Quantum field theory by Washington Soares Alves

Quantum Field Theory I Babis Anastasiou Institute for Theoretical Physics, ETH Zurich, 8093 Zurich, Switzerland E-mail: babis@phys.ethz.ch January 23, 2010

Contents 1 Quantum Field Theory. Why?

2 Theory of Classical Fields 2.1 The principle of least action . . . . . . . . . . . . . . . . . . . 2.2 Fields from a discretized space (lattice) . . . . . . . . . . . . . 2.3 Euler-Lagrange equations for a classical field from a Lagrangian density . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 2.4 Field Hamiltonian Density . . . . . . . . . . . . . . . . . . . . 2.5 An example: acoustic waves . . . . . . . . . . . . . . . . . . . 2.6 Noether’s theorem . . . . . . . . . . . . . . . . . . . . . . . . 2.6.1 Field symmetry transformations . . . . . . . . . . . . . 2.6.2 Space-Time symmetry transformations . . . . . . . . .

9 9 10

3 Quantization of the Schr¨ odinger field 3.1 The Schr¨odinger equation from a Lagrangian density . . . . . 3.2 Quantization of Fields . . . . . . . . . . . . . . . . . . . . . . 3.3 Quantized Schr¨odinger field . . . . . . . . . . . . . . . . . . . 3.4 Particle states from quantized fields . . . . . . . . . . . . . . . 3.5 What is the wave-function in the field quantization formalism?

30 30 32 33 35 39

4 The Klein-Gordon Field 4.1 Real Klein-Gordon field . . . . . . . . . . . . . . . . 4.1.1 Real solution of the Klein-Gordon equation . . 4.1.2 Quantization of the real Klein-Gordon field . . 4.1.3 Particle states for the real Klein-Gordon field 4.1.4 Energy of particles and “normal ordering” . . 4.1.5 Field momentum conservation . . . . . . . . . 4.1.6 Labels of particle states? . . . . . . . . . . . .

42 42 43 45 46 47 49 50

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14 16 17 19 19 20

4.2 Two real Klein-Gordon fields . . . . . . . . . . . . . . . . . . . 4.2.1 Two equal-mass real Klein-Gordon fields . . . . . . . . 4.2.2 Two real Klein-Gordon fields = One complex KleinGordon field . . . . . . . . . . . . . . . . . . . . . . . . 4.3 Conserved Charges as generators of symmetry transformations 4.4 Casimir effect: the energy of the vacuum . . . . . . . . . . . . 4.5 Can the Klein-Gordon field be an one-particle wave-function? 5 The Dirac Equation 5.1 Mathematical interlude . . . . . . . . . . . . 5.1.1 Pauli matrices and their properties . 5.1.2 Kronecker product of 2 × 2 matrices 5.2 Dirac representation of γ-matrices . . . . . . 5.3 Traces of γ− matrices . . . . . . . . . . . . 5.4 γ−matrices as a basis of 4 × 4 matrices . . . 5.5 Lagrangian for the Dirac field . . . . . . . .

50 52 55 57 58 62

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64 65 65 66 67 69 69 71

6 Lorentz symmetry and free Fields 6.1 Field transformations and representations of the Lorentz group 6.2 Scalar representation M(Λ) = 1 . . . . . . . . . . . . . . . . . 6.3 Vector representation M(Λ) = Λ . . . . . . . . . . . . . . . . . 6.4 How to find representations? . . . . . . . . . . . . . . . . . . . 6.4.1 Generators of the scalar representation . . . . . . . . . 6.4.2 Generators of the vector representation . . . . . . . . . 6.5 Spinor representation . . . . . . . . . . . . . . . . . . . . . . . 6.6 Lorentz Invariance of the Dirac Lagrangian . . . . . . . . . . . 6.7 General representations of the Lorentz group . . . . . . . . . . 6.8 Weyl spinors . . . . . . . . . . . . . . . . . . . . . . . . . . . . 6.8.1 The Dirac equation with Weyl spinors . . . . . . . . .

73 75 76 77 77 80 81 82 83 85 86 89

7 Classical solutions of the Dirac equation 7.1 Solution in the rest frame . . . . . . . . . . . . . 7.2 Lorentz boost of rest frame Dirac spinor along the 7.3 Solution for an arbitrary vector . . . . . . . . . . 7.4 A general solution . . . . . . . . . . . . . . . . . .

90 91 92 95 95

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. . . . z-axis . . . . . . . .

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8 Quantization of the Dirac Field 8.1 One-particle states . . . . . . . . . . . . . . . . . . 8.1.1 Particles and anti-particles . . . . . . . . . . 8.1.2 Particles and anti-particles of spin- 12 . . . . 8.2 Fermions . . . . . . . . . . . . . . . . . . . . . . . . 8.3 Lorentz transformation of the quantized spinor field 8.3.1 Transformation of the ladder operators . . . 8.3.2 Transformation of the quantized Dirac field 8.4 Parity . . . . . . . . . . . . . . . . . . . . . . . . . 8.5 Other discrete symmetries . . . . . . . . . . . . . .

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97 99 100 102 104 105 105 108 109 112

9 Quantization of the free electromagnetic field 114 9.1 Maxwell Equations and Lagrangian formulation . . . . . . . . 114 9.2 Quantization of the Electromagnetic Field . . . . . . . . . . . 118 10 Propagation of free particles 10.1 Transition amplitude for the Schr¨odinger field . . . . . . 10.2 Transition amplitude for the real Klein-Gordon field . . . 10.3 Time Ordering and the Feynman-St¨ uckelberg propagator the real Klein-Gordon field . . . . . . . . . . . . . . . . . 10.4 Feynman propagator for the complex Klein-Grodon field 10.5 Feynman propagator for the Dirac field . . . . . . . . . . 10.6 Feynman propagator for the photon field . . . . . . . . .

. . . . for . . . . . . . .

. . . .

128 130 131 132

11 Scattering Theory (S-matrix) 11.1 Propagation in a general field theory . . . . . . 11.1.1 A special case: free scalar field theory . . 11.1.2 “Typical” interacting scalar field theory 11.2 Spectral assumptions in scattering theory . . . . 11.3 “In” and “Out” states . . . . . . . . . . . . . . 11.4 Scattering Matrix-Elements . . . . . . . . . . . 11.5 S-matrix and Green’s functions . . . . . . . . . 11.6 The LSZ reduction formula . . . . . . . . . . . . 11.7 Truncated Green’s functions . . . . . . . . . . . 11.8 Cross-sections∗ . . . . . . . . . . . . . . . . . .

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133 134 138 139 141 142 144 145 147 150 151

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123 . 123 . 124

12 Perturbation Theory and Feynman Diagrams 12.1 Time evolution operator in the interaction picture . . . 12.2 Field operators in the interacting and free theory . . . 12.3 The ground state of the interacting and the free theory 12.4 Wick’s theorem . . . . . . . . . . . . . . . . . . . . . . 12.5 Feynman Diagrams for φ4 theory . . . . . . . . . . . . 12.6 Feynman rules in momentum space . . . . . . . . . . . 12.7 Truncated Green’s functions in perturbation theory . .

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152 . 154 . 156 . 158 . 160 . 164 . 167 . 171

13 Loop Integrals 13.1 The simplest loop integral. Wick rotation . 13.2 Dimensional Regularization . . . . . . . . 13.2.1 Angular Integrations . . . . . . . . 13.2.2 Properties of the Gamma function . 13.2.3 Radial Integrations . . . . . . . . . 13.3 Feynman Parameters . . . . . . . . . . . .

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186 . 186 . 189 . 192 . 193 . 194 . 198 . 201

14 Quantum Electrodynamics 14.1 Gauge invariance . . . . . . . . . . . . 14.2 Perturbative QED . . . . . . . . . . . 14.3 Dimensional regularization for QED . . 14.3.1 Gamma-matrices in dimensional 14.3.2 Tensor loop-integrals . . . . . . 14.4 Electron propagator at one-loop . . . . 14.5 Photon propagator at one-loop . . . . . 15 One-loop renormalization of QED

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. . . . . . . . . . . . . . . . . . . . . . . . regularization . . . . . . . . . . . . . . . . . . . . . . . .

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174 174 177 178 180 181 182

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Bibliography [1] The Quantum Theory of Fields, Volume I Foundations, Steven Weinberg, Cambridge University Press. [2] An introduction to Quantum Field Theory, M. Peskin and D. Schroeder, Addison-Wesley [3] Quantum Field Theory in a nutshell, A. Zee, Princeton University Press. [4] Quantum Field Theory, Mark Srednicki, Cambridge University Press. [5] An introduction to Quantum Field Theory, George Sterman, Cambridge University Press. [6] Classical Mechanics, Goldstein, Poole and Safko, Addison-Wesley [7] Lectures On Qed And Qcd: Practical Calculation And Renormalization Of One- And Multi-loop Feynman Diagrams, Andrea Grozin, World Scientific

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Chapter 1 Quantum Field Theory. Why? The goal of this lecture series is to introduce a beautiful synthesis of quantum mechanics and special relativity into a unified theory, the theory of quantized fields. We have already seen in the course of Quantum Mechanics II that quantized fields can be used to describe systems of many identical particles (fermions or bosons). In this formalism, a quantum-mechanical state of particles at fixed positions is generated from the vacuum state, containing no particles, after we act on it with field operators, |x~1 , x~2 , . . . , x~n i ∼ f † (x~1 , t)f † (x~2 , t) . . . f † (x~n , t) |0i .

(1.1)

We can also prove (QMII) that there is a wave-function for non-interacting identical particles which satisfies the Schr¨odinger equation, if the field operator itself f (~x, t) satisfies the Schr¨odinger equation. The power of the field theory formulation is that particle-exchange symmetry for bosons and antisymmetry for fermions are automatically satisfied, if we impose commutation quantization conditions for boson fields and anti-commutation quantization conditions for fermion fields: [f (x~1 , t), f (x~2 , t)]± = f † (x~1 , t), f † (x~2 , t) ± = 0 f (x~1 , t), f †(x~2 , t) ± = δ (3) (x~1 − x~2 ), (1.2)

where the commutator (“-” for bosons) and anticommutator (“+” for fermions) are defined as, [A, B]± = AB ± BA. (1.3) For example, we can easily verify that the Pauli exclusion principle for identical fermions is automatically respected. A state of two identical fermions 7

in the same position is, in quantum field theory, |~x, ~xi ∼ f † (~x, t)f † (~x, t) |0i =

1 † f (~x, t), f † (~x, t) + |0i = 0 |0i = 0, 2

(1.4)

which vanishes. In non-relativistic quantum mechanics, quantized fields offer an elegant framework which is convenient to describe a system of many identical particles. However, an equivalent description in terms of wave-functions yields the same physical results and it is always possible. Traditional quantum mechanics fails when we must merge it with special relativity. A number of new phenomena manifest themselves at small distances (quantum effects) and relativistic velocities; these are routinely studied at modern collider experiments. An example physical processes, exhaustivelly studied at the LargeElectron-Positron (LEP) collider in Geneva, is the production of a muon and its anti-particle out of the annihilation of an electron and a positron: e− + e+ → µ− + µ+ .

(1.5)

Such a phenomenon cannot be explained with a wave-function description. We can start with a wave-function for the electron/positron system a long time before they collide. A long time after the collision, we have a new wave-function for two different particles, the muon and the anti-muon. The Schrodinger equation, and the corresponding time evolution, does not predict wave-functions which are destroyed and replaced by others. Quantized fields allow for the creation and annihilation of particles. From our example reaction, we conclude that there exist an electron field and a muon field. As we shall see later, the quantization of these fields introduces creation and annihilation operators for the electron the muon and their antiparticles. In Quantum Field Theory, it is also possible to describe how fields interact, and to compute the probability for a state with a certain particle content to mutate to another state with a different particle content after some time. Field quantization is an essential part of quantum relativistic laws, and not an optional formalism.

Chapter 2 Theory of Classical Fields We start by introducing a Lagrangian formalism for fields at the classical level.

2.1

The principle of least action

The Lagrangian formalism has been introduced in classical mechanics for systems with a finite number of degrees of freedom. We review here the salient features of the formalism. We consider a physical system with ηi (t), i = 1 . . . N generalized coordinates. For any time interval t2 − t1 , the action is given by, Z t2

dtL [η1 (t), . . . ηN (t), η˙1 (t), . . . η˙N (t)] ,

(2.1)

where L [ηi , η˙ i ] is the Lagrangian function of the system. According to the principle of the least action, the physical generalized coordinates yield a stationary value for the action: h i h i S ηi = ηiphys = S ηi = ηiphys + aωi , (2.2)

where ωi (t1 ) = ωi (t2 ) = 0, and a is a small parameter deforming slightly the generalized coordinate functions from their physical solution ηiphys (t). Eq. 2.2 determines the dynamical behaviour of the physical coordinates

ηiphys (t). Performing a Taylor expansion in a through O(a), we obtain, " # Z t2 X N ∂L η phys , η˙ phys ∂L η phys , η˙ phys dt ωj + ω˙ j = 0 ∂ηjphys ∂ η˙jphys t1 j=1 " !# Z t2 X N ∂L η phys , η˙ phys d ∂L η phys , η˙ phys dt ; − ωj phys phys dt ∂η ∂ η ˙ t1 j j j=1 ! phys phys Z t2 N d X ∂L η , η˙ ωj (t) = 0 dt + dt j=1 ∂ η˙jphys t1 " phys phys !# Z t2 X N ∂L η phys , η˙ phys ∂L η , η˙ d ; dt − ωj = 0, dt ∂ηjphys ∂ η˙jphys t1 j=1

(2.3)

where the integral over the total derivative is zero due to ωj (t1 ) = ωj (t2 ) = 0. The functions ωj (t) are arbitrary smooth functions for t 6= t1 , t2 ; for the above integral to vanish for arbitrary choices of ωj , the physical generalized coordiates must obey the Euler-Lagrange equations phys phys ! ∂L η phys , η˙ phys ∂L η , η˙ d − = 0. (2.4) dt ∂ηjphys ∂ η˙jphys

2.2

Fields from a discretized space (lattice)

A classical field is a continuous function defined at every point in space-time. For example, the amplitude of a mechanical wave or an electromagnetic wave are fields. We may derive classical equations of motion for fields using the principle of least action, which we have postulated for a discrete number of generalized coordinates in the last section. An extension of the formalism to fields is made, if we first assume that a field function takes non-zero values only at discrete points of a lattice which spans all space, and then take the “continuous limit” of a zero distance between the lattice points. As a simple example, we shall study the vibration motion in an onedimensional elastic rod. The rod has a mass density µ. For an elastic rod, the force applied on it is proportional to the elongation or the compression per unit length (ξ) of the rod, F = −Y ξ, 10

(2.5)

where Y is a constant called Young’s modulus. A non-vibrational rod can be seen as an inifinite number of equally spaced particles at rest, all having the same mass m, with a relative distance a. For a small spacing a, we have dm m = lim . (2.6) a→0 dx a A vibration is created when these particles are displaced from their positions at rest. We will assume that each particle in the lattice can interact with its immediate neighbours only. When two neighboring particles are displaced to a larger relative distance, an attractive force tends to bring them back to their resting position, while a repulsive force is developed when they are found at a shorter relative distance. Such a force can be approximated (elastic rod) by considering an elastic spring between the particles, with a Hooke’s constant κ. The force required to elongate one spring is, µ=

F = −κ (yi+1 − yi ) = − (κa)

yi+1 − yi a

= − (κa) ξ.

(2.7)

We can then relate the microspopic Hooke’s constant to the macroscopic Young’s modulus: Y = lim (κa) . (2.8) a→0

The potential energy in the rod is equal to the sum of the potential energies in all springs, X1 X1 κ∆yi2 = κ(yi+1 − yi )2 , (2.9) V = 2 2 i i

where ∆yi = the expansion or contraction of the spring in between particle i and particle i + 1, and yi is the displacement of particle i from its position at rest. The kinetic energy of all particles is, X1 my˙i2 . (2.10) T = 2 i The Lagrangian of the system is simply given by,

L = T − V X 1 1 2 2 ; L= my˙i − κ (yi+1 − yi ) . 2 2 i 11

(2.11)

We can easily find the equations of motion for each particle j in the discretized rod, from the Euler-Lagrange equations 2.4, ∂L d ∂L − =0 dt ∂ y˙k ∂yk d ∂T ∂V ; + =0 dt ∂ y˙k ∂yk P P ∂ i 12 κ(yi+1 − yi)2 d ∂ i 21 my˙i2 + =0 ; dt ∂ y˙k ∂yk ∂ 12 κ(yk+1 − yk )2 + 21 κ(yk − yk−1 )2 d ∂ 12 my˙k2 ; + =0 dt ∂ y˙k ∂yk ; m¨ yk + κ(yk − yk−1 ) − κ(yk+1 − yk ) = 0. (2.12) Let us now consider the limit that the spacing in the discretized rod tends to zero. The vibration amplitude yk (t), defines a continuous function at each position x in the one-dimensional rod. yk (t) → y(x, t) yk+1(t) → y(x + a, t) yk−1(t) → y(x − a, t) yk+2(t) → y(x + 2a, t) yk−2(t) → y(x − 2a, t) ... The equations of motion are written as, m¨ y (x, t) − κ [(y(x + a, t) − y(x, t)) − (y(x, t) − y(x − a, t))] = 0 m y(x + a, t) − y(x, t) y(x, t) − y(x − a, t) ; =0 y¨(x, t) − κ − a a a m y¨(x, t) − (κa) ; a

y(x+a,t)−y(x,t) a

− a

y(x,t)−y(x−a,t) a

= 0.

(2.13)

Taking the limit a → 0, we obtain the equation of motion for the vibration field y(x, t), ∂ 2 y(x, t) ∂ 2 y(x, t) µ − Y = 0. (2.14) ∂t2 ∂x2 12

It is interesting to take the continous limit in the expression of Eq. 2.11 for the Lagrangian as well. We have, " 2 # X 1m 2 1 yi+1 − yi L = a . (2.15) y˙ − (κa) 2a i 2 a i In the continuoum limit, the summation turns into an integration over the position variable x, leading to " 2 2 # Z 1 ∂y(x, t) ∂y(x, t) 1 L = dx µ . (2.16) − Y 2 ∂t 2 ∂x The action is then written as an integral in time and space dimensions, Z S = dtdxL, (2.17) over a Lagrangian density, 2 2 ∂y(x, t) ∂y(x, t) 1 1 − Y . L= µ 2 ∂t 2 ∂x

(2.18)

Can we obtain the equations of motion (Eq. 2.14) from the action integral over the Lagrangian density of Eq. 2.17 using the principle of least action? If so, we can avoid the cumbersome discretization derivation and work using a direct formalism for continuoum systems. We require that Z ∂y ∂y = 0, (2.19) δ dtdxL y, , ∂t ∂x if we vary the field y(x, t) from its physical solution, y(x, t) → y(x, t) + δy(x, t).

(2.20)

We also require that the variation of the field vanishes at the boundaries of the integrations, y(x, t1 ) = y(x, t2) = 0, y(x1, t) = y(x2 , t) = 0.

(2.21)

We then have, â&#x2C6;&#x201A;y â&#x2C6;&#x201A;y Î´ dtdxL y, , =0 â&#x2C6;&#x201A;t â&#x2C6;&#x201A;x Z â&#x2C6;&#x201A;L â&#x2C6;&#x201A;L â&#x2C6;&#x201A;L ; dtdx Î´y + Î´(â&#x2C6;&#x201A;t y) + Î´(â&#x2C6;&#x201A;x y) = 0 â&#x2C6;&#x201A;y â&#x2C6;&#x201A;(â&#x2C6;&#x201A;t y) â&#x2C6;&#x201A;(â&#x2C6;&#x201A;x y) Z

(2.22)

The field variation commutes with space and time derivatives, and we can rewrite the above equation as, Z â&#x2C6;&#x201A;L â&#x2C6;&#x201A;L â&#x2C6;&#x201A;L dtdx Îý + â&#x2C6;&#x201A;t (Îý) + â&#x2C6;&#x201A;x (Îý) = 0 â&#x2C6;&#x201A;y â&#x2C6;&#x201A;(â&#x2C6;&#x201A;t y) â&#x2C6;&#x201A;(â&#x2C6;&#x201A;x y) Z â&#x2C6;&#x201A;L â&#x2C6;&#x201A;L â&#x2C6;&#x201A;L ; dtdx Îý â&#x2C6;&#x2019; â&#x2C6;&#x201A;t â&#x2C6;&#x2019; â&#x2C6;&#x201A;x â&#x2C6;&#x201A;y â&#x2C6;&#x201A;(â&#x2C6;&#x201A;t y) â&#x2C6;&#x201A;(â&#x2C6;&#x201A;x y) â&#x2C6;&#x201A;L â&#x2C6;&#x201A;L Îý + â&#x2C6;&#x201A;x Îý = 0. (2.23) +â&#x2C6;&#x201A;t â&#x2C6;&#x201A;(â&#x2C6;&#x201A;t y) â&#x2C6;&#x201A;(â&#x2C6;&#x201A;x y) Since the field variation Îý is generic and it vanishes at the boundaries, we obtain an Euler-Lagrange differential equation for y(x, t), â&#x2C6;&#x201A;L â&#x2C6;&#x201A;L â&#x2C6;&#x201A;L â&#x2C6;&#x2019; â&#x2C6;&#x201A;t â&#x2C6;&#x2019; â&#x2C6;&#x201A;x = 0. â&#x2C6;&#x201A;y â&#x2C6;&#x201A;(â&#x2C6;&#x201A;t y) â&#x2C6;&#x201A;(â&#x2C6;&#x201A;x y)

(2.24)

Substituting into this equation the Lagrangian density for the elastic rod, we find the following equation of motion: Âľ

â&#x2C6;&#x201A; 2 y(x, t) â&#x2C6;&#x201A; 2 y(x, t) â&#x2C6;&#x2019; Y = 0, â&#x2C6;&#x201A;t2 â&#x2C6;&#x201A;x2

(2.25)

which is the same as Eq. 2.12

2.3

Euler-Lagrange equations for a classical field from a Lagrangian density

We generalize readily the Lagrangian formalism developed for the one-dimensional elastic rod example, to fields which are defined in four space-time dimensions xÂľ = (x0 = ct, x1 = x, x2 = y, x3 = z). In this lecture series, we use natural units, setting h ÂŻ = c = 1, and the space-time metric is defined as 14

g µν = diag(1, −1, −1, −1). We consider a system of N fields φ(i) , i = 1, . . . , N with a Lagrangian density which depends only on the fields and their first (for simplicity) space-time derivatives L = L φ(i) , ∂µ φ(i) . (2.26) The action is given by the integral, Z Z t2 dx0 d3~xL φ(i) , ∂µ φ(i) . S≡ t1

(2.27)

and we consider small variations φ(i) (x, t) → φ(i) (x, t) + δφ(i) (x, t) of the fields from their physical solutions which are zero at the times t1 , t2 or at the boundary surface S(V ) of the volume V, but arbitrary otherwise. Applying the variational principle, we obtain

δSZ = 0 ; δ d4 xL = 0 Z ∂L ∂L 4 (i) (i) ; dx =0 δφ + δ ∂µ φ ∂φ(i) ∂ (∂µ φ(i) ) Z Z ∂L ∂L ∂L (i) 4 (i) 4 δφ + d x∂µ =0 − ∂µ δφ ; dx ∂φ(i) ∂ (∂µ φ(i) ) ∂ (∂µ φ(i) ) Z ∂L ∂L 4 δφ(i) = 0, (2.28) − ∂µ ; dx ∂φ(i) ∂ (∂µ φ(i) ) where we have used the conditions for vanishing field variations at the bounaries. Given that the functional form of the field variation δφ(i) is arbitrary, we obtain the Euler-Lagrange differential equations, ∂L ∂L − ∂µ = 0. (2.29) (i) ∂φ ∂ (∂µ φ(i) ) From the above derivation we can easily see that two Lagrangian densities, L and L + ∂ν F ν (φ(i) , ∂φ(i) ), which differ by a total divergence yield identical Euler-Lagrange identities. Indeed the variation of the total divergence term in the action integral is zero, Z Z Z 4 ν ν δ d x∂ν F = δ dσν F = dσν δF ν = 0. (2.30) V,T

S(V,T )

In the above we have used Stokes’ theorem and that the variation of a smooth function F of the fields and their derivatives vanishes at the boundaries of the action integral. 15

2.4

Field Hamiltonian Density

For a system with a finite number of degrees of freedom, we find the conjugate momentum of a generalized coordinate by differentiating the Lagrangian with respect to the time derivative of the coordinate: pj =

∂L ∂ q˙j

(2.31)

The Hamiltonian of the system is then given by, X H= pj q˙j − L.

(2.32)

Let us now start from the Lagrangian of a field (in one dimension for simplicity), Z L = dxL, (2.33) which after discretization becomes

aLi ,

(2.34)

where Li is the value of the Lagrangian density at the point i of the lattice (discretized line in our case). The field q(x, t) has a value qi (t), at the same point. The corresponding conjugate mometum is, pj =

∂L X ∂Li = a . ∂ q˙j ∂ q˙j i

(2.35)

The Hamiltonian of discretized system is then, H=

X i

X j

∂Li − Li q˙j ∂ q˙j

(2.36)

We now assume that the discretized Lagrangian at any point i, contains the time derivative of only one degree of freedom q˙i , and no other (e.g. q˙i+1 , q˙i−1 , . . .). This was indeed the case for the example of the elastic rod, and it is the case for all physical systems that we will study. It essentially means that each point in the lattice has its own self-determined kinetic energy. On 16

the other hand, the potential term in Li can depend on the coordinates of its neighbours. The Hamiltonian now becomes, X ∂Li H= a q˙i − Li . (2.37) ∂ q˙i i Finally, by taking the continuoum limit a → 0, we obtain a volume integral over a it Hamiltonian density, Z H = dxH (2.38) with H = q(x, ˙ t)π(x, t) − L,

(2.39)

with the field conjugate momentum defined as, π(x, t) =

∂L . ∂ q(x, ˙ t)

(2.40)

We notice that in the derivation of Euler-Lagrange equations (Eq. 2.29), derivatives with respect to space and time coordinates are treated on the same footing. In fact, the Lagrangian formulation is manifestly covariant. However, in computing the Hamiltonian (Eq. 2.36), differentiation with time maintains a special role and covariance is not manifest.

2.5

An example: acoustic waves

As an example, we use an elastic medium (e.g. air) for acoustic waves. The Lagrangian density is a generalization in three space dimensions of the elastic rod Lagrangian that we have already studied. 2 2 2 ρvsound ρ ∂y(~r, t) ~ ∇y(~r, t) , − (2.41) L= 2 ∂t 2 with vsound the speed of sound. Setting vsound = 1, we write ρ (∂µ y) (∂ µ y) 2 ρ ≡ (∂x0 y)2 − (∂x1 y)2 − (∂x2 y)2 − (∂x3 y)2 . 2

L =

(2.42)

We first find the corresponding Euler-Lagrange equations, where we need the derivatives ∂L = 0, (2.43) ∂y ∂L = ρ ∂ µ y, ∂ (∂µ y)

∂µ

∂L = ρ ∂µ ∂ µ y ≡ ∂ 2 y. ∂ (∂µ y)

(2.44)

The equations of motion are ∂µ

∂L ∂L = ∂ (∂µ y) ∂y ∂ 2 y = 0.

;

(2.45)

A solution of the equation of motion is a plane-wave µ

y = eikµ x ≡ eik·x ,

with

kµ k µ = k 2 = 0.

(2.46)

The conjugate momentum of the field y is, ∂L = ρ∂0 y ∂ (∂0 y) π ; ∂0 y = . ρ π≡

(2.47)

The Hamiltonian density is H ≡ π∂0 y − L ; H=

π 2 1 ~ 2 + ρ ∇y . 2ρ 2

(2.48)

R 3 In the theory of fields, we make a reference to the Lagrangian L = d xL R 3 and the Hamiltonian H = d xH rarely. In practice, we always work directly with the corresponding densities L and H. Thus, it has prevailed that we call H the “Hamiltonian” and L the “Lagrangian” assuming imlicitly that we refer to their densities. We shall adopt the same terminology from now on.

2.6

Noetherâ&#x20AC;&#x2122;s theorem

We can prove a very powerful theorem for Lagrangian systems wich exhibit symmetries, which is known as the Noether theorem. The theorem states that for every symmetry of the system, there exists a conserved current and a corresponding conserved charge.

2.6.1

Field symmetry transformations

Consider a Lagrangian density L Ď&#x2020;(i) , â&#x2C6;&#x201A;Î˝ Ď&#x2020;(i) ,

(2.49)

which we assume to be invariant under infinitesimal transformations of the fields Ď&#x2020;(i) (x) â&#x2020;&#x2019; Ď&#x2020;(i) (x) + Î´Ď&#x2020;(i) (x),

â&#x2C6;&#x201A;Î˝ Ď&#x2020;(i) (x) â&#x2020;&#x2019; â&#x2C6;&#x201A;Î˝ Ď&#x2020;(i) (x) + â&#x2C6;&#x201A;Î˝ Î´Ď&#x2020;(i) (x).

(2.50)

The change in the Lagrangian density can be computed as, X â&#x2C6;&#x201A;L â&#x2C6;&#x201A;L (i) (i) Î´L = Î´Ď&#x2020; + â&#x2C6;&#x201A; Î´Ď&#x2020; (i) (i) ) Î˝ â&#x2C6;&#x201A;Ď&#x2020; â&#x2C6;&#x201A; (â&#x2C6;&#x201A; Ď&#x2020; Î˝ i X â&#x2C6;&#x201A;L â&#x2C6;&#x201A;L â&#x2C6;&#x201A;L (i) (i) + â&#x2C6;&#x201A;Î˝ â&#x2C6;&#x2019; â&#x2C6;&#x201A;Î˝ Î´Ď&#x2020; Î´Ď&#x2020; = (i) (i) ) (i) ) â&#x2C6;&#x201A;Ď&#x2020; â&#x2C6;&#x201A; (â&#x2C6;&#x201A; Ď&#x2020; â&#x2C6;&#x201A; (â&#x2C6;&#x201A; Ď&#x2020; Î˝ Î˝ i X â&#x2C6;&#x201A;L (i) . (2.51) Î´Ď&#x2020; ; Î´L = â&#x2C6;&#x201A;Î˝ â&#x2C6;&#x201A; (â&#x2C6;&#x201A;Î˝ Ď&#x2020;(i) ) i Invariance of the Lagrangian under the symmetry transformation of Eq. 2.50 requires that X â&#x2C6;&#x201A;L (i) Î´L = â&#x2C6;&#x201A;Î˝ = 0. (2.52) Î´Ď&#x2020; â&#x2C6;&#x201A; (â&#x2C6;&#x201A;Î˝ Ď&#x2020;(i) ) i Therefore the currents

JÎ˝ =

â&#x2C6;&#x201A;L Î´Ď&#x2020;(i) , â&#x2C6;&#x201A; (â&#x2C6;&#x201A;Î˝ Ď&#x2020;(i) )

(2.53)

â&#x2C6;&#x201A;Î˝ J Î˝ = 0.

(2.54)

are conserved: The conserved currents J Î˝ are as many as the number of independent generators for the transformations Î´Ď&#x2020;(i) . 19

It may happen that a field symmetry transformation does not leave the Lagrangian invariant, but it only changes it by a total derivative, δL = ∂µ J µ .

(2.55)

In that case, we still obtain the same physical equations of motion since the action integral remains invariant, Z Z 4 d x δL = d4 x ∂µ J µ = 0. (2.56) From Eq. 2.55 and our general result of Eq. 2.51, we find that Jν =

∂L δφ(i) − J ν , ∂ (∂ν φ(i) )

(2.57)

are conserved currents ∂ν J ν = 0. To each conserved current, corresponds a conserved charge, i.e. a physical quantity which maintains the same value at all times. Indeed,

; ; ; ;

2.6.2

∂ν J ν = 0 ∂J 0 ∂J i + i =0 ∂t ∂x ∂J 0 ~ ~ +∇·J = 0 ∂t Z Z 0 3 ∂J ~ · J~ = 0 d ~x + d3~x∇ ∂t Z ∂Q = 0 with Q = d3~xJ 0 ∂t

(2.58)

Space-Time symmetry transformations

We now consider a more complicated version of the theorem, which deals with space-time symmetries. We consider a classical system of i = 1 . . . N physical fields φ(i) , and we assume that the Lagrangian density does not depend explicitly on space-time coordinates. This is a reasonble assumption if we require that physical laws which are valid in the present should also be valid after we perform a time translation, i.e. the same physical will hold in the future and they held in the past. Also, this assumption anticipates that experiments in two different positions discover the same physics. However, 20

the Lagrangian depends on the space-time coordinates implicitly, through the fields φ(i) (xµ ). L = L φ(i) , ∂ν φ(i) . (2.59)

We now require that if we perform a space-time symmetry transformation, such as a translation or a rotation or a boost µ

xµ → x′ , the action remains always the same: Z Z (i) 4 (i) S= d xL φ (x), ∂ν φ (x) = V

(2.60)

h i ′ (i) ′ ′ (i) ′ d x L φ (x ), ∂ν φ (x ) . 4 ′

V′

(2.61)

In this way, since the action is invariant under the transformation, the variational principle δS = 0 is automatically satisfied after the transformation if it was satisfied before the transformation. In general, the fields φ(i) also transform under general space-time tranformations. They may be scalar, vectors, etc. We write, (i)

φ′ (x′ ) = φ(i) (x) + ∆φ(i) (x) ∂ν φ

′ (i)

′

(i)

(2.62) (i)

(x ) = ∂ν φ (x) + ∆ ∂ν φ (x)

(2.63) (2.64)

The action after the transformation becomes, Z h i (i) (i) d4 x′ L φ′ (x′ ), ∂ν φ′ (x′ ) S = ′ ′ ZV ∂xµ 4 = d x det L φ(i) (x) + ∆φ(i) (x), ∂ν φ(i) (x) + ∆ ∂ν φ(i) (x) . ∂xν V (2.65) The variation of the Lagrangian density due to the symmetry transformation can be computed as, ∆L = L φ(i) (x) + ∆φ(i) (x), ∂ν φ(i) (x) + ∆ ∂ν φ(i) (x) − L φ(i) (x), ∂ν φ(i) (x) X ∂L ∂L (i) (i) (i) 2 ∆φ + ∆ ∂ φ + O ∆φ (2.66) ; ∆L = ν (i) (i) ) ∂φ ∂ (∂ φ ν i where the term inside the square brackets vanishes by requiring that φ(i) are physical fields satisfying the Euler-Lagrange equations. 21

We consider continuous transformations for which there exists a choice of the transformation parameters resulting to a unit transformation, i.e. no transformation. An example is a Lorentz boost with some velocity ~v, where for v = 0 we have xµ → x′ µ = xµ . There are examples of symmetry transformations where this does not occur. For example a parity transformation does not have this property, and the Noether theorem is not applicable then. For continuous symmetry transformations, we can consider that they are infinitesimally different than the unit transformation, xµ → x′µ = xµ + δxµ with δxµ very small. The Jacobian of such a transformation is, ′ ∂xµ ∂ [xµ + δxµ ] det = det = det δµν + ∂ ν δxµ . ∂xν ∂xν

(2.67)

(2.68)

We now need the determinant of a four by four matrix (µ, ν = 0, 1, 2, 3). We can verify with an explicit calculation that through order O(δx): ′ ∂xµ = 1 + ∂ µ (δxµ ) + O((δx)2 ). (2.69) det ∂xν [exercise: Verify the above explicitly] Substituting the expansions of Eq. 2.66 and Eq. 2.69 in the action integral (i) of Eq. 2.65, and keeping terms through O ∆φ , we find Z S = S + d4 x [∆L + L∂ µ δxµ ] ( ) Z X ∂L ∂L 4 (i) (i) µ ;0 = dx ∆φ + ∆ ∂ν φ + L∂ δxµ ∂φ(i) ∂ (∂ν φ(i) ) i

(2.70)

An untransformed field evaluated at a transformed coordinate is expanded as φ(i) (x′ ) = φ(i) (x + δx) ; φ(i) (x′ ) = φ(i) (x) + δxµ ∂µ φ(i) + O(δx2 ) ; δφ(i) (x) = δxµ ∂µ φ(i) 22

(2.71)

Similarly, ∂ν φ(i) (x′ ) = ∂ν φ(i) (x + δx) ; ∂ν φ(i) (x′ ) = ∂ν φ(i) (x) + δxµ ∂µ ∂ν φ(i) + O(δx2 ) ; δ∂ν φ(i) (x) = δxµ ∂ν ∂µ φ(i)

(2.72)

The difference of a transformed field evaluated at a transformed coordinate with respect to an untransformed field evaluated at a non-tranformed coordinate is: (i)

∆φ(i) = φ′ (x′ ) − φ(i) (x) (i)

; ∆φ(i) = φ′ (x′ ) − φ(i) (x′ ) + φ(i) (x′ ) − φ(i) (x) (i)

; ∆φ(i) = φ′ (x′ ) − φ(i) (x′ ) + δφ(i) (x) ; ∆φ(i) = δ∗ φ(i) (x) + δφ(i) (x) ; ∆φ(i) = δ∗ φ(i) (x) + δxµ ∂µ φ(i) ,

(2.73)

where we separate explicitly the variation of the field at a point (i)

δ∗ φ(i) (x) = φ′ (x′ ) − φ(i) (x′ ),

(2.74)

from the variation due to changing the space-time position. Similarly, for the variation of the field derivative, we have (i)

∆∂ν φ(i) = ∂ν φ′ (x′ ) − ∂ν φ(i) (x) (i)

; ∆∂ν φ(i) = ∂ν φ′ (x′ ) − ∂ν φ(i) (x′ ) + ∂ν φ(i) (x′ ) − ∂ν φ(i) (x) h i (i) ′ (i) ′ (i) ′ ; ∆∂ν φ = ∂ν φ (x ) − φ (x ) + ∂ν φ(i) (x + δx) − ∂ν φ(i) (x) ; ∆∂ν φ(i) = ∂ν δ∗ φ(i) (x) + δxµ ∂µ ∂ν φ(i) (x).

