Physics 523 & 524: Quantum Field Theory Iquantum.phys.unm.edu/523-18/sw.pdfPhysics 523 & 524: Quantum Field Theory I Kevin Cahill Department of Physics and Astronomy ... 2.8 Feynman

Physics 523 & 524: Quantum Field Theory I

Kevin Cahill

Department of Physics and Astronomy

University of New Mexico, Albuquerque, NM 87131

Contents

1 Quantum fields and special relativity page 1

1.1 States 1

1.2 Creation operators 1

1.3 How fields transform 3

1.4 Translations 7

1.5 Boosts 8

1.6 Rotations 9

1.7 Spin-zero fields 9

1.8 Conserved charges 12

1.9 Parity, charge conjugation, and time reversal 13

1.10 Vector fields 15

1.11 Vector field for spin-zero particles 18

1.12 Vector field for spin-one particles 18

1.13 Lorentz group 23

1.14 Gamma matrices and Clifford algebras 25

1.15 Dirac’s gamma matrices 27

1.16 Dirac fields 28

1.17 Expansion of massive and massless Dirac fields 41

1.18 Spinors and angular momentum 44

1.19 Charge conjugation 49

1.20 Parity 51

2 Feynman diagrams 53

2.1 Time-dependent perturbation theory 53

2.2 Dyson’s expansion of the S matrix 54

2.3 The Feynman propagator for scalar fields 56

2.4 Application to a cubic scalar field theory 60

2.5 Feynman’s propagator for fields with spin 62

ii Contents

2.6 Feynman’s propagator for spin-one-half fields 63

2.7 Application to a theory of fermions and bosons 66

2.8 Feynman propagator for spin-one fields 71

2.9 The Feynman rules 73

2.10 Fermion-antifermion scattering 74

3 Action 76

3.1 Lagrangians and hamiltonians 76

3.2 Canonical variables 76

3.3 Principle of stationary action in field theory 78

3.4 Symmetries and conserved quantities in field theory 79

3.5 Poisson Brackets 84

3.6 Dirac Brackets 85

3.7 Dirac Brackets and QED 86

4 Quantum Electrodynamics 93

4.1 Global U( 1) Symmetry 93

4.2 Abelian gauge invariance 94

4.3 Coulomb-gauge quantization 96

4.4 QED in the interaction picture 97

4.5 Photon propagator 98

4.6 Feynman’s rules for QED 100

4.7 Electron-positron scattering 101

4.8 Trace identities 104

4.9 Electron-positron to muon-antimuon 105

4.10 Electron-muon scattering 111

4.11 One-loop QED 112

4.12 Magnetic moment of the electron 124

4.13 Charge form factor of electron 132

5 Nonabelian gauge theory 134

5.1 Yang and Mills invent local nonabelian symmetry 134

5.2 SU(3) 135

6 Standard model 138

6.1 Quantum chromodynamics 138

6.2 Abelian Higgs mechanism 138

6.3 Higgs’s mechanism 140

6.4 Quark and lepton interactions 146

6.5 Quark and Lepton Masses 148

6.6 CKM matrix 153

6.7 Lepton Masses 156

6.8 Before and after symmetry breaking 157

Contents iii

6.9 The Seesaw Mechanism 161

6.10 Neutrino Oscillations 163

7 Path integrals 165

7.1 Path integrals and Richard Feynman 165

7.2 Gaussian integrals and Trotter’s formula 165

7.3 Path integrals in quantum mechanics 166

7.4 Path integrals for quadratic actions 170

7.5 Path integrals in statistical mechanics 175

7.6 Boltzmann path integrals for quadratic actions 180

7.7 Mean values of time-ordered products 183

7.8 Quantum field theory 185

7.9 Finite-temperature field theory 189

7.10 Perturbation theory 192

7.11 Application to quantum electrodynamics 196

7.12 Fermionic path integrals 200

7.13 Application to nonabelian gauge theories 208

7.14 Faddeev-Popov trick 208

7.15 Ghosts 211

7.16 Integrating over the momenta 212

8 Landau theory of phase transitions 214

8.1 First- and second-order phase transitions 214

9 Effective field theories and gravity 217

9.1 Effective field theories 217

9.2 Is gravity an effective field theory? 218

9.3 Quantization of Fields in Curved Space 221

9.4 Accelerated coordinate systems 228

9.5 Scalar field in an accelerating frame 229

9.6 Maximally symmetric spaces 233

9.7 Conformal algebra 236

9.8 Conformal algebra in flat space 237

9.8.1 Angles and analytic functions 241

9.9 Maxwell’s action is conformally invariant for d = 4 247

9.10 Massless scalar field theory is conformally invariant if

d = 2 248

9.11 Christoffel symbols as nonabelian gauge fields 249

9.12 Spin connection 254

10 Field Theory on a Lattice 256

10.1 Scalar Fields 256


iv Contents

10.3 The Propagator 261

10.4 Pure Gauge Theory 261

10.5 Pure Gauge Theory on a Lattice 264

References 266

Index 267

1

Quantum fields and special relativity

1.1 States

A Lorentz transformation Λ is implemented by a unitary operator U(Λ)

which replaces the state |p, σ〉 of a massive particle of momentum p and spin

σ along the z-axis by the state

U(Λ)|p, σ〉 =

√(Λp)0

p0

∑s′

D(j)s′σ(W (Λ, p)) |Λp, s′〉 (1.1)

where W (Λ, p) is a Wigner rotation

W (Λ, p) = L−1(Λp)ΛL(p) (1.2)

and L(p) is a standard Lorentz transformation that takes (m,~0) to p.

1.2 Creation operators

The vacuum is invariant under Lorentz transformations and translations

U(Λ, a)|0〉 = |0〉. (1.3)

A creation operator a†(p, σ) makes the state |p, σ〉 from the vacuum state

|0〉

|pσ〉 = a†(p, σ)|0〉. (1.4)

The creation and annihilation operators obey either the commutation rela-

tion

[a(p, s), a†(p′, s′)]− = a(p, s) a†(p′, s′)− a†(p′, s′) a(p, s) = δss′ δ(3)(p− p′)

(1.5)

2 Quantum fields and special relativity

or the anticommutation relation

[a(p, s), a†(p′, s′)]+ = a(p, s) a†(p′, s′) + a†(p′, s′) a(p, s) = δss′ δ(3)(p− p′).

(1.6)

The two kinds of relations are written together as

[a(p, s), a†(p′, s′)]∓ = a(p, s) a†(p′, s′)∓ a†(p′, s′) a(p, s) = δss′ δ(3)(p− p′).

(1.7)

A bracket [A,B] with no signed subscript is interpreted as a commutator.

Equations (1.1 & 1.4) give

U(Λ)a†(p, σ)|0〉 =

√(Λp)0

p0

∑s′

D(j)s′σ(W (Λ, p)) a†(Λp, s′)|0〉. (1.8)

And (1.3) gives

U(Λ)a†(p, σ)U−1(Λ)|0〉 =

√(Λp)0

p0

∑s′

D(j)s′σ(W (Λ, p)) a†(Λp, s′)|0〉. (1.9)

SW in chapter 4 concludes that

U(Λ)a†(p, σ)U−1(Λ) =

√(Λp)0

p0

∑s′

D(j)s′σ(W (Λ, p)) a†(Λp, s′). (1.10)

If U(Λ, b) follows Λ by a translation by b, then

U(Λ, b)a†(p, σ)U−1(Λ, b) = e−i(Λp)·a

√(Λp)0

p0

∑s′

D(j)s′σ(W (Λ, p)) a†(Λp, s′)

= e−i(Λp)·a

√(Λp)0

p0

∑s′

D†(j)s′σ (W−1(Λ, p)) a†(Λp, s′)

= e−i(Λp)·a

√(Λp)0

p0

∑s′

D∗(j)σs′ (W−1(Λ, p)) a†(Λp, s′)

(1.11)


The adjoint of this equation is

U(Λ, b)a(p, σ)U−1(Λ, b) = ei(Λp)·a

√(Λp)0

p0

∑s′

D∗(j)s′σ (W (Λ, p)) a(Λp, s′)

= ei(Λp)·a

√(Λp)0

p0

∑s′

D†(j)σs′ (W (Λ, p)) a(Λp, s′)

= ei(Λp)·a

√(Λp)0

p0

∑s′

D(j)σs′(W

−1(Λ, p)) a(Λp, s′).

(1.12)

These equations (1.11 & 1.12) are (5.1.11 & 5.1.12) of SW.

1.3 How fields transform

The “positive frequency” part of a field is a linear combination of annihila-

tion operators

ψ+` (x) =

∑σ

∫d3p u`(x; p, σ) a(p, σ). (1.13)

The “negative frequency” part of a field is a linear combination of creation

operators of the antiparticles

ψ−` (x) =∑σ

∫d3p v`(x; p, σ) b†(p, σ). (1.14)

To have the fields (1.13 & 1.14) transform properly under Poincare trans-

formations

U(Λ, a)ψ+` (x)U−1(Λ, a) =

∑¯

D`¯(Λ−1)ψ+

¯ (Λx+ a)

=∑

¯

D`¯(Λ−1)

∑σ

∫d3p u¯(Λx+ a; p, σ) a(p, σ)

U(Λ, a)ψ−` (x)U−1(Λ, a) =∑

¯

D`¯(Λ−1)ψ−¯ (Λx+ a)

=∑

¯

D`¯(Λ−1)

∑σ

∫d3p v¯(Λx+ a; p, σ) b†(p, σ)

(1.15)

the spinors u`(x; p, σ) and v`(x; p, σ) must obey certain rules which we’ll now

determine.


First (1.12 & 1.13) give

U(Λ, a)ψ+` (x)U−1(Λ, a) = U(Λ, a)

∑σ

∫d3p u`(x; p, σ) a(p, σ)U−1(Λ, a)

=∑σ

∫d3p u`(x; p, σ)U(Λ, a)a(p, σ)U−1(Λ, a)

(1.16)

=∑σ

∫d3p u`(x; p, σ) ei(Λp)·a

√(Λp)0

p0

∑s′

D(j)σs′(W

−1(Λ, p)) a(Λp, s′).

Now we use the identity

d3p

p0=d3(Λp)

(Λp)0(1.17)

to turn (1.16) into

U(Λ, a)ψ+` (x)U−1(Λ, a) =

∑σ

∫d3(Λp) u`(x; p, σ) ei(Λp)·a

×

√p0

(Λp)0

∑s′

D(j)σs′(W

−1(Λ, p)) a(Λp, s′). (1.18)

Similarly (1.11, 1.14, & 1.17) give

U(Λ, a)ψ−` (x)U−1(Λ, a) = U(Λ, a)∑σ

∫d3p v`(x; p, σ) b†(p, σ)U−1(Λ, a)

=∑σ

∫d3p v`(x; p, σ)U(Λ, a)b†(p, σ)U−1(Λ, a) (1.19)

=∑σ

∫d3p v`(x; p, σ) e−i(Λp)·a

√(Λp)0

p0

∑s′

D∗(j)σs′ (W−1(Λ, p)) b†(Λp, s′)

=∑σ

∫d3(Λp) v`(x; p, σ) e−i(Λp)·a

√p0

(Λp)0

∑s′

D∗(j)σs′ (W−1(Λ, p)) b†(Λp, s′).

So to get the fields to transform as in (1.15), equations (1.18 & 1.19) say


that we need

∑¯

D`¯(Λ−1)ψ+

¯ (Λx+ a) =∑

¯

D`¯(Λ−1)

∑σ

∫d3p u¯(Λx+ a; p, σ) a(p, σ)

=∑

¯

D`¯(Λ−1)

∑σ

∫d3(Λp) u¯(Λx+ a; Λp, σ) a(Λp, σ)

=∑σ

∫d3(Λp) u`(x; p, σ) ei(Λp)·a (1.20)

×

√p0

(Λp)0

∑s′

D(j)σs′(W

−1(Λ, p)) a(Λp, s′)

=∑s′

∫d3(Λp) u`(x; p, s′) ei(Λp)·a

×

√p0

(Λp)0

∑σ

D(j)s′σ(W−1(Λ, p)) a(Λp, σ)

and

∑¯

D`¯(Λ−1)ψ−¯ (Λx+ a) =

∑¯

D`¯(Λ−1)

∑σ

∫d3p v¯(Λx+ a; p, σ) b†(p, σ)

=∑

¯

D`¯(Λ−1)

∑σ

∫d3(Λp) v¯(Λx+ a; Λp, σ) b†(Λp, σ)

=∑σ

∫d3(Λp) v`(x; p, σ) e−i(Λp)·a (1.21)

×

√p0

(Λp)0

∑s′

D∗(j)σs′ (W−1(Λ, p)) b†(Λp, s′)

=∑s′

∫d3(Λp) v`(x; p, s′) e−i(Λp)·a

×

√p0

(Λp)0

∑σ

D∗(j)s′σ (W−1(Λ, p)) b†(Λp, σ).

Equating coefficients of the red annihilation and blue creation operators, we

find that the fields will transform properly if the spinors u and v satisfy the


rules

∑¯

D`¯(Λ−1) u¯(Λx+ a; Λp, σ) =

√p0

(Λp)0

∑s′

D(j)s′σ(W−1(Λ, p))u`(x; p, s′) ei(Λp)·a

(1.22)∑¯

D`¯(Λ−1)v¯(Λx+ a; Λp, σ) =

√p0

(Λp)0

∑s′

D∗(j)s′σ (W−1(Λ, p))v`(x; p, s′)e−i(Λp)·a

(1.23)

which differ from SW’s by an interchange of the subscripts σ, s′ on the

rotation matrices D(j). (I think SW has a typo there.) If we multiply both

sides of these equations (1.22 & 1.23) by the two kinds of D matrices, then

we get first

∑¯,`

D`′`(Λ)D`¯(Λ−1) u¯(Λx+ a; Λp, σ) = u`′(Λx+ a; Λp, σ)

=

√p0

(Λp)0

∑s′,`

D(j)s′σ(W−1(Λ, p))D`′`(Λ)u`(x; p, s′) ei(Λp)·a (1.24)

∑¯,`

D`′`(Λ)D`¯(Λ−1)v¯(Λx+ a; Λp, σ) = v`′(Λx+ a; Λp, σ)

=

√p0

(Λp)0

∑s′,`

D∗(j)s′σ (W−1(Λ, p))D`′`(Λ)v`(x; p, s′)e−i(Λp)·a (1.25)

1.4 Translations 7

and then with W ≡W (Λ, p)∑σ

D(j)σs (W )u`′(Λx+ a; Λp, σ)

=

√p0

(Λp)0

∑s′,σ,`

D(j)s′σ(W−1)D

(j)σs (W )D`′`(Λ)u`(x; p, s′) ei(Λp)·a

=

√p0

(Λp)0

∑`

D`′`(Λ)u`(x; p, s) ei(Λp)·a (1.26)∑σ

D∗(j)σs (W )v`′(Λx+ a; Λp, σ)

=

√p0

(Λp)0

∑σ,s′,`

D∗(j)s′σ (W−1)D

∗(j)σs (W )D`′`(Λ)v`(x; p, s′)e−i(Λp)·a

=

√p0

(Λp)0

∑`

D`′`(Λ)v`(x; p, s)e−i(Λp)·a (1.27)

which are equations (5.1.13 & 5.1.14) of SW:

∑s

u¯(Λx+ a; Λp, s)D(j)sσ (W (Λ, p)) =

√p0

(Λp)0

∑`

D ¯ (Λ)u`(x; p, σ) ei(Λp)·a

∑s

v¯(Λx+ a; Λp, s)D∗(j)sσ (W (Λ, p)) =

√p0

(Λp)0

∑`

D ¯ (Λ)v`(x; p, σ)e−i(Λp)·a.

(1.28)

These are the equations that determine the spinors u and v up to a few

arbitrary phases.

1.4 Translations

When Λ = I, the D matrices are equal to unity, and these last equations

(1.28) say that for x = 0

u`(a; p, σ) = u`(0; p, σ) eip·a

v`(a; p, σ) = v`(0; p, σ)e−ip·a.(1.29)

Thus the spinors u and v depend upon spacetime by the usual phase e±ip·x

u`(x; p, σ) = (2π)−3/2u`(p, σ) eip·x

v`(x; p, σ) = (2π)−3/2v`(p, σ)e−ip·x(1.30)


in which the 2π’s are conventional. The fields therefore are Fourier trans-

forms:

ψ+` (x) = (2π)−3/2

∑σ

∫d3p eip·xu`(p, σ) a(p, σ)

ψ−` (x) = (2π)−3/2∑σ

∫d3p e−ip·xv`(p, σ) b†(p, σ)

(1.31)

and every field of mass m obeys the Klein-Gordon equation

(∇2 − ∂20 −m2)ψ`(x) = (2−m2)ψ`(x) = 0. (1.32)

Since exp[i(Λp · (Λx + a))] = exp(ip · x + iΛp · a), the conditions (1.28)

simplify to

∑s

u¯(Λp, s)D(j)sσ (W (Λ, p)) =

√p0

(Λp)0

∑`

D ¯ (Λ)u`(p, σ)

∑s

v¯(Λp, s)D∗(j)sσ (W (Λ, p)) =

√p0

(Λp)0

∑`

D ¯ (Λ)v`(p, σ)

(1.33)

for all Lorentz transformations Λ.

1.5 Boosts

Set p = k = (m,~0) and Λ = L(q) where L(q)k = q. So L(p) = 1 and

W (Λ, p) ≡ L−1(Λp)ΛL(p) = L−1(q)L(q) = 1. (1.34)

Then the equations (1.33) are

u¯(q, σ) =

√m

q0

∑`

D ¯ (L(q))u`(~0, σ)

v¯(q, σ) =

√m

q0

∑`

D ¯ (L(q))v`(~0, σ).

(1.35)

Thus a spinor at finite momentum is given by a representation D(Λ) of

the Lorentz group (see the online notes of chapter 10 of my book for its

finite-dimensional nonunitary representations) acting on the spinor at zero

3-momentum p = k = (m,~0). We need to find what these spinors are.

1.6 Rotations 9

1.6 Rotations

Now set p = k = (m,~0) and Λ = R a rotation so that W = R. For rotations,

the spinor conditions (1.33) are∑s

u¯(~0, s)D(j)sσ (R) =

∑`

D ¯ (R)u`(~0, σ)∑s

v¯(~0, s)D∗(j)sσ (R) =

∑`

D ¯ (R)v`(~0, σ).(1.36)

The representations D(j)sσ (R) of the rotation group are (2j + 1) × (2j + 1)-

dimensional unitary matrices. For a rotation of angle θ about the ~θ = θ axis,

they are the ones taught in courses on quantum mechanics (and discussed

in the notes of chapter 10)

D(j)sσ (θ) =

[e−iθ·J

(j)]sσ

(1.37)

where [Ja, Jb] = iεabcJc. The representations D ¯ (R) of the rotation group

are finite-dimensional unitary matrices. For a rotation of angle θ about the~θ = θ axis, they are

D ¯ (θ) =[e−iθ·J

]¯

(1.38)

in which [Ja,Jb] = iεabcJ . For tiny rotations, the conditions (1.36) require

(because of the complex conjugation of the antiparticle condition) that the

spinors obey the rules∑s

u¯(~0, s)(J (j)a )sσ =

∑`

(Ja) ¯ u`(~0, σ)∑s

v¯(~0, s)(−Ja)∗(j)sσ ) =∑`

(Ja) ¯ v`(~0, σ)(1.39)

for a = 1, 2, 3.

1.7 Spin-zero fields

Spin-zero fields have no spin or Lorentz indexes. So the boost conditions

(1.210) merely require that u(q) =√m/q0u(0) and v(q) =

√m/q0v(0).

The conventional normalization is u(0) = 1/√

2m and v(0) = 1/√

2m. The

spin-zero spinors then are

u(p) = (2p0)−1/2 and v(p) = (2p0)−1/2. (1.40)

For simpicity, let’s first consider a neutral scalar field so that b(p, s) =


a(p, s). The definitions (1.13) and (1.14) of the positive-frequency and negative-

frequency fields and their behavior (1.30) under translations then give us

φ+(x) =

∫d3p√

(2π)32p0a(p) eip·x

φ−(x) =

∫d3p√

(2π)32p0a†(p) e−ip·x.

(1.41)

Note that [φ±(x)

]†= φ∓(x). (1.42)

Since [a(p), a(p′)]± = 0, it follows that

[φ+(x), φ+(y)]∓ = 0 and [φ−(x), φ−(y)]∓ = 0 (1.43)

whatever the values of x and y as long as we use commutators for bosons

and anticommutators for fermions.

But the commutation relation

[a(p, s), a†(q, t)]∓ = δst δ(3)(p− q) (1.44)

makes the commutator

[φ+(x), φ−(y)]∓ =

∫d3pd3p′

(2π)3√

2p02p′0eip·xe−ip

′·y δ3(p− p′)

=

∫d3p

(2π)32p0eip·(x−y) = ∆+(x− y)

(1.45)

nonzero even for (x− y)2 > 0 as we’ll now verify.

For space-like x, the Lorentz-invariant function ∆+(x) can only depend

upon x2 > 0 since the time x0 and its sign are not Lorentz invariant. So we

choose a Lorentz frame with x0 = 0 and |x| =√x2. In this frame,

∆+(x) =

∫d3p

(2π)32√p2 +m2

eip·x

=

∫p2dp d cos θ

(2π)22√p2 +m2

eipx cos θ

(1.46)

where p = |p| and x = |x|. Now∫d cos θ eipx cos θ =

(eipx − e−ipx

)/(ipx) = 2 sin(px)/(px), (1.47)

so the integral (1.46) is

∆+(x) =1

4π2x

∫ ∞0

sin(px) pdp√p2 +m2

(1.48)

1.7 Spin-zero fields 11

with u ≡ p/m

∆+(x) =m

4π2x

∫ ∞0

sin(mxu)udu√u2 + 1

=m

4π2xK1(mx2) (1.49)

a Hankel function.

To get a Lorentz-invariant, causal theory, we use the arbitrary parameters

κ and λ setting

φ(x) = κφ+(x) + λφ−(x) (1.50)

Now the adjoint rule (1.42) and the commutation relations (1.45 and 1.45)

give

[φ(x), φ†(y)]∓ = [κφ+(x) + λφ−(x), κ∗φ−(y) + λφ+(y)]∓

= |κ|2[φ+(x), φ−(y)]∓ + |λ|2[φ−(x), φ+(y)]∓

= |κ|2 ∆+(x− y)∓ |λ|2 ∆+(y − x)

[φ(x), φ(y)]∓ = [κφ+(x) + λφ−(x), κφ+(y) + λφ−(y)]∓

= κλ([φ+(x), φ−(y)]∓ + [φ−(x), φ+(y)]∓

)= κλ (∆+(x− y)∓∆+(y − x)) .

(1.51)

But when (x− y)2 > 0, ∆+(x− y) = ∆+(y − x). Thus these conditions are

[φ(x), φ†(y)]∓ =(|κ|2 ∓ |λ|2

)∆+(x− y)

[φ(x), φ(y)]∓ = κλ∆+(x− y)(1∓ 1).(1.52)

The first of these equations implies that we choose the minus sign and so

that we use commutation relations and not anticommutation relations for

spin-zero fields. This is the spin-statistics theorem for spin-zero fields.

SW proves the theorem for arbitrary massive fields in section 5.7.

We also must set

|κ| = |λ|. (1.53)

The second equation then is automatically satisfied. The common magnitude

and the phases of κ and λ are arbitrary, so we choose κ = λ = 1. We then

have

φ(x) = φ+(x) + φ−(x) = φ+(x) + φ+†(x) = φ†(x). (1.54)

Now the interaction density H(x) will commute with H(y) for (x− y)2 > 0,

and we have a chance of having a Lorentz-invariant, causal theory.

