∂µ ψ , ¯ µ ∂µ ψ ≈ (1 + 4ρ)ψγ ψ¯
(x
)γ µ ∂µ
ψ
(x
) = e4ρ ψγ

(5.19)

it follows that this transformation leaves the action unchanged. The Noether ¯ µ ψ − ixν ψγ ¯ µ ∂ν ψ + xµ L. current is j µ = − 32 iψγ

6 Green functions

6.1 The Green function of the Klein-Gordon equation satisfies the equation ( x + m2 )∆(x − y) = −δ (4) (x − y) .

(6.1)

Fourier transformations of the Green function and the δ-function in (6.1) gives



1 1 2 4 ˜ −ik·(x−y) ( x + m ) =− d k ∆(k)e d4 ke−ik·(x−y) . (6.2) (2π)4 (2π)4 From (6.2) follows ˜ ∆(k) =

k2

1 1 = 2 . 2 −m k0 − k2 − m2

Then, the Green function is defined by

1 d4 k e−ik·(x−y) . ∆(x − y) = (2π)4 k02 − k2 − m2

(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) convergent. 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 is from the upper side (Fig. 6.1). The exponential term in (6.3) for large energy k0 behaves as e(x0 −y0 )Imk0 , therefore the contour for x0 > y0 has to be closed from the lower side (Imk0 < 0), while in the case x0 < y0 we will close the integration contour on the upper side. By applying the Cauchy theorem we get

1 ∆(x − y) = − d3 keik·(x−y) 2πi(Resωk + Res−ωk )θ(x0 − y 0 ) . (6.4) (2π)4 From (6.4) follows

132

Solution

i ∆R = − (2π)3



0 0 d3 k ik·(x−y) −iωk (x0 −y0 ) e (e − eiωk (x −y ) )θ(x0 − y 0 ) . (6.5) 2ωk

∆R (x − y) is the retarded Green function. The solution of the inhomogeneous equation (  + m2 )φ = J is

φ(x) = − d4 y∆(x − y)J(y) + φ0 , (6.6) where φ0 is a solution of homogeneous equation. From the expressions (6.5) and (6.6) (because of θ−function), we conclude that we integrate over y 0 from −∞ to x0 . The value of the field φ at time x0 is determined by the source J at 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

3 0 0 i d k ik·(x−y) −iωk (x0 −y0 ) ∆A = e (e − eiωk (x −y ) )θ(y 0 − x0 ) . (6.7) 3 (2π) 2ωk The advanced Green function contributes nontrivially to the field φ(x) for 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

  i ∆F = d3 keik·(x−y) Res−ωk θ(y 0 − x0 ) − Resωk θ(x0 − y 0 ) (2π)3

3 d k ik·(x−y) ( −iωk (x0 −y0 ) i e e θ(x0 − y 0 ) (6.8) =− (2π)3 2ωk ! 0 0 +eiωk (x −y ) θ(y 0 − x0 ) . We can conclude that positive (negative) energy solutions propagate forward (backward) in spacetime. This is what we need in the relativistic quantum physics in contrast to the classical theory (for example in classical electrodynamics), 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. 6.2 From (6.5) and (6.8) it follows that (we take y = 0)

3 i d k i(ωk t+k·x) e , ∆F (x) − ∆R (x) = − 3 (2π) 2ωk since θ(t) + θ(−t) = 1. By applying (  + m2 ) on (6.9) we get (  + m2 )[∆F (x) − ∆R (x)] = 0. 6.3

I=

d4 kδ(k 2 − m2 )θ(k0 )f (k)

(6.9)

134

Solution

=

d4 kδ(k02 − ωk2 )θ(k0 )f (k)

1 [δ(k0 − ωk ) + δ(k0 + ωk )] θ(k0 )f (k) 2ωk 

3  d k f (k) . = 2ωk =

d3 kdk0

k0 =ωk

From this calculation it is clear that the expression d3 k/(2ωk ) is a Lorentz invariant measure. 6.5 Let 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

ωk −ρ

R

−ωk −ρ

+ + + + + =0. (6.10) Cρ−

−R

−ωk +ρ

Cρ+

ωk +ρ

CR

Fig. 6.5. The integration contour that defined the principal-part propagator

The integral along the half–circle, CR tends to zero for large R,  which can be seen if we take that limit in the integrand. If in the integral Cρ+ we take k0 = ωk + ρeiϕ , it becomes



= Cρ+

0

π

ie−ix0 (ωk +ρe

iϕ

)

1 dϕ . ρeiϕ + 2ωk

By taking ρ → 0 in (6.11) we get

iπ −iωk x0 =− e . + 2ω k Cρ In the same way we can show that

iπ iωk x0 = e . 2ωk Cρ− From (6.10), (6.12) and (6.13) we get (for x0 < 0)

(6.11)

(6.12)

(6.13)

Chapter 6. Green functions

¯ ∆(x) =

iπ (2π)4



 d3 k ik·x  −iωk x0 e e − eiωk x0 θ(−x0 ). 2ωk

The case x0 > 0 is analogous to the previous one. The result is

3  iπ d k ik·x  −iωk x0 ¯ ∆(x) =− e e − eiωk x0 θ(x0 ) . 4 (2π) 2ωk

135

(6.14)

(6.15)

By comparing equations (6.14) and (6.15) with the expressions for ∆R and ∆A we obtain 1 ¯ ∆(x) = (∆R (x) + ∆A (x)) . 2 6.6

3 i d k ik·x −iωk t ∆(x) = − e (e − eiωk t ) , (6.16) 3 (2π) 2ωk

3 i d k i(k·x∓ωk t) ∆± (x) = ∓ e . (6.17) (2π)3 2ωk 6.7 By using the expression for ∆ obtained in Problem 6.6 we get

3 i d k iki eik·x (e−iωk t − eiωk t ) = 0 , ∂i ∆(x) = − 3 (2π) 2ωk

(6.18)

since the integrand is an odd function of k. The second identity can be proven easily. 6.8 By applying the operator (  + m2 ) to the expression (6.16) we get

3 ! ( i d k 2 2 2 i(−ωk t+k·x) i(ωk t+k·x) , (−ω + k + m ) e − e (  + m2 )∆(x) = − k (2π)3 2ωk from which follows that (  + m2 )∆(x) = 0, as k 2 = m2 . 6.9 For m = 0 from (6.8) it follows that

3 ! d k ik·x ( −ikx0 i e ∆F |m=0 = − θ(x0 ) + eikx0 θ(−x0 ) e 3 (2π) 2k

∞ π i =− k sin θdkdθ 2(2π)2 0 0 ! ( × eik(−t+r cos θ) θ(t) + eik(t+r cos θ) θ(−t) ,

(6.19)

where in the second line we integrated over the polar angle ϕ. Integration over θ gives

∞ ( 1 ∆F (x)|m=0 = − dk (e−ik(t−r) − e−ik(t+r) )θ(t) 2(2π)2 r 0 ! (6.20) +(eik(t+r) − eik(t−r) )θ(−t) .

136

Solution

Now, we shall consider separately two cases: t > 0 and 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, where  → 0+ . In this way we ensure convergence of this integral. Then from (6.20) it follows that  1 1 i ∆F |m=0 = − (6.21) 2(2π)2 r t − r − i t + r − i 1 1 i i = . (6.22) = 2 2 2 2 2 (2π) t − r − i (2π) x − i By applying the formula 1 1 = P ∓ iπδ(z) , z ± i z

(6.23)

in expression (6.22) we get ∆F (x) |m=0 = −

1 i 1 δ(x2 ) + 2 P 2 . 4π 4π x

(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 θ and ϕ leads to

∞ ( ! 1 −ik(t−r) ik(t+r) −ik(t+r) ik(t−r) θ(t) . dk e − e − e + e ∆R (x) = − 2(2π)2 r 0 (6.25) The change of variable k
= −k in the third and the fourth integral in expression (6.25) gives

∞ 1 ∆R (x) = − dk(e−ik(t−r) − eik(t+r) )θ(t) . (6.26) 2(2π)2 r −∞ Note the change of the lower integration limit in the expression (6.26). From (6.26) follows ∆R |m=0 (x) = −

1 [δ(t − r) − δ(t + r)] θ(t) . 4πr

(6.27)

The second term in (6.27) has a ”wrong” sign but it is irrelevant as this term vanishes (t > 0 and r > 0). By changing this minus into a plus in (6.27) we finally obtain: ∆R |m=0 (x) = −

1 1 δ(t2 − r2 )θ(t) = − δ(x2 )θ(t) . 2π 2π

The case of advanced Green function is left for an exercise.

(6.28)

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 k 2 − m2 → k 2 − m2 + iηk0 in the propagator denominator, where η is a small positive number. Therefore,

e−ik·(x−y) d4 k . (6.29) ∆R (x − y) = (2π)4 k 2 − m2 + iηk0 Now the poles of the integrand in (6.29) are k0 = ±ωk − iη/2. From (6.6) and (6.29) we have



g e−ik·x 4 ik0 y0 φR (x) = − k e d dy d3 yδ (3) (y)e−ik·y . 0 (2π)4 k 2 − m2 + iηk0 (6.30) First in (6.30) we shall integrate over y0 , then over y and finally over k0 ; this gives

g eik·x 3 k . (6.31) φR (x) = d (2π)3 k 2 + m2 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
= −k )

∞ kdk g φR (x) = − e−ikr . (6.32) 2 2 (2π) ir −∞ k + m2

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) = which is the requested result.

g −mr e , 4πr

(6.33)

138

Solution

6.12 Apply i/ ∂ − m on S(x). 6.13 The Fourier transformation of the equation (i/ ∂ −m)S(x−y) = δ (4) (x−y) leads to



1 1 4 ˜ −ip·(x−y) (i/ ∂ − m) = (6.34) d pS(p)e d4 pe−ip·(x−y) . (2π)4 (2π)4 From (6.34) follows /+m p ˜ S(p) = 2 . p − m2 Therefore, the Green function is given by

/p + m d4 p e−ip·(x−y) . (6.35) S(x − y) = 2 4 (2π) p0 − p2 − m2 The poles of the integrand in (6.35) are p0 = ±Ep = ± p2 + m2 . The propagator is



p0 γ 0 + pi γi + m −ip0 (x0 −y0 ) 1 3 ip·(x−y) pe dp0 e , SF (x − y) = d 4 (2π) p20 − Ep2 CF (6.36) where the integration contour CF is defined in Problem 6.1. Applying the Cauchy theorem we get