(2.75)

Substituting Eq 2.73 and Eq 2.75 into Eq. 2.70, we obtain: Z X ∂L ∂L 4 (i) (i) 0= d x δ∗ φ + ∂ν δ∗ φ ∂φ(i) ∂ (∂ν φ(i) ) i ( ) Z X ∂L ∂L + d4 x L∂ µ δxµ + δxµ ∂ φ(i) + ∂ ∂ φ(i) (i) µ (i) ) µ ν ∂φ ∂ (∂ φ ν i

(2.76)

Applying the chain rule, we see that the term in the curly brackets is a total divergence, X ∂L ∂L µ µ µ (i) (i) ∂µ (Lδx ) = L∂µ δx + δx . (2.77) ∂ φ + ∂ ∂ φ (i) µ (i) ) µ ν ∂φ ∂ (∂ φ ν i The term in the square brackets of Eq. 2.76 can also be written as a total divergence, applying integration by parts and using the Euler-Lagrange equations, ∂L ∂L (i) (i) δ φ + ∂ δ φ ∗ ν ∗ ∂φ(i) ∂ (∂ν φ(i) ) ∂L ∂L ∂L (i) (i) = δ∗ φ + ∂ν − ∂ν δ∗ φ ∂φ(i) ∂ (∂ν φ(i) ) ∂ (∂ν φ(i) ) ∂L (i) δ∗ φ = ∂ν ∂ (∂ν φ(i) ) Eq. 2.76, with Eq. 2.78 and Eq. 2.78, becomes " # Z X ∂L 0 = d4 x ∂ν δ φ(i) + ∂µ (Lδxµ ) (i) ) ∗ ∂ (∂ φ ν # "i Z X ∂L δ φ(i) + Lδxν . ; 0 = d4 x∂ν (i) ) ∗ ∂ (∂ φ ν i

(2.78)

(2.79)

This is essentially the proof of Noether’s theorem. For simplicity, we restrict ourselves to fields φ(i) which transform as scalars under a space-time transformation, i.e. they do not transform: (i)

φ′ (x′ ) = φ(i) (x) ; ∆φ(i) = 0.

(2.80)

For this to happen, the variation of a field at a point δ∗ φ(i) must compensate the variation δφ(i) due to changing the space-time position: ∆φ(i) = 0 ; δ∗ φ(i) + δφ(i) = 0 ; δ∗ φ(i) = −δφ(i) ; δ∗ φ(i) = −δxµ ∂µ φ(i) . Substituting Eq. 2.80 into Eq 2.79 we obtain Z d4 x∂ν (T µν δxµ ) = 0. V

(2.81)

(2.82)

where we have defined: T µν ≡

∂L ∂ µ φ(i) − Lg µν (i) ∂ (∂ν φ )

X i

(2.83)

The tensor T µν is known as the energy-momentum tensor for reasons that will become obvious in a while. The name stress-energy tensor is also encountered very often. The volume of integration in equation Eq. 2.71 is considered to have boundaries which are arbitrary. In addition, the space-time variation δxµ depends on small but otherwise arbitrary parameters, as many of them as the generators of the symmetry transformation. Then, in order for Eq. 2.71 to hold, the integrand must vanish. We have therefore proven, that if the system possesses a space-time symmetry transformation, there are as many conserved currents as the generators of the symmetry transformation. ∂ν J ν = 0,

J ν = δxµ T µν .

(2.84)

We emphasize that we can write as many currents J ν as the number of the independent δxµ space-time symmetry transformations which are obeyed by the physical system. Translation symmetry transformations As a first application of the Noether theorem, we find the conserved currents and the corresponding charges for physical systems which are symmetric under space-time translations. An infinitesimal translation transformation is µ

xµ → x′ = xµ + ǫµ ; δxµ = ǫµ ,

(2.85)

where ǫµ is a small constant four-vector. The corresponding conserved current is J ν = T µν δǫν ,

(2.86)

satisfying the continuity equation ∂ν J ν = 0 ; ∂ν T µν ǫµ = 0. 25

(2.87)

The vector ǫµ is small but otherwise arbitrary. Then, we must have that the above equation is satisfied in general if ∂ν T µν = 0,

(2.88)

for every value of the index µ = 0, 1, 2, 3 seperately. To the four conserved currents correspond four “conserved charges” (Eq. 2.58), Z ν P = d3~x T 0ν . (2.89) Specifically, the “time-component” of P ν is Z 0 P = d3~x T 00 " # Z X ∂L = d3~x ∂0 φ(i) − g 00 L . (i) ) ∂ (∂ φ 0 i

(2.90)

We recall that the conjugate momentum for the field φ(i) is π (i) =

∂L , ∂ (∂0 φ(i) )

(2.91)

and thus P0 =

d3~x

" X i

π (i) ∂0 φ(i) − L .

(2.92)

In the square bracket, we recognize the Hamiltonian density (Eq. 2.39), Z 0 P = d3~x H. (2.93) concluding that the charge P 0 is the energy of the system, and it is conserved. Energy conservation is thus a pure consequence of time translation symmetry. Similarly, we can identify the charges P i, i = 1, 2, 3 as the momentum components of the system in the three space directions, and they are of course also conserved.

Lorentz symmetry transformations Lorentz transformations,

x′ = Λµν xν ,

(2.94)

ds2 = g µν dxµ dxν = g µν dx′ν dx′ν .

(2.95)

g µν dxµ dxν = g µν Λρµ Λσν dxρ dxσ = g ρσ Λµρ Λνσ dxµ dxν ; g µν = g ρσ Λµρ Λνσ .

(2.96)

preserve the distance

This leads to

Considering an infinitesimal transformation, Λµν = δνµ + ωνµ , +O(ω 2)

(2.97)

we obtain g µν = g ρσ Λµρ Λνσ ; g µν = g ρσ δρµ + ωρµ + . . . (δσν + ωσν + . . .)

; g µν = g µν + ω νµ + ω µν + O(ω 2) ; ω µν = −ω νµ .

(2.98)

The parameters ω µν for generating Lorentz transformations are therefore antisymmetric. A transformed space-time coordinate is given by x′

; δxµ

= = = ≡

Λµρ xρ (g µρ + ω µρ + . . .) xρ xµ + ω µρ xρ + O(ω 2 ) µ x′ − xµ = ω µρ xρ .

(2.99)

If Lorentz transformations are a symmetry of a physical system, then the corresponding conserved currents are 1 J ν = T µν δxµ = T µν xρ ωµρ = T µν xρ [ωµρ − ωρµ ] 2 1 1 µν ρ [T x ωµρ − T µν xρ ωρµ ] = [T µν xρ ωµρ − T ρν xµ ωµρ ] = 2 2 1 ; Jν = ωµρ J µνρ with J µνρ ≡ T µν xρ − T ρν xµ . (2.100) 2 27

The continuity equations for the conserved currents J ν are, ∂ν J ν = 0 ; ∂ν J µνρ = 0, and the correspoding conserved charges are: Z Z µρ 3 µ0ρ M = d ~xJ = d3~x T µ0 xρ − T ρ0 xµ .

(2.101)

(2.102)

Poincar´ e symmetry transformations We now require that a physical system possesses both space-time translation and Lorentz transformation symmetry, µ

xµ → x′ = ǫµ + Λµρ xρ .

(2.103)

The symmetry group of both translations and Lorentz transformations is called the Poincar´e group. The following two charge conservation equations should hold: ∂ν T µν = 0, due to translation symmetry, (2.104) ∂ν J µνρ = 0, due to Lorentz symmetry.

(2.105)

This is possible if the energy-momentum tensor T µν is symmetric. Indeed, 0 = = = = ;

∂ν J µνρ = ∂ν [T µν xρ − T ρν xµ ] xρ ∂ν T µν + T µν ∂ν xρ − xµ ∂ν T ρν − T ρν ∂ν xµ xρ 0 + T νµ δνρ − xµ 0 − T ρν δνµ T µρ − T ρµ T µρ = T ρµ .

(2.106)

It is not obvious that the energy-momentum tensor is symmetric. In fact, the definition of the tensor X ∂L µ (i) T µν ≡ ∂ φ − Lg µν (2.107) (i) ) ∂ (∂ φ ν i does not exhibit an explicit symmetry in the indices µ and ν. However, the energy momentum tensor is not uniquely defined. Let us consider a general tensor of the form ∂ρ f ρνµ , 28

which is antisymmetric in the indices ν, ρ: f ρνµ = −f νρµ . We can show that such a tensor is a conserved current, ∂ν (∂ρ f ρνµ ) =

1 [∂ν ∂ρ f ρνµ − ∂ν ∂ρ f νρµ ] 2

(antisymmetry)

1 [∂ν ∂ρ f ρνµ − ∂ρ ∂ν f ρνµ ] (relabeling) 2 1 = [∂ν ∂ρ f ρνµ − ∂ν ∂ρ f ρνµ ] = 0. (commuting derivatives) (2.108) 2 =

However it produces a null conserved charge, as we can verify easily: Z Q = d3~x ∂ρ f ρ0µ Z = d3~x ∂0 f 00µ + ∂i f i0µ Z = d3~x ∂0 0 + ∇i f i0µ Z = d3~x ∇i f i0µ (using Stokes’ theorem) = 0.

(2.109)

We can therefore add such a tensor to the energy-momentum tensor, T µν → T¯ µν = T µν + ∂ρ f ρνµ

(2.110)

without modifying the corresponding conserved charges (energy and momenta). The function f ρνµ can be chosen so that T¯ µν is manifestly symmetric in the indices µ, ν.

Chapter 3 Quantization of the Schr¨ odinger field We start our study of field quantization, by first reproducing the known results of Quantum Mechanics, for a system of many identical non-interacting particles.

3.1

The Schr¨ odinger equation from a Lagrangian density

As a first step, we introduce a Lagrangian that yields the Schr¨odinger equation with the Euler-Lagrange formalism. We can verify that the following density, ~ ~ ∗ ∇ψ ∇ψ ∂ψ − (3.1) L = iψ ∗ ∂t 2m does exactly this. ψ and ψ ∗ turn out to be complex conjugate when they take their physical solutions, but this is not an assumption that we have to do now ourselves. It will be a natural consequence of solving the Euler-Lagrange

equations, which are: ∂L ∂L − ∂ (∂µ ψ) ∂ψ ∂L ∂L ∂L ∂ψ ∗ ∂i ∂ i ψ ∗ = ∂0 + ∂i − =i + +0 ∂ (∂0 ψ) ∂ (∂i ψ) ∂ψ ∂t 2m ∇2 ∂ ψ∗, ;0 = i − ∂t 2m 0 = ∂µ

(3.2)

and ∂L ∂L − ∗ ∂ (∂µ ψ ) ∂ψ ∗ ∂L ∂L ∂L ∂i ∂ i ψ ∂ψ = ∂0 + ∂ − = 0 + −i i ∗ ∗ ∗ ∂ (∂0 ψ ) ∂ (∂i ψ ) ∂ψ 2m ∂t 2 ∇ ∂ ψ. ;0 = i + ∂t 2m 0 = ∂µ

(3.3)

Indeed, we obtain the Schr¨odinger equation for ψ and its complex conjugate for ψ ∗ . A general solution of the above equations is a superposition of planewave solutions, ψ(~x, t) =

~2 d3~k ~k)e−i(ωk t−~k~x) , with ωk = k . a( (2π)3 2m

(3.4)

It is also useful to compute the conjugate momenta for ψ and ψ ∗ as well as the Hamiltonian density corresponding to the Schr¨odinger Lagrangian. We find ∂L π= = iψ ∗ , (3.5) ∂ (∂0 ψ) and π∗ =

∂L = 0. ∂ (∂0 ψ ∗ )

(3.6)

The Hamiltonian density is H = π∂0 ψ + π ∗ ∂0 ψ ∗ − L ∗ ~ ~ ∇ψ ∇ψ (∂i ψ ∗ ) (∂ i ψ) ;H = − = . 2m 2m 31

(3.7)

Notice that this Hamiltonian density differs from ~ ~ ∗ ∇ψ ~ ψ ∗ ∇ψ ~ ∗ 2 ∇ψ ∇ ψ ∇ ψ − = − , 2m 2m 2m

(3.8)

by a total divergence which does not contribute to the Hamiltonian integral R H = d3~xH. Substituting the general solution of Eq. 3.4 in the expression for the Hamiltonian density, Eq. 3.7, we obtain H =

d3~k1 d3~k2 ~k1~k2 −i[(ωk −ωk )t−(~k2 −~k1 )~x] ∗ 2 1 a (k2 )a(k1 ) e (2π)3 (2π)3 2m

The Hamiltonian of the Schr¨odinger field is, Z H = d3~x H, and we can perform the volume integration easily using We find the following expression for the Hamiltonian, H=

3.2

d3~k ωk a∗ (k)a(k). 3 (2π)

(3.9)

(3.10) R

~ d3~xeik·~x = (2π)3 δ (3) (~k).

(3.11)

Quantization of Fields

For a system with a finite number of degrees of freedom L[qi q˙i ], the quantization procedure is the following. For each generalized coordinate we find first the corresponding conjugate momentum pj =

∂L . ∂ q˙j

(3.12)

Then we promote all qi , pi to operators, which satisfy commutation relations, [qi , pj ] = iδij ,

(3.13)

[pi , pj ] = [qi , qj ] = 0.

(3.14)

and

We can generalize this quantization procedure to systems of fields, with an infintite number of degrees of freedom. Viewing fields as the continuum limit of a discrete spectrum of degrees of freedom is particularly useful. In this way, we can make the “discrete to continuum” correspondece: i → ~x, qi (t) → φ(~x, t) pi (t) → π(~x, t).

(3.15)

The corresponding quantization conditions for fields are: [φ(~x1 , t), π(~x2 , t)] = iδ (3) (~x1 − ~x2 ),

(3.16)

[φ(~x1 , t), φ(~x2 , t)] = [π(~x1 t), π(~x2 , t)] = 0.

(3.17)

and

3.3

Quantized Schr¨ odinger field

We can apply the field quantization conditions of the previous section (Eqs 3.163.17) to the Schr¨odinger field ψ(~x, t) and its conjugate momentum π(~x, t) = iψ ∗ (~x, t). We obtain [ψ(~x1 , t), π(~x2 , t)] = iδ (3) (x1 − x2 ) ; [ψ(~x1 , t), ψ ∗ (~x2 , t)] = δ (3) (x1 − x2 ),

(3.18)

and [ψ(~x1 , t), ψ(~x2 , t)] = [π(~x1 , t), π(~x2 , t)] = 0 ; [ψ(~x1 , t), ψ(~x2 , t)] = [ψ ∗ (~x1 , t), ψ ∗ (~x2 , t)] = 0.

(3.19)

We now run into a problem if we impose a quantization condition for ψ ∗ and π ∗ , since we have found π ∗ = 0 in Eq. 3.6. The requirement, δ (3) (~x1 , ~x2 ) = [ψ ∗ (~x1 , t), π ∗~x2 , t] = [ψ ∗ (~x1 , t), 0] = 0,

(3.20)

is clearly inconsistent. However, we can write a different Lagrangian for the Schr¨odinger field than the one in Eq. 3.1, which is classically equivalent yielding identical equations of motion, ~ ~ ∗ ∇ψ ∇ψ ∗ i ∂ψ i ∂ψ Lnew = ψ ∗ − ψ − . (3.21) 2 ∂t 2 ∂t 2m 33

Exercise: Verify that this Lagrangian yields the Scr¨ odinger equation as Euler-Lagrange equations This Lagrangian differs from the one of Eq. 3.1 by a time derivative 2i ∂t (ψ ∗ ψ). With Lnew both π and π ∗ are non-vanishing, and all commutation relations that we can write lead to one of the Eqs 3.18-3.19. This is a good example showing that not all Lagrangian densities which are equivalent at the classical level, i.e. yield the same equations of motion, can be subjected to a self-consistent quantization. In our case, Lnew can be quantized while L not. The quantization conditions using the conjugate momenta π and π ∗ as derived from the Lagrangian density Lnew give the following three commutation relations: [ψ(~x1 , t), ψ(~x2 , t)] = [ψ ∗ (~x1 , t), ψ ∗ (~x2 , t)] = 0, [ψ(~x1 , t), ψ ∗ (~x2 , t)] = δ (3) (~x2 − ~x1 ) .

(3.22)

From these quantization conditions for the fields, it is easy to derive the corresponding quantization conditions for the operators a(~k) and a∗ (~k). First we observe, from Eq. 3.4 that the operators a(~k) and a∗ (~k) are Fourier transforms of ψ and ψ ∗ respectively, Z ~ ~ (3.23) a(k) = d3~xφ(~x, t)ei[ωk t−k·~x] . Exercise: Verify explicitly the above by substituting in the expression of Eq. 3.4 for the field φ. Then, we have h i ∗ ~ ~ a(k1 ), a (k2 ) Z ~ ~ = d3~x1 d3~x2 ei[ωk1 t−k1 ·~x1 ] e−i[ωk2 t−k2 ·~x2 ] [φ(~x1 , t), φ∗ (~x2 , t)] Z ~ ~ = d3~x1 d3~x2 ei[ωk1 t−k1 ·~x1 ] e−i[ωk2 t−k2 ·~x2 ] δ (3) (~x2 − ~x1 ) Z ~ ~ = d3~xei[(ωk1 −ωk2 )t−(k1 −k2 )·~x] h i ; a(~k1 ), a∗ (~k2 ) = (2π)3 δ (3) ~k1 − ~k2 . (3.24)

Similarly (exercise), we can prove that h i h i a(~k1 ), a(~k2 ) = a∗ (~k1 ), a∗ (~k2 ) = 0. 34

(3.25)

To emphasize that a∗ and ψ ∗ are now operators, we write a∗ → a† , ψ ∗ → ψ † .

3.4

Particle states from quantized fields

Let us now look at the so called “number density” operator N(k) ≡ a† (k)a(k).

(3.26)

We can prove that [N(k), a(p)] = a† (k)a(k), a(p) = a† (k) [a(k), a(p)] + a† (k), a(p) a(k)

; [N(k), a(p)] = − (2π)3 δ (3) (~k − p~)a(p). Also, in the same manner we find that N(k), a† (p) = (2π)3 δ (3) (~k − p~)a† (p).

(3.27)

(3.28)

It is also easy to prove ((exercise)) that two number density operators commute with each other, [N(k), N(p)] = N(k), a† (p)a(p) = N(k), a† (p) a(p) + a† (p) [N(k), a(p)] = (2π)3 δ (3) (~k − p~)a† (p)a(p) − (2π)3 δ (3) (~k − ~p)a† (p)a(p) ;

[N(k), N(p)] = 0.

We recall that the Hamiltonian is (Eq. 3.11), Z Z d3~k d3~k † H= ω a ωk N(k). (k)a(k) ; H = k (2π)3 (2π)3

(3.29)

(3.30)

Obviously, it commutes with the number density operator, " # Z Z d3~k d3~k [N(p), H] = N(p), ω N(k) = ωk [N(p), N(k)] = 0. k (2π)3 (2π)3 (3.31) 35

As a consequence, the eigenstates of the Hamiltonian can be found as eigenstates of the “number density” operator, or the “particle number operator”, defined as Z Z 3~ d k d3~k † ˆ≡ N N(k) = a (k)a(k), (3.32) (2π)3 (2π)3 and which also obviously commutes with the Hamiltonian. From Eqs 3.27-3.28, we derive that h i ˆ a† (p) = +a† (p), N, (3.33)

and

h i ˆ a(p) = −a(p). N,

(3.34)

ˆ |ni = n |ni , N

(3.35)

ˆ (a(p) |ni) = N

nh i o ˆ ˆ N, a(p) + a(p)N |ni

ˆ (a(p) |ni) = {−a(p) + a(p)n} |ni ; N ˆ (a(p) |ni) = (n − 1) (a(p) |ni) . ; N

(3.37)

We can prove that the eigenvalues of the number operator must always be positive, if the correspondinge eigestates are states of a Hilbert space with a positive norm. Consider the state |ωi = a(p) |ni, 0 ≤ || |ωi ||2 = hω| ωi = hn| a† (p)a(p) |ni = hn| N(p) |ni Z d3 p~ ˆ |ni = n hn| ni = n|| |ni ||2 ; 0 ≤ hn| N(p) |ni = hn| N (2π)3 ;

n ≥ 0.

(3.38) 36

This condition is in an apparent contradiction with Eq. 3.34, since a repeated application of the operator a(p) on a positive eigenvalue will eventually yield an eigenstate with a negative eigenvalue. This is true , if all eigenstates are not annihilated by the a(p) operator. Assume, however, that there is a state |0i which is annihilated by a(p), a(p) |0i = 0.

(3.39)

This state is an eigenstate of the number operator with a zero eigenvalue: a(p) |0i = 0 †

; a (p)a(p) |0i = 0 |0i ;

d3 ~p † a (p)a(p) |0i = 0 |0i (2π)3

ˆ |0i = 0 |0i ;N

(3.40)

Then, Eq. 3.39 only permits to act on the vacuum state with a† operators, producing eigenstates of the number operator with positive integer eigenvalues. For example, for a general state Z 3 d p~1 f (p1 )a† (p1 ) |0i , (3.41) |1particle i ≡ 3 (2π) we have (exercise): ˆ |1particle i = (+1) |1particle i . N Similarly, for a general state: Z 3 d p~1 d3 p~2 |2particles i ≡ f (p1 , p2 )a† (p1 )a† (p2 ) |0i , (2π)3 (2π)3

(3.42)

(3.43)

we have (exercise):

and so on:

ˆ |2particles i = (+2) |2particles i , N

(3.44)

ˆ |mparticles i = (+m) |mparticles i , N

(3.45)

with

Z Y m 3 d p~i † |mparticles i ≡ a (pi ) |0i f (p1 , . . . , pm ). 3 (2π) i=1

(3.46)

ˆ, Suggestively, we have labeled the eigenstates of the number operator N as states of a discrete number of particles. This becomes better justified when we find common eigenstates Q with the Hamiltonian H. The states of Eq. 3.46 with f (p1 , . . . , pm ) = i (2π)3 δ (3) (pi − ki ), i = 1 . . . m: ! m Y |mparticles , Ei ≡ a† (ki ) |0i (3.47) i=1

diagonalize the Hamiltonian, H |mparticles , Ei = E |mparticles , Ei ,

with E =

m X

ωki

(3.48)

i=1

Exercise: Show the above for m=1 and m=2. Get convinced that it is valid for arbitrary m. The Hamiltonian H is, according to Noether’s theorem, a conserved quantity due to the invariance of the action under time-translations. Symmetry under translations in the three space-time dimensions predicts the existence of a conserved field momentum P~ , with Z i P = d3 xT 0i . (3.49) Explicitly for the Schr¨ udinger field we find the field, P~ =

d3~k N(k)~k (2π)3

(3.50)

Exercise: Prove the above It is easy to see that a state |mparticles , Ei of Eq. 3.47 is also an eigenstate of the field momentum operator: P~ |mparticles , Ei = p~ |mparticles , Ei , 38

with ~p =

m X i=1

~ki

(3.51)

Exercise: Show the above for m=1 and m=2. Get convinced that it is valid for arbitrary m. To summarize, we have proven that the states of Eq. 3.47 are simultaneous eigenstates of the “number-operator”, the field Hamiltonian operator, and the field momentum operator. Their eigenvalues are the number of particles, the sum of their energies, and the number of their momenta respectively. We therefore see that the quantization of the Schr¨odinger field is a very elegant formalism to describe multi-particle states. It turns out that the “number-operator” is also a conserved quantity. The Lagrangian of the Schr¨odinger field has an additional symmetry which is responsible for this. The Lagrangian is invariant under the symmetry transformation: ψ(~x, t) → eia ψ(~x, t) ψ ∗ (~x, t) → e−ia ψ ∗ (~x, t).

(3.52)

You can prove that the corresponding charge to this symmetry transformation is Z ˆ N = d3~xJ 0 , with J 0 = ψ ∗ (~x, t) ∗ ψ(~x, t). (3.53) Thus, we have found that multi-particle systems where we have measured their number of particles at one moment, will always maintain the same number of particles in the future.

3.5

What is the wave-function in the field quantization formalism?

In quantum mechanics, the wave function y(~x, t)1 corrsponding to a single particle state is the probability amplitude to find a particle at a position ~x, y(~x, t) = h1particle | ~xi .

(3.54)

For a system of N particles the wave-function is the probability amplitude to find the N particles at positions x1 , x2 , . . . xN . y(~x1 , . . . ~xN , t) = h~x1 , . . . , ~xN | Nparticles i . 1

(3.55)

In this section, we use the letter y to denote the wave-function, and the letter ψ for the Schr¨odinger field operator.

The wave-function satisfies the Schr¨odinger equation, which for free particles is: " # N ~2 X ∇x i ∂ i + y(~x1 , . . . ~xN , t) = 0. (3.56) ∂t i=1 2m

Let us now look at a generic N-particle state of the previous section, Eq. 3.46, ! Z Y N d3 p~i † |Nparticles i = a (pi ) |0i f (p1 , . . . , pN ), (3.57) (2π)3 i=1 where we can express the creation operators a† (pi ) as Fourier transforms of the field operators φ† (~x, t). Substituting Eq. 3.23 into Eq. 3.57, we obtain ! Z Y N |Nparticles i = d3 x~i |S(x1 , . . . , xN )i f˜(x1 , . . . , xN , t), (3.58) i=1

with, |S(x1 , . . . , xN )i = and ˜ 1 , . . . , xN , t) = f(x

N Y i=1

ψ † (xi ) |0i ,

d3 p~i −i(ωpi t−~pi ~xi ) f (p1 , . . . , pN ). e (2π)3

By an explicit calculation (exercise), we can verify that " # N ~2 X ∇ ∂ xi i + f˜(~x1 , . . . ~xN , t) = 0, ∂t i=1 2m

(3.59)

(3.60)

(3.61)

i.e. it satisfies the Schr¨odinger equation. Therefore the function f˜, is identified as the wave-function y(x1 , . . . xN , t), y˜(x1 , . . . xN , t) = y(x1 , . . . xN , t) = h~x1 , . . . , ~xN | Nparticles i .

(3.62)

Using this observation, we write Eq.3.58 as ! Z Y N |Nparticles i = d3 x~i |S(x1 , . . . , xN )i h~x1 , . . . , ~xN | Nparticles i , ; |Nparticles i =

i=1 N Y

d3 x~i

i=1

; 1 = |Nparticles i

|S(x1 , . . . , xN )i h~x1 , . . . , ~xN | |Nparticles i N Y

d3 x~i

i=1

|S(x1 , . . . , xN )i h~x1 , . . . , ~xN | |Nparticles i (3.63)

Inside the square bracket we have a unit operator, which leads us to the conclusion that |S(x1 , . . . , xN )i = |~x1 , . . . , ~xN i . (3.64) We summarize the main result of this section, which connect the field operator formalism with the traditional wave-function description of non-relativistic Quantum Mechanics. • A quantum mechanical state of N particles at positions ~x1 , . . . ~xN is given by the action of field operators on the vacuum state: ! N Y † |~x1 , . . . ~xN i = ψ (xi ) |0i . (3.65) i=1

• A general N−particle state is a superposition, Z |Nparticles i = d3~x1 . . . d3~x1 f˜(x1 , . . . , xN ) |~x1 , . . . ~xN i .

(3.66)

• The kernel of the superposition integral is the wave-function, satisfying the Schr¨odinger equation # " N ~2 X ∇x i ˜ ∂ f (~x1 , . . . ~xN , t) = 0. (3.67) i + ∂t i=1 2m

Chapter 4 The Klein-Gordon Field In the study of the Schr¨odinger field, we demonstrated how Quantum Field theory can be made to construct quantum states of particles, with certain momentum and an energy momentum relation: E=

~p2 . 2m

(4.1)

This is the correct energy-momentum relation in the non-relativistic limit. Would it be possible to modify this procedure in such a way so that we obtain the relativistic energy-momentum relation E 2 = p2 + m2 ?

(4.2)

Such an energy-momentum relation is not a novelty of special relativity. In our classical mechanics example of acoustic waves, the Euler-Lagrange equations gave wave solutions of the form: ~

with

4.1

φ(~x, t) ∼ e−i(Et−k~x) ,

(4.3)

2 E 2 = ~k 2 vsound .

(4.4)

Real Klein-Gordon field

If we were to quantize acoustic waves, we would obtain particles which would always travel with the speed of sound. It is to be demonstrated that such 42

a quantization can be made consistently, but let us assume that this is possible. Then “particles” from the quantization of the acoustic field (usually called phonons) have the same energy-momentum relation as massless relativistic particles, if we were to identify the speed of sound as the speed of light (vsound = c). It is easy to modify the Lagrangian for acoustic waves so that the solutions for the field (Eq. 4.3), encode the correct relativistic energy-momentum relation of Eq. 4.2 for massive particles. The corresponding Lagrangian is 1 1 L = (∂µ φ) (∂ µ φ) − m2 φ2 . (4.5) 2 2 We will then attempt for the first time to write a quantum relativistic theory of free particles, by quantizing the field φ whose dynamics is described by this Lagrangian. In the case of acoustic waves, the field φ corresponds to the amplitude of an oscillation, and it is therefore real.

4.1.1

Real solution of the Klein-Gordon equation

The Euler-Lagrangian equation for the field φ of Eq. 4.5 is, ∂µ ;

∂L ∂L − =0 ∂ (∂µ φ) ∂φ 2 ∂ + m2 φ(x) = 0.

(4.6)

This is known as the Klein-Gordon equation. We shall find a general solution of the Klein-Gordon equation for a real field φ. We start with the ansatz: Z d4 k −ik·x ∗ +ik·x φ(x) = f (k)e + f (k)e , (4.7) (2π)4

which is manifestly real. In addition, it is invariant under Lorentz transformations, as it is anticipated for a scalar field. Substituting into the KleinGordon equation we find, Z d4 k m2 − k 2 f (k)e−ik·x + f ∗ (k)e+ik·x = 0. (4.8) 4 (2π) The above is satisfied if we have, f (k) = (2π)δ(k 2 − m2 )c(k),

f ∗ (k) = (2π)δ(k 2 − m2 )c∗ (k). 43

(4.9)

A general real solution of the Klein-Gordon equation is: Z d4 k µ φ(x ) = δ m2 − k 2 c(k)e−ik·x + c∗ (k)e+ik·x . 3 (2π)

(4.10)

q with ωk = ~k 2 + m2 .

(4.11)

δ(k0 − ωk ) + δ(k0 + ωk ) . 2ωk

(4.12)

The delta function requires that 2

k = kµ k = m ; k0 = ±ωk , The delta function can be written as, δ(k 2 − m2 ) = δ(k02 − ωk2 ) =

We can now perform the k0 integration in Eq. 4.10, obtaining Z d3~k h ~ ~ µ φ(x ) = c(ωk , ~k)e−i(ωk t−k~x) + c(−ωk , ~k)e+i(ωk t+k~x) 3 (2π) 2ωk i −i(ωk t+~k~ x) +i(ωk t−~k~ x) ∗ ~ ~ (4.13) + c(−ωk , k)e +c (ωk , k)e

We perform a change of integration ~k → −~k to the integrals corresponding to the second and fourth term of the sum inside the square bracket. Then, the four terms can be grouped together into a sum of only two exponentials, and we can write the general solution as: Z i d3~k h ~ −i(ωk t−~k~x) x) µ ∗ ~ i(ωk t−~k~ φ(x ) = , (4.14) + a (k)e a(k)e (2π)3 2ωk where a(~k) = c(ωk , ~k) + c∗ (−ωk , −~k). We remark that we often write the integration measure in an explicitly Lorentz invariant form, Z Z Z d4 k d4 k (+) 2 d3~k 2 2 = δ(k − m )Θ(k > 0) δ (k − m2 ), (4.15) 0 (2π)3 2ωk (2π)3 (2π)3 with δ (+) (k 2 − m2 ) ≡ δ(k 2 − m2 )Θ(k0 > 0).

(4.16)

The measure is invariant under orthochronous transformations which do not change the sign of the time-like component of the integration variable. 44

4.1.2

Quantization of the real Klein-Gordon field

The conjugate momentum of the Klein-Gordon field, φ(~x, t), is π=

∂L = ∂0 φ. ∂(∂0 φ)

(4.17)

The Hamiltonian of the Klein-Gordon field is, 2 Z ~ (∂0 φ)2 + ∇φ + m2 φ2 3 H = d ~x . 2

(4.18)

Substituting into the Hamiltonian integral the solution of Eq. 4.14, and performing the integration over the space volume, we find: Z d3~k ωk † † H= a (k)a(k) + a(k)a (k) , (4.19) (2π)3 2ωk 2

where we have promoted the field into an operator. We now impose commutation relations for the field and its conjugate momentum, [φ(~x1 , t), φ(~x2 , t)] = [π(~x1 , t), π(~x2 , t)] = 0,

(4.20)

[φ(~x1 , t), π(~x2 , t)] = iδ (3) (~x2 − ~x1 ) .

(4.21)

and With a straightforward calculation, we can verify (exercise) that the operators a and a† are written in terms of the φ and π operators as, Z ~ (4.22) a(k) = i d3~xei(ωk t−k~x) π(x,~ t) − iωk φ(~x, t) ,

and

†

a (k) = −i

~ d3~xe−i(ωk t−k~x) π(x,~ t) + iωk φ(~x, t) .

(4.23)

Then we can compute the commutators that can be formed with a and a† . We find (exercise), [a(k1 ), a(k2 )] = a† (k1 ), a† (k2 ) = 0, (4.24) and

†

a(k1 ), a (k2 ) = (2π) (2ωk1 )δ 45

(3)

~k1 − ~k2 .

(4.25)

4.1.3

Particle states for the real Klein-Gordon field

We can now construct the particle number operator, ˆ= N

d3~k a† (k)a(k), (2π)3 2ωk

(4.26)

for which we can prove that

and

h i ˆ a(p) = −a(p), N,

(4.27)

h i ˆ a† (p) = +a† (p). N,

(4.28)

Therefore, the operators a(p) and a† are ladder operators. In particular, if ˆ with eigenvalue n, a state |Si is an eigenstate of the number operator N † then the states a(p) |Si and a (p) |Si are also eigenstates with eigenvalues n − 1 and n + 1 respectively. The construction of particle states proceeds identically, as in our study of the quantization for the Schr¨odinger field, after we observe that the field Hamiltonian and the number operator commute (exercise): h i ˆ H = 0. N, (4.29)

The above equations lead to the existence of a vacuum state |0i, which is annihilated by the operator a(k), a(k) |0i = 0.

(4.30)

The vacuum state is an eigenstate of the number operator with a zero eigenvalue, ˆ |0i = 0 |0i . N (4.31) All other states can be produced from the vacuum state, by acting on it with creation operators a† (pi ). A generic state ! m Y |Ψm i ≡ a† (pi ) |0i , (4.32) i=1

contains m particles,

ˆ |Ψm i = m |Ψm i . N 46

(4.33)

4.1.4

Energy of particles and “normal ordering”

We consider the simplest case for |Ψim state, |pi = a† (p) |0i ,

(4.34)

which is an one-particle state, satisfying ˆ |pi = 1 |pi . N

(4.35)

We can compute the energy level of the state by acting on it with the field Hamiltonian operator, d3~k ωk † a (k)a(k) + a(k)a† (k) |pi 3 (2π) 2ωk 2 Z ωk † d3~k 2a (k)a(k) + a(k), a† (k) |pi = 3 (2π) 2ωk 2 Z d3~k ωk † 3 (3) = 2a (k)a(k) + (2π) 2ω δ (0) |pi k (2π)3 2ωk 2 Z (3) 3~ ωk ; H |pi = ωp + δ (0) dk |pi . (4.36) 2 H |pi =

To our surprise, we find an infinite eigenvalue! This is at first sight very embarrassing. We can make sense of this infinity, if we look at the energy of the vacuum state, which contains no particles. H |0i = = = ; ;

We find that the field Hamiltonian gives an energy eigenvalue for the vacuum state in Eq. 4.37 which is equal to the same infinite constant that appeared in the energy eigenvalue (Eq. 4.36) of the one-particle state. We can then re-write Eq. 4.36 in an apparently innocent manner, p H |pi = p~2 + m2 + Evacuum |pi . (4.38) Using the property (exercise) H, a† (pi ) = +ω(pi)a† (pi ),

we can prove that states with more particles, have an energy, ! Xp 2 2 (~pi + m ) |Ψm i . H |Ψm i = Evacuum +

(4.39)

(4.40)

All Hamiltonian eigenvalues contain and identical infinite constant which is equal to the vacuum energy. However, measurements of absolute energy levels are not possible. The vacuum energy is harmless if we want to measure the energy difference of the states |Ψm i from the vacuum. We can then “recalibrate” our energy levels (by an infinite constant) removing from the Hamiltonian operator the energy of the vacuum: H → : H : ≡ H − Evacuum = H − h0| H |0i .

(4.41)

Explicitly, the Hamiltonian operator after we subtract the vacuum constant can be written as Z Z † d3~k ωk † d3~k † : a (k)a(k) + a(k)a (k) : = ω a (k)a(k) . k (2π)3 2ωk 2 (2π)3 2ωk (4.42) A practical trick to remove from a conserved operator, such as the Hamiltonian, its vacuum expectation value: : O : =≡ O − h0| O |0i

(4.43)

is to perform what is known as “normal ordering”. As you can see in Eq. 4.42, this is simply achieved by putting to creation operators to the left of annihilation operators when these refer to the same momentum ~k. With this practical rule, : a† (k)a(k) + a(k)a† (k) : = a† (k)a(k) + a† (k)a(k) = 2a† (k)a(k). (4.44) 48

This is equivalent to using the commutation relation [a(k), a† (k)] = (2π)3 2ωk δ (3) (0) in order to put the creation operators to the right and dropping infinities δ (3) (0). These leaves annihilation operators to the left, and guarantees that the vacuum is automatically annihilated by the modified operator.

