Relativistic Quantum Field Theory, Volume 1: Canonical Formalism 1643276999, 9781643276991

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Relativistic Quantum Field Theory, Volume 1 Canonical formalism

Relativistic Quantum Field Theory, Volume 1 Canonical formalism Michael Strickland Kent State University, Kent, Ohio, USA

Morgan & Claypool Publishers

Copyright ª 2019 Morgan & Claypool Publishers All rights reserved. No part of this publication may be reproduced, stored in a retrieval system or transmitted in any form or by any means, electronic, mechanical, photocopying, recording or otherwise, without the prior permission of the publisher, or as expressly permitted by law or under terms agreed with the appropriate rights organization. Multiple copying is permitted in accordance with the terms of licences issued by the Copyright Licensing Agency, the Copyright Clearance Centre and other reproduction rights organizations. Rights & Permissions To obtain permission to re-use copyrighted material from Morgan & Claypool Publishers, please contact [email protected]. ISBN ISBN ISBN

978-1-64327-702-8 (ebook) 978-1-64327-699-1 (print) 978-1-64327-700-4 (mobi)

DOI 10.1088/2053-2571/ab30cc Version: 20191101 IOP Concise Physics ISSN 2053-2571 (online) ISSN 2054-7307 (print) A Morgan & Claypool publication as part of IOP Concise Physics Published by Morgan & Claypool Publishers, 1210 Fifth Avenue, Suite 250, San Rafael, CA, 94901, USA IOP Publishing, Temple Circus, Temple Way, Bristol BS1 6HG, UK

This book is dedicated to my wife, Dr Veronica Antocheviz Dexheimer Strickland, and our amazing daughter Emily.

Contents Preface

xi

Acknowledgements

xiv

Author biography

xv

Units and conventions

xvi

1

Classical field theory

1-1

1.1 1.2 1.3 1.4

Lagrangian formalism for fields The Klein–Gordon field The electromagnetic field Lorentz invariance 1.4.1 Rotations 1.4.2 Boosts 1.4.3 An example of a Lorentz-invariant quantity Transformation of fields under Lorentz transformations Noether’s theorem 1.6.1 Implications of Noether’s theorem Applications of Noether’s theorem 1.7.1 Translational invariance 1.7.2 The Klein–Gordon energy–momentum tensor 1.7.3 Angular momentum and Lorentz transformations The Hamiltonian formalism for fields 1.8.1 Hamilton’s equations for fields References

1.5 1.6 1.7

1.8

2

Quantization of free fields

2.1

The quantum linear chain and phonons 2.1.1 Going from classical to quantum 2.1.2 Construction of the vacuum and many phonon states Poisson brackets in classical field theory Quantization of a free scalar field theory 2.3.1 Quantized Hamiltonian operator 2.3.2 Vacuum renormalization and normal ordering 2.3.3 A note on dimensions 2.3.4 The vacuum state 2.3.5 Single particle states

2.2 2.3

1-1 1-2 1-4 1-6 1-6 1-6 1-7 1-7 1-9 1-10 1-11 1-11 1-12 1-13 1-15 1-16 1-17 2-1

vii

2-1 2-2 2-5 2-6 2-7 2-9 2-10 2-10 2-11 2-12

Relativistic Quantum Field Theory, Volume 1

2.4 2.5 2.6 2.7 2.8 2.9

Multi-particle states and Fock space Complex scalar fields Quantization of a complex scalar field Causality Propagators Propagators as Green’s functions References

3

Quantization of interacting field theories

3.1 3.2 3.3

Weakly-interacting scalar fields Two examples of interacting quantum field theories The interaction picture and Dyson’s equation 3.3.1 Dyson’s formula Interactions in scalar Yukawa theory The S-matrix 3.5.1 Example: pion decay in the scalar Yukawa model Beyond leading-order perturbation theory 3.6.1 Wick’s theorem 3.6.2 Example: nucleon–nucleon scattering in the scalar Yukawa model 3.6.3 Feynman diagrams 3.6.4 Feynman diagrams for scalar λϕ4 theory 3.6.5 Example: leading-order correction to the propagator 3.6.6 Wick rotation Decay rates and cross sections Examples using scalar Yukawa theory References

3.4 3.5 3.6

3.7 3.8

2-15 2-18 2-20 2-21 2-26 2-29 2-30 3-1 3-1 3-2 3-4 3-5 3-5 3-6 3-7 3-10 3-10 3-13 3-16 3-16 3-20 3-22 3-25 3-29 3-34

4

Quantum electrodynamics

4-1

4.1

Classical Dirac fields 4.1.1 The Dirac equation 4.1.2 Hamiltonian density 4.1.3 Energy–momentum tensor 4.1.4 Poincare invariance and generalized angular momentum Quantization of the Dirac field 4.2.1 The fermion field The Feynman propagator for Dirac fields

4-1 4-2 4-2 4-3 4-3 4-4 4-7 4-8

4.2 4.3

viii

Relativistic Quantum Field Theory, Volume 1

4.4

4.5

4.6

4.7

4.8

4.9

The electromagnetic field 4.4.1 Local gauge invariance 4.4.2 Gauss’ law 4.4.3 Gauge fixing Quantization of the electromagnetic field 4.5.1 Coulomb gauge 4.5.2 Poisson (Dirac) brackets for constrained systems 4.5.3 Coulomb-gauge mode expansion 4.5.4 General Lorenz (covariant) gauge 4.5.5 The photon propagator in general covariant gauge 4.5.6 The Gupta–Bleuler method Coupling the electron to the photon 4.6.1 Gauge invariance of the QED Lagrangian 4.6.2 Obtaining vertex functions directly from the QED Lagrangian QED Feynman rules 4.7.1 External lines 4.7.2 Internal lines 4.7.3 Interaction vertex 4.7.4 Special rules for fermions QED Feynman rules—Examples 4.8.1 Electron–positron scattering 4.8.2 Compton scattering The leading-order electron–positron scattering cross section References

5

Renormalization of quantum electrodynamics

5.1 5.2 5.3 5.4 5.5

Renormalization group flow Beta functions Renormalizable field theories Dimensional regularization in QED One-loop renormalization of QED 5.5.1 Electron self-energy 5.5.2 Photon polarization 5.5.3 Electron–photon vertex 5.5.4 Summary Schwinger–Dyson equations Photon wavefunction renormalization

5.6 5.7

ix

4-11 4-12 4-12 4-13 4-14 4-14 4-15 4-16 4-17 4-19 4-19 4-22 4-23 4-24 4-26 4-27 4-27 4-28 4-28 4-28 4-28 4-30 4-31 4-36 5-1 5-2 5-6 5-8 5-9 5-9 5-10 5-12 5-14 5-16 5-16 5-17

Relativistic Quantum Field Theory, Volume 1

5.8 5.9 5.10 5.11

Electron wavefunction and mass renormalization Vertex renormalization The renormalized QED Lagrangian The one-loop QED running coupling References

5-19 5-20 5-20 5-21 5-24

Appendices A Classical mechanics review

A-1

B Functionals and functional derivatives

B-1

C Tensor algebra

C-1

D Mandelstam variables

D-1

x

Preface In introductory quantum mechanics, one learns how to quantize a system given a fixed number of non-relativistic bosonic or fermionic particles. One uses a formalism in which the Hamiltonian is promoted from being a number to an operator resulting in the Schrödinger equation. The Hamiltonian operator itself is decomposed into kinetic and potential energy contributions and the potential energy form is typically taken as an external input. A question then arises: how does one arrive at the potential in the first place based on first principles? For example, how do we know (beyond the experiment) that the Coulomb potential is the appropriate potential for charged particles? Are there quantum corrections to this potential? In addition, since potentials are not well-defined relativistically (instantaneous interactions), how can we generalize quantum mechanics to the relativistic case and satisfy causality? In classical physics, fields are introduced in order to construct physical laws that are causal and local. Classical forces described by, for example Coulomb’s law or Newton’s universal gravitation, require action at a distance. As a result, the force felt by a body changes instantaneously if any other body’s position (or charge, etc) changes, no matter how far away the other object is. In the classical field theories of Maxwell (electromagnetic field) and Einstein (gravitational field), the interactions between objects are mediated by a field that acts locally and causality is restored. So it seems that, since they are causal and local, we should figure out how to quantize classical field theories. Another inconsistency in quantum mechanics was our somewhat haphazard treatment of wave-particle duality. For example, both electrons and photons act simultaneously like waves and particles and physically they share many common features. They both undergo wave-like diffraction from obstacles, but can also act like discrete particles (photoelectric effect, Compton scattering, etc). Despite this, in classical theory, electrons are simply postulated to exist as matter, while photons are interpreted as ripples in the electromagnetic field. Is it possible that electrons are themselves ripples in an ‘electron field’? As we will learn over the course of these three volumes, the answer is a definitive yes. In general, the field is the fundamental object and particles are derived concepts that appear only after the quantization of the field (e.g. the Higgs field gives birth to the Higgs boson). In quantum mechanics, we take classical number-valued quantities and promote them to operators acting in a Hilbert space. As we will see, at least in ‘canonical quantization’, the rules for quantizing a field are only slightly different. The basic degrees of freedom in quantum field theory (QFT) are operator-valued functions of space and time and, since space and time are continuous, we are dealing with an infinite number of degrees of freedom, so we will need to (re-)learn how to deal with systems with a large number of degrees of freedom (many-body theory). Once we are done, we will be able to properly define QFTs that can be used in a variety of different contexts, e.g. high energy theory, condensed matter, cosmology, quantum gravity, etc.

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Relativistic Quantum Field Theory, Volume 1

Beyond this, Dirac taught us that a consistent theory of relativistic electrons requires the existence of anti-electrons (aka positrons). As a consequence, it is possible to create particle-antiparticle pairs once the energy available exceeds twice the electron rest mass E > 2mc 2 in a process called pair production. And, of course, the reverse can happen, which is called pair annihilation. And one can easily see both of these types of events using modern particle detectors. The conclusion we must draw from this is that particles are not indestructible objects; they can be created and destroyed and may only live for a short amount of time. They are merely excitations surfing on a quantum field. But the story is more fantastical than this. If pair production has an energy threshold, then one could argue that as long as the energies available do not exceed this threshold (E > 2mc 2 ), then non-relativistic theories would be self-consistent; however, at this point the Heisenberg uncertainty principle comes into play. Let us say that we wanted to measure the position of a particle with a given spatial resolution L . The Heisenberg uncertainty principle tells us that the uncertainty in the momentum is Δp ≳ 1/L . In a relativistic setting, energy and momentum are connected, therefore, we also have an uncertainty in the energy ΔE ≳ 1/L . However, when the energy plus its uncertainty exceeds E + ΔE > 2mc 2 , then it is possible to create purely quantum-mechanical particle-antiparticle pairs. Equating the two, we obtain a threshold distance L 0 = 1/(2m ) = λ Compton /(4π ) with λ Compton = 2π /m. From this exercise, we learn that the spontaneous production of particle-antiparticle pairs is important when a particle of mass m is localized in space to a distance that is on the order of less than its Compton wavelength. A similar argument holds if one considers localizing particles in time. The energy uncertainty increases as the time interval is made shorter and one eventually turns on the possibility of pair production. The consequence of this is that as one considers shorter and shorter time (or space) intervals one sees more and more particles! Why is this relevant? Since microscopic particles interacting with one another can resolve the other particles’ microscropic dynamics (they behave like observers since they exchange quanta in order to interact), they ‘see’ their partners as being surrounded by an ensemble of particles and antiparticles that flit in and out of existence. In QFT, these ephemeral particles are called virtual particles. The ‘cloud’ of virtual particles that surround particles can modify its observable properties and must be taken into account to consistently understand them. Figure 0.1 shows a typical Feynman graph for the splitting of an electron (solid line with the arrows) into virtual photons (sinusoidal lines) and virtual electron– positron pairs (closed loops). In this figure, time progresses from the left to the right. Starting from the incoming electron line we see the radiation of a virtual photon, which then splits into a virtual electron–positron pair, and so forth. In QFT, during this time, all possible configurations of the various intermediate particles are sampled, but some configurations are more probable than others. We will learn how to quantify this over the course of these three volumes. A very similar phenomenon to pair production exists in condensed matter systems. When one considers the conduction of electrons in metals, for example, one finds that there is a valence band of electrons that are locally bound to atoms xii

Relativistic Quantum Field Theory, Volume 1

Figure 0.1. A typical Feynman graph for the splitting of an electron (solid line with the arrows) into virtual photons (sinusoidal lines) and virtual electron–positron pairs (closed loops).

and a conduction band in which electrons are able to move around. Because of the spin-statistics theorem, fermions obey the Pauli exclusion principle and ‘stack up’ to an energy called the Fermi energy. It is possible for a photon (or an energetic phonon) to excite one of the electrons that have an energy below the Fermi energy (in the Fermi sea) to above the Fermi sea. In metals, this process requires very little energy, however, in semiconductors the material is gapped. This introduces a lower limit on the amount of energy required to excite an electron into the conduction band, which is called the gap. The processes of electron–hole pair production and annihilation can be described using Feynman graphs similar to the relativistic case and the mathematical machinery used to describe many-body states in materials is very similar to what is encountered in relativistic QFT. This three-volume series began as lecture notes for a two-semester introductory course in quantum field theory and quantum chromodynamics and, as such, is written rather informally. The text is intended to be used in an introductory graduate-level course in quantum field theory or an advanced undergraduate course. Volume 1 introduces classical fields and the method of canonical quantization in order to have a bridge to the language and formalisms used by students focusing on condensed matter physics. Volume 2 builds upon what was learned in Volume 1, but starts anew using the modern path integral formalism and focuses on applications to quantum electrodynamics and chromodynamics. Volume 3 continues with discussions of applications to particle physics phenomenology, the weak interaction, the Higgs mechanism, and finite temperature field theory. Michael Strickland June 30, 2019

xiii

Acknowledgements I thank the students who provided feedback on the lectures notes that formed the basis for this book.

xiv

Author biography Michael Strickland Dr Michael Strickland is a professor of physics at Kent State University. His primary interest is the physics of the quark-gluon plasma (QGP) and high-temperature quantum field theory (QFT). The QGP is predicted by quantum chromodynamics (QCD) to have existed until approximately 10−5 s after the Big Bang. The QGP is currently being studied terrestrially by experimentalists at the Relativistic Heavy Ion Collider (RHIC) at Brookhaven National Laboratory and the Large Hadron Collider (LHC) at CERN. Dr Michael Strickland has published research papers on various topics related to the QGP, quantum field theory, relativistic hydrodynamics, and many other topics. In addition, he has co-written a classic text on the physics of neural networks.

xv

Units and conventions • I will use natural units in which ℏ = c = 1. In these units, masses, energies, momenta, and temperatures have the same ‘energy-like’ units. I will typically use eV, MeV, or GeV as the standard unit for these energy-like quantities. For example, the mass of an electron is me=0.511 MeV , where this is understood implicitly to denote me = 0.511 MeV/c 2 but, since c = 1, we typically suppress the division by c 2 when writing the mass. • Another consequence of using natural units is that the vacuum electric permeability and magnetic permeabilities are both equal to one, ε0 = μ0 = 1, which greatly simplifies the presentation of the formulas of electromagnetism. A similar simplification happens in thermodynamic formulas since kB = 1. • In natural units, spatial and temporal separations have units of inverse energy units (e.g. GeV −1). In some places in the notes we will use a bracket notation to indicate the dimension of a quantity, e.g. [m] = 1 tells us that masses have units of energy to the power one and lengths [L] = −1 indicating that length scales scale like energy to the power minus one. • If the argument of a function, e.g. exp, should be dimensionless, then the argument should be equivalent to an energy times a length scale (E × L ), since lengths have units of inverse energies in natural units. • In some cases, you might need to switch from energy scales to explicit distance scales or vice versa. For this purpose, the following combination is very useful

ℏc = 0.197326938 GeV fm, where fm stands for femtometers (1 fm = 10−15 m). For the purposes of estimation, one can use the approximate value ℏc ≈ 0.2 GeV fm . For example, if I were to tell you that a length (or time) scale is L = 0.2 fm, you can divide this by ℏc to find the equivalent in inverse GeV, e.g. L = (0.2 fm)/(ℏc ) ≃1 GeV−1. Similarly, we can work in reverse to convert energy (mass) scales to inverse distance scales. For example, from the electron mass me = 0.511 MeV we can determine the Compton wavelength λ Compton = 2π /m and then covert from GeV−1 to meters, i.e. λ Compton, e ≃ 12296 GeV−1 × ℏc ≃ 2.426 × 10−12m . • Finally, in the three volumes in this series, the Minkowski space metric is assumed to be mostly negative, which means that (g μν )Minkowski ≡ η μν = diag(1, −1, −1, −1).

xvi

IOP Concise Physics

Relativistic Quantum Field Theory, Volume 1 Canonical formalism Michael Strickland

Chapter 1 Classical field theory

In order to understand how field quantization is performed, it is necessary to have an understanding of the Lagrangian and Hamiltonian formalisms applied to classical fields. Since we are interested in theories that are manifestly consistent with special relativity, we will use the covariant (relativistic) formalism. For additional information and a mini review of classical mechanics, see appendix A. For more about classical fields, see refs. [1, 2].

1.1 Lagrangian formalism for fields In classical mechanics, we write the Hamiltonian in terms of generalized coordinates qa(t ) that are indexed by a label a . In field theory, we are interested in the dynamics of fields written as, for example, ϕa(x , t ) where both a and x can be considered as labels. As a result, we are dealing with an infinite number of degrees of freedom. The canonical fields from classical physics are the electric and magnetic fields. In this case, ϕ1, ϕ2 , and ϕ3 could correspond to the three vector components of the electric or magnetic fields, E(x , t ) or B(x , t ), respectively. The dynamics of fields can be obtained from a Lagrangian density that is a function of ϕa , ∂tϕa , and ∇ϕa . In the relativistic formalism, we notationally combine the time and space derivatives into a single 4-derivative ∂μϕa where ∂μ = (∂t, ∇). The Lagrangian itself is obtained from spatial integration of the Lagrangian density L

L (t ) =

∫ d 3xL(ϕa , ∂μϕa ).

(1.1)

The Lagrangian, L , is our first example of a functional. A functional is a map from continuous functions (in this case the fields and their derivatives) to a real or complex number. In this case, the Lagrangian functional takes the functions ϕa and ∂μϕa and integrates them over space to obtain a number. By integrating the Lagrangian in time we obtain the action associated with a particular field configuration doi:10.1088/2053-2571/ab30ccch1

1-1

ª Morgan & Claypool Publishers 2019

Relativistic Quantum Field Theory, Volume 1

S=

∫t

t2

dt L(t ) =

1

∫ d 4xL.

(1.2)

Note that in classical mechanics the Lagrangian depends on q and ∂tq = q ̇ but not on ∂ t2q = q .̈ In field theory, we also require that L only depends on ϕ and ϕ,̇ but not on ϕ.̈ In principle, one cannot rule out the appearance of higher-order spatial derivatives in the Lagrangian, however, as we will see, most fundamental theories only contain ∇ϕ and not higher derivatives, since otherwise it is not possible to maintain Lorentz invariance. Additionally, since we are looking for fundamental theories that are invariant under temporal and spatial translations, we will not consider Lagrangian densities that have an explicit dependence on space and time, i.e. L will not explicitly depend on x μ. The fields’ equations of motion follow from the principle of least action. One varies the path, keeping the end points fixed and requires that the variation of the action vanishes, δS = 0,

δS =

⎡

⎤

∫ d 4x ⎢⎣ ∂∂ϕL δϕa + ∂(∂∂μLϕ ) δ(∂μϕa )⎥⎦ a

=

∫

a

⎧ ⎡ ⎛ ∂L ⎞⎤ ⎛ ∂L ⎞⎫ ⎪ ∂L ⎪ ⎢ ⎥ ⎬ + ∂ − ∂ d 4x ⎨ δϕ δϕ ⎜ ⎟ ⎜ μ μ a a ⎟⎪ , ⎪ ⎝ ∂(∂μϕa ) ⎠⎥⎦ ⎝ ∂(∂μϕa ) ⎠⎭ ⎩⎢⎣ ∂ϕa

(1.3)

where we used δ (∂μϕa ) = ∂μ(δϕa ) and, in going from the first to the second lines, we have subtracted and added a term in order to construct a term that is a total derivative (the last term in curly brackets on the second line). The integral of the last term above vanishes for any variation δϕa that vanishes at spatial infinity and that also satisfies δϕa(x , t1) = δϕa(x , t2 ) = 0 (boundary condition for the variation). Requiring δS = 0 for such paths gives us the Euler–Lagrange equations of motion for the field(s) ϕa

⎛ ∂L ⎞ ∂L = 0. ∂μ⎜ ⎟− ⎝ ∂(∂μϕa ) ⎠ ∂ϕa

(1.4)

1.2 The Klein–Gordon field Let us start with a simple case and work our way towards reality by considering a non-interacting (free) real scalar field ϕ(x , t ). One can obtain the equation of motion for free relativistic scalar fields heuristically by writing the relativistic energy E 2 = p 2 + m2 as an operator acting on a field ϕ on the left- and right-hand sides. Using E = i ∂t and p = −i ∇ one obtains

(∂t2 − ∇2 )ϕ + m 2ϕ = 0.

1-2

(1.5)

Relativistic Quantum Field Theory, Volume 1

Making use of the fact that the relativistic 4-derivative is ∂μ = (∂t, ∇) and ∂ μ = (∂t, −∇), we see that ∂μ∂ μ = ∂ t2 − ∇2 , which allows us to write equation (1.5) using a compact relativistic notation1

(∂μ∂ μ + m 2 )ϕ = 0.

(1.6)

Equation (1.6) is called the Klein–Gordon equation. The Klein–Gordon field can be considered as the first candidate for a relativistic quantum field, since it incorporates the relativistic energy relation from the beginning. Now let us see how we can obtain the Klein–Gordon equation using the Lagrangian formalism. For this purpose, we consider the following Lagrangian2

1 1 L = ∂μϕ ∂ μϕ − m 2ϕ 2 2 2 1 αβ 1 = η ∂αϕ ∂βϕ − m 2ϕ 2 , 2 2

(1.7)

where η μν = diag(1, −1, −1, −1) is the Minkowski space metric tensor. Evaluating the two terms entering the Euler–Lagrange equations of motion (1.4), we obtain

∂L 1 1 = η αβ δαμ∂βϕ + η αβ δ βμ∂αϕ ∂(∂μϕ) 2 2 = η μβ ∂βϕ = ∂ μϕ ,

(1.8)

and

∂L = −m 2ϕ . ∂ϕ

(1.9)

Plugging the results above into the Euler–Lagrange equations of motion (1.4) we obtain

(∂μ∂ μ + m 2 )ϕ = 0,

(1.10)

which is precisely the Klein–Gordon equation obtained previously using our heuristic method. Note that, based on our classical mechanics understanding of Lagrangians, we can split the Lagrangian into kinetic energy and potential energy contributions. We begin by separating the temporal and spatial derivatives in the Lagrangian

Some authors also use the ‘box’ or d’Alembertian operator (□ ≡ ∂μ∂ μ ) as an abbreviation. Since the typesetting of the symbol makes it clear whether we are discussing the Lagrangian or the Lagrangian density, I will simply call both the Lagrangian from here on out. 1 2

1-3

Relativistic Quantum Field Theory, Volume 1

1 1 L = ∂μϕ ∂ μϕ − m 2ϕ 2 2 2 1 1 2̇ 1 = ϕ − ( ∇ϕ ) 2 − m 2 ϕ 2 . 2 2 2 Using

L = T − V,

we

obtain,

upon

inspection,

(1.11)

T=

1 2

∫ d 3xϕ2̇

and

1 2

V = ∫ d 3x[(∇ϕ )2 + m2ϕ2 ]. These are the kinetic energy and potential energy contributions for a free scalar field. Based on this, it is relatively straightforward to generalize this result to include field self-interactions. To accomplish this, we simply take V = ∫ d 3x⎡⎣ 12 (∇ϕ )2 + 12 m2ϕ2 + U (ϕ )⎤⎦, where U is an externally input function of the field that obeys all appropriate symmetries (e.g. the potential must be symmetric under ϕ → −ϕ if the theory is Z2 invariant). Taking interactions into account and going back to the relativistic form, we find the following Lagrangian for an interacting relativistic scalar field

L=

1 1 ∂μϕ ∂ μϕ − m 2ϕ 2 − U (ϕ), 2 2

(1.12)

which gives the general equation of motion

(∂μ∂ μ + m 2 )ϕ + ∂ϕU = 0.

(1.13)

1.3 The electromagnetic field A canonical example of fields from classical physics are the electric and magnetic fields, E(x , t ) and B(x , t ) [3]. The electric and magnetic fields are given by

E = −∇ϕ −

∂A , ∂t

(1.14)

B = ∇ × A, where ϕ and A are the scalar and vector potentials, respectively. Using these two definitions we can see that two of the four Maxwell equations, ∇ · B = 0 and d B/dt = −∇ × E, are automatically satisfied using standard vector calculus identities. In the covariant (relativistic) formalism we obtain both the electric and magnetic fields from a single 4-component vector potential field Aμ(x , t ) = (ϕ, A) where μ = 0, 1, 2, 3. One can then obtain Maxwell’s equations from the following compact Lagrangian

1 L = − FμνF μν − Aμ J μ, 4

(1.15)

where Fμν = ∂μAν − ∂νAμ is the electromagnetic field strength tensor and J μ = (ρ , J) is the 4-current which includes the charge density and currents in a single 4-vector.

1-4

Relativistic Quantum Field Theory, Volume 1

Expanding out the first term in (1.15), one obtains3

1 L = − [(∂μAν )(∂ μAν ) − (∂νAμ )(∂ μAν )] − Aμ J μ. 2

(1.16)

For the Euler–Lagrange equation of motion we need

∂L 1 = − ⎡⎣2δ μαδνβ ∂ μAν − δναδ μβ(∂ μAν ) − δ μαδνβ(∂ νAμ )⎤⎦ ∂(∂αAβ ) 2 1 = − [2∂ αAβ − 2∂ βAα ] 2 = − [∂ αAβ − ∂ βAα ] = − F αβ ,

(1.17)

∂L = −J β . ∂Aβ

(1.18)

⎛ ∂L ⎞ ∂L = 0, ⎟− ∂μ⎜ ∂ ∂ ∂ ( A ) Aβ ⎝ α β ⎠

(1.19)

and

Putting the pieces together using

one obtains a rather simple looking Euler–Lagrange equation of motion for electromagnetism

∂μF μν = J ν.

(1.20)

This is the covariant-formalism expression for the ‘non-trivial’ Maxwell equations. The ν = 0 component gives Gauss’ law and the ν = 1, 2, 3 components give the three vector components of the Ampere’s law. As mentioned earlier, the other two Maxwell equations follow from standard vector calculus identities applied to the definitions of the electric and magnetic fields in terms of the scalar and vector potentials. It is straightforward to prove that the current is automatically conserved, which in covariant formalism is expressed as ∂μJ μ = 0, by hitting both sides of equation (1.20) with a ∂ν . One finds, with some very straightforward manipulations, that ∂ν∂μF μν = 0 (alternatively one can see this from symmetry: F μν = −F νμ and the order of the partial derivatives does not matter). Exercise 1.1 The covariant form of Maxwell’s equations in vacuum is ∂μF μν = J ν where J ν = (ρ , J) is the 4-current.

3 One can integrate the second term in square brackets by parts twice to transform (∂νAμ )(∂ μAν ) → (∂μAμ )2 . This form can be useful in some situations.

1-5

Relativistic Quantum Field Theory, Volume 1

(a) Show that ∇ · E = ρ and ∇ × B − ∂tE = J follow from the zeroth [ν = 0] and spacelike [ν = (1, 2, 3)] components of the covariant form of Maxwell’s equations. (b) Show that the remaining two vacuum Maxwell equations follow from standard vector calculus identities applied to the definitions of the electric and magnetic fields in terms of the scalar and vector potentials.

1.4 Lorentz invariance One of the main motivations for the development of QFT is to find a self-consistent relativistic formulation of quantum mechanics. The postulates of special relativity (constancy of the speed of light and the equivalence of inertial frames) can be shown to require that four-vectors transform in the following way

(x′) μ = Λμν x ν ,

(1.21)

Λμα η αβ Λν β = η μν ,

(1.22)

where Λμ ν satisfies

or equivalently Λα μ ηαβ Λβ ν = ημν . I also note that ‘proper’ Lorentz transformations satisfy det (Λμ ν) = 1. ‘Improper’ Lorentz transformations satisfy det (Λμ ν) = −1 and involve either space inversion or time reversal, both of which are not by themselves symmetries of nature. As a result, from now on we will only consider proper Lorentz transformations. The Lorentz group includes three-dimensional rotations and relativistic boosts. Let us discuss them separately. 1.4.1 Rotations For a rotation by ϕ about the z -axis one has

⎛1 0 0 ⎜ 0 cos ϕ −sin ϕ Λμν = ⎜ ⎜⎜ 0 sin ϕ cos ϕ ⎝0 0 0

0⎞ 0⎟ ⎟. 0⎟ ⎟ 1⎠

(1.23)

For a general rotation one has

Λ00 = 1, Λ0 j = Λ j 0 = 0, i

Λ j = Rij , where Rij is a general 3d rotation matrix. 1.4.2 Boosts For a boost by v < 1 along the x -axis one has 1-6

(1.24)

Relativistic Quantum Field Theory, Volume 1

⎛γ ⎜ γv Λμν = ⎜ ⎜⎜ 0 ⎝0

γv γ 0 0

0 0 1 0

0⎞ ⎟ 0⎟ , 0 ⎟⎟ 1⎠

(1.25)

1

with γ = (1 − v2 )− 2 . The general form of a Lorentz boost by a three-velocity v = (v1, v2 , v3) can be expressed compactly as

Λ00 = γ , Λ j = Λ j 0 = γvj , 0

i

Λ j = δij +

vi vj v2

(1.26) (γ − 1),

1

where now γ = (1 − v2)− 2 . A general Lorentz transformation can be a combination of boosts and rotations. 1.4.3 An example of a Lorentz-invariant quantity As a simple example, let us prove that the contraction of two four-vectors is Lorentzinvariant

(x′)α (x′)α = (Λα μ x μ)ηαβ (Λβ ν x ν ) = (Λα μ ηαβ Λβ ν)x μx ν = ημνx μx ν = x μxμ,

(1.27)

where, in going from the second to third lines, I used equation (1.22). This proves that the contraction of two four vectors is the same in both the primed and unprimed coordinate systems and is therefore a Lorentz-invariant. In fact, it was one of the conditions used to derive the form of the Lorentz transformation in the first place.

1.5 Transformation of fields under Lorentz transformations Let us begin by considering the example of a simple scalar field theory. Under a Lorentz transformation x → Λx a scalar field transforms as

ϕ(x ) → ϕ′(x ) = ϕ(Λ−1x ).

(1.28)

It is the inverse Lorentz transformation that appears in the argument of the field because we are considering an active transformation in which the field is shifted in space rather than a passive transformation in which the coordinate system is shifted. For our scalar field theory to be Lorentz-invariant it must be true that if ϕ(x ) solves the equations of motion then ϕ(Λ−1x ) also solves the equations of motion. This will happen if the action for our theory is Lorentz-invariant. If the field is a vector instead of a scalar, then the transformation is more complicated. For example, under a Lorentz transform the electromagnetic vector potential transforms as 1-7

Relativistic Quantum Field Theory, Volume 1

Aμ (x ) → ΛμαAα (Λ−1x ).

(1.29)

Likewise, the 4-derivative of a scalar field transforms in a similar way

∂μϕ(x ) → (Λ−1)α μ∂αϕ(Λ−1x ).

(1.30)

Example: Lorentz invariance of the Klein–Gordon action We will now prove that the Klein–Gordon action is Lorentz-invariant. Recall that the Lagrangian for this simple scalar field theory is

L=

1 1 ∂μϕ(x )η μν∂νϕ(x ) − m 2ϕ 2(x ). 2 2

(1.31)

Using the transformation rules one finds

L→

1 1 ⎡ −1 α ⎣(Λ ) μ∂αϕ(y )⎤⎦η μν⎡⎣(Λ−1) β ν∂βϕ(y )⎤⎦ − m 2ϕ 2(y ), 2 2

(1.32)

where y = Λ−1x . Using equation (1.22) it follows that (Λ−1)α μη μν (Λ−1) β ν = η αβ and, as a result, one has

L→

1 1 ∂αϕ(y )η αβ ∂βϕ(y ) − m 2ϕ 2(y ), 2 2

(1.33)

which has the same form as our original Lagrangian, except with x → y . So the Lagrangian density changes, however, this is not a problem since what is supposed to be invariant is the action S, which is obtained from a 4-integral of the Lagrangian density

S=

∫ d 4x L(x) → ∫ d 4x L(y ) = ∫ d 4y L(y ) = S,

(1.34)

where, in the last step, we made a change of variables from x to y . In practice, one needs the Jacobian for this transformation, which is given by ∣det(Λ)∣, however, as we discussed in the beginning of this section, Lorentz transformations have ∣det(Λ)∣ = 1. From equation (1.34) we see that the Klein–Gordon action is Lorentz-invariant. Exercise 1.2 One can write an infinitesimal Lorentz transformation in the form

Λμν = δ μν + ω μν . Using equation (1.22) show that ωμν = −ωνμ. What does this imply about the number of independent infinitesimal Lorentz transformations in four dimensions?

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Relativistic Quantum Field Theory, Volume 1

Exercise 1.3 Show that the action S obtained from the electromagnetic field Lagrangian

1 L = − FμνF μν − Aμ J μ, 4 is Lorentz-invariant.

1.6 Noether’s theorem Symmetries play an important role in physics. Symmetries are even more important in relativistic quantum field theory: there are Lorentz symmetries, gauge symmetries, internal symmetries, etc. With this in mind it is useful to have a deeper understanding of Noether’s theorem, which relates continuous symmetries to conserved quantities. Noether’s theorem states that every continuous symmetry of the Lagrangian gives rise to a conserved current j μ (x ). In the relativistic formalism, current conservation is written compactly as

∂μj μ = 0.

(1.35)

Expanding out the four contraction using j μ = (j 0 , j) and ∂μ = (∂t, ∇), one obtains something that perhaps looks more familiar

∂j 0 + ∇ · j = 0. ∂t

(1.36)

This is nothing but the continuity equation. Here j 0 is the charge density and j is the spatial current. Noether’s theorem can be proven most easily by considering infinitesimal symmetry transformations. We begin by introducing an infinitesimal field transformation

δϕa (x ) = Xa(ϕ),

(1.37)

under which the Lagrangian, quite generally, becomes L → L + δ L. The transformation Xa will be a symmetry of the theory if the Lagrangian is invariant up to a total derivative

δ L = ∂μF μ,

(1.38)

for some set of functions F μ(ϕ ). This is related to the fact that a total derivative leaves the action invariant. We begin by considering an arbitrary field transformation δϕa for which one has

∂L ∂L δϕa + δ(∂μϕa ) ∂(∂μϕa ) ∂ϕa ⎡ ∂L ⎛ ∂L ⎞ ⎛ ∂L ⎞⎤ − ∂μ⎜ =⎢ δϕa ⎟ , ⎟⎥δϕa + ∂μ⎜ ⎢⎣ ∂ϕa ⎝ ∂(∂μϕa ) ⎠⎥⎦ ⎝ ∂(∂μϕa ) ⎠

δL =

1-9

(1.39)

Relativistic Quantum Field Theory, Volume 1

where, as in the previous, we have subtracted a term in order to construct a total derivative. The first term in square brackets in (1.39) vanishes if the equations of motion are satisfied and one obtains

⎛ ∂L ⎞ δ L = ∂μ⎜ δϕa ⎟ . ⎝ ∂(∂μϕa ) ⎠

(1.40)

As stated above, if the field transformation is a symmetry transformation δϕa = Xa(ϕ ), then one must have δ L = ∂μF μ. Requiring that this is satisfied by (1.40) gives

⎛ ∂L ⎞ ∂μ⎜ δϕa ⎟ = ∂μF μ, ⎝ ∂(∂μϕa ) ⎠

(1.41)

which upon moving the term on the right-hand side to the left allows us to identify a conserved current by comparing this to ∂μ j μ = 0. The resulting conserved current expressed generally is

jμ =

∂L Xa(ϕ) − F μ(ϕ). ∂(∂μϕa )

(1.42)

This is Noether’s theorem in mathematical form. We will look at some examples below, but first let us briefly discuss the deep implications of this result. 1.6.1 Implications of Noether’s theorem The existence of a conserved current is a strong statement because, as written above, this is a local condition: the current is conserved exactly at each point in spacetime. This can be seen by integrating the charge density4 over an arbitrary volume V

Q=

∫V

d 3x j 0 .

(1.43)

Taking a time derivative and using the continuity equation one obtains

∂Q = ∂t

∫V

d 3x

∂ j0 =− ∂t

∫V

d 3x ∇ · j = −

∫A j · dS,

(1.44)

where, in the last step, we applied the divergence theorem to change the integral of the divergence of the three-current over the volume V to a surface integral over the area A bounding V . The equation above states that the time rate of change of the charge contained in a volume V must come from three-current flowing out of the volume (through the bounding surface A). Since this is true for any volume, including an infinitesimal one, this holds at all points in spacetime. Note, in 4 Although we call it charge density, the conserved quantity need not be an actual charge. As we will see below it could be, e.g. the energy or three-momentum density, with the associated ‘charge’ in this case being the total energy or total three-momentum. For ease of discussion, however, we will simply call it a charge in analogy to charge conservation.

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Relativistic Quantum Field Theory, Volume 1

addition, that one could take the volume above to be all space. Requiring that j → 0 is sufficiently fast as ∣x∣ → ∞, one sees that the integral on the far right-hand side of equation (1.44) vanishes, and one has an explicitly conserved charge, i.e., dQ /dt = 0.

1.7 Applications of Noether’s theorem We will now look at some applications of Noether’s theorem. In the process, we will learn the field theory generalizations of energy, momentum, and angular momentum conservation. 1.7.1 Translational invariance In classical mechanics (and quantum mechanics for that matter), invariance under spatial translations results in conservation of momentum and invariance under temporal translations results in the conservation of energy. A similar thing occurs in field theory, as we will now demonstrate. To do this we consider a specific infinitesimal transformation

xμ → xμ − εν.

(1.45)

One can determine how the fields transform subject to this transformation by Taylor expanding the transformed fields

ϕa (x ) → ϕa (x ) + ε ν∂νϕa (x ),

(1.46)

which, upon comparing to the general variation ϕa(x ) → ϕa(x ) + δϕa(x ), allows us to identify

δϕ = ε ν∂νϕa .

(1.47)

Note that the sign on the right-hand side in the transformation of ϕa is related to the fact that we choose to make an active transformation of the field (the final result below is independent of this choice). Also note that, as usual, we have discarded higher-order terms since we are working in the limit ε → 0. Similarly, one can determine how the Lagrangian changes abstractly using Taylor expansion

L(x ) → L(x ) + ε ν∂νL(x ),

(1.48)

which allows us to identify

δ L = ε ν∂νL .

(1.49)

For concreteness, let us first consider a temporal translation, in which case one has ε μ = ε δ 0μ. This allows us to identify δ L = ε ∂tL and hence F μ = ε δ 0μ L and, since this is a total derivative, we can identify Xa = ε ∂tϕa . Plugging these into the general form of Noether’s theorem (1.42), we obtain the following conserved current (which I indicate as (j μ )0,ε since we have singled out the temporal direction and it depends on ε ) 1-11

Relativistic Quantum Field Theory, Volume 1

(j μ )0,ε =

∂L (ε ∂tϕa ) − (ε δ 0μ L), ∂(∂μϕa )

(1.50)

which obeys ∂μ(j μ )0,ε = 0. Since the overall factor of ε appears on both terms, we can eliminate it to define the quantity that is conserved in general

(j μ ) 0 =

∂L ∂tϕa − δ 0μL . ∂(∂μϕa )

(1.51)

If one repeats this exercise for a general four-translation, one finds that there is a conserved current associated with each component of ε ν . In this general case, one has F μ = ε δνμ L and Xa = ε ∂νϕa . Factoring out the arbitrary infinitesimal from each conserved current, we obtain

(j μ )ν =

∂L ∂νϕ − δνμL ≡ T μν. ∂(∂μϕa ) a

(1.52)

The quantity T μ ν defined above is called the energy–momentum tensor. To obtain T μν we contract equation (1.52) with a metric tensor to obtain

T μν =

∂L ∂ νϕ − η μνL . ∂(∂ μϕa ) a

(1.53)

Since it is conserved, each component indexed by ν satisfies

∂μT μν = 0.

(1.54)

Having determined the current, we can determine the globally conserved quantities (charges) corresponding to this current using equation (1.43) with the volume V taken to be all space. From the zeroth component we identify

E=

∫ d 3x T 00,

(1.55)

as the associated conserved quantity, which we identify as the total energy of the field configuration. From the spacelike ( μ = 1, 2, 3 ≡ i ) components we introduce

Pi =

∫ d 3x T 0i ,

(1.56)

as the conserved quantity, which we can easily identify as the total three-momentum of the field configuration. 1.7.2 The Klein–Gordon energy–momentum tensor To get a feeling for the field theory energy–momentum tensor, let us consider the simple Klein–Gordon theory presented previously. Recall, that for such a theory, the Lagrangian is

L=

1 α 1 ∂ ϕ∂αϕ − m 2ϕ 2 . 2 2

1-12

(1.57)

Relativistic Quantum Field Theory, Volume 1

To apply equation (1.52), we consider the first term on the right side, which requires us to compute

∂L = ∂ μϕ . ∂(∂μϕ)

(1.58)

As a result, we have T μ ν = ∂ μϕ ∂νϕ − δνμL. Using the Minkowski metric η μν to raise the ν index gives

T μν = ∂ μϕ ∂ νϕ − η μνL .