3 i d p ip·(x−y) e SF (x − y) = − (2π)3 2Ep ( (Ep γ 0 + pi γ i + m)e−iEp (x0 −y0 ) θ(x0 − y0 )+ ! +(−Ep γ 0 + pi γ i + m)eiEp (x0 −y0 ) θ(y0 − x0 )

3 ( i d p (/ p + m)e−ip·(x−y) θ(x0 − y0 )− =− (2π)3 2Ep ! −(/ p − m)eip·(x−y) θ(y0 − x0 ) . (6.37) The advanced Green function can be found in the same way. The result is

3 d p ip·(x−y) ( i e SA (x − y) = (Ep γ 0 + pi γ i + m)e−iEp (x0 −y0 ) − (2π)3 2Ep ! −(−Ep γ 0 + pi γ i + m)eiEp (x0 −y0 ) θ(y0 − x0 ) . (6.38) For simplicity we take y = 0 in (6.37) and (6.38). We have

3 i d p i(p·x−Ep x0 ) e (Ep γ 0 + pi γ i + m)(θ(x0 ) + θ(−x0 )) SF − S A = − (2π)3 2Ep

3 i d p i(p·x−Ep x0 ) =− e (Ep γ 0 + pi γ i + m) . (6.39) (2π)3 2Ep

Chapter 6. Green functions

Thus, SF − SA = −

i (2π)3



d3 p (Ep γ 0 + pi γi + m)e−ip·x . 2Ep

139

(6.40)

By applying i/ ∂ − m on (6.40) we get (i/ ∂ − m)(SF − SA ) = 0, since (/ p + m)(/ p − m) = p2 − m2 = 0. 6.14 The integration along the curve CF is equivalent to the integration along the real p0 –axis if we make the replacement p2 − m2 → p2 − m2 + i, where  is a small positive number in the propagator denominator. The simple poles are p0 = ±Ep ∓ i. So we get

ψ(x) =

g (2π)4





d4 y

dp0

  1 /p + m 3 −ip·(x−y) iq·y  0  e d p 2 δ(y0 )e   . 0 p − m2 + i 0

After the integration over the variables y0 and y we get

ψ(x) =

g 2π



  1 p / + m 0  e−i(p0 x0 −p·x) δ (3) (p − q)   . dp0 d3 p 2 2 0 p − m + i 0

(6.41)

Integration over the momentum p is simple and it gives

ψ(x) =

g iq·x e 2π

  1 p0 γ0 − q · γ + m −ip0 x0  0  e dp0 2   . 0 p0 − q2 − m2 + i −∞ 0



∞

(6.42)

Employing the Cauchy theorem we find that ig iq·x  (−Eq γ0 − q · γ + m)eiEq x0 θ(−x0 ) e 2Eq   1 0 −iEq x0 θ(x0 )   , +(Eq γ0 − q · γ + m)e 0 0

ψ(x) = −

(6.43)

which finally gives: ig iq·x e 2Eq       −Eq + m Eq + m 0 0      −iE x  × eiEq x0   θ(−x0 ) + e q 0   θ(x0 ) . (6.44) q3 q3 q1 + iq2 q1 + iq2

ψ(x) = −

140

Solution

6.15 The equation for the free massive vector field Aµ is given by (g ρσ   − ∂ ρ ∂ σ + m2 g ρσ )Aσ = 0 .

(6.45)

The Green function (it is in fact the inverse kinetic operator) is defined by (g ρσ   − ∂ ρ ∂ σ + m2 g ρσ )x Gσν (x − y) = δ (4) (x − y)δνρ . If we introduce Gσν

1 = (2π)4



(6.46)

˜ σν (k) , d4 ke−ik·(x−y) G

in (6.46), we get ˜ σν = δνρ . (−k 2 g ρσ + k ρ k σ + m2 g ρσ )G

(6.47)

˜ ρσ = Ak 2 gρσ + We shall assume that the solution of (6.47) has the form G Bkρ kσ , where A and B are scalars, i.e. they depend on k 2 and m2 . Inserting the solution into (6.47), after comparing of the appropriate coefficients, we get 1 1 . A= , B=− 2 2 −k 4 + k 2 m2 m (m − k 2 ) The final result takes the following form  1 kµ kν ˜ µν = G + −g . µν k 2 − m2 m2

(6.48)

6.16 Use the same procedure as in the previous problem. The result is ˜ µν = − gµν + 1 + λ kµ kν . G k2 λk 4

7 Canonical quantization of the scalar field

7.1 Starting from the expressions for scalar field φ and its canonical momen˙ tum π = φ,

  d3 k φ= a(k)e−ik·x + a† (k)eik·x , 3 2(2π) ωk

  d3 k φ˙ = i ωk −a(k)e−ik·x + a† (k)eik·x , (2π)3 2ωk we have


(2π)3/2  a(k
)e−iωk
t + a† (−k
)eiωk
t , (7.1) d3 xφ(x)e−ik ·x = √ 2ωk


 ωk
 † −ik
·x ˙ a (−k
)eiωk
t − a(k
)e−iωk
t . (7.2) = i(2π)3/2 d3 xφ(x)e 2

From (7.1) and (7.2) it follows that

( ! 1 1 3 ik·x ˙ √ ω xe φ(x) + i φ(x) , a(k) = d k (2π)3/2 2ωk

! ( 1 1 ˙ √ a† (k) = d3 xe−ik·x ωk φ(x) − iφ(x) . 3/2 (2π) 2ωk

(7.3) (7.4)

By using the expressions (7.3) and (7.4), we find:


1 i ˙ d3 xd3 yei(k·x−k ·y) −ωk [φ(x), φ(y)]+ [a(k), a† (k
)] = √ 3 2(2π) ωk ωk
 ˙ +ωk
[φ(x), φ(y)]


1 1 d3 xei(ωk −ωk
)t+i(k −k)·x (ωk + ωk
) = √ 2(2π)3 ωk ωk
= δ (3) (k − k
) .

(7.5)

142

Solutions

In the previous formula, we used the equal–time commutation relations for real scalar field (7.C) i.e. we took1 x0 = y 0 . We can do this because the creation and annihilation operators are time independent. This can be proved directly:

  1 1 da(k) √ = d3 xeik·x iωk2 φ + i∇2 φ − im2 φ . 3/2 dt (2π) 2ωk After two partial integrations in the second term we get

  i 1 da(k) √ = d3 xeik·x ωk2 − k2 − m2 φ . dt (2π)3/2 2ωk The dispersion relation, ωk2 = m2 + k2 gives da(k)/dt = 0. It is clear that a† (k) is also time independent. Similarly, we can prove that: [a(k), a(k
)] = [a† (k), a† (k
)] = 0 . 7.2 In this problem, φ(x) is a classical field, so that a(k) and a† (k) are the coefficients rather then operators. We can calculate them from the expressions ˙ = 0, x) = c: (7.3) and (7.4) inserting φ(t = 0, x) = 0 and φ(t

1 1 √ a(k) = d3 xe−ik·x ic (2π)3/2 2ωk ic = √ (2π)3/2 δ (3) (k) . 2m Then, the scalar field is φ(t, x) =

c sin(mt) . m

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) = [φ(x)∂µx ∆(x − y) − ∆(y − x)∂µ φ(x)]dΣ µ . σ

Solve this problem using the previous theorem. 7.3 The results are:



  d3 kωk a† (k)a(k) + b† (k)b(k) ,

  : Q : = q d3 k a† (k)a(k) − b† (k)b(k) ,

  : P : = d3 kk a† (k)a(k) + b† (k)b(k) .

:H:=

1

This will be done in the forthcoming problems, too.

(7.6) (7.7) (7.8)

Chapter 7. Canonical quantization of the scalar field

143

7.4 (up , uk ) = δ (3) (k − p), (up , u∗k ) = 0. 7.5 From (2.9), we have

1 d3 kωk 0| (a† (k)a(k) + a(k)a† (k)) |0 2

1 d3 kωk 0| a(k)a† (k) |0 = 2

1 = d3 kωk (δ (3) (0) − 0| a† (k)a(k) |0) 2

1 = δ (3) (0) d3 k k2 + m2 2

∞ (3) = 2πδ (0) dkk 2 k 2 + m2 .

0| H |0 =

0

√ By change of variable k = m t, the last integral becomes Euler’s beta function πm4 (3) 3 0| H |0 = πm4 δ (3) (0)B( , −2) = − δ (0)Γ (−2) . 2 4 7.6 Use the formulae from Problem 7.3 and the commutation relations (7.D). (a) Direct calculation yields

3 3
( ! 1 d kd k µ † µ
−ik
·x †
ik
·x √ a [P , φ] = k (k)a(k), a(k )e + a (k )e (2π)3/2 2ωk


3  1 d k µ √ k −a(k)e−ik·x + a† (k)eik·x = 3/2 (2π) 2ωk (7.9) = −i∂ µ φ . The same result can be obtained if we start from the transformation law of the field φ under translations (see Problem 7.20): φ(x + ) = ei·P φ(x)e−i·P = φ(x) + iµ [Pµ , φ(x)] + o(2 ) .

(7.10)

On the other hand, we have φ(x + ) = φ(x) + µ ∂µ φ + o(2 ) . From (7.11) and (7.10) the result (7.9) comes. (b) First, we calculate the commutator [P µ , φn (x)]: [P µ , φn (x)] = =

n  k=1 n 

φk−1 [P µ , φ]φn−k φk−1 (−i∂ µ φ)φn−k

k=1

= −i∂ µ φn .

(7.11)

144

Solutions

In the same way one can prove that [P µ , π n (x)] = −i∂ µ π n . As a consequence, [Pµ , φn (x)π m (x)] = −i∂µ (φn (x)π m (x)) . An arbitrary analytical function F (φ, π) can be expanded in series as  Cnm φn π m . F (φ, π) = nm

Then [Pµ , F (φ, π)] = −i∂µ F . †

(c) [H, a (k)a(q)] = (ωk − ωq )a† (k)a(q). (d) [Q, P µ ] = 0. (e) [H, N ] = 0.     ω (f) d3 x[H, φ(x)]e−ip·x = (2π)3/2 2p −a(p)e−iωp t + a† (−p)eiωp t 7.7 From the Baker–Hausdorff relation follows eiQ φe−iQ = φ + i[Q, φ] +

i2 [Q, [Q, φ]] + . . . . 2!