4.1.5

Field momentum conservation

Noether’s theorem predicts that field momentum is a conserved operator, due to the symmetry of the Klein-Gordon action under space translations. The momentum operator is given by, d3~k ~k † a (k)a(k) + a(k)a† (k) 3 (2π) 2ωk 2 Z Z δ (3) (0) d3~k ~ † = d3~k ~k k a (k)a(k) + 3 (2π) 2ωk 2 Z 3~ d k ~ † ; P~ = k a (k)a(k) . (2π)3 2ωk P~ =

(4.45)

Here, the infinity δ (3) (0) cancels because the momentum integral multiplying it is vanishing due to the antisymmetry of the integrand under the transformation ~k → −~k. P~ commutes with both the Hamiltonian, with which it has common eigenstates. The momentum of the vacuum state is zero, P~ |0i = 0.

(4.46)

The P~ eigenvalues of multi-particle states can be easily found by using, h i P~ , a† (p) = +~pa† (p). (4.47)

etc.

4.1.6

Labels of particle states?

The main outcome of quantizing the real Klein-Gordon field, is a spectrum of states which can be identified as states of many particles with definite energy and momentum. A general state, ! n Y |Ψn i ≡ a† (pi ) |0i , (4.49) i=1

describes m identical particles, ˆ |Ψn i = n |Ψn i , N

(4.50)

each of them carrying momentum ~pi , P~ |Ψn i =

n X i=1

p and relativistic energy + p~2i + m2 , H |Ψn i =

p~i

|Ψn i ,

(4.51)

(4.52)

n q X p~2i + m2 i=1

|Ψn i .

We have therefore arrived to a consistent combination of quantum mechanics and special relativity, where field quanta give rise to particles with the correct properties as anticipated from relativity.

4.2

Two real Klein-Gordon fields

Our task is to describe all known particles and their interactions. It is then interesting to study the quantization of a system with more than one fields. We can start simply, describing a system of two Klein-Gordon fields which they differ only in their mass parameter, 1 (∂µ φ1 ) (∂ µ φ1 ) − 2 1 (∂µ φ2 ) (∂ µ φ2 ) − + 2

L =

m21 2 φ 2 1 m22 2 φ. 2 2

(4.53)

This Lagrangian yields two Klein-Gordon equations as equations of motions for the fields φ1 and φ2 ., ∂ 2 + m21 φ1 = 0, ∂ 2 + m22 φ2 = 0, (4.54) with solutions

φ1 (x ) = µ

φ2 (x ) = and

i d3~k h ~ −ik·x † ~ ik·x , a ( k)e + a ( k)e 1 1 (2π)3 2ω1

(4.55)

i d3~k h ~ −ik·x † ~ ik·x , a ( k)e + a ( k)e 2 2 (2π)3 2ω2

(4.56)

q ω1 = ~k 2 + m21 ,

q ω2 = ~k 2 + m22 .

(4.57)

ˆ can be The Hamiltonian H, field momentum P~ and number operator N written as Z d3~k ˆ ˆ ˆ ˆ N = N1 + N2 Ni = a† (~k)ai (~k), (4.58) (2π)3 2ωi i Z d3~k ~k n † o ~ ~ ~ ~ a , ai , (4.59) P = P1 + P2 Pi = (2π)3 2ωi 2 i Z d3~k ωi n † o H = H1 + H2 Hi = ai , ai . (4.60) (2π)3 2ωi 2 We can construct particle states in the same fashion as with the Lagrangian of just a single Klein-Gordon field. Products of a†1 operators acting on the vacuum state create relativistic particles with mass m1 , while a†2 operators create particles with mass m2 . For example, the states |S1 i ≡ a†1 (p) |0i ,

|S1 i ≡ a†1 (p) |0i

(4.61)

have

H |S1 i =

ˆ |Si i N P~ |Si i p~2 + m21 |S1 i

= = and 51

+1 |Si i ,

+~p |Si i , H |S2 i =

(4.62) (4.63) q

p~2 + m22 |S2 i .

(4.64)

They are degenerate in that, they are single particle states with the same momentum p~, however they can be distinguished by measuring the energy of the particles as long as the masses m1 and m2 . The possibility of m1 = m2 is interesting; states created with the a†1 or a†2 operators are degenerate in both the energy and momentum.

4.2.1

Two equal-mass real Klein-Gordon fields

The special case of m1 = m2 = m worthies a detailed study, because a new rotation symmetry emerges in the space of fields φ1 and φ2 . This leads, according to Noether’s theorem, to a conserved quantity. We rewrite the Lagrangian in a form where the symmetry is manifest: 1 ~ µ ~ T L= ∂µ φ ∂ φ , (4.65) 2 where we have defined the field vector φ1 ~T = ~ with its transpose φ φ≡ φ2

φ1 φ2

(4.66)

Obviously, the Lagrangian is invariant (exercise) if we perform an orthogonal transformation, ~→φ ~ ′ = Rφ, ~ with RT = R−1 . φ

(4.67)

To compute the conserved current we need to find the change δφi of the fields under an infinitesimal symmetry transformation. We then expand, Rij = δij + θij + O(θ2 ).

(4.68)

Orthogonality of R implies that the matrix θij is antisymmetric, T −1 Rij = Rij ; δji + θji = δij − θij ; θij = −θji .

(4.69)

The variation of the fields is: φ1 → φ′1 = R1j φj = (δ1j + θ1j ) φj = φ1 + θ11 φ1 + θ12 φ2 = φ1 + θ12 φ2 ; δφ1 = θ12 φ2 , 52

(4.70)

and, similarly, δφ2 = θ21 φ1 = −θ12 φ1 .

(4.71)

The Noether theorem current is,

∂L ∂L δφ1 + δφ2 ∂ (∂µ φ1 ) ∂ (∂µ φ2 ) = θ12 [(∂ µ φ1 ) φ2 − (∂ µ φ2 ) φ1 ]

Jµ = ; Jµ

(4.72)

and the “charge” Q=

d3~x [(∂ µ φ1 ) φ2 − (∂ µ φ2 ) φ1 ] ,

(4.73)

is conserved. Substituting in the above expressions the physical solutions for the fields φi , and performing the ~x integration we obtain (exercise): Z i d3~k h † † a1 (k)a2 (k) − a2 (k)a1 (k) . (4.74) Q = −i (2π)3 2ωk A Noether charge is defined uniquely up to a constant multiplicative factor. Obviously, if Q is a time-independent (conserved) quantity, then also λQ, where λ is a number constant, is also conserved. The normalization in Eq. 4.74 is convenient yielding a Hermitian Q (exercise): Q† = Q.

(4.75)

We now proceed to find the eigenstates of the Q charge operator. This becomes easy, if we construct ladder operators for Q. We compute first the commutators of the charge with a1 and a2 (exercise). We find [Q, a1 (p)] = −ia2 (p),

(4.76)

[Q, a2 (p)] = +ia1 (p).

(4.77)

and Define the linear combinations, a(p) ≡

a1 (p) + ia2 (p) √ , 2

(4.78)

b(p) ≡

a1 (p) − ia2 (p) √ . 2

(4.79)

and

These are indeed ladder operators (exercise), satisfying the following commutation relations with the charge Q: [Q, a(p)] = −a(p),

[Q, b(p)] = +b(p).

(4.80)

By taking the hermitian conjugate of the above equations, and using Q† = Q, we find that Q, a† (p) = +a† (p), Q, b† (p) = −b† (p). (4.81)

From a state |Si with charge q,

Q |Si = q |Si ,

(4.82)

we can obtain states with charges q ± 1 by applying the ladder operators a† (p) and b† (p) (exercise), Q a† (p) |Si = (q + 1) a† (p) |Si , (4.83) Q b† (p) |Si = (q − 1) b† (p) |Si .

(4.84)

We remark that the operators a† , b† are also ladder operators for the Hamiltonian H and the field momentum operator P~ . This is a consequence of the fact that a† and b† are linear combinations of a†1 and a†2 which yield degenerate eigenstates for H and P . To find all the common eigenstates of the charge Q , it suffices to start from one common eigenstate and use the operators a† and b† to find all the others. The vacuum state is also an eigenstate of the charge operator (exercise): Q |0i = 0. (4.85) Repeated application of the a† operator on the vacuum, builds states of positive charge: ! n Y

(+)

Ψ ≡ a† (pi ) , (4.86) n i=1

with

ˆ Ψ(+) = n Ψ(+) N n n ! n X

(+)

P~ Ψn = p~i Ψ(+) n

H Ψ(+) = n

i=1 n q X

p~i 2 + m2

i=1

(4.87) (4.88) !

Q Ψ(+) = +n Ψ(+) n n

(+)

Ψn

(4.89) (4.90)

Repeated application of the b† operator on the vacuum, builds states of positive charge: ! n Y

(−)

Ψn ≡ b† (pi ) , (4.91) i=1

with

ˆ Ψ(−) N = n Ψn(−) n ! n X

(−)

= p~i Ψn(−) P~ Ψn

H Ψ(−) = n

i=1 n q X

p~i 2 + m2

i=1

. = −n Ψ(−) Q Ψ(−) n n

(4.92) (4.93) !

(−)

Ψ n

(4.94) (4.95)

To summarize the main results of this section, mass degeneracy resulted to a new O(2) symmetry of the Lagrangian. This gives rise to a new conserved quantity, the charge Q. A particle state, is now characterized by its momentum, the mass of the particle (or equivalently the energy), and its charge which can be either positive or negative. We have just demonstrated a method to describe physical systems of particles which are accompanied by their anti-particles.

4.2.2

Two real Klein-Gordon fields = One complex Klein-Gordon field

We can recast the original expressions for the operators whose eigenvalues ˆ P~ , H, Q) in terms of the charge ladder operators a, b, characterize particle (N, 55

rather than a1 , a2 . Substituting, a−b a2 = √ , 2−

a+b a1 = √ , 2

(4.96)

we find i d3~k h † ~ ~ † ~ ~k) , a ( k)a( k) + b ( k)b( (2π)3 2ωk Z d3~k ~k † † ~ P = a ,a + b ,b (2π)3 2ωk 2 Z d3~k ωk † † H= a ,a + b ,b . (2π)3 2ωk 2

ˆ = N

(4.97) (4.98) (4.99)

We can now re-write the field operators in terms of a and b. We find that φ1 =

φ + φ† φ2 = √ , 2i

φ + φ† √ , 2

(4.100)

with φ(x) =

d3~k −ik·c † ik·c a(k)e + b (k)e , (2π)3 2ωk

q k = ωk = ~k 2 + m2 . 0

(4.101) We can cast the Lagrangian representing the fields in terms of φ and φ† , rather than φ1 and φ2 . We find, 1 ~ T ~ 1 2 ~ T ∂µ φ ∂µ φ − m φ φ = (∂µ φ) ∂µ φ† − m2 φφ† , (4.102) L= 2 2

where

φ1 + iφ2 † φ1 − iφ2 √ ,φ = √ . (4.103) 2 2 The Lagrangian in the complex field representation, is more suggestive to the fact that it yields particle and anti-particle states. We could compute the Hamiltonian and momentum operators directly in terms of φ and φ† arriving to the same expressions as in the representation with two real fields. To compute the charge Q, we would need to identify what is the symmetry of this new Lagrangian. It is perhaps already obvious that the Lagrangian is invariant under a field phase-redefinition, ~= φ

φ1 φ2

, and φ =

φ → e−iα φ,

φ† → e+iα φ† . 56

(4.104)

This is the equivalent of the rotation symmetry transformation that we have found earlier, in the complex field representation. Let us verify this, by ~ performing a rotation on the vector φ, φ1 cos θ sin θ φ1 → φ2 − sin θ cos θ φ2 φ1 cos θ −i sin θ φ1 ; → iφ2 −i sin θ cos θ iφ2 ; φ1 + iφ2 → e−iθ (φ1 + iφ2 ) ; φ to e−iθ φ.

(4.105)

There is an “ambiguity” worth noting, when applying Noether’s theorem to find the conserved charged under the transformation φ → eiα φ and φ† → e−iα φ† . The corresponding field variations are, δφ = iαφ,

δφ† = −iαφ† .

Noether’s theorem states that the quantity, Z ∂L ∂L 3 † Q = d ~x δφ + δφ ∂(∂0 φ) ∂(∂0 φ† ) Z ;Q = d3~x πδφ + π † δφ†

(4.106)

(4.107)

is conserved. Obviously, if Q is conserved, then also every other operator λQ + c, with lambda, c constant numbers is also conserved. The expressions for Q are therefore unique up to a multiplicative and an an additive constant. The ambiguity on the additive constant is removed when we remove the contribution of the vacuum to the charge of particle states (as we have done for the energy). The operator : Q : = Q − h0| Q |0i ,

(4.108)

is ambiguous only up to a multiplicative factor, which essentially denotes the ”units” in which we measure the charge of a state.

4.3

Conserved Charges as generators of symmetry transformations

It may occur that we know of a conserved charge, without knowing the corresponding symmetry transformation. For example, in our previous discussion, 57

we found the charge from the Klein-Gordon Lagrangian in the representation of two-real fields before we “discovered” the U(1) symmetry of the Lagrangian in the complex field representation. A conserved charge is a generator of the symmetry transformation. Let us assume that a Lagrangian is symmetric under a field transformation, φi → φi + δφi leading to a conserved charge Z X Q = d3 x i

∂L δφi = ∂ (∂0 φi )

(4.109)

d3 x

πi δφi .

(4.110)

The commutator of the charge operator Q with a field operator φj is, Z X [Q, φj (x)] = d3 y [πi (y)δφi(y), φj (x)] i

= =

d3 y

([πi (y), φj (x)] δφi (y) + πi (y) [δφi (y), φj (x)])

d3 y

X i

iδij δ (3) (~y − ~x)δφj (x) + 0

; [Q, φj (x)] = iδφj (x).

(4.111)

Exercise: Verify this explicitly computing the commutators [Q, φ] † and Q, φ for the complex Klein-Gordon field.

4.4

Casimir effect: the energy of the vacuum

A remarkable consequence of the quantization of the Klein-Gordon field is that a state with no particles, the vacuum |0i, has energy. For the real Klein-Gordon field, for example, we find that the vacuum energy is: Z ωk (3) (4.112) Evacuum = δ (0) d3~k . 2 We can associate the infinity δ 3 (0), with the volume of our physical system. We have considered that the classical solutions for the fields extend to an 58

infinite volume, and we have performed all space integrations from −∞ to ∞. If we were to consider a finite volume, this delta function would be replaced by the volume of integration. Recall that, Z ~ d3~xe−ik·~x = (2π)3 δ (3) (~k). (4.113) For ~k = 0, we have V =

d3~x = (2π)3 δ (3) (~0),

(4.114)

and the density of the vacuum energy can be cast in the physically more appealing form, p Z Evacuum m2 + ~k 2 d3~k ǫ= = . (4.115) V (2π)3 2 A second source of infinity in the expression p for the vacuum energy is the integration over all frequency modes ωk = m2 + ~k 2 . The value of the energy of the vacuum, does not enter physical predictions and we may not worry that it comes out infinite. But we should be able to see that the vacuum energy is different if, for a reason, the fields vanish in some region of the space-volume if some frequencies ωk do not contribute to the vacuum. This can be achieved, if the field is forces to satisfy some boundary conditions. Let us assume, that the Klein-Gordon field is forced to vanish on the planes x = 0 and x = L, φ(x = 0, y, z, t) = φ(x = L, y, z, t) = 0.

(4.116)

We should also assume that there is no mass term, m = 0, for convenience. The boundary conditions filter only the wave-lengths with nπ kx = . (4.117) L One must make a careful promotion of the field into an operator. The general field solution which in addition satisfies the boundary conditions is: ∞ nπx Z X d2~k⊥ µ µ µ a(k)e−ikµ x + a† (k)eikµ x , (4.118) φ(x ) = sin 2 L (2π) 2ωk n=0 with

r n2 π 2 2 ωk = ~k⊥ + 2 . L 59

(4.119)

Notice that we only integrate over the perpendicular directions ky , kz , since kx is discretized. Consequently, the volume integral δ (3) (0) will be replaced by the surface of the planes δ (3) (0) → δ (2) (0) = S/(2π)2 . The vacuum energy in the new configuration with the two “field-free” planes is, q Z π 2 n2 2 ∞ + ~k⊥ Evacuum X d2~k⊥ L2 = , (4.120) S (2π)2 2 n=0 where we have defined ~k⊥ = (ky , kz ). S is now the surface of each planes. In polar coordinates, q π 2 n2 2 ∞ Z + k⊥ X dk⊥ k⊥ Evacuum L2 = . (4.121) S (2π) 2 n=0

This integral appears to be divergent in the limit k⊥ → ∞. Let us now work with a slightly modified integrand, q π 2 n2 2 ∞ Z 1−δ + k⊥ X dk⊥ k⊥ Evacuum L2 = . (4.122) S (2π) 2 n=0 There exists a value of δ for which the integral is well defined. We shall perform our calculation for such a value, and then we shall try to analytically continue the result to δ = 0. We change variables once again, k⊥ = l⊥

nπ , L

(4.123)

and we obtain, Evacuum 1 π 3−δ = S 4π L ;ǫ=

1 π 3−δ 8π L

∞ X

n3−δ

n=1

∞

n=1

∞ X

n−3+δ

∞

1−δ dl⊥ l⊥ δ

2 2 −2 dl⊥ (l⊥ )

2 1 + l⊥

(4.124)

We recognize the sum as the ζ-function, ∞ X n=1

1 n−3+δ

= ζ(δ − 3). 60

(4.125)

Performing the change of variables, 2 l⊥ →

x , 1−x

(4.126)

we find that the integral is, Z ∞ Z 1 q δ−5 δ 2 2 − δ2 2 dl⊥ (l⊥ ) 1 + l⊥ = dxx− 2 (1 − x) 2 = B(1 − δ/2, −3/2 + δ/2). 0

(4.127)

where we have recognized the beta function, B(x, y) =

Γ(x)Γ(y) . Γ(x + y)

(4.128)

The final result reads, Evacuum 1 π 3−δ = B(1 − δ/2, −3/2 + δ/2)ζ(δ − 3). S 8π L Amazingly 1 , the limit δ → 0 exists and it is finite. We find,

(4.129)

Evacuum π2 =− . (4.130) S 1440L3 The vacuum energy depends on the distance between the two planes, on which the field vanishes. Can we realize this in an experiment? The electromagnetic field is zero inside a conductor. If we place two “infinite” conducting sheets parallel to each other at a distance L, then we can reproduce the boundary conditions of the setup that we have just studied. The quantization of the electromagnetic field is rather more complicated than the Klein-Gordon field, but as far as the vacuum energy is concerned, we shall see that they are very similar. Our analysis, leads to an amazing prediction. Two electrically neutral conductors attract each other. This is known as the Casimir effect. Notice, that the energy of the vacuum Evacuum gets smaller when the conducting plates are closer. Therefore, there is an attractive force between them. This is an effect that it has been verified experimentally. The force per unit area (pressure or rather anti-pressure) between the two conductors is: ∂ Evacuum π2 S =− . (4.131) F =− ∂L 480L4 1

Many subtleties have been swept under the carpet in this calculation. They will become clear when we introduce properly the method of dimensional regularization.

4.5

Can the Klein-Gordon field be an oneparticle wave-function?

In the previous chapter, we performed the quantization of the Schr¨odinger field. We found out that field quantization, although an elegant formalism, it was not indispensable. We were able to connect this formalism with a traditional description where the Schr¨odinger field was taken to be directly, a wave-function of a single particle. It is a question worth answering, whether field quantization is truly necessary for a relativistic field or it is just a superfluous formalism. An one-particle wave-function, as a probability amplitude, is in general complex. We will then assume that the complex Klein-Gordon field is one. Recall that the Lagrangian is written as, LKG = (∂µ ψ) (∂µ ψ ∗ ) − m2 ψψ ∗ .

(4.132)

We denote the field with ψ in order to emphasize our intention to consider it a wave-function. As such, ψ ∗ is its complex conjugate and not a hermitian conjugate operator. The complex Klein-Gordon field Lagrangian and the Schr¨odinger field Lagrangian, |∇ψ|2 LS = ψ ∗ i∂t ψ + , (4.133) 2m are both symmetric under a phase transformation, as we have discussed in detail. In the Schr¨odinger case, this lead to a conserved charge, which we identified it with the probability for the particle to be anywhere in space, Z Total Probability = d3~x |ψ(~x, t)|2 = constant. (4.134) Therefore, we could also interpret the quantity, ρ ≡ |ψ(~x, t)|2 ,

(4.135)

as a probability density. An attempt to repeat the same steps for the case of the Klein-Gordon field fails badly. The “total probability” integral is now the charge, Z h i 3 ∗ ∗ ˙ ˙ Q = d ~x ψ(~x, t)ψ (~x, t) − ψ(~x, t)ψ (~x, t) , (4.136) 62

with a “density” ˙ x, t)ψ ∗ (~x, t), ρ ≡ ψ(~x, t)ψ˙ ∗ (~x, t) − ψ(~

(4.137)

which is not positive definite, and thus physically unacceptable. A second embarrassment from the Klein-Gordon equation as a relativistic wave-function equation, is that it predicts particles with a negative energy. This may be accepted for a free particle, if we assume that for whatever reason we can only “see” particles with positive energy. However, if we couple these particles to photons as it is required for all particles with electromagnetic interactions, spontaneous emission of photons will turn all particles into negative energy particles. Clearly, a single particle wave-function interpretation of the Klein-Gordon field is wrong and it cannot describe physical particles. Quantum Field Theory is the only consistent approach.

Chapter 5 The Dirac Equation Dirac identified the source of the problems in interpreting the Klein-Gordon equation as a wave -function equation the fact that it was a second order differential equation in time. The correspondence, i∂t → E, resulted to a second order polynomial equation for the energy of a freeparticle obeying, which of course has both a positive and a negative solution. Dirac’s idea was then to find a linear equation, whose differential operator represented a sort of a “square-root” of the differential operator in the KleinGordon equation. Explicitly, Dirac’s linear equation is, (iγ µ ∂µ − m) ψ = 0.

(5.1)

Multiplying the Dirac equation with the differential operator (−iγ µ ∂µ − m) from the left, ; (−iγ ν ∂ν − m) (iγ µ ∂µ − m) ψ = 0 ; γ µ γ ν ∂µ ∂ν + m2 ψ = 0 1 ν µ 1 µ ν 2 γ γ ∂µ ∂ν + γ γ ∂ν ∂µ + m ψ = 0 ; 2 2 1 µ ν 1 ν µ 2 ; γ γ ∂µ ∂ν + γ γ ∂µ ∂ν + m ψ = 0 2 2 1 µ ν 2 ; {γ , γ } ∂µ ∂ν + m ψ = 0, 2 64

(5.2)

which for {γ µ , γ ν } = 2g µν ,

(5.3)

gives the Klein-Gordon equation. In other words, if we can construct objects γ µ which satisfy the algebra of Eq. 5.3, known as Clifford algebra, then every solution ψ(~x, t) of the Dirac equation, will also be a solution of the Klein-Gordon equation. The objects γ µ are anti-commuting and they cannot be just numbers. They have to be matrices. The first dimensionality for which we can find four non-commuting matrices γ 0 , γ 1 , γ 2 , γ 3 is four. In the following we shall construct the so called “gamma-matrices” in four dimensions.

5.1

Mathematical interlude

It is useful to review here the properties of the 2 × 2 Pauli matrices, which we shall use as building blocks for constructing γ-matrices by taking their Kronecker product.

5.1.1

Pauli matrices and their properties

The Pauli matrices are, 0 1 0 1 1 0 σ1 = , σ2 = −i , σ3 = . 1 0 −1 0 0 −1

(5.4)

The Pauli matrices satisfy the following commutation and anti-commutation relations, [σi , σj ] = 2iǫijk σk (ǫ123 = 1), (5.5) and {σi , σj } = 2δij I2×2 .

(5.6)

σi2 = I2×2 , σ1 σ2 = iσ3 , σ3 σ2 = −iσ1 , . . .

(5.8)

det (σi ) = −1,

(5.9)

Adding the two equations together, we find that the product of two Pauli matrices is, σi σj = δij I2×2 + iǫijk σk . (5.7) For example, The determinant and trace of Pauli matrices are,

and Tr (σi ) = 0.

(5.10)

The Pauli matrices, together with the unit matrix I2×2 constitute a basis of general 2 × 2 matrices M2×2 . We write, M2×2 = a0 I2×2 +

3 X

ai σi .

(5.11)

i=1

From the above decomposition we have, Tr (M) = a0 (M) +

3 X

ai Tr (σi ) = 2a0 ,

(5.12)

i=1

and (exercise), Tr (σk M) = 2ak .

(5.13)

We can therefore write, 3

M2×2 =

5.1.2

X Tr (Mσi ) Tr (M) I2×2 + σi . 2 2 i=1

(5.14)

Kronecker product of 2 × 2 matrices

Let us define the Kronecker product of two 2 × 2 matrices b11 b12 a11 a12 , , B= A= b21 b22 a21 a22

(5.15)

as 

b11 a11  b11 a21 b11 A b12 A = A⊗B ≡  b21 a11 b21 A b22 A b21 a21

b11 a12 b11 a22 b21 a12 b21 a22

b12 a11 b12 a21 b22 a11 b22 a21

 b12 a12 b12 a22  . b22 a12  b22 a22

(5.16)

The following properties can be proved (exercise) explicitly, det (A ⊗ B) = (det A)2 (det B)2 ,

(5.17)

Tr (A ⊗ B) = Tr (A) Tr (B) ,

(5.18)

(A ⊗ B)† = A† ⊗ B † ,

(5.19)

(A + C) ⊗ B = A ⊗ B + C ⊗ B,

(5.20)

A ⊗ (B + C) = A ⊗ B + A ⊗ C,

(5.21)

(A1 ⊗ B1 ) (A2 ⊗ B2 ) = A1 A2 ⊗ B1 B2 .

(5.22)

and

5.2

Dirac representation of γ-matrices

The matrices γ 0 = I2×2 ⊗ σ3 ,

(5.23)

and γ j = σj ⊗ (iσ2 ) ,

(j = 1, 2, 3)

(5.24)

satisfy the Clifford algebra, {γ µ , γ ν } = 2g µν I4×4 .

(5.25)

We find, γ0

= (I2×2 ⊗ σ3 ) (I2×2 ⊗ σ3 ) 2 = I2×2 ⊗ σ32 = I2×2 ⊗ I2×2 = I4×4 0 0 ; γ ,γ = 2g 00 I4×4 . (5.26) Also, for the anti-commutator γ j , γ k with j, k = 1, 2, 3 we have j k γ ,γ = γj γk + γkγj = (σj ⊗ (iσ2 )) (σk ⊗ (iσ2 )) + (σk ⊗ (iσ2 )) (σj ⊗ (iσ2 )) = σj σk ⊗ (iσ2 )2 + σk σj ⊗ (iσ2 )2 = {σj , σk } ⊗ (−I2×2 ) = (2δjk I2×2 ) ⊗ (−I2×2 ) j k ; γ ,γ = −2δ jk I4×4 j k ; γ ,γ = 2g jk I4×4 . (5.27)

Finally, the anti-commutator {γ 0 , γ j } with j = 1, 2, 3 is 0 j γ ,γ = γ 0γ j + γ j γ 0 = (I2×2 ⊗ σ3 ) (σj ⊗ (iσ2 )) + (σj ⊗ (iσ2 )) (I2×2 ⊗ σ3 ) = σj ⊗ (iσ2 σ3 ) + σj ⊗ (iσ3 σ2 ) = σj ⊗ (i {σ2 , σ3 }) = σj ⊗ i2δ 23 I2×2 = σj ⊗ 02×2 = 04×4 ; γ 0, γ j = 2g 0j I4×4 . (5.28) We have therefore found a set of γ- matrices satisfying the Clifford algebra. This set is not unique. Let us define new matrices, −1 µ γ µ′ = M4×4 γ M4×4 ,

with M4×4 any 4 × 4 matrix with an inverse. Then, µ′ ν ′ −1 γ ,γ = M4×4 {γ µ , γ ν } M4×4 −1 = 2g µν M4×4 I4×4 M4×4 µν = 2g I4×4 .

(5.29)

(5.30)

The matrices γ µ′ satisfy the same algebra, furnishing a different representation. Let us now compute the hermitian conjugate of the γ−matrices. We find γ µ† = γ 0 γ µ γ 0 ,

(5.31)

which can also be written as, 2 γ 0 γ µγ 0 = γ 0 γ µ, γ 0 − γ 0 γ µ = 2g µ0 γ 0 − γ µ .

(5.32)

Explicitly, γ-matrices are hermitian for µ = 0, †

γ 0 = γ 0,

(5.33)

while they are anti-hermitian for µ = j = 1, 2, 3 †

γ j = −γ j .

(5.34)

† γ 0 = (I2×2 ⊗ σ3 )† = I2×2 ⊗ σ3† = I2×2 ⊗ σ3 = γ 0 ,

(5.35)

It is easy to verify this in the Dirac representation. We have †

and †

γ j = (σj ⊗ (iσ2 ))† = σj† ⊗ (−iσ2† ) = σj ⊗ (−iσ2 ) = −γ j . 68

(5.36)

5.3

Traces of γ− matrices

We can prove that γ−matrices are traceless in every representation. Let us start with Tr (γ µ γ ν γ µ ) = Tr γ ν (γ µ )2 = Tr (γ ν g µµ ) = g µµ Tr (γ ν ) (5.37) (No summation over the index µ) This is also equal to

Tr (γ µ γ ν γ µ ) = Tr (γ µ {γ ν , γ µ }) − Tr (γ µ )2 γ ν = 2g µν Tr (γ µ ) − g µµ Tr (γ ν ) . (5.38) For µ 6= ν, we then have g µµ Tr (γ ν ) = −g µµ Tr (γ ν ) ; Tr(γ ν ) = 0.

(5.39)

It is left as an exercise to prove a number of useful identities for the traces of gamma matrices (series 4).

5.4

γ−matrices as a basis of 4 × 4 matrices

The Clifford algebra restricts the number of independent matrices that we can construct by taking products or combinations of products of γ−matrices in a given representation. First, it imposes that the square of a γ−matrix is proportional to the unit 4 × 4 matrix, 1 µ µ +I4×4 , µ = 0 µ 2 µµ (γ ) = {γ , γ } = g I4×4 = (5.40) −I4×4 , µ = j = 1, 2, 3 2 This limits the number of terms in independent products of γ−matrices to at most four. • The only independent product of four γ−matrices is, γ5 = iγ 0 γ 1 γ 2 γ 3 .

(5.41)

Exercise: Show that you can write the matrix γ 5 as γ5 =

i µνρσ µ ν ρ σ ǫ γ γ γ γ , 4! 69

(ǫ0123 = 1).

(5.42)

• There are four independent products of three γ−matrices, γ 0 γ 1 γ 2 , γ 1 γ 2 γ 3 , γ 2 γ 3 γ 1 and γ 3 γ 1 γ 0 . These can be written as a linear combination of γ 5 γ µ, (5.43) and products of two only γ−matrices, by using the Clifford algebra. • Due to the Clifford algebra, only the six antisymmetric combinations of products of two γ−matrices are independent. These are the combinations σ 01 , σ 02 , σ 03 , σ 12 , σ 13 , σ 23 with σ µν =

i µ ν [γ , γ ] . 2

(5.44)

In the Dirac representation γ5 takes the form, iγ 0 γ 1 γ 2 γ 3 i (I2×2 ⊗ σ3 ) (σ1 ⊗ (iσ2 )) (σ2 ⊗ (iσ2 )) (σ3 ⊗ (iσ2 )) i (σ1 σ2 σ3 ) ⊗ (−iσ3 σ2 ) i iσ32 ⊗ (−σ1 ) = I2×2 ⊗ σ1 0 I2×2 = . I2×2 0

γ5 = = = = ; γ5

Exercise: Show that in the Dirac representation, σl 02×2 02×2 σj jk 0j , and σ = iǫjkl σ =i 02×2 σl σj 02×2

(5.45)

(5.46)

with j, k = 1, 2, 3. A general 4 × 4 matrix has 16 independent components. From a representation of γ−matrices we can construct 15 independent 4 × 4 matrices. Together with the unit matrix I4×4 , they constitute a basis. In other words, a general 4 × 4 matrix M can be written as, M4×4 = a0 I4×4 + aµ γ µ + a5µ γ5 γ µ + aµν σ µν .

(5.47)

Exercise: Find the coefficients a0 , a5µ , aµν for a given 4 × 4 matrix M.

5.5

Lagrangian for the Dirac field

After this survey of the properties of γ−matrices, we return to the Dirac equation, (i6∂ − mI4×4 ) ψ = 0. (5.48) We have introduced the “slash” notation: 6a ≡ γ µ aµ ,

6∂ ≡ γ µ ∂µ .

(5.49)

The Dirac differential operator i6∂ − m is a 4 × 4 matrix. The object ψ is a four dimensional “Dirac spinor” and it has its own Lorentz transformation, which we shall derive in detail later.   ψ1 (~x, t)  ψ2 (~x, t)   (5.50) ψ(~x, t) ≡ ψa (~x, t) =   ψ3 (~x, t)  . ψ4 (~x, t)

It is very important not to confuse a spinor ψ with a four dimensional vector, which has truly different Lorentz transformations. We can now take the Hermitian conjugate of the Dirac equation, (i6∂ − mI4×4 ) ψ = 0 ; −i (γ µ ∂µ ψ)† − mψ † = 0 ; −i ∂µ ψ † γ µ† − mψ † = 0 ; −i ∂µ ψ † γ 0 γ µ γ 0 − mψ † = 0

Multiplying this equation with γ 0 from the left, we obtain −i ∂µ ψ † γ 0 γ µ (γ 0 )2 − mψ † γ 0 = 0 ; −i ∂µ (ψ † γ 0 γ µ − m(ψ † γ 0 ) = 0.

We now define,

ψ¯ ≡ ψ † γ 0 .

(5.51)

(5.52) (5.53)

Then, the Hermitian conjugate of the Dirac equation reads ¯ µ − mψ¯ = 0. −i∂µ ψγ 71

(5.54)

We can obtain the Dirac equation and its conjugate from a Lagrangian density, L = ψ¯ (i6∂ − m) ψ. (5.55) Exercise: Derive the Euler-Lagrange equations for the spinor field components ψa and ψ¯a . This is a very important piece of information if we want to quantize the Dirac field ψ. We will postpone the quantization of the Dirac field until after we gain a deeper understanding of the origin of the Lagrangian, and the Lorentz transformation properties of spinor fields.

Chapter 6 Lorentz symmetry and free Fields A fundamental requirement for physical laws is that they must be valid for all relativistic observers. Let us consider fields φi (xµ ) describing a physical system, in a certain coordinate reference frame. Consider now a different relativistic observer. Space-time coordinates in the frame of the two observers are related with a Lorentz transformation, xµ → xµ′ = Λµν xν .

(6.1)

The new observer will have to transform appropriately both fields and coordinates when computing the Lagrangian, φi (x) → φ′i (x′ ),

L(x) → L′ (x′ ).

(6.2)

We will require that the action is the same for both observers. In this way, if it is an extremum for the physical values of φi (x) in the first reference frame, it will also be an extremum for the physical values of the field φ′i (x′ ) in the second relativistic frame. For proper Lorentz transformations (det (Λµν ) = 1) we have µ′ ∂x 4 ′ 4 d x = d x det = d4 x det (Λµν ) = d4 x. (6.3) ∂xν Then ′

S=S ;

d xL(x) =

d4 xL′ (x′ ) ; L(x) = L′ (x′ ). 73

(6.4)

We have just found that we can write relativistically invariant actions, and therefore physical laws for all relativistic observers, if the Lagrangian density is a scalar, i.e. it does not transform under a Lorentz transformation: L′ (x′ ) = L(x),

xµ → xµ′ = Λµν xν .

(6.5)

How many such Lagrangian densities can we find? Obviously, Lorentz symmetry is not sufficient on its own to determine all physical laws. We can restrict the problem to Lagrangians which describe free relativistic particles, which do not interact with each other. The Lagrangians that we are after may contain up to one only parameter, which is the mass m of the free particles. The equations of motion and field quantization should lead to the correct relativistic energy-momentum relation, p2 = m2 .

(6.6)

A Lagrangian of free particles must contain derivatives of the fields ∂µ φi . They are needed to generate the momentum ∂µ φi → pµ in the energymomentum relation. We include a generic label i, allowing for the possibility of a particle with a mass m and multiple degrees of freedom, such as charge, spin, etc. Our question has now become more specific: How can we combine the fields φi and their derivatives ∂µ φi into a scalar Lagrangian with one only mass parameter? We need two ingredients: • The transformation of derivatives ∂µ → ∂µ′ under xµ → Λµν xν , • The fields transformations φi (x) → φ′i (x′ ) under xµ → Λµν xν . It is straightforward to find the transformation of derivatives, ∂ ∂xν ∂ = ∂xµ′ ∂xµ′ ∂xν ν = Λ−1 µ ∂ν .

∂µ → ∂µ′ = ; ∂µ → ∂µ′ Similarly,

∂ µ → ∂ µ′ = Λ−1

µ ν

∂ν .

(6.7)

(6.8)

6.1

Field transformations and representations of the Lorentz group

In this section, we will examine the allowed possibilities for linear field transformations 1 , φi(x) → φ′i (x′ ) = M(Λ)ij φj (x) (6.9) under a space-time Lorentz transformation

xµ → Λµν xν .

