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ANDR E A S WE I L E R , T UM

R E L AT I V I T Y , PA R T I C L E S , F I E L D S

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Der wackre Schwabe forcht’ sich nit,

ging seines Weges Schritt vor Schritt.1 1 Ludwig Uhland oft zitiert von

Albert Einstein.

Copyright c© 2018 Andreas Weiler, TUM

Sommersemester 2017, Version vom December 31, 2018

mailto:[email protected]

Contents

1 Foreword 9

1.1 Units and conventions 9

2 Introduction 11

2.1 Why quantum field theory? 11

2.1.1 Necessity of the field viewpoint: causality 12

2.1.2 General features of relativistic quantum field theories 13

2.2 String Theory 13

2.2.1 The discrete chain 13

2.2.2 The continuous string 14

2.2.3 Quantum string theory 16

2.2.4 Review of the simple harmonic oscillator 17

2.2.5 Fock space of the discrete chain 18

3 Relativity 23

3.1 Galileo and Newton 23

3.1.1 Maxwell vs. Newton 24

3.1.2 Derivation of the Lorentz transformations 24

3.2 Consequences of special relativity 25

3.2.1 Relativity of simultaneity 26

3.2.2 Time dilation 27

3.2.3 Lorentz contraction 27

3.2.4 Addition of velocities 28

3.3 Minkowski space 29

3.3.1 Fourvectors 30

3.3.2 Minkowski Metric 31

3.3.3 Discrete transformations 32

3.3.4 Motion of a free particle 32

4 andreas weiler, tum

3.4 Indices upstairs and downstairs 33

3.4.1 A convenient notation and derivatives 33

3.4.2 The appearance of anti-matter 34

3.5 Relativistic mechanics 34

3.5.1 E = mc2 36

3.5.2 Application: weak decay of a pion 36

4 Classical Field Theory 37

4.1 Dynamics of fields 37

4.1.1 Example: the electro-magnetic field 37

4.1.2 The Lagrangian density 37

4.1.3 Example 1: the Klein-Gordon equation 39

4.1.4 Example 2: first order Lagrangians 39

4.1.5 Example 3: the Maxwell equations 40

4.1.6 Locality 40

4.1.7 Lorentz invariance 41

4.2 Symmetries: Noether’s Theorem and conserved currents 42

4.2.1 Conserved currents 43

4.2.2 Example 1: internal symmetries U(1) 43

4.2.3 Example 2: an internal SO(N) and a bit on Lie groups 44

4.2.4 Example 3: space-time translations 45

4.2.5 Summary of Noether’s theorem 47

4.3 Coulomb’s law and the propagator method 47

4.3.1 Fourier transformations 48

4.3.2 The Coulomb Potential 49

4.3.3 A non-linear example: the graviton Lagrangian 49

4.4 The Hamiltonian Formalism 51

4.4.1 Example: a real scalar field 52

5 Canonical Quantization 53

5.1 Free fields 53

5.1.1 The free scalar field 54

5.1.2 Infinities 55

5.1.3 Normal ordering 57

5.2 Particles 58

5.2.1 Multi-particle states and Bose statistics 59

relativity, particles, fields 5

5.3 Smeared operator valued distributions 60

5.3.1 Dimensional analysis and Coleman theorem 61

5.4 Some technicalities: relativistic normalization 62

5.4.1 Meaning of the state φ(x)|0〉 64

5.5 Complex scalar fields 64

5.6 Time dependent operators: the Heisenberg picture 66

5.7 Causal quantum fields 68

5.7.1 Propagators in quantum field theory 69

5.8 The Stuckelberg-Feynman propagator 71

5.8.1 Green’s Functions 73

5.8.2 Particle creation by a classical source 74

5.9 Non-relativistic fields 76

5.9.1 A special case: quantum mechanics 78

5.9.2 Non-relativistic interactions 80

6 Interacting quantum fields 81

6.1 Dimensional analysis of interactions 81

6.1.1 Some examples of weakly coupled theories 82

6.2 The interaction picture 84

6.2.1 Dyson’s formula 85

6.3 A first look at scattering 87

6.3.1 Scattering simplified 88

6.3.2 Example: Meson Decay φ→ ψψ 90

6.4 Wick’s Theorem 90

6.4.1 Example: nucleon scattering ψψ → ψψ 92

6.5 Diagrammatic Perturbation theory: Feynman Diagrams 94

6.5.1 Feynman rules 95

6.5.2 Revisiting ψψ → ψψ scattering 98

6.6 More scattering processes in the ψψφ theory 98

6.6.1 Nucleon anti-nucleon scattering: ψ(p1) + ψ(p2)→ ψ(q1) + ψ(q2) 98

6.6.2 Meson-meson scattering φφ→ φφ 99

6.6.3 Symmetry factors 99

6.7 Potentials 102

6.7.1 The Yukawa Potential 102


6.8 How to calculate observables 104

6.8.1 Fermi’s Golden Rule 104

6.8.2 Decay Rates 106

6.8.3 Example: two-body phase space 108

6.8.4 Examples: particle decays 108

6.9 Cross Sections 109

6.10 Green’s Functions 110

6.10.1 The true vacuum 110

6.10.2 Connected diagrams and vacuum bubbles 112

6.10.3 Outlook to S-matrices from Green’s functions 114

7 Lorentz Representations 117

7.1 The Lorentz algebra 118

7.1.1 Representations 121

7.1.2 Spinor representations 121

7.1.3 The Dirac equation 124

7.1.4 Lorentz-invariant actions 127

7.1.5 The Dirac Lagrangian 129

7.1.6 The slash 130

7.1.7 Chiral spinors and the Weyl equation 130

7.1.8 γ5 131

7.2 Discrete symmetries 132

7.2.1 Parity 132

7.3 Solutions of the free Dirac equation 133

7.3.1 Negative energy solutions 135

7.3.2 Examples 135

7.3.3 Helicity 136

7.3.4 Useful results: inner and outer products 136

7.3.5 Outer products 137

7.4 Symmetries of the Dirac Lagrangian 138

7.4.1 Internal Symmetries 139

7.5 Quantizing the Dirac Field 140

7.5.1 The Hamiltonian 140

7.5.2 Quantization of fermions 140

7.5.3 Fermi-Dirac Statistics 142

7.5.4 Causal Fermi Fields 142


7.6 Perturbation Theory for Spinors 143

7.7 The Fermion Propagator 144

7.7.1 The Feynman propagator for Dirac fields 145

7.7.2 Nucleon-Nucleon scattering with spin 146

7.8 Feynman rules for Dirac Spinors 148

7.8.1 Example: Nucleon scattering ΨΨ→ ΨΨ 149

7.8.2 Example: Nucleon-Anti-Nucleon scattering ΨΨ→ ΨΨ 149

7.8.3 Example: Meson Scattering 149

7.8.4 Example: the Yukawa potential 150

8 Quantum Electro Dynamics 153

8.1 The Feynman Rules for Quantum Electrodynamics 153

8.2 Maxwell’s equations 154

8.3 Gauge Symmetry 155

8.3.1 Gauges 156

8.4 Quantizing the Electro-magnetic field 157

8.4.1 Coulomb gauge 157

8.4.2 Lorentz Gauge 159

8.4.3 The propagator 164

8.5 Coupling gauge fields to matter 164

8.5.1 Coupling photons to Dirac fermions 164

8.5.2 Coupling photons to charged scalars 165

8.6 The electric charge 166

8.7 Discrete symmetries of QED 166

8.7.1 Parity 167

8.7.2 Charge conjugation 167

8.7.3 Majorana neutrino 169

8.7.4 Furrys theorem 170

8.8 Time reversal 170

8.8.1 Quantum mechanics 170

8.8.2 Scalar Field 171

8.8.3 Spinor Field 172

8.9 Electron scattering and the Coulomb potential 172

8.9.1 Coulomb potential 173

9 Outlook 175

1

Foreword

These notes are not original but mostly a combination of results

found in books and lecture notes. My contribution is here foremost

in the selection and the perspective provided. I have made liberal

use of the following references The ? indicates the technical level of

the book. It is roughly proportionalto the time needed per page for afull understanding. Note, that theZee book in particular is written in

a conversational tone, but coversmore concepts than any of the otherbooks.

• Peskin, Schroeder - An Introduction to Quantum Field Theory ??

• Schwartz - Quantum Field Theory and the SM ??

• Mandl, Shaw - Quantum Field Theory ?

• Zee - Quantum Field Theory in a Nutshell ?

• Srednicki - Quantum Field Theory ? ? ?

• Weinberg - Quantum Field Theory I ? ? ??

• Ramond - Field Theory: a Modern Primer ? ? ?

• Itzykson, Zuber - Quantum Field Theory ? ? ?

and many more. I strongly suggest that you find a book (or books)

you like from the ones above and study it as a complement to these

lecture notes.

This course provides a hopefully gentle introduction into the

beautiful world of quantum field theory while it familiarizes you

with relativity. It is the first real theory course in that it is the

first course to tackle the theoretical framework that underlies all of

nature. Let’s go.

1.1 Units and conventions

There are three fundamental dimensionful constants in nature: the

speed of light c, Planck’s constant (divided by 2π) ~ , and Newton’s

constant GN . Their dimensions are

[c] = length× time−1 (1.1)

[~] = length2 ×mass× time−1 (1.2)

[GN ] = length3 ×mass−1 × time−2 (1.3)


The whole point of units is that you can choose whatever units are

most convenient! In particle physics and cosmology, we use “natural

units”

~ = c = 1 (1.4)

and so

length = time = mass−1 = energy−1 (1.5)

we can express all dimensionful quantities in terms of a single scale

which we choose to be mass or, equivalently, energy.1 I will tem- 1 Since E = mc2 has become E = m.

porarily reintroduce ~ and c whenever convenient, mostly to show

the classical ~→ 0 limit or when we start with relativity. To convert

the unit of energy back to units of length or time, we have to insert

the relevant powers of c and ~.

Energies will be given in units of eV (the electron volt)2 or more 2 The electron-volt is a unit of energyequal to approximately 1.6 × 10−19

joules (J). It is the amount of energy

gained by the charge of a singleelectron moving across an electricpotential difference of one volt.

often GeV = 109 eV or TeV = 1012 eV , since we are often dealing

with high energies. Newton’s constant defines a mass scale

GN = M−2P , MP = 1.22× 1019 GeV (1.6)

this is the Planck scale. It corresponds to a length lP ≈ 10−33 cm.

Useful conversion factors are

(1 GeV)−1(~c) = 0.1973 fm (1.7)

(1 GeV)/c2 = 1.783× 10−24 g (1.8)

Quantity Mass Length

Cosmological constant Λ1/4 10−3 eV 10−4 m

Neutrino masses mν ≤ 0.12 eV ≥ 10−9 m

Atomic length (bohr a0 = 1meα

) 3.73 eV 5.29 · 10−11 m

Electron mass me− 0.511 keV 3 · 10−13 m

Muon mass mµ− 105.7 MeV 1.9 · 10−15 m

Proton mass mp 938.272 MeV 2.1 · 10−16 m

Z boson mass mZ 91.1876 GeV 2.16 · 10−18 m

Higgs mass mh 125.09 GeV 1.57 · 10−18 m

Top quark mass mPolet 172.44 GeV 1.14 · 10−18 m

Planck Mass MP 2.4 · 1018 GeV 8 · 10−35 m

Table 1.1: Overview over the relevant

fundamental scales of Nature.

2

Introduction

2.1 Why quantum field theory?

It is the theory of the world. Here are the main arguments:

• All elementary particles look the same. An electron at the begin-

ning of the universe has the exact same properties as an electron

which is part of the atoms of your body. An excellent explanation

for this is, if we assume that there is one field per elementary

particle permeating the whole universe

Ψa = Ψa(x, t) (2.1)

Elementary particles are then local, quantized excitations of this

universal field. Each elementary particle corresponds to exactly

one quantum field.

• Once we try to combine special relativity (SR) with quantum

mechanics (QM), we find that particle number is not conserved

anymore

QM + SR = QFT

We will study Dirac’s argument for anti-particles in detail later.

Let it just suffice to say, that if we try to localize a particle in a

box smaller than This is the so-called Compton

wavelength.

λ =~mc

(2.2)

then the energy uncertainty is of order of the mass of the particle.

To see this, start with the Heisenberg relation for the momentum

uncertainty of a particle in a box of size L

∆p ≥ ~/L

Relativistically, momentum and energy are on the same footing

∆E ≥ ~c/L.1 But if the uncertainty exceeds ∆E = 2mc2 then 1 E2 = (pc)2 + (mc2)2

we can create particle anti-particle pairs out of the vacuum.

The Compton wavelength is always smaller than the de Broglie

wavelength λdB = h/p. If you like, the de Broglie wavelength is

the distance at which the wavelike nature of particles becomes

apparent; the Compton wavelength is the distance at which

the concept of a single point like particle breaks down completely.


2.1.1 Necessity of the field viewpoint: causality

We will show now that causality makes a multi-particle theory

necessary. We start with the amplitude for a particle propagating

from x0 to x during the time t

U(t) = 〈x|e−iHt|x0〉 (2.3)

The energy of a free particle in non-relativistic quantum mechanics is

E = p2/(2m) and so2 2 Inserting the identity

1 =

∫d3p

(2π)3|p〉〈p|

and

〈x|p〉 = eip·x

U(t) = 〈x|e−ip2/(2m)t|x0〉 (2.4)

=

∫d3p

(2π)3〈x|e−ip2/(2m)t|p〉〈p|x0〉 (2.5)

=1

(2π)3

∫d3p e−ip

2/(2m)t〈x|p〉〈p|x0〉 (2.6)

=1

(2π)3

∫d3p e−ip

2/(2m)teip·(x−x0) (2.7)

=( m

2πit

)3/2

exp

(im(x− x0)2

2t

)(2.8)

where in the last step we have used a general gaussian integral3 with 3 A very useful formula:∫dnx e

− 12

n∑i,j=1

Aijxixj+n∑i=1

Bixi

=

∫dnx e−

12xTAx+BTx

=

√(2π)n

detAexp

(1

2BTA−1B

)

A = it/m 1 and B = i(x − x0) This amplitude is non-zero for all x

and t, which implies that a particle can propagate between any two

points in an arbitrarily short time. In a relativistic theory, no signal

can propagate faster than light4 and this would mean that causality

4 More on that later.

is violated! We can try to amend the above formula by using the

relativistic Pythagoras for the energy

E2 = (pc)2 + (mc2)2 (2.9)

and taking the positive root5 5 Why?

U(t) = 〈x|e−i√p2+m2t|x0〉

=

∫d3p

(2π)3〈x|e−i

√p2+m2t|p〉〈p|x0〉

=1

(2π)3

∫d3p e−i

√p2+m2teip·(x−x0)

=1

(2π)3

∫ ∞

0

dpp2

∫ 2π

0

dφ

∫ π

0

dθ sin θ e−i√p2+m2teip|x−x0| cos θ

=1

(2π)2

∫ ∞

0

dpp2 e−i√p2+m2t

∫ 1

−1

d cos θeip|x−x0| cos θ

=1

2π2

1

|x− x0|

∫ ∞

0

dpp e−i√p2+m2t sin(p|x− x0|)

where we have chosen the z-axis along the x− x0 direction. We now The main idea of stationary phasemethods relies on the cancellationof sinusoids with rapidly varying

phase. If many sinusoids have thesame phase and they are addedtogether, they will add constructively.

If, however, these same sinusoidshave phases which change rapidlyas the frequency changes, they willadd incoherently, varying between

constructive and destructive additionat different times.https://en.wikipedia.org/wiki/

Stationary_phase_approximation

want an approximate solution for this integral for x2 t2 (far out-

side the light-cone). We will use the stationary phase approximation.

The phase function

φ(p) ≈ px− t√p2 +m2 (2.10)

https://en.wikipedia.org/wiki/Stationary_phase_approximation

https://en.wikipedia.org/wiki/Stationary_phase_approximation


has a stationary point at

φ′(p)

∣∣∣∣p0

= 0 p0 = imx/√x2 − t2 (2.11)

Substituting this value back for p, we find that up to a rational

function of x and t, that You can also solve this integral

exactly in terms of Bessel-functionsand convince yourself of the veracity

of this statement.U(t) ∼ eiφ(p0) ∼ e−m

√x2−t2 (2.12)

We find a small, exponentially suppressed amplitude but still non-

zero outside the light-cone, causality is still violated.

We see from the exponential suppression that causality is violated

if we get to distances as small as x ∼ 1/m, for distances x 1/m

there is a negligible chance to find the particle outside the light-

cone6, so at distances much greater than the Compton wavelength 6 Light-cone: x2 = c2t2, withtemporarily reintroduced c.of a particle, the single-particle theory will not lead to measurable

violations of causality. This is in accordance with our earlier ar-

guments based on the uncertainty principle: multi-particle effects

become important when you are working at distance scales of order

the Compton wavelength of a particle.

Quantum field theory finds a miraculous solution to the causality

problem: the propagation across a space-like interval is indistin-

guishable from the propagation of an anti-particle in the opposite

direction. We will find that the amplitudes for particle and anti-

particle propagation exactly cancel outside the light-cone – causality

is preserved!

2.1.2 General features of relativistic quantum field theories

• CPT

• Spin-statistics

• Very constraining for higher spins, only S ≤ 2 theories.

• Massless particles (m = 0): S = 1 gauge theory (like electro-

magnetism), S = 2 gravity For the massless spin-2 field, the

requisite gauge principle can be shown to be general covariance,

which leads to Einstein’s theory, if you want to couple S = 3/2 to

gravity: unique theory super-gravity

2.2 String Theory

2.2.1 The discrete chain

We end up with a quantum field by quantizing a classical field. A

simple example is the classical string. We will define our string

as a long-wavelength limit (= low energy limit) of a harmonically

coupled discrete chain. We take N + 2 masses with a quadratic

nearest-neighbor coupling

L(q, q) = T − V =

N+1∑

j=0

[m2q2j −

κ

2(qj − qj+1)2

](2.13)


The coordinate qj(t) is the longitudinal displacement of the j-th

mass along the one-dimensional chain. The equilibrium distance a

between masses sets the rest location of the masses

xj ≡ j · a (2.14)qn1 qn qn+1

. . .. . .

Figure 2.1: One-dimensional discrete

chain.

At the endpoints, the chain is fixed:

q0(t) = qN+1(t) ≡ 0 (2.15)

The remaining N masses satisfy the equation of motion (EOM)

mqj − κ(qj+1 − 2qj + qj−1) = 0 (2.16)

for j = 1, . . . , N . The normal modes diagonalize the EOMs and have

the form

qj(t) = cos(ω t) sin(j p) (2.17)

The boundary-conditions in Eq. (2.15) require p to be

pn =πn

N + 1(2.18)

for n = 1, . . . , N . Substituting this into the EOMs, we get the

dispersion relations for the N independent normal frequencies ωn:

ωn = ω0 sin

(πn

2(N + 1)

), ω0 = 2

√κ

m(2.19)

The frequency ω0 is the cut-off frequency since the modes with

n > N are merely repeating the lower ones. For N = 4, there are 4 independentmodes. Above, n = 5 is trivial andequivalent to n = 0, since sin(jp5) =

sin(jπ) = sin(p0j) = sin(0) = 0. Andn = 6 e.g. is the same as n = 4, andso on.

2.2.2 The continuous string

The total length of the chain is

R = N a (2.20)

We find the continuum limit by taking the mass distance a to zero

while keeping the the total length R fixed

a→ 0, N →∞, R = N a = const. (2.21)

In this limit the discrete chain approaches a continuous string and

the labeled coordinates qj(t) approach a classical field defined as

q(x, t) ≡ qj(t) (2.22)

We can take the limit also for the Lagrangian by replacing

(qj − qj+1)2 = a2

(qj − qj+1

a

)2

→ a2

(∂q(x, t)

∂x

)2

(2.23)

∑

j

=1

a

∑

j

a → 1

a

∫ R

0

dx (2.24)

Keeping the mass density ρ and string tension σ finite, we define

ρ =m

a, σ = κ a (2.25)


to obtain the Lagrangian

L =1

2

∫ R

0

dx

[ρ

(∂q(x, t)

∂t

)2

− σ(∂q(x, t)

∂x

)2]

(2.26)

and the EOM1

c2∂2q(x, t)

∂t2− ∂2q(x, t)

∂x2= 0 (2.27)

which is of course a wave-equation with a speed of sound of

c2 =σ

ρ(2.28)

General solutions can be expanded in

q(x, t) = a · ei(kx−ωt) + b · e−i(kx−ωt) (2.29)

with a dispersion law

ω = c k (2.30)

As above, the boundary conditions of fixed ends

q(0, t) = q(R, t) = 0 (2.31)

give us a form of the normal modes

qn(x, t) = cos(ωn t) sin(kn x) (2.32)

with ωn = c kn and

kn =πn

R(2.33)

The normal-modes reproduce the discrete chain as long as n/N

0 2 4 6 8

0.0

0.2

0.4

0.6

0.8

1.0

1.2

1.4

!0

discrete chain

continuous string

Figure 2.2: Normal modes of aclassical string ωn ∼ n and adiscrete chain for N = 8 which has

ωn ∼ sin(n/N).

1. We find however that the number of modes of the continuous

string is infinite! Only the first N modes correspond to those of the

discrete chain. Therefore, we see that there is a cut-off frequency

ωC which is

ωC ≡ ωN =πc

a∼ 1/a (2.34)


This is of the same parametric size as the maximum frequency of the

discrete chain ω0 = 2√

κm = 2 c/a, see Eq. (2.19). Our continuum

model q(x, t) is a good representation of the discrete system only for

ω ωC .

This is a first encounter with the important concept of renormalization. The cut-off frequency

ωC ∼ 1/a

is a theoretical necessity. The cut-off tells us until which scale, we can trust our continuum theory. A continuous

string at frequencies ω ωC is not a good description of the discrete chain: the minimal wavelength is set by the

atomic distance, below which there is nothing to oscillate. Without a cut-off the specific heat of the string would

diverge since at a temperature T each mode (equipartition theorem in statistical physics) contributes an amount

kT , where k is the Boltzmann constant.

Here, we cannot determine the value of the cut-off from the long-wavelength theory alone, because only the

combination

c ∼ aωCappears. We can absorb the cut-off in measurable parameters, as in Eq. (2.25). This is called renormalization. If

this can be done, a theory is called renormalizable. You will discuss this at length in the chapter of loop correc-

tions in quantum field theories.

If a theory is non-renormalizable, the behavior can be sensitive to short-distance physics, which in this case would

be the atomic motion. This behavior does appear random on macroscopic scales, as in the propagation of a crack

or the nucleation of a raindrop.

2.2.3 Quantum string theory

We quantize the classical continuum theory to obtain a first example

of a quantum field theory. The Hamiltonian of the discrete chain is

H(p, q) = T + V =

N∑

j=1

[p2j

2m+κ

2(qj − qj+1)2

](2.35)

with pj = mqj . We can quantize the system, if we can replace pj and

qj with hermitian operators which satisfy the commutation In the following we will mostly set~ = 1.

[pj , qk] = −i~δjk (2.36)

We will switch to periodic boundary conditions now, Think of the system as being on atorus. The spacing of the frequenciesis doubled but each frequency ap-

pears two-fold. We are not interestedin the behaviour at the endpoints

and we will be at large N .

qj+N (t) = qj(t) (2.37)

since they allow us to use just the exponentials

qj(t) =1√N

N/2∑

n=−N/2Qn e

i2πnj/N (2.38)

pj(t) =1√N

N/2∑

n=−N/2Pn e

i2πnj/N (2.39)

where the operators Pn and Qn satisfy

[P †n, Qm] = −iδnm, (2.40)

P †n = P−n (2.41)

Q†n = Q−n (2.42)


Substituting back into the Hamilto-

nian, showing just the p2j -term.

N∑j=1

p2j =1

N

N∑j=1

N/2∑n=−N/2

N/2∑m=−N/2

Pn ei2πnj/N Pm ei2πmj/N

=

N/2∑n=−N/2

N/2∑m=−N/2

Pn Pm1

N

N∑j=1

ei2π(n+m)j/N

︸︷︷︸δn,−m

=

N/2∑n=−N/2

Pn P−n =

N/2∑n=−N/2

Pn P†n

=

N∑n=1

P †n Pn

where in the last step we have usedthat P †n and Pn commute and that

the product P †n Pn therefore is

symmetric n→ −n.The orthogonality relation follows

from the geometric series.

N∑j=1

ei2π(n+m)j/N =1− ei2π(n+m)j

1− ei2π(n+m)j/N= 0

because we have ei2π(n+m)j = 1since n,m ∈ N and ei2π(n+m)j/N 6= 1for n+m 6= 0 . The case n+m = 0 is

trivial.

Ex: Show the result for the qjsubstitution!

The Hamiltonian is now simplified to a sum of independent

harmonic oscillators

H =

N∑

n=1

[1

2mP †nPn +

1

2mω2

n Q†nQn

](2.43)

with eigen-frequencies

ω2n =

4κ

msin2

(πn

N

)(2.44)

As you remember from your QM course, we can write the energy-

eigenvalues of the harmonic oscillator with frequency ω as

εn = ~ω(

1

2+ n

), n = 0, 1, . . . (2.45)

where n is the occupation number. We can therefore write the energy-

eigenvalues of H as a set of occupation numbers

Eα =

N/2∑

n=−N/2ωn

(1

2+ αn

)(2.46)

with αn = 0, 1, 2, . . ..

2.2.4 Review of the simple harmonic oscillator

Consider the quantum mechanical Hamiltonian

H =1

2p2 +

1

2ω2q2 (2.47)

with the canonical commutation relations

[q, p] = i (2.48)

To find the spectrum we define the creation and annihilation opera-

tors (also known as raising/lowering operators, or sometimes ladder

operators)

a =

√ω

2q +

i√2ωp, (2.49)

a† =

√ω

2q − i√

2ωp (2.50)

which we can invert to

q =1√2ω

(a+ a†), (2.51)

p = −i√ω

2(a− a†) (2.52)

Substituting the expressions we find

[a, a†] = 1 (2.53)


while the Hamiltonian is

H =1

2ω(aa† + a†a) (2.54)

H = ω(a†a+1

2) (2.55)

One can easily confirm that the commutators between the Hamilto-

nian and the creation and annihilation operators are given by

[H, a†] = ωa†, (2.56)

[H, a] = −ωa (2.57)

Let |E〉 be an eigenstate with energy E, so that

H|E〉 = E|E〉 (2.58)

Then we can construct more eigenstates by operating with a and a†,

Ha†|E〉 = (E + ω)a†|E〉, (2.59)

Ha|E〉 = (E − ω)a|E〉 (2.60)

So we find that we can generate a ladder of states for each k with

energies

. . . , E − ω,E,E + ω,E + 2ω, . . . (2.61)

The energy is bounded from below and there must be a ground state

|0〉 which satisfies

a|0〉 = 0 (2.62)

This has the ground-state energy (also called zero-point energy),

H|0〉 =1

2ω|0〉 (2.63)

2.2.5 Fock space of the discrete chain

For this simple system, the Hilbert space can be identified with

the tensor product of the Hilbert spaces of the individual atoms

or masses. Thus, if we identify with |Ψ〉n an arbitrary state in the

Hilbert space of the n-th atom, the states of the chain can then be

written as

|Ψ〉 = |Ψ〉1 ⊕ |Ψ〉2 ⊕ . . .⊕ |Ψ〉N (2.64)

This is a first example of the Fock space of multi-particle systems.

We can further diagonalize the Hamiltonian by making use of the

creation and annihilation operator formalism for the harmonic

oscillator. We will drop the ”ˆ” for operatorsfrom now on since it will be clear

from the context.ak ≡

√mωk

2

(Qk +

i

mωkPk

)(2.65)

a†k ≡√mωk

2

(Q−k −

i

mωkP−k

)(2.66)


With this definition we find that the ladder operators obey the

commutation relations.

[ak , a†k′ ] = δkk′

[ak , ak′ ] = [a†k , a†k′ ] = 0

(2.67)

(2.68)

Proof: Plugging the definitions of ak and a†k into the commutator

and using the fact that the commutator is bilinear, we can calculate7 7 Thanks to Nepomuk Ritz!

[ak, a

†k′

]=

[√mωk

2

(Qk +

i

mωkPk

),

√mωk′

2

(Q−k′ −

i

mωk′P−k′

)]

=m√ωkωk′

2

[Qk, Q−k′ ]︸︷︷︸

=0

− i

mωk′[Qk, P−k′ ] +

i

mωk[Pk, Q−k′ ] +

1

m2ωkωk′[Pk, P−k′ ]︸︷︷︸

=0

=m√ωkωk′

2

i

mωk′

[P †k′ , Qk

]

︸︷︷︸=−iδkk′

− i

mωk

([P †k , Qk′

])†

=m√ωkωk′

2

(δkk′

mωk′+δkk′

mωk

)=mωk

2

2

mωkδkk′ = δkk′

where we have also used that P−k = P †k , Q−k = Q†k and that[P †k′ , Qk

]= −iδkk′ .

[ak, ak′ ] =

[√mωk

2

(Qk +

i

mωkPk

),

√mωk′

2

(Qk′ +

i

mωk′Pk′

)]

=m√ωkωk′

2

[Qk, Qk′ ]︸︷︷︸

=0

+i

mωk[Pk, Qk′ ] +

i

mωk′[Qk, Pk′ ]−

1

m2ωkωk′[Pk, Pk′ ]︸︷︷︸

=0

Since the commutator is antisymmetric, we have [Qk, Pk′ ] =

− [Pk′ , Qk] and therefore, upon exchanging the indices k and k′

for this term (which we can do, because [Pk′ , Qk] = −iδk,−k′), the

remaining terms cancel and we have

[ak, ak′ ] = 0.

If we take the hermitian conjugate of this equation, we immediately

get

[ak, ak′ ]†

=[a†k′ , a

†k

]= −

[a†k, a

†k′

]= 0

and therefore also [a†k, a

†k′

]= 0.

which completes the proof.

The Hamiltonian becomes

H =

N/2∑

k=−N/2ωk

(a†kak +

1

2

)(2.69)


Further we can show that

[H, ak] = −ωkak (2.70)

[H, a†k] = ωka†k (2.71)

and therefore these operators take us between energy eigenstates.

Let |E〉 be an eigenstate with energy E, so that

H|E〉 = E|E〉 (2.72)

Then we can construct more eigenstates by operating with ak and

a†k,

Ha†k =∑q

ωq(a†qaq + 1/2)a†k

=∑q

ωq(a†qaqa

†k + 1/2a†k)

=∑q

ωq(a†q(a†kaq + δkq) + 1/2a†k)

= a†k(H + ωk)

Ha†k|E〉 = (E + ωk)a†k|E〉, (2.73)

Hak|E〉 = (E − ωk)ak|E〉 (2.74)

So we find that we can generate a ladder of states for each k with

energies

. . . , E − ωk, E,E + ωk, E + 2ωk, . . . (2.75)

The energy is bounded from below and there must be a ground state

|0〉 which satisfies

ak|0〉 = 0 for all k (2.76)

This has the ground-state energy (also called zero-point energy),

H|0〉 =1

2

∑

k

ωk|0〉 (2.77)

Excited states then can be generated by repeated application of a†k,

|n1, n2, . . . , nN 〉 = (a†1)n1(a†2)n2 · · · (a†N )nN |0〉 (2.78)

where we have not been careful about normalization, so 〈n|n〉 6= 1. Ex: Derive the correct normalization

factor.

We now again move to the continuum limit with x = j · a to find

Hcont =

∫ R

0

dx

[1

2ρπ2(x, t) +

σ

2

(∂φ(x, t)

∂x

)](2.79)

and the conjugate momentum operator

π(x, t) =pj(t)

a= ρ

∂qj(t)

∂t= ρ

∂φ(x, t)

∂t(2.80)

The quantum field φ(x, t) and its conjugate momentum π(x, t)

satisfy the equal-time commutation relation This is the generalization of the

canonical commutation relations ofone-particle quantum mechanics,

[xa, pb] = iδab.[φ(x, t), π(x′, t)] = i~ δ(x− x′) (2.81)

where I have temporarily reintroduced ~ to show how to recover the

classical limit ~→ 0. Also for quantum string we have to introduce a

cut-off frequency ωC .


We have seen that in particular for distances smaller than x ∼1/m or energies above E ∼ m, we need to extend the QM frame-

work because of causality violation and the possibility of particle

creation. These energies are however also energies where relativistic

effects become important. Therefore we will spend some time on the

basics of relativity in the next chapter.

We started with a discrete chain for pedagogical reasons. Of course we do not believe that the fields observed

in Nature, such as the electron field or the photon field, are actually constructed of point masses tied together

with springs. A modern picture which was first used by Landau-Ginzburg and fully understood by S. Wein-

berg, is that we start with the desired symmetry, say Lorentz invariance or a conserved U(1) charge (more on

both later) decide on the fields we want by specifying how they transform under the symmetry (in this case we

decided on a scalar field φ(x, t) as the local displacement) and then write down the action involving no more than

two time derivatives – because we don’t know how to quantize actions with more than two time derivatives (we

usually end up with ghosts). See also the discussion in the statistical physics script (in particular the chapter:

Landau-Ginzburg Theorie)

3

Relativity

3.1 Galileo and Newton

Two observers, one on a moving train (x, y, z) with velocity u and

one on the ground (x′, y′, z′), can be related by

t′ = t (3.1)

x′ = x+ u · t (3.2)

y′ = y (3.3)

z′ = z (3.4)

which are known as the Galilean transformations. The differen-

tial form is

dt′ = dt (3.5)

dx′ = dx+ u · dt (3.6)

We can add velocities trivially: e.g. a ball with 3m/s on a train

with 20m/s will be observed on the ground as 3m/s + 20m/s =

23m/s. We see this also with the Galilean transformations

v′ =dx′

dt′=

d

dt′(x+ u t) =

dx

dt+ u = v + u (3.7)

We now see that the invariance of Newtonian mechanics under

Galilean transformations follows merely because Newton’s law

involves 2nd derivatives, so

md2x′

dt′2= m

d

dt′

(dx

dt+ u

)= m

d2x

dt2(3.8)

We also see importantly where Galilean invariance in Newtonian

mechanics fails: if u changes in magnitude or direction with time

then we find1 1 All inertial reference frame is aframe in which Newton’s first axiom

holds, i.e. every object on which noforce is acting either stays at rest ormoves uniformly.

F ′ = ma′ = ma+mdu

dt(3.9)

we feel this as an additional force, e.g. as a centrifugal force in an

accelerated frame of reference. Let us check that the Newtonian

action is invariant, too

S =

∫dt′

1

2m

(dx′

dt′

)2

=

∫dt

1

2m

(dx

dt+ u

)2

(3.10)

=

∫dt

1

2m

(dx

dt

)2

+ u

∫dtm

dx

dt+ u2

∫dt

1

2m (3.11)


for fixed initial and final conditions, the second term is just an

irrelevant additional constant, like the last term. We can generalize

this to a many particle system with interactions

S =

∫dt

∑

a

1

2ma

(dxadt

)2

−∑

a6=bV (xa − xb)

(3.12)

Note, that it is necessary for the interaction potential to just de-

pend on the difference xa − xb. We also see that all observers have

to agree on the same mass m, there is no m′ otherwise Galilean

transformations would not work.

3.1.1 Maxwell vs. Newton

Figure 3.1: Michelson and Morley’sinterferometric setup, mounted on a

stone slab that floats in an annular

trough of mercury.

Maxwell’s equations are not invariant under Galilean transforma-

tions. In particular, the velocity of light is the same in all inertial

systems

c′ = c (3.13)

contradicting Eq. (3.7). Various eminent physicists in the late 19th

century tried to reconcile this fact by postulating that light – like

sound – had to propagate in a medium, the aether. In 1887 (when

Einstein was eight years old) Michelson and Morley performed a

famous experiment to detect the aether and failed.

The Michelson–Morley experiment was performed over the spring and summer of 1887 by Albert A. Michel-

son and Edward W. Morley at what is now Case Western Reserve University in Cleveland, Ohio, and published

in November of the same year. It compared the speed of light in perpendicular directions, in an attempt to detect

the relative motion of matter through the stationary luminiferous aether (‘’aether wind”). The result was negative,

in that the expected difference between the speed of light in the direction of movement through the presumed

aether, and the speed at right angles, was found not to exist; this result is generally considered to be the first

strong evidence against the then-prevalent aether theory, and initiated a line of research that eventually led to

special relativity, which rules out a stationary aether.

Michelson–Morley type experiments have been repeated many times with steadily increasing sensitivity. These

include experiments from 1902 to 1905, and a series of experiments in the 1920s. More recent optical resonator

experiments confirmed the absence of any aether wind at the 10−17 level. Together with the Ives–Stilwell and

Kennedy–Thorndike experiments, Michelson–Morley type experiments form one of the fundamental tests of special

relativity theory. https://en.wikipedia.org/wiki/Michelson-Morley_experiment

3.1.2 Derivation of the Lorentz transformations

We consider two inertial systems and assume that they coincide for

t = 0 = t′ with a relative velocity v.

Let us know see how we can modify the Galilean transformations

so that a spherical light wave emanating from the origin

(ct)2 = x2 + y2 + z2 (3.14)

does not depend on the observer and is the same in the moving

inertial system

(ct′)2 = x′2 + y′2 + z′2 (3.15)

https://en.wikipedia.org/wiki/Michelson-Morley_experiment


Let us assume that the origin is moving along the x-axis. We know Three useful hyperbolic function

formulae

cosh2 η − sinh2 η = 1, (H1)

sinh η =tanh η√

1− tanh2 η, (H2)

cosh η =1√

1− tanh2 η(H3)

H2 and H3 can be easily shown byplugging in the definition of tanh η

and use of H1.

then

y = y′, z = z′ (3.16)

The most general linear solution is, as can be verified by direct

substitution using (H1)

(x

ct

)=

(cosh η sinh η

sinh η cosh η

)(x′

ct′

)(3.17)

This should remind you of an euclidean rotation which leaves x2 + y2

invariant (instead of (ct)2 − x2).

To find the role of η in the physical setting, record the origin’s

progression, i.e. x′ = 0, x = vt. The equations become (using first

x′ = 0),

x = ct′ sinh η, (3.18)

ct = ct′ cosh η. (3.19)

Now divide :

x

ct= tanh η =

v

c= β ⇒

sinhη =vc√

1− v2

c2

, cosh η =1√

1− v2

c2

= γ,

where x = vt was used in the first step, (H2) and (H3) in the second,

which, when plugged back in (3.17), gives

x =x′ + vt′√

1− v2

c2

, t =t′ + v

c2x′

√1− v2

c2

, (3.20)

or, with the usual abbreviations,

x = γ(x′ + vt′),

t = γ

(t′ +

vx′

c2

),

(3.21)

(3.22)

We see that the transformations for time and space are now symmet-

ric2! With 2 Very clear if you set c = 1.

Or,

x′ = γ(x− vt), (3.23)

t′ = γ(t−

vx

c2

). (3.24)

γ =1√

1−(vc

)2 (3.25)

3.2 Consequences of special relativity

All reference frames in this section are inertial frames.


ct

x

future

past

light-coneworld line

space-like

time-like

Figure 3.2: Light-cones and a world-line in Minkowski-Space.

3.2.1 Relativity of simultaneity

Two events happening in two different locations x1 and x2 that

occur simultaneously E.g. by sending light signals back

and forth. https://en.wikipedia.

org/wiki/Einstein_synchronisationt1 = t2 ∆t = 0 (3.26)

in the reference frame of one observer A, may occur non-simultaneously

in the reference frame of another observer B′. Let’s move in the

boosted frame B′ traveling with v relative to A.

t′1 = γ(t− vx1

c2

), t′2 = γ

(t− vx2

c2

)(3.27)

∆t′ = t′1 − t′2 = γv

c2(x2 − x1) 6= 0 (3.28)

This is counter-intuitive and seems to contradict our usual notion of

cause and effect. Can we change the order of two events? If t2−t1 > 0,

we want that

0 < t′2 − t′1 = γ(t2 − t1 −

v

c2(x2 − x1)

)(3.29)

We have to have that

t2 − t1 >v

c

x2 − x1

c(3.30)

Since v < c, the order of the events is kept for

t2 − t1 ≥x2 − x1

c(3.31)

If two events are to be causally connected, the cause can only be

transmitted with finite speed v ≤ c. This means

t2 − t1 =x2 − x1

v≥ x2 − x1

c(3.32)

https://en.wikipedia.org/wiki/Einstein_synchronisation

https://en.wikipedia.org/wiki/Einstein_synchronisation


We find that cause and effect cannot be exchanged.

However, the order of causally disconnected events can very well

be changed.

x

t

x’

t’

A BtA tB = 0

t0 > 0

Figure 3.3: Relativity of simultaneityin a space-time diagram (c = 1). The

coordinate lines show unit-ticks andunit-distances in the restframe (x, t)

and a moving inertial frame (x′, t′)with v = 0.3. Two simultaneousevents A and B are happening at

different times in (x′, t′).

3.2.2 Time dilation

Let us assume that we have a clock in A which ticks every ∆t sec-

onds. In B′ we will find the ticks at

∆t′ = γ∆t =∆t√

1−(vc

)2 ≥ ∆t (3.33)

The stationary clock ticks faster! Why is the situation not symmet-

ric? An observer in B′ will have to use two clocks to compare to the

for him moving clock of A.

We can now define the important quantity of proper time ∆τ :

it is the time measured by a clock which is at a fixed place. We can

observe time dilation in (almost) everyday experiments like the decay

of unstable particles, like the muons generated in the atmosphere by

cosmic radiation. See Ex1!

3.2.3 Lorentz contraction

Suppose we measure a length as ∆x = l in A. How did we perform

the measurement? We put a standard ruler on the distance and

compare the endpoints simultaneously ∆t = 0. A measurement

of the length in B′ moving with v relative to A is characterized by

∆t′ = 0. The positions of the endpoints in B′ are

∆x′ = γ(∆x− v∆t) (3.34)


where ∆t = t2 − t1 =?. We have to measure simultaneously in B′:

t1 −v

c2x1

!= t2 −

v

c2x2 (3.35)

Therefore

t1 − t2 =v

c2(x1 − x2) (3.36)

Plugging back

l′ = x′1 − x′2 = γ

(x1 − x2 −

v2

c2(x1 − x2)

)(3.37)

which means

l′ = l

√1− v2

c2(3.38)

A stick of length l at rest in A will appear contracted in B′ by a

factor√

1− v2

c2 < 1. Crucial for the length measurement is the

requirement of simultaneity.

x

t

x’

t’

time dilationt = t0 > t0

Figure 3.4: Time dilation in aspace-time diagram (c = 1). Thecoordinate lines show unit-ticks andunit-distances in the restframe (x, t)

and a moving inertial frame (x′, t′)with v = 0.3. We see that we needtwo clocks in the rest frame to

measure the moving system’s clock.