(7.12)

The first commutator in the previous expansion is given by

[Q, φ] = iq d3 y[φ† (y)π † (y) − φ(y)π(y), φ(x)]

= −q d3 yδ (3) (x − y)φ(y) = −qφ(x) . Then [Q, [Q, φ]] = (−q)2 φ , Finally, iQ

e φe

−iQ

[Q, [Q, [Q, φ]]] = (−q)3 φ , . . .

 (−iq)2 + . . . φ = e−iq φ . = 1 − iq + 2

(7.13)

(7.14)

7.8 The angular momentum of a scalar field has the form

M µν = d3 x(xµ T 0ν − xν T 0µ ) . (a) By inserting the previous formula in the commutator, we have

˙ ν φ − g0ν L) − yν (φ∂ ˙ µ φ − g0µ L), φ(x)] . (7.15) [Mµν , φ(x)] = d3 y[yµ (φ∂ 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)∂µ φ(y), φ(x)] = −i∂µ φδ (3) (x − y) − iδµ0 π(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, e 2 ωµν M φ(x)e− 2 ωµν M i

µν

i

µν

= φ(Λ−1 (ω)x) .

(b) We first calculate the commutator [Mµν , P0 ]:

[Mµν , P0 ] = d3 x[xµ T0ν − xν T0µ , P0 ]

= d3 x (xµ [T0ν , P0 ] − xν [T0µ , P0 ])

= i d3 x (xµ ∂0 T0ν − xν ∂0 T0µ )

  = i d3 x −xµ ∂i T iν + xν ∂i T iµ

  = i d3 x gµi T iν − giν T iµ

  = i d3 x Tµν − gµ0 T 0ν − Tνµ + g0ν T 0µ = −i(gµ0 Pν − gν0 Pµ ) .

(7.17)

In (7.17), we used the results of Problem 7.6 (b), the continuity equation ∂µ T µν = 0 and integrated by parts. In the case λ = i we can use of a partial integration. The result is [Mµν , Pi ] = −i(giµ Pν − giν Pµ ). Thus, [Mµν , Pλ ] = i(gλν Pµ − gλµ Pν ) . (c) Let us calculate firstly the commutator [Mij , Mkl ].

(7.18)

146

Solutions



( ˙ ˙ d3 xd3 y xi φ(x)∂ j φ(x) − xj φ(x)∂i φ(x), ! ˙ ˙ yk φ(y)∂ l φ(y) − yl φ(y)∂k φ(y)

 ˙ ˙ = d3 xd3 y xi yk [φ(x)∂ j φ(x), φ(y)∂l φ(y)]−

[Mij , Mkl ] =

˙ ˙ −xi yl [φ(x)∂ j φ(x), φ(y)∂k φ(y)] ˙ ˙ −xj yk [φ(x)∂ i φ(x), φ(y)∂l φ(y)]

 ˙ ˙ . +xj yl [φ(x)∂ i φ(x), φ(y)∂k φ(y)]

(7.19)

Applying the equal–time commutation relations, we obtain2

 ( x ˙ [Mij , Mkl ] = i d3 xd3 y xi yk φ(x)∂ l φ(y)∂j  y (3) ˙ − φ(y)∂ (x − y) j φ(x)∂l δ   y x (3) ˙ ˙ − xi yl φ(x)∂ (x − y) k φ(y)∂j − φ(y)∂j φ(x)∂k δ   y x (3) ˙ ˙ − xj yk φ(x)∂ (x − y) l φ(y)∂i − φ(y)∂i φ(x)∂l δ   ! y x (3) ˙ ˙ + xj yl φ(x)∂ (x − y) . k φ(y)∂i − φ(y)∂i φ(x)∂k δ If we use the relation x (3) y (3) ∂m δ (x − y) = −∂m δ (x − y)

we obtain



[Mij , Mkl ] = −i d3 xd3 y    y (3) x (3) ˙ ˙ (x − y) − φ(y)∂ (x − y) xi yk φ(x)∂ l φ(y)∂j δ j φ(x)∂l δ   y (3) x (3) ˙ ˙ (x − y) − φ(y)∂ (x − y) −xi yl φ(x)∂ k φ(y)∂j δ j φ(x)∂k δ   y (3) x (3) ˙ ˙ (x − y) − φ(y)∂ (x − y) −xj yk φ(x)∂ l φ(y)∂i δ i φ(x)∂l δ   y (3) x (3) ˙ ˙ +xj yl φ(x)∂k φ(y)∂i δ (x − y) − φ(y)∂i φ(x)∂k δ (x − y) . By performing partial integrations in the last expression, we obtain 2

We have used the following notation: x ∂m =

∂ ∂ ; ∂xm = . ∂xm ∂xm

Chapter 7. Canonical quantization of the scalar field

[Mij , Mkl ] = −i

147

 3 ˙ ˙ d x gjk (xl φ(x)∂ i φ(x) − xi φ(x)∂l φ(x))

˙ ˙ +gil (xk φ(x)∂ j φ(x) − xj φ(x)∂k φ(x)) ˙ ˙ +gik (xj φ(x)∂ l φ(x) − xl φ(x)∂j φ(x))  ˙ ˙ +gjl (xi φ(x)∂k φ(x) − xk φ(x)∂i φ(x)) = i(gjk Mil + gli Mjk − gik Mjl − gjl Mik ) .

(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

1 a b d3 xd3 yτij τmn [Qa , Qb ] = − 4 ( ! × φ˙ †i (x)φj (x) − φ†i (x)φ˙ j (x), φ˙ †m (y)φn (y) − φ†m (y)φ˙ n (y) . Recall that as the charges are time-independent we can work with the equal–time commutators and we have

  i a b d3 x φ˙ † [τ a , τ b ]φ − φ† [τ a , τ b ]φ˙ . [Q , Q ] = − 4 By using [τ a , τ b ] = 2iabc τ c , we get [Qa , Qb ] = iabc Qc . The second case is similar to the previous one:



[Qi , Qj ] = imn jpq d3 x d3 y[φm (x)φ˙ n (x), φp (y)φ˙ q (y)]

= i d3 x(−imn jnq φm φ˙ q + imn jpm φp φ˙ n )

= i d3 x(δij φm φ˙ m − φj φ˙ i − δij φm φ˙ m + φi φ˙ j )

d3 x(φi φ˙ j − φj φ˙ i )

= iijk kmn d3 xφm φ˙ n =i

= iijk Qk . 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  symbols.

148

Solutions

(b) The commutator between the charges Qa and the field φm can be found similarly:

i a ˙† [Qa , φm (x)] = − [φi (y)φj (y) − φ†i (y)φ˙ j (y), φm (x)] d3 yτij 2

i a = − τij d3 y[φ˙ †i (y), φm (x)]φj (y) 2

1 a = − τij d3 yδ (3) (x − y)δim φj (y) 2 1 a φj (x) . = − τmj 2 In the same way, we find: [Qa , φ†m (x)] =

1 a † τ φ . 2 im i

The previous two results can be rewritten in the form [θa Qa , φm (x)] = iδ0 φm (x) , †

†

[θa Qa , φm (x)] = iδ0 φm (x) . In the case of SO(3) symmetry, the calculation is the same as above. The result is [Qk , φm (x)] = ikmj φj (x) . 7.11 The dilatation current is j µ = −φ∂ µ φ − xν ∂ µ φ∂ν φ + Lxµ . (a) The dilatation generator is 

1 0 ˙2 3 i ˙ i ˙ D = − d x φφ + x φ∂i φ + x (φ − ∂i φ∂ φ) . 2 (b) The commutator between the generator D and the field φ(x) is given by

[D, φ(y)] = − d3 x[φ(x)π(x) + xi π(x)∂i φ(x) 1 0 2 1 x π (x) − x0 ∂i φ(x)∂ i φ(x), φ(y)] 2

2  3 = − d x φ(x)[π(x), φ(y)] + x0 π(x)[π(x), φ(y)]  + xi [π(x), φ(y)]∂i φ(x) .

+

By using the commutation relations (7.C), we have

Chapter 7. Canonical quantization of the scalar field

ρ[D, φ(y)] = iρ(φ(y) + y 0 π(y) + y i ∂i φ) = iρ(φ(y) + y µ ∂µ φ(y)) = iδ0 φ . In the same way, we obtain: ρ[D, π(x)] = iρ(2π + xµ ∂µ π) = iδ0 π . (c) By applying the previous result, we easily get ρ[D, φ2 ] = ρ([D, φ]φ + φ[D, φ]) = i((δ0 φ)φ + φδ0 φ) = iδ0 (φ2 ) , and generally ρ[D, φa ] = iδ0 (φa ) . Similarly, one can show that ρ[D, π a ] = iδ0 (π a ) . An arbitrary analytic function can be expanded in the following form  cab φa π b , F (φ, π) = ab

so that ρ[D, F ] = ρ

 a,b

=ρ

 a,b

=i



cab [D, φa π b ]   cab [D, φa ]π b + φb [D, π b ]

  cab δ0 (φa )π b + φa δ0 (π b )

a,b

   cab φa π b  = iδ0  a,b

= iδ0 F . (d) We first consider the case µ = i:

i

[D, P ] = d3 x[D, π∂ i φ]

  = d3 x π[D, ∂ i φ] + [D, π]∂ i φ . By using part (b) of this problem, we obtain

149

150

Solutions



 d3 x (2π + x0 ∂0 π + xj ∂j π)∂ i φ  + π(2∂ i φ + x0 ∂ i π + xj ∂ i ∂j φ) .

[D, P i ] = i

(7.21)

The second term in this expression is transformed in the following way



1 3 0 k i 3 0 k i d xx ∂k ∂ φ∂ φ = − d xx ∂ φ∂k ∂ φ = − d3 x∂ i (x0 ∂k φ∂ k φ) , 2 where we used the Klein-Gordon equation, ∂0 π = −∂ i ∂i φ and then performed a partial integration. Thus, we conclude  that the second term can be dropped as a surface term. The expression d3 xπx0 ∂ i π is also a surface term. Similarly, one can show that



d3 xxj ∂j π∂ i φ = −3 d3 xπ∂ i φ − d3 xxj π∂j ∂ i φ . Inserting these results in the formula (7.21) we obtain [D, P i ] = iP i . The commutator [D, P 0 ] = iP 0 can be calculated in the same way. 7.12 In the expression for the vacuum expectation value, express the fields φf in terms of the creation and annihilations operators. From four terms, only one, which is proportional to 0| a(k)a† (k
) |0 = δ (3) (k − k
), is nonzero. Then, we have 1 1 0| φf (t, x)φf (t, x) |0 = 2 3 (a π) (2π)3



d3 k 2ωk



3

d ye

−(x−y)2 /a2 +ik·(x−y)

2 .

