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Quantum Field Theory Demystified

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Quantum Field Theory Demystified
David McMahon
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DOI: 10.1036/0071543821

ABOUT THE AUTHOR
David McMahon works as a researcher at Sandia National Laboratories. He has advanced degrees in physics and applied mathematics, and is the author of Quantum Mechanics Demystified, Relativity Demystified, MATLAB® Demystified, and several other successful books.
Copyright © 2008 by The McGraw-Hill Companies, Inc. Click here for terms of use.

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CONTENTS AT A GLANCE

CHAPTER 1 Particle Physics and Special Relativity

1

CHAPTER 2 Lagrangian Field Theory

23

CHAPTER 3 An Introduction to Group Theory

49

CHAPTER 4 Discrete Symmetries and Quantum Numbers

71

CHAPTER 5 The Dirac Equation

85

CHAPTER 6 Scalar Fields

109

CHAPTER 7 The Feynman Rules

139

CHAPTER 8 Quantum Electrodynamics

163

CHAPTER 9 Spontaneous Symmetry Breaking and

the Higgs Mechanism

187

CHAPTER 10 Electroweak Theory

209

CHAPTER 11 Path Integrals

233

CHAPTER 12 Supersymmetry

245

Final Exam

263

Solutions to Quizzes and Final Exam

281

References

289

Index

291

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For more information about this title, click here
CONTENTS

Preface

xv

CHAPTER 1 Particle Physics and Special Relativity

1

Special Relativity

5

A Quick Overview of Particle Physics

12

Elementary Particles

14

The Higgs Mechanism

18

Grand Unification

18

Supersymmetry

19

String Theory

19

Summary

20

Quiz

20

CHAPTER 2 Lagrangian Field Theory

23

Basic Lagrangian Mechanics

23

The Action and the Equations of Motion

26

Canonical Momentum and the Hamiltonian

29

Lagrangian Field Theory

30

Symmetries and Conservation Laws

35

Conserved Currents

38

The Electromagnetic Field

39

Gauge Transformations

43

Summary

47

Quiz

47

x

Quantum Field Theory Demystified

CHAPTER 3 An Introduction to Group Theory

49

Representation of the Group

50

Group Parameters

52

Lie Groups

52

The Rotation Group

54

Representing Rotations

55

SO(N)

58

Unitary Groups

62

Casimir Operators

67

Summary

68

Quiz

68

CHAPTER 4 Discrete Symmetries and Quantum Numbers

71

Additive and Multiplicative Quantum Numbers

71

Parity

72

Charge Conjugation

76

CP Violation

78

The CPT Theorem

80

Summary

82

Quiz

83

CHAPTER 5 The Dirac Equation

85

The Classical Dirac Field

85

Adding Quantum Theory

87

The Form of the Dirac Matrices

89

Some Tedious Properties of the

Dirac Matrices

91

Adjoint Spinors and Transformation Properties

94

Slash Notation

95

Solutions of the Dirac Equation

95

Free Space Solutions

99

Boosts, Rotations, and Helicity

103

Weyl Spinors

104

Summary

107

Quiz

108

Contents

xi

CHAPTER 6 Scalar Fields

109

Arriving at the Klein-Gordon Equation

110

Reinterpreting the Field

117

Field Quantization of Scalar Fields

117

States in Quantum Field Theory

127

Positive and Negative Frequency

Decomposition

128

Number Operators

128

Normalization of the States

130

Bose-Einstein Statistics

131

Normal and Time-Ordered Products

134

The Complex Scalar Field

135

Summary

137

Quiz

137

CHAPTER 7 The Feynman Rules

139

The Interaction Picture

141

Perturbation Theory

143

Basics of the Feynman Rules

146

Calculating Amplitudes

151

Steps to Construct an Amplitude

153

Rates of Decay and Lifetimes

160

Summary

160

Quiz

160

CHAPTER 8 Quantum Electrodynamics

163

Reviewing Classical Electrodynamics Again

165

The Quantized Electromagnetic Field

168

Gauge Invariance and QED

170

Feynman Rules for QED

173

Summary

185

Quiz

185

CHAPTER 9 Spontaneous Symmetry Breaking and

the Higgs Mechanism

187

Symmetry Breaking in Field Theory

189

xii

Quantum Field Theory Demystified

Mass Terms in the Lagrangian

192

Aside on Units

195

Spontaneous Symmetry Breaking and Mass

196

Lagrangians with Multiple Particles

199

The Higgs Mechanism

202

Summary

207

Quiz

207

CHAPTER 10 Electroweak Theory

209

Right- and Left-Handed Spinors

210

A Massless Dirac Lagrangian

211

Leptonic Fields of the Electroweak Interactions

212

Charges of the Electroweak Interaction

213

Unitary Transformations and the Gauge Fields

of the Theory

215

Weak Mixing or Weinberg Angle

219

Symmetry Breaking

220

Giving Mass to the Lepton Fields

222

Gauge Masses

224

Summary

231

Quiz

231

CHAPTER 11 Path Integrals

233

Gaussian Integrals

233

Basic Path Integrals

238

Summary

242

Quiz

243

CHAPTER 12 Supersymmetry

245

Basic Overview of Supersymmetry

246

Supercharge

247

Supersymmetric Quantum Mechanics

249

The Simplified Wess-Zumino Model

253

A Simple SUSY Lagrangian

254

Summary

260

Quiz

260

Contents
Final Exam Solutions to Quizzes and Final Exam References Index

xiii
263 281 289 291

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PREFACE
Quantum field theory is the union of Einstein’s special relativity and quantum mechanics. It forms the foundation of what scientists call the standard model, which is a theoretical framework that describes all known particles and interactions with the exception of gravity. There is no time like the present to learn it—the Large Hadron Collider (LHC) being constructed in Europe will test the final pieces of the standard model (the Higgs mechanism) and look for physics beyond the standard model. In addition quantum field theory forms the theoretical underpinnings of string theory, currently the best candidate for unifying all known particles and forces into a single theoretical framework.
Quantum field theory is also one of the most difficult subjects in science. This book aims to open the door to quantum field theory to as many interested people as possible by providing a simplified presentation of the subject. This book is useful as a supplement in the classroom or as a tool for self-study, but be forewarned that the book includes the math that comes along with the subject.
By design, this book is not thorough or complete, and it might even be considered by some “experts” to be shallow or filled with tedious calculations. But this book is not written for the experts or for brilliant graduate students at the top of the class, it is written for those who find the subject difficult or impossible. Certain aspects of quantum field theory have been selected to introduce new people to the subject, or to help refresh those who have been away from physics.
After completing this book, you will find that studying other quantum field theory books will be easier. You can master quantum field theory by tackling the reference list in the back of this book, which includes a list of textbooks used in the development of this one. Frankly, while all of those books are very good and make fine references, most of them are hard to read. In fact many quantum field theory books are impossible to read. My recommendation is to work through this book first, and then tackle Quantum Field Theory in a Nutshell by Anthony Zee. Different than all other books on the subject, it’s very readable and is packed with great
Copyright © 2008 by The McGraw-Hill Companies, Inc. Click here for terms of use.

xvi

Quantum Field Theory Demystified

physical insight. After you’ve gone through that book, if you are looking for mastery or deep understanding you will be well equipped to tackle the other books on the list.
Unfortunately, learning quantum field theory entails some background in physics and math. The bottom line is, I assume you have it. The background I am expecting includes quantum mechanics, some basic special relativity, some exposure to electromagnetics and Maxwell’s equations, calculus, linear algebra, and differential equations. If you lack this background do some studying in these subjects and then give this book a try.
Now let’s forge ahead and start learning quantum field theory.

David McMahon

Quantum Field Theory Demystified

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CHAPTER 1
Particle Physics and
Special Relativity
Quantum field theory is a theoretical framework that combines quantum mechanics and special relativity. Generally speaking, quantum mechanics is a theory that describes the behavior of small systems, such as atoms and individual electrons. Special relativity is the study of high energy physics, that is, the motion of particles and systems at velocities near the speed of light (but without gravity). What follows is an introductory discussion to give you a flavor of what quantum field theory is like. We will explore each concept in more detail in the following chapters.
There are three key ideas we want to recall from quantum mechanics, the first being that physical observables are mathematical operators in the theory.
Copyright © 2008 by The McGraw-Hill Companies, Inc. Click here for terms of use.

2

Quantum Field Theory Demystified

For instance, the Hamiltonian (i.e., the energy) of a simple harmonic oscillator is the operator

Hˆ =

ω

⎛ ⎝⎜

aˆ†aˆ

+

1⎞ 2 ⎠⎟

where aˆ†, aˆ are the creation and annihilation operators, and is Planck’s constant.
The second key idea you should remember from quantum mechanics is the uncertainty principle. The uncertainty relation between the position operator xˆ and the momentum operator pˆ is

Δxˆ Δpˆ ≥

(1.1)

2

There is also an uncertainty relation between energy and time.

ΔE Δt ≥

(1.2)

2

When considering the uncertainty relation between energy and time, it’s important to remember that time is only a parameter in nonrelativistic quantum mechanics, not an operator.
The final key idea to recall from quantum mechanics is the commutation relations. In particular,

[xˆ, pˆ ] = xˆpˆ − pˆxˆ = i

Now let’s turn to special relativity. We can jump right to Einstein’s famous equation that every lay person knows something about, in order to see how special relativity is going to impact quantum theory. This is the equation that relates energy to mass.

E = mc2

(1.3)

What should you take away from this equation? The thing to notice is that if there is enough energy—that is, enough energy proportional to a given particle’s mass as described by Eq. (1.3)—then we can “create” the particle. Due to conservation laws, we actually need twice the particle’s mass, so that we can create a particle and its antiparticle. So in high energy processes,

• Particle number is not fixed. • The types of particles present are not fixed.

These two facts are in direct conflict with nonrelativistic quantum mechanics. In nonrelativistic quantum mechanics, we describe the dynamics of a system with the

CHAPTER 1 Particle Physics and Special Relativity 3

Schrödinger equation, which for a particle moving in one dimension with a potential V is

−

2
2m

∂2ψ ∂x 2

+ Vψ

=

i

∂ψ ∂t

(1.4)

We can extend this formalism to treat the case when several particles are present. However, the number and types of particles are absolutely fixed. The Schrödinger equation cannot in any shape or form handle changing particle number or new types of particles appearing and disappearing as relativity allows.
In fact, there is no wave equation of the type we are used to from nonrelativistic quantum mechanics that is truly compatible with both relativity and quantum theory. Early attempts to merge quantum mechanics and special relativity focused on generating a relativistic version of the Schrödinger equation. In fact, Schrödinger himself derived a relativistic equation prior to coming up with the wave equation he is now famous for. The equation he derived, which was later discovered independently by Klein and Gordon (and is now known as the Klein-Gordon equation) is

1 ∂2ϕ − ∂2ϕ = m2c2 ϕ

c2 ∂t2 ∂x2

2

We will have more to say about this equation in future chapters. Schrödinger discarded it because it gave the wrong fine structure for the hydrogen atom. It is also plagued by an unwanted feature—it appears to give negative probabilities, something that obviously contradicts the spirit of quantum mechanics. This equation also has a funny feature—it allows negative energy states.
The next attempt at a relativistic quantum mechanics was made by Dirac. His famous equation is

i ∂ψ = −i cα ⋅ ∇ψ + βmc2ψ ∂t

Here, α and b are actually matrices. This equation, which we will examine in detail in later chapters, resolves some of the problems of the Klein-Gordon equation but also allows for negative energy states.
As we will emphasize later, part of the problem with these relativistic wave equations is in their interpretation. We move forward into a quantum theory of fields by changing how we look at things. In particular, in order to be truly compatible with special relativity we need to discard the notion that j and y in the

4

Quantum Field Theory Demystified

Klein-Gordon and Dirac equations, respectively describe single particle states. In their place, we propose the following new ideas:
• The wave functions ϕ and ψ are not wave functions at all, instead they are fields.
• The fields are operators that can create new particles and destroy particles.
Since we have promoted the fields to the status of operators, they must satisfy commutation relations. We will see later that we make a transition of the type

[xˆ, pˆ ] → [ϕˆ(x,t),πˆ(y,t)]
Here, πˆ (y,t) is another field that plays the role of momentum in quantum field theory. Since we are transitioning to the continuum, the commutation relation will be of the form

[ϕˆ(x,t),πˆ(y,t)] = i δ (x − y)

where x and y are two points in space. This type of relation holds within it the notion of causality so important in special relativity—if two fields are spatially separated they cannot affect one another.
With fields promoted to operators, you might wonder what happens to the ordinary operators of quantum mechanics. There is one important change you should make sure to keep in mind. In quantum mechanics, position xˆ is an operator while time t is just a parameter. In relativity, since time and position are on a similar footing, we might expect that in relativistic quantum mechanics we would also put time and space on a similar footing. This could mean promoting time to an operator tˆ. This is not what is done in ordinary quantum field theory, where we take the opposite direction—and demote position to a parameter x. So in quantum field theory,
• Fields ϕ and ψ are operators.
• They are parameterized by spacetime points (x, t).
• Position x and time t are just numbers that fix a point in spacetime—they are not operators.
• Momentum continues to play a role as an operator.
In quantum field theory, we frequently use tools from classical mechanics to deal with fields. Specifically, we often use the Lagrangian

L =T −V

(1.5)

CHAPTER 1 Particle Physics and Special Relativity 5

The Lagrangian is important because symmetries (such as rotations) leave the form of the Lagrangian invariant. The classical path taken by a particle is the one which minimizes the action.

S = ∫ L dt

(1.6)

We will see how these methods are applied to fields in Chap. 2.

Special Relativity
The arena in which quantum field theory operates is the high energy domain of special relativity. Therefore, brushing up on some basic concepts in special relativity and familiarizing ourselves with some notation is important to gain some understanding of quantum field theory.
Special relativity is based on two simple postulates. Simply stated, these are:
• The laws of physics are the same for all inertial observers. • The speed of light c is a constant.
An inertial frame of reference is one for which Newton’s first law holds. In special relativity, we characterize spacetime by an event, which is something that happens at a particular time t and some spatial location (x, y, z). Also notice that the speed of light c can serve in a role as a conversion factor, transforming time into space and vice versa. Space and time therefore form a unified framework and we denote coordinates by (ct, x, y, z).
One consequence of the second postulate is the invariance of the interval. In special relativity, we measure distance in space and time together. Imagine a flash of light emitted at the origin at t = 0. At some later time t the spherical wavefront of the light can be described by
c2t2 = x2 + y2 + z2 ⇒ c2t2 − x2 − y2 − z2 = 0
Since the speed of light is invariant, this equation must also hold for another observer, who is measuring coordinates with respect to a frame we denote by (ct′, x′, y′, z′). That is,
c2t′2 − x′2 − y′2 − z′2 = 0

6

Quantum Field Theory Demystified

It follows that

c2t2 − x2 − y2 − z2 = c2t′2 − x′2 − y′2 − z′2

Now, in ordinary space, the differential distance from the origin to some point (x, y, z) is given by

dr 2 = dx2 + dy2 + dz2

We define an analogous concept in spacetime, called the interval. This is denoted by ds2 and is written as

ds2 = c2dt2 − dx2 − dy2 − dz2

(1.7)

From Eq. (1.7) it follows that the interval is invariant. Consider two observers in two different inertial frames. Although they measure different spatial coordinates (x, y, z) and (x′, y′, z′) and different time coordinates t and t′ to label events, the interval for each observer is the same, that is,
ds2 = c2dt2 − dx2 − dy2 − dz2 = c2dt′2 − dx′2 − dy′2 − dz′2 = ds′2

This is a consequence of the fact that the speed of light is the same for all inertial observers.
It is convenient to introduce an object known as the metric. The metric can be used to write down the coefficients of the differentials in the interval, which in this case are just +/−1. The metric of special relativity (“flat space”) is given by

⎛1 0 0 0⎞

η μν

=

⎜ ⎜

0

−1

0

0

⎟ ⎟

⎜ 0 0 −1 0 ⎟

(1.8)

⎝⎜ 0 0 0 −1⎠⎟

The metric has an inverse, which in this case turns out to be the same matrix. We denote the inverse with lowered indices as

⎛1 0 0 0⎞

ημν

=

⎜ ⎜

0

⎜0

−1 0 0 −1

0

⎟ ⎟

0⎟

⎝⎜ 0 0 0 −1⎠⎟

CHAPTER 1 Particle Physics and Special Relativity 7

The symbol ημν is reserved for the metric of special relativity. More generally, the metric is denoted by gμν . This is the convention that we will follow in this book. We
have

gμν gνρ

=

δ

ρ μ

(1.9)

where

δ

ρ μ

is

the

Kronecker

delta

function

defi ned

by

δ

ρ μ

=

⎧1 ⎨⎩0

if μ if μ

= ≠

ρ ρ

Hence Eq. (1.9) is just a statement that

gg−1 = I

where I is the identity matrix.
In relativity, it is convenient to label coordinates by a number called an index. We take ct = x0 and (x, y, z) → (x1, x2, x3 ). Then an event in spacetime is labeled by the
coordinates of a contravariant vector.

xμ = (x0, x1, x2, x3 )

(1.10)

Contravariant refers to the way the vector transforms under a Lorentz transformation, but just remember that a contravariant vector has raised indices. A covariant vector has lowered indices as

xμ = (x0 , x1, x2 , x3 )

An index can be raised or lowered using the metric. Specifically,

xα = gαβ xβ

xα = gαβ xβ

(1.11)

Looking at the metric, you can see that the components of a covariant vector are related to the components of a contravariant vector by a change in sign as

x0 = x0

x1 = − x1

x2 = −x2

x3 = − x3

We use the Einstein summation convention to represent sums. When an index is repeated in an expression once in a lowered position and once in a raised position, this indicates a sum, that is,

3
∑ sα sα ≡ sα sα = s0s0 + s1s1 + s2s2 + s3s3 α=0

8

Quantum Field Theory Demystified

So for example, the index lowering expression in Eq. (1.11) is really shorthand for
xα = gαβ xβ = gα 0 x0 + gα1x1 + gα 2 x2 + gα 3x3
Greek letters such as α, β, μ, and ν are taken to range over all spacetime indices, that is, μ = 0, 1, 2, and 3. If we want to reference spatial indices only, a Latin letter such as i, j, and k is used. That is, i = 1, 2, and 3.