The field (1.54)

φ(x) =

∫d3p√

(2π)32p0

[a(p) eip·x + a†(p) e−ip·x

](1.55)


obeys the Klein-Gordon equation

(∇2 − ∂20 −m2)φ(x) ≡ (2−m2)φ(x) = 0. (1.56)

1.8 Conserved charges

If the field φ adds and deletes charged particles, an interaction H(x) that is

a polynomial in φ will not commute with the charge operator Q because φ+

will lower the charge and φ− will raise it. The standard way to solve this

problem is to start with two hermitian fields φ1 and φ2 of the same mass.

One defines a complex scalar field as a complex linear combination of the

two fields

φ(x) =1√2

(φ1(x) + iφ2(x))

=

∫d3p√

(2π)32p0

[1√2

(a1(p) + ia2(p)) eip·x +1√2

(a†1(p) + ia†2(p)

)e−ip·x

].

(1.57)

Setting

a(p) =1√2

(a1(p) + ia2(p)) and b†(p) =1√2

(a†1(p) + ia†2(p)

)(1.58)

so that

b(p) =1√2

(a1(p)− ia2(p)) and a†(p) =1√2

(a†1(p)− ia†2(p)

)(1.59)

we have

φ(x) =

∫d3p√

(2π)32p0

[a(p) eip·x + b†(p) e−ip·x

](1.60)

and

φ†(x) =

∫d3p√

(2π)32p0

[b(p) eip·x + a†(p) e−ip·x

]. (1.61)

Since the commutation relations of the real creation and annihilation oper-

ators are for i, j = 1, 2

[ai(p), a†j(p′)] = δij δ

3(p− p′) and [ai(p), aj(p′)] = 0 = [a†i (p), a

†j(p′)]

(1.62)

the commutation relations of the complex creation and annihilation opera-

tors are

[a(p), a†(p′)] = δ3(p− p′) and [b(p), b†(p′)] = δ3(p− p′) (1.63)

1.9 Parity, charge conjugation, and time reversal 13

with all other commutators vanishing.

Now φ(x) lowers the charge of a state by q if a† adds a particle of charge

q and if b† adds a particle of charge −q. Similarly, φ†(x) raises the charge of

a state by q

[Q,φ(x)] = − qφ(x) and [Q,φ†(x)] = qφ†(x). (1.64)

So an interaction with as many φ(x)’s as φ†(x)’s conserves charge.

1.9 Parity, charge conjugation, and time reversal

If the unitary operator P represents parity on the creation operators

Pa†1(p)P−1 = η a†1(−p) and Pa†2(p)P−1 = η a†2(−p) (1.65)

with the same phase η. Then

Pa1(p)P−1 = η∗a1(−p) and Pa2(p)P−1 = η∗a2(−p) (1.66)

and so both

Pa†(p)P−1 = ηa†(−p) and Pa(p)P−1 = η∗a(−p) (1.67)

and

Pb†(p)P−1 = ηb†(−p) and Pb(p)P−1 = η∗b(−p). (1.68)

Thus if the field

φ1(x) =

∫d3p√

(2π)32p0

[a1(p) eip·x + a†1(p) e−ip·x

](1.69)

or φ2(x), or the complex field (1.60) is to go into a multiple of itself under

parity, then we need η = η∗ so that η is real. Then the fields transform under

parity as

Pφ1(x)P−1 = η∗φ1(x0,−x) = ηφ1(x0,−x)

Pφ2(x)P−1 = η∗φ(x0,−x) = ηφ2(x0,−x)

Pφ(x)P−1 = η∗φ(x0,−x) = ηφ(x0,−x).

(1.70)

Since P2 = I, we must have η = ±1. SW allows for a more general phase

by having parity act with the same phase on a and b†. Both schemes imply

that the parity of a hermitian field is ±1 and that the state

|ab〉 =

∫d3p f(p2) a†(p) b†(−p) |0〉 (1.71)

has even or positive parity, P|ab〉 = |ab〉.


Charge conjugation works similarly. If the unitary operator C represents

charge conjugation on the creation operators

Ca†1(p)C−1 = ξa†1(p) and Ca†2(p)C−1 = − ξa†2(p) (1.72)

with the same phase ξ. Then

Ca1(p)C−1 = ξ∗a1(p) and Ca2(p)C−1 = − ξ∗a2(p) (1.73)

and so since a = (a1 + ia2)/√

2 and b = (a1 − ia2)/√

2

Ca(p)C−1 = ξ∗ b(p) and Cb(p)C−1 = ξ∗a(p) (1.74)

and since a† = (a†1 − ia†2)/√

2 and b† = (a†1 + ia†2)/√

2

Ca†(p)C−1 = ξ b†(p) and Cb†(p)C−1 = ξa†(p). (1.75)

Thus under charge conjugation, the field (1.60) becomes

Cφ(x)C−1 =

∫d3p√

(2π)32p0

[ξ∗ b(p) eip·x + ξ a†(p) e−ip·x

](1.76)

and so if it is to go into a multiple of itself or of its adjoint under charge

conjugation then we need ξ = ξ∗ so that ξ is real. We then get

Cφ(x)C−1 = ξ∗ φ†(x) = ξ φ†(x). (1.77)

Since C2 = I, we must have ξ = ±1. SW allows for a more general phase

by having charge conjugation act with the same phase on a and b†. Both

schemes imply that the charge-conjugation parity of a hermitian field is ±1

and that the state

|ab〉 =

∫d3p f(p2) a†(p) b†(p) |0〉 (1.78)

has even or positive charge-conjugation parity, C|ab〉 = |ab〉.The time-reversal operator T is antilinear and antiunitary. So if

Ta1(p)T−1 = ζ∗a1(−p) and Ta2(p)T−1 = − ζ∗a2(−p)

Ta†1(p)T−1 = ζa†1(−p) and Ta†2(p)T−1 = − ζa†2(−p)(1.79)

then

Ta(p)T−1 = T1√2

(a1(p) + ia2(p)

)T−1 =

1√2

(Ta1(p)T−1 − iTa2(p)T−1

)= ζ∗

1√2

(a1(−p) + ia2(−p)

)= ζ∗ a(−p)

(1.80)


and

Tb†(p)T−1 = T1√2

(a†1(p) + ia†2(p)

)T−1 =

1√2

(Ta†1(p)T−1 − iTa†2(p)T−1

)= ζ

1√2

(a†1(−p) + ia†2(−p)

)= ζ b†(−p)

(1.81)

then one has

Tφ(x)T−1 = T

∫d3p√

(2π)32p0

[a(p) eip·x + b†(p) e−ip·x

]T−1

=

∫d3p√

(2π)32p0

[Ta(p)T−1 e−ip·x + Tb†(p)T−1 eip·x

]=

∫d3p√

(2π)32p0

[ζ∗ a(−p) e−ip·x + ζ b†(−p) eip·x

].

(1.82)

So if ζ is real, then after replacing −p by p, we get

Tφ(x)T−1 = ζ∗∫

d3p√(2π)32p0

[a(p) eip

0x0+ip·x + b†(p) e−ip0x0+ip·x

]= ζ∗φ(−x0,x) = ζφ(−x0,x).

(1.83)

Since T2 = I, the phase ζ = ±1. SW lets ζ be complex but defined only for

complex scalar fields and not for their real and imaginary parts.

1.10 Vector fields

Vector fields transform like the 4-vector xi of spacetime. So

D ¯ (Λ) = Λ¯

` (1.84)

for ¯, ` = 0, 1, 2, 3. Again we start with a hermitian field labelled by i =

0, 1, 2, 3

φ+i(x) = (2π)−3/2∑s

∫d3p eip·xui(p, s) a(p, s)

φ−i(x) = (2π)−3/2∑s

∫d3p e−ip·xvi(p, s) a†(p, s).

(1.85)


The boost conditions (1.210) say that

ui(p, s) =

√m

p0

∑k

L(p)ikuk(~0, s)

vi(p, s) =

√m

p0

∑k

L(p)ikvk(~0, s).

(1.86)

The rotation conditions (1.39) give∑s

ui(~0, s)(J (j)a )ss =

∑k

(Ja)ikuk(~0, s)

−∑s

vi(~0, s)(J∗(j)a )ss =∑k

(Ja)ikvk(~0, s).(1.87)

The (2j+ 1)× (2j+ 1) matrices (J(j)a )ss are the generators of the (2j+ 1)×

(2j+ 1) representation of the rotation group. (See my online notes on group

theory.) You learned that

3∑a=1

[(J (j)a )2

]ss′

=3∑

a=1

j∑s=−j

(J (j)a )ss (J (j)

a )ss′ = j(j + 1)δss′ (1.88)

and that

J1 =1√2

0 1 0

1 0 1

0 1 0

, J2 =1√2

0 −i 0

i 0 −i0 i 0

, J3 =

1 0 0

0 0 0

0 0 −1

(1.89)

in courses on quantum mechanics.

For k = 1, 2, 3, the three 4 × 4 matrices (Jk)ij are the generators of

rotations in the vector representation of the Lorentz group. Their nonzero

components are

(Jk)ij = − iεijk (1.90)

for i, j, k = 1, 2, 3, while (Jk)00 = 0, (Jk)0

j = 0, and (Jk)i0 = 0 for i, j, k =

1, 2, 3. So

(J 2)ij = 2δij (1.91)

with (J 2)00 = 0, (J 2)0

j = 0, and (J 2)i0 = 0 for i, j = 1, 2, 3. Apart from a

factor of i, the Jk’s are the 4 × 4 matrices Ja = iRa of my online notes on


the Lorentz group

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

.

(1.92)

Since (Ja)0k = 0 for a, k = 1, 2, 3, the spin conditions (1.87) give for i = 0∑

s

u0(~0, s)(J (j)a )ss = 0 and −

∑s

v0(~0, s)(J∗(j)a )ss = 0. (1.93)

Multiplying these equations from the right by (J(j)a )ss′ while summing over

a = 1, 2, 3 and using the formula (1.88) [(J (j))2]ss′ = j(j + 1) δss′ , we find

j(j + 1)u0(~0, s) = 0 and j(j + 1) v0(~0, s) = 0. (1.94)

Thus u0(~0, σ) and v0(~0, σ) can be anything if the field represents particles of

spin j = 0, but u0(~0, σ) and v0(~0, σ) must both vanish if the field represents

particles of spin j > 0.

Now we set i = 1, 2, 3 in the spin conditions (1.87) and again multiply

from the right by (J(j)a )ss′ while summing over a = 1, 2, 3 and using the

formula (1.88) (J (j))2 = j(j + 1). The Lorentz rotation matrices generate a

j = 1 representation of the group of rotations.

3∑ka=1

(Ja)ik (Ja)k` = j(j + 1) δi` = 2δi`. (1.95)

So the remaining conditions on the fields are

j(j + 1)ui(~0, s′) =∑ssa

ui(~0, s)(J (j)a )ss (J (j)

a )ss′ =∑ksa

(Ja)ikuk(~0, s) (J (j)a )ss′

=∑kà

(Ja)ik (Ja)k` u`(~0, s′) =∑k

2 δi` u`(~0, s′) = 2ui(~0, s′)

j(j + 1) vi(~0, s′) =∑ssa

vi(~0, s)(J∗(j)a )ss (J∗(j)a )ss′ =∑ksa

(Ja)ikvk(~0, s) (J (j)a )ss′

=∑kà

(Ja)ik (Ja)k` v`(~0, s′) =∑k

2 δi` v`(~0, s′) = 2 vi(~0, s′).

(1.96)

Thus if j = 0, then for i = 1, 2, 3 both ui(~0, s) and vi(~0, s) must vanish,

while if j > 0, then since j(j + 1) = 2, the spin j must be unity, j = 1.


1.11 Vector field for spin-zero particles

The only nonvanishing components are constants taken conventionally as

u0(~0) = i√m/2 and v0(~0) = − i

√m/2. (1.97)

At finite momentum the boost conditions (1.210) give them as

uµ(~p) = ipµ/√

2p0 and vµ(~p) = − ipµ/√

2p0. (1.98)

The vector field φµ(x) of a spin-zero particle is then the derivative of a scalar

field φ(x)

φµ(x) = ∂µφ(x) =

∫d3p√

(2π)32p0

[ipµ a(p) eip·x − ipµ b†(p) e−ip·x

](1.99)

1.12 Vector field for spin-one particles

We start with the s = 0 spinors ui(~0, 0) and vi(~0, 0) and note that since

(J(j)3 )s0 = 0, the a = 3 rotation conditions (1.87) imply that

(J3)ik uk(~0, 0) = iR3 u

i(~0, 0) = 0 and (J3)ik vk(~0, 0) = iR3 v

i(~0, 0) = 0.

(1.100)

Referring back to the explicit formulas for the generators of rotations and

setting u, v = (0, x, y, z) we see that

J3 u =

0 0 0 0

0 0 −i 0

0 i 0 0

0 0 0 0

0

x

y

z

=

0

−iyx

0

=

0

0

0

0

(1.101)

and

J3 v =

0 0 0 0

0 0 −i 0

0 i 0 0

0 0 0 0

0

x

y

z

=

0

−iyx

0

=

0

0

0

0

. (1.102)

Thus only the 3-component z can be nonzero. The conventional choice is

uµ(~0, 0) = vµ(~0, 0) =1√2m

0

0

0

1

. (1.103)

We now form the linear combinations of the rotation conditions (1.87)


that correspond to the raising and lowering matrices J(1)± = J

(1)1 ± iJ (1)

2

J(1)+ =

√2

0 1 0

0 0 1

0 0 0

and J(1)− =

√2

0 0 0

1 0 0

0 1 0

. (1.104)

Their Lorentz counterparts are

J (1)± = J (1)

1 ± iJ (1)2 =

0 0 0 0

0 0 0 ∓1

0 0 0 −i0 ±1 i 0

. (1.105)

In these terms, the rotation conditions (1.87) for the j = 1 spinors ui(~0, s)

are ∑s

ui(~0, s)(J(1)± )ss =

∑k

(J±)ikuk(~0, s). (1.106)

But

J∗(1)1 ± iJ∗(1)

2 = J(1)1 ∓ iJ (1)

2 = J∓. (1.107)

So the rotation conditions (1.87) for the j = 1 spinors vi(~0, s) are

−∑s

vi(~0, s)(J(1)∓ )ss =

∑k

(J±)ikvk(~0, s). (1.108)

So for the plus sign and the choice s = 0, the condition (1.106) gives ui(~0, 1)

as

∑s

ui(~0, s) J(1)+s0 =

√2ui(~0, 1) = (J+)iku

k(~0, 0) =

0 0 0 0

0 0 0 −1

0 0 0 −i0 1 i 0

1√2m

0

0

0

1

(1.109)

or

ui(~0, 1) =1

2√m

0

−1

−i0

. (1.110)


Similarly, the minus sign and the choice s = 0 give for ui(~0,−1)

∑s

ui(~0, s) J(1)−s0 =

√2ui(~0,−1) = (J−)iku

k(~0, 0) =

0 0 0 0

0 0 0 1

0 0 0 −i0 −1 i 0

1√2m

0

0

0

1

(1.111)

or

ui(~0,−1) =1

2√m

0

1

−i0

. (1.112)

The rotation condition (1.108) for the j = 1 spinors vi(~0, s) with the minus

sign and the choice s = 0 gives

−∑s

vi(~0, s)J(1)−s0 = −

√2 vi(~0,−1) = (J+)ikv

k(~0, 0) =

0 0 0 0

0 0 0 −1

0 0 0 −i0 1 i 0

1√2m

0

0

0

1

(1.113)

or

vi(~0,−1) =1

2√m

0

1

i

0

. (1.114)

Similarly, the plus sign and the choice s = 0 give

−∑s

vi(~0, s)J(1)+s0 = −

√2 vi(~0, 1) = (J−)ikv

k(~0, 0) =

0 0 0 0

0 0 0 1

0 0 0 −i0 −1 i 0

1√2m

0

0

0

1

(1.115)

or

vi(~0, 1) =1

2√m

0

−1

i

0

. (1.116)

The boost conditions (1.210) now give for i, k = 0, 1, 2, 3

ui(~p, s) = v∗i(~p, s) =√m/p0 Lik(~p)u

k(~0, s) = ei(~p, s)/√

2p0 (1.117)


where

ei(~p, s) = Lik(~p) ek(~0, s) (1.118)

and

e(~0, 0) =

0

0

0

1

, e(~0, 1) = − 1√2

0

1

i

0

, and e(~0,−1) =1√2

0

1

−i0

.

(1.119)

A single massive vector field is then

φi(x) = φ+i(x)+φ−i(x) =1∑

s=−1

∫d3p√

(2π)32p0ei(~p, s) a(~p, s)eip·x+e∗i(~p, s) a†(~p, s)e−ip·x.

(1.120)

The commutator/anticommutator of the positive and negative frequency

parts of the field is

[φ+i(x), φ−k(y)]∓ =

∫d3p√

(2π)32p0eip·(x−y) Πik(~p) (1.121)

where Π is a sum of outer products of 4-vectors

Πik(~p) =1∑

s=−1

ei(~p, s)e∗k (~p, s). (1.122)

At ~p = 0, the matrix Π is the unit matrix on the spatial coordinates

Π(~0) =

1∑s=−1

ei(~0, s)e∗k (~0, s) =

0 0 0 0

0 1 0 0

0 0 1 0

0 0 0 1

. (1.123)

So Π(~p) is

Π(~p) = LΠ(0)LT = LηLT + L

1 0 0 0

0 0 0 0

0 0 0 0

0 0 0 0

LT (1.124)

or

Π(~p)ik = ηik + pipk/m2. (1.125)


This equation lets us write the commutator (1.126) in terms of the Lorentz-

invariant function ∆+(x− y) (1.45) as

[φ+i(x), φ−k(y)]∓ = (ηik − ∂i∂k/m2)

∫d3p√

(2π)32p0eip·(x−y)

= (ηik − ∂i∂k/m2) ∆+(x− y).

(1.126)

As for a scalar field, we set

vi(x) = κφ+i(x) + λφ−i(x) (1.127)

and find for (x− y)2 > 0 since ∆+(x− y) = ∆+(y − x) for x, y spacelike

[v(x), v†(y)]∓ =(|κ|2 ∓ |λ|2

)(ηik − ∂i∂k/m2) ∆+(x− y)

[v(x), v(y)]∓ = (1∓ 1)κλ(ηik − ∂i∂k/m2) ∆+(x− y).(1.128)

So we must choose the minus sign and set |κ| = |λ|. So then

vi(x) = v+i(x) + v−i(x) = v+i(x) + v+i†(x) (1.129)

is real. This is a second example of the spin-statistics theorem.

If two such fields have the same mass, then we can combine them as we

combined scalar fields

vi(x) = v+i1 (x) + iv−i2 (x). (1.130)

These fields obey the Klein-Gordon equation

(2−m2)vi(x) = 0. (1.131)

And since both

pi = Lijkj and ek(~p) = Lkè

`(0) (1.132)

it follows that

p · e(~p) = k · e(0) = 0. (1.133)

So the field vi also obeys the rule

∂i vi(x) = 0. (1.134)

These equations (1.133) and 1.134 are like those of the electromagnetic

field in Lorentz gauge. But one can’t get quantum electrodynamics as the

m → 0 limit of just any such theory. For the interaction H = Ji vi would

lead to a rate for v-boson production like

JiJkΠik(~p) (1.135)

which diverges as m → 0 because of the pipk/m2 term in Πik(~p). One can

1.13 Lorentz group 23

avoid this divergence by requiring that ∂iJi = 0 which is current conserva-

tion.

Under parity, charge conjugation, and time reversal, a vector field trans-

forms as

Pva(x)P−1 = − η∗Pabvb(Px)

Cva(x)C−1 = ξ∗va†(x)

Tva(x)T−1 = ζ∗Pabvb(−Px).

(1.136)

1.13 Lorentz group

The Lorentz group O(3, 1) is the set of all linear transformations L that

leave invariant the Minkowski inner product

xy ≡ x · y − x0y0 = xTηy (1.137)

in which η is the diagonal matrix

η =

−1 0 0 0

0 1 0 0

0 0 1 0

0 0 0 1

. (1.138)

So L is in O(3, 1) if for all 4-vectors x and y

(Lx) T ηL y = xTLT η Ly = xTη y. (1.139)

Since x and y are arbitrary, this condition amounts to

LTη L = η. (1.140)

Taking the determinant of both sides and recalling that detAT = detA and

that det(AB) = detAdetB, we have

(detL)2 = 1. (1.141)

So detL = ±1, and every Lorentz transformation L has an inverse. Multi-

plying (1.140) by η, we get

ηLTηL = η2 = I (1.142)

which identifies L−1 as

L−1 = ηLTη. (1.143)


The subgroup of O(3, 1) with detL = 1 is the proper Lorentz group SO(3, 1).

The subgroup of SO(3, 1) that leaves invariant the sign of the time compo-

nent of timelike vectors is the proper orthochronous Lorentz group

SO+(3, 1).

To find the Lie algebra of SO+(3, 1), we take a Lorentz matrix L = I +ω

that differs from the identity matrix I by a tiny matrix ω and require L to

obey the condition (1.140) for membership in the Lorentz group(I + ωT

)η (I + ω) = η + ωTη + η ω + ωTω = η. (1.144)

Neglecting ωTω, we have ωTη = −η ω or since η2 = I

ωT = − η ω η. (1.145)

This equation implies that the matrix ωab is antisymmetric when both in-

dexes are down

ωab = −ωba. (1.146)

To see why, we write it (1.145) as ωea = − ηab ωbc ηce and the multiply both

sides by ηde so as to get ωda = ηde ωea = −ηab ωbc ηce ηde = −ωac δcd = −ωad.

The key equation (1.145) also tells us that under transposition the time-

time and space-space elements of ω change sign, while the time-space and

spacetime elements do not. That is, the tiny matrix ω is for infinitesimal θ

and λ a linear combination

ω = θ ·R+ λ ·B (1.147)

of three antisymmetric space-space matrices

R1 =

0 0 0 0

0 0 0 0

0 0 0 −1

0 0 1 0

R2 =

0 0 0 0

0 0 0 1

0 0 0 0

0 −1 0 0

R3 =

0 0 0 0

0 0 −1 0

0 1 0 0

0 0 0 0

(1.148)

and of three symmetric time-space matrices

B1 =

0 1 0 0

1 0 0 0

0 0 0 0

0 0 0 0

B2 =

0 0 1 0

0 0 0 0

1 0 0 0

0 0 0 0

B3 =

0 0 0 1

0 0 0 0

0 0 0 0

1 0 0 0

(1.149)

all of which satisfy condition (1.145). The three R` are 4× 4 versions of the

familiar rotation generators; the three B` generate Lorentz boosts.

1.14 Gamma matrices and Clifford algebras 25

If we write L = I + ω as

L = I − iθ` iR` − iλ` iB` ≡ I − iθ`J` − iλ`K` (1.150)

then the three matrices J` = iR` are imaginary and antisymmetric, and

therefore hermitian. But the three matrices K` = iB` are imaginary and

symmetric, and so are antihermitian. The 4×4 matrix L = exp(iθ`J`−iλ`K`)

is not unitary because the Lorentz group is not compact.