Calculating the Poisson integral in this formula, we obtain

3 1 d k −k2 a2 /2 0| φf (t)φf (t) |0 = e 2(2π)3 ωk

∞ 2 2 k 2 dk 1 √ = e−k a /2 . 2 2 (2π)2 0 k +m By the change of variable k 2 = t, the last integral becomes

∞ √ tdt −ta2 /2 1 √ 0| φf (t)φf (t) |0 = e 2 8π 0 t + m2   m2 m2 a2 /4 m2 a2 m2 a2 ) − K0 ( ) , (7.22) = e K1 ( 16π 2 4 4 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

K1 (x) =

151

1 , x

K0 (x) = −(log(x/2) + 0, 5772) for x  1, we obtain in the limit m → 0 0| φf (t)φf (t) |0 =

1 . 4π 2 a2

µ and use the commu7.13 Express the operators Lm and Ln in terms of αm tation relations.

7.14 After a very simple calculation, we find that

∞  0 0 i 1 −k 0| {φ(x), φ(y)} |0 = eik(y −x −|x−y|) lim dke →0 2 2(2π) |x − y| 0 − eik(y

0

−x0 +|x−y|)

0

+ eik(x  0 0 − eik(x −y +|x−y|) .

−y 0 −|x−y|)

(7.23)

The integrals in the previous expression are regularized by introducing  as a regularization parameter. At the end we have to take the limit  → 0. The result is 1 1 0| {φ(x), φ(y)} |0 = − 2 . 2π (x − y)2 7.15 The vacuum expectation value φ(x)φ(y) is given by   φ(x)φ(y) = φ+ (x)φ− (y)

d3 q d3 k √ ei(q·y−k·x) δ (3) (k − q) = (2π)3/2 2ωk (2π)3/2 2ωq

3 d k −ik·(x−y) 1 e , = 3 (2π) 2ωk where we split the field φ into 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:   φ(x1 )φ(x2 )φ(x3 )φ(x4 ) = φ+ (x1 )φ+ (x2 )φ− (x3 )φ− (x4 )   + φ+ (x1 )φ− (x2 )φ+ (x3 )φ− (x4 ) . (7.24) The first term in the last expression is 

4

  φ (x1 )φ (x2 )φ (x3 )φ (x4 ) = +

+

−

−

i=1

/ 0 d3 qi † † √ a a a a 1 2 3 4 (2π)3/2 2ωi

× ei(−q1 ·x1 −q2 ·x2 +q3 ·x3 +q4 ·x4 ) ,

152

Solutions

where ai = a(qi ). Using the relation / 0 / 0 a1 a2 a†3 a†4 = a1 (δ (3) (q2 − q3 ) + a†3 a2 )a†4 / 0 = δ (3) (q2 − q3 )δ (3) (q1 − q4 ) + a1 a†3 (δ (3) (q2 − q4 ) − a†4 a2 ) = δ (3) (q2 − q3 )δ (3) (q1 − q4 ) + δ (3) (q1 − q3 )δ (3) (q2 − q4 ) , we obtain

3 1 d q1 d3 q2 −iq2 ·(x2 −x3 )−iq1 ·(x1 −x4 ) e (2π)6 2ω1 2ω2

3 1 d q1 d3 q2 −iq2 ·(x2 −x4 )−iq1 ·(x1 −x3 ) + e (2π)6 2ω1 2ω2 = φ(x2 )φ(x3 ) φ(x1 )φ(x4 ) + φ(x1 )φ(x3 ) φ(x2 )φ(x4 ) .

 +  φ (x1 )φ+ (x2 )φ− (x3 )φ− (x4 ) =

The following result can be derived in the same way:  +  φ (x1 )φ− (x2 )φ+ (x3 )φ− (x4 ) = φ(x1 )φ(x2 ) φ(x3 )φ(x4 ) . By adding two last expressions, we get φ(x1 )φ(x2 )φ(x3 )φ(x4 ) = φ(x1 )φ(x3 ) φ(x2 )φ(x4 ) + φ(x1 )φ(x4 ) φ(x2 )φ(x3 ) + + φ(x1 )φ(x2 ) φ(x3 )φ(x4 ) . This result is a special case of Wick’ s theorem. 7.16 Scalar field in two dimensional spacetime can be represented as

∞ ( ! µ µ dk a(k)e−ikµ x + a† (k)eikµ x , φ(x) = (2π)2ωk −∞ so that φ(x)φ(y) =

1 4π



∞

−∞

dk i|k|(y0 −x0 )−ik(y−x) e . |k|

(7.25)

If we introduce the notation y0 − x0 = τ , y − x = r, the previous integral becomes

∞  dk  ik(τ −r) 1 (7.26) e + eik(τ +r) . φ(x)φ(y) = 4π 0 k Denoting the integral in (7.26) by I and introducing the regularization parameter , we get:

∞   ∂I i = lim→0 dke−k eik(τ −r) + eik(τ +r) ∂τ 4π 0 τ 1 . (7.27) =− 2π τ 2 − r2

Chapter 7. Canonical quantization of the scalar field

153

From (7.27), it follows that φ(x)φ(y) = −

τ 2 − r2 (x − y)2 1 1 log log = − , 4π µ2 4π µ2

where µ is an integration constant which has the dimension of length. 7.17 By taking partial derivative of the expression 0| T (φ(x)φ(y)) |0 with respect to x0 , we get: ∂x0 0| T (φ(x)φ(y)) |0 = δ(x0 − y0 ) 0| [φ(x), φ(y)] |0 + + θ(x0 − y0 ) 0| ∂x0 φ(x)φ(y) |0 + θ(y0 − x0 ) 0| φ(y)∂x0 φ(x) |0 . 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: ∂x20 0| T (φ(x)φ(y)) |0 = δ(x0 − y 0 )[π(x), φ(y)] + θ(x0 − y 0 ) 0| ∂x20 φ(x)φ(y) |0 + + θ(y 0 − x0 ) 0| φ(y)∂x20 φ(x) |0 . In the first term, we use the equal–time commutation relation, and finally get the result ∂x20 0| T (φ(x)φ(y)) |0 = −iδ (4) (x − y) + + θ(x0 − y 0 ) 0| ∂x20 φ(x)φ(y) |0 + + θ(y 0 − x0 ) 0| φ(y)∂x20 φ(x) |0 , which implies ( x + m2 ) 0| T (φ(x)φ(y)) |0 = −iδ (4) (x − y) + + θ(x0 − y0 ) 0| ( x + m2 )φ(x)φ(y) |0 + + θ(y0 − x0 ) 0| φ(y)( x + m2 )φ(x) |0 . The last two terms vanish since the field φ satisfies the Klein–Gordon equation. Therefore, ( x + m2 ) 0| T (φ(x)φ(y)) |0 = −iδ (4) (x − y) . (7.28) 7.18 (a) Applying the variational principle to the given action leads to the equations:  1 ∂ψ = − ∆ + V (r) ψ i ∂t 2m  † † 1 ∂ψ = − ∆ + V (r) ψ . −i ∂t 2m 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−iEk t+ik·r , where Ek = k 2 /2m so that the general solution is

d3 k ψ(t, r) = a(k)e−iEk t+ik·r . (7.29) (2π)3/2 The negative energy solutions are not present in previous expression since Ek > 0 in nonrelativistic quantum mechanics. The field ψ † is

d3 k † ψ † (t, r) = a (k)eiEk t−ik·r . (7.30) (2π)3/2 In the quantum theory these classical fields are replaced by operators in the Hilbert space. The field conjugate to ψ is π=

∂L = iψ † . ∂ ψ˙

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

1 iEk t a(k) = e d3 xψ(t, x)e−ik·x (2π)3/2

1 † −iEk t a (k) = e d3 xψ † (t, x)eik·x . (2π)3/2 From (7.31) and previous relations one easily gets the commutation relations: (7.32) [a(k), a† (p)] = δ (3) (p − k) , [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 0| ψ(x0 , x)ψ (y0 , y) |0 θ(x0 − y0 )

i =− d3 kd3 pe−i(Ek x0 −k·x−Ep y0 +p·y) (2π)3 × 0| a(k)a† (p) |0 θ(x0 − y0 )

i =− d3 kd3 pe−i(Ek x0 −k·x−Ep y0 +p·y) (2π)3 × δ (3) (p − k)θ(x0 − y0 )

k2 i 3 −i 2m (x0 −y 0 )+ik·(x−y) =− ke θ(x0 − y0 ) d (2π)3  3/2 im(x−y)2 m = −i e 2(x0 −y0 ) θ(x0 − y0 ) . 2πi(x0 − y0 )

Chapter 7. Canonical quantization of the scalar field

(d) The eigenfunctions are

 uk =

155

2 sin(kx) , π

hence the (nonrelativistic) field operators are  ∞ k2 2 ψ= dka(k)e−i 2m t sin(kx) , π 0  ∞ k2 2 † ψ = dka† (k)ei 2m t sin(kx) . π 0

(7.34)

(7.35)

We shall leave to the reader to prove that  G(x0 , x, y0 , y) = −i

m 2πi(x0 − y0 )

 1/2  im(x−y)2 im(x+y)2 2(x0 −y0 ) 2(x0 −y0 ) −e e 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 e−iEn (x0 −y0 ) un (x)u∗n (y)θ(x0 − y0 ) . (7.37) n

(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 components of the conserved currents. For the spatial translations the time component of the current is j0 = −

∂L ∂i ψi ∂(∂0 ψ) †

= −iψ † ∂i ψi = −iψ ∇ψ ·  . The conserved charge is the linear momentum

P = − d3 xψ † (i∇)ψ .

The Hamiltonian H=

d3 xψ † (−

1 )∆ψ 2m

is generator of time translations. The angular momentum

J = −i d3 xψ † (x × ∇)ψ

(7.38)

(7.39)

(7.40)

(7.41)

is generator of rotations. Under Galilean boosts we have δxi = −vi t, δψ = −imv · xψ so that

156

Solutions

j0 = v · j0 = mv · xψ † ψ + ivtψ † ∇ψ. Consequently, the boost generator is

G = d3 xψ † (mx + it∇)ψ .