LORENTZ TRANSFORMATIONS

A Lorentz transformation Λ allows us to transform between different inertial reference frames. For simplicity, consider an inertial reference frame x′μ moving along the x axis with respect to another inertial reference frame xμ with speed
v < c. If we define

β=v c

γ= 1 1− β2

(1.12)

Then the Lorentz transformation that connects the two frames is given by

Specifically,

⎛ γ −βγ /c 0 0⎞

Λμν

=

⎜ ⎜

−βγ

/c

⎜0

γ 0

0 0⎟⎟ 1 0⎟

⎝⎜ 0

0 0 1⎠⎟

(1.13)

x′0

=

γ

⎛ ⎝⎜

x0

−

β c

x1

⎞ ⎠⎟

x ′1

=

γ

⎛ ⎝⎜

x1

−

β c

x

0

⎞ ⎠⎟

x′2 = x2

x′3 = x3

(1.14)

We can write a compact expression for a Lorentz transformation relating two sets of coordinates as

x′μ

=

Λ

μ ν

xν

(1.15)

CHAPTER 1 Particle Physics and Special Relativity 9

The rapidity φ is defined as

tanhφ = β = v c

(1.16)

Using the rapidity, we can view a Lorentz transformation as a kind of rotation (mathematically speaking) that rotates time and spatial coordinates into each other, that is,

x′0 = −x1 sinh φ + x0 cosh φ x′1 = x1 cosh φ − x0 sinh φ
Changing velocity to move from one inertial frame to another is done by a Lorentz transformation and we refer to this as a boost.
We can extend the shorthand index notation used for coordinates to derivatives. This is done with the following definition:

∂ ∂t

→

∂ ∂x 0

=

∂0

∂ ∂y

→

∂ ∂x 2

=

∂2

∂ ∂x

→

∂ ∂x1

=

∂1

∂ ∂z

→

∂ ∂x3

=

∂3

We can raise an index on these expressions so that

∂μ = gμν ∂ν

∂0 = ∂0

∂i = −∂i

In special relativity many physical vectors have spatial and time components. We call such objects 4-vectors and denote them with italic font (sometimes with an index) reserving the use of an arrow for the spatial part of the vector. An arbitrary 4-vector Aμ has components
Aμ = ( A0 , A1, A2 , A3 ) Aμ = ( A0 , − A1, − A2 , − A3 ) Aμ = (A0, A)
Aμ = ( A0 , − A)

10

Quantum Field Theory Demystified

We denote the ordinary vector part of a 4-vector as a 3-vector. So the 3-vector part of Aμ is A. The magnitude of a vector is computed using a generalized dot product, like
A ⋅ A = Aμ Aμ = A0 A0 − A1A1 − A2 A2 − A3 A3 = gμν Aμ Aν
This magnitude is a scalar, which is invariant under Lorentz transformations. When a quantity is invariant under Lorentz transformations, all inertial observers agree on its value which we call the scalar product. A consequence of the fact that the scalar product is invariant, meaning that x′μ xμ′ = xμ xμ, is

Λαβ

Λαμ

=

δ

β μ

(1.17)

Now let’s consider derivatives using relativistic notation. The derivative of a field is written as

∂ϕ ∂x μ

=

∂μϕ

(1.18)

The index is lowered because as written, the derivative is a covariant 4-vector. The components of the vector are

⎛ ⎝⎜

∂ϕ ∂x 0

,

∂ϕ ∂x1

,

∂ϕ ∂x 2

,

∂ϕ ∂x3

⎞ ⎠⎟

We also have

∂ϕ = ∂μϕ ∂xμ

which is a contravariant 4-vector. Like any 4-vector, we can compute a scalar

product, which is the four-dimensional generalization of the Laplacian called the

D’Alembertian operator which using ordinary notation is

1 c2

∂2 ∂t 2

− ∇2

≡

. Using the

relativistic notation for derivatives together with the generalized dot product we

have

∂μ∂μ

=

1 c2

∂2 ∂t 2

− ∇2

≡

(1.19)

CHAPTER 1 Particle Physics and Special Relativity 11

One 4-vector that is of particular importance is the energy-momentum 4-vector which unifies the energy and momentum into a single object. This is given by

pμ = (E, − p) = (E, − p1, − p2 , − p3 ) ⇒ pμ = (E, p) = (E, p1, p2 , p3 )

(1.20)

The magnitude of the energy-momentum 4-vector gives us the Einstein relation connecting energy, momentum, and mass.

E2 = p2c2 + m2c4

(1.21)

We can always choose a Lorentz transformation to boost to a frame in which the 3-momentum of the particle is zero p = 0 giving Einstein’s famous relation between energy and rest mass, like

E = mc2

Another important 4-vector is the current 4-vector J. The time component of this vector is the charge density ρ while the 3-vector part of J is the current density J .
That is,

J μ = (ρ, Jx , Jy, Jz )

(1.22)

The current 4-vector is conserved, in the sense that

∂μJ μ = 0

(1.23)

which is nothing other than the familiar relation for conservation of charge as shown here.

∂μJ μ = ∂tρ + ∂xJ x + ∂yJ y + ∂zJ z = 0 ⇒ ∂ρ + ∇ ⋅ J = 0
∂t

12

Quantum Field Theory Demystified

A Quick Overview of Particle Physics
The main application of quantum field theory is to the study of particle physics. This is because quantum field theory describes the fundamental particles and their interactions using what scientists call the standard model. In this framework, the standard model is believed to describe all physical phenomena with the exception of gravity. There are three fundamental interactions or forces described in the standard model:
• The electromagnetic interaction
• The weak interaction
• The strong interaction
Each force is mediated by a force-carrying particle called a gauge boson. Being a boson, a force-carrying particle has integral spin. The gauge bosons for the electromagnetic, weak, and strong forces are all spin-1 particles. If gravity is quantized, the force-carrying particle (called the graviton) is a spin-2 particle.
Forces in nature are believed to result from the exchange of the gauge bosons. For each interaction, there is a field, and the gauge bosons are the quanta of that field. The number of gauge bosons that exist for a particular field is given by the number of generators of the field. For a particular field, the generators come from the unitary group used to describe the symmetries of the field (this will become clearer later in the book).

THE ELECTROMAGNETIC FORCE
The symmetry group of the electromagnetic field is a unitary transformation, called U(1). Since there is a single generator, the force is mediated by a single particle, which is known to be massless. The electromagnetic force is due to the exchange of photons, which we denote by γ . The photon is spin-1 and has two polarization states. If a particle is massless and spin-1, it can only have two polarization states. Photons do not carry charge.

THE WEAK FORCE
The gauge group of the weak force is SU(2) which has three generators. The three physical gauge bosons that mediate the weak force are W +, W −, and Z. As we will see, these particles are superpositions of the generators of the gauge group. The gauge bosons for the weak force are massive.

CHAPTER 1 Particle Physics and Special Relativity 13

• W + has a mass of 80 GeV/c2 and carries +1 electric charge. • W − has a mass of 80 GeV/c2 and carries −1 electric charge. • Z has a mass of 91 GeV/c2 and is electrically neutral.
The massive gauge bosons of the weak interaction are spin-1 and can have three polarization states.

THE STRONG FORCE
The gauge group of the strong force is SU(3) which has eight generators. The gauge bosons corresponding to these generators are called gluons. Gluons mediate interactions between quarks (see below) and are therefore responsible for binding neutrons and protons together in the nucleus. A gluon is a massless spin-1 particle, and like the photon, has two polarization states. Gluons carry the charge of the strong force, called color. Since gluons also carry color charge they can interact among themselves, something that is not possible with photons since photons carry no charge. The theory that describes the strong force is called quantum chromodynamics.

THE RANGE OF A FORCE
The range of a force is dictated primarily by the mass of the gauge boson that mediates this force. We can estimate the range of a force using simple arguments based on the uncertainty principle. The amount of energy required for the exchange of a force mediating particle is found using Einstein’s relation for rest mass as
ΔE ≈ mc2
Now we use the uncertainty principle to determine how long the particle can exist as shown here.

Δt ≈ ΔE = mc2
The special theory of relativity tells us that nothing travels faster than the speed of light c. So, we can use the speed of light to set an upper bound on the velocity of the force-carrying particle, and estimate the range it travels in a time Δt, that is,

distance Velocity =
time

⇒

Δx

=

cΔt

=

c mc2

=

mc

14

Quantum Field Theory Demystified

This is the range of the force. From this relation, you can see that if m → 0, Δx → ∞. So the range of the electromagnetic force is infinite. The range of the weak force, however, is highly constrained because the gauge bosons of the weak force have large masses. Plugging in the mass of the W as 80 GeV/c2 you can verify that the range is
Δx ≈ 10−3 fm
This explains why the weak force is only felt over nuclear distances. This argument does not apply to gluons, which are massless, because the strong force is more complicated and involves a concept known as confinement. As stated above, the charge of the strong force is called color charge, and gluons carry color. Color charge has a strange property in that it exerts a constant force that binds colorcarrying particles together. This can be visualized using the analogy of a rubber band. The stronger you pull on the rubber band, the tighter it feels. If you don’t pull on it at all, it hangs loose. The strong force acts like a rubber band. At very short distances, it is relaxed and the particles behave as free particles. As the distance between them increases, the force gets them back in stronger pulling. This limits the range of the strong force, which is believed to be on the order of 10−15 m, the dimension of a nuclear particle. As a result of confinement, gluons are involved in mediating interactions between quarks, but are only indirectly responsible for the binding of neutrons and protons, which is accomplished through secondary particles called mesons.

Elementary Particles
The elementary particles of quantum field theory are treated as mathematical pointlike objects that have no internal structure. The particles that make up matter all carry spin-1/2 and can be divided into two groups, leptons and quarks. Each group comes in three “families” or “generations.” All elementary particles experience the gravitational force.
LEPTONS
Leptons interact via the electromagnetic and weak interaction, but do not participate in the strong interaction. Since they do not carry color charge, they do not participate in the strong interaction. They can carry electric charge e, which we denote as −1

CHAPTER 1 Particle Physics and Special Relativity 15
(the charge of the electron), or they can be electrically neutral. The leptons include the following particles:
• The electron e carries charge −1 and has a mass of 0.511 MeV/c2. • The muon μ− carries charge −1 and has a mass of 106 MeV/c2. • The tau τ − carries charge −1 and has a mass of 1777 MeV/c2.
Each type of lepton described above defines one of the three families that make up the leptons. In short, the muon and tau are just heavy copies of the electron. Physicists are not sure why there are three families of particles. The muon and tau are unstable and decay into electrons and neutrinos.
Corresponding to each particle above, there is a neutrino. It was thought for a long time that neutrinos were massless, but recent evidence indicates this is not the case, although experiment puts small bounds on their masses. Like the electron, muon, and tau, the three types of neutrinos come with masses that increase with each family. They are electrically neutral and are denoted by
• Electron neutrino νe • Muon neutrino νμ • Tau neutrino ντ
Since they are electrically neutral, the neutrinos do not participate in the electromagnetic interaction. Since they are leptons, they do not participate in the strong interaction. They interact only via the weak force.
To each lepton there corresponds an antilepton. The antiparticles corresponding to the electron, muon, and tau all carry charge of +1, but they have the same masses. They are denoted as follows:
• The positron e+ carries charge +1 and has a mass of 0.511 MeV/c2. • The antimuon μ+ carries charge +1 and has a mass of 106 MeV/c2. • The antitau τ + carries charge +1 and has a mass of 1777 MeV/c2.
In particle physics, we often indicate an antiparticle (a particle with the same properties but opposite charge) with an overbar; so if p is a given particle, we can indicate its corresponding antiparticle by p. We will see later that charge is not the only quantum number of interest; a lepton also carries a quantum number called lepton number. It is +1 for a particle and −1 for the corresponding antiparticle. The antineutrinos νe,νμ , and ντ , like their corresponding particles, are also electrically neutral, but while the neutrinos νe,νμ , and ντ all have lepton number +1, the antineutrinos νe,νμ , and ντ have lepton number −1.

16

Quantum Field Theory Demystified

In particle interactions, lepton number is always conserved. Particles that are not leptons are assigned a lepton number 0. Lepton number explains why there are antineutrinos, because they are neutral like ordinary neutrinos. Consider the beta decay of a neutron as shown here.
n → p + e +νe
A neutron and proton are not leptons, hence they carry lepton number 0. The lepton number must balance on each side of the reaction. On the left we have total lepton number 0. On the right we have
0 + ne + nνe
Since the electron is a lepton, ne = 1. This tells us that the neutrino emitted in this decay must be an antineutrino, and the lepton number is nνe = −1 allowing lepton number to be conserved in the reaction.

QUARKS
Quarks are fundamental particles that make up the neutron and proton. They carry electrical charge and hence participate in the electromagnetic interaction. They also participate in the weak and strong interactions. Color charge, which is the charge of the strong interaction, can come in red, blue, or green. These color designations are just labels, so they should not be taken literally. There is also “anticolor” charge, antired, antiblue, and antigreen. Color charge can only be arranged such that the total color of a particle combination is white. There are three ways to get white color charge:
• Put three quarks together, one red, one blue, and one green.
• Put three quarks together, one antired, one antiblue, and one antigreen.
• Put two quarks together, one colored and one anticolored, for example a red quark and an antired quark.
The charge carried by a quark is −1/3 or +2/3 (in units of electric charge e). There are six types or “flavors” of quarks:
• Up quark u with charge +2/3 • Down quark d with charge −1/3 • Strange quark s with charge +2/3

CHAPTER 1 Particle Physics and Special Relativity 17
• Charmed quark c with charge −1/3 • Top quark t with charge +2/3 • Bottom quark b with charge −1/3 Like the leptons, the quarks come in three families. One member of a family has charge +2/3 and the other has charge −1/3. The families are (u,d), (s,c), and (t,b). With each family, the mass increases. For example, the mass of the up quark is only
mu ≈ 4 MeV/c2
while the mass of the top quark is a hefty
mt ≈ 172 GeV/c2
which is as heavy as a single gold nucleus. Like the leptons, there are antiparticles corresponding to each quark.
Bound states of quarks are called hadrons. Bound states of observed quarks consist of two or three quarks only. A baryon is a hadron with three quarks or three antiquarks. Two famous baryons are
• The proton, which is the three-quark state uud • The neutron, which is the three-quark state udd Bound states consisting of a quark and antiquark are called mesons. These include: The pion π 0 = uu or dd The charged pion π + = ud or π − = ud
SUMMARY OF PARTICLE GENERATIONS OR FAMILIES
The elementary particles come in three generations: • The first generation includes the electron, electron neutrino, the up quark, the down quark, and the corresponding antiparticles. • The second generation includes the muon, muon neutrino, strange quark, and charmed quark, along with the corresponding antiparticles. • The third generation includes the tau, the tau neutrino, the top quark, and the bottom quark, along with the corresponding antiparticles.