3.2.4 Addition of velocities

We can now derive the correct law of addition of velocities. Consider

again the simple case of an object moving in the x direction. The

Lorentz transformation reads

x′ =1√

1−(vc

)2 (x+ vt), (3.39)

t′ =1√

1−(vc

)2(t+

vx

c2

). (3.40)


x

t

x’

t’

Lorentz contraction

l = l0/ < l0

Figure 3.5: Length contraction in aspace-time diagram (c = 1). Thecoordinate lines show unit-ticks and

unit-distances in the restframe (x, t)and a moving inertial frame (x′, t′)with v = 0.3. We see that when we

measure the length of the movingmeasuring stick, we do so at thesame time t = 0, corresponding totwo different times t′1 6= t′2 in the

moving system.

Let the velocity seen in A be u = dxdt and in B′ be u′ = dx′

dt′ . Then

dividing dx′ by dt′. We obtain

u′ =dx′

dt′=vdt+ dx

dt+ vdxc2

=v + dx

dt

1 +v dxdtc2

=v + u

1 + vuc2

(3.41)

Instead of the Galilean u′ = u+ v, the correct law contains a crucial

denominator

u′ =u+ v

1 + vuc2

(3.42)

This function is remarkable: it is symmetric under v ↔ u. If the

object is slowly moving u c, then u′ ≈ u + v, in accordance with

everyday intuition. But if a particle happens to be moving at the

speed of light (e.g. a photon), u = c, then u′ = c+v1+ vc

c2= c.

3.3 Minkowski spaceWe really, really set c = 1 from now

on.In Euclidean space, the invariance of the combination

dl2 = dx2 + dy2 + dz2 (3.43)

allows us to define dl as the distance between two points. Two ob-

servers whose coordinate systems are rotated to each other measure

the same distance.3 3 Indeed, the invariance of dl2 definesrotations and the Lie group SO(3).With profound insight, Minkowski realized4 that the invariance of4 In 1907, a mere 2 years after SR.

the combination

ds2 = dt2 − dx2 − dy2 − dz2 (3.44)


allows us to talk about distance in space-time. This invariance

determines the Lorentz transformation5. Similar to the case of 5 Or the SO(1, 3) rotations

rotations, two observers in uniform motion relative to each other can

now agree on the space-time distance between two points.

We say that the separation between two nearby points in space-

time is

ds2 > 0 dt2 > dx2 + dy2 + dz2 timelike

ds2 = 0 dt2 = dx2 + dy2 + dz2 lightlike

ds2 < 0 dt2 < dx2 + dy2 + dz2 spacelike

The first two are causally connected. Light-like is also sometimes

called ’null’.

We see that we can identify ds with a proper-time intervall dτ

since it is the time measured by a resting particle ds2 = dt2. Since

ds2 is Lorentz-invariant, we can determine it in any inertial system

by calculating ds2 = dt2 − (dx2 + dy2 + dz2), or

dτ2 = ds2 = dt2(

1− dx2 + dy2 + dz2

dt2

)(3.45)

= dt2(1− v2(t)

)(3.46)

which is exactly the already discussed time dilation.

For an arbitrarily moving particle x(t) with a velocity v(t) =

dx(t)/dt, we can calculate

τB − τA =

∫ τB

τA

dτ =

∫ tB

tA

dt√

1− v2(t) ≤ tB − tA (3.47)

3.3.1 Fourvectors

We define space-time points X using the components of a four-vector

xµ ≡

x0

x1

x2

x3

≡

ct

x

y

z

(3.48)

The components are defined relative to basis vectors eµ

X = xµeµ =

3∑

µ=0

xµeµ (3.49)

where we have used the Einstein summation convention, when

an index variable appears twice in a single term, it implies summa-

tion of that term over all the values of the index. In the relativistic

context, we will always require one ’upstairs’ and one ’downstairs’

index. In the following, we will therefore not show the ”∑3µ=0”

anymore.

A Lorentz transformation can then be written as

x′µ = Λµνxν (3.50)


where we have moved one index to the right to make obvious, which

one is the first and which one is the second index of the matrix.

Since we have only changed the coordinates and the physical vectors

has to remain the same, we know

X = xµeµ = x′µe′µ = Λµν xνe′µ (3.51)

we find for the basis vectors

eν = Λµν e′µ (3.52)

and a Lorentz transformation along the x-direction would look like

(Λµν) =

γ γβ 0 0

γβ γ 0 0

0 0 1 0

0 0 0 1

(3.53)

with γ = 1/√

1− v2/c2 and β = v/c. This is called a boost. We

can of course also parametrize ordinary rotations within a Lorentz

transformation, say a rotation around the z-Axis

(Λµν) =

1 0 0 0

0 cos θ − sin θ 0

0 sin θ cos θ 0

0 0 0 1

(3.54)

3.3.2 Minkowski Metric

Lorentz transformations leave the relativistic distance ds2 invariant.

This defines a metric of the Minkowski-space

ds2 = dt2 − dx2 − dy2 − dz2 = ηµν dxµdxν (3.55)

with

(ηµν) =

1 0 0 0

0 −1 0 0

0 0 −1 0

0 0 0 −1

(3.56)

The invariance of ds2 then implies

ds2 = ηµν dxµdxν = ηµν dx

′µdx′ν = ηµν ΛµαΛνβ dxαdxβ (3.57)

and we derive the condition

ηαβ = ηµν ΛµαΛνβ (3.58)

which we will use extensively later. This defines the set of all

Lorentz transformations.6 6 To get the representations of theLorentz group SO(1, 3).


3.3.3 Discrete transformations

Space-time distances as measured by the Minkowski metric, or

Minkowski scalar products like

VµWµ (3.59)

are left invariant not only by transformations which can be continu-

ously connected to the the identity, but also by discrete transfor-

mations like parity Using

ηαβ = ηµν ΛµαΛνβ

and taking the determinant, we find

det Λ = ±1

P : (t, x, y, z) −→ (t,−x,−y,−z) (3.60)

or time reversal

T : (t, x, y, z) −→ (−t, x, y, z) (3.61)

Parity and time reversal are special because they cannot be written

as a product of rotations and boosts7. We can write them as 7 Because rotations and boosts

reduce to 14×4 for v → 0 or θ → 0.

P =

1 0 0 0

0 −1 0 0

0 0 −1 0

0 0 0 −1

, T =

−1 0 0 0

0 1 0 0

0 0 1 0

0 0 0 1

(3.62)

As you will learn in the discussion of the Standard Model of particle

physics, neither symmetry is conserved in nature!8 8 The weak interaction SU(2)L ×U(1)Y are explicitly parity violating

because the gauge charges distinguishbetween left-handed and right-

handed fermions. T is violated in

flavor changing transitions, since theYukawa couplings of the quarks have

a complex phase.

3.3.4 Motion of a free particle

Newton enunciated that in the absence of external forces, a particle

will maintain a constant velocity. Hence the equation of motion

d2X

dt2= 0 (3.63)

or in other words dXdt stays unchanged.

How does a free particle move in special relativity? Again, the

answer had to be enunciated (or guessed) by Einstein. He knew that

it must reduce to Newton’s equation for a slowly moving particle.

Also, all inertial frame observers have to agree that the particle is

freely moving. Both requirements are satisfied by

d2Xµ

dτ2= 0 (3.64)

where Xµ = Xµ(τ) denotes the location of a particle with τ being

the proper time. In another inertial frame X ′µ = ΛµνXν and so

d2X ′µ

dτ2= Λµν

d2Xν

dτ2= 0 (3.65)

and that dτ is the same for both observers. This motivates the

introduction of the four-velocity

uµ ≡ dxµ

dτ(3.66)


which because of dt/dτ = γ(v) can be written in terms of the usual

three-velocity v as

uµ =dxµ

dt

dt

dτ= γ(v)

(c

v

)(3.67)

The (Lorentz) square of the four-velocity is9 9 See Ex for a four-acceleration

u2 = uµuµ = γ(v)2(c2 − v2) = c2 (3.68)

3.4 Indices upstairs and downstairs

For now ηµν is the only object with lower indices. When we want to

sum over indices, the rule is that we multiply by ηµν and invoke the

Einstein summation convention, e.g. for two vectors pµqν

p · q ≡ ηµν pµqν (3.69)

To save ourselves from constantly writing the metric ηµν , we define,

when we are given a vector pµ a vector with lower index

pν ≡ ηµνpµ (3.70)

Or in other words,

pµ = (p0,p), and pµ = (p0,−p) (3.71)

and therefore

p · q = pµqµ = p0q0 − p · q (3.72)

I hope you are not confused by this trivial act of notational sloth

and the fact that we have ”two kinds of vectors”, contravariant

vectors pµ and covariant vectors pµ. There is nothing profound

going on here. Just a convenient notation.10 10 See however box below!

Indeed, we can also define an inverse for ηµν , (ηµν)−1 = ηµν

ηµνηνλ = δλµ (3.73)

As you can see it is numerically the same matrix11 and we will use 11 In general curved space-times themetric can be a function of xµ and

the inverse will in general not be thesame matrix.

the same symbol η with two covariant indices: ηµν . From this we can

see there is no such thing as δµν or δµν , neither does ηµν exist !

3.4.1 A convenient notation and derivatives

It follows that the short-hand ∂µ for ∂∂xµ has to carry a lower index

because

∂µxν =

∂xν

∂xµ= δνµ (3.74)

We will use this fact repeatedly. You can also argue that a variation

of a Lorentz invariant scalar field φ(xµ) should still be a scalar

δφ =∂φ

∂xµδxµ (3.75)


and therefore

∂µ =∂

∂xµ=

(∂

∂t,∂

∂x,∂

∂y,∂

∂z

)(3.76)

and

∂µ =∂

∂xµ=

(∂

∂t,− ∂

∂x,− ∂

∂y,− ∂

∂z

)(3.77)

A useful mnemonic is to think of xµ carrying an upper index but

since in ∂∂xµ it appears in the denominator, it should carry a lower

index.

We will also use

∂µAµ = ∂0A

0 + ∂jAj (3.78)

and

∂µ∂µ =

∂2

∂t2−∇2 = (3.79)

Finally, we will make use of the completely anti-symmetric tensor

εµναβ , known as the Levi-Civita tensor. It is defined as

εµναβ =

1, if µναβ is an even permutation of (0, 1, 2, 3)

−1, if µναβ is an odd permutation of (0, 1, 2, 3)

0, if µναβ is not a permutation of (0, 1, 2, 3)

(3.80)

So e.g.ε0123 = −ε1023 = 1 and ε0223 = 0.

3.4.2 The appearance of anti-matter

An actual physics process (event A), a proton turns into a neu-

tron by emitting a pion, the π+, which by conservation of charge

necessarily carries positive charge. In event B, the π+ is absorbed

which turns a neighboring neutron into a proton. This generates an

attraction between the proton and the neutron and is an effective

description of the strong force.

The same process as seen by two observers. Since the pion prop-

agates over a spacelike intervall, it is possible to see a temporal

order in which A occurs after B. This something, by charge conser-

vation, has to carry negative charge which implies the existence of

anti-matter.

3.5 Relativistic mechanics

We define the four-momentum of a particle of mass m with

pµ ≡ muµ (3.84)

which we can write in an inertial system where the particle has the

three-velocity v(t)

pµ =

(p0

p

)= mγ(v)

(c

v

)(3.85)


A+

B

n

np

p

AB

n

np

p

?

Figure 3.6: The need for anti-matter. The same process as seenby two observers. Since the pion

propagates over a spacelike intervall,it is possible to see a temporalorder in which A occurs after B.

This something ”?”, by chargeconservation, has to carry negativecharge

The dual space: The Minkowski metric defines a (not positive definite) norm for four vectors

∆s2 = 〈∆X ,∆X〉 = ∆xµ∆xν 〈eµ, eν〉︸︷︷︸ηµν

(3.81)

and a Lorentz-invariant scalar product for arbitrary vectors A,B

〈A,B〉 = aµbνηµν (3.82)

Every vector space V has a dual space, often denoted V ∗, which is simply the set of all linear maps from the

vector space into the real (or complex) numbers. We can find a basis eµ for the dual space with

eµ[eν ] = δµν (3.83)

where eµ[·] takes a vector to produce a number.

The metric gives us a canonical isomorphism between the space of covariant vectors V and its dual space V ∗,

allowing us to be sloppy and conflate the two. More general objects with n upper indices and m lower indices can

be thought of as living in a tensor product of copies of V and V ∗, i.e. V ⊗m ⊗ V ∗⊗n.

For us contra-variant and co-variant vectors are just a convenient notation but if you are a mathematician-want-to-

be you can read about this in more profound books, or here https://en.wikipedia.org/wiki/Dual_space.

The three-momentum is therefore

p = mγ(v)v (3.86)

This coincides with the usual definition of momentum if we introduce

a velocity dependent mass m(v) = γ(v)m. We call m = m(0) the

rest mass.

The relativistic energy12 E = cp0 is proportional to the 0- 12 See Ex for a derivation from the

relativistic Lagrangian

S = −mc∫ds

of a free point-particle.

component

E = cp0 = mγ(v)c2 =mc2√1− v2

c2

≈ mc2(

1 +1

2

v2

c2+ . . .

)(3.87)

where a new term, the rest energy, appeared. We find for the

square of the momentum

p2 = pµpµ = (p0)2 − p2 = m2c2 (3.88)

where we have used uµuµ = c2.

https://en.wikipedia.org/wiki/Dual_space


3.5.1 E = mc2

The energy and the momentum of a particle of mass m form a four-

vector

pµ =

(E

p

)(3.89)

whose square is

p2 = E2 − p2 = m2 (3.90)

This is the relativistic Pythagoras which defines a Lorentz-

invariant relationship between the energy, the rest-mass and the

momentum of a particle:

E2 = m2 + p2 (3.91)

3.5.2 Application: weak decay of a pion

+µ

µ+

Figure 3.7: Feynman diagram of a π+

decay.

A charged π± can decay to a muon µ± and a muon anti-neutrino νµ.

The muon has about 3/4 the mass of a pion13 and the neutrino is

13mπ± = 139.57 MeV and mµ± =105.658 MeV.

almost massless. We can calculate the energy and the velocity of the

decay products using the four-momentum conservation

pµini = pµfinal (3.92)

which is(Eµini

pini

)=

(Eπ+

pπ+

)=

(Eµ+ + Eν

pµ+ + pν

)=

(Eµfinal

pfinal

)(3.93)

In the rest-system of the pion we derive from energy conservation

Eπ+ = mπ+ = Eµ+ + Eν (3.94)

The almost mass-less neutrino has Eν = |pν | and because the initial

three-momentum was zero in the rest system of the pion, we get

|pν | = |pµ+ | =m2π+ −m2

µ+

2mπ+

(3.95)

For the decay to an electron and an electron anti-neutrino, the

electron mass is negligible: we find the decay products back-to-back

and the rest-mass has been converted entirely into kinetic energy

with |pν | = |pµ+ | = mπ+/2.

4

Classical Field Theory

We will first introduce various aspects of classical fields. We will

return to the subject at several later stages when we need new

concepts.

4.1 Dynamics of fields

In QM, observables are constructed out of q’s and p’s. Classical

fields are defined at each point in space-time.

φa(xµ) (4.1)

Thus we are dealing with a system with infinitely many degrees of

freedom,1 at least one for each point x in space. Notice, that the 1 Compare the discrete chain to the

continous string!concept of position has been demoted from a dynamical variable in

particle mechanics to a mere label in field theory.

4.1.1 Example: the electro-magnetic field

You are very familiar with E(x, t) and B(x, t) from classical electro-

dynamics. In a more advanced treatment of electromagnetism, we

derive the spatial 3-vectors from a single 4-component field

Aµ(xµ) =

(φ

A

)(4.2)

where the vector potential Aµ is a vector in space-time. The original

electric and magnetic fields are related by

E = −∇φ− ∂A

∂t, and B = ∇×A (4.3)

which have two of Maxwell’s equations as an immediate consequence

∇ ·B = 0, and∂B

∂t= −∇×E (4.4)

4.1.2 The Lagrangian density

The dynamics of fields is determined by a Lagrangian which is a

function of φ(x, t), φ(x, t), ∇φ(x, t). All the systems in this course

will have the form

L(t) =

∫d3xL(φa, ∂µφa) (4.5)


where the L is the Lagrangian density2. The action is 2 We will not be very careful in

distinguishing the Lagrangian densityand the Lagrangian.S =

∫ t2

t1

dt

∫d3xL (4.6)

Remember that in classical mechanics, L depends on q and q. In Interactions with multiple derivatives,

e.g.

L =1

M2φ2φ

may occur due to quantum effects in

all but the simplest renormalizablefield theories. The are generic in all

effective field theories, as we may

discuss at the end of the course, timepermitting.

field theory we analogously restrict to Lagrangians depending on

φ and φ and not on φ. We could in principle write Lagrangian’s

which depend on ∇φ, ∇2φ, ∇3φ and so on. But anticipating Lorentz-

invariance, we will only consider ∇φ and no higher derivatives. Also

we will not include an explicit xµ dependence, all such dependence

comes only implicitly through φ and its derivatives.

In the following, we will use functional derivatives. A functional F [φ] for us is a map from a vector space of

functions into the real numbers, like e.g. the action S[φ]. A functional derivative is denoted by δF [φ]δφ(x)

and it is

defined as the linear term in an expansion

F [φ+ η] = F [φ] +

∫dx′

δF [φ]

δφ(x′)η(x′) + . . . (4.7)

or ∫dx

δF

δφ(x)η(x) = lim

ε→0

F [φ+ εη]− F [φ]

ε(4.8)

=

[d

dεF [φ+ εη]

]ε=0

, (4.9)

where η is an arbitrary function. In particular, for the functional F [φ] = φ(x), we find

δF [φ]

δφ=

δφ(x)

δφ(x′)= δ(x− x′) (4.10)

The quantity εη is called the variation of φ. The differential (or variation or first variation) of the functional F [φ]

is

δηF (φ) =

∫dxδF

δφ(x) η(x) (4.11)

This is similar in form to the total differential of a function. You should familiarize yourself with the mechanics of

functional derivatives on the next exercise sheet.

We can derive the equations of motion by the principle of least

action. We vary the field at each space-time point, keeping the

end-points fixed and requiring

δS = 0 (4.12)

with

δS =

∫d4x

[∂L∂φa

δφa +∂L

∂(∂µφa)δ(∂µφa)

](4.13)

=

∫d4x

[∂L∂φa

− ∂µ(

∂L∂(∂µφa)

)]δφa + ∂µ

(∂L

∂(∂µφa)δφa

)

The last term is a total derivative and therefore its integral only

depends on the field values at the spatial and temporal boundaries.

It vanishes for any δφa that decays at spatial infinity and obeys Note, we will always make the

assumption that we can drop suchtotal derivatives from the Lagrangian.It let’s us integrate by parts without

consequence. We will thereforeequate

φ∂µψ = −(∂µφ)ψ

in a Lagrangian.

δφa(x, t1) = δφa(x, t2) = 0 (4.14)


Requiring δS = 0 then yields the Euler-Lagrange equations of

motions for the fields

∂L∂φa

− ∂µ(

∂L∂(∂µφa)

)= 0 (4.15)

4.1.3 Example 1: the Klein-Gordon equation

Consider a real scalar field φ(xµ) and the Lagrangian

L =1

2ηµν∂µφ∂νφ−

1

2m2φ2 (4.16)

=1

2(φ)2 − 1

2(∇φ)2 − 1

2m2φ2 (4.17)

Comparison with a the usual structure L = T − V we see, that we

have a kinetic energy

T =

∫d3x

1

2(φ)2 (4.18)

and the potential energy of the field

V =

∫d3x

[1

2(∇φ)2 +

1

2m2φ2

](4.19)

where the first term is called gradient energy, while we usually call

the second term the ”potential”. Using

∂L∂φ

= −m2φ, and∂L

∂(∂µφ)= ∂µφ (4.20)

we get the Euler-Lagrange equation for the Klein-Gordon field

∂µ∂µφ+m2φ = 0 (4.21)

If we add an arbitrary potential V (φ) we can generalize the Klein-

Gordon equation to

φ+∂V

∂φ= 0 (4.22)

where we have used = ∂µ∂µ for the Laplacian in Minkowski space.

4.1.4 Example 2: first order Lagrangians

Alternatively we can write a Lagrangian which is linear in time

derivatives. Take a complex scalar field ψ whose dynamics is defined

by a real Lagrangian

L =i

2(ψ∗∂tψ − (∂tψ

∗)ψ)− 1

2m∇ψ∗ · ∇ψ − mω2

2ψ∗ψ (4.23)

We derive the EOMs by treating the ψ and ψ∗ as independent

variables

∂L∂ψ∗

=i

2∂tψ −

mω2

2ψ (4.24)

∂L∂(∂tψ∗)

= − i2ψ (4.25)

∂L∂(∇ψ∗) = − 1

2m∇ψ (4.26)


from which we obtain

i∂ψ

∂t= − 1

2m∇2ψ +

mω2

2ψ (4.27)

This looks like the Schrodinger equation except for the harmonic

potential, but its interpretation is very different. The field ψ is a

classical field and we have no probability interpretation attached to

it.

The plane which gives the initial conditions that determine the

future (and the past) uniquely is called a Cauchy surface. It differs

between the two examples above. If L ∝ (∂tφ)2, we must specify

both φ and ∂tφ to fix the future evolution. For L ∝ ψ∗∂tψ, we only

need ψ and ψ∗.

4.1.5 Example 3: the Maxwell equations

The Maxwell equations without sources are given by the Lagrangian

L = −1

2(∂µAν)(∂µAν) +

1

2(∂µA

µ)2 (4.28)

Note that the minus signs are crucial to ensure positive kinetic terms

for the Ai

L =1

2(∂tA

i)2 +1

2(∇A0)2 − 1

2(1− δij)(∂iAj)2 (4.29)

Observe, that there is no kinetic term for A0 (no L ∝ (∂tA0)2),

which will have important consequences in QED. With

∂L∂(∂µAν)

= −∂µAν + (∂αAα)ηµν (4.30)

we get the Euler-Lagrange-equations

∂µ∂L

∂(∂µAν)= −∂2Aν + ∂ν(∂αA

α) = −∂µ(∂µAν − ∂νAµ) (4.31)

≡ −∂µFµν (4.32)

where the field strength is defined as3 3 You should check that this repro-

duces Maxwell’s equations in thevacuum.

Fµν = ∂µAν − ∂νAµ (4.33)

The Maxwell Lagrangian therefore can be written in the compact

form

L = −1

4FµνF

µν (4.34)

where we’ve used an integration by parts.

4.1.6 Locality

All examples above are local Lagrangians. We have not written any

terms that look like

L =

∫d3xd3yφ(x)φ(y)


which would couple a field at x to a field at y. The closest we get is Again, effective field theories are

special and can describe the correctphysics until some scale while being

non-local.

a coupling between φ(x) and φ(x + δx)through the gradient term

(∇φ(x))2. This property of locality is, as far as we know, a key

feature of all theories of Nature. Locality has a deep connection

to unitarity and one finds that non-local theories quickly violate

unitarity.

4.1.7 Lorentz invarianceAn example for an active transfor-mation: start with a temperature

field φ(x) which has a hotspot atx = (1, 0, 0). After a rotation

x→ Rx

The new field φ′(x) will have the

hotspot at x = (0, 1, 0). If we wantto express φ′(x) in terms of the oldfield φ, we need to place ourselves atx = (0, 1, 0) and ask ourselves what

the old field looked like where we’vecome from at

R−1x = R−1(0, 1, 0) = (1, 0, 0)

See next comment on passive trans-formations.

Lorentz transformations have a representation on the fields. The

simplest example is the scalar field which, under the Lorentz trans-

formation xµ → Λµνxν , transforms as

φ(x)→ φ′(x) = φ(Λ−1x) (4.35)

There is an inverse Λ−1 in the argument because we are dealing with

an active transformation in which we truly shift the field.

We call a theory Lorentz invariant if φ(x) solves the EOMs then

also φ(Λ−1x) solves the EOMs. We can ensure that this holds by

Continuing with the temperature

example: This R−1 is the reason forthe inverse transformation in theformula.

If we were instead dealing with apassive transformation in which werelabel our choice of coordinates, we

would have instead

φ(x)→ φ′(x) = φ(Λx)

requiring the action to be Lorentz invariant. Let’s look at our three

examples.

1. Klein-Gordon equation: For our real scalar field we have

φ(x) → φ′(x) = φ(Λ−1x) = φ(y). The contraction of deriva-

tives is obviously invariant using Eq. (3.58). The potential terms

transform similarly, with φ2(x) → φ2(y). Putting this all to-

gether, we find that the action is indeed invariant under Lorentz

transformations,

S =

∫d4xL(x) =

∫d4x

[∂µφ(x)∂µφ(x)−m2φ2(x)

]

−→∫d4y

[∂′µφ(y)∂′µφ(y)−m2φ2(y)

]=

∫d4yL(y) = S

where we’ve denoted ∂′µ = ∂∂y and used the fact that the Jacobian

is trivial

d4x → det(Λ) d4y = d4y

with4 4 Ex: Show that

| det(Λ)| = 1

using Eq. (3.58).

det(Λ) = 1

for Lorentz transformations connected to the identity, which we

are using for now.

2. First order equation: The theory is not Lorentz invariant since

space and time are not on the same footing. In practice, it’s easy

to see if the action is Lorentz invariant: just make sure all the

Lorentz indices µ, ν, . . . are contracted with Lorentz invariant

objects, such as the metric ηµν .

3. Maxwell’s equations: Under a Lorentz transformation we

obtain

Aµ(x)→ ΛµνAν(Λ−1x)


and you can see that Maxwell’s Lagrangian is indeed invariant.

Historically of course, it was Maxwell’s Lagrangian which first

show the concept of Lorentz invariance.

4.2 Symmetries: Noether’s Theorem and conserved currents

Symmetries in field theory are probably even more important than

in particle mechanics. Every additional symmetry constraints and

reduces our freedom to pick terms and their coefficients in the

Lagrangian. There are Lorentz symmetries, internal symmetries (like

U(1) or SU(3) in QCD), gauge redundancies, supersymmetry and so

on.

They are the real point of the Lagrangian formalism: the La-

grangian provides a natural framework for the quantum mechanical

implementation of symmetry principles. Noether’s theorem is useful

because it allows you to make exact statements about the solutions

of a theory without solving it explicitly. Since in quantum field the-

ory we won’t be able to solve almost anything exactly, symmetry

arguments will be extremely important!

Noether’s theorem: Every continuous symmetry of the action

gives rise to a conserved current jµ(x) such that the equations of

motion imply

∂µjµ(x) = 0 (4.36)

which in components is ∂tj0 +∇ · j = 0. We will now prove this.

Since we are assuming a continuous symmetry, we may prove it by

working infinitesimally. The transformation

δφa(x) = Xa(x) (4.37)

is a symmetry of the action. This means that the Lagrangian

changes at most by a total derivative

δXa(x)L = ∂µBµ (4.38)

For an arbitrary variation δφa(x), we know that

δL =

[∂L∂φa

− ∂µ(

∂L∂(∂µφa)

)]δφa + ∂µ

(∂L

∂(∂µφa)δφa

)(4.39)

When the equations of motion are satisfied, the first term [. . .]

vanishes! We then have

δL∣∣EOM

= ∂µ

(∂L

∂(∂µφa)δφa

)(4.40)

For a transformation which is a symmetry of the action, we know

however that δXa(x)L = ∂µBµ. If we restrict ourselves to variations

that satisfy the EOMs, we obtain for the transformations which leave

the action invariant

δXa(x)L∣∣EOM

= ∂µ

(∂L

∂(∂µφa)Xa(x)

)= ∂µB

µ (4.41)


or

∂µ

(∂L

∂(∂µφa)Xa(x)−Bµ

)= 0 (4.42)

This proves the theorem and we find Noether’s theorem was provedfor classical fields and one usually

extends it to quantum theory byreplacing the fields by the corre-sponding quantized fields. This does

however not always give a conservedcurrent in the quantum theory! Anon-zero divergence is called an

current anomaly. In the SM, e.g.baryon and lepton number are globalsymmetries which have a quantumanomaly.

∂µjµ(x) = 0 with jµ(x) =

∂L∂(∂µφa)

Xa(x)−Bµ (4.43)

4.2.1 Conserved currents

A conserved current implies a conserved charge Q with

Q ≡∫d3xj0 (4.44)

where the integration is over spatial coordinates only. We can see

this by taking the time-derivative

dQ

dt=

∫d3x

∂j0

∂t= −

∫d3x∇ · j = 0 (4.45)

where we have assumed that j → 0 fast enough as x→∞.Noether’s theorem does not applyto discrete symmetries, such asφ→ −φ of

L =1

2φφ−m2φ2 − λφ4

with φ real. Discrete symmetries canstill lead to selection rules.

There is one such conserved current jµ and one constant of the

motion Q for each independent infinitesimal symmetry transforma-

tion.

A conserved current is a stronger statement than a global con-

served charge (i.e. integrated over all of R3), since a conserved

current implies a local conservation of charge! With the analysis

above for a finite volume V in R3

QV =

∫

V

d3x j0 (4.46)

we find5 5 ∂V is the boundary area of V .

dQVdt

= −∫

V

d3x ∇ · j = −∫

∂V

dS · j (4.47)

with Stokes theorem. Any charge leaving V per time interval must

be equal to the flow of three-current out of the volume. This holds in

any local field theory.

4.2.2 Example 1: internal symmetries U(1)

A global internal symmetry only involves a transformation of

the fields and acts the same at every point in space-time. A simple

example can be obtained using a complex scalar field ψ(x)

ψ(x) =1√2

(φ1(x) + i φ2(x))

with φi real fields. We assume the real Lagrangian

L = ∂µψ∗∂µψ − V (ψ∗ψ) (4.48)


where the potential is a general polynomial in |ψ|2 = ψ∗ψ. Instead

of expanding in φ1, φ2 and deriving the equations of motion, we can

equivalently treat ψ∗, ψ as independent variables. We obtain the

EOM

∂µ∂µψ +

∂V (ψ∗ψ)

∂ψ∗= 0 (4.49)

after varying ψ∗. The Lagrangian is invariant under phase rotations

of ψ

ψ(x)→ eiαψ(x), or δψ = iαψ (4.50)

This is called a global U(1) symmetry. We say the internal sym-

metry is G = U(1). This implies a conservation law, since the

Lagrangian remains invariant6 under this transformation δL = 0, 6 Not even a total derivative.

and so Bµ = 0. We obtain for the conserved current7 7 We drop the constant α and we

perform the sum over internal indices

1, 2 by including ψ and ψ∗.jµ =∂L

∂(∂µφa)Xa(x) =

∂L∂(∂µψ)

δψ +∂L

∂(∂µψ∗)δψ∗ (4.51)

and find

jµ = i(∂µψ∗)ψ − iψ∗(∂µψ) (4.52)

Once we quantize this field theory, we will see that it can be in-

terpreted as electric charge and implies a conservation of particle

number.

4.2.3 Example 2: an internal SO(N) and a bit on Lie groups

We can rewrite this Lagrangian using a basis of real fields8 8 A summation over repeated indicesis implied here, too.

L = ∂µφa∂µφa − V (φaφa) (4.53)

We see that all the terms are polynomials of the form φaφa = φ21 +φ2

2,

similar to r2 = x2 + y2 which is invariant under rotations. We

reproduce the complex transformation with(φ1(x)

φ2(x)

)→(

cosα sinα

− sinα cosα

)(φ1(x)

φ2(x)

)(4.54)

or(δφ1(x)

δφ2(x)

)→(

0 α

−α 0

)(φ1(x)

φ2(x)

)(4.55)

which we can write as Remember only greek indices like

µ, ν, α, . . . are Lorentz contractedwith ηµν , here we just sum over

repeated indices.δφi(x) = αεijφj(x) (4.56)

with εij = −εji and ε12 = 1. This is a two dimensional, real rotation

or G = O(2). If we restrict ourselves to transformations which can

be continuously connected to the identity, we would have G = SO(2)

which has the additional restriction that the group elements have

unit determinant or

detO = 1 (4.57)


This is called the special orthogonal group in 2 dimensions.9 9 We will always factor G into

a direct product of semi-simplegroups and discrete factors. For

the conservation laws, the discrete

factors are not important.

Infinitesimal group elements can be written as

O = 1 + αεij + . . . = 1 + iαT ij + . . . (4.58)

with O ∈ G = SO(2). We call the matrix T ij parameterizing

an infinitesimal rotation, a (or ”the” since there is only one here)

generator of the group. We could have written the transformation

of φa as

δφi(x) = iαT ijφj(x) (4.59)

and find the conserved current as

jµ = (∂µφ1)φ2 − (∂µφ2)φ1 (4.60)

or in terms of the generator

jµ = iT ij(∂µφi)φj (4.61)

Generally, we obtain one conserved current for each infinitesimal

transformation, or generator, that leaves the action invariant.

We can generalize this symmetry to a = 1, 2, . . . , N . This is a

non-Abelian symmetry group G = SO(N). Similarly, we could

sum a complex field ψa over an index a and build a Lagrangian

with polynomials of ψ∗aψa (and derivatives thereof) to obtain a

G = SU(N) invariant theory. SU(N) is complex non-abelian group

and contains all special unitary transformations in N dimensions.

These groups are called non-abelian, because their generators in

general do not commute, e.g. for SO(3)

[T a, T b] = iεabcT c (4.62)

where we have suppressed the obvious matrix indices of the genera-

tors. We find 3 conserved currents:

jµa = iT ija (∂µφi)φj (4.63)

We will return to currents and conserved charges once we have

introduced quantized fields.10 10 Anticipating the result, we statethat the non-abelian nature of the

algebra implies that it is not possibleto simultaneously measure more than

one of the SO(3) charges of a state:

the charges are non-commutingobservables.

4.2.4 Example 3: space-time translations

Let us now study the case in which we transform space and time.

Recall that in classical mechanics, invariance under spatial transla-

tions gives rise to the conservation of momentum, while invariance

under time translations leads to the conservation of energy. Consider

an infinitesimal transformation11 11 Also called the Poincare transfor-mations.

xν → xν − εν (4.64)


φa(x)→ φa(x) + εν∂νφa(x) (4.65)


Note the ”+” sign because we are performing an active transforma-

tion. The Lagrangian itself will transform as

L(x)→ L(x) + εν∂νL(x) (4.66)

Noether’s theorem tells us that there are four conserved currents

(jµ)ν for each one of the four space-time translations εν . Together

with the total derivative (Bµ)ν = δµν εµL, we find

(jµ)ν =∂L

∂(∂µφa)∂νφa − δµνL ≡ Tµν (4.67)

This defines the energy-momentum-tensor Tµν . The four con-

served quantities are the total energy E

E =

∫d3xT 00 (4.68)

and

P i =

∫d3xT 0i (4.69)

where P i is the total physical momentum12 carried by the field. 12 Which is not to be confused withthe canonical momentum, which we

will introduce below.For the Klein-Gordon Lagrangian, we can e.g. compute13

13 Ex: Verify that indeed

∂µTµν = 0

using the equations of motion.

Tµν = ∂µφ∂νφ− ηµνL (4.70)

We find for E and P i

E =

∫d3x

[1

2(∂tφ)2 +

1

2(∇φ2) +

1

2m2φ2

](4.71)

P i =

∫d3x(∂tφ)∂iφ (4.72)

or with ∂i = −∂i = −∇i which we will use later:

P = −∫d3x(∂tφ)∇φ (4.73)

Note that the energy momentum tensor accidentally came out

symmetric here, such that Tµν = T νµ. This might not always be the

case, but we can always find a symmetric form!14 The trick is that 14 In general relativity, we need a

symmetric energy-momentum tensor

Rµν − 12Rgµν = −

8πG

c4Tµν ,

since the metric and the Ricci tensorare symmetric.

we add an extra term which is anti-symmetric in the first two indices

Γρµν = −Γµρν

Θµν = Tµν + ∂ρΓρµν

because the anti-symmetry of Γρµν guarantees that ∂µ∂ρΓρµν = 0,15 15 Thanks to Schwarz’s theorem or

the symmetry of second order partialderivatives.

which means that ∂µΘµν = 0 still holds.

In general relativity, the metric gµν is a field and we can expand it

as gµν = ηµν +√GNhµν + . . . . Once we enter this into the Einstein

action and expand, we will find for the linear terms in hµν a coupling

hµνΘµν .


4.2.5 Summary of Noether’s theorem

We summarize Noether’s theorem:

1. There is one conserved current jµ and one constant of motion Q

for each independent infinitesimal symmetry transformation.

2. The symmetry must be continuous, otherwise the δφa has no

meaning.

3. The current is conserved on-shell, that is when the equations of

motions are satisfied.

4. It applies to global symmetries, parametrized by numbers α, e.g.

φ→ eiαφ.16 16 As we shall see later: also forlocal symmetries parametrized byα = α(x)Currents will appear as sources for vector fields, e.g. appearing in

the Lagrangian as

L = . . .− jµ(x)Aµ(x). (4.74)

This can be an explicit external current such as the charge current or

just a formal place-holder for the composite expression

jµ = i(∂µψ∗)ψ − iψ∗(∂µψ) (4.75)

We will discuss this issue in more detail once we tackle the issue of

gauge invariance.

4.3 Coulomb’s law and the propagator method

Let us do some calculations to better understand aspects of classical

field theory and introduce Fourier transformation techniques along

the way. We will derive Coulomb’s law from relativistic electro-

dynamics.

A charge e at the origin can be represented by the current jµ

jµ =

(ρ(x)

vi(x)

)=

(e δ3(x)

0

)(4.76)

Adding the source term to the Lagrangian, we get

L = −1

4FµνF

µν − jµ(x)Aµ(x). (4.77)

The Euler-Lagrange equations are

0 =∂L∂Aν

− ∂µ∂L

∂(∂µAν)= −jµ + ∂2Aν − ∂ν(∂αA

α) (4.78)

= −jµ + ∂µFµν (4.79)

where we have used Eq. (4.31). As you may remember from your

electro-dynamics lectures, we can freely choose a gauge for Aµ and

we will use Lorenz gauge ∂αAα = 0 to obtain

Aν = jµ (4.80)


which has the formal solution

Aν =1

jµ (4.81)

where 1/ is just the inverse of which we will define more pre-

cisely later. This is a very common procedure in quantum field

theory: we say that the field Aν is determined by the source jν after

it propagates with the propagator

ΠA =1

(4.82)

We will discuss propagators in great detail. For our example above

we have to solve the following equations

Ai(x) = 0 (4.83)

A0(x) =e

δ3(x) (4.84)

The homogeneous solutions Aν = 0 , which are the electro-

magnetic waves, we will ignore here since they have nothing to

do with our source.

4.3.1 Fourier transformations

We will determine the solution of the equation of motion by taking a

Fourier transformation. Generally, we have We will not be very careful in distin-guishing the field from its Fourier

transformation. It will be clear fromthe variables what we mean. We’lluse p or k for the transformed field,e.g.

φ(x) =

∫d3p

(2π)3φ(p) eix·p (4.85)

f(x) =

∫d3p

(2π)3f(p) eix·p (4.86)

Recall, that the Fourier transformation of a δ-function (or δ tempered-

distribution) is just

Note the choice of normalization!The D-dimensional momentum

integral is accompanied with a factor1

(2π)D

δ3(p) =

∫d3x δ3(x) e−ix·p = 1 (4.87)

and

δ3(x) =

∫d3p

(2π)3eix·p (4.88)

We obtain for the derivatives of δ-functions

∆ δ3(x) =

∫d3p

(2π)3∆ eix·p =

∫d3p

(2π)3(ip)2 eix·p (4.89)

This also works for four-vectors

δ4(x) =

∫d4k

(2π)4eix

µkµ (4.90)

and so

δ4(x) =

∫d4k

(2π)4(−kµkµ) eix

µkµ (4.91)

We will use this all the time: for you as a field theorist, means

”−k2”.


4.3.2 The Coulomb Potential

Back to our problem: since δ3(x) is time-independent, we simplify

A0(x) =e

δ3(x) = − e

∆δ3(x) (4.92)

We solve this using a Fourier transformation

A0(x) =

∫d3p

(2π)3(− e

∆) eix·p

= e

∫d3p

(2π)3

1

p2eix·p

=e

(2π)3

∫ ∞

0

dp p2

∫ 1

−1

d cos θ

∫ 2π

0

dφ1

p2eipr cos θ

=e

(2π)2

∫ ∞

0

dpeipr − e−ipr

ipr

=e

8π2

1

ir

∫ ∞

−∞dp

eipr − e−iprp

Note that the integrand does not diverge at p→ 0 since the numera-

tor is ∝ sin(pr) ≈ pr + . . .. So we can introduce a small shift in the

denominator iε and simplify

∫ ∞

0

dpeipr − e−ipr

p= limε→0

(∫ ∞

0

dpeipr − e−iprp+ iε

)

Assuming ε > 0, we see that the pole is on the negative imaginary

axis. We will now use the residue theorem to calculate the integral:

for eipr we need to close the contour in the upper half-plane to get

exponential suppression of the integrand ∼ e−|p|r, which misses the

pole. For e−ipr, we close the contour in the lower half plane and

obtain∫ ∞

−∞dp−e−iprp+ iε

= (−2πi) Resp→−iε

[−e−iprp+ iε

]= (2πi) e−εr (4.93)

We therefore find

A0(x) =e

8π2

1

ir(2πi) e−εr =

e

4πre−εr (4.94)

and so for ε→ 0, we obtain finally

A0(x) =e

4πr(4.95)

which is Coulomb’s law.

4.3.3 A non-linear example: the graviton Lagrangian

Propagators are a useful tool to solve similar types of equations. Let

us assume that our Lagrangian has a non-linear term A3 (or as we

will see: an interaction). Electro-magnetism does not have these

terms and the equations of motion are linear, but there are many

examples in nature, like the gluon or electro-weak theory. Another is

the graviton. A heuristic Lagrangian for the graviton h which is the

fluctuation around a flat background metric gµν(x) = ηµν+hµν(x)+. . .

can be written as See Ex!