1.14 Gamma matrices and Clifford algebras

In component notation, L = I + ω is

Lab = δab + ωab, (1.151)

the matrix η is ηcd = ηcd, and ωT = − η ω η is

ωab = (ωT) ab = − (ηωη) a

b = − ηbc ωcd ηda = − ωbd ηda = − ω ab . (1.152)

Lowering index a we get

ωeb = ηea ωab = −ωbd ηda ηea = −ωbd δde = − ωbe (1.153)

That is, ωab is antisymmetric

ωab = − ωba. (1.154)

A representation of the Lorentz group is generated by matrices D(L) that

represent matrices L close to the identity matrix by sums over a, b = 0, 1, 2, 3

D(L) = 1 +i

2ωab J ab. (1.155)

The generators J ab must obey the commutation relations

i[J ab,J cd] = ηbc J ad − ηac J bd − ηda J cb + ηdb J ca. (1.156)

A remarkable representation of these commutation relations is provided

by matrices γa that obey the anticommutation relations

γa, γb = 2 ηab. (1.157)

One sets

J ab = − i

4[γa, γb] (1.158)

where η is the usual flat-space metric (1.138). Any four 4× 4 matrices that


satisfy these anticommutation relations form a set of Dirac gamma matrices.

They are not unique. Is S is any nonsingular 4×4 matrix, then the matrices

γ′a = S γa S−1 (1.159)

also are a set of Dirac’s gamma matrices.

Any set of matrices obeying the anticommutation relations (1.157) for

any n × n diagonal matrix η with entries that are ±1 is called a Clifford

algebra.

As a homework problem, show that

[J ab, γc] = − i γa ηbc + i γb ηac. (1.160)

One can use these commutation relations to derive the commutation rela-

tions (1.156) of the Lorentz group.

The gamma matrices are a vectors in the sense that for L near the identity

D(L) γcD−1(L) ≈ (I +i

2ωab J ab) γc (I − i

2ωab J ab)

= γc +i

2ωab [J ab, γc]

= γc +i

2ωab (−i γa ηbc + i γb ηac)

= γc +1

2ωab γ

a ηbc − 1

2ωab γ

b ηac

= γc − 1

2ηcb ωba γ

a − 1

2ηca ωab γ

b

= γc − 1

2ωca γ

a − 1

2ωcb γ

b

= γc − ωca γa

= γc + ω ca γa

= (δ ca + ω c

a ) γa

= L ca γa

(1.161)

in which we used (1.152) to write − ωca = ω ca . The finite ω form is

D(L) γaD−1(L) = L ac γc. (1.162)

The unit matrix is a scalar

D(L) I D−1(L) = I. (1.163)

The generators of the Lorenz group form an antisymmetric tensor

D(L)J abD−1(L) = L ac L b

d J cd. (1.164)

1.15 Dirac’s gamma matrices 27

Out of four gamma matrices, one can also make totally antisymmetric ten-

sors of rank-3 and rank-4

Aabc ≡ γ[a γb γc] and Babcd ≡ γ[a γb γc γd] (1.165)

where the brackets mean that one inserts appropriate minus signs so as to

achieve total antisymmetry. Since there are only four γ matrices in four

spacetime dimensions, any rank-5 totally antisymmetric tensor made from

them must vanish, Cabcde = 0.

Notation: The parity transformation is

β = iγ0. (1.166)

It flips the spatial gamma matrices but not the temporal one

β γi β−1 = − γi and β γ0 β−1 = γ0. (1.167)

It flips the generators of boosts but not those of rotations

β J i0 β−1 = − J i0 and β J ik β−1 = J ik. (1.168)

1.15 Dirac’s gamma matrices

Weinberg’s chosen set of Dirac matrices is

γ0 = − i(

0 1

1 0

)= −γ0† and γi = −i

(0 σi

− σi 0

)= γi† (1.169)

in which the σ’s are Pauli’s 2× 2 hermitian matrices

σ1 =

(0 1

1 0

), σ2 =

(0 −ii 0

), and σ3 =

(1 0

0 −1

), (1.170)

which are the gamma matrices of 3-dimensional spacetime. With this choice

of γ’s, the matrix β is

β = i γ0 =

(0 1

1 0

)= β†. (1.171)

In spacetimes of five dimensions, the fifth gamma matrix γ4 which tradi-

tionally is called γ5 = γ5 is

γ5 = γ5 = −iγ0γ1γ2γ3 =

(1 0

0 −1

). (1.172)

It anticommutes with all four Dirac gammas and its square is unity, as it

must if it is to be the fifth gamma in 5-space:

γa, γb = 2ηab (1.173)


for a, b = 0, 1, 2, 3, 4 with η44 = 1 and η40 = η04 = 0.

With Weinberg’s choice of γ’s, the Lorentz boosts are

J i0 = − i

4[γi, γ0] = − i

4

[−i(

0 σi

− σi 0

),−i

(0 1

1 0

)]=i

4

[(σi 0

0 −σi)−(−σi 0

0 σi

)]=i

2

(σi 0

0 −σi).

(1.174)

The Lorentz rotation matrices are

J ik = − i

4[γi, γk] = − i

4

[−i(

0 σi

− σi 0

),−i

(0 σk

−σk 0

)]=i

4

[(−σiσk 0

0 −σiσk)−(−σkσi 0

0 −σkσi)]

=−i4

([σi, σk] 0

0 [σi, σk]

)=−i4

(2iεikjσ

j 0

0 2iεikjσj

)=

1

2εikj

(σj 0

0 σj

).

(1.175)

The Dirac representation of the Lorentz group is reducible, as SW’s choice

of gamma matrices makes apparent. The Dirac rotation matrices are

Ji =1

2

(σi 0

0 σi

). (1.176)

Some useful relations are

β γa† β = − γa, β J ab† β = J ab and β D(L)† β = D(L)−1 (1.177)

as well as

β γ†5 β = − γ5 and β(γ5γa)† β = − γ5γ

a. (1.178)

1.16 Dirac fields

The positive- and negative-frequency parts of a Dirac field are

ψ+` (x) = (2π)−3/2

∑s

∫d3p u`(~p, s) e

ip·x a(~p, s)

ψ−` (x) = (2π)−3/2∑s

∫d3p v`(~p, s) e

−ip·x b†(~p, s).

(1.179)


The rotation conditions (1.39) are∑s

u¯(~0, s)(J(j)i )ss =

∑`

(Ji) ¯ u`(~0, s)∑s

v¯(~0, s)(−J (j)i )∗ss) =

∑`

(Ji) ¯ v`(~0, s).(1.180)

The Dirac rotation matrices (1.176) are

Ji =1

2

(σi 0

0 σi

)(1.181)

so we set the four values `, ¯= 1, 2, 3, 4 to ` = (m,±) with m = ±12 . And we

consider u`(s) to be u+m(s) stacked upon u−m(s) and similarly take v`(s) to

be v+m±(s) above v−m±(s) where u±m(s) and v±m±(s) are, a priori, 2× (2j+ 1)-

dimensional matrices with indexes m = ±1/2 and s = −j, . . . , j. That is,u1(s)

u2(s)

u3(s)

u4v

=

u+

+1/2(s)

u+−1/2(s)

u−+1/2(s)

u−−1/2(s)

and

v1(s)

v2(s)

v3(s)

v4(s)

=

v+

+1/2(s)

v+−1/2(s)

v−+1/2(s)

v−−1/2(s)

. (1.182)

We then have four equations∑s

u+m(~0, s)(J

(j)i )ss =

∑m

12 σ

immu

+m(~0, s)∑

s

u−m(~0, s)(J(j)i )ss =

∑m

12 σ

immu

−m(~0, s)∑

s

v+m(~0, s)(−Ji)∗(j)ss ) =

∑m

12 σ

immv

+m(~0, s)∑

s

v−m(~0, s)(−Ji)∗(j)ss ) =∑m

12 σ

immv

−m(~0, s).

(1.183)

SW defines the four 2× (2j + 1) matrices

U+ms = u+

m(~0, s) and U−ms = u−m(~0, s)

V +ms = v+

m(~0, s) and V −ms = v−m(~0, s).(1.184)

in terms of which the four Dirac rotation conditions (1.183) are

U+ J(j)i = 1

2 σi U+ and U− J

(j)i = 1

2 σi U−

V + (−J∗(j)i ) = 12 σi V

+ and V − (−J∗(j)i ) = 12 σi V

−.(1.185)


Taking the complex conjugate of the second of these equations, we get

−J (j)i = V +∗−1(1

2σi∗)V +∗ = V +∗−1(−1

2σ2 σi σ2)V +∗

−J (j)i = V −∗−1(1

2σi∗)V −∗ = V −∗−1(−1

2σ2 σi σ2)V −∗

(1.186)

or more simply

J(j)i = (σ2V

+∗)−1 12σ

i (σ2V+∗)

J(j)i = (σ2V

−∗)−1 12σ

i (σ2V−∗).

(1.187)

The 2 × 2 Pauli matrices ~σ and the (2j + 1) × (2j + 1) matrices ~J (j) both

generate irreducible representations of the rotation group. So by writing

U+ J(j)i J

(j)k = 1

2 σi U+ J

(j)k = = 1

2 σi12 σk U

+ (1.188)

and similar equations for U−, V +, V −, we see that

U+D(j)(~θ) = U+ e−iθ·~J(j)

= e−i~θ·~σ U+ = D(1/2)(~θ)U+

U−D(j)(~θ) = U− e−iθ·~J(j)

= e−i~θ·~σ U− = D(1/2)(~θ)U−

(1.189)

and similar equations for V ±.

σ2V+∗D(j)(~θ) = σ2V

+∗ e−iθ·~J(j)

= e−i~θ·~σ σ2V

+∗ = D(1/2)(~θ)σ2V+∗

σ2V−∗D(j)(~θ) = σ2V

−∗ e−iθ·~J(j)

= e−i~θ·~σ σ2V

−∗ = D(1/2)(~θ)σ2V−∗.

(1.190)

Now recall Schur’s lemma (section 10.7 of PM):

Part 1: If D1 and D2 are inequivalent, irreducible representations of a group G, andif D1(g)A = AD2(g) for some matrix A and for all g ∈ G, then the matrix A mustvanish, A = 0.

Part 2: If for a finite-dimensional, irreducible representation D(g) of a group G,we have D(g)A = AD(g) for some matrix A and for all g ∈ G, then A = cI. That is,any matrix that commutes with every element of a finite-dimensional, irreduciblerepresentation must be a multiple of the identity matrix.

Part 1 tells us that D(j)(~θ) and D(1/2)(~θ) must be equivalent. So j = 1/2

and 2j + 1 = 2. A Dirac field must represent particles of spin 1/2.

Part 2 then says that the matrices U± must be multiples of the 2 × 2

identity matrix

U+ = c+ I and U− = c− I (1.191)

and that the matrices σ2V±∗ must be multiples of the 2× 2 identity matrix

σ2V+∗ = d′+ I and σ2V

−∗ = d′− I (1.192)


or more simply

V + = − id+ σ2 and V − = − id− σ2. (1.193)

That is,

v+m(~0, s) = d+

(0 −1

1 0

)and v−m(~0, s) = d−

(0 −1

1 0

). (1.194)

Going back to ` = (m,±) by using the index code (1.182)u1(s)

u2(s)

u3(s)

u4(s)

=

u+

+1/2(s)

u+−1/2(s)

u−+1/2(s)

u−−1/2(s)

=

c+ δ1/2,s

c+ δ−1/2,s

c− δ1/2,s

c− δ−1/2,s

and

v1(s)

v2(s)

v3(s)

v4(s)

=

v+

+1/2(s)

v+−1/2(s)

v−1/2(s)

v−−1/2(s)

=

d+ δ−1/2,s

d+ δ1/2,s

d− δ−1/2,s

d− δ+1/2,s

.(1.195)

we have for the u’s

u+1/2(1/2) = c+ and u+

−1/2(1/2) = 0 (1.196)

u−1/2(1/2) = c− and u−−1/2(1/2) = 0 (1.197)

u+1/2(−1/2) = 0 and u+

−1/2(−1/2) = c+ (1.198)

u−1/2(−1/2) = 0 and u−−1/2(−1/2) = c− (1.199)

v+1/2(1/2) = 0 and v+

−1/2(1/2) = d+ (1.200)

v−1/2(1/2) = 0 and v−−1/2(1/2) = d− (1.201)

v+1/2(−1/2) = − d+ and v+

−1/2(−1/2) = 0 (1.202)

v−1/2(−1/2) = − d− and v−−1/2(−1/2) = 0 (1.203)

So

u(~0,m = 12) =

u+

1/2(1/2)

u+−1/2(1/2)

u−1/2(1/2)

u−−1/2(1/2)

=

c+

0

c−0

and u(~0,m = −12) =

u+

1/2(−1/2)

u+−1/2(−1/2)

u−1/2(−1/2)

u−−1/2(−1/2)

=

0

c+

0

c−

,

v(~0,m = 12) =

v+

1/2(1/2)

v+−1/2(1/2)

v−1/2(1/2)

v−−1/2(1/2)

=

0

d+

0

d−

and v(~0,m = −12) =

v+

1/2(−1/2)

v+−1/2(−1/2)

v−1/2(−1/2)

v−−1/2(−1/2)

= −

d+

0

d−0

.(1.204)


To put more constraints on c± and d±, we recall that under parity

Pa(~p, s)P−1 = η∗a a(−~p, s) and Pb†(~p, s)P−1 = ηb β†(−~p, s) (1.205)

and so

Pψ+` (x)P−1 = (2π)−3/2

∑s

∫d3p u`(~p, s) e

ip·x η∗a a(−~p, s)

= (2π)−3/2∑s

∫d3p u`(−~p, s) eip·Px η∗a a(~p, s)

Pψ−` (x)P−1 = (2π)−3/2∑s

∫d3p v`(~p, s) e

−ip·x ηb b†(−~p, s)

= (2π)−3/2∑s

∫d3p v`(−~p, s) e−ip·Px ηb b†(~p, s).

(1.206)

We recall the relations (1.177)

β γa† β = − γa, β J ab† β = J ab, and β D(L)† β = D(L)−1 (1.207)

and in particular, since J 0i† = − J 0i, the rule

βJ 0iβ = J 0i† = −J 0i. (1.208)

We also have the pseudounitarity relation

β D†(L)β = D−1(L). (1.209)

In general spinors at finite momentum are related to those at zero momen-

tum by

u¯(q, s) =

√m

q0

∑`

D ¯ (L(q))u`(~0, s)

v¯(q, s) =

√m

q0

∑`

D ¯ (L(q))v`(~0, s)

(1.210)

which for Dirac spinors is

u(p, s) =

√m

p0D(L(p))u(~0, s)

v(p, s) =

√m

p0D(L(p)) v(~0, s).

(1.211)

So now by using the boost rule (1.208) we have

u`(−~p, s) =√m/p0D(L(−~p))u(0, s) =

√m/p0D(L(~p))−1u(0, s) (1.212)

=√m/p0 β D(L(~p))β u(0, s) (1.213)


and

v`(−~p, s) =√m/p0D(L(−~p))v(0, s) =

√m/p0D(L(~p))−1v(0, s) (1.214)

=√m/p0 β D(L(~p))β v(0, s). (1.215)

So under parity

Pψ+(x)P−1 = (2π)−3/2∑s

∫d3p

√m/p0 β D(L(~p))β u(0, s) eip·Px η∗a a(~p, s)

Pψ−(x)P−1 = (2π)−3/2∑s

∫d3p

√m/p0 β D(L(~p))β v(0, s) e−ip·Px ηb b

†(~p, s).

(1.216)

So to have Pψ±` (x)P−1 ∝ ψ±` (x), we need

β u(0, s) = bu u(0, s) and β v(0, s) = bv u(0, s). (1.217)

We then get

Pψ+` (t, ~x)P−1 = bu β η

∗a ψ

+` (t,−~x) and Pψ−` (t, ~x)P−1 = bv β ηb ψ

−` (t,−~x).

(1.218)

Here since P2 = 1, these factors are just signs, b2u = b2v = 1. The eigenvalue

equations (1.217) tell us that c− = bu c+ and that d− = bv d+. So rescaling

the fields we get

u(~0,m = 12) =

1√2

1

0

bu0

and u(~0,m = −12) =

1√2

0

1

0

bu

,

v(~0,m = 12) =

1√2

0

1

0

bv

and v(~0,m = −12) =

−1√2

1

0

bv0

.(1.219)

If the annihilation and creation operators a(p, s) and a†(p, s) obey the rule

[a(p, s), a†(p′, s′)]∓ = δss′δ3(~p− ~p′) (1.220)

and if the field is the sum of the positive- and negative-frequency parts (2.55)

ψ+` (x) = (2π)−3/2

∑s

∫d3p u`(~p, s) e

ip·x a(~p, s)

ψ−` (x) = (2π)−3/2∑s

∫d3p v`(~p, s) e

−ip·x b†(~p, s)

(1.221)


with arbitrary coefficients κ and λ

ψ`(x) = κψ+` (x) + λψ−` (x) (1.222)

then

[ψ`(x), ψ†`′(y)]∓ = [κψ+` (x) + λψ−` (x), κ∗ψ+†

`′ (y) + λ∗ψ−†`′ (y)]

=

∫d3p

(2π)3

∑s

[|κ|2 u`(~p, s)u∗`′(~p, s)eip·(x−y)

∓|λ|2 v`(~p, s) v∗`′(~p, s)e−ip·(x−y)]

=

∫d3p

(2π)3

∑s

[|κ|2N``′(p) e

ip·(x−y) ∓ |λ|2N``′(p) e−ip·(x−y)

](1.223)

where

N``′(p) =∑s

u`(~p, s)u∗`′(~p, s)

M``′(p) =∑s

v`(~p, s) v∗`′(~p, s).

(1.224)

When ~p = 0, these matrices are

N``′(0) =∑s

u`(~0, s)u∗`′(~0, s)

N(0) =1

2

1

0

bu0

[1 0 bu 0]

+1

2

0

1

0

bu

[0 1 0 bu]

=1

2

1 0 bu 0

0 0 0 0

bu 0 1 0

0 0 0 0

+1

2

0 0 0 0

0 1 0 bu0 0 0 0

0 bu 0 1

=1 + bu β

2

(1.225)


and

M``′(0) =∑s

v`(~0, s) v∗`′(~0, s)

M(0) =1

2

0

1

0

bv

[0 1 0 bv]

+1

2

1

0

bv0

[1 0 bv 0]

=1

2

0 0 0 0

0 1 0 bv0 0 0 0

0 bv 0 1

+1

2

1 0 bv 0

0 0 0 0

bv 0 1 0

0 0 0 0

=1 + bv β

2.

(1.226)

So then using the boost relations (1.211) we find

N(~p) =∑s

u`(~p, s)u∗`′(~p, s) =

m

p0D(L(p))

∑s

u`(~0, s)u∗`′(~0, s)D

†(L(p))

=m

2p0D(L(p)) (1 + bu β)D†(L(p))

M(~p) =∑s

v`(~p, s) v∗`′(~p, s) =

m

p0D(L(p))

∑s

v`(~0, s) v∗`′(~0, s)D

†(L(p))

=m

2p0D(L(p)) (1 + bv β)D†(L(p)).

(1.227)

The pseudounitarity relation (1.209)

β D†(L)β = D−1(L). (1.228)

gives

β D†(L) = D−1(L)β (1.229)

which implies that

D(L)β D†(L) = β. (1.230)

The pseudounitarity relation also says that

D†(L) = β D−1(L)β (1.231)

so that

D(L)D†(L) = D(L)β D−1(L)β. (1.232)

Also since the gammas form a 4-vector (1.162)

D(L) γaD−1(L) = L ac γc (1.233)


and since β = iγ0, we have

D(L(p))β D−1(L(p)) = D(L(p)) iγ0D−1(L(p)) = iL 0c (p) γc = −iLc0γc.

(1.234)

Now

pa = Lab(p)kb = La0(p)m (1.235)

so

D(L(p))β D−1(L(p)) = −i pcγc/m (1.236)

which implies that

D(L)D†(L) = − i (pcγc/m)β. (1.237)

Thus

N(~p) =m

2p0D(L(p)) (1 + bu β)D†(L(p)) =

m

2p0[(−i pcγc/m)β + bu β]

=1

2p0[−i pcγc + bum] β (1.238)

and

M(~p) =m

2p0D(L(p)) (1 + bv β)D†(L(p)) =

m

2p0[(−i pcγc/m)β + bv β]

=1

2p0[−i pcγc + bvm] β. (1.239)

We now put the spin sums (1.238) and (1.239) in the (anti)commutator

(1.223) and get

[ψ`(x), ψ†`′(y)]∓ =

∫d3p

(2π)32p0

[|κ|2 [(− i pcγc + bum)β]``′ e

ip·(x−y)

∓ |λ|2 [(− i pcγc + bvm)β]``′ e−ip·(x−y)

]= |κ|2 [(− ∂cγc + bum)β]``′

∫d3p


∓ |λ|2 [(− ∂cγc + bvm)β]``′

∫d3p

(2π)32p0e−ip·(x−y)

= |κ|2 [(− ∂cγc + bum)β]``′ ∆+(x− y)

∓ |λ|2 [(− ∂cγc + bvm)β]``′ ∆+(y − x).

(1.240)

Recall that for (x− y)>0, i.e. spacelike, ∆+(x− y) = ∆+(y− x). So its first


derivatives are odd. So for x− y spacelike

[ψ`(x), ψ†`′(y)]∓ = |κ|2 [(− ∂cγc + bum)β]``′ ∆+(x− y)

∓ |λ|2 [(∂cγc + bvm)β]``′ ∆+(x− y)

=(|κ|2 ± |λ|2

)[(− ∂cγc)β]``′∆+(x− y)

+(|κ|2bu ∓ |λ|2 bv

)mβ``′ ∆+(x− y).

(1.241)

To get the first term to vanish, we need to choose the lower sign (that is,

use anticommutators) and set |κ| = |λ|. To get the second term to be zero,

we must set bu = − bv. We may adjust κ and and bu so that

κ = λ and bu = − bv = 1. (1.242)

In particular, a spin-one-half field must obey anticommutation relations

[ψ`(x), ψ†`′(y)]+ ≡ ψ`(x), ψ†`′(y) = 0 for (x− y)2 > 0. (1.243)

Finally then, the Dirac field is

ψ`(x) =∑s

∫d3p

(2π)3/2

[u`(~p, s) e

ip·x a(~p, s) + v`(~p, s) e−ip·x b†(~p, s)

].

(1.244)

The zero-momentum spinors are

u(~0,m = 12) =

1√2

1

0

1

0

and u(~0,m = −12) =

1√2

0

1

0

1

,

v(~0,m = 12) =

1√2

0

1

0

−1

and v(~0,m = −12) =

1√2

−1

0

1

0

.(1.245)

The spin sums are[N(~p)

]`m

=∑s

u`(~p, s)u∗m(~p, s) =

[ 1

2p0(−i pcγc +m) β

]`m[

M(~p)]`m

=∑s

v`(~p, s)v∗m(~p, s) =

[ 1

2p0(−i pcγc −m) β

]`m.

(1.246)

The Dirac anticommutator is

[ψ`(x), ψ†`′(y)]+ ≡ ψ`(x), ψ†`′(y) = [(− ∂cγc + m)β]``′ ∆+(x− y). (1.247)


Two standard abbreviations are

β ≡ iγ0 =

[0 1

1 0

]and ψ ≡ ψ†β = iψ†γ0 =

[ψ∗` ψ∗r

] [0 1

1 0

]=[ψ∗r ψ∗`

].

(1.248)

A Majorana fermion is represented by a field like

ψ`(x) =∑s

∫d3p

(2π)3/2

[u`(~p, s) e

ip·x a(~p, s) + v`(~p, s) e−ip·x a†(~p, s)

].