(1.59)

One can verify explicitly using the equation of motion for ϕ obtained previously (1.10) that this quantity is conserved, i.e. ∂μT μν = 0. Based on this, we can identify the total energy and three-momentum of the Klein–Gordon field

E=

1 2

∫ d 3x[ϕ2̇ + (∇ϕ)2 + m2ϕ2 ], Pi =

∫ d 3x ϕ̇ ∂iϕ.

(1.60) (1.61)

Notice that, for this specific case, the energy–momentum tensor T μν is symmetric, i.e., T μν = T νμ. This is required, e.g. for use in Einstein’s equations of general relativity, but is not always guaranteed. If one requires a symmetric energy– momentum tensor it is possible to modify the definition of the energy–momentum tensor by adding an extra term that does not affect its conservation. The method, originally introduced by Belinfante, is as follows [4]: one defines a new quantity, which here I will call T˜ μν , via μν T˜ ≡ T μν + ∂λΓλμν ,

(1.62)

where T μν is the ‘canonical energy–momentum tensor’ introduced above and Γλμν is a function of the fields that is anti-symmetric in the first two indices, i.e., Γλμν = −Γμλν . The antisymmetry of Γλμν guarantees that ∂μ∂λΓλμν = 0, so that the modified energy– momentum tensor, T˜ μν , is also conserved. 1.7.3 Angular momentum and Lorentz transformations As mentioned previously, general Lorentz transformations include rotations and boosts. In addition, we discussed the fact that all fundamental theories should be Lorentzinvariant. We will now consider what the field theoretical implications of this in terms of conserved quantities are. We can anticipate that we will be able to derive something akin to angular momentum conservation for fields, but at first sight it is unclear what the implications of Lorentz boost invariance are. Luckily, the mathematical machinery developed so far will tell us without us having to guess. We will develop this for a single component scalar field, but the technique can be easily generalized. To begin, let us consider an infinitesimal Lorentz transformation

Λμν = δ μν + ω μν .

1-13

(1.63)

Relativistic Quantum Field Theory, Volume 1

Using equation (1.22), one can show that ω μν must be an anti-symmetric tensor, i.e. ω μν = −ω νμ. The transformation of a scalar field subject to this transformation is

ϕ(x ) → ϕ′(x ) = ϕ(Λ−1x ) ≃ ϕ(x μ − ω μνx ν ) ≃ ϕ(x ) − ω μνx ν∂μϕ(x ).

(1.64)

From this we learn that δϕ = −ω μ νx ν ∂μϕ. Similarly one finds that the variation of the Lagrangian subject to this transformation is

δ L = −ω μνx ν∂μL = −∂μ(ω μνx νL),

(1.65)

where the last equality holds because ∂μω μ ν = 0 since (a) it has no coordinate dependence and (b) because the antisymmetry of ω μ ν implies that ω μ μ = 0. As we see for the last relation, when subject to a Lorentz transformation, the Lagrangian changes by a total derivative. This means that there will be an associated conserved current and we can identify X = δϕ = −ω μ νx ν ∂μϕ. To complete the application Noether’s theorem, from (1.65) we identify F μ = −ω μ νx ν L. Plugging these into the general expression for the conserved (Noether) current (1.42) we obtain

∂L ωα νx ν∂αϕ + ω μνx νL ∂(∂μϕa ) ⎡ ∂L ⎤ ∂αϕ − δαμL⎥ = − ω α νx ν⎢ ⎣ ∂(∂μϕ) ⎦

jμ = −

(1.66)

= − ω α ν x νT μ α . Since there are six independent components of the anti-symmetric tensor ω μ ν , there will be six conserved currents associated with each choice of ω μ ν . As in the case of translations we can divide out the actual infinitesimals in each case to give the physical currents. The result is5

M μνρ = x νT μρ − x ρT μν.

(1.67)

The quantity above obeys the conservation law ∂μM μνρ = 0. For ν, ρ = 1, 2, 3 the Lorentz transformation is a rotation and the three conserved charges resulting from equation (1.43) give the total angular momentum of the field

Q ij =

∫ d 3x (xiT 0j − x jT 0i ).

(1.68)

From this expression one sees immediately that Q ii = 0 and Q ij = −Q ji , so there are three independent conserved quantities. Additionally, based on our previous discussion of the energy–momentum tensor, one can identify T 0i as the three-momentum density. The expression is simple for our scalar field example because T μν is symmetric for this theory. In general, the expression is more complicated unless one uses the symmetrized form T˜ μν , in which case the general formula looks like the one obtained here but with T μν → T˜ μν . 5

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Relativistic Quantum Field Theory, Volume 1

Upon inspection one sees that Q ij is field theoretical analog of the conservation of the three vector components of L = r × p as anticipated in the preliminary discussion. We are now in a position to consider the Lorentz boosts. In this case, we can choose either ν or ρ to be zero and the other index to be 1, 2, 3 to obtain

Q 0i =

∫ d 3x (x 0T 0i − xiT 00).

(1.69)

This one is harder to interpret immediately. However, since Q 0i corresponds to a conserved charge we can take a derivative of this expression with respect to time (t = x 0) to obtain a somewhat more transparent relation

dQ 0i = 0= dt

∫ d 3x T 0i + t dtd ∫ d 3x T 0i − dtd ∫ d 3x xiT 00

dP i d =P +t − dt dt i

∫ d xxT 3

i

00

(1.70)

.

Since P i is a conserved quantity (proven earlier), one has dP i /dt = 0. As a consequence one finds that

d dt

∫ d 3x xiT 00 = Pi = constant.

(1.71)

Since T 00 is the energy density, the integral appearing on the left-hand side is something like ‘center of energy’. With this in mind we see that the equation above states that the center of energy moves with a constant momentum. This is the fieldtheoretical version of Newton’s first law emerging from a conservation law! Exercise 1.4 Using (1.10), show that the Klein–Gordon energy–momentum tensor (1.59) is conserved, i.e. that ∂μT μν = 0.

1.8 The Hamiltonian formalism for fields In modern relativistic QFT, one could proceed solely with the Lagrangian formalism and, by using the path integral formalism, one can obtain a properly quantized theory. In this volume, I will concentrate on the so-called method of canonical quantization, for which we need the Hamiltonian formalism for a field theory. We will cover the method of path integral quantization in the second volume. To start, we introduce the canonical momentum π a(x )6, which is the conjugate variable to the field ϕa(x ),

π a (x ) =

∂L . ∂ϕȧ

(1.72)

In the condensed matter context, one does not usually need to indicate ‘up’ or ‘down’ indices, however, with an eye towards relativistic applications, I will try to keep the labels in the right place in preparation for when they are Lorentz indices. 6

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Relativistic Quantum Field Theory, Volume 1

The Hamiltonian density, which is a function of π a and ϕa (and their gradients, ∇π a and ∇ϕa ), is given by

H(π a , ϕa ) = π aϕȧ − L .

(1.73)

Similarly to classical mechanics, one eliminates ϕȧ in favor of π a . The Hamiltonian itself is obtained by spatial integration of the Hamiltonian density

H [π a , ϕa ] =

∫ d 3x H(π a, ϕa ).

(1.74)

1.8.1 Hamilton’s equations for fields The Hamilton’s equations arise in a similar manner as in classical mechanics, with the result being expressed in terms of functional derivatives of the Hamiltonian with respect to the field and its conjugate momentum7

ϕȧ (x ) =

δH [π , ϕ ] , δπ a(x )

π ̇ a (x ) = −

δH [π , ϕ ] . δϕa (x )

(1.75)

(1.76)

Based on the discussion in appendix B (see in particular equation (B.9)), we see that, since the Hamiltonian density is an ordinary function of π a and ϕa and their gradients, ∇π a and ∇ϕa , the functional derivatives above can be expressed explicitly in terms of the Hamiltonian density

⎡ ∂H ⎤ ∂H δH ⎥, − ∇⎢ = ∂ϕa δϕa ⎣ ∂(∇ϕa ) ⎦

(1.77)

⎡ ∂H ⎤ ∂H δH = a − ∇⎢ ⎥. a ∂π δπ ⎣ ∂(∇π a ) ⎦

(1.78)

Note that, although the Hamiltonian formalism looks like it manifestly breaks Lorentz invariance since space and time are not treated on equal footing, it is in fact Lorentz-invariant. Example: The Hamiltonian for a simple interacting scalar field theory To begin, we expand out the Lorentz contractions appearing in the interacting scalar field Lagrangian in order to easily identify the canonical momentum

7

See appendix B for an introduction/review of functional derivatives.

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Relativistic Quantum Field Theory, Volume 1

1 1 L = ∂μϕ ∂ μϕ − m 2ϕ 2 − U (ϕ) 2 2 1 1 2̇ 1 = ϕ − (∇ϕ)2 − m 2ϕ 2 − U (ϕ). 2 2 2

(1.79)

From the second line, it is easy to see that π = ϕ ̇ and one straightforwardly obtains

H=

1 1 2 1 π + (∇ϕ)2 + m 2ϕ 2 + U (ϕ). 2 2 2

(1.80)

In order to determine the field equation of motion, we can evaluate the necessary functional derivatives in equations (1.75) and (1.76) with the help of equations (1.77) and (1.78). The result is

δH = π = ϕ ̇, δπ

(1.81)

δH = m 2ϕ + ∂ϕU − ∇2 ϕ = −π ̇ = −ϕ .̈ δϕ

(1.82)

The first equation above simply returns the definition of π as usual. Upon rearrangement, the second equation gives

(∂t2 − ∇2 )ϕ + m 2ϕ + ∂ϕU = 0,

(1.83)

which can be written compactly as

(∂ μ∂μ + m 2 )ϕ + ∂ϕU = 0,

(1.84)

which is the Klein–Gordon equation of motion obtained previously using the Lagrangian formalism (1.13). Exercise 1.5 Show that equations (1.77) and (1.78) follow from equation (B.8). Hint: Recall that H depends on ϕa , ∇ϕa , πa , and ∇πa . Also take a look at equation (2.29).

References [1] Landau L D and Lifshitz E M 1980 The Classical Theory of Fields vol 2 4th ed (Oxford: Butterworth-Heinemann) [2] Goldstein H, Poole C P and Safko J L 2001 Classical Mechanics 3rd ed (Reading, MA: Addison-Wesley) [3] Jackson J D 1998 Classical Electrodynamics 3rd ed (New York: Wiley) [4] Belinfante F 1940 Physica 7 449–74

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IOP Concise Physics

Relativistic Quantum Field Theory, Volume 1 Canonical formalism Michael Strickland

Chapter 2 Quantization of free fields

We now turn to the quantization of free field theories, the result of which will be a launching point for full quantum field theory. I will focus on a simple free scalar field theory (U (ϕ ) = 0) but, first, I would like to discuss how one quantizes a quantum mechanical system consisting of N equal masses connected by identical springs with spring constant κ , with the masses being confined to move in only one dimension. We will refer to this as the ‘quantum linear chain.’ This example will provide the background necessary for the leap to quantum field theory.

2.1 The quantum linear chain and phonons To prepare for the quantum linear chain, we must first consider the classical problem: a linear ‘chain’ of equal masses connected by ideal springs with spring constant. We will assume that (1) the masses in the chain can only move in one dimension, (2) all springs have the same spring constant κ , and (3) all springs have zero equilibrium (rest) length and obey periodic boundary conditions. In the static equilibrium case, all masses are separated by a distance ε . We introduce coordinates qi which are the displacement of each oscillator from its equilibrium position. The classical Hamiltonian for this system is N

pn2

N

1 H=∑ + κ ∑(qn+1 − qn)2 . 2m 2 n=1 n=1

(2.1)

In figure 2.1, I sketch the equilibrium configuration in the top row and a particular ‘perturbed’ state in the bottom row. For the details of the classical analysis see appendix A.4 where I show that the classical solution can be written in terms of complex-valued normal coordinates bk . The normal coordinates bk are coefficients in a Fourier expansion of the generalized coordinates of the masses, qn with n ∈ (1, ⋯ , N ). The solution for the generalized coordinates obtained in section A.4 is

doi:10.1088/2053-2571/ab30ccch2

2-1

ª Morgan & Claypool Publishers 2019

Relativistic Quantum Field Theory, Volume 1

Figure 2.1. Setup for the linear chain discussion. The top line shows the equilibrium configuration and the bottom line shows a particular ‘perturbed’ state.

qn(t ) =

∑⎡⎣bk e−iω tun,k + b k*e iω tu n*,k ⎤⎦, k

k

(2.2)

k

where k is the discrete wavenumber of the basis element, ω k2 = (4κ /m ) sin2(kε /2) is the energy of the k th mode (dispersion relation), and un,k = e ikεn / N is the normalized spatial Fourier basis function. In the classical case, the complex-valued coefficients bk are fixed by specifying the 2N initial conditions qn(0) and qṅ (0). Based on equation (2.2) and the relationship between the generalized coordinates and momenta, pn = mqṅ , one obtains

pn (t ) =

∑( −imωk )⎡⎣bk e−iω tun,k − b k*e iω tu n*,k ⎤⎦. k

k

(2.3)

k

As discussed in appendix A, one can use the Poisson bracket formalism to derive some suggestive relations between the generalized coordinates and momenta

{qn, pn ′ }PB = δnn ′, {qn, qn ′}PB = {pn , pn ′ }PB = 0.

(2.4)

From these two relations and the expansions of qn and pn in terms of the normal coordinates bk , one obtains the Poisson bracket relations for the normal coordinates

{bk , b k*′}PB =

−i δkk ′, 2mωk

{bk , bk ′}PB = {b k*, b k*′}PB = 0.

(2.5) (2.6)

2.1.1 Going from classical to quantum The bridge from here to the quantum world is actually quite straightforward. We simply transform the generalized coordinates and momenta into operators and 2-2

Relativistic Quantum Field Theory, Volume 1

replace the Poisson brackets by commutators (up to a factor of i on the right-hand side)

[qˆn, pˆn ′ ] = iδnn ′, [qˆn, qˆn ′] = [pˆn , pˆn ′ ] = 0.

(2.7)

Note that there is no factor of ℏ on the right-hand side of the top line since in natural units ℏ = 1. The Hamiltonian in this case is obtained from a straightforward promotion of the c-number variables to operators N

N

1 1 Hˆ = pˆn2 + κ ∑(qˆn+1 − qˆn)2 . ∑ 2m n = 1 2 n=1

(2.8)

As in the classical case, we will once again introduce normal coordinates, expanding the operators qˆn and pˆn in terms of a basis uk,n . However, in the quantum case, the coefficients bˆk become operators. Therefore, in analogy to (2.2), we obtain

qˆn(t ) =

∑⎡⎣bˆk (t )un,k + bˆ k (t )u n*,k ⎤⎦, †

(2.9)

k

where here we have absorbed the explicit time dependence into the coefficients. Likewise, the generalized momenta operators become

pˆn (t ) =

∑( −imωk )⎡⎣bˆk (t )un,k − bˆ k (t )u n*,k ⎤⎦. †

(2.10)

k

Both the coordinate and momenta operators are manifestly Hermitian, i.e., qˆn† = qˆn and pˆn† = pˆn , which guarantees that expectation values are real valued. Based on these results, one can obtain an explicit expression for the normal coordinate operators N ⎡ ⎤ 1 i ˆ pˆn (t )⎥ . bk (t ) = ∑u k*,n⎢qˆn(t ) + ⎦ 2 n=1 ⎣ mωk

(2.11)

The time dependence of the normal coordinate operators bˆk (t ) is simple

bˆk (t ) = bˆk (0)e−iωkt ,

(2.12)

which can be verified by plugging (2.11) and (2.8) into the Heisenberg time-evolution equation i dtd bˆk (t ) = [bˆk (t ), Hˆ ]1. Equation (2.11) can be used to calculate the commutators of the normal coordinate operators, with the result being

1 This time-dependence could also have been anticipated from the fact that we absorbed a factor of e−iωkt into the coefficients in going from the classical to quantum-mechanical notation.

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Relativistic Quantum Field Theory, Volume 1

1 ⎡ ˆ ˆ †⎤ δkk ′, ⎢⎣bk , b k ′⎥⎦ = 2mωk

(2.13)

⎡ † †⎤ [bˆk , bˆk ′] = ⎢⎣bˆ k , bˆ k ′⎥⎦ = 0,

(2.14)

where I have made use of the completeness relations (A.27) and (A.30). I would like you to take a moment and reflect on the similarity of the result above and the classical result presented in equations (2.5) and (2.6). Once again, we see that the quantum commutator relations can be obtained (up to multiplication by a factor of i on the right-hand side) from the classical Poisson brackets. This suggests a straightforward path to field quantization: We first determine the necessary theory and Poisson brackets for classical fields. We then promote them to operator-valued functions with the Poisson brackets replaced by commutators via {A, B} → 1i [Aˆ , Bˆ ]. In fact, this is what we will do for bosons next. For fermions, the commutators will become anti-commutators, but we will cross that bridge when we come to it. However, before proceeding, we need to clean up some things. As defined so far, the normal coordinate operators have dimensions of length, so the commutators have dimensions of length squared. Looking forward, we will obtain a somewhat simpler/more transparent formalism if we define new dimensionless operators aˆk via

aˆk =

2mωk bˆk .

(2.15)

Written in terms of these new operators, the generalized coordinate and momenta operators become

qˆn(t ) =

1 ⎡⎣aˆk(t )u n,k + aˆ †(t )u n*,k ⎤⎦ , k 2mωk

∑ k

pˆn (t ) = −i∑ k

mωk ⎡ † ⎣aˆk(t )u n,k − aˆ k (t )u n*,k ⎤⎦ . 2

(2.16)

(2.17)

The commutation relations become simple

⎡⎣ aˆk , aˆ † ⎤⎦ = δkk ′, k′

(2.18)

[aˆk , aˆk ′] = ⎡⎣ aˆ k†, aˆ k†′⎤⎦ = 0.

(2.19)

Using (2.16) and (2.17) one can follow similar steps as in the classical case (see appendix A) to express the Hamiltonian in terms of the normal coordinate operators, with the exception being that one needs to take care with the fact that the coefficients no longer commute in some cases. The result is

(

)

† † Hˆ = ∑mω k2 bˆk bˆ k + bˆ k bˆk =

k

⎛ 1⎞ = ∑ωk⎜aˆ k†aˆk + ⎟ , ⎝ 2⎠ k 2-4

ω

∑ k ( aˆkaˆ k† + aˆ k†aˆk ) 2 k

(2.20)

Relativistic Quantum Field Theory, Volume 1

where, in going from the first to second lines, we have used the commutation relation (2.18). Remarkably, this is the Hamiltonian of an ensemble of uncoupled harmonic oscillators with energies ωk . The resulting quantized vibrations of the system are called phonons. We recognize aˆ k† as the creation operator for a phonon with wavenumber k and aˆk as the destruction operator for a phonon with wavenumber k . At this point, however, we should take a step back, since worrisomely, the final expression for the Hamiltonian includes a zero-point energy contribution 12 ωk for each and every oscillator mode2. As we take the number of oscillators to infinity (continuum limit), this could be troublesome since we will obtain an infinity. The way out of this box is to recognize that this contribution is a constant and, since we can shift the energy of a system by an overall constant without affecting the dynamics, we can simply get rid of this (potentially infinite) constant. We will return to this issue shortly. 2.1.2 Construction of the vacuum and many phonon states The eigenstates of the Hamiltonian operator can be constructed in the same manner as is done with a single harmonic oscillator, however, now the creation/destruction operators have an index k associated with them, which labels the spatial wavenumber of the phonon. Since the states are uncoupled and creation/destruction operators with different wavenumbers commute, the full quantum state is given by a direct product of quantum harmonic oscillator states. Generalizing from the case of a single harmonic oscillator, the many-particle ground state (vacuum), ∣0〉, is defined by requiring that any mode-number destruction operator applied to it returns zero, i.e.,

aˆk 0 = 0.

(2.21)

Physically, this means that the ground state contains no phonons. Multi-phonon states can be constructed in a similar manner to a single harmonic oscillator. First, we define the multiple-particle state

n ≡ ∣n1, n2 , ⋯〉 =

∏ ∣n k 〉 ,

(2.22)

k

where nk indicates the number of phonons with wavenumber k . Note that in this notation ∣0〉 = ∣0, 0, 0, ⋯〉 = ∏k ∣0k 〉. We can construct ∣nk 〉 in the usual manner by applying multiple creation operators

∣n k 〉 =

1 n aˆ k†) k ∣0k 〉, ( nk !

(2.23)

where the square root of nk ! guarantees that 〈nk ∣nk 〉 = 1. 2 In quantum field theory, we lump all of these zero-point contributions together into something we call the vacuum energy.

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Relativistic Quantum Field Theory, Volume 1

Finally, in this context, I note that these states satisfy the stationary Schrödinger equation

(Hˆ − En) n = 0,

(2.24)

with

En =

⎛

∑ωk⎜⎝nk + k

1⎞ ⎟. 2⎠

(2.25)

2.2 Poisson brackets in classical field theory At the end of the last chapter we discussed the Hamiltonian formalism for classical fields. We will now discuss the mathematical background associated with Poisson brackets for continuous fields in preparation for quantizing fields. Given two functionals A[ϕ, π ] and B[ϕ, π ], in analogy to (A.12), we define

{A , B}PB =

⎛

⎞

∫ d 3x⎜⎝ δϕδA(x) δπδB(x) − δπδA(x) δϕδB(x) ⎟⎠.

(2.26)

If we evaluate the Poisson bracket between a functional that has no explicit time dependence and the Hamiltonian functional, we obtain, similarly to (A.17),

dA = dt

⎡

⎤

∫ d 3x⎢⎣ δϕδA(x) ϕ(̇ x) + δπδA(x) π (̇ x)⎥⎦ = {A, H }PB ,

(2.27)

where in the last step, we used the Hamilton equations of motion (1.76) and (1.77). In the case that the functionals are simply functions, things simplify a bit. For example, consider the Poisson bracket between ϕ(x , t ) and the Hamiltonian

ϕ(̇ x , t ) = {ϕ(x , t ), H }PB ⎡ δϕ(x , t ) δH δϕ(x , t ) δH ⎤ = d 3x′⎢ − ⎥ δπ (x′ , t ) δϕ(x′ , t ) ⎦ ⎣ δϕ(x′ , t ) δπ (x′ , t ) δϕ(x , t ) δH = d 3x′ δϕ(x′ , t ) δπ (x′ , t ) δH = d 3x′δ 3(x − x′) δπ (x′ , t ) δH = , δπ (x , t )

∫ ∫

(2.28)

∫

which is a result we have already obtained, but gives an example of the manipulation of functional derivatives. In going from the second to third lines, we used the fact that ϕ and π are independent functionals and hence

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Relativistic Quantum Field Theory, Volume 1

δϕ(x , t ) = 0. δπ (x′ , t )

(2.29)

In going from the third to fourth lines, we used the fact that

δϕ(x , t ) = δ 3(x − x′). δϕ(x′ , t )

(2.30)

This formula can be obtained from the defining relation for the functional derivative (B.3) generalized to three dimensions and taking F → ϕ to obtain

δϕ(x , t ) =

δϕ(x , t ) δϕ(x′ , t ). ∫ d 3x′ δϕ ( x′ , t )

(2.31)

Upon inspection, this immediately implies equation (2.30). Alternatively, one could use (B.6), generalized to three dimensions, to obtain equation (2.30). Likewise, one has

δπ (x , t ) = δ 3(x − x′). δπ (x′ , t )

(2.32)

Based on the discussion above, we can now derive the Poisson brackets of the field and the conjugate momentum

{ϕ(x , t ), π (x′ , t )}PB = =

δϕ(x , t ) δπ (x′ , t ) ∫ d 3x″ δϕ (x″ , t ) δπ (x″ , t )

∫ d 3x″δ 3(x − x″)δ 3(x′ − x″)

(2.33)

= δ 3(x − x′). Similarly, one can evaluate the Poisson brackets between two ϕ’s or two π ’s. Summarizing, the final results for the equal time field Poisson brackets are

{ϕ(x , t ), π (x′ , t )}PB = δ 3(x − x′),

(2.34)

{ϕ(x , t ), ϕ(x′ , t )}PB = {π (x , t ), π (x′ , t )}PB = 0.

(2.35)

2.3 Quantization of a free scalar field theory The discussion in the preceding two sections has set the stage for the quantization of a free scalar field theory with U (ϕ ) = 0. Following the example of the quantum linear chain, we write the field and the canonical momentum as a linear superposition of creation and annihilation operators, indexed by the 3-momentum, which is now continuous since we are dealing with infinite 3-space. I remind you that the Lagrangian density in this case is

=

1 μ 1 ∂ ϕ∂μϕ − m 2ϕ 2 , 2 2

2-7

(2.36)

Relativistic Quantum Field Theory, Volume 1

and the corresponding Hamiltonian density is

=

1 1 2 1 π + ( ∇ϕ ) 2 + m 2 ϕ 2 . 2 2 2

(2.37)

The equations of motion can be obtained from the Euler–Lagrange equation (or Hamilton’s equations) resulting in

(∂μ∂ μ + m2 )ϕ = 0.

(2.38)

This is solved by μ

ϕ = Ae ±ikμx ,

(2.39)

where A is an arbitrary constant and k 2 = k μkμ = m2 . Writing k μ = (E , k) we find E 2 = k2 + m2 , as expected. The general classical solution for the field is given by a superposition of the form

ϕ(x ) =

∫k

μ

μ

[e ikμx a*(k) + e−ikμx a(k)],

(2.40)

where, for notational convenience, I have introduced

∫k ≡ ∫

d 3k (2π )3

1 . 2E (k)

(2.41)

From this, one can obtain the canonical momentum via π = ϕ ̇

π (x ) =

∫k

μ

μ

(iEk )[e ikμx a*(k) − e−ikμx a(k)].

(2.42)

After quantization of the field and canonical momentum, one obtains

ϕˆ (x ) = πˆ (x ) =

∫k

∫k

μ

μ

[e ikμx aˆ †(k) + e−ikμx aˆ(k)], μ

μ

(iEk )[e ikμx aˆ †(k) − e−ikμx aˆ(k)].

(2.43) (2.44)

The creation (aˆ†) and annihilation (aˆ ) operators obey the commutation relations

[aˆ(k), aˆ †(k′)] = (2π )3δ 3(k − k′),

(2.45)

[aˆ(k), aˆ(k′)] = [aˆ †(k), aˆ †(k′)] = 0.

(2.46)

The factor of (2π )3 appearing in the first line stems from the factor of 1/(2π )3 appearing in the integration measure (2.41). These commutation relations guarantee that the following equal-time commutation relations between the field and canonical momentum operators are satisfied

[ϕˆ (x , t ), πˆ (x′ , t )] = iδ 3(x − x′),

2-8

(2.47)

Relativistic Quantum Field Theory, Volume 1

[ϕˆ (x , t ), ϕˆ (x′ , t )] = [π (x , t ), π (x′ , t )] = 0.

(2.48)

These are analogous to the classical Poisson brackets (2.34) and (2.35). Exercise 2.1 Show that equations (2.47) and (2.48) follow from equations (2.43), (2.44), (2.46), and (2.45).

2.3.1 Quantized Hamiltonian operator Using the quantum mode decompositions given in equations (2.43) and (2.44), the Hamiltonian density (2.37) in operator form becomes ⎧ ˆ = 1⎨  2⎩ +

∫p ∫k

( − E (p)E (k))[e ipμ x aˆ †(p) − e−ipμ x aˆ(p)][e ikμx aˆ †(k) − e−ikμx aˆ(k)]

∫p ∫k

( − p · k)[e ipμ x aˆ †(p) − e−ipμ x aˆ(p)][e ikμx aˆ †(k) − e−ikμx aˆ(k)]

+ m2

μ

μ

μ

∫p ∫k

μ

μ

μ

μ

μ

(2.49)

⎫ μ μ μ μ [e ipμ x aˆ †(p) + e−ipμ x aˆ(p)][e ikμx aˆ †(k) + e−ikμx aˆ(k)]⎬ . ⎭

Integrating over space ∫ ≡ ∫ d 3x using x

∫x

μ

e ±i (pμ +kμ )x = (2π )3δ 3(p + k)e ±2iEkt ,

(2.50)

∫x

(2.51)

μ

e ±i (pμ −kμ )x = (2π )3δ 3(p − k),

ˆ one obtains the Hamiltonian operator Hˆ = ∫ H x

{

1 − Hˆ = Ek[aˆ †( −k)aˆ †(k)e 2iEkt − aˆ †(k)aˆ(k) 4 k⃗ − aˆ(k)aˆ †(k) + aˆ( −k)aˆ(k)e−2iEkt ] +

∫

∫k⃗ Ek[aˆ†(−k)aˆ†(k)e 2iE t + aˆ†(k)aˆ(k) k

(2.52)

}

+ aˆ(k)aˆ †(k) + aˆ( −k)aˆ(k)e−2iEkt ] , where ∫ ⃗ ≡ ∫ d 3k/(2π )3 and I have introduced a compact notation Ek = E (k). k Simplifying this expression, one finds

1 Hˆ = 2

∫k⃗ Ek[aˆ†(k)aˆ(k) + aˆ(k)aˆ†(k)].

(2.53)

Once again, I would like you to take a step back and appreciate the similarities between the derivation above and the case of the classical linear chain presented in 2-9

Relativistic Quantum Field Theory, Volume 1

section A.4. Once again, the primary difference in the classical and quantum treatments is that the Fourier coefficients are now operators that have non-trivial commutation relations. 2.3.2 Vacuum renormalization and normal ordering We will close this section with a conundrum. We can use the commutation relation (2.45) to rewrite the field Hamiltonian operator above in a similar form as was obtained for the quantum linear chain (2.20). The result is

Hˆ =

⎡

⎤

∫k⃗ Ek⎢⎣aˆ†(k)aˆ(k) + 12 (2π )3δ 3(0)⎥⎦.

(2.54)

As we can see, the second term on the right-hand side is (doubly) infinite! This is related to the comments I made earlier in this chapter about the continuum limit of the quantum linear chain. This infinity is worrisome, but since only energy differences matter, one can subtract this particular infinity to obtain a finite field Hamiltonian operator and, hence, energy. This is refered to as vacuum energy subtraction. The standard notational prescription for doing this subtraction is called normal ordering. This operation is indicated by wrapping operators with colons before and after them. The notation means that we should take all creation operators and move them to the left of all annihilation operators. For example, the normal-ordered version of the field Hamiltonian operator above is indicated as :Hˆ : and applying the rules, one finds

1 : Hˆ : =: 2

∫k⃗ Ek[aˆ†(k)aˆ(k) + aˆ(k)aˆ†(k)]: =∫k⃗ Ekaˆ†(k)aˆ(k).

(2.55)

As we advance in quantum field theory, we will develop a much less ad hoc way to deal with the subtraction of infinities which is related to renormalization of the vacuum energy. Exercise 2.2 Show that equation (2.53) follows from equation (2.49). Note: Some of the connecting steps are already provided. You should provide, for example, the explicit details that were not shown when going from (2.49) to (2.52). 2.3.3 A note on dimensions I would like to briefly comment on the dimensionality of the field operators in order to get you thinking about this topic in general. We can determine the dimensionality of the fields by first considering the dimensionality of the Lagrangian density. In natural units, the action (S ) is dimensionless and, assuming four spacetime dimensions, the spacetime integral ∫ d 4x of the Lagrangian density gives the action. This implies that, in four spacetime dimensions, the Lagrangian density has

(

)

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Relativistic Quantum Field Theory, Volume 1

units of (length)−4 or, equivalently, (energy)4 . In the notation introduced in the ‘Units and Conventions’ preface, one can then write

[] = 4.

(2.56)

For a real scalar field we can use this to determine the dimension of the field operators. Since all terms in the Lagrangian density (2.36) have the same dimension, we can consider either term. In the first term, we have the product of two derivatives which each have dimension

[∂μ] = 1,

(2.57)

therefore, in four dimensions, one has

[ϕ ] = 1.

(2.58)

The same conclusion would be reached if we looked instead at the second term in the scalar Lagrangian. Looking forward, I note that in D-dimensions one would find instead

[] = D ,

(2.59)

and, since the derivatives have the same dimensionality, one would have, in general,

[ϕ ] = (D − 2)/2.

(2.60)

For now, however, let us focus on the case that D = 4, in which case we have [ϕ ] = 1. Looking at equation (2.41) we see that the integral ∫ has dimensions ⎡⎣∫ ⎤⎦ = 52 . The k k creation and annihilation operators in equation (2.43) must compensate this so that the right-hand side has the correct dimensions. As a consequence,

[aˆ ] = [aˆ † ] = −3/2.

(2.61)

This is consistent with equation (2.45), since on the right-hand side we have [δ 3(k − k′)] = −3 and on the left-hand side we have the product of two creation/ annihilation operators. 2.3.4 The vacuum state Similar to the case of the quantization of the quantum linear chain, once we have determined the quantum Hamiltonian operator, we can specify the vacuum state and multi-particle states that are constructed using the creation and annihilation operators introduced previously. We define the vacuum state ∣0〉 by requiring that an annihilation operator corresponding to any (continuous) wavenumber k returns zero, i.e.

aˆ(k) 0 = 0.

(2.62)

The vacuum state ∣0〉 is a vector consisting of vacuum states for every possible value of k , i.e. ∣0〉 = ∣0 k1, 0 k2 , ⋯〉, but now there are infinitely many values possible for k . 2-11

Relativistic Quantum Field Theory, Volume 1

As a result, our vacuum state is infinite-dimensional3. The vacuum state is normalized such that

〈0∣0〉 = 1.

(2.63)

2.3.5 Single particle states We started with a continuous scalar field and, upon quantization, we expressed the Hamiltonian in terms of creation and annihilation operators. As a result, we now have discrete excitations which we call particles. These particles can have continuous 3-momentum, but they are discrete in number (their number is integer-valued). To see how this arises, let us begin by looking at the basics. Using equation (2.53) and the commutation relations (2.45) and (2.46), one can show that

[Hˆ , aˆ †(k)] = Ekaˆ †(k),

(2.64)

[Hˆ , aˆ(k)] = −Ekaˆ(k),

(2.65)

where I remind you that Ek = E (k) = k2 + m2 . As a result of the commutation relations (2.64) and (2.65), we can construct energy eigenstates by acting on the vacuum ∣0〉 with the creation operator aˆ†(k). We define a single-particle state via

k = aˆ †(k) 0 .

(2.66)

Using the commutation relations (2.45) and (2.46), one can show that

aˆ(k′) k = aˆ(k′)aˆ †(k) 0 = (2π )3δ 3(k − k′) 0 .

(2.67)

Note, if we were to set k = k′ above we would obtain an infinity in the form of (2π )3δ 3(0) on the right-hand side above. This is related to the fact that the particle states we have created are not localized in space, they are instead perfectly localized in momentum space – they are momentum eigenstates. This is nothing new: in non-relativistic quantum mechanics, free-particle position and momentum eigenstates are also not normalizable (they also normalize to delta functions). To construct well-defined normalizable states we have to, for example, create a Gaussian wavepacket by creating a linear superposition of the momentum eigenstates. We will return to this issue in the future. For now let us proceed with the setup and defining relations. Let us next consider the action of the Hamiltonian operator on the vacuum state

3 Notationally, I will no longer indicate the vacuum (ground) state with a bold-faced zero; from now on, this is implicit.

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Relativistic Quantum Field Theory, Volume 1

Hˆ ∣0〉 =

{

1 2

}

∫→k Ek[aˆ†(k)aˆ(k) + aˆ(k)aˆ†(k)]

1 Ekaˆ(k)aˆ †(k)∣0〉 2 → k ⎛1 ⎞ = ⎜ δ 3(0) d 3kEk⎟ ∣0〉 = ∞∣0〉, ⎝2 ⎠  =E vacuum =

∫

∣0〉 (2.68)

∫

where, in going from the first to second lines, we have used equation (2.62) and I remind you that ∫ ⃗ = ∫ d 3k/(2π )3. Unfortunately, as alluded to previously, we see k that the vacuum energy E vacuum is doubly infinite: there is a singular delta function multiplying a divergent integral of the form ∫ d 3k k 2 + m2 . The prescription we will use is to define the real energy of a state via a shifted operator Hˆ ′ = Hˆ − E vacuum . Using this modified operator we would find Hˆ ′∣0〉 = 04. In practice, this is equivalent to always using the normal ordered version of the Hamiltonian operator

: Hˆ : 0 =

{∫

ˆ → Eka k

†

(k)aˆ(k)} 0 = 0,

(2.69)

where we have, once again, used equation (2.62). For the rest of this chapter, the Hamiltonian will implicitly be given by the normal-ordered version, but I will suppress the colons before and after in order to keep the formulas relatively compact. Using the commutation relations above we have

Hˆ ∣k〉 = =

{∫

→ k′

}

Ek ′aˆ †(′k)aˆ(′k) aˆ †(k)∣0〉

∫→k ′ Ek′aˆ†(′k)(2π )3δ 3(k − ′k)∣0〉

(2.70)

= Ekaˆ †(k)∣0〉 = Ek∣k〉. We interpret the state ∣k〉 as a momentum eigenstate of a single particle with mass m. We will verify this explicitly in the next section. Exercise 2.3 Show that equations (2.64) and (2.65) follow from equations (2.53), (2.45), and (2.46).

To make this mathematically sound, one should work in a finite-sized box and, additionally, put an upper limit (ultraviolet cutoff) on the k -integration. Only at the end of the calculation can we take the box size to infinity and the upper limit to infinity. Luckily, the end result is the same as our simple prescription. 4

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Relativistic Quantum Field Theory, Volume 1

Total linear momentum and angular momentum To further explore the particle interpretation, let us look at the other quantum numbers of the single particle state. Taking the classical total momentum determined using Noether’s theorem (1.61) and promoting the field to an operator-valued function we obtain

∫ d 3x[∂tϕˆ(∂iϕˆ ) + (∂iϕˆ )∂tϕˆ ] = − 12 ∫ d 3x[πˆ (∇i ϕˆ ) + (∇i ϕˆ )πˆ ],

1 i Pˆ = 2

(2.71)

where we have symmetrized the classical integrand prior to quantization since the order of the operators is irrelevant in the classical case but is important in quantum i field theory. This construction guarantees that Pˆ is a Hermitean operator. Using (2.43), we have

∇i ϕˆ (x ) =

∫k

μ

μ

( −ik i )[e ikμx aˆ †(k) − e−ikμx aˆ(k)],

(2.72)

where I remind you that ∫ = ∫ ⃗ (2Ek )−1/2 . Using this and equation (2.44), we obtain k k

1⎧ i Pˆ = − ⎨ 2⎩ +

∫x ∫k ∫p k iEp[eik x aˆ†(k) − e−ik x aˆ(k)][eip x aˆ†(p) − e−ip x aˆ(p)] μ

μ

∫x ∫k ∫p p Ek[e i

μ

ikμx μ ˆ †

a ( k) − e

μ

μ

−ikμx μ ˆ

a(k )][e

μ

ipμ x μ ˆ †

a ( p) − e

μ

μ

⎫ a(p)]⎬ . ⎭

(2.73)

−ipμ x μ ˆ

Using equations (2.50) and (2.51), we can perform the integral over x to obtain 1 i Pˆ = − 4

=

{∫

k⃗

k i [aˆ†(k)aˆ†( − k)e 2iEkt + aˆ(k)aˆ( − k)e−2iEkt − aˆ†(k)aˆ(k) − aˆ(k)aˆ†(k)]

+

∫k⃗ k i[−aˆ†(k)aˆ†(−k)e 2iE t − aˆ(k)aˆ(−k)e−2iE t − aˆ†(k)aˆ(k) − aˆ(k)aˆ†(k)] } (2.74)

1 2

∫k⃗ k i[aˆ†(k)aˆ(k) + aˆ(k)aˆ†(k)].

k

k

Similarly to the Hamiltonian operator, using the commutation relations we will once again contain an infinite contribution which is not physically relevant. Normal ordering the total momentum operator, we obtain the result

∫k⃗ kaˆ†(k)aˆ(k).

ˆ: = :P

(2.75)

From here on I will drop the colons as I did with the Hamiltonian operator, but it should be understood that the operator is the normal-ordered version. Acting with the total momentum operator on our single-particle state ∣k〉, one finds

ˆ k =k k , P

(2.76)

which demonstrates that our single-particle state is, in fact, a momentum eigenstate.

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Relativistic Quantum Field Theory, Volume 1

Following a similar (but even more laborious) procedure, one can construct the angular momentum operator for our scalar field. For this purpose, one must use the symmetric form of the energy momentum tensor discussed previously. I will not present the gory details in this case, and instead simply present the final result, which is

∫k⃗ aˆ†(k)(k × i∇k)aˆ(k).

: Jˆ : =

(2.77)

For more details see [1]. Using this operator and the commutation relations for the creation and annihilation operators, one can verify explicitly that a scalar particle at rest (k = 0) satisfies Jˆ ∣p = 0〉 = 0. This means that our quantized scalar field theory gives a spin-0 particle.