(7.42)

(7.43)

The commutation relations can be found using the commutation relations (7.31). Let us start with [Pi , Gj ]:

[Pi , Gj ] = i d3 xd3 y[−ψ † (y)∂iy ψ(y), ψ † (x)(mxj + it∂j )ψ(x)]

 = −im d3 xd3 y ψ † (y)[∂i ψ(y), ψ † (x)xj ψ(x)]  + [ψ † (y), ψ † (x)xj ψ(x)]∂i ψ(y)

= −im d3 x(−∂i ψ † xj ψ(x) − xj ψ † ∂i ψ) = −iM δij , (7.44)  3 † where M = m d xψ ψ is the mass operator. It appears since the representation 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: [Gi , Gj ] = [H, P] = [H, J] = 0 [Ji , Jj ] = iijk Jk [Ji , Gj ] = iijk Gk [Ji , Pj ] = iijk Pk [H, Gi ] = −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

d3 q [a(p), a† (q)]f˜(q) [a(p), a† ] = C 2ωq

d3 q ˜ =C f (q)δ (3) (p − q) 2ωq 1 ˜ = C f (p) . 2ωp

(7.45)

The second commutator can be evaluated in the same way. The result is [a† (p), a] = −C

1 ˜∗ f (p) . 2ωp

(7.46)

Chapter 7. Canonical quantization of the scalar field

157

(b) Using (7.45), we have a(p)(a† )n = C

1 ˜ f (p)(a† )n−1 + a† a(p)(a† )n−1 . 2ωp

(7.47)

By repeating this procedure n times, we get a(p)(a† )n = C

1 nf˜(p)(a† )n−1 + (a† )n a(p) . 2ωp

(7.48)

Hence, nf˜(p) † n−1 [a(p), (a† )n ] = C (a ) . 2ωp

(7.49)

(c) This calculation is straightforward: a(p) |z = e−|z|

2

/2

= e−|z|

2

/2

a(p)

∞  z n (a† )n |0 n! n=0

∞ 

C z n f˜(p) † n−1 (a ) |0 2ωp (n − 1)! n=1

C ˜ = f (p)z |z . 2ωp

(7.50)

(d) By using the previous relation and the property z|z = 1, we have

  d3 p z| a(p) |z e−ip·x + z| a† (p) |z eip·x z| φ |z = 3/2 (2π) 2ωp

  3 d p ˜(p)e−ip·x + z ∗ f˜∗ (p)eip·x z f =C (2π)3/2 2ωp C = (zf (x) + z ∗ f ∗ (x)) . (7.51) (2π)3/2 In the same manner we have

 d3 q d3 p 2 z| : φ : |z = z| a(p)a(q) |z e−i(p+q)·x (2π)3/2 2ωp (2π)3/2 2ωq + z| a† (q)a(p) |z ei(q−p)·x + z| a† (p)a(q) |z ei(p−q)·x + z| a† (p)a† (q) |z ei(q+p)·x

 d3 q d3 p 2 =C f˜(p)f˜(q)z 2 e−i(p+q)·x (2π)3/2 2ωp (2π)3/2 2ωq + f˜(p)f˜∗ (q)|z|2 e−i(p−q)·x  + f˜∗ (p)f˜(q)|z|2 ei(p−q)·x + f˜∗ (p)f˜∗ (q)(z ∗ )2 ei(p+q)·x =

C2 2 (zf (x) + z ∗ f ∗ (x)) . (2π)3



(7.52)

158

Solutions

Hence, (∆φ)2 = 0 .

(7.53)

(e) It is easy to see that

z| H |z = C 2 |z|2

d3 p|f˜(p)|2 .

(7.54)

7.20 (a) By substituting the expression for φ in the relation U (Λ, a)φ(x)U −1 (Λ, a) = φ(Λx + a) we obtain

  d3 k √ U (Λ, a) a(k)e−ik·x + a† (k)eik·x U −1 (Λ, a) (2π)3/2 2ωk

  d3 k

−ik
·(Λx+a) †
ik
·(Λx+a) √ a(k . (7.55) = )e + a (k )e (2π)3/2 2ωk


In the integral on the right hand side we make the changing of variables k
µ Λµν = k ν . In Problem 6.3, we proved that d3 k/(2ωk ) is a Lorentz invariant measure, so that  d3 k
ωk
d3 k √ = . 2 ωk 2ωk
By performing the inverse Fourier transformation, we obtain the requested result. (b) It is easy to see that †

U (Λ, a) |k1 , . . . , kn  = U (Λ, a)a (k1 )U −1 (Λ, a)U (Λ, a) · · · †

· · · U (Λ, a)a (kn )U −1 (Λ, a) |0  ωk1
· · · ωkn
iaµ Λµ (kν +...+kν ) ν 1 n |Λk , . . . , Λk  . = e 1 n ωk 1 · · · ωk n (c) From the expressions (7.6) and (7.8) and the first part of this problem, we have

µ −1 U (Λ)P U (Λ) = d3 kk µ U (Λ)a† (k)a(k)U −1 (Λ)

ωk
† = d3 kk µ a (Λk)a(Λk) ωk

= Λνµ d3 k
k
ν a† (k
)a(k
) = Λνµ P ν , where we made the change of variables k µ = Λνµ k
ν in 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 = d3 kωk a† (k)a(k) . The Backer–Hausdorff relation reads 1 P HP −1 = eA He−A = H + [A, H] + [A, [A, H]] + . . . (7.56) 2   3  † where A = − iπ d q a (q)a(q) − ηp a† (q)a(−q) . The first commutator 2 in this expression is

  iπ [A, H] = − ηp d3 kωk a† (k)a(−k) − a† (−k)a(k) . 2 By changing k → −k in 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 τ Pτ −1 = −P, τ Hτ −1 = H †

†

7.23 The first step is to show that Cφ C −1 = ηc∗ φ, CπC −1 = ηc π and † Cπ C −1 = ηc π.

8 Canonical quantization of the Dirac field

8.1 If we use the anticommutation relation (8.E) the anticommutator iSab (x− y) = {ψa (x), ψ¯b (y)}, where a, b = 1, . . . , 4 are Dirac indices, becomes  1

m {ψa (x), ψ¯b (y)} = δrs δ (3) (p − q) d3 pd3 q 3 (2π) E E p q r,s  ub (q, s)ei(q·y−p·x) × ua (p, r)¯  + va (p, r)¯ vb (q, s)e−i(q·y−p·x) . Applying the solution of Problem 4.4 we have

3 ( ! 1 d p −ip·(x−y) ip·(x−y) iSab = (/ p + m) . e + (/ p − m) e ab ab (2π)3 2Ep

(8.1)

The last expression can be easily transformed into the following form

3 ( ! 1 d p −ip·(x−y) µ x ip·(x−y) ¯ {ψa (x), ψb (y)} = (iγ ∂µ + m)ab e . (8.2) − e (2π)3 2Ep From (8.2) we see that ∆(x − y) is given by

3 ( ! i d p −ip·(x−y) ip·(x−y) ∆(x − y) = − e . − e (2π)3 2Ep The function ∆(x − y) was defined in Problem 6.6. In the special case x0 = y0 we shall make change p → −p in the second term of expression (8.1) and obtain

d3 p ip·(x−y) e = (γ 0 )ab δ (3) (x − y) . (8.3) {ψa (x), ψ¯b (y)}|x0 =y0 = (γ 0 )ab (2π)3

162

Solutions

8.2 (a) Substituting (8.A,B) in the expression for charge Q we obtain

Q = −e d3 x : ψ † ψ : 

m  † cr (p)cs (p)u†r (p)us (p) = −e d3 p E p r,s + : dr (p)d†s (p) : vr† (p)vs (p) + c†r (p)d†s (−p)u†r (p)vs (−p)e2iEp t  (8.4) +dr (p)cs (−p)vr† (p)us (−p)e−2iEp t . From (4.52) and (8.4) we get 

  d3 p c†r (p)cr (p) − d†r (p)dr (p) . Q = −e

(8.5)

r

(b) As ψ satisfy the Dirac equation, (−iγ i ∂i + m)ψ = iγ0 ∂0 ψ the Hamiltonian is

H = i d3 x : ψ † ∂0 ψ :    1

m m  † 3 3 3 xd pd q : ur (p)c†r (p)eip·x d = 3 (2π) E E p q r,s    +vr† (p)dr (p)e−ip·x Eq us (q)cs (q)e−iq·x − vs (q)d†s (q)eiq·x : 

  = (8.6) d3 pEp c†r (p)cr (p) + d†r (p)dr (p) . r

(c) P=



  d3 pp c†r (p)cr (p) + d†r (p)dr (p) .

(8.7)

r

8.3 (a) It is easy to see that

  1 m 3 3 [H, ψ] = pd qE d p 3/2 E (2π) q r,s   † × cr (p)cr (p) + d†r (p)dr (p), cs (q)us (q)e−iq·x + d†s (q)vs (q)eiq·x

  1 m 3 3 pd qE δrs δ (3) (p − q) d = p 3/2 E (2π) q r,s   × −cr (p)us (q)e−iq·x + d†r (p)vs (q)eiq·x  d3 p   mEp −cr (p)ur (p)e−ip·x + d†r (p)vr (p)eip·x = 3/2 (2π) r = −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) = −δrs δ (3) (p − q)cr (p) , and the similar expression for d−operators. (b) If we had used commutation relations instead of anticommutation relations in the quantization process we would have obtained:   

† H= d3 pEp c†r (p)cr (p) − dr (p)dr (p) . r

From here we conclude that the energy spectrum would have been unbounded from below, which is physically unacceptable. 8.4 [H, c†r (p)cr (p)] =



s

= + =



d3 qEq [c†s (q)cs (q) + d†s (q)ds (q), c†r (p)cr (p)]  d3 qEq [c†s (q)cs (q), c†r (p)]cr (p)

s † cr (p)[c†s (q)cs (q), cr (p)]



 d3 qEq c†s (q){cs (q), c†r (p)}cr (p)

s

− {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))   = Ep c†r (p)cr (p) − c†r (p)cr (p) = 0 8.5 The form variation of a spinor field is δ0 ψ = δψ − δxµ ∂µ ψ = i = − ω µν σµν ψ − ω µν xν ∂µ ψ 4  i 1 µν = ω xµ ∂ν − xν ∂µ − σµν ψ . 2 2 On the other hand we have δ0 ψ = − 2i ω µν Mµν ψ . Comparing these results we conclude that the generators are given by 1 Mµν = i(xµ ∂ν − xν ∂µ ) + σµν . 2 8.6 (a) Applying the formula [AB, C] = A{B, C} − {A, C}B we obtain