18

Quantum Field Theory Demystified

The Higgs Mechanism
As the standard model of particle physics is formulated, the masses of all the particles are 0. An extra field called the Higgs field has to be inserted by hand to give the particles mass. The quantum of the Higgs field is a spin-0 particle called the Higgs boson. The Higgs boson is electrically neutral.
The Higgs field, if it exists, is believed to fill all of empty space throughout the entire universe. Elementary particles acquire their mass through their interaction with the Higgs field. Mathematically we introduce mass into a theory by adding interaction terms into the Lagrangian that couple the field of the particle in question to the Higgs field. Normally, the lowest energy state of a field would have an expectation value of zero. By symmetry breaking, we introduce a nonzero lowest energy state of the field. This procedure leads to the acquisition of mass by the particles in the theory.
Qualitatively, you might think of the Higgs field by imagining the differences between being on land and being completely submerged in water. On dry land, you can move your arm up and down without any trouble. Under water, moving your arm up and down is harder because the water is resisting your movement. We can imagine the movement of elementary particles being resisted by the Higgs field, with each particle interacting with the Higgs field at a different strength. If the coupling between the Higgs field and the particle is strong, then the mass of the particle is large. If it is weak, then the particle has a smaller mass. A particle like the photon with zero rest mass doesn’t interact with the Higgs field at all. If the Higgs field didn’t exist at all, then all particles would be massless. It is not certain what the mass of the Higgs boson is, but current estimates place an upper limit of ≈140 GeV/c2. When the Large Hadron Collider begins operation in 2008, it should be able to detect the Higgs, if it exists.

Grand Unification
The standard model, as we have described above, consists of the electromagnetic interaction, the weak force, and quantum chromodynamics. Theorists would like to unify these into a single force or interaction. Many problems remain in theoretical physics, and in the past, many problems have been solved via some kind of unification. In many cases two seemingly different phenomena are actually two sides of the same coin. The quintessential example of this type of reasoning is the discovery by Faraday, Maxwell, and others that light, electricity, and magnetism are all the same physical phenomena that we now group together under electromagnetism.

CHAPTER 1 Particle Physics and Special Relativity 19
Electromagnetism and the weak force have been unified into a single theoretical framework called electroweak theory. A grand unified theory or GUT is an attempt to bring quantum chromodynamics (and hence the strong force) into this unified framework.
If such a theory is valid, then there is a grand unification energy at which the electromagnetic, weak, and strong forces become unified into a single force. There is some support for this idea since the electromagnetic and weak force are known to become unified at high energies (but at lower energies than where unification with the strong force is imagined to occur).
Supersymmetry
There exists yet another unification scheme beyond that tackled by the GUTs. In particle physics, there are two basic types of particles. These include the spin-1/2 matter particles (fermions) and the spin-1 force-carrying particles (bosons). In elementary quantum mechanics, you no doubt learned that bosons and fermions obey different statistics. While the Pauli exclusion principle prevents two fermions from inhabiting the same state, there is no such limitation for bosons.
One might wonder why there are these two types of particles. In supersymmetry, an attempt is made to apply the reasoning of Maxwell and propose that a symmetry exists between bosons and fermions. For each fermion, supersymmetry proposes that there is a boson with the same mass, and vice versa. The partners of the known particles are called superpartners. Unfortunately, at this time there is no evidence that this is the case. The fact that the superpartners do not have the same mass indicates either that the symmetry of the theory is broken, in which case the masses of the superpartners are much larger than expected, or that the theory is not correct at all and supersymmetry does not exist.
String Theory
The ultimate step forward for quantum field theory is a unified theory known as string theory. This theory was originally proposed as a theory of the strong interaction, but it fell out of favor when quantum chromodynamics was developed. The basic idea of string theory is that the fundamental objects in the universe are not pointlike elementary particles, but are instead objects spread out in one dimension called strings. Excitations of the string give the different particles we see in the universe.
String theory is popular because it appears to be a completely unified theory. Quantum field theory unifies quantum mechanics and special relativity, and as a result is able to describe interactions involving three of the four known forces.

20

Quantum Field Theory Demystified

Gravity, the fourth force, is left out. Currently gravity is best described by Einstein’s general theory of relativity, a classical theory that does not take quantum mechanics into account.
Efforts to bring quantum theory into the gravitational realm or vice versa have met with some difficulty. One reason is that interactions at a point cause the theory to “blow up”—in other words you get calculations with infinite results. By proposing that the fundamental objects of the theory are strings rather than point particles, interactions are spread out and the divergences associated with gravitational interactions disappear. In addition, a spin-2 state of the string naturally arises in string theory. It is known that the quantum of the gravitational field, if it exists, will be a massless spin-2 particle. Since this arises naturally in string theory, many people believe it is a strong candidate for a unified theory of all interactions.

Summary
Quantum field theory is a theoretical framework that unifies nonrelativistic quantum mechanics with special relativity. One consequence of this unification is that the types and number of particles can change in an interaction. As a result, the theory cannot be implemented using a single particle wave equation. The fundamental objects of the theory are quantum fields that act as operators, able to create or destroy particles.

Quiz
1. A quantum field (a) Is a field with quanta that are operators (b) Is a field parameterized by the position operator (c) Commutes with the Hamiltonian (d) Is an operator that can create or destroy particles
2. The particle generations (a) Are in some sense duplicates of each other, with each generation having increasing mass (b) Occur in pairs of three particles each (c) Have varying electrical charge but the same mass (d) Consists of three leptons and three quarks each

CHAPTER 1 Particle Physics and Special Relativity 21
3. In relativistic situations (a) Particle number and type is not fixed (b) Particle number is fixed, but particle types are not (c) Particle number can vary, but new particle types cannot appear (d) Particle number and types are fixed
4. In quantum field theory (a) Time is promoted to an operator (b) Time and momentum satisfy a commutation relation (c) Position is demoted from being an operator (d) Position and momentum continue to satisfy the canonical commutation relation
5. Leptons experience (a) The strong force, but not the weak force (b) The weak force and electromagnetism (c) The weak force only (d) The weak force and the strong force
6. The number of force-carrying particles is (a) Equivalent to the number of generators for the fields gauge group (b) Random (c) Proportional to the number of fundamental matter particles involved in the interaction (d) Proportional to the number of generators minus one
7. The gauge group of the strong force is: (a) SU(2) (b) U(1) (c) SU(3) (d) SU(1)
8. Antineutrinos (a) Have charge -1 and lepton number 0 (b) Have lepton number +1 and charge 0 (c) Have lepton number -1 and charge 0 (d) Are identical to neutrinos, since they carry no charge

22

Quantum Field Theory Demystified

9. The lightest family of elementary particles is (a) The electron, muon, and neutrino (b) The electron, up quark, and down quark (c) The electron, electron neutrino, up quark, and down quark (d) The electron and its antiparticle
10. The Higgs field (a) Couples the W and Z bosons to each other (b) Is a zero mass field (c) Has zero mass and charge +1 (d) Gives mass to elementary particles

CHAPTER 2
Lagrangian Field Theory

We begin our study of quantum field theory by building up some fundamental mathematical tools that can be applied to any fundamental physical theory. The main quantity of interest in this chapter is a function known as the Lagrangian, which is constructed by taking the difference of the kinetic and potential energies. In classical mechanics the Lagrangian is an equivalent method to Newtonian mechanics that can be used to derive the equations of motion. When Lagrangian methods are applied to fields, we can use the same techniques to derive the field equations.

Basic Lagrangian Mechanics

For now, we work in one spatial dimension x and consider the motion of a single

particle. Let T be the kinetic energy of the particle moving in a potential V. The

Lagrangian L is defined as

L = T − V

(2.1)

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24

Quantum Field Theory Demystified

The Lagrangian is a foundational concept which captures all the dynamics of the system and allows us to determine many useful properties such as averages and dynamic behavior.
Given L we can find the equations of motion from the Euler-Lagrange equations. For a single particle moving in one dimension, these are given by the single equation.

d ∂L − ∂L = 0

(2.2)

dt ∂x ∂x

where we have used a dot to denote differentiation with respect to time, that is,

x = dx dt

EXAMPLE 2.1

Consider

a

particle

of

mass

m

with

kinetic

energy

T

=

1 2

mx 2

moving

in

one

dimension

in a potential V(x). Use the Euler-Lagrange equations to find the equation of motion.

SOLUTION

We follow five basic steps. (1) First we write down the Lagrangian L, then we

calculate the derivatives we need for the Euler-Lagrange equation. (2) The first is

the derivative of L with respect to

x,

∂L ∂x

.

(3)

We

take

the

time

derivative

of

this

quantity,

d dt

(

∂L ∂x

).

(4)

Next

is

the

derivative

of

L

with

respect

to

x,

∂L ∂x

.

(5)

We

are

finally able to form the equation

d dt

(

∂L ∂x

)

=

∂L ∂x

.

Begin by writing the Lagrangian [step (1)] which is

L = 1 mx2 − V (x) 2

Next is the derivative of L with respect to x. When doing calculations involving the

Lagrangian, we treat x just as though it were an independent variable. For example,

∂ ∂x

( x )2

=

2 x

and

∂ ∂x

V

(

x

)

=

0.

Applying

this

[step

(2)]

we

get

∂L ∂x

=

∂ ∂x

⎡1 ⎣⎢ 2

mx 2

− V (x)⎤⎦⎥

=

mx

which is a momentum term: mass times velocity. Now we take the time derivative of this momentum [step (3)] which is

d dt

⎛ ⎝⎜

∂L ∂x

⎞ ⎠⎟

=

d dt

(mx)

=

mx

+

mx

=

mx

CHAPTER 2 Lagrangian Field Theory

25

and we have mass times acceleration. The remaining derivative is now a simple calculation [step (4)], that is,

∂L ∂x

=

∂ ∂x

⎡1 ⎣⎢ 2

mx 2

− V (x)⎤⎦⎥

=

−

∂V ∂x

Next [step (5)] is writing the equation which describes the dynamical behavior of the system, like

d dt

⎛ ⎝⎜

∂L ∂x

⎞ ⎠⎟

−

∂L ∂x

=

0

⇒

d dt

⎛ ⎝⎜

∂L ∂x

⎞ ⎠⎟

=

∂L ∂x

⇒

mx

=

−

∂V ∂ x

This elegant result is quite familiar. From classical mechanics, we recall that if a force F is conservative then

F = −∇V

In one dimension this becomes

F = − ∂V ∂ x

Therefore,

∂L ∂x

=

− ∂V ∂x

=

F. Using our

calculation

we

have

mx = − ∂V = F ∂ x

Hence we arrive at Newton’s second law,

F = mx = ma

where the acceleration a is given by a = d 2x /dt2 = x.
EXAMPLE 2.2 Consider a particle of mass m undergoing simple harmonic motion. The force on the particle is given by Hooke’s law F(x) = −kx. Determine the equation of motion by using the Euler-Lagrange equation.
SOLUTION Once again, the kinetic energy is given by
T = 1 mx2 2

26

Quantum Field Theory Demystified

We integrate the force F(x) = −kx to compute the potential and find that V = 1 k x2 2
Using Eq. (2.1) the Lagrangian is [step (1)]

L = T − V = 1 mx2 − 1 k x2

2

2

[step

(2)]

∂ ∂x

(L)

=

∂ ∂x

(

1 2

mx 2

−

1 2

kx2 )

=

mx.

As

above,

[step

(3)]

the

time

derivative

of

the

momentum

is

a

force

d dt

(mx)

=

mx

and

the

last

derivative

[step

(4)]

is

∂L ∂x

=

∂ ∂x

⎡1 ⎣⎢ 2

mx 2

−

1 2

kx 2

⎤ ⎦⎥

=

−kx

To obtain the equation of motion [step (5)] for the particle we use Eq. (2.2). We have

mx = −kx

This is the familiar equation of a simple harmonic oscillator, that is,

d2x dt 2

+

ω

2 0

x

=

0

where

the

natural

frequency

ω

2 0

=

k/m.

The Action and the Equations of Motion

If we integrate the Lagrangian with respect to time, we obtain a new quantity called the action which we denote by S

S = ∫ L dt

(2.3)

The action is functional because it takes a function as argument and returns a number. Particles always follow a path of least action. By varying (minimzing the variance of) the action, we can determine the path actually followed by a particle. Consider two fixed points x(t1) and x(t2 ). There are an infinite number of paths connecting these points. This means that there are an infinite number of paths for the particle to follow between these two points. The actual path that the particle follows is the path of least action. The path of least action represents a minimum. To find this path we

CHAPTER 2 Lagrangian Field Theory

27

minimize the variance of the action. We do this by describing the unknown action as minimal term and a variation.

S → S +δS

then the path followed by the particle is the one for which

δS = 0

(2.4)

This is the path with zero variation. This is the path of least action. Computing the variation δS leads to the equations of motion for the system. To
see how this works, we start with a simple example, returning to the derivation of Newton’s second law. We consider a small change in coordinates given by

x → x+ε

where ε is small. This variation is constrained to keep the end points fixed, that is,

ε(t1) = ε(t2 ) = 0

(2.5)

Using a Taylor expansion, the potential can be approximated as

The action then becomes

V (x + ε) ≈ V (x) + ε dV dx

∫ ∫ ∫ S =

t2 Ldt =
t1

t2 1 mx2 − V (x)dt = 2 t1

t2 t1

1 2

m(x

+

ε)2

−

⎛⎝⎜ V

(x)

−

ε

dV dx

⎞⎠⎟

dt

Now we expand out the first term, the kinetic energy term, giving

(x + ε)2 = x2 + 2xε + ε2 ≈ x2 + 2xε
We dropped the ε 2 term, since by assumption ε is small, so squaring it gives a term we can neglect. That is, we only keep the leading order terms. So we have

∫ S =

t2 t1

1 2

m(x2

+

2xε)

−

⎛⎝⎜V (x)

+

ε

dV dx

⎞⎠⎟

dt

This expression can be written in a more useful fashion with some manipulation. The idea is to isolate the terms containing ε. We can do this by applying integration

28

Quantum Field Theory Demystified

by parts to the 2xε term, transferring the time derivative from ε to x. First let’s recall the formula for integration by parts

∫ ∫ t2

dg f (t) dt

=

f (t)g(t)

t2

−

t2 g(t) df dt

t1

dt

t1

t1

dt

(2.6)

In our case,

f (t) = x and

dg dt

=

ε.

Recalling

Eq.

(2.5),

the

fact

that

the

variation

vanishes at the end points of the interval, the boundary term vanishes in our case.