L = −1

2hh+

1

3λh3 + J · h (4.96)

where we have dropped the indices. This defines the non-linear

equations of motion. We solve perturbatively and we start with the

non-interacting theory with λ = 0

h0 =1

J (4.97)

We obtain the first oder correction h = h0 + h1 where h1 = O(λ)

(h0 + h1)− λ(h0 + h1)2 − J = 0 (4.98)

and using the 0th order result above,

h1 = λh20 +O(λ2) (4.99)

This means that

h1 = λ1

h2

0 = λ1

[(

1

J)(

1

J)

](4.100)

In summary, the first order solution is

h = h0 + h1 =1

J + λ

1

[(

1

J)(

1

J)

]+O(λ2) (4.101)

This is the so-called Green’s function method. We will call the

Π = − 1

(4.102)

the 2-point Green’s function or the propagator. If we define x =

∂2 with respect to xµ and

xΠ(x, y) = −δ4(x− y) (4.103)

Note, Π(x, y) = Π(y, x). We can use Fourier transformations to show

Π(x, y) =

∫d4p

(2π)4

1

p2ei(x−y)·p (4.104)

which we can check

xΠ(x, y) = −∫

d4p

(2π)4ei(x−y)·p = −δ4(x− y) (4.105)

We ignore subtleties with boundary conditions here but we will come

back to this later on.

We can write the field as

h(x) =

∫d4y δ4(x− y)h(y) = −

∫d4z (yΠ(x, y))h(y) (4.106)

= −∫d4y Π(x, y)yh(y) (4.107)

where we have used integration by parts. We can solve the free field

with λ = 0 equation, by inserting yh0(y) = J(y) to obtain

h0(x) = −∫d4yΠ(x, y)J(y) (4.108)


and the next order as

wh1(w) = λh0(w)2 (4.109)

= λ

∫d4y Π(w, y)J(y)

∫d4z Π(w, z)J(z) (4.110)

Substituting again as above in Eq. (4.107) we find for the perturba-

tive solution

h(x) = −∫d4y Π(x, y)J(y)− λ

∫d4s

∫d4z

∫d4yΠ(x, s)Π(s, y)Π(s, z) J(y)J(z) +O(λ2)

which is the meaning of Eq. (4.101).

There is a nice pictorial representation of this solution in the form

of Feynman diagrams. This pictorial representation allows us to

J(y)(x, y)

h(x)

J(y)

h(x) = +

J(z)s

+ . . .

Figure 4.1: Feynman diagrams in a

classical field theory.

derive the most general λn expansion for a classical field with the

following classical Feynman rules:

1. Draw a line from a point x to a new point xj .

2. Truncate with a source J(xi) or branch into two lines and multi-

ply with a factor λ.

3. Repeat the previous step.

4. The final result for h(x) is given by all graphs up to some order

λn with the ends truncated by currents J(xi) and the lines re-

placed by propagators Π(xi, xj). All internal points are integrated

over.

As we will see, in quantum field theory the Feynman rules are almost

identical except for ~ → 0 (classical limit), lines can not close

in on themselves.17 In classical general relativity, these diagrams 17 No loops!

describe the post-Newtonian effects of the Sun on Mercury! The

lines represent gravitons and the sources are in this case the Sun.

The first diagram is the well-known Newtonian potential while the

second order represents the self-interaction λ ∼ √GN which is a

prediction of general relativity.

Figure 4.2: The perihelion precession

of Mercury. https://en.wikipedia.org/wiki/

Two-body_problem_in_general_relativity

4.4 The Hamiltonian Formalism

In this course we will not discuss path integrals, and focus instead on

canonical quantization. For this we need the Hamiltonian formalism

of field theory, since canonical commutators are defined between the

field and its canonically conjugate momentum.

We first define the conjugate momentum πa(x)

πa(x) =∂L∂φa

(4.111)

https://en.wikipedia.org/wiki/Two-body_problem_in_general_relativity

https://en.wikipedia.org/wiki/Two-body_problem_in_general_relativity


The Hamilton density is obtained by a Legendre transformation The conjugate momentum πa(x)

should not be confused with thetotal momentum P i which is just

a number characterizing a field

configuration.

H = πa(x)φa(x)− L(x) (4.112)

As in classical mechanics, we replace φa(x) in favor of πa(x) every-

where. The Hamiltonian is simply

H =

∫d3xH (4.113)

4.4.1 Example: a real scalar field

For the Lagrangian

L =1

2(∂tφ)2 − 1

2(∇φ2)− 1

2m2φ2

we obtain π = φ and the Hamiltonian

H =1

2π2 +

1

2(∇φ2) +

1

2m2φ2

The advantage of the Lagrangian formalism is its explicit Lorentz

invariance. In contrast, the Hamiltonian formalism is not mani-

festly Lorentz invariant, since we have picked a preferred time. The

equations of motions do not look Lorentz invariant

∂tφ(x, t) =∂H

∂π(x, t), ∂tπ(x, t) = − ∂H

∂φ(x, t)

The physics is of course unchanged (and still Lorentz invariant). In

the canonical formalism, we will have to check at various points that

that the invariance is indeed preserved.

5

Canonical Quantization

5.1 Free fields

You are very familiar with the transition from classical physics

to quantum mechanics by using canonical quantization: it is a

recipe based on the Hamiltonian formalism. We take the generalized

coordinates and their conjugate momenta qa, pa and promote them

to hermitian operators. This recipe is closest to classical mechanics,

when we formulate the dynamics using Poisson brackets qb, pa = δaband morph them into commutation relations between operators

[qa, pb] = i~ δab

[qa, qb] = 0

[pa, qb] = 0

As in the introductory example of the discrete chain, we will do the

same now for the field φa(x) and its momentum conjugate πb(x).

A quantum field operator valued function of space obeys the

commutation relations This is also often confusingly referredto as second quantization. How-

ever, the fact that there are discretemodes is a classical phenomenon(think of the eigenmodes of a string).

The two steps are in fact (1) inter-pret these modes as having energy

E = ~ω and (2) quantize each mode

as a harmonic oscillator. In thatsense we are only quantizing once.

[φa(x), πb(y)] = i~ δab δ3(x− y)

[φa(x), φb(y)] = 0

[πa(x), πb(y)] = 0

(5.1)

(5.2)

(5.3)

We are working in the Schrodinger picture so that the operators

φa(x), πb(x) do not depend on time at all. Note, that we have lost

track of Lorentz invariance since we have separated space and time.

All time dependence in the Schrodinger picture is encapsulated in

the states |ψ〉 which evolve according to the Schrodinger equation

i~d

dt|ψ〉 = H |ψ〉 (5.4)

This is exactly the same as in quantum mechanics, we are just

applying the formalism to the fields. Warning: the notation |ψ〉is deceiving since the wave function in quantum field theory is a

functional of every possible configuration of the field φa.

In quantum mechanics we would determine the spectrum of the

Hamiltonian H and obtain the time evolution this way. This is too


hard in quantum field theories (most of the time). Clearly, dealing

with an infinite number of degrees of freedom1 complicates the 1 At least one for each point x inspace.problem somewhat. However for free theories, we can solve the

dynamics by going to a basis where we can separate the dynamics of

the degrees of freedom.2 Typically, these will be Lagrangians which 2 Recall the eigen-modes of thediscrete chain.are quadratic in the fields.

The simplest relativistic field theory is the Klein-Gordon equation

for a real scalar field φ(x, t)

∂µ∂µφ+m2φ = 0

To decouple the degrees of freedom, we only need to take the Fourier

transformation

φ(x, t) =

∫d3p

(2π)3φ(p, t) eix·p (5.5)

Then [∂2

∂t2+ (p2 +m2)

]φ(p, t) = 0 (5.6)

For each p, φ(p, t) solves the harmonic oscillator equation with a

frequency Reminder:

H =1

2p2 +

1

2ω2q2

and the commutation relations

[q, p] = i

Define

a =

√ω

2q +

i√

2ωp,

a† =

√ω

2q −

i√

2ωp

which we can invert to

q =1√

2ω(a+ a†), (5.7)

p = −i√ω

2(a− a†) (5.8)

Substituting the expressions we find

[a, a†] = 1

and H = ω(a†a+ 12

).

ωp = +√p2 +m2

The most general solution is a linear superposition of simple har-

monic oscillators, each vibrating at a different frequency with a

different amplitude. To quantize this theory we must quantize these

infinitely many harmonic oscillators. You should read Section 2.2.3

again.

5.1.1 The free scalar field

We write the field and its conjugate momentum as a linear sum of

infinitely many creation a†p and annihilation ap operators labeled by

their 3-momentum p

φ(x) =

∫d3p

(2π)3

1√2ωp

[ap e

ix·p + a†p e−ix·p] (5.9)

and for the conjugate momentum3 3 Inspired by Eq. (2.51) or Eq. (5.8).

You can check that this will lead tothe correct commutation relations for

the fields consistent with the ladder

operators.π(x) =

∫d3p

(2π)3(−i)

√ωp2

[ap e

ix·p − a†p e−ix·p]

(5.10)

Later, when we will consider time-

dependent field operators (Heisen-berg picture), you will see that this isconsistent with

π =∂L

∂(∂tφ)= ∂tφ

As in the case of the quantized string, the commutation rela-

tions for φ(x) and π(x) of Eq. (5.1) are equivalent to commutation

relations for ap and a†p. Quantum field theory is just quantum me-

chanics with an infinite number of harmonic oscillators.

[ap , a†q] = (2π)3 δ(3)(p− q)

[ap , aq] = [a†p , a†q] = 0

(5.11)

(5.12)


We will show this one way

[φ(x), π(y)] =

∫d3p

(2π)3

∫d3q

(2π)3

−i2

√ωqωp

− [ap, a

†q]

︸︷︷︸(2π)3 δ3(p−q)

eix·p−iy·q + [a†p, aq]︸︷︷︸

−(2π)3 δ3(p−q)

e−ix·p+iy·q

=

∫d3p

(2π)3

−i2

[−ei(x−y)·p − e−i(x−y)·p

]

= i

∫d3p

(2π)3ei(x−y)·p

= i δ(3)(x− y)

We obtain for the Hamiltonian in terms of ap , a†q Ex: Assuming canonical com-

mutation relations for the fields,show that the creation andannihilation operators satisfy

[ap , a†q ] = (2π)3 δ(3)(p− q).

H =1

2

∫d3x

[π(x)2 + (∇φ(x))2 +m2φ(x)2

]

=1

2

∫d3x

d3p

(2π)3

d3q

(2π)3(−1)

√ωpωq

4

[ap e

ix·p − a†p e−ix·p] [aq e

ix·q − a†q e−ix·q]

+1√

4ωpωq

[ip ap e

ix·p − ip a†p e−ix·p] [iq aq e

ix·q − iq a†q e−ix·q]

+m2 1√4ωpωq

[ap e

ix·p + a†p e−ix·p] [ aq eix·q + a†q e

−ix·q]

We will now collect the terms, integrate over d3x and replace

e−ix·(p+p) → δ3(p− p) and e−ix·(p−p) → δ3(p+ p) which allows us to

remove one d3q integral.

H =

∫d3p

(2π)3

1

4ωp

[apa−p + a†pa

†−p]

(−ω2p + p2 +m2)

+1

4ωp

[apa†p + a†pap

](ω2p + p2 +m2)

The first line vanishes because of ω2p = p2 +m2 and we obtain

H =1

2

∫d3p

(2π)3ωp[apa†p + a†pap

](5.13)

=

∫d3p

(2π)3ωp

[a†pap +

1

2(2π)3 δ(3)(0)

](5.14)

What is this!? We clearly have to work to interpret this properly.

There are two infinities: the delta-function has an infinite spike at 0

and the integral over ωp also diverges.4 4 The latter infinity is closely related

to the infinity of the specific heat

mentioned in the introduction.5.1.2 Infinities

We will separate the infinities into two classes: infrared diver-

gencies which are caused by assuming infinitely large scales (or

arbitrarily low energies) and ultraviolet divergencies which come

from extrapolating to infinitely small scales (or arbitrarily high

energies).

We will better understand the problem by discussing the ground

state or the vacuum |0〉. We define |0〉 as the state being annihi-

lated by all the ap

ap|0〉 ≡ 0 for all p (5.15)


We obtain for the ground-state energy

H|0〉 = E0|0〉 =

∫d3p

(2π)3ωp

[a†pap +

1

2(2π)3 δ(3)(0)

]|0〉 (5.16)

=

∫d3p

(2π)3ωp

1

2(2π)3 δ(3)(0)|0〉 (5.17)

= ”∞2” |0〉 (5.18)

This is common for quantum field theories: infinities pop up left and

right. They tell us that we might be asking the wrong question or

are doing something wrong.

The infrared divergence or long-distance divergence can be taken

care of by putting the field in a box of length L or volume V = L3

and assuming periodic boundary conditions. We obtain

(2π)3 δ(3)(0) = limL→∞

∫ L2

−L2dx

∫ L2

−L2dy

∫ L2

−L2dz eix·p

∣∣∣∣p=0

= V (5.19)

This tells us: we should be calculating the energy density E0instead of the total energy! We therefore divide by V to find

E0 =E

V=

∫d3p

(2π)3

1

2ωp =

4π

(2π)3

1

2

∫ ∞

0

dp p2√m2 + p2 →∞ (5.20)

which is however still infinite.5 5 Although a little less infinite:”∞” |0〉...

The ultraviolet divergence or short-distance divergence can be

understood by recognizing this as the integral over all ground state

energies of the harmonic oscillators. In the limit that |p| → ∞ we

find E0 → ∞. This divergence is caused by us assuming that our

quantum field theory is valid down to the smallest distances (or

highest energies)6. This is certainly not true. We need to cut-off the 6 Even above the Planck scale!Compare to the discrete chain,where this would correspond to

extrapolating to distances muchsmaller than the average distancebetween the mass points

integral at high energy (or high momentum) to include the fact that

we are ignorant above a certain scale.

We could introduce an explicit cut-off and stop integrating at some

energy Λ since we do not want to extrapolate beyond this point.

E0(Λ) =1

4π2

∫ Λ

0

dp p2√m2 + p2 ∝ aΛ4 + . . . (5.21)

which is called regularization.7 7 Note, that this is not a Lorentz-

invariant regulator: Lorentz invari-ance is broken when we impose acut-off on the spatial momentum

only. To meaningfully evaluate the

zero-point energy density, you mustuse a regularization scheme that

does not break Lorentz invariance.A possible choice is dimensionalregularization which will be covered

in more advanced QFT courses.

We can then renormalize the vaccuum energy to its observed

value by realizing that we could have added a constant V0 to the

Lagrangian

Lnew = L − V0 (5.22)

This constant will not affect the equations of motion and we usually

just drop it. V0 is a bare parameter which we choose to cancel the

divergence of zero-point energies

V0 = −E0(Λ) + χ (5.23)


this diverges as we take the cut-off to infinity V0 = −aΛ4 + . . .. But

now the observed energy-density is finite even for Λ→∞ as

1

VH |0〉 = (E0(Λ) + V0)|0〉 (5.24)

= (E0(Λ)− E0(Λ) + χ)|0〉 (5.25)

= Eobserved |0〉 (5.26)

the step is called renormalisation. Note that the finite piece

χ = Eobserved is completely arbitrary. It must be determined ex-

perimentally by measuring an observable – in this case the vacuum

energy.8 Since Λ only appears in the ”bare” Lagrangian and interme- 8 Vacuum energy is a meaningfulobservable only in the presence of

gravity – see below.diate steps of the calculation, but not in physical observables, we can

remove it at the end.

5.1.3 Normal ordering

Alternatively, we can follow a more practical approach called nor-

mal ordering. Since we are only interested in energy-differences9, 9 Again, as long as we ignore gravity,which couples directly to the energymomentum tensor.

the practical approach is good enough. We will just subtract off the

infinity and redefine our Hamiltonian as

H ′ =

∫d3p

(2π)3ωp a

†pap (5.27)

and now

H ′|0〉 = 0 (5.28)

If you are a bit taken aback by the lack of rigor implied by an ar-

bitrary redefinition of our Hamilton, note the following: there is

an inherent ambiguity of ordering transitioning from classical to

quantum. If we had defined the classical Hamiltonian as

H ′classical =1

2(ωq − ip)(ωq + ip) =

1

2p2 +

ω

2q2 = Hclassical

which is classically equivalent H ′classical = Hclassical. We would

naturally get after quantization10 10 Compare to the definition

a =

√ω

2q +

i√

2ωp,

a† =

√ω

2q −

i√

2ωp =

.

H ′QM = ω a† a

This method of exploiting the order ambiguity is called normal

ordering: we define it as the ordering in which all the creation

operators stand to the left of the annihilation operators.

(apa†q)normal ordered = : apa

†q : = a†qap

(a†paq)normal ordered = : a†paq : = a†paq

In general, we define a string of operators as normal ordered

: φa(x1)φb(x2) . . . φf (xn) : (5.29)


The Cosmological Constant : As mentioned before, gravity can indeed measure energy differences. Gravity

sees everything! We would expect the sum of a all zero-point energies to contribute to the cosmological constant

Λ = E0/V and appear on the right side of the Einstein-Hilbert equations

Rµν − 12Rgµν = −8πGTµν + Λ gµν ,

Cosmological observations tell us that ca. 70% of the energy density of the universe today is in a term that looks

like the cosmological constant with

Λ ∼ (10−3 eV)4

This is many orders of magnitude smaller than all the other scales in particle physics (the SM is valid to ∼ 200

GeV = 2 · 1011 eV). Why don’t zero-point energies contribute to Λ? What cancels them to such accuracy? We do

not know the answer. In fact, we don’t even know how to ask the right question.

if all the annihilation operators are placed to the right. Another

consequence of normal-ordering is that vacuum matrix elements

vanish

〈0| : φa(x1)φb(x2) . . . φf (xn) : |0〉

and the only normal-ordered expressions that do not vanish are

c-number functions11 which satisfy 11 The term c-number (or classicalnumber) is an old nomenclature

used by Paul Dirac which refers to

real and complex numbers. It isused to distinguish from operators

(q-numbers or quantum numbers) inquantum mechanics.

〈0| : f(x) : |0〉 = f(x)

Our Hamiltonian then has the normal-ordered form

: H : =

∫d3p

(2π)3ωp a

†pap (5.30)

5.2 Particles

What are the excitations of the field with respect to the vacuum |0〉?We know that

[H, a†p] = ωpa†p (5.31)

[H, ap] = −ωpap (5.32)

As in the case of the simple harmonic oscillator, we can construct Ex: Verify the relation

[H, a†p] = ωpa†p

energy eigenstates by applying the creation operators a†p on the

vacuum |0〉. We define

a†p |0〉 =1√2Ep

|p〉 (5.33)

The factor√

2Ep constitutes a relativistic normalization – see

below.12 The energy of this state is 12 The√

2 is for convenience only.

: H : |q〉 =

∫d3p

(2π)3ωp a

†pap

√2Eq a

†q |0〉

=√

2Eq

∫d3p

(2π)3ωp a

†p

(a†qap + (2π)3 δ(3)(p− q)

)|0〉

= ωq√

2Eq a†q |0〉

= ωq |q〉


with

ω2q = q2 +m2 (5.34)

This is the relativistic dispersion relation (or relativistic Pythagoras)

for a particle of mass m and momentum q. We therefore interpret

this as a momentum eigenstate of a single particle of mass m.

Its energy is E2q = q2 +m2 and we can interchange Eq and ωq.

What about the other quantum numbers? The total momen-

tum P of a classical field configuration which we have derived in

Eq. (4.73), can be turned into a (normal ordered) operator

: P : = −∫d3x : π(x)∇φ(x) :

= −∫d3x

d3p

(2π)3

d3q

(2π)3(−i)

√ωp4ωq

:[ap e

ix·p − a†p e−ix·p]∇[aq e

ix·q + a†q e−ix·q] :

= −∫d3x

d3p

(2π)3

d3q

(2π)3(−i)

√ωp4ωq

:[ap e

ix·p − a†p e−ix·p]iq[aq e

ix·q − a†q e−ix·q]

:

=

∫d3p

(2π)3(−1)

1

2p[apa−p + a†pa

†−p]

+ p a†pap

The first term vanishes because the term in the brackets is sym-

metric under p → −p which together with p makes this part of the

integrand anti-symmetric and therefore vanishes after integrating

over a symmetric interval. We find

: P : =

∫d3p

(2π)3p a†pap (5.35)

In summary we find

: Pµ : =

∫d3p

(2π)3pµ a†pap (5.36)

We can do the same for the classical expression of the angular

momentum13 to obtain the operator 13 See Ex and solution!

J i = εijk∫d3x(J 0)jk (5.37)

Acting on a one-particle state with zero-momentum, we find

J i|p = 0〉 = 0

which we understand as telling us that the particle does not carry

internal angular momentum. Quantizing a scalar field gives rise to a

spin-0 particle!

5.2.1 Multi-particle states and Bose statistics

Similarly, we can see that the state

a†p1a†p2

. . . a†pn |0〉 (5.38)


has momentum

p1 + p2 + . . .+ pn (5.39)

We interpret this as an n-particle state. Note that these are not localizedin space; a†pn creates a momentumeigenstate.|p1, . . . ,pn〉 = Na†p1

a†p2. . . a†pn |0〉 (5.40)

where N is a normalization. Since the creation operators commute

[a†p1, a†p2

] = 0, these states are symmetric under exchange of any

two particles, e.g.

|p1,p2〉 = |p2,p1〉 (5.41)

and we know that these particles are bosons. We can construct the

Fock space as the Hilbert space spanned by all possible combina-

tions of the a†pi

|0〉, a†p1|0〉, a†p1

a†p2|0〉, a†p1

a†p2a†p3|0〉, . . . (5.42)

As we have seen in the introduction, this is simply the direct sum of

the n-particle Hilbert spaces

|Ψ〉 = |Ψ〉1 ⊕ |Ψ〉2 ⊕ . . . (5.43)

We can count the number of particles in a given state in Fock space

using the number operator N defined as

N =

∫d3p

(2π)3a†pap (5.44)

which satisfies

N |p1, . . . ,pn〉 = n |p1, . . . ,pn〉 (5.45)

and it commutes with Hamiltonian,

[H,N ] = 0

which it should since in our free theory, particle number is conserved.

This will no longer hold once we include interactions (e.g. L ∼ φ3).

Interactions can destroy and annihilate particles, taking us between

different sectors in Fock space.

5.3 Smeared operator valued distributions

As already mentioned, |p〉 is not really a particle and not localized in

space but |p〉 is a momentum eigenstate.

Crucially, neither the operators φ(x) nor a†p are good operators

acting on the Fock space. 14 14 This is in direct analogy with the

position and momentum eigenstates

of quantum mechanics which are notgood elements of the Hilbert space

because they are not normalizable.

They normalize to delta-functions.

〈p|p〉 = 2Ep〈0|apa†p|0〉= 2Ep〈0|a†pap + (2π)3δ(3)(p− p)|0〉= (2π)32Ep δ

(3)(0) 〈0|0〉︸︷︷︸=1


and so Recall, ap|0〉 = 0 and therefore by

hermitian conjugation 〈0|a†p = 0.

〈p|p〉 = 2Ep (2π)3 δ(3)(0) (5.46)

Similarly

〈0|φ(x)φ(x)|0〉 = 〈x|x〉 = δ(3)(0) (5.47)

These are therefore operator valued distributions, rather than

functions. Even though we obtain a finite vacuum expectation value

〈0|φ(x)|0〉 = 0 (5.48)

The quadratic fluctuations of the operator at a point are infinite

〈0|φ(x)φ(x)|0〉 = ∞. We can obtain finite results by smearing the

distributions over space (or momentum) using test functions,

φf (x) =

∫d3y φ(x)f(x− y)

with e.g. the (normalized) Gaussian smearing

f(x) =1

(a2π)32

exp

(−x

2

a2

)

We can obtain 〈0|φf (x)φf (x)|0〉 by observing that of the four terms

after expanding φ(x)2 only the one proportional to 〈0|apa†q|0〉 ∼δ3(p− q) will be non-zero. We have

〈0|φf (x)φf (x)|0〉 =1

8(aπ)6

∫d3p

2Ep

[∫d3y exp

(− (x− y)2

a2+ ip(x− y)

)]2

The Gaussian integral can be easily calculated by completing the

squares15 and we get 15 Or by using Mathematica as theauthor did.

〈0|φf (x)φf (x)|0〉 =1

(2π)3

∫d3p

2Epe−p

2a2/2

=1

(2π)2

∫ ∞

0

dpp2

√p2 +m2

e−p2a2/2

For a 1/m we can ignore the m-dependence16 to obtain 16 Define q = ap and approximate√q2 + (am)2 ≈ q

〈0|φf (x)φf (x)|0〉 =1

4π2

1

a2

We see that the fluctuations diverge without bounds in the limit of

a→ 0.

5.3.1 Dimensional analysis and Coleman theorem

We could have in fact guessed the result for the fluctuations easily.

In the limit a → 0, there is no scale17 other than a. Similarly, we 17 As a → 0, we can ignore the onlyother scale (m) since 1/a m.know by counting the dimensions in the expansion of 〈0|φ(x)φ(y)|0〉

and realizing again that the only scale is |x− y|, we have to have Up to factors of order one, like π etc.

〈0|φ(x)φ(y)|0〉 ∼ 1

|x− y|2


This dimensional analysis is equivalent to saying that the field has

dimensions of inverse length. We could have already realized that a

scalar field in 3 + 1 Dimensions has canonical mass dimension18 1, or 18 The mass dimension of a field ψis defined to be the exponent

[ψ] = n ψ ∼ (mass)n

which means that ψ has dimension(mass)n ∼ (energy)n ∼ (length)−n.

[φ] = 1 by looking at the kinetic term of the action

S =

∫d4x ∂µφ(x)∂µφ(x) + . . .

Since the action is dimensionless [S] = [~] = 0 and since [d4x] = −4

and [∂µ] = 1, we find

[φ(x)] = 1

So the growth of the fluctuations with |x− y|−2 is really just dimen-

sional analysis. We can obtain similar results in any dimension. In D

space-time dimensions

S =

∫dDx ∂µφ(x)∂µφ(x) + . . .

from which we get

[φ(x)] =D − 2

2

and so in D−dimensions we obtain

〈0|φ(x)φ(y)|0〉D ∼1

|x− y|D−2

We immediately see an interesting feature by going to lower dimen-

sions. The divergence of the fluctuations is less rapid. In particular

for D = 2, that is in 1+1 space-time dimensions, the power vanishes.

This is really a log and we get19 19 See Exercise 5.2.

〈0|φ(x)φ(y)|0〉D=2 ∼ ln

(1

|x− y|

)

The fluctuations diverge logarithmically slow as |x − y| → 0. The

flip side of this is that also for large |x− y|, the fluctuations diverge

logarithmically.

So long-range fluctuations in D ≤ 2 are then actually very

important now and if you follow this through you find that these

fluctuations are large enough to prevent the formation of long-

range order.20 This is a form of the Coleman-Mermin-Wagner 20 which in particle physics language

is called spontaneous symmetrybreaking.

theorem21 which holds in quantum field theory and in statistical

21 Continuous symmetries cannot bespontaneously broken in dimensions

D ≤ 2, https://en.wikipedia.org/wiki/

Mermin-Wagner_theorem

mechanics. They hold in both since we can get statistical mechanics

from quantum field theory by analytic continuation – but this is for

another lecture.

5.4 Some technicalities: relativistic normalization

The vacuum |0〉 is defined as normalized 〈0|0〉 = 1. Our one-particle

|p〉 =√

2Ep a†p|0〉 states then have

〈p|q〉 = (2π)3√

4EpEqδ(3)(p− q)

https://en.wikipedia.org/wiki/Mermin-Wagner_theorem

https://en.wikipedia.org/wiki/Mermin-Wagner_theorem


This is a Lorentz invariant normalization. How do we show this? In

a quantum theory, we are always looking for unitary representa-

tions of our symmetries, such that

pµ → pµ = Λµνpν |p〉 → |p〉 = U(Λ)|p〉

and therefore leave the normalizations unchanged

〈p|p〉 = 〈p|p〉

but since p is a three-vector, this is not at all clear. The problem is

that spatial volumes are not invariant under boosts V → V/γ! We

show that we have a Lorentz invariant normalization by proving that

another object is Lorentz invariant.

Claim: We claim that∫

d3p

2Ep(5.49)

is the Lorentz invariant measure.

Proof: We show this by using that∫d4p is obviously Lorentz

invariant and by using that

pµpµ = m2 p20 = E2

p = p2 +m2

is a Lorentz invariant statement. We choose the positive branch p0 =

+Ep, but Lorentz transformations (the ones which are continuously

connected to the 1) also leave this choice invariant. We conclude

that the following combination must be Lorentz invariant∫d4p δ(p2

0 − p2 −m2)

∣∣∣∣p0>0

=

∫d3p

2p0

∣∣∣∣p0=Ep

which completes the proof. Now we know that the Lorentz-invariant Note, that

δ(g(x)) =∑i

δ(x− xi)|g′(xi)|

where the sum extends over all rootsof g(x) which are assumed to be

simple. For example

δ(x2 − α2

)=

1

2|α|

[δ (x+ α)+δ (x− α)

].

δ(3)-function for 3 vectors is:

2Ep δ(3)(p− q)

this holds because∫

d3p

2Ep2Ep δ

(3)(p− q) = 1

This finally explains our normalization for the momentum states,

|p〉 =√

2Ep a†p |0〉

These states now satisfy

〈q|p〉 = (2π)3 2Ep δ(3)(p− q) (5.50)

The identity on one-particle states is therefore

1 =

∫d3p

(2π)3

1

2Ep|p〉〈p|

In the following, we will only use relativistically invariant normaliza-

tions.


5.4.1 Meaning of the state φ(x)|0〉Let us consider the meaning of φ(x)|0〉. Using the expansion

Eq. (5.9) and Eq. (5.33), we obtain

φ(x)|0〉 =

∫d3p

(2π)3

1

2Epe−ip·x|p〉

which is a linear superposition of single particle states with well-

defined momentum. This should remind you of the non-relativistic

expression for the eigen-state of position |x〉, except for the factor

of 1/2Ep. We observe that for non-relativistic p the factor is nearly

a constant 1/2Ep ≈ 1/2m.

We will therefore put forward the same interpretation: the ope-

rator φ(x) acting on the vacuum creates a particle at position x.

Additional evidence comes from computing

〈0|φ(x)|p〉 = 〈0|∫

d3p′

(2π)3

1√2Ep′

[ap′ e

ix·p′ + a†p′ e−ix·p′

]√2Ep a

†p |0〉

= eix·p (5.51)

This can be interpreted as the position-space representation of the

single-particle wavefunction of the state |p〉, exactly like in non-

relativistic quantum mechanics 〈x|p〉 ∝ eix·p is the wavefunction of

the state |p〉.

5.5 Complex scalar fields

Let us consider a complex scalar field ψ(x) with

L = ∂µψ∗∂µψ −m2ψ∗ψ

with the equation of motion

∂µ∂µψ +m2ψ = 0

and the same equation of motion for ψ∗. We expand the complex

field operator in the Schroedinger picture

ψ(x) =

∫d3p

(2π)3

1√2Ep

[bp e

ix·p + c†p e−ix·p]

ψ†(x) =

∫d3p

(2π)3

1√2Ep

[b†p e−ix·p + cp e

ix·p]

(5.52)

(5.53)

The classical field is not real: the corresponding quantum field ψ

is not hermitian. This explains the appearance of the different

operators b and c†. The conjugate momentum is

π =∂L∂∂tψ

= ∂tψ∗

and22 22 Recall, that we define π(x) inanalogy with the result for theharmonic oscillator and that it

satisfies the canonical commutationrelations.


π(x) =

∫d3p

(2π)3(+i)

√Ep2

[b†p e−ix·p − cp eix·p

](5.54)

π†(x) =

∫d3p

(2π)3(−i)

√Ep2

[bp e

ix·p − c†p e−ix·p]

(5.55)

with the commutation relations

[ψ(x), π(y)] = iδ(3)(x− y)

all other commutators are vanishing

0 = [ψ(x), π(y)†] = [ψ(x), ψ(y)] = [π(x), π(y)] = . . .

You can check these commutation relations imply

[bp, b†q] = (2π)3δ(3)(p− q) (5.56)

[cp, c†q] = (2π)3δ(3)(p− q) (5.57)

and the rest vanishes

0 = [bp, bq] = [c†p, c†q] = [bp, cq] = [bp, c

†q]

naturally including their hermitian conjugates, e.g. [cp, cq] = 0.

Interpretation: The quantization of a complex scalar field

gives rise to two types of creation operators b†q and c†q. These are

particles and anti-particles. Compare to a real scalar field.

There is only one type of particle: itis its own anti-particle.

The complex scalar field has an internal U(1) symmetry, which as

we saw in Sec. 4.2.2 gives rise to a conserved current

jµ = i(∂µψ∗)ψ − iψ∗(∂µψ) (5.58)

and therefore a conserved quantum number

Q = i

∫d3x (ψ∗ψ − ψ∗ψ) = i

∫d3x (πψ − ψ∗π∗) (5.59)

The normal ordered quantum operator is

: Q : = i

∫d3x : (πψ − ψ∗π∗) :

= −∫d3x

∫d3p

(2π)3

∫d3q

(2π)3

√Eq4Ep

:[bp e

ix·p + c†p e−ix·p] [ b†q e−ix·q − cq eix·q

]: + h.c.

= −∫

d3p

(2π)3

[b†pbp − c†pcp

]−[b−pcp − b†−pc†p

]+[b−pcp − b†−pc†p

]

where we’ve used that hermitian conjugation of the first bracket

leaves it unchanged with normal-ordering. The second and third

bracket cancel. We finally obtain for the normal-ordered charge

operator

: Q : =

∫d3p

(2π)3

[c†pcp − b†pbp

]= Nc −Nb (5.60)


which counts the number of anti-particles (created by c†p) minus the

number of particles (created by b†p). We know that23 23 Ex: Show this!

[H,Q] = 0

which tells us that Q is a conserved quantity. We saw that in the

free field theory this is trivially true with Eq. (5.44), since [N,H] =

0. Later, we will see that in interacting field theories, new particles

can be created, e.g. L ∼ φ3 can branch a particle into two new ones:

Nc and Nb will not be individually conserved. However, Q = Nc −Nbsurvives as a conserved quantity.

5.6 Time dependent operators: the Heisenberg picture

We have been working in the Schroedinger picture, where the op-

erators depend on space but not on time. We will move to the

Heisenberg picture and make the operators time-dependent in

the usual way

φH(x) = φH(x, t) = eiHt φS(x) e−iHt

and In the Schroedinger picture, statesevolve in time

i~d

dt|p(t)〉S = H |p(t)〉S = Ep |p(t)〉S

which means

|p(t)〉S = e−iEpt |p(0)〉S

States in the Heisenberg picture are

related by the unitary transformation

|ψ(t)〉S = e−iHt |ψ〉H

and similarly operators

OS = e−iHtOH(t) e+iHt

This means that the matrix-elementsare the same

H〈ψ|OH(t)|φ〉H = S〈ψ(t)|OS |φ(t)〉S

We will assume that at t = 0 bothpictures coincide.

|ψ(0)〉S = |ψ〉HOS = OH(0)

π(x) = eiHt π(x) e−iHt

In general

OH(t) = eiHtOS e−iHt

and sod

dtOH(t) = i[H,OH(t)]

In field theory we will drop the subscripts ”S” or ”H” and indicate

using the arguments if we are in the Schrodinger picture φ(x) or in

the Heisenberg picture φ(x, t) = φ(x).

We assume that say at t = 0 the two pictures coincide and we find

that Eq. (5.1) become equal-time commutation relations

[φa(x, t), πb(y, t)] = i~ δab δ3(x− y)

[φa(x, t), φb(y, t)] = 0

[πb(x, t), πb(y, t)] = 0

(5.61)

(5.62)

(5.63)

We can now discuss the time-evolution of the operator φ(x, t), e.g.24 24 We use

[A2, B] = A[A,B] + [A,B]A = 2fA

if the commutator is just a c-valued

function

[A,B] = f

e.g. f = δ(x− y).

∂tφ(y) = i[H(t), φ]

=i

2

[∫d3xπ(x, t)2 + (∇φ(x, t))2 +m2φ(x, t)2, φ(y)

]

= i

∫d3xπ(x)(−i)δ(3)(x− y)

= π(y)


And

∂tπ(y) = i[H,π] =i

2

[∫d3xπ(x)2 + (∇xφ(x))2 +m2φ(x)2, π(y)

]

= i

∫d3x∇x[φ(x), π(y)]∇xφ(x) + im2φ(x)δ3(x− y)

= −(∫

d3x (∇xδ3(x− y))∇xφ(x)

)−m2φ(y)

= ∇2yφ(y)−m2φ(y)

which when combined with ∂tφ(y) = π(y) gives the Klein-Gordon

equation

∂µ∂µφ+m2φ = 0

We find the important result: Field operators satisfy the same

equation as the classical field. This starts to resemble a relativistic

theory.

How does the Heisenberg field operator expansion look like?

We want to transform

φS(x) → φ(x) = eiHt φS(x) e−iHt

and therefore need to transform the creation and annihilation opera-

tors in the expansion

eiHt ap e−iHt = e−iEpt ap (5.64)

eiHt a†p e−iHt = e+iEpt a†p (5.65)

which we can prove using the Baker-Campbell-Hausdorff relation

eXY e−X =

∞∑

m=0

1

m![X,Y ]m (5.66)

with the general commutator

[[X,Y ]m = [X, [X,Y ]m−1] and [X,Y ]0 = Y

We know that

[H, ap] = −Epap, and so [iHt, ap]m = (−iEpt)map

which is all we need to show

eiHt ap e−iHt =

∞∑

m=0

1

m![iHt, ap]m = e−iEpt ap

An analogous calculation (or hermitian conjugation) allows us to

obtain the Heisenberg transformation of a†p.

Plugging this into Eq. (5.9) means that the field operator of the real

scalar in the Heisenberg picture φ(x, t) = eiHt φ(x) e−iHt has the

following expansion

φ(x) =

∫d3p

(2π)3

1√2Ep

[ap e

−ip·x + a†p eip·x] (5.67)


where the exponent has the opposite sign, because we have used the

Lorentz-vectors

p · x = pµxµ = Ept− p · x

You can easily check that Eq. (5.67) indeed satisfies the Klein-

Gordon equation with E2p = p2 +m2.

For completeness, we show the straight-forward result for the

complex scalar field of Eq. (5.52) in the Heisenberg picture

ψ(x) =

∫d3p

(2π)3

1√2Ep

[bp e−ip·x + c†p e

ip·x]

ψ†(x) =

∫d3p

(2π)3

1√2Ep

[b†p e

ip·x + cp e−ip·x]

(5.68)

(5.69)

5.7 Causal quantum fields

We already know from Eq. (5.61) that φ(x) and π(x) satisfy equal

time commutation relations, e.g.

[φ(x, t), π(y, t)] = i δ(3)(x− y) (5.70)

We would like to study now: how does the commutator look like at

arbitrary space-time distances? This brings us back to the discussion

of causality in the introduction. Recall that we must require that

space-like separated events (x− y)2 < 0 do not overlap. Or in other

words, we should be able to find a basis in which matrix elements of

operators can be brought into diagonal form (for x and y space-like).

This is equivalent to requiring that space-like separated operators

commute This is the very important causalitycondition.

[O1(x),O2(y)] = 0 if (x− y)2 < 0 (5.71)

which ensures that a measurement at x will not affect a measure-

ment at y if they are not causally connected. Let us check if we

satisfy this property. We define

∆(x− y) ≡ [φ(x), φ(y)] (5.72)

By direct substitution we find that in the free theory the ∆(x− y) is

simply a c-number function with the integral expression25 25 Thanks to Nepomuk Ritz forsending in the solution!

∆(x− y) =

∫d3p

(2π)3

1

2Ep

[e−i(x−y)·p − ei(x−y)·p

]

Plugging the expression for φ(x) from Eq. (5.67) into the definition of ∆(x − y)

yields

∆(x− y) ≡ [φ(x), φ(y)] =

∫d3p

(2π)3

∫d3q

(2π)31√

4EpEq

[ap e

−ip·x + a†p eip·x, aq e

−iq·y + a†q eiq·y]

=

∫d3p

(2π)3

∫d3q

(2π)31√

4EpEq

[ap, aq ] e−ip·x−iq·y +

[ap, a

†q

]e−ip·x+iq·y

+[a†p, aq

]eip·x−iq·y +

[a†p, a

†q

]eip·x+iq·y

.


Using the commutation relations

[ap, aq ] = 0[ap, a

†q

]= (2π)3δ(3)(p− q)[

a†p, a†q

]= 0

[a†p, aq

]= −(2π)3δ(3)(q − p).

this expression simplifies to

∆(x− y) =

∫d3p

(2π)31

2Ep

[e−i(x−y)·p − ei(x−y)·p

]which is the desired result.

Properties of ∆(x− y):

1. ∆(x − y) is Lorentz-invariant. The exponentials are functions of

contracted four-vectors and we have just shown the integration

measure∫d3p/(2Ep) to be Lorentz-invariant.

2. For time-like distances ∆(x − y) does not vanish: if we take

zµ = xµ − yµ = (t, 0, 0, 0)µ, we obtain

∆(z) = [φ(x, t), φ(x, 0)] ∼∫d3p

1

2Epsin(Ept)

which is non-vanishing.

3. For space-like distances ∆(x− y) vanishes. We first show it for

a space-like distance at equal times

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

We obtain in this case

[φ(x, t), φ(y, t)] =

∫d3p

(2π)3

1

2√p2 +m2

[ei(x−y)·p − e−i(x−y)·p

]

This vanishes. We can e.g. replace p → −p in the second term

since it is just an integration variable and we are integrating over

a symmetric interval, which changes the sign in the exponent and

causes the sum to identically vanish. Since ∆(x − y) is Lorentz-

invariant, it only depends on (x− y)2 and must therefore vanish for

all (x− y)2 < 0 (and not only for equal times).

We conclude that we have defined a theory of causal quantum

fields, which solves this problem of relativistic quantum mechanics

as described in the introduction. This property will continue to hold

in interacting theories. Even though [φ(x), φ(y)] will not be

a c-function anymore, but will beoperator valued.

5.7.1 Propagators in quantum field theoryConsider a purely space-like distance:

x0 − y0 = 0 and x− y = r and repeat

steps after Eq. (4.92)

D(x− y) =

∫d3p

(2π)31

2Epe−ip·(x−y)

=2π

(2π)3

∫ ∞0

dpp2

2√p2 +m2

eipr − e−ipr

ipr

=−i

2(2π)2r

∫ ∞−∞

dppeipr√p2 +m2

The integrand can be considered a

complex function with branch-cutson the imaginary axis (from the√. . .) starting at ±im. We evaluate

the integral by pushing (see Fig. 5.1)the contour to wrap around the

upper branch but. With ρ = −ip, weget

D(x− y) =1

4π2r

∫ ∞m

dρρ e−ρr√ρ2 −m2

r→∞∼ e−mr (5.73)

Outside the light-cone, the propa-

gation amplitude is exponentiallysuppressed but still non-zero!