2.4 Multi-particle states and Fock space One can construct multi-particle states by repeatedly acting with creation operators. Similarly to the quantum linear chain, since by equations (2.45) and (2.46) creation/ annihilation operators with different wavenumbers commute, multi-particle states are direct products of single-wavenumber states. As a result, we can define a multiparticle via N

∣n〉 ≡ ∣n k1, n k1, ⋯ , n kN 〉 =

∏

∣n ki〉,

(2.78)

i=1

where n k1 indicates the number of particles with wavenumber k1. Each of the states that enter the direct product on the right-hand side are constructed in analogy to harmonic oscillator states

∣n k〉 =

1 (aˆ †(k))nk ∣0 k〉, n k!

(2.79)

where, as before, the square root of nk ! guarantees that 〈nk ∣nk 〉 = 1. If the momenta are distinct, i.e. all n ki = 1, we have, e.g.

∣k1, k2 , ⋯ , kN 〉 = aˆ †(k1)aˆ †(k2)⋯aˆ †(kN )∣0〉.

(2.80)

Since the creation operators aˆ†(ki) commute among each other by equation (2.46), the resulting states are symmetric under the exchange of any two particles. For example,

∣k1, k2〉 = ∣k2 , k1〉.

(2.81)

This means that our spin-0 scalar particles are bosons. In fact, that we were even able to put an arbitrary number of particles in a single wavenumber state above, implies that they are bosons.

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Relativistic Quantum Field Theory, Volume 1

In general, the full Hilbert space for our theory is spanned by acting on the vacuum state with all possible combinations of creation operators with various wave numbers, e.g.

∣0〉, aˆ †(k1)∣0〉, aˆ †(k1)aˆ †(k2)∣0〉, aˆ †(k1)aˆ †(k2)aˆ †(k3)∣0〉, ….

(2.82)

Note that, as mentioned above, since the states are bosonic, we can put multiple particles into one state. In practice, this means that the k ’s above can be degenerate. The full state is a linear combination of the basis states above and the constructed state space is called a Fock space. The Fock space is the sum of the n-particle Hilbert spaces. Since our states can now contain multiple particles, it is useful to introduce an operator that counts the number of particles in a given state. The operator is, quite naturally, called the number operator

Nˆ =

∫k⃗ aˆ†(k)aˆ(k).

(2.83)

One can verify by explicit calculation that the number operator satisfies

Nˆ ∣k1, k2 , ⋯ , kN 〉 = N ∣k1, k2 , ⋯ , kN 〉.

(2.84)

For the free field theory we are considering, the number operator can be shown to commute with the Hamiltonian operator, i.e. [Nˆ , Hˆ ] = 0. In practice, this means that, if we start with a given number of particles, the time evolution that occurs via the Hamiltonian operator will conserve this number, i.e. Ṅ = −i [Nˆ , Hˆ ] = 0. This is special to free field theories5. This may no longer be true when we include interactions, since interactions can create/destroy particles, thereby moving the state between different ‘sectors’ of Fock space. Exercise 2.4 Show that the normal-ordered number operator (2.83) commutes with the normal-ordered Hamiltonian (2.55). A note regarding relativistic normalization So far we have defined the vacuum normalization such that 〈0∣0〉 = 1. Using this and the commutation relations for the creation/annihilation operator, we have

〈k1∣k2〉 = 〈0∣aˆ(k1)aˆ †(k2)∣0〉 = (2π )3δ 3(k1 − k2).

(2.85)

With this normalization, one can also define a complete set of states that, when summed (integrated) over, give identity 3

∫ (2dπk)3 5

k k = 1.

This could also happen if the interactions are always number conserving.

2-16

(2.86)

Relativistic Quantum Field Theory, Volume 1

This definition is Lorentz invariant, but this is not obvious. The integration measure d 3k by itself is not Lorentz invariant and neither is the integrand, but it turns out that the entire expression is Lorentz invariant. To make the Lorentz invariance of our setup more obvious, in some texts you will encounter a different ‘relativistic normalization’ for the states, so I feel it is my duty to alert you to this. To define the ‘relativistic normalization’, we start by defining a momentum-space integration measure that is manifestly Lorentz-invariant. We then work out the consequences of this. We start by considering the 4-integral over momentum, restricted to on-shell states that have positive energy. This can be written as

∫ dK ≡ ∫ =

∫

d 4k 2πδ(k 02 − k2 − m 2 )θ (k 0) (2π )4 ∞ d 3k dk 0δ(k 02 − k2 − m 2 )θ (k 0), (2π )3 −∞

(2.87)

∫

where we have introduced the shorthand notation ∫ dK which will allow us to write many formulas compactly in the future. The delta function enforces k 0 = Ek = p2 + m2 and the theta function enforces energy positivity k 0 > 0. The delta function is obviously Lorentz invariant since the argument k 02 − k2 − m2 = kμk μ − m2 is Lorentz invariant. The theta function is also invariant since a Lorentz boost cannot change a negative energy state to a positive energy state or vice versa. Finally the 4-integration measure d 4k is invariant under proper Lorentz transformations since, in that case, det(Λ) = 1. Therefore, ∫ dK is Lorentz invariant. To simplify this further, we can factorize the argument of the delta function and use

δ(f (x )) =

∑ xi

1 δ(x − xi ), ∣f ′(x )∣x=xi ∣

(2.88)

where xi are the zeros of the function f (x ). The argument of the delta function in equation (2.87) can be factorized into (k 0 − k2 + m2 )(k 0 + k2 + m2 ). For realvalued k and m, only the first factor gives a positive solution, with k 0 = k2 + m2 = Ek . Taking the derivative of the argument and inserting this first solution gives f ′(x )∣x=x1 = 2Ek . Putting the pieces together we obtain

∫ dK = ∫

d 3k 1 . (2π )3 2Ek

(2.89)

Requiring ∫ dK to be our default 3-integration instead of ∫ ⃗ , we introduce a new set k of relativistic one-particle states indicated by ∣k 〉6, which satisfy

6

In this case the k is not typeset in bold-face to distinguish it from our original states.

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Relativistic Quantum Field Theory, Volume 1

∫ dK

k k = 1.

(2.90)

Expanding this out explicitly we obtain 3

∫ (2dπk)3 2E1 k

k k = 1.

(2.91)

Comparing this with equation (2.86), we can relate the relativistically-normalized states to the ones we have already obtained

∣k 〉 =

2Ek ∣k〉 =

2Ek aˆ †(k)∣0〉.

(2.92)

With this definition, the normalization condition for the relativistically-normalized states is

〈k1∣k2〉 = 2Ek(2π )3δ 3(k1 − k2).

(2.93)

In some texts you may also see relativistically normalized creation operators defined as aˆ†(k ) = 2Ek aˆ†(k). You have been warned, so do not be confused when you pick up another text and nothing makes sense at first glance. I will note, however, that in addition to being an impetus for a word of caution, during this discussion above we have learned something that will come up again, namely the definition of the invariant momentum-space measure given in equation (2.89).

2.5 Complex scalar fields We now turn to a generalization of the simple scalar field theory we have been considering thus far. The generalization is to make the fields complex-valued. As we will see below, this seemingly innocuous change has a profound effect since, in a complex scalar field theory, there is a conserved charge that remains conserved even when we add interactions. This is useful because it provides insight into how one might construct quantum field theories that conserve, e.g., electric charge, lepton number, baryon number, etc. We begin by considering the following classical Lagrangian density for a complex scalar field

 = ∂μψ *∂ μψ − m 2ψ *ψ − U (∣ψ ∣2 ),

(2.94)

where the potential is a general function of ∣ψ ∣2 = ψ *ψ . At this point one could expand ψ into two real-valued scalar fields ψ (x ) = 1 [ψ1(x ) + iψ2(x )] and treat these 2 two as independent fields, however, in practice, one can also treat the field ψ and its complex-conjugate ψ * as the independent degrees of freedom. For example, using the Euler–Lagrange equations of motion (1.4) we obtain the following equation of motion for ψ by varying ψ *

∂μ∂ μψ + m 2ψ +

∂U = 0. ∂ψ *

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(2.95)

Relativistic Quantum Field Theory, Volume 1

Likewise, one can obtain the following equation of motion for ψ * by varying ψ

∂μ∂ μψ * + m 2ψ * +

∂U = 0. ∂ψ

(2.96)

Expanding out the Lorentz-contractions in the Lagrangian it is straightforward to identify the two canonical momenta

π=

∂ = ψ ̇ *, ∂ψ ̇

(2.97)

π* =

∂ = ψ̇ . ∂ψ ̇ *

(2.98)

Using these, we can construct the Hamiltonian

∫x =∫ x

H=

(π ∂tψ + π *∂tψ * − ) (2.99) (π *π + ∇ψ * · ∇ψ + m 2ψ *ψ + U (∣ψ ∣2 )).

The non-trivial classical Poisson brackets for this theory are

{ψ (x , t ), π (′x , t )}PB = δ 3(x − ′x),

(2.100)

{ψ (x , t ), π *(′x , t )}PB = 0.

(2.101)

Finally, and probably most importantly, we note that the Lagrangian (and Hamiltonian) is invariant under a global phase change of the field, i.e. if we make the transformation

ψ → e iαψ ,

(2.102)

where α is independent of position and time7, then the Lagrangian is completely unchanged. This is an example of an internal symmetry. In this case, the symmetry corresponds to a symmetry under rotation of the complex-valued field ψ in the complex plane. As one might expect already, by Noether’s theorem, this implies that there is a conserved current. In this case, the infinitesimal field transform corresponds to δψ = iαψ and δL = 0. The associated locally conserved current is

j μ = i (∂ μψ *)ψ − iψ *(∂ μψ ).

(2.103)

Exercise 2.5 Show that the equations of motion (2.95) and (2.96) are consistent with those obtained if one instead splits ψ (x ) = 1 [ψ1(x ) + iψ2(x )] and derives equations 2 of motion for ψ1 and ψ2 instead. 7

This is what the word ‘global’ above implies.

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Relativistic Quantum Field Theory, Volume 1

Exercise 2.6 Show that equation (2.103) follows from the fact that the Lagrangian is invariant under a global phase rotation of the complex field ψ via equation (2.102).

2.6 Quantization of a complex scalar field We now turn to the quantization of the complex scalar field. We can now simply follow the rules and promote the fields to field operators and the Poisson brackets to commutators. The resulting field operators are

ψˆ (x ) =

∫k

ψˆ †(x ) =

∫k

μ μ [e ikμx aˆ †(k) + e−ikμx bˆ(k)],

μ † μ [e ikμx bˆ (k) + e−ikμx aˆ(k)],

(2.104) (2.105)

where, since there are two independent degrees of freedom in our complex field, there are two different types of creation/annihilation operators, which we call a and b. Based on these one can easily obtain the canonical momenta π = ψ ̇ and π † = ψ ̇ †. The equal-time commutation relations for the fields are

[ψˆ (x , t ), πˆ (x′ , t )] = iδ 3(x − x′),

(2.106)

[ψˆ (x , t ), πˆ †(x′ , t )] = 0.

(2.107)

† [aˆ(k), aˆ †(k′)] = [bˆ(k), bˆ (k′)] = (2π )3δ 3(k − k′),

(2.108)

† † [aˆ(k), aˆ(k′)] = [bˆ(k), bˆ(k′)] = [aˆ †(k), aˆ †(k′)] = [bˆ (k), bˆ (k′)] = 0,

(2.109)

† † [aˆ(k), bˆ(k′)] = [aˆ(k), bˆ (k′)] = [aˆ †(k), bˆ(k′)] = [aˆ †(k), bˆ (k′)] = 0.

(2.110)

These imply that

Based on the conserved classical current (2.103), we can define a conserved charge Q = ∫ j 0 (x ). The conserved charge can be expressed in terms of the canonical x momenta and field operators

: Qˆ : = −i

∫x

(πψ ˆ ˆ †). ˆ ˆ † − ψπ

(2.111)

Inserting the canonical momenta and field operators expanded in terms of the creation and annihilation operators one finds

: Qˆ : =

∫k

† (a ) (b ) [aˆ †(k)aˆ(k) − bˆ (k)bˆ(k)] ≡ Nˆ − Nˆ ,

(2.112)

where, in the final step, we defined the number operators for particles of type a and type b. One can easily verify that this charge is conserved, i.e. Qˆ ̇ = −i [Qˆ , Hˆ ] = 0, (a ) (b ) or Nˆ are however, when including interactions we will find that neither Nˆ 2-20

Relativistic Quantum Field Theory, Volume 1

individually conserved8. Note that the sign appearing in front each of the number operators above can be interpreted as their charge. We see from equation (2.112) that particles of type a have positive charge (particles) and particles of type b have negative charge (anti-particles). Earlier in this chapter, we covered the quantization of non-interacting scalar and complex scalar fields. We learned that we can straightforwardly transition from classical field theory to quantized quantum fields by rewriting the field and canonical momentum as operators that are expanded in a basis of creation and annihilation operators. For the real-valued scalar field theory, for example, the field and canonical momentum operators obeyed equal-time commutation relations of the form

[ϕˆ (x , t ), πˆ (x′ , t )] = iδ 3(x − x′),

(2.113)

[ϕˆ (x , t ), ϕˆ (x′ , t )] = [π (x , t ), π (x′ , t )] = 0.

(2.114)

This prescription is not obviously Lorentz invariant since time is singled out. This might be a problem since, if Lorentz invariance is broken, then the theory may not be causal. For a theory to be causal, it must be true that, if an event (e.g. measurement) occurs at a point x1 = (t1, x1) in space-time, it can only affect physics at a point x2 = (t2, x2) if any signal that propagates from x1 to x2 moves at a speed that is less than or equal to the speed of light. We will now formalize this statement and check to see if our construction obeys this constraint.

2.7 Causality To begin, we consider an event that occurs at point x1 in spacetime. The signal from this event propagates outward in all directions at a fixed velocity, which must be less than or equal to the speed of light. In figure 2.2, I show a two-dimensional slice of space-time in which I show four events. The origin of the coordinate system is chosen to coincide with the spacetime location of x1 and the points x2 , x3, and x4 are timelike, lightlike, and spacelike separated from the point x1, respectively. The dashed lines indicate the path that would be followed by signals traveling at the speed of light from point x1. In this figure, the timelike region with t > 0 is the future light cone9 and the region with t < 0 is called the past light cone. If an event occurs at x1, it can only affect physics in the forward light cone (this includes the light cone itself). Space-time points that are in the past light cone or the spacelike regions are causally disconnected regions of spacetime. The spacelike region is causally disconnected since, in order for a signal to propagate into that region from x1, the signal would have to travel faster than the speed of light (superluminal propagation). In general, one can classify the space-time separation of two points x and y into three categories

8 9

For a free complex scalar field theory, they are individually conserved, but this is a very special case. This is sometimes called the forward light cone.

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Relativistic Quantum Field Theory, Volume 1

Figure 2.2. Space–time digram showing four points x1, x2 , x3, and x4 (black filled circles). The origin of the coordinate system is chosen to coincide with the spacetime location of x1 and the points x2 , x3, and x4 are timelike, lightlike, and spacelike separated from the point x1, respectively. The dashed lines indicate the path that would be followed by signals traveling at the speed of light from point x1.

(x − y ) 2 > 0 (x − y ) 2 = 0 (x − y ) 2 < 0

timelike separated, lightlike separated, spacelike separated,

(2.115)

where x and y represent 4-vectors and I have used the shorthand notation (x − y )2 = (xμ − yμ )(x μ − y μ ). For our theory to be causal, it must be true that if we perform two measurements using operator Oˆ 1(x ) at space-time point x and operator Oˆ 2(y ) at space-time point y , then these two operators should commute (the measurement of Oˆ 1 cannot influence the measurement of Oˆ 2 ) if x and y are spacelike separated, i.e.

[Oˆ 1(x ), Oˆ 2(y )] = 0

if (x − y )2 < 0.

(2.116)

If this condition were violated, it would mean that a signal had propagated at a superluminal speed. Let us check to see if the real scalar field theory we introduced previously obeys this constraint. To do this, we consider an object Δ(x − y ) which is defined by the commutator of two field operators evaluated at different points in space-time

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Relativistic Quantum Field Theory, Volume 1

Δ(x , y ) = [ϕˆ (x ), ϕˆ (y )].

(2.117)

For a real scalar field theory, the fields are observable, so we can use this as our test observable. We can compute Δ(x , y ) by plugging in our general expansion of the field in terms of creation and annihilation operators (2.43) and performing some algebra Δ(x, y ) = =

=

∫p ∫k

μ

μ

μ

μ

[e ikμx aˆ †(k) + e−ikμx aˆ(k), e ipμ y aˆ †(p) + e−ipμ y aˆ(p)]

1 i (k x μ+p y μ ) † † −i (k x μ+p y μ ) {e μ μ [aˆ (k), aˆ (p)] + e μ μ [aˆ(k), aˆ(p)] 2 EkEp (2.118) μ−p y μ ) i (kμx μ−pμ y μ ) ˆ † i ( k x − † μ μ +e [a (k), aˆ(p)] + e [aˆ(k), aˆ (p)]}

∫k⃗ ∫p⃗ ∫

d 3k 1 −ikμ(x μ−y μ ) μ μ − e ikμ(x −y )] ≡ D(x − y ) − D(y − x ), [e (2π )3 2Ek

where we have introduced a function D, called the propagator, in the last line upon noticing that the result splits into two very similar terms. Properties of Δ(x , y ) • It is a classical-number (c -number) valued function. • It only depends on the separation of the two points x − y and, hence, in what follows we can write it as a single-argument function Δ(x , y ) → Δ(x − y ). • It is Lorentz invariant since the integration measure ∫ d 3k /(2Ek ) and the integrand are separately Lorentz invariant10. • It is antisymmetric under x ↔ y, i.e. Δ(x − y ) = −Δ(y − x ) consistent with equation (2.117). • It has support (does not vanish in general) for timelike separations. For example, consider x − y = (t , 0, 0, 0) which gives

[ϕ(x , t ), ϕ(x , 0)] = ∼

t →∞

∫

d 3k 1 −iEkt [e − e iEkt ] (2π )3 2Ek m [sin(mt ) − cos(mt )], t3

(2.119)

where I have discarded multiplicative constants that are not important for the discussion. To find this result, consider, for example, the first term in the integral above

D(x − y ) =

∫

1 d 3k 1 −iEkt e = 3 (2π ) 2Ek 4π 2

∫0

where k = ∣k∣. Changing variables to Ek = and k = E k2 − m2 which gives 10

See the discussion leading up to equation (2.89).

2-23

∞

k2d k −i e Ek

k 2+m 2 t

,

(2.120)

k2 + m2 we have k d k = Ek dEk

Relativistic Quantum Field Theory, Volume 1

D(x − y ) =

∞

1 4π 2

∫m

dEk E k2 − m 2 e−iEkt =

1 m K1(imt ), 4π 2 it

(2.121)

which, in the large time limit, becomes

im −imt e . t3

lim D(x − y ) ∼ t →∞

(2.122)

The same method can be used to evaluate the second term in the integral. • It is zero for spacelike separations. To see this, consider, for example, two points with equal time such that x − y = (0, x − y) and (x − y )2 = −(x − y)2 < 0. In this case, one has

[ϕ(x , t ), ϕ(y , t )] =

d 3k 1 i k·(x−y) [e − e−i k·(x−y)]. (2π )3 2Ek

∫

(2.123)

We can flip the sign in the second exponential11 and we immediately see that the second term then identically cancels the first. To see this another way, we can evaluate the integrals explicitly. As above, we begin by considering the first term in the integral. Defining r = x − y we have

d 3k 1 i k·r e (2π )3 2Ek ∞ 1 1 i kr cos θ 1 k2 d k d (cos θ ) e = 2 −1 2Ek (2π ) 0 ∞ k2 e i kr − e−i kr 1 k d = 2Ek (2π )2 0 i kr ∞ ke i kr i , dk =− 2 2(2π ) r −∞ k2 + m 2

D(x − y ) =

∫

∫

∫

(2.124)

∫

∫

where, in the last step, we took k → −k and then combined the two terms in to a single integral over positive and negative k . To evaluate the integral appearing finally in equation (2.124), we notice that the integrand has no poles. It does, however, possess two cuts associated with the square root in the denominator that run along the imaginary axis as shown in figure 2.3. As shown in figure 2.3, to evaluate this integral, we can deform the integration contour so that it wraps around the upper cut giving

D(x − y ) =

1 4π 2r

∞

∫m

du

ue−ur 2

u −m

2

=

1 m K1(mr ) 4π 2 r

∼

r →∞

m −mr e . r

(2.125)

If one repeats this analysis for the second term in equation (2.123) one finds the same result, namely

11

Change variables k → −k and use the fact that Ek = E−k .

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Relativistic Quantum Field Theory, Volume 1

Figure 2.3. Contour deformation going from (a) to (b) that can be used to evaluate the final integral appearing in equation (2.124).

D(y − x ) =

∫

1 m d 3k 1 −i k·r K1(mr ) e = (2π )3 2Ek 4π 2 r

∼

r →∞

m −mr e . r

(2.126)

As a result, the two terms contributing to Δ(x − y ) cancel and one finds Δ(x − y ) as argued previously. We have now proven two different ways that Δ(x − y ) is zero for spacelike separated points that have equal time. To generalize beyond this specific case, we now invoke the Lorentz invariance of Δ(x − y ). Since Lorentz transformations can mix space and time, but preserve the fact that the separation remains spacelike, we can cover the entire spacelike region using this result and, hence, Δ(x − y ) = 0 for all (x − y )2 < 0. Therefore, all spacelike-separated measurements are causally disconnected. Note, however, that each of the spacelike propagators D(x − y ) = D(y − x ) are non-zero for all ∣r∣. This means that the quantum fields themselves are not restricted to the light cone; however, as we showed above, any physical measurement will be causally disconnected since Δ(x − y ) = D(x − y ) − D(y − x ) = 0. Physically, the two different propagators, D(x − y ) and D(y − x ), correspond to forward and backwards propagation of the fields between two points x and y . For spacelike-separated points, these two contributions exactly cancel12.

12 For the real-valued scalar field considered in this discussion, the scalar particle is its own antiparticle, and the backwards propagation from y to x can be viewed as the propagation of an antiparticle from x to y . With complex scalar fields, this becomes more transparent since there are two types of charged particles. For complex scalar fields, one finds that the two types of fields are required to have equal masses and opposite charges in order for this cancellation to occur. The take-home message is profound: causality requires the existence of anti-particles.

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Relativistic Quantum Field Theory, Volume 1

2.8 Propagators We have already introduced the notation D(x − y ) for the propagator above. Now let us discuss the physics of this function. As I will now demonstrate, if we compute the vacuum expectation value of the product of two field operators evaluated at two different points, we will obtain the propagator

D(x − y ) = 〈0∣ϕˆ (x )ϕˆ (y )∣0〉 =

∫p ∫k

μ

μ

〈0∣[e ikμx aˆ †(k) + e−ikμx aˆ(k)] μ

(2.127)

μ

× [e ipμ y aˆ †(p) + e−ipμ y aˆ(p)]∣0〉. Expanding the right-hand side, we obtain four terms, but only the one involving

〈0∣aˆ(k)aˆ †(p)∣0〉 = (2π )3δ 3(p − k),

(2.128)

is non-vanishing. Using this delta function to eliminate the integral over p, we obtain

D(x − y ) = 〈0∣ϕˆ (x )ϕˆ (y )∣0〉 =

∫

d 3k 1 −ikμ(x μ−y μ ) e , (2π )3 2Ek

(2.129)

which matches the assignment we made in the preceding discussion surrounding equation (2.118). As we can see from equation (2.127), the propagator is the amplitude for a particle to propagate from y to x . In practice, there are three important types of propagators that will appear repeatedly in quantum field theory. They are called the Feynman propagator, the retarded propagator, and the advanced propagator. I will now give some details concerning these three different types of propagators. The Feynman propagator The Feynman propagator is defined as the time-ordered vacuum expectation value of two field operators

⎧ D(x − y ) DF(x − y ) ≡ 〈0∣Tϕˆ (x )ϕˆ (y )∣0〉 = ⎨ ⎩ D(y − x )

x0 > y0, x0 < y0,

(2.130)

where T stands for the time ordering operator. Its action is to place all operators in order from left to right with those with the later time appearing to the left

⎧ ϕˆ (x )ϕˆ (y ) Tϕˆ (x )ϕˆ (y ) = ⎨ ⎪ ⎩ ϕˆ (y )ϕˆ (x ) ⎪

x0 > y0, x0 < y0.

(2.131)

One can write the Feynman propagator in terms of a four-dimensional integral of the form d 4k 1 μ μ e−ikμ(x −y ), D F(x − y ) = i (2.132) 4 2 2 CF (2π ) k − m

∫

where ∫ indicates a k 0 contour integration along the contour shown in figure 2.4. CF

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Relativistic Quantum Field Theory, Volume 1

Figure 2.4. The contour CF for the integral definition of the Feynman propagator given in equation (2.132). The × marks indicate the positions of the poles in the integrand.

To show that equation (2.132) gives equation (2.130) we need to evaluate the dk 0 integral along the contour shown in figure 2.4. To start, we factorize

1 1 1 = 02 = 0 . 2 2 k −m (k − Ek )(k 0 + Ek ) (k ) − E k 2

(2.133)

This shows that there are two poles corresponding to k 0 = ±Ek with residues ±1/(2Ek ), respectively. The other term in the integrand is exp( −ikμ(x μ − y μ )) = exp( −ik 0(x 0 − y 0 ))exp( +ik i (x i − yi )). To evaluate the contour integral in equation (2.132) we can close the contour in either the lower or upper half of the complex plane. Whether we close in lower or upper half plane depends on the sign of x 0 − y 0. If x 0 > y 0, we must close the contour in the lower half plane, since for x 0 > y 0, lim k0→−i∞ exp( −ik 0(x 0 − y 0 )) = 0 and we can ignore the contribution from semicircle as we take its size to infinity. As a result, we will only pick up a contribution from the positive pole k 0 = +Ek . Since, in this case the resulting loop around the pole goes in the clockwise direction, the p0 integration returns −2πi /(2Ek ) giving

D F(x − y ) = i =

∫

∫

d 3k −2πi −ikμ(x μ−y μ ) e (2π )4 2Ek d 3k 1 −ikμ(x μ−y μ ) e = D(x − y ) (2π )3 2Ek

(2.134) x0 > y0.

If y 0 > x 0 , we must close the contour in the upper half plane, in which case we only obtain a contribution from the negative pole k 0 = −Ek . Since now the loop around the pole is counter-clockwise, the p0 integration returns 2πi /( −2Ek ) giving

D F(x − y ) = i =

∫

∫

=

∫

=

∫

d 3k 2πi +iEk(x 0−y 0)+ik i(xi−yi ) e (2π )4 −2Ek d 3k 1 −iEk(y 0 −x 0)−ik i(yi −xi ) e (2π )3 2Ek d 3k 1 −iEk(y 0 −x 0)+ik i(yi −xi ) e (2π )3 2Ek d 3k 1 −ikμ(y μ −x μ ) e = D(y − x ) (2π )3 2Ek

2-27

(2.135)

y 0 > x 0,

Relativistic Quantum Field Theory, Volume 1

where, in going from the second to the third lines, we changed integration variables from k → −k and used the fact that all other quantities appearing in the integrand only depend on k2 . This completes the proof that equation (2.132) gives equation (2.130). The ‘iε prescription’ One can achieve the same result for the Feynman propagator by defining

D F(x − y ) = i

∫

d 4k 1 μ μ e−ikμ(x −y ), 4 2 (2π ) k − m 2 + iε

(2.136)

with ε > 0 being an infinitesimal, which is taken to zero at the end. The addition of this infinitesimal in the denominator of the integrand shifts the poles slightly off the real k 0 axis. As a consequence, one can then simply integrate directly along the k 0-axis from −∞ to ∞ without encountering singularities. For an illustration of this see figure 2.5. The advanced and retarded propagators We can define two other contours that can be used to define the so-called advanced (A) and retarded (R) propagators using the contours shown in figure 2.6. The corresponding propagators are defined via

DR,A(x − y ) ≡ i

∫C

R,A

d 4k 1 μ μ e−ikμ(x −y ). (2π )4 k 2 − m 2

(2.137)

Figure 2.5. Illustration of how the poles in the k0 integration are shifted off the real k 0 -axis when using the iε prescription.

Figure 2.6. The contours CR and CA for the integral definition of the retarded (left) and advanced (right) propagators.

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Evaluating the contour integration along CR, one finds that the retarded propagator is given by

⎧ D(x − y ) − D(y − x ) DR(x − y ) = ⎨ ⎩0

x0 > y0, x0 < y0.

(2.138)

Likewise, evaluating the contour integration along CA , one finds that the advanced propagator is given by

⎧0 DA(x − y ) = ⎨ ⎩ D(x − y ) − D(y − x )

x0 > y0, x0 < y0.

(2.139)

Exercise 2.7 Show that using the iε prescription expression for the Feynman propagator given by equation (2.136), one obtains equation (2.130).

2.9 Propagators as Green’s functions Although we have introduced them as the amplitude for quantum field vacuum expectation values of products of two field operators, the propagators in general can also be seen from a different perspective as Green’s functions of the classical field equation. For the case considered so far, namely a real-valued scalar field, one can show that, irrespective of the contour chosen, the propagator is the Green’s function for the Klein–Gordon equation. The Green’s function of a given dynamical equation is the solution for a source that is point-like in space and time. For example, for the Klein–Gordon equation one sees that

d 4k 1 μ μ e−ikμ(x −y ) 4 2 2 (2 ) k m π − R,A,F d 4k −k 2 + m 2 −ikμ(x μ−y μ ) e =i 4 (2.140) k 2 − m2 CR,A,F (2π ) 4 d k −ikμ(x μ−y μ ) e =−i 4 CR,A,F (2π ) = − iδ 4(x − y ).

(∂ μ∂μ + m 2 )DR,A,F(x − y ) = i (∂ μ∂μ + m 2 )

∫C

∫

∫

In practice, one uses the different contours to address different types of boundary conditions. For example, in classical field theory one uses the retarded propagator (Green’s function) if one knows the initial condition of the field and wants to evolve the configuration forward in time in the presence of some fixed source function J (x ), e.g.

(∂ μ∂μ + m 2 )ϕ(x ) = J (x ).

(2.141)

One uses the advanced propagator, if instead one knows the end point (final time) field configuration and wants to figure out the field configuration at earlier times. 2-29

Relativistic Quantum Field Theory, Volume 1

The Feynman propagator is a bit harder to interpret in this way, but as we will see, it is the one that naturally arises when one starts to consider interacting quantum field theories.

Reference [1] Greiner W 2000 Relativistic Quantum Mechanics - Wave Equations (Berlin: Springer)

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IOP Concise Physics

Relativistic Quantum Field Theory, Volume 1 Canonical formalism Michael Strickland

Chapter 3 Quantization of interacting field theories

In the prior chapter, we covered the quantization of non-interacting (free) real and complex scalar field theories. We were able to obtain particle and anti-particle excitations, but the result is rather boring since the particles/anti-particles do not interact. We will now discuss how to include field interactions. As usual we will start with simple cases and add more complexity as the chapter progresses. For further reading on this topic see references [1–4].

3.1 Weakly-interacting scalar fields Consider the following general expansion of the Lagrangian for a weakly interacting real scalar field

L=

1 1 ∂μϕ∂ μϕ − m 2ϕ 2 − 2 2

λn n ϕ . n! n⩾3

∑

(3.1)

Note above, that I have dropped the ‘hats’ on the operators in order to make the notation more manageable. From now on, it will be implicitly assumed that our quantum fields etc are, in fact, operator-valued functions. The coefficients λ n are the coupling constants of the theory. We will now focus on the case that all coupling constants in the sum above are ‘small’, in which case we might try to treat the corrections implied by them using perturbation theory. However, the first question is: what are the coupling constants small in relation to? For example, if we were considering a problem in mechanics and I told you that I was going to take the limit of small momentum in order to take the non-relativistic limit, I would have to say that the momentum was small compared to the mass scale, i.e. that the mass dominates the relativistic energy and I can then expand E = p2 + m2 = m 1 + p2 /m2 in a Taylor series to obtain E ≃ m(1 + p2 /(2m2 ) + ⋯) ≃m + p2 /(2m ) + ⋯. As we can see from this example, we need to be precise when making series expansions and define ‘small’ as being relative to some other quantity if the thing we are expanding in has non-trivial dimensions. doi:10.1088/2053-2571/ab30ccch3

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ª Morgan & Claypool Publishers 2019

Relativistic Quantum Field Theory, Volume 1

To figure this out, we start by looking at the dimensions of a generic coupling constant appearing in the sum above. As we discussed in the previous chapter, the action is dimensionless in natural units, i.e. [S ] = 0, and, therefore, in D dimensions, the Lagrangian density has units of [L] = D. Since every term in the Lagrangian must have the same dimension and [∂μ] = [m ] = 1, based on the first two terms one has [ϕ ] = (D − 2)/2. Since all terms in the sum must also have dimension D, this implies

[λn ] = 4 −

⎛D ⎞ ⎜ − 1⎟n . ⎝2 ⎠

(3.2)

In four-dimensional space-time one has [λ n ] = 4 − n . Relevant couplings: If we look at the first term in the sum in equation (3.1), we find [λ3] = 1. This means that this coupling constant has dimensions of energy and therefore the necessary dimensionless parameter that we will expand in will be of the form λ3 /E , where E is some quantity with units of energy. Typically, the scale E is set by the energy scale of the process we are considering. This means that the cubic correction in the potential is a small perturbation at high energies E ≫ λ3, but at small energies E ≪ λ3 is a large perturbation. Couplings that dominate in the low energy limit are called relevant couplings. This terminology means that, if we look at low energies/momentum, which correspond to long time/distances scales, e.g. the times scales/sizes relevant for everyday human existence, the relevant couplings are the ones that are large and, therefore, cannot be ignored. In the relativistic case one has E ⩾ m and we can make the cubic perturbation small if λ3 ≪ m. Irrelevant couplings: If we look at the terms with n ⩾ 5 in the sum in equation (3.1), we find [λ n ] < 0. The corresponding dimensionless parameter is λ nE n−4 which is small at low-energies and large at high energies. This kind of perturbation is called irrelevant. This does not mean that one should always ignore this type of perturbation, it simply means that in the low-energy limit they are much less important than the relevant couplings. Marginal couplings: Finally, if we look at the second term in the sum in equation (3.1), we find [λ 4 ] = 0. This term is already dimensionless, so such perturbations will be small simply if λ 4 ≪ 1. These types of couplings are called marginal since they are on the dividing line between relevant and irrelevant. I will note, however, that quantum corrections can change couplings from e.g. being marginal to being relevant or vice versa. Also, one can have an irrelevant operator being promoted to being marginal and so on. Also note that the classification or the type of coupling depends on the theory one is considering. The classifications above are specific to scalar field theories. In other theories, e.g. electromagnetism, the fields will have a different dimension and the counting above has to be reconsidered. Additionally, if we have a theory with D ≠ 4, we will also need to reconsider the dimensional counting and classification of operators.

3.2 Two examples of interacting quantum field theories Having introduced this terminology, we can now consider some concrete examples of interacting quantum field theories. In this text, we focus on weakly coupled field 3-2

Relativistic Quantum Field Theory, Volume 1

Figure 3.1. The fundamental ‘vertex’ in λϕ4 theory gives 2 → 2 , 3 → 1, and 1 → 3 processes.

theories, i.e. ones in which the couplings can be treated as perturbations of the free version of the field theory at all energies. We will start by specifying two types of interacting field theories and use these as examples to develop general methods. λϕ4 theory: The canonical interacting field theory one usually encounters first is a real scalar field theory with only a quartic interaction. This is colloquially referred to as ‘λϕ4 theory’. The Lagrangian density for this theory is

L=

1 1 λ ∂μϕ∂ μϕ − m 2ϕ 2 − ϕ4 , 2 2 4!

(3.3)

where, for purposes of our ongoing discussion, one can assume λ ≪ 1. If we expand out the ϕ4 term in terms of creation and annihilation operators using equation (2.43) we will obtain sixteen terms of the form

a†(k1)a†(k2)a†(k3)a†(k 4), a†(k1)a†(k2)a†(k3)a(k 4), a†(k1)a†(k2)a(k3)a(k 4), … ,

(3.4)

where ki above are the momenta associated with each of the powers of ϕ in ϕ4 . These various operators will create and destroy particles. The ϕ4 interaction can be visualized as a four-particle vertex which can be flipped around to give either 2 → 2, 3 → 1, or 1 → 3 interactions as shown in figure 3.1. Scalar Yukawa theory: Another simple example that can be used to understand the rules of the game is ‘scalar Yukawa theory’. Hideki Yukawa (1907–1981) was a Nobelprize winning Japanese theoretical physicist who was the first to write down a semi1 realistic theory for nuclear interactions [5]. The Yukawa theory involved spin- 2 nucleons exchanging a light spin-0 pseudoscalar particle that represents the ‘pion’ field1. Our version is a toy model because our nucleons/anti-nucleons are represented by a complex scalar field instead of a fermionic field and we will have scalar mesons (ϕ ) instead of pseudoscalar mesons. The Lagrangian for our scalar Yukawa theory toy model is

L = ∂μψ †∂ μψ +

1 1 ∂μϕ∂ μϕ − M 2ψ †ψ − m 2ϕ 2 − gψ †ψϕ , 2 2

(3.5)

where, as before, we will assume weak coupling, which means that g ≪ M , m. As mentioned above, this theory takes a complex scalar field (ψ ) and a real-valued scalar field (ϕ ) and couples them together via the last term above. The complex field ψ is our toy model for a nucleon field (it includes positively charged particles and negatively charged anti-particles) and the real field ϕ represents our scalar pion field. The fundamental vertices all come from the final term in the 1

A pseudoscalar is a scalar particle that is odd (field flips sign) under parity transformations.

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Relativistic Quantum Field Theory, Volume 1

Figure 3.2. The fundamental vertices in scalar Yukawa theory. Time moves from left to right. Dotted lines represent pions and solid lines represent nucleons if the arrow points to the right and anti-nucleons if the arrow points point to the left.

Lagrangian above and are depicted in figure 3.2. As we can see graphically, the individual particle number of pions and (anti-)nucleons is not conserved (in the leftmost diagram a pion transforms into a nucleon/anti-nucleon pair); however, the number of nucleons minus the number of anti-nucleons is conserved in this theory.

3.3 The interaction picture and Dyson’s equation Before looking at specific examples using the toy models above, we need to remind ourselves of some technology of time-dependent perturbation theory and scattering theory from quantum mechanics. A key ingredient in the the quantum mechanical treatment was the introduction of a new ‘picture’ for wavefunctions and operators, where it is the interaction part of the Hamiltonian that evolves the wavefunction in time and not the full Hamiltonian. I will briefly review it in the context of quantum mechanics for a finite number of degrees of freedom. In the Schrödinger picture, states evolve in time via

i

d ψ dt

S

= H ψ S,

(3.6)

and Schrödinger-picture operators OS are time-independent. In the Heisenberg picture, the states are time-independent and it is the operators that are time-dependent

ψ

H

= e iHt ψ S ,

O H(t ) = e iHtO Se−iHt .

(3.7) (3.8)

So far, we have been working in the Heisenberg picture so we will need this relation below. Note, also that HH = HS since the Hamiltonian commutes with itself. In the interaction picture2 we land somewhere in the middle with both the states and operators depending on time. The interaction picture starts by splitting the Hamiltonian up into ‘free’ and ‘interacting’ parts

H = H0 + Hint.

(3.9)

In practice, this split should be made in such a way that the eigenstates corresponding to H0 should be analytically obtainable, but this is not absolutely necessary. States and operators in the interaction picture are indicated with a subscript ‘I’ 2

This is sometimes referred to as the Dirac picture.

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Relativistic Quantum Field Theory, Volume 1

ψ

I

= e iH0t ψ S ,

(3.10)

O I(t ) = e iH0tO Se−iH0t .

(3.11)

Using the second equation above, we can determine the Hint in the interaction picture

HI(t ) ≡ (Hint )I = e iH0tHinte−iH0t .

(3.12)

Starting from equation (3.6) and using equations (3.9) and (3.6) we obtain

i

d −iH0t ψ I ) = (H0 + Hint )e−iH0t ψ I . (e dt

(3.13)

Expanding and simplifying this expression, we obtain

i

d ψ dt

I

(3.14)

= HI(t ) ψ I .

3.3.1 Dyson’s formula Equation (3.14) is solved by a unitary evolution operator U (t , t0 )

ψ (t )

I

= U (t , t 0 ) ψ (t 0 ) I ,

(3.15)

with

⎛ U (t , t0) = T exp ⎜ −i ⎝

∫t

t 0

⎞ dt′HI(t′)⎟ , ⎠

(3.16)

where T is the time-ordering operator which satisfies

⎧O1(t1)O2(t2 ) TO1(t1)O2(t2 ) = ⎨ ⎩O2(t2 )O1(t1)

t1 > t2, t2 > t1.

(3.17)

The solution given by equations (3.15) and (3.16) taken together is called Dyson’s formula. Exercise 3.1 Show that equation (3.14) is solved by equation (3.15).

3.4 Interactions in scalar Yukawa theory It is straightforward to construct the Hamiltonian density based on equation (3.5). The result is

H = ∂μψ †∂ μψ +

1 1 ∂μϕ∂ μϕ + M 2ψ †ψ + m 2ϕ 2 + gψ †ψϕ . 2 2

3-5

(3.18)

Relativistic Quantum Field Theory, Volume 1

The first four terms above correspond to free propagation of our complex and real scalar fields ψ and ϕ. The last term is the interaction term from which we can construct the interaction Hamiltonian

Hint(t ) = g

∫ d 3x ψ †ψϕ.