Hence,

∫ ∫ ∫ t2

1 m(x

+

ε)2 dt

=

1

m

t2 (x + ε)2 dt ≈ 1 m

t2 (x2 + 2xε)dt

2 t1

2 t1

2 t1

∫ ∫ ∫ t2 1 m(x2 + 2xε) dt = t2 1 mx2dt − t2 m xε dt

2 t1

2 t1

t1

We can now collect the ε terms

∫ ∫ ∫ S =

t2 1 mx2dt − 2 t1

t2 m xε dt −
t1

t2 t1

⎛⎝⎜V (x)

+

ε

dV dx

⎠⎟⎞

dt

∫ ∫ =

t2 1 mx2 − V(x)dt + 2 t1

t2 t1

⎝⎜⎛

−mx

−

dV dx

⎞⎠⎟

ε dt

= S +δS

The requirement that δS = 0 can be satisfied only if the integral of the second term is 0. Since the end points are arbitrary, the integrand must be identically 0, that is,

mx + dV = 0 ⇒ mx = − dV

dx

dx

Now let’s consider a more general situation where there are N generalized coordinates qi (t) where i = 1,…, N. Considering a Lagrangian expressed in terms of these coordinates and their first derivatives only, we have an action of the form
∫ S = t2 L(q, q) dt t1
In this case the system evolves from some initial point q1 = q(t1) to some final point q2 = q(t2 ). To find the trajectory followed by the system, we apply the principle of

CHAPTER 2 Lagrangian Field Theory

29

least action and solve δ S = 0. Once again, the end points of the trajectory are fixed so that
δq(t1) = δq(t2 ) = 0

Also note that

δq(t) = d (δq) dt

(2.7)

Then,

∫ δ S = δ t2 L(q, q) dt t1

∫ ∑ =

t2 t1

i

⎡ ⎢ ⎣

∂L ∂qi

δ

qi

+

∂L ∂qi

δ qi

⎤ ⎥ ⎦

dt

∫ ∑ =

t2 t1

i

⎡ ∂L

⎢ ⎣

∂qi

δ qi

+

∂L ∂qi

d dt

(δ

qi

)

⎤ ⎥ ⎦

dt

To move from the second to the third line, we applied Eq. (2.7). Now we use integration by parts on the second term, giving

∫ ∑ δS =

t2 t1

i

⎡ ⎢ ⎣⎢

∂L ∂qi

δ

qi

−

d dt

⎛ ⎝⎜

∂L ∂qi

⎞ ⎠⎟

δ

qi

⎤ ⎥ ⎦⎥

dt

For this expression to vanish, since δqi is arbitrary each coordinate index must satisfy

∂L ∂qi

−

d dt

⎛ ⎝⎜

∂L ∂qi

⎞ ⎠⎟

=

0

(2.8)

which are of course the Euler-Lagrange equations and i = 1,…, N. Therefore, we see that the principle of least action gives rise to the Euler-Lagrange equations. Hence the Lagrangian satisfies the Euler-Lagrange equation independently for each coordinate.

Canonical Momentum and the Hamiltonian

In Examples 2.1 and 2.2 we saw that the derivative of the kinetic energy with respect to velocity was the momentum: a completely classical result, that is,

∂ ∂x

(L)

=

∂ ∂x

(T

−V

)

=

∂ ∂x

⎛⎝⎜

1 2

mx2 ⎞⎠⎟

=

mx

30

Quantum Field Theory Demystified

This result can be made more general and more useful. The canonical momentum is defined as

pi

=

∂L ∂qi

(2.9)

This allows us to define a Hamiltonian function, which is given in terms of the Lagrangian and the canonical momentum as

∑ H( p, q) = piqi − L
i

(2.10)

Lagrangian Field Theory

Now that we have reviewed the basics of the Lagrangian formalism, we are prepared to generalize these techniques and apply them to fields, that is, functions on spacetime ϕ(x,t) which we write more compactly as ϕ(x). In the continuous case, we actually work with the Lagrangian density.

L = T −V = ∫ L d3x

(2.11)

The action S is then the time integral of this expression.

S = ∫ dt L = ∫ L d 4x

(2.12)

Typically, the Lagrangians encountered in quantum field theory depend only on the fields and their first derivatives.

L = L(ϕ, ∂μϕ) L → L(ϕ,∂μϕ)

(2.13)

Moreover we are interested in fields that are local, meaning that at a given spacetime
point x, the Lagrangian density depends on the fields and their first derivatives
evaluated at that point.
Now we apply the principle of least action to Eq. (2.12). We vary the action with respect to the field ϕ(x) and with respect to the first derivative of the field ∂μϕ (x) as follows:

0 =δS

=δ∫d4x L

∫ =

d

4

x

⎧ ⎨ ⎩

∂L ∂ϕ

δϕ

+

∂L ∂[∂uϕ

]

δ

(∂μϕ

⎫ )⎬ ⎭

CHAPTER 2 Lagrangian Field Theory

31

Now we use the fact that
δ (∂μϕ) = ∂μ (δϕ)
and apply integration by parts to the second term in this expression. The boundary terms vanish because the end points are fixed and the second term becomes

∫ ∫ d

4

x

∂L ∂[∂μϕ

]

δ

(∂

μϕ

)

=

d

4

x

∂L ∂[∂μϕ

]

∂

μ

(δϕ

)

∫ = −

d

4

x∂μ

⎛ ⎝⎜

∂L ∂[∂μϕ

⎞ ] ⎠⎟

δϕ

All together, the variation of the action is then

∫ 0 = δ S =

d

4

x

⎧⎪ ⎨ ⎩⎪

∂L ∂ϕ

−

∂μ

⎛ ⎝⎜

∂L ∂[∂μϕ

⎞ ] ⎠⎟

⎫⎪ ⎬ ⎭⎪

δϕ

What does it mean when an integral is 0? There are two cases: Either there is cancellation because the integrand takes positive and negative values or the integrand is 0 over the entire domain of integration. In this case the domain of integration can vary, so we can’t rely on cancellation and we know the integrand must be 0. That is, the term inside the braces vanishes. This gives us the Euler-Lagrange equations for a field j.

∂L ∂ϕ

−

∂μ

⎛ ⎝⎜

∂L ⎞ ∂[∂ μϕ ] ⎠⎟

=

0

(2.14)

The Einstein summation convention is in force, so there is an implied sum on the second term. That is,

∂μ

⎛ ⎝⎜

∂L ⎞ ∂[∂ μϕ ] ⎠⎟

=

∂t

⎛ ⎝⎜

∂L ⎞ ∂[∂tϕ ] ⎠⎟

−

∂x

⎛ ⎝⎜

∂L ⎞ ∂[∂ xϕ ] ⎠⎟

−

∂y

⎛ ⎝⎜

∂L ⎞ ∂[∂ yϕ ] ⎠⎟

−

∂z

⎛ ⎝⎜

∂L ⎞ ∂[∂ zϕ ] ⎠⎟

The canonical momentum density for the field is given by π (x) = ∂L ∂ϕ

(2.15)

32

Quantum Field Theory Demystified

just as in the classical case. The Hamiltonian density is then H = π (x)ϕ (x) − L
To obtain the Hamiltonian, we integrate this density over space
H = ∫ Hd3x

(2.16) (2.17)

EXAMPLE 2.3 Find the equation of motion and the Hamiltonian corresponding to the Lagrangian

{ } L = 1 2

(∂μϕ)2 − m2ϕ 2

SOLUTION We can find the field equations with a straightforward application of Eq. (2.14). Let’s follow the five-step process from the previous examples. Since we are given the Lagrangian, step (1) is done. The remaining steps require some insight.

The trick in doing the problems is that we treated ∂μϕ as a variable. If you need a

crutch to get used to thinking this way, as an analogy think of ϕ as x and ∂μϕ as y.

Then

∂ ∂ϕ

[

1 2

(∂μϕ)2 ]

is

like

calculating

[∂ 1
∂x 2

(y)2 ] =

0.

Now

we

are

ready

for

step

(2),

compute

∂ ∂ (∂μϕ )

(L).

Begin by expanding the first term in the Lagrangian:

(∂μϕ)2 = (∂μϕ)(∂μϕ) = (∂μϕ)gμν (∂νϕ)

Hence, we have

{ } ∂L
∂[∂ μϕ ]

=

∂⎛ ∂[∂μϕ] ⎝⎜

1 2

(∂μϕ)2 − m2ϕ 2

⎞⎠⎟

=

1 2

∂ ∂[∂ μϕ ]

(∂μϕ )g μv

(∂vϕ )

−

1 2

∂ ∂[∂ μϕ ]

(m2ϕ 2 )

CHAPTER 2 Lagrangian Field Theory

33

Using our observation that we simply act like ∂μϕ is a variable when computing these derivatives, it is clear that

∂ (m2ϕ 2 ) = 0 ∂[∂ μϕ ]

This means we are left with

∂L ∂[∂ μϕ ]

=

1 2

∂ ∂[∂μϕ

]

(∂μϕ

)g

μν

(∂νϕ )

But

∂ ∂[∂ μϕ ]

(∂ μϕ )g μν

(∂νϕ )

=

g μν

(∂νϕ )

+

g μν

(∂μϕ )

=

2∂μϕ

Meaning that

∂L = ∂μϕ ∂[∂ μϕ ] Step (3) entails taking the derivative of this result.

∂μ

⎛ ⎝⎜

∂L ⎞ ∂ [∂ μϕ ] ⎠⎟

=

∂μ (∂ μϕ)

Next, step (4) is the easiest.

{ } ∂L
∂ϕ

=

∂ ∂ϕ

⎡ ⎣⎢

1 2

(∂μϕ)2 − m2ϕ 2

⎤ ⎦⎥

=

∂ ∂ϕ

⎡1 ⎣⎢ 2

(∂μϕ)2

⎤ ⎦⎥

−

∂ ∂ϕ

1 (m2ϕ 2) 2

= − ∂ 1 (m2ϕ 2 ) = −m2ϕ ∂ϕ 2

34

Quantum Field Theory Demystified

Finally step (5) is to form the equations of motion.

0

=

∂L ∂ϕ

−

∂μ

⎛ ⎝⎜

∂L ⎞ ∂[∂ μϕ ] ⎠⎟

= −m2ϕ − ∂μ (∂μϕ)

Using

∂μ∂μ

=

∂2 ∂t 2

− ∇2,

the

field

equations

corresponding

to

the

Lagrangian

given

in

this problem can be written as

∂2φ ∂t 2

−

∇2ϕ

+

m 2ϕ

=

0

EXAMPLE 2.4

Derive the sine-Gordon equation

∂2ϕ ∂t 2

−

∂2ϕ ∂x 2

+ sinϕ

=

0

from the Lagrangian

{ } L = 1 2

(∂tϕ)2 − (∂xϕ)2

+ cosϕ

SOLUTION First we calculate

{ } ∂L
∂ϕ

=

∂ ∂ϕ

⎡ ⎣⎢

1 2

(∂tϕ)2 − (∂xϕ)2

+

cos

ϕ

⎤ ⎦⎥

{ } =

∂ ∂ϕ

⎡ ⎣⎢

1 2

(∂tϕ)2 − (∂xϕ)2

⎤ ⎦⎥

+

∂ ∂ϕ

cos

ϕ

= ∂ (cosϕ) = − sinϕ ∂ϕ

The Lagrangian only has one spatial coordinate, so

∂μ

⎛ ⎝⎜

∂L ⎞ ∂[∂ μϕ ] ⎠⎟

=

∂t

⎛ ⎝⎜

∂L ⎞ ∂[∂tϕ ] ⎠⎟

−

∂x

⎛ ⎝⎜

∂L ⎞ ∂[∂ xϕ ] ⎠⎟

Now tackling the time and space derivatives separately leads us to

{ } ∂L
∂[∂tϕ ]

=

∂ ∂[∂tϕ ]

⎡1 ⎣⎢ 2

(∂tϕ)2 − (∂xϕ)2

+

cosϕ

⎤ ⎦⎥

=

∂tϕ

{ } ∂L
∂[∂ xϕ ]

=

∂ ∂[∂ xϕ ]

⎡1 ⎣⎢ 2

(∂tϕ)2 − (∂xϕ)2

+

cosϕ

⎤ ⎦⎥

=

∂xϕ

CHAPTER 2 Lagrangian Field Theory

35

and so

∂μ

⎛ ⎝⎜

∂L ⎞ ∂[∂ μϕ ] ⎠⎟

=

∂t

⎛ ⎝⎜

∂L ⎞ ∂[∂tϕ ] ⎠⎟

−

∂x

⎛ ⎝⎜

∂L ⎞ ∂[∂ xϕ ] ⎠⎟

=

∂t

(∂tϕ )

−

∂x

(∂ xϕ )

= ∂2ϕ − ∂2ϕ = ϕ ∂t2 ∂x2

Therefore, the equation of motion in this case is

ϕ + sinϕ = ( + sin)ϕ = 0

This equation is called the sine-Gordon equation, due to its resemblance to the Klein-Gordon equation.

Symmetries and Conservation Laws

A symmetry is a change in perspective that leaves the equations of motion invariant. For example, the change could be a translation in space, a change in time or a rotation. These are external symmetries, that is, they depend upon changes in spacetime. There also exist internal symmetries, changes in the fields that do not involve changes with respect to spacetime at all.
A famous theorem in classical mechanics which is very important is known as Noether’s theorem. This theorem allows us to relate symmetries to conserved quantities like charge, energy, and momentum. Mathematically, a symmetry is some kind of variation to the fields or the Lagrangian that leaves the equations of motion invariant. We will see how to make such a change and then deduce a conserved quantity.
Two of the most fundamental results of physics, conservation of energy and momentum, are due to a symmetry that results from a small displacement in spacetime. That is we let the spacetime coordinates vary according to

xμ → xμ + aμ

(2.18)

where aμ is a small and arbitrary parameter describing a displacement in spacetime. Expanding in Taylor, the field changes according to

ϕ(x) → ϕ(x + a) = ϕ(x) + aμ∂μϕ

(2.19)

Under a small variation (perturbation), the field can be described as ϕ → ϕ + δϕ

36

Quantum Field Theory Demystified

This means we can write the variation in the field explicitly as

δϕ = aμ∂μϕ

(2.20)

Now let’s reconsider the variation of the Lagrangian. We have, in the case of a Lagrangian depending only on the field and its first derivatives,

δL

=

∂L ∂ϕ

δϕ

+

∂L ∂(∂ μϕ )

δ (∂μϕ)

From the Euler-Lagrange equations [Eq. (2.14)] we know that

∂L ∂ϕ

=

∂μ

⎛ ⎝⎜

∂L ⎞ ∂[∂μϕ ] ⎠⎟

Hence the variation in the Lagrangian can be written as

δ

L

=

∂μ

⎛ ⎝⎜

∂L ∂[∂μϕ

⎞ ]⎠⎟

δϕ

+

∂L ∂(∂ μϕ )

δ

(∂μϕ

)

=

∂μ

⎛ ⎝⎜

∂L ⎞ ∂[ ∂μϕ]⎠⎟

δϕ

+

∂L ∂( ∂μϕ)

∂

(δϕ)
μ

Now we have an expression that can be written as a total derivative. Remember the product rule from ordinary calculus that ( fg)′ = f ′g + g′f . We take

f = ∂L ∂ [∂μϕ ]

g = δϕ

Allowing us to write

δL

=

∂μ

⎛ ⎝⎜

∂

∂L [∂μϕ

]

δϕ

⎞ ⎠⎟

=

(

fg )′

Now we can apply Eq. (2.20). Note that the same index can only be used twice in a single
expression, so we need to change the label used in Eq. (2.20) to another dummy index, say δϕ = aμ∂μϕ = aν∂νϕ. The quantities involved are just ordinary scalars, so we can also move them around and write δϕ = ∂νϕaν . So, we arrive at the following expression

δ

L

=

∂μ

⎛ ⎝⎜

∂L ∂[∂μϕ

]

∂νϕ

⎞ ⎠⎟

a ν

CHAPTER 2 Lagrangian Field Theory

37

Equivalently, we can also write the variation of the Lagrangian as δ L = ∂μ (L)aμ = δνμ∂μ (L)aν

That is, we consider how it varies directly with respect to the displacement Eq. (2.18). Equating these two results gives

δL

=

δνμ∂μ (L)aν

=

∂μ

⎛ ⎝⎜

∂L ∂[∂μϕ

]

∂νϕ

⎞ ⎠⎟

a ν

Moving both terms to the same side of the equation gives

∂μ

⎛ ⎝⎜

∂L ∂[∂μϕ

]

∂νϕ

− δνμL⎞⎠⎟

aν

=

0

Now, recall that aν is arbitrary. So in order for this expression to vanish, the derivative must be zero. That is,

∂μ

⎛ ⎝⎜

∂L ∂[∂ μϕ ]

∂νϕ

− δνμL⎞⎠⎟

=

0

This expression is so important that we give this quantity its own name. It turns out this is the energy-momentum tensor. We write this as

Tνμ

=

∂L ∂[∂μϕ

]

∂νϕ

− δνμL

(2.21)

Hence the conservation relation expressed by zero total divergence is

∂μTνμ = 0

(2.22)

Notice that

T00

=

∂L ϕ ∂ϕ

−

L

=

H

(2.23)