We have already seen the usefulness of the Green’s function method

in Sec. 4.3. We will encounter a similar object in quantum field

theories. Continuing with the causality problem, we could ask the

question: what is the amplitude to create a particle at x and observe

it at y while leaving the vacuum |0〉 undisturbed? We calculate


〈0|φ(x)φ(y)|0〉 =

∫d3p

(2π)3

d3q

(2π)3

1√4EpEq

〈0|apa†q|0〉 e−ip·x+iq·y

=

∫d3p

(2π)3

1

2Epe−ip·(x−y)

≡ D(x− y) (5.74)

We call D(x − y) the propagator. For space-like separations (x −y)2 < 0, we can show, see Eq. (5.73), that

D(x− y) ∼ e−m|x−y|

Although it decays exponentially quickly outside the light-cone, it

is non-vanishing! The quantum field leaks out of the light-cone, vio-

lating causality? How do we reconcile this with the fact, Eq. (5.71),

that all space-like separated measurements commute? We can write

the commutator Eq. (5.72) as

[φ(x), φ(y)] = D(x− y)−D(y − x) = 0 if (x− y)2 < 0 (5.75)

Let us describe what this equation means: we know that for space-

like separation (x− y)2 < 0 of x and y, there is no Lorentz invariant

ordering of events: if a particle can travel from x −→ y, it can just

as easily travel from y −→ x. In any measurement, the amplitudes

for the two processes cancel.

+im

im

branch-cut

I bran

ch

cu

t

Iradial

IR

Figure 5.1: Since the integrand

has no poles or branch-cuts insidethis contour, the integral has to

vanish. The integrand on the radialboundary vanishes Iradial → 0. Wecan therefore ’push the contour’ to

wrap around the branch cut. Weknow IR + Ibranch−cut + Iradial = 0and therefore IR = −Ibranch−cut

Let us consider this for a complex scalar-field, which has distinct

particle and anti-particle excitations, see Eq. (5.52). When the

complex scalar field ψ(x) is quantized,

• ψ(x) will create positive charged particles and destroy negatively

charged ones

• ψ†(x) will create negative charged particles and destroy positively

charged ones

The commutator [ψ(x), ψ†(y)] will be generally non-zero but most

delicately cancel outside of the light-cone to preserve causality.

The interpretation of the two terms in Eq. (5.75) now has charges

attached. Since

ψ(x)|0〉 ∼ (. . .) c†p |0〉

produces an anti-particle at x and

ψ†(y)|0〉 ∼ (. . .) b†p |0〉

produces a particle at y, we can interpret the commutator

[ψ(x), ψ†(y)] = D(x− y)−D(y − x)

= 〈0|ψ(x)ψ†(y)|0〉 − 〈0|ψ†(y)ψ(x)|0〉= ”particle y → x”− ”anti-particle x→ y”

In oder for these two processes to cancel, both particles must exist

and they must have the same mass.


In quantum field theory, causality requires that every particle has

a corresponding anti-particle with the same mass and opposite

quantum numbers (here it would be electric charge). What is the

anti-particle of the real Klein-Gordon field? The particle of the real

field is its own anti-particle.26 26 And by the property of anti-particles of having opposite quantumnumbers, we found an alternative

way to show that real scalar fieldsare neutral.

5.8 The Stuckelberg-Feynman propagator

We will now discover one of the most important quantities in inter-

acting field theories, the Stuckelberg-Feynman propagator. We

will define the product of fields so that an interpretation in terms of

the causal propagation of particles is always possible Its importance will become clearerin the next chapter and we will call

it only the Feynman propagator, as

it is unfortunately common. One ofthe many cases of the Matthew effect

in science, https://en.wikipedia.

org/wiki/Matthew_effect oder”Der Teufel scheisst immer auf den

großten Haufen.”

∆F (x− y) = 〈0|Tφ(x)φ(y)|0〉 =

D(x− y) x0 > y0

D(y − x) y0 > x0

(5.76)

where T is the time ordering, which places all operators evaluated

at later times to the left

Tφ(x)φ(y) =

φ(x)φ(y) x0 > y0

φ(y)φ(x) y0 > x0(5.77)

Claim: We can express the Feynman propagator of Eq. (5.76) in

terms of a Lorentz-invariant expression

∆F (x− y) =

∫

contour

d4p

(2π)4

i

p2 −m2e−ip·(x−y)

The meaning of the ”contour” will become clear soon. This is the

first integral over 4-momenta. Until now, we have integrated only

over 3-momenta and kept the p0 on the mass shell with p0 = Ep.

Here, we have no such condition. The integral for ε = 0 is ill-defined,

because for each p the denominator has a pole at

p2 −m2 = (p0)2 − p2 −m2 p0 = ±Ep = ±√p2 +m2

We need a prescription to avoid these singularities in the p0 integra-

tion. The Feynman propagator follows from the contour

Ep

+Ep

Figure 5.2: The contour C for theFeynman propagator in the complex

p0-plane.

https://en.wikipedia.org/wiki/Matthew_effect

https://en.wikipedia.org/wiki/Matthew_effect


Proof: We first rewrite the numerator

1

p2 −m2=

1

(p0)2 − E2p

=1

(p0 − Ep)(p0 + Ep)

The residue is Recall the residue for a simple pole is

Resz=c

[f(z)] = limz→c

(z − c)f(z)

and that for a positively oriented

simple closed curve C∮Cf(z)dz = 2πi

∑i

Resz=ci

[f(z)]

where the sum is over the residues

which are enclosed by C. Clockwiseencircled residues contribute with the

opposite sign.

Resp0=±Ep

[1

(p0 − Ep)(p0 + Ep)

]= ± 1

2Ep

For x0 > y0: we can close the contour in the lower half plane

p0 = −i|ρ|, since here

e−ip0(x0−y0) = e−|ρ|(x

0−y0) → 0

for p0 → −i∞ and therefore the radial integration will not con-

tribute. We pick up the residue at p0 = +Ep which gives∮

Cdp0 i

p2 −m2e−ip

0(x0−y0) = −2πi1

2Epe−iEp(x0−y0)

where the minus sign is due to us taking a clockwise contour. There-

fore when x0 > y0, we get

∆F (x− y) =

∫d3p

(2π)4

−2πi

2Epie−iEp(x0−y0)+ip·(x−y)

=

∫d3p

(2π)3

1

2Epe−ip·(x−y)

= D(x− y)

which is the Feynman propagator for x0 > y0.

For x0 > y0: we will close the contour in the upper-half plane. We

catch the residue at −Ep in an anti-clockwise direction and find

∆F (x− y) =

∫d3p

(2π)4

2πi

(−2Ep)ie+iEp(x0−y0)+ip·(x−y)

=

∫d3p

(2π)3

1

2Epe−iEp(y0−x0)−ip·(y−x)

=

∫d3p

(2π)3

1

2Epe−ip·(y−x)

= D(y − x)

where in the step from the 2nd to the 3rd line, we have flipped the

integration variable p → −p which is valid since the rest of the

expression (including the boundaries) is symmetric in p. Also in this

case we reproduce the Feynman propagator. QED.

Instead of explicitly showing the contour, we usually write for the

Feynman propagator the beautiful expression This is the most important result of

this chapter.

∆F (x− y) =

∫d4p

(2π)4

i

p2 −m2 + iεe−ip·(x−y) (5.78)

with ε > 0 and infinitesimal. This shifts the poles slightly off the real

axis, such that the integral along p0 is equivalent to the contour C.This form of the propagator is called the iε-prescription.


+Ep i"

Ep + i"

<(p0)

=(p0)Figure 5.3: The iε-prescription forthe Feynman contour. Note that theε is not exactly the same as in the

propagator, they are proportional toeach other since εprop = 2εplotEp

5.8.1 Green’s Functions

We will now see that the objects we’ve been discussing above are

closely related to the Green’s functions which we have discussed in

Sec. 4.3. In fact, if we stay away from the singularities, we obtain

(+m2)∆F (x− y) =

∫d4p

(2π)4(−p2 +m2)

i

p2 −m2e−ip·(x−y)

= −i∫

d4p

(2π)4e−ip·(x−y)

= −iδ(4)(x− y) (5.79)

Note, that we did not need the contour in the derivation. For other

applications it is useful to pick alternative contours, which also result

in Green’s functions.

We can define a retarded Green’s function ∆R(x− y) with

∆R(x− y) =

D(x− y)−D(y − x) x0 > y0

0 y0 > x0

which corresponds to the contour in Fig. 5.4 (left) This is useful if

Ep +Ep

retarded

Ep +Ep

advanced

Figure 5.4: Contours for the retardedand advanced Green’s function in the

complex p0-plane.we know the initial value of a field configuration and want to know

what it evolves into in the presence of a source, e.g.

φ+m2φ = J(x)

Along the same lines, we can define an advanced Green’s function,

see contour in Fig. 5.4 (right)

∆A(x− y) =

0 x0 > y0

−D(x− y) +D(y − x) y0 > x0

which is helpful if we know the final field configuration and want to

figure out where it came from.


Using Heaviside θ-functions we can express the three types of

Green’s functions as

∆F (x− y) = 〈0|Tφ(x)φ(y)|0〉= θ(x0 − y0) 〈0|φ(x)φ(y)|0〉+ θ(y0 − x0) 〈0|φ(y)φ(x)|0〉

and

∆R(x− y) = θ(x0 − y0) 〈0| [φ(x), φ(y)]|0〉∆A(x− y) = −θ(y0 − x0) 〈0| [φ(x), φ(y)]|0〉

We are still a long way from being able to do any real calculation,

since so far we have only talked about the free Klein-Gordon theory,

where the field equations are linear and there are no interactions.

On the other hand, the formalism here is extremely important since

the free theory forms the basis for doing perturbative calculations in

interacting theories.

5.8.2 Particle creation by a classical source

Even though we have no interactions yet, we can discuss what

happens if we disturb the field by a classical, external source J(x)

(+m2)φ(x) = J(x) (5.80)

which follows from the Lagrangian

L =1

2∂µφ∂µφ−

1

2m2φ2 + J · φ (5.81)

We will now turn on J(x) for a finite time. Before we turn on J(x),

the field has the form of a free field, see Eq. (5.67)

φ0(x) =

∫d3p

(2π)3

1√2Ep

[ap e

−ip·x + a†p eip·x]

without a source, this would be the solution for all time. We can con-

struct the solution with a source, by using the retarded Green’s

function

(+m2)∆R(x− y) = −iδ(4)(x− y) (5.82)

∆R(x− y) = 0, (x0 < y0) (5.83)

The second requirement, that ∆R be the retarded Green function, is

required so that the boundary condition φ(x) → φ0(x) as x0 → −∞is satisfied. You can check that this is the correct

Ansatz, by plugging the first line into

Eq. (5.80) and using the propertiesof the retarded Green’s function

Eq. (5.82)

(+m2)φ(x)

=(+m2)φ0(x) + i

∫d4y(+m2)x∆R(x− y)J(y)

=0 + i(−i)J(x) = J(x)

which clearly satisfies Eq. (5.80).

φ(x) = φ0(x) + i

∫d4y∆R(x− y)J(y)

= φ0(x) + i

∫d4y

∫d3p

(2π)3

1

2Epθ(x0 − y0)

[e−ip·(x−y) − eip·(x−y)

]J(y)

x0→+∞= φ0(x) + i

∫d4y

∫d3p

(2π)3

1

2Ep

[e−ip·(x−y) − eip·(x−y)

]J(y)

We have used the fact that if we wait until all of J(y) is in the past,

the the θ(x0 − y0) equals 1 over the whole domain of integration


and may be dropped. Then the solution involves only the Fourier

transform of J(y)

J(p) =

∫d4y eip·yJ(y)

evaluated at pµ such that

p2 = m2 (on-shell) (5.84)

and with (J(p))∗ = J(−p) because J(y) is real, we obtain

φ(x)x0→+∞

= φ0(x) + i

∫d3p

(2π)3

1

2Ep

[e−ip·xJ(p)− eip·x(J(p))∗

]

We can now group positive frequency terms together with ap and

negative frequency terms together with a†p:

φ(x) =

∫d3p

(2π)3

1√2Ep

[(ap +

i√2Ep

J(p)

)e−ip·x +

(a†p −

i√2Ep

(J(p))∗)eip·x

]

Since all observables are built out of the fields, we have solved the

theory. We can now guess (or derive) the Hamiltonian after J(x) has

been switched on and off. The free Hamiltonian (with J(x) ≡ 0) is

H =

∫d3p

(2π)3Ep a

†pap

and

〈0|H|0〉 = 0

If we act with J(y) and study the system in the far future (x0 →+∞), we can get HJ looking at the above discussion of the retarded

Green’s function by just replacing

ap → ap +i√2Ep

J(p)

a†p → a†p −i√2Ep

(J(p))∗

to obtain

HJ =

∫d3p

(2π)3Ep

(a†p −

i√2Ep

(J(p))∗)(

ap +i√2Ep

J(p)

)

The energy of the system in the far future after the source has been

switched off, is therefore

〈0|HJ |0〉 =

∫d3p

(2π)3Ep|J(p)|2

2Ep(5.85)

Note that because we are in the Heisenberg representation, we

are still in the ground state of the free theory – the state has not

evolved.

How can we interpret Eq. (5.85)? In the far future, we are in the

free theory again and the spectrum of the Hamiltonian is just free


particles. This means that the expectation value of the total number

of particles created with momentum p is

dN(p) =|J(p)|2

2Ep

and each Fourier component of J(p) produces particles27 with a 27 This can be made more explicitonce we discuss interacting fields

and have developed the relevant

formalism. Here you are asked tointuit this interpretation by starring

long enough at the form of the free

Hamiltonian in the absence of asource.

probability density |J(p)|2/(2Ep) for creating a particle in the mo-

mentum eigenstate p. The total number of particles produced

is∫dN =

∫d3p

(2π)3

1

2Ep|J(p)|2

We see from the condition Eq. (5.84), that only the Fourier modes

of J(x) which are in resonance (p2 = m2) with the on-shell Klein-

Gordon waves are effective at creating particles. This is just the

classical phenomenon of resonance occurring in the quantum field

theory setting.

5.9 Non-relativistic fields

We will now derive the non-relativistic limit of the Klein-Gordon

equation to see if we can recover the Schrodinger equation. Let us

start with a classical complex scalar field which satisfies the Klein-

Gordon equation. We decompose

ψ(x, t) = e−imtψ(x, t) (5.86)

Plugging this into the Klein-Gordon equation

0 = (∂2t −∇2 +m2)ψ = e−imt

[∂2t − 2im∂t −∇2

]ψ(x, t) (5.87)

where the m2 term cancelled against the time derivatives. In the

non-relativistic limit

|p| m

and so after a Fourier transformation28 28 Recall

φ(x) =

∫d3p

(2π)3

[φ(p) e−ip·x

]with p2 = m2, such that (+m2)φ =0

∂tψ(x, t)→ −iEpψ(p), ∂tψ(x, t)→ −i(Ep −m)ψ(p)

in the non-relativistic limit, Ep −m m and we find

|∂2t ψ(x, t)| m|∂tψ(x, t)|.

We therefore drop the term with two time derivatives to obtain

i∂ψ

∂t= − 1

2m∇2ψ (5.88)

This looks like the Schrodinger equation, except it does not have

a probabilistic interpretation – it is just a free classical field, evolving

according to an equation of first order in the time derivatives.

In Sec. 4.1.4, we discussed a Lagrangian which was first order

in the time derivatives. Once again, we can derive Eq. (4.23) from


the relativistic Klein-Gordon Lagrangian using ∂tψ mψ and

ψ(x, t) = e−imtψ(x, t)

L = ψ∗(−−m2)ψ

= |ψ|2 − |∇ψ|2 −m2|ψ|2

= | − imψ + ∂tψ|2 − |∇ψ|2 −m2|ψ|2

≈ im(ψ∗ (∂tψ)− ψ (∂tψ

∗))− |∇ψ|2

Once we divide by 1/2m, replace ψ → ψ and partially integrate in

time, we recover Eq. (4.23) which also was first order in the time

derivatives

L = i ψ∗ (∂tψ)− 1

2m(∇ψ∗)(∇ψ) (5.89)

Like its relativistic origin, this Lagrangian has a U(1) symmetry

ψ → eiαψ. The associated conserved current is Recall,

jµ(x) =∂L

∂(∂µψ)δψ +

∂L∂(∂µψ∗)

δψ∗

jµ(x) =

(−ψ∗ψ

i2m (ψ∗∇ψ − ψ∇ψ∗)

)(5.90)

Let us compute the Hamiltonian with the goal of quantizing the

theory. The conjugate momentum is

π =∂L

∂(∂tψ)= iψ∗

Interestingly, the momentum does not depend on time derivatives,

which is consistent for a theory that only is first order in the time

derivatives.29 We compute the Hamiltonian H = πψ − L 29 The trajectory is determined

by ψ and ψ∗ at a time t, no time-derivatives on the initial slice re-

quired.H =1

2m(∇ψ∗)(∇ψ)

where the time derivatives drop out. As before we quantize by

imposing in the Schrodinger picture the canonical commutation

relations

[ψ(x), π(y)] = iδ(3)(x− y)

or

[ψ(x), ψ†(y)] = δ(3)(x− y)

and all others vanishing

[ψ(x), ψ(y)] = [ψ†(x), ψ†(y)] = 0

We can Fourier expand the field operator ψ(x)

ψ(x) =

∫d3p

(2π)3ap e

ip·x

The commutation relations imply

[ap, a†q] = (2π)3 δ(3)(p− q)


As before the vacuum (or ground-state) satisfies ap|0〉 = 0, the

multi-particle excitations are proportional to a†p1. . . a†pn |0〉. They are

eigenstates of the Hamiltonian

H|p〉 =p2

2m|p〉

The single particle states satisfy the non-relativistic dispersion

relation.

Comments

• Quantizing the first order Lagrangian above gives rise to non-

relativistic particles of mass m.

• We started with a complex field but found only a single type of

particle. The anti-particle is not part of the spectrum. 30 Only 30 which is present in the quantized

relativistic Klein-Gordon theory.with relativity does the existence of anti-particles follow.

• The conserved charge implied by Eq. (5.90) is equal to particle

number

Q =

∫d3x : ψ†ψ :

Even if we introduce interactions ∆L = V (ψ∗ψ), particle number

will still be conserved. Only with relativity, anti-particles appear

and particle number can change.

• We cannot find a non-relativistic limit of a real scalar field. In

the relativistic theory the particles were their own anti-particles

in this case. There can be no way to construct a multi-particle

theory which conserves particle number!

5.9.1 A special case: quantum mechanics

How do we recover quantum mechanics? In quantum mechanics,

we describe physics in terms of the momentum operator P and the

position operator X. We already have the momentum operator

P =

∫d3p

(2π)3p a†pap

In the non-relativistic limit we can also construct a position operator,

see discussion in Sec. 5.4.1. The operator

ψ†(x) =

∫d3p

(2π)3a†p e

−ip·x (5.91)

creates a particle at x and we write

|x〉 = ψ†(x)|0〉

The best candidate for a position operator is then

X =

∫d3xxψ†(x)ψ(x)


so that

X |x〉 = x |x〉

Let us now construct a state |ϕ〉 which we would usually call the

Schrodinger wavefunction (in the position representation). We

obtain it by taking superpositions of one-particle states |x〉

|ϕ〉 =

∫d3xϕ(x) |x〉

Let us check if it has the right properties. The position operator has

the right action, as we can see

Xi |ϕ〉 =

∫d3xxi ϕ(x) |x〉 (5.92)

The momentum operator acts as See Exercise!

P i |ϕ〉 =

∫d3x

(−i∂ϕ(x)

∂xi

)|x〉 (5.93)

So we learn that when acting on one-particle states, P and X act as

position and momentum operators in quantum mechanics, with

[Xi,P j ] |ϕ〉 = iδij |ϕ〉

What about time-evolution? The wave-function ϕ(x, t) evolves in

time according to the Hamiltonian

H =

∫d3x

1

2m∇ψ∗∇ψ =

∫d3p

(2π)3

p2

2ma†pap

and we find

i∂ϕ

∂t= Hϕ = − 1

2m∇2ϕ (5.94)

This is the same equation obeyed by the original (full quantum) field.

Only this time, it is really the Schrodinger equation with the usual

probabilistic interpretation for the wavefunction ϕ. In particular, the

total probability which is conserved in quantum mechanics arises

as the conserved charge of the Noether current:

Q =

∫d3x |ϕ(x)|2

It is useful to know that if we treat the one-particle Schrodinger

equation as the equation for a quantum field then it will give the

proper generalization to multi-particle quantum field theory. His-

torically, the fact that the equation for the classical field Eq. (5.88)

and the one-particle wavefunction Eq. (5.94) gave rise to consider-

able confusion and people thought that perhaps we are quantizing

the wavefunction itself and called it ”second quantization”. This is

clearly incorrect, we are only quantizing the classical field once!


5.9.2 Non-relativistic interactions

Often we are interested in some fixed background potential

V (x). This can be incorporated into field theory by working with a

Lagrangian with

L = i ψ∗ (∂tψ)− 1

2m(∇ψ∗)(∇ψ)− V (x)ψ∗ψ (5.95)

We have broken translational symmetry and there will not be an

associated, conserved energy-momentum tensor. Such Lagrangians

are useful in condensed-matter physics but we almost never

encounter them in high-energy physics, where everything is Poincare

invariant, that is translational and Lorentz invariant.

We can also have interactions between particles, which require

n particle states with n ≥ 2. They can arise from Lagrangians of the

form

∆L = λψ∗(x)ψ∗(x)ψ(x)ψ(x)

which destroys two particles before creating two new ones. In the

following chapter we will explore interactions like there in detail for

relativistic theories.

6

Interacting quantum fields

We were able to completely solve the free scalar quantum field

theory, but nothing interesting happens in this theory. Most impor-

tantly, the particle excitations do not interact with each other. Now

we will discuss more complicated theories with interaction terms,

which in particular will lead to non-linear field equations.

6.1 Dimensional analysis of interactions

We can add the following terms to the free Lagrangian

L =1

2∂µφ∂

µφ− 1

2m2φ2 − λ3 φ

3 − λ4 φ4 − λ5 φ

5 + . . .

We call the coefficients of the higher order terms λn coupling

constants. The equations of motion now contain non-linear terms

and they cannot be solve by Fourier analysis as the free Klein-

Gordon equation.

When are these additional terms small perturbations? Which

terms should we include? What about λ15 φ15 ? I have restricted the expansion

to powers of φ but we could haveequally added terms of the form

1

M4−4n−m (∂µφ∂µφ)nφm

with integer n,m. In particular, wecould have added

∆L =1

M4(∂µφ∂

µφ)2

Here you can see explicitly how the

’typical energy scale’ E appears if

you replace ∂µ ∼ pµ ∼ E. We obtainfor the typical size of the interaction

∼E4

M4

that it is clearly an irrelevant pertur-bation.

Recall our discussion in Sec. 5.3.1, where we derived the mass

dimension of [φ] = 1 and of the action [S] = 0 from which we

immediately conclude [L] = 4. We obtain for the coupling constants

[λn] = 4− n, e.g. [λ3] = 1, [λ4] = 0, [λ5] = −1

So for all but λ4 we cannot just ask for λn 1, since this only

makes sense for dimensionless quantities. We will introduce mass

scales Mi for the dimensionful couplings to write

L =1

2∂µφ∂

µφ− 1

2m2φ− λ3M3 φ

3 − λ4φ4 − λ5

M5φ5 + . . .

Now the statement of λn 1 makes sense but it has be accom-

panied by discussion of the typical energy E of the process under

consideration vs. the mass scale Mi of the interaction.

We distinguish three classes of interactions Note here, that the meaning of

relevant, marginal and irrelevantis tied to the number of space-timedimensions. In D = 2 for example,

[φ]=0 and all the φn terms arerelevant!

• Relevant perturbation: ∆L = −λ3 φ3 = −λ3M3 φ

3 and

[λ3] = 1. The dimensionless parameter is

λ3

E= λ3

M3

E


What is the meaning of the scale E? We typically take E to

be the energy scale of the process of interest.1 This means that 1 Think of the center of mass energyof a scattering, or the mass of

particle if you consider decays of

particles.

when we test the theory a very high energies E λ3 or E M3,2 λ3 φ

3 is a small perturbation, suppressed by M3/E. If we2 Always assuming that λn = O(1).test at low energies E λ3 it is a large perturbation. We call

terms like this relevant because they are most relevant at low

energies, which is where most of the physics we see takes place.

In relativistic theories, E > m, and we can make this a small

perturbation by taking λ3 m or M3 m.

• Marginal perturbation: ∆L = λ4 φ4 with [λ4] = 0. The

effects of terms with d = 4 are independent of energy, and these

interactions are called marginal. This term is small if λ4 1.

• Irrelevant perturbation: ∆L = λ5 φ5 with [λ5] = −1 (and

λn φn for n ≥ 5). The dimensionless parameter is

λ5 = λ5E

M5

This interaction is small at low energies E M5 and large at

high energies. Such perturbations are called irrelevant.

Note, that his naive categorization of relevant, marginal and irrele-

vant is sometimes subject to change due to quantum corrections.3 3 One example are the pions of QCD,

they are scalars [φ] = 1 but emergeas strongly coupled bound states oftwo quarks [ψψ] = 3 at low energy.

Fermions in a Dirac spinor havemass-dimension [ψ] = 3/2.

Further, we see that irrelevant terms can be problematic in QFT

since, as we have seen, we often have to integrate to very high

energies. These terms lead to so-called ’non-renormalizable’ field

theories and which require some additional work to make sense of

the infinities at arbitrarily high energies.

6.1.1 Some examples of weakly coupled theories

Here we will only study theories where the interaction terms can be

considered small perturbations. You might discuss strongly interact-ing theories in a QFT2 course.

• φ4 theory: This is a subset of the marginal and relevant interac-

tions above

L =1

2∂µφ∂

µφ− 1

2m2φ2 − λ

4!φ4

with small λ 1. Is there a reason why we should not have

a ∆L ∼ φ3 coupling? This could be explained by a discrete

symmetry φ(x) → −φ(x) which allows only even powers in the

field.

We can guess the effects of the extra term. If we expand out φ4 in

creation and annihilation operators, we find terms like

a†p1a†p2

a†p3a†p4

, a†p1a†p2

a†p3ap4 , . . .

These terms create and destroy particles and particle number will

in general not be conserved and we could calculate [H,N ] 6= 0

(even though the discrete parity would be conserved.)


Why QFT is so simple

Typically we consider only relevant and marginal couplings, since irrelevant couplings become small at low energies.

This simplifies QFT a lot, since out of the infinite number of interaction terms we can write down, only a small

number is needed. In the case of a scalar field, we would have

L =1

2∂µφ∂

µφ− 1

2m2φ− λ3φ

3 − λ4φ4 +O(Λ/E)

Let us assume, we someday discover the true Lagrangian of the universe which is valid up to very high energies,

say Λ with [Λ] = 1. This could be the GUT scale (Λ ≈ 1016 GeV) or the Planck scale (Λ ≈ 1018 GeV). If we now

investigate nature at the small energies E accesible to us E Λ and if we assume she is described by a scalar field

φ (think of the Higgs). This scalar field will then have some complicated couplings λnφn which depend on the UV

completion of our amazing theory of everything at Λ. In terms of the dimensionless couplings λn, we expect them

to scale like

λn =λn

Λn−4

Now for experiments at E Λ (think of E ≈ 104 GeV at the LHC), the interaction therms of φn with n > 4 will

be suppressed by very small powers of (E

Λ

)n−4

This simple argument using dimensional analysis tells us that we can usually focus on the first few terms in the

interactions. It also tells us that it will be very difficult to figure out the high energy theory. These considerations

are a simplified version of what is called effective field theory and Wilson’s renormalization group. Due to the

unknown UV physics completing our QFT, we are generally required to write down all the allowed marginal and

relevant interactions to capture the IR behaviour of a theory with a scalar field.

This discussion is the modern viewpoint of what people previously discussed under the notion of ”renormalizable”

quantum field theories.

• Scalar Yukawa theory: We introduce a real scalar φ and a

complex scalar ψ with the Lagrangian We will use this theory in the sec-

tions below!

L = ∂µψ∗∂µψ −M2ψ∗ψ +

1

2∂µφ∂

µφ− 1

2m2φ2 − g ψ∗ψ φ

The coupling g has [g] = 1 and is a relevant coupling. For the per-

turbation to be small, we require g M,m. Particle number is

again not conserved anymore but since the Lagrangian is invariant

under a continuous phase symmetry (U(1))

ψ → eiαψ

a conserved charge Q exists, with [Q,H] = 0. The number of

ψ-particles minus the number of ψ-anti-particles is conserved. We

will usually denote the antiparticle as ψ.

• Quantum electro-dynamics This theory involves Dirac 4-

spinors4 Ψ and the electro-magnetic vector potential Aµ. 4 Which we will discuss in the nextchapter

L = Ψ(iγµ∂µ −m)Ψ − 1

4FµνF

µν − e ΨγµΨ Aµ

You can easily determine the canonical dimensions of the fields to

find that [e] = 0. You know e = −|e| as the electron charge. In

our units |e| ≈ 0.3 and the interaction term is small.


• Fermion Yukawa theory This theory involves spinors Ψ and a

real scalar φ

L = Ψ(iγµ∂µ −m)Ψ +1

2∂µφ∂

µφ− 1

2M2φ2 − g ΨΨ φ

You can also determine the canonical dimensions of the fields

to find that [g] = 0. Yukawa invented this theory to describe

nucleons (Ψ) and pions (φ). Nowadays we use it in the Standard

Model (SM), since this is the interaction between the Higgs

and the matter fermions (quarks and leptons). Most of the free

parameters in the SM are Yukawa coupling constants.

In our now interacting quantum field theory, we impose the equal-

time commutation relations

[φa(x, t), πb(y, t)] = iδab δ(3)(x− y)

which are unaffected by Lint.5 You can straight-forwardly check 5 Note, that if Lint contained a

∂µφ, the definition of the conjugatemomentum π(x) would change.

then that the field operator in the Heisenberg picture satisfies the

equation of motion including interaction term.

6.2 The interaction picture

We have discussed the Schroedinger (states time-dependent, oper-

ators not) and the Heisenberg picture (operators time-dependent,

states not) in Sec. 5.6 and we will now introduce a hybrid called the

interaction picture. We write the Hamiltonian as the free theory

part H0 plus the interaction Hint

H = H0 +Hint (6.1)

for the φ4-example that would be

H = HKlein-Gordon +

∫d3x

λ

4!φ4(x) (6.2)

We now define states and operators in the interaction picture as It is a hybrid: informally speaking itis the Heisenberg picture for the freeH0 and the Schroedinger picture for

the Hint.|ψ(t)〉I = eiH0t |ψ(t)〉SOI(t) = eiH0tOS e−iH0t

(6.3)

(6.4)

when the interaction is small, e.g. λ 1, then the most important

time dependence generated by H0 is already taken care off. The

interaction Hamiltonian in the interaction picture is

HI = (Hint)I = eiH0t (Hint) e−iH0t (6.5)

States in the interaction picture evolve according to (starting with

the Schroedinger picture result)

id

dt|ψ(t)〉S = HS |ψ(t)〉S


plugging in the definition of the interaction state |ψ(t)〉I , we get

id

dt(e−iH0t |ψ〉I) = (H0 +Hint)S e

−iH0t |ψ〉I (6.6)

id

dt|ψ〉I = HI |ψ〉I (6.7)

where we have first canceled H0 on both sides, multiplied the equa-

tion by eiH0t and used Eq. (6.5).

6.2.1 Dyson’s formulaNo relation to the vacuum applianceinventor.At a fixed time t0, we can expand φ(x) as before in terms of ladder

operators

φ(x, t0) =

∫d3p

(2π)3

1√2Ep

[ap e

ix·p + a†p e−ix·p]

We now want a perturbative expansion that captures the full time-

dependence of the quantum field φ. In the Heisenberg picture, we

would get

φH(x, t) = eiH(t−t0) φH(x, t0) e−iH(t−t0) (6.8)

For λ = 0, we have H = H0 and this reduces with Eq. (6.4) to

φ(x, t)∣∣λ=0

= eiH0(t−t0) φ(x, t0) e−iH0(t−t0) = φI(x, t) (6.9)

When λ is small, this will give the most important part of the time-

dependence. Since H0 is diagonalized by the Fourier expansion, we

get

φI(x) =

∫d3p

(2π)3

1√2Ep

[ap e

−ip·x + a†p eip·x]

∣∣∣∣x0=t−t0

(6.10)

which is just the familiar result of Eq. (5.67). We now want to

express the Heisenberg picture field operator φ in terms of φI(x),

again as before to have the full time-dependence in the operators

(while keeping the states time-independent). We can write this as

φ(x, t) = eiH(t−t0)e−iH0(t−t0) φI(x, t) eiH0(t−t0)e−iH(t−t0) (6.11)

≡ U†(t, t0)φI(x, t)U(t, t0) (6.12)

with the unitary operator

U(t, t0) = eiH0(t−t0)e−iH(t−t0) (6.13)

which is also called the interaction picture propagator or time-

evolution operator. States in the interaction picture evolve as From the definition of a state in the

interaction picture

|ψ(t)〉I = eiH0t|ψ(t)〉S= eiH0te−iHt|ψ(0)〉S = U(t, 0)|ψ(0)〉S

where we have use the time-evolutionin the Schroedinger picture.

|ψ(t)〉I = U(t, t0)|ψ(t0)〉I (6.14)

We want to express U(t, t0) entirely in terms of φI for which we have

an explicit expression in terms of ladder operators. We note, that


U(t, t0) is the unique solution, with initial condition U(t0, t0) = 1 of

a simple differential equation

i∂

∂tU(t, t0) = eiH0(t−t0)(H −H0)e−iH(t−t0)

= eiH0(t−t0)(Hint)e−iH(t−t0)

= eiH0(t−t0)(Hint)e−iH0(t−t0)eiH0(t−t0)e−iH(t−t0)

= HI(t)U(t, t0)

with HI(t) as defined in Eq. (6.5).6 The solution of this equation 6 E.g.

HI(t) =

∫d3x

λ

4!φ4I(x)

should naively look something like U ∼ exp(−i∫ tdt′HI(t

′)), but that

ignores the fact that [HI(t), HI(t′)] 6= 0 when t 6= t′.

A more careful treatment shows that the actual solution is

U(t, t0) = 1+(−i)∫ t

t0

dt1HI(t1) + (−i)2

∫ t

t0

dt1

∫ t1

t0

dt2HI(t1)HI(t2)

+(−i)3

∫ t

t0

dt1

∫ t1

t0

dt2

∫ t2

t0

dt3HI(t1)HI(t2)HI(t3) + . . .

To verify, that this indeed the correct solution, take the derivative:

you find that each term gives the previous one ×(−i)HI(t). The For the first two terms:

∂

∂tU(t, t0)

= (−i)HI(t) + (−i)2HI(t)

∫ t

t0

dt2HI(t2) + . . .

= −iHI(t)

[1 + (−i)

∫ t

t0

dt1HI(t1) + . . .

]

= −iHI(t)U(t, t0)

initial condition is U(t0, t0) = 1 is trivially satisfied.

We observe that that the factors of HI(ti) stand in time order,

with the later operators to the left. We can now simplify using the

time ordering T . Starting with H2I

∫ t

t0

dt1

∫ t1

t0

dt2HI(t1)HI(t2) =1

2

∫ t

t0

dt1

∫ t

t0

dt2 T HI(t1)HI(t2)

The time-ordered double integral on the RHS just counts everything

twice, since in the t1 − t2 plane, the integrand T HI(t1)HI(t2) is

symmetric about the t1 = t2 line. Similarly, one can show

∫ t

t0

dt1 . . .

∫ tn−1

t0

dtnHI(t1) . . . HI(tn) =1

n!

∫ t

t0

dt1 . . .

∫ t

t0

dtn T HI(t1) . . . HI(tn)

It’s not easy to draw this but you should be able to convince yourself

that this is in fact correct. We can now write the time evolution

operator in a very compact form:

U(t, t0) = 1+(−i)∫ t

t0

dt1HI(t1) +(−i)2

2!

∫ t

t0

dt1

∫ t

t0

dt2 T HI(t1)HI(t2)

+(−i)3

3!

∫ t

t0

dt1

∫ t

t0

dt2

∫ t

t0

dt3 T HI(t1)HI(t2)HI(t3)+ . . .

and we obtain Dyson’s formula

U(t, t0) = T exp

(−i∫ t

t0

dt′HI(t′)

)(6.15)

It is usually very hard (impossible) to compute time ordered expo-

nentials in practice. The power of Dyson’s formula is that just keep

the first couple of terms in the expansion when the interaction is a

small perturbation.


t2

t1t0 t

t0

t

Z t

t0

dt1

Z t

t0

dt2

Z t

t0

dt1

Z t1

t0

dt2

Figure 6.1: Dyson integration.

6.3 A first look at scattering

We will now apply what we have just learned to a the interaction in

the scalar Yukawa theory

HI = g

∫d3xψ†ψφ

where all the fields are in the interaction picture.7 This HI allows 7 Which means the field operators are

in the Heisenberg picture of the free

theory.particles to morph into each other. To understand why that is,

follow the time evolution of a state

|ψ(t)〉I = U(t, t0)|ψ(t0)〉I

where U(t, t0) is given by Dyson’s formula, which contains an expan-

sion in powers of HI . Now, HI contains annihilation and creation

operators for different types of particles. This is a simplified descrip-

tion of meson-nucleon interactions.8 We can have 8 Nucleons are neutron and the

proton, wich are composed of down

and up quarks. We will now take thehistorical perspective and consider

them elementary particles. In our

toy model we will describe them asscalars. In reality nucleons are spin

1/2 particles, and therefore fermions.

• ψ† ∼ b† + c : This operator creates ψ particles trough b† and

destroys anti-particles through c. Let us call these ψ-particles

nucleons.

• ψ ∼ b+ c† : This operator destroys ψ particles (nucleons) trough

b and creates anti-particles (anti-nucleons) through c†.

• φ ∼ a+a† We will call the φ particles mesons and φ can therefore

create and destroy mesons.

Due to the U(1) symmetry of ψ in L, we know that the charge

Q = Nc −Nb will be conserved even in the presence of LI = −HI .


The interaction ψ†ψφ contains a collection of creation and annihi-

lation operators, such as

c†b†a

which annihilates a φ particle (meson) and creates ψ particle and

antiparticle (nucleon and anti-nucleon): this corresponds to the

decay of the process φ→ ψψ.

Another example

cc†a ψ + φ→ ψ

which corresponds to the absorption of a meson by a nucleon.

At second order in perturbation theory, ∼ H2I , which produces

more complicated processes like

ψ + ψ → φ→ ψ + ψ

nucleon-antinucleon scattering through the creation of an intermedi-

ary meson.

6.3.1 Scattering simplified

We will now discuss scattering processes. They are particularly

convenient since in many cases the initial and final states look

like non-interacting particles. What does that mean? In a scat-

tering process, we start with an initial state |i〉 of isolated particles.

The particles are widely separated and do not feel the effects of the

interaction: they look like free plane wave states9. Once the particles 9 Eigenstates of H0 and so eigen-

states also of N , even though ingeneral [H,N ] 6= 0

approach each other, they begin to feel the interaction (and the

resulting potential) and they evolve in complicated non-linear way

as in Eq. (6.14). In this intermediate stage the system will look ex-

tremely complicated if expressed in terms of free particles. Particles

will be created and destroyed, since in general

[HI , N ] 6= 0

We will not just have two colliding protons but a very complicated

mess of protons, pions, photons, gluons, etc.

The result of the scattering process can have several outcomes.

Multiple initial particles could form a bound state.10 In this case, 10 e.g. p + p → 2D, two protonsforming a deuterium nucleus.no matter how long we wait after the scattering, the final state will

never look like the eigenstate of the free Hamiltonian H0, because

the interaction is responsible for the bound state.11 The formalism 11 Without the interaction the boundstate will fly apart.here will not be useful in this case.

We will now consider a process where no bound state is formed.

After some long time the system will just be a couple of widely

separated (non-interacting) particles. It will look simple again.

Before we go any further, you need to know that this is a bit of a

fake.12 Despite this, our quick and dirty scattering model will still 12 No matter how far in the future

we go, we never end up with acollection of free particles. Youalready know this from electro-

magnetism: the electron alwayscarries its electromagnetic field

with it (which corresponds to avirtual cloud of photons around the

electron). Similarly, nucleons alwayshave a cloud of mesons around them.Turning off our interaction, the stateswill change, and our simple picture is

not quite right.

work in the processes we consider here. We can see this by imagining

a theory like

L = Lφ + Lψ − g f(t)ψ†ψφ (6.16)


f(t)

1

t

T

Figure 6.2: Turning on and off ofthe function f(t) in Eq. (6.16). Thescattering process occurs near t = 0.

with f(t) like in Fig. 6.2. The function f(t) = 0 for large |t| and

f(t) = 1 around 0 where the scattering happens. A long time

after the scattering has occurred, we turn the interaction off very

slowly (adiabatically) over a period ∆: we expect that the simple

states in the real theory slowly turn into eigenstates of H0 with unit

probability.

There will be a one-to-one correspondence between asymptotic

simple eigenstates of the full Hamiltonian H and the eigenstates of

the free H0. 13 We should recover the full theory if 13 Again, we can’t consider bound

states, which are not eigenstates ofthe free Hamiltonian.T →∞, ∆→∞, ∆/T → 0

where the last limit is required to ensure no edge effects occur

(switching on and off of HI is adiabatic). We can justify this ap-

proach in a more rigorous manner but we do not have time for the

technical details in this course. Look for the LSZ reduction for-

mula in your favorite QFT textbook, named after the German

physicists Harry Lehmann, Kurt Symanzik, and Wolfhart Zimmer-

mann14. 14 Wolfhart Zimmermann was hon-

orary professor of TUM and directorat the Heisenberg MPI in Freimann.