(3.19)

Following the rules introduced chapter 2, we quantize the fields appearing above in terms of different types of creation and annihilation operators

ψ (x ) =

∫k

[e ikμx a†(k) + e−ikμx b(k)],

ψ †(x ) =

∫k

[e ikμx b†(k) + e−ikμx a(k)],

ϕ(x ) =

∫k

[e ikμx c†(k) + e−ikμx c(k)],

μ

μ

(3.20)

μ

μ

(3.21)

μ

μ

(3.22)

where, in our toy model, – a† creates positively charged nucleons, – b† creates negatively charged anti-nucleons, – c† creates neutral pions. Therefore, roughly speaking we see that ψ ∼ a† + b creates a nucleon and destroys an anti-nucleon, ψ † ∼ b† + a creates an anti-nucleon and destroys a nucleon, and finally ϕ ∼ c† + c creates and destroys pions. At first order in perturbation theory, the relevant terms are linear in Hint ∼ ψ †ψϕ. As a result, we will find terms of the form

ψ †ψϕ → abc , abc†, aa†c , aa†c†, b†bc , b†bc†, b†a†c , b†a†c†,

(3.23)

where I have suppressed the momentum labels for compactness. These operator strings should be read from right to left. The term involving b†a†c , for example, annihilates a pion and creates a nucleon/anti-nucleon pair. This term maps to the leftmost diagram in figure 3.2, which is the pion decay process ϕ → ψψ † in our toy model3. At second order in perturbation theory, (Hint )2 will generate more complicated terms, e.g. (b†a†c )(c†ba ) that map to scattering processes like ψψ † → ϕ → ψψ †. Given this setup, we now need to compute the quantum amplitude for these processes.

3.5 The S-matrix In order to calculate quantum amplitudes, we start with an assumption: that the initial and final states are eigenstates of the free (non-interacting) theory. On the surface this assumption seems natural, since particles will only significantly interact with one another when they are relatively close and, once they are sufficiently far 3

In the diagram, the process is read from left to right.

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Relativistic Quantum Field Theory, Volume 1

away from one another, we can ignore their interactions. Let us work this assumption for the time being and then return to the caveats. Mathematically, the assumption above means that we assume that (a) at some far distant time in the past lim t → −∞ we are given an incoming state ∣i 〉 which is an eigenstate of H0 and (b) we additionally assume that at some far distant time in the future lim t → +∞ the final state ∣f 〉 is also an eigenstate of H0. These states are called asymptotic states. The amplitude to go from the initial state to the final state is

lim 〈f ∣U (t f , ti )∣i〉 ≡ 〈f ∣S∣i〉.

ti →−∞ t f →+∞

(3.24)

The operator S is called the S -matrix (scattering matrix). We will return to how this is formally obtained shortly, but first I would like the reader to get a feeling for where we are going and how, once the rules are specified, one can calculate things. Before going forward, I remind the reader that the assumption that we can formulate processes in terms of asymptotic states is questionable in some cases, for example: (1) if a bound state forms in the process of colliding (e.g. e−p → H ), then our underlying model would also need to include these bound states (hydrogen atoms) in the fundamental theory. As we advance in quantum field theory, we will see that there is a way to deal with this, because bound states appear as poles in the S -matrix; (2) if particles interact strongly and are ‘confined’ (like quarks and gluons) they cannot appear as asymptotic states; (3) particles are never really ‘alone’ and are instead surrounded by virtual particles that flit in and out of the vacuum. As we advance in quantum field theory, we will see that taking this final point seriously leads to the development of something called renormalization. For now, however, we will use this assumption since we can learn many useful things in the process. 3.5.1 Example: pion decay in the scalar Yukawa model I would like to run through how to calculate the rate at which pions decay into nucleon/anti-nucleon pairs in our toy model to leading order in perturbation theory. In the end, I will attempt to write down some general rules that will make the job much less tedious when considering higher-order graphs. We begin by requiring that our incoming state is a single pion and that our outgoing state contains a single nucleon and a single anti-nucleon. Using relativistically normalized states (see chapter 2.4) we have4

i = f =

2Ek ck† 0 ,

(3.25)

4EpEq a p†bq† 0 .

(3.26)

The incoming pion has 4-momentum k = (Ek , k) and (figure 3.3) the outgoing nucleon and anti-nucleon have 4-momenta, p = (Ep, p) and q = (Eq, q), respectively. To leading order (first order in g ), the quantum transition amplitude is 4 As we will see, the particular normalization is irrelevant for the physical decay rate, but this one makes the final result we obtain in this section for the S -matrix more ‘pretty’.

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Relativistic Quantum Field Theory, Volume 1

Figure 3.3. Diagram for the incoming state (‘pion’ with 4-momentum k ) and the outgoing state (‘nucleon’ with 4-momentum p and ‘anti-nucleon’ with 4-momentum q ).

f S i ≃ f

⎛ ⎜1 − i ⎝

∞

⎞

∫−∞ dt′HI(t′) + ⋯⎠ ⎟

i ,

(3.27)

where I have expanded the Dyson formula to first order. If the initial and final states do not have the same particle content, then 〈 f ∣i 〉 = 0. This is true for this example, so the first term above vanishes. Using

HI (t ) = e iH0t(Hint )Se−iH0t ,

(3.28)

and the fact that the Schrödinger representation operator (Hint )S is time-independent, we can perform the time integration in the second term above to obtain ∞

∫−∞ dt′

′⎞ ⎛ ∞ f HI (t′) i = ⎜ dt′e i (Ei−Ef )t ⎟ f (Hint )S i ⎝ −∞ ⎠ = 2πδ(Ei − Ef ) f (Hint )S i ,

∫

(3.29)

where, for the given initial and final states, we have Ei = Ek and Ef = Ep + Eq . We have now reduced the problem to evaluating the matrix elements of a timeindependent operator. One complication is that thus far our field operators are defined in the Heisenberg representation. To do this calculation properly we need to convert them to the Schrödinger picture. However, we can perturbatively expand this conversion and, at least, at leading order we obtain a simple relationship

(Hint )S = e−iHt(Hint )H e iHt , = e−i (H0+Hint )t(Hint )H e i (H0+Hint )t ≃ e−iH0t(Hint )H e iH0t + O((Hint )2 ).

(3.30)

As a result, to leading order in Hint

f S i ≃ −i 2πδ(Ei − Ef ) f (Hint )H i ,

(3.31)

where we have used the fact that the delta function forces Ei = Ef so that the overall phase factor that comes out is equal to one. We now just need to evaluate the final matrix element

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Relativistic Quantum Field Theory, Volume 1

f (Hint )H i = g

∫ d 3x

f ϕψ †ψ i

∫x ∫p ∫p ∫p

= g 2Ek 2Ep 2Eq

1

2

0 a(p)b(q)

3

(3.32)

× ⎡⎣e ip1 ·xc†(p1) + e−ip1 ·xc(p1)⎤⎦ × [e ip2 ·xb†(p2) + e−ip2 ·xa(p2)][e ip3 ·xa†(p3) + e−ip3 ·xb(p3)]c†(k) 0 , where ∫ = ∫ d 3pi (2E pi )−1/2(2π )−3 and I have introduced a shorthand notation for p i

four-vector contractions pi · x = (pi )μx μ. Expanding this expression and using the commutation relations for the various creation and annihilation operators, one finds that only the following term will survive f (Hint ) H i = g 2Ek 2Ep 2Eq

∫x, p , p , p 1

= g (2π )9

2

0 a(p)b(q)[c(p1)b†(p 2)a†(p 3)]c†(k) 0 e−i (p1 −p2 −p3 )·x

3

∫x, p , p , p 1

2

δ 3(p1 − k)δ 3(p 2 − q)δ 3(p 3 − p)e−i (p1 −p2 −p3 )·x

3

= g

e−i (Ek −Ep−Eq )t 2Ek 2Ep 2Eq

= g

e−i (Ek −Ep−Eq )t (2π )3δ 3(k − p − q). 2Ek 2Ep 2Eq

∫ d 3xei(k−p−q)·x

(3.33)

Eliminating matching factors on the left and right, we obtain finally

f (Hint )H i = g(2π )3δ 3(k − p − q)e−i (Ek−Ep−Eq )t .

(3.34)

Inserting this into equation (3.31), we obtain our expression for the leading order ‘pion’ decay S -matrix amplitude in our toy model

f S i = − ig(2π )4δ(Ei − Ef )δ 3(k − p − q) = − ig(2π )4δ 4(k − p − q ),

(3.35)

where, in the final step we used, Ei − Ef = Ek − Ep − Eq = k 0 − p0 − q 0 , and then combined the delta functions into a four-dimensional delta function. We also used the fact that the energy delta function implies that exp( −i (Ek − Ep − Eq )t ) → 1. This result hints at a very simple ‘Feynman rule’ for our toy model: for every vertex, we insert a factor of ‘−ig’ and require that energy and momentum are conserved by the 4-momenta entering/leaving the vertex. Exercise 3.2 Show that the first line of equation (3.33) follows from equation (3.32).

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3.6 Beyond leading-order perturbation theory So far we have learned how one can include weakly-coupled interactions and worked through an example of how to compute the quantum amplitudes for pion decay in the scalar Yukawa model. In the process of that example calculation, we saw some general features emerge: (1) the time-dependence of the amplitude was simple and resulted, in the end, in the energy conservation delta function appearing in equation (3.31); (2) after expanding the interaction Hamiltonian we saw that, when projected with the initial and final states, only a single term survived5; (3) the final result automatically conserved energy and 3-momentum. These features will emerge in general, but in order to see this, we need to come up with some tricks to make the calculations easier since, as we go to higher orders in perturbation, things could quickly become unmanageable. 3.6.1 Wick’s theorem We learned in chapter 3.3 that in order to perform perturbative calculations, we need to expand the Dyson formula order-by-order in Hint . Generally, the problem will be reduced to calculating matrix elements of the form

f THI(x1)⋯HI(xn) i ,

(3.36)

where ∣i 〉 and ∣f 〉 are asymptotic free states. The order of the operators in this matrix element are set by the time-ordering operator T. Since HI contains combinations of creation and annihilation operators, it will be useful to move all annihilation operators to the right, where they can eliminate particles in the initial state ∣i 〉. If we move all of the annihilation operators to the right (creation operators to the left) then the final form will be normal ordered. Wick’s theorem tells us how to transform time-ordered strings of operators to normal-ordered strings of operators. Before looking at the general case, let us start with the specific case of two field operators. Time-ordered products of two fields and the propagator Returning to the case of a scalar field, we can decompose our scalar field operator into two terms

ϕ(x ) = ϕ(+)(x ) + ϕ(−)(x ),

(3.37)

ϕ(+)(x ) =

∫k

a(k)e−ik·x ,

(3.38)

ϕ(−)(x ) =

∫k

a†(k)e+ik·x ,

(3.39)

where

and k · x = kμx μ. Choosing x 0 > y 0, we have 5

This happened in the transition from equation (3.32) to equation (3.33).

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Relativistic Quantum Field Theory, Volume 1

Tϕ(x )ϕ(y ) = ϕ(x )ϕ(y ) = [ϕ(+)(x ) + ϕ(−)(x )][ϕ(+)(y ) + ϕ(−)(y )] = ϕ(+)(x )ϕ(+)(y ) + ϕ(−)(x )ϕ(+)(y ) + ϕ(+)(x )ϕ(−)(y ) + ϕ(−)(x )ϕ(−)(y ) (3.40) = ϕ(+)(x )ϕ(+)(y ) + ϕ(−)(x )ϕ(+)(y ) + ϕ(−)(y )ϕ(+)(x ) + ϕ(−)(x )ϕ(−)(y ) + [ϕ(+)(x ), ϕ(−)(y )].

Upon inspection we see that the first four terms on the last line are normal ordered since in all cases the creation operators are always on the left. In fact, the first four terms can be written compactly as ‘: ϕ(x )ϕ(y ): ’. The final commutator can be seen to be the propagator introduced in equation (2.118): ′

(+) (−) [ϕˆ (x ), ϕˆ (y )] =

∫k ∫k′ e−ik·xeik ·y[a(k), a†(k′)] = ∫ ∫ e−ik·(x−y )(2π )3δ 3(k − k′) k k′

(3.41)

1 −ik·(x−y ) e = ⃗ k 2Ek = D(x − y ).

∫

With this, we have for x 0 > y 0

Tϕ(x )ϕ(y ) = : ϕ(x )ϕ(y ): +D(x − y )

if x 0 > y 0 .

(3.42)

if x 0 < y 0 .

(3.43)

Similarly, we can determine that for y 0 > x 0

Tϕ(x )ϕ(y ) = : ϕ(x )ϕ(y ): +D(y − x )

We can combine this into one compact formula by invoking the Feynman propagator DF defined in equation (2.130)

Tϕ(x )ϕ(y ) = : ϕ(x )ϕ(y ): +DF(y − x ),

(3.44)

where I remind you that

D F(x − y ) =

∫

d 4k ie−ik·(x−y ) . (2π )4 k 2 − m 2 + iε

(3.45)

Field contraction notation To proceed, we introduce some notation called the contraction of a pair of field operators. We define the contraction of a pair of fields in a string of operators ⋯ϕ(x1)⋯ϕ(x2 )⋯ as replacing the two contracted operators with a Feynman propagator connecting the two spacetime points at which the fields are evaluated, leaving all other operators unaffected. The notation for this is to connect the two field operators using a horizontal square bracket, e.g.

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Relativistic Quantum Field Theory, Volume 1

denotes the contraction of ϕ(x1) and ϕ(x2 ). The simplest example is the product of two fields

Note that there is a similar procedure for complex fields, in which case one has

Tψ (x )ψ †(y ) = : ψ (x )ψ †(y ): +DF(y − x ),

(3.48)

and

Statement of Wick’s theorem Wick’s theorem states that, for any collection of fields ϕ1 = ϕ(x1), ϕ2 = ϕ(x2 ), etc, one has

Tϕ1⋯ϕn = : ϕ1⋯ϕn : +: all possible contractions: .

(3.51)

To see this in action, let us consider the simplest case first, namely two operators

This is the same as we obtained in the previous subsection. Next, let us consider four operators

The proof of Wick’s theorem is made by induction. It is true for n = 2 as shown above. One then supposes that it is true for some given n and proves that, if it is, then it must be true for n + 1. Then we can bootstrap up from the case n = 2 to an arbitrary number of operators. Next, I point out a very useful fact. Since the vacuum expectation value (VEV) of any normal-order operator string vanishes, i.e. 〈0∣:O: ∣0〉 = 0, if we take the vacuum expectation value of the result above, we will find that all terms except the ‘fully

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Relativistic Quantum Field Theory, Volume 1

contracted’ terms (last three terms above) vanish and we obtain a rather simplelooking result

〈0∣Tϕ1ϕ2ϕ3ϕ4∣0〉 = DF(x1 − x2 )DF(x3 − x4) + DF(x1 − x3)DF(x2 − x4) + DF(x1 − x4)DF(x2 − x3).

(3.54)

Exercise 3.3 Prove Wick’s theorem by induction. 3.6.2 Example: nucleon–nucleon scattering in the scalar Yukawa model The method introduced above becomes very useful when we have to go beyond leading order in the expansion of Dyson’s formula. At second order, for example, we will encounter terms that contribute to nucleon–nucleon scattering. Let us look first at ψψ → ψψ scattering. In this case, the relativistically normalized initial states and final states are

i = f =

2E p1 2E p2 a†(p1)a†(p2) 0 ≡ ∣p1 , p2 〉,

(3.55)

2E q1 2E q2 a†(q1)a†(q 2) 0 ≡ ∣q1, q2〉,

(3.56)

where p1 and p2 are the incoming nucleons’ 4-momenta and q1 and q2 are the outgoing nucleons’ 4-momenta. We now look at the expansion of 〈f ∣S∣i 〉. To leading order we get S = 1, which is the case of no scattering. To get rid of this trivial contribution, we can evaluate instead 〈f ∣S − 1∣i 〉. For the set of initial and final states above, the linear contribution in g vanishes since the initial and final states have the same particle content. At second order, we will find a contribution of the form

( −ig )2 2

∫ d 4x ∫ d 4yTψ †(x)ψ (x)ϕ(x)ψ †(y )ψ (y )ϕ(y ).

(3.57)

We are now in a good position to make use of Wick’s theorem since there is a timeordered string of operators above

When we ‘sandwich’ this between the initial and final states above, we see that only the last term in the expansion above survives since the other terms have additional terms that annihilate pions that do not exist in our initial or final states. As a result, one has 3-13

Relativistic Quantum Field Theory, Volume 1

f Tψ †(x )ψ (x )ϕ(x )ψ †(y )ψ (y )ϕ(y ) i = DFϕ(x − y ) f : ψ †(x )ψ (x )ψ †(y )ψ (y ): i .

(3.59)

We are now left with the task of evaluating the matrix element on the right-hand side

f : ψ †(x )ψ (x )ψ †(y )ψ (y ): i = 〈q1, q2∣ : ψ †(x )ψ (x )ψ †(y )ψ (y ): ∣p1 , p2 〉.

(3.60)

Let us do it explicitly and then find a shortcut to make it not as painful

∫x, y =

〈q1, q2∣ψ †(x )ψ †(y )ψ (x )ψ (y )∣p1 , p2 〉 2E p1 2E p2 2E q1 2E q2

∫x, y DFϕ(x − y ) ∫k , k , k , k 1

2

3

0 a(q1)a(q 2) 4

(3.61)

: [e ik1·xb†(k1) + e−ik1·xa(k1)] × [e ik2·yb†(k2) + e−ik2·ya(k2)][e ik3·xa†(k3) + e−ik3·xb(k3)] × [e ik 4·ya†(k 4) + e−ik 4·yb(k 4)]: a†(p1)a†(p2) 0 , where here ∫ = ∫ d 4x and ∫ = ∫ d 3k (2π )−3(2Ek )−1/2 . Looking at the rightmost x k term in square brackets above, we see that the second term will give zero since the annihilation operator for an anti-particle b(k 4) commutes with the two creation operators for the particles and therefore gives zero. Therefore, only the first term involving an a†(k 4) survives. Looking to the next term on the right we see again that b(k3) can be moved to the right and also gives zero. The same analysis works for the b† operators in the two sets of square brackets on the left if one acts with them on the vacuum state on the left. As a result, only the following term survives

∫x, y =

〈q1, q2∣ψ †(x )ψ †(y )ψ (x )ψ (y )∣p1 , p2 〉 2E p1 2E p2 2E q1 2E q2

∫x, y DFϕ(x − y ) ∫k , k , k , k 1

2

3

e−i (k1−k3)·xe−i (k2−k 4)·y 4

× 〈0∣a(q1)a(q 2): a(k1)a(k2)a†(k3)a†(k 4): a†(p1)a†(p2)∣0〉, =

∫x, y DFϕ(x − y ) ∫k , k , k , k 1

2

3

(3.62)

e−i (k1−k3)·xe−i (k2−k 4)·y 4

× 〈0∣a(q1)a(q 2)a†(k3)a†(k 4)a(k1)a(k2)a†(p1)a†(p2)∣0〉. To evaluate the remaining vacuum expectation value, we apply the commutation relations repeatedly to move the factors a†(k3)a†(k 4) to the left so that they eventually hit the outgoing vacuum state and return 0 and move the factors a(k1)a(k2) to the right so that they hit the vacuum, once again giving zero. The end result is

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Relativistic Quantum Field Theory, Volume 1

0 a(q1)a(q 2)a†(k3)a†(k 4)a(k1)a(k2)a†(p1)a†(p2) 0 = (2π )12 [δ 3(q1 − k3)δ 3(q 2 − k 4) + δ 3(q1 − k 4)δ 3(q 2 − k3)]

(3.63)

3

3

3

3

× [δ (p1 − k1)δ (p2 − k2) + δ (p1 − k2)δ (p2 − k1)]. All terms above contain a product of four delta functions that can be used to perform the three-momentum integrations over k1, k2, k3, and k 4 . Performing these now-trivial integrals and simplifying, we obtain

∫x, y 〈q1, q2∣ψ †(x)ψ †(y )ψ (x)ψ (y )∣p1 , p2 〉 =2

∫x, y DFϕ(x − y )[e−i(p −q )·xe−i(p −q )·y + e−i(p −q )·ye−i(p −q )·x ]

= 2i ×

1

∫

1

2

2

1

2

2

1

d 4k 1 (2π )4 k 2 − m 2 + iε

(3.64)

∫x, y e−ik·(x−y) [e−i(p −q )·xe−i(p −q )·y + e−i(p −q )·ye−i(p −q )·x ]. 1

1

2

2

1

2

2

1

We can now perform the integrals over x and y to obtain

∫x, y 〈q1, q2∣ψ †(x)ψ †(y )ψ (x)ψ (y )∣p1 , p2 〉 d 4k [δ 4(k − q1 + p1 )δ 4(k + q2 − p2 ) k 2 − m 2 + iε + δ 4(k + q2 − p1 )δ 4(k − q1 + p2 )].

= 2(2π )4i

∫

(3.65)

Finally, using the delta functions to perform the 4-momentum integral over k , we obtain

∫x, y 〈q1, q2∣ψ †(x)ψ †(y )ψ (x)ψ (y )∣p1 , p2 〉 ⎡ ⎤ 1 1 ⎥ = 2i ⎢ + (p1 − q2 )2 − m 2 + iε ⎦ ⎣ (p1 − q1)2 − m 2 + iε

(3.66)

× (2π )4δ 4(p1 + p2 − q1 − q2 ). Multiplying by the overall factor of ( −ig )2 /2 from equation (3.57) we obtain our final expression for the leading order (first non-vanishing perturbative) contribution to nucleon-nucleon scattering in the scalar Yukawa theory

⎡ ⎤ 1 1 ⎥ f S − 1 i ≃ i ( −ig )2 ⎢ + (p1 − q2 )2 − m 2 ⎦ ⎣ (p1 − q1)2 − m 2

(3.67)

4 4

× (2π ) δ (p1 + p2 − q1 − q2 ), where I have taken ε → 0 since there are no singularities in our final expression. As in the example of pion decay, we find an overall four-dimensional Dirac delta function that expresses the conservation of energy and momentum during the

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Relativistic Quantum Field Theory, Volume 1

Figure 3.4. Diagrammatic representation of the result equation (3.67).

process of scattering. The two terms in the square brackets are harder to interpret immediately, but as we will see, they correspond to a ‘t-channel’ pion exchange and a ‘u-channel’ pion exchange, respectively. To visualize this, I have drawn two diagrams corresponding to these terms above in figure 3.4. This is our first official Feynman diagram. As we have learned, even with Wick’s theorem to help us eliminate many terms at the beginning of the calculation, the calculation above was still rather tedious. The final result, however, suggests that there is an easier way that involves drawing diagrams and then inferring the expressions from them. After doing a few calculations of quantum amplitudes using Wick’s theorem, one starts to see patterns emerge that can be formed into a set of rules for how to perturbatively compute the scattering matrix. This allows one to formulate a compact set of rules that can be used to compute quantum amplitudes. To see how these emerge, we begin by considering the space–time structure of expectation values of time-ordered strings of operators. Exercise 3.4 Show that equation (3.63) is correct.

3.6.3 Feynman diagrams We begin by considering the case of four real scalar fields. The time-ordered expectation value of four fields was evaluated previously using Wick’s theorem and the result was presented in equation (3.54). I will repeat it here for easier reference

0 Tϕ1ϕ2ϕ3ϕ4 0 = DF(x1 − x2 )DF(x3 − x4) + DF(x1 − x3)DF(x2 − x4) + DF(x1 − x4)DF(x2 − x3).

(3.68)

The right-hand side of equation (3.68) can be depicted as a sum of three Feynman diagrams as shown in figure 3.5. The picture suggests a physical interpretation: two particles are generated at two space-time points and then each one propagates to other points, where they are both annihilated. For a real scalar field this can happen in three possible ways corresponding to the three graphs shown in figure 3.5. The total amplitude for this process is given by the sum of these three diagrams. 3.6.4 Feynman diagrams for scalar λϕ4 theory Let us consider the leading-order expansion of the interacting propagator (2-point function) using Dyson’s formula 3-16

Relativistic Quantum Field Theory, Volume 1

Figure 3.5. Space-time Feynman diagrams for the time-ordered product of four scalar fields. The open circles labelled 1, 2, 3, and 4 correspond to the space-time points x1, x2 , x3, and x4 , respectively. Solid lines indicate a scalar Feynman propagator DF(xi − xj ).

⎡ 0 TϕI(x )ϕI(y ) 0 = 0 T ϕ(x )ϕ(y ) + ϕ(x )ϕ(y )⎣⎢ −i

{

∫ dt HI(t )⎤⎦⎥ + ⋯} (3.69)

0 , for the case of a scalar field theory with Hint = λϕ4 /4!. The first term above gives the non-interacting (free field) result, 〈0∣T ϕ(x )ϕ(y )∣0〉 = DF(x − y ). The second term is explicitly given by ⎧ ⎫ ⎪ ⎪ ⎪ ⎪ ⎛ −iλ ⎞ 3 4 ⎜ ⎟ dt d z ϕ (z )⎬∣0〉. (3.70) 〈0∣T⎨ ϕ(x )ϕ(y ) ⎝ ⎠ 4!  ⎪ ⎪   ⎪ ⎪ ∫ d 4z ⎩ ⎭

∫ ∫

If we apply Wick’s theorem, we obtain one term for each possible way to contract the six different ϕ’s with each other in pairs. There are 15 ways to contract them, but luckily only two of these ways are actually different. If we contract ϕ(x ) with ϕ(y ), there are three unique ways to contract the remaining ϕ(z )’s. If we contract ϕ(x ) with one of the ϕ(z ) (four choices), followed by contracting ϕ(y ) with one of the remaining ϕ(z )’s (three choices remaining), we are left with only one possibility for which ϕ(z )’s to contract. Taken together, in this case, there are 4 · 3 = 12 different contractions which all give the same final result. As a consequence, we have

⎧ ⎛ −iλ ⎞ ⎟ 〈0∣T⎨ϕ(x )ϕ(y )⎜ ⎝ 4! ⎠ ⎩

⎫

∫ d 4zϕ4(z)⎬⎭∣0〉

⎛ −iλ ⎞ ⎟D (x − y ) d 4zDF(z − z )DF(z − z ) = 3⎜ ⎝ 4! ⎠ F ⎛ −iλ ⎞ ⎟ + 12⎜ d 4zDF(x − z )DF(z − y )DF(z − z ). ⎝ 4! ⎠

∫

(3.71)

∫

We can associate both of the terms above with Feynman diagrams. As in figure 3.5, we connect spacetime points with Feynman propagators. In this case, however, we need to distinguish between external points x and y and the internal point z , which is an integration variable. In this way, we find that the two terms above correspond to the graphs shown in figure 3.6. As can be seen from this example, the first term factorizes into two contributions that are not connected to one another (they are disconnected graphs). As we will see, only connected graphs like the second one in figure 3.6 contribute to S − 1. 3-17

Relativistic Quantum Field Theory, Volume 1

Figure 3.6. Space-time Feynman diagrams corresponding to the integrals written in equation (3.71).

Momentum-space Feynman rules for the λϕ4 theory Although it is possible to continue along the lines above and specify all Feynman diagrams as space–time integrals, it turns out to be more efficient to work in momentum-space by introducing the Fourier-transform of the Feynman propagator. With each propagator we associate a 4-momentum ki and we enforce energymomentum conservation at every vertex. In the end, we are left with the following simple Feynman rules: 1. For each purely internal propagator line we assign a four-momentum ki and insert a Feynman propagator in momentum-space

Above, I have introduced the momentum-space representation of the Feynman propagator. Some texts indicate this with a tilde over it, but in this text whether we are in position space or momentum space, this will be implicit. 2. For every vertex we associate a factor of −iλ and require that the 4-momentum entering the vertex sums to zero (energy-momentum conservation).

3. For every purely internal propagator line, we integrate over all possible 4momenta with ∫ d 4ki /(2π )4 . 4. For each external (uncontracted) point we add (multiplicatively) a plane wave.

5. Divide by the symmetry factor for the graph. Symmetry factors The last step above mentions the symmetry factor. In general, this is the number of ways one can change parts of the diagram without changing the integral representation of the Feynman graph. There are lots of mnemonic devices for calculating these but I think it is best to calculate them explicitly since, in the end, the rules have some special cases. Here are my rules for computing symmetry factors in λϕ4 theory:

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Relativistic Quantum Field Theory, Volume 1

• Identify the order n of perturbation theory you are working at. Start with n! (4! ) n , which accounts for the factorial coming from the Taylor expansion of the exponential and the factors of 4! that come from each of the vertices. • If there are external lines, count the number of ways that one can connect the external lines to the vertices to generate the target graph. Let us call this factor Next . • Next count the number of ways that you can connect the internal lines to generate the target graph. Let us call this factor Nint . • The final symmetry factor is obtained by dividing the number from step 1 by the product of Next and Nint , i.e. Nsym = n! (4! ) n /(NextNint ). It is best to learn these rules with some examples. We being by considering the case of the leading-order (connected) correction to the scalar propagator. The process is illustrated in figure 3.7. In the first line of this figure, we have a factor of 4! since we are working at leading order (one vertex). We then imagine connecting one of the external lines to the vertex. Since there are four possible legs to connect, we pick up a factor of 4. We then need to connect the second external line, but now there is one less leg available, so we pick up a factor of 3. This tells us that Next = 4 · 3 = 12. Once the external lines are connected, the graph will look like the middle row of figure 3.7. At this point we consider the possibilities for closing the internal lines. In this case, there is only one way to join the two lines, so Nint = 1. Putting the results together we obtain Nsym = 4! /12 = 2. As the next example let us consider the so-called ‘basketball diagram’ which is obtained by contracting all legs of two ϕ4 vertices together. The process is illustrated in figure 3.8. In this case, since we need to go to second order in perturbation theory, we start with a factor of 2! (4! )2 . Since there are no external lines Next = 1. To connect all legs from one vertex there are 4 · 3 · 2 distinct ways to connect them. The final symmetry factor for this graph, Nsym = 48, is shown in the lower right corner of

Figure 3.7. The step-by-step process for determining the symmetry factor for a simple graph. See the text for a description.

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Relativistic Quantum Field Theory, Volume 1

Figure 3.8. The step-by-step process for determining the symmetry factor for a slightly more complicated graph.

figure 3.8 along with a ‘pretty’ version of the resulting graph6. You can practice your skills at computing symmetry factor by seeing if you can reproduce the symmetry factors listed in figure 3.9. Exercise 3.5 Calculate the symmetry factors for the following Feynman diagrams in λϕ4 theory.

3.6.5 Example: leading-order correction to the propagator As an example, let us consider the Feynman diagrams that contribute to the leadingorder, O(λ ), modification of the free propagator (2-point function) in λϕ4 theory. Our building blocks consist of one vertex and Feynman propagators that can be used to connect two of the four ‘legs’ of the vertex in all possible ways that leave two unmatched legs. We can start by picking one of four lines entering the vertex, e.g. the line corresponding to k1. In order to construct a graph with one line incoming and one outgoing, we can only contract two of the lines in the four-vertex using the Feynman propagator. If we, e.g. choose k1 to begin with, there are three possible lines corresponding to those labeled as k2 , k3, and k4 that we might connect to. When we are done, the diagram generated by this procedure is the one shown in the bottom row of figure 3.7. In this graph, there is one ‘internal’ line corresponding to the closed loop on the top, therefore we must integrate over it. The symmetry factor for the graph is 2. Calling the incoming 4-momentum p, the outgoing 4-momentum q , 6

This is our first example of a vacuum graph. Vacuum graphs have no external lines.

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Relativistic Quantum Field Theory, Volume 1

Figure 3.9. Examples of symmetry factors for some Feynman diagrams.

and the internal momentum k as shown in figure 3.10, and applying the momentumspace Feynman rules, we obtain

1 2

d 4k −ip·x (e )[( −iλ)(2π )4δ 4(p − k + k − q )]DF(k)(e iq·y ) (2π )4 ⎡ iλ ⎤ d 4k i = (2π )4δ 4(p − q )e−ip·(x−y )⎢ − ⎥, (2π )4 k 2 − m 2 + iε ⎦ ⎣ 2

∫

(3.72)

∫

where I have used the delta function to set q = p in the overall phase factor. The nontrivial part of this expression is contained inside the square brackets. The factors multiplying this are merely a delta function that enforces conservation of 4-momentum and a trivial phase factor. Now let us compare this with the result one obtains by evaluating the second term on the right-hand side of equation (3.71) which was obtained by constructing the position-space Feynman diagram ⎛ − iλ ⎞ ⎟ 12⎜ ⎝ 4! ⎠ iλ 2 iλ =− 2 =−

=

∫z

D F (x − z )D F (z − z )D F (z − y )

ie −ip·(x−z ) i ie −iq·(z−y ) 2 2 2 2 − m + iε k − m + iε q − m 2 + iε ie −ip·x ie iq·y i e −i (p−q )·z 2 2 2 2 2 2 p q p − m + iε q − m + iε z k k − m + iε ⎧ ⎤⎫ ⎡ iλ i DF(p )DF(q )⎨(2π )4δ 4(p − q )e −ip·(x−y )⎢ − ⎥⎬ , 2 2 ⎣ ⎩ 2 k k − m + iε ⎦⎭ q

∫z ∫p ∫k ∫q

p2

{∫

∫ ∫

∫p ∫

∫

∫

3-21

}

(3.73)

Relativistic Quantum Field Theory, Volume 1

Figure 3.10. A Feynman diagram with momentum labels for the leading-order correction to the scalar propagator.

where above integrals over space are of the form ∫ = ∫ d 4z and integrals over z momenta are of the form ∫ = ∫ d 4p/(2π )4 . As before, I have used the delta function p to set q = p in the overall phase factor. As we can see from this comparison, the momentum-space Feynman rules give the result in curly brackets above. The outside integrals over p and q together with DF(p ) and DF(q ) correspond to the ‘legs’ of the diagram, which describe the free propagation of the particle from the external point x to the vertex and then from the vertex to y. As we see, the momentum-space rules give us a quick path to the non-trivial part of the quantum-amplitude. For this particular diagram, the benefit is small; however, for more complicated diagrams it is much more efficient to construct the necessary expressions using the momentum-space Feynman rules. Exercise 3.6 (a) Write down the position-space Feynman diagram integral corresponding to the upper-left diagram in figure 3.9. (b) Write down the momentum-space Feynman diagram integral corresponding to the upper-left diagram in figure 3.9. (c) Show that the answers obtained in the previous two problems are consistent with one another. 3.6.6 Wick rotation Continuing with the example above, let us now turn to the evaluation of the integral that appears on the right-hand side of equation (3.72)

I≡

∫

d 4k i = 4 2 (2π ) k − m 2 + iε

∞

∫−∞

dk 0 2π

∫

d 3k i . 2 3 2 (2π ) k 0 − k − m 2 + iε

(3.74)

There are many ways to attack this integral, but here I would like to introduce a method that is often used in quantum field theory called Wick rotation [6]. We begin by taking a look at figure 2.5 and we notice that, since there are only two poles in the propagator, we can deform the integration contour by rotating it (in the direction that the integration contour never hits the poles) so that the k 0 integral runs along the imaginary k 0 axis, as shown in figure 3.11. After this rotation one has i∞

I=

∫−i∞

dk 0 2π

∫

d 3k i , 2 3 (2π ) k 0 − k2 − m 2

3-22

(3.75)

Relativistic Quantum Field Theory, Volume 1

Figure 3.11. Rotation of the integration contour used when transforming to a Euclidean metric.

where I have dropped the iε since the contour does not hit the poles when ε → 0 any longer. The next step is to change variables for the k 0 integration to k4 ≡ −ik 0 such that dk 0 = idk4 and k 02 = −k 42 . Simplifying the resulting integral, one finds ∞

I=

∫−∞

dk4 2π

∫

d 3k 1 . 2 3 (2π ) k 4 + k2 + m 2

(3.76)

In fact, we can simplify this even further by introducing a ‘Euclidean’ four-vector kEμ = (k , k4 ) and requiring that contractions of this vector utilize a Euclidean metric, i.e. kE · kE = k2 + k 42 . Using this, we can write the integral above compactly as

I=

∫

d 4kE 1 . 2 4 (2π ) k E + m 2

(3.77)

We will now see the benefit of this trick. Since the space for kE is Euclidean, if one uses four-dimensional spherical coordinates to parameterize kE , then k E2 is simply the four-dimensional radius. As a result, the angular integrals become trivial, and one can immediately write

I=

S4 (2π )4

∫0

∞

dkE k E3

k E2

1 , + m2

(3.78)

where S4 is the surface area of four-dimensional unit-sphere (sphere with radius of 1). However, at this point we run into a problem: the remaining integral is divergent! To see this without even evaluating the integral, we look at the asymptotic behavior of the integrand in the limit kE → ∞, in which case one has

lim k E3

kE →∞

k E2

1 = kE . + m2

(3.79)

As a result, the integral will be quadratically divergent, i.e. if we were to put an upper limit ΛUV on the (radial) kE integration, we would find that the integral diverges like Λ 2UV . Quantities that diverge like a power of the cutoff are called power-law divergent. In this case that the divergence comes from large momenta so it is a called an ultraviolet (UV) divergence. If instead the integral were divergent for small momenta, it would be called an infrared (IR) divergence. In practice, one could

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simply leave a finite ultraviolet cutoff, ΛUV , on this integral and treat the theory as an effective theory that only describes a finite range of momenta. However, putting cutoffs on integrals in this naive way breaks gauge invariance, so we cannot use this method in general. There are better methods to regulate this integral that, when we start dealing with more complicated theories like electromagnetism, QED, and QCD, will respect gauge invariance. The most common (and, in my opinion, best) way to regulate integrals in quantum field theory is to use dimensional regularization (dimreg) [7, 8]. In this method, one imagines that, instead of living in a four-dimensional space time, we live in a d = 4 − 2ε dimensional spacetime and then we only take ε to zero at the very end of the calculation. Following this method, we can generalize the I integral above to

⎛ e γE μ2 ⎞ε I=⎜ ⎟ ⎝ 4π ⎠ d

∫ d

ε d d kE 1 Sd ⎛ e γE μ2 ⎞ = ⎜ ⎟ (2π )d k E2 + m 2 (2π )d ⎝ 4π ⎠

∫0

∞

dkE

k Ed −1 , k E2 + m 2

(3.80)

∞

where Sd = 2π 2 /Γ( 2 ) with Γ(n ) = ∫ u n−1e−udu = (n − 1)! being the Euler gamma 0 function, the irrational number

⎛ n 1 ⎞ γ E = lim ⎜⎜ ∑ − ln n⎟⎟ = 0.577215 … n →∞ ⎝ k ⎠ k=1

(3.81)

is the Euler–Mascheroni constant, and μ is called the MS renormalization scale, which has dimensions of energy, i.e. [μ] = 1. The collection of constants multiplying the integral raised to the power ε go to 1 in the limit ε → 0 and the specific form chosen specifies what is called the renormalization scheme. The standard scheme is the MS (MS bar) scheme [9–12]. In the end, physical results are independent of the particular scheme chosen. The MS scheme gives intermediate results that are more compact and easier to work with. The appearance of the scale μ is more subtle. It is related to the fact that, when we change the number of dimensions, the energy dimension of field operators also change and, as a result, a new dimensionful scale must be introduced to compensate for this. In a loose sense, you can think of μ as the dimreg equivalent of the cutoff ΛUV on the integral. We will return to this in the final chapter when we properly discuss the renormalization of QED. For now, let us take this prescription at face value. Evaluating the remaining integral we obtain finally d ⎛ e γE μ2 ⎞ε 2π 2 I=⎜ ⎟ ⎝ 4π ⎠ (2π )d Γ

() d 2

⎡ π csc(dπ /2) ⎤ ⎢⎣ ⎥. 2m 2−d ⎦

(3.82)

If we Taylor expand the result above around ε = 0 we obtain

⎤ ⎛ m2 ⎞ ⎛ m ⎞2 ⎡ 1 I = −⎜ ⎟ ⎢ + 1 − ln ⎜ 2 ⎟ + O(ε )⎥ . ⎝ 4π ⎠ ⎣ ε ⎝μ ⎠ ⎦

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(3.83)

Relativistic Quantum Field Theory, Volume 1

As we can see from this result, the leading term is divergent as ε → 0. This is the manifestation of the quadratic UV divergence mentioned earlier. In dimensional regularization, this divergence becomes a pole in ε . Next semester, we will learn how to deal with this divergence using mass renormalization. For now, we can press forward and focus on scattering/decay diagrams that are not divergent. Consider this an advanced warning that loop corrections like the one shown in figure 3.10 are nontrivial to handle properly.

3.7 Decay rates and cross sections In quantum field theory, the probability of a transition from a given initial state to a given final state occurring is related to the modulus squared of the corresponding quantum amplitude. For non-trivial transitions in which the initial and final states are different, the relevant object is the expectation value of S − 1. One subtle aspect, however, is that S − 1 comes with a multiplicative delta function. If we square this naively, we will get an infinity. As I will discuss below, this particular infinity is related to the fact that we assumed, from the beginning, that we are working in an infinite space. Decay rates We now compute the probability for a single-particle initial state ∣i 〉 with momentum pi and rest mass m to decay into a final state ∣f 〉 consisting of n particles with total n momentum pf ≡ ∑ j =1qj as shown in figure 3.12. The n-particle decay probability is given by

Pn =

∣〈f ∣S∣i〉∣2 . 〈f ∣f 〉〈i∣i〉

(3.84)

The states ∣i 〉 and ∣f 〉 are assumed to satisfy the relativistic normalization convention, i.e.