That is, T00 is nothing other than the Hamiltonian density: it’s the energy density and the equation ∂0T00 = 0 reflects conservation of energy. The components of momentum density are given by Ti0 where i runs over the spatial indices. The components of
momentum of the field are found by integrating each of these terms over space, that is,

∫ Pi = d 3x Ti0

(2.24)

38

Quantum Field Theory Demystified

Conserved Currents
Now let’s go through the process used in the last section again to see how Noether’s theorem can be applied to derive a conserved current and an associated conserved charge. We let the field vary by a small amount

ϕ → ϕ + δϕ

(2.25)

We then start from the premise that under this variation, the Lagrangian does not change. The variation in the Lagrangian due to Eq. (2.25) will be of the form

L→ L+δL

(2.26)

So what we mean is that

δL = 0

(2.27)

Now following the usual procedure the variation of the Lagrangian due to a variation in the field will be

δL

=

∂L ∂ϕ

δϕ

+

∂L ∂(∂ μϕ )

∂μ

(δϕ )

Once again, from the Euler-Lagrange equations [Eq. (2.14)] we can write

Therefore, we have

∂L ∂ϕ

=

∂μ

⎛ ⎝⎜

∂L ⎞ ∂[∂μϕ ] ⎠⎟

δ

L

=

∂μ

⎛ ⎝⎜

∂L ∂[∂μϕ

⎞ ]⎠⎟

δϕ

+

∂L ∂[∂μϕ

]

∂μ

(δϕ )

=

∂μ

⎛ ⎝⎜

∂L ∂[∂ μϕ ]

δϕ

⎞ ⎠⎟

Since we are operating under the premise that the variation in the field does not change the Lagrangian [Eq. (2.27)], this leads us to the result

∂μ

⎛ ⎝⎜

∂L ∂[∂μϕ

]

δϕ

⎞ ⎠⎟

=

0

(2.28)

CHAPTER 2 Lagrangian Field Theory

39

We call the quantity in the parentheses a conserved current. In analogy with electrodynamics we denote it with the letter J and write

J μ = ∂L δϕ ∂[∂ μϕ ]

(2.29)

The conservation law Eq. (2.28) then can be written as

∂μJμ = 0

(2.30)

This is the central result of Noether’s theorem:
• For every continuous symmetry of the Lagrangian—that is, a variation in the field that leaves form of the Lagrangian unchanged—there is a conserved current whose form can be derived from the Lagrangian using Eq. (2.29).
There is a conserved charge associated with each conserved current that results from a symmetry of the Lagrangian. This is found by integrating the time component of J:

∫ Q = d 3x J 0

(2.31)

We see that the translation symmetry in spacetime worked out in the previous section that led to the energy-momentum tensor is a special case of Noether’s theorem, with the conserved “charges” being energy and momentum.

The Electromagnetic Field

The Maxwell or electromagnetic field tensor is given by

⎛ 0 −Ex −Ey −Ez ⎞

F μν

= ∂μ Aν

− ∂ν Aμ

=

⎜ ⎜

Ex

⎜ Ey

0 Bz

− Bz

By

⎟ ⎟

0 −Bx ⎟

⎝⎜ Ez − By Bx 0 ⎠⎟

(2.32)

The Aμ is the usual vector potential, but this is a 4-vector whose time component is
the scalar potential and whose spatial component is the usual vector potential used to write down the magnetic field, that is, Aμ = (ψ , A). It can be shown that F μν

40

Quantum Field Theory Demystified

satisfies or leads to Maxwell’s equations. Note that F μν is antisymmetric; the sign flips when the indices are interchanged. That is,

F μν = −Fνμ

(2.33)

Without sources, the homogeneous Maxwell’s equations can be written in terms of the electromagnetic field tensor as

∂λ F μν + ∂ν F λμ + ∂μ Fνλ

(2.34)

Meanwhile, the inhomogeneous Maxwell’s equations can be written as

∂μ F μν − Jν = 0

(2.35)

where Jν are the current densities. It is an instructive exercise to derive Maxwell’s equations using a variational procedure so that you can learn how to work with tensors of higher order, that is, vector fields. In the next example we derive Eq. (2.35) from a Lagrangian.

EXAMPLE 2.5

Show

that

the

Lagrangian

L

=

−

1 4

Fμν F μν

−

Jμ

Aμ

leads

to

the

inhomogeneous

Maxwell’s equations [Eq. (2.35)] if the potential Am is varied, leaving the current

density constant.

SOLUTION We begin as usual by writing the action as an integral of the Lagrangian density. The action in this case is

∫ S =

d4x

⎛⎝⎜

−

1 4

Fμν F μν

−

Jμ

Aμ

⎞⎠⎟

The variation we will compute is δ Aμ . We have

∫ δ S =

d4x

⎛⎝⎜

−

1 4

(δ Fμν

)F μν

−

1 4

Fμν

(δ F μν

)

−

J μδ

Aμ

⎞⎠⎟

Let’s consider the first term. Using the definition of the electromagnetic field tensor Eq. (2.32), we can write Fμν = ∂μ Aν − ∂ν Aμ and so we have

−

1 4

(δ

Fμν

)F

μν

=

−

1 4

(∂μδ

Aν

− ∂νδ Aμ )F μν

CHAPTER 2 Lagrangian Field Theory

41

Now we integrate by parts, transferring the derivatives from the δ Aν terms to F μν. This allows us to write

−

1 4

(δ Fμν )F μν

=

−

1 4

(∂μδ Aν

−

∂νδ Aμ )F μν

=

1 4

(∂μ F μνδ

Aν

−

∂ν F μνδ

A μ

)

But, repeated indices are dummy indices. So let’s swap μ and ν in the second term, and write this as

1 4

(∂μ F μνδ

Aν

−

∂μ Fνμδ

Aν )

Now we use the antisymmetry of the electromagnetic tensor under interchange of the indices in Eq. (2.23). This will get rid of the minus sign on the second term, giving

1 4

(∂μ F μνδ

Aν

−

∂μ Fνμδ

Aν )

=

1 4

(∂μ F μνδ

Aν

+

∂μ F μνδ

Aν )

=

1 2

∂μ F μνδ Aν

So we have the result

−

1 4

(δ

Fμν

)F

μν

=

1 2

∂

μ

F

μνδ

A ν

Now

let’s

tackle

the

next

term

−

1 4

Fμν (δ F μν ).

In

this

case

we

have

(2.36)

−

1 4

Fμν (δ F μν )

=

−

1 4

Fμνδ (∂μ Aν

−

∂ν

Aμ )

=

−

1 4

Fμν (∂μδ

Aν

−

∂νδ

Aμ )

We’re going to have to lower and raise some indices using the metric (see Chap. 1) to get this in the form of Eq. (2.36). The first step is to raise the indices on the field tensor term

−

1 4

Fμν (∂μδ

Aν

−

∂νδ

Aμ )

=

−

1 4

gμρ gνσ F ρσ

(∂μδ Aν

−

∂νδ

Aμ )

42

Quantum Field Theory Demystified

Now let’s move F ρσ inside the parentheses and integrate by parts to transfer the derivative onto it

−

1 4

gμρ gνσ F ρσ

(∂μδ

Aν

−

∂νδ

Aμ )

=

1 4

gμρ gνσ [∂μ (F ρσ

)δ

Aν

−

∂ν (F ρσ

)δ

Aμ ]

=

1 4

gμρ gνσ ∂μ (F ρσ

)δ

Aν

−

1 4

gμρ gνσ ∂ν (F ρσ

)δ

A μ

Now we lower the indices on the derivative operators to give

=

1 4

gμρ gνσ ∂μ (F ρσ

)δ

Aν

−

1 4

gμρ gνσ ∂ν (F ρσ

)δ

Aμ

=

1 4

gνσ ∂ρ (F ρσ

)δ

Aν

−

1 4

gμρ ∂σ

(F ρσ

)δ

Aμ

Next, do the same thing to the vector potential terms

1 4

gνσ ∂ρ (F ρσ

)δ

Aν

−

1 4

gμρ ∂σ

(F ρσ

)δ

Aμ

=

1 4

∂ρ (F ρσ

)δ

Aσ

−

1 4

∂σ

(F ρσ

)δ

A ρ

Once again, repeated indices are dummy indices so we can change the labels. Focusing on the second term, let’s change ρ → ν, σ → μ. We obtain

1 4

∂ρ (F ρσ )δ Aσ

−

1 4

∂σ (F ρσ )δ Aρ

=

1 4

∂ρ (F ρσ )δ Aσ

−

1 4

∂μ (Fνμ )δ Aν

Now apply the antisymmetry of the electromagnetic tensor to the second term, to get rid of the minus sign as shown here.

1 4

∂ρ (F ρσ )δ Aσ

−

1 4

∂μ (Fνμ )δ Aν

=

1 4

∂ρ (F ρσ )δ Aσ

+

1 4

∂μ (Fνμ )δ Aν

CHAPTER 2 Lagrangian Field Theory

43

The relabeling procedure can also be applied to the first term. This time we let ρ → μ, σ → ν and we get

1 4

∂ρ (F ρσ )δ Aσ

+

1 4

∂μ (F μν )δ Aν

=

1 4

∂μ (F μν )δ Aν

+

1 4

∂μ (F μν )δ Aν

=

1 2

∂μ

(F μν

)δ

Aν

We can combine this result with Eq. (2.36), and we see that the variation in the action becomes

∫ δ S =

d4x

⎛⎝⎜

1 2

∂μ F μνδ

Aν

+

1 2

∂μ F μνδ

Aν

−

J μδ

Aμ

⎞⎠⎟

∫= d 4 x(∂μ F μν − J μ )δ Aν

We require that the variation in the action vanish, that is, δ S = 0. Since the variation is arbitrary, δ Aν cannot vanish. Once more we arrive at the conclusion that the integral will be 0 only if the integrand is 0 everywhere in the domain. This means
the action will only vanish if Maxwell’s equations are satisfied, that is,

∂μ F μν − J μ = 0

Gauge Transformations
In this section we will consider an extension of the idea of invariance, by introducing what is known as a gauge transformation. Here we’re only going to provide a brief introduction to these ideas; they will be elaborated as we proceed through the book.
The idea of a gauge transformation follows from studies of electricity and magnetism where we can make changes to the scalar and vector potentials ψ and A without changing the field equations and hence the physical fields E and B themselves. For example, recall that the magnetic field B can be defined in terms of the vector potential A using the curl relation
B=∇×A
A rule from vector calculus tells us that ∇i(∇ × F) = 0 for any vector field F. Hence the Maxwell’s equation ∇i B = 0 is still satisfied when we make the

44

Quantum Field Theory Demystified

definition B = ∇ × A. Now let f be some scalar function and define a new vector potential A′via
A′ = A + ∇f
We also know from vector calculus that ∇ × ∇f = 0. Hence we can add a term of the form ∇f to the vector potential with impunity if it’s mathematically convenient. The magnetic field B is unchanged since

B = ∇ × A′ = ∇ × (A + ∇f ) = ∇ × A + ∇ × ∇f = ∇ × A

Therefore the magnetic field, the physical quantity of interest, is unchanged by a

transformation of the form A′ = A + ∇f. We call this type of transformation in

electrodynamics a gauge transformation. There are different choices that can be

made when implementing a gauge transformation. For example, if we impose the

requirement that ∇iA = 0, we call this the Coulomb gauge. On the other hand, if

∇iA =

− μ0ε0

∂ψ ∂t

,

we

have

what

is

known

as

the

Lorentz

gauge.

In field theory, we arrive at a similar notion by considering transformations to the

field that leave the Lagrangian invariant. To see how this works in field theory, let’s

consider a simple example, the Klein-Gordon Lagrangian with a complex field.

L = ∂μϕ †∂μϕ − m2ϕ†ϕ

(2.37)

Let U be a unitary transformation applied to the fields such that U does not, in any way, depend on spacetime. That is, we let

ϕ → Uϕ

(2.38)

Then

ϕ † → ϕ †U †

(2.39)

Since the transformation is unitary, we also know that UU† = U†U = 1. Let’s see how this transformation affects the Lagrangian [Eq. (2.37)]. Looking at each term individually, we start with the first term where we have

∂μϕ†∂μϕ → ∂μ (ϕ†U † )∂μ (Uϕ)

CHAPTER 2 Lagrangian Field Theory

45

But U does not depend on spacetime in any way, so the derivative operators do not affect it. Hence
∂μ (ϕ†U † ) ∂μ (Uϕ) = ∂μ (ϕ† )(U †U )∂μ (ϕ) = ∂μϕ†∂μϕ
Similarly for the second term in Eq. (2.37), we have
m2ϕ†ϕ → m2 (ϕ†U † )(Uϕ) = m2ϕ† (U †U )ϕ = m2ϕ†ϕ
Therefore we see that under the transformation [Eq. (2.38)], the Lagrangian [Eq. (2.37)] is invariant. Since U is a constant, we can write it in the form
U = eiΛ
where Λ is a constant. However, in certain contexts, Λ can also be a matrix so long as it is Hermitian. Since it is a constant we say that the gauge transformation in this case is global, it does not depend on spacetime in any way.
LOCAL GAUGE TRANSFORMATIONS
The gauge transformations that are of interest are local transformations that do depend on spacetime. This type of transformation satisfies the requirements of special relativity—that no signal can travel faster than the speed of light.
Let’s return to the transformation ϕ → Uϕ. In the following, we still consider U to be a unitary transformation, however now we let it depend on spacetime so that U = U(x). This means that terms like ∂μU will not vanish. Next consider how the Lagrangian is changed by the transformation ϕ → Uϕ. We use the same Lagrangian we considered in the previous section, namely L = ∂μϕ†∂μϕ − m2ϕ†ϕ. This time the second term in the Lagrangian remains invariant as shown here.
m2ϕ†ϕ → m2ϕ†U † (x)U(x)ϕ = m2ϕ†ϕ
The first term will change due to the spacetime dependence of U = U(x), that is,
∂μϕ † → ∂μ (ϕ †U † ) = (∂μϕ † )U † + ϕ †∂μ (U † )
Similarly we find that
∂μϕ → ∂μ (Uϕ) = (∂μU)ϕ + U∂μ (ϕ)

46

Quantum Field Theory Demystified

We can write this in a more useful form by exploiting the fact that U is unitary, like

∂μϕ → ∂μ (Uϕ) = (∂μU)ϕ + U∂μ (ϕ) = UU † (∂μU )ϕ + U∂μ (ϕ) = U[∂μϕ + (U †∂μU )ϕ ]
To maintain invariance, we would like to cancel the extra term that has shown up here. In other words we want to get rid of
(U †∂μU)ϕ
We can do this by introducing a new object, a spacetime dependent field Aμ = Aμ (x) called the gauge potential. It is given the label Aμ due to the analogy with electrodynamics—we are introducing a hidden field to keep the form of the Lagrangian invariant. As we will see later this has dramatic consequences, and this is one of the most important techniques in quantum field theory.
We introduce a covariant derivative that acts on the field as

Dμϕ = ∂μϕ − iAμϕ

(2.40)

(for readers of Relativity Demystified, notice the similarity to general relativity). Remember what the word covariant means—the form of the equations won’t change. We introduce this derivative operator to keep the Lagrangian invariant under a local gauge transformation. Recalling the effect of a global gauge transformation ϕ → Uϕ, we found that

∂μϕ → U∂μϕ
The covariant derivative Eq. (2.40) will allow us to recover this result in the case of the local gauge transformation. That is, ϕ → U(x)ϕ will lead to Dμϕ → U(x)Dμϕ. This can be accomplished if we define Aμ such that it obeys the similarity transformations.

Aμ → UAμU † + iU∂μU †

(2.41)

Note that some authors use the semicolon notation Dμϕ = ϕ ;μ to represent the covariant derivative.

EXAMPLE 2.6
Consider a charged scalar particle of mass m with charge q and describe a suitable modification of the derivative operator ∂μ → ∂μ + qAμ that will yield a Lorentz invariant, real Lagrangian.