(6.10)

Lorentz transformations are a group, i.e if we perform two successive Lorentz transformations, x → x′ = Λx, and

x′ → x′′ = Λ′ x′ ,

this is equivalent to performing a direct Lorentz transformation x → x′′ = Λ′′ x, with Λ′′ = Λ′ Λ.

(6.11)

How about the matrices M(Λ)ij ? Consider again the transformation x → x′′ = Λ′′ x, where the field is transformed as, φ(x) → φ′′ (x′′ ) = M(Λ′′ )φ(x).

(6.12)

φ′′ (x′′ ) = M(Λ′ )φ′ (x′ ) = M(Λ′ )M(Λ)φ(x).

(6.13)

We can arrive to the frame x′′ by performing successive Lorentz transformations, x → x′ = Λx → x′′ = Λ′ x′ = Λ′ Λx. Then we have that, For the last two equations to be consistent with each other, field transformations M(Λ) must obey: M(Λ′ Λ) = M(Λ′ )M(Λ).

(6.14)

In group theory terminology, this means that the matrices M(Λ) furnish a representation of the Lorentz group. Field Lorentz transformations are therefore not random, but they can be found if we find all (finite dimension) representations of the Lorentz group. 1

Non-linear transformations “can be made” linear, and are not necessary to consider (eg BRST symmetry in QFTII)

6.2

Scalar representation M (Λ) = 1

The simplest representation M(Λ) is the scalar representation, where a field φ(x) does not change under Lorentz transformations, φ(x) → φ′ (x′ ) = φ(x).

(6.15)

A Lagrangian of a scalar field and its derivatives must have the transformations of the derivatives canceled among themselves. A single derivative term, ∂µ φ is not a scalar. The minimum number of derivatives that are needed for a scalar is two, (∂µ φ) (∂ µ φ) , φ∂ 2 φ. The second term is actually equivalent to the first up to a total divergence. Let us assume that a Lagrangian contains this term, L = (∂µ φ) (∂ µ φ) + . . .

(6.16)

Then we can find the mass dimension of the field φ, by requiring that the action is dimensionless (in natural units where c = h ¯ = 1 ). Z [mass]0 = [S] = d4 xL = [x]4 [L] = [mass]−4 [L] (6.17)

which means that all Lagrangian densities must have mass dimensionality 4, [L] = [mass]4 .

(6.18)

From this we find that a scalar field φ has a mass dimensionality one, [(∂µ φ) (∂ µ φ)] = [mass]4 ; [∂µ ]2 [φ]2 = [mass]4 ; [mass]2 [φ]2 = [mass]4 ; [φ] = [mass]1 .

(6.19)

A relativistically invariant Lagrangian may include terms with no derivatives, of the type φn . Notice that the Klein-Gordon equation 1 m2 2 LKlein−Gordon = (∂µ φ) (∂ µ φ) − φ, 2 2 is, indeed, invariant under Lorentz transformations. 76

(6.20)

6.3

Vector representation M (Λ) = Λ

Another representation of the Lorentz group which is easy to figure out is the vector representation, corresponding to M(Λµν ) = Λµν .

(6.21)

A field in the vector representation transforms as, Aµ (x) → Aµ′ (x′ ) = Λµν Aν (x),

xµ′ = Λµν xν .

(6.22)

Terms where all Lorentz indices are contracted are invariant under Lorentz transformations, Exercise: Show that the following terms are invariant under Lorentz transformations (∂µ Aµ ) , (∂µ Aν ) (∂ µ Aν ) , (∂µ Aν ) (∂ ν Aµ ) .

(6.23)

Lorentz invariance is not the only symmetry that is required for free fields. For, example, the Lagrangian for the electromagnetic field is (as we shall see in detail) 1 (6.24) LEM = − [(∂µ Aν ) (∂ µ Aν ) − (∂µ Aν ) (∂ ν Aµ )] , 2 where only a certain combination out of two dimension four operators with two derivatives is present in the Lagrangian. This is because of an additional symmetry under gauge transformations which constrains further the Lagrangian form. Invariance of the Lagrangian under Lorentz transformations is a definite requirement for relativistic theory, but this is not sufficient in order to determine its exact form.

6.4

How to find representations?

Are the scalar and the vector representation the only representations that we can find? How do we find the rest of them? This question simplifies enormously if we look at small Lorentz transformations in arbitrary representations and the corresponding generators. A small Lorentz transformation is, Λµν = δνµ + ωνµ , 77

(6.25)

and any n × n matrix representation of this transformation, Mab (Λµν ) = Mab (δνµ + ωνµ ) ,

a, b = 1 . . . n

(6.26)

can be expanded in the small parameters ωνµ , Mab (Λµν )

Mab (δνµ )

+ ωµν

∂Mab (Λ)

+ O(ω 2 ). ∂ωµν ω=0

(6.27)

The n−dimensional representation Mab (δνµ ) of the unit transformation is the n × n unit matrix, Mab (δνµ ) = In×n . (6.28) The derivatives of the representation matrices M with respect to the small transformation parameters define the so called generators of the representation,

1 ∂Mab (Λµν )

µν . (6.29) (JM )ab ≡ i ∂ωµν ω=0 The generators form a basis, such that the representation of every small Lorentz transformation can be written as a linear combination of the generators plus the unit, µν Mab (Λ) = In×n + iωµν JM .

(6.30)

Finite Lorentz transformations are also generated by exponentiating linear combinations of the generators. If Λµν is not expandable in small parameters, we can use the representation property, M(Λ) = M(Λ1 ) . . . M(ΛN ),

with Λ = Λ1 Λ2 . . . ΛN

(6.31)

and construct it from the infinite product of infinitesimal Lorentz transformations with small parameters ωij /N, µν ωµν µν N M(Λ) = lim In×n + i JM = eiωµν JM . (6.32) N →∞ N The generators of a group representation is all we ever need to determine completely the representation. The full set of commutators of the generators define what is known as the Lie algebra of the group. Consider a Lorentz transformation Λ made up from the following sequence of small transformations, −1 Λ = Λ1 Λ2 Λ−1 1 Λ2 .

(6.33)

The representation of Λ is then −1 M(Λ) = M(Λ1 )M(Λ2 )M(Λ−1 1 )M(Λ2 ) (2) µν (1) µν (2) µν (1) µν JM JM In×n − iωµν JM In×n − iωµν = In×n + iωµν JM In×n + iωµν

µν ρσ (1) (2) ; M(Λ) = In×n + ωµν ωρσ [JM , JM ]+ ...

(6.34)

Now we take seriously the statement that representations M(Λ) of any small transformation Λ can be written as a linear combination of the generators X of the representation plus the unit matrix. There must exist parameters ωµν such that (X) αβ (6.35) M(Λ) = In×n + iωαβ JM + . . .

From the last equations we have that

(X)

µν ρσ αβ (1) (2) ωµν ωρσ [JM , JM ] = iωαβ JM .

(6.36)

Notice that the number of generators JM is finite and always the same for every representation of the group. Specifically for the Lorentz group we can Λ write only six independent derivatives limωµν →0 ∂M for a given representa∂ωµν tion M (recalling the antisymmetry of ωµν = −ωνµ ). Eq. 6.36 tells us that the commutator of two generators is a linear combination of generators, µν ρσ αβ αβ [JM , JM ] = iCµνρσ JM .

(6.37)

αβ The coefficients Cµνρσ as known as structure constants. Powerful theorems from group theory state the following:

1. The structure constants are indeed constants. The same ones irrespective of the choices of group parameters ω in Eq. 6.36. 2. The structure constants are characteristic of the group and they are the same for all representations M(Λ). We now have a better strategy in order to find representations of the Lorentz group: • First, we will use any of the representations that we already know, the scalar or vector representation, in order to find the structure constants αβ Cµνρσ of the Lorentz group. 79

µν • Then we shall attempt to find matrices Jn×n for each finite dimension n which satisfy the same commutation relations (known as Lie algebra) of Eq. 6.37. If we succeed, then we have found an n × n representation of the generators, and by exponentiation, an n × n representation of the group.

6.4.1

Generators of the scalar representation

We shall use the known scalar representation in order to find the structure constants of the Lorentz group. In this representation, under a Lorentz transformation, xµ → xµ′ = Λµν xν (6.38) a scalar field remains invariant φ′ (xµ′ ) = φ(xµ ) ; φ′ (Λµν xν ) = φ(xµ ).

(6.39)

Setting x → Λ−1 x in the above equation we have, µ

φ′ (xµ ) = φ(Λ−1 ν xν ),

(6.40)

which for small Lorentz transformations becomes φ′ (xµ ) = φ(xµ − ωνµ xν ) = φ(xµ ) − ωµν xν ∂ µ φ(xµ ) + O(ω 2 ) 1 = φ(xµ ) − (ωµν − ωνµ ) xν ∂ µ φ(xµ ) + O(ω 2 ) 2 1 = φ(xµ ) + ωµν (xµ ∂ ν − xν ∂ µ ) φ(xµ ) + O(ω 2) 2 i µν φ(x) ; φ′ (x) = 1 − ωµν Jscalar 2

(6.41)

with the generators of the Lorentz group in the scalar representation being: µν Jscalar = −i (xµ ∂ ν − xν ∂ µ ) .

(6.42)

. In an exponentiated form we have, i

µν

φ′ (x) = e− 2 ωµν Jscalar φ(x).

(6.43)

With one representation of the group generators at hand we can find the structure constants. An explicit calculation (exercise) yields the identity, µ1 µ2 µ3 µ4 [Jscalar , Jscalar ]= µ2 µ3 µ1 µ3 µ2 µ4 µ2 µ3 µ1 µ4 i (g Jscalar + g µ1 µ4 Jscalar − g µ2 µ4 Jscalar − g µ1 µ3 Jscalar ).

(6.44)

This is the Lie algebra of the Lorentz group, and we expect it to be satisfied by every other representation of the generators.

6.4.2

Generators of the vector representation

We can verify that the Lie algebra is universal to all generator representations by repeating the same analysis for the generators of the vector representation. A vector field Aα (xµ ), transforms as Aα (xµ ) → Aα′ (xµ′ ) = Λαβ Aβ (xµ ) = δβα + ωβα Aβ (xµ ) + O(ω 2 )

= Aα (xµ ) + ωβαAβ (xµ ) + O(ω 2 ) 1 αβ = Aα (xµ ) + ω − ω βα Aβ (xµ ) + O(ω 2 ) 2 µν ω δµα δνβ − δνα δµβ Aβ (xµ ) + O(ω 2) = Aα (xµ ) + 2 i α µν α′ µ′ αβ ; = A (x ) = δ − ωµν (Jvector )β Aβ (xµ ), 2

with

µν (Jvector )αβ = i δαµ δβν − δαν δβµ .

(6.45)

(6.46)

A Lorentz transformation on a vector field changes its orientation by mixing the indices of Aα . The generators of the transformation for the vector orientation are given in Eq. 6.46. In an exponentiated form, we have α′

′

µν −iωµν (Jvector )

A (x ) = e

α β

Aβ (x).

(6.47)

Exercise: Prove that a Lorentz transformation of a vector field “at a point” is: i h α µν µν −iωµν (Jvector )β +δβα Jscalar α′ Aβ (x). (6.48) A (x) = e

We can now compute explicitly (exercise) the commutator, µ1 µ2 µ3 µ4 [Jvector , Jvector ]αβ = i h µ1 µ4 µ2 µ3 µ1 µ3 µ2 µ4 i g µ2 µ3 (Jvector )αβ + g µ1 µ4 (Jvector )αβ − g µ2 µ4 (Jvector )αβ − g µ1 µ3 (Jvector )αβ .

(6.49)

We find that the commutators of the vector representation satisfy the same Lie algebra as in the scalar representation (Eq. 6.44), which is of course what we expect from the theory of Lie groups.

6.5

Spinor representation

We can find all possible generators for the representations of the Lorentz with a finite dimension in a systematic manner. Of particular interest to us is the Dirac or spinor representation of the Lorenz group, which has the same dimensionality 4 × 4 s the vector representation. We can verify, after some algebra, that the 4 × 4 matrices, S µν =

i µ ν [γ , γ ] , 4

(6.50)

satisfy the same commutation relation as in Eq. 6.44. Explicitly, [S µ1 µ2 , S µ3 µ4 ] = i (g µ2 µ3 S µ1 µ4 + g µ1 µ4 S µ2 µ3 − g µ2 µ4 S µ1 µ3 − g µ1 µ3 S µ2 µ4 ) .

(6.51)

Therefore, they are generators of a new representation for the Lorentz group, which is neither the scalar nor the vector representation. We call this new representation “spinor” representation. A spinor field   ψ1 (x)  ψ2 (x)   (6.52) ψ(x) =   ψ3 (x)  , ψ4 (x) transforms under as,

ψ ′ (x′ ) = Λ 1 ψ(x), 2

(6.53)

with a finite spinor transformation being i

µν

Λ 1 = e− 2 ωµν S . 2

(6.54)

We find that (exercise) the commutator of the generator S µν with a γ−matrix is [γ µ , S ρσ ] = i δµρ γ σ − δµσ γ ρ , (6.55)

or, in the more suggestive form,

ρσ [γ µ , S ρσ ] = (Jvector )ν γ ν .

(6.56)

The above equation can be exponentiated into a general result, µ ν µ Λ−1 1 γ Λ 1 = Λν γ . 2

(6.57)

Exercise: Prove Eq. 6.57 in the case of very small Lorentz transformations, by expanding in the transformation parameters ωµν both sides of the equation and using Eq 6.56. In the same manner, we verify that: Λ 1 γ µ Λ−1 = (Λ−1 )µν γ ν . 1 2

6.6

(6.58)

Lorentz Invariance of the Dirac Lagrangian

¯ and ψ6¯∂ ψ, in the Dirac Lagrangian are It is easy to verify that both terms, ψψ invariant under Lorentz transformations. As in many previous instances, we shall use the mathematical properties of Lie groups, which allow us to find transformation rules for representations for small Lorentz transformations, and simply exponentiate the results at the end in order to write down the general transformations. Starting from the transformation of a spinor field, we have: ψ → Λ1 ψ 2

; ψ † → ψ † Λ†1 2 ≃ ψ† 1 − † ≃ ψ 1+ 83

† i µν ωµν S 2 i † µν ωµν S 2

(6.59)

We also have, S

† µν

† i µ ν i 0 µ 0 0 ν 0 i γ γ γ ,γ γ γ [γ , γ ] = − γ ν † , γ µ† = 4 4 4 ; S µν † = γ 0 S µν γ 0 . (6.60)

Thus, we have for the transformation of the conjugate spinor field, i µν † † 0 γ 0, ψ → ψ γ 1 + ωµν S 2

(6.61)

and for finite Lorentz transformations, 0 ψ † → ψ † γ 0 Λ−1 1 γ .

(6.62)

We observe that ψ † does not really transform with the inverse transformation of ψ. However, ψ¯ does. Multiplying from the left with γ 0 both sides of the above transformation rule, we obtain 2 ψ † γ 0 → ψ † γ 0 Λ−1 γ0 1 2

¯ −1 ; ψ¯ → ψΛ 1 .

(6.63)

It is now obvious that the mass term in the Dirac Lagrangian is Lorentz invariant, ¯ ¯ → mψΛ ¯ −1 (6.64) mψψ 1 Λ 1 ψ = mψψ. 2

The derivative term of the Dirac equation is also Lorentz invariant, ν µ ¯ µ ∂µ ψ → ψΛ ¯ −1 ψγ Λ−1 µ Λ 1 ∂ν ψ 1 γ 2 2 ν µ = ψ¯ Λ−1 Λ−1 µ ∂ν ψ 1 γ Λ1 2 2 ρ µ −1 ν ¯ = ψγ Λρ Λ µ ∂ν ψ ¯ ρ δ ν ∂ν ψ = ψγ ρ ¯ = ψ6∂ ψ. (6.65) We see that the Dirac Lagrangian contains only the two simplest scalar terms that may be formed out of fields transforming in the spinor representation of the Lorentz group. The relative sign between these two terms is easy to determine by requiring that the equations of motion will also satisfy the Klein-Gordon equation, which gives the relativistically correct energy momentum relations, p2 = m2 . 84

6.7

General representations of the Lorentz group

The scalar, vector, and spinor representations are the simplest representations of the Lorentz group. However, there is an infinite number of n × n representations. All of them can be found easily by observing that the Lie algebra of the Lorentz group is the same as for an SU(2) × SU(2) group. Let us consider the generators in an arbitrary representation, which satisfy the commutation relationship, [J µ1 µ2 , J µ3 µ4 ] = i (g µ2 µ3 J µ1 µ4 + g µ1 µ4 J µ2 µ3 − g µ2 µ4 J µ1 µ3 − g µ1 µ3 J µ2 µ4 ) . (6.66) We now define the linear combinations, Ki ≡ J 0i ,

(6.67)

and

1 Li ≡ ǫijk J jk . (6.68) 2 We can change the basis of generators comprising from the six 2 generators J µν with the linear combinations {Li , K i }. The generators Li are generators of rotations around the axis−i, and the generators K i are generators of boosts in the direction−i (exercise: demonstrate this for the generators in the vectorial representation). Rotations are a subgroup of the Lorentz group, which satisfy an SU(2) algebra. We find the commutation relations, [Li , Lj ] = iǫijk Lk .

(6.69)

However, boosts do not form a subgroup of the Lorentz group. Two boosts in two different directions are not a boost in a third direction, but rather a rotation. We find that the commutators of boost generators are, [Ki , Kj ] = −iǫijk Lk .

(6.70)

The commutator of a boost and a rotation generator is, [Li , Kj ] = iǫijk Kk . 2

they are antisymmetric

(6.71)

Define now the linear combinations, Ji± ≡

Li ± iKi . 2

Then the Lie algebra breaks into two independent SU(2) algebras, + + Ji , Jj = iǫijk Jk+ , − − Ji , Jj = iǫijk Jk− ,

(6.72)

(6.73) (6.74)

and

+ − Ji , Jj = 0.

(6.75)

We have already learned about the representations of the SU(2) group in Quantum Mechanics. These can be labeled according to a “spin” number, 3 1 j = 0, , 1, , . . . 2 2 Each representation acts on (2j + 1) objects ψm with m = −j, −j + 1, . . . j − 1, j,

(6.76)

transforming them into each other under SU(2). A general representation of the Lorentz group acts on (2j + + 1)(2j − + 1) objects ψm+ ,m− , and is labeled by the numbers (j + , j − ), e.g. 1 1 1 1 (0, 0), ( , 0), (0, ), (0, 1), ( , ), . . . . 2 2 2 2

(6.77)

The massive Dirac field transforms in the mixed representation, 1 1 ( , 0) ⊕ (0, ). 2 2

6.8

Weyl spinors

As we have already noted, γ−matrices are not unique, and there are many equivalent representations. We introduce here a representation which shows better the decomposition of the Lorentz group into two SU(2) subgroups. This is called the chiral or Weyl representation, defined as γ 0 = I2×2 ⊗ σ1 ,

γ i = σi ⊗ (iσ2 ). 86

(6.78)

As an exercise, you can prove that indeed this set of gamma-matrices satisfies the Clifford algebra, {γ µ , γ ν } = 2g µν I4×4 .

(6.79)

In the chiral representation, the generators of the spinorial Lorentz group representation i S µν = [γ µ , γ ν ] , (6.80) 4 take a block-diagonal form, i i σj 02×2 0j S = − σj ⊗ σ3 = − , (6.81) 2 2 02×2 −σj and S

1 1 = ǫjkiσl ⊗ I2×2 = ǫjki 2 2

A Lorentz transformation for a spinor is i

σl 02×2 02×2 σl

µν

Λ 1 = e− 2 ωµν S , 2

(6.82)

(6.83)

and the exponent takes the explicit form, i i i i −i ωµν S µν = − ω0j S 0j − ωj0S j0 − ωjk S jk 2 2 2 2 ~ σ ~ σ = −β~ · ⊗ σ3 − i~θ · ⊗ I2×2 , 2 2

(6.84)

with

1 ω0j and θl = ǫjkl ω jk . 2 2 Under a small Lorentz transformation a spinor ψ transforms as, βj =

(6.85)

µν

ψ → ψ ′ = e− 2 ωµν S ψ i µν ≃ 1 − i ωµν S ψ 2  h i ψ  1 i ~ ~  1 − 2 θ − iβ · ~σ  ψ 2    =   h  i   ψ3 1 − 2i ~θ + iβ~ · ~σ ψ4 87

(6.86)

The two-dimensional objects, ψ3 ψ1 , and ψR ≡ ψL ≡ ψ4 ψ2

(6.87)

are called “Weyl spinors” and transform independently under a Lorentz transformation, i ~ ′ ~ ψL → ψL = 1 − θ − iβ · ~σ ψL , (6.88) 2 and

ψR → In an exponential form,

ψR′

i ~ = 1− θ + iβ~ · ~σ ψR . 2

ψL → ψL′ = ΛL ψL and ψL → ψR′ = ΛR ψR , with and

ΛL = e− 2 (θ−iβ )·~σ , i

ΛR = e− 2 (θ+iβ )·~σ . i

(6.89)

(6.90) (6.91) (6.92)

The two-transformations are connected via complex conjugation. Exercise: First we prove the identity, σ2~σ ∗ = −~σ σ2 .

(6.93)

ψLc ≡ σ2 ψL∗ ,

(6.94)

The spinor transforms as,

i ~ θ − iβ~ · ~σ )∗ ψL∗ 2 i ~ = σ2 (1 + θ + iβ~ · ~σ ∗ )ψL∗ 2 i ~ θ + iβ~ · ~σ )σ2 ψL∗ = (1 − 2

σ2 ψL∗ → σ2 (1 −

; ψLc → ΛR ψLc

(6.95)

Similarly, ψRc → ΛL ψRc . 88

(6.96)

6.8.1

The Dirac equation with Weyl spinors

We now write the Dirac equation in terms of Weyl spinors ψL and ψR , 0 = (i6∂ − m) ψ = iγ 0 ∂0 + iγ j ∂j − m14×4 ψ ~ − m12×2 ⊗ 12×2 ψ = i(I2×2 ⊗ σ1 )∂0 − i (~σ ⊗ (iσ2 )) ∇ 

−m

;  ~ i ∂0 − ~σ · ∇

~ i ∂0 + ~σ · ∇ −m

 

ψL ψR

= 0.

(6.97)

The mass m mixes the two Weyl spinors ψL , ψR . However, for a massless particle the Dirac equation is diagonalized into two independent equations for the Weyl spinors, ~ R, ∂0 ψR = −~σ · ∇ψ (6.98) and

~ L. ∂0 ψL = ~σ · ∇ψ

These are known as Weyl equations.

(6.99)

Chapter 7 Classical solutions of the Dirac equation In this chapter, we shall solve classically the equations of motion for the Dirac field, i.e. the classical Dirac equation. As we have already demonstrated, a field ψ which satisfies the Dirac equation it also satisfies the Klein-Gordon equation. So, it must admit a plane-wave solution. We consider first the general case ψ(x) = u(p)e−ip·x , (7.1) with p2 = m2 .

(7.2)

For now, we shall restrict ourselves to the cases with positive energy solutions, p0 > 0. Substituting into the Dirac equation, we obtain additional restrictions for the spinor u(p), (i6∂ − m) u(p)e−ip·x = 0 ; (6p − m) u(p) = 0.

(7.3)

Using the Weyl representation for γ−matrices, γ 0 = I2×2 ⊗σ1 , γ J = σj ⊗(iσ2 ), we obtain −mI2×2 p0 I2×2 − ~σ · p~ ξ 0 = , (7.4) 0 p I2×2 + ~σ · ~p −mI2×2 η 0

where we write the four-component spinor u(p) in terms of two Weyl-spinors ξ, η with two components each: ξ u(p) ≡ . (7.5) η It is of course possible to find solutions of the above equation directly. However, we may also want to use some cleverness, and exploit the fact that the field ψ transforms as a spinor. Consequently, ψ ′ (x′ ) = Λ 1 ψ(x),

(7.6)

and equivalently ′

u′ (p′ )e−ip ·x ′

′

= Λ 1 u(p)e−ip·x 2

−i(Λp)·(Λx)

= Λ 1 u(p)e−ip·x

; u (Λp)e

; u′ (Λp)e−ip·x = Λ 1 u(p)e−ip·x 2

u′ (Λp) = Λ 1 u(p).

(7.7)

and u(p) also transforms as a spinor, under a Lorentz transformation, pµ → pµ′ = Λµν pν .

(7.8)

The strategy for finding a general solution u(p) is then simple. We first solve the Dirac equation for an especially simple value of the momentum pµ , and then we perform a Lorentz transformation to obtain the result for a general value of the momentum.

7.1

Solution in the rest frame

The Dirac equation (Eq. 7.4) assumes its simplest form in the rest frame, pµ = m(1, ~0).

(7.9)

yielding,

−mI2×2 mI2×2 mI2×2 −mI2×2

ξ η

0 0

(7.10)

and, equivalently, ξ = η.

(7.11)

The general solution for the spinor u in the rest frame is, √ ξ µ u p = (m, ~0) = m . (7.12) ξ √ The factor m is a convenient normalization for future purposes, and the Weyl spinor ξ is arbitrary. We can choose it to be a linear combination, X ξ= cs ξ s , (7.13) s=+,−

with +

ξ =

1 0

−

, and ξ =

Notice that this basis diagonalizes σ3 , with

0 1

σ3 ξ ± = ±ξ ± .

(7.14)

(7.15)

From this solution of the Dirac equation in the rest frame, we can go to any other momentum value with a boost.

7.2

Lorentz boost of rest frame Dirac spinor along the z-axis

First, we boost any vector along the z-axis, with a very small Lorentz transformation, pµ → pµ′ = Λµν pν ≃ pµ + ωνµ pν , (7.16) where, ωµν = 0, except ω03 = −ω30 = Y.

Notice that,

ω30 = g 00 ω03 = Y,

ω03 = g 33 ω30 = −(−Y ) = Y.

(7.17)

(7.18)

The transformed momenta are, ′

′

p0 = p0 + Y p3 , p1 = p1 , p2 = p2 , p3 = p3 + Y p0 , 92

(7.19)

which is conveniently written as, 0′ 0 p p . = (I2×2 + Y σ1 ) 3′ p3 p For a finite boost of a vector along the z-direction, we have 0 0′ p p Y σ1 . =e 3′ p3 p

(7.20)

(7.21)

We can compute the exponential easily, noting that σ12 = I2×2 , eY σ1 = =

∞ X Yn

n=0 ∞ X k=0

σ1n ∞

Y 2k 2k X Y 2k+1 2k+1 σ + σ1 (2k)! 1 (2k + 1)! k=0

∞ ∞ X X Y 2k Y 2k+1 = I2×2 + σ1 (2k)! (2k + 1)! k=0 k=0

; eY σ1 = I2×2 cosh Y + σ1 sinh Y. We apply this transformation for the momentum vector at rest, m E m = (I2×2 cosh Y + σ1 sinh Y ) → 0 p3 0 m cosh Y E . = ; 3 m sinh Y p

(7.22)

(7.23)

We can then invert the above equations and compute the parameter Y which generates the Lorentz boost, known as rapidity, in terms of the components of the boosted momentum. We find (exercise:) eY =

E + p3 , m

e−Y =

E − p3 , m

Y =

1 E + p3 ln . 2 E − p3

(7.24)

Let us now find the transformation of a spinor, under the same Lorentz

boost. The transformation matrix is, i

Λ 1 = e− 2 ωµν S

µν

∞ X − Y2 = n! n=0 ∞ X − Y2 = n! n=0

n n

03 − i ω S 30 2 30

= e− 2 σ3 ⊗σ3

(σ3 ⊗ σ3 )n σ3n ⊗ σ3n

n − Y2 = ⊗ + σ3n ⊗ σ3n n! n=even n=odd n ∞ ∞ X −Y n X −Y 2 2 σ3 ⊗ σ3 + I2×2 ⊗ I2×2 = n! n! n=even n=odd Y Y I2×2 ⊗ I2×2 − sinh σ3 ⊗ σ3 . = cosh 2 2 ∞ X

; Λ1

= e− 2 ω03 S

− Y2 n!

σ3n

∞ X

(7.25)

Applying this spin boost transformation to our solution of the Dirac equation at the rest frame, we obtain s √ ξ u(p) = Λ 1 m 2 ξs i   hp p (1−σ3 ) (1+σ3 ) 3 3 ξs E+p 2 + E −p 2 i  (7.26) ; us (p) =  hp p 3) 3 (1−σ3 ) ξ s + E + p3 (1+σ E − p 2 2

where we have used the expression for the rapidity, derived in Eq. 7.24. We observe that, 2 1 ± σ3 1 ± σ3 , (7.27) = 2 2 or, equivalently, r 1 ± σ3 1 ± σ3 = . (7.28) 2 2 We can then collect all terms in Eq. 7.26 under a common square root,   q 3 ) (1−σ3 ) + (E − p3 ) (1+σ3 ) ξ s (E + p 2 2  us (p) =  q (1+σ ) (1−σ 3) s 3 3 3 (E + p ) 2 + (E − p ) 2 ξ (7.29)

which simplifies to p 3σ ξ s E − p 3 u (p) = p E + p3 σ3 ξ s s

(7.30)

This is the general solution for the spinor u(p), when pµ = (E, 0, 0, p3).

7.3

Solution for an arbitrary vector

We can obtain the spinor solution u(p) of the Dirac equation for an arbitrary vector pµ , by performing a rotation of the solution that we found for pµ = (E, 0, 0, p3). In fact, things are simpler now and there is not much work to do, if we can write down the previous result in a covariant form. We define, σ µ ≡ (1, ~σ ) , (7.31) and σ ¯ µ ≡ (1, −~σ ) .

Then the expression for the spinor us (p), Eq. 7.30, is written as √ σ · p ξs s u (p) = √ σ ¯ · p ξs

(7.32)

(7.33)

You can verify (exercise) that this is actually a solution of the Dirac equation for an arbitrary vector pµ . For this, you will need the generally useful identity, (σ · p)(¯ σ · p) = p2 = m2 , and to rewrite the Dirac differential operator as −m p · σ 6p − m = p·σ ¯ −m

(7.34)

(7.35)

in Eq. 7.4.

7.4

A general solution

We have found a general solution for the Dirac equation, of the form ψ(x) = us (p)e−ip·x , with p0 > 0. 95

(7.36)

Another possibility is, ψ(x) = v(p)e+ip·x , with p0 > 0.

(7.37)

Substituting into the Dirac equation we find that the spinor v s (p) satisfies (6p + m) v(p) = 0.

(7.38)

Repeating the same steps as for u(p), we find this equation is satisfied by spinors √ σ · p ξs s √ v (p) = . (7.39) ¯ · p ξs − σ A general solution for the classical Dirac equation is, XZ d3 ~p as (p)us (p)e−ip·x + b∗s (p)v s (p)e+ip·x , ψ(x) = 2 (2π) 2ωp s

where the as (p) and bs (p) are arbitrary coefficients, and p ωp = p0 = p~2 + m2 .

(7.40)

(7.41)

We have not discussed explicitly solutions with p0 < 0. At the classical level, these are included in the above integral, as you can easily verify by repeating here the analysis of Section 4.1.1. Dirac interpreted these negative energy solutions with an ingenious theory (Dirac sea), which lead to the discovery of antimatter. We do not need to bother with this interpretation, except out of historic interest. As with the Klein-Gordon equation, field quantization gives consistently rise to physical states which have a positive energy.

Chapter 8 Quantization of the Dirac Field To quantize the Dirac field, we promote the field into an operator, Z d3 ~p X −ip·x † ip·x a (p)u (p)e + b (p)v (p)e . ψ(~x, t) = s s s s (2π)3 2ωp s

(8.1)

The conjugate momentum of the field is, π(~x, t) =

∂L = iψ † (~x, t), ˙ ∂ψ

(8.2)

where the Lagrangian of the Dirac field is, L = ψ¯ (i6∂ − m) ψ. The conjugate field is given by, Z d3 p~ X † † +ip·x † ip·x † a (p)u (p)e + b (p)v (p)e . ψ (~x, t) = s s s s (2π)3 2ωp s

(8.3)

(8.4)

We also find, d3 p~ X † as (p)¯ us (p)e+ip·x + bs (p)¯ vs† (p)eip·x . 3 (2π) 2ωp s (8.5) ¯ In terms of ψ and ψ, the operators aa , bs are written as, Z as (p) = d3~xu¯s (p)eip·x γ 0 ψ(x), (8.6) ¯ x, t) ≡ ψ † (~x, t)γ 0 = ψ(~

and bs (p) =

0 −ip·x ¯ d3~xψ(x)γ e vs (p).

To prove the above you will need the spinor summations, X X us (p)¯ us (p) = 6p + m, and vs (p)¯ vs (p) = 6p − m, s

(8.7)

(8.8)

which can be derived easily from the explicit expressions for us (p), vs (p) of the last chapter. Using Noether’s theorem, we can compute the Hamiltonian of the Dirac field, which is Z X d3~k H= ω a†s (k)as (k) − bs (k)b†s (k) . (8.9) k 3 (2π) (2ωk ) s

In the above, we have been careful to preserve the order of operators as they appear. We can now proceed with the quantization procedure. It is left as an exercise to show that imposing commutation relations, leads to particle states b†s (p) |0i which have a negative energy. This is due to the minus sign in the second term of the Hamiltonian integral in Eq. 8.9. To obtain positive energy eigenvalues for such states, we are forced to quantize by requiring anti-commutation relations, {ψ(~x1 , t), π(~x2 , t)} = iδ (3) (~x1 − ~x2 ),

(8.10)

{ψ(~x1 , t), ψ(~x2 , t)} = {π(~x1 , t), π(~x2 , t)} = 0.

(8.11)

{as (~p1 ), as′ (~p2 )} = {bs (~p1 ), bs′ (~p2 )} = 0, n o n o † † † † as (~p1 ), as′ (~p2 ) = bs (~p1 ), bs′ (~p2 ) = 0.

(8.13)

and For the operators as , bs this results to the anticommutation relations, o o n n as (~p1 ), a†s′ (~p2 ) = bs (~p1 ), b†s′ (~p2 ) = δs′ s (2π)3 (2ωp1 )δ (3) (~p2 − ~p1 ), (8.12)

(8.14)

Using the anticommutation relation of Eq. 8.12, we cast the Hamiltonian of Eq. 8.9 as, Z X d3~k † † H= ω a (k)a (k) + b (k)b (k) + h0| H |0i , (8.15) k s s s s (2π)3 (2ωk ) s 98

with (3)

h0| H |0i = −δ (0)

d3~kωk .

(8.16)

Exercise: Notice that the sign of the vacuum energy is now negative. 1. What would be the Casimir force in a gedanken experiment where we could impose that the Dirac field vanishes on twoparallel plates? 2. In supersymmetric theories, for each fermion field exists a symmetric boson field, with the same mass. What is the Casimir force? As in the Klein-Gordon field, we can redefine the field Hamiltonian to have zero vacuum expectation value, : H : = H − h0| H |0i .

(8.17)

The subtraction of the vacuum energy is implemented at the operator level with normal ordering, which in the case of anticommuting fields requires an additional (-) sign, : bs (k)b†s (k) : = −b†s (k)bs (k). (8.18) Again with the use of N¨other’s theorem, we find that the momentum is conserved and takes the form, P~ =

8.1

X d3~k † † ~k a (k)a (k) + b (k)b (k) . s s s s (2π)3 (2ωk ) s

(8.19)

One-particle states

It is left as an exercise to repeat the steps that we followed in the case of the Schr¨odinger and the Klein-Gordon fields, and to to prove that • there exists a vacuum state with as (p) |0i = bs (p) |0i = 0, • the operators as (p), bs (p), a†s (p), b†s (p) ladder operators. 99

(8.20)

We can build two types of one-particle states, using either the a†s creation operator |p, sia ≡ a†s (p) |0i , (8.21)

or the b†s creation operator

|p, sib ≡ b†s (p) |0i .

(8.22)

These states are degenerate eigenstates of the Hamiltonian and momentum operators. We define the field momentum four-vector, P µ ≡ (: H : , P~ ).

(8.23)

P µ |p, sia = pµ |p, sia ,

(8.24)

P µ |p, sib = pµ |p, sib ,

(8.25)

Then and with µ

p =

p~2

m2 , p~

(8.26)

These are states of positive energy, and the correct relativistic energy-momentum relation, p2 = m2 . (8.27)

8.1.1

Particles and anti-particles

The Dirac Lagrangian is symmetric under the U(1) transformation, ′

(8.28)

δψ = ψ ′ − ψ = iθψ,

(8.29)

ψ(x) → ψ ′ (x) = eiθ ψ(x),

ψ † (x) → ψ † (x) = e−iθ ψ † (x).

For small values of θ, we find

and or, equivalently,

′

δψ † = ψ † − ψ † = −iθψ † ,

(8.30)

¯ δ ψ¯ = ψ¯′ − ψ¯ = −iθψ.

(8.31)

100

Using Noether’s theorem, we find that the current, ∂L ∂L + δψ, ¯ ∂(∂µ ψ) ∂(∂µ ψ) ¯ µ ψ, = ψγ

J µ = δ ψ¯ Jµ

is conserved, ∂µ J µ = 0. The corresponding conserved charge is, Z Z Z 3 0 3 ¯ 0 Q = d ~xJ = d ~xψγ ψ = d3~xψ † ψ.

(8.32)

(8.33)

We can express Q in terms of creation and annihilation operators, Q=

d3~k X † † a (k)a (k) + b (k)b (k) . s s s s (2π)3 2ωk s

(8.34)

We calibrate the charge of the vacuum state to be zero with normal ordering, : Q : = Q − h0| Q |0i Z d3~k X † : as (k)as (k) : + : bs (k)b†s (k) : = 3 (2π) 2ωk s :Q: =

d3~k X † † a (k)a (k) − b (k)b (k) . s s s s (2π)3 2ωk s

(8.35)

The states and |p, sia and |p, sib are non-degenerate eigenstates of the charge operator. We find, : Q : |p, sia = (+1) |p, sia ,

(8.36)

: Q : |p, sib = (−1) |p, sib .