〈i∣i〉 = 2Ei (2π )3δ 3(0) → 2EiV .

(3.85)

In the last step we have regulated the infinity implied by the δ 3(0) but imagining, for the time being, that our system is a box with size L on each side L /2

∫ L →∞ −L /2

(2π )3δ 3(0) = lim

L /2

dx

L /2

= lim L →∞

∫−L/2

∫−L/2

L /2

dy

L /2

dx

∫−L/2

∫−L/2

dz e i k·x k=0

L /2

dy

∫−L/2

dz

(3.86)

= lim L3 = V , L →∞

where, in the last step, I made the limit L → ∞ implicit. Similarly to the singleparticle initial state we can regulate the normalization of the final state n

〈f ∣f 〉 =

∏

2EjV .

j=1

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(3.87)

Relativistic Quantum Field Theory, Volume 1

Figure 3.12. Schematic representation of a general 1 → n decay.

The transition general amplitude can be expressed, in general as,

〈f ∣S − 1∣i〉 ≡ i (2π )4δ 4(pf − pi )Mfi ,

(3.88)

where Mfi is the i → f transition amplitude, which requires us to specify the model, e.g. scalar Yukawa theory, to compute, and the overall factor of i is a convention that allows us to match the non-relativistic quantum mechanics formalism. Note that, in this case, the 1 in S − 1 does not matter since the initial and final states have different particle content and are, therefore, orthogonal. If we boost into the frame in which the initial particle is at rest, we have Ei = m and pi = 0. With this, we can write an explicit expression for the transition probability

⎞ ⎛ n 1 ⎜ 1 ⎟ Pn = [(2π )4δ 4(pf − pi )]2 ∣Mfi ∣2 . ∏ 2mV ⎜⎝ j = 1 2EjV ⎟⎠ ⎞ ⎛ n 1 ⎟ 1 4 4 2 ⎜ = (2π ) δ (pf − pi )∣Mfi ∣ Vt⎜∏ ⎟, 2mV ⎝ j = 1 2E j V ⎠

(3.89)

where, in going from the first to second lines, we replaced one of the 4d delta functions which evaluates to (2π )4δ 4(0) by the space-time volume Vt . We now divide the left and right by t to get the probability per unit time. To obtain the total transition rate, we integrate over all possible momentum for the final state particles by multiplying by V ∫ d 3qj /(2π )3 on the right, which is dimensionless, to produce the Lorentz-invariant measures. The result is the partial decay width, Γn , for our initial particle decaying into the particular n-particle final state we have considered

Γn =

1 2m

⎞ 1 ⎟ 4 4 2 ⎟(2π ) δ (pf − pi ) ∣Mfi ∣ . 3 (2 ) 2 π E j ⎠ ⎝ j=1  ⎛

n

∫ ⎜⎜∏

d 3qj

(3.90)

d Πn

Above we have introduced d Πn , which is called the density of final states. We can see by inspection that, since the integration measure d 3qj /(2Ej ) is Lorentz invariant for

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all final particles, the density of final states is a Lorentz-invariant quantity. This allows us to write this formula compactly as

Γn =

1 2m

∫ d Πn∣Mfi ∣2 .

(3.91)

If we want the total decay width for decay into all possible n-particle final states, we should sum over n, and within each n, over all possible different particle-type combinations C

Γ=

1 ∑∑ 2m n C

∫ d Πn∣Mfi ∣2 .

(3.92)

For example, in the three-particle final state sector, a τ − lepton can decay into μ−νμντ or e−νeντ , so if we wanted to compute the total decay width, we would have to include (at least) these two different combinations when n = 3. The total decay width is equal to the inverse of the particle’s half-life, i.e. τ1/2 = 1/Γ . If the decaying particle is not at rest, the decay rate can be obtained by taking into account the effect of time dilation. This leads to an increased half-life. Another quantity that one will see in particle data tables is the branching ratio (or branching fraction) for the n-particle decay i → f . It is obtained by taking the n-particle decay width and dividing by the total width

BR n(i → f ) ≡

Γn . Γ

(3.93)

Of course, BRn obeys 0 ⩽ BRn(i → f ) ⩽ 1 and ∑n BRn(i → f ) = 1. Cross sections Next we consider an initial state consisting of two particles which can scatter into n particles in the final state. We will take the initial state to be ∣i 〉 = ∣pA , pB 〉 and the final state to be ∣f 〉 = ∣q1, q2, …, qn〉 as shown in figure 3.13. The differential transition probability per unit time and flux, which is the analog of equation (3.92) without the integral, is given by

dσ =

1 d Πn ∣Mfi ∣2 . F 4EAEBV

Figure 3.13. Schematic representation of a general 2 → n scattering.

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(3.94)

Relativistic Quantum Field Theory, Volume 1

The quantity above is usually refered to as the differential cross section. In the expression above F stands for the flux of the incoming beam of particles, which can be written as

∣p / EA − pB / EB∣ ∣v ∣ ∣v − vB∣ F = rel = A = A = V V V

(pA · pB )2 − m A2m B2 EAEBV

.

(3.95)

In the last step above we have used pA2 = m A2 = E A2 − p2A and pB2 = m B2 = E B2 − p2B . Putting the pieces together, we obtain

dσ =

1

d Πn∣Mfi ∣2 .

(3.96)

∫ d Πn∣Mfi ∣2 .

(3.97)

4 (pA · pB )2 − m A2m B2

Integrating, we obtain the total cross section

1

σ=

2

4 (pA · pB ) −

m A2m B2

Example: 2 → 2 scattering cross section of equal mass particles Let us work out some more details in the case of 2 → 2 scattering of equal mass particles of mass m. To simplify the calculation, it is convenient to work in the center-of-momentum (COM) frame, which is defined as being the frame in which the total initial 3-momentum vanishes, i.e. pA + pB = 0. In this frame, momentum conservation implies that q1 + q2 = 0. Since the particles have equal masses, one has pA2 = pB2 = q12 = q22 = m2 . Four-momentum conservation implies that s ≡ (pA + pB )2 = (q1 + q2 )2 , where s is the total center of mass energy squared. Note that the squares appearing in the definition of s are shorthand for the full four-vector contractions, i.e. s = (pA + pB )2 = (pA + pB ) · (pA + pB ) = (pAμ + pBμ )(pA,μ + pB,μ ). Using the equivalence (pA + pB )2 = (q1 + q2 )2 and the fact that the particles all have equal mass, one finds that pA · pB = q1 · q2 . Using this, we can rewrite the square root coming from the flux factor in terms of the outgoing momentum of particle 1

(pA · pB )2 − m 4 = (q1 · q2 )2 − m 4 = (E1E2 − q1 · q 2)2 − (E12 − q12)(E 22 − q 22) =

(E1E2 + q12)2 − (E12 − q12)(E 22 − q 22)

(3.98)

= (E1 + E2 )2 q12 = s ∣q1∣ , where, in going from the second to third lines, I used the fact that q2 = −q1. In the COM frame, we can also simplify the two-body density of states integral

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Relativistic Quantum Field Theory, Volume 1

d 3q1 d 3q2 1 1 (2π )4δ 4(q1 + q2 − pA − pB ) (2π )3 (2π )3 2E1 2E2 d 3q1 1 1 2πδ(E1 + E2 − EA − EB ) = (2π )3 2E1 2E2 d ∣q1∣∣q1∣2 = d Ω q1 δ(E1 + E2 − EA − EB ) 16π 2E1E2 ∣q1∣2 ⎛ ∣q1∣ ∣q1∣ ⎞−1 = d Ω q1 + ⎜ ⎟ 16π 2E1E2 ⎝ E1 E2 ⎠ ∣q ∣ 1 = d Ω q1 1 , 2 16π s

∫ d Π2 = ∫

∫ ∫

∫

(3.99)

∫

∫

where, in going from the third to fourth lines, we have used the general rule for evaluating an integral involving a delta function of a function. Putting the pieces together using (3.96) and dividing the left and right by d Ω q1, we obtain the scattering cross section in the COM frame

dσ (pA + pB → q1 + q2 ) 1 ∣M(pA + pB → q1 + q2 )∣2 = . d Ω q1 64π 2 s

(3.100)

Our final result is for the differential cross section per unit solid angle for 2 → 2 scattering of equal mass particles. Note that the expression for dσ in (3.94) is also valid for identical particles in the final state. However, if one has identical particles in the final state, one has to restrict the two-particle phase-space integration to a region where e.g. ∣q2∣ < ∣q1∣ in order to avoid double-counting. The end of the effect of this is to reduce the scattering cross section by a factor of 2, i.e. σ →

1 2

∫ dσ .

3.8 Examples using scalar Yukawa theory The momentum-space Feynman rules for scalar Yukawa theory are: • Pions are represented by dashed lines. • Nucleons/anti-nucleons are represented by solid lines with arrows. • Add a momentum ki for each internal line. • To each vertex, include a factor of

⎛ ⎞ ( −ig )(2π )4δ 4⎜⎜∑qi ⎟⎟ , ⎝ i ⎠

(3.101)

where ∑i qi is the sum of all momenta entering the vertex. • For each internal dotted line (internal = not connected to initial or final states) with momentum ki , include a factor of

∫

d 4k i . (2π )4 k 2 − m 2 + iε

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(3.102)

Relativistic Quantum Field Theory, Volume 1

• For each internal solid line with momentum ki , include a factor of

∫

d 4k i . (2π )4 k 2 − M 2 + iε

(3.103)

Example: pion decay width in scalar Yukawa theory In a previous example, we obtained the S -matrix for pion decay with the final result being given in equation (3.35). Let us compute this again quickly using the momentum-space Feynman rules specified above. The diagram is shown in figure 3.3. Since there are no internal lines, the result is simply

〈f ∣S∣i〉 = −ig(2π )4δ 4(k − p − q ).

(3.104)

That was painless. You should like Feynman rules now. From this result we can obtain the transition amplitude by using its definition given in equation (3.88), which results in

Mfi = −g.

(3.105)

Plugging this into equation (3.91), we can obtain the leading-order approximation for the pion two-body decay width in the local rest frame (LRF)

Γ LRF = 2

g2 2m

∫

d 3p d 3q 1 1 (2π )4δ 4(p + q − k ) (2π )3 (2π )3 2Ep 2Eq

=

g2 32π 2m

∫

d 3p δ(Ep + Eq − Ek ) EpEq

=

g2 32π 2m

∫ d Ωp ∫

=

g2 32π 2m

∫ d Ωp E∣ppE∣ q ⎜⎝ ∣Epp∣ + ∣Epq∣ ⎟⎠

=

g 2∣p∣ , 8πm 2

d ∣p∣∣p∣2 δ(Ep + Eq − Ek ) EpEq 2

⎛

(3.106)

⎞−1

where, in the last step, I used the fact that in the LRF one has Ep + Eq = Ek = m. The LRF constraint Ep + Eq = Ek = m also fixes ∣p∣ above, since this implies m = p 2 + M 2 + p 2 + M 2 = 2 p 2 + M 2 . This allows us to express ∣p∣ solely in terms of the masses of the ‘meson’ and the ‘(anti)-nucleon’

∣p∣ =

⎛ m ⎞2 ⎜ ⎟ − M2 . ⎝2⎠

(3.107)

Since the argument of the square root above should be positive, this implies that M ⩽ m/2. This condition is called the threshold for this decay. For the case M = m/2

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(‘at threshold’) the two particles would be produced at rest in the LRF, but as we can see from the decay rate expression above as ∣p∣ → 0, one has Γ L2 RF → 0. Example: nucleon–nucleon scattering in scalar Yukawa theory Building upon our toy model, the scattering matrix for nucleon–nucleon scattering in scalar Yukawa theory, was computed in equation (3.67). It is straightforward to check that the momentum-space Feynman rules stated above applied to the sum of the two diagrams in figure 3.4 give the right-hand side of equation (3.67), which I will repeat here for easier reference

⎡ ⎤ 1 1 ⎥ 〈f ∣S − 1∣i〉 ≃ i ( −ig )2 ⎢ + (p1 − q2 )2 − m 2 ⎦ ⎣ (p1 − q1)2 − m 2

(3.108)

4 4

× (2π ) δ (p1 + p2 − q1 − q2 ). From this result we can obtain the transition amplitude by using its definition given in equation (3.88) which gives

⎡ ⎤ 1 1 ⎥. Mfi = ( −ig )2 ⎢ + (p1 − q2 )2 − m 2 ⎦ ⎣ (p1 − q1)2 − m 2

(3.109)

Using this result and equation (3.100), taking into account the factor of 1/2 due to the fact that the outgoing particles are identical, we obtain the following expression for the leading-order differential cross section in the COM frame

dσ g4 1 1 = ∣ + ∣2 . 2 2 2 dΩ 128π s (p1 − q1) − m (p1 − q2 )2 − m 2

(3.110)

Choosing the incoming particle to be propagating in the x -direction, we have from figure 3.14

p1 = (E , ∣p∣ , 0, 0),

(3.111)

p2 = (E , −∣p∣ , 0, 0),

(3.112)

q1 = (E , ∣p∣ cos θ , ∣p∣ sin θ , 0),

(3.113)

q2 = (E , −∣p∣ cos θ , −∣p∣ sin θ , 0),

(3.114)

Figure 3.14. Setup for the calculation of the nucleon–nucleon scattering cross section.

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where I have used the fact that, since p1 = −p2 and the particles have equal mass, one has E p1 = E p2 ≡ E in the COM frame. Likewise, since q1 = −q2 and they have same mass, we have Eq1 = Eq2 and, since p1 + p2 = q1 + q2 , one has Eq1 = Eq2 = E . Finally, since all particles have the same mass and same energy E , this implies that the length of their three-momentum is the same, i.e. ∣p1∣ = ∣p2∣ = ∣q1∣ = ∣q2∣ ≡ ∣p∣. With this, we have

(p1 − q1)2 − m 2 = p12 + q12 − 2p1 · q1 − m 2 = −2p 2(1 − cos θ ) − m 2 ,

(3.115)

(p1 − q2 )2 − m 2 = p12 + q22 − 2p1 · q2 − m 2 = −2p 2(1 + cos θ ) − m 2 .

(3.116)

With this we have

1 1 + (p1 − q1)2 − m 2 (p1 − q2 )2 − m 2 ⎡ ⎤ 1 1 = −⎢ 2 + ⎥. ⎣ 2p (1 − cos θ ) + m 2 2p 2(1 + cos θ ) + m 2 ⎦

(3.117)

Before proceeding in general, let us consider a special case, namely that m = 0. In this case one has ⎤ ⎡ ⎤ 1 1 1 ⎡ 1 1 + + lim ⎢ ⎥ = − 2⎢ ⎥ 2 2 2 2 m → 0 ⎣ (p1 − q1) − m (p1 − q2 ) − m ⎦ 2p ⎣ 1 − cos θ 1 + cos θ ⎦ = −

1 , p 2 sin2 θ

(3.118)

and

⎞2 1 ⎛ ⎞2 dσ g4 ⎛ 1 g2 = lim ⎜ ⎟ = ⎜ ⎟ , m→0 d Ω 128π 2s ⎝ p 2 sin2 θ ⎠ 2 ⎝ 16πE p 2 sin2 θ ⎠

(3.119)

where we have used the fact s = (2E )2 = 4E 2 . The expression above is essentially the relativistic version of the Rutherford scattering cross section (we have ignored spin etc. though). As we see from this expression, the cross section is peaked for very forward (θ = π ) and backward (θ = 0) scattering of particle 1. Returning to the case that m is finite, we find

⎡ ⎤ 1 1 + −⎢ 2 ⎥ 2 2 2 2p (1 + cos θ ) + m ⎦ ⎣ 2p (1 − cos θ ) + m 4p 2 + 2m 2 . = 2 (2p + m 2 )2 − 4p4 cos2 θ

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(3.120)

Relativistic Quantum Field Theory, Volume 1

This gives the final result for the general differential cross section

⎤2 dσ g4 ⎡ 2p 2 + m 2 = ⎢ ⎥ . dΩ 128π 2E 2 ⎣ (2p 2 + m 2 )2 − 4p4 cos2 θ ⎦

(3.121)

Note that the denominator of this expression can never go to zero since (2p 2 + m2 )2 > 4p 2 for m > 0. From this expression, we can obtain the total scattering cross section σ=

g4 128π 2E 2

g4 = 64πE 2 =

∫0 1

∫−1

2π

1

dϕ

⎡

4p 2 + 2m2

⎤2

∫−1 d (cos θ )⎢⎣ (2p2 + m2)2 − 4p4 cos2 θ ⎥⎦

⎡ ⎤2 2p 2 + m2 d (cos θ )⎢ 2 ⎥ ⎣ (2p + m2 )2 − 4p4 cos2 θ ⎦

g4 1 64πE 2 m2(4E 2 + m2 − 4M 2 ) ⎡ ⎛ ⎞⎤ m2 ⎢ m2(4E 2 + m2 − 4M 2 )tanh−1 ⎜1 − ⎟⎥ 2 2 2 ⎝ 2E + m − 2M ⎠ ⎥ ⎢ × ⎢1 + ⎥. 2(E 2 − M 2 )(2E 2 + m2 − 2M 2 ) ⎥ ⎢ ⎦ ⎣

(3.122)

In the high energy E ≫ m, M limit this reduces to

lim σ = E ≫ m, M

g4 1 . 256πm 2 E 4

(3.123)

Exercise 3.7 By adding the correction from the Feynman diagram shown below to the leading-order contribution shown in figure 3.3, write down an expression for the scalar Yukawa theory π → nn decay rate including contributions through order g 4.

. Exercise 3.8 Using the Feynman diagram shown below, calculate the leading-order nucleon–antinucleon scattering cross section in scalar Yukawa theory. Analyze your

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Relativistic Quantum Field Theory, Volume 1

result and determine if it is possible to have a singularity in the scattering cross section. If a singularity exists, what would this imply?

References [1] [2] [3] [4] [5] [6] [7] [8] [9] [10] [11] [12]

Kleinert H 2016 Particles and Quantum Fields (Singapore: World Scientific) Srednicki M 2007 Quantum Field Theory (Cambridge: Cambridge University Press) Ryder L H 2013 Quantum Field Theory (Cambridge: Cambridge University Press) Tong D 2006 Lectures on Quantum Field Theory (Cambridge: Cambridge University Press) Yukawa H 1935 Proc. of the Physico-Mathematical Society of Japan 3rd Series 17 48–57 Wick G C 1954 Phys. Rev. 96 1124–34 Bollini C G and Giambiagi J J 1972 Il Nuovo Cimento B (1971–1996) 12 20–6 ’t Hooft G and Veltman M 1972 Nucl. Phys. B 44 189–213 ’t Hooft G 1973 Nucl. Phys. B 61 455–68 Weinberg S 1973 Phys. Rev. D 8 3497–509 Bardeen W A, Buras A J, Duke D W and Muta T 1978 Phys. Rev. D 18 3998–4017 Collins J C 1984 Renormalization: An Introduction to Renormalization, the Renormalization Group and the Operator-Product Expansion Cambridge Monographs on Mathematical Physics (Cambridge: Cambridge University Press)

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IOP Concise Physics

Relativistic Quantum Field Theory, Volume 1 Canonical formalism Michael Strickland

Chapter 4 Quantum electrodynamics

In the last chapter we focused on simple scalar field theories. Now we will develop the QFT formalism for quantum electrodynamics (QED). In this theory, we have spinor-valued fields ψ that describe the fermions (electrons) and four-vector fields Aμ that describe the gauge field (photons). Let us start with the fermions since they are fundamentally different than what we have covered thus far.

4.1 Classical Dirac fields We begin by writing down the Lagrangian density for a free relativistic fermion field [1–6]

L = ψ (i ∂ − m )ψ ,

(4.1)

where ψ is a four-component Dirac bi-spinor, ψ = ψ †γ 0, and a = γμa μ with γ μ being the Dirac gamma matrices1 i⎞ ⎛ ⎛ ⎞ γ 0 = ⎜1 0 ⎟ γ i = ⎜ 0 i σ ⎟ , ⎝ 0 − 1⎠ ⎝− σ 0 ⎠

(4.2)

where it is understood that each entry is a 2 × 2 matrix, ‘1’ standards for an identity matrix, and σ i are the Pauli spin matrices

⎛ ⎞ σ1 = ⎜ 0 1 ⎟ ⎝1 0 ⎠

⎛ ⎞ σ2 = ⎜ 0 − i ⎟ ⎝i 0 ⎠

⎛ ⎞ σ3 = ⎜ 1 0 ⎟ , ⎝ 0 − 1⎠

which obey the commutation relation [σi , σj ] = 2iεijkσk and the anti-commutation relation {σi , σj} = 2δij 2. The γ -matrices obey 1 This is the ‘Dirac representation’ of the γ -matrices. Other representations are possible, such as the Weyl representation. In all representations, however, the algebra is the same. 2 The index position on the Pauli spin matrices is irrelevant, e.g. σ1 = σ1 = σx , since they are not components of a four-vector.

doi:10.1088/2053-2571/ab30ccch4

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ª Morgan & Claypool Publishers 2019

Relativistic Quantum Field Theory, Volume 1

{γ ν, γ μ} = 2η μν ,

(4.3)

where η μν is the Minkowski space metric tensor. Note that the dimension of the Dirac field is [ψ ] = 3/2. 4.1.1 The Dirac equation Similarly to the case of complex scalar fields, we can treat ψ and ψ as independent variables. The equations of motion for ψ can be determined using

⎡ ∂L ⎤ ∂L ⎥ = 0, − ∂μ⎢ ∂ψ ⎣ ∂(∂μψ ) ⎦

(4.4)

which results in the Dirac equation

(i ∂ − m)ψ = 0.

(4.5)

Alternatively, one can obtain an equation of motion for ψ

ψ (i ∂⃖ + m) = 0,

(4.6)

where the left arrow of the derivative indicates that it should act to the left. Multiplying equation (4.5) by ψ from the left and equation (4.6) from the right by ψ and then adding the results, one obtains

ψ ∂ ψ + ψ ∂⃖ ψ = ∂μ(ψ γ μψ ) = 0.

(4.7)

From the expression above we can identify the conserved Dirac current

j μ ≡ −eψ γ μψ ,

(4.8)

where we have introduced a multiplicative number, the charge −e which is appropriate for an electron, that does not affect the statement of charge conservation. Note also, for future reference, that the Lagrangian vanishes when evaluated at a solution to Dirac’s equation. To see this, multiply (4.5) from the left by ψ which gives the equation (4.1) on the left and hence L → 0 when a ψ is a solution of the Dirac equation. 4.1.2 Hamiltonian density The Hamiltonian density is obtained in the usual manner, starting from

∂L = iψ †, ∂ψ ̇ ∂L π ψ† = † = 0. ∂ψ ̇ πψ =

After performing the Legendre transformation necessary, one obtains

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(4.9)

Relativistic Quantum Field Theory, Volume 1

H = π ψ ψ ̇ + π ψ†ψ ̇ † − L = ψ †( −iα · ∇ + βm)ψ ,

(4.10)

where β = γ 0 and α i = γ 0γ i . Exercise 4.1 Show that equation (4.10) is correct.

4.1.3 Energy–momentum tensor The energy–momentum tensor for the fermions is

∂L ∂ψ ∂L ∂ψ − η μνL + ∂(∂μψ ) ∂xν ∂(∂μψ ) ∂xν i = [ψ γ μ∂ νψ − (∂ νψ )γ μψ ] − η μνL 2 i = [ψ γ μ∂ νψ − (∂ νψ )γ μψ ]. 2

T μν =

(4.11)

The spatial integral of the 0-component gives the conserved four-momentum

Pν =

∫ d 3x T 0ν = ∫ d 3x [ψ iγ 0∂ νψ − η0νψ (iγ · ∂ − m)ψ ].

(4.12)

Using P ν = (P0, P) we obtain the conserved energy

P0 =

∫ d 3x ψ [iγ 0∂ 0 − iγ 0∂ 0 − iγ · ∇ + m]ψ = ∫ d 3x H,

(4.13)

and conserved three-momentum

P=

∫ d 3x ψ [iγ 0∂i]ψ = −i ∫ d 3x ψ †∇ψ .

(4.14)

Note that equation (4.11) is not symmetric. One can use the Belinfante procedure (or the trick of working a curved space mentioned earlier) to construct a symmetric energy–momentum tensor. The symmetrized result is

Tˆ

μν

=

i ψ (γ μD ν + γ νD μ)ψ , 2

(4.15)

where Dμ ≡ 12 (∂μ⃗ − ∂μ⃖ ). 4.1.4 Poincare invariance and generalized angular momentum Using the invariance under three-dimensional rotations, we can find the conserved angular momentum of the Dirac field. The transformation of Dirac spinors under infinitesimal Lorentz transformations is given by

4-3

Relativistic Quantum Field Theory, Volume 1

i δωμνσ μνψ (x ), 4

ψ ′(x′) = ψ (x ) −

(4.16)

where

σ μν =

i μ ν [γ , γ ]. 2

(4.17)

Using this, one finds that the generalized angular-momentum tensor for the Dirac field is of the form

Lμν =

Mμν = Lμν + Sμν,

(4.18)

∫ d 3x ( T 0νxμ − T 0μxν),

(4.19)

Sμν =

1 2

∫ d 3x ψ †σμνψ .

(4.20)

The three-dimensional orbital and spin angular momentum vectors are

L = −i S=

∫ d 3x ψ †(x × ∇)ψ ,

(4.21)

1 2

(4.22)

∫ d 3x ψ †Sψ ,

where

⎛

∑ = ⎜⎝ σ0

0 ⎞⎟ . σ⎠

(4.23)

4.2 Quantization of the Dirac field We begin by looking for classical plane wave solutions to the Dirac equation of the form μ

ψ (x , p ) = e−ixμp u(p ),

(4.24)

( p − m)u(p ) = 0.

(4.25)

which gives

Using pμ = (E , −p) and expanding in terms of the γ -matrices one finds

4-4

Relativistic Quantum Field Theory, Volume 1

p − m = γ 0E − γ · p − m ⎛1 0 ⎞ ⎛ 0 σ⎞ ⎛1 0 ⎞ ⎟E − ⎜ ⎟ · p − ⎜ ⎟m =⎜ ⎝ 0 −1⎠ ⎝−σ 0 ⎠ ⎝0 1 ⎠ ⎛ E − m −σ · p ⎞ =⎜ ⎟. ⎝ σ · p −E − m⎠

(4.26)

Based on the 2 × 2 block structure of the matrices, it is natural to write the fourcomponent bi-spinor u as a two-component vector of spinors

u u = uA , B

( )

(4.27)

which upon substitution into the Dirac equation gives

(E − m)uA = (σ · p)uB , (E + m)uB = (σ · p)uA.

(4.28)

Using the second equation to solve for uB one obtains

σ·p uA E+m px − ipy ⎞ 1 ⎛ pz ⎜⎜ ⎟u . = −pz ⎟⎠ A E + m ⎝ px + ipy

uB =

(4.29)

Particle spinors Choosing

⎛ ⎞ uA = ⎜1 ⎟ or ⎝0⎠

⎛ ⎞ uA = ⎜ 0 ⎟ , ⎝1 ⎠

(4.30)

⎛ ⎞ 0 ⎜ ⎟ 1 ⎜ ⎟ px − ipy ⎟ ⎜ (2) u = A2 , ⎜E+m ⎟ ⎜ −p ⎟ z ⎜ ⎟ ⎝E+m ⎠

(4.31)

gives our first two bi-spinors

⎛ ⎜ ⎜ (1) u = A1⎜ E ⎜ ⎜ px ⎜ ⎝E

⎞ 1 ⎟ 0 pz ⎟ ⎟ + m ⎟ and + ipy ⎟ ⎟ +m ⎠

where A1 and A2 are constants to be fixed below. Anti-particle spinors Choosing instead

⎛ ⎞ uB = ⎜1 ⎟ or ⎝0⎠

⎛ ⎞ uB = ⎜ 0 ⎟ , ⎝1 ⎠

4-5

(4.32)

Relativistic Quantum Field Theory, Volume 1

gives our final two negative energy bi-spinors

v(1)

⎛ ⎜E ⎜ p = B1⎜ x ⎜E ⎜ ⎜ ⎝

pz ⎞ + m ⎟⎟ + ipy ⎟ and +m ⎟ ⎟ 1 ⎟ ⎠ 0

v(2)

⎛ px − ipy ⎞ ⎜ ⎟ ⎜E+m ⎟ −pz ⎟ , = B2⎜ ⎜E+m ⎟ ⎜ ⎟ 0 ⎜ ⎟ ⎝ ⎠ 1

(4.33)

where B1 and B2 are constants to be fixed below. Note that above, we have already implemented the Feynman–Stückelberg prescription [7, 8] by identifying the second two solutions u (3,4) and obtained positive energy solutions by inverting p → −p and E → −E , i.e. v(1,2)(E , p) = u (3,4)( −E , −p). In this prescription, the negative energy solutions, which propagate backwards in time, are interpreted as positive energy anti-particles, which propagate forwards in time. Normalization In both cases, the normalization is fixed by requiring

u (r )†(p )u (s )(p ) = 2Eδ rs ,

(4.34)

v(r )†(p )v(s )(p ) = 2Eδ rs .

(4.35)

Evaluating u (1)†(p )u (1)(p ) = A12 (1 + p 2 /(E + m )2) gives A1 = find A2 = B1 = B2 = E + m .

E + m . Likewise, we

Orthogonality and completeness relations One can prove that the following useful relations hold

u (s )u (s ) = 2m ,

(4.36)

v (s )v(s ) = −2m ,

(4.37)

and

∑ u (s )(p)u (s )(p) =

p + m,

(4.38)

p − m.

(4.39)

s = 1,2

∑ v(s )(p)v(s )(p) = s = 1,2

Exercise 4.2 Show that the relations listed in equations (4.34)-(4.39) are satisfied by the spinor solutions listed in equations (4.31) and (4.33).

4-6

Relativistic Quantum Field Theory, Volume 1

Exercise 4.3 Defining γ 5 ≡ iγ 0γ 1γ 2γ 3. (a) Show that (γ 5)† = γ 5. (b) Show that (γ 5)2 = 1. (c) Show that {γ 5, γ μ} = 0.

4.2.1 The fermion field With the above choice of normalization, the quantized Dirac field can be written in terms of these basis spinors

ψ (x ) =

∫k ∑ [bs(k )u (s )(k )e−ik·x + d s†(k )v(s )(k )eik·x ],

(4.40)

∫k ∑ [bs†(k )u (s )(k )eik·x + ds(k )v(s )(k )e−ik·x ],

(4.41)

s = 1,2

ψ (x ) =

s = 1,2

where the operator b† creates particles, d † creates anti-particles, s labels the spinstate with s ∈ {1, 2}, and

∫k ≡ ∫

d 3k (2π )3

1 . 2Ek

(4.42)

The Dirac field operators defined above satisfy equal-time anti-commutation relations

{ψ (x, t ), ψ (x′, t )} = δ α

† β

αβδ

3

(x − x′),

(4.43)

}

(4.44)

{

{ψα(x , t ), ψβ(x′ , t )} = ψα†(x , t ), ψ β†(x′ , t ) = 0,

where α and β are spinor component indices. As a consequence, the creation and annihilation operators can be shown to obey the following anti-commutation relations † {br (p), bs (k)} = (2π )3δrsδ 3(p − k), † {d r(p), d s (k)} = (2π )3δrsδ 3(p − k),

(4.45)

with all other anti-commutators vanishing, e.g.

{br , bs } = {d r , ds} = {br , d s†} = {br , ds} = ⋯ = 0.

(4.46)

The Hamiltonian density is obtained by identifying π = ∂L/∂ψ ̇ = iψ † as the canonical momentum

H = πψ ̇ − L = iψ †∂tψ =

i † [ψ ∂tψ − (∂tψ †)ψ ], 2

4-7

(4.47)

Relativistic Quantum Field Theory, Volume 1

where we have used the Dirac equation in going from the first to second lines and we symmetrized in the end to make the Hamiltonian manifestly Hermitian. As a final exercise for this section, let us compute the Hamiltonian H = ∫ H x

H =i =

∫x

ψ †∂tψ

i ∑ 2 r, s

∫x ∫p ∫k { ⎡⎣ br†(p)u†(r )(p)eip·x + dr(p)v†(r )(p)e−ip·x⎤⎦

× ⎡⎣ bs (k)u (s )(k)( −ik 0)e−ik·x + d s†(k)v(s )(k)(ik 0)e ik·x⎤⎦

(4.48)

− ⎡⎣br†(p)u †(r )(p )(ip0 )e ip·x + d r(p)v†(r )(p)( −ip0 )e−ip·x⎤⎦ × ⎡⎣ bs (k)u (s )(k)e−ik·x + d s†(k)v(s )(k)e ik·x⎤⎦ .

}

Putting together the details one obtains

H=

1 ∑ 2 r, s

∫k⃗ ⎡⎣ br†(k)bs(k)u†(r )(k)u (s )(k) − dr(k)d s†(k)v†(r )(k)v(s )(k)⎤⎦,

(4.49)

where ∫ ⃗ = ∫ d 3k/(2π )3. Next, we can use the inner product formulas (4.34) and k (4.35)

u †(r )(k)u (s )(k) = v†(r )(k)v(s )(k) = 2Ekδ rs ,

(4.50)

to obtain

H=

∑∫ ⃗ s

k

Ek[bs†(k)bs (k) − ds(k)d s†(k)].

(4.51)

Finally, normal order and apply the anti-commutation relation to obtain

: H : =∑ s

∫k⃗ Ek⎡⎣ bs†(k)bs(k) + d s†(k)ds(k)⎤⎦.

(4.52)

Note that above we used the fact that one picks up a minus sign when normalordering anti-commuting fermionic fields. Exercise 4.4 Show that equation (4.49) follows from equation (4.48).

4.3 The Feynman propagator for Dirac fields As was the case with scalar fields, the Feynman propagator is defined as the vacuum expectation value of the time-ordered product of field operators taken at different points in spacetime

i (SF (x,y ))αβ = 〈0∣Tψα(x )ψβ(y )∣0〉,

4-8

(4.53)

Relativistic Quantum Field Theory, Volume 1

where α and β are spinor indices. For fermionic fields, the time-ordering operator is slightly different:

⎧ ⎪ ψα(x )ψβ (y ) Tψα(x )ψβ(y ) = ⎨ ⎪ ⎩− ψβ(y )ψα(x )

x0 > y0, x0 < y0,

(4.54)

where the minus sign in the second case stems from the anti-commuting nature of the Dirac field. The Feynman propagator can also be written as

i (SF (x,y ))αβ = θ (x 0 − y 0)〈0∣ψα(x )ψβ(y )∣0〉 − θ (y 0 − x 0)〈0∣ψβ(y )ψα(x )∣0〉.

(4.55)

Inserting the Fourier expansions (4.40) and (4.41) into the first vacuum expectation value appearing above, we obtain

〈0∣ψα(x )ψβ(y )∣0〉 =

μ μ 〈0∣⎡⎣ br (p)u α(r )(p)e−ipμ x + d r†(p)vα(r )(p)e ipμ x ⎤⎦

∫p,k ∑

r, s = 1,2

μ μ × ⎡⎣bs†(k)u β(s )(k)e ikμy + ds(k)v β(s )(k)e−ikμy ⎤⎦∣0〉

=

∫p,k e−ip x eik y ∑

u α(r )(p)u β(s )(k)〈0∣br (p)bs†(k)∣0〉

∫p,k e−ip x eik y ∑

u α(r )(p)u β(s )(k)(2π )3δrsδ 3(p − k)

μ

μ

μ

μ

r, s = 1,2

=

μ

μ

μ

μ

(4.56)

r, s = 1,2

=

∫→p

1 −ipμ (x μ−y μ ) e u α(r )(p)u β(s )(p), ∑ 2E p s = 1,2

where ∫ = ∫ d 3p/(2π )3. To simplify this further we can make use of the identity p⃗ (4.38)

∑ u α(r )(p)u β(s )(p) = ( p

+ m)αβ ,

(4.57)

d 3p 1 −ip (x μ−y μ ) e μ ( p + m)αβ . (2π )3 2Ep

(4.58)

s = 1,2

which gives

〈0∣ψα(x )ψβ(y )∣0〉 =

∫p⃗

Likewise, using one can use the identity (4.39)

∑ vα(r )(p)v β(s )(p) = ( p s = 1,2

to obtain

4-9

− m)αβ ,

(4.59)

Relativistic Quantum Field Theory, Volume 1

〈0∣ψβ(y )ψα(x )∣0〉 =

∫p⃗

d 3p 1 ip (x μ−y μ ) e μ ( p − m)αβ . (2π )3 2Ep

(4.60)

By turning the factors of p + m and p − m into operators, we can write, e.g.

d 3p 1 −ip (x μ−y μ ) e μ (2π )3 2Ep = (i ∂ + m)αβ D(x − y ),

〈0∣ψα(x )ψβ(y )∣0〉 = (i ∂ + m)αβ

∫p⃗

(4.61)

where the ∂ is understood to be with respect to x − y and in the last step we have recognized that the integral is simply the scalar two-point function introduced in equation (2.118). Likewise, one finds

〈0∣ψβ(y )ψα(x )∣0〉 = −(i ∂ + m)αβ D(y − x ).

(4.62)

Inserting these results into equation (4.55), we obtain i (SF (x , y ))αβ = θ (x 0 − y 0 )(i ∂ + m )αβ D(x − y ) + θ (y 0 − x 0 )(i ∂ + m )αβ D(y − x ) = (i ∂ + m )αβ [θ (x 0 − y 0 )D(x − y ) + θ (y 0 − x 0 )D(y − x )]

(4.63)

= (i ∂ + m )αβ DF (x − y ),

where DF is the scalar Feynman propagator introduced in equation (2.130)3. Summarizing (and dropping the spinor indices for compactness), we found that the Feynman propagator for a Dirac field can be expressed in the spatial representation as

iSF (x − y ) = (i ∂ + m)DF (x − y ),

(4.64)

where I have written the argument on the left as x − y since SF only depends on this difference. One can also obtain a compact representation of the Feynman propagator by expressing it directly in momentum space. To do this most easily, one can use the iε prescription

⎡ iSF (x − y ) = (i ∂ + m)⎢ ⎣

∫

=

∫

⎤ d 4p i −ipμ (x μ−y μ ) e ⎥ (2π )4 p 2 − m 2 + iε ⎦

d 4p i ( p + m) −ip (x μ−y μ ) e μ . (2π )4 p 2 − m 2 + iε

(4.65)

One might be concerned about the derivatives acting on the θ functions in going from the first to second lines in equation (4.63). In practice, one finds that the extra terms cancel between the first and second contributions listed in the second line of equation (4.63). 3

4-10

Relativistic Quantum Field Theory, Volume 1

To proceed further, we make use of the identity aa = a μaμ = a 2 to rewrite p2 − m2 + iε = ( p + m + iε )( p − m + iε ) where 2 p ε = ε. Adding an infinitesimal iε in the numerator of (4.65), we see that the factors of ( p + m + iε ) cancel in the numerator and denominator, i.e.

p +m p + m + iε 1 = → , 2 p − m + iε ( p + m + iε )( p − m + iε ) p − m + iε 2

(4.66)

giving

iSF (x − y ) =

∫

d 4p i μ μ e−ipμ (x −y ). 4 (2π ) p − m + iε

(4.67)

From this result, we can read off the Fourier transform as the argument of the integral on the right-hand side, i.e.

iSF (p ) =

i . p − m + iε

(4.68)

Exercise 4.5 Show that the projection operators PL = 12 (1 − γ 5) and PR = 12 (1 + γ 5) obey PL2,R = PL,R and PLPR = 0. Also show that, for massless fermions the projection operators PL and PR , project out the left-handed (negative helicity) and right-handed (positive helicity) parts of the fermion spinor, respectively. Exercise 4.6 Show that the following ‘trace identities’ are true. Here ‘Tr’ acts in spinor space, summing the diagonal elements of the argument. For a 4 × 4 identity matrix, one has trivially Tr(1) = 4. For γ μ one has, e.g. Tr(γ μ ) = 0. (a) Tr( a b ) = 4aμb μ = 4a · b. (b) Tr( a b c d ) = 4[(a · b )(c · d ) − (a · c )(b · d ) + (a · d )(b · c )]. (c) Tr(γ 5) = 0. (d) Tr(γ 5 a b ) = 0. (e) Tr(γ 5 a b c d ) = −4iεμνρσ a μb νc ρd σ .