CHAPTER 2 Lagrangian Field Theory

47

SOLUTION The field equations for a complex field corresponding to a charged scalar particle are obtained using the covariant derivative in Eq. (2.40), in this case with the form
i∂μ → i∂μ − qAμ
The Lagrangian is L = (i∂μ + qAμ )ϕ* (i∂μ − qAμ )ϕ − m2ϕ*ϕ
Variation of this Lagrangian leads to the equation of motion as shown here.
(i∂μ − qAμ )2ϕ − m2ϕ = 0

Summary
The Lagrangian is the difference of kinetic and potential energy given by L = T − V. It can be used to obtain the equations of motion for a system by applying variational calculus to the action S which is the integral of the Lagrangian. When extended to continuous systems, these techniques can be applied to fields to obtain the field equations. For problems with expected symmetries or conserved quantities, we require that the Lagrangian remains invariant under corresponding transformations. When a given transformation leaves the form of the Lagrangian invariant, we call that transformation a symmetry. Noether’s theorem allows us to derive conservation laws, including conserved charges and currents from symmetries in the Lagrangian. Symmetries can be local, meaning that they are spacetime dependent, or they can be internal symmetries that are intrinsic to the system at hand. To maintain covariance of the equations, a covariant derivative must be introduced which necessitates the use of a gauge potential.

Quiz

1. Find the equation of motion for a forced harmonic oscillator with Lagrangian

L = 1 mx2 − 1 mω 2x2 + α x

2

2

Here α is a constant.

48

Quantum Field Theory Demystified

2. Consider a Lagrangian given by

L

=

−

1 2

∂μϕ

∂μϕ

−

1 2

m2ϕ

2

−

V

(ϕ )

(a) Write down the field equations for this system. (b) Find the canonical momentum density π (x).

(c) Write down the Hamiltonian.
3. Consider a free scalar field with Lagrangian L = ∂μϕ ∂μϕ and suppose that the field varies according to ϕ → ϕ + α, where α is a constant. Determine the conserved current.

4. Refer to the Lagrangian for a complex scalar field Eq. (2.37). Determine the equations of motion obeyed by the fields ϕ and ϕ†.

5. Refer to Eq. (2.37) and calculate the conserved charge.

∫ 6.

Consider the action

S

=

1 4

Fμν F μνd 4 x. Vary the potential according to

Aμ → Aμ + ∂μϕ where ϕ is a scalar field. Determine the variation in

the action.

CHAPTER 3
An Introduction to Group Theory
An abstract branch of mathematics called group theory plays a fundamental role in modern particle physics. The reason it does so is because group theory is related to symmetry. For example, an important group in physics is the rotation group, which is related to the fact that the laws of physics don’t change if you rotate your frame of reference. In general what we are after is a set of equations, or laws of physics, that keep the same mathematical form under various transformations. Group theory is related to this type of symmetry.
Defi nitions
We begin our discussion of group theory in a somewhat abstract manner, but later we will get to some more concrete material. Let’s state what a group is and the four properties it must have.
Copyright © 2008 by The McGraw-Hill Companies, Inc. Click here for terms of use.

50

Quantum Field Theory Demystified

A group G is a set of elements {a, b, c, . . .} which includes a “multiplication” or composition rule such that if a ෈ G and b ෈ G, then the product is also a member of the group, that is,

ab ෈ G

(3.1)

We call this the closure property if ab = ba, we say that the group is abelian. On the other hand if ab ≠ ba, the group is nonabelian. The multiplication or composition rule is meant to convey a product in the abstract sense, the actual implementation of this rule can vary from group to group.
A group G must satisfy four axioms. These are
1. Associativity: The multiplication rule is associative, meaning (ab)c = a(bc).
2. Identity element: The group has an identity element e that satisfies ae = ea = a. The identity element for the group is unique.
3. Inverse element: For every element a ෈ G, there exists an inverse which we denote by a–1 such that aa–1 = a–1 a = e.
4. Order: The order of the group is the number of elements that belong to G.

Representation of the Group
In particle physics we are often interested in what is called a representation of the group. Let’s denote a representation by F. A representation is a mapping that takes group elements g ෈ G into linear operators F that preserve the composition rule of the group in the sense that
• F(a)F(b) = F(ab). • The representation also preserves the identity, that is, F(e) = I. Suppose that a,b ෈ G and f ෈ H where H is some other group. If the composition rule satisfies
f (a) f (b) = f (ab)
we say that G is homomorphic to H. This is a fancy way of saying that the two groups have a similar structure.
EXAMPLE 3.1A Does the set of all integers form a group under addition?

CHAPTER 3 An Introduction to Group Theory

51

SOLUTION The set of all integers forms a group under addition. We can take the composition rule to be addition. Let z1 ෈ Z and z2 ෈ Z. Clearly the sum
z +z ෈Z 1 2
is another integer, so it belongs to the group. For the identity, we can take e = 0 since
z+0=0+z=z
for any z ෈ Z. Addition is commutative, that is,
z1 + z2 = z2 + z1 Therefore the group is abelian. The inverse of z is just −z, since
z + (–z) = e = 0
which satisfies aa–1 = a–1 a = e.

EXAMPLE 3.1 Does the set of all integers form a group under multiplication?

SOLUTION

The set of all integers does not form a group under multiplication. We can take the

composition

rule

to

be

multiplication.

Let

z 1

෈

Z

and

z 2

෈

Z.

Clearly

the

product

z1z2 ෈ Z

is another integer, so it belongs to the group. For the identity, we can take e = 1 since

z×1=1×z=z for any z ෈ Z. Multiplication is commutative, that is,

z ×z =z ×z 1 2 2 1
Therefore the group is abelian. The problem is the inverse: 1/z is not an integer.

z×1 =e=1 z

52

Quantum Field Theory Demystified

So even though an inverse exists, the inverse is not in the group, and therefore the set of all integers does not form a group under multiplication.

Group Parameters

An ordinary function of position is specified by an input x, that is, we have y = f(x). In an analogous way, a group can also be a function of one or more inputs that we call parameters.
Let a group G be such that individual elements g ෈ G are specified by a finite set of parameters, say n of them. If we denote the set of parameters by

{q1,

q2,

.

.

.

,q } n

The group element is then written as

g

=

G

(q1,

q2,

.

.

.

,q ) n

The identity is the group element where the parameters are all set to 0.

e = G (0, 0, . . . ,0)

Lie Groups

While there are discrete groups with a finite number of elements, most of the groups
we will be concerned with have an infinite number of elements. However, they have
a finite set of continuously varying parameters. In the expression g = G (q1, q2, . . . ,qn) we have suggestively labeled the
parameters as angles, since several important groups in physics are related to rotations. The angles vary continuously over a finite range 0 . . . 2π. In addition, the group is parameterized by a finite number of parameters, the angles of rotation.
So, if a group G

•

Depends

on

a

finite

set

of

continuous

parameters

q i

• Derivatives of the group elements with respect to all the parameters exist

we call the group a Lie group. To simplify the discussion we begin with a group with a single parameter q. We obtain the identity element by setting q = 0

g(θ) θ=0 = e

(3.2)

CHAPTER 3 An Introduction to Group Theory

53

By taking derivatives with respect to the parameters and evaluating the derivative at q = 0, we obtain the generators of the group. Let us denote an abstract generator by X. Then

X = ∂g ∂θ

θ=0

(3.3)

More generally, if there are n parameters of the group, then there will be n generators such that each generator is given by

Xi

=

∂g ∂θi

θi =0

(3.4)

Rotations have a special property, in that they are length preserving (that is, rotate a vector and it maintains the same length). A rotation by –q undoes a rotation by q,
hence rotations have an orthogonal or unitary representation. In the case of quantum
theory, we seek a unitary representation of the group and choose the generators Xi to be Hermitian. In this case

Xi

=

−i

∂g ∂θi

θi =0

(3.5)

For some finite q, the generators allow us to define a representation of the group. Consider a small real number e > 0 and use a Taylor expansion to form a representa-
tion of the group (which we denote by D)

D (εθ) ≈ 1 + iεθ X

If q = 0, then clearly the representation gives the identity. You will also recall that the exponential function has a series expansion

ex = 1+ x + 1 x2 + 2!

So we can define the representation of the group in terms of the exponential using

D (θ)

=

lim
n→∞

⎛ ⎝⎜

1

+

i

θX n

⎞ ⎠⎟

n

=

eiθ X

(3.6)

54

Quantum Field Theory Demystified

Notice that if X is Hermitian, X = X† and the representation of the group is unitary, since
D† (θ ) = (eiθX )† = e−iθX† = e−iθX ⇒ D† (θ)D(θ) = (e−iθX ) (eiθX ) = 1

One reason that the generators of a group are important is that they form a vector space. This means we can add two generators of the group together to obtain a third generator, and we can multiply generators by a scalar and still have a generator of the group. A complete vector space can be used as a basis for representing other vector spaces, hence the generators of a group can be used to represent other vector spaces. For example, the Pauli matrices from quantum mechanics can be used to describe any 2 × 2 matrix.
The character of the group is defined in terms of the generators in the following sense. The generators satisfy a commutation relation we write as

[Xi , X j ] = ifijk Xk

(3.7)

This is called the Lie algebra of the group. The quantities f are called the structure ijk
constants of the group. Looking at the commutation relation for the Lie algebra, Eq. (3.7), you should recognize the fact that you’ve already been working with group generators in your studies of non-relativistic quantum mechanics. You’ll see this explicitly when we discuss the Pauli spin matrices later on.

The Rotation Group
The rotation group is the set of all rotations about the origin. A key feature is that rotations preserve the lengths of vectors. This mathematical property is expressed by saying the matrices are unitary. It is easy to see that the set of rotations forms a group. Let’s check off each of the basic properties that a group must have.
The first is a group composition rule. Remember that if a ෈ G and b ෈ G, if G is a group then ab ෈ G as well. Now let R1 and R2 be two rotations. It is clear that the composition of these rotations, say by performing the rotation R first followed
1
by the rotation R2, is itself just another rotation, as shown here.
R3 = R2 R1
That is, performing the two rotations as described is the same as doing the single rotation R3. Hence R3 is a member of the rotation group. Next, we illustrate how two

CHAPTER 3 An Introduction to Group Theory

55

rotations in sequence are the same as a single rotation that is the sum of the two angles:

q1

+

q2
=

q1 + q2

It is not the case that rotations commute. That is, in general
R1 R2 ≠ R2 R1 To this, put a book on the table in front of you. Then rotate it about two different axes, and then try the experiment again doing the same rotations but in different order. You will see that the end results are not the same. Therefore, the rotation group is nonabelian. However, it’s easy to see that rotations are associative, like
R1 (R2 R3) = (R1 R2) R3
The rotation group has an identity element—this is, simply doing no rotation at all. The inverse of a rotation is simply the rotation carried out in the opposite direction.

Representing Rotations

Let x be the coordinates of a two-dimensional vector and let x ′ be the coordinates

i

i

of the vector rotated by an angle q in the plane. The components of the two vectors

are related by a transformation as

xj′ = Rij xi
where R is a rotation matrix. This is a representation of the rotation group. ij
Specifically, a rotation by an angle q (in two dimensions) can be represented by the matrix

R

(θ )

=

⎛ ⎝⎜

cosθ − sinθ

sinθ ⎞ cos θ ⎠⎟

56

Quantum Field Theory Demystified

So that

x1′ = cosθ x1 + sinθ x2

x2′ = − sinθ x1 + cosθ x2

⎛ ⎝⎜

x1′⎞ x2′ ⎠⎟

=

⎛ cosθ ⎝⎜ − sinθ

sinθ ⎞ cosθ ⎠⎟

⎛ ⎝⎜

x1 ⎞ x2 ⎠⎟

=

⎛ x1 cosθ + x2 sinθ ⎞ ⎝⎜ −x1 sinθ + x2 cosθ⎠⎟

Let’s write down the transpose of the rotation matrix as

RT

(θ

)

=

⎛ ⎝⎜

cosθ − sinθ

sinθ ⎞ cos θ ⎠⎟

Notice that

R

(θ

)

RT

(θ

)

=

⎛ ⎝⎜

cosθ − sinθ

sinθ ⎞ ⎛ cosθ cosθ⎠⎟ ⎝⎜ sinθ

− sinθ⎞ cosθ ⎠⎟

RRT

=

⎛ cosθ ⎝⎜ − sinθ

sinθ ⎞ ⎛ cosθ cosθ⎠⎟ ⎝⎜ sinθ

− sinθ⎞ cosθ ⎠⎟

=

⎛ ⎝⎜

−

cos2 θ + sin2 θ sinθ cosθ + cosθ

sinθ

− cosθ sinθ + cosθ sinθ⎞ cos2 θ + sin2 θ ⎠⎟

=

⎛ ⎝⎜

1 0

0⎞ 1 ⎠⎟

This tells us that the transpose of the matrix is the inverse group element, since
multiplying two matrices together gives the identity. You can see why this is true using basic trigonometry: sin(–q ) = –sin(q ) and cos(–q ) = cos(q ).

RT

(θ

)

=

R(−θ )

=

⎛ ⎝⎜

cosθ sinθ

− sinθ⎞ cosθ ⎠⎟

So, the inverse of R(q ) is R(–q ): sin(q – q ) = 0 and cos(q – q ) = 1.

R

(θ ) RT

(θ )

=

R

(θ)R (−θ)

=

R

(θ ) R−1 (θ )

=

⎛ ⎝⎜

1 0

0⎞ 1 ⎠⎟

=

I

CHAPTER 3 An Introduction to Group Theory

57

In group theory, various groups are classified according to the determinants of the matrices that represent the groups. Notice that in this case

det

R (θ

)

=

det

⎛ ⎝⎜

cosθ − sinθ

sinθ ⎞ cos θ ⎠⎟

= cos2 θ + sin2 θ = 1

In general, the determinant will not be +1. However, when this condition is satisfied, the rotation matrix corresponds to a proper rotation.
A more readable notation is to make the angle implicit and to use a subscript. That is,
R(θ1) → R1
This is a better way to discuss problems to multiple angles. Now, if det R1 = 1 and det R2 = 1, then the product is unity. We see this from the
property of multiplication for determinants.
det (R1R2 ) = det R1 det R2 = (1) (1) = 1
The product of two rotations also has an inverse, since

R1R2 (R1R2 )T = R1R2 R2T R1T = R1R1T = I

Another way to look at this is to use the closure property of this group. Call the

successive rotations R and R . Taken together they form the group element R :

1

2

3

R1R2 = R3

As before, the transpose is the inverse:

R3T = R3−1

In terms of the components, we have

R3T = (R1R2 )T = R2T R1T and R3−1 = (R1R2 )−1 = R2−1R1−1

We now see

as above.

R3R3T = I

58

Quantum Field Theory Demystified

So the matrix representation preserves the properties of the rotation group. Now that we’ve introduced the notion of a group, we will explore groups important for particle physics.

SO(N)

The group SO(N ) are special orthogonal N × N matrices. The term special is a reference to the fact that these matrices have determinant +1. A larger group, one that contains SO(N ) as a subgroup, is the group O(N) which are orthogonal N × N matrices that can have arbitrary determinant. Generally speaking, rotations can be
represented by orthogonal matrices, which themselves form a group. So the group
SO(3) is a representation of rotations in three dimensions, and the group consists of 3 × 3 orthogonal matrices with determinant +1.
A matrix O is called orthogonal if the transpose OT is the inverse, that is,

OOT = OTO = I ⇒ O is orthogonal

(3.8)

As we stated above, a special orthogonal matrix is one with positive unit determinant

detO = +1

(3.9)

Now let’s turn to a familiar case, SO(3). This group has three parameters, the three
angles defining rotations about the x, y, and z axes. Let these angles be denoted by V, f, and q. Then

⎛1

Rx

(ς

)

=

⎜ ⎜

0

⎝0

0 cosς − sinς

0⎞

sin

ς

⎟ ⎟

cos ς ⎠

⎛ cosφ 0 sinφ ⎞

Ry

(φ

)

=

⎜ ⎜

0

1

0

⎟ ⎟

⎝ − sinφ 0 cosφ⎠

⎛ cosθ sinθ 0⎞

Rz

(θ

)

=

⎜ ⎜

−

sin

θ

cosθ

0⎟⎟

⎝0

0 1⎠

(3.10) (3.11) (3.12)

CHAPTER 3 An Introduction to Group Theory

59

These matrices are the representation of rotations in three dimensions. Rotations in three or more dimensions do not commute, and it is an easy although tedious exercise to show that the rotation matrices written down here do not commute either.
The task now is to find the generators for each group parameter. We do this using Eq. (3.5). Starting with Rx (V ), we have

⎛0

dRx dς

=

⎜ ⎜

0

⎝0

0 − sinς − cosς

0⎞

cosς

⎟ ⎟

sinς ⎠

Now we let V → 0 to obtain the generator

⎛0 0 0⎞

Jx

= −i dRx dς

ς =0

=

⎜ ⎜

0

0

−i⎟⎟

⎝0 i 0⎠

(3.13)

Next, we compute the generator for rotations about the y axis evaluated at f = 0.