After this long motivation, we finally get to the meat.15 We want

15 Or the tofu if you are so inclined.to solve Eq. (6.7)

id

dt|ψ〉I = HI |ψ〉I

with the boundary condition I will drop the subscript ”I” on

states from now on, since we will be

in the interaction picture from nowon if not otherwise stated.

|ψ(−∞)〉 = |i〉 (6.17)

and

|ψ(+∞)〉 = |f〉 (6.18)

The amplitude to go from |i〉 to |f〉 is

limt±→±∞

〈f |U(t+, t−)|i〉 ≡ 〈f |S|i〉 (6.19)

where the unitary operator S is called the S-matrix. Alternatively,

we can with Eq. (6.15) write

S = T exp

(−i∫d4xHI(x)

)(6.20)


6.3.2 Example: Meson Decay φ→ ψψ

We take the (relativistically) normalized initial φ and final states ψψ

|i〉 = |pφ〉 =√

2Ep a†p |0〉 (6.21)

|f〉 = |q1 q2〉 =√

4Eq1Eq2 b†q1c†q2 |0〉 (6.22)

where q1,2 are the momenta of the decay products, a nucleon anti-

nucleon pair (ψψ). We can compute the amplitude for the decay

with Eq. (6.15) to leading order in g

〈f |S|i〉 = 〈f |i〉 − i〈f |∫ ∞

−∞dtHI(t)|i〉+ . . . (6.23)

= 0− ig〈f |∫d4xψ†(x)ψ(x)φ(x) |i〉 (6.24)

We first expand φ ∼ a + a†. We will only need the a piece which The a† term turns |i〉 into a two

meson state with zero overlap withthe two nucleon state in |f〉.

annihilates the initial meson state |i〉 into something ∼ |0〉. We

obtain,

〈f |S|i〉 = i〈f |∫ ∞

−∞dtHI(t) |i〉+ . . .

= −ig〈f |∫d4xψ†(x)ψ(x)

∫d3k

(2π)3

√2Ep√2Ek

aka†pe−ik·x|0〉

= −ig〈f |∫d4xψ†(x)ψ(x)e−ip·x|0〉

where we have used the commutation relation of the ak. To get non-

zero overlap with 〈f |, only the b† and c† contribute, since we need to

create a nucleon and anti-nucleon final state from |0〉. We have,

〈f |S|i〉 = −ig〈0|∫d4xd3k1d

3k2

(2π)6

√4Eq1Eq2√4Ek1

Ek2

e−i(p−k2−k1)·x bq1cq2 b†k1c†k2|0〉

= −ig (2π)4 δ(4)(p− q1 − q2) (6.25)

and we have found our first quantum field theory amplitude! Notice

the δ-function which encapsulates the 4-vector conservation. In

particular, the decay will only take place if mφ ≥ 2Mψ.16 16 Ex: Show this!

We post-pone the steps to turn this into a something observable,

like a life-time of the meson. You can already see that it might in-

volve some additional insights since once we calculated probabilities,

we will get the square of a δ-function.

6.4 Wick’s Theorem

We now want to find a systematic way to compute quantities like

〈f |THI(x1) . . . HI(xn)|i〉

from Dyson’s formula. The time-ordering T fixes the order of the

operators, but we would have a much simpler calculation if we could

move all annihilation operators to the right, where they can remove

particles in |i〉. What we want find is therefore: how to go from time


ordered products to normal order products of fields. This is called

Wick’s theorem.

We define the contraction of two fields

A(x)B(y) ≡ TA(x)B(y) − : A(x)B(y) : (6.26)

We can easily see that A(x)B(y) is a c-number. Let us consider the

example of a real scalar field

φ(x) = φ+(x) + φ−(x)

decomposed as Note, that these are either freefields in the Heisenberg picture, or

interaction picture fields φI(x) for

H = H0 +HI .φ+(x) =

∫d3p

(2π)3

1√2Ep

ap e−ip·x

φ−(x) =

∫d3p

(2π)3

1√2Ep

a†p eip·x

where I would have used the opposite assignment of + and − but we

have Pauli and Heisenberg to blame.17 Let us consider x0 > y0, we 17 It is due to the sometimes usednotion of positive energies in thetime evolution of φ+ ∼ e−iEpt andthe negative energies in the time

evolution of φ− ∼ e−i(−Ep)t.

obtain

Tφ(x)φ(y) = φ(x)φ(y)

= (φ+(x) + φ−(x))(φ+(y) + φ−(y))

= φ+(x)φ+(y) + φ−(x)φ+(y) + φ+(x)φ−(y) + φ−(x)φ−(y)

= φ+(x)φ+(y) + φ−(x)φ+(y) + φ−(y)φ+(x) + [φ+(x), φ−(y)] + φ−(x)φ−(y)

= : φ(x)φ(y) : + [φ+(x), φ−(y)]

= : φ(x)φ(y) : + D(x− y)

where we have achieved normal ordering at the price of an ex-

tra term [φ+(x), φ−(y)] = D(x − y), which is the propagator of

Eq. (5.74).18 Similarly for y0 > x0, 18 Recall,

D(x− y) = 〈0|φ(x)φ(y)|0〉

= 〈0| : φ(x)φ(y) : + [φ+(x), φ−(y)]|0〉

= 〈0|[φ+(x), φ−(y)]|0〉

= [φ+(x), φ−(y)]

because in the last line, the commu-tator is a c-number.

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

= : φ(x)φ(y) : + D(y − x)

Putting both together, we finally obtain

Tφ(x)φ(y) = : φ(x)φ(y) : + ∆F (x− y) (6.27)

where ∆F (x − y) is the Feynman propagator! As derived in

Sec. 5.8, it has the integral representation

∆F (x− y) =

∫d4p

(2π)4

i

p2 −m2 + iεe−ip·(x−y)

An alternative derivation proceeds as follows: we have shown above,

that the difference between normal ordering and time ordering

φ(x)φ(y) is just a c-number.19 So we can equally switch it between 19 The operator structure is lost oncewe use the commutator for a and a†.


the vacuum

φ(x)φ(y) = 〈0|φ(x)φ(y)|0〉= 〈0|Tφ(x)φ(y)|0〉 − 〈0| : φ(x)φ(y) : |0〉= 〈0|Tφ(x)φ(y)|0〉= ∆F (x− y)

For charged fields, it is straight-forward to show that the propaga-

tor is

ψ(x)ψ†(y) = ψ†(x)ψ(y) =

∫d4p

(2π)4

i

p2 −M2 + iεe−ip·(x−y)

(6.28)

Note, that the only difference is the mass M of ψ. All other contrac-

tions vanish

ψ(x)ψ(y) = ψ†(x)ψ†(y) = 0

as they should. The last equation is easy to show since ψ ∼ c† + b

and ψ therefore only creates c-type particles and annihilates b-type

particles, ergo 〈0|Tψ(x)ψ(y)|0〉 = 0.

We can now state Wick’s theorem. For any collection of fields

φ1 ≡ φa1(x1), φ2 ≡ φa2(x2), . . ., the T -product of the fields has the

expansion

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

For n = 2, it is identical to Eq. (6.26). The term ”all possible

contractions” means, there will be one term for each possible con-

traction of the n fields in pairs. The proof that shows this for all n is

by induction and so not very illuminating. With this theorem, we Proof: Suppose it is true forφ2 . . . φn and we now add φ1. We

choose x01 > x0k for all 2 ≤ k ≤ n. Wemove φ1 out to the left of the time

ordered product

Tφ1φ2 . . . φn

= (φ+1 + φ−1 )(: φ2 . . . φn + contractions :)

The φ−1 ∼ a† is already normalordered. To have the RHS as a

normal ordered product, we need to

commute φ+1 past the φ−k operators.

Each time φ+1 moves past one φ−k , wepick up a factor

φ1φk = ∆F (x1 − xk)

Try it!

This proves Wick’s theorem by

induction.

obtain thus for n = 4,

Tφ1φ2φ3φ4 = :φ1φ2φ3φ4 + φ1φ2φ3φ4 + φ1φ2φ3φ4 + φ1φ2φ3φ4

+φ1φ2φ3φ4 + φ1φ2φ3φ4 + φ1φ2φ3φ4

+φ1φ2φ3φ4 + φ1φ2φ3φ4 + φ1φ2φ3φ4 :

where a contraction across operators which are not adjacent, still

gives a factor of ∆F , e.g.

: φ1φ2φ3φ4 : = ∆F (x1 − x3) : φ2φ4 :

6.4.1 Example: nucleon scattering ψψ → ψψ

We are scattering ψψ → ψψ and therefore have the in- and outgoing

states

|i〉 = |p1p2〉 =√

2Ep1

√2Ep2 b

†p1b†p2|0〉 (6.29)

|f〉 = |q1q2〉 =√

2Eq1√

2Eq2 b†q1b†q2 |0〉 (6.30)


We now want to obtain the expansion of 〈f |S|i〉. In fact, we are

interested in 〈f |S − 1|i〉, since we do not care about amplitudes where

nothing happens.20 At second order in g, we get for 〈f |S − 1|i〉 20 The two nucleons in this casejust continue to fly without any

interaction as free particles.(−ig)2

2!

∫d4x1d

4x2 ψ†(x1)ψ(x1)φ(x1)ψ†(x2)ψ(x2)φ(x2) (6.31)

Wick’s theorem can relate this to a sum of normal-ordered products.

The interesting piece for us is

: ψ†(x1)ψ(x1)ψ†(x2)ψ(x2) : φ(x1)φ(x2) (6.32)

which contributes to ψψ → ψψ scattering, because the two ψ fields

annihilate the initial nucleons and the two ψ†’s creates the outgoing

nucleons (or ψ-particles). Any other contraction will give a zero

result. We obtain with Eq. (5.68)

〈q1q2| : ψ†(x1)ψ(x1)ψ†(x2)ψ(x2) : |p1p2〉= 〈q1q2|ψ†(x1)ψ†(x2) |0〉〈0|ψ(x1)ψ(x2) |p1p2〉=[eiq1·x1+iq2·x2 + e+iq1·x2+iq2·x1

] [e−ip1·x1−ip2·x2 + e−ip1·x2−ip2·x1

]

=[eix1·(q1−p1)+ix2·(q2−p2) + eix1·(q2−p1)+ix2·(q1−p2)

]+ (x1 ↔ x2)

Let us plug this into Eq. (6.31) to obtain

(−ig)2

2!

∫d4x1d

4x2 . . .+ (x1 ↔ x2)∫

d4k

(2π)4

i

k2 −m2 + iεe−ik·(x1−x2)

where the . . . corresponds to the sum of exponentials above result-

ing from the matrix element of the normal-ordered operator. Since

the propagator is symmetric under x1 ↔ x2 we can drop this in

exchange for the 1/2 in front. The xi integrals result in δ-functions,

e.g. for the first term we get for the exponentials

∫d4x1d

4x2 eix1·(q1−p1)+ix2·(q2−p2)−ik·(x1−x2)

= (2π)8 δ(4)(q1 − p1 − k)δ(4)(q2 − p2 + k)

and we find

i(−ig)2

∫d4k

(2π)4

(2π)8

k2 −m2 + iε

[δ(4)(q1 − p1 − k)δ(4)(q2 − p2 + k)

+ δ(4)(q1 − p2 + k)δ(4)(q2 − p1 − k)]

We finally obtain

i(−ig)2

[1

(p1 − q1)2 −m2+

1

(p1 − q2)2 −m2

](2π)4δ(4)(p1 + p2 − q1 − q2)

Notice that performing the final integral over δ-functions leaves us

with a factor of

(2π)4δ(4)(p1 + p2 − q1 − q2)

which just enforces 4-momentum conservation for the scattering

process. Since we usually have Poincare invariant Lagrangians21, it 21 Poincare invariant = space-timetranslation (and Lorentz) invariant.For the δ(4) function the space-

time translation invariance is thecrucial part, since it guarantees total

4-momentum conservation.


is customary to define the invariant Feynman amplitude Afi by22

22 We will get back to why we can

remove the δ-function later. We willsee it has to do with our assumptionof free particles being in plane wave

states.

〈f |S − 1|i〉 = iAfi (2π)4δ(4)(pf − pi)

where the factor i is by convention, since it reproduces the phase

conventions for scattering in non-relativistic quantum mechanics.

Further, pµf =∑

n pµf,n and pµi =

∑n p

µi,n. We can ignore the iε,

because the denominator is never zero: in the center of mass frame,

we can write the momenta as23 23 Careful: p = |p| and not a 4-vector!

p1 = (√p2 +M2, p, 0, 0) (6.33)

p2 = (√p2 +M2,−p, 0, 0) (6.34)

q1 = (√p2 +M2, p cos θ, p sin θ, 0) (6.35)

q1 = (√p2 +M2,−p cos θ,−p sin θ, 0) (6.36)

which immediately gives24 24 We see that e.g. (p1 − q1)2 < 0 andso (p1 − q1)2 6= m2 which justifies

dropping iε.(p1 − q1)2 = −2p2(1− cos θ), (p1 − q2)2 = −2p2(1 + cos θ)

and so

A = g2

[1

2p2(1− cos θ) +m2+

1

2p2(1 + cos θ) +m2

](6.37)

Note that the two terms are required because of Bose statistics. Scat-

tering into two identical particles at an angle θ is indistinguishable

from scattering at an angle θ − π, and so the probability must be

symmetrical under the interchange of the two processes. Since these

particles are bosons, the amplitude must also be symmetric.

This was a bit tedious. There is a much simpler way which we

will discuss in a little bit using Feynman diagrams.

6.5 Diagrammatic Perturbation theory: Feynman Diagrams

Our final result for ψψ → ψψ scattering in Eq. (6.37) was remark-

ably simple, even though the intermediate steps were a bit messy.

ψ†

ψ

φ

ψ†ψφ

Figure 6.3: Interaction vertex. Each

field in HI is represented by a lineemanating from the vertex. Theψ†-line creates a nucleon or destroys

an anti-nucleon, the ψ-line creates an

anti-nucleon or destroys a nucleon,and the φ line creates or annihilatesa meson.

This motivates a diagrammatic short-hand called Feynman dia-

grams. These are pictures of the fields and contractions which we

need to evaluate to give the matrix element. At the nth order in per-

turbation theory HI will act n times and so a Feynman diagram will

contain n interaction vertices. For our toy model the interaction

vertex looks like in Fig. 6.3.

We need to distinguish ψ from ψ† and we will therefore draw an

arrow on the line.

Next, contractions are the lines joining different vertices. Every

time there is a contraction, we join the lines of the contracted fields.

The example of Eq. (6.32) corresponds to the diagram Fig. 6.4.φ1φ2

Figure 6.4: Diagrammatic representa-tion of Eq. (6.32).

Arrows will line up because ψ(x)ψ(y) = ψ†(x)ψ†(y) = 0 and

the arrows indicate the flow of the U(1) charge. Now, any fields


which are left uncontracted must either annihilate particles from the

incoming state or create particles from the outgoing state. If there

are different ways of doing this, they correspond to indistinguishable

processes, so we must add the corresponding amplitudes: we write

down a separate Feynman diagram for each distinct labeling of the

external legs. For ψψ → ψψ scattering, this results in two Feynman

diagrams

p1

p2

q1

q2

k

p1

p2 q1

q2

k

Figure 6.5: Feynman diagramscontributing at order g2 to ψψ → ψψscattering, see Eq. (6.32). We show

the explicit momentum assignmentfor the external and internal lines.There are two distinct options.

We will revisit them below.

6.5.1 Feynman rules

1. For each particle in |i〉 and |f〉, draw an external line. Assign a

directed momentum pi for each line, and add an arrow to denote

its charge.

2. Join the external lines together with vertices as in Fig. 6.3.

3. At each vertex, write a factor of

(−ig)(2π)4δ(4)(∑

i

ki)

with∑i ki is the sum of all momenta flowing into the vertex.

4. For each internal φ line, we write a factor of

∫d4k

(2π)4

i

k2 −m2 + iε

and for each internal ψ line, we write

∫d4k

(2π)4

i

k2 −M2 + iε

5. Divide the final result by the overall energy-momentum conserv-

ing δ-function, (2π)4δ(4)(pf − pi).

In fact, we can simplify things further and get rid of the trivial

delta-functions: we impose energy-momentum conservation on the

momenta flowing into each vertex and so we change

3.’ At each vertex, write a factor

(−ig)

4.’ Each contracted internal line represents a propagator (or M2 for

internal ψ lines)

i

k2 −m2 + iε

This is ok for graphs like the ones we have been considering.

However, there are also diagrams with closed loops where energy-

momentum conservation at the vertices does not fully fix all the

internal momenta! For example the matrix element obtained from p p+k

k

p

Figure 6.6: Diagrammatic repre-sentation of Eq. (6.38) where theblue arrows show the momentum

assignments.


〈p| : ψ†(x1)ψ(x2)ψ(x1)ψ†(x2)φ(x1)φ(x2) : |p〉 (6.38)

This means, we need to keep the factor

∫d4k

(2π)4

Similarly for the fully contracted term

〈0| : ψ†(x1)ψ(x2)ψ(x1)ψ†(x2)φ(x1)φ(x2) : |0〉 (6.39)

p+k

k

p

Figure 6.7: Diagrammatic repre-

sentation of Eq. (6.39) where theblue arrows show the momentumassignments.

neither p nor k is constrained (not fixed by external momenta) and

we must integrate over both momenta. Thus we must add a rule that

we integrate with∫

d4k(2π)4 over each internal line whose momentum

k is unconstrained .


Let us summarize our simplified and final Feynman rules to calcu-

late iAfi which we defined as 〈f |S − 1|i〉 = iAfi (2π)4δ(4)(pf − pi)

1. For each particle in |i〉 and |f〉, draw an external line. Assign a

directed momentum pi for each line, and add an arrow to denote

the flow of charge.

2. Join the external lines together with vertices as in Fig. 6.3.

3. At each vertex, write a factor of

(−ig)

and impose 4-momentum conservation.

4. For each internal φ line, we write a factor of

i

k2 −m2 + iε

and for each internal ψ line, we write

i

k2 −M2 + iε

5. For each loop with momentum k unconstrained write a factor of

∫d4k

(2π)4


6.5.2 Revisiting ψψ → ψψ scattering

Let us apply the above Feynman rules to compute the amplitude of

ψψ → ψψ scattering at order g2. The two diagrams contributing to

this process at this order are and we obtain using the Feynman rules

p1

p2

q1

q2

kp1

p2 q1

q2

k

Figure 6.8: Feynman diagramscontributing at order g2 to ψψ → ψψscattering. We show the explicit

momentum assignment for theexternal and internal lines. There aretwo distinct options.

iA = (−ig)2

[i

(p1 − q1)2 −m2+

i

(p1 − q2)2 −m2

](6.40)

The interpretation of these diagrams is the following: the nucleons ψ

exchange a meson φ which has momentum k = p1− q1 = p2− q2. This

meson does not satisfy the usual energy momentum relation, since as

we saw above k2 = −2p2(1 − cos θ) < 0 and therefore in particular

k2 6= m2! The meson φ is called a virtual particle and which is

off-shell.25 Heuristically we can say, it cannot live long enough for 25 k2 = m2 would be on-shell.

its energy to be measured to great accuracy.

6.6 More scattering processes in the ψψφ theory

Let us follow Feynman’s motto and let us ’shut up and calculate’ a

couple of sample processes.

6.6.1 Nucleon anti-nucleon scattering: ψ(p1)+ ψ(p2)→ ψ(q1)+

ψ(q2)

There are two diagrams contributing to this process. Note the

opposite flow of the arrows on the Feynman-lines indicating the

difference between a nucleon and an anti-nucleon. We obtain,

p1

p2

q1

q2

k

p1

p2

q1

q2

kFigure 6.9: Feynman diagramscontributing at order g2 to ψ(p1) +ψ(p2) → ψ(q1) + ψ(q2) scattering.We show the explicit momentum

assignment for the external andinternal lines. There are two distinct

options.

iA = (−ig)2

[i

(p1 − q1)2 −m2+

i

(p1 + p2)2 −m2

](6.41)

Note, since the diagrams are simply a shorthand for matrix elements

of operators in the Wick expansion, the orientation of the lines

inside the graphs have absolutely no significance. In the second

contribution, the virtual meson can go on-shell for certain values of


the external momenta. In the center of mass frame p1 = −p2 and

the 2nd denominator is

1

(p1 + p2)2 −m2=

1

4(M2 + p21)−m2

which can lead to a divergence if m > 2M for certain values of p1.

This will be fixed if one adds higher order corrections to the Feyn-

man graph in the form of an imaginary contribution (the width) and

we will find a bump if we scan the momenta p1 in an experiment.

6.6.2 Meson-meson scattering φφ→ φφ

For φφ → φφ, the simplest diagram appears only at g4. One of the

momenta remains unconstrained

p1

p2

q1

q2

k

k + q2

k q1

k q1 + p1

Figure 6.10: Feynman diagram

contributing at order g4 to φφ → φφ

scattering. We show the explicitmomentum assignment for the

external and internal lines.

and we need to integrate over it,

iA =

∫d4k

(2π)4

i

k2 −M2 + iε

i

(k − q1)2 −M2 + iε

× i

(k − q1 + p1)2 −M2 + iε

i

(k + q2)2 −M2 + iε

This is a one-loop diagram and the integration of kµ can require

some additional insights as diagrams can be divergent. This was a

serious problem in the early days of quantum field theory but we

now understand it along the lines of the discussion in Sec. 5.1.2. In

the example at hand we are fine though, since the integration is

convergent, since iA ∼∫d4k/k8 for very large kµ.

6.6.3 Symmetry factors

Are there combinatoric considerations in associating operators with

Feynman diagrams? Write the expansion of Eq. (6.20)

S = T exp

(−i∫d4xHI(x)

)

in powers of the interaction

S =

∞∑

n=0

S(n) (6.42)

=

∞∑

n=0

(−i)nn!

∫d4x1 . . . d

4xn THI(x1) . . .HI(xn) (6.43)


Generally, we might expect the factor 1/n! to get canceled when

we carry out the Wick expansion, because we can permute the

vertices to find a different contraction that gives the same normal

ordered operator. But this counting can change if the diagrams have

symmetries, e.g. permutations of the vertices that do not give rise

to a new contraction!

Let us consider three Klein-Gordon scalar fields φ1, φ2, φ3 with

masses m1,m2,m3 and an interaction

HI = λφ1(x)φ2(x)φ3(x)

and discuss the diagram associated with the contraction

(φ1φ2φ3)(φ1φ2φ3)(φ1φ2φ3) (6.44)

There are 3! contractions associated with this diagram, and the 1/3!

1

3

2

1

32

1

Figure 6.11: A λ3 diagram in theλφ1(x)φ2(x)φ3(x)-theory.

in front of the S(3) gets completely canceled.

We now need to add to our list of rules associating operators with

diagrams, we may need to add one more: Divide by the symme-

try factor of the graph. The symmetry factor is the number of

permutations of the vertices that leave the contraction unchanged. We will give a precise definition

below.In the λφ1(x)φ2(x)φ3(x)-theory, this consideration will give the

complete symmetry factor. In general, another contribution to the

symmetry factor of a graph occurs if there are identical fields at a

single vertex.

Let us illustrate this with in λφ3(x)-theory, where we consider a

single real KG-scalar field of mass m with the interaction

HI =λ

3!φ3(x)

The factor 1/3! is included since there are usually 3! different ways

of contracting the fields of a vertex with the fields of neighboring

vertices, and so the 1/3! gets canceled.

However, the 1/3! does not always get completely canceled be-

cause permutations of the lines might not give a new contraction.

We must therefore also include in the symmetry factor of a graph

the number of permutations of lines that leave the contraction

unchanged.


You can convince yourself, that we can put these considerations in a

simple formula for the symmetry factor S(G) of a graph G

S(G) =n!(η)n

r

where n is the number of vertices, η comes from the coupling con-

stant and is η = 4! in λ4!φ

4 theory and η = 3! in λ3!φ

3 theory, and r is

the multiplicity of the diagram. The multiplicity r is determined by

labeling all vertices with three (or four) lines emerging. We assume

all these lines to be distinguishable. The multiplicity r is then the

total number of ways to connect external points and vertices to form

the diagram .26 26 If a diagram is direction sensitive(charge flow), the we take thisinto account by only including the

number of ways to draw the diagramgiven the directional conditions ofthe external points.

To get the correct overall constant for a diagram, we divide by its

symmetry factor.

Examples: We will use the λ3! φ

3(x) and λ4! φ

4(x) theory

• No symmetry since all three external lines are distinct:

S(G) = 1

which comes about from

〈q1q2| :λ

3!φ(x)φ(x)φ(x) : |p3〉

=λ

3!3!〈q1q2| : φ(x)φ(x)φ(x) : |p3〉

we can contract each of the three field operators with each of the

external fields. Let us check by calculating

S(G) =1!(3!)1

3!= 1 (6.45)

where η = 3! and r = 3! since there are 3! ways to connect the

external points to the 3 lines emerging from the vertex.

• This is a 2nd order graph in λ3!φ

3 theory:

S(G) =2!(3!)2

(3 · 2)22= 1


4 theory.

S(G) =2!(4!)2

8 · 3 · 4 · 3 · 2 = 2

with r = 8 · 3 · 4 · 3 · 2 because we can connect the first external

point to 8 different lines emerging from the two vertices, after


that we can use the remaining 3 lines and the 4 · 3 lines from the

other vertex. The internal propagator is then fixed apart from a

two-fold ambiguity.


4 with two loops

S(G) =2!(4!)2

8 · 4 · 3 · 2 = 6

with r = 8 · 4 · 3 · 3 because there are 8 · 4 ways to connect to the

external points and once we choose one of the internal lines we

have first 3 options, then 2 to connect it to the other vertex.

• A 4th order λ3!φ

3 graph:

S(G) =4!(3!)4

12 · 2 · 9 · 2 · 6 · 2 · 3 · 2 = 1

Most people never need to evaluate a diagram with a symmetry

factor greater than 2, so you should not worry too much about this

technicality.

6.7 Potentials

Let us look at the non-relativistic limit of our scattering amplitude

with the goal to treat it as a potential in non-relativistic quantum

mechanics.

6.7.1 The Yukawa Potential

We first re-examine the problem of a classical Klein-Gordon field

in the presence a point source. We will just repeat the steps of

Sec. 4.3.2 where we now have to solve the static Klein-Gordon

equation

(+m2)φ(x) = δ3(x)

we can invert this and simplify since δ3(x) is time-independent

φ(x) =1

−∇2 +m2δ3(x)

We solve again using the Fourier transformation of the δ-function

φ(x) =

∫d3p

(2π)3

1

−∇2 +m2eix·p

=

∫d3p

(2π)3

1

p2 +m2eix·p

=1

4π2

1

ir

∫ ∞

0

dp peipr − e−iprp2 +m2

=1

2πr

∫ ∞

−∞

dp

2πip

eipr

p2 +m2


where we have skipped in the 2nd and 3rd line the steps which we

have derived carefully in Sec. 4.3.2. We close the contour in the

upper half plane p → +i∞ where we pick up the pole at p = +im,

which finally gives

φ(x) =1

4πre−mr (6.46)

If you compare this to the ∼ 1/r of Coulomb’s law Eq. (4.95) you see

that the field now dies off exponentially at distances > 1/m, which is

the Compton wavelength of the meson.

Can we now understand the profile of the φ field as a force be-

tween ψ particles? Recall, that we treat the electro-static potential

sourced by a δ-function as the potential energy for another charged

(test) particle moving in this background.

Is there a classical limit of the scalar Yukawa theory with the ψ

particles as δ-function sources for φ, creating the Yukawa potential?

The answer is yes, at least in the limit M m, that is, when the

nucleons are much heavier than the mesons and act as quasi static

sources.

Recall the Born approximation27 in non-relativistic quantum me- 27 See e.g. Cohen-Tannoudji, Quan-

tum Mechanics, Vol. II, ChapterVIII, especially section B.4

chanics: at first order in perturbation theory, we find the amplitude

to scatter an incoming state with momentum p of a potential U(r)

into an outgoing state with momentum q

AQM (p→ q) = 〈q|U(r)|p〉 = −i∫d3r U(r)e−i(q−p)·r

We will now compare the non-relativistic potential scattering with

that from the first diagram in Eq. (6.40) in the center of mass frame

iA = −g2 i

(p1 − q1)2 −m2= g2 i

|p1 − q1|2 +m2(6.47)

where we have used the fact that in the center of mass frame the

energies of the incoming and scattered particles are the same, see

Eq. (6.33).

We need to account for the different normalizations of the rel-

ativistic and non-relativistic amplitudes and divide the result by

(2M)2 to account for the difference between relativistic and non-

relativistic normalization of states, see Sec. 5.4. We define the dimen-

sionless quantity

λ ≡ g

2M

and obtain

∫d3r U(r)e−i(q1−p1)·r = − λ2

|p1 − q1|2 +m2

Inverting the Fourier-transform gives

U(r) = −λ2

∫d3p

(2π)3e−ip·r

1

p2 +m2


This is exactly the integral, that we already calculated in the begin-

ning of the section and we find

U(r) = − λ2

4πre−mr

This is the Yukawa potential, where the minus sign tells us that it

is attractive. Yukawa made this potential28 the basis for his theory 28 To be precise: he used the fermion-scalar version of this interaction

which however gives the same poten-

tial.

of the nuclear force and worked backwards from the range of the

force (about 1 fm) to predict the mass (about 200 MeV) of the

required meson, the pion.

We can directly from the scattering amplitudes for ψψ → ψψ, Ex: Check this!

ψψ → ψψ, and ψψ → ψψ that the sign of the spin-0 generated

Yukawa term is always positive – it leads to a universally attractive

potential. Compare this to the Coulomb potential which is mediated

by the exchange of a spin-1 boson, which can be either attractive

and repulsive. Gravity, which is mediated by a spin-2 field is again

universally attractive.

Notice that quantum field theory has given us an entirely new

view of forces between particles. Rather than being a fundamental

concept, the force arises from the virtual exchange of other particles,

in this case the meson.

6.8 How to calculate observables

We are now able to calculate amplitudes for a set of processes by

evaluating Feynman diagrams,

〈f |S − 1|i〉 = iAfi (2π)4δ(4)(pF − pI) (6.48)

but we have yet to calculate a probability, which would be measur-

able. For this, we need to square the amplitudes and sum over all

observed final states. But it looks like we are in trouble here, since

|〈f |S − 1|i〉|2 = |Afi|2[(2π)4 δ(4)(pF − pI)

]2(6.49)

Squaring a δ-function makes no sense. What happened?

As we will see, the problem is that we are not working with

square-integrable states. Instead, our states are normalized to δ(3)-

functions, since they are plane waves which exist at every point in

space-time. Thus, the scattering process is occuring at all points in

space for all time. No suprise that we got divergent non-sense.

6.8.1 Fermi’s Golden Rule

Let us derive the familiar Fermi’s golden rule from Dyson’s formula

of Eq. (6.15). Consider two energy eigenstates |n〉 and |m〉 of H0,

with En 6= Em. We find in leading order in the perturbation Recall from Eq. (6.5), that theinteraction Hamiltonian in theinteraction picture is

HI = (Hint)I = eiH0t (Hint) e−iH0t


〈m|U(t)|n〉 = −i〈m|∫ t

0

dt′HI(t′)|n〉

= −i〈m|Hint|n〉∫ t

0

dt′ei(Em−En)t′

= −i〈m|Hint|n〉∫ t

0

dt′eiωt′

= −〈m|Hint|n〉eiωt − 1

ω

with ω = Em − En. We obtain for the probability Pn→m(t) for the

transition |m〉 → |n〉 after a time t,

Pn→m(t) = |〈m|U(t)|n〉|2 = |〈m|Hint|n〉|2[4

sin2(ωt/2)

ω2

]

t

t2

!

4sin2(!t/2)

!2

Figure 6.12: 4[sin2(ωt/2)

ω2

]vs ω.

We show the function in brackets in Fig. 6.12. As you can see, after

a time t most of the transitions happen in a region between energy

eigenstates separated by

∆E ≈ 2π

t

As we take t → ∞, the function in brackets starts to approach a

δ-function. We get the normalization by

∫ ∞

−∞dω

[4

sin2(ωt/2)

ω2

]= 2πt

and therefore

4sin2(ωt/2)

ω2→ 2πt δ(ω), for t→∞

We consider now a transition to a cluster of states with density

ρ(E). In the limit t→∞, we get

Pn→m(t) =

∫dEm ρ(Em)|〈m|Hint|n〉|2 4

sin2(ωt/2)

ω2

→ 2πt |〈m|Hint|n〉|2ρ(En)

We finally obtain a constant transition probability per unit time

for states around the same energy En ∼ Em = E. This is Fermi’s

golden rule for the transition rate Γn→m = Pn→m Note however, according to

Wikipedia: “Although named af-

ter Enrico Fermi, most of the workleading to the Golden Rule is due

to Paul Dirac who formulated 20

years earlier a virtually identicalequation, including the three com-

ponents of a constant, the matrix

element of the perturbation andan energy difference. It was given

this name because, on account of its

importance, Fermi dubbed it ’GoldenRule No. 2.’ ”. This seems to show

how to avoid the Matthew principle,but you need to be yourself famous

enough. ’Golden Rule No. 1.’ states

the second-order contributions to thetransition probability.

Pn→m = 2π |〈m|Hint|n〉|2 ρ(E) (6.50)

This was a useful exercise, since it showed us how to carefully

take the limit t → ∞. If we chose to compute the amplitude for the

state |n〉 at t → −∞ to transition to |m〉 at t → +∞ we would get

the amplitude

−i〈m|∫ ∞

−∞dt′HI(t

′)|n〉 = −i〈m|Hint|n〉 2π δ(ω)

Squaring this, we wold run into the same problem as for the Feynman-

amplitude, since Pn→m(t) = |〈m|Hint|n〉|2 (2π)2 δ(ω)2, a non-sensical


infinity. Comparing to the result above, we realize now that the

extra infinity is coming because Pn→m is the probability for the

transition to happen in infinite time t → ∞. We can write these

δ-functions as

(2π)2 δ(ω)2 = (2π) δ(ω)T (6.51)

where T signals t → ∞. We can now divide by T , to derive the

probability per unit time as We used a similar trick, when we

discussed the vacuum energy inSec. 5.1.2.Pn→m = 2π |〈m|Hint|n〉|2 δ(ω)

which after integrating over the density of final states gives us

Fermi’s Golden rule again. We will now interpret the squares of the

δ(4)-functions in Eq. (6.49) as space-time volume factors.

6.8.2 Decay Rates

We now want to obtain the probability for a single particle |i〉 of

momentum pI to decay into some number of particles |f〉 with

momentum pF =∑l pfl . This probability is given by

P =|〈f |S|i〉|2〈f |f〉〈i|i〉

We have to be careful with the normalization since we work with

states obeying the relativistic normalization convention, see Sec. 5.4.

In particular, with Eq. (5.50), we obtain

〈i|i〉 = (2π)3 2Ep δ(3)(0) = 2EpIV

and

〈f |f〉 =∏

final states

2EplV

Let us place our initial particle at rest pI = 0 and EpI = mi, we

obtain for the probability for the decay

P = |〈f |S|i〉|2 × 1

〈f |f〉〈i|i〉 (6.52)

= |Afi|2 (2π)4δ(4)(pI − pF )V T × 1

2mV

∏

final states

1

2EplV

where like in the derivation of the Golden Rule we have replaced

one of the δ(4)-functions with a space-time factor (2π)4δ(4) = V T .

We have been already computing the amplitude Afi. We can now Recall the example of meson decayφ→ ψψ of Eq. (6.25)

〈f |S|i〉 = −ig (2π)4 δ(4)(p− q1 − q2)

= iAfi (2π)4 δ(4)(p− q1 − q2)

and therefore

Afi = −g

in leading order.

get rid of T by calculating the transition rate, which we get by

taking the T derivative (or just dividing by T ). One worry is still the

sum over all final states. We proceed in two steps: first we put our

theory in a box with side L assuming periodic boundary conditions

which allows us to count states. Then we integrate over all possible

momenta of all final states.


As we discussed in the introduction, in a box of dimension L3

with periodic boundary conditions, the allowed momentum values

are of the form

ki =2πniL

, with ni = 0,±1,±2, . . . , and i = x, y, z

which implies a density of states (L/(2π))3 in momentum space.

In the real world, no experimentalist can measure a cross-section

with exact momenta assignments. We can only measure all states

about some small region ∆k in momentum space. In a region of size

∆kx∆ky∆kz, there are

∆kx∆ky∆kz ×(L

2π

)3

=V

(2π)3∆kx∆ky∆kz

states. The normalization of our momentum

states and the field expansion alsochanges since for the discrete system

δ(3)(x) =1

V

∑k

e−ik·x

which means

φ(x) =

∫d3p

(2π)31√2Ep

[ap e

−ip·x + a†p eip·x

]→

1√V

∑k

1√2Ep

[ap e

−ip·x + a†p eip·x

]The reason, the 1√

Vappears, is

because we have chosen a discretenormalization

[ak, a†p] = δk,p

and φ therefore satisfies

[φ(x, t), φ(y, t)] = iδ(3)(x− y)

We recover the continuum as in the

introduction with∑k1

→L

2π

∫dk1

∑k

→V

(2π)3

∫d3k

ak →(

1

V

)1/2

ak

The number of final one-particle states in an infinitesimal momen-

tum interval is

V

(2π)3d3k

and we find for the density of final states that the factors of V

cancel

dΠ = (2π)4δ(4)(pI − pF )∏

final states

d3pl(2π)3

1

2Epl(6.53)

and find a Lorentz-invariant quantity, since the 1/(2Epl) conspire

to give us a Lorentz-invariant measure. This gives us our final

expression for the decay probability per unit time Γ = P ,

Γ =1

2m

∑

final states

∫|Afi|2 dΠ (6.54)

where the sum is over final states with different numbers and types

of particles.After a time t, the probability that the particle has not

decayed is just e−Γt. We call Γ the width of the particle. It is the

inverse of the half-life τ = 1/Γ. If we consider the uncertainty

principle, we can see this: the particle exists for a time τ , any mea-

surement of its energy (or mass in the rest-frame) must be uncertain

by ∼ 1/τ = Γ. Therefore, a series of measurements of the particle

mass will have a characteristic spread of order Γ.

Since we will need it later, we will also derive a general expression

for the probability per unit time using Eq. (6.52)

P = |Afi|2 V ·∏

initial states

1

2EpiV· dΠ (6.55)

where we have used Eq. (6.53) for the phase space of the final-states.


6.8.3 Example: two-body phase space

For two particles, there are six integrals in d3p1d3p2 but four of these

variables are constrained by 4-vector conservation δ(4)(p1 + p2 − pI),leaving only two variables to integrate over.

dΠ =d3p1

(2π)3

1

2Ep1

d3p2

(2π)3

1

2Ep2

(2π)4δ(4)(p1 + p2 − pI)

In the center of mass frame, pI = 0 and EI ≡ ET which is total

energy in the process.

dΠ =d3p1

(2π)3

1

2Ep1

d3p2

(2π)3

1

2Ep2

(2π) δ(E1 + E2 − ET )(2π)3 δ(3)(p1 + p2)

=d3p1

(2π)3

1

4Ep1Ep2

(2π) δ(E1 + E2 − ET )

=1

(2π)34Ep1Ep2

p21dp1dΩ1 (2π) δ(E1 + E2 − ET )

We now need to convert the δ-function of energy into a constraint

on p1. Using the notation of the margin note, we have g(p1) = We use again that

δ(g(x)) =∑i

δ(x− xi)|g′(xi)|

where the sum extends over all roots

of g(x) which are assumed to be

simple. For example

δ(x2 − α2

)=

1

2|α|

[δ (x+ α)+δ (x− α)

].

E1(p1) + E2(p2) =∑i

√p2i +m2 and

g′(p1) =

2∑

i=1

1√p2

1 +m2

1

2(2p1) =

p1

E1+p1

E2

=p1(E1 + E2)

E1E2=p1

E2

ETE2

where we have used p1 = −p2.

We obtain finally for the two-body phase space in the center of

mass frame,

dΠ2 =1

16π2

p1

ETdΩ (6.56)

Here we have assumed that the particles A and B in the final state

are distinguishable and we treated the final states |A(p1), B(p2)〉 and

|A(p2), B(p1)〉 as distinct. Note, that if the final states are instead

identical, we would have double-counted by 2!. In general, we need

to multiply an n-body phase space of n identical particles in the final

states by a factor of

dΠn →1

n!dΠn

6.8.4 Examples: particle decays

Let us apply this to an example. We assume that m > 4M2 and the

pion to two nucleon decay

φ→ ψψ

is kinematically allowed. There is only one diagram at leading order

in perturbation theory, see Eq. (6.25),

ψ†

ψ

φ

ψ†ψφ

Figure 6.13: Pion decay into two

nucleons. This is kinematicallyallowed in our toy model if m > 4M2.Note, in nature the pion is the

lightest QCD resonance and wouldnot be able to decay to two nucleons.

iA = −ig


and the decay width of φ is therefore

Γ(φ→ ψψ) =1

2m

∫dΠ |A|2

=g2

2m

1

16π2

p1

ET

∫dΩ

=g2

2m

p1

8πm

where we have used∫dΩ = 4π and ET = m. We adapt the result of

Sec. 3.5.2, and find p1 = 12

√m2 − 4M2 to obtain

Γ(φ→ ψψ) =g2

32π

1

m

√1− (2M)2

m2(6.57)

6.9 Cross Sections

In physical experiments, we often collide two beams of particles or a

beam hits a target. From time to time, they hit each other. We then

measure the resulting number of particles incident on a detector.

The fraction of the times (per unit flux) they collide is called

the cross section and we denote it as σ. The incoming flux F is

defined as the number of incoming particles per area per unit time.

The total number of scattering events N per unit time is then given

by

N = F · σ (6.58)

We can use quantum field theory to calculate σ. In fact, we can

extract even more and we can calculate dσ, which is the differential

cross-section. This dσ is related the probability for a given scatter-

ing process into a solid angle (θ, φ). With E1 and E2 as the energies

of the two incoming particles, we obtain with Eq. (6.55)

dσ =differential probability

unit time · unit flux

=|Afi|2

2E1 2E2V· 1

F· dΠ

We now need an expression for the unit flux F . Let us assume for

simplicity that we are in the center of mass frame of the collision.

We have so far considered a single particle per spatial volume V such

that the flux is given in terms of the 3-velocities as This is easy to see: consider a beamof particles perpendicular to a plane

of area A moving with v. If thedensity of particles is n, then after atime t, the total number of particlespassing through the plane is

N = |v|A · t · n

The flux is therefore N/(At) = |v| · n.

With our normalization, there is one

particle in the box with volume V , son = 1/V and the flux is

F = |v|/V

If we have two colliding beams, theprobability of finding either particle

is 1/V but since the collision can

occur anywhere in the box, the totalflux is

|v1 − v2|/V 2 × V = |v1 − v2|/V

F =|v1 − v2|

V

We therefore get

dσ =1

4E1E2

1

|v1 − v2|· |Afi|2 · dΠ (6.59)

You can now use Eq. (6.59) and take your favorite scattering

amplitude to compute the probability for particles to end up at

various angles.