4.4 The electromagnetic field Having covered the electron, now let us turn to the photon. In the first chapter of this volume we learned that the classical Lagrangian for electromagnetism can be expressed in terms of the field strength tensor F μν = ∂ μAν − ∂ νAμ where Aμ = (ϕ, A), with the result being

1 L = − FμνF μν, 4 where, for now, I have taken the source J μ to zero.

4-11

(4.69)

Relativistic Quantum Field Theory, Volume 1

4.4.1 Local gauge invariance The Lagrangian density (4.69) is invariant under a local gauge transformation

Aμ (x ) → Aμ (x ) + ∂μλ(x ),

(4.70)

where λ(x ) is an arbitrary function that dies off sufficiently rapidly as ∣x∣ → ∞. The invariance of the Lagrangian density under this transformation is called a gauge symmetry. Physical observables such as the electric and magnetic fields, energy–momentum tensor, etc do not depend on the choice of the gauge (choice of λ(x ) above). As a result, our theory has a infinite number of symmetries associated with the choice of λ(x ). However, unlike global symmetries that can be used in conjunction with Noether’s theorem, a local gauge transformation does not change the physical state we are describing. Since physically equivalent states can be labeled by an infinite number of possible gauge choices, this tells us that there is a very large redundancy in our description. In order to remove this redundancy, we can choose a particular gauge to work in. This is similar in some ways to the choice of coordinate system that one will use to attack the problem and, as with coordinate systems, some gauge choices allow you to more easily solve a problem than others. In practice, one can use a class or classes of different gauge choices and verify explicitly that the results for physical observables are independent of the gauge choice. 4.4.2 Gauss’ law Since our massless vector field Aμ has four components, it seems like the photon in our theory has four degrees of freedom. However, we know experimentally that, in vacuum, the photon has only two independent degrees of freedom (polarization states). How can we make a consistent quantization of the vector field Aμ that ensures that there are only two physical degrees of freedom in vacuum? Part of the answer lies in Gauss’ law and the other part lies in gauge symmetry, so let us discuss these now. If one expands the Lagrangian (4.69), you will find that there is no term involving A0̇ = ϕ.̇ As a result, one cannot define a canonical momentum conjugate to the electric potential ϕ. This means that this variable is not dynamical. If we are given initial values for Ai and Ai̇ at some time t0 , then A0 can be determined using Gauss’ law ∇ · E = 0. To see how this works in practice, using E = −∇A0 − ∂tA , we see that Gauss’ law implies that

−∇2 A0 − ∇ · ∂tA = 0.

(4.71)

This equation has a solution of the form

A0 (x) =

1 4π

∫ d 3y (∇∣x· ∂−tAy)(∣ y)

=

1 ∂t 4π

∫ d 3y (∇∣x· −A)(y∣y) .

(4.72)

This means that, at any given time t , A0 is completely determined by the divergence of Ai̇ . As a result, there are really only three independent degrees of freedom in Aμ. We have reduced four degrees of freedom to three. 4-12

Relativistic Quantum Field Theory, Volume 1

4.4.3 Gauge fixing To remove the remaining extra degree of freedom we must specify (fix) a gauge. To fix the gauge we must supply an additional constraint equation to be satisfied by the vector potential. The two most commonly used gauges are the Lorenz gauge (∂μAμ = 0) and the Coulomb gauge (∇ · A = 0) [9], however, there are many other possibilities, e.g. Landau gauge, axial gauges, light-cone gauge, etc. For simplicity, we will focus on the two canonical (classes of) gauges. Lorenz or covariant gauge, ∂μAμ = 0 If we are given a vector potential A˜ μ that has ∂μA˜ μ ≠ 0, we can make a general gauge μ transformation Aμ ≡ A˜ + ∂ μλ such that

∂μAμ = ∂μ(A˜ + ∂ μλ) = 0.

μ

(4.73)

μ ∂μ∂ μλ = −∂μA˜ .

(4.74)

This is solved by

Note that since this is a partial differential linear equation, there are homogeneous solutions that satisfy

∂μ∂ μλ = 0.

(4.75)

Any gauge transformation that satisfies this homogeneous partial differential equation, leaves ∂μAμ = 0 invariant. This means that we have not completely fixed the gauge. The remaining freedom in the choice of λ(x ) is called a residual gauge freedom. Coulomb or radiation gauge, ∇ · A = 0 We can use the residual gauge freedom in the Lorenz gauge to additionally enforce ∇ · A = 0 using a similar method as above. Since ∇ · A = 0, from equation (4.72) we have A0 = 04. As a result, Coulomb gauge is a special case of the general Lorenz gauge and still satisfies the Lorenz gauge condition ∂μAμ = 0. Note that, by singling out the spacelike components in the specification of the Coulomb gauge, we have broken Lorentz invariance. As a result, one should be careful when using the Coulomb gauge. However, the reduction of the number of degrees of freedom is quite clear with this gauge choice. In the Coulomb gauge, the three independent components of A satisfy a single constraint equation ∇ · A = 0, allowing us to write one of the components in terms of the other two. This means that we only have two real degrees of freedom remaining that can then be identified with the two polarization states of the photon.

4

If we include charges (sources), then A0 ≠ 0 since ∇ · E = ρ .

4-13

Relativistic Quantum Field Theory, Volume 1

4.5 Quantization of the electromagnetic field Using the Lagrangian density (4.69), we can compute the Hamiltonian density. We begin, as usual, by computing the canonical momentum conjugate to Aμ

∂L = 0, ∂A0̇ ∂L = − F 0i ≡ E i . πi = ∂Ai̇

π0 =

(4.76)

As mentioned above, the fact that π 0 = 0 tells us that A0 is not a dynamical field. Using the results above, we can obtain the Hamiltonian density

H = π iAi̇ − L 1 = (E 2 + B 2 ) − A0 (∇ · E). 2

(4.77)

As we can see from the result above, A0 acts as a Lagrange multiplier multiplying Gauss’ law. The ‘equation of motion’ for A0 simply imposes Gauss’ law

∇ · E = 0,

(4.78)

which is a constraint on the system that must be imposed if one treats the components of A as the dynamical degrees of freedom. Exercise 4.7 Show that equation (4.77) follows from equation (4.76).

4.5.1 Coulomb gauge Let us begin by considering quantization in the Coulomb gauge. In the Coulomb gauge, one obtains the following vacuum wave equation for the vector potential [9]

□A = ∂ μ∂μA = 0,

(4.79)

which is solved by

A=

∫

d 3k ε ke ±ik·x , (2π )3

(4.80)

as long as k 02 = E k2 = ∣k∣2 . Above, εk is a momentum-dependent three-vector that represents the polarization of the wave. Evaluating ∇ · A we obtain

∇ · A = ∓i

∫

d 3k (ε k · k) e ±ik·x . (2π )3

(4.81)

Requiring the Coulomb gauge condition ∇ · A = 0 then implies that

ε k · k = 0,

4-14

(4.82)

Relativistic Quantum Field Theory, Volume 1

which tells us that εk is always transverse to the direction of the wave specified by k . We have a choice concerning the precise vector basis that is used to describe εk . In general, we can take εk to be a linear combination of two orthonormal vectors εkr with r = 1, 2 where the basis vectors both satisfy εkr · k = 0 (i.e. they are both in the transverse plane relative to k ). The polarization basis vectors satisfy

εkr · εks = δrs ,

(4.83)

for r ∈ {1, 2}. The solutions (4.80) will form the basis of plane wave states we use to decompose the quantum electromagnetic field. However, before making this decomposition, we should consider what happens to Poisson brackets subject to the Coulomb gauge constraint. 4.5.2 Poisson (Dirac) brackets for constrained systems One would naively guess that the correct Coulumb-gauge equal-time commutation relations for the electromagnetic field should be

[Ai (x , t ), Aj (y , t )] = [Ei (x , t ), Ej (y , t )] = 0, [Ai (x , t ), Ej (y , t )] = iδijδ 3(x − y).

(4.84)

However, this does not work because it is not consistent with the constraints that ∇ · A=∇ · E = 0. To see this, consider the commutator

[∇ · A(x , t ), ∇ · E(y , t )] = ⎡⎣∇ix Ai (x , t ), ∇ yj Ej (y , t )⎤⎦ = ∇ix ∇ yj [Ai (x , t ), Ej (y , t )] x

= i∇ ·

∇y δ 3(x

(4.85)

− y) ≠ 0.

The problem encountered above also occurs in the classical theory and one finds that it is necessary to modify the Poisson bracket structure in order to account for the constraints [10]. The resulting modified Poisson brackets are called Dirac brackets and translate into the following non-trivial Coulomb-gauge commutation relation

⎛ ∂i∂j ⎞ [Ai (x , t ), Ej (y , t )] = i ⎜δij − 2 ⎟δ 3(x − y). ⎝ ∇ ⎠

(4.86)

To see how this satisfies the constraints, it is easier to work in momentum–space. Writing the 3d delta function above as a momentum–space integral, we obtain

[Ai (x , t ), Ej (y , t )] = i

⎛

∫k⃗ ⎜⎝δij −

kikj ⎞ i k·(x−y) ≡i ⎟e k2 ⎠

∫k⃗ PijT ei k·(x−y),

(4.87)

where I have introduced the transverse projector

PijT ≡ δij −

ki kj , k2

which obeys PijT ki = PijT kj = 0 and PikT PkjT = PijT . This gives, for example,

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(4.88)

Relativistic Quantum Field Theory, Volume 1

[∇ix Ai (x , t ), Ej (y , t )] = −

∫k⃗

PijT ki e ik·(x−y) = 0.

⏟

(4.89)

=0

Similarly, one can show that using equation (4.87) all necessary constraints are satisfied. 4.5.3 Coulomb-gauge mode expansion As a result of the previous discussion, we are now in a position to make our quantum mode expansion of the vector potential and electric field (conjugate momentum) 2

A(x ) =

∫k ∑er(k)⎡⎣ ar(k)e−ik·x + ar†(k)eik·x⎤⎦,

(4.90)

r=1

2

E(x ) =

∫k ∑εr(k)(−iEk )⎡⎣ ar(k)e−ik·x − ar†(k)eik·x⎤⎦,

(4.91)

r=1

where ∫ ≡ ∫ d 3k (2π )−3(2Ek )−1/2 and, since photons are massless, one has Ek = ∣k∣. k The polarization vectors above satisfy

εkr · k = 0

εkr · εks = δrs 2

∑εkr,iεkr,j = δij − r=1

(transversality),

(orthogonality), k ik j k2

(completeness).

(4.92)

(4.93) (4.94)

One can show that the creation and annihilation operators introduced above obey

[ar(k), as(p)] = ⎡⎣ a r†(k), a s†(p)⎤⎦ = 0,

(4.95)

⎡ a (k), a †(p)⎤ = (2π )3δ δ 3(k − p). rs s ⎣ r ⎦

(4.96)

These guarantee that the field commutation relations (4.87) are satisfied. Using equations (4.90) and (4.91), one can obtain an expression for the quantized Coulomb-gauge Hamiltonian. After some algebra and normal ordering, we obtain the following simple expression 2

∫k⃗ ∣k∣∑ar†(k)ar(k),

:H : =

(4.97)

r=1

where, as previously, ∫ ⃗ ≡ ∫ d 3k/(2π )3. k The Coulomb gauge is appealing because the physical degrees of freedom are clear in this gauge, however, this gauge is not Lorentz invariant. This can be seen by looking at, for example, the Ai propagator 4-16

Relativistic Quantum Field Theory, Volume 1

DijT (x − y ) ≡ 〈0∣TAi (x )Aj (y )∣0〉

=

∫

⎛ ⎞ ⎜ ⎟ d k i ⎜ δ − kikj ⎟e−ik·(x−y ). ij 2 ⎟ (2π )4 k 2 + iε ⎜ k ⎜ ⎟ T Pij ⎝ ⎠ 4

(4.98)

This form turns out to be very inconvenient when including interactions in the theory, so let us turn back to the Lorenz gauge, which is manifestly Lorentzinvariant, for some guidance. 4.5.4 General Lorenz (covariant) gauge In the Lorenz gauge one imposes ∂μAμ = 0. The corresponding vacuum equations of motion for the vector potential are

□Aν = ∂μ∂ μAν = 0.

(4.99)

In order to enforce our gauge choice, in this case we will introduce a Lagrange multiplier λ = 1/ξ into the Lagrangian

1 1 (∂μAμ )2 . L = − FμνF μν − 4 2ξ

(4.100)

The equation of motion for the Lagrange multiplier λ = 1/ξ gives the Lorenz gauge condition. In constructing the Lagrangian above, we have generalized the covariant Lorenz gauge. The end result is that different values of ξ will map to specific gauge choices in the family of Lorenz (covariant) gauges. Taking ξ = 1 is the standard Lorenz gauge which is referred to as Feynman gauge. Taking ξ = 0 gives the Landau gauge and taking ξ = 3 gives the, not often used, Yennie gauge. For now, we will take ξ = 1 (Feynman gauge) and then generalize the result to arbitrary ξ at the end. Using the modified Lagrangian with ξ = 1 we obtain

∂L = −∂ μAμ , ∂A0̇ ∂L = ∂ iA0 − Ai̇ . πi = ∂Ai̇

π0 =

(4.101)

Promoting Aμ and π μ to operators, we obtain the following Feynman-gauge commutation relations

[Aμ (x , t ), Aν (y , t )] = [πμ(x , t ), πν(y , t )] = 0, [Aμ (x , t ), πν(y , t )] = iημνδ 3(x − y).

(4.102)

Using this, we can make a mode expansion involving four polarization four-vectors εμ(k , λ ) with λ ∈ {0, 1, 2, 3} as follows:

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Relativistic Quantum Field Theory, Volume 1

3

Aμ (x ) =

∫k ∑ε μ(k, λ)[a(k, λ)e−ik·x + a†(k, λ)eik·x ],

(4.103)

λ= 0

3

π μ(x ) = i

∫k

∣k∣ ∑ ε μ(k , λ)[a(k , λ)e−ik·x − a†(k , λ)e ik·x ].

(4.104)

λ= 0

The relevant commutation relations are ′

[a(k , λ), a†(p , λ′)] = − η λλ (2π )3δ 3(k − p), [a(k , λ), a(p , λ′)] = [a†(k , λ), a†(p , λ′)] = 0.

(4.105)

There are now four polarization vectors that are each four-vectors. We choose one of these ε(k , 0) to be timelike and the others ε(k , i ) to be spacelike. The normalization of the polarization vectors is taken to be ′

ε μ(k , λ)εμ(k , λ′) = η λ λ ,

(4.106)

and, as a result, we have ′

εμ(k , λ)εν(k , λ′)η λλ = ημν ,

(4.107)

where the sum over λ and λ′ is implicit due to the repeated indices above. We choose to make two of the spacelike polarizations, corresponding to ‘1’ and ‘2’ orthogonal to the four-momentum, i.e.

k μεμ(k , 1) = k μεμ(k , 2) = 0.

(4.108)

The remaining spacelike polarization vector εμ(k , 3) is the longitudinal polarization. If, for example, the momentum lies in the x 3 direction, we have k = (E , 0, 0, E ) and we can choose

⎛1 ⎞ ⎜ ⎟ ε 0 = ⎜ 0 ⎟ ε1 = ⎜0⎟ ⎝0⎠

⎛0⎞ ⎛0⎞ ⎜0⎟ ⎜1 ⎟ ⎜ ⎟ ε 2 = ⎜ ⎟ ε3 = ⎜1 ⎟ ⎜0⎟ ⎝0⎠ ⎝0⎠

⎛0⎞ ⎜0⎟ ⎜ ⎟. ⎜0⎟ ⎝1 ⎠

(4.109)

For other momenta, we can use a Lorentz transformation to boost/rotate these vectors to the appropriate frame. Note that the normalization (4.106) implies some strange consequences like the existence of negative norm states. We will return to this issue in the next section, where we will use a method invented by Gupta and Bleuler that requires that the expectation value of the operator-valued Lorenz gauge condition between any two physical states returns zero, i.e. 〈ψ ′∣∂μAμ∣ψ 〉 = 0. We will see that for physical observables there will be a delicate cancellation between the timelike- and longitudinally-polarized modes, leaving only the two transverse degrees of freedom as the physical degrees of freedom.

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Relativistic Quantum Field Theory, Volume 1

Exercise 4.8 Show that equation (4.107) is consistent with equation (4.106).

4.5.5 The photon propagator in general covariant gauge The Feynman propagator is obtained from a time-ordered expectation value of fields

iD Fμν(x − y ) ≡ 〈0∣TAμ (x )Aν (y )∣0〉 3 ⎞ 1 ⎛ ⎜⎜ ∑ ηλλε μ(k , λ)ε ν(k , λ)⎟⎟ = − k ⃗ 2Ek ⎝ ⎠ λ= 0

∫

× [θ (x 0 − y 0)e−ik·(x−y ) + θ (y 0 − x 0)e ik·(x−y )] 1 [θ (x 0 − y 0)e−ik·(x−y ) + θ (y 0 − x 0)e ik·(x−y )] = − η μν k ⃗ 2Ek = − η μνDF (x − y ),

(4.110)

∫

where I used equation (4.107) in going from the second to third lines. This can be simplified using the iε prescription for the scalar Feynman propagator appearing on the last line above

iD Fμν(x − y ) =

∫

d 4k −iη μν −ik·(x−y ) e . (2π )4 k 2 + iε

(4.111)

This allows us easily identify the Feynman propagator in momentum–space

iD Fμν(k ) =

−iη μν . k 2 + iε

(4.112)

The result above is specific to the Feynman gauge (ξ = 1). In a general covariant gauge, the result is

iD Fμν(k ) =

−i ⎡ μν k μk ν ⎤ + − ( 1) η ξ ⎢ ⎥. k 2 + iε ⎣ k 2 + iε ⎦

(4.113)

4.5.6 The Gupta–Bleuler method In this section, I will discuss the Gupta–Bleuler method for imposing gauge fixing constraints in QFT [11, 12]. However, before doing this, we need to set up the problem. In the end of the previous section we discussed how to quantize the electromagnetic field in a Lorentz-invariant manner using the Lorenz gauge. Although we were able to do this, the final expression contained a sum over four different polarization states when there should only be two physical states in

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Relativistic Quantum Field Theory, Volume 1

vacuum. Even worse, the commutation relation obtained for the photon creation/ annihilation operators seemed to imply that negative norm states exist. To see this explicitly, I remind you of the result obtained in the last section (4.105). ′

[a(k , λ), a†(p , λ′)] = −η λλ (2π )3δ 3(k − p).

(4.114)

Let us try to construct a Fock space for photons using this. As usual, we assume that there exists a vacuum state ∣0∣ which satisfies

a(k , λ) 0 = 0,

(4.115)

〈0∣0〉 = 1.

(4.116)

which is normalized as

Likewise, one can construct eigenstates of the number operator by applying creation operators to the vacuum state. We will focus on a single field mode. In this case one has

n(k , λ) =

1 [a†(k , λ)]n(k,λ ) 0 . n(k , λ)!

(4.117)

Using this, we can attempt to compute the norm of a one-photon state

〈1(k , λ)∣1(k , λ)〉 = 〈0∣a(k , λ)a†(k , λ)∣0〉 = 〈0∣−η λλ(2π )3δ 3(k − k) + a†(k , λ)a(k , λ)∣0〉 = − η λλ(2π )3δ 3(0).

(4.118)

This result is infinite, but this is nothing new since, as discussed previously, plane wave states are not normalizable in an infinite volume and in a finite volume (2π )3δ 3(0) → V . The disturbing thing about the result above is that, if we consider the timelike (sometimes called the scalar) polarization λ = 0, one has η λλ = 1 and one finds that the norm is negative. This is a fundamental problem since this destroys our ability to interpret things probabilistically. Similarly, one finds that the photon energy, defined by

〈1(k , λ)∣H ∣1(k , λ)〉 = Ek〈1(k , λ)∣1(k , λ)〉,

(4.119)

is also negative for the scalar polarization λ = 0 because the norm on the right-hand side is negative. Clearly, our naive Fock space construction is flawed. This is because we have not yet enforced the gauge constraint ∂μAμ = 0. The results obtained above point to a problem with scalar photons in Lorenz gauge quantization. So far, we have not applied the constraint implied by ∂μAμ = 0. As we will see, this constraint will guarantee that expectation values evaluated with physical states will not depend on the unphysical scalar and longitudinal polarization contributions. If one tries to naively apply ∂μAμ = 0 as an operator identity using (4.103) one finds that ∂μAμ ≠ 0. As explained by Gupta and Blueler, the solution is to instead use the Lorenz gauge constraint as a condition on the state vectors in the photon Hilbert space ∣Ψ∣. Only state vectors that satisfy

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Relativistic Quantum Field Theory, Volume 1

〈Ψ′∣∂μAμ ∣Ψ〉 = 0,

(4.120)

will be part of the physical Hilbert space. In fact, we can require a somewhat stronger condition that guarantees that the equation above is satisfied. To do this we decompose the vector potential into positive- and negative-frequency parts

Aμ (x ) = A μ(+)(x ) + A μ(−)(x ),

(4.121)

where A μ(+) and A μ(−) contain only annihilation and creation operators, respectively, and (A μ(+) )† = A μ(−). This is the same decomposition we used to prove Wick’s theorem. Gupta and Bleuler postulated the following condition for the annihilation part of the Lorenz-gauge constraint

∂ μA μ(+) Ψ = 0,

(4.122)

where Ψ is an arbitrary state. Taking the Hermitian conjugate, this also implies

Ψ ∂ μA μ(−) = 0.

(4.123)

These conditions are sufficient to guarantee that equation (4.120) is satisfied

〈Ψ′∣∂ μAμ ∣Ψ〉 = 〈Ψ′∣ ∂ μA μ(+)∣Ψ〉 + 〈Ψ′∣∂ μA μ(−) ∣Ψ〉 = 0.   =0

(4.124)

=0

Let us now consider the implications of (4.122) in terms of photon creation/ annihilation operators. Extracting the positive-frequency contribution from equation (4.103), we find the constraint

⎡ ⎛ ⎢∂μ⎜ ⎢⎣ ⎜⎝

3

⎞⎤

λ=0

⎠⎦

∫k ∑ε μ(k, λ)a(k, λ)e−ik·x⎟⎟⎥⎥

= −i

∫k

Ψ

⎤ ⎡ 3 e−ik·x⎢ ∑ k · ε(k , λ) a(k , λ)⎥ Ψ = 0. ⎢⎣ λ = 0 ⎦⎥

(4.125)

Since this has to be true for all Fourier modes, this implies that the quantity in square brackets acting on the state has to vanish. To simplify this further, we can make use of equations (4.108) and (4.109), which tell us that k · ε(k , 1) = k · ε(k , 2) = 0, k · ε(k , 1) = E , and k · ε(k , 3) = −E 5. As a result of this simplification, we have

−i

∫k

e−ik·xE [a(k , 0) − a(k , 3)] Ψ = 0.

(4.126)

Again, since this has to apply to all Fourier modes, this implies that

5 Although we used the forms (4.108) and (4.109), the conclusions reached for the dot products are true in any frame since these are Lorentz invariant.

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Relativistic Quantum Field Theory, Volume 1

[a(k , 0) − a(k , 3)] Ψ = 0,

(4.127)

Ψ [a†(k , 0) − a†(k , 3)] = 0.

(4.128)

and

This demonstrates that the Lorenz gauge condition implies that there is a relationship between scalar and longitudinal photons. The relations above tell us that the expectation values of the scalar and longitudinal number operators are equal for all physically admissible states, i.e.

〈Ψ∣a†(k , 0)a(k , 0)∣Ψ〉 = 〈Ψ∣a†(k , 3)a(k , 3)∣Ψ〉.

(4.129)

As a result, we can now demonstrate that the energy is positive as well by evaluating the expectation value of the Hamiltonian between physical states. To do this, we begin with the Hamiltonian density in Lorenz gauge, which is

1 1 H = − π μπμ − ∇Aμ · ∇Aμ . 2 2

(4.130)

Computing the Hamiltonian operator, one finds 3

H=

∫k⃗

Ek ∑ ( −ηλλ )a†(k , λ)a(k , λ).

(4.131)

λ= 0

Using this result, one has

〈Ψ∣H ∣Ψ〉 =

∫k

Ek

∑ 〈Ψ∣a†(k, λ)a(k, λ)∣Ψ〉 λ = 1,2

+ 〈Ψ∣a†(k , 3)a(k , 3) − a†(k , 0)a(k , 0)∣Ψ〉 =

∫k

Ek

(4.132)

∑ n(k, λ), λ = 1,2

where n(k , λ ) is the expectation value of the number operator. As we see, only the transverse photons contribute to the energy in the end. The negative energy scalar photon contribution is canceled exactly by the contribution from the longitudinal photon. The same thing will happen for all observable quantities, e.g. momentum, angular momentum, etc.

4.6 Coupling the electron to the photon We have now built two separate non-interacting field theories, one for the electron/ anti-electron and one for the photon. We would now like to couple the photon to the electron. Luckily, we already have all of the pieces we need. Recall, that when including sources, the electromagnetic Lagrangian is (1.15)

1 LMaxwell = − FμνF μν − Aμ J μ. 4

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(4.133)

Relativistic Quantum Field Theory, Volume 1

Identifying the current appearing above with the electron current derived in equation (4.8)

J μ = (j μ )electron = −eψ γ μψ ,

(4.134)

we obtain

1 FμνF μν + eAμ ψ γ μψ 4 1 = − FμνF μν + eψ A ψ . 4

LMaxwell = −

(4.135)

To complete the QED Lagrangian, we should add the contribution from the Dirac fields (4.1)

LQED = LDirac + LMaxwell 1 = ψ (i ∂ − m)ψ − FμνF μν + eψ A ψ . 4

(4.136)

This can be simplified a bit by introducing the gauge covariant derivative

Dμ ≡ ∂μ − ieAμ ,

(4.137)

which allows us to write the QED Lagrangian compactly as

LQED = ψ (iD − m)ψ −

1 FμνF μν. 4

(4.138)

4.6.1 Gauge invariance of the QED Lagrangian One can verify almost immediately that the QED Lagrangian (4.138) is invariant under a global phase transformation ψ → e iαψ , which is related to charge conservation; however, there is a much larger class of transformations that leave the QED Lagrangian invariant. These correspond to local gauge transformations under which the vector potential and Dirac field transform as

Aμ → Aμ +

1 ∂μλ(x ), e

ψ → e iλ(x )ψ ,

(4.139) (4.140)

where λ(x ) is a arbitrary function. We can verify this explicitly

⎡ ⎤ ⎛ 1 ⎞ 1 LQED → [e−iλψ ]⎢iγ μ∂μ + eγ μ⎜Aμ + ∂μλ⎟ − m⎥[e iλψ ] − FμνF μν, ⎝ ⎠ ⎣ ⎦ 4 e

(4.141)

where I have used the fact that FμνF μν is invariant under local gauge transformations. Expanding out the derivative

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Relativistic Quantum Field Theory, Volume 1

i ∂μ[e iλψ ] = −[∂μλ ]e iλψ + e iλ[i ∂μψ ].

(4.142)

This first term above cancels the extra term generated by the gauge transform of Aμ

LQED → ψ [ −( ∂ λ)ψ + i ∂ ψ + e A ψ + ( ∂ λ)ψ − mψ ] − → ψ (i ∂ + e A − m )ψ − → ψ (i D − m )ψ −

1 FμνF μν 4

1 FμνF μν 4

(4.143)

1 FμνF μν = LQED . 4

This demonstrates that QED is invariant under local gauge transformations. I note, in closing, one can turn this on its head and, instead of coupling the current in the way we did above, one can simply add the free Dirac and Maxwell Lagrangians and then require local gauge invariance and one will find that one needs to add a term of the form eψ Aψ to the sum of the free Lagrangians in order to guarantee local gauge invariance. We will use this method in the second volume to obtain the QCD Lagrangian which, in the end, is strikingly similar to the QED Lagrangian. Exercise 4.9 Show that the electromagnetic field strength tensor can be expressed as a commutator of gauge covariant derivatives as follows −ieF μν = [D μ, D ν ]. (Hint: consider this an operator relation with the both the right and left acting on an arbitrary scalar function f .)

4.6.2 Obtaining vertex functions directly from the QED Lagrangian In the last chapter, we derived a systematic and somewhat tedious procedure for determining the Feynman rules for a given theory. Here I would like to demonstrate a quick method that can be used to read off the Feynman rules for any given model Lagrangian. To do this we will extract the vertex functions for the theory. The twopoint vertex function for Dirac fields will give the inverse Dirac propagator, the twopoint vertex function for the vector potential will give the photon propagator, and the three-point function involving a photon and two Dirac fields will give the interaction vertex. To start we expand out each term of the QED Lagrangian explicitly

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Relativistic Quantum Field Theory, Volume 1

1 FμνF μν 4 = ψ (iγ μ∂μ − m)ψ + eψ γ μψAμ 1 − (∂μAν − ∂νAμ )(∂ μAν − ∂ νAμ ) 4 = ψ (iγ μ∂μ − m)ψ + eψ γ μψAμ 1 − [(∂μAν )(∂ μAν ) − (∂νAμ )(∂ μAν )]. 2

LQED = ψ (i D − m)ψ −

(4.144)

Focusing on the third term, we can write μ

μ

(∂μAν )(∂ μAν ) − (∂νAμ )(∂ μAν ) = Aα (∂μ⃖ ∂⃗ η αβ )Aβ + Aμ (∂ν⃖ ∂⃗ )Aν .

(4.145)

Since what matters is the action, we can integrate by parts to transform all left derivatives to right derivatives times a minus sign μ μ μ μ Aα (∂μ⃖ ∂⃗ η αβ )Aβ + Aμ (∂ν⃖ ∂⃗ )Aν → − Aα (∂μ⃗ ∂⃗ η αβ )Aβ − Aμ (∂⃗ν∂⃗ )Aν

→ − Aα (η αβ □)Aβ − Aμ (∂ μ∂νAν ) → − Aμ (η μν□)Aν ,

(4.146)

where, in going from the second to third lines, I have used the fact that we will work in Lorenz gauge and hence ∂νAν = 0 and I relabeled α → μ and β → ν . With this simplification, we can write the Lorenz gauge (LG) QED Lagrangian as

LQED,LG = ψ (iγ μ∂μ − m)ψ + eψ γ μψAμ +

1 Aμ (η μν□)Aν . 2

(4.147)

We now transform to momentum space by taking ∂μ → −ipμ to obtain

˜ QED,LG = ψ (p − m)ψ + eψ γ μψAμ − 1 Aμ (η μνp 2 )Aν . L 2

(4.148)

To obtain the Feynman rules we take functional derivatives with respective to the fields. For example, the inverse electron propagator is obtained by evaluating

(SF (p ))−1 ≡

δ 2L δψ δψ

= p − m.

(4.149)

ψ = ψ =Aμ =0

Adding an iε to take care of the time-ordering and inverting, we obtain

SF (p ) =

1 . p − m + iε

(4.150)

This is precisely the same result we obtain previously (4.68) by considering the timeordered vacuum expectation value of Dirac fields. Note that in the above analysis I have discarded all of the trivial delta functions that enforce energy and momentum conservation. I will do this for all terms below for simplicity.

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Relativistic Quantum Field Theory, Volume 1

Next, let us consider the photon propagator

δ 2L δAμ δAν

(D Fμν(p ))−1 ≡

= −η μνp 2 .

(4.151)

ψ = ψ =Aμ =0

Adding an iε to take care of the time-ordering as before and inverting using the fact that (η μν )−1 = η μν , we obtain

D Fμν(p ) =

−η μν . p 2 + iε

(4.152)

Once again, this is precisely the Feynman gauge (ξ = 1) photon propagator obtained previously (4.112). Finally, we can obtain the electron–photon vertex by evaluating

e Γμ(p ) ≡

δ 3L δAμ δψ δψ

= e Γμ ,

(4.153)

ψ = ψ =Aμ =0

where the e on the left-hand side is a convention for defining the vertex function Γμ. The end result is that, at leading-order in the e , one has

Γ μ = γ μ.

(4.154)

The associated Feynman rule is that we insert a factor of ie Γμ for every electron– photon vertex. Having obtained the Lagrangian for QED we can now present the Feynman rules for QED and some example calculations. These are a bit more involved than our previous toy models. Hopefully, you will now appreciate why we did not jump straight to this theory from the beginning. Exercise 4.10 Show that

1 4π

i q·x

∫ d 3x e∣x∣

=

1 . ∣q∣2

This illustrates the relation between the Fourier transform of the Coulomb potential and the 00-component of the free photon propagator.

4.7 QED Feynman rules As with our two examples considered so far (scalar Yukawa and λϕ4 theories), one can calculate time-ordered expectation values of operator strings using an appropriately generalized Wick’s theorem. In the end, one finds that, similarly to the case of scalar fields, there is an efficient way to calculate interaction corrections to a given quantity, namely by using a set of Feynman rules that are formulated in momentum space. There are three different types of lines that correspond to the propagation of photons, electrons, and anti-electrons (positrons). Photons are indicated by a wavy 4-26

Relativistic Quantum Field Theory, Volume 1

line, electrons are indicated by a solid line with an arrow directed with the flow of time, and anti-electrons are indicated by a solid line with an arrow directed against the flow of time. 4.7.1 External lines • Incoming electron with spin s

• Outgoing electron with spin s

• Incoming anti-electron with spin s

• Outgoing anti-electron with spin s

• Incoming photon with polarization εμ

• Outgoing photon with polarization εμ

4.7.2 Internal lines • For each internal photon line, we assign a four-momentum k and two Lorentz indices μ and ν (incoming and outgoing). This maps to the photon Feynman propagator in momentum space

In all examples below, we will use Feynman gauge ξ = 1. • For each internal (anti-)electron line, we assign a four-momentum pi . This maps to (i times) the electron Feynman propagator in momentum space

• As usual, we integrate over all internal momenta.

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Relativistic Quantum Field Theory, Volume 1

4.7.3 Interaction vertex • Make sure that energy–momentum is conserved at every vertex. Previously, we had included explicit four dimensional delta functions, but, in practice these delta functions always reduce to overall energy–momentum conservation if applied correctly. • For each vertex we include

4.7.4 Special rules for fermions • Fermionic ordering Order matters for the fermions. The rule is to move along each fermionic line along the direction of the arrow and, as you proceed, left multiply by the spinors or Dirac matrices encountered along the line. If there are multiple fermions, follow each ‘line’ separately. The ordering for the different ‘lines’ themselves does not matter. • Identical fermions in the incoming or outgoing states If more than one diagram contributes, include a relative minus sign between diagrams that differ only in the exchange of two incoming (or outgoing) electrons or in the exchange of an incoming electron with an outgoing positron (or vice versa). For an example of this rule in practice, see the discussion surrounding figure 4.3. • Fermionic loops For every closed fermionic loop, multiply the result by –1.

4.8 QED Feynman rules—Examples The best way to learn how to apply the Feynman rules is to see some examples. 4.8.1 Electron–positron scattering We will begin by considering electron–positron scattering, e +e−→e +e−. • Scattering (t -channel) The first diagram we will consider is shown in figure 4.1. In this case there are two outgoing fermionic lines, the ‘top line’ and the ‘bottom line’. According to the Feynman rules we can evaluate each line separately

top line: bottom line:

u (p3 )(ieγ μ)u(p1 ), v (p2 )(ieγ ν )v(p4 ),

(4.155)

where I have suppressed the spin labels on the spinors. They should be implicit from here on. Next, we have to include the photon propagator (Feynman gauge) 4-28

Relativistic Quantum Field Theory, Volume 1

Figure 4.1. e +e−→e +e− scattering diagram.

[u (p3 )(ieγ μ)u(p1 )]

−iη μν [v (p2 )(ieγ ν )v(p4 )], q 2 + iε

(4.156)

where q = p1 − p3. To turn this into the invariant amplitude, we multiply by i

Mscattering = i [u (p3 )(ieγ μ)u(p1 )]

−iη μν [v (p2 )(ieγ ν )v(p4 )] (p1 − p3 )2 + iε

=−

e2 [u (p3 )γ μu(p1 )][v (p2 )γμv(p4 )] (p1 − p3 )2 + iε

=−

e2 [u (p3 )γ μu(p1 )][v (p2 )γμv(p4 )], t

(4.157)

where, t = (p1 − p3 )2 , and I have suppressed the iε for compactness. • Annihilation (s -channel) Our next example is electron–positron annihilation, which is depicted in figure 4.2. Following the Feynman rules, we obtain

Mannhilation = i [v (p2 )(ieγ μ)u(p1 )]

−iη μν [u (p3 )(ieγ ν )v(p4 )] q 2 + iε

e2 = − [v (p2 )γ μu(p1 )][u (p3 )γμv(p4 )], s

(4.158)

where s = (p1 + p2 )2 . • Total invariant amplitude For electron–positron scattering, both s - and t -channel scattering occur. Naively, we would simply add the the amplitudes from each process, however, this is a case where one of the ‘special rules for fermions applies’. If we exchange the incoming electron with the outgoing positron in figure 4.2, we can deform it into figure 4.1, which means that there should be a relative minus sign between the two amplitudes. I demonstrate how this can be visualized in figure 4.3. In the first step, we exchange an incoming electron with an outgoing positron; we then deform the diagram pulling all incoming (blue) lines to the left and all outgoing (red) lines to the right. As a result of the discussion above, we have

Mtotal = Mscattering − Mannhilation .

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(4.159)

Relativistic Quantum Field Theory, Volume 1

Figure 4.2. Drell–Yan scattering diagram.

Figure 4.3. Deformation of the annihilation graph into the scattering graph.

Figure 4.4. The two diagrams that contribute to Compton scattering, e−γ → e−γ .

Note that we could have equally well have chosen Mtotal = Mannhilation − Mscattering . The only thing that matters is that there is a relative sign between the two contributions. Since the two choices differ by an overall phase (e ±iπ ), this choice does not affect observables. 4.8.2 Compton scattering The diagram on the left in figure 4.4 gives

⎛ ⎞ i (ieγ μ)u(p1 )⎟εν(p2 )ε μ*(p3 ) M1 = i ⎜u (p4 )(ieγ ν ) q −m ⎝ ⎠ 1 ε *(p3 )u(p1 ) = e 2u (p4 ) ε (p2 ) q −m q +m ε *(p3 )u(p1 ) = e 2u (p4 ) ε (p2 ) 2 q − m2 e2 u (p4 ) ε (p2 )( p1 − p 3 + m) ε *(p3 )u(p1 ), = t − m2 where, as usual, t = (p1 − p3 )2 .

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(4.160)

Relativistic Quantum Field Theory, Volume 1

The diagram on the right in figure 4.4 gives

M2 =

e2 u (p4 ) ε (p3 )( p1 + p 2 + m) ε *(p2 )u(p1 ), s − m2

(4.161)

where s = (p1 + p2 )2 . As before, the total amplitude is given by the sum

MCompton = M1 + M2 .

(4.162)

4.9 The leading-order electron–positron scattering cross section Let us return to electron–positron scattering. By applying the Feynman rules to figures 4.1 and 4.2 we obtained the invariant amplitudes listed in equations (4.157) and (4.158), respectively. According to equation (4.159), the total invariant amplitude is

M=−

e2 [u (p3 )γ μu(p1 )] [v (p2 )γμv(p4 )]    t  μ =a

= bμ

2

+

e [v (p2 )γ μu(p1 )] [u (p3 )γμv(p4 )]    s  μ =c

(4.163)

= dμ

e2 e2 = − a μbμ + c μdμ. s t Note that the quantities a μ, b μ, c μ, and d μ are four-vectors of numbers (all of the spinor indices are contracted). For the cross section, we need

∣M∣2 =

e4 μ e4 μ ν *+ [ ][ ] [c dμ ][c νd ν ]* a b a b μ ν s2 t2 e4 − ([a μbμ ][c νd ν ]* +[a μbμ ]*[c νd ν ]). st

(4.164)

To simplify things, we will specialize to the case of scattering of spin-unpolarized incoming states. In this case we average over the incoming spin projections s1, s2 = 1, 2. If the experiment does not measure the cross section for each possible outgoing spin combination between the electron and positron, then we should, in addition, sum the invariant amplitude over all possible outgoing spin combinations

〈∣M∣2 〉spin−averaged =

1 ∑∑ ∣M∣2 , 4 s ,s s ,s

(4.165)

1 2 3 4

where the leading factor of 1/4 divides by the number of spin combinations possible in the incoming state (average over ↑↑, ↑↓, ↓↑, ↓↓).