(8.37)

and Therefore the operators a†s (p) give rise to particles with momentum pµ and charge (+1), while the operators b†s (p) give rise to anti-particles with the same momentum pµ and opposite charge (−1). To emphasize this, we shall display explicitly the charge of one-particle states, |p, s, +i ≡ |p, sia = a†s (p) |0i , and |p, s, −i ≡ |p, sib = b†s (p) |0i . 101

(8.38)

8.1.2

Particles and anti-particles of spin- 12

We have demonstrated in previous chapters that the Dirac Lagrangian is invariant under Lorentz transformations. The spinor field transforms “at a point” as, ψ(x) → ψ ′ (x) = Λ 1 ψ(Λ−1 x). (8.39) 2

The variation of the field “at a point” is

δ∗ ψ ≡ ψ ′ (x) − ψ(x) = Λ 1 ψ(Λ−1 x) − ψ(x) 2 i µν = 1 − ωµν S ψ(xµ − ωνµ xν ) − ψ(xµ ) + O(ω 2 ) 2 i = ψ(xµ − ωνµ xν ) − ψ(xµ ) − ωµν S µν ψ(xµ − ωνµ xν ) + O(ω 2). 2 (8.40) We can expand, ψ(xµ − ωνµ xν ) = ψ(xµ ) − ωµν xν ∂µ ψ = ψ µ (x) − i

ωµν µν J ψ(xν ), 2 scalar

(8.41)

where we remind that, µν Jscalar = i (xµ ∂ ν − xν ∂ µ ) .

(8.42)

Eq. 8.40 becomes,

i µν δ∗ ψ = − ωµν (Jscalar + S µν ) ψ. (8.43) 2 Noether’s theorem, Eq. 2.79, leads to the result that the quantity Z ∂L µν (J µν + S µν ) ψ, (8.44) J = d3~x ∂(∂0 ψ) scalar

is conserved. Explicitly, J

µν

¯ 0 (J µν + S µν ) ψ. d3~x ψγ scalar

(8.45)

µν d3~x ψ † (Jscalar + S µν ) ψ.

(8.46)

and, equivalently, J

µν

102

This is very close to what we find for the total angular momentum of a scalar field. But not exactly. The S µν is novel, and it arises here for the first time. This term is responsible to giving spin, an intrinsic angular momentum, to Dirac particle states. How can we figure out if a particle has spin? We can do the following test. We put the particle at rest, and measure its total angular momentum. If we find it zero, then the particle has an angular momentum only when it moves and therefore has no intrinsic spin. If it is not zero besides having a zero momentum, then it is all due to the intrinsic spin of the particle. First, we would like to use normal ordering, as with every conserved quantity, to make sure that the vacuum has a zero total angular momentum: : J µν : |0i = 0.

(8.47)

Let us consider a state of a positively charged Dirac particle which has a zero momentum. |0, s, +i = a†s (0) |0i . (8.48)

Does it have any angular momentum? The action of the total-angular momentum on the state is, : J µν : |0, s, +i = : J µν :, a†s (0) |0i+a†s (0) : J µν : |0i = : J µν :, a†s (0) |0i . (8.49) We now specialize on a rotation around the z−axis, ω12 = −ω21 = θ, otherwise ωµν = 0.

(8.50)

After some algebra, we find that J 3 |0, s, +i ≡

(8.51)

(8.52)

which diagonalizes σ3 , σ3 ξ (1) = +ξ (1) ,

σ3 ξ (2) = −ξ (2) .

103

(8.53)

Then we have, 1 1 J |0, s, +i = δs,(1) − δs,(2) |0, s, +i . 2 2 3

(8.54)

We now have a physical interpretation of the quantum number s. A particle at rest with charge + and s = (1) has total angular momentum + 12 in the z−direction, 1 1 3 |0, s, +i . (8.55) J |0, s, +i = δs,(1) − δs,(2) 2 2 A particle at rest with charge + and s = (2) has total angular momentum − 12P in the z−direction. We have then discovered that a general superposition of s cs |p, s, +i describes a particle with total spin- 12 . P Repeating the same steps, we also conclude that a general anti-particle state s cs |p, s, −i is also a state with spin- 21 . In summary, the quantization of the Dirac field yields quantum states with spin− 21 . Spin emerged simply as a component of the total angular momentum, due to the Dirac fields transforming in the spinorial representation of the Lorentz group. This theory predicts also the existence of both matter and anti-matter. Of course, the theory is highly successful. The electron and positron are states of the quantized Dirac field. The same is valid for other known spin- 12 particles, the muon and tau leptons, the quarks and their antiparticles.

8.2

Fermions

Particles of the Dirac field are fermions. A state with two positively charged particles is |p1 , s1 , +; p2 , s2 , +i ≡ a†s1 (p1 )a†s2 (p2 ) |0i 1 † as1 (p1 )a†s2 (p2 ) − a†s1 (p1 )a†s2 (p2 ) |0i using Eq. 8.14 ; |p1 , s1 , +; p2 , s2 , +i = 2 (8.56) The state where all quantum numbers of the two particles are identical p1 = p2 and s1 = s2 is forbidden, |p, s, +; p, s, +i =

1 † as (p)a†s (p) − a†s (p)a†s (p) |0i = 0. 2 104

(8.57)

which is Pauli’s exclusion principle for fermions. Using the anti-commutation of creation operators we can write a state of N identical particles or antiparticles as, |p1 , s1 , ±; p2 , s2 , ±; . . . ; pN , sN , ±;i =

1 X sgn(σ)a†sσ(1) a†sσ(2) . . . a†sσ(N) |0i , n! σinS n

(8.58) which is manifestly anti-symmetric. The quantization of the Dirac field explains another mystery of quantum mechanics., i.e. that elementary particles with spin− 12 are fermions.

8.3

Lorentz transformation of the quantized spinor field

A spinor field transforms classically, as ψ(x) → ψ ′ (x′ ) = Λ 1 ψ(x). 2

(8.59)

What is the transformation of the corresponding quantum field? It does not have to be the same, and in fact it is not, as for the classical field. Relativistic invariance at the classical level is cast as the requirement that the Lagrangian is a scalar, L′ (x′ ) = L(x). This is sufficient to guarantee that observers in different reference frames find the same minimum value for the action. However, this is not the same criterion that we must introduce for Lorentz invariance of the quantum system.

8.3.1

Transformation of the ladder operators

We consider a quantum one-particle state, e.g. |p, s, +i . Under a Lorentz transformation Λ, a state transforms |p, s, +i → U(Λ) |p, s, +i . 105

(8.60)

What is the new state, i.e. what is the transformation representation U(Λ)? It is logical to require that the action of U(Λ) yields a particle state with a momentum Λp ≡ Λµν pν , |p, s, +i → U(Λ) |p, s, +i = N(Λ, p) |Λp, s, +i .

(8.61)

Here we encounter an ambiguity. Consider a four vector pµ = (m, 0). An SO(3) rotation leaves the vector pµ invariant. In general, for any vector k µ there are Lorentz transformations W , forming what is called the “little group”, which leave it invariant, Wνµ pν = pµ .

(8.62)

Therefore the representation U(Λ) is not unique, since, for example, a different one U(Λ)U(W ) could also be used for transforming a state with a momentum p to a state with a momentum Λp. It is beyond our scope to discuss how this ambiguity is removed. We will assume here for simplicity that there exists a method to choose uniquely one of the “little group” transformations. The vacuum state has zero momentum and all Lorentz transformations belong to the little group. We then have that U(Λ) |0i = |0i .

(8.63)

Requiring that the normalization of a state is Lorentz invariant, we obtain ||p, s, +i|2 = |U(Λ) |p, s, +i|2 ; U † = U −1 .

(8.64)

We express Eq. 8.61 in terms of creation operators, U(Λ) |p, s, +i = N(Λ, p) |Λp, s, +i ; U(Λ)a†s (p) |0i = N(Λ, p)a†s (Λp)U(Λ) |0i ; U(Λ)a†s (p)U −1 (Λ) = N(Λ, p)a†s (Λp),

(8.65)

where N(Λ, p) is a normalization factor that we need to determine. Conjugating Eq. 8.65 we find that, U(Λ)as (p)U −1 (Λ) = N(Λ, p)∗ as (Λp).

106

(8.66)

; hp1 , s1 , +| p2 , s2 , +i = (2π)3 2ωp1 δ (3) (p2 − p1 )δs1 s2 .

(8.67)

Two different particles in a reference frame should remain distinct in any other reference frame. If we perform a Lorentz transformation, Eq. 8.61, we have hp1 , s1 , +| U −1 U |p2 , s2 , +i = h0| Uas1 (p1 )U −1 Ua†s2 (p2 )U −1 |0i = h0| Uas1 (p1 )U −1 Ua†s2 (p2 )U |0i

= |N(Λ, p1 )|2 (2π)3 2ωΛp1 δ (3) (Λp2 − Λp1 )δs1 s2 (8.68)

We can now prove that 2ωΛp1 δ (3) (Λp1 − Λp2 ) = 2ωp1 δ (3) (p1 − p2 ). Proof: Consider a scalar function f (~p). Then, Z 3 d p~ ′ 2ωp δ (3) (~p − p~′ ). f (~p ) = 2ωp The measure

(8.69)

(8.70)

Z d3 p~ = d4 pδ(p2 − m2 )Θ p0 > 0 2ωp is invariant under Lorentz transformations. Given that f is a scalar, we must have that ωp δ (3) (~p − ~p′ ) Z

is also a scalar, which proves Eq. 8.69. The orthogonality of the states is therefore preserved under Lorentz transformations. Eq. 8.68 becomes hp1 , s1 , +| U −1 U |p2 , s2 , +i = |N(Λ, p1 )|2 (2π)3 2ωp1 δ (3) (p2 − p1 )δs1 s2 , (8.71) which for N(Λ, p) = 1, guarantees that hp1 , s1 , +| U −1 U |p2 , s2 , +i = hp1 , s1 , +| p2 , s2 , +i . 107

(8.72)

8.3.2

Transformation of the quantized Dirac field

Quantum field theories provide us with states and operators which are all made out of creation and annihilation operators. The latter transform under Lorentz transformations in a representation U(Λ) of the Lorentz group, such that (explicitly written for the Dirac field ladder operators) U(Λ)a†s (p)U −1 (Λ) U(Λ)as (p)U −1 (Λ) U(Λ)b†s (p)U −1 (Λ) U(Λ)bs (p)U −1 (Λ)

= = = =

a†s (Λp), as (Λp), b†s (Λp), bs (Λp).

(8.73) (8.74) (8.75) (8.76)

Measurable observables are obtained from matrix-elements of the type, ˆ |STATE2 i . hSTATE1 | O

(8.77)

We measure first one such matrix-element, hp, s, +| ψ(x) |p, s, +i ,

(8.78)

in a certain reference system. Then we Lorentz transform the state, |p, s, +i → |p, s, +i′ = U(Λ) |p, s, +i ,

(8.79)

and the quantum field, ψ(x) → ψ ′ (x′ ).

(8.80)

We should not hurry into claiming that ψ ′ (x′ ) = Λ 1 ψ(x). This is the trans2 formation of the classical field. It does not have to transform in the same representation as the quantum field, which is now an operator. What is then the quantum ψ ′ (x′ )? It should be such that the two measurements in the two frames are the same, hp, s, +| U −1 (Λ)ψ ′ (x′ )U(Λ) |p, s, +i = hp, s, +| ψ(x) |p, s, +i ; U −1 (Λ)ψ ′ (x′ )U(Λ) = ψ(x) ; ψ ′ (x′ ) = U(Λ)ψ(x)U −1 (Λ).

108

(8.81)

We can compute the right hand side, knowing the transformation of the creation and annihilation operators from Eq. 8.73, Z −1 d3 k X −1 −ik·x † +ik·x Uψ(x)U = U a (k)u(k)e + b (k)v(k)e U s s (2π)3 2ωk s Z d3 k X −ik·x † +ik·x a (k)u(Λk)e + b (k)v(Λk)e . = s s (2π)3 2ωk s

(8.82)

Now we use that, d3 k d3 Λk = , (2π)3 2ωk (2π)3 2ωΛk

k · x = Λk · Λx,

(8.83)

and u(k) = Λ−1 1 u(Λk),

and v(k) = Λ−1 1 v(Λk),

(8.84)

which yields, setting p = Λk, Z d3 p X −1 −1 −ip·Λx † +ip·Λx Uψ(x)U = Λ1 a (p)u(Λp)e + b (p)v(Λp)e s s 2 (2π)3 2ωp s = Λ−1 1 ψ(Λx).

(8.85)

We have found the transformation of the quantum Dirac field. It reads, ψ(x) → ψ ′ (x′ ) = U(Λ)ψ(x)U −1 (Λ) = Λ−1 1 ψ(Λx).

(8.86)

8.4

Parity

We now derive the transformation properties of a spinor field under parity. This is a discrete transformation, (t, ~x) → (t, −~x).

(8.87)

Under parity, the momentum of a particle transforms as, ~p → −~p 109

(8.88)

The representation of the parity transformation on an one-particle state should be, for example, UP |~p, s, +i = λ |−~p, s, +i ,

(8.89)

where λ is such that 1 = h~p, s, +| UP† UP |~p, s, +i = |λ|2 h−~p, s, +| −~p, s, +i ; |λ|2 = 1.

(8.90)

This yield the following transformation for the creation and annihilation operators, UP as (~p)UP−1 = λa as (−~p), (8.91) and UP bs (~p)UP−1 = λb bs (−~p).

(8.92)

Under a parity transformation, a Dirac quantum field transforms as Z i h d3~k X −1 −ik·x † ~ +ik·x ~ ~ ~ UP ψ(~x, t)UP = UP−1 U a ( k)u( k)e + b ( k)v( k)e P s s (2π)3 2ωk s Z d3~k X ~~ −ik·x ∗ † +ik·x ~ ~ λa as (−k)u(k)e + λb bs (−k)v(k)e = (2π)3 2ωk s

(8.93)

Now we use that the measure is invariant under a parity transformation,

and we rewrite, Eq. 8.93 becomes Z

d3~k d3 (−~k) = , (2π)3 2ωk (2π)3 2ω−k

(8.94)

k · x = Et − ~k~x = Et − (−~k)(−~x).

(8.95)

UP ψ(~x, t)UP−1 = d3 ~p X −i(Et−~ p·(−x)) ∗ † +i(Et−~ p·(−x)) λ a (~ p )u(~ p )e + λ b (~ p )v(−~ p )e (, 8.96) a s b s (2π)3 2ωp s

where we set p~ = −~k. Recall now that, in the Weyl representation, √ E − ~ p ~ σ 0 I 2×2 0 us (p) = √ . and γ = I2×2 0 E + ~p~σ 110

(8.97)

We then have γ 0 us (~p) = us (−~p),

(8.98)

γ 0 vs (~p) = −vs (−~p).

(8.99)

λa = −λ∗b = 1.

(8.100)

and similarly, We can now fix the arbitrary normalization phases λa , λb to

Then, Eq. 8.96 yields a simple parity transformation for the field, UP ψ(~x, t)UP−1 = γ 0 ψ(−~x, t).

(8.101)

It is easy to show that we can take UP to be its own inverse. We have, UP ψ(−~x, t)UP−1 = γ 0 ψ(~x, t) ; ψ(−~x, t) = γ 0 UP−1 ψ(~x, t)UP ; γ 0 ψ(−~x, t) = (γ 0 )2 UP−1 ψ(~x, t)UP ; UP ψ(~x, t)UP−1 = UP−1 ψ(~x, t)UP ,

(8.102)

which is satisfied for UP = UP−1 .

(8.103)

We can study the transformation of composite operators made up from ¯ which transforms as a scalar under two Dirac fields. The combination ψψ Lorentz transformations, transforms under parity as a scalar too. ¯ ~x)ψ(t, ~x) → ψ¯′ (t, −~x)ψ ′ (t, −~x) = UP ψ(t, ¯ ~x)UP UP ψ(t, ~x)UP ψ(t, ¯ −~x)γ 0 γ 0 ψ(t, −~x, t) = ψ(t, ¯ −~x)ψ(t, −~x). = ψ(t, (8.104) ¯ 5 ψ which transforms as a scalar under However, there is a combination ψγ Lorentz transformations (exercise: prove that it is indeed so), but transforms differently under parity. ¯ ~x)γ 5 ψ(t, ~x) → ψ(t, = = =

¯ ~x)UP γ 5 UP ψ(t, ~x)UP ψ¯′ (t, −~x)γ 5 ψ ′ (t, −~x) = UP ψ(t, ¯ −~x)γ 0 γ 5 γ 0 ψ(t, −~x, t) ψ(t, ¯ −~x)γ 5 γ 0 γ 0 ψ(t, −~x, t) −ψ(t, ¯ −~x)γ 5 ψ(t, −~x). −ψ(t, (8.105) 111

This is called a “pseudoscalar”. ¯ µ ψ which transforms as a vector under Lorentz transThe combination ψγ formations (exercise), transforms also as a vector under parity. ¯ ~x)γ µ ψ(t, ~x) → ψ¯′ (t, −~x)γ µ ψ ′ (t, −~x) = UP ψ(t, ¯ ~x)UP γ µ UP ψ(t, ~x)UP ψ(t, ¯ −~x)γ 0 γ µ γ 0 ψ(t, −~x, t) = ψ(t, ¯ −~x) 2g 0µ γ 0 − γ µ (γ 0 )2 ψ(t, −~x) = ψ(t, ¯ −~x)γ µ ψ(t, −~x) = 2g 0µ − 1 ψ(t, (8.106)

For µ = 0,

¯ ~x)γ 0 ψ(t, ~x) → ψ(t, ¯ −~x)γ 0 ψ(t, −~x). ψ(t,

(8.107)

¯ ~x)γ i ψ(t, ~x) → ψ(t, ¯ −~x)γ i ψ(t, −~x). ψ(t,

(8.108)

For µ = i = 1, 2, 3,

This is exactly how a vector xµ = (t, ~x) → (t, −~x) transforms under parity. ¯ µ γ 5 ψ which transforms as a scalar under Lorentz The combination ψγ transformations (exercise: prove that it is indeed so), transforms oppositely under parity. ¯ ~x)γ m uγ 5 ψ(t, ~x) → ψ(t, = = = =

¯ ~x)UP γ µ γ 5 UP ψ(t, ~x)UP ψ¯′ (t, −~x)γ µ γ 5 ψ ′ (t, −~x) = UP ψ(t, ¯ −~x)γ 0 γ µ γ 5 γ 0 ψ(t, −~x, t) ψ(t, ¯ −~x)γ 0 γ µ γ 0 γ 5 ψ(t, −~x, t) −ψ(t, ¯ −~x) 2g 0µ γ 0 − γ µ (γ 0 )2 γ 5 ψ(t, −~x) −ψ(t, ¯ −~x)γ µ γ 5 ψ(t, −~x) − 2g 0µ − 1 ψ(t, (8.109)

This is called a “pseudo-vector”. The Dirac Lagrangian, ¯ iψ6¯∂ ψ − mψψ,

is composed from true scalar terms, and it transforms as a scalar under parity. Exercise: Prove that the classical Dirac field has the same parity transformation as the quantum Dirac field.

8.5

Other discrete symmetries

Exercise: Prove the following, 112

1. Under time-reversal (t, ~x) → (−t, ~x), UP as (~p)UP = a−s (−~p),

ξ −s = −iσ2 (ξ s )∗ .

(8.110)

and UP ψ(t, ~x)UP = ψ(−t, ~x).

(8.111)

2. Under charge conjugation, exchanging a particle with its antiparticle, UC as (~p)UC = bs (~p), and

UC bs (~p)UC = as (~p),

¯ ~x)γ 0 γ 2 )T . UC ψ(t, ~x)UC = −i(ψ(t,

113

(8.112) (8.113)

Chapter 9 Quantization of the free electromagnetic field Light is travelling with the maximum possible relativistic velocity. The quantization of the electromagnetic fields and the realization that light is packaged in photons, are one of the greatest advances ever made in the history of physics.

9.1

Maxwell Equations and Lagrangian formulation

We first review the equations of Maxwell, ~ ·B ~ = 0, ∇ ~ ~ ×E ~ + ∂ B = 0, ∇ ∂t ~ ·E ~ = ρ, ∇

~ ~ ×B ~ − ∂ E = ~j. ∇ ∂t We now introduce the four-vector, ~ , Aµ ≡ φ, A 114

(9.1) (9.2) (9.3) (9.4)

(9.5)

where the components are chosen such as, ~ =∇ ~ × A, ~ B

(9.6)

and

~ ~ = − ∂ A − ∇φ. ~ E (9.7) ∂t Then, the first two Maxwell equations (Eq. 9.1, Eq. 9.2) are automatically satisfied, ~ ·B ~ =∇ ~ · ∇ ~ ×A ~ = 0, ∇ (9.8) and

~ × ∇

~ ∂A ~ − ∇φ − ∂t

∂ ~ ~ ~ ~ = ǫijk ∂i ∂j φ = 0. + ∇ × A = ∇ × ∇φ ∂t

(9.9)

We now define the tensor, F µν = ∂ µ Aν − ∂ ν Aµ .

(9.10)

The components of F µν are, F

=∂ A −∂ A =

~ ∂A ~ + ∇φ ∂t

!(i)

= −E i ,

(9.11)

and F ij = ∂ i Aj − ∂ j Ai = −ǫijk B k .

(9.12)

All other components are zero, since F µν is anti-symmetric. In a matrix form we have,   0 −E 1 −E 2 −E 3  E1 0 −B 3 B 2   (9.13) F µν =   E2 B3 0 −B 1  E 3 −B 2 B 1 0 This is called the Electromagnetic field tensor. Define now the current four-vector, j ν ≡ ρ, ~j . 115

(9.14)

We can easily verify that the remaining Maxwell equations are given by the compact equation, ∂µ F µν = j ν . (9.15) For ν = 0, ∂µ F µ0 = φ 3 X 00 ; ∂0 F + ∂i F i0 = φ i=1

~ ·E ~ = φ. ; ∇

(9.16)

For ν = 1, ∂µ F µ1 = j 1 3 X 01 ∂i F i1 = j 1 ; ∂0 F + i=1

; −

∂E ∂B 2 ∂B 3 −+ = j1, + ∂t ∂x2 ∂x3

(9.17)

which is the first component of the vector equation −

~ ∂E ~ ×B ~ = ~j. +∇ ∂t

(9.18)

Maxwell equations, expressed in terms of the vector potential Aµ , take the form, ∂µ F µν = 0 ; ∂µ (∂ µ Aν − ∂ ν Aµ ) = 0 ; ∂ 2 Aν − ∂ ν (∂µ Aµ ) = 0.

(9.19)

Classical gauge Invariance and gauge-fixing Maxwell equations, Eq. 9.15, are invariant under gauge transformations, where the vector potential transforms as Aµ → Aµ′ = Aµ + ∂ µ χ, 116

(9.20)

or, in components, φ → φ′ = φ + χ˙ ~→A ~′ = A ~ − ∇χ. ~ A

(9.21) (9.22)

χ is an arbitrary function of space-time coordinates. Indeed, the field tensor F µν is invariant under this transformation, ν

F µν → F µν ′ = ∂ µ A′ − ∂ ν A′ = ∂ µ (Aν + ∂ ν χ) − ∂ ν (Aµ + ∂ µ χ) = F µν .

(9.23)

We can often derive simplify our calculations if we exploit gauge invariance. Suppose that we are given a field Aµ satisfying Maxwell equations Eq. 9.19. We can “fix the gauge”, i.e. perform a special gauge transformation, Aµ → Aµ ′ , such that the second term in Eq. 9.19 disappears, 0 = ∂µ Aµ′ = ∂µ Aµ + ∂ 2 χ ; ∂ 2 χ = −∂µ Aµ .

(9.24)

This particular choice of gauge is known as the “Lorentz gauge”. In the Lorentz gauge, Maxwell equations take the very simple form (dropping the prime) ∂ 2 Aµ = j µ . (9.25) In the vacuum, j ν = 0, we have ∂ 2 Aµ = 0.

(9.26)

This equation is particularly easy to solve as plane-waves, and we write the general real classical field solution as an integral Z d3~k X (λ) Aµ (x) = ǫ (k) aλ (k)e−ik·x + a∗λ (k)eik·x , (9.27) 3 (2π) 2ωk λ=0

µ ~ with k = ωk , k and ωk = ~k . We have introduced a basis of four-vectors which determine the direction of the electromagnetic field (polarization),         1 0 0 1         0  (1)  1  , ǫ(2) =  0  , ǫ(3) =  0  . (9.28) , ǫ = ǫ(0) =   0   1   0   0  1 0 0 0 117

The polarization vectors are normalized as, ′

′

ǫ(λ) · ǫ(λ ) = g λλ .

(9.29)

The “Lorentz gauge condition”, ∂µ Aµ = 0, imposes in addition that, Z d3~k X (λ) −ik·x ∗ ik·x k · ǫ (k) a (k)e − a (k)e = 0. (9.30) λ λ (2π)3 2ωk λ=0 Lagrangian of the electromagnetic field The following Lagrangian, 1 L = − Fµν F µν , 4

(9.31)

gives rise to the Maxwell equations as classical equations of motion. Explicitly, 1 (9.32) L = − [(∂µ Aν ) (∂ µ Aν ) − (∂µ Aν ) (∂ ν Aµ )] . 2 It is easy to verify that the Euler-Lagrange equations are ∂L ∂L − =0 ∂ (∂µ Aν ) ∂Aν ; ∂ 2 Aν − ∂ ν (∂µ Aµ ) = 0. ∂µ

9.2

(9.33)

Quantization of the Electromagnetic Field

The conjugate momentum for the field components Aµ are, πµ =

∂L = F µ0 . ∂ (∂0 Aµ )

(9.34)

One would like to impose the bosonic quantization condition, [Aµ (x), π ν (x′ )] = g µν δ (3) (~x − ~x′ ) .

(9.35)

Notice the presence of g µν , which is needed in order to maintain covariance on the indices µ, ν of the quantization condition. We immediately come across a problem. For µ = 0, π 0 = F 00 = 0, 118

we encounter an inconsistency 0 = A0 (x), π 0 (x′ ) = −δ (3) (~x − ~x′ ) 6= 0.

(9.36)

Our first attempt to quantizing the classical Lagrangian of the electromagnetic field has failed. However, it is possible to quantize a different Lagrangian which nevertheless gives rise to the same physics as the Lagrangian of Eq. 9.31. We can exploit the fact that classical physics is the same in every gauge for Aµ . Consider a modified Lagrangian,

1 λ L = − Fµν F µν − (∂µ Aµ )2 . (9.37) 4 2 This yields the following Euler-Lagrange equations, which is not invariant under gauge transformations, ∂ 2 Aµ − (1 − λ) ∂ µ (∂ν Aν ) = 0.

(9.38)

which are generally different than Maxwell equations. Notice however, that for a special value λ = 1, the equations of motion are identical to Maxwell equations, ∂ 2 Aµ = 0, in the Lorentz gauge. For the new Lagrangian and with λ = 1, we find ∂L = −∂µ Aµ , (9.39) π0 = ∂ A˙ 0 which is not zero, as long as we do not impose the Lorentz condition ∂µ Aµ = 0. This is at first irreconcilable with classical physics, however there is a way out. We may think the classical electromagnetic field as the expectation value of a quantum field in a physical quantum state |ψi, i.e. Aµclassic ∼ hψ| Aµoperator |ψi .

(9.40)

We therefore only need to guarantee that the expectation value of ∂µ Aµ on physical states |ψi, hψ| ∂µ Aµ |ψi = 0, (9.41)

which may be done in a way such that ∂µ Aµ 6= 0 as an operator identity. We can now carry on with the quantization of the Lagrangian of Eq. 9.37. The quantization conditions of Eq. 9.35, substituting in the expressions of , yield the conjugate momenta π µ = ∂(∂∂L 0 Aµ ) h i A˙ µ (~x, t) , Aν (~x′ , t) = igµν δ (3) (~x − ~x′ ) . (9.42) 119

We now substitute the solution of Eq. 9.27 into Eq. 9.43. We obtain, h i 3 (3) ~ † ′ λλ′ ′ ~ (9.43) aλ (k), aλ′ (k ) = −g 2k0 (2π) δ k−k .

We can safely interpret the ai (k) and a†i (k) for i = 1, 2, 3 as annihilation and creation operators correspondingly, of photons with space-like polarizations ǫµ(i) . But this is problematic for the operators a0 (k), a†0 (k), which satisfy i h (9.44) aλ (k), a†0 (k ′ ) = −2k0 (2π)3 δ (3) ~k − ~k ′ . Consider a general one-photon state with a time-like polarization, Z d3~k |1i = f (k)a†0 (k) |0i . 3 (2π) (2k0 )

(9.45)

The norm of the state is, h1| 1i = −

d3~k |f (k)|2 h0| 0i , (2π)3 (2k0 )

(9.46)

which is negative. In general, a state with nt photons with a time-like polarization has a norm hnt | nt i = (−1)nt . (9.47)

This is a serious problem. Let us calculate the energy of such states. After a standard, by now, calculation, we find that the Hamiltonian is given by the expression: " 3 # Z X † d3~k H= k0 (9.48) aλ (k)aλ (k) − a†0 (k)a0 (k) (2π)3 2k0 λ=1

with k0 = ~k > 0. The energy of the state |1i is E1 =

h1| H |1i , h1| 1i

(9.49)

which yields (exercise), E1 = −

d3~k k |f (k)|2 (2π)3 (2k0 ) 0 R d3~k |f (k)|2 (2π)3 (2k0 )

120

< 0.

(9.50)

We have found that this state has a negative energy. This problem is removed if we implement properly a quantum Lorentz gauge condition. We need Eq. 9.41 to hold for any physical state. We will now describe how this can be achieved following the prescription of Gupta and Bleuler. We separate the electromagnetic field into a positive and a negative frequency component, (−) Aµ (x) = A(+) (9.51) µ (x) + Aµ (x), with (+) µ

d3~k X µ ǫλ aλ (k)e−ik·x , 3 (2π) 2k0 λ

(9.52)

d3~k X µ † ǫ a (k)e+ik·x . (2π)3 2k0 λ λ λ

(9.53)

and (−) µ

Physical states |ψi are required that they satisfy µ

∂µ A(+) |ψi = 0.

(9.54)

This is satisfied automatically by the vacuum state, |ψi = |0i, since A(+) contains only annihilation operators. We also find that, hψ| ∂µ Aµ |ψi µ

= hψ| ∂µ A(+) |ψi + hψ| ∂µ A(−) |ψi

= hψ| ∂µ A(+) |ψi + hψ| ∂µ A(+) |ψi∗ µ

= 2Re hψ| ∂µ A(+) |ψi = 0.

The Bleuler-Gupta condition takes the explicit form Z 3 X d3~k −ik·x (+) µ k · ǫλ (k)aλ (k) |ψi = 0. e ∂µ A |ψi = (2π)3 2k0 λ=0

(9.55)

(9.56)

It is easy to verify that the above is not satisfied for photon-states with a purely longitudinal polarization, |ψi = |1i, as in Eq. 9.45. What is a physical state for a single photon? Consider the general onephoton state with a momentum pµ = (1, 0, 0, 1), choosing its momentum along the z-axis. This state can be written as, |γi =

3 X λ=0

cλ a†λ (p) |0i ,

121

(9.57)

where cλ are arbitrary coefficients determining the mixture of polarizations in the photon physical state. We must have, µ

0 = ∂µ A(+) |γi ! X ;0 = cλ p · ǫλ = 0.

(9.58)

From the expressions for the vector polarizations in Eq. 9.28 we derive that, p · ǫ1 = p · ǫ2 = 0

(9.59)

p · ǫ0 = −p · ǫ3 .

(9.60)

and Therefore, for a physical photon state we need that c1 = −c3 . In other words, a component of the photon state with a time-like polarization must be accompanied with an “opposite” component with a longitudinal polarization. The contributions of these components cancel each other in all physical quantities.

122

Chapter 10 Propagation of free particles In previous chapters we studied the particle states from the quantization of free fields, such as the Schr¨odinger, the Klein-Gordon, the spinor, and the electromagnetic field. We are now ready to study the simplest transition amplitude, for the transition of a free particle from a space-time position y µ ≡ (y 0, ~y ) to a position xµ ≡ (x0 , ~x), with x0 > y 0. The transition amplitude is given by, Mx→y = hx| yi|x0 >y0 . (10.1) where |xi , |yi are states of a single particle at positions x, y correspondingly.

10.1

Transition amplitude for the Schr¨ odinger field

We shall compute the transition amplitude for the case of the Scr¨odinger field. This will give us an opportunity to compare with other theories, which are relativistic, and appreciate the merits of the latter. We recall that the Schr¨odinger field is given by, 0

ψ(x , ~x) =

d3~k −i(Ek x0 −~k~ x) , a(k)e (2π)3

where Ek =

~k 2 , 2m

123

(10.2)

(10.3)

and the operator a(k) satisfies the quantization condition, a(k), a† (k ′ ) = (2π)3 δ (3) (~k − ~k ′ ).

(10.4)

A state of a particle at a position x is given by the action of a field on the vacuum (see Section 3.5), |xi = ψ † (x0 , ~x) |0i ,

|yi = ψ † (y 0 , ~y ) |0i .

(10.5)

The transition amplitude of Eq. 10.1 is then,

with

My→x = h0| ψ(x0 , ~x)ψ † (y 0 , ~y ) |0i Z d3~k −i(Ek ∆t−~k∆~x) = , e (2π)3 ∆t = x0 − y 0,

∆~x = ~x − ~y .

(10.6) (10.7)

We can compute the above integral exactly. We first change to spherical coordinates, d3~k = dkk 2 dcosθ dφ, (10.8) and write

~k∆~x = k ∆x cos θ.

(10.9)

Then it is easy to perform successively the integrations in φ, cos θ and k. The result reads, i h m 32 2 −i∆t 21 m( ∆x ∆t ) . (10.10) My→x = e 2πi∆t This is an oscillatory function which is defined for arbitrary values of the pareven when its value is not smaller than the speed of light. Particle speed ∆x ∆t ticles which are quanta of the Schr¨odinger field, can propagate at space-time intervals anywhere outside the light-cone, which is in blatant disagreement with special relativity.

10.2

Transition amplitude for the real KleinGordon field

We write the real Klein-Gordon field as a sum of a “positive frequency” and a “negative frequency” term, φ(x) = φ(+) (x) + φ(−) (x), 124

(10.11)

with φ

(x) =

d4 k δ(k 2 − m2 )Θ(k 0 )e−ik·x a(k), 3 (2π)

(10.12)

(x) =

d4 k δ(k 2 − m2 )Θ(k 0 )e+ik·x a† (k). (2π)3

(10.13)

(+)

and φ

(−)

We notice that,

φ(+) (x) |0i = 0,

(10.14)

h0| φ(−) (x) = 0.

(10.15)

and The operator a(k) satisfies the quantization condition,

and

a(k), a† (k ′ ) = (2π)3 2ωk δ (3) (~k − ~k ′ ), q ωk = ~k 2 + m2 .

(10.16)

(10.17)

The amplitude for the transition y µ → xµ , with x0 > y 0, is then My→x = h0| φ† (x)φ(y) |0i = h0| φ(x)φ(y) |0i = h0| φ(+) (x) + φ(−) (x) φ(+) (y) + φ(−) (y) |0i

= h0| φ(+) (x)φ(+) (y) + φ(+) (x)φ(−) (y) + φ(−) (x)φ(+) (y) + φ(−) (x)φ(−) (y) |0i = h0| φ(+) (x)φ(−) (y) |0i = h0| φ(+) (x), φ(−) (y) + φ(−) (x)φ(+) (y) |0i = h0| φ(+) (x), φ(−) (y) |0i . (10.18)

The commutator gives,

(+) φ (x), φ(−) (y) =

d4 k Θ(k 0 )e−ik(x−y) δ(k 2 − m2 ), (2π)3

(10.19)

which is a c−number. We then have for the transition amplitude, My→x = h0| φ(+) (x), φ(−) (y) |0i = φ(+) (x), φ(−) (y) h0| 0i = φ(+) (x), φ(−) (y) , (10.20) 125

and explicitly, My→x =

d4 k Θ(k 0 )e−ik(x−y) δ(k 2 − m2 ). (2π)3

(10.21)

We would like to investigate whether the above expression for the transition amplitude respects the expectation from special relativity, which restricts particles to transitions within the light-cone. It is now a bit more tedious to evaluate the integral of Eq. 10.21. For simplicity, we shall evaluate it for two special transitions. • Transition A: where we take ∆~x = ~x − ~y = 0 and ∆t = x0 − y 0 > 0, which is a time-like transition (within the light-cone) • Transition B: where we take |∆~x| = |~x − ~y | > 0 and ∆t = x0 −y 0 = 0, which is a space-like transition (within the light-cone). Time-like transition A The transition amplitude becomes, Z d4 k 0 Θ(k 0 )e−ik ∆t δ(k 2 − m2 ) My→x = 3 (2π) Z √ d3~k 2 ~2 p = e−i∆t k +m (2π)3 2 ~k 2 + m2 Z ∞ √ k2 1 −i∆t k 2 +m2 √ dk e = (2π)2 0 k 2 + m2 √ Setting E = k 2 + m2 , we find that Z ∞ √ 1 E 2 − m2 e−iE∆t dE My→x = (2π)2 m

(10.22)

(10.23)

For large times, ∆t → inf ty, we can prove (exercise) that lim ∼ e−im∆t .