Let us look in more detail at 2 → 2 scattering in the center of

mass frame. The 3-velocities are v1 = p1/E1 and v2 = p2/E2 =

−p1/E2 and so

|v1 − v2| = |p1|(

1

E1+

1

E2

)= p1

E1 + E2

E1E2=p1ETE1E2

which leads together with the two-body phase space of Eq. (6.56) to

dσ =1

4E1E2

1

|v1 − v2|· |Afi|2 · dΠ2

=1

4E1E2

E1E2

piET|Afi|2

1

16π2

pfET

dΩ

=1

4piET

1

16π2

pf |Afi|2ET

dΩ


dσ

dΩ=

1

64π2

1

E2T

pfpi|Afi|2 (6.60)

where pi and pf are the magnitudes of the three-momenta of the

incoming and outgoing particles.

6.10 Green’s Functions

Often times, we are not interested in questions that can be answered

by calculating scattering cross-sections. We might want to figure out

the non-Gaussianity of density perturbations arising in the CMB

from novel models of inflation or calculate the optical properties of

strange metals. These questions are best addressed with the help of

correlation functions, which we will now learn how to calculate

with Feynman diagrams.

6.10.1 The true vacuum

So far we have approximated the ground state with the ground state

of the free theory H0|0〉 = 0. We will now define the true vacuum

of the interacting theory |Ω〉. The energy is chosen such that for

H = H0 +Hint

H|Ω〉 = 0 (6.61)

and we normalize the vacuum to

〈Ω|Ω〉 = 1

We define correlation functions as

G(n)(x1, . . . , xn) ≡ 〈Ω|T φH(x1) . . . φH(xn)|Ω〉 (6.62)

where φH is in the Heisenberg picture of the full theory. We also

call G(n) the Green’s functions of the full theory. How can we

compute these objects using Feynman Diagrams?


Claim: The n-point correlation function is given as

G(n)(x1, . . . , xn) = 〈Ω|T φH(x1) . . . φH(xn)|Ω〉 (6.63)

=〈0|T φI(x1) . . . φI(xn)S|0〉

〈0|S|0〉 (6.64)

where we the interaction picture operators on the right-hand side

are applied on the vacuum of the free theory |0〉 and the usual time

evolution operator

S = T exp

(−i∫d4xHI(x)

)= lim

t+→+∞t−→−∞

U(t+, t−)

Proof: The time evolution operator in the interaction picture

U(t, t0) = T exp

(−i∫ t

t0

dt′∫d3xHI(x, t

′)

)

has the following properties

U†(t, t0) = U(t0, t), U(t0, t2) = U(t0, t1)U(t1, t2), U(t0, t0) = 1

The numerator of the RHS of Eq. (6.64) can be written with

φH(x) = U†(t, t0)φI(x, t)U(t, t0) (see Eq. (6.12)) as

〈0|T φI(x1) . . . φI(xn)U(t+, t−)|0〉= 〈0|T U(t0, t1)φH(x1)U(t1, t0)U(t0, t2)φH(x2)U(t2, t0) . . .

U(t0, tn)φH(xn)U(tn, t0)U(t+, t−)|0〉

Since we are allowed to commute operators under time ordering T ,

using U(t0, t1)U(t1, t0) = 1, we get

〈0|T U(t+, t0)φH(x1)φH(x2) . . . φH(xn)U(t0, t−)|0〉

Since t± → ±∞ the ordering of the remaining U is independent

of the time-ordering T and they will always appear to the left and

right of the string of operators. We therefore need to figure out the

meaning of the two boundary U operators, e.g.

limt−→∞

U(t0, t−)|0〉

We consider an arbitrary state |Ψ〉,

〈Ψ|U(t0, t−)|0〉 = 〈Ψ|UH(t0, t−)|0〉

Where we have used that H0|0〉 = 0 and UH is the time evolution

operator of the full theory. Let us insert a complete set of states, Recall from Eq. (6.13), that in theinteraction picture

U(t, t0) = eiH0(t−t0)e−iH(t−t0)

and the full time-evolution is

UH = e−iH(t−t0)

which we take as energy eigenstates of the full Hamiltonian H =

H0 +Hint,

〈Ψ|UH(t,−∞)|0〉 = limt−→−∞

〈Ψ|UH(t0, t−)

|Ω〉〈Ω|+

∑

n 6=0

|n〉〈n|

|0〉

= limt−→−∞

〈Ψ|UH(t0, t−)|Ω〉︸︷︷︸|Ω〉

〈Ω|0〉+ limt−→−∞

∑

n 6=0

eiEn(t−−t)〈Ψ|n〉〈n|0〉

= 〈Ψ|Ω〉〈Ω|0〉+ limt−→−∞

∑

n6=0

eiEn(t−−t)〈Ψ|n〉〈n|0〉


where we have used Eq. (6.61) in the first term. The second term

can also be simplified. In fact, it vanishes!

Figure 6.14: The Riemann-Lebesgue

lemma. The function f(x) is multi-

plied by a rapidly oscillating function.The product integrates to zero in the

limit of infinite oscillation frequency.

Proof:∣∣∣∣∫ f(x)e−iωx dx

∣∣∣∣ =

∣∣∣∣∫ 1

iωf ′(x)e−iωx dx

∣∣∣∣≤

1

|ω|

∫|f ′(x)| dx→ 0 as ω → ±∞.

The sum∑n is actually an integral

∫d3p . . . since the momentum

states form a continuum. We can now use the Riemann-Lebesgue

lemma, which states that for any well-behaved function f(x),

limω→∞

∫ β

α

dx f(x) eiωx = 0

We can alternatively derive this by shifting the integration slightly

into the imaginary t→ −∞(1 + iε), which exponentially suppresses all

energy eigenstates except for the ground-state.29 The above result 29 See e.g. Peskin & Schroeder,chapter 4.2therefore gives,

〈Ψ|UH(t,−∞)|0〉 = 〈Ψ|Ω〉〈Ω|0〉 (6.65)

In particular, we see that the vacuum of the free theory and the

vacuum of the full theory are proportional to each other Since the result is true for any state〈Ψ|, we have

UH(t,−∞)|0〉 = |Ω〉〈Ω|0〉

|0〉 = U†H(t,−∞)|Ω〉〈Ω|0〉|0〉 = |Ω〉〈Ω|0〉

|0〉 = 〈Ω|0〉 · |Ω〉

which justifies the somewhat hand-waving argument in Sec. 6.3.1 to

use free states, or creation operators acting on the free vacuum |0〉,in scattering processes.

If we insert the result of Eq. (6.65) in the RHS of Eq. (6.64), we find With

〈0|S|0〉 = limt±→±∞

〈0|U(t+, t−)|0〉

= limt±→±∞

〈0|U(t+, t)U(t, t−)|0〉

= limt±→±∞

〈0|UH(t+, t)UH(t, t−)|0〉

= 〈0|Ω〉〈Ω|Ω〉〈Ω|0〉

〈0|T φI(x1) . . . φI(xn)S|0〉〈0|S|0〉

=〈0|Ω〉〈Ω|T φH(x1)φH(x2) . . . φH(xn)|Ω〉〈Ω|0〉

〈0|Ω〉〈Ω|Ω〉〈Ω|0〉= 〈Ω|T φH(x1)φH(x2) . . . φH(xn)|Ω〉

where we have used 〈Ω|Ω〉 = 1. This completes the proof of this very

important result which we can write in slightly more explicit form

〈Ω|T φH(x1) . . . φH(xn)|Ω〉 =〈0|T φI(x1) . . . φI(xn) e−i

∫d4xHI(x)|0〉

〈0| e−i∫d4xHI(x)|0〉

(6.66)

This expression for the n-point Green’s function G(n)(x1, . . . , xn)

is ideally suited to doing perturbative calculations. We need only

keep as many terms as needed in the Taylor series expansions of the

exponentials.

6.10.2 Connected diagrams and vacuum bubbles

We will now sketch some further results before we move on to

greener pastures30. With Dyson’s formula and Wick’s theorem, 30 Spin 12

representations!

we know how to calculate the expressions appearing in Eq. (6.66).

Feynman diagrams greatly simplify these calculations. But what is

the meaning of the factor (〈0|S|0〉)−1? What are we dividing the


expression by? We will now sketch that the denominator removes

all vacuum bubbles and that we only need to consider connected

Feynman graphs, in words:

〈Ω|T φH(x1) . . . φH(xn)|Ω〉 =∑

connected Feynman graphs

(6.67)

where a connected diagram is defined by the requirement that every

line is joined to an external leg.

Let us explain this in φ4 theory. The diagramatic expansion for

〈0|S|0〉 is

None of these diagrams connect to external particles and we there-

fore call them vacuum bubbles. We will not show this here31, 31 See however Sheet 7, Ex 2!

but the pre-factors of the diagrams consisting of combinatoric and

symmetry factors conspire such that the whole series can be expo-

nentiated

Quantum Field Theory: Example Sheet 4

Dr David Tong, November 2007

1. A real scalar field with φ4 interaction has the Lagrangian

L =1

2∂µφ∂

µφ− 1

2m2φ2 − λ

4!φ4 (1)

Use Dyson’s formula and Wick’s theorem to show that the leading order contribution

to 3-particle → 3-particle scattering includes the amplitude

p3

2p

p1

p12p

p3

//

/

= (−iλ)2 i

(p1 + p2 + p3)2 − m2(2)

Check that this result is consistent with the Feynman rules for the theory. What other

diagrams also contribute to this process?

2. Examine ⟨0|S|0⟩ to order λ2 in φ4 theory. Identify the different diagrams with

the different contributions arising from an application of Wick’s theorem. Confirm

that to order λ2, the combinatoric factors work out so that the the vacuum to vacuum

amplitude is given by the exponential of the sum of distinct vacuum bubble types,

⟨0|S |0⟩ = exp ( + + + ... ) (3)

3. Consider the Lagrangian for 3 scalar fields φi, i = 1, 2, 3, given by

L =3!

i=1

1

2(∂µφi)(∂

µφi) − 1

2m2(

3!

i=1

φ2i ) − λ

8(

3!

i=1

φ2i )

2 (4)

Show that the Feynman propagator for the free field theory (i.e. λ = 0) is of the form

⟨0|Tφi(x)φj(y)|0⟩ = δijDF (x − y) (5)

where DF (x − y) is the usual scalar propagator. Write down the Feynman rules of the

theory. Compute the amplitude for the scattering φiφj → φkφl to lowest order in λ.

1

and we find that the amplitude for the vacuum of the free theory to

evolve into itself (the numerator of the RHS of Eq. (6.66))

〈0|S|0〉 = exp(all distinct vacuum bubbles)

The same combinatoric simplification takes place for the numerator

evaluating of Eq. (6.66) generic correlation functions. One can show See e.g. Schwartz 7.2.5 or Peskin 4.4.

Compare also to the exponentiationof the cluster expansion in statisticalphysics, see script 3.4.1 of SS17 of

the TH4A lecture by yours truly.

〈0|T φI(x1) . . . φI(xn)S |0〉 = 〈0|S|0〉 ·∑

connected diagrams

where again connected means no vacuum bubbles, or that all the

lines are joined to an external leg. We can now conclude that the

division by 〈0|S|0〉 has a very cute interpretation in terms of Feyn-

man diagrams. We only need to calculate the connected Feynman

graphs and we do not have to care for the vacuum bubbles. We

therefore obtain as promised Eq. (6.67)

〈Ω|T φH(x1) . . . φH(xn)|Ω〉 =∑

connected Feynman graphs

Here we show another example for HI(x) = g3!φ

3(x) theory for a

two-point function

〈0|T φI(x1)φI(x2) e−i∫d4xHI(x)|0〉 =

D12 − g2

∫ (1

4D1xDxxDyyDy2 +

1

8D12DxxDxyDyy +

1

12D12D

3xy + . . .

)=


+

+=

D12D1xDxyDyxDy2

where integration over repeated indices is implied and DF (x1, x2) =

D12. The sum over all graphs (disconnected and connected) in the

numerator is then the sum over all graphs with no bubbles multiply-

ing the sum over the bubbles. In the calculation of 〈Ω|T φH(x1) . . . φH(xn)|Ω〉,we therefore can ignore all bubble diagrams, they are automatically

included by using the real vacuum |Ω〉 instead of the free one |0〉.

6.10.3 Outlook to S-matrices from Green’s functions

How can we related the Grenn’s functions back to S-matrix elements

relevant for scattering processes? The Feynman rules for the Green’s

functions include the propagators for the external legs, which we

have to amputate. Also, the 4-momenta assignment to the external

legs in G(n) is arbitrary and generally not on-shell. In order to

achieve both we can relate the Green’s function in momentum space

G(n)(p1, . . . , pn) to S-matrix elements

〈q1, . . . , qn|S − 1|p1, . . .pn′〉

= (−i)n+n′n∏

i=1

(q2i −m2 + iε)

n′∏

j=1

(p2j −m2 + iε) G(n+n′)(−q1, . . . ,−qn, p1, . . . , pn′)

where we take the limit p0 → Ep in all of the external legs. These

prefactors therefore vanish and we only get a non-zero answer for

diagrams contributing to Gn which have the same propagators for

each external leg.

The important step that we have taken here, is that this provides

a systematic approach to dealing with true particle states in an

interacting theory. This formula still applies once we take into

account virtual particles in asymptotic states. This is the formally

correct way to consider scattering and is known as the LSZ reduction

formula. It will be worked out and shown in more detail in your

quantum field theory lecture.


I(a) I(b) I(c) I(d) I(e)

I(f) I(g) I(h) I(i) I(j)

II(a) II(b) II(c) II(d) II(e)

II(f) III(a) III(b) III(c) IV

V VI(a) VI(b) VI(c) VI(d) VI(e)

VI(f) VI(g) VI(h) VI(i) VI(j) VI(k)

Figure 6.15: Self-energy-like di-

agrams representing 32 gauge-

invariant subsets contributing to themuon g − 2 at the tenth order, which

leads to prediction for the anomalous

magnetic dipole moment a = g−22

as

aµ(SM) = 116 591 840 (59)× 10−11

and compare to the measurement

aµ(exp)− aµ(SM) = 249 (87)× 10−11

which is a terrifically precise pre-diction. From Phys.Rev.Lett. 109(2012) 111808.

7

Lorentz Representations

In non-relativistic quantum mechanics we learn that the spin ± 12

states of the electron along a specific direction, are conveniently

described by a complex doublet of states

|ψ〉 =

∣∣∣∣∣ψ↑ψ↓

⟩(7.1)

The dynamics of the doublet ψ = 〈x|ψ〉 is governed by the Schrodinger-

Pauli equation

i∂tψ =

[(1

2m(i∇− eA)2 − eA0

)12×2 + µBB · σ

]ψ (7.2)

with the vector potential Aµ, B = ∇×A, and the Bohr magneton

µB =e

2me

describing the strength of the coupling between the B-field and the

electron’s magnetic moment.1 1 Recall the Stern-Gerlach experi-ment.The σ = (σ1, σ2, σ3) are the Pauli-matrices which we have

combined into a vector:

σ1 =

(0 1

1 0

), σ2 =

(0 −ii 0

), σ3 =

(1 0

0 1

)(7.3)

Now it would seem that we have a problem with rotational invari-

ance, since the σ-vector will not change under rotations contrary

to ψ and B. How do we make sure that the equation is rotationally

invariant? It works because

[σi, σj ] = 2i εijkσk

are the exact same algebraic relations that the generator of infinites-

imal rotations satisfy, see Eq. (7.8) and the discussion below. We

anticipate the result and conclude that the changes induced by

rotations cancel in

B · σ ψ

or any other vector field, like ∂i which transforms also as a 3-vector.

So we could have written down a rotationally invariant equation for

ψ with

(1 ∂t − ∂iσi)ψ = 0 (7.4)


It turns out, this equation is Lorentz-invariant, too. We combine

σµ = (1, σ1, σ2, σ3) (7.5)

to obtain This is the equation for a Weyl

2-spinor, which is a slightly differ-

ent form of the equation which weusually associate with the Dirac equa-

tion, which describes the dynamics of

a Dirac 4-spinor.

σµ∂µψ = 0 (7.6)

This is the Dirac equation! We could try to be bold and add a

mass term with

(σµ∂µ +m)ψ = 0 (7.7)

but this is not a Lorentz invariant equation. Please forget Eq. (7.7)

immediately (and remember only Eq. (7.6)).

In conclusion, we need to understand the intricate Lorentz prop-

erties of spin- 12 particles, discuss their representations. The Dirac

equation and its non-relativistic limit, the Schrodinger-Pauli equa-

tion, will follow.

7.1 The Lorentz algebra

We will now be a bit more rigorous and introduce groups, algebras

and representations. A group consists of a set of elements gi and

a rule

gi × gj = gk

The rule is required to be associative and have an identity element.

A representation is a particular embedding of the gi into operators

that act in a vector space. For finite-dimensional representations this Think of e.g. matrices acting onvectors. Technically, the matrixembedding is the representation,

not the vectors, even though we arenot careful usually in making thisdistinction.

means an embedding of the gi into matrices.

Every group has a trivial representation r : gi → 1. We dis-

tinguish this from a faithful representation, where each group

element has its own matrix.

For the Lorentz-group, which we recall from Eq. (3.58) leaves

ηαβ = ηµν ΛµαΛνβ

invariant, we have only discussed the fundamental representation

which acts on 4-vectors

xµ → Λµνxν

and if you go back to Sec. 3.3.1, you’ll find examples of this embed-

ding as the set of 4× 4 rotation and boost matrices. Our goal is now

to find all the representations.

Our goal is now to extract the properties of the group inde-

pendent of one particular representation. We will now focus on

continuously connected groups, which can be described by a set

of coordinates which are real numbers. The easiest approach is to A non-continuous group, is e.g. the

reflection group Z2, which has theelements 1,−1

consider infinitesimal transformations: for any group G, group

elements g ∈ G can be written as

g = exp(iαgi λi)


where the αgi are just numbers and the λi are group generators. Gen-

erators λi form an algebra, because we can add and multiply them,

whereas group elements can only be multiplied. They can be ex-

tracted from the group elements using infinitesimal transformations

g = exp(iαgi λi) = 1 + iαgi λi + . . .

or through

λi =1

i

dg

dαgi

∣∣∣∣αgi=0

Lie groups are a special class, with an infinite number of ele-

ments but with a finite number of generators. The generators of a Think of the rotation group in3D, which contains infinitely manyrotations matrices depending on

the 3 continuous Euler rotationangles, but there are only 3 different

infinitesimal rotations in the x − y,

x− z, and y − z plane.

Lie group form the Lie algebra. Lie groups are crucial for the SM

as it contains the SU(3)× SU(2)× U(1) special unitary groups. The

Lorentz group is also a (non-compact) Lie group and is sometimes

called O(1, 3), an orthogonal group that preserves vector products

with a metric with the (1, 3) signature ηµν = (1,−1,−1,−1).

Let us start with a reminder of the rotation group O(3).2 The 2 We will denote the Lie group with

capital letters, like SO(3), and thecorresponding Lie algebra with so(3)

three generators Ji of the rotation group satisfy

[Ji, Jj ] = iεijkJk (7.8)

when we act on a space-time vector xµ they are represented as

J1 =

0 0 0 0

0 0 0 0

0 0 0 −i0 0 i 0

, J2 =

0 0 0 0

0 0 0 i

0 0 0 0

0 −i 0 0

, J3 =

0 0 0 0

0 0 −i 0

0 i 0 0

0 0 0 0

and so e.g. for the group element G1 = exp(iθJ1)

G1 =

1 0 0 0

0 1 0 0

0 0 cos θ sin θ

0 0 − sin θ cos θ

(7.9)

You can obtain J2 and J3 by straight-forward permutations. How Check that the commutator in

Eq. (7.8) is indeed satisfied!about the Lorentz-boosts? We recall Eq. (3.17)

(x

t

)=

(cosh η sinh η

sinh η cosh η

)(x′

t′

)(7.10)

which means infinitesimally

x→ x+ η t

t→ t+ η x

The generator of a Lorentzboost in the x direction

t

x

y

z

=

t

x

y

z

+ iK1

t

x

y

z


is therefore given by the hermitian matrix

iK1 =

0 1 0 0

1 0 0 0

0 0 0 0

0 0 0 0

and analogously

iK2 =

0 0 1 0

0 0 0 0

1 0 0 0

0 0 0 0

, iK3 =

0 0 0 1

0 0 0 0

0 0 0 0

1 0 0 0

We can now check the commutator between the generators of rota-

tions Ji and the boost generators Ki. We find

[Ji,Kj ] = iεijkKk (7.11)

What does that mean? This just reflects that the boost generators

Ki transform as a 3-vector under rotations, as expected since the

boost direction is a 3−vector characterized by the relative velocity vi.

We will now do one of the most important calculations in history. Under a rotation we would get for

the boost group elements U =

exp(iηlKl) which are 4× 4 matrices,

U → G−1i UGi = (1− iJi)(1 + iηlKl)(1− iJi) + . . .

= 1 + iηlKl + ηl[Kl, Ji] + . . .

and so

Kl → i[Ji,Kl]

Compute [K1,K2]. What do we get?

K1K2 −K2K1 =

0 0 0 0

0 0 −1 0

0 0 0 0

0 0 0 0

−

0 0 0 0

0 0 0 0

0 −1 0 0

0 0 0 0

=

0 0 0 0

0 0 −1 0

0 1 0 0

0 0 0 0

= −i J3

Two boosts produce a rotation! Summarizing, we find the Lorentz-

algebra SO(1, 3)

[Ji, Jj ] = iεijkJk

[Ji,Kj ] = iεijkKk

[Ki,Kj ] = −iεijkJk

(7.12)

(7.13)

(7.14)

Please note the very important negative sign in the commutator of

two boosts!

Now use a crucial observation: the algebra falls apart in two

independent sets if complexify it using

J+i ≡

1

2(Ji + iKi), J−i ≡

1

2(Ji − iKi)

which satisfies

[J+i , J

+j ] = iεijkJ

+k (7.15)

[J−i , J−j ] = iεijkJ

−k (7.16)

[J+i , J

−j ] = 0 (7.17)


This shows that the complexified Lie-algebra for the Lorentz-group

has two commuting sub-algebras. The algebra generated by J±

is the 3D rotation algebra or so(3) = sl(2,R) = su(2). So we have

shown that

so(1, 3) = su(2)⊕ su(2)

which means that the Lorentz group SO(1, 3) is locally isomorphic to

two independent SU(2) groups. As one can show, the fact that the

complexifications of the corresponding Lie algebras coincide implies

that the representation theory of both groups coincides.

7.1.1 Representations

We can now simply use our knowledge of SU(2) representations

to determine all representations of SO(1, 3). We know that the

representations are labeled by their spin

j = 0, 12 , 1,

32 , . . . (7.18)

and each SU(2) representation consists of (2j + 1) objects

ψm with m = −j,−j + 1, . . . , j − 1, j

which are rotated into each other by elements of SU(2). We there-

fore know that the irreducible representations of SO(1, 3) are

labeled by (j+, j−) with j± taking on the values of Eq. (7.18).

Each SO(1, 3) representation contains (2j+ + 1)(2j− + 1) objects

Ψm+,m− with

Ψm+,m− with m± = −j±,−j± + 1, . . . , j± − 1, j±

SU(2)⊗ SU(2) (0, 0) ( 12 , 0) (0, 1

2 ) ( 12 ,

12 ) (1, 0) (1, 1)

SO(3) 0 12

12 0⊕ 1 1 0⊕ 1⊕ 2

Table 7.1: Decomposition of ir-reducible representations of the

Lorentz SU(2) × SU(2) into irre-

ducible representations of its SO(3)subgroup describing spin.

We recognize the trivial, one-dimensional representation (0, 0)

as the Lorentz-scalar. Counting the size of the representation, we

conclude that the 4-dimensional representation ( 12 ,

12 ) is the Lorentz

vector, which is the defining representation of the Lorentz group.

7.1.2 Spinor representations

How can we understand the representations ( 12 , 0)? We write The name ”left-handed” and ”right-

handed” from the way the spinprecesses as a massless fermionmoves: we will see this in Sec. 7.3.4.

ψL = ψα, with α = 1, 2 (7.19)

How about the notation ( 12 , 0) ? This says that J+

i = 12 (Ji + iKi)

acting on ψα is represented by

1

2σi


while J−i = 12 (Ji − iKi) is represented by 0. This means that the

representations of the generators (before complexification) are

Ji =1

2σi

iKi =1

2σi

after solving the simple linear system. The equality here means

”represented by”.

We will similarly denote the ( 12 , 0) with the slightly exotic symbol We use dotted and undotted indices

to make sure we don’t accidentally

contract the α of ( 12, 0) with an

α = 1, 2 of ( 12, 0), which would not be

Lorentz invariant. This is called vander Waerden notation. It will prove

extremely useful in supersymmetry

and in superstring theory.

ψR = χα, with α = 1, 2 (7.20)

Do not be confused by the bar. For now it is just part of the typo-

graphic symbol.

The two-component spinors ψα and χα are called Weyl spinors

and furnish perfectly good representations of the Lorentz group.

What about the Dirac spinor? If you have already encountered

it, you know that it has also 4 components. Why? The reason is

parity. We know that under parity, see XXXX, that

x→ −xt→ t

and similarly for other 3-vectors like the 3-momentum or the boosts

p→ −pK → −K

Rotations are pseudo-vectors and they transform as Think of the angular momentum as

L = r × v or imagine a clock-wise

rotation in a mirrored coordinatesystem.

J → J

This means for the complexified generators

J+i =

1

2(Ji + iKi) →

1

2(Ji − iKi) = J−i

In other words J+ ↔ J− and so under parity we exchange

( 12 , 0) ↔ (0, 1

2 )

In order to describe an electron in QED which conserves parity,

we need to use both 2-dimensional representations, which we can

describe mathematically

( 12 , 0)⊕ (0, 1

2 )

This reducible representation is thus given by stacking two Weyl QED and QCD are parity invari-

ant, which means the equations ofmotions and the Lagrangian areinvariant under P transformations. If

had chosen to write the theory using

just the ( 12, 0) spinor, we would have

explicitly violated P. This is in fact,

what happens in the electro-weak

sector: SU(2)L gauge bosons onlycouple to ( 1

2, 0) spinors and not to

the (0, 12

) representation.

spinors to form a Dirac spinor

Ψ =

(ψL

ψR

)=

(ψα

χα

)


The spinor Ψ(p) is (like the scalar) a function of 4-momentum, or a

function of xµ if you Fourier-transform. We know its transformation

properties under rotations

JDirac =

(12σ 0

0 12σ

)

where the equality means ”represented by”. Similarly for boosts

iKDirac =

(12σ 0

0 − 12σ

)

Note, the very important minus sign, which distinguishes the two

irreducible representations entering a Dirac spinor.

Parity forced us to us a 4-spinor but we know that a electron has

only two physical degrees of freedom. We must project out two of

four the components described by Ψ(p). Let us go to the rest-frame

(pµ)r ≡ (m,0)

Anticipating the result, we will understand fully later, we write the

projection operator as You can already guess that γ0 willturn out to be one of the γ matrices.

P =1

2(1− γ0)

For now γ0 is just a 4× 4 matrix. A projection applied twice should

be the same as just applying it once

P2 = P (γ0)2 = 1

and the eigen-values of γ0 are therefore ±1. Since we know that

under parity

ψL ↔ ψR

we cannot simply use the projection operator to set one of the fields

(e.g. χα) to 0. Parity means that the two irreducibles should be on

the same footing. We choose the following representation of γ0

γ0 =

(0 12×2

12×2 0

)

or explicitly We could have used also a different

basis choice.

γ0 =

0 0 1 0

0 0 0 1

1 0 0 0

0 1 0 0

So in the rest-frame, we can write the projection to two degrees of

freedom as

P Ψ(pr) = 0

which is the same as

ψL − ψR = 0 (7.21)


Since this is an equation equating the two-spinors, it removes two of

the four degrees of freedom contained in the Dirac spinor.

Eq. (7.21) does not look Lorentz invariant since it equates states

in two different representations. Note however, that we required a

specific frame (the rest frame) and therefore should not respect the

resulting equation to be covariant. Once we boost to a generic frame,

we will find that the projection equation is covariant.

7.1.3 The Dirac equation

We have ”derived” the Dirac equation above. Why? We know now

how to obtain the equation which is satisfied by Ψ(p) of a general pµ

by Lorentz-transforming the spinor! We boost

Ψ(p) = e−iη·KΨ(pr)

which implies for the projection equation for a general pµ

(e−iη·Kγ0eiη·K − 1

)Ψ(p)

we define

γµpµm≡ e−iη·Kγ0eiη·K

to obtain the Dirac-equation

Figure 7.1: A memorial stone inWestminster Abbey in the nave near

Newton’s monument. According to a

physics legend, Dirac was staring intoa fire on evening in 1928 when the

equation occurred to him.

(γµpµ −m)Ψ(p) = 0 (7.22)

which we can write in coordinate space as

(iγµ∂µ −m)Ψ(x) = 0 (7.23)

which is a first-order differential equation for the Dirac 4-spinor.

We can expand on the last steps by noting that Ex: Show this!

e−iη·Kγ0eiη·K =

(0 exp(−η · σ)

exp(η · σ) 0

)

and with the unit vector η/|η| we get Ex: Show this, too! Use σi, σj =

σiσj + σjσi = 2δij

exp(η · σ) = cosh |η|+ η

|η| · σ sinh |η|

If we identify the relation between boost and the 3-momentum and

the relevant boost as Ex: Show this, too! You can imme-diately convince yourself, that this is

the correct result in 1 + 1 dimensions,

since

pr = (m, 0)→ (p0, p1)

which requires a boost: p1 =m sinh(η) + 0 · cosh(η) which isEq. (7.24) in 1+1 space-time dimen-

sions.

p = mη

|η| sinh |η| (7.24)

We can then easily see that

γi =

(0 σi

−σi 0

)

The Dirac matrices satisfy the anti-commutator relations A,B = AB + BA


γµ, γν = 2gµν 14×4 (7.25)

The γ-matrices generate the Dirac algebra, which is a special

case of a so-called Clifford algebra. This particular basis is the the

Weyl-representation.

We additionally define

σµ ≡ (1,σ), σµ ≡ (1,−σ) (7.26)

and we can write the Dirac matrices in the Weyl or chiral basis as

γµ =

(0 σµ

σµ 0

)(7.27)

We have found a non-trivial representation of the Lorentz group.

Note however that the representation is non-unitary! The complexi-

fied algebra is unitary but it will come with imaginary coefficients.

You can see that most easily by looking at an example of a boost

in the z-direction for ( 12 , 0)-representation which generated by K3

iK3 =σ3

2 K3 =

1

2

(−i 0

0 i

)

which means that the group g3(η3) ∈ SO(1, 3) element is

g3(η3) = exp(iη3K3) =

(eη2 0

0 e−η2

)

This is not a unitary matrix, since g†3g3 6= 1. This is expected since

the Lorentz group is not compact.3 and therefore there are no 3 The rotational subgroup iscompact since it is parametrized by

angles θi which lie on a circle S1

with 0 and 2π identified (there is

more to this but since the topology

of SO(3) is a ball with antipodalsurface points identified). Boosts

however are parametrized by η = v/c

with 0 ≤ η < 1 where η = 0 andη = 1 are not the same and the

group is therefore not compact.

finite dimensional unitary representations of the Lorentz group. We

will have to make use of projective representations and work with a

so-called little group, but this is for your next QFT course.4

4 See e.g. S. Weinberg, QFT1, chap-

ter 2.

We can find a Lorentz-covariant form of the Lorentz-algebra in

Eq. (7.14) by generalizing the rotation group generators and writing

them as differential operators

J = x× p = x× (−i∇)

which we can also write as an anti-symmetric tensor with J3 = J12

and so on:

J ij = −i(xi∇j − xj∇i)

The generalization to 4 dimensions is naturally

Jµν = i(xµ∂ν − xν∂µ) (7.28)

which is an anti-symmetric 4×4 matrix, which contains N(N−1)/2 =

3 · 4/2 = 6 elements. We can identify them with the previously defined

Ji and Ki by

J i → J ij and Ki → J0i


again with J3 = J12 and so on. A general Lorentz transformation is

parametrized by the anti-symmetric tensor

ωµν = −ωνµ

and

Λ = e−i2ωµνJ

µν

The commutation rules of the generators can be determined simply

from the commutators of the differential operators5 5 See Ex Sheet 9!

[Jµν , Jρσ] = i(ηνρJµσ − ηµρJνσ − ηνσJµρ + ηµσJνρ). (7.29)

You can see that the fundamental representation for Jµν acting on

4-vectors is given by

(Jµν)αβ = i(δµαδνβ − δµβδνα) (7.30)

Compare with the definition of Ki and Ji above to confirm. We can

understand it better by looking at an infinitesimal transformation of

xµ

xα →(δαβ −

i

2ωµν(Jµν)αβ

)xβ

which becomes e.g. in the case of ω23 = −ω32 = θ

xα →

1

1

1 −θθ 1

x

ν

or in exponentiated form e−iω23J23

, we reproduce Eq. (7.9)

G1 =

1 0 0 0

0 1 0 0

0 0 cos θ sin θ

0 0 − sin θ cos θ

(7.31)

and you can also verify that ω10 = −ω01 = η gives K1.

The Lorentz-algebra for Dirac spinors can be represented by Or with the other common Diraccombination

σµν =i

2[γµ, γν ]

we trivially have

Sµν =1

2σµν

Sµν =i

4[γµ, γν ] (7.32)

which can be written with Eq. (7.25) as

Sµν =i

4[γµ, γν ] =

0 µ = ν

i2γ

µγν µ 6= ν(7.33)

=i

2γµγν − i

2ηµν (7.34)

you can check that Eq. (7.32) is a representation of the Lorentz-

algebra satisfying Eq. (7.29)

[Sµν , Sρσ] = i(ηνρSµσ − ηµρSνσ − ηνσSµρ + ηµσSνρ). (7.35)


by using Eq. (7.25) repeatedly. So for a given Lorentz-transformation

Λµν defined by the parameters ωµν , we have the Dirac-spinor repre-

sentation

S[Λ] = exp

(− i

2ωµνS

µν

)(7.36)

with the transformation

Ψ(x)→ S[Λ]Ψ(Λ−1x)

We can again see, that the Lorentz-transformations are not unitary,

since the the exponent is not hermitian6 because 6 For general unitary matrix

U = exp(iA)

in order to satisfy

U†U = 1

we see

U†U = exp(−iA†) exp(iA)

= 1− i(A† −A) + . . .

we need

A = A†

(Sµν)† = − i4

[(γν)†, (γµ)†]

=i

4[(γµ)†, (γν)†]

which would be hermitian if all the γµ are hermitian (γµ)† = γµ,

or all are anti-hermitian (γµ)† = −γµ. However this never happens

because of Eq. (7.25)

(γ0)2 = 1 real eigenvalues

(γi)2 = −1 imaginary eigenvalues

So we could pick γ0 to be hermitian, but we can only chose γi to be

anti-hermitian. Indeed, you can check that the Dirac matrices in the

Weyl or chiral basis of Eq. (7.27) exactly satisfy this

(γ0)† = γ0, (γi)† = −γi

We see that there is no way to pick the γµ such that the exponent is

hermitian and the representation is therefore non-unitary.

7.1.4 Lorentz-invariant actions

We have now a new field that we can work with, the Dirac spinor Ψ.

We now want to construct a Lorentz invariant action, which will lead

to a Lorentz invariant equation of motion.

Let us try something naive, which will not work. We define the

adjoint spinor as

Ψ†(x) = (Ψ∗)T (x)

Can we build a Lorentz scalar by contracting Ψ†Ψ =∑3l=0(Ψ∗)lΨl ?

How does this behave under Lorentz transformations

Ψ(x)→ S[Λ]Ψ(Λ−1x)

Ψ†(x)→ Ψ†(Λ−1x)S[Λ]†


Ψ†(x)Ψ(x)→ Ψ†(Λ−1x)S[Λ]†S[Λ]Ψ(Λ−1x)


This would be invariant if only S[Λ] was unitary but we in general

have S[Λ]†S[Λ] 6= 1. This means Ψ†Ψ is not a Lorentz-invariant

scalar, and certainly cannot be used in an action.

But we can use our insight in why it fails to find a solution. We

pick a representation of the Clifford algebra which like the Weyl

basis satisfies (γ0)† = γ0 and (γi)† = −γi, then we have

γ0γµγ0 = (γµ)†

which means that

(Sµν)† =i

4[(γµ)†, (γν)†] = γ0Sµνγ0

such that

S[Λ]† = exp

(i

2ωµν(Sµν)†

)= γ0(S[Λ]−1)γ0 (7.37)

This allows us to define the Dirac adjoint

Ψ(x) = Ψ†(x)γ0 (7.38)

We now see that ΨΨ is a Lorentz scalar since with Eq. (7.37)

Ψ(x)Ψ(x) = Ψ†(x)γ0Ψ(x)

→ Ψ†(Λ−1x)S[Λ]†γ0S[Λ] Ψ(Λ−1x)

= Ψ†(Λ−1x)γ0Ψ(Λ−1x)

= Ψ(Λ−1x)Ψ(Λ−1x)

which is indeed the tranformation law for a Lorentz-scalar field, see

e.g. Eq. (4.35).

Let us show another important bi-linear: ΨγµΨ is a Lorentz-

vector. We claim, it therefore transforms as

Ψ(x)γµΨ(x)→ ΛµνΨ(Λ−1x)γνΨ(Λ−1x)

and we treat the µ = 0, 1, 2, 3 label on the γ matrices as a true

vector index. We can use this bi-linear to form Lorentz-scalars

by contracting it with other Lorentz-vectors. We suppress the x-

argument for notational simplicity.

ΨγµΨ→ ΨS[Λ]−1γµS[Λ] Ψ

To show that ΨγµΨ transforms as a vector, we need

S[Λ]−1γµ S[Λ] = Λµνγν (7.39)

which we show by working infinitesimally

Λ = 1− i

2ωρσJ

ρσ + . . .

S[Λ] = 1− i

2ωρσS

ρσ + . . .


From Eq. (7.39) we find that we have to show

[Sρσ, γµ] = −(Jρσ)µνγν (7.40)

which holds if we plug in the definition of Jρσ.We use

γµγν = −γνγµ + γµ, γν= −γνγµ + 2ηµν

to obtain for ρ 6= σ with Eq. (7.33) for the LHS

[Sρσ, γµ] =i

2[γργσ, γµ]

=i

2(γργσγµ − γµγργσ)

=i

2(γργσ, γµ − γργµγσ − γµ, γργσ + γργµγσ)

= i(γρησµ − γσηµρ) (7.41)

For the RHS we plug in Eq. (7.30) with (Jρσ)µν = i(ηρµδσν − ησµδρν)

to finally obtain

−(Jρσ)µνγν = −i(ηρµγσ − ησµγρ)

which equals Eq. (7.41) and therefore proves Eq. (7.40).

Similarly, ΨγµγνΨ transforms as a Lorentz-tensor. Ex: Show this!

7.1.5 The Dirac Lagrangian

We can now use the Lorentz-covariant bilinears ΨΨ, ΨγµΨ, and

ΨγµγνΨ to build Lorentz-invariant action. We choose

S =

∫d4xΨ(x)(iγµ∂µ −m)Ψ(x) (7.42)

This is the Dirac action. If we vary with respect to Ψ, we get

(iγµ∂µ −m)Ψ(x) = 0

which is the beautiful Dirac equation. It is first order in deriva-

tives but miraculously Lorentz invariant. If we tried a similar thing

with a scalar field it would look like vµ∂µφ = . . . which necessarily in-

volves a privileged vector vµ in space-time and would not be Lorentz

invariant.

The Dirac equation mixes the various components of Ψ through

the γµ matrices. Note, each individual component itself solves the

Klein-Gordon equation. We write

(iγν∂ν +m)(iγµ∂µ −m)Ψ(x) = (−γνγµ∂ν∂µ +m2)Ψ(x) = 0

with

γνγµ∂ν∂µ =1

2γν , γµ ∂ν∂µ = ∂ν∂ν

and so we get

−(∂ν∂µ +m2)Ψα(x) = 0

which is the Klein-Gordon equation. We show the spinor index α

explicitly to show that it applies to each component.


7.1.6 The slash

We will introduce some useful notation. Often we contract 4-vectors

with gamma matrices and we write

Aµγµ = /A

and the Dirac equation reads

(i/∂ −m)Ψ = 0

7.1.7 Chiral spinors and the Weyl equation

We write our spinor in the Weyl basis as

Ψ(x) =

(ψL(x)

ψR(x)

)

which is in the ( 12 , 0)⊕(0, 1

2 ) representation. From XXX, we know that

the Weyl spinors ψL(x) in (12 , 0) and ψR(x) in (0, 1

2 ) transform the

same way under rotations

S[Λrot] =

(exp( i2θ · σ) 02

02 exp( i2θ · σ)

)(7.43)

but differently under boosts

S[Λboost] =

(exp(+ 1

2η · σ) 02

02 exp(− 12η · σ)

)(7.44)

or in components for rotations

ψL/R → exp(i

2θ · σ)ψL/R

and for boosts

ψL/R → exp(±1

2η · σ)ψL/R

which decomposes the Dirac Lagrangian into

L = Ψ(iγµ∂µ −m)Ψ

= iψ†Lσµ∂µψL + iψ†Rσ

µ∂µψR −m(ψ†LψR + ψ†RψL) (7.45)

We see that a massive fermion seems to require both ψL and ψR

since they couple trough the mass term.7 In the massless case the 7 You can also write a mass term

with only one of them called a

Majorana mass, see below.two spinors decouple and we get the the Weyl equations

iσµ∂µψL = 0

iσµ∂µψR = 0

We have already seen in the ”derivation” of the Dirac equation in

Sec. ??, that the equation of motion projects out half the degrees of

freedom. The crucial aspect was, that the equation of motion was

first order rather than second order as the Klein-Gordon equation.


In particular the conjugate momentum to the Dirac Lagrangian is

given by

πΨ =∂L∂Ψ

= iΨ†

which is not proportional to the time-derivative of Ψ. Therefore, the

phase space of the spinor is parametrized by Ψ and Ψ†, while for a

scalar it is φ and π = φ. So the phase space is described by 8 real

dimensions and consequently the Dirac spinor has 4 real degrees of

freedom.

7.1.8 γ5

In our choice of basis for the γµ matrices, the Lorentz-transformations

S[Λ] came out block-diagonal. What happens if we choose a different

representation of the Clifford algebra?

γµ → UγµU−1, and Ψ→ UΨ

S[Λ] will now likely not be block-diagonal. Can we define the Weyl

spinors in an invariant way? We introduce a ”fifth” gamma-matrix8 8 You can convince yourself that wecan also write γ5 as

γ5 =i

4!εµνσργ

µγνγσγρ

using the totally anti-symmetric four

index tensor with ε0123 = 1.