164

Solutions

   1 † [Mµν , ψa (x)] = d y ψb (y) i(yµ ∂ν − yν ∂µ ) + σµν ψc (y), ψa (x) 2 bc 

1 † 3 = − d y{ψb (y), ψa (x)} i(yµ ∂ν − yν ∂µ ) + σµν ψc (y) 2 bc 1 = −[i(xµ ∂ν − xν ∂µ ) + σµν ]ac ψc (x) , 2

3

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 d3 xd3 y    1 × ψa† (x) i(xµ ∂ν − xν ∂µ ) + σµν ψb (x), ψc† (y)∂ρ ψc (y) . 2 ab First we suppose that all indices are the spatial: µ = i, ν = j, ρ = k. Then,

[Mij , Pk ] = i d3 xd3 y   1 1 † † × ψa (x) i(xi ∂j − xj ∂i ) + σij ψb (x), ψc (y) ∂k ψc (y) 2 ab  1 † † ψb (x) − ψc (y){ψa (x), ∂k ψc (y)} i(xi ∂j − xj ∂i ) + σij 2 ab

= i d3 xd3 y   1 × ψa† (x) i(xi ∂j − xj ∂i ) + σij δ (3) (x − y)∂k ψb (y) 2 ab  1 y (3) † − ψc (y)∂k δ (x − y)δac i(xi ∂j − xj ∂i ) + σij ψb (x) , 2 ab where we used the equal-time anticommutation relations (8.C,D). The integration over y leads to

  † † [Mij , Pk ] = i d3 x igjk ψ ∂i ψ − igik ψ ∂j ψ , or [Mij , Pk ] = i(gjk Pi − gik Pj ). Now we take µ = 0, ν = i, and ρ = k, i.e. we calculate the commutator [M0i , Pk ]. In order to do it first we are going to compute anticommutator ¯ {∂x0 ψ(x), ψ(y)}| x0 =y 0 .

Chapter 8. Canonical quantization of the Dirac field

165

Taking partial derivative of (8.1) with respect to x0 and substituting x0 = y 0 we get

( i 3 {∂x0 ψa (x), ψ¯b (y)}|x0 =y0 = p (−Ep γ 0 + p · γ − m)ab eip·(x−y) d 2(2π)3 ! + (Ep γ 0 − p · γ − m)ab e−ip·(x−y)

i d3 p(p · γ − m)ab eip·(x−y) = (2π)3 = γ ab ∇x δ (3) (x − y) − imδab δ (3) (x − y) . Then



[M0i , Pk ] = i d3 xd3 y   1 1 × ψa† (x) i(x0 ∂i − xi ∂0 ) + σ0i ψb (x), ψc† (y) ∂k ψc (y) 2 ab  1 † † − ψc (y){ψa (x), ∂k ψc (y)} i(x0 ∂i − xi ∂0 ) + σ0i ψb (x) 2 ab

 3 3 † x (3) = i d xd y ix0 ψ (x)∂i δ (x − y)∂k ψ(y) − ixi ψa† (x)(γγ0 ∇x − imγ0 )ac δ (3) (x − y)∂k ψc (y) 1 + ψa† (x) (σ0i )ab δ (3) (x − y)∂k ψb (y) 2 − ix0 ψ † (y)∂ky δ (3) (x − y)∂i ψ(x) + ixi ψ † (y)∂ky δ (3) (x − y)∂0 ψ(x) 1 − ψa† (y) (σ0i )ab ∂ky δ (3) (x − y)ψb (x) 2

  = i d3 x −ixi ψ † γγ 0 ∂k ∇ψ − mxi ψ † γ0 ∂k ψ − ixi ∂k ψ † ∂0 ψ

  ¯ 0 ∂0 + iγ∇ − m)∂k ψ . = i d3 x igik ψ † ∂0 ψ + xi ψ(iγ The second term in the last line vanishes since ψ satisfies the Dirac equation. Then we get [M0i , Pk ] = igik P0 . The remaining commutators [M0i , P0 ] and [Mij , P0 ] can be computed in the same way. 8.7 The helicity operator is 1 Sp = 2



d3 x : ψ †

Σ·p ψ: . |p|

(8.8)

166

Solutions †

Inserting expressions for fields ψ and ψ in the previous formula and using the fact that ur (p) and vr (p) are eigenspinors of Σ · p/|p| with eigenvalues (−1)r+1 and (−1)r , respectively (see Problem 4.7) we get

2

 1 m 3 x d d3 pd3 q 2(2π)3 E p Eq r,s=1 ( × c†r (q)cs (p)(−1)s+1 u†r (q)us (p)ei(q−p)·x

Sp =

+ c†r (q)d†s (p)(−1)s u†r (q)vs (p)ei(q+p)·x + dr (q)cs (p)(−1)s+1 vr† (q)us (p)e−i(q+p)·x ! − d†s (p)dr (q)(−1)s vr† (q)vs (p)ei(p−q)·x .

(8.9)

Performing the x integration 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 q gives 1 Sp = 2 r=1 2



  d3 p(−1)r+1 c†r (p)cr (p) + d†r (p)dr (p) .

(8.10)

Let us emphasize that we have used the expansion of the fields with respect to helicity basis. 8.8 The two-particle state given in the problem is eigenstate of the operators H, Q, and Sp . Using the explicit form of the Hamiltonian from Problem 8.2 we have 

 † † d3 pEp c†r (p)cr (p) Hcr1 (p1 )cr2 (p2 ) |0 = r

 + d†r (p)dr (p) c†r1 (p1 )c†r2 (p2 ) |0 .

(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 = δr1 r δ (3) (p − p1 )cr (p)c†r2 (p2 ) |0 − c†r (p)c†r1 (p1 )cr (p)c†r2 (p2 ) |0 .

(8.12)

Repeating once more we get †

c†r (p)cr (p)c†r1 (p1 )c†r2 (p2 ) |0 = δr1 r δ (3) (p − p1 )cr (p)c†r2 (p2 ) |0 − c†r (p)c†r1 (p1 )δrr2 δ (3) (p − p2 ) |0 . (8.13) It is easy to see that d†r (p)dr (p)c†r1 (p1 )c†r2 (p2 ) |0 = 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 p we obtain (8.15) Hc†r1 (p1 )c†r2 (p2 ) |0 = (Ep1 + Ep2 )c†r1 (p1 )c†r2 (p2 ) |0 . Similar as before we have: Qc†r1 (p1 )c†r2 (p2 ) |0 = −2ec†r1 (p1 )c†r2 (p2 ) |0 ,

(8.16)

for charge and Sp c†r1 (p1 )c†r2 (p2 ) |0  1 (−1)r1 +1 + (−1)r2 +1 c†r1 (p1 )c†r2 (p2 ) |0 = 2

(8.17)

for helicity. To summarize: energy, charge and helicity of the two–particle state |p1 , r1 ; p2 , r2  are Ep1 + Ep2 ,

−2e,

 1 (−1)r1 +1 + (−1)r2 +1 , 2

(8.18)

respectively. 8.9 The commutator is

1 a b [Qa , Qb ] = τkl [ψi† (x)ψj (x), ψk† (y)ψl (y)] d3 xd3 yτij 4

1 a b = τkl (ψi† (x)ψl (y)δjk − ψk† (y)ψj (x)δil )δ (3) (x − y) d3 xd3 yτij 4

1 a b b a τjl ψl − ψk† τkl τlj ψj ) = d3 x(ψi† τij 4

1 = d3 xψ † [τ a , τ b ]ψ 4

i = abc d3 xψ † τ c ψ = iabc Qc . 2 The generators Qa satisfy the commutation relations of SU(2) algebra as we expected. 8.10 The charges are



1 b Qb = d3 xj0b = d3 x(abc π˙ a π c + Ψi† τij Ψj ) . 2 (a) The commutator is

 [Qb , Qe ] = d3 xd3 y abc def [π˙ a (x)π c (x), π˙ d (y)π f (y)] b e τij τmn † [Ψi† (x)Ψj (x), Ψm + (y)Ψn (y)] 2 2

(8.19)

168

Solutions

=

 d3 xd3 y abc def (iδ (3) (x − y)δ cd π˙ a (x)π f (y)

− iδ (3) (x − y)δ af π˙ d (y)π c (x)) b e  τij τmn (3) † + δ (x − y)(δjm Ψi† (x)Ψn (y) − δin Ψm (y)Ψj (x))

2 2  i bed † d 3 e b b e = d x i(π˙ π − π˙ π ) +  Ψ τ Ψ 2 

1 bed 3 adc a c † d = i d x  π˙ π + Ψ τ Ψ 2 = ibed Qd . (b) The results are [Qb , π a (x)] = −iabc π c (x) , [Qb , ψi (x)] = −

b τin ψn (x) , 2 b

τ [Qb , ψ¯i (x)] = −ψ¯n (x) ni . 2 8.11 The conserved charge for dilatation is 



3 † 3 0 3 j † 0¯ j ψ ψ + x ψ ∂j ψ − x ψγ ∂j ψ . D = d xj = −i d x 2

(8.20)

Let us find the commutator between the operator D and momentum P i 

3 [D, P i ] = d3 xd3 y [ ψ † (x)ψ(x) + xj ψ † (x)∂j ψ(x), ψ † (y)∂ i ψ(y)] 2  j ¯ − [x0 ψ(x)γ ∂j ψ(x), ψ † (y)∂ i ψ(y)] . We decompose the previous expression on three commutators. The first one is [ψ † (x)ψ(x), ψ † (y)∂ i ψ(y)] = [ψa† (x)ψa (x), ψb† (y)]∂ i ψb (y) + ψb† (y)[ψa† (x)ψa (x), ∂ i ψb (y)] = ψa† (x){ψa (x), ψb† (y)}∂ i ψb (y) − ψb† (y){ψa† (x), ∂ i ψb (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)∂ i ψ(y)] = ψ † (x)∂ i ψ(y)δ (3) (x − y) − ψ † (y)ψ(x)∂yi δ (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)∂j ψ(x), ψ † (y)∂ i ψ(y)] = ψ † (x)∂ i ψ(y)∂jx δ (3) (x − y) − ψ † (y)∂j ψ(x)∂yi δ (3) (x − y) , j j i ¯ ¯ ∂j ψ(x), ψ † (y)∂ i ψ(y)] = ψ(x)γ ∂ ψ(y)∂jx δ (3) (x − y) [ψ(x)γ j ¯ − ψ(y)γ ∂j ψ(x)∂ i δ (3) (x − y) . y

(8.22)

(8.23)

Inserting (8.21), (8.22) and (8.23) in (8.21) and applying ∂xk δ (3) (x − y) = −∂yk δ (3) (x − y) , we get

[D, P i ] = −

d3 xψ † ∂ i ψ = iP i .