∆t→∞

(10.24)

The transition amplitude for a particle to remain within the light-cone, such as in our exemplary case A, is then finite. 126

Space-like transition B It is interesting to find out whether transitions outside the light-cone, such as in case B, are forbiden in a relativistic field theory. In this case, the transition amplitude becomes Z d4 k ~ My→x = Θ(k 0 )e+ik∆~x δ(k 2 − m2 ) 3 (2π) Z k 2 e+ik(∆x) cos θ dk √ = dcosθ (2π)2 k 2 + m2 Z ∞ k2 eik(∆x) − eik(∆x) 1 √ dk = (2π)2 0 ik(∆x) 2 k 2 + m2 Z +∞ −i keik(∆x) √ . (10.25) = dk 2(2π)2 (∆x) −∞ k 2 + m2 The above integrals has branch cuts at k = ±im, as in Figure 10.1. For

Figure 10.1: Branch cuts of the transition amplitude integral for a space-like transition (∆x) > 0 we can wrap the contour of integration around the upper branch cut. The integral is then becoming, after setting k = iρ, Z ∞ 1 ρe−ρ(∆x) My→x = 2 . (10.26) dρ p 4π (∆x) m ρ2 − m2 127

Asymptotically, for a large space-like interval ∆x → ∞, the transition integral becomes lim My→x ∼ e−m(∆x) . (10.27) (∆x)→∞

We find that the transition amplitude vanishes exponentially the further away we move from the light-cone (∆x = 0). Is this a contradiction with special relativity? Actually it is not when we remember that our theory is in addition a quantum theory. Essentially, a particle can escape from the lightcone by a distance ∆x which is not observable to us due to the uncertainty principle. The transition amplitude is only significant for m(∆x) ∼ 1,

(10.28)

which is the uncertainty we expect for determining its position for a momentum uncertainty ∆p ∼ m. Essentially, to observe that a particle of a certain mass has escaped the light-cone we must first measure the momentum of the particle with an accuracy as good as its mass (in order to identify the particle from its mass). Then, the uncertainty in its position will be as large as the distance that we expect a finite probability for the particle to escape away from the light-cone.

10.3

Time Ordering and the Feynman-St¨ uckelberg propagator for the real Klein-Gordon field

In order for the “correlation function”, h0| φ(x)φ(y) |0i , to be a transition amplitude where a particle is created at point y µ and it is destroeyd at a point xµ , the time x0 must occur after y 0, i.e. x0 > y 0 . However, h0| φ(x)φ(y) |0i = h0| [φ(x), φ(y)] |0i + h0| φ(y)φ(x) |0i = h0| 0 |0i + h0| φ(y)φ(x) |0i = h0| φ(y)φ(x) |0i .

(10.29)

We can write the same correlation function with the space-time points x and y interchanged. How can we interpret this? If we insist on a causal 128

interpretation, where a particle is first created at a point x and it is destroyed at a point y, we must have y 0 > x0 . The Feynman-St¨ uckelberg propagator is defined to account for both cases, 0 0 0 0 x > y and x < y , for arbitrary x, y. This is written as, h0| T φ(x)φ(y) |0i = Θ(x0 − y 0) h0| T φ(x)φ(y) |0i +Θ(y 0 − x0 ) h0| T φ(y)φ(x) |0i .

(10.30)

The “time-ordering” symbol T is defined to order the operators following it from the later to the earlier times. The Feynman-St¨ uckelberg propagator has a very nice integral representation. Explicitly, from the above definition and the result of Eq. 10.21, we find Z d3~k 1 p h0| T φ(x)φ(y) |0i = Θ(x0 − y 0 )e−ik·(x−y) + Θ(y 0 − x0 )e−ik·(y−x) 3 (2π) 2 ~k 2 + m2 Z +∞ 4 dk i = lim e−ik·(x−y) . (10.31) 4 2 δ→0 −∞ (2π) k − m2 + iδ We can verify that the four-dimensional integral of the last line is equal to the line before explicitly, by using Cauchy’s theorem. In Fig. 10.2 we plot the poles of the integrand on the energy k 0 complex-plane. These are located

Figure 10.2: Poles of the integrand for the Feynman-St¨ uckelberg propagator

129

at k 0 = −ω + iδ and k 0 = ω − iδ, with ω = exponential in the integrand behaves as, ~

p ~k 2 + m2 . At k 0 = ±i∞, the

e−ik·(x−y) → e+ik(~x−~y) e±∞(x

0 −y 0 )

(10.32)

If x0 > y 0 we can close the contour of integration to the lower complex haplfplane, guaranteeing that the integrand vanishes at −i∞. If x0 < y 0 we must close the contour to the upper half-plane. In each case, we pick up one of the residue at k 0 = ±ω, recovering the two terms of the first line in Eq. 10.31. The expression for the Feynman propagator, Z +∞ 4 dk i h0| T φ(x)φ(y) |0i = lim e−ik·(x−y) , (10.33) 4 2 δ→0 −∞ (2π) k − m2 + iδ is of paramount importance for the computation of generic transition amplitudes in quantum field theory. An important property is that it is a Green’s function of the Klein-Gordon equation. We can act on it with the differential operator of the Klein-Gordon equation, Z +∞ d4 k i 2 2 2 2 ∂ + m h0| T φ(x)φ(y) |0i = ∂ + m e−ik·(x−y) 4 2 2 + iδ (2π) k − m −∞ Z +∞ 4 d k −ik·(x−y) e = −i 4 −∞ (2π) ; ∂ 2 + m2 h0| T φ(x)φ(y) |0i = −iδ 4 (x − y). (10.34)

10.4

Feynman propagator for the complex KleinGrodon field

We will now compute the transition amplitude, or the Feynman propagator, for the complex Klein-Gordon field, Z d3~k φ(~x, t) = a(k)e−ik·x + b† (k)eik·x . (10.35) 3 (2π) 2ωk Notice that the hermitian conjugate of the field acting on the vacuum creates a particle with charge +1 at a certain position x, |particle at position xi = φ† (~x, t) |0i , 130

(10.36)

while a field acting on the vacuum creates an anti-particle with charge -1, |anti-particle at position xi = φ(~x, t) |0i .

(10.37)

Consider the correlation function, h0| φ(y)φ†(x) |0i

(10.38)

which we can interprete it as a causal sequence where a particle is first created at xµ ≡ (x0 , ~x) and it is destroyed at y µ ≡ (y 0 , ~y). We must then have y 0 > x0 . It is easy to prove that h0| φ(y)φ†(x) |0i = h0| φ† (x)φ(y) |0i

(10.39)

The inerpretation of the rhs is that an anti-particle is first created at y µ ≡ (y 0, ~y ) and it is destroyed at xµ ≡ (x0 , ~x). We must then have x0 > y 0 . We define the Feynman propagator to account for the two equivalent possibilities for the forward in time propagation of either a particle or an anti-particle. h0| T φ(x)φ† (y) |0i = Θ(x0 − y 0 ) h0| φ(x)φ† (y) |0i + Θ(y 0 − x0 ) h0| φ† (y)φ(x) |0i (10.40) We can repeat the same computation as for the Feynamn propagator of the real Klein-Gordon field. The result is identical: Z +∞ 4 dk i † h0| T φ(x)φ (y) |0i = lim e−ik·(x−y) . (10.41) 4 2 δ→0 −∞ (2π) k − m2 + iδ

10.5

Feynman propagator for the Dirac field

Let us now compute the correlation function of two Dirac spinor fields, Z d3 ~p X s ¯ ua (p)¯ usb (p)e−ip·(x−y) h0| ψa (x)ψb (y) |0i = 3 (2π) 2ωp s Z d3 ~p (6p + m)ab e−ip·(x−y) = 3 (2π) 2ωp Z ∂ d3 p~ µ ; h0| ψa (x)ψ¯b (y) |0i = iγ e−ip·(x−y) + m1 3 2ω ∂xµ (2π) p ab (10.42) 131

Similarly, we find (exercise) Z ∂ d3 p~ µ h0| ψ¯a (y)ψb (x) |0i = − iγ + m1 e−ip·(y−x) (10.43) 3 2ω ∂xµ (2π) p ab Notice the overall (-) sign for the transition amplitude of an anti-particle with respect to the corresponding amplitude in the case of a particle. We can combine the two forward in time propagations by including this relative minus sign into the definition of the time ordering for fermions. h0| T ψa (x)ψ¯b (y) |0i = Θ(x0 −y 0 ) h0| ψa (x)ψ¯b (y) |0i−Θ(y 0 −x0 ) h0| ψ¯a (y)ψb (x) |0i (10.44) From the above, we find for: Z ∂ d3 p~ µ h0| T ψa (x)ψ¯b (y) |0i = iγ Θ(y 0 − x0 )e−ip·(y−x) + m1 µ 3 ∂x (2π) 2ωp ab +Θ(x0 − y 0 )e−ip·(x−y) Z +∞ 4 dk i µ ∂ = iγ + m1 e−ik·(x−y) , (10.45) µ 4 2 2 + iδ ∂x (2π) k − m ab −∞ which yields for the fermion Feynman propagator the integral representation, Z +∞ 4 d k i (6k + m) −ik·(x−y) ¯ |0i = e . (10.46) h0| T ψ(x)ψ(y) 4 2 2 −∞ (2π) k − m + iδ

10.6

Feynman propagator for the photon field

132

Chapter 11 Scattering Theory (S-matrix) We have studied the quantum field theories of free elementary particles with spin-0 (Klein-Gordon field), spin- 21 (Dirac spinor field), and spin-1 (Electromagnetic photon field). In all these cases, we could identify a field four momentum P µ = H, P~ which was a conserved quantity. It was also relatively easy to find a complete set of eigenstates for P µ . In all these “free field theories” the Hamiltonian eigenstates happened to be, in addition, eigenstates of a time-independent “number operator”. States with definite energy and momentum were, therefore, also states of a conserved number of indestructible particles 1 . In reality, however, particles interact and may also be destroyed or created in a scattering process. Realistic field theories contain additional terms in their Lagrangian which do not make it possible to find a constant particlenumber operator. In a free field theory, conservation of energy and momentum is a consequence of the fact that the theory is invariant under time and space translations. This is a property which will also hold for interacting field theories. Conservation of the particle number was, on the other hand, only a consequence of an “accident” that a field free theory was rather simple. This “accident” does not occur anymore in an interacting field theory, where we expect it to describe successfully particle creation and annihilation. In this chapter we will consider interacting field theories in general, without specifying the details of their Lagrangian. We will only require that they yield physical laws which are invariant under space-time translations, and therefore their is a conserved field momentum operator. We will also make 1

In mathematical parlance, the Hilbert space of states had a Fock-state representation.

133

one further assumption, which we will explain later in its detail, that particle interactions happen at short distances, and that particles when are at far distances they behave as if they were free. These two considerations will lead us to general expressions for the probability amplitude of a scattering process.

11.1

Propagation in a general field theory

We start our analysis by studying the propagator of fields in a general theory, as we have specified it above. We shall also assume that there is a ground state, which we denote by |Ωi . For simplicity we consider here the propagator of a scalar (interacting) field φ(x), which we write as hΩ| T {φ(x)φ(y)} |Ωi = hΩ| T {φ(x)1φ(y)} |Ωi .

(11.1)

Let us form a unit operator 1 out of a complete set of states. Space-time translation invariance dictates that P µ ≡ (H, P~ ), the field four momentum of the Lagrangian system is a conserved quantity. Notice that we do not need the explicit expression for P µ ; Noether’s theorem guarantees that such an operator does exist. We denote with |pi the corresponding eigenstates of the momentum-operator, P µ |pi = pµ |pi , (11.2) or, in components,

H |pi = Ep |pi ,

P~ |pi = ~p |pi ,

(11.3)

with pµ = (Ep , ~p) .

(11.4)

As we have remarked, the states |pi, are not , necessarily, states with a definite number of particles. To simplify the discussion, we also assume here that all momentum eigenvalues are not light-like, having p2 6= 0. Then, for each state |pi with momentum pµ , we can perform a Lorentz boost which renders ~p = 0. Conversely, all states |pi can be produced by applying Lorentz transformations to states |λ0 i with P~ |λ0 i = 0. 134

(11.5)

(11.6)

The sum in λ runs over all possible states with a zero space momentum |λ0 i. Let us now assume that x0 > y 0 . The propagator is then written as, hΩ| T {φ(x)φ(y)} |Ωi|x0 >y0 = ! XZ d3 p~ hΩ| φ(x) |Ωi hΩ| + |λp i hλp | φ(y) |Ωi (2π)3 2Ep (λ) λ

= hΩ| φ(x) |Ωi hΩ| φ(y) |Ωi XZ d3 ~p + hΩ| φ(x) |λp i hλp | φ(y) |Ωi . (2π)3 2Ep (λ)

(11.7)

Consider now the matrix-element hΩ| φ(x) |λp i , and recall that the system is invariant under space-time translations xµ → xµ +ǫµ . The field four-momentum is the generator of space-time translations. If we know the value of the quantum field at a point y µ , then we can find its value at a different space-time point by applying, φ(xµ ) = eiP ·(x−y) φ(y)e−iP (x−y).

(11.8)

It is very convenient to express the field in the matrix-element in terms of the value of the quantum field at the origin, hΩ| φ(x) |λp i = hΩ| eiP ·x φ(0)e−iP ·x |λp i .

(11.9)

The ground state is invariant under space-time translations, e−iP ·x |Ωi = |Ωi . Also, a state |λp i has a definite momentum pµ = (Ep (λ), ~p). Thus,

e−iP ·x |λp i = |λp i e−ip·x p0 =Ep (λ) . 135

(11.10)

(11.11)

We can then write,

hΩ| φ(x) |λp i = hΩ| φ(0) |λp i e−ip·x p0 =Ep (λ) .

(11.12)

|λp i = U (Λp ) |λ0 i .

(11.13)

We now recall that the state |λp i is produced from a state with a zero spacemomentum with a Lorentz boost,

We can also exploit that the vacuum state is invariant under Lorentz transformations, U (Λp ) |Ωi = U −1 (Λp ) |Ωi = |Ωi , (11.14)

and cast Eq. 11.12 as

hΩ| φ(x) |λp i = hΩ| U −1 (Λp ) φ(0)U (Λp ) |λ0 i e−ip·x p0 =Ep (λ) .

(11.15)

For a scalar field,

U (Λp ) φ(xµ )U −1 (Λp ) = φ (Λµν xν ) ,

(11.16)

which for a position at the origin xµ = 0 yields U (Λp ) φ(0)U −1 (Λp ) = U −1 (Λp ) φ(0)U (Λp ) = φ(0).

(11.17)

We then find that the space-time dependence of the matrix-element in Eq. 11.15 is a simple exponential,

hΩ| φ(x) |λp i = hΩ| φ(0) |λ0 i e−ip·x p0 =Ep (λ) . (11.18) Similarly, Finally,

hλp | φ(y) |Ωi = hλ0 | φ(0) |Ωi e+ip·y p0 =Ep (λ) .

hΩ| φ(y) |Ωi = hΩ| eiP ·y φ(0)e−iP ·y |Ωi = hΩ| φ(0) |Ωi .

(11.19)

(11.20)

Using Eqs 11.18-11.20, we find that the propagator in Eq. 11.7 becomes, hΩ| T {φ(x)φ(y)} |Ωi|x0 >y0 = hΩ| φ(0) |Ωi2 Z X

d3 ~p 2 |hΩ| φ(0) |λ0 i| + e−ip·(x−y) p0 =Ep (λ) . (11.21) 3 (2π) 2Ep (λ) λ 136

In the same fashion, we obtain for the second time-ordering possibility hΩ| T {φ(x)φ(y)} |Ωi|y0 >x0 = hΩ| φ(0) |Ωi2 Z X

d3 ~p 2 −ip·(y−x)

+ |hΩ| φ(0) |λ0 i| e . (11.22) p0 =Ep (λ) (2π)3 2Ep (λ) λ

The propagator is,

hΩ| T {φ(x)φ(y)} |Ωi = hΩ| T {φ(x)φ(y)} |Ωi|x0 >y0 Θ(x0 − y 0 ) + hΩ| T {φ(x)φ(y)} |Ωi|x0 >y0 Θ(x0 − y 0 ),

(11.23)

which yields, hΩ| T {φ(x)φ(y)} |Ωi = hΩ| φ(0) |Ωi2 + ×

d p~ (2π)3 2Ep (λ)

X λ

|hΩ| φ(0) |λ0 i|2 ×

e−ip·(x−y) Θ(x0 − y 0 ) + e−ip·(y−x) Θ(y 0 − x0 )

p0 =Ep (λ)

(11.24)

We recognize that the above integral is the Feynman-St¨ uckelberg propagator of Eq. 10.31, of the free scalar field theory. We denote by m2λ the eigen-value of the squared field-momentum P 2 corresponding to the state |λ0 i, P 2 |λ0 i = m2λ |λ0 i . Then, for the boosted state |λp i, we have P 2 |λp i = H 2 − P~ 2 |λp i = Ep (λ)2 − ~p2 |λp i .

(11.25)

(11.26)

Given that |λp i = U(Λp ), and P 2 is invariant under a Lorentz transformation, we obtain that Ep (λ)2 − p~2 = m2λ . (11.27)

Then, using the result of Eq. 10.31, we write

hΩ| T {φ(x)φ(y)} |Ωi = hΩ| φ(0) |Ωi2 Z X i d4 p 2 e−ip·(x−y) + |hΩ| φ(0) |λ0 i| 2 4 2 (2π) p − mλ + iδ λ 137

(11.28)

This is a general result for the propagator of any scalar field in any quantum field theory for which is symmetric under Poincare’ (Lorentz and translation) transformations. Let us now assume that there exists a density of states |λ0 i, with P 2 |λ0 i = 2 mλ |λ0 i. We can replace the sum in the above expression with an integral, hΩ| T {φ(x)φ(y)} |Ωi = hΩ| φ(0) |Ωi2 Z Z 2 2 2 2 + dM ρ(M − mλ ) |hΩ| φ(0) |λ0 i|

d4 p i e−ip·(x−y) 4 2 (2π) p − M 2 + iδ (11.29)

This is the K¨allenLehmann representation for the scalar field propagator. It is an amazing result for its simplicity, given its generality. Up to constants (hΩ| φ(0) |Ωi2 , |hΩ| φ(0) |λ0 i|2 ), the propagator is determined fully by the density of the energy eigenstates with zero space-momentum.

11.1.1

A special case: free scalar field theory

Our result for the Feynman-St¨ uckelberg can be now rederived as a special case of the K¨allen-Lehmann propagator of Eq. 11.29 when considering the free Klein-Gordon field. In this case, we are able to find a simple expression for the quantum field φ(x), Z d3 ~p φ(x) = a(p)e−ip·x + a† (p)e+ip·x , (11.30) 2 (2π) 2ωp p with ωp = p~2 + m2 . The vacuum state is, |Ωi = |0i ,

(11.31)

and it is annihilated by the ladder operator a(p), a(p) |0i = h0| a† (p) = 0.

(11.32)

h0| φ(0) |0i = 0.

(11.33)

We can easily see that We also need to compute the density of states |λ0 i, with H |λ0 i = mλ |λ0 i and P~ |λ0 i = 0. In the case of the free Klein-Gordon field we can identify a number of particles for each eigenstate of the P µ operator. We can build 138

states of higher and higher energy with p~ = 0, by applying repeatedly the creation operator a† (0) on the vacuum. We have H a† (0) |0i = m a† (0) |0i , H a† (0)2 |0i = 2m a† (0)2 |0i , ... † n H a (0) |0i = nm a† (0)n |0i , ... The density is then 2

ρ(M −

m2λ )

∞ X n=1

δ(M 2 − n2 m2 ).

(11.34)

Finally, we require the constants † n h0| φ(0) |λ0 i = h0| φ(0) a (0) |0i = δn,1 .

(11.35)

Exercise: prove the above. We now substitute Eqs 11.33, 11.34 and Eq. 11.35 into Eq. 11.29. We recover Z d4 p i h0| T {φ(x)φ(y)} |0i = e−ip·(x−y) , (11.36) 4 2 (2π) p − m2 + iδ which is the familiar Feynman-St¨ uckelberg propagator for the free KleinGordon field.

11.1.2

“Typical” interacting scalar field theory

It is not possible to compute the density ρ(M 2 −m2λ ) of zero-momentum ~p = 0 states |λ0 i in realistic field theories where the Lagrangian contains additional potential terms to the ones of the free scalar Klein-Gordon field. We can make, however, some reasonable assumptions which agree with observations on the systems that we would like to describe. We have emphasized several times that in interacting theories it is not, in general, possible to identify a number of particles corresponding to a certain 139

energy-momentum eigenstate. Nevertheless, we do observe particles which act as if they were free and isolated. Such a state is characterized by the mass m of the particle. If we now consider a state of two “approximately free” particles the energy of the state when the combined momentum of the two particles is p~ = 0 is also “approximately” twice the mass of one-particle 2 m. However, their interaction potential will not allow for this exact value, and in general, their energy will be larger. States other than one-particle states form a continuum of masses M 2 which is larger than the minimum energy (corresponding to free particles) required to get two particles in the same state. An exception to this occurs when the two-particles form a bound state, where their binding energy is negative and their effective mass is generally smaller than 2 m. Although a very important issue, we will not analyze in this course bound state problems. Then, a typical density is, ρ(M 2 − m2λ ) = δ(M 2 − m21−particle ) + Θ(M 2 − 4m2 )ρcont (M 2 − m2λ ). (11.37) where ρcont a continuous function. Substituting this density functional form into the general expression of Eq. 11.29 for the scalar propagator we find, hΩ| T {φ(x)φ(y)} |Ωi = Z

2 d4 p i

hΩ| φ(0) 1particle

e−ip·(x−y) 0 2 4 2 (2π) p − m1particle + iδ Z Z i d4 p 2 2 2 2 e−ip·(x−y) + dM ρcont (M − mλ ) |hΩ| φ(0) |λ0 i| 4 2 (2π) p − M 2 + iδ + hΩ| φ(0) |Ωi2 (11.38) The above result is very important, stating that one-particle states are responsible for poles in the propagator of an interacting field. If a theory has states which behave as isolated then we anticipate to find poles in the propagator for squared momenta p2 = m1particle which correspond to the physical mass of the particles. Suggestive to the above, it is then worth to remember in practice that the scalar propagator is hΩ| T {φ(x)φ(y)} |Ωi =

d4 p iZ˜φ e−ip·(x−y) + continuum, (2π)4 p2 − m2phys + iδ 140

(11.39)

where

2 Z˜φ = hΩ| φ(0) 1particle 0 ,

(11.40)

a constant, and mphys the physical mass of a particle.

11.2

Spectral assumptions in scattering theory

We are now ready to study using quantum field theory the scattering of elementary particles. Scattering processes occur in Lagrangian systems with interaction potentials, i.e. additional terms in the Lagrangian other than the ones that we have found in the quantization of free fields. It has never been possible to find an exact solution for the Hamiltonian of any field theory which describes realistic particle interactions in four space-time dimensions. However, it is possible under some reasonable assumptions to develop a theory for scattering probability amplitudes. For simplicity, we shall carry out our analysis for the case of interacting scalar fields. This is sufficient to demonstrate the salient features of the scattering theory. Our basic set of assumptions is the following, • The theory has a ground state |Ωi • There are “first excited” states |pi with P 2 |pi = m2phys |pi ,

(11.41)

corresponding intuitively to “single-particle” states. We also assume that m2phys ≥ 0, which means that there are no tachyons (particles which travel faster than the light) • All remaining eigenstates |qi of the field momentum operator have P 2 |qi = m2phys |qi ,

(11.42)

M 2 ≥ 4m2phys .

(11.43)

with These states form a continuum, and they correspond, intuitively, to multi-particle excitations. 141

• The expectation value of the field operator in the ground state is zero, hΩ| φ(x) |Ωi = 0.

(11.44)

Roughly, φ(x) |Ωi should correspond to an one-particle excited state. A single particle should not disappear spontaneously into a state with no particles |Ωi. The corresponding probability amplitude should therefore vanish, hΩ| 1particle i ∼ hΩ| φ(x) |Ωi = 0.

11.3

“In” and “Out” states

As we have discussed earlier, states in interacting field theories do not longer describe systems of particles with a necessarily constant number of them. Free field theories have a Hilbert-space of states with states of definite particle number, but Hamiltonian eigenstates in interacting theories are presumably. Nevertheless, we do observe systems of free electrons and photons when they are sufficiently separated. By the uncertainty principle, such states describing particles which are isolated at certain times must be wave-packets. Two classes of wave-packets are interesting in order to describe the scattering of freely moving particles. • Wave-packets which are isolated at a time very far in the past, t = −∞, but may be overlapping at some finite time t • Wave-packets which are isolated at in the far future, t = +∞, but may be overlapping at some finite time t. We should remember that in our development of the field theory formalism, we are using the Heisenberg picture for time evolution. In this picture, field operators depend on time, while states are time independent 2 . Time-independent states are perhaps counter-intuitive, and can mislead us to believe that they describe non-evolving systems. But this is of course not 2

For example, in the free scalar field case the field has a solution which is explicitly time-dependent, Z d3 p~ a(p)e−ip·x + a† (p)e+ip·x , φ(x) = 2 (2π) 2ωp

while states, |0i , |pi = a† (p) |0i , . . ., are manifestly time-independent.

142

true. Measurements of expectation values of field operators on a state are time-dependent, and in this sense, states encapsulate the time-evolution of the physical system that they describe. We define a state as an |ini state, if it describes a system of freely moving in the very far past, which may or may not be interacting at a later time. We define a state as an |outi state, if it describes a system of particles which may or may not be interacting at a finite time but they are freely moving in the far future. Consider an |ini state which describes a single particle with momentum p in the far past. We will assume that free particles are indestructible, stable, in the absence of other particles. Then this state will also describe a free single particle in the far future, and it will also be an state |ini: |in, single − particle, pi = |out, single − particle, pi .

(11.45)

More complicated “in” and “out” states containing many free particles in the very far past or future respectively, allow for particle interactions at a finite time. What are the possible “in” and “out” states? We will make the following assumption: “ Every possible state is a linear combination of either |ini or |outi states.” In physical terms, this means that every physical system will behave as a system of freely moving particles which are getting increasingly isolated if we wait long enough. Conversely, if we look further back in the far past, every physical system originates from a system of freely moving isolated particles. The above assumption is in accordance with the particle interactions occur when particles are brought very close together. Short range interactions cannot keep particles together perpetually. There is a striking important “exception” to this assumption: elementary particles may form bound states. In practice, we know from experimentation which are the bound states that may be formed (hadrons, atoms, molecules, etc ). These are stable, and we can account for them by including them in a generalized definition of |ini and |outi states. We adopt the following notation. We denote by |{pi }, ini an “in” state where the ith particle has momentum pi . We denote by |{kj }, outi 143

an “out” state where the j th particle has momentum kj . These states are complete. Every state can be written as a linear combination of either of them, and we have: X X |{pi }, ini hin, {pi}| = |{kj }, outi hout, {kj }| = 1 (11.46)

11.4

Scattering Matrix-Elements

Obviously, an |{pi }, ini state can be written as a superposition of |{kj }, outi states. This physically means that a system of originally free particles will evolve to a system of ultimately free particles. If these particles never interact, the “in” state is the same as the “out” state. If they do, then we may have isolated particles after the interaction with different momenta or even a different number of them. We write, X |{pi }, ini = c ({kj }, {pi }) |{kj }, outi . (11.47) {kj }

The probability amplitude for the transition of a system of i = 1 . . . N incoming particles with momenta {pi } to a system of j = 1 . . . M outgoing particles with momenta {kj } is known as the “scattering matrix-element”, Sji ≡ c ({kj }, {pi }) |{kj }i = hout, {kj }| {pi }, ini

(11.48)

The scattering probability is P ({pi } → {kj }) = |Sji|2 = |hout, {kj }| {pi }, ini|2 .

(11.49)

The scattering matrix-elements constitute the S-matrix, which is a matrix acting on the space of “in” states and transforms them into “out” states, S ≡ Sij ,

|α, outi = Sαβ |β, ini .

(11.50)

The S-matrix is unitary, X X † SS † αγ = Sαβ Sβγ = hα, in| out, βi hβ, out| γ, ini β

= hα, in|

X β

|β, outi hout, β| |γ, ini = δαγ . 144

(11.51)

We are usually not interested in computing the probability of not having a scattering. The matrix-elements of the S-matrix which are relevant for a true scattering are the non-diagonal. We then define the so called “transitionmatrix” T , which is given by S ≡ 1 + iT.

(11.52)

In the rest of this chapter we shall prove a fundamental relation of Smatrix elements to expectation values of field operators in the ground state of the interacting theory.

11.5

S-matrix and Green’s functions

In the last section, we established very few properties of the S-matrix which, at a first sight, offer little help in computing it explicitly. It is possible, however, to extract S-matrix elements from Green’s functions, which are defined as G(x1 , x2 , . . . , xN ) ≡ hΩ| T {φ(x1 )φ(x2 ) . . . φ(xN )} |Ωi .

(11.53)

In the simplest case of one space-time point, hΩ| φ(x) |Ωi = 0,

(11.54)

we have assumed that the corresponding Green’s function vanishes, as part of our spectral assumptions for an interacting field theory. The two-point Green’s function is our familiar propagator, for which we have found the K¨allen-Lehmann representation G(x, y) =

d4 p iZ˜φ e−ip·(x−y) + continuum. (2π)4 p2 − m2phys + iδ

(11.55)

Consider the explicit case with x0 > τ > −τ > y 0 with τ → ∞, for two space-time points where one is very far in the future and the second very far

145

in the past. Then, G(x, y)| x0 →+∞

y 0 →−∞

= hΩ| T {φ(x)φ(y)} |Ωi = hΩ| {φ(x)φ(y)} |Ωi = hΩ| {φ(x)1 1 φ(y)} |Ωi ! X = hΩ| φ(x) |outi hout| out

; G(x, y)| x0 →+∞

y 0 →−∞

XX in

out

X in

|ini hin| φ(y) |Ωi

hΩ| φ(x) |outi hout| ini hin| φ(y) |Ωi

(11.56)

We have defined an |outi as any state which behaves as a state of any number of free isolated particles in the far future. The the sum over all “out” states corresponds to X hΩ| φ(x) |outi hout| → hΩ| φ(x) |Ωi hΩ| out

We now make use the property of “out” states behaving as states of free particles at large times x0 > τ , and compute hΩ| φ(x) |outi in free field theory. In free field theory, the state φ(x)Ω corresponds to a state with exactly oneparticle, and it has no overlap with states of a different particle multiplicity. We then have, hΩ| φ(x) |Ωi = hΩ| φ(x) |p1 , p2 ; outi = hΩ| φ(x) |p1 , p2 , p3 ; outi = . . . = 0, (11.58) and only hΩ| φ(x) |p1 ; outi = 6 0. (11.59) We have already computed this non-vanishing already up to a constant. From Eq. 11.18 and Eq. 11.40 we have q hΩ| φ(x) |p; outi = Z˜φ e−ip·x (11.60) 146

We compute the sum over “in” states in Eq. 11.56 in an identical fashion. Then we arrive to a simple result Z d3 p~ d3~q ˜ G(x, y)| x0 →+∞ = Zφ e−i(p·x−q·y) hp; out| q; ini . (11.61) (2π)3 2ωp (2π)3 2ωq y 0 →−∞ Eq. 11.61 relates the Green’s function function for two well separated in time points to the transition amplitude from an “in” single-particle state to an “out” single-particle state. How about more complicated Green’s functions and transition amplitudes? We can repeat the same procedure for a general Green’s function G(x1 , . . . , y1 . . .)| x0i →+∞ = hΩ| T {φ(x1 ) . . . φ(yy ) . . .} |Ωi x0i →+∞ . yi0 →−∞

(11.62)

yi0 →−∞

with Nout points at far future times x0i and Nin points yj at far past times. We find that ! N ! Z NY out in Y d3 p~i e−ipi ·xi d3 ~qj e+iqj ·yj G(x1 , . . . , y1 . . .)| x0i →+∞ = (2π)3 2ωpi (2π)3 2ωqj y 0 →−∞ i=1 j=1 i

Nin +Nout 2

×Z˜φ

hp1 , . . . , pNout ; out| q1 , . . . qNin ; ini

(11.63)

We arrive to a very useful result. Green’s functions with Nin + Nout , with Nin of them chosen in the far past and Nout in the far future are, up to an overall constant, some sort of a Fourier transform of S-matrix elements for the scattering of Nin particles to Nout particles. In the next section we shall use Eq. 11.63 to compute the matrix-elements.

11.6

The LSZ reduction formula

We will compute scattering matrix elements by inverting Eq. 11.63. This would require a type of a Fourier transformation on the space-time coordinates xµi and yjµ . The integration over the space components, ~xi and ~yj , can be unrestricted (from −∞ to +∞), but we need to be careful concerning the time integrations, since we have made assumptions about x0i being far future and yj0 far past times.

147

Let us try to integrate xµi and yjµ over the maximum space-time region where we do not contradict our time-ordering assumptions. Consider the integral, ! ! Z Z YZ ∞ Y Z −∞ I [{ki}, {lj }] ≡ dx0i d3~xi eiki ·xi dyj0 d3~yj e−ilj ·yj τ

−τ

×G(x1 , . . . , y1, . . .).

(11.64)

We now substitute the expression for the Green’s function that we have found in Eq. 11.63, and perform the space integrations using the identity Z ~ d3~xeik~x = (2π)3 δ (3) (~k). (11.65) The integral of Eq. 11.64 becomes, I [{ki }, {lj }] ≡

∞

i(ki0 −ω(~ki ))x0i 0e dxi 2ω(~ki)

Nin +Nout 2

×Z˜φ

YZ j

−∞ −τ

−i(lj0 −ω(~lj ))yj0 0e dyj 2ω(~lj )

hk1 , . . . , kNout ; out| l1 , . . . lNin ; ini

The integrals over the time variables are also easy to perform, Z ∞ i h 0 0 0 −i ~ ~ ~ lim eiT (l −ω(l)) − eiτ (l −ω(l)) dteit(l −ω(l)) = l0 − ω(~l) T →∞ τ

(11.66)

(11.67)

The above is problematic due to the term which requires a not well defined limit at infinity. Here we pay a price for not having considered proper localized wave-packets for “in” and “out” states. This undefined limit can be evaded in a rigorous proof. An “easy” way out is to consider an alternative integration over a time which is slightly complex 3 , Z ∞ 0 i 0 ~ ~ dteit(1+iδ)(l −ω(l)) = lim (11.68) eiτ (l −ω(l)) 0 ~ δ→0 τ l − ω(l) 3 Certainly, our analysis here is very sloppy with exchanging limits, as well as with not having considered proper wave-packets for “in” and “out” states. Careful and also lengthier proofs can be found in the original literature and the book of Weinberg

148

Given the above caveats, we can write I [{ki }, {lj }] ≡

Y i

−i

− ω(~k)

Nin +Nout 2

×Z˜φ

eiτ (

k 0 −ω(~k)

)

Y j

lj0

−i

− ω(l~j )

hk1 , . . . , kNout ; out| l1 , . . . lNin ; ini

eiτ (

lj0 −ω(l~j )

)

(11.69)

We have now achieved to relate S−matrix elements for the probability of a transition of a system with Nin particles with momenta {ljµ } to a system with Nout particles with momenta {kj } to a Fourier-type integrals over Green’s functions. Let us analyze Eq. 11.69 further. We observe that a “Fourier transformed” Green’s function on the lhs has poles at q 0 ~ li = ω(li) = ~li2 + m2phys , due to every particle in the “in” and “out” states. We also notice that the poles are independent of the time τ , since the τ dependent exponential becomes one at the pole position l0 = ω. Indeed , we can expand 1 iτ eiτ (l0 −ω) = + + O((l0 − ω)τ 2 ) 2ω(l0 − ω) 2ω(l0 − ω) 2ω

(11.70)

This means that had we decided to put no restrictions in the time integrations for I [{ki}, {lj }] the coefficient of the poles (their residue) would be unchanged. We can then write, Regular at li0 , kj0 → ω(~li ), ω(~li)    1 1 2 2 ˜ ˜ Y Y iZφ iZφ   hk1 , . . . , kNout ; out| l1 , . . . lNin ; ini +  ~ ~ ~ ~ 2ω(li)(l0 − ω(li )) 2ω(kj )(k0 − ω(kj )) ˜ (k1 , . . . , kNout ; l1 , . . . lN ) , = G in

(11.71)

˜ in momentum space is defined as a normal where the Green’s function G Fourier transform of a Green’s function in x−space, Z +∞ P P ˜ G (k1 , . . . , kNout ; l1 , . . . lNin ) = d4 x1 . . . d4 xNout d4 y1 . . . d4 yNin ei[ i xi ·ki − j yj ·lj ] −∞

× hΩ| T {φ(x1 ) . . . φ(xNout )φ(y1 ) . . . φ(yNin )} |Ωi (11.72) 149

A comment is in order concerning the time integrations in the above integral. The exponential factors must be slightly deformed in the complex plane, 0 0 ~ eik·x → eik x (1+iδ)−ik~x . (11.73)

Alternatively, we can leave intact the exponentials and perform a real timeintegration in Eq. 11.72, but then we have to evaluate the kernel Green’s function at deformed space-time points, Z +∞ P P i[ i xi ·ki − j yj ·lj ] 4 4 4 4 ˜ (k1 , . . . , kNout ; l1 , . . . lN ) = G d y . . . d y e d x . . . d x 1 Nin 1 Nout in −∞

y1 ) . . . φ(˜ yNin )} |Ωi × hΩ| T {φ(˜ x1 ) . . . φ(˜ xNout )φ(˜ (11.74)

where x˜µ = (x0 (1 − iδ), ~x).