(1.249)

Since C = γ2β it follows that C−1 = βγ2 and so that J ∗ab = −βCJ abC−1β =

−βγ2βJ abβγ2β. But βγ2β = − ββγ2 = −i2γ20γ2 = −γ2. So J ∗ab = −

γ2J abγ2. Thus

D∗(L) = e−iωabJ∗ab

= e−iωab(−γ2J abγ2) = γ2eiωabJ abγ2 = γ2D(L) γ2.

(1.250)

Now with SW’s γ’s,

γ2 u(~0,±12) = v(~0,±1

2) and γ2 v(~0,±12) = u(~0,±1

2). (1.251)

Thus the hermitian conjugate of a Majorana field is

ψ∗(x) =∑s

∫d3p

(2π)3/2

[u∗(~p, s) e−ip·x a†(~p, s) + v∗(~p, s) eip·x a(~p, s)

]=∑s

∫d3p

(2π)3/2

[v∗(~p, s) eip·x a(~p, s) + u∗(~p, s) e−ip·x a†(~p, s)

]=∑s

∫d3p

(2π)3/2

[D∗(L(p)) v∗(~0, s) eip·x a(~p, s) +D∗(L(p))u∗(~0, s) e−ip·x a†(~p, s)

]= γ2

∑s

∫d3p

(2π)3/2

[D(L(p))γ2v(~0, s) eip·x a(~p, s) +D(L(p))γ2u(~0, s) e−ip·x a†(~p, s)

]= γ2

∑s

∫d3p

(2π)3/2

[D(L(p))u(~0, s) eip·x a(~p, s) +D(L(p))v(~0, s) e−ip·x a†(~p, s)

]= γ2

∑s

∫d3p

(2π)3/2

[u(~p, s) eip·x a(~p, s) + v(~p, s) e−ip·x a†(~p, s)

]= γ2 ψ(x).

(1.252)

This works because the spinors obey the Majorana conditions

u(p, s) = γ2 v∗(p, s) and v(p, s) = γ2 u∗(p, s). (1.253)


The parity rules (1.254) now are

Pψ+` (t, ~x)P−1 = β η∗a ψ

+` (t,−~x) and Pψ−` (t, ~x)P−1 = − β ηb ψ−` (t,−~x).

(1.254)

So to have a Dirac field survive a parity transformation, we need the phase

of the particle to be minus the complex conjugate of the phase of the an-

tiparticle

η∗a = − ηb or ηb = − η∗a. (1.255)

So the intrinsic parity of a particle-antiparticle state is odd. So negative-

parity bosons like π0, ρ0, J/ψ can be interpreted as s-wave bound states of

quark-antiquark pairs. Under parity a Dirac field goes as

Pψ(t, ~x)P−1 = η∗ β ψ(t,−~x). (1.256)

If a Dirac particle is the same as its antiparticle, then its intrinsic parity

must be odd under complex conjugation, η = −η∗. So the intrinsic parity of

a Majorana fermion must be imaginary

η = ± i. (1.257)

But this means that if we express a Dirac field ψ as a complex linear com-

bination

ψ =1√2

(χ1 + iχ2) (1.258)

of two Majorana fields with intrinsic parities η∗1 = ±i and η∗2 = ±i, then

under parity

Pψ(t, ~x)P−1 =1√2

(η∗1 β φ1(t,−~x) + iη∗2 β φ2(t,−~x)

)(1.259)

so we need η∗1 = η∗2 to have

Pψ(t, ~x)P−1 =1√2

(η∗1 β φ1(t,−~x) + iη∗2 β φ2(t,−~x)

)= η∗ β ψ(t,−~x).

(1.260)

But in that case the Dirac field has intrinsic parity η = ±i.The equation (1.236) that shows how beta goes under D(L(p))

D(L(p))β D−1(L(p)) = − i pcγc/m (1.261)

tells us that the spinors (1.211)

u(p, s) =

√m

p0D(L(p))u(~0, s) and v(p, s) =

√m

p0D(L(p)) v(~0, s)

(1.262)


are eigenstates of −i pcγc/m with eigenvalues ±1

(− i pcγc/m)u(p, s) = D(L(p))β D−1(L(p))

√m

p0D(L(p))u(~0, s)

=

√m

p0D(L(p))β u(~0, s) =

√m

p0D(L(p)) u(~0, s) = u(p, s)

(− i pcγc/m) v(p, s) = D(L(p))β D−1(L(p))

√m

p0D(L(p)) v(~0, s)

=

√m

p0D(L(p))β v(~0, s) = −

√m

p0D(L(p)) v(~0, s) = − v(p, s).

(1.263)

So the spinors u and v obey Dirac’s equation in momentum space

(i pcγc +m)u(p, s) = 0 and (−i pcγc +m)v(p, s) = 0 (1.264)

which implies that a Dirac field obeys Dirac’s equation

(γa ∂a +m)ψ(x) = (γa ∂a +m)∑s

∫d3p

(2π)3/2

[u(~p, s) eip·x a(~p, s) + v(~p, s) e−ip·x b†(~p, s)

]=∑s

∫d3p

(2π)3/2

[(γa ∂a +m)u(~p, s) eip·x a(~p, s)

+ (γa ∂a +m)v(~p, s) e−ip·x b†(~p, s)]

=∑s

∫d3p

(2π)3/2

[(iγa pa +m)u(~p, s) eip·x a(~p, s)

+ (−iγa pa +m)v(~p, s) e−ip·x b†(~p, s)]

= 0.

(1.265)

The Majorana conditions (1.253) imply that

u∗(p, s) = − βCv(p, s) and v∗(p, s) = − βCu(p, s). (1.266)

So for a Dirac field to survive charge conjugation, the particle-antiparticle

phases must be related

ξb = ξ∗a. (1.267)

Then under charge conjugation a Dirac field goes as

Cψ(x)C−1 = − ξ∗ βC ψ∗(x). (1.268)

If a Dirac particle is the same as its antiparticle, then ξ must be real (and


η imaginary), ξ = ±1, and must satisfy the reality condition

ψ(x) = − β C ψ∗(x). (1.269)

Suppose a particle and its antiparticle form a bound state

|Φ〉 =∑ss′

∫d3p d3p′χ(p, s; p′, s′) a†(p, s) b†(p′s′) |0〉. (1.270)

Under charge conjugation

C |Φ〉 = ξaξb∑ss′

∫d3p d3p′χ(p, s; p′, s′) b†(p, s) a†(p′s′) |0〉

= − ξaξb∑ss′

∫d3p d3p′χ(p, s; p′, s′) a†(p′s′) b†(p, s) |0〉

= − ξaξb∑ss′

∫d3p d3p′χ(p′, s′; p, s) a†(p, s) b†(p′, s′) |0〉

= − ξaξb |Φ〉 = −ξaξ∗a |Φ〉 = −|Φ〉.

(1.271)

The intrinsic charge-conjugation parity of a bound state of a particle and

its antiparticle is odd.

1.17 Expansion of massive and massless Dirac fields

A Dirac field is a sum over s = ±1/2 of

ψa(x) =∑s

∫d3p

(2π)3/2ua(~p, s) e

ip·x a(~p, s) + va(~p, s) e−ip·x b†(~p, s) (1.272)

in which

u(p, s) =m− i/p√

2p0(p0 +m)u(0, s) and v(p, s) =

m+ i/p√2p0(p0 +m)

v(0, s)

(1.273)

and the zero-momentum spinors (1.245) are

u(0, s = 12) =

1√2

1

0

1

0

and u(0, s = −12) =

1√2

0

1

0

1

,

v(0, s = 12) =

1√2

0

1

0

−1

and v(0, s = −12) =

1√2

−1

0

1

0

.(1.274)


These static spinors and the matrix formulas (1.273) give explicit expressions

for the spinors at arbitrary momentum p:

u(p, 12) =

1

2√p0(p0 +m)

m+ p0 − p3

−p1 − ip2

m+ p0 + p3

p1 + ip2

(1.275)

u(p,−12) =

1

2√p0(p0 +m)

−p1 + ip2

m+ p0 + p3

p1 − ip2

m+ p0 − p3

(1.276)

v(p, 12) =

1

2√p0(p0 +m)

−p1 + ip2

m+ p0 + p3

−p1 + ip2

−m− p0 + p3

(1.277)

v(p,−12) =

1

2√p0(p0 +m)

−m− p0 + p3

p1 + ip2

m+ p0 + p3

p1 + ip2

. (1.278)

In the m→ 0 limit with p = p0z in the 3-direction, they are

u(p0z, 12) =

0

0

1

0

and u(p0z,−12) =

0

1

0

0

(1.279)

v(p0z, 12) =

0

1

0

0

and v(p0z,−12) =

0

0

1

0

. (1.280)

They tell us that the upper two components which are those of the left-

handed Dirac field represent particles of negative helicity u(p0z,−12) and

antiparticles of positive helicity v(p0z, 12). They also say that the lower two

components which are those of the right-handed field represent particles of

positive helicity u(p0z, 12) and antiparticles of negative helicity v(p0z,−1

2).


In the m→ 0 limit, the spinors are for the particles

u(p, 12) =

1

2p0

0

0

p0 + p3

p1 + ip2

and u(p,−12) =

1

2p0

−p1 + ip2

p0 + p3

0

0

(1.281)

and for the antiparticles

v(p, 12) =

1

2p0

−p1 + ip2

p0 + p3

0

0

and v(p,−12) =

1

2p0

0

0

p0 + p3

p1 + ip2

. (1.282)

The massless Dirac field is

ψa(x) =

12∑

s=−12

∫d3p

(2π)3/2

[ua(p, s)e

ip·x a(p, s) + va(p, s) e−ip·x b†(p, s)

].

(1.283)

The left-handed field is

ψà(x) =

∫d3p

(2π)3/2

[uà(p,−1

2)eip·x a(p,−12) + và(p,

12) e−ip·x b†(p, 1

2)]

=

∫d3p

(2π)3/22p0

[(−p1 + ip2

p0 + p3

)eip·x a(p,−1

2)

+

(−p1 + ip2

p0 + p3

)e−ip·x b†(p, 1

2)].

(1.284)

The right-handed field is

ψra(x) =

∫d3p

(2π)3/2

[ura(p,

12)eip·x a(p, 1

2) + vra(p,−12) e−ip·x b†(p,−1

2)]

=

∫d3p

(2π)3/22p0

[( p0 + p3

p1 + ip2

)eip·x a(p,+1

2)

+

(p0 + p3

p1 + ip2

)e−ip·x b†(p,−1

2)].

(1.285)

The massless left-handed field and the massless right-handed field create

and absorb different particles. These are different fields. That is why the

unbroken gauge group SU(2)` can act on ψ` without affecting ψr.


1.18 Spinors and angular momentum

Under a Lorentz transformation a Dirac field goes as

U(L)ψa(x)U−1(L) =∑b

Dab(L−1)ψb(Lx) (1.286)

whereD(L) is the (12 , 0)⊕(0, 1

2) representation of the Lorentz group. If L = R

is an active rotation of |~θ| radians about the vector ~θ, then the matrix is

D(R−1) =

(ei~θ·~σ/2 0

0 ei~θ·~σ/2

). (1.287)

In particular, if ~θ = θz, then

D(R−1) =

(eiθσ3/2 0

0 eiθσ3/2

)=

eiθ/2 0 0 0

0 e−iθ/2 0 0

0 0 eiθ/2 0

0 0 0 e−iθ/2

. (1.288)

So since the field is

ψa(x) =∑s

∫d3p

(2π)3/2ua(~p, s) e

ip·x a(~p, s) + va(~p, s) e−ip·x a†c(~p, s),

(1.289)

we need

U(R)ψa(x)U−1(R) = D(R−1)abψb(Rx)

=∑s

∫d3p

(2π)3/2

[ua(~p, s)e

ip·xU(R)a(~p, s)U−1(R)

+ va(~p, s)e−ip·xU(R)a†c(~p, s)U

−1(R)]. (1.290)

A state of momentum ~p, kind n, and spin s = ±1/2 in the z direction

transforms under a right-handed rotation R(θ, z) of θ radians about the z

axis as

U(R(θ, z))|pz, s, n〉 = e−iθJz |pz, s, n〉 = e−iθs|pz, s, n〉. (1.291)

The state |pz, s, n〉 is made from the vacuum by the creation operator for a

particle of momentum ~p, kind n, and spin s = ±1/2 in the z direction

|pz, s, n〉 = a†(pz, s, n)|0〉. (1.292)

The creation operator transforms as

U(R(z, θ))a†(pz, s, n)U(R(z, θ))−1 = e−iθsa†(pz, s, n). (1.293)


The antiparticle creation operator goes as

U(R)a†c(~p, s)U−1(R) =

∑s′

(e−i

~θ·~σ/2)ss′a†c(R~p, s

′), (1.294)

and so the particle annihilation operator goes as

U(R)a(~p, s)U−1(R) =∑s′

(e−i

~θ·~σ/2)∗ss′a(R~p, s′) =

∑s′

(ei~θ·~σ/2

)s′sa(R~p, s′).

(1.295)

Our requirement (1.290) is

U(R)ψa(x)U−1(R) = D(R−1)abψb(Rx)

=∑s

∫d3p

(2π)3/2

[ua(~p, s)e

ip·x∑s′

(ei~θ·~σ/2

)s′sa(R~p, s′)

+ va(~p, s)e−ip·x

∑s′

(e−i

~θ·~σ/2)ss′a†c(R~p, s

′)]. (1.296)

But

D(R−1)abψb(Rx) =

(ei~θ·~σ/2 0

0 ei~θ·~σ/2

)ab

ψb(Rx)

=

(ei~θ·~σ/2 0

0 ei~θ·~σ/2

)ab

∑s

∫d3p

(2π)3/2

[ub(~p, s)e

ip·Rxa(~p, s)

+ vb(~p, s)e−ip·Rxa†c(~p, s)

]. (1.297)

Equivalently since Rp ·Rx = p · x,

D(R−1)abψb(Rx) =

(ei~θ·~σ/2 0

0 ei~θ·~σ/2

)ab

ψb(Rx)

=

(ei~θ·~σ/2 0

0 ei~θ·~σ/2

)ab

∑s

∫d3p

(2π)3/2

[ub(~Rp, s)e

ip·xa(R~p, s)

+ vb(~Rp, s)e−ip·xa†c(R~p, s)

]. (1.298)

Let’s focus on the zero-momentum terms. We need∑s′

(ei~θ·~σ/2

)ss′ua(~0, s

′) =

(ei~θ·~σ/2 0

0 ei~θ·~σ/2

)ab

ub(~0, s) (1.299)

and ∑s′

(e−i

~θ·~σ/2)s′sva(~0, s

′) =

(ei~θ·~σ/2 0

0 ei~θ·~σ/2

)ab

vb(~0, s). (1.300)


The spinors

u(~0, 12) =

1√2

1

0

1

0

and u(~0,−12) =

1√2

0

1

0

1

(1.301)

satisfy the condition (1.299) on the spinor ua(~0, s). Let’s look at the condition

(1.300) on va(~0, s) for the simpler case of ~θ = θz. We set s = 12 or equivalently

s = + and also a = 1. Since the rotation is about z,(e−i

~θ·~σ/2)s′s

=(e−iθσ3/2

)s′s

= e−iθsδs′s. (1.302)

So the condition (1.300) says that

e−iθ/2v1(~0, 12) = eiθ/2v1(~0, 1

2). (1.303)

So v1(~0, 12) = 0. Now we set a = 2. Then the condition says that

e−iθ/2v2(~0, 12) = e−iθ/2v2(~0, 1

2). (1.304)

So v2(~0, 12) can be anything. Now we set a = 3. Then the same condition

(1.300) says that

e−iθ/2v3(~0, 12) = eiθ/2v3(~0, 1

2) (1.305)

which implies that v3(~0, 12) = 0. Finally, we set a = 4. Now the same condi-

tion (1.300) says that

e−iθ/2v4(~0, 12) = e−iθ/2v4(~0, 1

2). (1.306)

So v4(~0, 12) can be anything. That is, v(~0, 1

2) must be of the form

v(~0, 12) =

0

α

0

β

(1.307)

in which α and β are arbitrary.

Now we set s = −12 or equivalently s = − and also a = 1. Then the

condition (1.300) says that

eiθ/2v1(~0,−12) = eiθ/2v1(~0,−1

2). (1.308)

So v1(~0,−12) can be anything. But the condition (1.300) for a = 2 says

eiθ/2v2(~0,−12) = e−iθ/2v2(~0,−1

2) (1.309)


which means that v2(~0,−12) = 0. For a = 3, we get

eiθ/2v3(~0,−12) = eiθ/2v3(~0,−1

2), (1.310)

so v3(~0,−12) is unrestricted. Finally, for a = 4, we have

eiθ/2v4(~0,−12) = e−iθ/2v4(~0,−1

2) (1.311)

which makes v4(~0,−12) vanish. So

v(~0,−12) =

γ

0

δ

0

(1.312)

in which γ and δ are arbitrary. To make the v’s suitably normalized and

orthogonal to the u’s, one may set

v(~0, 12) =

1√2

0

1

0

−1

and v(~0,−12) =

1√2

−1

0

1

0

. (1.313)

The state a†(0, s, nc)|0〉 is (summed over a and s′)

1

(2π)3/2

∫d3x va(0, s)ψa(x)|0〉 = va(0, s)

∫d3xd3p

(2π)3v(~p, s′) e−ip·x a†(~p, s′, nc)|0〉

= va(0, s)

∫d3p δ3(p)v(~p, s′) a†(~p, s′, nc)|0〉

= va(0, s)v(0, s′) a†(0, s′, nc)|0〉 (1.314)

= δss′ a†(0, s′, nc)|0〉 = a†(0, s, nc)|0〉.

So

|0, s, nc〉 =1

(2π)3/2

∑a

∫d3x va(0, s)ψa(x)|0〉. (1.315)

To determine its spin, we act on it with the operator e−iθJ3 that rotates

states about the z axis by angle θ in a right-handed way.

e−iθJ3 |0, s, nc〉 = e−iθJ31

(2π)3/2

∑a

∫d3x va(0, s)ψa(x)|0〉 (1.316)

=1

(2π)3/2

∑a

∫d3x va(0, s)e

−iθJ3ψa(x)eiθJ3e−iθJ3 |0〉.


Since the vacuum is invariant, this is

e−iθJ3 |0, s, nc〉 =1

(2π)3/2

∑a

∫d3x va(0, s)e

−iθJ3ψa(x)eiθJ3 |0〉

=1

(2π)3/2

∑a

∫d3x va(0, s)D(R−1)abψb(Rx)|0〉. (1.317)

Since the jacobian of a rotation is unity, we have

e−iθJ3 |0, s, nc〉 =1

(2π)3/2

∑a

∫d3x va(0, s)D(R−1)abψb(x)|0〉 (1.318)

in which

D(R−1)ab =

eiθ/2 0 0 0

0 e−iθ/2 0 0

0 0 eiθ/2 0

0 0 0 e−iθ/2

. (1.319)

So this is

e−iθJ3 |0, s, nc〉 =1

(2π)3/2

∑a

∫d3x va(0, s)e

iσ(a)θ/2ψa(x)|0〉 (1.320)

in which σ(a) = 1 for a = 1 & 3, and σ(a) = −1 for a = 2 & 4. For s = ±1/2,

the spinors v(0, s) are

v(~0, 12) =

1√2

0

1

0

−1

and v(~0,−12) =

1√2

−1

0

1

0

. (1.321)

So for s = 1/2, slots a = 2 & 4 are nonzero, and

e−iθJ3 |0, 1/2, nc〉 = e−iθ/21

(2π)3/2

∑a

∫d3x va(0, s)ψa(x)|0〉 (1.322)

which means that the state has spin 1/2 in the z direction. For s = −1/2,

slots a = 1 & 3 are nonzero, and

e−iθJ3 |0, 1/2, nc〉 = eiθ/21

(2π)3/2

∑a

∫d3x va(0, s)ψa(x)|0〉 (1.323)

which means that the state has spin −1/2 in the z direction.

1.19 Charge conjugation 49

1.19 Charge conjugation

Under charge conjugation, a massive Dirac field

ψ(x) =∑s

∫d3p

(2π)3/2u(~p, s) eip·x a(~p, s) + v(~p, s) e−ip·x a†c(~p, s) (1.324)

transforms as

Cψ(x)C−1 = −ξ∗βγ2βψ∗(x) = ξ∗γ2ψ

∗(x) (1.325)

in which ξ is the charge-conjugation parity of the particle annihilated by a

(which is the complex conjugate of that of the particle annihilated by ac)

and

βγ2β =

(0 1

1 0

)(−i)

(0 σ2

−σ2 0

)(0 1

1 0

)= i

(0 σ2

−σ2 0

)= −γ2. (1.326)

Under charge conjugation, the creation operators for particles and their

antiparticles, transform as

Ca†(~p, s)C−1 = ξa†c(~p, s) and Ca†c(~p, s)C−1 = ξca

†(~p, s). (1.327)

So

Ca(~p, s)C−1 = ξ∗ac(~p, s) and Cac(~p, s)C−1 = ξ∗ca(~p, s). (1.328)

So, suppressing explicit mention of the sum over spins, we have

Cψ(x)C−1 =

∫d3p

(2π)3/2u(~p, s) eip·x Ca(~p, s)C−1 + v(~p, s) e−ip·x Ca†c(~p, s)C

−1

=

∫d3p

(2π)3/2u(~p, s) eip·x ξ∗ac(~p, s) + v(~p, s) e−ip·x ξca

†(~p, s)C−1.

Under charge conjugation, the field changes to

Cψ(x)C−1 = − iξ∗(

0 σ2

−σ2 0

)ψ∗(x) = ξ∗γ2ψ

∗(x). (1.329)

Explicitly and with ξc = ξ∗, we get

Cψ(x)C−1 = − iξ∗(

0 σ2

−σ2 0

)∑s

∫d3p

(2π)3/2

[u∗(~p, s) e−ip·x a†(~p, s)

+ v∗(~p, s) eip·x ac(~p, s)]

(1.330)


So leaving aside the factor −iξ∗, the spinors for the ac particles (aka, an-

tiparticles) are

ub(~p, s) =

(0 σ2

−σ2 0

)v∗(~p, s) (1.331)

while the antispinors for the a particles are

va(~p, s) =

(0 σ2

−σ2 0

)u∗(~p, s). (1.332)

Let’s focus on the zero-momentum spinors. The zero-momentum spinors are

u(~0, 12) =

1√2

1

0

1

0

and u(~0,−12) =

1√2

0

1

0

1

,

v(~0, 12) =

1√2

0

1

0

−1

and v(~0,−12) =

1√2

−1

0

1

0

.(1.333)

The spinor ub(0,+) is

ub(~0,+) =

(0 σ2

−σ2 0

)v∗(~0,+) (1.334)

or in all its components,

ub(~0,+) =

0 0 0 −i0 0 i 0

0 i 0 0

−i 0 0 0

1√2

0

1

0

−1

=i√2

1

0

1

0

= i u(~0,+) (1.335)

which is what we expect. Similarly,

ub(~0,−) =

(0 σ2

−σ2 0

)v∗(~0,−) (1.336)


ub(~0,−) =

0 0 0 −i0 0 i 0

0 i 0 0

−i 0 0 0

1√2

−1

0

1

0

=i√2

0

1

0

1

= i u(~0,−) (1.337)

1.20 Parity 51

which is what we expect. Similarly, the spinor va(0,+) is

va(~0,+) =

(0 σ2

−σ2 0

)u∗(~0,+) (1.338)


va(~0,+) =

0 0 0 −i0 0 i 0

0 i 0 0

−i 0 0 0

1√2

1

0

1

0

=i√2

0

1

0

−1

= i v(~0,+) (1.339)

which is what we expect. Finally, the spinor va(0,−) is

va(~0,−) =

(0 σ2

−σ2 0

)u∗(~0,−) (1.340)


va(~0,−) =

0 0 0 −i0 0 i 0

0 i 0 0

−i 0 0 0

1√2

0

1

0

1

=i√2

−1

0

1

0

= i v(~0,−) (1.341)

which is what we expect.