⎛ 0 0 −i⎞

Jy

=

−i dRy dφ

=

⎜ ⎜

0

0

0

⎟ ⎟

⎝i 0 0⎠

(3.14)

And finally, for rotations about the z axis we find

⎛ 0 −i 0⎞

Jz

= −i dRz dθ

θ=0

=

⎜ ⎜

i

0

0⎟⎟

⎝ 0 0 0⎠

These matrices are, of course, the familiar angular momentum matrices. So we’ve discovered the famous result that the angular momentum operators are the generators of rotations. We can use the generators to build infinitesimal rotations. For example, an infinitesimal rotation about the z axis by an angle eq, where e is a small positive parameter, is written as
Rz (εθ) = 1 + i Jzε θ

60

Quantum Field Theory Demystified

From quantum mechanics, you already know the algebra of the group. This is just the commutation relations satisfied by the angular momentum operators.

[Ji , J j ] = iεijk Jk

(3.15)

The structure constants in this case are given by the Levi-Civita tensor, which takes on values of +1, –1, or 0 according to

ε123 = ε312 = ε231 = +1 ε132 = ε321 = ε213 = −1

(3.16)

with all other combinations of the indices giving 0.

EXAMPLE 3.2 Show that the representation of the rotation group is of the form eiJyφ .

SOLUTION What we need to show is that Ry (φ) = eiJyφ . This is easy to do by just writing down the first few terms in the Taylor series expansion of the exponential. Remember
Euler’s formula which tells us that

eiθ = cosθ + i sinθ From this we can extract the expansions for the sine and cosine.

cosθ = 1 − 1 θ 2 + 1 θ 4 + 2 4!
sinθ = θ − 1 θ 3 + 1 θ 5 + 3! 5!

Now,

eiJyφ

=1+

iJ yφ

−

1 2

Jy2φ 2

−

i

1 3!

J

y3φ

3

+

What

are

the

powers

of

the

J y

matrix?

A

quick

calculation

shows

that

Jn y

takes

one

of two values depending upon whether n is odd or even.

Jy

=

J

3 y

=

J

5 y

=

⎛ 0 0 −i⎞

=

⎜ ⎜

0

0

0

⎟ ⎟

and

Jy2

=

J

4 y

=

J

6 y

=

⎝i 0 0⎠

⎛ 1 0 0⎞

=

⎜ ⎜

0

0

0⎟⎟

⎝ 0 0 1⎠

CHAPTER 3 An Introduction to Group Theory

61

The expansion becomes

⎛ 1 0 0⎞ ⎛ 0 0 i ⎞

⎛ 0 0 i⎞2

⎛ 0 0 i⎞3

eiJyφ

=

⎜ ⎜

0

1

0⎟⎟

+

i

⎜ ⎜

0

0

0⎟⎟

φ

−

1 2

⎜ ⎜

0

0

0⎟⎟

φ2

−

i

1 3!

⎜ ⎜

0

0

0⎟⎟ φ3 +

⎝ 0 0 1⎠ ⎝ −i 0 0⎠ ⎝ −i 0 0⎠

⎝ −i 0 0⎠

⎛ 1 0 0⎞ ⎛ 0 0 −φ⎞ ⎛φ2 0 0 ⎞ ⎛ 0 0 φ3⎞

=

⎜ ⎜

0

1

0⎟⎟

+

⎜ ⎜

0

0

⎝ 0 0 1⎠ ⎝φ 0

0 0

⎟ ⎟ ⎠

−

1⎜ 2 ⎝⎜⎜

0 0

0 0

0 φ2

⎟ ⎠⎟⎟

−

1 3!

⎜ ⎝⎜⎜

0 φ3

0 0

0 0

⎟ ⎠⎟⎟

+

⎛ ⎜ ⎜

1−

1 2

φ2

+

=⎜

0

⎜ ⎜φ

−

1

φ3

+

⎝ 3!

0

−

⎛ ⎝⎜

φ

−

1 3!

φ

3

+

⎞⎞ ⎠⎟ ⎟ ⎟

0

0

⎟

0

1− 1φ2 +

⎟ ⎟

2

⎠

⎛ cosφ 0 − sinφ⎞

=

⎜ ⎜

0

0

0

⎟ ⎟

⎝ sinφ 0 cosφ ⎠

So, rotations about the x, y, and z axes are represented by

Rx (ς ) = eiJxς

Ry (φ) = eiJyφ

Rz (θ ) = eiJzθ

(3.17)

A rotation about an arbitrary axis defined by a unit vector n is given by

Rn (θ ) = eiJ⋅θ

(3.18)

As stated earlier, orthogonal transformations (rotations) preserve the lengths of
vectors. We say that the length of a vector is invariant under rotation. This means that given a vector x with length x 2 = x2 + y2 + z2 , when we transform it under a rotation x′ = Rx we have

x′2 = x2 ⇒ x′2 + y′2 + z′2 = x2 + y2 + z2

(3.19)

The preservation of the lengths of vectors by orthogonal transformations will be important in establishing a correspondence between the unitary transformation SU(2) (see in the next section, Unitary Groups) and rotations in three dimensions represented by SU(3).

62

Quantum Field Theory Demystified

Unitary Groups
In particle physics unitary groups play a special role. This is due to the fact that unitary operators play an important role in quantum theory. Specifically, unitary operators preserve inner products—meaning that a unitary transformation leaves the probabilities for different transitions among the states unaffected. That is, quantum physics is invariant under a unitary transformation. As a result unitary groups play a special role in quantum field theory.
When the physical predictions of a theory are invariant under the action of some group, we can represent the group by a unitary operator U. Moreover, this unitary operator commutes with the Hamiltonian as shown here.
[U, H] = 0
The unitary group U(N) consists of all N × N unitary matrices. Special unitary groups, denoted by SU(N) are N × N unitary matrices with positive unit determinant. The dimension of SU(N) and hence the number of generators, is given by N2 – 1. Therefore,
• SU(2) has 22 – 1 = 3 generators. • SU(3) has 32 – 1 = 8 generators.
The rank of SU(N) is N – 1. So
• The rank of SU(2) is 2 – 1 = 1. • The rank of SU(3) is 3 – 1 = 2.
The rank gives the number of operators in the algebra that can be simultaneously diagonalized.
The simplest unitary group is the group U(1). A “1 × 1” matrix is just a complex number written in polar representation. In another way, we can say a U(1) symmetry has a single parameter q and is written as
U = e−iθ
where q is a real parameter. It is completely trivial to see that U(1) is abelian, since
U1U2 = e e −iθ1 −iθ2 = e e −iθ2 −iθ1 = U2U1
U1U2 = e e −iθ1 −iθ2 = e−i(θ1+θ2 ) = e−i(θ2+θ1) = e e −iθ2 −iθ1 = U2U1

CHAPTER 3 An Introduction to Group Theory

63

We will see that many Lagrangians in field theory are invariant under a U(1) transformation. When you look at the problem in the complex plane this invariance becomes obvious. Consider the arbitrary complex number z = reia. When we multiply z by eiq we get
eiθ z = eiθreiα = rei(θ+α )
The new complex number has the same length, r, and the angle is increased by q. For example, the Lagrangian for a complex scalar field
L = ∂μϕ*∂μϕ − m2ϕ*ϕ

is invariant under the transformation ϕ → e−iθϕ

As described in Chap. 2, when a Lagrangian is invariant under a transformation there is a symmetry, and in this case there is a U(1) symmetry. Force-mediating particles, called gauge bosons, will be associated with unitary symmetries like this one. We will see later that when considering electrodynamics, the gauge boson associated with the U(1) symmetry of quantum electrodynamics is the photon.
We will see in later chapters that a U(1) symmetry also manifests itself in terms of the conservation of various quantum numbers. If there is a U(1) symmetry associated with a quantum number a, then
U = e−iaθ
The importance of the U(1) symmetry is that the Hamiltonian H is invariant under the transformation e−iaθ Heiaθ , that is,

UHU† = H

Again we see that adjoint U† (the Hermitian conjugate) is also the inverse.

U(θ )U†(θ ) = U(θ )U(−θ ) = 1

We will see that such symmetries are present in quantum field theory with conservation of lepton and baryon number, for example.
In summary, an element of the group U(1) is a complex number of unit length written as

U = e−iθ where q is a number. This is the familiar unit circle.

(3.20)

64

Quantum Field Theory Demystified

Moving right along, the next non-trivial unitary group is U(2), which is the set of all 2 × 2 unitary matrices. Being unitary these matrices satisfy

UU † = U †U = I

(3.21)

For physics, we are interested in a subgroup of U(2), which is the set of all 2 × 2 unitary matrices with determinant +1. This group is called SU(2). The generators of SU(2) are the Pauli matrices, which we reproduce here for your convenience

σ1

=

⎛ ⎝⎜

0 1

1⎞ 0⎠⎟

σ2

=

⎛ ⎝⎜

0 i

−i⎞ 0 ⎠⎟

σ3

=

⎛ ⎝⎜

1 0

0⎞ −1⎠⎟

(3.22)

Now we see how the rank of a unitary group comes into play. The rank of SU(2)

is 1, and there is one diagonalized operator s in the basis we have chosen. The 3

generators of SU(2) are actually taken to be

1 2

σ

i

and the Lie algebra is the familiar

commutation relations that are satisfied by the Pauli matrices

⎡ ⎣⎢

σi 2

,σ j 2

⎤ ⎦⎥

=

iεijk

σk 2

(3.23)

The similar algebraic structure between SU(2) as indicated in Eq. (3.23) and SO(3)
as indicated by Eq. (3.5) indicates that there will be a correspondence between
these two groups. Since the Pauli matrices do not commute, SU(2) is nonabelian,
as seen in Eq. (3.23). Recalling that the rank of SU(2) is 1, there is one diagonal generator which we have chosen to be s3.
An element of SU(2) can be written as

U = eiσ jα j /2

(3.24)

where

s i

is

one

of

the

Pauli

matrices

and

a j

is

a

number.

To

understand

the

presence

of the factor ½ in Eqs. (3.23) and (3.24), let’s explore the correspondence between

SU(2) and SO(3). We saw in Eq. (3.19) that SO(3) preserves the lengths of vectors.

Sinceitisunitary,SU(2) preservesthelengthsofvectorsaswell.Let r = x xˆ + y yˆ + z zˆ.

We construct a matrix of the form σ ⋅ r .

σ ⋅r =σxx +σyy+σzz

=

⎛ ⎝⎜

0 x

x⎞ 0⎠⎟

+

⎛ ⎝⎜

0 iy

−iy⎞ 0 ⎠⎟

+

⎛ ⎝⎜

z 0

0⎞ − z ⎠⎟

=

⎛ ⎝⎜

x

z +

iy

x − iy⎞ −z ⎠⎟

(3.25)

CHAPTER 3 An Introduction to Group Theory

65

If we take the determinant of this matrix, we obtain the length of the vector as shown here.

det

⎛ ⎝⎜

x

z +

iy

x − iy⎞ −z ⎠⎟

=

−x2

−

y2

−

z2

=

−x2

Also note that the matrix σ ⋅ r is Hermitian and has zero trace. Now consider a unitary transformation on this matrix. For example, we can take

U (σ ⋅ r )U† = σ x (σ ⋅ r )σ x

=

⎛ ⎝⎜

0 1

1⎞ ⎛ z 0⎠⎟ ⎝⎜ x + iy

x − iy⎞ ⎛ 0 −z ⎠⎟ ⎝⎜ 1

1⎞ 0⎠⎟

=

⎛ ⎝⎜

x

−z − iy

x + iy⎞ z ⎠⎟

The transformed matrix still has zero trace, and is still Hermitian. Moreover, the determinant is preserved, and it again gives the length of the vector, that is,

det

⎛ ⎝⎜

−z x − iy

x

+ z

iy⎞ ⎠⎟

=

−x2

−

y2

−

z2

=

−x2

The conclusion is that like SO(3), SU(2) preserves the lengths of vectors as shown here.

x′2 = x2 ⇒ x′2 + y′2 + z′2 = x2 + y2 + z2

The correspondence works by considering an SU(2) transformation on a two component spinor, like

ψ

=

⎛α⎞ ⎝⎜ β⎠⎟

where

x = 1 (β2 − α 2) 2

y = − i (α 2 + β 2 ) 2

z = αβ

66

Quantum Field Theory Demystified

) Then
( ) on x

( =anxzyS.UA(s2)ytoruancsafnorsmeeaftrioonmotnhiψs e=quiαβvailsenecqeu,ivanalSenUt(t2o)

an SO(3) transformation transformation has three

real parameters that correspond to the three angles of an SO(3) transformation.

Label the “angles” for the SU(2) transformation by a, b, and g . Half the rotation

angle generated by SU(2) corresponds to the rotation generated by SO(3). For

arbitrary angle a, a transformation generating a rotation about x in SU(2) is

given by

U

=

⎛ ⎝⎜

cosα/2 i sinα/2

i sinα/2⎞ cosα/2 ⎠⎟

(see quiz problem 1). This transformation corresponds to the rotation Eq. (3.10), a rotation about the x axis. Next, consider an SU(2) transformation generating a rotation around the y axis. The unitary operator is

U

=

⎛ ⎝⎜

cos β/2 − sin β/2

sin β/2 ⎞ cos β/2⎠⎟

This corresponds to the SO(3) transformation given in Eq. (3.11). Finally, for a rotation about the z axis we have the SU(2) transformation
⎛ eiγ /2 0 ⎞ U = ⎝⎜ 0 e−iγ /2 ⎠⎟

which corresponds to Eq. (3.12). All 2 × 2 unitary matrices are specified by two parameters, complex numbers
a and b where

U

=

⎛a ⎝⎜ −b*

b⎞ a* ⎠⎟

For an element of SU(2), det U = +1 so we require that a 2 + b 2 = 1. Later we will seek to define Lagrangians that are invariant under an SU(2)
symmetry. This symmetry will be of particular importance in the case of electroweak interactions. The gauge bosons corresponding to the SU(2) symmetry will be the W and Z bosons that carry the weak interaction.
Next, we consider the unitary group SU(3), which will be important in the study of quarks and the theory of quantum chromodynamics. Earlier we indicated

CHAPTER 3 An Introduction to Group Theory

67

that SU(3) has eight generators. These are called the Gell-Mann matrices and are given by

⎛ 0 1 0⎞

λ1

=

⎜ ⎜

1

0

0⎟⎟

⎝ 0 0 0⎠

⎛ 0 −i 0⎞

λ2

=

⎜ ⎜

i

0

0⎟⎟

⎝ 0 0 0⎠

⎛ 1 0 0⎞

λ3

=

⎜ ⎜

0

−1

0⎟⎟

⎝ 0 0 0⎠

⎛ 0 0 1⎞

λ4

=

⎜ ⎜

0

0

0⎟⎟

⎝ 1 0 0⎠

⎛ 0 0 −i⎞

λ5

=

⎜ ⎜

0

0

0

⎟ ⎟

⎝i 0 0⎠

⎛ 0 0 0⎞

λ6

=

⎜ ⎜

0

0

1⎟⎟

⎝ 0 1 0⎠

⎛0 0 0⎞

λ7

=

⎜ ⎜

0

0

−i⎟⎟

⎝0 i 0⎠

⎛1 0 0 ⎞

λ8 =

1 3

⎜ ⎜

0

1

0

⎟ ⎟

⎝ 0 0 −2⎠

Notice two of the matrices are diagonal, l and l , as we would expect from the

3

8

rank of the group. The Gell-Mann matrices are traceless and they satisfy the

commutation relations

8
∑ ⎡⎣λi , λ j ⎤⎦ = 2i fijkλk k=1

(3.26)

This defines the algebraic structure of SU(3). The nonzero structure constants are

f123 = 1

f147

=

f165

=

f246

=

f257

=

f345

=

f376

=

1 2

f458 = f678 =

3 2

(3.27)

We will see more of SU(3) when we examine the standard model.