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Relativistic Quantum Field Theory, Volume 1

To proceed, let us consider the first term in equation (4.164)

〈∣M∣2 〉spin−averaged,1 =

1 e4 ∑ [a μbμ ][a νbν ]* 4 t 2 s ,s ,s ,s 1 2 3 4

4

=

=

1e 4 t2

∑

[a μa ν*][bμbν*]

(4.166)

s1, s2, s3, s4

⎛ ⎞⎛ ⎞ e4 ⎜ 1 μ ν*⎟⎜ 1 *⎟ . a a b b ∑ ∑ μ ν ⎟⎜ 2 ⎟ t 2 ⎜⎝ 2 s , s ⎠⎝ s , s ⎠ 1 3

2 4

In going from the second line to the third line above, I have used the fact that a μa ν* only depends on particles 1 and 3 and bμbν* only depends on particles 2 and 4 to separate the sums. Consider the first term in parentheses above

1 1 a μa ν* = ∑[u (p3 )γ μu(p1 )][u (p3 )γ μu(p1 )]*. ∑ 2 s ,s 2 s ,s 1 3

(4.167)

1 3

To proceed let us introduce a more general form for the tensor above which will be useful when we encounter similar terms in the future

L≡

1 ∑ [u (pa )Γ1u(pb )][u (pa )Γ2u(pb )]* , 2 s ,s a

(4.168)

b

where Γ1 and Γ2 are two general 4 × 4 spinor matrices. One can show that

[u (pa )Γ2u(pb )]* = u (pb )Γ2u(pa ),

(4.169)

where Γ2 ≡ γ 0 Γ †2γ 0. This gives

L=

1 ∑ [u (pa )Γ1u(pb )][u (pb )Γ2u(pa )] 2 s ,s a

=

b

⎡ ⎤ 1 ⎢ ( ) ( ) ( ) u p Γ u p u p ∑ a 1 ⎢∑ b b ⎥⎥ Γ2u(pa ) 2 s ⎣ sb ⎦ a  

(4.170)

= pb + mb

=

1 ∑u (pa )Γ1( p b + mb)Γ2u(pa ). 2 s a

Introducing G ≡ Γ1( p b + mb )Γ2 , which is the matrix in spinor space, and making the spinor indices explicit, we can write

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Relativistic Quantum Field Theory, Volume 1

L=

1 ∑∑ui(pa )Gijuj(pa ) 2 s ij a

1 = ∑Gij∑uj (pa )ui (pa ) 2 ij s a

⎡ ⎤ 1 = ∑Gij ⎢∑u(pa )u (pa )⎥ ⎢⎣ ⎥⎦ 2 ij sa ji =

1 ∑Gij( p a + ma )ji 2 ij

=

1 ∑[G(p p a + ma )]ii , 2 i

(4.171)

where above i and j are spinor indices. In the last line above we see that the expression looks like two matrices that have been multiplied and summed over the diagonal elements. We can write the sum over the diagonal components as the trace ‘Tr’. Plugging back in the definition of G , we obtain finally

L=

1 Tr[Γ1( p b + mb)Γ2( p a + ma )]. 2

(4.172)

Using this method we can evaluate all spin-averaged terms of this form. Summarizing, we have

1 1 [u (pa )Γ1u(pb )][u (pa )Γ2u(pb )]* = Tr[Γ1( p b + mb)Γ2( p a + ma )], ∑ 2 2 s ,s

(4.173)

1 1 [u (pa )Γ1v(pb )][u (pa )Γ2v(pb )]* = Tr[Γ1( p b − mb)Γ2( p a + ma )], ∑ 2 2 s ,s

(4.174)

1 1 [v (pa )Γ1u(pb )][v (pa )Γ2u(pb )]* = Tr[Γ1( p b + mb)Γ2( p a − ma )], ∑ 2 2 s ,s

(4.175)

1 1 [v (pa )Γ1v(pb )][v (pa )Γ2v(pb )]* = Tr[Γ1( p b − mb)Γ2( p a − ma )]. ∑ 2 2 s ,s

(4.176)

a

a

a

a

b

b

b

b

Returning to the specific case at hand, we have Γ1 = γ μ and Γ2 = γ ν . Note that Γ2 = γ 0γ ν†γ 0 = γ ν , giving

L aμν =

1 1 a μa ν* = Tr[γ μ( p 3 + m)γ ν( p1 + m)]. ∑ 2 s ,s 2

(4.177)

2 4

We have reduced the problem of evaluating L μν to evaluating traces of some Dirac matrices. Expanding the result we obtain

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Relativistic Quantum Field Theory, Volume 1

1 1 L aμν = Tr[γ μ p 3γ ν p1] + mTr[γ μγ ν p1] 2 2 1 1 + mTr[γ μ p 3γ ν ] + m 2 Tr[γ μγ ν ]. 2 2

(4.178)

Using the fact that the trace of an odd number of γ matrices vanish, the second and third terms vanish. Using the trace identities proven in homework

Tr(γ μγ αγ νγ β ) = 4(η μαη νβ − η μνη αβ + η μβ η αν ),

(4.179)

Tr(γ μγ ν ) = 4η μν ,

(4.180)

and

we obtain

1 1 L aμν = Tr[γ μ p 3γ ν p1] + m 2 Tr[γ μγ ν ] 2 2 1 1 μ α ν β = p3α p1β Tr[γ γ γ γ ] + m 2 Tr[γ μγ ν ] 2 2 μα νβ μν αβ = 2p3α p1β (η η − η η + η μβ η αν ) + 2m 2η μν

(4.181)

= 2[p3μ p1ν + p1μ p3ν + (m 2 − p1 · p3 )η μν ]. We now turn to the evaluation of the second term in parentheses in equation (4.166)

L bμν =

1 1 b μb ν* = ∑ [v (p2 )γ μv(p4 )][v (p2 )γ νv(p4 )]* ∑ 2 s ,s 2 s ,s 2 4

2 4

1 = Tr[γ μ( p 4 − m)γ ν( p 2 − m)] 2 = 2[p4μ p2ν + p2μ p4ν + (m 2 − p2 · p4 )η μν ].

(4.182)

With this equation, (4.166) becomes e 4 μν L a Lb, μν t2 ⎤ e4 ⎡ = 4 2 ⎢p3μ p1ν + p1μ p3ν + (m 2 − p1 · p3 )η μν ⎥ ⎦ t ⎣

〈∣M∣2 〉spin−averaged,1 =

× [p4μ p2ν + p2μ p4ν + (m 2 − p2 · p4 )ημν ] = 4

e4 t2

(4.183)

[2(p1 · p2 )(p3 · p4 ) + 2(p2 · p3 )(p1 · p4 ) + 2(m 2 − p2 · p4 )(p1 · p3 )

+ 2(m 2 − p1 · p3 )(p2 · p4 ) + 4(m 2 − p1 · p3 )(m 2 − p2 · p4 )].

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Relativistic Quantum Field Theory, Volume 1

Using the relationships between the dot products of the four vectors and the Mandelstam variables (s, t, and u) found in appendix D

p1 · p2 = p3 · p4 =

1 (s − 2m 2 ), 2

(4.184)

p1 · p3 = p2 · p4 =

1 (2m 2 − t ), 2

(4.185)

p1 · p4 = p2 · p3 =

1 (2m 2 − u ), 2

(4.186)

one can write the result as

〈∣M∣2 〉spin−averaged,1 =

2e 4 [24m 4 − 8m 2(s + u ) + s 2 + u 2 ], t2

(4.187)

where I have used s + t + u = 4m2 to eliminate t . In the high-energy limit, one can take m → 0 to obtain

〈∣M∣2 〉spin−averaged,1 ≃ 2e 4

s2 + u2 . t2

(4.188)

Using similar methods, one can evaluate the second and third terms in equation (4.164), with the results in the high-energy limit being

〈∣M∣2 〉spin−averaged,2 ≃ 2e 4

t 2 + u2 , s2

(4.189)

u2 . ts

(4.190)

and

〈∣M∣2 〉spin−averaged,3 ≃ 4e 4

Putting all three pieces together, we obtain the high-energy limit of the spin-averaged amplitude squared

⎛ s2 + u2 2u 2 ⎞ t 2 + u2 〈∣M∣2 〉spin−averaged ≃ 2e 4⎜ + + ⎟. ⎝ t2 s2 ts ⎠

(4.191)

Just to remind you, the three terms above come from scattering, annihilation, and interference between the two processes. Finally I note that it is, of course, possible to write down the full result including the masses, but it is not very enlightening. Exercise 4.11 Show that equation (4.169) is correct. Exercise 4.12 Show that equations (4.188) and (4.189) are correct. You can assume that the electron is massless.

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Relativistic Quantum Field Theory, Volume 1

Exercise 4.13 What are the QED Feynman diagrams contributing to scattering of electrons with muons (e− + μ− → e− + μ−) at leading order in e? Write an expression for the incoming spin-averaged and outgoing spin-summed modulus-squared of the invariant amplitude for this process in terms of Mandelstam variables. Exercise 4.14 What are the QED Feynman diagrams contributing to annihilation of an electron and an anti-electron into two photons (e− + e + → γ + γ ) at leading order in e ? Write an expression for the incoming spin-averaged and outgoing polarizationsummed modulus-squared of the invariant amplitude for this process in terms of Mandelstam variables. For this problem, you can assume that the electron is massless.

References [1] Dirac P 1981 The Principles of Quantum Mechanics (Oxford: Clarendon) [2] Itzykson C 2005 Quantum Field Theory (New York: Dover) [3] Peskin M and Schroeder D 2015 An Introduction to Quantum Field Theory (Boca Raton, FL: CRC Press) [4] Ryder L H 1996 Quantum Field Theory (Cambridge: Cambridge University Press) [5] Kleinert H 2016 Particles and Quantum Fields (Singapore: World Scientific) [6] Srednicki M 2007 Quantum Field Theory (Cambridge: Cambridge University Press) [7] Stueckelberg E 1941 Helv. Phys. Acta 14 322–3 [8] Feynman R P 1948 Rev. Mod. Phys. 20 367–87 [9] Jackson J 1999 Classical Electrodynamics (New York: Wiley) [10] Dirac P 1950 Can. J. Math. 2 014 [11] Gupta S 1950 Proc. Phys. Soc. A 63 681–91 [12] Bleuler K 1950 Helv. Phys. Acta 23 567–86

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IOP Concise Physics

Relativistic Quantum Field Theory, Volume 1 Canonical formalism Michael Strickland

Chapter 5 Renormalization of quantum electrodynamics

Having covered the Feynman rules for QED and how to compute invariant amplitudes based on these, I would now like to turn to the topic of renormalization [1–12]. As we have seen in the case of a scalar field theory, when one considers certain corrections to quantities, like the propagator, one obtains answers that are divergent. The same thing happens in QED. For example, if we consider the leadingorder correction to the electron propagator as shown in figure 5.1, one finds that the diagram corresponding to SF(1)(p ) is divergent. The type of correction shown in figure 5.1 is called a self-energy correction because it represents the interaction of the electron with itself. In order to obtain a finite result, one must impose a regulator and then remove the regulator in a systematic manner. In practice, in order to regulate the integral, one introduces a finite parameter or set of parameters that render the necessary integral convergent (figure 5.2). Some commonly used methods are: • Introduce momentum-space cut-offs (cut-off regularization). • Change the number of dimensions for the integral (dimensional regularization) [13, 14]. • Introduce an artificial mass (Pauli–Villars regularization) [15]. • Zeta function regularization (related to dimensional regularization) [16, 17]. • Schwinger proper-time regularization [18, 19]. For gauge theories, one should not use cut-off regularization since this breaks gauge invariance, however, momentum cut-offs are perfectly suitable to use for a scalar field theory. The other methods listed can all be used in gauge theories and do not break gauge invariance. The method most commonly used these days is dimensional regularization, which we have already seen applied in a scalar field theory. Below, I will present the one-loop renormalization of QED. Before proceeding to QED, however, I need to introduce the concept of the renormalization group flow (RG flow), which will form the conceptual underpinning for what we will do in

doi:10.1088/2053-2571/ab30ccch5

5-1

ª Morgan & Claypool Publishers 2019

Relativistic Quantum Field Theory, Volume 1

Figure 5.1. Feynman diagram for the leading-order correction to the electron propagator. The first term is the bare (free) propagator and the second term is the (divergent) leading-order correction.

Figure 5.2. A sharp cut-off (notch band pass) in momentum-space for use in setting up renormalization group flow.

QED. For the discussion of RG flow, we will consider the simpler case of λϕ4 , in which case we can use momentum-space cut-offs as our regulators.

5.1 Renormalization group flow To begin this discussion, we assume that we are given a Lagrangian density that is specified for momenta restricted to a particular momentum scale, which I will call Λ . This momentum scale maps to a particular distance scale and we assume that we know the Lagrangian at that scale. Typically, this scale is associated with a microscopic (high momentum) scale where we think we know the fundamental interactions in our field theory. For the purposes of our discussion, we will assume that the Lagrangian density is that of λϕ4 theory

=

λ Z0 μ 1 ∂ ϕ∂μϕ − m 02ϕ 2 − 0 ϕ4 , 2 2 4!

(5.1)

where Z0 is called the bare wavefunction renormalization, m 0 is the bare mass, and λ 0 is the bare coupling. The word bare in this context means that the quantity does not include the effect of quantum fluctuations. In order to make this more formal, we introduce the concept of the smeared field, which is a band-passed version of the full quantum field

ϕk,Λ(x ) =

∫y

ρk,Λ (x − y )ϕ(y ) ,

5-2

(5.2)

Relativistic Quantum Field Theory, Volume 1

Figure 5.3. Depiction of the process of lowering the infrared cut-off k and raising the ultraviolet cut-off Λ .

where ϕ(y ) is the original field, ρk,Λ (x − y ) is a smearing function that selects only certain Fourier modes, ∫ = ∫ d 4x , and ϕk,Λ(x ) is the smeared field. The subscripts k x and Λ above indicate the momentum range present in the smeared field, with k being the infrared cut-off and Λ being the ultraviolet cut-off. After the application of the smearing function, the field only has Fourier-modes in the range k ≲ p ≲ Λ . In momentum-space, the action of this smearing is to modify the kinetic part of bare propagator for our scalar field

Δ(p ) =

ρ˜k,Λ (p ) 1 − ρk,Λ (p ) 1 → Δk,Λ(p ) = = , 2 2 p p p2

(5.3)

where ρk,Λ (p ) is the Fourier-transform of the spatial smearing function

ρk,Λ (p ) =

∫x

ρk,Λ (x )e ip·x ,

(5.4)

and ρ˜k,Λ (p ) ≡ 1 − ρk,Λ (p ). In practice, any momentum-space regulator that suppresses modes with momentum p ≲ k and p ≳ Λ suffice, but for simplicity it is easier to consider a sharp cut-off function that maps to a notch band pass filter for the fields1. This is illustrated in figures 5.2 and 5.3. As can be seen in the figure, the notch filter completely eliminates any mode with p ⩽ k or p ⩾ Λ . This is reflected in the kinetic part of the smeared effective propagator (sometimes called the blocked propagator) appearing on the right-hand side of equation (5.3). As we can see from this expression, modes that do not have k ⩽ p ⩽ Λ have a vanishing propagator and hence do not contribute to any physical process. For the bare Lagrangian presented in equation (1.1), we imagine that we take the k → Λ so that only modes with p = Λ can propagate in the system. This defines the bare Lagrangian, however, we are interested in the theory including the effects of all quantum fluctuations. In order to do this, we will take k → 0 and, eventually, Λ → ∞. However, we will not do this all at once, but instead we will include an infinitesimal shell by, for example, lowering k from Λ to Λ − dk . In this way, we will

1

One could, for example, consider instead a smooth regulator such as a plateau with Gaussian edges, etc.

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Relativistic Quantum Field Theory, Volume 1

include the effect of quantum fluctuations in the range Λ − dk ⩽ p ⩽ Λ . This will give us the dressed propagator, dressed vertex function, etc., which include the effect of quantum fluctuations in this (infinitesimal) momentum range. We can then consider the next infinitesimal shell and include modes in the range Λ − 2 dk ⩽ p ⩽ Λ and so on until we have reached k = 0. This is illustrated graphically in figure 5.3. In terms of Feynman diagrams, we construct them as usual, but we use the smeared propagator for all internal propagators. This is illustrated for the propagator (2-point function) and the four-vertex (4-point function) in figure 5.4. In this figure, I depict the inclusion of a thin shell of momentum Λ − dk < p < Λ in the loop integrals appearing in these two n-point functions. However, the story does not end there. We can generate a six-field vertex (6-point function) by combining three four-vertices as shown in figure 5.5. And, of course, we can generate a eight-field vertex by combining four fourvertices and so on. As a result, we see that in the process of integrating out quantum fluctuations we actually modify the effective Lagrangian density for our field theory with more and more (higher-order) interactions induced. Of course, in perturbation theory each of the higher induced n -point functions would be suppressed by powers of the coupling constant, but if we were not making a strict perturbative expansion, then we would need to take into account all of these vertices. So far, we have considered integrating out one infinitesimal shell. What happens when we integrate out the next shell so that we have included quantum fluctuations in the range Λ − 2 dk < p < Λ ? This is depicted in figure 5.6. As can be seen from figure 5.6, as we integrate out successive shells in momentum space, we include more and corrections due to quantum fluctuations in the n-point functions and the effective couplings now depend on the infrared cut-off k . All of this can be systematized by introducing the concept of the effective action (sometimes called the blocked action)

Figure 5.4. Integration of the a single thin shell in momentum-space for the loop corrections to the 2- and 4point functions in λϕ4 theory.

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Relativistic Quantum Field Theory, Volume 1

Figure 5.5. Generation of a six-point function from the quantum fluctuations.

Figure 5.6. Diagrams included in the integration of the second thin shell in momentum-space. Here I only show the 2- (top) and 4-point (bottom) functions. The momentum in all loops now ranges over Λ − 2 dk < p < Λ .

e iS˜k,Λ[Φ(x )] =

∫ Dϕ ∏ δ(ϕk,Λ(x) − Φ(x)) eiS[ϕ],

(5.5)

x

where the integral above is a path integral that integrates over all possible functional values of ϕ. On the left-hand side S˜k,Λ[Φ(x )] is the effective action, which describes fields with momentum in the range k ⩽ p ⩽ Λ and in the argument of the integrand on the right-hand side S[ϕ ] is the microscopic action that depends on the fundamental fields. The delta function in the integrand restricts the fields contributing to the path integral to be smeared (blocked) fields. The resulting effective action describes the physics of modes with a spatial scale (resolution) λIR = 2π /k . It includes the effects of quantum fluctuations from scales down to a scale λUV = 2π /Λ . Taking a partial derivative of equation (5.5) with respect to k gives us a partial differential equation, which should be satisfied by the effective action [8, 12]

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Relativistic Quantum Field Theory, Volume 1

−1⎤ ⎡ ∂S˜k,Λ δ 2S˜k,Λ ⎞ ⎥ 1 ⎢ 1 ⎛ ∂Δk,Λ ⎞⎛ ⎜k ⎟⎜1 + Δk,Λ 2 ⎟ = − Tr k ∂k δ Φ ⎠ ⎥⎦ 2 ⎣⎢ Δk,Λ ⎝ ∂k ⎠⎝ ⎡ ρ˜k,Λ δ 2S˜k,Λ ⎞−1⎤ 1 ⎢ −1 ⎛ ∂ρ˜k,Λ ⎞⎛ ⎜ = − Tr ρ˜ k, Λ ⎜k ⎟ 1 + 2 2 ⎟ ⎥, 2 ⎢⎣ p δ Φ ⎠ ⎥⎦ ⎝ ∂k ⎠⎝

(5.6)

where the trace (Tr) above refers to a normalized expectation value evaluated using the smeared path integral

∫ Dϕ ∏ δ(ϕk,Λ(x) − Φ(x)) A eiS[ϕ] Tr[A] =

x

∫ Dϕ ∏ δ(ϕk,Λ(x) − Φ(x)) eiS[ϕ]

.

(5.7)

x

The partial differential equation in equation (5.6) tells us how the effective action for our quantum field theory changes as we lower the infrared cutoff k which corresponds to including more and more quantum fluctuations.

5.2 Beta functions If one Taylor-expands the effective Lagrangian emerging from the RG flow equations and truncates the series at m-th order, one can write down RG flow equations for the various coupling constants2

∂λ1 = β1(k , λ1, λ2 , … , λm ) , ∂k 2 ∂λ k 2 22 = β2(k , λ1, λ2 , … , λm ) , ∂k ⋮ ∂ λ k 2 m2 = βm(k , λ1, λ2 , … , λm ) . ∂k k2

(5.8)

The functions β1, β2 , …, βm are called the beta functions for the theory and tell us how to evolve the coupling constants appearing in the Lagrangian as we follow the RG flow [4, 5]. The action of the RG flow on the coupling constants and resulting effective potential is visualized in figure 5.7. In the plot on the left in this figure, the axes in the diagram are the various coupling constants in the Lagrangian. The flow starts with the bare action defined at k = Λ and terminates with the infinite volume limit when k = 0. The three possible different trajectories shown correspond to three different choices for smearing function ρ˜k,Λ . Note, importantly, that if the RG flow is computed exactly, the final result at k = 0 is independent of the choice of the 2

Note that, if the bare Lagrangian only contains a single bare coupling in the effective potential, e.g. λ 4(Λ), then all other induced couplings can be expressed in terms of this coupling constant and, as a result, there is only one beta function that depends only on this coupling.

5-6

Relativistic Quantum Field Theory, Volume 1

Figure 5.7. The RG flow in coupling constant space is shown on the left. The flow starts with the bare action defined at k = Λ and terminates with the infinite volume limit k = 0 . The three different trajectories shown correspond to three different choices for regulator function ρ˜k,Λ . On the right I show how the effective potential might evolve under the RG flow using a sharp cut-off function.

Figure 5.8. Dependence of the effective Φ4 coupling λ 4(k ) on the infrared cut-off k .

smearing function. The plot on the right shows how the effective potential might evolve under the RG flow using a typical smearing function. If we select one particular coupling, say the coefficient of Φ 4 in the effective action we can plot its dependence on the infrared cut-off k . As an example, in figure 5.8 I show the dependence of the effective Φ 4 coupling λ 4(k ) on the infrared cut-off k . As we can see from this figure, the coupling constant depends on the scale. This is our first example of a running coupling constant, which is a direct consequence of quantum fluctuations. To close this section, I would like to mention that for condensed matter theories it suffices to keep the ultraviolet cut-off fixed since the microscopic theories do not themselves describe fundamental particles like quarks and gluons, but instead describe physics at atomic scales. For QED and QCD, however, which purport to

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Relativistic Quantum Field Theory, Volume 1

describe the fundamental constituents of all matter, one also has to take the ultraviolet cut-off Λ to infinity at the same time that the infrared cut-off is taken to zero. That being said, the nice thing about the RG flow equations is that they can also be used to determine how the theory ‘flows’ as the UV scale Λ is increased as well. Having covered the conceptual basis for the fact that coupling constants can depend on the scale due to quantum fluctuations, I will now turn to the technicalities involved in the renormalization of QED. Since we need to work in a manifestly gauge-invariant scheme, instead of using momentum cut-off regularization in the remainder of this chapter we will use dimensional regularization. The issue, however, is that it is less clear physically what is going on, hence the lead-up discussion in scalar field theory. Similarly to the preceding discussion, however, one finds that the fundamental quantities that need to be considered are the one-loop corrections in the field theory. In QED, however, there are three different types of divergent graphs that will give rise to wavefunction renormalization, mass renormalization, and vertex renormalization.

5.3 Renormalizable field theories The basic method for renormalization of a general QFT is as follows: • Evaluate loop corrections in the QFT. • Identify any divergent quantities and use your regulator of choice to make them finite. • Insert extra (eventually infinite) terms into the bare Lagrangian as ‘counterterms’ Δ ct . These will cancel the divergences as the regulator is removed. For example, the renormalized QED Lagrangian will be of the form

 QED = ψ (i D − m)ψ −

1 FμνF μν +  gf + Δ ct, 4

(5.9)

where gf is the gauge fixing term. • If only a finite number of counterterms are needed in Δ ct , which are in the same form as the original Lagrangian, then the theory is said to be renormalizable. • Otherwise, the theory is unrenormalizable. It is acceptable for effective theories which do not claim to describe the fundamental constituents of matter to be unrenormalizable (e.g. condensed matter theories), however, fundamental theories must be renormalizable, otherwise, one would require an infinite number of counterterms and hence an infinite number of terms in the Lagrangian. This would lead to zero predictability for the theory. Thankfully, QED and QCD are both renormalizable QFTs, as is the theory of weak-interactions which unified the electromagnetic and weak interactions. At this moment it time, in the general case, quantum gravity appears to be unrenormalizable. This is one of the outstanding problems in QFT/quantum gravity.

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Relativistic Quantum Field Theory, Volume 1

5.4 Dimensional regularization in QED Since we are going to use dimensional regularization with d = 4 − 2ε space-time dimensions, we first need to consider the dimensions of the fields and the electron charge e . By examining the QED Lagrangian, one finds

d−1 , 2 d [Aμ ] = − 1. 2 [ψ ] =

(5.10)

This means that we must multiply the electron charge e by a factor of μ2−d /2 , where μ is the renormalization scale in dimensional regularization. So, when applying the Feynman rules, one must take e → eμ2−d /2 = eμε at every vertex. Next, we need to write down the rules for γ -matrices in d dimensions

{γ μ, γ ν} = 2η μν ,

(5.11)

γ μγμ = d,

(5.12)

γμγ νγ μ = (2 − d )γν ,

(5.13)

Tr[odd # of γ matrices] = 0,

(5.14)

Tr[I ] = f (d ) ,

(5.15)

Tr[γ μγ ν ] = f (d ) η μν ,

(5.16)

Tr[γ μγ νγ λγ σ ] = f (d )(η μνη λσ − η μλη νσ − η μσ η νλ ) ,

(5.17)

where f (d ) is an arbitrary function that satisfies f (4) = 4.

5.5 One-loop renormalization of QED In order to prove the renormalizability of QED it suffices to consider the primitive one-loop divergences. In the second volume, we will prove that this is true. The summary is that, once the primitive one-loop divergences are obtained, then all graphs can be regulated using only this information. For the purposes of this course, we will take this statement as a given and proceed to obtain the one-loop running coupling in QED3. There are three primitive divergent one-loop graphs that need to be accounted for in order to properly renormalize the QED Lagrangian: the electron self-energy, the photon polarization, and electron–photon vertex.

3

It can be proven that this suffices. For the proof I refer you to refs. [20] and [21].

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Relativistic Quantum Field Theory, Volume 1

Figure 5.9. One-loop electron self-energy.

5.5.1 Electron self-energy We begin by considering the one-loop electron self-energy shown in figure 5.9. Note that, for this purpose we do not add the external legs of the graph since these are the same for all self-energy graphs at all orders, and can, therefore, be factored out. Using the QED Feynman rules in the Feynman gauge, this diagram becomes

−i Σ(p ) = (ie )2

∫

d 4k i −iη μν . γ γ μ ν (2π )4 p − k − m k 2

(5.18)

Simplifying and generalizing the integral to d = 4 − 2ε dimensions using the MS scheme, we obtain

⎛ e γ E ⎞ε Σ(p ) = − ie 2⎜ ⎟ μ4−d ⎝ 4π ⎠ ⎛ e γ E μ2 ⎞ε = − ie 2 ⎜ ⎟ ⎝ 4π ⎠ 

∫

∫

1 1 d dk γ γμ (2π )d μ p − k − m k 2 1 1 d dk γ γμ (2π )d μ p − k − m k 2

(5.19)

≡ μ 2ε

= − ie 2μ 2ε

∫

d dk 1 1 γ γμ 2 , d μ p − k −m k (2π )

where, in going from the second to third lines, I have introduced a compact notation for the overall scale factor, which will help to keep the formulas somewhat more manageable going forward. To proceed, we use p p = p2 to write this as

Σ(p ) = −ie 2μ 2ε

∫

μ d d k γμ( p − k + m)γ . (2π )d [(p − k )2 − m 2 ]k 2

(5.20)

At this point we will have to use a trick introduced by Schwinger/Feynman, which is to introduce a dummy integration variable in order to separate the two terms in the denominator4. The trick for the two terms of the form above follows from ⎛1 1 1 1⎞ 1 = − ⎜ ⎟= A1A2 A2 − A1 ⎝ A1 A2 ⎠ A2 − A1

4

∫A

A2

1

du = u2

This trick is commonly called Feynman parameter integration.

5-10

∫0

1

dz , (5.21) [A1z + A2 (1 − z )]2

Relativistic Quantum Field Theory, Volume 1

which is a specific case of the general formula

Γ(a1 +⋯+an) 1 a1 an = Γ(a1)⋯Γ(an) A1 ⋯An

∫0

1

∫0

dz1⋯

1

dzn

δ(1 − z1 −⋯−zn)z1a1−1⋯znan−1 . (5.22) (z1A1 +⋯+znAn )a1+⋯+an

Using the specific case for n = 2 with A1 = (p − k )2 − m2 and A2 = k 2 , we obtain

∫0

Σ(p ) = −ie 2μ 2ε

1

dz

γμ(p − k + m)γ μ d dk . (2π )d [(p − k )2 z − m 2z + k 2(1 − z )]2

∫

(5.23)

Changing variables to k′ ≡ k − pz , we obtain

Σ(p ) = −ie 2μ 2ε

∫0

1

dz

∫

μ d d k′ γμ( p − p z − k ′ + m)γ . (2π )d [k′2 − m 2z + p 2 z(1 − z )]2

(5.24)

The term that is linear in k′ integrates to zero since the denominator is even in k′, leaving

Σ(p ) = − ie 2μ 2ε

∫0

= − ie 2μ 2ε

∫0

×

∫

1

dz

∫

γμ(p − pz + m)γ μ d d k′ (2π )d [k′2 − m 2z + p 2 z(1 − z )]2

1

(5.25)

dz γμ[p(1 − z ) + m ]γ μ

d d k′ 1 . 2 2 d (2π ) [k′ − m z + p 2 z(1 − z )]2

To proceed, we must evaluate the integral appearing on the right-hand side

Id ≡

∫

d d k′ 1 . d 2 2 (2π ) [k′ − m z + p 2 z(1 − z )]2

(5.26)

Following the discussion in chapter 3.6.4, after Wick rotation, one finds

Id = i

∫

d d kE 1 (2π )d ⎡ k 2 + m 2z − p 2 z(1 − z )⎤2 ⎣ E ⎦

S =i dd (2π )

∫0

k Ed −1

∞

dkE

⎡ k 2 + m 2z − p 2 z(1 − z )⎤2 ⎣ E ⎦

(5.27) ,

where Sd = 2π d /2 /Γ(d /2) is the surface area of a d -dimensional unit sphere. Using

∫0

∞

dx

Γ(b − d /2)Γ(d /2) d /2−b x d −1 a , = 2 b (x + a ) 2Γ ( b )

(5.28)

we obtain

Id = i

Sd Γ(2 − d /2)Γ(d /2) 2 [m z − p 2 z(1 − z )]d /2−2 . (2π )d 2

5-11

(5.29)

Relativistic Quantum Field Theory, Volume 1

Inserting this result back into the expression for the electron self-energy (5.25) and simplifying, we obtain

Γ(2 − d /2) 1 dz γμ[ p (1 − z ) + m ]γ μ[m 2z − p 2 z(1 − z )]d /2−2 0 (4π )d /2 1 (5.30) Γ( ε ) = e 2μ 2ε dz [( −2 + 2ε ) p (1 − z ) + (4 − 2ε )m ] (4π )2−ε 0 × [m 2z − p 2 z(1 − z )]−ε ,

∫

Σ( p ) = e 2 μ 2 ε

∫

where we have used equations (5.12) and (5.13) to simplify the spinor contractions. For the purposes of renormalization, we need to only consider the leading-order term by Taylor-expanding around ε = 0. For this purpose, the following expansions are useful

(4πμ 2 )ε Γ(ε ) =

⎛ 1 π2 ⎞ + 2 log(μ) + ε⎜2 log 2(μ) + ⎟ + O(ε 2 ), ⎝ ε 12 ⎠

(5.31)

a −ε = 1 − log(a )ε + O(ε 2 ).

(5.32)

and

Inserting these expansions and keeping only the divergent terms, we obtain 1 e2 dz [ −2 p (1 − z ) + 4m ] + O(ε 0) 2 16π ε 0 e2 = ( − p + 4m) + O(ε 0). 16π 2ε

Σ( p ) =

∫

(5.33)

I note that one can see that the two terms in parentheses are in the same form as the terms that appear in the inverse electron propagator in the bare Lagrangian. This means that the counterterm necessary can be included through a redefinition of the fermionic kinetic energy coefficient and mass. 5.5.2 Photon polarization At one-loop level the photon propagator receives a correction from the production of virtual electron–positron pairs as shown in the diagram shown in figure 5.10. Note that the fermionic loop in this graph is summed over the spin states, which results in the overall trace and, since there is a closed fermionic loop, one will have to multiply by an overall factor of −1. Using the Feynman rules, we obtain

i Πμν(k ) = −(ie )2 μ 2ε

∫

⎡ ⎤ d dp i i μ ν ⎢ ⎥. Tr γ γ (2π )d ⎣ p − m p − k − m ⎦

5-12

(5.34)

Relativistic Quantum Field Theory, Volume 1

Figure 5.10. One-loop photon polarization. I show the photon legs in grey because they are not really part of this diagram.

Solving for Πμν(k ) and simplifying a bit, we obtain

Πμν(k ) = ie 2μ 2ε

d d p Tr[γ μ( p + m)γ ν( p − k + m)] . (2π )d (p 2 − m 2 )((p − k )2 − m 2 )

∫

(5.35)

Using the two-factor Feynman parameter integral formula (5.21) and then changing variables to p′ = p − kz , we obtain Πμν(k ) = ie 2μ 2ε

∫0

1

dz

∫

d d p′ Tr[γ μ( p ′ + k z + m )γ ν ( p ′ − k (1 − z ) + m )] . (5.36) (2π )d [ p ′2 − m 2 + k 2z (1 − z )]2

Focusing on the numerator above, we see that all terms that are odd in p will integrate to zero since the denominator is an even function of p′. As a result, the only terms necessary to evaluate in the numerator of the integrand are

[p′α p′ β − kαkβz(1 − z )]Tr[γ μγ αγ νγ β ] + m 2Tr[γ μγ ν ].

(5.37)

Using equations (5.16) and (5.17), this becomes

f (d ){2p′μ p′ν − 2z(1 − z )(k μk ν − k 2η μν ) − η μν[p′2 − m 2 − k 2z(1 − z )]},

(5.38)

which gives

Πμν(k ) = ie 2f (d )μ 2ε

∫0

1

dz

∫

d dp ⎧ 2p μ p ν ⎨ (2π )d ⎩ [p 2 − m 2 + k 2z(1 − z )]2

⎫ 2z(1 − z )(k μk ν − η μνk 2 ) η μν ⎬, − − 2 [p − m 2 + k 2z(1 − z )]2 [p 2 − m 2 + k 2z(1 − z )] ⎭

(5.39)

where I have relabeled p′ → p. The first and last terms cancel upon integration5. The middle term remains and gives

Πμν(k ) = − 2ie 2(k μk ν − η μνk 2 )f (d )μ 2ε 1 d dp 1 dz z(1 − z ) . 0 (2π )d [p 2 − m 2 + k 2z(1 − z )]2

∫

5

∫

This is not obvious. If you are not convinced, I encourage you to check it for yourself.

5-13

(5.40)

Relativistic Quantum Field Theory, Volume 1

Figure 5.11. One-loop vertex correction. As before, I show the legs in grey because they are not really part of this diagram.

The momentum integral appearing on the right-hand side is of the form encountered previously (5.26). Using that example, and after a quick bit of algebra, we obtain μ 2ε Γ(2 − d /2) 1 dz z(1 − z )[m 2 − k 2z(1 − z )]d /2−2 0 (4π )d /2 μ 2ε Γ(ε ) 1 = 2e 2(k μk ν − η μν k 2 )f (4 − 2ε ) dz z(1 − z )[m 2 − k 2z(1 − z )]−ε (4π ) 2−ε 0 (5.41) 1 e2 μ ν μν 2 0 = 2 (k k − η k ) dz z(1 − z ) + O(ε ) 0 2π ε e2 = (k μk ν − η μν k 2 ) + O(ε 0). 12π 2ε

∫

Πμν(k ) = 2e 2(k μk ν − η μν k 2 )f (d )

∫

∫

Once again, I note that the second divergent term above has the same form as the Feynman-gauge photon propagator and, as a result, can be absorbed via rescaling as I will explicitly demonstrate below. The first term will result in a modification of the gauge fixing term in the Lagrangian which, while interesting on its own, we will not delve into in this volume. 5.5.3 Electron–photon vertex We now consider the one-loop modification of the QED vertex function, which is shown in figure 5.11. Using the Feynman rules, one obtains

ieμ ε Λμ(p , q − p , q ) = (ieμ ε )3

∫

d dk i i −iη λν . (5.42) γ γ γ ν μ λ (2π )d q − k − m p − k − m k2

Simplifying and solving for Λμ(p, q − p, q ), one obtains

Λμ(p , q − p , q ) = − ie 2μ 2ε = − ie 2μ 2ε

∫

1 1 1 d dk γ γ γν (2π )d ν q − k − m μ p − k − m k 2

∫

ν d d k γν( q − k + m)γμ( p − k + m)γ . (2π )d k 2[(q − k )2 − m 2 ][(p − k )2 − m 2 ]

5-14

(5.43)

Relativistic Quantum Field Theory, Volume 1

Using the Feynman parameter integral based on equation (5.22), one has

1 =2 A1A2 A3 =2

∫0 ∫0

1

dx 1

dx

∫0 ∫0

1

1

δ(1 − x − y − z ) [xA1 + yA2 + zA3]3 1 . dy [xA1 + yA2 + (1 − x − y )A3]3

dy 1 −x

∫0

dz

(5.44)

This gives

Λμ(p , q − p , q ) = − ×

2ie 2μ 2ε (2π )d

∫0

1

dx

∫0

1 −x

dy

γν( q − k + m)γμ( p − k + m)γ ν

(5.45)

∫ d k [k 2 − m2(x + y ) − 2k(px + qy ) + p2 x + q 2y ]3 . d

Changing variables to k′ = k − px − qy and then relabeling k′ → k gives 1 −x 2ie 2μ 2ε 1 dx dy 0 (2π )d 0 γν[ q (1 − y ) − p x − k + m ]γμ[ p (1 − x ) − q y − k + m ]γ ν

∫

Λμ(p , q − p , q ) = −

×

∫

(5.46)

∫ d dk [k 2 − m2(x + y ) + p2 x(1 − x) + q 2y(1 − y ) − 2(p · q)xy ]3 .

The resulting integral over the momentum above contains a divergent and convergent contributions. The divergent contributions correspond to the terms in the numerator that are proportional to k 2 . The divergent piece is related to renormalization and the convergent piece is related to the anomalous magnetic moment. We can split them up as (2) Λμ(p , q − p , q ) = Λ (1) μ (p , q − p , q ) + Λ μ (p , q − p , q ) ,

(5.47)

(2) where Λ (1) μ is the divergent part and Λ μ is the convergent part. Here we will focus on the divergent part

Λ(1) μ (p , q − p , q ) = − ×

2ie 2μ 2ε (2π )d

∫0

1

dx

∫0

1 −x

dy

γ k γ k γν

∫ d dk [k 2 − m2(x + y ) + p2 x(1 ν− xμ) + q 2y(1 − y ) − 2(p · q)xy ]3

2ie 2μ 2ε =− (2π )d

∫0

1

dx

∫0

1 −x

dy γνγαγμγβγ

ν α β

×

∫ d dk [k 2 − m2(x + y ) + p2 x(1 −kxk) + q 2y(1 − y ) − 2(p · q)xy ]3 .

5-15

(5.48)

Relativistic Quantum Field Theory, Volume 1

The resulting momentum-space integral vanishes unless α = β by symmetry, leaving

Λ (1) μ (p , q − p , q ) = − ×

∫

2ie 2μ 2ε (2π )d

∫0

1

dx

∫0

1 −x

dy γνγαγμγ αγ ν

k2 . d k 2 [k − m 2(x + y ) + p 2 x(1 − x ) + q 2y(1 − y ) − 2(p · q )xy ]3

(5.49)

d

The product of Dirac matrices remaining can be evaluated, with the result being

γνγαγμγ αγ ν = (2 − d )2 γμ = (2 − 2ε )2 γμ.

(5.50)

The remaining momentum–space integral can be evaluated using 2

∫ d dk (k 2 k+ a)b

=

idπ d /2 Γ(b − 1 − d /2) , 2Γ(b) a b−1−d /2

(5.51)

which gives Λ (1) μ (p , q − p , q ) =

e 2μ 2ε Γ(ε ) (2 − 2ε )2 γμ 2(4π )2−ε ×

∫0

1

dx

∫0

1 −x

dy

1 . [m 2(x + y ) − p 2 x (1 − x ) − q 2y(1 − y ) + 2(p · q )xy ]ε

(5.52)

In the limit ε → 0, one obtains

Λ (1) μ (p , q − p , q ) =

e2 γμ + O(ε 0). 16π 2ε

(5.53)

And once again we see that, since this is proportional to γμ, it has a form that can be absorbed into the bare coupling by a rescaling. 5.5.4 Summary Summarizing, we have isolated all of the primitive one-loop divergences in QED. We found

Σ( p ) = Πμν(k ) =

e2 ( −p + 4m) + O(ε 0), 16π 2ε e2 (k μk ν − η μνk 2 ) + O(ε 0), 12π 2ε

Λ (1) μ (p , q − p , q ) =

e2 γμ + O(ε 0). 16π 2ε

(5.54) (5.55) (5.56)

5.6 Schwinger–Dyson equations So what is the connection of these quantities to the propagators, etc.? Well, unfortunately, at this point, I have to present something that we have not proven 5-16

Relativistic Quantum Field Theory, Volume 1

thus far (we will do this in the next volume). By analyzing the structure of the perturbative series one can show that the full propagators and vertex functions can be expressed into terms of the self-energy Σ , polarization function Πμν , and vertex function Λμ as follows

S˜F (p )−1 = SF (p )−1 − Σ(p ) ,

(5.57)

μν D˜ F (k )−1 = D Fμν(k )−1 − Πμν(k ),

(5.58)

Γ˜μ(p , q − p , q ) = γμ + Λμ(p , q − p , q ),

(5.59)

μν where S˜F (p ), D˜ F (k ), and Γ˜μ(p, q − p, q ) are the full electron propagator, the full photon propagator, and the full electron–photon vertex function. The propagators appearing on the right-hand side are the bare (free) propagators. The self-energy, polarization function, and vertex function appearing on the right-hand side contain all one-particle irreducible graphs6. These equations are called the Schwinger–Dyson equations. Note that, if we imagine that Σ is a small quantity, we can expand equation (5.57) in a Taylor series. To see how this emerges, we can write

S˜F (p )−1 = SF (p )−1 − Σ(p ), S˜F (p )−1SF (p ) = 1 − Σ(p )SF (p ), ⎛ ⎞ (5.60) 1 S˜F (p ) = SF (p )⎜ ⎟, ⎝ 1 − Σ(p )SF (p ) ⎠ ˜ SF (p ) = SF (p ) + SF (p )Σ(p )SF (p ) + SF (p )Σ(p )SF (p )Σ(p )SF (p ) + ⋯. This is depicted graphically in figures 5.12 and 5.13. What we see from this is that equation (5.57) sums an infinite set of Feynman graphs. Similar expansions can be made from equations (5.58) and (5.59).