1.20 Parity

Under parity the field ψ(x) (1.324) changes to

Pψ(x)P−1 = η∗βψ(Px) (1.342)

where η is the parity of the particle annihilated by a, which is minus the

complex conjugate of that of the particle annihilated by b. The annihilation

and creation operators transform as

Pa†(~p, s)P−1 = ηa†(−~p, s) and Pa†c(~p, s)P−1 = ηca

†c(−~p, s) (1.343)

and

Pa(~p, s)P−1 = η∗a(−~p, s) and Pa(~p, s)P−1 = η∗ca(−~p, s). (1.344)

So

Pψ(x)P−1 =

∫d3p

(2π)3/2u(~p, s) eip·x Pa(~p, s)P−1 + v(~p, s) e−ip·x Pa†c(~p, s)P

−1

=

∫d3p

(2π)3/2u(~p, s) eip·x η∗a(−~p, s) + v(~p, s) e−ip·x ηca

†c(−~p, s)


But ηc = −η∗, so

Pψ(x)P−1 = η∗∫

d3p

(2π)3/2u(~p, s) eip·x a(−~p, s)− v(~p, s) e−ip·x a†c(−~p, s).

Now

Pψ(x)P−1 = η∗βψ(Px),

that is,

Pψ(x)P−1 = η∗β

∫d3p

(2π)3/2u(~p, s) eiPp·x a(~p, s) + v(~p, s) e−iPp·x a†c(~p, s)

= η∗β

∫d3p

(2π)3/2u(−~p, s)eip·xa(−~p, s) + v(−~p, s)e−ip·xa†c(−~p, s)

So we need

βu(−~p, s) = u(~p, s) and βv(−~p, s) = − v(~p, s). (1.345)

These rules are easy to check for ~p = 0

βu(~0, s) = u(~0, s) and βv(~0, s) = − v(~0, s) (1.346)

which the spinors (1.333) satisfy.

2

Feynman diagrams

2.1 Time-dependent perturbation theory

Most physics problems are insoluble, and we must approximate the un-

known solution by numerical or analytic methods. The most common ana-

lytic method is called perturbation theory and is based on an assumption

that something is small compared to something simple that we can analyze.

In time-dependent perturbation theory, we write the hamiltonian H as

the sum H = H0 + V of a simple hamiltonian H0 and a small, complicated

part V which we give the time dependence induced by H0

V (t) = eiH0t/~ V e−iH0t/~. (2.1)

We alter the time dependence of states by the same simple exponential (and

set ~ = 1)

|ψ, t〉 = eiH0te−iHt|ψ〉 (2.2)

and find as its time derivative

id

dt|ψ, t〉 = eiH0t(H −H0)e−iHt|ψ〉 = eiH0t V e−iH0t eiH0te−iHt|ψ〉

= V (t) |ψ, t〉.(2.3)

We can formally solve this differential equation by time-ordering factors of

V (t) in expansion of the exponential

|ψ, t〉 = T[e−i

∫ t0 V (t′) dt′

]|ψ〉

=

1− i∫ t

0V (t′) dt′ − 1

2!

∫T[V (t′)V (t′′)

]dt′dt′′ + . . .

|ψ〉

(2.4)

so that V ’s of later times occur to the left of V ’s of earier times, that is, if

t > t′ then T [V (t)V (t′)] = V (t)V (t′) and so forth.

54 Feynman diagrams

2.2 Dyson’s expansion of the S matrix

The time-evolution operator in the interaction picture is the time-ordered

exponential of the integral over time of the interaction hamiltonian

T[e−i

∫V (t) dt

]. (2.5)

Time ordering has V (t>) to the left of V (t<) if the time t> is later than the

time V (t<). The interaction hamiltonian is

V (t) =

∫H(t, x) d3x. (2.6)

The density of the interaction hamiltonian is (or is taken to be) a sum of

terms

H(t, x) =∑i

giHi(t, x) (2.7)

each of which is a monomial in the fields ψ`(x) and their adjoints ψ†`(x). A

generic field is an integral over momentum

ψ`(x) =∑s

∫d3p

(2π)3/2

[u`(p, s, n) a(p, s, n) eip·x+v`(p, s, n) b†(p, s, n) e−ip·x

](2.8)

in which n labels the kind of field.

The elements of the S matrix are amplitudes for an initial state |p1, s1, n1; . . . ; pk, sk, nk〉to evolve into a final state |p′1, s′1, n′1; . . . ; p′k′ , s

′k′ , n

′k′〉. The case considered

most often is when k = 2 incoming particles and k′ = 2 or 3 outgoing par-

ticles, but in high-energy collisions, k′ can be much larger than 2 or 3. An

S-matrix amplitude is

Sp′1,s′1,n′1;...;p1,s1,n1;... = 〈p′1, s′1, n′1; . . . ; p′k′ , s′k′ , n

′k′ |T

[e−i

∫H(x) d4x

]|p1, s1, n1; . . . ; pk, sk, nk〉

= 〈0|a(pk′ , sk′ , nk′ ; . . . ; a(p1, s1, n1) (2.9)

×∞∑N=0

(−i)n

N !

∫d4x1 . . . d

4xNT[H(x1) . . . H(xN )

]× a†(p1, s1, n1; . . . ; a†(pk, sk, nk)|0〉

in which |0〉 is the vacuum state. It is the mean value in the vacuum state

of an infinite polynomial in creation and annihilation operators.

To make sense of it, the first step is to normally order the time-evolution

operator by moving all annihilation operators to the right of all creation

operators. We assume for now that H(x) itself is already normally ordered.

2.2 Dyson’s expansion of the S matrix 55

To do that we use these commutation relations with plus signs for bosons

and minus signs for fermions:

a(p, s, n) a†(p′, s′, n′) = ± a†(p′, s′, n′) a(p, s, n) + δ3(~p− ~p′) δss′ δnn′a(p, s, n) a(p′, s′, n′) = ± a(p′, s′, n′) a(p, s, n) (2.10)

a†(p, s, n) a†(p′, s′, n′) = ± a†(p′, s′, n′) a†(p, s, n).

Since

a(p, s, n) |0〉 = 0 and 〈0| a†(p, s, n) = 0, (2.11)

the S-matrix amplitude is just the left-over pieces proportional to products

of delta functions multiplied by u’s and v’s and Fourier factors all divided

by 2π’s.

Each such piece arises from the pairing of one annihilation operator with

one creation operator. Such pairs arise in six ways:

1. A particle p′, s′, n′ in the final state can pair with a field adjoint ψ†`(x) in

H(x) and yield the factor

[a(p′, s′, n′), ψ†`(x)]∓ = u∗` (p′, s′, n′)e−ip

′·x(2π)−3/2. (2.12)

2. An antiparticle p′, s′, n′ in the final state can pair with a field ψ`(x) in

H(x) and yield the factor

[b(p′, s′, n′), ψ`(x)]∓ = v`(p′, s′, n′)e−ip

′·x(2π)−3/2. (2.13)

3. A particle p, s, n in the initial state can pair with a field ψ`(x) in H(x)

and yield the factor

[ψ`(x), a†(p, s, n)]∓ = u`(p, s, n)eip·x(2π)−3/2. (2.14)

4. An antiparticle p, s, n in the initial state can pair with a field adjoint

ψ†`(x) in H(x) and yield the factor

[ψ†`(x), b†(p, s, n)]∓ = v∗` (p, s, n)eip·x(2π)−3/2. (2.15)

5. A particle (or antiparticle) p′, s′, n′ in the final state can pair with a

particle (or antiparticle) p, s, n in the initial state and yield

[a(p′, s′, n′), a†(p, s, n)]∓ = δ3(~p− ~p′) δss′ δnn′ . (2.16)

6. A field ψ`(x) = ψ+` (x) + ψ−` (x) in H(x) can pair with a field adjoint in

H ′(y), ψ†m(y) = ψ+†m (y) +ψ−†m (y). But for ψ+

` (x) to cross ψ+†m (y) the time

x0 must be later than y0, and for ψ−†m (y) to cross ψ−` (x), the time y0 must

56 Feynman diagrams

be later than x0. If for instance H(x) = ψ†(x)ψ(x)φ(x), then in Dyson’s

expansion these terms would appear

θ(x0−y0)ψ†(x)ψ(x)φ(x)ψ†(y)ψ(y)φ(y)+θ(y0−x0)ψ†(y)ψ(y)φ(y)ψ†(x)ψ(x)φ(x).

(2.17)

The resulting pairings are the propagator

θ(x0 − y0) [ψ+` (x), ψ+†

m (y)]∓ ± θ(y0 − x0) [ψ−†m (y), ψ−` (x)]∓ ≡ −i∆`m(x, y)

(2.18)

in which the ± signs will be explained later.

One then integrates over N spacetimes and the implicit momenta. The

result is defined by a set of rules and Feynman diagrams. In general, the

1/N ! in Dyson’s expansion is cancelled by the N ! ways of labelling the xi’s.

For example, in the term

1

2!

∫d4x1d

4x2T[H(x1)H(x2)

](2.19)

one can have H(x1) absorb an incoming electron of momentum p and have

H(x2) absorb an incoming electron of momentum p′ or the reverse.

But some processes require special combinatorics. So people often write

H(x) =g

3!φ3(x) or H(x) =

g

4!φ4(x) (2.20)

to compensate for multiple possible pairings. But these factorials don’t al-

ways cancel. Fermions introduce minus signs. The surest way to check the

signs and factorials in each process until one has gained sufficient experience.

2.3 The Feynman propagator for scalar fields

Adding ±iε to the denominator of a pole term of an integral formula for a

function f(x) can slightly shift the pole into the upper or lower half plane,

causing the pole to contribute if a ghost contour goes around the upper half-

plane or the lower half-plane. Such an iε can impose a boundary condition

on a Green’s function.

The Feynman propagator ∆F (x) is a Green’s function for the Klein-

Gordon differential operator (Weinberg, 1995, pp. 274–280)

(m2 −2)∆F (x) = δ4(x) (2.21)

in which x = (x0,x) and

2 = 4− ∂2

∂t2= 4− ∂2

∂(x0)2(2.22)


is the four-dimensional version of the laplacian 4 ≡ ∇·∇. Here δ4(x) is the

four-dimensional Dirac delta function

δ4(x) =

∫d4q

(2π)4exp[i(q · x− q0x0)] =

∫d4q

(2π)4eiqx (2.23)

in which qx = q · x − q0x0 is the Lorentz-invariant inner product of the

4-vectors q and x. There are many Green’s functions that satisfy Eq.(2.21).

Feynman’s propagator ∆F (x)

∆F (x) =

∫d4q

(2π)4

exp(iqx)

q2 +m2 − iε=

∫d3q

(2π)3

∫ ∞−∞

dq0

2π

eiq·x−iq0x0

q2 +m2 − iε. (2.24)

is the one that satisfies boundary conditions that will become evident when

we analyze the effect of its iε. The quantity Eq =√q2 +m2 is the energy

of a particle of mass m and momentum q in natural units with the speed of

light c = 1. Using this abbreviation and setting ε′ = ε/2Eq, we may write

the denominator as

q2 +m2 − iε = q · q−(q0)2

+m2 − iε =(Eq − iε′ − q0

) (Eq − iε′ + q0

)+ ε′2

(2.25)

in which ε′2 is negligible. Dropping the prime on ε, we do the q0 integral

I(q) = −∫ ∞−∞

dq0

2πe−iq

0x0 1

[q0 − (Eq − iε)] [q0 − (−Eq + iε)]. (2.26)

As shown in Fig. 2.1, the integrand

e−iq0x0 1

[q0 − (Eq − iε)] [q0 − (−Eq + iε)](2.27)

has poles at Eq − iε and at −Eq + iε. When x0 > 0, we can add a ghost

contour that goes clockwise around the lower half-plane and get

I(q) = ie−iEqx0 1

2Eqx0 > 0. (2.28)

When x0 < 0, our ghost contour goes counterclockwise around the upper

half-plane, and we get

I(q) = ieiEqx0 1

2Eqx0 < 0. (2.29)

Using the step function θ(x) = (x+ |x|)/2, we combine (2.28) and (2.29)

− iI(q) =1

2Eq

[θ(x0) e−iEqx0

+ θ(−x0) eiEqx0]. (2.30)

58 Feynman diagrams

Figure 2.1 In equation (2.27), the function f(q0) has poles at ±(Eq − iε),and the function exp(−iq0x0) is exponentially suppressed in the lower halfplane if x0 > 0 and in the upper half plane if x0 < 0. So we can add aghost contour (. . . ) in the LHP if x0 > 0 and in the UHP if x0 < 0.

In terms of the Lorentz-invariant function

∆+(x) =1

(2π)3

∫d3q

2Eqexp[i(q · x− Eqx0)] (2.31)

and with a factor of −i, Feynman’s propagator (2.24) is

− i∆F (x) = θ(x0) ∆+(x) + θ(−x0) ∆+(x,−x0). (2.32)


The integral (2.31) defining ∆+(x) is insensitive to the sign of q, and so

∆+(−x) =1

(2π)3

∫d3q

2Eqexp[i(−q · x+ Eqx

0)] (2.33)

=1

(2π)3

∫d3q

2Eqexp[i(q · x+ Eqx

0)] = ∆+(x,−x0).

Thus we arrive at the Standard form of the Feynman propagator

− i∆F (x) = θ(x0) ∆+(x) + θ(−x0) ∆+(−x). (2.34)

The annihilation operators a(q) and the creation operators a†(p) of a

scalar field φ(x) satisfy the commutation relations

[a(q), a†(p)] = δ3(q − p) and [a(q), a(p)] = [a†(q), a†(p)] = 0. (2.35)

Thus the commutator of the positive-frequency part

φ+(x) =

∫d3p√

(2π)32p0exp[i(p · x− p0x0)] a(p) (2.36)

of a scalar field φ = φ+ + φ− with its negative-frequency part

φ−(y) =

∫d3q√

(2π)32q0exp[−i(q · y − q0y0)] a†(q) (2.37)

is the Lorentz-invariant function ∆+(x− y)

[φ+(x), φ−(y)] =

∫d3p d3q

(2π)32√q0p0

eipx−iqy [a(p), a†(q)]

=

∫d3p

(2π)32p0eip(x−y) = ∆+(x− y) (2.38)

in which p(x− y) = p · (x− y)− p0(x0 − y0).

At points x that are space-like, that is, for which x2 = x2−(x0)2 ≡ r2 > 0,

the Lorentz-invariant function ∆+(x) depends only upon r = +√x2 and has

the value (Weinberg, 1995, p. 202)

∆+(x) =m

4π2rK1(mr) (2.39)

in which the Hankel function K1 is

K1(z) = −π2

[J1(iz) + iN1(iz)] =1

z+z

2

[ln(z

2

)+ γ − 1

2

]+ . . . (2.40)

where J1 is the first Bessel function, N1 is the first Neumann function, and

γ = 0.57721 . . . is the Euler-Mascheroni constant.

60 Feynman diagrams

The Feynman propagator arises most simply as the mean value in the

vacuum of the time-ordered product of the fields φ(x) and φ(y)

T φ(x)φ(y) ≡ θ(x0 − y0)φ(x)φ(y) + θ(y0 − x0)φ(y)φ(x). (2.41)

The operators a(p) and a†(p) respectively annihilate the vacuum ket a(p)|0〉 =

0 and bra 〈0|a†(p) = 0, and so by (2.36 & 2.37) do the positive- and negative-

frequency parts of the field φ+(z)|0〉 = 0 and 〈0|φ−(z) = 0. Thus the mean

value in the vacuum of the time-ordered product is

〈0|T φ(x)φ(y) |0〉 = 〈0|θ(x0− y0)φ(x)φ(y) + θ(y0− x0)φ(y)φ(x)|0〉= 〈0|θ(x0− y0)φ+(x)φ−(y) + θ(y0− x0)φ+(y)φ−(x)|0〉= 〈0|θ(x0− y0)[φ+(x), φ−(y)]

+ θ(y0− x0)[φ+(y), φ−(x)]|0〉. (2.42)

But by (2.38), these commutators are ∆+(x− y) and ∆+(y − x). Thus the

mean value in the vacuum of the time-ordered product of two real scalar

fields

〈0|T φ(x)φ(y) |0〉 = θ(x0 − y0)∆+(x− y) + θ(y0 − x0)∆+(y − x)

= −i∆F (x− y) =

∫d4q

(2π)4eiq·x

−iq2 +m2 − iε

(2.43)

is the Feynman propagator (2.32) multiplied by −i.

2.4 Application to a cubic scalar field theory

The action density

L = − 1

2∂νφ∂

νφ− 1

2m2φ2 − g

3!φ3, (2.44)

describes a scalar field with cubic interactions. We let

V (t) =g

3!

∫φ(x)3 d3x (2.45)

in which the field φ has the free-field time dependence

φ(x) =

∫d3p√

(2π)32p0

[a(p)eip·x + a†(p)e−ip·x

](2.46)

in which p · x = ~p · x − p0t and p0 =√p2 +m2. The amplitude for the

scattering of bosons with momenta p and k to momenta p′ and k′ is to

2.4 Application to a cubic scalar field theory 61

lowest order in the coupling constant g

A = 〈p′, k′| − 1

2!

∫T[V (t′)V (t′′)

]dt′ dt′′ |p, k〉

= − g2

2!(3!)2〈p′, k′|

∫T[φ3(x)φ3(y)

]d4xd4y |p, k〉.

(2.47)

This process happens in three ways, so A = A1 +A2 +A3.

The first way is for the incoming particles to be absorbed at the same

vertex x or y and for the outgoing particles to be emitted at the other vertex

y or x. We cancel the 2! by choosing the initial particles to be absorbed at y

and the final particles to be emitted at x. Then using the expansion (2.46)

of the field, we find

A1 = − g2

4〈0|a(k′)a(p′)

∫d3p′′d3k′′d3p′′′d3k′′′

(2π)6√

2p′′02k′′02p′′′02k′′′0a†(p′′′)a†(k′′′)

× e−i(p′′′+k′′′)·xT[φ(x)φ(y)

]a(p′′)a(k′′)ei(p

′′+k′′)·yd4xd4y a†(p)a†(k)|0〉(2.48)

after canceling two factors of 3 because each of the 3 fields φ3(x) and each

of the 3 fields φ3(y) could be the one to remain in the time-ordered product

T [φ(x)φ(y)]. The commutation relations [a(p), a†(k)] = δ3(p− k) now give

A1 = − g2

∫d4xd4y

(2π)6√

2p′02k′02p02k0ei(p+k)·y−i(p′+k′)·x〈0|T

[φ(x)φ(y)

]|0〉

(2.49)

in which the mean value in the vacuum of the time-ordered product

〈0|T[φ(x)φ(y)

]|0〉 = − i∆F (x− y) = −i

∫d4q

(2π)4

exp(iqx)

q2 +m2 − iε(2.50)

is Feynman’s propagator (2.24 & 2.43). Thus the amplitude A1 is

A1 = ig2

∫d4xd4yd4q

(2π)10√

2p′02k′02p02k0

ei(p+k)·y−i(p′+k′)·x+iq·(x−y)

q2 +m2 − iε. (2.51)

Thanks to Dirac’s delta function, the integrals over x, y, and q are easy, and

in the smooth ε→ 0 limit A1 is

A1 = ig2

∫d4q

(2π)2√

2p′02k′02p02k0

δ4(p+ k − q)δ4(q − p′ − k′)q2 +m2 − iε

=δ4(p+ k − p′ − k′)

(2π)2√

2p′02k′02p02k0

ig2

(p+ k)2 +m2.

(2.52)

62 Feynman diagrams

The sum of the three amplitudes is

A =δ4(p+ k − p′ − k′)16π2

√p′0k′0p0k0

[ ig2

(p+ k)2 +m2+

ig2

(p− k′)2 +m2+

ig2

(p− p′)2 +m2

](2.53)

in which the delta function conserves energy and momentum.

2.5 Feynman’s propagator for fields with spin

The time-ordered product for fields with spin is defined so as to compen-

sate for the minus sign that arises when a Fermi field is moved past a

Fermi field. Thus the mean value in the vacuum of the time-ordered product

Tψ`(x)ψ†m(y) is

〈0|Tψ`(x)ψ†m(y)

|0〉 = 〈0|θ(x0− y0)ψ`(x)ψ†m(y)± θ(y0− x0)ψ†m(y)ψ`(x)|0〉

= 〈0|θ(x0− y0)ψ+` (x)ψ†−m (y)± θ(y0− x0)ψ†+m (y)ψ−` (x)|0〉

= 〈0|θ(x0− y0)[ψ+` (x), ψ†−m (y)]∓

± θ(y0− x0)[ψ†+m (y), ψ−` (x)]|0〉∓. (2.54)

in which the upper signs are used for bosons and the lower ones for fermions.

The expansions

ψ+` (x) = (2π)−3/2

∑s

∫d3p u`(~p, s) e

ip·x a(~p, s)

ψ−` (x) = (2π)−3/2∑s

∫d3p v`(~p, s) e

−ip·x b†(~p, s)

(2.55)

give for the (anti)commutators

[ψ+` (x), ψ†−m (y)]∓ =

[(2π)−3/2

∑s

∫d3p u`(~p, s) e

ip·x a(~p, s),

(2π)−3/2∑r

∫d3k u∗m(~k, r)e−ik·y a†(~k, r)

]∓

=∑s

∫d3p

(2π)3u`(~p, s)u

∗m(~p, s) eip·(x−y)

(2.56)


and

[ψ†+m (y), ψ−` (x)]∓ =[(2π)−3/2

∑s

∫d3p v∗m(~p, s) eip·y b(~p, s),

(2π)−3/2∑r

∫d3k v`(~k, r)e

−ik·x b†(~k, r)]∓

=∑s

∫d3p

(2π)3v`(~p, s) v

∗m(~p, s) eip·(y−x).

(2.57)

Putting these expansions into the formula (2.54) for the time-ordered

product, we get for its mean value in the vacuum


|0〉 = θ(x0 − y0)

∑s

∫d3p

(2π)3u`(~p, s)u

∗m(~p, s) eip·(x−y)

± θ(y0− x0)∑s

∫d3p

(2π)3v`(~p, s) v

∗m(~p, s) eip·(y−x).

(2.58)

2.6 Feynman’s propagator for spin-one-half fields

For fields of spin one half, the spin sums are

[N(~p)

]`m

=∑s

u`(~p, s)u∗m(~p, s) =

[ 1


]`m[

M(~p)]`m

=∑s

v`(~p, s)v∗m(~p, s) =

[ 1


]`m,

(2.59)

so for spin-one-half fields Feynman’s propagator is


|0〉 = θ(x0 − y0)

∫d3p

(2π)3

[ 1


]`meip·(x−y)

− θ(y0− x0)

∫d3p

(2π)3

[ 1


]`meip·(y−x).

(2.60)

64 Feynman diagrams

Using the derivative term −∂cγc to generate −ipcγc, we get


|0〉 = θ(x0 − y0)

[(− ∂cγc +m) β

]`m

∫d3p


− θ(y0− x0)[

(∂cγc −m) β

]`m

∫d3p

(2π)32p0eip·(y−x)

= θ(x0 − y0)[

(− ∂cγc +m) β]`m

∆+(x− y)

+ θ(y0 − x0)[

(− ∂cγc +m)β]`m

∆+(y − x).