Casimir Operators
A casimir operator is a nonlinear function of the generators of a group that commutes with all of the generators. The number of casimir operators for a group is given by the rank of the group.

68

Quantum Field Theory Demystified

Considering SU(3) as an example, the generators are the angular momentum operators. A casimir operator in this case is

J2

=

J x2

+

J

2 y

+

J

2 z

A casimir operator is an invariant. In this case, the invariance suggests J2 is a multiple of the group identity element.

Summary
Group theory plays an important role in physics because groups are used to describe symmetries. The structure of a group is defined by the algebra among its generators. If two groups have the same algebra, they are related. Unitary transformations preserve the probabilities of state transitions in quantum theory. As a result, the most important groups in quantum field theory are the unitary groups, specifically U(1), SU(2), and SU(3).

Quiz
1. Consider an element of SU(2) given by U = eiσxα/2. By writing down the power series expansion, write U in terms of trigonometry functions.
2. Consider SU(3) and calculate tr (λiλ j ). 3. How many casimir operators are there for SU(2)?
4. Write down the casimir operators for SU(2).
A Lorentz transformation can be described by boost matrices with rapidity defined by tanhφ = v /c. A boost in the x direction is represented by the matrix

⎛ coshφ sinhφ 0 0⎞

⎜ ⎜

sinh

φ

cosh φ

0

0⎟⎟

⎜0

0 1 0⎟

⎝⎜ 0

0 0 1⎠⎟

5. Find the generator Kx .

CHAPTER 3 An Introduction to Group Theory

69

6. Knowing that [Kx , Ky ] = −iJz, where Jz is the angular momentum operator written in four dimensions as

⎛ 0 0 0 0⎞

⎜ 0 0 1 0⎟

Jz

=

−i ⎜ ⎜0

−1

0

⎟ 0⎟

⎝⎜ 0 0 0 0⎠⎟

find Ky. 7. Do pure Lorentz boosts constitute a group?

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CHAPTER 4
Discrete Symmetries and Quantum Numbers
In Chap. 3 we examined continuous symmetries, that is, symmetries with continuously varying parameters such as rotations. Now we consider a different kind of symmetry, a discrete symmetry. There are three important discrete symmetries in particle physics: parity, charge conjugation, and time reversal.
Additive and Multiplicative Quantum Numbers
A quantum number is some quantity (a quantized property of a particle) that is conserved in a particle reaction (decay, collision, etc.). An additive quantum number
Copyright © 2008 by The McGraw-Hill Companies, Inc. Click here for terms of use.

72

Quantum Field Theory Demystified

n is one such that if n1, n2 ,…, ni ,… are the quantum numbers before the reaction, and m1, m2 ,…, mi ,… are the quantum numbers after the reaction, then the sum is
preserved.

∑ ni = ∑ mi

i

i

(4.1)

Or, if we have a composite system with quantum numbers n1, n2 ,…,ni ,…, and if the quantum number is additive, then the quantum number of the composite
system is

∑ ni i

Amultiplicative quantum number is one such that if n1, n2 ,…, ni ,… are the quantum numbers before the reaction, and m1, m2 ,…, mi ,… are the quantum numbers after the reaction, then the product is preserved.

∏ ni = ∏ mi

i

i

(4.2)

Or, if we have a composite system, and if a quantum number is multiplicative, then

∏ ni i
is the quantum number for the composite system. If a quantum number is conserved, then it represents a symmetry of the system.

Parity

We begin our discussion of parity by examining nonrelativistic quantum mechanics. Consider a potential V that is symmetric about the origin and therefore V (−x) = V (x). This implies that if ψ (x) is a solution of the Schrödinger equation, then so is ψ (−x), and it solves the equation with the same eigenvalue. This is because

−

2
2m

d2ψ (−x) dx 2

+

V

(−

x)ψ

(−x)

=

Eψ

(−x)

⇒

−

2
2m

d2ψ (−x) dx 2

+

V

(x)ψ

(−x)

=

Eψ

(−x)

CHAPTER 4 Discrete Symmetries

73

when V (−x) = V (x). If ψ (x) and ψ (−x) both solve the Schrödinger equation with the same eigenvalue E, then they must be related to each other as

ψ (x) = αψ (−x)

(4.3)

If we let x → −x, then we obtain ψ (−x) = αψ (x)

Inserting this into Eq. (4.3) gives
ψ (x) = αψ (−x) = α [αψ (x)] = α 2ψ (x)
⇒1=α2 This tells us that α = ±1. Then either

ψ (−x) = ψ (x)

in which case we say that the wave function has even parity or

ψ (−x) = −ψ (x)

in which case we say the wave function has odd parity. This leads us to the concept of the parity operator P. The parity operator causes a change in sign when x → −x

in the wave function.

Pψ (x) = ψ (−x)

(4.4)

As seen here in the illustrations, even powers lead to functions with even parity
[ψ (−x) = ψ (x)] while odd powers lead to functions with odd parity [ψ (−x) = −ψ (x)].

An example of an even function where y(–x) = y(x). Monomials with even powers (x2, x4, x6, …) are even functions.

An example of an odd function where y (–x) = –y(x). Monomials with odd powers (x1, x3, x5, …) are odd functions.

74

Quantum Field Theory Demystified

Obviously, applying the parity operator twice in succession gives the original wave function back as shown here.

P2ψ (x) = Pψ (−x) = ψ (x)

Think of the parity operator as a reflection through the y axis. If y is even, we see the same function values; if y is odd, we see the negative of the function values. In either case, another reflection through the y axis brings us back to our initial state. The reflection of the reflection is the image.
It follows that

P2 = I

(4.5)

The eigenstates of parity are ±1:

Pψ =±ψ

(4.6)

As we have seen, true reflections preserve length. If ψ is an angular momentum state with angular momentum L, that is, ψ =
L, mz , then the parity operator acts as

P L,mz = (−1)L L,mz

(4.7)

We showed above that α = ±1 when ψ (x) and ψ (−x) both solve the Schrödinger equation with the same eigenvalue E. We can generalize this by saying that if the Parity operator and Hamiltonian commute

[P, H ] = 0 = PH − HP
then parity is conserved. A consequence of this is that a state with parity a cannot evolve into a state with parity −a since the Hamiltonian governs the time evolution of the states. Even parity states remain even during time evolution and odd parity states remain odd. Now if ψ is a nondegenerate eigenstate of H with eigenvalue E, then
P(H ψ ) = P(E ψ ) = EP ψ

But if [P, H ] = 0 = PH − HP, then
P(H ψ ) = H(P ψ ) = E(P ψ )

CHAPTER 4 Discrete Symmetries

75

So the eigenstates of H are also eigenstates of the parity operator. Also, it follows that the eigenvalues of P are α = ±1. Notice that this precludes states of mixed, or indefinite parity. This is a powerful constraint mathematically and physically it means that particles are either fermions or bosons.
In quantum field theory, the eigenvalue of parity a is a property of particles called the intrinsic parity of the particle. Following our discussion of parity and wave functions at the beginning of this section, if α = +1for a given particle, we say that the particle has even parity. If α = −1, the particle has odd parity.
Parity for fermions is assigned as follows:
• Particles with spin-1/2 have positive parity. Hence an electron and a quark both have α = +1.
• Antiparticles with spin-1/2 have negative parity. Therefore a positron has α = −1.
Bosons have the same intrinsic parity for both particles and antiparticles. Parity is a multiplicative quantum number. Let ψ = a b define a composite
system. If the parities of a and b are Pa and Pb respectively, then the parity of the composite system is the product of the individual parities, that is,
Pψ = PaPb
We can construct new parity operators by combining P with one of the conserved charges of the standard model. These are
• The electric charge operator Q
• Lepton number L
• Baryon number B
Earlier we said something about the conservation of parity. Parity is not always conserved, and there are specific cases, like
• Parity is conserved in the electromagnetic and strong interactions.
• Parity is not conserved in the weak interaction.

Particles are often labeled as follows

J P ≡ spinparity

(4.8)

A spin-0 particle with negative parity is called a pseudoscalar. Examples of
pseudoscalar particles include the π and K mesons. Using the notation in Eq. (4.8) we write 0− to indicate a pseudoscalar particle.
A spin-0 particle with positive parity is called a scalar. We denote a scalar by 0+.
An example of a scalar particle is the Higgs boson, the particle corresponding to the

76

Quantum Field Theory Demystified

field believed to be responsible for mass generation. The elusive Higgs hasn’t been detected at the time of writing, but may be found soon when the Large Hadron Collider (LHC) begins operation.
A vector boson has spin-1 and negative parity (1−). The most famous vector boson is the photon. A pseudovector has unit spin and positive parity 1+.
Parity is conserved in the electromagnetic and strong interactions, so the total parity of a system before an electromagnetic or strong interaction is the same as the parity after the interaction. In the 1950s, two physicists named Lee and Yang proposed that parity conservation is violated in the weak interaction.
This is called parity violation. This was demonstrated experimentally by observing weak decays of Cobalt-60, leading to the Nobel prize for Lee and Yang. Parity violation also became apparent in the weak decays of two particles called the θ and τ mesons. They decay as
θ + → π +π 0
τ + → π +π −π +
The final states of these two decays have opposite parity and therefore physicists believed the q + and the t + to be different particles. However, successively refined measurements of the q + and t + mass and lifetimes suggested they were actually the same particle. The discovery of parity violation in the weak interaction resolved this dilemma and today we call this particle the K+ meson.

Charge Conjugation

We now consider charge conjugation C, an operator which converts particles into antiparticles. Let ψ represent a particle state and ψ represent the antiparticle state. Then the charge conjugation operator acts as

Cψ =ψ

(4.9)

Charge conjugation also acts on antiparticle states, turning them into particle states.

It follows that

Cψ =ψ C2 ψ = CC ψ = C ψ = ψ

(4.10)

We can use this relation to determine the eigenvalues of charge conjugation. It is apparent that like parity, they must be C = ±1. Charge conjugation is also like parity

CHAPTER 4 Discrete Symmetries

77

in that it is a multiplicative quantum number. Since charge conjugation converts particles into antiparticles and vice versa, it reverses the sign of all quantum numbers (and also changes the sign of magnetic moment). Consider a proton p . It has positive charge q
Q p =q p
and Baryon number B = +1. If we operate on the proton with the charge conjugation operator, then C p = p and since
Q p = −q p
the charge has been reversed. The baryon number has also been changed to B = −1. Notice that the proton state cannot be an eigenstate of charge conjugation, since the result of C p = p is a state with different quantum numbers—a different quantum state. The eigenstates of the charge conjugation operator are eigenstates with 0 charge, that is, neutral particles. More generally, the eigenstates of C must have all additive quantum numbers equal to 0. An example is a neutral pion π 0. In this case, π 0 is its own antiparticle and so
C π0 =α π0
for some a. Applying charge conjugation twice
C2 π 0 = αC π 0 = α2 π 0 ⇒ α = ±1
We can find the charge conjugation properties of the photon in the following way, and hence determine the eigenvalue α for the π 0. First charge conjugation will reverse the sign of the charge density J as shown here.
CJC−1 = −J
Now, the interaction part of the electromagnetic Lagrangian can be used to determine the charge conjugation properties of the photon. We need to find the action of C on Jμ Aμ . We have
CJμ AμC −1 = CJμC −1CAμC −1 = −JμCAμC −1

78

Quantum Field Theory Demystified

This can only be invariant if
CAμC −1 = − Aμ ⇒ CJμ AμC −1 = Jμ Aμ Since Aμ is the electromagnetic vector potential, this tells us that the eigenvalue of charge conjugation for the photon is α = −1. Therefore if there are n photons, the charge conjugation is (−1)n. The π 0 decays into two photons as
π0 →γ +γ The two photon state has α = (−1)2 = +1, therefore we conclude that
C π 0 = (+1) π 0
⇒ α = +1 for the π 0
Charge conjugation proceeds in the same way as parity, that is,
• Charge conjugation C is conserved in the strong and electromagnetic interactions. • Charge conjugation is not conserved in the weak interaction.

CP Violation
The fact that charge conjugation and parity are each individually violated in weak interactions led to the hope that the combination of charge conjugation and parity would be conserved. It almost is, but there is a slight violation that can be seen from the decay of neutral K mesons.
The neutral K 0 meson is an interesting particle which is observed in a linear combination of states with its antiparticle. This is because the K 0 spontaneously transitions into its antiparticle and vice versa as shown here.
K0 ↔ K0
The K 0 and its antiparticles are pseudoscalars 0− and so have negative parity, that is
P K0 = − K0
P K0 = − K0

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79

Charge conjugation, is of course the operation that transforms the K 0 into its antiparticle, that is,
C K0 = K0 C K0 = K0
Taken together, CP acts on the states as CP K 0 = −C K 0 = − K 0 CP K 0 = −C K 0 = − K 0

We see from this relation that K 0 and K 0 are not eigenstates of CP. To see if CP is violated, we need to construct states that are eigenstates of CP out of K 0 and K 0 . The states that do this are

K0 − K0

K1 =

2

K0 + K0

K2 =

2

It is helpful to think of particle states in terms of vector spaces. In terms of a rotation by π /4 we see that

⎛ ⎝⎜

K1 K2

⎞ ⎠⎟

=

R

⎛⎝⎜

−

π 4

⎞⎠⎟

⎛ ⎜ ⎝

K0 K0

⎞ ⎟ ⎠

which demonstrates an advantage of the vector space view of particle. So we have CP K1 = + K1
CP K2 = − K2

Fortunately, these states can be created in the laboratory. It turns out that they both decay into π mesons. Now, if CP is conserved, then each of these states will decay into a state with the same value of CP. That is,

K1 → decays into a state with CP = +1 K2 → decays into a state with CP = −1

80

Quantum Field Theory Demystified

A neutral K meson K 0 can decay into a state with two a mesons or into a state with three π mesons. The charge conjugation and parity eigenvalues of these states are
2π mesons : C = +1, P = +1, ⇒ CP = +1 3π mesons : C = +1, P = −1, ⇒ CP = −1
If CP is conserved then, we would have K1 → only decays into 2π mesons
K1 → only decays into 3π mesons This is not what is observed experimentally. It is found that a small fraction of the time

K2 → decays into 2π mesons

Hence we have a transition from CP = −1 → CP = +1. It turns out that the long lived K meson state is

KL

=

K2 + ε K1 1+ ε 2

The parameter ε is a measure of the amount of violation of CP conservation. Experimental evidence indicates that

ε = 2.3 × 10−3

A small number indeed, but not zero. CP violation happens because a small fraction
of the time, the long lived neutral K meson state, is found in K1 , giving the unexpected decays.

The CPT Theorem
To restore invariance, we have to bring in one more symmetry, time reversal. This is another discrete transformation on states, turning a state ψ into a state ψ ′ that evolves with time flowing in the negative direction. Momentums change sign:

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linear momentum p → − p and angular momentum L → −L, but all other quantities maintain the same sign. The time-reversal operator acts to transform the states as
T ψ = ψ′
The time-reversal operator is antiunitary and antilinear. To say that it is antilinear, we mean that
( ) T α ψ + β φ = α* ψ ′ + β* φ′
While a unitary operator U preserves inner products,
Uφ Uψ = φ ψ
an antiunitary operator does not, but instead gives the complex conjugate as shown here.
Aφ Aψ = φ ψ *
The time-reversal operator is antiunitary and can be written as a product of an operator K that converts states into their complex conjugates (note that K does not refer to the K meson of the previous section, but in this context is an operator)
Kψ = ψ *
and a unitary operator U
T = UK
If the time-reversal operator commutes with the Hamiltonian [T, H ] = 0, and if
ψ satisfies the Schrödinger equation, T ψ will also satisfy the Schrödinger equation when t → −t. Hence the name time-reversal operator. If the laws of physics are unchanged under time reversal, then they are a symmetry of the system.
The CPT theorem considers the three symmetries C, P, and T taken together. According to this theorem, if charge conjugation, parity reversal, and time reversal are taken together we have an exact symmetry and so the laws of physics are invariant. More colloquially, the theorem means that if matter is replaced by antimatter (charge conjugation), momentum is reversed with spatial inversion (parity conjugation), and time is reversed, the result would be a universe indistinguishable from the one we live in. For the CPT theorem to be valid, all three symmetries must