(11.75)

We now multiply Eq. 11.71 with a factor l2 − m2phys q i Z˜φ

for each “in” state and “out” state particle, and take the limit l2 = (l0 − ω)(l0 + ω) → m2phys . The regular term vanishes in this limit from the lhs of the equation. We are left with the LSZ reduction formula,    Y kj2 − m2phys Y li2 − m2phys   q q lim  hk1 , . . . , kNout ; out| l1 , . . . lNin ; ini = l2 =m2 ˜ ˜ i phys j i i Zφ i Zφ 2 2 kj =mphys

˜ (k1 , . . . , kNout ; l1 , . . . lN ) . ×G in

(11.76)

which is due to Lehmann, Symanzin and Zimmermann.

11.7

Truncated Green’s functions

Let us recall here the K¨ahlen-Lehman representation of the propagator of Eq. 11.39, which we can invert by applying a Fourier transformation. We find iZ˜φ 2 2 ˜ G(p) = 2 + Regular at p → m (11.77) phys p − m2phys 150

where we have defined the propagator in momentum space as Z ˜ G(p) ≡ d4 xeip·x hΩ| T {φ(x)φ(0)} |Ωi .

(11.78)

We can then cast the S-matrix from the LSZ formula as Nin +Nout 2

hk1 , . . . , kNout ; out| l1 , . . . lNin ; ini = Z˜φ

˜ G (k1 , . . . , kNout ; l1 , . . . lNin )

Q ˜ i ) Q G(l ˜ j )

ki2 →m2phys G(k i j 2 2

lj →mphys

(11.79) A Green’s function which is divided by a propagator for each external particle in the “in” and “out” states is called truncated: ˜ Nout ; l1 , . . . lNin ) ˜ trunc (k1 , . . . , kNout ; l1 , . . . lN ) ≡ G (k1 , . . . , k Q G in Q ˜ ˜ i G(ki ) j G(lj )

(11.80)

The LSZ formula takes the form

Nin +Nout 2

hk1 , . . . , kNout ; out| l1 , . . . lNin ; ini = Z˜φ

˜ trunc (k1 , . . . , kNout ; l1 , . . . lN )

k2 →m2 G in i

phys

lj2 →m2phys

(11.81) We now have a relation of probability amplitudes for the scattering of free particles in Qunatum Field theory, in terms of truncated Green’s functions. As we have discussed already, there is no realistic field theory in four dimensions where all such probabilities can be computed exactly. In the next chapter, we shall resort to perturbation theory, which turns to be an amazingly powerful tool.

11.8

Cross-sections∗

151

Chapter 12 Perturbation Theory and Feynman Diagrams We have been able to solve exactly Hamiltonian systems for the simplest quantum field theories which describe free particles. For these, we could find all eigenstates and eigenvalues of the Hamiltonian. However, in general, when interactions among particles are present, we must resort to perturbation theory. As a concrete example, we consider a simple new term in the Hamiltonian of a real Klein-Gordon field, 1 λ 1 L = (∂µ φ)(∂ µ φ) − m2 φ2 − φ4 . 2 2 4! The Hamiltonian density is, H = π φ˙ − L,

φ=

∂L , ∂ φ˙

(12.1)

(12.2)

The Hamiltonian is then found to be H = H0 + H ′ , where H0 =

d ~x

~ π + ∇φ 2

(12.3) + m2 φ2

2 is the Hamiltonian a free real Klein-Gordon field, and Z λ ′ H = d3~x φ4 , 4! 152

(12.4)

(12.5)

is a new term. For a generic λ 6= 0, it is not possible to find an exact expression for the quantum field φ. In this chapter, we will learn how to construct a solution using perturbation theory, by expanding around the known result for λ = 0. It is only necessary to know the field φ(~x, t) at a specific time value t = τ . We can use symmetry under time translations in order to determine the field µ µ µ at any other moment t. The generator of space-time translations x → x +ǫ is the field momentum P µ = H, P~ . Specifically for time translations, the generator is the Hamiltonian H. Fields at two different times are related via, φ(x, t) = e+iH(t−τ ) φ(~x, τ )e−iH(t−τ )

(12.6)

(this is an exponentiated form of an equation such as Eq. 4.111, with Q = P µ δxµ , δxi = 0). As you may have noticed field operators evolve with time, while states do not. This is usually termed as the Heisenberg picture. Let us now compute the time evolution of the field φ(~x, τ ) in the hypothetical case that there is no interaction (λ = 0). We find that φI (x, t) = e+iH0 (t−τ ) φ(~x, τ )e−iH0 (t−τ ) ,

(12.7)

and solving for φ(~x, τ ), we find φ(x, τ ) = e−iH0 (t−τ ) φI (~x, t)e+iH0 (t−τ ) .

(12.8)

φI (~x, τ ) is the field in the “interaction picture”. The fields in the interaction picture and the Heisenberg picture are identical if H = H0 . Combining Eq 12.6 and Eq. 12.8 we find the identity φ(~x, t) = U −1 (t, τ )φI (~x, t)U(t, τ )

(12.9)

where we have defined the time-evolution operator in the interaction picture U(t, τ ) ≡ e+iH0 (t−τ ) e−iH(t−τ ) .

(12.10)

We now make a very important assumption. We require that there is a time, for example in the far past, for which the field of the full Hamiltonian H is a solution of the free Hamiltonian H0 . This is a reasonable approximation if we can identify a time that particles are far from each other and they do not feel their interaction can be neglected. We can formally implement this by requiring for example that the interaction switches on at a certain time. Given the existence of such a special time, the interaction field φI (~x, t) will continue to be a solution of the free Hamiltonian at any time t. Then, Eq. 12.9 is a transformation of the field φI in the free theory (H0 ) to the field φ in the full “interacting” theory (H). 153

12.1

Time evolution operator in the interaction picture

The calculation of the field operator in the full theory proceeds through the evaluation of the time-evolution operator U(t, τ ) given that we can determine the field operator in the free theory. We shall first derive a differential equation for U(t, τ ). Differentiating with respect to time, we have: i

∂U(t, τ ) = i iH0 eiH0 (t−τ ) e−iH(t−τ ) − ieiH0 (t−τ ) He−iH(t−τ ) ∂t = eiH0 (t−τ ) (H − H0 )e−iH(t−τ ) = eiH0 (t−τ ) (H − H0 )e−iH0 (t−τ ) eiH0 (t−τ ) e−iH(t−τ ) (12.11)

We write the above in the form, ∂U(t, τ ) = VI (t − τ )U(t, τ ), ∂t with the interaction potential defined as, i

VI (t − τ ) ≡ eiH0 (t−τ ) (H − H0 )e−iH0 (t−τ ) .

(12.12)

(12.13)

We can cast the general solution of the differential equation Eq. 12.12 as a time ordered exponential (as with an ordinary Schr¨odinger equation). Integrating both sides of the equation with respect to time, we obtain Z 1 t U(t, τ ) = U(τ, τ ) + dt1 VI (t1 − τ )U(t1 , τ ). (12.14) i τ Let us rewrite the above equation, replacing t with t1 and t1 with t2 . We have, Z 1 t U(t1 , τ ) = U(τ, τ ) + dt2 VI (t2 − τ )U(t2 , τ ). (12.15) i τ We can substitute Eq. 12.15 into Eq. 12.14, obtaining Z 1 t dt1 VI (t1 − τ ) U(τ, τ ) U(t, τ ) = 1 + i τ 2 Z t Z t1 1 + dt1 dt2 VI (t1 − τ )VI (t2 − τ )U(t2 , τ ). i τ τ (12.16) 154

Obviously, we can repeat inserting Eq. 12.14 to itself as many times as we wish. After an infinite number of iterations we obtain " Z t1 Z tn−1 ∞ n Z t X 1 U(t, τ ) = 1 + dt1 dt2 . . . dtn i τ τ τ n=1 # VI (t1 − τ )VI (t2 − τ ) . . . VI (tn − τ ) U(τ, τ ).

(12.17)

Time-Ordering Consider one of the simplest integrals in the above series, Z t Z t1 I[t, τ ] = dt1 dt2 VI (t1 − τ )VI (t2 − τ ). τ

It can be written as Z t Z t I[t, τ ] = dt1 dt2 VI (t1 − τ )VI (t2 − τ )Θ(t1 − t2 ). τ

(12.19)

By changing integration variables, t1 → t2 and t2 → t1 , we obtain Z t Z t I[t, τ ] = dt1 dt2 VI (t2 − τ )VI (t1 − τ )Θ(t2 − t1 ). τ

(12.18)

(12.20)

Adding Eq. 12.19 and Eq. 12.20, we obtain Z Z t Z t1 1 t dt1 dt2 T {VI (t1 − τ )VI (t2 − τ )} , dt2 VI (t1 − τ )VI (t2 − τ ) = dt1 2! τ τ τ (12.21) where the symbol T {. . .} is our familiar time-ordering symbol, ordering operators from the largest to the smallest times, T {VI (t1 − τ )VI (t2 − τ )} = VI (t1 − τ )VI (t2 − τ )Θ(t1 − t2 ) +VI (t2 − τ )VI (t1 − τ )Θ(t2 − t1 ).

(12.22)

It is not hard to convince ourselves that in general, Z t Z t1 Z tn−1 dt1 VI (t1 − τ ) dt2 VI (t2 − τ ) . . . dtn VI (tn − τ ) τ

1 n!

t τ

dt1 . . . dtn T {VI (t1 − τ ) . . . VI (tn − τ )} . 155

(12.23)

Time-Ordered Exponentiated Integral We define a time-ordered exponentiated integral as the time-ordering of its Taylor series expansion, hR in t ′ ′ ∞ dt A(t ) X τ Rt ′ ′ T e τ dt A(t ) ≡ T n! n=0 Z ∞ 1 X t = 1+ dt1 . . . dtn T {A(t1 ) . . . A(tn )} . (12.24) n! n=1 τ For times t > ta > to we have the identity, R t ′ ′ R ta ′ ′ Rt Rt ′ R ta ′ ′ ′ ′ ′ T e ta dt A(t ) × T e τ dt A(t ) = T e ta dt A(t ) e τ dt A(t ) = T e τ dt A(t ) , (12.25) where we can put the two exponentials under a common time-ordering symbol given that the times of the operators in the first exponent are always larger than the times of the operators in the second exponent. We can now write a compact expression for the time-evolution operator of Eq. 12.17 as a time-ordered exponential, by using Eq. 12.23 and the definition of Eq. 12.25. We obtain that, U(t, τ ) = T e−i

12.2

Rt τ

dt′ V (t′ −τ )

U(τ, τ ).

(12.26)

Field operators in the interacting and free theory

Let us summarize here what we have achieved. • The field in the full theory φ(~x, t) is related to the field in the free theory φI (~x, t) at any time t, via a simple relation, φ(~x, t) = U(t, τ )−1 φI (~x, t)U(t, τ ).

(12.27)

• The time evolution operator, U(t, τ ), is a time ordered exponential given by Eq. 12.26.

156

To determine the operator U(t, τ ) at an arbitrary time, we require a boundary value U(τ, τ ) at a time τ . We can select τ to be any time for which the field in the full theory is essentially equal to the field in the free theory, φ(~x, τ ) = φI (~x, τ ).

(12.28)

Such as special time τ is not hard to find, if the assumptions that we have made for the S−matrix of the theory are correct. Essentially, fields are most of the time free, except for the short duration of particle interactions. We can then select τ to be a time in the far past or the far future of the scattering event that we would like to describe. For this time, we can set U(τ, τ ) = 1.

(12.29)

A time-ordered product of fields in the full theory can be written as, T {φ(x1 )φ(x2 ) . . . φ(xn )} = o −1 0 0 0 0 0 0 0 0 ˜ ˜ ˜ T U (x1 , τ )φI (x1 )U (x1 , x2 )φI (x2 )U(x2 , x3 ) . . . U (xn−1 , xn )φI (xn )U(xn , τ ) , n

(12.30)

where we define the operator U˜ (t2 , t1 ) ≡ U(t2 , τ )U −1 (t1 , τ ).

(12.31)

The above operator is independent of the reference time τ . It is given by (exercise): Rt ˜ 2 , t1 ) = T ei t12 dt′ VI (t′ ) , (12.32) U(t where t2 > t1 . It is easy to prove that U˜ (t3 , t2 )U(t2 , t1 ) = U(t3 , t1 ),

(12.33)

with t3 > t1 . Notice that, for t2 > x0 > t1 , we have n o n o T . . . U˜ (t2 , t1 )φI (x) . . . = T . . . U˜ (t2 , x0 )φI (x)U˜ (x0 , t1 ) . . . . (12.34)

157

12.3

The ground state of the interacting and the free theory

Our goal is to develop a formalism for the evaluation of Green’s functions hΩ| T {φ(xi ) . . . φ(xn )} |Ωi , in the interacting theory. As we have seen, their Fourier transforms will give us, with the LSZ reduction formula, the probability amplitudes for physical scattering processes. We remind that the points xµi in the above need to have time components which are slightly complex (Eq. 11.75). From the discussion of the previous section we have that hΩ| T {φ(xi ) . . . φ(xn )} |Ωi = n o ˜ 0 , x0 )φI (xi ) . . . φI (xn )U(x0 , τ ) |Ωi (12.35) hΩ| T U −1 (x01 , τ )U(x 1 n n

(12.36)

and |ψ0 i ≡ |Ωi. Eigenstates of the full Hamiltonian form a complete set, X 1 = |Ωi hΩ| + |ψn i hψn | . (12.37) n6=0

Let us now act with an operator to the vacuum state |0i of the free theory H0 , e−iH(τ +T )(1−iδ) e+iH0 (τ +T )(1−iδ) |0i , choosing T a large time in the future and δ a very small “dumbing” parameter δ → 0+ . This is in anticipation of needing to evaluate Green’s functions with slightly complex times. We have, e−iH(τ +T )(1−iδ) e+iH0 (τ +T )(1−iδ) |0i = e−iH(τ +T )(1−iδ) e+i0(τ +T )(1−iδ) |0i = e−iH(τ +T )(1−iδ) |0i (12.38)

158

−iE0 (τ +T )−E0 (τ +T )δ

= |Ωi e

hΩ| 0i +

X n6=0

|ψn i e−iEn (τ +T )−En (τ +T )δ hψn | 0i . (12.39)

Assuming a hierarchy En > E0 for the first energy level of the full Hamiltonian with respect to the ground energy, the factor e−En δ(τ +T ) vanishes faster than e−E0 δ(τ +T ) , as δ → 0+ . In this limit, we can write a simple relation between the ground state in the full theory and the vacuum in the free theory, e−iH(τ +T )(1−iδ) e+iH0 (τ +T )(1−iδ) |0i = |Ωi e−iE0 (1−iδ)(τ +T ) hΩ| 0i .

(12.40)

We now observe that the operator on the left side of the above equation is nothing else than U −1 (−T (1 − iδ), τ (1 − iδ)). We then have |Ωi = N U −1 (−T (1 − iδ), τ (1 − iδ)) |0i ,

(12.41)

hΩ| = N ∗ h0| U (T (1 − iδ), τ (1 − iδ)) ,

(12.42)

where N is a normalization constant, which we determine from the normalization condition hΩ| Ωi = 1, |N |2 = hΩ| U˜ (T (1 − iδ), −T (1 − iδ)) |Ωi

(12.43)

Substituting into Eq. 12.35, we have that hΩ| T {φ(xi ) . . . φ(xn )} |Ωi = |N |2 h0| U(T, τ ) n o −1 0 0 0 0 ˜ T U (x , τ )U(x , x )φI (xi ) . . . φI (xn )U(x , τ ) 1

U −1 (−T, τ ) |0i

159

(12.44)

Given that the times T > x0i > −T , we can put all operators under the timeordering symbol. We then obtain, the final result for the Green’s function hΩ| T {φ(xi ) . . . φ(xn )} |Ωi = lim δ→0

T →∞

h0| T φI (x1 ) . . . φI (xn )e−i h0| T e−i

−T

dt′ VI (t′ ) −T

dt′ VI (t′ )

|0i

, (12.45)

where the points xµi are computed in slightly complex times x0i (1 − iδ). Eq. 12.45 is an exact result, but it can serve as our basis for perturbation theory. The exponential can be written as a function of the perturbation Lagrangian with free fields. For our example interaction e−i

R∞

−∞

dt′ VI (t′ )

= e−i

λ d4 x 4! φI (x)4

(12.46)

and it can be expanded as a Taylor series in λ.

12.4

Wick’s theorem

In the previous section, we found that Green’s functions can be cast in a form suitable for applying the method of perturbation theory. For the example Hamiltonian of Eq. 12.5, we can write the result, hΩ| T {φ(x1 ) . . . φ(xn )} |Ωi = lim δ→0

T →∞

n R 4 P −iλ d xφI (x)4 |0i h0| T φI (x1 ) . . . φI (xn ) ∞ n=0 4! R , P −iλ 4 xφ (x)4 n |0i d h0| ∞ I n=0 4! (12.47) To evaluate the terms of the right hand side of Eq. 12.47, we need to be able to compute expectation values in the free-field theory of time-ordered products of field operators. This can be achieved by means of Wick’s theorem. We decompose the free scalar field φI into a term with a creation operator and a term with an annihilation operator, φI (x) = φ+ (x) + φ− (x), 160

(12.48)

with φ− (x) =

d3~k e−ik·x a(k), (2π)3 2ωk

(12.49)

φ+ (x) =

d3~k e+ik·x a† (k), (2π)3 2ωk

(12.50)

and

The time-ordered product of two scalar fields is, T {φI (x)φI (y)} = Θ(x0 − y 0 )φI (x)φI (y) + (xµ ↔ y µ) (

= Θ(x0 − y 0 ) φ+ (x)φ+ (y) + φ− (x)φ− (y) + φ+ (x)φ− (y) +φ− (x)φ+ (y)

)

+ (xµ ↔ y µ )

(

= Θ(x0 − y 0 ) φ+ (x)φ+ (y) + φ− (x)φ− (y) + φ+ (x)φ− (y) +φ+ (y)φ−(x) + [φ+ (x), φ− (y)] =

Θ(x0 − y 0 ) + Θ(y 0 − x0 )

)

+ (xµ ↔ y µ)

(

+φ+ (x)φ− (y) + φ+ (y)φ−(x)

φ+ (x)φ+ (y) + φ− (x)φ− (y)

)

+ [φ− (x), φ+ (y)] Θ(x0 − y 0 ) + [φ− (y), φ+(x)] Θ(y 0 − x0 ) (12.51) The sum of the theta functions in the first term is equal to one. We observe that in the curly bracket all field products appear with creation operators preceding annihilation operators. The curly bracket is then just the normal ordering of the product of the two field operators ( ) ...

=: φI (x)φI (y) :

(12.52)

The sum of the two terms in the last line is a known object, the FeynmanSt¨ uckelberg propagator, which is not an operator but a c−number. Using 161

that the relation [a(k), a† (k ′ )] = (2π)3 2ωk δ (3) (~k − ~k ′ ), we find that

(12.53)

d3 k e−ik(x−y) . (12.54) (2π)3 2ωk Combining the two terms together we have [φ− (x), φ+ (y)] Θ x0 − y 0 + [φ− (y), φ+(x)] Θ y 0 − x0 Z d3 k −ik(x−y) 0 0 −ik(y−x) 0 0 = e Θ(x − y ) + e Θ(y − x ) (2π)32ωk = h0| T {φI (x)φI (y)} |0i . (12.55) [φ− (x), φ+ (y)] =

We can then write that, T {φI (x)φI (y)} =: φI (x)φI (y) : + h0| T {φI (x)φI (y)} |0i .

(12.56)

We have proved that the time-ordered product of two free-field operators is the normal-ordering of the same product plus a c−function which is the propagator of the two-fields. This result generalizes easily to the time-ordered product of an arbitrary number of field operators. For three fields we have T {φI (x1 )φI (x2 )φI (x3 )} = + + +

: φI (x1 )φI (x2 )φ(x3 ) : h0| T {φI (x1 )φI (x2 )} |0i φI (x3 ) h0| T {φI (x1 )φI (x3 )} |0i φI (x2 ) h0| T {φI (x2 )φI (x3 )} |0i φI (x1 ) (12.57)

For four fields we have T {φI (x1 )φI (x2 )φI (x3 )φ(x4 )} = + + + + + +

: φI (x1 )φI (x2 )φ(x3 )φI (x4 ) : h0| T {φI (x1 )φI (x2 )} |0i : φI (x3 )φ(x4 ) : h0| T {φI (x1 )φI (x3 )} |0i : φI (x2 )φ(x4 ) : h0| T {φI (x1 )φI (x4 )} |0i : φI (x2 )φ(x3 ) : h0| T {φI (x1 )φI (x2 )} |0i h0| T {φI (x3 )φ(x4 )} |0i h0| T {φI (x1 )φI (x3 )} |0i h0| T {φI (x2 )φ(x4 )} |0i h0| T {φI (x1 )φI (x4 )} |0i h0| T {φI (x2 )φ(x3 )} |0i . (12.58) 162

In general, Wick’s theorem states that T {φI (x1 ) . . . φ(xn )} = = : φI (x1 ) . . . φ(xn ) + all contractions :

(12.59)

where a “contraction” means to replace one or more pairs of fields with their propagator. The theorem can be proved easily by induction. Notice that the normal ordered products in the expressions produced with Wick’s theorem are vanishing when bracketed with the vacuum, Y h0| : φI (xj ) : |0i = 0, (12.60) j

given that creation operators are placed before annihilation operators in the normal ordering and h0| a† (k) = a(k) |0i = 0. From Eqs 12.56-12.58 we derive the tautology h0| T {φI (x)φI (y)} |0i = h0| T {φI (x)φI (y)} |0i ,

(12.61)

and the more informative equations, h0| T {φI (x1 )φI (x2 )φI (x3 )} |0i = 0,

(12.62)

and h0| T {φI (x1 )φI (x2 )φI (x3 )φ(x4 )} |0i = h0| T {φI (x1 )φI (x2 )} |0i h0| T {φI (x3 )φ(x4 )} |0i + h0| T {φI (x1 )φI (x3 )} |0i h0| T {φI (x2 )φ(x4 )} |0i + h0| T {φI (x1 )φI (x4 )} |0i h0| T {φI (x2 )φ(x3 )} |0i . (12.63) Such equations admit a graphical representation. Let us represent the “two-point” correlation function with a straight line h0| T {φI (x1 )φI (x2 )} |0i =

163

(12.64)

The four point function is represented as 1

h0| T {φI (x1 )φI (x2 )φI (x3 )φ(x4 )} |0i =

contracted 2

(12.65)

It is then very easy to apply Wick’s theorem pictorially, simply by drawing all possible pairing of the points which appear in h0| T {φI (x1 )φI (x2 ) . . .} |0i.

12.5

Feynman Diagrams for φ4 theory

The pictorial application of Wick’s theorem yields to the representation of Green’s functions in the full theory as a perturbative expansion in terms of Feynman diagrams. As a concrete example, we shall consider the two-point function in the full theory, through order O(λ) in the coupling parameter λ. From Eq. 12.47 we have hΩ| T {φ(x1 )φ(x2 )} |Ωi R 4 4 d xφ (x) |0i + O(λ2 ) h0| T φI (x1 )φI (x2 ) 1 + (−iλ) I 4! = (−iλ) R 4 4 d xφI (x) |0i + O(λ2 ) h0| T 1 + 4! Z (−iλ) d4 x h0| T φI (x1 )φI (x2 )φI (x)4 |0i = h0| T {φI (x1 )φI (x2 )} |0i + 4! Z (−iλ) − h0| T {φI (x1 )φI (x2 )} |0i d4 x h0| φI (x)4 |0i + O(λ2 ) (12.66) 4! Let us compute pictorially the second term in the expansion, Z (−iλ) d4 x h0| T φI (x1 )φI (x2 )φI (x)4 |0i 4! 164

Figure 12.1: The term h0| T {φI (x1 )φI (x2 )φI (x)4 } |0i as the sum of all contractions of the graph above The corresponding Green’s function is equal to the contractions of the graph in Fig 12.5. We find two different types of contractions which lead to two different “Feynman diagrams”. • Diagram I: The two fields defined at the “external” points x1 and x2 get contracted with each other. The four fields defined at the “internal” point x contract then among themselves.

This Feynman diagram is classified as “disconnected”, meaning that the internal point x is not connected to any of the external points x1 and x2 . This configuration occurs in more than one ways. The external fields at x1 and x2 can be contracted in 1 way. Take now one of the four fields at the internal point x. This field can be contracted with any of the remaining internal fields with 3 ways. Finally, the remaining two internal fields can be contracted in 1 way only. Therefore, this Feynman diagram appears 1 × 3 × 1 times. • Diagram II: The two external fields defined at x1 and x2 contract with internal fields at x.

165

This is a “connected” diagram, meaning that internal points are connected to external points by following the lines of the graph. We can now compute the multiplicity of the Feynman diagram. There are 4 ways that we can contract the field at the first external point with one of the fields at the internal point x. The second external point can then be contracted to another internal field in three ways. Finally, two remaining internal fields can be contracted with each other in one way. Therefore, this Feynman diagram occurs 4×3×1=

4! times. 2

Now we examine the last term (−iλ) h0| T {φI (x1 )φI (x2 )} |0i − 4!

d4 x h0| φI (x)4 |0i

in Eq. 12.66, which is the contribution of the denominator in Eq. 12.47 to the λ Taylor expansion. It is easy to convince ourselves that this term yields the same Feynman diagram as the disconnected Diagram I. The two external fields are contracted together. The four internal fields are also contracted among themselves. However, this terms carries an overall minus sign. As a result, the disconnected Feynman diagram drops out from the result. This observation holds at all orders in perturbation theory. Disconnected Feynman diagrams do not contribute to the perturbative expansion of Green’s functions 1 . These factorize in the numerator of Eq. 12.47 and cancel exactly against the denominator. In summary, the perturbative expansion of the two-point Green’s function through order O(λ) can be represented very simply with the following Feynman diagrams. hΩ| T {φ(x1 )φ(x2 )} |Ωi = 2 +

1 1 2

+ ...

(12.67)

We can obtain a concrete mathematical expression from the above diagrams by following very simple rules, pioneered by Feynman: 1

The general proof of this statement is easier with the path integral formalism, and we postpone it for QFTII.

166

1. We associate a propagator to each line 1

→ h0| T {φI (x1 )φI (x2 )} |0i

(12.68)

2. We associate a factor of (−iλ) and a space-time integration to each vertex Z → (−iλ) d4 x (12.69)

These rules give us a simple pictorial representation in terms of Feynman diagrams of the perturbative series for any Green’s function G(x1 , x2 , . . . , xN ). We draw all possible graphs with N external and with no more vertices than the maximum order in the perturbative expansion λn that we require. Then we translate the Feynman diagrams into mathematical expressions using the rules above. One difficulty in this procedure is to determine the combinatorial rational factor that multiplies the diagram, as for example the factor 1/2 in the second diagram of Eq. 12.67. This number can always be obtained following the procedure in Eq. 12.67, which is a rather brute force method. For everything practical, this method is sufficient and for more complicated cased we can easily program it in a computer code. The combinatorial factor, also known as symmetry factor, can be determined cleverly as the inverse of the independent exchange symmetry operations of the Feynman diagram (exchanging vertices and propagators) which leave it intact. This is a nice method for everyone who is confident in spotting all symmetries without double counting equivalent ones. It is, however, prone to human errors.

12.6

Feynman rules in momentum space

The LSZ reduction formula expresses scattering amplitudes as truncated Green’s functions in momentum space. We can develop simple Feynman rules for computing them directly. We start with the Fourier transform of the propagator from the origin to a point x. Z ˜ G(p) = d4 xeip·x hΩ| T {φ(x)φ(0)} |Ωi (12.70)

167

At leading order in perturbation theory we have, Z ˜ G(p) = d4 xeip·x h0| T {φI (x)φI (0)} |0i + O(λ) Z Z d4 k i 4 ip·x = d xe e−ikx + O(λ) 4 2 2 (2π) k − m + iδ Z Z 4 dk i = d4 xei(p−k)·x + O(λ) (2π)4 k 2 − m2 + iδ Z i d4 k (2π)4 δ (4) (p − k) + O(λ) = 4 2 (2π) k − m2 + iδ ˜ ; G(p) =

i + O(λ) p2 − m2 + iδ

(12.71)

It is interesting to look at the transition amplitude for a particle with momentum p to a particle with momentum p′ . The corresponding Green’s function is Z Z ′ ·y ′ 4 +ip ˜ (p; p ) = d ye d4 xe−ip·x hΩ| T {φ(x)φ(y)} |Ωi , (12.72) G At leading order in perturbation theory gives (exercise) ˜ (p; p′ ) = (2π)4 δ (4) (p − p′ ) G

i + O(λ). − m2 + iδ

(12.73)

Notice the delta function, which simply imposes momentum conservation. Execrise: What is the corresponding “truncated” Green’s function? A more complicated case is the scattering of four particles. The Green’s function we require is G(x1 , x2 , x3 , x4 ) =

1 2

4 + 3 Ga

+ 2

3 Gb

2 Gc

+...

1 2 x Gd

+ 3

1 2

(12.74) 168

Let us now compute the Green’s function in momentum space, Z ˜ 3 , p4 ; p1 , p2 ) = d4 x1 e−i(x1 ·p1 +x2 ·p2 −x3 ·p3 −x4 ·p4 ) G(x1 , x2 , x3 , x4 ), (12.75) G(p where we consider p1 , p2 in the initial state and p3 , p4 in the final state. We write, ˜ 3 , p4 ; p1 , p2 ) = Ga + Gb + Gc + Gd + Ge + . . . , G(p

(12.76)

where Ga,b,c are the contributions of the first three diagrams and Gd , Ge are the contributions of the fourth and fifth diagram at order O(λ) and O(λ2 ). The first three diagrams give, i i (2π)4 δ 4 (p2 − p3 ) 2 2 − m + iδ p2 − m2 + iδ i i 4 4 + (2π)4 δ 4 (p1 − p3 ) 2 (2π) δ (p − p ) 2 4 2 p1 − m2 + iδ p2 − m2 + iδ i i (2π)4 δ 4 (p3 + p4 ) 2 + (2π)4 δ 4 (p1 + p2 ) 2 2 p1 − m + iδ p3 − m2 + iδ + (12.77)

Ga + Gb + Gc = (2π)4 δ 4 (p1 − p4 )

p21

The fourth diagram is converted into a mathematical expression following again the Feynman rules of the previous section. We have Z Gd = d4 x1 d4 x2 d4 x3 d4 x4 e−i(x1 ·p1+x2 ·p2−x3 ·p3 −x4 ·p4 ) Z × (−iλ) d4 x ( vertex ) × × × ×

h0| T h0| T h0| T h0| T

{φI (x1 )φI (x)} |0i {φI (x2 )φI (x)} |0i {φI (x3 )φI (x)} |0i {φI (x4 )φI (x)} |0i

( ( ( (

contraction contraction contraction contraction

of of of of

x1 x2 x3 x4

and and and and

x x x x

) ) ) ).(12.78)

Substituting the expression for h0| T {φI (x3 )φI (x)} |0i Performing the x and xi integrations we find, Gd = (−iλ)

p21

i i i i (2π)4 δ 4 (p1 +p2 −p3 −p4 ). 2 2 2 − m2 + iδ p2 − m2 + iδ p3 − m2 + iδ p4 − m2 + iδ (12.79) 169

The diagram Ge is a “loop diagram”. Let us write the corresponding expression, using the Feynman rules in position space. We have Z Ge = d4 x1 d4 x2 d4 x3 d4 x4 e−i(x1 ·p1 +x2 ·p2 −x3 ·p3 −x4 ·p4) Z × (−iλ) d4 xd4 y ( vertices ) × × × × ×

h0| T {φI (x1 )φI (x)} |0i h0| T {φI (x2 )φI (x)} |0i h0| T {φI (x3 )φI (y)} |0i h0| T {φI (x4 )φI (y)} |0i h0| T {φI (x)φI (y)} |0i2

( contraction of x1 and x ) ( contraction of x2 and x ) ( contraction of x3 and y ) ( contraction of x4 and y ) ( two contractions of x and y ) (12.80)

Performing all x−space integrations we find, Ge =

i i i i 2 2 2 − m2 + iδ p2 − m2 + iδ p3 − m2 + iδ p4 − m2 + iδ ×(2π)4 δ 4 (p1 + p2 − p3 − p4 ) ×(−iλ)2 Z +∞ 4 dk i i × . 4 2 2 2 2 −∞ (2π) k − m + iδ (k + p1 + p2 ) − m + iδ p21

(12.81)

Observe our final expressions for Ga, b, c, Gd and Ge. We can make easy rules (Feynman rules) to produce them from the Feynman diagrams. • In every vertex we pick up a factor (−iλ), • With each propagator comes a factor p2

i . − m2 + iδ

• At each vertex, we have delta functions guaranteeing that momentum is conserved. As a result, for all particles that are connected with each 170

other there is an overal factor (2π)4 δ (4)

X in

pin −

X out

pout ,

which forces the sum of incoming momenta in a connected diagram to be equal to the sum of the outcoming momenta. • The loop in diagram Ge introduces an integration over a “loop momentum” and a measure Z +∞ 4 d k 4 −∞ (2π) .

12.7

Truncated Green’s functions in perturbation theory

Figure 12.2: Feynman diagrams which are correction to external lines and do not contribute to physical scattering amplitudes. We are interested in computing amplitudes for physical scattering processes, as given by the LSZ formula. For the scattering of two particles with momenta p1 , p2 to two particles with momenta p3 , p4 we then need a

171

truncated Green’s function

˜ trunc (p3 , p4 ; p1 , p2 )

hp3 , p4 | p1 , p2 i = Z˜φ2 G

(12.82)

p2i =m2phys

compute for squared external momenta equal to the physical mass of the particles. The truncated Green’s function is defined as ˜ trunc (p3 , p4 ; p1 , p2 ) = G

˜ 3 , p4 ; p1 , p2 ) G(p , ˜ 1 )G(p ˜ 2 )G(p ˜ 3 )G(p ˜ 4) G(p

(12.83)

where the Green’s function in momentum space is divided with the propagator of each of the particles in the initial and final states. The K¨ahlen-Lehmann representation for the pole of the propagator is ˜ G(p) =

iZ˜ + regular terms p2 − m2phys

(12.84)

In the previous section we computed the perturbative result, ˜ G(p) =

i + O(λ). p2 − m2

(12.85)

At leading order in perturbation theory, we can then identify the physical mass of a particle mphys with the mass parameter m of our Lagrangian, and determine the normalization Z˜φ = 1. We then find that the diagrams Ga , Gb , Gc which do not connect all eternal particles with each other do not contribute to the scattering amplitude hp3 , p4 | p1 , p2 i. We find for example that, at leading order in λ,

2 2 2 2 = 0. ∼ (p3 − m )(p4 − m )δ(p4 − p1 )δ(p3 − p2 )

˜ ˜ ˜ ˜

2 2 G(p1 )G(p2 )G(p3 )G(p4 ) pi =m

(12.86) i The diagrams Gd and Ge connect all external particles and have poles p2 −m 2 i for each one of them. These contribute to the scattering amplitude. We can perform the division with the external propagators as it is demanded by the LSZ formula we need three more rules in our set of Feynman rules. 172

• Lines starting from an external point contribute a factor 1 (and not i/(p2 − m2 )). We must be, though, a bit more careful. Eq. 12.85 has perturbative corrections of order λ. We need to divide our Feynman diagrams with the full perturbative expansion of the propagator and not just the leading order contribution. We can account for this division entirely with one more clever trick. • Through away all Feynman diagrams which “correct” external lines. For example, the diagrams of Fig. 12.2 should all be disregarded when computing truncated Green’s functions for physical scattering amplitudes.

173

Chapter 13 Loop Integrals In order to compute the perturbative expansion of any Green’s function, we must perform unrestricted integrations over the momenta of particles circulating in loops of Feynman diagrams. The exact evaluation of loop integrals is a difficult and often prohibitive task. Many loop integrals are divergent! This is a very unpleasant surprise if we aim towards a realistic description of physical phenomena with finite probabilities. The problem of divergences is a very serious one, and it jeopardizes the mathematical foundation of the perturbative method.

13.1

The simplest loop integral. Wick rotation

Let us consider the simplest loop-integral in field theory, Z d4 k 1 I≡ 4 2 (2π) k − m2 + iδ

(13.1)

This is a formidable integral, given that it requires four integrations for each space-time dimension. We notice that only the combination k 2 = k02 − ~k 2 appears in the integrand. The integral would be easily solvable had we had a Euclidean metric for kE2 = k02 + ~k 2 , by using spherical coordinates in four dimensions. With a Minkowski metric, we ought to treat the time integration specially. We write Z +∞ 3~ Z +∞ dk 1 I = lim , (13.2) dk0 4 δ→0+ −∞ (2π) (k0 − ωk + iδ) (k0 + ωk − iδ) −∞ 174

with

q ωk = ~k 2 + m2 .

(13.3)

The integrand has two poles, at k0 = ωk − iδ and k0 = −ωk + iδ, which are found in the lower-right and upper-left quarter planes in the (ℜk0 , ℑk0 ) plane We can perform the so called Wick rotation, rotating the integration

Figure 13.1: Wick’s rotation: The integrand has no poles on the upper-right and lower-left k0 complex plane. We can rotate the integration axis by 90o , and integrate along the imaginary axis ko ∈ [−i∞, +i∞]. axis by 90 degrees and integrating over the imaginary axis, Z

+∞ −∞

d3~k (2π)4

+i∞

dk0

−i∞

1 k02 − ~k 2 − m2 + iδ

(13.4)

Setting k0 = ik0′ , the integral becomes I=i

+∞

−∞

d3~k (2π)4

+i∞

−i∞

dk0′ 175

1 −(k0′ )2

− ~k 2 − m2 + iδ

(13.5)

Notice that after Wick’s rotation we can consider the four-vector kE ≡ (k0′ , ~k) in Euclidean space, and express kE in four-dimensional spherical coordinates: dkE = d|k| |k|3dΩ4 .