(2.61)

Since the derivative of the step function is a delta function

∂0θ(x0 − y0) = δ(x0 − y0), (2.62)

we can write Feynman’s propagator as


|0〉 =

[(−∂cγc +m)β

]`m

×(θ(x0 − y0) ∆+(x− y) + θ(y0 − x0) ∆+(y − x)

)+ ∂0γ

0βθ(x0 − y0) ∆+(x− y)

− ∂0γ0βθ(y0 − x0) ∆+(y − x)

(2.63)

and so also as


|0〉 =

[(−∂cγc +m)β

]`m

×(θ(x0 − y0) ∆+(x− y) + θ(y0 − x0) ∆+(y − x)

)+γ0βδ(x0 − y0) ∆+(x− y)− γ0βδ(y0 − x0) ∆+(y − x)

=[

(− ∂cγc +m) β]`m

×(θ(x0 − y0) ∆+(x− y)− θ(y0 − x0) ∆+(y − x)

)+γ0βδ(x0 − y0)

(∆+(x− y)−∆+(y − x)

).

(2.64)

But at equal times ∆+(x− y) = ∆+(y−x). So the ugly final term vanishes,

and Feynman’s propagator for spin-one-half fields is


|0〉 =

[(− ∂cγc +m) β

]`m

×(θ(x0 − y0)∆+(x− y) + θ(y0 − x0)∆+(y − x)

).

(2.65)


But this last piece is Feynman’s propagator for real scalar fields

〈0|T φ(x)φ(y) |0〉 = θ(x0 − y0)∆+(x− y) + θ(y0 − x0)∆+(y − x)

= − i∆F (x− y) =

∫d4q

(2π)4

−iq2 +m2 − iε

eiq·(x−y).

so Feynman’s propagator for spin-one-half fields is


|0〉 ≡ − i∆`m(x− y)

=[

(− ∂cγc +m) β]`m

∫d4q

(2π)4

−iq2 +m2 − iε

eiq·(x−y).

(2.66)

Letting the derivatives act on the exponential, we get


|0〉 ≡ − i∆`m(x− y)

= − i∫

d4q

(2π)4

[(− iqaγa +m)β

]`m

q2 +m2 − iεeiq·(x−y).

(2.67)

Since

(−iqaγa +m)(iqaγa +m) = q2 +m2, (2.68)

people often write this as


|0〉 ≡ − i∆`m(x− y)

= − i∫

d4q

(2π)4

[(− iqaγa +m)β

]`m

q2 +m2 − iεeiq·(x−y)

= − i∫

d4q

(2π)4

[ 1

iqaγa +m− iεβ]`meiq·(x−y).

(2.69)

Feynman was a master of notation (and of everything else). He set /p = paγa

and wrote


|0〉 ≡ − i∆`m(x− y)

= − i∫

d4q

(2π)4

[(− i/q +m)β

]`m


= − i∫

d4q

(2π)4

( 1

i/q +m− iεβ)`meiq·(x−y).

(2.70)

66 Feynman diagrams

In the common notation ψ = ψ†β, his propagator is

〈0|Tψ`(x)ψm(y)

|0〉 ≡ − i∆`m(x− y)β

= − i∫

d4q

(2π)4

[(− i/q +m)

]`m


= − i∫

d4q

(2π)4

( 1

i/q +m− iε

)`meiq·(x−y).

(2.71)

2.7 Application to a theory of fermions and bosons

Let us consider first the theory

H(x) = g`m ψ†`(x)ψm(x)φ(x) (2.72)

where g`m is a coupling constant, ψ(x) is the field of a fermion f , and φ(x) =

φ†(x) is a real boson b. Let’s compute the amplitude A for f + b→ f ′ + b′.

The lowest-order term is

A = − 1

2!

∫d4xd4y 〈0|a(p′, s′)b(k′)T

[H(x)H(y)

]b†(k)a†(p, s)|0〉 (2.73)

= − g`mg`′m′

2!

∫d4xd4y〈0|a(p′, s′)b(k′)T

[ψ†`(x)ψm(x)φ(x)ψ†`′(y)ψm′(y)φ(y)

]× b†(k)a†(p, s)|0〉.

Here the operators b(k′) and b†(k) are the boson deletion and addition op-

erators. Either the boson field φ(y) deletes the boson from the initial state

and the boson field φ(x) deletes the boson from the final state or the boson

field φ(x) deletes the boson from the initial state and the boson field φ(y)

deletes the boson from the final state. These give the same result. So we

cancel the 2! and let φ(y) delete the boson from the initial state and have


φ(x) add the boson of the final state. We then get

A = − g`mg`′m′∫d4xd4y〈0|a(p′, s′)b(k′)T

[ψ†`(x)ψm(x)φ−(x)ψ†`′(y)ψm′(y)φ+(y)

]× b†(k)a†(p, s)|0〉

= − g`mg`′m′∫d4xd4y 〈0|a(p′, s′)b(k′)T

[ψ†`(x)ψm(x)

×∫

d3k′′√(2π)32k′′0

b†(k′′)e−ik′′·xψ†`′(y)ψm′(y)φ+(y)

]b†(k)a†(p, s)|0〉

= − g`mg`′m′∫d4xd4y 〈0|a(p′, s′)T

[ψ†`(x)ψm(x)

×∫

d3k′′√(2π)32k′′0

δ3(~k′ − ~k′′) e−ik′′·xψ†`′(y)ψm′(y)φ+(y)]b†(k)a†(p, s)|0〉

(2.74)

= − g`mg`′m′∫d4xd4y 〈0|a(p′, s′)T

[ψ†`(x)ψm(x)

1√(2π)32k′0

e−ik′·x

× ψ†`′(y)ψm′(y)φ+(y)]b†(k)a†(p, s)|0〉.

We now let φ+(y) delete the boson from the initial state.

A = − g2

∫d4xd4y 〈0|a(p′, s′)T

[ψ†`(x)ψm(x)

1√(2π)32k′0

e−ik′·x

× ψ†`′(y)ψm′(y)

∫d3k′′√

(2π)32k′′0b(k′′)eik

′′·y]b†(k)a†(p, s)|0〉

= − g`mg`′m′∫d4xd4y 〈0|a(p′, s′)T

[ψ†`(x)ψm(x)

1√(2π)32k′0

e−ik′·y

× ψ†`′(y)ψm′(y)

∫d3k′′√

(2π)32k′′0δ3(k′′ − k)eik

′′·y]a†(p, s)|0〉 (2.75)

= − g`mg`′m′∫d4xd4y 〈0|a(p′, s′)T

[ψ†`(x)ψm(x)

1√(2π)32k′0

e−ik′·x

× ψ†`′(y)ψm′(y)1√

(2π)32k0eik·y

]a†(p, s)|0〉

= − g`mg`′m′∫

d4xd4y√(2π)62k′02k0

eik·y−ik′·x〈0|a(p′, s′)T

[ψ†`(x)ψm(x)ψ†`′(y)ψm′(y)

]a†(p, s)|0〉.

68 Feynman diagrams

Now there are two terms. In one the initial fermion is deleted at y and the

final fermion is added at x

A1 = − g`mg`′m′∫

d4xd4y√(2π)62k′02k0


[ψ+†` (x)ψm(x)ψ†`′(y)ψ+

m′(y)]a†(p, s)|0〉

= − g`mg`′m′∫

d4xd4y√(2π)62k′02k0


[∫ d3p′′√(2π)3

e−ip′′·xa†(p′′, s′′)u∗` (p

′′, s′′)

× ψm(x)ψ†`′(y)

∫d3p′′′√(2π)3

eip′′′·ya(p′′′, s′′′)um′(p

′′′, s′′′)]a†(p, s)|0〉

(2.76)

= − g`mg`′m′∫

d4xd4y√(2π)62k′02k0

eik·y−ik′·x〈0|T

[∫ d3p′′√(2π)3

e−ip′′·xu∗` (p

′′, s′′)δ3(~p′ − ~p′′)δs′s′′

× ψm(x)ψ†`′(y)

∫d3p′′′√(2π)3

eip′′′·yum′(p

′′′, s′′′)δ3(~p− ~p′′′)δss′′′]|0〉

= − g`mg`′m′u∗` (p′, s′)um′(p, s)∫

d4xd4y√(2π)122k′02k0

eik·y−ik′·xe−ip

′·xeip·y〈0|T[ψm(x)ψ†`′(y)

]|0〉

= − g`mg`′m′u∗` (p′, s′)um′(p, s)∫

d4xd4y√(2π)122k′02k0

ei(k+p)·y−i(k′+p′)·x〈0|T[ψm(x)ψ†`′(y)

]|0〉.

In terms of SW’s definition (6.2.31) of the fermion propagator, A1 is

A1 = −g`mg`′m′u∗` (p′, s′)um′(p, s)∫

d4xd4y√(2π)122k′02k0

ei(k+p)·y−i(k′+p′)·x(−i∆m`′(x, y)).

(2.77)

In the other term, the initial fermion is deleted at x and the final fermion

is added at y

A2 = − g`mg`′m′∫

d4xd4y√(2π)62k′02k0


[ψ†`(x)ψ+

m(x)ψ+†`′ (y)ψm′(y)

]a†(p, s)|0〉

= − g`mg`′m′∫

d4xd4y√(2π)62k′02k0


[ψ†`(x)ψ+†

`′ (y)ψm′(y)ψ+m(x)

]a†(p, s)|0〉

(2.78)

= g`mg`′m′

∫d4xd4y√

(2π)62k′02k0eik·y−ik

′·x〈0|a(p′, s′)T[ψ+†`′ (y)ψ†`(x)ψm′(y)ψ+

m(x)]a†(p, s)|0〉

in which the minus sign arises from the transposition ψ†`(x)ψ+†(y)→ ψ+†(y)ψ†`(x).

The earlier transposition ψ+(x)ψ+†(y)ψm′(y) → ψ+†(y)ψm′(y)ψ+(x) pro-

duced two minus signs or one plus sign. Inserting the expansions of ψ+†(y)


and ψ+(x), we have

A2 = g`mg`′m′

∫d4xd4y√

(2π)62k′02k0eik·y−ik

′·x〈0|a(p′, s′)T[ ∫ d3p′′√

(2π)3u∗`′(p

′′, s′′)a†(p′′, s′′)e−ip′′·y

× ψ†`(x)ψm′(y)

∫d3p′′′√(2π)3

um(p′′′, s′′′)a(p′′′, s′′′)eip′′′·x]a†(p, s)|0〉

= g`mg`′m′u∗`′(p′, s′)um(p, s)

∫d4xd4y√

(2π)122k′02k0ei(k−p

′)·y−i(k′−p)·x〈0|T[ψ†`(x)ψm′(y)

]|0〉.

(2.79)

In terms of SW’s definition (6.2.31) of the fermion propagator, A2 is

A2 = − g`mg`′m′u∗`′(p′′, s′′)um(p, s)

∫d4xd4y√

(2π)122k′02k0ei(k−p

′)·y−i(k′−p)·x〈0|T[ψm′(y)ψ†`(x)

]|0〉

= − g`mg`′m′u∗`′(p′, s′)um(p, s)

∫d4xd4y√

(2π)122k′02k0ei(k−p

′)·y−i(k′−p)·x (− i∆m′`(y, x)).

(2.80)

The full amplitude for f + b→ f ′ + b′ is the sum A = A1 +A2 of the two

amplitudes

A = − g`mg`′m′u∗` (p′, s′)um′(p, s)∫

d4xd4y√(2π)122k′02k0

ei(k+p)·y−i(k′+p′)·x(− i∆m`′(x, y))

− g`mg`′m′u∗`′(p′, s′)um(p, s)

∫d4xd4y√

(2π)122k′02k0ei(k−p

′)·y−i(k′−p)·x (− i∆m′`(y, x)).

(2.81)

Interchanging x and y, m and m′, and ` and `′ in A1, we get

A = − g`′m′g`mu∗`′(p′, s′)um(p, s)

∫d4xd4y√

(2π)122k′02k0ei(k+p)·x−i(k′+p′)·y(− i∆m′`(y, x)

)− g`mg`′m′u∗`′(p′, s′)um(p, s)

∫d4xd4y√

(2π)122k′02k0ei(k−p

′)·y−i(k′−p)·x (− i∆m′`(y, x)).

(2.82)

Combining terms, we get

A = − g`′m′g`mu∗`′(p′, s′)um(p, s)

∫d4xd4y√

(2π)122k′02k0eip·x−ip

′·y(− i∆m′`(y, x))

×(eik·x−ik

′·y + eik·y−ik′·x)

(2.83)

which agrees with SW’s (6.1.27) when his boson is restricted to a single

scalar field.

70 Feynman diagrams

We now replace mean value in the vacuum of the time-ordered product

by its value (2.70)


|0〉 ≡ − i∆`m(x− y) = − i

∫d4q

(2π)4

[(− iqaγa +m)β

]`m

q2 +m2 − iεeiq·x.

(2.84)

Replacing `,m, x, y by m′, `, y, x, we get

〈0|Tψm′(y)ψ†`(x)

|0〉 ≡ − i∆m′`(y − x) = − i

∫d4q

(2π)4

[(− iqaγa +m)β

]m′`

q2 +m2 − iεeiq·(x−y).

(2.85)

We then find

A = − g`′m′g`mu∗`′(p′, s′)um(p, s)

∫d4xd4y

(2π)6√

2k′02k0eip·x−ip

′·y

×(eik·x−ik

′·y + eik·y−ik′·x)

(−i)∫

d4q

(2π)4

[(− i/q +m)β

]m′`

q2 +m2 − iεeiq·(y−x).

(2.86)

In matrix notation, this is

A = −∫

d4xd4y

(2π)6√

2k′02k0

(ei(p+k)·x−i(p′+k′)·y + ei(k−p

′)·y+i(p−k′)·x)

× (−i)∫

d4q

(2π)4

u†(p′, s′) g[(− i/q +m)β

]g u(p, s)

q2 +m2 − iεeiq·(y−x). (2.87)

The d4x and d4y integrations give

A = i

∫d4q

(2π)2√

2k′02k0

(δ(q − k′ − p′)δ(p+ k − q) + δ(q + k − p′)δ(p− k′ − q)

)×u†(p′, s′) g

[(− i/q +m)β

]g u(p, s)

q2 +m2 − iε(2.88)

which is

A =iδ(p′ + k′ − p− k)

8π2√k′0k0

u†(p′, s′)g

[−i(/p+ /k) +m

(p+ k)2 +m2+−i(/p+ /k′) +m

(p− k′)2 +m2

]βgu(p, s).

(2.89)

2.8 Feynman propagator for spin-one fields 71

2.8 Feynman propagator for spin-one fields

The general form of the propagator is

〈0|T[ψa(x)ψ†b(y)

]|0〉 = θ(x0 − y0)

∑s

∫d3p

(2π)3ua(~p, s)u

†b(~p, s) e

ip·(x−y)

± θ(y0− x0)∑s

∫d3p

(2π)3va(~p, s) v

∗b (~p, s) e

ip·(y−x).

(2.90)

We use the upper (+) sign for spin-one fields because they are bosons. For

single real massive vector field

ψa(x) =1∑

s=−1

∫d3p√

(2π)32p0

[ea(~p, s)a(~p, s)eip·x + e∗a(~p, s)a†(~p, s)e−ip·x

],

(2.91)

the mean value in the vacuum of its time-ordered product is


]|0〉 = θ(x0 − y0)

∑s

∫d3p

(2π)32p0ea(~p, s) e

∗b(~p, s) e

ip·(x−y)

+ θ(y0− x0)∑s

∫d3p

(2π)32p0ea(~p, s) e

∗b(~p, s) e

ip·(y−x).

(2.92)

The spin-one spin sum is

∑s

ea(~p, s) e∗b(~p, s) = ηab + pa pb/m

2. (2.93)

So the mean value (2.58) is


]|0〉 = θ(x0 − y0)

∫d3p

(2π)32p0

(ηab +

pa pbm2

)eip·(x−y)

+ θ(y0− x0)

∫d3p

(2π)32p0

(ηab +

pa pbm2

)eip·(y−x)

(2.94)

72 Feynman diagrams

in which the momentum pa is physical (aka, on the mass shell) in that

p0 =√~p2 +m2 and p2

0 = ~p2 +m2. In terms of derivatives, we have


]|0〉 = θ(x0 − y0)

(ηab −

∂a ∂bm2

)∫d3p


+ θ(y0− x0)

(ηab −

∂a ∂bm2

)∫d3p

(2π)32p0eip·(y−x).

(2.95)

In this formula, we may interpret the double time derivative as ∂20 = ∇2−m2.

For spatial values of a and b, we can move the derivatives to the left of the

step functions. And we can move the product ∂0 ∂i of one spatial and one

time derivative by the argument we used for spin one-half. We find


]|0〉 =

(ηab −

∂a ∂bm2

)[θ(x0 − y0)

∫d3p


+ θ(y0− x0)

∫d3p

(2π)32p0eip·(y−x)

]=

(ηab −

∂a ∂bm2

)[θ(x0 − y0) ∆+(x− y) + θ(y0 − x0) ∆+(y − x)

]=

(ηab −

∂a ∂bm2

)[−i∆F (x− y)]

= − i(ηab −

∂a ∂bm2

)∫d4q

(2π)4

eiq·(x−y)

q2 +m2 − iε(2.96)

in which ∂20 = ∇2 −m2.

We can relax the rule ∂20 = ∇2 − m2 if we add an extra term to the

propagator. The extra term is

i

m2

(∇2 −m2 − ∂2

0

) ∫ d4q

(2π)4

eiq·(x−y)

q2 +m2 − iε=

i

m2

∫d4q

(2π)4

−~q2 −m2 + q20


= − i

m2δ4(x− y). (2.97)

So the Feynman propagator for spin-one fields is


]|0〉 = − i

∫d4q

(2π)4

(ηab + pa pb/m2)

q2 +m2 − iεeiq·(x−y) − i

m2δ0a δ

0b δ

4(x− y).

(2.98)

Suppose the vector field ψa(x) interacts with a current ja(x) thru a term

ψa(x)ja(x) in H(x). Then in Dyson’s expansion, the awkward second term

2.9 The Feynman rules 73

in the Feynman propagator (2.98) contributes the term

− iH2(x) =1

2

∫[−ija(x)][−ijb(y)]

[− i

m2δ0a δ

0b δ

4(x− y)

]d4y =

i[j0(x)]2

2m2.

(2.99)

So if H(x) also contains the term

HC(x) =[j0(x)]2

2m2, (2.100)

then the effect of the awkward second term in the Feynman propagator is

cancelled. This actually happens in a natural way as SW explains in chapter

7.

2.9 The Feynman rules

1) Draw all diagrams of the order you are working at. Label each internal

line with its (unphysical) 4-momentum considered to flow in the direction

of the arrow or in either direction if the particle is uncharged.

2) For each vertex of type i, include the factor

− i(2π)4giδ4(∑

p+ q − p′ − q′)

(2.101)

which makes the sum of the 4-momenta p + q entering each vertex add up

to sum of the 4-momenta p′ + q′ leaving each vertex.

For each outgoing line include the factor

u∗` (~p′, s′, n′)

(2π)3/2or

v`(~p′, s′, n′)

(2π)3/2(2.102)

for arrows pointing out (u∗` (~p′, s′, n′)) or pointing in (v`(~p

′, s′, n′)).

For each incoming line include the factor

u`(~p, s, n)

(2π)3/2or

v∗` (~p, s, n)

(2π)3/2(2.103)

for arrows pointing in (u∗` (~p′, s′, n′)) or pointing out (v`(~p

′, s′, n′)).

For each internal line of a spin-zero particle carrying momentum qa include

the factor−i

(2π)4

1

q2 +m2` − iε

. (2.104)

For each internal line of a spin-one-half particle with ends labelled by `

and m and carrying momentum qa include the factor

−i(2π)4

[(− i/q +m)β

]`m

q2 +m2` − iε

. (2.105)

74 Feynman diagrams

For each internal line of a spin-one particle with ends labelled by ` and m

and carrying momentum qa include the factor

−i(2π)4

η`m + q`qm/m2

q2 +m2` − iε

(2.106)

and keep in mind the delta-function term in (2.98).

3). Integrate the product of all these factors over all the internal momenta

and sum over ` and m, etc.

4) Add the results of all the Feynman diagrams.

2.10 Fermion-antifermion scattering

We can watch Feynman’s rules emerge in fermion-antifermion scattering.

We consider a fermion interacting with a neutral scalar boson (2.72)

H(x) = g`m ψ†`(x)ψm(x)φ(x). (2.107)

The initial state is |p, s; q, t〉 = a†(p, s)b†(q, t)|0〉 and the final state is |p′, s′; q′, t′〉 =

a†(p′, s′)b†(q′, t′)|0〉. The lowest-order term is

A = − 1

2!

∫d4xd4y 〈0|b(q′, t′) a(p′, s′)T

[H(x)H(y)

]a†(p, s) b†(q, t)|0〉

(2.108)

= − g`mg`′m′

2!

∫d4xd4y〈0|b(q′, t′)a(p′, s′)T

[ψ†`(x)ψm(x)φ(x)ψ†`′(y)ψm′(y)φ(y)

]× a†(p, s)b†(q, t)|0〉

in which the operators a and b delete fermions and antifermions. There is

only one Feynman diagram (Fig. 2.2).

We cancel the 2! by choosing to absorb the incoming fermion-antifermion

pair at vertex y and to add the outgoing fermion-antifermion pair at vertex

x. We are left with

A = − g`mg`′m′

(2π)6

∫d4xd4y〈0|u∗` (p′, s′) vm(q′, t′) e−ix·(p

′+q′) T[φ(x)φ(y)

]× um′(p, s) v∗`′(q, t) eiy·(p+q)|0〉. (2.109)

Adding in the scalar propagator (2.50), we get

A = − g`mg`′m′

(2π)6

∫d4x d4y u∗` (p

′, s′) v∗m(q′, t′) e−ix·(p′+q′)

× um′(p, s) v∗`′(q, t) eiy·(p+q)∫

d4k

(2π)4

−ieik·(x−y)

k2 +m2 − iε. (2.110)

2.10 Fermion-antifermion scattering 75

Fermion-antifermion scattering

Figure 2.2 Fermion-antifermion scattering p+q → p′+q′ (arbitrary colors).

The integration over y conserves 4-momentum at vertex y, and the integra-

tion over x conserves 4-momentum at vertex x. We then have

A = ig`mg`′m′

(2π)2

∫d4k u∗` (p

′, s′) vm(q′, t′) δ(k − p′ − q′) δ(p+ q − k)

× um′(p, s) v∗`′(q, t)1

k2 +m2 − iε(2.111)

or

A = i δ(p+ q − k′ − q′) g`mg`′m′

(2π)2

u∗` (p′, s′) vm(q′, t′)um′(p, s) v

∗`′(q, t)

(p+ k)2 +m2.

(2.112)

3

Action

3.1 Lagrangians and hamiltonians

A transformation is a symmetry of a theory if the action is invariant or

changes by a surface term. So we choose to work with actions that are sym-

metrical. The action is normally an integral over spacetime of an action

density often called a lagrangian. Often the action density itself is invari-

ant under the transformation of the symmetry.

There are procedures, sometimes clumsy procedures, for computing the

hamiltonian from the lagrangian. The hamiltonian often is not invariant

under the transformation of the symmetry. So it’s very hard to find a suitably

symmetrical theory by starting with a hamiltonian. But once one has a

hamiltonian, one can compute scattering amplitudes energies, and states

with these energies.