γ5 = iγ0γ1γ2γ3

which satisfies which in the Weyl basis looks like

γ5 is called the ”fifth” gamma-matrix, because with ΓA = γA for

A = 0, 1, 2, 3 and Γ4 = iγ5 it satisfiesthe five-dimensional Clifford algebra

ΓA,ΓB = 2ηAB

γ5 =

(−12 02

02 12

)(7.46)

γµ, γ5 = 0, and (γ5)2 = +1

As you will show in the exercises

[Sµν , γ5] = 0

which means that γ5 is a scalar under rotations and boosts. This

is another manifestation of the fact, that the Dirac representation

is reducible, since eigenvectors of γ5 whose eigenvalues are different

transform without mixing9 9 This criterium for reducibility is

also known as Schur’s lemma.We can therefore define Lorentz-invariant projection operators

PL/R =1

2(1∓ γ5)

such that

P 2L = PL, P 2

R = PR, PLPR = 04

with Eq. (7.46), we see that PL/R projects into the Weyl-basis

ψL = PLΨ

ψR = PRΨ


7.2 Discrete symmetries

7.2.1 Parity

We already saw that under parity P the left-handed and right-

handed spinors are exchanged

P : ψL ↔ ψR

Using this, we can write the action of parity on the Dirac spinor

itself as

P : Ψ(x, t)→ γ0Ψ(−x, t)

Notice that the Dirac Lagrangian conserves parity as you can easily

see from Eq. (7.45). Let us now discuss how our interaction terms

change under parity. The ”mass” bi-linear transforms as

P : Ψ(x, t)Ψ(x, t)→ Ψ(−x, t)Ψ(−x, t)

thanks to (γ0)2 = 1. This is the transformation of a scalar field. How

about the vector ΨγµΨ? We split in temporal

P : Ψ(x, t)γ0Ψ(x, t)→ Ψ(−x, t)Ψ(−x, t)

and spatial components

P : Ψ(x, t)γiΨ(x, t)→ Ψ(−x, t)γ0γiγ0Ψ(−x, t)= −Ψ(−x, t)γiΨ(−x, t)

which shows that ΨγµΨ is a vector. We can use γ5 in the bilinear to

form another Lorentz scalar and vector

Ψγ5Ψ, and Ψγ5γµΨ

We find with γ5, γµ = 0 that

P : Ψ(x, t)γ5Ψ(x, t)→ Ψ(−x, t)γ0γ5γ0Ψ(−x, t)= −Ψ(−x, t)Ψ(−x, t)

it is a pseudo-scalar and for the Ψγ5γµΨ bilinear

P : Ψ(x, t)γ5γµΨ(x, t)→

−Ψ(−x, t)γ5γ0Ψ(−x, t)Ψ(−x, t)γ5γiΨ(−x, t)

we find that it is a pseudo-vector or axial-vector. Like the rotation vector Ji or theangular momentum L = x× p.

In summary we find the following bi-linears in Table 7.2.1 and their

properties with

Ψ(x) Γ Ψ(x)

for various Γ matrices. The total number of bi-linears is 16, which What about Γ = γ[µγνγσ] orΓ = γ[µγνγσγρ]? The brackets

indicate anti-symmetrization, sincewe do not need to consider the

symmetric part which is ∼ ηµν (anti-

symmetrization also explains why weconsider at most 4-tensors). One can

show that γ[µγνγσ] = −iεµνσργργ5and γ[µγνγσγρ] = −iεµνσργ5 andso they are linearly related to the

simpler bilinears with Γ = γργ5 and

Γ = γ5.

is exactly what we could hope for from a 4× 4 component object.

We have now a set of terms to build covariant Lagrangians with.

If we add γ5 to our terms then these terms will typically break


Γ type number of bi-linears

1 scalar 1

γµ vector 4

σµν = i2 [γµ, γν ] tensor 6

γµγ5 pseudo-vector 4

γ5 pseudo-scalar 1

16

parity. The electro-weak sector of the SM treats the ψL and ψRfields differently and they have different quantum numbers. They

do not describe the same particle, e.g. the left-handed electron eL

which couples to W -bosons of the SU(2)L gauge interactions is

only distantly related to the right-handed electron eR which does

not. A theory which treats ψL and ψR on the same footing is called In fact they are only related through

a scalar coupling via the Higgs h

L ⊃ e†LheRa vector-like theory. A Lagrangian in which ψL and ψR appear

differently, is called a chiral theory. QCD and QED are vector-like,

and the electro-weak sector including the Higgs are chiral theories.

7.3 Solutions of the free Dirac equation

Let us study the solutions of the free Dirac equation

(/p−m)Ψ(p) =

(−m σµpµ

σµpµ −m

)Ψ(p) = 0 (7.47)

We start with a simple ansatz

Ψ = u(p)e−ip·x, with p2 = m2

where the on-shell condition is motivated because the components

satisfy the Klein-Gordon equation. For now we focus on positive

frequencies, such that p0 > 0. The column vector u(p) needs to

satisfy the additional constraint

(γµpµ −m)u(p) = 0

We can easily analyze this equation in the rest frame, where pr =

(m,0). We get the general solution by boosting with S[Λ]. In the

rest frame

(mγ0 −m)u(pr) = m

(−12 12

12 −12

)u(pr) = 0

the solutions are

u(pr) =√m

(ξ

ξ

)


for any two-component spinor ξ which is just a number. We conven-

tionally normalize

ξ†ξ = 1

and we added√m for future convenience. The interpretation for the

ξ two spinor is given by the transformation under rotations, where

ξ transforms as an ordinary two-component spinor of the rotation

group, see Eq. (7.43). We have

ξ → exp( i2 θ · σ) ξ

and ξ therefore determines the spin orientation of the Dirac solution,

as usual. E.g. when

ξ =

(1

0

)

then the particle has spin up (↑) along the z-direction. Notice again,

that after applying the Dirac equation we are free to choose only 2 of

the four components u(p).

Let us now apply a boost in the z-direction(E

p3

)=

(m cosh η

m sinh η

)

Now apply the same boost to u(p). With Eq. (7.44) we get Show this!

u(p) =

(√p · σ ξ√p · σ ξ

)(7.48)

with p = (E, 0, 0, p3) and where the square-root of a matrix is

understood as taking the positive root of each eigenvalue. Let us

show that this satisfies the Dirac equation. With u(p)T = (u1, u2) we

can write the Dirac equation Eq. (7.47)as

(p · σ)u2 = mu1 (7.49)

(p · σ)u1 = mu2 (7.50)

We can see that either of the two equations implies the other using

(p · σ)(p · σ) = p20 − pipjσiσj = p2

0 − pipjδij = pµpµ = m2

Let us try the ansatz u1 = p ·σξ′ for some ξ′ then Eq. (7.50) immedi-

ately gives

u2 = mξ′

We find that any spinor of the form

u(p) = A

((p · σ)ξ′

mξ′

)

with constant A would be solution. To make this more symmetric,

we choose A = 1/m and ξ′ =√p · σ ξ with constant ξ. Then

u1 = (p · σ)√p · σ = m

√p · σξ


7.3.1 Negative energy solutions

We get another solution to the Dirac equation with the ansatz

Ψ = v(p) e+ip·x

The u(p) solution oscillates in time as Ψ ∼ e−iEt and is called a

positive frequency solution. The v(p) solutions oscillates with

Ψ ∼ e+iEt and are called negative frequency solutions. Both are

solutions to the classical Dirac equation and note that both have

positive energy. The Dirac equation requires that the 4-component

spinor v(p) satisfies

(m σµpµ

σµpµ m

)v(p) = 0

which with the above we see is solved by

v(p) =

( √p · σ χ

−√p · σ χ

)(7.51)

with some normalized (χ†χ = 1) 2-component spinor, taken to be a

constant.

7.3.2 Examples

We take again the positive frequency solution with mass m at rest

p = 0

u(pr) =√m

(ξ

ξ

)

Consider ξT = (1, 0) we boost to (E, 0, 0, p3) to get

u(p) =

√p · σ

(1

0

)

√p · σ

(1

0

)

=

√E − p3

(1

0

)

√E + p3

(1

0

)

This expression in fact also makes sense for a massless particle with

E = p3. For a massless spinor we therefore obtain We picked the normalization such

that this limit would make sense.

u(p) =√

2E

0

0

1

0

Similarly, if we boost ξT = (0, 1) (spin down), we get

u(p) =

√p · σ

(0

1

)

√p · σ

(0

1

)

=

√E + p3

(0

1

)

√E − p3

(0

1

)

m→0−→√

2E

0

1

0

0


7.3.3 Helicity

We finally explain the meaning of ”left-handed” for ψL and ”right-

handed” for ψR. The massless solutions are eigenstates of the helic-

ity operator

h ≡ p · S =1

2pi

(σi 0

0 σi

)(7.52)

where pi = p/|p|. In Ex9, you will discuss the Pauli-

Lubanski polarization vector,

Wµ ≡ (1/2)εµνσρPνJσρ

which is a Lorentz covariant operator.

Using the commutation relations forthe generators of the Poincare group,one can show that P 2 = PµPµ

and W 2 = WµWµ, are Casimiroperators for the Poincare algebra.Since the Poincare group has rank2, P 2 and W 2 are its only Casimir

operators. Therefore a massive state

can be labelled by two numbers, itsmass and spin. A massless state

can be labeled by only one number

(called helicity), since Pµ and Wµ formassless particles are proportional to

each other.

A particle with h = +1/2 is called right-handed, while one with

h = −1/2 is called left-handed.

The helicity of a massive particle depends on the reference frame:

we can always boost to a frame in which its momentum is in the

opposite direction (but its spin stays unchanged). 10

10 For a massless particle movingwith v = c, we cannot perform such a

boost.

The Lorentz-invariance of helicity (for mass-less particles) is

manifest in the representation of Weyl spinors ψL and ψR which live

in different representations of the Lorentz group.

7.3.4 Useful results: inner and outer products

It is convenient to introduce a basis ξs and χs with s = 1, 2 for the

two-component spinores such that

ξr†ξs = δrs, and χr†χs = δrs

which e.g. is satisfied for

ξ1 =

(1

0

), and ξ2 =

(0

1

)

and similarly for χs. The two independent solutions to the positive

frequency ansatz are

us(p) =

(√p · σ ξs√p · σ ξs

)

We can take the inner product in two ways, either u† · u or u · u. Only

u · u will be Lorentz-invariant of course, but we will need u† · u to

quantize the theory later. Let us determine the results.

(ur(p))† · us(p) = (ξr†√p · σ, ξr†√p · σ)


)

= ξr†p · σξs + ξr†p · σξs

= 2ξr†p0ξs

= 2E δrs

and for the Lorentz-invariant inner product

ur(p) · us(p) = (ξr†√p · σ, ξr†√p · σ)

(0 1

1 0

)(√p · σ ξs√p · σ ξs

)= 2mδrs


Similarly for the negative frequency solutions we have

vs(p) =

( √p · σ χs

−√p · σ χs

)

with

(vr(p))† · vs(p) = 2p0δrs

vr(p) · vs(p) = −2mδrs

What about the inner products of u and v ? We compute

ur(p) · vs(p) = (ξr†√p · σ, ξr†√p · σ)

(0 1

1 0

)( √p · σ χs

−√p · σ χs

)

= ξr†(−√p · σp · σ)χs + ξr†√p · σp · σχs = 0

and similarly

vr(p) · us(p) = 0

For the adjoint inner product, there is another combination which

has useful properties. With (p′)µ = (p0,−p), we have

(ur(p))† · vs(p′) = (ξr†√p · σ, ξr†√p · σ)

( √p′ · σ χs

−√p′ · σ χs

)(7.53)

= ξr†√

(p · σ)(p′ · σ)χs − ξr†√

(p · σ)(p′ · σ)χs

(7.54)

The terms under the square root are

(p · σ)(p′ · σ) = (p0 + piσi)(p0 − piσi) = p2

0 − p2 = m2

and the same holds for (p · σ)(p′ · σ). This means the two terms in

Eq. (7.54) cancel and we get

(ur(p))† · vs(−p) = (vr(p))† · us(−p) = 0

7.3.5 Outer products

A final spinor identity will prove very useful before we turn to

quantizing everything. We claim

2∑

s=1

us(p)us(p) = /p+m (7.55)

Note, the two spinors are not contracted, but placed back to back to

give a 4× 4 matrix Similarly Think of |a〉〈b| instead of 〈b|a〉.

2∑

s=1

vs(p)vs(p) = /p−m

Now let us show this.

2∑

s=1

us(p)us(p) =

2∑

s=1


)(ξs†√p · σ, ξs†√p · σ) (7.56)


Now

2∑

s=1

ξsξs† = 12

which gives for Eq. (7.56)

2∑

s=1

us(p)us(p) =

(m p · σp · σ m

)= /p+m (7.57)

You can now easily prove∑2s=1 v

s(p)vs(p) = /p−m.

The Dirac Sea:

A historical detour: if we interpret Ψ as a single-particle wave function, the plane wave solutions to the Dirac

equation are energy eigenstates, with

Ψ = u(p)e−ip·x i∂Ψ

∂t= +EpΨ

Ψ = v(p)e+ip·x i∂Ψ

∂t= −EpΨ

which look like positive and negative energy solutions. The spectrum is unbounded below! This seems disas-

trous. Dirac invented the ’Dirac see’, where all negative energy states are filled up thanks to the Pauli-exclusion

principle. He postulated only charge differences are observable (similar to the energy difference argument mo-

tivating normal ordering). He postulated that if you excite a negative energy state, it leaves a hole with all the

properties of an electron, except that it carries positive charge! This is the prediction of anti-matter.

We now understand that the interpretation of the Dirac spinor as a single-particle wavefunction is not really

correct. The picture of a Dirac sea relies on the fact, that the particles are fermions – which breaks down for a

charged scalar, a boson, which also predicts anti-particles.

What we can still learn from Dirac’s analysis is that there is no consistent way to interpret the Dirac equation

as describing a single particle. We need to think of it as a classical field equation with only positive energy solu-

tions (the Hamiltonian is positive definite!). Once we quantize the theory, we encounter particle and anti-particle

excitations.

7.4 Symmetries of the Dirac Lagrangian

We can proceed as in the scalar Lagrangian and derive the space-

time symmetries

δΨ = εµ∂µΨ

and use the standard formula Eq. (4.67) for the energy-momentum

tensor Recall

(jµ)ν =∂L

∂(∂µφa)∂νφa − δµνL ≡ TµνTµν = iΨγµ∂νΨ− L

using the fact that L does not depend on ∂µΨ. Since the current

is only conserved on-shell, we can impose the equations of motions

directly on Tµν and we won’t lose anything. Since the equations of

motions are first order, this affects the energy-momentum tensor. We

use (iγµ∂µ −m)Ψ(x) = 0 which means, we can set L = 0, to obtain

Tµν = iΨγµ∂νΨ


which e.g. leaves for the total energy

E =

∫d3xT 00 =

∫d3x iΨγ0Ψ (7.58)

=

∫d3x iΨ†γ0(−iγi∂i +m)Ψ (7.59)

In the last equality we have again used the equations of motion.

7.4.1 Internal Symmetries

We can rotate the Dirac spinor by a phase

Ψ→ eiαΨ

which leaves the Dirac Lagrangian invariant. This gives rise to a

current

jµV = ΨγµΨ (7.60)

where we specify that it is a vector symmetry which transforms

ψL and ψR in the same way. We can easily obtain jµV using the

Ψ→ eiαΨ

ψL → eiαψL

ψR → eiαψR

Noether procedure or we can simply check that it is conserved on-

shell:

∂µjµV = (∂µΨ)γµΨ + Ψγµ(∂µΨ)

= imΨΨ− imΨΨ = 0

where we have used i/∂Ψ = mΨ and the Dirac adjoint equation

Ψi←−/∂ = −mΨ, where the arrow points to the term we take the

derivative of. The conserved charge

Q =

∫d3xΨγ0Ψ =

∫d3xΨ†Ψ (7.61)

has the interpretation of electric charge or particle number for

fermions.

For a massless Dirac spinor m = 0, the left-handed ψL and

right-handed ψR spinors decouple and we can rotate the left and

right-handed fermions in opposite directions

Ψ→ eiαγ5Ψ

ψL → e−iαψL

ψR → e+iαψRΨ→ eiα γ

5

(7.62)

which gives the conserved axial current

jµA = Ψγ5γµΨ (7.63)

This is only conserved when m = 0, as we can see by taking the

divergence

∂µjµA = (∂µΨ)γµγ5Ψ + Ψγµγ5(∂µΨ)

= (∂µΨ)γµγ5Ψ−Ψγ5γµ(∂µΨ)

= 2imΨγ5Ψ

which vanishes only for m = 0. Once we couple an axial current to

a gauge field (similar to the photon field Aµ), axial symmetries do

not survive the quantization process, they develop anomalies. An

Anomaly is a symmetry of the classical theory which is not preserved

in the quantum theory.11 11 Another symmetry which isusually anomalous is scale symmetryxµ → e−αxµ, see Ex4-2.


7.5 Quantizing the Dirac Field

If we were to use the same approach for the free Dirac Lagrangian

L = Ψ(x)(iγµ∂µ −m)Ψ(x)

as for the Klein-Gordon scalar theory, we would run into serious

trouble: the quantized free Hamiltonian would seem to be able

to lower the energy by creating particles (The Hamiltonian is not

bounded below). If we tried anti-commutation relations for a scalar

field, the cancellation between negative and positive frequencies to

guarantee causality in Eq. (5.71) and Eq. (5.72) would not work

anymore. This is the gist of the spin-statistics theorem. See Ex10-1.

7.5.1 The Hamiltonian

We will need the Hamiltonian. With the conjugate momentum

π = iΨ†

we obtain

H = πΨ− L = Ψ(−iγi∂i +m)Ψ (7.64)

which is the same as the result we obtained deriving the energy-

momentum tensor in Eq. (7.59).

We now want to turn this into an operator. We will expand Ψ(x)

into a basis of eigenfunctions of the free Dirac equation

ψ(x) =

2∑

s=1

∫d3p√2Ep

(bsp u

s(p)e+ip·x + cs†p vs(p)e−ip·x

)

ψ†(x) =

2∑

s=1

∫d3p√2Ep

(bs†p u

s†(p)e−ip·x + csp vs†(p)e+ip·x )

where the operators bs†p create particles associated with the spinor

state us(p), while the cs†p create anti-particles associated to vs(p).

Except for the us(p) and vs(p), this has the exact same form as the

expansion of a complex scalar field, see Sec. 5.5.

7.5.2 Quantization of fermions

We need to account for the Fermi-Dirac-statistics for fermions,

meaning that the state picks up a minus sign upon the interchange

of any two particles. The spin-statistics theorem says that integer

spin must be quantized as bosons, while half-integer spin must be

quantized as fermions. If we try to do otherwise (as you should in

the exercise), we will find inconsistencies, such as unboundedness of

the free Hamiltonian.

How do we quantize fermionic fields? Recall for bosons

[a†p, a†q] = 0 a†p a

†q |0〉 ≡ |p, q〉 = |q,p〉

If we want Fermi-Dirac statistics, we will need anti-commutation

relations A,B = AB +BA to hold.


Recall that for a scalar field φ, we could interpret a†p and ap as

creation an annihilation operators because of their commutation

relation with the number operator12 12 ... or equivalently with the freeHamiltonian

H =

∫d3k

(2π)3Ek a

†kak

Nφ =

∫d3k

(2π)3a†kak

Recall, that this worked because of the following identity for commu-

tators

[AB,C] = A[B,C] + [A,C]B

which immediately gives13 13 ... and also

[N, ak′ ] = −ak′

Recall that this means for eigenstates

of the number operator

N |n〉 = n|n〉

and so

N a†k|n〉 = ([N, a†k] + a†kN)|n〉

= (a†k + a†kN)|n〉

= (1 + n)a†k|n〉

and so

a†k|n〉 ∝ |n+ 1〉

up to normalization.

[N, a†k′

]=

∫d3k

(2π)3

[a†kak , a

†k′

]= a†k′

Therefore a†k′ acting on a state raises the eigenvalue of N by one

and the energy by Ek, as expected for a creation operator. However

there is another useful relation for commutators

[AB,C] = AB,C − A,CB (7.65)

with the anticommutator A,B = AB + BA. This means that if

we were to impose anticommutation relations on the a†k and ak, then

they could still have an interpretation as creation an annihilation

operators!

We will therefore ask the spinor field to satisfy

Ψα(x),Ψβ(y) = Ψ†α(x),Ψ†β(y) = 0 (7.66)

and

Ψα(x),Ψ†β(y)

= δαβ δ

(3)(x− y) (7.67)

Using the expansion for Ψ and Ψ† gives

brp, bsq = crp, csq = brp, cs†q = . . . = 0

and the canonical non-zero relations

brp, b

s†q

= (2π)3 δrs δ(3)(p− q)

crp, c

s†q

= (2π)3 δrs δ(3)(p− q)

(7.68)

(7.69)

The calculation of the Hamiltonian is straight-forward but somewhat

tedious and we get Send me the LATEXand I’ll post ithere.

H =

∫d3p

(2π)3Ep

(bs†p b

sp − cspcs†p

)

=

∫d3p

(2π)3Ep

(bs†p b

sp + cs†p c

sp − (2π)3δ(3)(0)

)

where the anti-commutation relation saved us from the indignity of

an unbounded Hamiltonian! Interestingly the contribution to the

cosmological constant now has the opposite sign compared to the

bosonic result.14 14 This anticipates a result of su-persymmetry, which is a symmetrybetween fermions and bosons. Exact

supersymmetry predicts H|0〉 =((2π)3δ(3)(0) − (2π)3δ(3)(0))|0〉 = 0.

This holds after quantization andeven in the presence of interactions!


7.5.3 Fermi-Dirac Statistics

We define the (free) vacuum as usual

bsp|0〉 = csp|0〉 = 0

for all creation operators. The Hamiltonian still has nice commuta-

tion relations with the anti-commuting b and c operators. You can

easily see with Eq. (7.65) that

[H, bs†p ] = Epbs†p

[H, bsp] = −Epbsp

and exactly the same for b ↔ c. We can treat them therefore again

as ladder operators creating a tower of energy-eigenstates by acting

with bs† and cs† on the vacuum, just as in the bosonic case. The (not

relativistically normalized) one-particle states are

|p, r〉 = br†p |0〉

and the (not relativistically normalized) two-particle states statisfy

|p1, r1;p2, r2〉 = (br1p1)†(br2p2

)†|0〉 = −|p2, r2;p1, r1〉

exactly as we wanted, confirming that the particles do indeed obey

Fermi-Dirac statistics. In particular, we obtain the Pauli-Exclusion

principle

|p1, r1;p1, r1〉 = 0

As for the scalar field (which had spin 0), we could act with the

angular momentum operator and confirm that a particle at rest

|p = 0, r〉

does indeed carry intrinsic angular momentum 1/2 as expected.

7.5.4 Causal Fermi Fields

It is not clear that a theory with Fermi-Dirac statistics will lead to

causal predictions. For bosons, we said that

[φ(x), φ(y)] = 0, for (x− y)2 < 0.

For Fermi fields we now have the relations

Ψα(x),Ψβ(y) = 0

and

Ψα(x),Ψ†β(y) = 0, for (x− y)2 < 0

How do we see that this guarantees causality in the quantized the-

ory?


The reason is that observables are necessarily always bilinear in

the fields, e.g. the energy Eq. (??), momentum and fermion number

Eq. (7.61) (or charge) are given by

E =

∫d3x iΨ†∂tΨ

P i =

∫d3x iΨ†∂iΨ

Q =

∫d3xΨ†Ψ

This is not entirely surprising: we know that spinors form a double-

valued representation of the Lorentz group since they change

sign under 2π rotations. Observables, however, are unaffected by

a rotation by 2π and so must be composed of an even number of

spinor fields.

7.6 Perturbation Theory for Spinors

We will consider a simple nucleon-meson theory15 15 The nucleons are fermions now andwe are graduating from a toy model

to something closer to Nature, eventhough it does not in fact describenucleon-meson interactions at low

energy correctly.

L = Ψ(iγµ∂µ −m)Ψ +1

2(∂µφ)2 − µ2

2φ2 − λΨΓΨφ (7.70)

with Γ = 1 in case of φ being a scalar or Γ = iγ5 for a pseudo-

scalar.16 16 We need the i for hermiticity of L.

Note, the dimensions of the fields, as before [φ] = 1, but the

kinetic term of the spinor requires [Ψ] = 3/2. This turns the in-

teraction to a marginal term [λ] = 0, compare to the dimensionful

(relevant) interaction in the case of a trilinear coupling between

scalars.

Dyson’s formula and Wick’s theorem can be used in almost

the same way for Fermi fields as for scalars. However, the an-

ticommutation relations lead to a crucial difference. Remem-

ber that for space-like distances (x − y)2 < 0, time-ordering is

not Lorentz-invariant17 Still, the T product of two scalar fields is 17 Since we can boost to a reference

frame,where the temporal orderx0 > y0 is reversed to x0 < y0.

Lorentz-invariant because the fields commute for space-like distances

(x− y)2 < 0, so

φ(x)φ(y) = φ(y)φ(x)

and the order here doesn’t matter. For fermions this no longer

holds! If (x− y)2 < 0, Fermi fields anticommute. For (x− y)2 < 0 and

if we are in a frame where x0 > y0

TΨ(x)Ψ(y) = Ψ(x)Ψ(y) (7.71)

in the Lorentz-transformed frame where y0 > x0, we have

TΨ(x)Ψ(y) = Ψ(y)Ψ(x) = −Ψ(x)Ψ(y) (7.72)

We conclude that we must modify the definition of the T ordering of

Fermi-fields to make it Lorentz invariant.


The solution is straight-forward. We define the T -product to in-

clude a factor of (−1)n where n is the number of exchanges required

to time order the Fermi fields.

For two fields

TΨ(x)Ψ(y) =

Ψ(x)Ψ(y) for x0 > y0

−Ψ(y)Ψ(x) for y0 > x0(7.73)

in particular we have

TΨ1(x)Ψ2(y) = −TΨ2(y)Ψ1(x)

The normal-ordering product is defined as before and we define

with

Ψ = Ψ(+) + Ψ(−)

where Ψ(+) multiplies an annihilation operator, and Ψ(−) multiplies

a creation operator. Remember the strange historicconvention stemming from the

positive/negative energy in theexponent.

: Ψ1Ψ2 : = : Ψ(+)1 Ψ

(+)2 + Ψ

(+)1 Ψ

(−)2 + Ψ

(−)1 Ψ

(+)2 + Ψ

(−)1 Ψ

(−)2 :

= Ψ(+)1 Ψ

(+)2 −Ψ

(−)2 Ψ

(+)1 + Ψ

(−)1 Ψ

(+)2 + Ψ

(−)1 Ψ

(−)2

where the second picked up a (−1) because we had to interchange to

Fermi fields. Just as for the T product, we therefore have Recall that for scalar fields, theirorder in T and Nproducts was

unimportant.: Ψ1Ψ2 : = − : Ψ2Ψ1 :

With these modifications, we can use Dyson’s formula and Wick’s

theorem as before. We just need to be careful with contractions in

Wick’s theorem, e.g.

: Ψ1Ψ2Ψ3Ψ4 : = − : Ψ1Ψ3Ψ2Ψ4 : = −Ψ1Ψ3 : Ψ2Ψ4 :

7.7 The Fermion Propagator

We will now move to the Heisenberg picture and define spinor

operators at every point in spacetime Ψ(x, t) such that they evolve

according to

∂Ψ

∂t= i[H,Ψ]

which is the usual operator time evolution. As before, we can solve

this by expanding (replacing ix · p→ −ix · p in the exponent)

Ψ(x) =

2∑

s=1

∫d3p√2Ep

(bsp u

s(p)e−ip·x + cs†p vs(p)eip·x

)(7.74)

Ψ†(x) =

2∑

s=1

∫d3p√2Ep

(bs†p u

s†(p)eip·x + csp vs†(p)e−ip·x

)(7.75)


We define the fermionic propagator to be

iSαβ ≡ Ψα,Ψβ (7.76)

Inserting the expansion Eq. (7.74) we obtain We will in the following often dropthe indices, but you should rememberthat S(x − y) = Ψ(x),Ψ(y) is a

4× 4 matrix.iSαβ(x− y) =

∫d3pd3q

(2π)6

1√4EpEq

[bsp, br†q usα(p)urβ(q)e−i(p·x−q·y)

+cs†p , crq vsα(p)vrβ(q)e+i(p·x−q·y)]

We use Eq. (7.55) and the anticommutation relations for b and c of

Eq. (7.68) to obtain

iSαβ(x− y) =

∫d3p

(2π)3

1

2Ep

[usα(p)usβ(p)e−ip·(x−y) + vsα(p)vsβ(p)e+ip·(x−y)

]

=

∫d3p

(2π)3

1

2Ep

[(/p+m)αβe

−ip·(x−y) + (/p−m)αβe+ip·(x−y)

]

We can therefore write

iS(x− y) = (i/∂x +m)(D(x− y)−D(y − x))

in terms of the propagator for the real scalar field D(x−y). If we stay Recall, we can write

D(x− y) =

∫d3p

(2π)31

2Epe−ip·(x−y)

with pµ = (Ep,p) or p2 = m2.

away from singularities, the propagator satisfies

(i/∂x −m)S(x− y) = 0

which directly follows because

(i/∂x −m)S(x− y) = (i/∂x −m)(i/∂x +m)(D(x− y)−D(y − x))

= −(x +m2)(D(x− y)−D(y − x)) = 0

using the properties of the scalar propagator for p2 = m2 away from

the singularities.

7.7.1 The Feynman propagator for Dirac fields

Similarly, we obtain first for the vacuum expectation value

〈0|Ψα(x)Ψβ(y)|0〉 =

∫d3p

(2π)3

1

2Ep(/p+m)αβe

−ip·(x−y)

= (i/∂x +m)D(x− y)

and

〈0|Ψβ(y)Ψα(x)|0〉 =

∫d3p

(2π)3

1

2Ep(/p−m)αβe

+ip·(x−y)

= (−i/∂x −m)D(y − x)

from which we define the Feynman propagator Note, it is also a 4× 4 matrix.

SF (x− y) = Ψ(x)Ψ(y)


or

SF (x− y) = 〈0|T Ψ(x)Ψ(y)|0〉 ≡

〈0|Ψ(x)Ψ(y)|0〉 x0 > y0

〈0| −Ψ(y)Ψ(x)|0〉 y0 > x0

Note, the minus sign! As we have argued around Eq. (7.72) it is

necessary for Lorentz-invariance. Let us relate the spinor Feynman

propagator to the scalar Feynman propagator

Ψ(x)Ψ(y) = θ(x0 − y0)(i/∂x +m)D(x− y) + θ(y0 − x0)(i/∂x +m)D(y − x)

= (i/∂x +m) [ θ(x0 − y0)D(x− y) + θ(y0 − x0)D(x− y) ]

= (i/∂x +m)φ(x)φ(y)

where the Feynman propagator for scalar fields is If you are worried about the time-

derivative of the θ function, you cansee that

∂tθ(x0 − y0)D(x− y) = δ(x0 − y0)D(x− y) = 0

using the properties of the scalar

propagator.

φ(x)φ(y) = ∆F (x− y) =

∫d4p

(2π)4e−ip·(x−y) i

p2 −m2 + iε

The integral representation of the Feynman propagator for Dirac

spinors is thereforeWe can write Eq. (7.77) also as

SF (x− y) = i

∫d4p

(2π)4e−ip·(x−y)

/p−m+ iε

which is a commonly used short

hand. We see that like for the scalarfield, the Feynman propagator is justthe inverse of the Dirac equation

(with suitable ε pole structure).

SF (x− y) = i

∫d4p

(2π)4e−ip·(x−y) /p+m

p2 −m2 + iε(7.77)

which satisfies

(i/∂x −m)SF (x− y) = iδ(4)(x− y)

thanks to −(x + m2)φ(x)φ(y) = iδ(x − y) derived in Eq. (5.79).

We conclude that SF (x − y) is a Green’s function of the Dirac

operator.

7.7.2 Nucleon-Nucleon scattering with spin

Let us study ΨΨ→ ΨΨ scattering again. We have already discussed

this process in Sec. 6.4.1, except that the nucleons now have spin.

The initial state is

|i〉 = |p1, s;p2, r〉 =√

2Ep1

√2Ep2

bs†p1br†p2|0〉 (7.78)

and the final state is

|f〉 = |q1, s′; q2, r

′〉 =√

2Eq1√

2Eq2 bs′†q1 b

r′†q2 |0〉 (7.79)

We need to be careful with signs, since the b† now anti-commute.

For the adjoint we therefore have

〈f | =√

4Eq1Eq2〈0|br′

q2bs′

q1 (7.80)

We are interested in 〈f |S − 1|i〉 at order λ2,

(−iλ)2

2

∫d4x1d

4x2 T[

Ψ(x1)Ψ(x1)φ(x1)Ψ(x2)Ψ(x2)φ(x2)]

(7.81)


where as usual, all field operators are in the interaction picture.

As in the pure-scalar calculation, the contribution to Nucleon-

Nucleon scattering is due to the contraction

: Ψ(x1)Ψ(x1)Ψ(x2)Ψ(x2) : φ(x1)φ(x2)

We now have to be careful about how the spinor indices are con-

tracted. Let us start with the Ψ and deal with Ψ later. Since we will

only find no non-vanishing contributions for the c† at this order, we

can focus on

: Ψ(x1)Ψ(x1)Ψ(x2)Ψ(x2) : bs†p1br†p2|0〉

= −∫d3k1d

3k2

(2π)6(Ψ(x1) · um(k1))(Ψ(x2) · un(k2))

e−ik1·x1−ik2·x2

√4Ek1Ek2

bmk1bnk2

bs†p1br†p2|0〉

where the − sign is from commuting Ψ past the Ψ. Let us now

anticommute the b’s past the b†’s 18 to obtain 18 Using bi, b†j = δij and bj |0〉 = 0,we find

b1b2b†3b†4|0〉 =(−b1b†3b2b

†4 + b1δ23b

†4)|0〉

=(−b1b†3δ24 + δ23δ14)|0〉=(−δ13δ24 + δ23δ14)|0〉.

= − 1√4Ep1

Ep2

[(Ψ(x1) · ur(p2))(Ψ(x2) · us(p1))e−ip2·x1−ip1·x2

−(Ψ(x1) · us(p1))(Ψ(x2) · ur(p2))e−ip1·x1−ip2·x2]|0〉

Note again the relative minus sign between the two terms. What

happens when we contract this with 〈f |? Let’s see the first term

〈0|br′q2bs′

q1(Ψ(x1) · ur(p2))(Ψ(x2) · us(p1)) |0〉 =

e+iq2·x1+iq1·x2

2√Eq1Eq2

(us′(q1) · ur(p2))(ur

′(q2) · us(p1))

−e+iq1·x1+iq2·x2

2√Eq1Eq2

(ur′(q2) · ur(p2))(us

′(q1) · us(p1))

the second term contributes the same and cancels the 1/2 in front

of Eq. (7.81). The factors of 1/√E cancel the relativistic state

normalization. Combining everything, we obtain

−(−iλ)2

∫d4x1d

4x2d4k

(2π)4

iek·(x1−x2)

k2 − µ2 + iε

[(us′(q1) · ur(p2))(ur

′(q2) · us(p1))e+i(q2−p1)·x1+i(q1−p2·x2

−(ur′(q2) · ur(p2))(us

′(q1) · us(p1))e+i(q1−p1)·x1+i(q2−p2·x2

]

where have used the φ-propagator. Integrating over x1 and x2, we

obtain

−∫d4k i(−iλ)2(2π)4

k2 − µ2 + iε

[(us′(q1) · ur(p2))(ur

′(q2) · us(p1))δ(4)(p1 − q1 + k)δ(4)(p2 − q2 + k)

−(ur′(q2) · ur(p2))(us

′(q1) · us(p1))δ(4)(p1 − q2 + k)δ(4)(p2 − q1 + k)

]

Almost done. We use the usual normalization

〈f |S − 1|i〉 = iA (2π)4δ(4)(p1 + p2 − q1 − q2)

we find

iA = (−iλ2)

[(ur′(q2) · ur(p2))(us

′(q1) · us(p1))

(p2 − q1)2 − µ2 + iε− (us

′(q1) · ur(p2))(ur

′(q2) · us(p1))

(p1 − q1)2 − µ2 + iε

]

which is the final result.


7.8 Feynman rules for Dirac Spinors

The calculation above is really cumbersome and you will hopefully

never have to redo it this way ever again. Luckily introducing Feyn-

man rules simplifies the calculation significantly and we will see that

we will be able to derive the final result in a couple of easy steps.

1. Propagators:

φ(x)φ(y) =i

p2 − µ2 + iε=

Ψ(x)Ψ(y) =i(/p+m)

p2 −m2 + iε=

2. Vertices:

−iλ =

3. External legs incoming

p

= us(p)

p

= vs(p)

4. External legs outgoing

p

= us(p)

p

= vs(p)

5. Impose momentum conservation at each vertex.

6. Integrate over each undetermined momentum k with∫

d4k(2π)4 .

7. Figure out the overall sign of the diagram.

8. A closed fermion loop always gives a factor of −1 and the trace of

a product of Dirac matrices.19 19 We will see an example of thisbelow.

We draw arrows on fermion lines to represent the direction of

particle-number flow. The momentum assigned to a fermion

propagator flows in the direction of this arrow. For external antipar-

ticles, however, the momentum flows opposite to the arrow. We show

this explicitly by drawing a second arrow next to the line.


7.8.1 Example: Nucleon scattering ΨΨ→ ΨΨ

The two diagrams are

p

q

p′

q′

+

p

q

p′

q′

(7.82)

iA = (−ig)2

[u(q′) · u(q)

i

(q − q′)2 − µ2u(p′) · u(p)

−u(p′) · u(q)i

(p′ − q)2 − µ2u(q′) · u(p)

](7.83)

Where we have suppressed the spinor spin labels. The minus sign

between the diagrams is a reflection of Fermi-Dirac statistics.

7.8.2 Example: Nucleon-Anti-Nucleon scattering ΨΨ→ ΨΨ

The two diagrams are

p

q

p′

q′

+ p

q

p′

q′

(7.84)

with the Feynman rules we have

iA = (−iλ)2

[− u(p′) · u(p) v(q) · v(q′)

(p− p′)2 − µ2+v(q) · u(p) u(p′) · v(q′)

(p+ q)2 − µ2

]

There is a relative minus sign between the two diagrams, which is a

bit subtle to figure out. You can convince yourself that it is correct,

by repeating the relevant steps using the explicit Wick technology

above.

7.8.3 Example: Meson Scattering

Finally, in φφ→ φφ scattering, the leading contribution appears first

at 1-loop. The amplitude for the diagram is

p

p′

q

q′


iA = −(−iλ)4

∫d4k

(2π)4Tr

[/k + m

k2 −m2

/k + /p′ + m

(k + p′)2 −m2

/k + /p′ − /q′ + m

(k + p′ − q′)2 −m2

/k − /p+ m

(k + p)2 −m2

]

where we have suppressed the +iε piece in the denominator. Note,

that the integral diverges logarithmically20 In the k → ∞ limit, the 20 Compare to the pure scalar case

which was convergent.integral behaves like

iA ∼∫d4k

k4∼ ln Λ

You will have to come to the QFT course next semester to learn how

to make sense of results like these. There is an overall minus sign Like for the cosmological constant in

Sec. ?? we will have to regularize andrenormalize the result in a systematic

way.

sitting in front the of the amplitude.

In complicated diagrams, we can often find the minus signs, by

noting that the product (ΨΨ) or any other pair of fermions is a

commuting object. So,

. . . (ΨΨ)x(ΨΨ)y(ΨΨ)z . . . = . . . (+1)(ΨΨ)x(ΨΨ)z(ΨΨ)y . . .

. . . SF (x− z)SF (z − y) . . .

In a closed loop of n fermion propagators we have

which involves the contraction

Ψα(x)Ψα(x)Ψβ(y)Ψβ(y) = −Ψβ(y)Ψα(x)Ψα(x)Ψβ(y)

= −Tr[SF(y − x)SF(x− y)]

This is a general result: a closed fermion loop results in a factor of

(−1) and a trace of a string of γ matrices.

7.8.4 Example: the Yukawa potential

We saw in Sec. ?? that the exchange of a real scalar between to

charged scalars gives rise to always attractive Yukawa potential

between two spin zero particles. Do we still find the same for the

scalar force between two spin 1/2 particles?

We will again take the non-relativistic limit p m which

implies for the spinors

us(p) =


)→ √

m

(ξs

ξs

)

and

vs(p) =

( √p · σ ξs

−√p · σ ξs

)→ √

m

(ξs

−ξs

)

If the two interacting particles are distinguishable only the first one

in Eq. (7.82) contributes. In this limit, the spinor contractions in the

ψψ → ψψ scattering are

us′(q′) · us(q) = 2mδss′


and the amplitude becomes

iA = −i(−iλ)2(2m)

[δs′sδr′r

(p− p′)2 + µ2

]

We see that the spin is conserved at each vertex in the non-relativistic

limit (due to the δ factors) and that the momentum dependence is

exactly the same as in the bosonic case Eq. (6.47), telling us that the

particles feel an attractive Yukawa potential,

U(r) = −λ2e−µr

4πr

If we repeat the calculation for ΨΨ → ΨΨ, two minus signs can-

cel and the Yukawa interaction once again leads to an attractive

potential.

8

Quantum Electro Dynamics

8.1 The Feynman Rules for Quantum Electrodynamics

We replace the scalar particle φ with a vector particle Aµ and re-

place the Yukawa interaction Hamiltonian with

Hint =

∫d3x eΨγµ ΨAµ

What are the Feynman rules? We can easily guess, even though the

theory behind them is quite involved.

1. Interaction vertex:

−ieγµ =

2. Propagators:

Aµ(x)Aν(y) =−iηµνq2 + iε

=

3. External photon lines:

Outgoing

p

= ε∗µ(p)

Incoming

p

= εµ(p)

We draw photons as wavy lines. The symbol εµ(p) stands for

the polarization vectors of the initial or final state photons. The

polarization vectors are always of the form

εµ =

(0

ε

), with p · ε = 0 (8.1)


If p is along the z-axis, the right- and left-handed polarization

vectors are

εµ =1√2

0

1

±i0

8.2 Maxwell’s equations

The Maxwell-Lagrangian in the absence of any sources is simply

L = −1

4FµνF

µν (8.2)

with the field strength

Fµν = ∂µAν − ∂νAµ (8.3)

and the equations of motions are

∂µ

(∂L

∂(∂µAν)

)= −∂µFµν = 0

Additionally, the field strength satisfies the Bianchi identities

∂λFµν + ∂µF

νλ + ∂νFλν = 0

The version of Maxwell’s equation you learn in pre-school is recov-

ered atopwithdelims

Aµ =

(φ

A

)

with the electric and magnetic fields E and B

E = −∇φ− ∂A

∂t

and

B = ∇×A

which in terms of the field strength is

Fµν =

0 Ex Ey Ez

−Ex 0 −Bz By

−Ey Bz 0 −Bx−Ez −By Bx 0

and the Bianchi identity results in two Maxwell equations which are

unchanged even in the presence of sources

∇ ·B = 0

∂B

∂t= −∇×E

The equations of motions give the two remaining Maxwell equations

∇ ·E = 0

∂E

∂t= ∇×B

Note, the interesting electro-magneticduality E → B and B → −E!