(8.24)

(8.25)

Similarly one can show that [D, P 0 ] = iP 0 .

(8.26)

8.12 (a) Using the expression (5.G) the energy–momentum tensor is ¯ α ∂β ψ − gαβ (iψ/ ¯∂ ψ − gx2 ψψ) ¯ . Tαβ = iψγ Taking derivative of the previous expression we get ¯ , ∂ α Tαβ = 2gxβ ψψ where we have used the equations of motion: i/ ∂ ψ − gx2 ψ = 0 , ¯ µ + gx2 ψ¯ = 0 . i∂µ ψγ The result ∂ α Tαβ = 0 shows that there is no translation symmetry in the theory. As a consequence, the energy and momentum are not conserved in this theory. (b) From the expression for the four-momentum (5.6) we have

¯ j ∂j ψ + gx2 ψψ) ¯ , P 0 (t) = d3 x(−iψγ

i

P (t) = i so

†

d3 xψ ∂ i ψ ,

170

Solutions



[P 0 (t), P i (t)] =

d3 xd3 y

 ¯ x)γ j ∂j ψ(t, x), ψ † (t, y)∂ i ψ(t, y)] × [ψ(t,  ¯ x)ψ(t, x), ψ † (t, y)∂ i ψ(t, y)] + igx2 [ψ(t,



= d3 xd3 y  † † × (γ 0 γ j )ab [ψa (t, x)∂j ψb (t, x), ψc (t, y)∂ i ψc (t, y)]  † † 0 + igx2 γab [ψa (t, x)ψb (t, x), ψc (t, y)∂ i ψc (t, y)] . The commutators in the previous expression can be found in the same way as in the previous problem

 ¯ j ∂j ∂i ψ ¯ j ∂ i ψ − ψγ [P 0 (t), P i (t)] = d3 x −∂j ψγ  ¯ i ψ + (∂ i ψ)ψ) ¯ + igx2 (ψ∂

  ¯ j ∂ i ψ) + igx2 ∂ i (ψψ) ¯ = d3 x −∂j (ψγ

¯ , = −2ig d3 xxi ψψ where we dropped the surface terms. (c) It is easy to show that ∂µ M µνρ = 0, which is a consequence of the Lorentz symmetry of the Lagrangian density. 8.13 (a) Under the Lorentz transformation the commutator [J µ (x), J ν (y)] transforms in the following way U (Λ)[J µ (x), J ν (y)]U −1 (Λ) µ ν = U (Λ)[ψ¯a (x)γab ψb (x), ψ¯c (y)γcd ψd (y)]U −1 (Λ) (8.27) −1 µ −1 −1 ν −1 ¯ ¯ = [U ψa (x)U γab U ψb (x)U , U ψc (y)U γcd U ψd (y)U ] . Taking the adjoint of (8.G) and multiplying by γ 0 we obtain −1 ¯ ¯ U (Λ)ψ(x)U (Λ) = ψ(Λx)S(Λ) .

(8.28)

By using (8.G), last expression and S −1 γ µ S = Λµν γ ν in (8.27) we get U (Λ)[J µ (x), J ν (y)]U −1 (Λ) = Λρµ Λσν [J ρ (Λx), J σ (Λy)] .

(8.29)

From the last result we see that the commutator [J µ (x), J ν (y)] is a covariant 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 = y 0 = t, x = y. We get [Jµ (t, x), Jν (t, y)] = (γ0 γµ )ab (γ0 γν )cd [ψa† (t, x)ψb (t, x), ψc† (t, y)ψd (t, y)]  = (γ0 γµ )ab (γ0 γν )cd ψa† (t, x){ψb (t, x), ψc† (t, y)}ψd (t, y)  − ψc† (t, y){ψa† (t, x), ψd (t, y)}ψb (t, x) .

(8.30)

Using the anticommutation relation (8.D) in (8.30) gives [Jµ (t, x), Jν (t, y)]   ¯ y)γν γ0 γµ ψ(t, x) δ (3) (x − y) . ¯ x)γµ γ0 γν ψ(t, y) − ψ(t, = ψ(t, Since x = y then δ (3) (x − y) = 0 and the commutator is equal to zero in the special frame we have chosen. Because of the covariance it follows that it is equal to zero for (x − y)2 < 0. Therefore, microcausality principle is valid. 8.14 First show that 

 ψa (x)ψ¯b (y) =

1 (2π)3



 ψ¯a (x)ψb (y) =

1 (2π)3



d3 p (/ p + m)ab e−ip·(x−y) , 2Ep

(8.31)



d3 p (/ p − m)ba e−ip·(x−y) . (8.32) 2Ep   If in the expression ψ¯a (x1 )ψb (x2 )ψc (x3 )ψ¯d (x4 ) , we substitute the expansions (8.A–B), we obtain   ψ¯a (x1 )ψb (x2 )ψc (x3 )ψ¯d (x4 ) " 4

#   m2 d3 pi = (2π)6 i=1 Epi r1 ,...,r4 0 / ¯4d ei(−p1 ·x1 −p2 ·x2 +p3 ·x3 +p4 ·x4 ) × d1 c2 d†3 c†4 v¯1a u2b v3c u 0  / + d1 d†2 c3 c†4 v¯1a v2b u3c u ¯4d ei(−p1 ·x1 +p2 ·x2 −p3 ·x3 +p4 ·x4 ) , 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 solutions from Problem 4.4 and using / 0 d1 c2 d†3 c†4 = −δr1 r3 δr2 r4 δ (3) (p1 − p3 )δ (3) (p2 − p4 ) ,

172

Solutions

/ 0 d1 d†2 c3 c†4 = δr1 r2 δr3 r4 δ (3) (p1 − p2 )δ (3) (p3 − p4 ) we have 

 ψ¯a (x1 )ψb (x2 )ψc (x3 )ψ¯d (x4 )

3 d p1 d3 p2 1 (/ p1 − m)ca (/ p2 + m)bd e−ip1 ·(x1 −x3 )−ip2 ·(x2 −x4 ) =− (2π)6 4Ep1 Ep2

3 1 d p1 d3 p3 + (/ p1 − m)ba (/ p3 + m)cd e−ip1 ·(x1 −x2 )−ip3 ·(x3 −x4 ) . 6 (2π) 4Ep1 Ep3

By using (8.31) and (8.32) the last expression takes the form      ψ¯a (x1 )ψb (x2 )ψc (x3 )ψ¯d (x4 ) = − ψ¯a (x1 )ψc (x3 ) ψb (x2 )ψ¯d (x4 )    + ψ¯a (x1 )ψb (x2 ) ψc (x3 )ψ¯d (x4 ) . The previous formula is special case of the Wick theorem. 8.15 Substituting (8.A-B) in the commutator we obtain

1 ¯ µ 1  m [ψ, γ ψ] = [¯ ur (p)γ µ us (q) d3 pd3 q 2 2(2π)3 r,s Ep Eq †

†

× (cr (p)cs (q) − cs (q)cr (p))ei(p−q)·x †

†

†

†

+u ¯r (p)γ µ vs (q)(cr (p)ds (q) − ds (q)cr (p))ei(p+q)·x + v¯r (p)γ µ us (q)(dr (p)cs (q) − cs (q)dr (p))e−i(p+q)·x ! † † + v¯r (p)γ µ vs (q)(dr (p)ds (q) − ds (q)dr (p))ei(q−p)·x . (8.33) Using the anticommutation relations (8.E) we obtain 1 ¯ µ ¯ µψ : − [ψ, γ ψ] = : ψγ 2

µ  1 3 p − d p (¯ ur (p)ur (p) + v¯r (p)vr (p)) , 2(2π)3 Ep r 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 0| T (ψ¯a (x)ψb (y)) |0 = −iSF ba (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



d3 p ( (/ p − m)ba eip·(y−x) θ(x0 − y0 ) 2Ep ! − (/ p + m)ba eip·(x−y) θ(y0 − x0 ) . (8.34)

0| T (ψ¯a (x)ψb (y)) |0 =

1 (2π)3

With a help of Problem 6.13 we see that right hand side of the expression (8.34) is −iSF ba (y − x) and we have ¯ 0| T (ψ(x)Γ ψ(y)) |0 = Γab 0| T (ψ¯a (x)ψb (y)) |0 = −iΓab SF ba (y − x) = −i tr [Γ SF (y − x)]

d4 p e−ip·(y−x) tr [(/ p + m)Γ ] . = −i (2π)4 p2 − m2 + i Using the identities from the Problems 3.6(b),(d),(e) and (i) we obtain tr [(/ p + m)γ5 ] = tr [(/ p + m)γ5 γµ ] = 0, tr [(/ p + m)γµ γν ] = 4mgµν . From here the requested result follows. 8.17 (a) In the Weyl representation for γ–matrices the charge conjugate spinor is ψc = C ψ¯T   σ2 0 0 =i 1 0 −σ2  χ = . −iσ2 ϕ∗

1 0



ϕ∗ −iσ2 χ



c gives ϕ = χ. The condition ψM = ψM (b) If   χ ϕ ψM = = and φ , M −iσ2 χ∗ −iσ2 ϕ∗

then † ψ¯M φM = −iχ σ2 ϕ∗ + iχT σ2 ϕ

= −iσ2ab χ∗a ϕ∗b + iσ2ab χa ϕb = −iσ2ba ϕ∗b χ∗a + iσ2ba ϕb χa † = −iϕ σ2 χ∗ + iϕT σ2 χ = φ¯M ψM . In the last expression we used that ϕ and χ are Grassmann variables. The other identities can be proved in the same way. For the second one the following identity is useful: σ2 σ µ σ2 = σ ¯ µT .