3.2 Canonical variables

In quantum mechanics, we use the equal-time commutation relations

[qi, pk] = iδik, [qi, qk] = 0, and [pi, pk] = 0. (3.1)

In general, the operators qi and qi(t) = eiHtqie−iHt do not commute. Quan-

tum field theory promotes these equal-time commutation relations to ones

in which the indexes i and k denote different points of space

[qn(~x, t), pm(~y, t)]∓ = iδ(~x− ~y)δnm,

[qn(~x, t), qm(~y, t)]∓ = 0, and [pn(~x, t), pm(~y, t)]∓ = 0.(3.2)

The commutator of a real scalar field

φ(x) =

∫d3p√

(2π)32p0

[a(~p) eip·x + a†(~p) e−ip·x

](3.3)

3.2 Canonical variables 77

is

[φ(x), φ(y)] = ∆(x− y) =

∫d3p

(2π)32p0

(eip·(x−y) − e−ip·(x−y)

). (3.4)

At equal times, one has

∆(~x− ~y, 0) = 0,∂

∂x0∆(x− y)|x0=y0 = − i δ3(~x− ~y), and

∂2

∂x0∂y0∆(x− y)|x0=y0 = 0.

(3.5)

So a real field φ and its time derivative φ satisfy the equal-time commutation

relations (3.60)

[φ(~x, t), φ(~y, t)]− = i δ3(~x− ~y),

[φ(~x, t), φ(~y, t)]− = 0, and [φ(~x, t), φ(~y, t)]− = 0.(3.6)

A complex scalar field

φ(x) =1√2

[φ1(x) + iφ2(x)

](3.7)

obeys the commutation relations

[φ(x), φ†(y)]− =1

2[φ1(x) + iφ2(x), φ1(x)− iφ2(x)]

=1

2([φ1(x), φ1(x)] + [φ2(x), φ2(x)]) = ∆(x− y)

(3.8)

and

[φ(x), φ(y)]− =1

2[φ1(x) + iφ2(x), φ1(x) + iφ2(x)]

=1

2([φ1(x), φ1(x)]− [φ2(x), φ2(x)]) = 0.

(3.9)

So the complex scalar fields φ(x) and φ†(x) obey the equal-time commuta-

tion relations (3.6)

[φ(~x, t), φ†(~y, t)]− = i δ3(~x− ~y) and [φ(~x, t), φ†(~y, t)]− = 0 (3.10)

[φ(~x, t), φ(~y, t)] = 0 and [φ†(~x, t), φ†(~y, t)] = 0 (3.11)

[φ(~x, t), φ(~y, t)] = 0 and [φ†(~x, t), φ†(~y, t)] = 0. (3.12)

78 Action

3.3 Principle of stationary action in field theory

If φ(x) is a scalar field, and L(φ) is its action density, then its action S[φ] is

the integral over all of spacetime

S[φ] =

∫L(φ(x)) d4x. (3.13)

The principle of least (or stationary) action says that the field φ(x) that

satisfies the classical equation of motion is the one for which the first-order

change in the action due to any tiny variation δφ(x) in the field vanishes,

δS[φ] = 0. And to keep things simple, we’ll assume that the action (or La-

grange) density L(φ) is a function only of the field φ and its first derivatives

∂aφ = ∂φ/∂xa. The first-order change in the action then is

δS[φ] =

∫ [∂L

∂φδφ+

∂L

∂(∂aφ)δ(∂aφ)

]d4x (3.14)

in which we sum over the repeated index a from 0 to 3. Now δ(∂aφ) =

∂a(φ+δφ)−∂aφ = ∂aδφ. So we may integrate by parts and drop the surface

terms because we set δφ = 0 on the surface at infinity

δS[φ] =

∫ [∂L

∂φδφ+

∂L

∂(∂aφ)∂a(δφ)

]d4x

=

∫ [∂L

∂φ− ∂a

∂L

∂(∂aφ)

]δφ d4x.

(3.15)

This first-order variation is zero, for arbitrary δφ only if the field φ(x) sat-

isfies Lagrange’s equation

∂a

(∂L

∂(∂aφ)

)≡ ∂

∂xa

[∂L

∂(∂φ/∂xa)

]=∂L

∂φ(3.16)

which is the classical equation of motion.

Example 3.1 (Theory of a scalar field) The action density of a scalar field

φ of mass m is

L = 12 (φ)2 − 1

2 (∇φ)2 − 12 m

2φ2 = −12 ∂aφ∂

aφ− 12 m

2φ2. (3.17)

Lagrange’s equation (3.16) for this action density is

∇2φ− φ = ∂a ∂aφ = m2φ (3.18)

which is the Klein-Gordon equation (1.56).


In a theory of several fields φ1, . . . , φn with action density L(φk, ∂aφk),

the fields obey n copies of Lagrange’s equation

∂

∂xa

(∂L

∂(∂aφk)

)=

∂L

∂φk(3.19)

one for each k.

3.4 Symmetries and conserved quantities in field theory

An action density L(φi, ∂aφi) that is invariant under a transformation of the

coordinates xa or of the fields φi and their derivatives ∂aφi is a symmetry

of the action density. Such a symmetry implies that something is conserved

or time independent.

Suppose that an action density L(φi, ∂aφi) is unchanged when the fields

φi and their derivatives ∂aφi change by δφi and by δ(∂aφi)

0 = δL =∑i

∂L

∂φiδφi +

∂L

∂∂aφiδ (∂aφi). (3.20)

For many transformations (but not for Lorentz transformations), we can set

δ(∂aφi) = ∂a(δφi). In such cases by using Lagrange’s equations (3.19) to

rewrite ∂L/∂φi, we find

0 =∑i

(∂a

∂L

∂∂aφi

)δφi +

∂L

∂∂aφi∂aδφi = ∂a

∑i

∂L

∂∂aφiδφi (3.21)

which says that the current

Ja =∑i

∂L

∂∂aφiδφi (3.22)

has zero divergence

∂aJa = 0. (3.23)

Thus the time derivative of the volume integral of the charge density J0

QV =

∫VJ0 d3x (3.24)

is the flux of current ~J entering through the boundary S of the volume V

QV =

∫V∂0J

0 d3x = −∫V∂kJ

k d3x = −∫SJkd2Sk. (3.25)

80 Action

If no current enters V , then the charge Q inside V is conserved. When the

volume V is the whole universe, the charge is the integral over all of space

Q =

∫J0 d3x =

∫ ∑i

∂L

∂φiδφi d

3x =

∫ ∑i

πi δφi d3x (3.26)

in which πi is the momentum conjugate to the field φi

πi =∂L

∂φi. (3.27)

Example 3.2 (O(n) symmetry and its charge) Suppose the action density

L is the sum of n copies of the quadratic action density (3.17)

L =

n∑i=1

12 (φi)

2 − 12 (∇φi)2 − 1

2 m2φ2

i = −12 ∂aφ∂

aφ− 12 m

2φ2, (3.28)

and Aij is any constant antisymmetric matrix, Aij = − Aji. Then if the

fields change by δφi = ε∑

j Aij φj , the change (3.20) in the action density

δL = − εn∑

i,j=1

[m2φiAijφj + ∂aφiAij∂aφj

]= 0 (3.29)

vanishes. Thus the charge (3.26) associated with the matrix A

QA =

∫ ∑i

πi δφi d3x = ε

∫ ∑i

πiAij φj d3x (3.30)

is conserved. There are n(n− 1)/2 antisymmetric n×n imaginary matrices;

they generate the group O(n) of n× n orthogonal matrices.

An action density L(φi, ∂aφi) that is invariant under a spacetime transla-

tion, x′a = xa + δxa, depends upon xa only through the fields φi and their

derivatives ∂aφi

∂L

∂xa=∑i

(∂L

∂φi

∂φi∂xa

+∂L

∂∂bφi

∂2φi∂xb∂xa

). (3.31)

Using Lagrange’s equations (3.19) to rewrite ∂L/∂φi , we find

0 =

(∑i

∂b

(∂L

∂∂bφi

)∂aφi +

∂L

∂∂bφi

∂2φi∂xb∂xa

)− ∂L

∂xa

0 = ∂b

[∑i

(∂L

∂∂bφi

∂φi∂xa

)− δbaL

] (3.32)


that the energy-momentum tensor

T ba =∑i

∂L

∂∂bφi

∂φi∂xa− δbaL (3.33)

has zero divergence, ∂bTba = 0.

Thus the time derivative of the 4-momentum PaV inside a volume V

PaV =

∫V

(∑i

∂L

∂∂0φi

∂φi∂xa− δ0

aL

)d3x =

∫VT 0a d

3x (3.34)

is equal to the flux entering through V ’s boundary S

∂0PaV =

∫V∂0T

0a d

3x = −∫V∂kT

ka d

3x = −∫ST ka d

2Sk. (3.35)

The invariance of the action density L under spacetime translations implies

the conservation of energy P0 and momentum ~P .

The momentum πi(x) that is canonically conjugate to the field φi(x) is

the derivative of the action density L with respect to the time derivative of

the field

πi =∂L

∂φi. (3.36)

If one can express the time derivatives φi of the fields in terms of the fields φiand their momenta πi, then hamiltonian of the theory is the spatial integral

of

H = P0 = T 00 =

(n∑i=1

πi φi

)− L (3.37)

in which φi = φi(φ, π).

Example 3.3 (Hamiltonian of a scalar field) The hamiltonian density

(3.37) of the theory (3.17) is

H = 12 π

2 + 12 (∇φ)2 + 1

2 m2φ2. (3.38)

A Lorentz transformation (1.15) of a field is like this

U(Λ, a)ψ`(x)U−1(Λ, a) =∑

¯

D`¯(Λ−1)ψ¯(Λx+ a). (3.39)

The action is invariant under Lorentz transformations. The action density

82 Action

is constructed so as to be invariant when the fields transform this way:

ψ′`(x) =∑

¯

D`¯(Λ−1)ψ¯(x) (3.40)

The action density is constructed so as to be invariant when the fields and

their derivatives of the fields transform under infinitesimal Lorentz transfor-

mations as

δψ`(x) =i

2ωab(Jab)

`m ψ

m(x)

δ(∂k ψ`)(x) =

i

2ωab(Jab)

`m ∂k ψ

m(x) + ω jk ∂jψ

`(x).

(3.41)

The invariance of the action density says that

0 =∂L

∂ψ`δψ` +

∂L

∂∂kψ`δ(∂k ψ

`) (3.42)

=∂L

∂ψì

2ωab(Jab)

`m ψ

m +∂L

∂∂kψ`

[ i2ωab(Jab)

`m ∂k ψm + ω j

k ∂jψ`].

Remembering that ωab = −ωba and ωab = −ωba, we rewrite the last bit as

ω bk ∂bψ` = ηka ω

ab∂b =1

2

(ηka ω

ab∂bψ` − ηkb ωab∂a)ψ`. (3.43)

We then have

0 =∂L

∂ψì

2ωab(Jab)

`m ψ

m

+∂L

∂∂kψ`

[ i2ωab(Jab)

`m ∂k ψm +

1

2

(ηka ω

ab∂b − ηkb ωab∂a)ψ`

](3.44)

or

0 =∂L

∂ψì

2(Jab)

`m ψ

m

+∂L

∂∂kψ`

[ i2

(Jab)`m ∂k ψm +

1

2

(ηka ∂b − ηkb ∂a

)ψ`

]. (3.45)

The equations of motion now give

0 =

(∂k

∂L

∂∂kψ`

)i

2(Jab)

`m ψ

m

+∂L

∂∂kψ`

[ i2

(Jab)`m ∂k ψm +

1

2


)ψ`

](3.46)

or

0 = ∂k

(∂L

∂∂kψì

2(Jab)

`m ψ

m

)+

∂L

∂∂kψ`1

2


)ψ`

]


or

0 = ∂k

(∂L

∂∂kψì

2(Jab)

`m ψ

m

)+

1

2

(∂L

∂∂aψ`∂b −

∂L

∂∂bψ`∂aψ`

).

We recall (3.33) the energy-momentum tensor

T ba =∑i

∂L

∂∂bφi

∂φi∂xa− δbaL (3.47)

or

Tba =∂L

∂∂bψ`

∂ψ`∂xa− ηbaL (3.48)

which has zero divergence, ∂bTba = 0. In terms of it, we have

0 = ∂k

(∂L

∂∂kψì

2(Jab)

`m ψ

m

)− 1

2(Tab − Tba) .

So Belinfante defined the symmetric energy-momentum tensor as

Θab = T ab − i

2∂k

[ ∂L

∂∂kψ`(Jab)`m ψ

m

− ∂L

∂∂aψ`(Jkb)`m ψ

m − ∂L

∂∂bψ`(Jka)`m ψ

m]. (3.49)

The quantity in the square brackets is antisymmetric in a, k. So

∂aΘab = ∂aT

ab − i

2∂a∂k

[ ∂L

∂∂kψ`(Jab)`m ψ

m

− ∂L

∂∂aψ`(Jkb)`m ψ

m − ∂L

∂∂bψ`(Jka)`m ψ

m]

= ∂aTab = 0.

(3.50)

The Belinfante energy-momentum tensor is symmetric

Θab = Θba (3.51)

and so it is the one to use when gravity is involved.

The quantity

Mabc = xbΘac − xcΘab (3.52)

is a conserved current

∂aMabc = ∂a

(xbΘac − xcΘab

)= Θbc −Θcb = 0. (3.53)

So the angular-momentum operators

Jbc =

∫M0bc d3x =

∫ (xbΘ0c − xcΘ0b

)d3x (3.54)

84 Action

are conserved.

3.5 Poisson Brackets

In a classical theory with coordinates qn and conjugate momenta

pn =∂L

∂qn, (3.55)

the Poisson bracket is

[A,B]P =∑n

∂A

∂qn

∂B

∂pn− ∂B

∂qn

∂A

∂pn. (3.56)

One of Dirac’s many insights was to say that in the quantum version of such

a theory the equal-time commutation relations are

[A,B] = i~ [A,B]P . (3.57)

In field theories, with coordinates φa = φn(x) and conjugate momenta

πa = πn(x) =∂L(x)

∂φn(x), (3.58)


[A,B]P =∑n

∫d3x

∂A

∂φn(x)

∂B

∂πn(x)− ∂B

∂φn(x)

∂A

∂πn(x). (3.59)

In the quantum theory, the equal-time commutator is

[A,B] = i~ [A,B]P . (3.60)

When these equations and definitions (3.55–3.60) make sense, one some-

times can express the velocities φn(x) in terms of the coordinates φn(x) and

their conjugate momenta πn(x). In such cases, one can make a hamiltonian

density

H = φn(x)πn(x)− L(x) (3.61)

in which L(x) is the action density. And the action principle gives us the

equations of motion

∂i∂L

∂∂iφn=

∂L

∂φn. (3.62)

But things can go wrong. For instance, the Lagrange’s equations for the

action densities L = q and L = φn lead to the inconsistent equation

0 = 1. (3.63)

3.6 Dirac Brackets 85

Similarly,

So Nature doesn’t care about our ideas. The theory of a massive vector

field V

L = − 14FikF

ik − 12ViV

i − JiV i (3.64)

provides another example in which the momentum π0 is constrained to van-

ish

π0 =∂L(x)

∂V0(x)= 0 (3.65)

and the equation of motion for the field V 0 (3.62)

∂i∂L

∂∂iV 0=

∂L

∂V 0(3.66)

is

− ∂iF i0 = ∂iFi0 = −m2V0 − J0 (3.67)

or with πi = − F 0i = F i0

∂iFi0 = ∇ · π = m2V 0 + J0. (3.68)

This last equation constrains the ∇ · π to be m2V 0 + J0 at each point x, t.

To cope with such problems and thereby to extend the variety of quantum

field theories, Dirac invented a kind of bracket that is different from his 〈a|b〉bracket.

3.6 Dirac Brackets

First, one makes a list of all the constraints

χ = 0 (3.69)

that occur due to the definitions of the conjugate momenta and to the equa-

tions of motion.

A constraint χf = 0 is first class if its Poisson bracket with all the other

constraints vanishes

[χf , χa]P = 0. (3.70)

One satisfies these constraints by choosing a gauge.

One then makes a matrix of the Poisson brackets of the second-class

constraints

Cab = [χa, χb]P . (3.71)

86 Action

In computing this matrix C, one does not apply the constraints (3.69) until

after one has computed their Poisson brackets. Otherwise C would vanish.

Dirac defined his new bracket in terms of the second-class constraints as

[A,B]D = [A,B]P − [A,χa]P (C−1)ab [χb, B]P . (3.72)

This new Dirac bracket inherits the algebraic properties of Poisson’s bracket

[A,B]D = − [B,A]D,

[A,BC]D = [A,B]D C +B [A,C]D,

[A, [B,C]D]D + [B, [C,A]D]D + [C, [A,B]D]D = 0.

(3.73)

Dirac then extended his rule that the commutator of two operators should

be i~ times their Poisson bracket to his new rule that it should be i~ times

their Dirac bracket

[A,B] = i~ [A,B]D. (3.74)

3.7 Dirac Brackets and QED

The action density is

L = − 14FikF

ik − JiAi. (3.75)

The usual rules give the momenta as

πi =∂L

∂(∂0Ai)(3.76)

and the commutation relations as

[Ai(x, t), πk(y, t)] = iδki δ

3(x− y). (3.77)

These equations contain the first-class constraints

π0 = 0

∇ · π − J0 = 0.(3.78)

We first choose a gauge that satisfies these first-class constraints. One

such gauge is the radiation or Coulomb gauge

∇ ·A = 0. (3.79)

In this gauge, the first-class constraints say that

− ∂iF i0 = −∇2A0 +∇ · ∇A = −∇2A0 = J0. (3.80)


So we can solve for A0

A0(x, t) =

∫d3y

J0(y, t)

4π|x− y|. (3.81)

Once we have eliminated A0 and π0 from the theory, we list the second-

class constraints

χ1(x, t) = ∇ ·A(x, t) = 0

χ2(x, t) = ∇ · π(x, t) + J0(x, t) = 0. (3.82)

Their Poisson bracket is

[χ1(x, t), χ2(y, t)]P =∑i

∫d3z

∂∇ ·A(x, t)

∂Ai(z)

∂∇ · π(y, t) + J0(y, t)

∂πi(z)

− ∂∇ · π(y, t) + J0(y, t)

∂Ai(z)

∂∇ ·A(x, t)

∂πi(z). (3.83)

The second terms vanish as do terms with J0. Since

∂∇ ·A(x, t)

∂Ai(z)= ∂iδ

3(x− z) and∂∇ · π(y, t)

∂πi(z)= ∂iδ

3(y− z), (3.84)


[χ1(x, t), χ2(y, t)]P = −∇2δ3(x− y). (3.85)

The other Poisson brackets vanish, so

C1x,2y = −∇2δ3(x− y) = −C2y,1x

C1x,1y = 0 (3.86)

C2x,2y = 0.

The matrix C is nonsingular, detC 6= 0. Its inverse is

(C−1)1x,2y = −∫

d3k

(2π)3

eik·(x−y)

k2= − 1

4π|x− y| = − (C−1)2y,1x (3.87)

in which the last minus sign is the one I missed in class. (Incidentally, there

are typos in the b-ok.org versions of Weinberg’s QTFI which I put on the

class website. Does anyone have a pdf of the 2005 paperback edition?)

88 Action

To check the formula (3.87) for the inverse C−1, we do the integral

∫d3z C1x,2z(C

−1)2z,1y = −∫d3z C1x,2z(C

−1)1y,2z

= −∫d3z [−∇2δ3(x− z)][−

∫d3k

(2π)3

eik·(y−z)

k2]

= −∫d3z δ3(x− z)∇2

∫d3k

(2π)3

eik·(y−z)

k2

=

∫d3z δ3(x− z)

∫d3k

(2π)3eik·(y−z)

=

∫d3z δ3(x− z)δ(y− z) = δ(x− y). (3.88)

To evaluate the Dirac bracket [Ai(x, t), πk(y, t)]D we need to compute the

Poisson brackets

[Ai(x, t), χ2y]P =∑k

∫d3z

[∂Ai(x, t)

∂Ak(z, t)

∂∇ · π(y, t) + J0(y, t)

∂πk(z, t)

− ∂∇ · π(y, t) + J0(y, t)

∂Ak(z, t)

∂Ai(x, t)

∂πk(z, t)

]=∑k

∫d3z

∂Ai(x, t)

∂Ak(z, t)

∂∇ · π(y, t)

∂πk(z, t)(3.89)

=∑k

∫d3z δikδ

3(x− z) ∂kδ3(y− z)

=∂

∂yiδ3(x− y) = − ∂

∂xiδ3(x− y)


and

[πi(x, t), χ1y]P =∑k

∫d3z

[∂πi(x, t)

Ak(z, t)

∂∇ ·A(y, t)

∂πk(z, t)

− ∂∇ ·A(y, t)

Ak(z, t)

∂πi(x, t)

∂πk(z, t)

]= −

∫d3z

∂∇ ·A(y, t)

Ak(z, t)

∂πi(x, t)

∂πk(z, t)

= −∫d3z

∂

∂ykδ(y− z) δki δ(x− z) (3.90)

= −∫d3z

∂

∂yiδ(y− z) δ(x− z)

= − ∂

∂yiδ(y− x)

=∂

∂xiδ(y− x) =

∂

∂xiδ(x− y).

So finally we compute the Dirac bracket (3.72) of Ai(x) with πk(y) at

equal times

[Ai(x, t),πk(y, t)]D = [Ai(x, t), πk(y, t)]P (3.91)

−∫d3z d3w [Ai(x, t), χ2(z, t)]P (C−1)2z,1w[χ1(w, t), πk(y, t)]P .

The first Poisson bracket is

[Ai(x, t), πk(y, t)]P =∑j

∫d3z

[∂Ai(x, t)

∂Aj(z)

∂πk(y, t)

∂πj(z)− ∂πk(y, t)

∂Aj(z)

∂Ai(x, t)

∂πj(z)

]

=∑j

∫d3z

∂Ai(x, t)

∂Aj(z)

∂πk(y, t)

∂πj(z)(3.92)

=∑j

∫d3z δijδ

3(x− z)δjkδ3(y− z) = δik δ

3(x− y).

The matrix element (3.87) of C−1 is

(C−1)2z,1w =1

4π|z−w| . (3.93)

The Poisson brackets (3.89) and (3.90) are

[Ai(x, t), χ2y]P = − ∂

∂xiδ3(x− y)

[πi(x, t), χ1y]P =∂

∂xiδ(x− y).