(13.6)

The integral then becomes I = −iΩ4

∞

d|k|

|k|3 . |k|2 + m2 − iδ

(13.7)

Notice that the integral diverges as |k| → ∞, which is an embarrassment. Infinities were historically the biggest puzzle in the development of quantum field theory. But, we now know what to do with them for many Lagrangian systems which are realistic descriptions of nature. We will learn later that we can “sweep infinities under the carpet”, exploiting certain “freedoms” that we have. We are allowed to absorb infinities in the the normalization constants Zφ of the LSZ formula, or in the definition of physical mass and coupling parameters in terms of the corresponding parameters appearing in the Lagrangian. These are very few “freedoms”, and it is amazing that they suffice to solve the problem. The method of “sweeping infinities under the carpet” is called renormalization. A prerequisite to it is to quantify the infinities with the help of a regulator. We shall introduce a parameter which renders the integrations well defined and finite except at a certain limit. For example, we can restrict the integration in Eq. 13.7 below a certain cut-off value Λ for |k|, 2 Z Λ |k|3 Λ I(Λ) ≡ −iΩ4 = −iΩ4 + ... (13.8) d|k| 2 |k| + m2 − iδ 2 0 The regularized integral I(Λ) is equal to the original integral I in the limit of an infinite Λ, lim I(Λ) = I. (13.9) Λ→∞

What is the benefit of introducing a regularization method? It will give us the possibility to compute the divergent parts of all loop-integrals that enter the evaluation of S−matrix elements for values of the regulator Λ where they are all well-defined. Then, we can perform “renormalization” and express the parameters of the Lagrangian in terms of physical mass and coupling parameters. Then we shall take the limit of the regulator which corresponds to the original loop integrals. It will turn out that for many field theories the final renormalized S-matrix elements are finite. 176

13.2

Dimensional Regularization

A method to regularize loop integrals is not unique. A well chosen regularization procedure can be very beneficial for practical computations, since the difficulty in carrying out the integrations may vary significantly. What is more important, some regularization methods may violate some of the symmetries of the theory. For example, a “cut-off” regularization violates Lorentz invariance. When symmetries are broken due to the regularization method it may be required to pay additional efforts in order to relate S-matrix elements to physical observables. The most elegant regularization method known to date was developed by t’Hooft and Veltman in the 70’s. It is called dimensional regularization. It has revolutionized the evaluation of loop integrals for its simplicity. It is also known to preserve most of the symmetries which are found in physical Lagrangian systems. In dimensional regularization, the regulator is the number of space-time dimensions d. We shall explain how this work, by revisiting the simplest example of a loop integral in Eq. 13.7. Let us now compute this integral in an arbitrary number of dimensions, Z 1 dd k , (13.10) Id = d 2 (2π) k − m2 + iδ which after Wick’s rotation becomes Z dd k 1 Id = i , (2π)d −k 2 − m2 + iδ and

k 2 = k02 + ~k 2 .

(13.11)

(13.12)

In spherical coordinates, iΩd Id = (2π)d

∞

k d−1 . −k 2 − m2 + iδ

For infinitely large loop momenta it behaves as, Z ∞ iΩd dkk d−3 . Id → − (2π)d 177

(13.13)

(13.14)

We observe that the integral has a finite ultraviolet limit (k → ∞) for values of the dimension d < 3. In dimensional regularization we compute the integral for a generic value of the dimension d which is allowed for it in in order to be well-defined. At the end of our computation, we shall perform an “analytic continuation” to a physical dimension value d = 4.

13.2.1

Angular Integrations

For d = 2, the solid angle is dΩ2 = dφ, for d = 3, we have dΩ3 = sin θdθdφ, and so on. We are used to defining solid angles for an integer number of dimensions. What is the value of the solid angle Ωd for an arbitrary real valued dimension d? Our generalization of the dimension from integer to real values for the solid angle is mathematically similar to the generalizing of the factorial to the complex-valued Γ function, n! → Γ(z), and relies on finding a defining recurrence relation. Let us take a d−dimensional vector ~k (d) = ~k (d−1) , kd ,

(13.15)

where kd is the component corresponding to the dth dimension. We shall assume that we know how to express the ~k (d−1) in terms of (d − 1) polar coordinates, i.e. a radial rd−1 coordinate and (d − 2) angular coordinates. Then Z +∞ Z +∞ Z ∞ Z d−2 dk1 . . . dkd = dkd dr(d−1) r(d−1) dΩd−1 . (13.16) −∞

−∞

The d−dimensional vector is now expressed in cylindrical coordinates, ~k (d) = kd−1 uˆ(d−1) (θ1 , θ2 , . . . θd−2 ) , kd , (13.17) where uˆ(d−1) is a unit vector. Now we change variables to polar coordinates in d−dimensions by performing the transformation, r(d−1) = l sin θd−1 . kd = l cos θd−1 178

(13.18)

This gives for the integration measure Z

+∞

dkd

−∞

∞

d−2 dr(d−1) r(d−1)

dΩd−1 =

dd k, ∞

dll

d−1

d−2

dθd−1 sin

θd−1

dΩd−1 .

(13.19)

We have then arrived to the recurrence relation, Z Z π Z d−2 dΩd = dθd−1 sin θd−1 dΩd−1 .

(13.20)

Recall that for d = 2, Z

dΩ2 =

2π

dθ1 .

(13.21)

Using this as a boundary condition for Eq. 13.20, we obtain the solid-angle in arbitrary integer dimensions, Z Z π Z π Z 2π d−2 dΩd = dθd−1 sin θd−1 . . . dθ2 sin θ2 dθ1 . (13.22) 0

All integrals can be performed easily using that, Z π Z 1 x−1 x dθ sin θ = d(cos θ)(1 − cos θ) 2 0 −1 Z 1 x+1 1 = daa 2 −1 (1 − a) 2 −1

(a = cos2 θ)

;

dθ sinx θ = B 0

1 1+x , 2 2

The the solid angle in d−dimensions is

1 2

x+1 2 x+2 2

(13.23)

π2 Ωd = 2 d . Γ 2

(13.24)

On the rhs we have a function which is defined, as we shall discuss immediately, to arbitrary complex values of the dimension parameter d, with the exception of non-positive integers. We thus have derived a generalization of the solid angle to an arbitrary such value of d. 179

A quicker derivation exponential integral,

of Eq. 13.24 proceeds as follows. Consider the Z

∞

dxe−x =

√

π.

(13.25)

−∞

For d such integrals we have, Z Z Z Pd d 2 d − x d xe i=1 i = dΩd π2 =

∞

dxxd−1 e−x .

(13.26)

d . 2

(13.27)

Setting t = x2 , we find 1 π = Ωd 2 d 2

13.2.2

∞

dtt

d −1 2

−t

1 = Ωd Γ 2

Properties of the Gamma function

In the above, the Beta function is defined as Z 2 B(x, y) = 1dξxix−1(1 − ξ)y−1 .

(13.28)

The Γ−function is defined through the integral representation, Z ∞ Γ(z) ≡ dxxz−1 e−x .

(13.29)

By using integration by parts, we can easily show that Γ(z + 1) = zΓ(z),

(13.30)

which is the same recursion relation as for the factorial of integers. For z = n a positive integer, we have Γ(n) = (n − 1)!

(13.31)

The function Γ(z) is analytic except for z = 0, −1, −2, . . .. From the integral representation we conclude that the Γ function is convergent for Re z > 0. But, the recursion relation of Eq. 13.30 defines it for all complex values of z, with the exception of non-positive integers. 1

shown in class

180

A useful value is

√ 1 Γ = π. 2

(13.32)

We can also prove the identity, x x + 1 1 x−1 =2 Γ Γ . Γ(x)Γ 2 2 2

(13.33)

We shall often need to expand a Gamma function around an integer value for its argument. We have Z ∞ Γ(1 + ǫ) = xǫ e−x Z0 ∞ = eǫ ln x e−x 0 Z ∞ ǫ2 2 = 1 + ǫ ln x + ln x + . . . e−x 2 0 = 2 γ π2 2 ; Γ(1 + ǫ) = 1 − ǫγ + ǫ + O(ǫ3 ). (13.34) + 2 12 The “Euler-gamma” constant is defined as Z ∞ γ≡− dx log(x) ≈ 0.544 . . .

(13.35)

13.2.3

Radial Integrations

We now continue with the remaining integration of Eq. 13.14 over the magnitude of the Euclidean four-momentum. We have Z k d−2 i Ωd dk . (13.36) Id = − 2 (2π)d k 2 + m2 − iδ We perform the change of variables x k2 → m2 − iδ , 1−x

(13.37)

which yields

d −1 i Ωd m2 − iδ 2 Id = − d 2 (2π)

181

dxx 2 −1 (1 − x)− 2

(13.38)

which yields the final result, Z d d2 −1 d k 1 d 2 m − iδ = −Γ 1 − d 2 iπ 2 k 2 − m2 + iδ

(13.39)

Notice that the integral is divergent for d = 2, 3, 4, . . ., but it is well-defined for any other value of d. Exercise: Prove that Z d d d2 −ν Γ ν − d k 1 2 ν 2 m − iδ = (−1) . (13.40) ν d Γ(ν) iπ 2 [k 2 − m2 + iδ] An interesting case is when m2 = 0. In dimensional regularization, where the dimension parameter is considered non-integer, we have that lim md = 0d = 0.

m→0

(13.41)

We then have that integrals with no mass scales vanish, and in our examplary case, Z d d k 1 (13.42) ν = 0. d 2 iπ 2 [k + iδ]

13.3

Feynman Parameters

Consider a more complicated integral. For example, Z d 1 d k . I2 = d 2 2 iπ 2 [k − m + iδ] [(k + p)2 − m2 + iδ] The integrand has now four poles for q q 2 2 2 2 ~ ~ k + m − iδ , ± (k + p~) + m − iδ . k0 = ±

(13.43)

(13.44)

It is now more difficult to perform the integration over k0 , as well as the angular integrations since the denominator depends explicilty on angles, (k + p)2 − m2 = k 2 + p2 − m2 + 2k0 p0 − 2|~k||~p| cos θ.

(13.45)

The method of Feynman parameters allows to integrate out the loop momentum k easily, applying Wick’s rotation and integrating over angles in 182

exactly the same manner as for the simplest integral of Eq. 13.40. The price that we have to pay is to introduce new integration variables over “Feynman parameters”. It is easy to prove that, Z 1 1 1 . (13.46) = dx AB [xA + (1 − x)B]2 0 We can use such identity in order to combine denominators in loop intergrals under one term. For example, the integral of Eq. 13.43 can be written as Z d d k 1 I2 = d 2 2 2 iπ 2 [x ((k + p) − m ) + (1 − x) (k 2 − m2 ) + iδ] Z d d k 1 = d 2 2 2 iπ 2 [k − m + 2xk · p + xp2 + iδ] Z d 1 d k (13.47) = d 2 iπ 2 [(k + px)2 − m2 + x(1 − x)p2 + iδ] We perform a shift in the integration variable k, usually called “loop momentum”, defining K µ = k µ + xpµ . (13.48) We then have, I2 =

dd k iπ

d 2

[K 2

−

(m2

1 − x(1 − x)p2 − iδ)]2

(13.49)

The integral over K µ can be solved using the general result of Eq. 13.40, Z 1 d −2 d (13.50) I2 = Γ 2 − dx m2 − x(1 − x)p2 − iδ 2 2 0 The above integral is only one-dimensional. We still have the task of performing the task of integrating over Feynman parameters. Feynman parameter integrals are generalized hypergeometric functions, which are only partially understood in the mathematical science. Said differently, the mathematics of hypergeometric functions is far from sufficient in order to tackle the Feynman integrals that appear in the perturbative expansion of Green’s functions. This limits computations to simple cases. Nevertheless, even with the mathematical understanding of loop integrations being in its infancy, we 183

can still derive rather precise theoretical predictions for S-matrix elements of interesting scattering processes. For physics predictions, we are interested in the value of loop integrals as a Laurent expansions around the physical value of the dimension parameter, d = 4. It is customary to write, d = 4 − 2ǫ,

(13.51)

and expand around ǫ = 0. The (−) sign is because we usually think of the dimension parameter as being smaller than four, in order to make an integral convergent in the UV limit |k| → ∞. The factor of 2 in front of ǫ is convenient due to that typically the combination d2 emerges as result of loop-integrations. The integral of Eq. 13.50 becomes Z 1 ǫ I2 = Γ(ǫ) dx m2 − x(1 − x)p2 − iδ 0 Z 1 1 −γ+ dx log m2 − x(1 − x)p2 − iδ + O (ǫ) . (13.52) ǫ 0 Exercise: Solve the above integral, assuming that p2 < 4m2 . For p2 > 4m2 the integral develops an imaginary part. Compute it, by performing an analytic coninuation of the result for p2 < 4m2 . The procedure of combining denominators in a single term using Feynman parameters and making complete squares of the loop momenta can be performed in general. For an arbitrary number of denominators in a loopintegral we can use the formula, ! Z X 1 Γ (ν1 + . . . νn ) 1 xν11 −1 . . . xνnn −1 P Q = dx . . . dx δ 1 − x P 1 n i νi Aν11 . . . Aνnn Γ(νi ) [ x A ] 0 i i i (13.53) Proof: Let us perform the change of variables xi =

ai 1 + an

in the rhs of Eq. 13.53. The delta function becomes, ! Pn−1 ! n X 1 − i=1 ai ai =δ = (1+an )δ δ 1− 1 + a 1 + a n n i=1 184

(13.54)

1−

n−1 X i=1

. (13.55)

The parameters ai are constrained to be in the interval [0, 1], due to the δ function for i = 1 . . . n − 1, while the parameter an ranges in between an = 0(xn = 0) and an = ∞(xn = ∞). The Jacobian of the transformation is dai dxi = , i 6= n, (13.56) 1 + an and dan dxn = . (13.57) (1 + an )2 All factors conspire in Eq. 13.53, so that the integrand is almost the same as the original, replacing xi with ai . The only difference is that the delta fucntion does not contain an anymore, and that the range of this variable is from 0 to infinity. Explicitly, Γ (ν1 + . . . νn ) Q Γ(νi ) ! Z ∞ Z 1 n−1 X aν11 −1 . . . aνnn −1 P (13.58) × dan da1 . . . dan−1 δ 1 − ai P νi [ a A ] 0 0 i i i=1

rhs of Eq. (13.53) =

We now change once more variables to Pn−1 x i=1 ai Ai an = , 1−x An

(13.59)

and perform the integration over x, which ranges from 0 to 1. The result is 1 × Aνnn

rhs of Eq. (13.53) = Γ (ν1 + . . . νn−1 ) × Γ(ν1 ) . . . Γ(νn−1 )

dx1 . . . dxn δ 1 −

X i

x1ν1 −1 . . . xnνn −1 P P [ xi Ai ] νi (13.60)

We have now factorized the term 1/Aνnn out of an integral, which is now cast in the same form as the original but with n − 1 Feynman parameters. Obviously, we can repeat the same procedure as many times as needed in order to factorize all A1νi terms, as in the lhs of Eq. 13.53. i

185

Chapter 14 Quantum Electrodynamics One of the most amazing successes of Quantum Field theory, is the precision in which it describes the interaction of radiation and matter. The rules governing these interactions are very simple, and are given by the theory of Quantum Electrodynamics (QED). What is also astonishing, emerges naturally by combining two very simple principles: symmetry and localilty. The QED Lagrangian is symmetric under phase-redefinitions (U(1) transformations) of the electron field which are local, i.e. we can choose the symmetry trnsformation parameters differently at each space-time point. Local symmetry transformations govern all quantum field theories, QED, QCD and the Standard Model of electroweak interactions, which describe the three forces of nature other than gravity. These theories describe physical phenomena extremelly well within the experimentally accessible accuracy.

14.1

Gauge invariance

We consider a free electron field: ¯ (i6∂ − m) ψ(x). L = ψ(x)

(14.1)

As we have discussed already, this Lagrangian is invariant under a U(1) transformation, ψ(x) → ψ ′ (x) = exp(ieθ)ψ(x), (14.2) if the phase θ is global, i.e. it is chosen to be the same at avery point in ∂θ =0 . space-time ∂x 186

Let us now take a bold step and ask whether we can have a a different phase θ at every space-time point xµ , U(x) = exp (ieθ(x)) .

(14.3)

The Lagrangian is no longer invariant, ieθ ¯ −ieθ ∂ ψ6¯∂ ψ → ψ6¯∂ ψ + ψe 6 e ψ.

(14.4)

The problem is that the space-time derivative does not transform simply under the local U(1) transformation any more, 6∂ ψ → eieθ 6∂ ψ + something else . If we insist on locality, we shall need to modify the derivative so that it transforms more conveniently. We will look for a new derivative which, under a local U(1) transformation transforms as: 6D ψ → U(x)6D ψ.

(14.5)

The simplest modification we can think of, is to add to the space-time derivative a space-time function, ∂ µ → D µ = ∂ µ + functionµ

(14.6)

which transforms in the same way as the space-time derivative under Lorentz transformations, i.e. it transforms as vector. A function of space-time, in field theory, is nothing else but a field. We are then proposing to modify the definition of a derivative by adding to it a vector field. We write a “covarian derivative”, Dµ = ∂µ − ieAµ (x), (14.7) where the −ie factor is conventional. The vector field Aµ (x) must have a very specific trasformation which we can find, in order for the modified derivative to transform simply. We require that

Dµ ψ(x) → Dµ′ ψ ′ = U(x)Dµ ψ ; ∂µ − ieA′µ (U(x)ψ) = U(x) (∂µ − ieAµ ) ψ ; U(x)∂µ ψ + [∂µ U(x)] ψ − ieA′µ U(x)ψ = U(x)∂µ ψ − ieAµ U(x)ψ i ; A′µ = Aµ − U −1 (x)∂µ U(x) (14.8) g 187

This is an astonishing result. We find that the vector field transforms as, A′µ = Aµ + g∂µ θ,

(14.9)

which is a gauge transformation, the transformation of the photon field. In other words, if we would like that a Lagrangian of electrons is invariant under local U(1) transformations, then the same Lagrangian must describe a photon field. The covariant derivative transforms as:

= = = =

Dµ → Dµ′ = ∂µ − ieA′µ i −1 ∂µ − ie Aµ − U ∂µU g −1 ∂µ − ieAµ − U (∂µ U) ∂µ − ieAµ + U ∂µ U −1 U(x) (∂µ − ieAµ ) U −1 (x)

(14.10)

Therefore: Dµ → Dµ′ = U(x)Dµ U −1 (x)

(14.11)

We can now replace the free Lagrangian of the spin-1/2 field with a new Lagrangian which is symmetric locally, L = ψ¯ [i6D − m] ψ ¯ −1 U [i6D − m] U −1 Uψ → ψU ψ¯ [i6D − m] ψ. If Aµ is a physical field, we need to introduce a kinetic term in the Lagrangian for it. We will insist on constructing a fully gauge invariant Lagrangian. To this purpose, we can use the covariant derivative as a building block. Consider the gauge transformation of the product of two covariant derivatives: Dµ Dν → Dµ′ Dν′ = UDµ U −1 UDν U −1 = UDµ Dν U −1 .

This is not a gauge invariant object. Now look at the commutator: [Dµ , Dν ] → Dµ′ , Dν′ = U [Dµ , Dν ] U −1 188

(14.12)

This is gauge invariant. To convince ourselves we write the commutator explicitly: [Dµ , Dν ] = (∂µ − ieAµ ) (∂ν − ieAν ) − [µ ↔ ν] = ∂µ ∂ν − ie (∂µ Aν ) − ieAν ∂µ − ieAµ ∂ν + (ie)2 Aµ Aν − [µ ↔ ν] = −ie [∂µ Aν − ∂ν Aµ ] . (14.13) Inserting Eq. 14.13 into Eq. 14.12, we find that the commutator of covariant derivatives (in the abelian case) is gauge invariant. We have also found that it is proportional to the field strength tensor of the gauge (photon) field: Fµν =

i [Dµ , Dν ] = ∂µ Aν − ∂ν Aµ e

(14.14)

We now have invariant terms for a Lagrangian with an “electron” and a “photon” field. The classical Lagrangian for QED reads 1 L = ψ¯ (i6D − m) ψ − Fµν F µν . 4

(14.15)

This Lagrangian accounts for the majority of the phenomena that we experience in nature. It is a beautiful synthesis of locality and symmetry! Exercise: Find the Noether current and conserved charge due to the invariance under the U(1) gauge transformation of the QED Lagrangian.

14.2

Perturbative QED

Let us take the QED Lagrangian of Eq. 14.15 and substitute the expression for the covariant derivative. We have, L = ψ¯ [i6∂ − m] ψ (free electron field) 1 + − Fµν F µν ( free photon field) 4 + eψ6¯Aψ (electron-photon interaction) (14.16) The last term, which couples the electron and fermion fields, differentiates the QED Lagrangian from the Lagrangian of two free electron and photon fields. Can we solve the energy eigenstates of the QED Hamiltonian? No, 189

unless we resort to perturbation theory. The strength of the photon-electron interaction is very small. We can consider the last term a small perturbation, and expand probability amplitudes around e = 0. For e = 0, we should recover the results from the quantization of the free photon and free electron fields. We then encounter the same problems as when we quantized the free electromagnetic field, which required to fix the gauge for the field Aµ . We then add to the QED Lagrangian a term, Lgauge−f ix = −

1 (∂µ Aµ )2 . 2ξ

(14.17)

In this way, we can at least be sure that the theory is correctly defined for e = 0. It is not obvious that this modification will solve the problem of obtaining a consistently quantized theory at higher orders in perturbation theory. It actually does! But the justification of this statement will be a very important topic of study in QFTII. As a primitive diagnostic, we shall consider ξ an arbitrary parameter in our calculations. We will then check that higher order corrections do not modify it. We then have, LQED = Le + Lγ + Lint , (14.18) with

and

Le = ψ¯ [i6∂ − m] ψ,

(14.19)

1 (∂µ Aµ )2 , Lγ = − Fµν F µν − 4 2ξ

(14.20)

Lint = eψ6¯Aψ.

(14.21)

We can develop a similar formalism for perturbation theory as in the case of a scalar field theory. This leads to the derivation of Feynman rules for S−matrix elements for QED transition amplitudes. Essentially, we can picture QED processes as photons and electrons propagating in between interactions, where a photon is absobed or emmitted by electrons. For every photon absorption or emission we need a factor of e, which is a number representing the strenght of the interaction. We then have to include all possibilities for a scattering process to happen, with so many interaction vertices as the maximum order in our perturbative series. The rules are:

190

• For the propagation of a photon with momentum p, assign the factor pµ pν i −gµν + (1 − ξ) 2 . (14.22) p2 + iδ p + iδ • For the propagation of an electron, assign the factor i . p6 − m + iδ

(14.23)

• For photon absorption or emission, assign a factor

− ieγ µ

(14.24)

In addition, • For each loop in a Feynman diagram, we perform an integration over the loop momentum Z dd k , (14.25) (2π)d where d is the dimension.

• For each fermion loop we multiply with a factor (−1)

(14.26)

• We divide each Feynman diagram with a “symmetry factor”, accounting for the equivalent field permutations. The LSZ reduction formula is slightly modified for fermions and photons, where the constants Z˜ 1/2 are replaced by Z˜ 1/2 times a “spin factor” which is different from one. For truncated external lines, we have the rules • multiply with a factor

u¯(p),

for each outgoing fermion,

191

• a factor

v(p),

for each outgoing anti-fermion, • a factor

u(p),

for each incoming fermion, • a factor

v¯(p),

for each incoming anti-fermion, • and a factor

ǫµ (p)

for each external photon.

14.3

Dimensional regularization for QED

QED is plagued by divergent loop integrals. We shall use dimensional regularization as our regularization method. In this Section, we discuss some new features that arise in dimensional regularization in gauge theories. First, we consider the QED action integral in arbitrary d dimensions. Z S = dd xLQED . (14.27) The action has no mass dimensions, [S] = dd xLQED = [mass]0 .

(14.28)

Since the space-time coordinates have an inverse mass dimension we find that all terms in the Lagrangian must have mass dimensionality d, [LQED ] = [mass]d .

(14.29)

From the kinetic terms of the photon and fermion fields we can deternime that their mass dimensionalities ought to be, d (∂µ Aν )2 = [mass]d ; [Aν ] = [mass] 2 −1 , (14.30) 192

and

d−1 ψ6¯∂ ψ = [mass]d ; [ψ] = [mass] 2 .

(14.31)

[ξ] = [mass]0 ,

(14.32)

From the gauge fixing term, we find that the parameter ξ

is dimensionless. Finally, the interaction term gives, 4−d eψ6¯Aψ = [mass]d ; [e] = [mass] 2 .

(14.33)

The coupling constant e is a dimensionful parameter in dimensions other than four. In dimensional regularization, we take d = 4 − 2ǫ, which translates into

[e] = [mass]ǫ .

(14.34)

We can then substitute in the Lagrangian and the Feynman rule for photon absorption or emission, e → eµǫ , (14.35) introducing an arbitrary mass scale µ in the Lagrangian. Naively, given that we shall take the ǫ → 0 after renormalization, the arbitrary scale µ seems innocuous. In practice, we cannot get rid of µ easily, unless we are able to compute the perturbative series at all orders. Loop integrals will produce poles in ǫ of the form, µǫ 1 = + log(µ) + O(ǫ). ǫ ǫ

(14.36)

The procedure of renormalization will eliminate the residues of the poles in ǫ in the expressions for scattering matrix-elements. However the log(µ) terms can only eliminated by means of a resummation of all perturbative orders.

14.3.1

Gamma-matrices in dimensional regularization

How do we treat γ-matrices in dimensional regularization? The prescription that is followed in conventional dimensional regularization for the Clifford algebra is {γ µ , γ ν } = 2g µν 14×4 , (14.37) 193

where the metric is taken in D = 4 − 2ǫ dimensions, gµµ = 4 − 2ǫ,

(14.38)

and the dimensionality of the gamma matrices is four by four, tr14×4 = 4.

(14.39)

Let us do some simple calculations with this algebra. γ µ γµ =

1 µ {γ , γµ} = gµµ14×4 = (4 − 2ǫ)14×4 . 2

(14.40)

Also, γ µ γ ν γµ = {γ µ , γ ν } γµ − γ ν γ µ γµ = 2g µν γµ − (4 − 2ǫ)γ ν ; γ µ γ ν γµ = −2(1 − ǫ)γ ν .

(14.41)

and so on. A difficulty arises in defining the matrix γ5 in 4 − 2ǫ dimensions. In four dimensions this is defined as, γ5 = iγ 0 γ 1 γ 2 γ 3 .

(14.42)

It is often just fine to use a prescription of an anti-commuting γ5 , {γ µ , γ5} = 0, in performing the gamma-matrix algebra for QED amplitudes. But a discussion of this issue lies beyond the scope of this course.

14.3.2

Tensor loop-integrals

In computing QED scattering amplitude, we need to calculate loop integrals where tensors of the loop-momentum appear in the numerator. These arise due to the Feynman rule for the fermion propagator, which is written as i i (γ µ kµ + m) = . 6k − m k 2 − m2 194

(14.43)

Particles propagating in loops, called virtual, generate loop-momenta in the numerators of loop integrands. At one-loop we can always express tensor integrals in terms of scalar integrals. The technique is known as Passarino-Veltman technique and we can illustrate it with a couple of examples. Consider the following tensor one-loop integral, Z kµ µ I [k ] ≡ dd k 2 , (14.44) k (k + p)2

where the +iδ term in the denominator is implicit and we take for simplicity m = 0. The integrand is a rank one vector in the loop momentum. Due to Lorentz convariance, the result of the integration will also be a rank one tensor. We can then write, I [k µ ] = Apµ , (14.45) where on the rhs we have written the most general rank-one tensor that we can construct from the momenta in the integral other than the loop momentum. The coefficient A is as yet undetermined. We can express it in terms of integrals which have scalar numerators. We multiply Eq. 14.45 with pµ and rearrange 1 (14.46) A = 2 I [k · p] . p

Notice that the scalar product in the numerator of the integral above can be expressed in terms of the denominators of the integral, k 2 p2 1 − . k · p = (k + p)2 − 2 2 2 We then have that, Z Z Z 1 1 1 1 1 d d 1 d A= − d k 2 + d k 2− 2 d k . 2 k (k + p)2 p2 k p (k + p)2

(14.47)

(14.48)

Notice, that we only find integrals with a constant numerator, which is independent of the loop momentum. The last two integrals are identical, as we can see by performing the shift k → k + p in one of them. In this particular example, they cancel against each other, and we have Z Z 1 pµ kµ d dd k 2 =− . (14.49) d k 2 2 k (k + p) 2 k (k + p)2 195

Consider now a more complicated example, Z kµ kν µ ν d I [k k ] ≡ d k 2 . k (k + p)2

(14.50)

The most general rank two tensor that we can write using the external momenta of the integral is, pµ pν (14.51) I [k µ k ν ] = A1 g µν + A2 2 . p Contracting with gµν and g µν gµν g µν pµp2pν

pµ pν , p2

we obtain the system of equations, ! ! pµ pν I [k 2 ] g A 2 µν 1 p . = I [(k·p)2 ] pµ pν pµ pν A 2 2 2 2 2 p p

(14.52)

(p )

In arbitrary dimensions, gµν g µν = d = 4 − 2ǫ, and the above system of equations becomes, 4 − 2ǫ 1 A1 = A2 1 1 with a solution, 1 1 −1 A1 = A2 3 − 2ǫ −1 4 − 2ǫ

I [k 2 ] I [(k·p)2 ] (p2 )2

(14.53)

(14.54)

The two integrals on the rhs can be expressed in terms of two scalar “master integrals”, Z 1 Ia ≡ dd k 2 , (14.55) k and Z 1 Ib ≡ dd k 2 . (14.56) k (k + p)2 Specifically, we have Z Z 2 1 k2 d = dd k I k = d k 2 2 k (k + p) (k + p)2 Z Z 1 1 = dd k 2 = dd (k + p) 2 (k + p) k = Ia . (14.57) 196

Ia is a scaleless integral and, as we have discussed, it is zero in dimensional regularization. We let it as an exercise to prove that p2 (p2 )2 I (k · p)2 = Ia + Ib . (14.58) 2 4 We can express all one-loop tensor integrals in terms of four master integrals, Z d d k 1 MA ≡ , (14.59) d 2 iπ 2 k 2 − m1 + iδ Z d d k 1 , (14.60) MB ≡ d 2 2 2 iπ 2 (k − m1 + iδ) [(k + p)2 − m2 + iδ] Z d d k 1 MC ≡ , d 2 2 2 2 2 iπ 2 (k − m1 + iδ) [(k + p1 ) − m2 + iδ] [(k + p2 )2 − m3 + iδ] (14.61) and Z d d k MD ≡ d × iπ 2 1 . × 2 2 2 2 (k − m1 + iδ) [(k + p1 ) − m2 + iδ] [(k + p2 )2 − m23 + iδ] [(k + p2 )2 − m24 + iδ] (14.62) Exercise: Express the following tensor integrals in terms of master integrals, Z kµ dd k 2 k − m2 Z kµkν dd k 2 k − m2 Z kµ dd k 2 (k − m21 ) [(k + p)2 − m22 ] Z kµ kν dd k 2 (k − m21 ) [(k + p)2 − m22 ] (14.63) The above master integrals are known analytically since many years. Their computation is involved for the general case of arbitrary masses mi and momenta pi . We shall need the following two special cases, Z d d k 1 2 1−ǫ = −Γ(−1 + ǫ) m , (14.64) d iπ 2 k 2 − m2 197

and

where

dd k

1 cΓ 2 −ǫ −p , = ǫ(1 − 2ǫ) iπ k 2 (k + p)2 d 2

cΓ =

Γ(1 + ǫ)Γ(1 − ǫ)2 . Γ(1 − 2ǫ)

(14.65)

(14.66)

Exercise: Prove the above using the method of Feynman parameters

14.4

Electron propagator at one-loop

As an application of the above, we can compute the electron propagator through one-loop order. Applying the QED Feynman rules we have, Z ¯ |Ωi = d4 xeip·x hΩ| T ψ(x)ψ(0)

i i i + (−iΣ2 (p)) + ... 6p − m 6p − m 6p − m

(14.67)

with dd k kµ kν i ǫ ν ǫ µ i −iΣ2 = (−ieµ γ ) (−ieµ γ ) 2 −gµν + (1 − ξ) 2 , (2π)d 6k + 6p − m k k (14.68) where the iδ terms in the denominators are implicit. As we have discussed, in d = 4 − 2ǫ dimensions, we need to introduce a mass scale factor µǫ for every coupling factor e. There are many ways that one can decide to compute this integral. Perhaps, the easiest is with applying directly Feynman parameters. I choose a bit of a longer method here, the details are not important, but the outcome is, in the fact that it contains ultraviolet divergences. We first make all denominators quadratic in the loop momentum by writing, Z

1 6k + 6p + m 6k + 6p + m , = = 6 + 6p − m k (6k + 6p − m) (6k + 6p + m) (k + p)2 − m2 198

(14.69)

and then contract the Lorentz indices and use the d-dimensional Clifford algebra. We obtain Z dd k (2 − d)(6k + 6p) + md 2 2ǫ −iΣ2 = −e µ (2π)d k 2 [(k + p)2 − m2 ] Z dd k 6k (6k + 6p + m)6k 2 2ǫ (14.70) + e µ (1 − ξ) (2π)d (k 2 )2 [(k + p)2 − m2 ] We can simplify the second integral if we use the identity 1 1 6k 6k = (6k − p6 + m) + (6p − m) (6p − m) , (14.71) 6k + 6p − m 6k + 6p − m which is proved easily by writing 6k = (6k + 6p − m) − (6p − m). The first term in the rhs of the identity leads to a scaleless integral which is zero in dimensional regularization. We then have Z dd k (2 − d)(6k + 6p) + md 2 2ǫ −iΣ2 = −e µ (2π)d k 2 [(k + p)2 − m2 ] Z dd k (6k + 6p + m) 2 2ǫ + e µ (1 − ξ) (6p − m) (6p − m) , 2 d 2 (2π) (k ) [(k + p)2 − m2 ] (14.72) and by changing variables to k → k − p, we write Z dd k (2 − d)6k + md 2 2ǫ −iΣ2 = −e µ (2π)d (k − p)2 [k 2 − m2 ] Z dd k (6k + m) 2 2ǫ + e µ (1 − ξ) (6p − m) (6p − m) . d (2π) [(k − p)2 ]2 (k 2 − m2 ) (14.73) Now we can write 6k = γ µ kµ ,

and reduce the emerging tensor integrals to master integrals (exercise). However, there is an easier method. First, we simplify further the gamma matrices in the numerators and we write Z dd k (2 − d)6k + md 2 2ǫ −iΣ2 = −e µ (2π)d (k − p)2 [k 2 − m2 ] Z dd k −6k (p2 − m2 ) + 26p(p · k − m2 ) + m (k 2 − m2 ) − (k − p)2 2 2ǫ . − e µ (1 − ξ) (2π)d [(k − p)2 ]2 (k 2 − m2 ) (14.74) 199

We anticipate the result to be written in terms of the vector pµ , contracted with gamma matrices. We can write the general ansatz, −iΣ2 =

∞ X

(6p)n fn (p2 , m2 ).

(14.75)

n=0

Given that 6p2 = p2 , 6p3 = p2 6p etc, the series terminates after the second term, and we cast the general functional form of the one-loop electron propagator as, −iΣ2 = −iΣ2,V (p2 , m2 )6p − imΣ2,S (p2 , m2 )1. (14.76) We can find expressions for the functions Σ2,S and Σ2,V by taking the traces

and

1 −imΣ2,S (p2 , m2 ) = tr (−iΣ2 ) , 4

(14.77)

1 −iΣ2,V (p2 , m2 )p2 = tr (−iΣ2 6p) . 4

(14.78)

Explicitly, 2

2 2ǫ

Σ2,S (p , m ) = −ie µ [d − (1 − ξ)]

and 2

Σ2,V (p , m )p

dd k 1 , d 2 (2π) (k − p) [k 2 − m2 ]

(14.79)

(2 − d)k · p dd k d (2π) (k − p)2 [k 2 − m2 ] Z dd k k · p(p2 + m2 ) − 2p2 m2 2 2ǫ − ie µ (1 − ξ) . (2π)d [(k − p)2 ]2 (k 2 − m2 ) (14.80) 2 2ǫ

= −ie µ

Both functions Σ2,S and Σ2,V are divergent due to the ultraviolet limit k → ∞. In this limit, we can ignore the electron mass in the above expressions. We then have Z 1 dd k 2 2 2ǫ , (14.81) Σ2,S (p , 0) = −ie µ [d − (1 − ξ)] d (2π) (k − p)2 [k 2 ]

which has a single pole in ǫ,

e 1 0 2 ǫ 0 Σ2,S (p , m ) = Σ2,S (p , 0) UVpole +O(ǫ ) = 2 (4πµ ) (3+ξ) − γ + O(ǫ ) 4π ǫ (14.82) 2

200

with d = 4 − 2ǫ. Similarly, the UV poles of Σ2,V can be computed from 2

Σ2,V (p , 0)p

dd k (2 − d)k · p (2π)d (k − p)2 k 2 Z dd k k·p 2 2ǫ 2 . − ie µ (1 − ξ)p (2π)d [(k − p)2 ]2 k 2 2 2ǫ

= −ie µ

(14.83)

Writing

1 2 k + p2 − (k − p)2 , 2 and setting to zero scale-less integrals we have k·p=

Σ2,V (p2 , 0)p2 =

14.5

Photon propagator at one-loop

201

(14.84)

Chapter 15 One-loop renormalization of QED

202