5.7 Photon wavefunction renormalization Let us look at the photon propagator first. If we use the one-loop polarization function (5.55) in equation (5.58), we obtain

e2 (k μk ν − η μνk 2 ) + O(ε 0) 12π 2ε e 2 μν 2 e2 = D Fμν(k )−1 + η − k k μk ν + O(ε 0). 12π 2ε 12π 2ε

μν D˜ F (k )−1 = D Fμν(k )−1 −

(5.61)

Recalling that the inverse photon propagator is the two-point function in the Lagrangian, we see that this implies that there will be an infinite correction to our 6 One-particle reducible graphs can be split into two subgraphs by cutting only one internal line. Conversely, one-particle irreducible graphs cannot be split into two subgraphs by cutting only one internal line.

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Relativistic Quantum Field Theory, Volume 1

Figure 5.12. Graphical depiction of equations (5.57)–(5.59). Double lines indicate full propagators and the shaded circle in the last line indicates the full vertex function.

Figure 5.13. Graphical depiction of the power series expansion of equation (5.57).

Lagrangian, but we can now determine the counterterm necessary to remove this divergence. In the Feynman gauge, the electromagnetic (EM) part of the QED Lagrangian is

1 1  EM = − F μνFμν − (∂μAμ )2 . 4 2

(5.62)

If the theory is to be renormalizable, the EM counterterm contribution should have the same form

 EM,ct = −

A μν B F Fμν − (∂μAμ )2 . 4 2

(5.63)

The sum of these two terms is

 EM +  EM,ct = −

1 (1 + A) F μνFμν+ gauge fixing term 4  ≡ Z3

(5.64)

Z = − 3 F μνFμν + gauge fixing term, 4 where we have absorbed the counterterm into an overall factor multiplying the first term7. From equation (5.61) we see that the second term modifies the photon propagator (the third term modifies the gauge-fixing term). In order to cancel this term, we need

7 Note that we will not concern ourselves with the gauge fixing term here. Just be warned that there some subtleties hiding here.

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Relativistic Quantum Field Theory, Volume 1

A=−

e2 , 12π 2ε

(5.65)

and, hence,

e2 . 12π 2ε

Z3 = 1 −

(5.66)

5.8 Electron wavefunction and mass renormalization If we use the one-loop electron self-energy (5.54) in equation (5.57), we obtain

e2 ( − p + 4m) + O(ε 0) 16π 2ε ⎛ ⎛ e2 ⎞ e2 ⎞ = p ⎜1 + − + m 1 ⎟ ⎜ ⎟ + O(ε 0). ⎝ ⎝ 16π 2ε ⎠ 4π 2ε ⎠

S˜F (p )−1 = SF (p )−1 −

(5.67)

Similarly to the purely electromagnetic contribution, we can separate out the purely fermionic contribution from the QED Lagrangian

 Dirac = ψ ( p − m)ψ ,

(5.68)

and we require that the electron contribution counterterm has the same form, but with some adjustable coefficients

 Dirac,ct = ψ (C p − Dm)ψ .

(5.69)

Adding these together, we obtain

 Dirac +  Dirac,ct =  (1 + C (1 + D ) ψ pψ −  ) mψ ψ ≡ Z2

≡ Zm

(5.70)

= Z2ψ pψ −mZmψ ψ . Compared with equation (5.67), we see that

C=−

e2 , 16π 2ε

(5.71)

e2 , 4π 2ε

(5.72)

D=− and, hence,

Z2 = 1 −

e2 , 16π 2ε

(5.73)

Zm = 1 −

e2 . 4π 2ε

(5.74)

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Relativistic Quantum Field Theory, Volume 1

Importantly, we notice that there is a wavefunction renormalization Z2 and a mass renormalization Zm which are different. The fact that there is a mass renormalization implies that the physical mass is not necessarily the same as the mass that appears in the bare Lagrangian.

5.9 Vertex renormalization Finally, let us turn to the vertex renormalization. Using equation (5.56) in equation (5.59), we obtain

⎛ e2 ⎞ Γ˜μ(p , q − p , q ) = γμ⎜1 + ⎟ + O(ε 0). ⎝ 16π 2ε ⎠

(5.75)

Separating out the electron–photon interaction from the QED Lagrangian, we have

 int = eψ γ μψAμ .

(5.76)

Once again, the counterterm should have the same form, but with an adjustable coefficient

 int,ct = eFψ γ μψAμ .

(5.77)

Combining these, we obtain

 int +  int,ct = e  (1 + F) ψ γ μψAμ ≡ Z1

(5.78)

= eZ1ψ γ μψAμ . Compared with equation (5.76) we find that F has to be

F=−

e2 , 16π 2ε

(5.79)

and, hence,

Z1 = 1 −

e2 . 16π 2ε

(5.80)

5.10 The renormalized QED Lagrangian Combining the three pieces determined above, we obtain the full bare (B) Lagrangian for QED  QED,B = Z2ψ pψ − mZmψψ + eZ1ψγ μψAμ −

Z3 μν F Fμν + gauge fixing term. (5.81) 4

In order to absorb these infinities in the bare Lagrangian we introduce rescaled fields

ψB =

Z2 ψ ,

5-20

(5.82)

Relativistic Quantum Field Theory, Volume 1

ABμ =

Z3 Aμ ,

(5.83)

and write the Lagrangian in terms of these

⎛ eμ ε Z ⎞ ⎛Z m⎞ 1 ⎟ ψBγ μψBA μB L QED,B = ψBpψB − ⎜ m ⎟ ψ ψ + ⎜ ⎝ Z2 ⎠ Z Z ⎝ ⎠  23 ≡ mB

−

≡ eB

(5.84)

1 μν B FB F μν + gauge fixing term 4

= ψBpψB − mB ψ ψ + eBψBγ μψBA μB −

1 μν B FB F μν + gauge fixing term, 4

where we have introduced the scaled coupling and mass

eB ≡

eμ ε Z1 = eμ ε Z3−1/2, Z2 Z3

(5.85)

Zmm . Z2

(5.86)

mB ≡

The fact that we were able to use these multiplicative rescalings and the Lagrangian ended up in the same form as the original QED Lagrangian, proves the renormalizability of QED! Note that, it may seem strange that we can rescale the fields in this way but, as we will see in the Volume 2, all physical quantities are expectation values of quantum fields defined in terms of a normalized path integral and the field rescaling cancels exactly between the numerator and denominator (normalization term) and, as a result, physical quantities independent of this field rescaling.

5.11 The one-loop QED running coupling Using equation (5.86), we can determine the scale dependence of the physical coupling e . We start from −1/2 ⎛ ⎛ e2 ⎞ e2 ⎞ ε eB = eμ ε Z3−1/2 = eμ ε ⎜1 − = e μ 1 + ⎟ ⎜ ⎟ + O(e 4). ⎝ ⎝ 12π 2ε ⎠ 24π 2ε ⎠

(5.87)

To proceed, we express this in compact form as

⎛ a⎞ eB = μ ε ⎜e + 1 ⎟ + O(e 4), ⎝ ε⎠

(5.88)

with a1 ≡ e 3 /(24π 2 ). Taking a derivative with respect to the scale μ on the left and right and using the fact that the bare coupling is independent of the scale (∂ μ eB = 0), we obtain

⎛ ∂e ⎛ a⎞ 1 ∂a1 ∂e ⎞ + 0 = εμ ε−1⎜e + 1 ⎟ + μ ε ⎜ ⎟ + O(e 4). ⎝ ε⎠ ε ∂e ∂μ ⎠ ⎝ ∂μ

5-21

(5.89)

Relativistic Quantum Field Theory, Volume 1

Simplifying a bit, we obtain

μ

⎛ 1 ∂a1 ⎞ a⎞ ∂e ⎛ ⎜1 + ⎟ = −ε⎜e + 1 ⎟ + O(e 4). ⎝ ⎝ ⎠ ∂μ ε ∂e ε⎠

(5.90)

∂e

Solving for μ ∂μ to leading-order in e , we obtain

μ

⎛ ⎛ 1 ∂a1 ⎞ ∂a ⎞ a ⎞⎛ ∂e ⎟ = − ⎜a1 − e 1 ⎟ = −ε⎜e + 1 ⎟⎜1 − ⎝ ⎠ ⎝ ∂e⎠ ∂μ ε ⎝ ε ∂e ⎠  

+ O(e 4).

(5.91)

e 3 − 3e 3 =− e 3 24π 2 24π 2 12π 2

From this, we obtain our final expression for the one-loop running of the QED coupling constant

μ

e3 ∂e = + O(e 4). ∂μ 12π 2

(5.92)

I remind you at this point that μ = e γ μ /(4π ) and, as a result, this reduces to

μ

e3 ∂e = + O(e 4) , ∂μ 12π 2

(5.93)

which shows that the running of the coupling constant is independent of the renormalization scheme, which is defined by the overall constant scale in front of μ . This differential equation tells us how the coupling constant changes with the scale. It can be solved analytically subject to a boundary condition at a reference scale μ0. The result is

e 2 (μ ) =

e 2 (μ 0 ) . e 2 (μ 0 ) μ2 log 2 1− 12π 2 μ0

(5.94)

As we can see from this expression, the QED coupling constant is a monotonically increasing function of μ8. Note that one can express this in terms of the fine structure constant α ≡ e 2 /(4π )

μ2

∂α 1 2 = α 2 ∂μ 3π

↔

α( μ ) =

⏟ ≡ β1(α )

α( μ 0 ) , α( μ 0 ) μ2 log 2 1− 3π μ0

(5.95)

where I have introduced the one-loop beta function β1(α ) for QED above. Note that one can go beyond one-loop order in the running by considering higher-order 8 Actually, there is the possibility of a pole at large μ, which is called the Landau pole [22]. In practice, this energy scale is higher than anything we would ever possibly experience, so it is not a practical concern. In addition, numerical studies have shown that the Landau pole lies in a region of parameter space which is made inaccessible by spontaneous chiral symmetry breaking [23].

5-22

Relativistic Quantum Field Theory, Volume 1

corrections to the photon polarization, electron self-energy, and electron–photon vertex function. In general, one can express the n-loop running coupling in the form

μ2

∂α = ∂μ2

N

∑ bnα n+2 . n = 0  

(5.96)

≡ βN (α )

The first few coefficients in the QED beta function are

1 , 3π 1 b1 = 2 , 4π 31 , b2 = − 288π 3 b0 =

(5.97)

which map to the one-, two-, and three-loop running of the coupling constant. The QED running coupling is currently known up to five loops [24]. To attach some numbers to this, let us assume that α = 1/137.036 at μ0 = 1 eV = 10−90 GeV , and evaluate the one-loop running. Figure 5.14 shows the result. As we can see from this figure, the fine structure constant has a small (but important!) scale variation. At modern high-energy colliders it is crucial to take into the QED running coupling when computing electromagnetic processes. Note that the correction to α (μ = 1 GeV ) in going from the one- to two-loops is 6.0661 × 10−7 and the correction in going from two- to three-loops at the same scale is −6.20216 × 10−10 . This is consistent with the rapid convergence of perturbation theory applied to QED. As a result, unless you are doing an ultra-high-precision QED experiment, it suffices to use the one-loop running. When we come to QCD,

Figure 5.14. One-loop running coupling α(μ) multiplied by 137.036 as a function of the renormalization scale μ in GeV.

5-23

Relativistic Quantum Field Theory, Volume 1

we will see that it is imperative to include the higher loop corrections to the beta function when comparing with data.

References [1] [2] [3] [4] [5] [6] [7] [8] [9] [10] [11] [12] [13] [14] [15] [16] [17] [18] [19] [20] [21]

Stueckelberg E C G and Petermann A 1953 Helv. Phys. Acta 26 499–520 Gell-Mann M and Low F E 1954 Phys. Rev. 95 1300–12 Kadanoff L P 1966 Phys. Phys. Fizika 2 263–72 Callan C G 1970 Phys. Rev. D 2 1541–7 Symanzik K 1970 Commun. Math. Phys. 18 227–46 Wilson K G 1971 Phys. Rev. B 4 3174–83 Wilson K G and Fisher M E 1972 Phys. Rev. Lett. 28 240–3 Wegner F J and Houghton A 1973 Phys. Rev. A8 401–12 Wilson K G and Kogut J B 1974 Phys. Rept. 12 75–199 Wilson K G 1983 Rev. Mod. Phys. 55 583–600 Polchinski J 1984 Nucl. Phys. B231 269–95 Wetterich C 1993 Phys. Lett. B301 90–4 Bollini C G and Giambiagi J J 1972 Il Nuovo Cimento B (1971–1996) 12 20–6 ‘t; Hooft G and Veltman M 1972 Nucl. Phys. B 44 189–213 Pauli W and Villars F 1949 Rev. Mod. Phys. 21 434–44 Dowker J S and Critchley R 1976 Phys. Rev. D13 3224 Hawking S W 1977 Comm. Math. Phys. 55 133–48 Schwinger J S 1951 Phys. Rev. 82 664–79 Liao S B 1996 Phys. Rev. D53 2020–36 Ward J C 1950 Phys. Rev. 78 182 Jauch J M and Rohrlich F 2011 The Theory of Photons and Electrons: The Relativistic Quantum Field Theory of Charged Particles with Spin One-halfTheoretical and Mathematical Physics (Berlin: Springer) [22] Pauli W, Rosenfeld L and Weisskopf V 1955 Niels Bohr and the Development of Physics (London: Pergamon) [23] Göckeler M, Horsley R, Linke V, Rakow P, Schierholz G and Stüben H 1998 Phys. Rev. Lett. 80 4119–22 [24] Baikov P A, Chetyrkin K G, Kuhn J H and Rittinger J 2012 JHEP 07 017

5-24

IOP Concise Physics

Relativistic Quantum Field Theory, Volume 1 Canonical formalism Michael Strickland

Appendix A Classical mechanics review In this appendix, I present a brief review of classical mechanics and explain how to extend the usual few-body formalism to many bodies through a specific example of a linear chain of coupled harmonic oscillators.

A.1 Lagrangian mechanics Recall that in the Lagrangian formalism for non-relativistic classical mechanics, one deals with functions of the generalized coordinates

L(q , q )̇ = T − V =

1 2 mq ̇ − V (q ), 2

(A.1)

where q is a function of t only. As can be see from this, the Lagrangian is a function of q and q .̇ The action is defined as an integral of the Lagrangian

S [q ] =

∫t

t2

dt L(q , q ). ̇

(A.2)

1

Above, the notation with square brackets, S [q ], indicates that the action is a functional of q(t ). The Euler–Lagrange equations of motion from the principle of least action

δS =

⎛

⎞

∫ dt⎜⎝ ∂∂Lq δq + ∂∂Lq̇ δq⎟⎠̇ = 0.

(A.3)

After adding and subtracting a term to create a total derivative that vanishes upon integration, one finds the following equation

doi:10.1088/2053-2571/ab30ccch6

A-1

ª Morgan & Claypool Publishers 2019

Relativistic Quantum Field Theory, Volume 1

d ⎛ ∂L ⎞ ∂L = 0, ⎜ ⎟− dt ⎝ ∂q ̇ ⎠ ∂q

(A.4)

where we have used the fact that δq is arbitrary to require that the integrand of the action variation vanishes.

A.2 Hamiltonian mechanics In classical mechanics one also encounters the Hamiltonian formalism. The Hamiltonian is obtained via a Legendre transform of the Lagrangian

H = pq ̇ − L(q , q ), ̇

(A.5)

where we have introduced the canonical momentum

p=

∂L . ∂q ̇

(A.6)

One can use the last relation to eliminate q ̇ in favor of p. Also note that we can rewrite the Euler–Lagrange equations of motion compactly as

ṗ =

∂L . ∂q

(A.7)

By evaluating the variation of the Hamiltonian, we can determine the Hamilton equations of motion

⎛ ∂L ∂L ⎞ δH = qδ̇ p + pδq ̇ − ⎜ δq + δq⎟̇ ∂q ̇ ⎠ ⎝ ∂q = qδ̇ p + pδq ̇ − (p ̇ δq + pδq )̇ = qδ̇ p − p ̇ δq .

(A.8)

Comparing this with the general expression for the variation of the Hamiltonian

δH =

∂H ∂H δp + δq, ∂q ∂p

(A.9)

we obtain the Hamilton equations

∂H = q,̇ ∂p

(A.10)

∂H = −p ̇ . ∂q

(A.11)

A.3 Poisson brackets For quantities A and B we define the Poisson bracket as

A-2

Relativistic Quantum Field Theory, Volume 1

{A , B}PB ≡

∂A ∂B ∂A ∂B . − ∂p ∂q ∂q ∂p

(A.12)

From this definition, it is straightforward to see that

{p , p }PB = 0,

(A.13)

{q , q}PB = 0,

(A.14)

{q , p }PB = 1.

(A.15)

In addition, if we evaluate the full derivative of a quantity, we obtain

dA ∂A ∂A ∂A q̇ ṗ + + = dt ∂q ∂p ∂t ∂A ∂H ∂A ∂H ∂A . + − = ∂q ∂p ∂p ∂q ∂t

(A.16)

Writing this in terms of the Poisson bracket results in a formula that looks very similar to the time evolution formula from quantum mechanics!

dA ∂A + {A , H }PB . = dt ∂t

(A.17)

A.4 Classical mechanics with many degrees of freedom In order to review the formalism, we will consider the canonical example of a classical linear ‘chain’ of equal masses connected by ideal springs. For simplicity, we will assume that (1) the masses in the chain can only move in one dimension, (2) all springs have the same spring constant, and (3) all springs have zero equilibrium (rest) length. In the static equilibrium case, all masses are separated by a distance ε . We introduce the coordinate qi as the displacement of each oscillator from its equilibrium position. In figure A1, I sketch the equilibrium configuration in the top row and a particular ‘perturbed’ state in the bottom row. We will use periodic boundary conditions, which means that the (N + 1)th oscillator is identified with oscillator 1, i.e., qN +1(t ) = q1(t ). Physically, this means that we are constructing a kind of ring. In the limit that there are large number of oscillators, this assumption should not matter. In practice, it makes the math a bit simpler. Based on the setup specified above, it is easy to write down the Lagrangian N

L=T−V=

N

1 1 m∑qṅ2 − κ ∑(qn+1 − qn)2 , 2 n=1 2 n=1

(A.18)

where κ is the spring constant. From the Euler–Lagrange equations of motion we obtain

mqn̈ = κ (qn+1 + qn−1 − 2qn).

A-3

(A.19)

Relativistic Quantum Field Theory, Volume 1

N

1

q2

q3

q4

Figure A1. Setup for the linear chain discussion. The top line shows the equilibrium configuration and the bottom line shows a particular ‘perturbed’ state.

We can then determine the canonical momentum from the Lagrangian

∂L = mqṅ , ∂qṅ

pn =

(A.20)

which allows us to obtain the Hamiltonian N

H=

N

∑pn qṅ − L =

∑

n=1

n=1

pn2 2m

N

+

1 κ ∑(q − qn)2 . 2 n = 1 n +1

(A.21)

I also note that one can evaluate the Poisson bracket of qi and pi for this many-body system

⎛ ∂q ∂p ∂q ∂p ⎞ n′ − n n ′ ⎟ = δnn ′. ∂pk ∂qk ⎠ k ∂pk k = 1⎝ N

{qn, pn ′ }PB =

∑ ⎜ ∂qn

(A.22)

Exercise A.1 Show that equation (A.19) follows from equation (A.18). A.4.1 Normal coordinates We can introduce normal coordinates by expanding the position of each of the oscillators in an orthogonal basis

qn(t ) =

∑ak(t )un,k ,

(A.23)

k

where k is the wavenumber and the basis functions un,k are

u n,k =

e ikεn , N

A-4

(A.24)

Relativistic Quantum Field Theory, Volume 1

where the normalization factor is chosen to make the orthnormality condition simple (see equation (A.27)). Due to the discreteness of the system, the allowed values of k are restricted to −π /ε ⩽ k ⩽ π /ε (the first Brillouin zone). Additionally, the periodic boundary condition requires that q1(t ) = qN +1(t ) and hence

1 ∑ak(t )e ikε = N k

1 ∑ak(t )e ikε(N +1), N k

(A.25)

which requires that exp (ikεN ) = 1 and hence kεN = 2πℓ where ℓ is an integer. This gives the following restriction on the possible values of k

2π ℓ, Nε

k=

(A.26)

and hence −N /2 ⩽ ℓ ⩽ N /2. In addition to the label k , the basis functions un,k (t ) have dimension N corresponding to the number of oscillators. The basis functions obey N

orthnormality:

∑u n*,k ′un,k = δkk ′,

(A.27)

n=1 k max

completeness:

∑

u n*,ku n ′,k = δnn ′,

(A.28)

k = k min

symmetry: u n*,k = u n,−k .

(A.29)

As a consequence of the last line together with the fact that the qn are real valued, one has a k*(t ) = a−k (t ). In addition, the final symmetry relation can be used to derive two alternative versions of the orthonormality relation, which will come in handy below: N

N

∑un,k ′un,k =

∑u n*,k ′u n*,k = δk,−k ′.

n=1

n=1

(A.30)

Plugging equation (A.23) into the Euler–Lagrange equations of motion (A.19) and performing some algebra gives

ak̈ (t ) = −ω k2ak(t ),

(A.31)

where

ω k2 = 4

⎛ kε ⎞ κ sin2 ⎜ ⎟ . ⎝2⎠ m

A-5

(A.32)

Relativistic Quantum Field Theory, Volume 1

1.0 0.8 Ωk 2

0.6 Κ m

0.4 0.2 0.0

Π

Π 2

0

Π 2

Π

k Figure A2. The dispersion relation (A.32) for modes in the classical linear oscillator chain. Only the first Brillouin zone is shown. For a discrete system (finite ε ), the dispersion relation is periodic in k with period 2π /ε .

Exercise A.2 Show that equations (A.31) and (A.32) follow from equation (A.23). This is the dispersion relation, which tells us how to relate the energy and the wavenumber. It is plotted in figure A2. In the limit of a continuous system one has κ lim ω k2 = k 2ε 2 . (A.33) ε→ 0 m The set of differential equations (A.31) shows that the coupled system of oscillators looks like an ensemble of uncoupled oscillators when expressed in terms of ak (t ). The solution to (A.31) can be expressed as

ak(t ) = bk e−iωkt + b−*k e iωkt ,

(A.34)

where we have used the fact that a k*(t ) = a−k (t ). The coefficients bk and b k* are the normal coordinates. As a result, the solution for the generalized coordinates can be written as

qn(t ) = ∑[bk e−iωkt + b−*k e iωkt ]u n,k k

= ∑⎡⎣bk e−iωktu n,k + b k*e iωktu n*,k ⎤⎦ ,

(A.35)

k

where in going from the first to second lines we have taken k → −k in the second term and used the symmetry property of the basis elements. From the above expression, it is straightforward to see that qn is real valued. Using the relation between qn and pn given in equation (A.20), one obtains the following expression for the generalized momenta

pn (t ) =

∑( −imωk )⎡⎣bk e−iω tun,k − b k*e iω tu n*,k ⎤⎦. k

k

A-6

k

(A.36)

Relativistic Quantum Field Theory, Volume 1

Using the explicit expressions for q and p in equation (A.21) we can evaluate the Hamiltonian N

N

1 1 H= ∑pn2 + κ∑(qn+1 − qn)2 2m n = 1 2 n=1 N ⎞2 m ⎛ i t ω ω i t − ⎡ ⎤ * * k k = − ∑⎜⎜∑ωk⎣bk e u n,k − b k e u n,k ⎦⎟⎟ 2 n = 1⎝ k ⎠ N κ ⎛ + ∑⎜⎜∑⎡⎣bk e−iωktu n+1,k(t ) + b k*e iωktu n*+1,k 2 n = 1⎝ k

(A.37)

⎞2

− bk e

−iωk t

u n,k(t ) −

b k*e iωktu n*,k ⎤⎦ ⎟⎟ ⎠

.

Expanding the kinetic energy term we find N

T=−

m ∑∑ωkωk ′⎡⎣bk bk ′e−i(ωk +ωk′)tun,kun,k ′ + b k*b k*′e i(ωk +ωk′)tu n*,ku n*,k ′ 2 n = 1 k, k ′ −

bk b k*′e−i (ωk −ωk′)tu n,ku n*,k ′

− b k*bk ′e

i (ωk −ωk ′ )t

(A.38)

u n*,ku n,k ′⎤⎦ .

Using the orthonormality relations (A.27) and (A.30) and the fact that ωk = ω−k gives m T = − ∑ω k2⎡⎣ b−k bk e−2iωkt + b−*k b k*e 2iωkt − bk b k* − b k*bk ⎤⎦ . (A.39) 2 k The calculation of the potential energy is a bit more involved. One expands it out and then uses the fact that (exp(−ikε ) − 1)(exp(ikε ) − 1) = 4 sin2(kε /2) = mω k2 /κ to obtain m V = ∑ω k2⎡⎣ b−k bk e−2iωkt + b−*k b k*e 2iωkt + bk b k* + b k*bk ⎤⎦ . (A.40) 2 k Combining these two expressions we see that the first two terms in the kinetic and potential energies cancel, leaving us with the following simple result

H=T+V=

∑mω k2(bk b k* + b k*bk ) = ∑2mω k2b k*bk . k

(A.41)

k

Note that the coefficients (normal coordinates) bk can be obtained from the initial conditions for q and q ̇ with the result being

bk =

N ⎡ ⎤ 1 i u n*,k⎢qn(0) + qṅ (0)⎥ . ∑ ⎦ 2 n=1 ⎣ ωk

A-7

(A.42)

Relativistic Quantum Field Theory, Volume 1

In this way, the initial conditions (2N real numbers) are encoded in the set of N complex coefficients bk . In fact, one can generalize (A.42) to arbitrary time to obtain N ⎡ ⎤ 1 i pn (t )⎥ . bk = ∑u n*,ke iωkt⎢qn(t ) + ⎣ ⎦ 2 n=1 mωk

(A.43)

One can verify by explicit calculation that the right-hand side does not depend on time and, therefore, the Hamiltonian (A.41) is constant in time (energy conservation). Exercise A.3 Show that equation (A.41) follows from equation (A.35). A.4.2 Poisson brackets for the linear chain Finally, I note that using (A.43) and (A.22) one can evaluate the Poisson bracket for the normal coordinates

⎛ ∂b ∂b *′ ∂b ∂b * ⎞ k − k k ′ ⎟, ∂pn ∂qn ⎠ n ∂pn n=1

(A.44)

−i δkk ′, 2mωk

(A.45)

N

{ bk , }PB = ∑⎜⎝ ∂qk b k*′

to obtain the following

{bk , b k*′}PB =

{

}PB = 0.

{bk , bk ′}PB = bk , b k*′

(A.46)

Exercise A.4 Show that equations (A.45) and (A.46) follow from equations (A.43) and (A.44).

A-8

IOP Concise Physics

Relativistic Quantum Field Theory, Volume 1 Canonical formalism Michael Strickland

Appendix B Functionals and functional derivatives Since we will find them appearing at various points, in this appendix I review the concept of functionals and functional derivatives. The notation F [ϕ ] indicates that F is a functional of ϕ(x ), i.e., a mapping from a normal linear space of functions M = {ϕ(x ): x ∈ } to real or complex numbers, F : M →  or  . An example of a functional is, e.g. ∞

F [ϕ(x )] =

∫−∞ dx ϕ2(x) .

(B.1)

As we can see from this example, in this case the functional F [ϕ ] takes the function ϕ(x ), defined on the entire real axis, and returns a number. The functional derivative

δF [ϕ ] , δϕ(x )

(B.2)

tells us how the value of the functional changes if the function ϕ(x ) is changed at the point x . The defining relation for a functional derivative is

δF [ϕ ] =

∫ dx δδϕF([xϕ)] δϕ(x) .

(B.3)

As in ordinary differentiation, the functional derivative can also be represented as the limit of divided differences. One can construct a variation of the ‘independent variable’, i.e., the function ϕ(x ) which is localized at the point y and has strength ε :

δϕ(x ) = εδ(x − y ) .

doi:10.1088/2053-2571/ab30ccch7

B-1

(B.4)

ª Morgan & Claypool Publishers 2019

Relativistic Quantum Field Theory, Volume 1

Inserting this into the defining relation above one finds δF [ϕ] = F [ϕ(x ) + εδ(x − y )] − F [ϕ(x )] =

δF [ ϕ ]

δF [ ϕ ]

∫ dx δϕ(x) ε δ(x − y ) = ε δϕ(y ) ,

(B.5)

which, in the limit of vanishing ε becomes

δF [ϕ ] F [ϕ(x ) + εδ(x − y )] − F [ϕ(x )] = lim . ε→ 0 δϕ(y ) ε

(B.6)

Note that the order of operations is important! The rule is that the limit ε → 0 must be taken first, before the evaluation of other operations/limits. Most of the rules of ordinary differential calculus apply to functional derivatives. Obviously, the functional derivative is linear. From the product of two functionals G[ϕ ] and F [ϕ ] the product rule applies. If F [ϕ ] = G [ϕ ]H [ϕ ] one has

δH [ϕ ] δF [ϕ ] δG [ϕ ] . H [ϕ ] + G [ϕ ] = δϕ(x ) δϕ(x ) δϕ(x )

(B.7)

Similarly, the chain rule can be applied to a functional of a functional:

δ F [G [ϕ ]] = δϕ(x )

∫ dx δδGF ([ϕx)] δδϕG([xϕ)] .

(B.8)

The proofs of these two relations follow almost directly from the analogous proofs in ordinary differential calculus. In the case that the function G[ϕ ] is an ordinary function g(ϕ ), e.g. ϕ2(x ), then the integral in the last relation collapses and one has the simpler relation

δF [ϕ ] dg[ϕ ] δ . F [g[ϕ ]] = δg(x ) dϕ(x ) δϕ(x )

(B.9)

Examples Let us get some practice with functional derivatives in the form of some examples. In all integrals listed below, the integration domain extends over the entire real axis x ∈ ( −∞ , ∞). (a) For F [ϕ ] = ϕ(x ) one has

δF [ϕ ] = δ (x − y ) . δϕ(x )

(B.10)

This follows trivially from (B.6). (b) For F [ϕ ] = ∂xϕ(x ) one has

δF [ϕ ] = ∂x[δ(x − y )] . δϕ(x )

B-2

(B.11)

Relativistic Quantum Field Theory, Volume 1

This also follows trivially from (B.6). (c) For F [ϕ ] = ∫ dx′ K (y, x′)ϕ(x′) one has

δF [ϕ ] = K (y , x ) . δϕ(x )

(B.12)

Proof: we apply (B.6) to obtain δF [ϕ] = lim δϕ(x ) ε→ 0 =

∫ dx′ K (y, x′)[ϕ(x′) + εδ(x′ − x)] − ∫ dx′ K (y, x′)ϕ(x′) (B.13)

ε

∫ dx′ K (y, x′)δ(x′ − x) = K (y, x).

(d) Show that if F [ϕ ] = ∫ dx′ [ϕ(x′)]n one has

δF [ϕ ] = n[ϕ(x )]n−1 . δϕ(x )

(B.14)

Proof: we apply (B.6) to obtain

δF [ϕ ] = lim δϕ(x ) ε→ 0 =n

∫ dx′ [ϕ(x′) + εδ(x′ − x)]n − ∫ dx′ [ϕ(x′)]n (B.15)

ε

∫ dx′ [ϕ(x′)]n−1δ(x′ − x) = n[ϕ(x)]n−1 ,

where, in going from the first to second lines, we have discarded all terms that vanish in the limit ε → 0. dϕ(x ′) n (e) Show that if F [ϕ ] = ∫ dx′ ⎡⎣ dx ′ ⎤⎦ one has

d ⎡ dϕ(x′) ⎤n−1 δF [ϕ ] = −n ⎢ ⎥ dx′ ⎣ dx′ ⎦ δϕ(x )

.

(B.16)

x ′→x

Proof: we apply (B.6) to obtain ⎛

δF [ϕ] = lim δϕ(x ) ε→ 0 =n

−

⎡

⎤n

∫ dx′ ⎢⎣ dϕdx(x′′) ⎥⎦

ε ⎡ dϕ(x′) ⎤ ⎥ dx′ ⎦

∫ dx′ ⎢⎣

= −n

⎞n

∫ dx′ ⎜⎝ dxd ′ [ϕ(x′) + εδ(x′ − x)]⎟⎠

∫

dx′

n −1

d [δ(x′ − x )] dx′

d ⎡ dϕ(x′) ⎤n−1 d ⎡ dϕ(x′) ⎤n−1 ( ) x x n ′ − = − δ ⎢ ⎥ ⎢ ⎥ dx′ ⎣ dx′ ⎦ dx′ ⎣ dx′ ⎦

B-3

(B.17)

, x ′→x

Relativistic Quantum Field Theory, Volume 1

where, in going from the first to second lines, we have once again discarded all terms that vanish in the limit ε → 0, and, in going from the second to the third lines, we have integrated by parts and used the fact that the delta function has no support at infinity.

B-4

IOP Concise Physics

Relativistic Quantum Field Theory, Volume 1 Canonical formalism Michael Strickland

Appendix C Tensor algebra Here I will attempt to summarize the rules of tensor algebra. Since, in this course, we will work exclusively in flat Minkowski space, I will state the rules in terms of the Minkowski space metric tensor (g μν )Minkowski ≡ η μν with

η μν

⎛1 0 0 0 ⎞ ⎜0 − 1 0 0 ⎟= = ημν = ⎜ ⎟ diag(1, −1, −1, −1). ⎜0 0 − 1 0 ⎟ ⎝0 0 0 − 1⎠

(C.1)

Note that, since the metric tensor is diagonal in Minkowski space, it is also symmetric, i.e. η μν = η νμ. A general tensor of rank-N has N indices, which can be mixed in the up and down positions, e.g. a rank-3 tensor might be written as T μνλ or T μ ν λ , however, these two objects are not the same since the position of the indices (up vs down) are different but that does not change the fact that they are both rank-3 tensors. Indices that are up are called contravariant indices and indices that are in the down position are called covariant indices. We can convert between the two types of indices using the metric tensor, e.g.

η μαxα = x μ,

(C.2)

ημαx α = xμ.

(C.3)

If x μ = (t , x , y, z ) then xμ = (t , −x , −y, −z ). One can think of the 4-vectors on the right above as column vectors that multiply the matrix above. 4-vectors are examples of rank-1 tensors. It is obvious from the matrix form that the square of the metric tensor returns an identity matrix. In index notation, this translates into the following identity

doi:10.1088/2053-2571/ab30ccch8

C-1

ª Morgan & Claypool Publishers 2019

Relativistic Quantum Field Theory, Volume 1

ημαη αν = Iμν = δμν,

(C.4)

where I is an identity matrix and δμ ν is a ‘fancy notation’ for a Kronecker delta symbol that simply requires μ to be equal to ν , if μ is summed over. Sometimes the distinction between up and down indices for the Kronecker delta is ignored since it does not matter in practice. Note that the metric tensor with one index up and one down is also a Kronecker delta

ημα = ημα , η ημα = η ναημα , να

(C.5)

η μν = δ ν μ = δ μν, where, on the left, I used the action of the metric tensor to change indices from covariant to contravariant and, on the right, I used equation (C.4) followed by the fact that the Kronecker delta is symmetric. One can express the action of the metric tensor on a general rank-N tensor as

ηαμi T μ1μ2⋯μi⋯μN = T μ1μ2⋯ α⋯μN ,

(C.6)

for the case of lowering indices, or

η αμi Tμ1μ2⋯μi⋯μN = Tμ1μ2⋯ α⋯μN ,

(C.7)

for the case of raising indices. Note that the up or down positions of the indices other than the one corresponding to μi in the example above do not matter. Whatever they are on the left, they will remain in that position on the right, e.g.

ηαμi Tμ1μ2⋯μi⋯μN = Tμ1μ2⋯ α⋯μN .

(C.8)

One can construct Lorentz-invariant products of tensor of various ranks if one makes sure that there is always one up index and one down index with a matching label, e.g.

xμx μ, ημνT μν = Tν ν = T μ μ, and x μxνTμν, are all examples of Lorentz-invariant tensor contractions.

C-2

(C.9)

IOP Concise Physics

Relativistic Quantum Field Theory, Volume 1 Canonical formalism Michael Strickland

Appendix D Mandelstam variables In figure D1, I have drawn a general two-to-two scattering digram with the lines labeled with each of the incoming/outgoing particle’s 4-momentum. The kinematics is subject to the constraints

pi 2 = mi2 ,

(D.1)

p1 + p2 = p3 + p4 ,

(D.2)

where i = 1, 2, 3, 4. In this case we have four 4-momenta at our disposal, so in addition to the invariant masses pi2 = mi2 , we can construct six possible invariants: p1 · p2 , p1 · p3, p1 · p4 , p2 · p3, p2 · p4 , and p3 · p4 . Of these additional six invariants, only two are linearly independent. However, since pi2 = mi2 and p1 + p2 = p3 + p4 , only two of these additional invariants are actually independent. One could proceed with the remaining independent dot products, but it turns out to be more convenient to introduce three Mandelstam variables that are related to these dot products

s ≡ (p1 + p2 )2 = (p3 + p4 )2 ,

(D.3)

t ≡ (p1 − p3 )2 = (p2 − p4 )2 ,

(D.4)

u ≡ (p1 − p4 )2 = (p1 − p3 )2 ,

(D.5)

which satisfy the constraint

s+t+u=

∑mi2 .

(D.6)

i

doi:10.1088/2053-2571/ab30ccch9

D-1

ª Morgan & Claypool Publishers 2019

Relativistic Quantum Field Theory, Volume 1

p1

p3

p2

p4

Figure D1. Two-to-two general scattering diagram.

y

p3 θ

p1

p2 p4

x

Figure D2. Setup for COM scattering.

Going to the center-of-momentum frame and using the setup indicated in figure D2 one can take the four vectors to be

p1 = (E1, ∣p∣ , 0, 0),

(D.7)

p2 = (E2, −∣p∣ , 0, 0),

(D.8)

p3 = (E3, ∣p′∣ cos θ , ∣p′∣ sin θ , 0),

(D.9)

p4 = (E4, −∣p′∣ cos θ , −∣p′∣ sin θ , 0).

(D.10)

Using these 4-vectors, one obtains

s = (p1 + p2 )2 = (E1 + E2 )2 = (E3 + E4)2 ,

(D.11)

and

t = (p1 − p3 )2 = (E1 − E3)2 − [(∣p∣ − ∣p′∣ cos θ )2 + ∣p′∣2 sin2 θ ] = E12 + E 22 − 2E1E2 − p∣2 − ∣p′∣2 + 2∣p∣∣p′∣ cos θ =

m12

+

m32

(D.12)

− 2E1E2 + 2∣p∣∣p′∣ cos θ .

In the limit of equal mass particles m1 = m2 = m3 = m4 = m, we have E1 = E3 = E and ∣p∣ = ∣p′∣ = m12 + m32 − 2E1E2 + 2∣p∣∣p′∣ cos θ and, as a result, t = 2m2 − 2E 2 + 2∣p∣2 cos θ = −2∣p 2∣ + 2∣p∣2 cos θ = −2∣p 2∣(1 − cos θ ). For u , one obtains

u = (p1 − p4 )2 = (E1 − E4)2 − [(∣p∣ + ∣p′∣ cos θ )2 + ∣p′∣2 sin2 θ ] = E12 + E 42 − 2E1E4 − ∣p∣2 − ∣p′∣2 − 2∣p∣∣p′∣ cos θ =

m12

+

m 42

− 2E1E4 − 2∣p∣∣p′∣ cos θ .

D-2

(D.13)

Relativistic Quantum Field Theory, Volume 1

In the limit of equal mass particles m1 = m2 = m3 = m4 = m, we have E1 = E4 = E and ∣p∣ = ∣p′∣ and, as a result, u = 2m2 − 2E 2 − 2∣p∣2 cos θ = −2∣p 2∣ − 2∣p∣2 cos θ = −2∣p 2∣(1 + cos θ ). It is also useful to express the dot products between the various vectors in terms of s , t , and u

p1 · p2 =

1 1⎡ 2 2 2 s − m12 − m 22 , ⎣(p1 + p2 ) − p1 − p2 ⎤⎦ = 2 2

(D.14)

p3 · p4 =

1 1⎡ 2 2 s − m32 − m 42 , ⎣(p3 + p4 )2 − p3 − p4 ⎤⎦ = 2 2

(D.15)

p1 · p3 =

1 1⎡ 2 2 m12 + m32 − t , ⎣p1 + p3 − (p1 − p3 )2 ⎤⎦ = 2 2

(D.16)

p2 · p4 =

1 1⎡ 2 2 m 22 + m 42 − t , ⎣p2 + p4 − (p2 − p4 )2 ⎤⎦ = 2 2

(D.17)

p1 · p4 =

1 1⎡ 2 2 m12 + m 42 − u , ⎣p1 + p4 − (p1 − p4 )2 ⎤⎦ = 2 2

(D.18)

p2 · p3 =

1 1⎡ 2 2 m 22 + m32 − u . ⎣p2 + p3 − (p2 − p3 )2 ⎤⎦ = 2 2

(D.19)

(

( (

(

(

(

)

)

)

)

)

)

In the limit of equal mass particles m1 = m2 = m3 = m4 = m, these relations simplify to

p1 · p2 = p3 · p4 =

1 (s − 2m 2 ) , 2

(D.20)

p1 · p3 = p2 · p4 =

1 (2m 2 − t ) , 2

(D.21)

p1 · p4 = p2 · p3 =

1 (2m 2 − u ) . 2

(D.22)

D-3