8.3 Gauge Symmetry

We have used Aµ to describe a massless vector field, which would

seem to imply that since the gauge field has 4 components, we

naively count 4 degrees of freedom. Yet we know has only two E.g. from counting states and

calculating the thermodynamics of a

blackbody in thermodynamics.degrees of freedom, which are the polarization states we’ve used

in the Feynman rules. How can this be resolved? There are two

mechanisms at work which ensure this.

• The field A0 is not dynamical: It has no kinetic term A0 in the

Lagrangian. If we fix Ai and Ai at initial some time t0, then A0 is

fully determined by the equation of motion ∇ ·E = 0 which we can

expand out to obtain

∇2A0 +∇ · ∂A∂t

= 0

which has the solution

A0(x) =

∫d3x′

∇x′ · ∂A(x′)∂t

4π|x− x′| (8.4)

which you can convince yourself is correct using the arguments

around Eq. (4.95) and the Green’s function method. So A0 is

clearly dependent and we do not get to specify it fixing the initial

conditions. This reduces our the degrees of freedom from 4 to 3.

gauge fixing

gauge orbits

field space

Figure 8.1: Schematic picture of

the gauge redundancy. The gauge

orbits are families of gauge-equivalentconfigurations. The physically

independent states are in a gauge

fixed slice that intersects each orbitonce.

• The Maxwell Lagrangian has a very large redundancy,

Aµ(x)→ Aµ(x) + ∂µα(x) (8.5)

for any function α(x). We will require that α(x) vanishes fast

enough at x → ∞. This is a gauge symmetry. The field

strength is invariant under it

Fµν → ∂µ(Aν + ∂να)− ∂ν(Aµ + ∂µα) = Fµν

We seem to have a theory with an infinite number of symmetries,

each for each local choice of function α(x). All our other internal

symmetries where global and acted the same at every point in

spacetime, e.g.

ψ → eiβψ

with β = const.. These symmetries lead to selection rules or

conservation laws, thanks to Noether’s theorem. Do we now have

infinitely many conserved currents?

No! A symmetry takes a physical state to different physical

state with the same properties, the gauge symmetry however is a

redundancy in our description. The two states connected by a

gauge symmetry have to be identified, they are the same physical

state. You can see this also in the Maxwell’s equation which are

not sufficient to specify the time evolution of Aµ

(ηµν(∂λ∂λ)− ∂µ∂ν)Aν = 0


Now, the differential operator (ηµν(∂λ∂λ) − ∂µ∂ν) is not invert-

ible, since it has a zero eigenvector for functions of the form ∂µλ.

Fixing initial data, we can not uniquely determine Aµ since there

no way to distinguish between

Aµ, and Aµ + ∂µλ

If we are willing to identify Aµ and Aµ + ∂µλ as the same

physical state, then this is not a problem anymore.

How about formulating the theory solely in E and B? Lorentz-

invariance would not be manifest, but we might be able to deal

with this. However, this is not advisable given the relevance of Aµ

for quantum mechanical effects like the Aharonov-Bohm effect.

Further, we would like to describe charged gauge fields, which also

requires generalizations of Aµ.

Figure 8.2: The the Aharonov-

Bohm effect is a quantum mechan-ical phenomenon in which an elec-

trically charged particle is affected

by an electromagnetic potential Aµ,despite being confined to a region in

which both the magnetic field B and

electric field E are zero. The underly-ing mechanism is the coupling of the

electromagnetic potential with thecomplex phase of a charged particle’s

wave function,

ϕ =q

~

∫P

A · dx

and the Aharonov-Bohm effect is ac-cordingly illustrated by interferenceexperiments. https://en.wikipedia.

org/wiki/Aharonov-Bohm_effect.

We discover an enlarged phase space, foliated by gauge orbits.

To remove the redundancy, we need to pick one representative from

each equivalent gauge orbit. We call these different configuration

of a physical state, different gauges. We can pick gauges as we

please, but as for coordinate systems, there are better and less suited

choices depending on the problem.

8.3.1 Gauges

We start with the Lorentz gauge1

1 Other popular choices are

A0 = 0 (temporal gauge)

A3 = 0 (axial gauge)

∂µAµ = 0 (8.6)

This does not yet pick a unique representation from the gauge orbit,

since we are free to make further gauge transformations which are

functions that satisfy

∂µ∂µλ(x) = 0

which certainly has non-trivial solutions. The advantage of the

Lorentz gauge condition is clear from its name: it is manifestly

Lorentz invariant.

Another useful gauge, is the Coulomb gauge

∇ ·A = 0 (8.7)

there is still options to choose harmonic functions which satisfy

∇2λ = 0. Since Eq. (8.4) fixes A0, we find that now2 2 This will not hold in the presence

of sources, as you know from electro-statics, since in this case, e.g. a

charge induces a potential for A0 = φA0 = 0

Coulomb gauge is not manifestly Lorentz invariant, but it clearly

shows the degrees of freedom of the physical system: The 3 degrees

of freedom satisfy a single constraint ∇ ·A = 0, leaving us with just 2

independent degrees of freedom.

https://en.wikipedia.org/wiki/Aharonov-Bohm_effect

https://en.wikipedia.org/wiki/Aharonov-Bohm_effect


8.4 Quantizing the Electro-magnetic field

We will now try to quantize the electric field in Coulomb and in

Lorentz gauge. A common subtlety to both methods is due to the

absence of a canonical momentum for A0:

π0 =∂L

∂(∂tA0)= 0 (8.8)

πi =∂L

∂(∂tAi)= −F 0i = Ei (8.9)

We again see that A0 is not a dynamical field. The Hamiltonian is

H =

∫d3x(πiAi − L)

=

∫d3x(

1

2E ·E +

1

2B ·B −A0(∇ ·E))

We see that A0 acts as a Lagrange multiplier, which has the equation

of motion

∇ ·E = 0

which is just enforcing Gauss’ law. What do the different gauge

fixing conditions mean for the system?

8.4.1 Coulomb gauge

In Coulomb gauge, the equation of motion for A is

∂µ∂νA = 0

which we solve as usual by a Fourier-ansatz

A(x) =

∫d3p

(2π)3ξ(p)eip·x

with p2 = 0 or p20 = |p|2. The Coulomb gauge condition ∇ ·A = 0

requires

ξ(p) · p = 0

which means that ξ(p) is orthogonal to the direction of motion of

the gauge field p. We can choose it to be a linear combination of two

ortho-normal vectors εr with r = 1, 2 which satisfy

εr(p) · p = 0

and

εr(p) · εs(p) = δrs

These are the two polarization states of the photon. Now we

attempt to quantize and we turn the classical Poisson brackets into

commutator for operator valued fields. Naively (and wrongly, we try

for the equal-time commutator

[Ai(x), Ej(y)] = iδji δ(3)(x− y) (wrong)


and

[Ai(x), Aj(y)] = [Ei(x), Ej(y)] = 0

This is not correct! Why? It does not work with the gauge and

Gauss constraints. We need to have imposed on the operators

∇ ·A = ∇ ·E = 0

but if we evaluate the operator relations above, we find

[∇x ·A(x),∇y ·E(y)] = i∇2δji δ(3)(x− y) 6= 0

You can see this problem already in the classical Poisson bracket

structure.3 The correct Poisson bracket structure implies different 3 See e.g. P. Ramond - QFT, orDirac ”Lectures on Quantum me-

chanics”.commutation relations in Coulomb gauge

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

(δji −

∂i∂j∇2

)δ(3)(x− y) (8.10)

Let us check if this is consistent with the constraints. We write the δ

function in momentum space

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

∫d3p

(2π)3

(δji −

pipjp2

)eip·(x−y)

Applying e.g. the Coulomb gauge fixing

[∂xi Ai(x), Ej(y)] = i

∫d3p

(2π)3ipi

(δji −

pipjp2

)eip·(x−y)

= i

∫d3p

(2π)3i

(pj −

p2pjp2

)eip·(x−y) = 0

We expand in modes again as usual

A(x) =

2∑

r=1

∫d3p

(2π)3

1√2|p|

εr(p)[arpe

ip·x + ar †p e−ip·x]

E(x) =

2∑

r=1

∫d3p

(2π)3

(−i)√|p|√

2εr(p)

[arpe

ip·x − ar †p e−ip·x]

with the before define polarization vectors.

εr(p) · p = 0 and εr(p) · εs(p) = δrs

You should convince yourself that the commutation relations imply

for the ladder operators

[arp, as †q ] = (2π)3 δrs δ(3)(p− q)

and

[arp, asq] = [ar †p , a

s †q ] = 0


where you will have to use the outer product (or completeness

relation) for the polarization vectors

2∑

i=1

εir(p)εjr(p) = δij − pipj

p2

We can confirm this equation by acting on both sides with a com-

plete basis (ε1(p), ε2(p),p).

We substitute the expansion into the Hamiltonian and obtain

after normal ordering

H =

∫d3p

(2π)3|p|

2∑

r=1

ar †p arp

Quantizing in the Coulomb gauge gave us direct insight into

the physical degrees of freedom. Lorentz invariance is however not

manifest anymore. We can see this looking at the propagator in the

Heisenberg picture

Dtrij (x− y) ≡ 〈0|TAi(x)Aj(y)|0〉

=

∫d4p

(2π)4

i

p2 + iε

(δji −

pipjp2

)e−ix·p

Where explicitly denote with tr, the transverse part of the photon.

Let us therefore explore the manifestly invariant Lorentz gauge.

8.4.2 Lorentz Gauge

The Lorentz gauge condition is

∂µAµ = 0

which implies for the equations of motion

∂µ∂µAν = 0

We will depart from the procedure enforcing Coloumb gauge.4 and 4 This can be motivated much more

cleanly using the path integral

formalism and a trick by Fadeev andPopov.

will modify the Lagrangian such that the equations of motion are

directly in Lorentz gauge. With

L = −1

4FµνF

µν − 1

2(∂µA

µ)2

we obtain

∂µFµν + ∂ν(∂µA

µ) = ∂µ∂µAν = 0

We can be more general and choose

L = −1

4FµνF

µν − 1

2α(∂µA

µ)2

with arbitrary and constant α. We confusingly call α = 1 the

Feynman gauge and α = 0 the Landau gauge.

We will now firstly quantize the theory and only secondly enforce

the gauge constraint. Thankfully, both conjugate momenta are


non-vanishing

π0 =∂L

∂(∂tA0)= −∂µAµ (8.11)

πi =∂L

∂(∂tAi)= ∂iA0 − ∂0Ai (8.12)

We impose the equal-time commutation relations in Lorentz-gauge

[Aµ(x), πν(y)] = i ηµνδ(3)(x− y) (8.13)

and

[Aµ(x), Aν(y)] = [πµ(x), πν(y)] = 0

Expanding again

Aµ(x) =

3∑

λ=0

∫d3p

(2π)3

1√2|p|

ελµ(p)[aλpe

ip·x + aλ †p e−ip·x]

πµ(x) =

3∑

λ=0

∫d3p

(2π)3

+i√|p|√

2ελµ(p)

[aλpe

ip·x − aλ †p e−ip·x]

Note that we here have (+i) in the momentum, rather than the

usual (−i). The reason is that the conjugate momentum is

πµ = −Aµ + . . .

We now have to deal with four polarization vectors! This is much

worse than the two physical polarizations in Coloumb gauge. We

choose ε0 time-like and εi to be space-like. We choose the normaliza-

tion

ελ · ελ′ = ηλλ′

which implies

(εµ)λ(εν)λ′ηλλ′ = ηµν

We choose two of the space-like polarizations to be transverse to the

momentum pµ = (|p|,p)

ε1 · p = ε2 · p = 0 (8.14)

The third vector is a longitudinal polarization. If pµ(z) ∝ (1, 0, 0, 1),

we would have

ε0 =

1

0

0

0

, ε1 =

0

1

0

0

, ε2 =

0

0

1

0

, ε3 =

0

0

0

1

,

We define for a general four-momentum pµ ∝ (|p|,p),

ε0(p) =

1

0

0

0

, ε3(p) =

1

|p|

0

p1

p2

p3

, εr(p) =

0

εr1εr1εr1

,

with εr(p) · p = εr(p) · p = 0, r = 1, 2


For later use, we note for the time-like and longitudinal polarizations

that

ε0(p) · p = |p| ε3(p) · p = −|p| (8.15)

We derive the ladder operator commutation relations using the field

expansion to obtain

[aλp, aλ′ †q ] = −ηλλ′(2π)3 δ(3)(p− q)

with all other commutators vanishing as usual. The minus sign

is problematic! For space-like λ, λ′ = 1, 2, 3 we are in well-know

territory

[aλp, aλ′ †q ] = (2π)3 δλλ

′δ(3)(p− q), λ, λ′ = 1, 2, 3

but for time-like ladder operators, we have

[a0p, a

0 †q ] = −(2π)3 δ(3)(p− q)

This means trouble. Let us see what it implies e.g. for one-particle

states. With the vacuum

aλp|0〉 = 0

we can define (yet to be normalized) one-particle states as usual

|p, λ〉 = aλ †q |0〉

For space-like λ = 1, 2, 3 we are in the green, but for the time-like

polarization λ = 0, the state |p, λ = 0〉 has negative norm

〈p, 0|q, 0〉 = 〈0|a0pa

0 †q |0〉 = −(2π)3δ(3)(p− q)

This strange. Negative norm would imply negative probabilities! We

put ourselves in danger already from the start as we can see from the

kinetic term for A0 which has the wrong sign

L =1

2A2 − 1

2A2

0 + . . .

How can we makes sense of this?

We have yet to impose the gauge constraint ∂µAµ = 0 on the theory.

It will be responsible to remove the time-like negative norm states

and remove all but two polarizations. How can we implement the

constraint in a quantum theory?

• Let us start with the strongest possible condition. We impose

∂µAµ = 0 as a operator equation. This does not work, because

it is not compatible with the commutation relations Eq. (8.13),

since the conjugate momentum Eq. (8.11)

π0 =∂L

∂(∂tA0)= −∂µAµ

would vanish universally.


• A weaker option would be to impose it on a subset of the Hilbert

space instead of on the operators. We could imagine splitting

the Hilbert space into ”good” states |ψ〉 and a ”bad” states which

somehow are not part of physical observables. How can we define

”good” states. We could impose

∂µAµ|ψ〉 = 0 (8.16)

on all the ”good” states. However this does not work either. It is

still too strong, since we can e.g. decompose Aµ = A+µ +A−µ with

A+µ (x) =

3∑

λ=0

∫d3p

(2π)3

1√2|p|

ελµ(p) aλpe−ip·x

A−µ (x) =

3∑

λ=0

∫d3p

(2π)3

1√2|p|

ελµ(p) aλ †p e+ip·x

The vacuum satisfies

A+µ |0〉 = 0

by definition. But we get

∂µA−µ |0〉 6= 0

which shows that the vacuum as a physical state isn’t even one of

the good states.

• The solution: We want the vacuum as one of the good physical

states. We therefore ask physical states |ψ〉 to only satisfy the

Gupta-Bleuler condition5 5 The problem with the Gupta-Bleuercondition is that is not suited fornon-abelian gauge theories due to

the non-linearity of the free fieldequations (they contain terms∼ A3, A4). The modern approach

replacing this QED-only ’hack’ iscalled BRST symmetry (Becchi,Rouet, Stora and Tyutin) and will

be developed in the QFT lectures.One identifies a nilpotent symmetry

generator which commutes with the

Hamiltonian H which allows to splitthe eigenstates of H into unphysicaland physical sub-sectors which havezero inner product with each other.

∂µA+µ |ψ〉 = 0 (8.17)

and the adjoint equation

〈ψ|∂µA−µ = 0

With this we make sure that

〈ψ′|∂µA−µ + ∂µA+µ |ψ〉 = 〈ψ′|∂µAµ|ψ〉 = 0

holds. Since the constraint is linear, the physical states span the

physical Hilbert space Hphys. How does the physical Hilbert space

look like?

We consider a basis for the Fock space. We decompose any ele-

ment of this basis into

|ψ〉 = |ϕT 〉|ϕL−0〉

where |ϕT 〉 contains the transverse photons (created by a1 †p and a2 †

p )

and |ϕL−0〉 contains time-like and longitudinal photons (created by

a0 †p and a3 †

p ).


The Gupta-Bleuer condition Eq. (8.17) therefore reads with

the polarization condition for the transverse states Eq. (8.14) and

Eq. (8.15) as

(a0p − a3

p) |ϕL−0〉 = 0 (8.18)

What does this mean? Condition Eq. (8.18) requires all physical

states to contain pairs of timelike and longitudinal photons: for each

timelike photon with p, it must also contain a longitudinal photon

with the same p. We can write |ϕL−0〉 as a sum over states |ϕ(n)L−0〉

containing n pairs of timelike and longitudinal photons

|ϕL−0〉 =

∞∑

n=0

cn |ϕ(n)L−0〉 (8.19)

where |ϕ(0)L−0〉 = |0〉 is just the vacuum.

One can show that Eq. (8.18) indeed decouples the negative norm

states and the all the remaining timelike and longitudinal photons

have zero norm

〈ϕ(n)L−0|ϕ

(n)L−0〉 = δn,0δm,0

We have therefore defined an inner product on the physical Hilbert

space which is positive semi-definite. This is clearly a step in the

right direction. We will now have to find a way to deal with the

zero-norm states.

We understand the zero-norm states if we treat them as gauge-

equivalent to the vacuum. If two states that differ only by pairs

of timelike and longitudinal photons, |ϕ(n)L−0〉 with n ≥ 1, they

describe the same physical state.

We need to check then that they give the same result for all

physical observables. Let us start with the Hamiltonian, which you

can easily compute

H =

∫d3p

(2π)3|p|[

3∑

i=1

ai †p aip − a0 †

p a0p

]

The Gupta-Bleuer condition Eq. (8.18) guarantees that

〈ψ|a3 †p a

3p|ψ〉 = 〈ψ|a0 †

p a0p|ψ〉

and the contributions of of timelike and longitudinal photons cancel

each other in the Hamiltonian. Further, the Hamiltonian evaluated

on physical states satisfying the Gupta-Bleuer condition is posi-

tive definite and contains only the contributions from transverse

photons, as it should be.

One can show that expectation values of gauge-invariant operators

evaluted on physical states are independent of the coefficients cn of

Eq. (8.19).


8.4.3 The propagator

The propagator for the photon in Lorentz gauge is given by

〈0|TAµ(x)Aν(x)|0〉 = Aµ(x)Aν(y) =

∫d4p

(2π)4

−iηµνq2 + iε

e−ip·(x−y)

Alternatively, you can show that for general gauges with arbitrary α

in Eq. (??), we would obtain in momentum space Keeping the α dependence and

testing that it cancels out in the finalresult is a useful check.−i

q2 + iε

(ηµν + (α− 1)

pµpνp2

)

8.5 Coupling gauge fields to matter

We now want to couple the gauge field Aµ to other fields, like scalars

or fermions (or more poetically, coupling light to matter). Let us see

how we can do this. We know that the simplest coupling has to have

the form

L = −1

4FµνF

µν −AµRµ (8.20)

What are the conditions on Rµ? We know that we need the gauge

redundancy to find the correct number of degrees of freedom. The

equations of motions are

∂µFµν = Rν

Let us apply ∂ν on both sides

∂ν∂µFµν = 0 = ∂νR

ν

In other words Rν must be a conserved current! We write Rν = jν

and recongnize this as the usual conservation equation

∂µjµ = 0

We have already encountered a large number of these conserved

currents jµ. Which are the Noether currents related to the gauge

symmetry for the photon field?

8.5.1 Coupling photons to Dirac fermions

Let us try with the U(1) global, internal symmetry of the Dirac

Lagrangian

Ψ→ eiαΨ

which with Eq. (7.60) gives rise to a current

jµ = ΨγµΨ (8.21)

Now let us couple this to the gauge field L ⊃ Aµjµ to obtain to-

gether with the kinetic term of the Dirac spinor

L = −1

4FµνF

µν + iΨ/∂Ψ− eΨγµΨAµ (8.22)


where I have introduced a yet to be specified constant e. We can

rewrite this as

L = −1

4FµνF

µν + iΨ /DΨ (8.23)

with the covariant derivative The ”covariant” is with respect to

it’s behaviour under gauge transfor-

mations as we will see below.Dµ = ∂µ + ieAµ (8.24)

Under a gauge transformation Aµ(x)→ Aµ(x) + ∂µλ(x) we find that

the Lagrangian transforms as

L → L− eΨ′γµ(∂µλ(x))Ψ′

we see that we need to cancel the transformation of Aµ with a

transformation of Ψ → Ψ′. Let us try a local generalization of the

global U(1) transformation

Ψ→ e−ie λ(x) Ψ

The covariant derivative transforms as

DµΨ = [∂µ + ieAµ]Ψ→ [∂µ + ieA′µ]Ψ′

= [∂µ + ie(Aµ + ∂µλ(x))]e−ie λ(x)Ψ

= e−ie λ(x)[∂µ − ie∂µλ(x) + ie(Aµ + ∂µλ(x))]Ψ

= e−ie λ(x)DµΨ

which shows that the covariant derivative has the convenient prop-

erty that it only picks up a factor exp(−ieλ(x)), with the derivative

canceling the transformation of the gauge field. The whole La-

grangian Eq. (8.26) is then gauge invariant, since

Ψ→ Ψe+ie λ(x)

Let us summarize now the gauge symmetry of QED

Aµ(x)→ Aµ(x) + ∂µλ(x), Ψ→ e−ie λ(x) Ψ (8.25)

which leaves the Lagrangian and in particular iΨ /DΨ invariant.

8.5.2 Coupling photons to charged scalars

We will write ϕ for the charged scalar field and not ψ as previously

in order no to confuse it with the spinors.

How would we couple the charged scalar to a gauge boson? The

linear coupling to the U(1) current would not work6 but we can use 6 Try it!

the covariant derivative Dµ

Dµ = ∂µ + ieAµ

and replace the ∂µ everywhere. This gives

L = −1

4FµνF

µν + (Dµϕ)∗Dµϕ−m2ϕ∗ϕ (8.26)


a simultaneous gauge transformation

Aµ(x)→ Aµ(x) + ∂µλ(x), ϕ(x)→ e−ie λ(x) ϕ(x) (8.27)

This tricks works for any theory. Replacing the derivatives by co-

variant derivatives in suitable representations works in almost any

theory. This procedure is called minimal coupling.

8.6 The electric charge

Back to QED and the gauge-spinor coupling: he factor e has the

interpretation of the coupling strength between the charged field and

the photon. We know from electro-magnetism that j0 is the charge

density, as can be seen from the equations of motion, ∂µFµν = jν or

XXX. The total charge Q is then given by

Q = e

∫d3xΨγ0Ψ

Inserting the Heisenberg expansion of the free spinor, we get a

similar expression as for the charged scalar field Eq. (5.60)

Q = e

∫d3q

(2π)3

2∑

r=1

(bs†q b

sq − cs†q csq

)

We find again that this is the number of particles minus the num-

ber of antiparticles. The value of the coupling constant in QED is

usually written in terms of the fine-structure constant There is a subtlety here, in that

these couplings are scale dependent

α = α(µ)

Once you include loop corrections,the value of the fine-structure con-stant grows logarithmically as

the energy scale is increased. Thevalue here is associated with theenergy scale of the electron mass, soα(µ = me) ≈ 1/137, however e.g. atthe mass of the Z boson

α(µ = MZ) ≈ 1/129

The scale dependence is encoded in

the renormalization group equationwhich you will discuss in the QFTlecture

∂e(µ)

∂ lnµ≡ β(e) =

e(µ)3

12π2+ . . .

where just show the leading order

dependence. The running of the

coupling is a loop effect as you seefrom its e(µ)3 dependence.

α =e2

4π~c≈ 1

137(8.28)

which in natural units means for e =√

4πα ≈ 0.3.

8.7 Discrete symmetries of QED

In addition to the Lorentz transformations which are continuously

connected to the 1, there are two other space-time operations which

can be symmetries of the Lagrangian: parity P and time-reversal

T . Additionally, there can be a particle-antiparticle symmetry called

charge conjugation C. We have already encountered parity as

P : (t,x)→ (t,−x)

which reverses handedness. Time reversal acts as

T : (t,x)→ (−t,x)

both leave the Minkowski interval x2 = t2 − x2 invariant. Formally,

we can say that the full Lorentz group breaks into four disconnected

subsets. We call the continuous Lorentztransformation the proper,

orthochronous Lorentz group L↑+

L↑+ L↑−

L↓+ L↓−

P

T T

P


Although a relativistic field theory must be invariant under L↑+, it

is not necessarily invariant under C, P, or T . The gravitational,

electromagnetic and the strong force respect all three, but weak

interactions violate C and P separately, and certain suppressed pro-

cesses violate CP and T violation. The combination CPT is observed

in all processes so far in nature. There is also the famous CPT the-

orem which states that any Lorentz invariant local quantum field

theory with a Hermitian Hamiltonian must have CPT symmetry.7 7 It is very difficult to write a con-

sistent quantum field theory whichwould lead to observable CPT viola-tion.

In all the quantum field theories that you will encounter in this lec-

ture (and in QFT1/2) you can therefore assume that CPT is good

symmetry and as a consequence that CP = T , which means that

the measured violation of CP symmetry implies that nature is not

time-reversal symmetric.

8.7.1 Parity

We have already discussed parity extensively. Let us just add the

result for the gauge field and any four-vector

P : Aµ(x) = (A0,A)→ (A0,−A)

the Dirac spinor in the Weyl basis exchanges ψL and ψR, whose

representation is

P : Ψ(x, t)→ γ0Ψ(−x, t)

8.7.2 Charge conjugation

We now know what the meaning of charge is in an interaction and

hence we can talk about charge conjugation. We will try to flip the

charge e in

iγµ(∂µ + ieAµ)Ψ = 0

Start with the complex conjugation of the equation of motion

−iγµ∗(∂µ − ieAµ)Ψ = 0

If we complex conjucate the Clifford algebra γµ, γν = 2gµν 1,

we see that (−γµ∗) also satisfies it. Therefore (−γµ∗) must be the

gamma matrices expressed in a different basis, which means there

exists a matrix Cγ0 such that Including γ0 in the definition is

convention.

−γµ∗ = (Cγ0)−1γµCγ0 (8.29)

We plug this into the conjugate equation of motion to find

iγµ(∂µ − ieAµ)ΨC = 0

with

ΨC ≡ Cγ0Ψ∗


Let us find the specific form of the charge conjugation matrix C.

We can write Eq. (8.29) with (γ0)2 = 1 and so (γ0)−1 = γ0 as

Cγ0γµ∗γ0C−1 = −γµ

We know that the hermitian conjugate is (γµ)† = γ0γµγ0 and

taking the complex conjugate gives (γµ)T = γ0(γµ∗)γ0 if γ0 is real.

Therefore we have

(γµ)T = −C−1γµC

which explains why C is defined with a γ0 attached. In both bases

(Weyl and Dirac from the Ex-sheet), γ2 is the only imaginary matrix,

so γµ∗ = γµ for µ 6= 2 and γ2∗ = −γ2. Then the defining equation Recall, that

γ2Weyl =

(0 σ2

−σ2 0

)with

σ2 =

(0 −ii 0

)Eq. (8.29) just says,

Cγ0, γµ = 0, for µ 6= 2,

[Cγ0, γ2] = 0

which means

C = −iγ2γ0

up to an arbitrary phase8. So we find the simple relation for the 8 We choose −i for convenience.

charge conjugate spinor where we multiply by (−i) for conve-

nience

ΨC = −iγ2Ψ∗ (8.30)

What does it mean in terms of the Weyl spinors? We find

ΨC = −iγ2

(ψ∗Lψ∗R

)=

(−iσ2ψ∗Riσ2ψ∗L

)

This is interesting! So the charge conjugate of a left-handed field is

right-handed and vice versa. In terms of the Lorentz representations, Experimentally, we know that theneutrino is left-handed. Thus, weare now able to predict that the

anti-neutrino is right-handed!

we have found an important relation, if

ψL : ( 12 , 0), then iσ2ψ∗L : (0, 1

2 )

and vice versa. We can explicitly check that ΨC transforms as a This is due to the fact the represen-tations of SU(2) are pseudo-real.A representation and its complex

conjugate are related to each otherby a simple transformation S, e.g.

for the complex doublet 2

2∗ = S−12S

in this case with εij = iσ2ij it is just

2∗i = εij2j

spinor with the known Lorentz transformation, Eq. (7.36), Ψ →e−

i4ωµνσ

µν

Ψ we have for the complex conjugate

Ψ∗ → e+ i4ωµν(σµν)∗Ψ∗

hence

Since for µ, ν 6= 2

(σµν)∗ =

(i

2[γµ, γν ]

)∗= −

i

2[γ∗µ, γ

∗ν ]

= −i

2[γµ, γν ]

= −σµν

since these γ matrices are real (only

γ2 is purely imaginary). If one of theindices µ = 2 or ν = 2, then e.g.

(σ2ν)∗ = −i

2[(−)γ2, (±)γν ]

= +σ2ν

ΨC → −iγ2e+ i4ωµν(σµν)∗Ψ∗

= e−i4ωµν(σµν) ΨC

where we have used the fact, that [γ2, γµγν ] = 0 for µ, ν 6= 2

and γ2, γ2γν = 0 for ν 6= 2. This shows that ΨC transforms

like a regular Dirac spinor and confirms our statement about the

representations of ψL and ψR being truly exchanging roles.


8.7.3 Majorana neutrino

Because ΨC transforms as a spinor, Majorana understood that

Lorentz invariance no only allows the Dirac equation i/∂Ψ = mΨ but

also the Majorana equation

i/∂Ψ = mΨC (8.31)

This type of mass term is called a Majorana mass. We can obtain

it from the Lagrangian

L = Ψi/∂Ψ− 1

2m(

ΨTCΨ + ΨCΨT)

Since Ψ and ΨC carry opposite charge, we can use the Majorana

equation, unlike the Dirac equation, only for electrically neutral

fields. You can also see this from the Lagrangian: the mass term

breaks the U(1) symmetry Ψ → eiαΨ. We can therefore also not

couple it to gauge fields. How does it look in components? The mass

contains terms like

ΨTCΨ = CαβΨαΨβ

but since

C = −iγ2γ0 =

(−iσ2 0

0 iσ2

)

and the anti-symmetric matrix tensor for i, j = 1, 2

i(σ2)ij = εij

we have Cαβ = −Cαβ this seems to indicate that the term has

to vanish if the spinor fields Ψα commute. In your QFT lecture,

you will learn that spinors Ψ have to be treated as anticommuting

Grassmannian numbers,

Ψ1Ψ2 = −Ψ2Ψ1

which we have already (secretly) realised when defining time-ordered

or normal ordered products of Fermion fields.

In terms of the Weyl two-spinors Ψ = (ψL, ψR) the mass term

looks like

1

2mΨTCΨ + h.c. = −1

2m(ψTL iσ2ψL −mψTRiσ2ψR + h.c.)

We see that we could have been more economical and started with

just one Weyl fermion, ψL since the ψL and ψR fields are now not

mixing through the mass term.9 A minimal real Majorana field 9 As we know, they are irreduciblerepresentations of the Lorentz groupand will be separately Lorentz

covariant.

would be

L = iψLγµψL +m(ψTL iσ2ψL + h.c.) (8.32)

which brings us back to the initial mystery of how to give a single

Weyl fermion a Lorentz-invariant mass term. All that we needed to

save the naive guess in Eq. (7.7), was an anti-symmetric matrix.

iσµ∂µψL −mψ†Liσ2 = 0


What happens to spinor bilinears under charge conjugation? You

can easily check that

C : CΨγµΨC → −ΨγµΨ

This means that the current jµ of QED changes sign.

8.7.4 Furrys theorem

We can start with the photon one-point function. We note that the

external photon must be attached to a QED vertex. Neglecting the

external propagator, the amplitude is hence

318 Chapter 10 Systematics of Renormalization

(a)

(c)

(e)

() . . . D=4 (b)

(d)

D =0 ( f)

(g)

D=3

D= 1

D= 1

D =0

Figure 10.2. The seven QED amplitudes whose superficial degree of di-vergence (D) is > 0. (Each circle represents the sum of all possible QED diagrams.) As explained in the text, amplitude (a) is irrelevant to scattering processes, while amplitudes (b) and ( d) vanish because of symmetries. Am-plitude ( e) is nonzero, but its divergent parts cancel due to the Ward identity. The remaining amplitudes ( c, f, and g) are all logarithmically divergent, even though D > 0 for (c) and (f).

photon propagator, this amplitude is therefore

-q = -ie J d4x e-iq·x (OJ T Jµ(x) JO), (10.5)

-where jl' = 'ljryl''lj; is the electromagnetic current operator. But the vacuum expectation value of jl' must vanish by Lorentz invariance, since otherwise it would be a preferred 4-vector.

The photon one-point function also vanishes for a second reason: charge-conjugation invariance. Recall that C is a symmetry of QED, so C JO) = JO). But jl'(x) changes sign under charge conjugation, Cjl'(x)Ct = -jl'(x), so its vacuum expectation value must vanish:

The same argument applies to any vacuum expectation value of an odd num-ber of electromagnetic currents. In particular, the photon three-point function, Fig. 10.2d, vanishes. (This result is known as Furry's theorem.) It is not hard to check explicitly that the photon one- and three-point functions vanish in the leading order of perturbation theory (see Problem 10.1).

The remaining amplitudes in Fig. 10.2 are all nonzero, so we must analyze their structures in more detail. Consider, for example, the electron self-energy

= −ie∫d4xe−iq·x〈Ω|Tjµ(x)|Ω〉 (8.33)

where jµ = ΨγµΨ is the electro-magnetic current operator. The

vacuum expection value of jµ has to vanish because of Lorentz-

invariance: otherwise jµ would be a preferred direction in Minkowski

space.

The photon one-point function has to vanish also for other rea-

sons, namely charge conjugation invariance. Since C is symmetry of

QED, we know that the vacuum |0〉 and the vacuum of the interact-

ing theory

C|Ω〉 = |Ω〉

are C invariant. A diagram with n external photons (and no external

fermions) is proportional to

〈Ω|Tjµ1(x1) . . . jµn(xn)|Ω〉

Figure 8.3: Example of a vanishing

3-point function.

If n is odd then

〈Ω|Tjµ1(x1) . . . jµn(xn)|Ω〉 = 〈Ω|TCCjµ1

(x1)CC . . . CCjµn(xn)CC|Ω〉= 〈Ω|T (Cjµ1(x1)C)(C . . . C)(Cjµn(xn)C)|Ω〉= (−1)n〈Ω|Tjµ1

(x1) . . . jµn(xn)|Ω〉

This means that the a pure photon n-point function vanishes, for n

odd. This is known as Furry’s theorem.

8.8 Time reversal

8.8.1 Quantum mechanics

We will now complete our discussion of the discrete symmetries of

QED and discuss time reversal invariance. Time reversal symmetry

is a bit subtle as you might already know from your quantum me-

chanics course. Let us investigate this first using the Schrodinger

equation of QM We are suppressing the trivial xdependence in this section.

i∂

∂tψ(t) = Hψ(t) (8.34)


with for concreteness think H = − 12m∇

2 − V (x). We now consider

the time reversal transformation

t→ t′ = −t

Our goal is now to find ψ′(t′) which satisfies

i∂

∂t′ψ(t′) = Hψ′(t′) (8.35)

We write for the transformed field

ψ′(t′) = Tψ(t)

where T is an operator which we will determine in the following. Let Again, we can fix it up to an arbi-trary phase.us plug this into Eq. (8.35), we obtain

i∂

∂(−t)Tψ(t) = HTψ(t)

T−1i∂

∂(−t)Tψ(t) = T−1HTψ(t)

Since H is time-independent by assumption, the time-inversion

operator has no effect and

T−1H = H T−1,

which results in

T−1(−i)T ∂

∂tψ(t) = Hψ(t)

Comparing to Eq. (8.34), we are no forced to conclude

T−1(−i)T = i (8.36)

We can define

T ≡ U K

where K complex conjugates everything to the right. Applied to For real φ1, φ2 we have

K(φ1 + iφ2) = φ1 − iφ2Eq. (8.36) we get with T−1 = KU−1

KU−1(−i)U K = U−1∗(+i)U∗K2 = U−1∗(+i)U∗

We see that it does the job, if U−1iU = i, that is, if U−1 is just an

ordinary (unitary) operator that leaves i be.10 The contribution of 10 We will determine U below

K to T makes it antiunitary.

8.8.2 Scalar Field

Let us check how T acts on a scalar state solving the free Klein-

Gordon equation φ(t) = e−ip·x = ei(k·x−Et). We have

φ′(t′) = Tφ(t) = UKφ(t) = Uφ∗(t) = Ue−i(k·x−Et)

Since φ has just one component, U is a trivial phase-factor.11 We 11 Since we require |φ|2 = |φ′|2, we

choose the phase factor here to be

U = 1.can rewrite this as

φ′(t) = e−i(k·x+Et) = ei(−k·x−Et)

from which we see that φ′ describes a wave moving in the opposite

direction. The energy is still positive since ψ ∝ e−iEt. We see that

two time-reversals are equal to the identity

T 2 = UKUK = UU∗K2 = +1


8.8.3 Spinor Field

Now we need to consider how T acts on spinors. The Dirac equa-

tion is after multiplying by γ0 from the left

i∂

∂tΨ(t) = HΨ(t)

with H = −iγ0γi∂i + γ0m. As above, we want

i∂

∂t′Ψ′(t′) = HΨ(t′)

with

Ψ′(t′) = TΨ(t)

Following our discussion above, we can make similar arguments if

T−1HT = H or

KU−1HUK = H

which means we require

KU−1γ0UK = γ0

and

KU−1(iγ0γi)UK = iγ0γi

If we move K to the RHS, we see that we can determine U with

U−1γ0U = γ0∗, and U−1γiU = −γi∗

Restricting ourselves to the Weyl (and Dirac) basis, we know that γ2

is the only imaginary matrix. So we need to find a U which flips γ1

and γ3 but not γ0 and γ2 ? This choice works12 12 Again, with an arbitrary phase

which we set to 1.

U = γ1γ3

and so

Ψ(t,x)→ γ1γ3Ψ(−t,x) =

0 1

−1 0

0 1

−1 0

Ψ(−t,x)

Thus T flips the spin of particles.

8.9 Electron scattering and the Coulomb potential

We describe electron scattering e−e− → e−e− with the two tree-level

Feynman diagrams as

p

q

p′

q′

+

p

q

p′

q′

(8.37)


Figure 8.4: Summary of C, P, T .

iA = (−ig)2

[u(q′)γµu(q)

(−iηµν)

(q − q′)2 − µ2u(p′)γνu(p)

− u(p′)γµu(q)(−iηµν)

(p′ − q)2 − µ2u(q′)γνu(p)

](8.38)

Compared to the scalar mediator we seem to find an overall minus

sign, but remember that the the expression is really positive for

µ, ν = 1, 2, 3.

8.9.1 Coulomb potential

We take the first diagram of e−e− → e−e− and calculate the non-

relativistic limit. The steps proceed as in Sec. 6.7.1 and Sec. 7.8.4

for the Yukawa potential. The only difference is that we now need to

evaluate uγµu in Eq. (8.38) using the non-relativistic spinor

u(p)→ √m(ξ

ξ

)

We see that

ur(p)γ0us(q)→ 2mδrs

ur(p)γius(q)→ 0

Comparing the scattering amplitude in the non-relativistic limit

to the quantum mechanical expression again, we find an effective

potential between two electrons

U(r) = +e2

∫d3p

(2π)3

eip·r

|r|2 = +e2

4πr

Which is the familiar repulsive Coulomb potential between two

electrons. What about e−e+ → e−e+ scattering? The relevant

diagram in the non-relativistic limit is

p

q

p′

q′


which gives

iA = −(−ig)2

[v(q′)γµv(q)

(−iηµν)

(q − q′)2u(p′)γνu(p)

](8.39)

The overall + sign comes from treating the fermions correctly: we

saw the same minus sign when studying scattering in Yukawa theory.

The non-relativistic limit is now difference however, since vγ0v → 2m

which gives us Compare to the Yukawa potential

where the limit is vv → −2m.

U(r) = −e2

∫d3p

(2π)3

eip·r

|r|2 = − e2

4πr

As expected, we find an attractive force between an electron and a

positron. We can see the same effect without all the spinor-ology in

scalar QED. The interaction is (Dµφ)∗Dµφ and so the relevant term

for the scalar-photon interaction is

−ieAµ(φ∗∂µφ− φ∂µφ∗) + e2AµAµφ∗φ

the two vertices are

p

q

= −ie(p+ q)µ , = +2ie2ηµν

(8.40)

The momentum dependence comes from the ∂µ term and the factor

of 2 is because of the identical particles AµAµ appearing in the

Lagrangian Similar to the cancellation of the 1/4!

in the φ4 vertex.We can see that the difference in sign comes from the A0 piece

of the propagator −iηµν/p2, we have in the non-relativistic limit of

scalar e−e− → e−e− scattering

= −iηµν(−ie)2 (p+ p′)µ(q + q′)ν(p′ − p)2

→ −i(−ie)2 (2m)2

−(p− p′)2

where the numerator selects the A0 piece since (p + p′)µ(q + q′)µ ≈(p+ p′)0(q + q′)0 ≈ (2m)2. The Coulomb potential for spin 0 particles

is again repulsive. If switch to scalar e + e− → e + e− scattering,

one of the arrows on the legs changes direction and the amplitude

picks up an additional minus sign because since the charge arrows

are correlated with momentum arrows. This leads as expected to an

attractive potential

9

Outlook

If you found this interesting you should continue and attend the

QFT course in the next semester. You have yet to encounter many

of the truly awesome and deep insights that quantum field theory

contains. It is the language in which the laws of Nature are written.

As Sidney Coleman said: ”Not only God knows, I know, and by the

end of the semester, you will know.”

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