174

Solutions

(c) The Majorana field operator is 1 ψM = √ (ψ + ψc ) 2 

 d3 p m  cr (p) + dr (p) √ = ur (p)e−ip·x (2π)3 Ep r 2 c† (p) + d† (p) vr (p)eip·x . + r √ r 2 The annihilation and creation operators can easily be read off: bM (p, r) =

cr (p) + dr (p) √ , 2

†

bM (p, r) =

†

†

cr (p) + dr (p) √ . 2

The anticommutation relations are derived from (8.E): {bM (p, r), b†M (q, s)} = δrs δ (3) (p − q) , †

{bM (p, r), bM (q, s)} = {bM (p, r), b†M (q, s)} = 0 . (d) The Dirac spinor is ψD = ψ1 + iψ2 where ψ1,2 are Majorana spinors. The Lagrangian density is ∂ ψ1 + iψ¯2 / ∂ ψ2 − m(ψ¯1 ψ1 + ψ¯2 ψ2 ) + ie(ψ¯1 /Aψ2 − ψ¯2 /Aψ1 ) . L = iψ¯1 / ¯ 8.18 Under Lorentz transformations the operator Vµ (x) = ψ(x)γ µ ψ(x) transforms in the following way: −1 ¯ U (Λ)Vµ (x)U −1 (Λ) = U (Λ)ψ(x)U (Λ)γµ U (Λ)ψ(x)U −1 (Λ) −1 ¯ = ψ(Λx)S(Λ)γ (Λ)ψ(Λx) = Λνµ Vν (Λx) , µS

(8.35) ¯ since Sγµ S −1 = Λν µ γν . The other operator Aµ (x) = ψ(x)γ 5 ∂µ ψ(x) transforms as −1 ¯ U (Λ)Aµ (x)U −1 (Λ) = U (Λ)ψ(x)U (Λ)γ5 ∂µ U (Λ)ψ(x)U −1 (Λ) ¯ = ψ(Λx)γ 5 ∂µ ψ(Λx) ,

where we used well known relation Sγ5 S −1 = γ5 (see Problem 4.38). Since ∂µ = Λρ µ ∂ρ
we have U (Λ)Aµ (x)U −1 (Λ) = Λρ µ Aρ (Λx) . Under parity vector Vµ transforms as follows: Vµ (x) → P Vµ (x)P −1 = ψ † (t, −x)γµ γ0 ψ(t, −x)  V0 (t, −x), for µ = 0 = −Vi (t, −x), for µ = i = V µ (t, −x) ,

(8.36)

Chapter 8. Canonical quantization of the Dirac field

175

since −1 ¯ = (P ψ(x)P −1 )† γ0 = (γ0 ψ(t, −x))† γ0 = ψ † (t, −x) . P ψ(x)P

In the similar way we get ¯ −x)γ5 ∂µ ψ(t, −x) P Aµ (x)P −1 = −ψ(t,  ¯ −x)γ5 ∂
ψ(t, −x), −ψ(t, 0 = ¯ ψ(t, −x)γ5 ∂i
ψ(t, −x), = −Aµ (t, −x) .

for µ = 0 for µ = i

From τ ψ(t, x)τ −1 = T ψ(−t, x), where τ is an antiunitary operator of time reversal follows ¯ x)τ −1 = τ ψ † (t, x)τ −1 γ ∗ = ψ † (−t, x)T † γ ∗ . τ ψ(t, 0 0 From the previous expressions we get τ Vµ (t, x)τ −1 = ψ † (−t, x)T † (γ0 γµ )∗ T ψ(−t, x) .

(8.37)

With a help of T γµ T −1 = γ µ∗ we get ¯ x)γ µ ψ(−t, x) = V µ (−t, x) . τ Vµ (x)τ −1 = ψ(−t,

(8.38)

We would suggest to reader to prove the previous result by taking T = iγ 1 γ 3 . The identity (8.39) (iγ 1 γ 3 )† γ0∗ γµ∗ iγ 1 γ 3 = γ 0 γ µ , has to be shown. Under time reversal the operator Aµ (x) transforms as ¯ x)γ5 ∂
µ ψ(−t, x) = −Aµ (−t, x) . τ Aµ (x)τ −1 = −ψ(−t,

(8.40)

−1 From Cψa (x)C −1 = (Cγ0T )ab ψb† (x) follows C ψ¯a C −1 = −ψb Cba , where C is a unitary charge conjugation operator while C is a matrix. It is easy to see −1 µ γab Cbd ψ¯d CV µ C −1 = −ψc Cca µ T ¯ = ψc (γ )cd ψd = ψc (γ µ )dc ψ¯d µ = −ψ¯d γdc ψc µ = −V .

The minus sign in the forth line of the previous calculation appears since the fields ψ and ψ¯ anticommute. An infinity constant is ignored. Compare this result with result of Problem 4.37. In the similar way result CAµ C −1 = ¯ 5 ψ is derived. ∂µ ψγ 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) −1 −1 ¯ ¯ = iU (Λ)ψ(x)U (Λ)γ µ ∂µ U (Λ)ψ(x)U −1 (Λ) − mU (Λ)ψ(x)ψ(x)U (Λ) µ −1 −1 ¯ ¯ = iψ(Λx)Sγ ∂µ S ψ(Λx) − mψ(Λx)SS ψ(Λx) ν ρ
¯ ¯ = i(Λ−1 )µ ψ(Λx)γ Λ ∂ ψ(Λx) − ψ(Λx)ψ(Λx) ν

µ ρ

µ
¯ ¯ = iψ(Λx)γ ∂µ ψ(Λx) − mψ(Λx)ψ(Λx) = L(Λx) .

Under the parity L transforms as follows P LP −1 = iψ † (t, −x)γ µ ∂µ γ 0 ψ(t, −x) − ¯ −x)ψ(t, −x) . − mψ(t, From we get

γ µ γ 0 ∂µ = γ 0 γ 0 ∂0
+ γ 0 γ i ∂i
= γ 0 γ µ ∂µ
, P L(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 T and C. Here, we use explicit expressions for them. Starting from τ ψ(t, x)τ −1 = iγ 1 γ 3 ψ(−t, x) ,

(8.41)

we obtain ¯ x)τ −1 = τ ψ † (t, x)τ −1 γ ∗ τ ψ(t, 0 = −iψ † (−t, x)(γ 3 )† (γ 1 )† (γ 0 )∗ ¯ = −iψ(−t, x)γ 3 γ 1 . Further, ¯ τ Lτ −1 = −iψ(−t, x)γ 3 γ 1 (γ µ )∗ γ 1 γ 3 ∂µ ψ(−t, x) ¯ − mψ(−t, x)γ 3 γ 1 γ 1 γ 3 ψ(−t, x) . Applying (γ 0 )∗ = γ 0 , (γ 1 )∗ = γ 1 , (γ 2 )∗ = −γ 2 , (γ 3 )∗ = γ 3 , the anticommutation relation among γ–matrices and introducing derivatives with respect to new coordinates t
= −t, x
= x instead of the old ones gives

Chapter 8. Canonical quantization of the Dirac field

177

¯ ¯ τ Lτ −1 = iψ(−t, x)γ µ ∂µ
ψ(−t, x) − mψ(−t, x)ψ(−t, x) = L(−t, x) . The transformation law for field ψ under charge conjugation Cψa C −1 = i(γ 2 )ab ψb† induces

C ψ¯a C −1 = iψb (γ 2 γ 0 )ba .

Then Lagrangian density transforms as CLC −1 = −iψc (γ 2 γ 0 γ µ γ 2 )ca ∂µ ψa† + mψb (γ 2 γ 0 γ 2 )ba ψa† . Since γ 2 γ 0 γ µ γ 2 ∂µ = (−γ 0 ∂0 + γ 1 ∂1 − γ 2 ∂2 + γ 3 ∂3 )γ0 , then the kinetic term becomes   −iψc −γ 0 ∂0 + γ 1 ∂1 − γ 2 ∂2 + γ 3 ∂3 ) cd ψ¯d . In the Dirac representation of γ–matrices the following relations are satisfied: (γ 1 )T = −γ 1 ,

(γ 0 )T = γ 0 ,

(γ 2 )T = γ 2 ,

(γ 3 )T = −γ 3 ,

and the kinetic term is µ ψc . iψc (γ µT )cd ∂µ ψ¯d = −i∂µ ψ¯d γdc

As in the previous problem we anticommute the fields ψ¯ and ψ, and ignore the infinity constant δ (3) (0). At the end we obtain ¯ , ¯ µ ψ − mψψ CLC −1 = −i∂µ ψγ which is the starting Lagrangian density up to four divergence. 8.20 From follows

S(Λ)σµν S −1 (Λ) = Λρ µ Λσ ν σρσ ,

(8.42)

U (Λ)Tµν U −1 (Λ) = Λρ µ Λσ ν Tρσ (Λx) ,

(8.43)

and therefore Tµν is a second rank tensor. Under parity the transformation rule is: P T0i (t, x)P −1 = −T0i (t, −x) , P Tij (t, x)P −1 = Tij (t, −x) . Charge conjugation act on a Tµν tensor according to CTµν (x)C −1 = −Tµν (x) .

(8.44)

178

Solutions

In order to confirm the previous result you should to prove that

The identity

C −1 σµν C = −(σµν )T .

(8.45)

T σµν T −1 = −(σ µν )∗ ,

(8.46)

can be derived easily. Consequently, τ T0i (t, x)τ −1 = T0i (−t, x) , τ Tij (t, x)τ −1 = −Tij (−t, x) .

9 Canonical quantization of the electromagnetic field

9.1 The commutator is [A (t, x), A˙ ν (t, y)] = µ

 λ,λ


i (2π)3



d3 kd3 q ωq µλ (k)νλ
(q) √ 2 ωk ωq

 † × [aλ (k), aλ
(q)]ei(k·x−q·y) †

− [aλ (k), aλ
(q)]e−i(k·x−q·y)

 .

Using the commutation relations (9.G) as well as orthogonality relations (9.D) we obtain

  i µν 3 ik·(x−y) ik·(y−x) [Aµ (t, x), A˙ ν (t, y)] = − g k e + e d 2(2π)3 = −ig µν δ (3) (x − y) . 9.2 Using the commutation relations (9.G) and the completeness relation (9.D) we get

3   1 d k −ik·(x−y) iDµν = [Aµ (x), Aν (y)] = −g µν − eik·(x−y) . (9.1) e 3 (2π) 2|k| In order to calculate the integral (9.1) we shall use spherical coordinates (using notation x0 − y0 = t, |x − y| = r)

π

∞ 1 iDµν (x − y) = −g µν kdk dθ sin θ 2(2π)2 0 0   × e−i(kt−kr cos θ) − ei(kt−kr cos θ)

 1 1 ∞  −ikt ikr dk e (e − e−ikr ) + eikt (e−ikr − eikr ) = −g µν 2 2(2π) ir 0

180

Solutions

  1 1 ∞ = −g dk e−ikt+ikr − e−ikt−ikr 2(2π)2 ir −∞ 1 (δ(t − r) − δ(t + r)) = −g µν 4πir 1 = ig µν (t)δ(t2 − r2 ) , 2π µν

where

2 1, (t) =

−1, 0,

(9.2)

t>0 t