(3.94)

90 Action

So the second term T2 in the Dirac bracket (3.91) is

T2 = −∫d3z d3w [Ai(x, t), χ2(z, t)]P

1

4π|z−w| [χ1(w, t), πk(y, t)]P

=

∫d3z d3w

∂

∂xiδ3(x− z)

1

4π|z−w|∂

∂wkδ(w− y)

=

∫d3z d3w δ3(x− z)

∂

∂zi1

4π|z−w|∂

∂wkδ(w− y)

=

∫d3z d3w δ3(x− z)

∂2

∂zi∂zk1

4π|z−w| δ(w− y)

=∂2

∂xi∂xk1

4π|x− y| . (3.95)

So Dirac’s rule of quantization says that the commutator of Ai with πk is

the transverse delta function

[Ai(x, t), πk(y, t)] = i[Ai(x, t), πk(y, t)]D

= iδik δ3(x− y) + i

∂2

∂xi∂xk1

4π|x− y| (3.96)

and that

[Ai(x, t), Ak(y, t)] = 0 and [πi(x, t), πk(y, t)] = 0. (3.97)

Apart from matter terms, the hamiltonian for QED is

H =

∫d3x[π⊥ · A− L

](3.98)

in which π⊥ is the transverse part of π, that is, the part that has no diver-

gence

∇ · π⊥ = 0. (3.99)

The longitudinal part π‖ of π does not contribute because it is the gradient

of a scalar S which when dotted into a transverse vector vanishes as an

integration by parts shows∫d3x∇S · A = −

∫d3xS∇ · A = 0. (3.100)

So we can take the conjugate momentum π to be transverse

∇ · π⊥ = 0. (3.101)

in the term π · A The transverse part π⊥ is

π⊥ =∂L

∂A−∇A0 = A. (3.102)


The hamiltonian for QED is then

H =

∫d3x[π⊥ · A− L

]=

∫d3x[π⊥

2 − L]

=

∫d3x[A

2 − L]

(3.103)

in which L is the action density

L = − 14FikF

ik + JiAi = 1

2(A+∇A0)2 − 12(∇×A)2 + JiA

i. (3.104)

So H is

H =

∫d3x

[π⊥

2 − 12(π⊥ +∇A0)2 + 1

2(∇×A)2 − J ·A+ J0A0]

(3.105)

apart from terms due to the matter fields. The integral of π⊥ ·∇A0 vanishes

so H is

H =

∫d3x

[12π⊥

2 + 12(∇×A)2 − J ·A+ J0A0 − 1

2(∇A0)2]. (3.106)

By Gauss’s law and the Coulomb gauge condition ∇ ·A = 0

− ∂iF i0 = −∇2A0 = J0, (3.107)

so the last term in H is∫d3x − 1

2(∇A0)2 =

∫d3x 1

2A0∇2A0 =

∫d3x − 1

2A0J0. (3.108)

The hamiltonian then is

H =

∫d3x

[12π⊥

2 + 12(∇×A)2 − J ·A+ 1

2J0A0

](3.109)

in which the last term is the Coulomb energy

VC = 12

∫d3xJ0A0 =

1

2

∫d3x d3y

J0(x, t)J0(y, t)

4π|x− y| . (3.110)

One may satisfy the commutation relations (3.96 & 3.97) by setting

A(x) =

∫d3p√

(2π)32p0

∑s

[eip·x e(p, s) a(p, s) + e−ip·x e∗(p, s) a†(p, s)

](3.111)

and using π⊥(x) = A

π⊥(x) = − i∫ √

p0 d3p√(2π)32

∑s

[eip·x e(p, s) a(p, s)− e−ip·x e∗(p, s) a†(p, s)

](3.112)

in both of which x = (x, t) and

[a(p, s), a†(q, t)] = δst δ3(p− q) and [a(p, s), a(q, t)] = 0. (3.113)

92 Action

One gets the transverse delta function because the two polarization 3-vectors

e(p, s) are perpendicular to p and obey the sum rule∑s

ei(p, s) ek∗(p, s) = δik −pipkp2

. (3.114)

The 3-vectors e(p, s) can be thought of as the vector parts of 4-vectors with

vanishing time components.

4

Quantum Electrodynamics

4.1 Global U(1) Symmetry

In most theories of charged fields, the action density is invariant when the

charged fields change by a phase transformation

ψ′`(x) = eiθ`ψ`(x). (4.1)

If the phase θ is independent of x, then the symmetry is called a global

U(1) symmetry.

Section 3.4 describes Noether’s theorem according to which a current

Ja =∑`

∂L

∂∂aψ`δψ` (4.2)

has zero divergence

∂aJa = 0. (4.3)

The charge

Q =

∫J0 d3x (4.4)

is conserved due to the global U(1) symmetry. Here for tiny θ`, δψ` = iθ`ψ`,

and the charge density J0 is

J0 =∑`

∂L

∂∂0ψ`δψ` =

∑`

∂L

∂∂0ψìθ`ψ` =

∑`

πìθ`ψ`. (4.5)

One imagines that the angle is proportional to the charge of the field ψ`,

θ` = q`θ.

94 Quantum Electrodynamics

4.2 Abelian gauge invariance

Quantum electrodynamics is a theory in which the action density is invariant

when the charged fields change by a phase transformation that varies with

the spacetime point x

ψ′`(x) = eiθ`(x)ψ`(x). (4.6)

Such a symmetry is called a local U(1) symmetry. A theory with a local

U(1) symmetry also has a global U(1) symmetry and so it conserves charge.

Quantities like ψ†`(x)ψ`(x) are intrinsically invariant under local U(1)

symmetries because

(ψ†`(x))′ ψ′`(x) = ψ†`(x)e−iθ`(x) eiθ`(x)ψ(x) = ψ†`(x)ψ(x). (4.7)

Derivatives of fields present problems, however, because

∂a(eiθ`(x)ψ`(x)

)= eiθ`(x)

[∂a(ψ`(x)

)+ i(∂aθ(x))ψ`(x)

]6= eiθ`(x)∂a

(ψ`(x)

)(4.8)

so things like ψ†`(x) ∂aψ`(x) and like (∂aψ†`(x)) ∂aψ`(x) are not invariant

under local phase transformations.

The trick is to introduce a field Aa(x) that transforms so as to cancel the

awkward term eiθì(∂aθ)ψ` in the derivative (4.8). We want a new derivative

Daψ` that transforms like ψ` .

(Daψ`)′ = eiθ`Daψ`. (4.9)

So we set

Da = ∂a + iAa (4.10)

and require that

(Daψ`)′ = (∂a + iA′a)ψ

′a = eiθ`Daψ` = eiθ`(∂a + iAa)ψ`. (4.11)

That is, we insist that

(∂a + iA′a)ψ′a = (∂a + iA′a)e

iθ`ψa = eiθ`(∂a + iAa)ψ`. (4.12)

So we need

i∂aθ` + iA′a = iAa (4.13)

or

A′a(x) = Aa(x)− ∂aθ`. (4.14)

This is what we need except that the field Aa does not carry the index `.

4.2 Abelian gauge invariance 95

The solution to this problem is to define the symmetry transformation

(4.6) so that the angle θ` is proportional to the charge of the field ψ`

ψ′`(x) = eiq`θ(x)ψ`(x). (4.15)

This definition has the advantage that the charge density (4.5) becomes

J0 = iθ∑`

q` π`ψ` (4.16)

which makes more sense than the old formula (4.5). More importantly, the

definition (4.15) means that equations (4.9–4.22) change to

(Daψ`)′ = eiq`θDaψ`. (4.17)

So we make the covariant derivative

Daψ` = (∂a + iqÀa)ψ` (4.18)

depend upon the field ψ` and require that

(Daψ`)′ = (∂a + iqÀ

′a)ψ′a = eiq`θDaψ` = eiq`θ(∂a + iqÀa)ψ`. (4.19)

That is, we insist that

(∂a + iqÀ′a)ψ′a = (∂a + iqÀ

′a)e

iq`θψa = eiq`θ(∂a + iqÀa)ψ`. (4.20)

So we need

iq`∂aθ + iqÀ′a = iqÀa (4.21)

or

A′a(x) = Aa(x)− ∂aθ(x) (4.22)

which is much better than the old rule (4.14). The twin rules

ψ′`(x) = eiq`θ(x)ψ`(x)

A′a(x) = Aa(x)− i∂aθ(x)(4.23)

constitute an abelian gauge transformation.

Note that a field ψ` couples to the electromagnetic field Aa in a way

qÀaψ` that is proportional to its charge q`.

To insure that the right charge q` appears in the right place, we can

introduce a charge operator q such that q ψ` = q`ψ` and redefine the abelian

gauge transformation (4.23) as

ψ′`(x) = eiqθ(x)ψ`(x) = eiq`θ(x)ψ`(x)

A′a(x) = Aa(x)− ∂aθ(x).(4.24)


So

∂bA′a = ∂bAa − ∂b∂aθ (4.25)

is not invariant, but the antisymmetric combination

∂bA′a − ∂aA′b = ∂bAa − ∂b∂aθ − ∂aAb + ∂a∂bθ = ∂bAa − ∂aAb (4.26)

is invariant. Maxwell introduced this combination

Fba = ∂bAa − ∂aAb. (4.27)

Thus

L = −1

4FbaF

ba (4.28)

is a Lorentz-invariant, gauge-invariant action density for the electromagnetic

field.

4.3 Coulomb-gauge quantization

The first step in the canonical quantization of a gauge theory is to pick a

gauge. The most physical gauge for electrodynamics is the Coulomb gauge

defined by the gauge condition

0 = ∇ · ~A. (4.29)

If the action density (4.28) is modified by an interaction with a current Ja

L = −1

4FbaF

ba +Aa Ja (4.30)

then the equation of motion is

∂bFba = − Ja (4.31)

while the homogeneous equations

0 = ∂aFbc + ∂bFca + ∂cFab (4.32)

follow from the antisymmetry of Fab.

Because of the antisymmetry of Fab, the derivative A0 does not appear

in the action density (4.30). So for a = 0, the equation of motion (4.31)

actually is a constraint

∂iFi0 = − ∂iEi = − J0 (4.33)

known as Gauss’s law

∇ · ~E = ρ = J0. (4.34)

4.4 QED in the interaction picture 97

This constraint together with the Coulomb gauge condition (4.29) lets us

express A0 in terms of the charge density ρ = J0. We find

∇2A0 − ∂0∇ · ~A = ∇ · E = J0 (4.35)

or

∇2A0 = J0. (4.36)

The solution is

A0(~x, t) =

∫J0(~y, t)

4π|~x− ~y|d3y. (4.37)

The quantum fields in this gauge are the transverse parts of ~A and their

conjugate momentum ~E.

4.4 QED in the interaction picture

The Coulomb-gauge hamiltonian in the interaction picture is

H = H0 + V

H0 =1

2

∫~E2 + ~B2 d3x+Hψ,0

V = −∫

~J · ~Ad3x+ VC + Vψ

(4.38)

in which B = ∇× ~A and both ~E and ~A are travsverse, that is

∇ · ~E = 0 and ∇ · ~A = 0 (4.39)

and the Coulomb potential energy is

VC =1

2

∫J0(~x, 0)J0(~y, 0)

4π|~x− ~y|d3xd3y. (4.40)

The electromagnetic field is

Ab(x) =

∫d3p√

(2π)32p0

∑s

[eip·xeb(p, s)a(p, s) + e−ip·xeb∗(p, s)a†(p, s)

].

(4.41)

The polarization vectors may be chosen to be

eb(p,±1) = R(p)1√2

0

1

±i0

(4.42)


where R(p) rotates z into p. Thus

~p · ~e(p, s) = 0 (4.43)

and

e0(p, s) = 0 (4.44)

because A0 is a dependent variable (4.37). The commutation relations are

[a(p, s), a†(p′, s′)] = δss′ δ3(~p− ~p′) and [a(p, s), a(p′, s′)] = 0. (4.45)

The term H0 in the hamiltonian (4.38) is

H0 =

∫ ∑s

1

2p0[a(p, s), a†(p, s)]+ d

3p

=

∫ ∑s

p0

(a†(p, s)a(p, s) +

1

2δ3(~p− ~p)

)d3p. (4.46)

The interaction is

V (t) = eiH0t[−∫~J(~x, 0)· ~A(~x, 0)d3x+

∫J0(~x, 0)J0(~y, 0)

8π|~x− ~y|d3xd3y+Vm(0)

]e−iH0t

(4.47)

in which Vm(0) is the non-electromagnetic part of the matter interaction.

Since A0 = 0, ~J · ~A = J ·A.

4.5 Photon propagator

The photon propagator is

− i∆ab(x− y) = 〈0|T [Aa(x), Ab(y)]|0〉. (4.48)

Inserting the formula for the electromagnetic field (4.41), we get

−i∆ij(x− y) =

∫d3p

(2π)32p0

(δij −

pipj~p2

)×[eip·(x−y)θ(x0 − y0) + e−ip·(x−y)θ(y0 − x0)

]=

∫d4q

(2π)4

−iq2 − iε

(δij −

qiqj~q2

)eiq·(x−y) (4.49)

with the understanding that this propagator vanishes when a or b is zero. If

nc = (1, 0, 0, 0) has only a time component, one can write the δij − qiqj/~q2

as (δab −

qaqb~q2

)θ(ab) = ηab +

q0(qanb + qbna)− qaqb + q2nanb~q2

(4.50)

4.5 Photon propagator 99

in which q0 is arbitrary and may be chosen to be determined by conservation

of energy. The terms qanb, qbna, and qaqb act like ∂anb, ∂bna, and ∂a∂b so they

appear in the S-matrix as ∂aJanb, and ∂bJ

bna, etc., which vanish because

of current conservation. The remaining term

q2nanb~q2

−iq2 − iε

=−inanb~q2

(4.51)

gives in the S-matrix a term

T =1

2

∫d4xd4y[−iJ0(x)][−iJ0(y)]

−i(2π)4

∫d4q

~q2eiq·(x−y)

= i1

2

∫d4xd4yJ0(x)J0(y)

d3q

(2π)3

δ(x0 − y0)

~q2ei~q·(~x−~y) (4.52)

= i1

2

∫d3x d3y dt

J0(~x, t)J0(~y, t)

4π|~x− ~y|(4.53)

which cancels the Coulomb term

TC = −i12

∫d3x d3y dt

J0(~x, t)J0(~y, t)

4π|~x− ~y|. (4.54)

The bottom line is that the effective photon propagator is

− i∆ab(x− y) =

∫d4q

(2π)4

−iηabq2 − iε

eiq·(x−y) = −i∆ab(y − x). (4.55)

Dirac’s action density is

Lψ = − ψ(γa∂a +m)ψ (4.56)

with ψ = iψ†γ0. The conjugate momentum is

π =∂L

∂ψ= −ψγ0 = −iψ†γ0γ0 = iψ†. (4.57)

The hamiltonian is

H =

∫(πψ − L)d3x =

∫[iψ†ψ + ψ(γa∂a +m)ψ] d3x

=

∫ψ(~γ · ~∇+m)ψ d3x.

(4.58)

The free Dirac field aka the Dirac field in the interaction picture is

ψ(x) =

∫d3p

(2π)3/2

∑s

[u(p, s)eipxa(p, s) + v(p, s)e−ipxb†(p, s)

](4.59)


The addition and deletion operators obey the anticommutation relations

[a(p, s), a†(p′, s′)]+ = [b(p, s), b†(p′, s′)]+ = σss′ δ3(~p− ~p′) (4.60)

with the other anticommutators equal to zero. Putting ψ(x) into H gives

H =

∫ ∑s

p0[a†(p, s)a(p, s) + b†(p, s)b(p, s)− δ3(~p− ~p)

]d3p. (4.61)

4.6 Feynman’s rules for QED


L = −1

4FabF

ab − ψ[γa(∂a + ieAa) +m]ψ. (4.62)

The electric current is

Ja =∂L

∂Aa= −ieψγaψ. (4.63)

The interaction is

V (t) = ie

∫ψ(~x, t)γaψ(~x, t)Aa(~x, t) d

3x+ VC(t) (4.64)

Draw all appropriate diagrams.

Label each vertex with α = 1, 2, 3, or 4 on an electron line of momentum

p entering the vertex and β = 1, .., 4 on an electron line of momentum p′

leaving the vertex, and an index a on the photon line of momentum q. The

vertex carries the factor

(2π)4eγaβαδ4(p− p′ + q). (4.65)

An outgoing electron line gives

uβ(p, s)

(2π)3/2. (4.66)

An outgoing positron line gives

vα(p, s)

(2π)3/2. (4.67)

An incoming electron line gives

uα(p, s)

(2π)3/2. (4.68)


An incoming positron line gives

vβ(p, s)

(2π)3/2. (4.69)

An outgoing photon gives

e∗a(p, s)√(2π)32p0

. (4.70)

An incoming photon gives

ea(p, s)√(2π)32p0

. (4.71)

An internal electron line of momentum p from vertex β to vertex α gives

−i(2π)4

(−i/p+m)αβ

p2 +m2 − iε. (4.72)

An internal photon line of momentum q linking vertexes a and b gives

−i(2π)4

ηabq2 − iε

. (4.73)

Integrate over all momenta, sum over all indexes.

Add up all terms and get the combinatorics and minus signs right.

4.7 Electron-positron scattering

There are two Feynman diagrams for electron-positron scattering at order

e2 where α = e2/(4πε0~c) = 0.00729735256 ≈ 1/137 is the fine-structure

constant. The direct or s-channel diagram is represented in Fig. 4.1.


S-channel electron-positron scattering

Figure 4.1 Direct or s-channel diagram for electron-positron scatteringe− + e+ → e− + e+ via electron-positron annihilation into an electron-positron pair (arbitrary colors).

Feynman’s rules give for the annihilation diagram

As =

∫d4q (2π)4eγaβαδ

4(p+ k − q)(2π)4eγbγδδ4(−p′ − k′ + q)

× uγ(p′, s′)

(2π)3/2

vδ(k′, t′)

(2π)3/2

uα(p, s)

(2π)3/2

vβ(k, t)

(2π)3/2

−i(2π)4

ηabq2 − iε

= (2π)4eγaβα(2π)4eγbγδδ4(p+ k − p′ − k′)

× uγ(p′, s′)

(2π)3/2

vδ(k′, t′)

(2π)3/2

uα(p, s)

(2π)3/2

vβ(k, t)

(2π)3/2

−i(2π)4

ηab(p+ k)2

(4.74)

= eγaβαeγbγδδ

4(p+ k − p′ − k′)

× uγ(p′, s′)vδ(k′, t′)uα(p, s)vβ(k, t)

−i(2π)2

ηab(p+ k)2

= e2vβ(k, t)γaβαuα(p, s)uγ(p′, s′)γbγδvδ(k′, t′)δ4(p+ k − p′ − k′)

× −i(2π)2

ηab(p+ k)2

= − i e2

(2π)2

δ4(p+ k − p′ − k′)(p+ k)2

vβ(k, t)γaβαuα(p, s)uγ(p′, s′)γaγδvδ(k′, t′).


T-channel electron-positron scattering

Figure 4.2 Exchange or t-channel diagram for electron-positron scattering(arbitrary colors).

Suppressing indexes, we get

As = − i e2

(2π)2δ4(p+ k − p′ − k′) vγ

au u′γav′

(p+ k)2. (4.75)

The exchange diagram is

At = ± i e2

(2π)2δ4(p+ k − p′ − k′) u

′γau vγav′

(p− p′)2, (4.76)

and is illustrated in Fig. 4.2 and we must check our minus signs.

The initial state is |p, s; k, t〉 = a†(p, s)b†(k, t)|0〉 and the final state is

|p′, s′; k′, t′〉 = a†(p′, s′)b†(k′, t′)|0〉. The direct diagram comes from

T [ψ(x)γaψ(x)Aa(x)ψ(y)γbψ(y)Ab(y)]

= (ψ)−(x)γaψ−(x)T [Aa(x)Ab(y)](ψ)+(y)γbψ+(y) (4.77)

∼ a†(p′s′)γab†(k′, t′)T [Aa(x)Ab(y)]b(k, s)γba(p, s),


while the exchange diagram illustrated in Fig. 4.2comes from

T [ψ(x)γaψ(x)Aa(x)ψ(y)γbψ(y)Ab(y)]

= (ψ)+(x)γaψ−(x)T [Aa(x)Ab(y)](ψ)−(y)γbψ+(y)

∼ b(k, t)γab†(k′, t′)T [Aa(x)Ab(y)]a†(p′, s′)γba(p, s) (4.78)

∼ a†(p′, s′)b(k, t)γab†(k′, t′)T [Aa(x)Ab(y)]γba(p, s)

∼ − a†(p′, s′)γab†(k′, t′)T [Aa(x)Ab(y)]b(k, t)γba(p, s)

which differs by a minus sign. So the total amplitude is

A = As +At

= − i e2

(2π)2δ4(p+ k − p′ − k′) vγ

au u′γav′

(p+ k)2

−[− i e2

(2π)2δ4(p+ k − p′ − k′) u

′γbu vγbv′

(p− p′)2

](4.79)

= − i e2

(2π)2δ4(p+ k − p′ − k′)

[vγau u′γav

′

(p+ k)2− u′γbu vγbv

′

(p− p′)2

]].

The probability is the square |A|2. To compute such things, we need to know

some trace identities.

4.8 Trace identities

Since γ5 is a fourth spatial gamma matrix (1.173), its square is unity, γ5γ5 =

1, and it anticommutes with the ordinary gammas. So since the trace is

cyclic,

Tr(γa) = Tr(γ5γ5γa) = −Tr(γ5γaγ5) = −Tr(γ5γ5γa) = −Tr(γa) (4.80)

the trace of a single gamma matrix is zero.

The trace of two gammas

Tr(γaγb) = Tr(2ηab − γbγa) (4.81)

is

Tr(γaγb) = ηabTr(1) = 4ηab. (4.82)

The trace of an odd number of gammas vanishes because

Tr(γa1 · · · γa2n+1) = Tr(γ5γ5γa1 · · · γa2n+1) = − Tr(γ5γa1 · · · γa2n+1γ5)

= − Tr(γ5γ5γa1 · · · γa2n+1) = −Tr(γa1 · · · γa2n+1) (4.83)


which implies that

Tr(γa1 · · · γa2n+1) = 0. (4.84)

To find the trace of four gammas, we use repeatedly the fundamental rule

γaγb = 2ηab − γbγa. (4.85)

We thus find

Tr(γaγbγcγd) = Tr[(2ηab − γbγa)γcγd] = Tr[2ηabγcγd − γbγaγcγd]= Tr[2ηabγcγd − γb(2ηac − γcγa)γd]= Tr[2ηabγcγd − 2ηacγbγd + γbγcγaγd] (4.86)

= Tr[2ηabγcγd − 2ηacγbγd + 2ηadγbγc − γbγcγdγa]

or

Tr(γaγbγcγd) = Tr[ηabγcγd − ηacγbγd + ηadγbγc]

= 4(ηabηcd − ηacηbd + ηadηbc

). (4.87)

Reversing completely the order of the gammas changes nothing:

Tr(γaγb · · · γyγz

)= Tr

(γzγy · · · γbγa

). (4.88)

4.9 Electron-positron to muon-antimuon

To avoid the extra complications that the sum of two amplitudes As + Atentail, let’s consider the process e−(p) + e+(p′) → µ−(k) + µ+(k′) which

requires only the direct amplitude

A = − i e2

(2π)2δ4(p+ p′ − k − k′) v(p′, s′)γau(p, s) u(k, t)γav(k′, t′)

(p+ p′)2

= − 2πi δ4(p+ p′ − k − k′)M(4.89)

where

M =e2

(2π)3

v(p′, s′)γau(p, s) u(k, t)γav(k′, t′)

(p+ p′)2(4.90)

in which I have relabeled the momenta so that k& k′ belong to the muons.

The probability |A|2 includes the term

vγau uµγavµ (vγbu uµγbvµ)∗ (4.91)

in which µ’s label muons. Since v = iv†γ0 = v†β as well as β2 = (iγ0)2 = 1,

and since (1.177) βγa†β = −γa, we get

(vγau)∗ = (v†βγau)∗ = u†γa†βv = u†ββγa†βv = u†βγav = uγav. (4.92)


So the electron part of the term (4.91) is

vγau (vγbu)∗ = vγau uγbv. (4.93)

The spin sums are∑s

u`(~p, s)u∗m(~p, s) =

[ 1


]`m∑

s

v`(~p, s)v∗m(~p, s) =

[ 1


]`m.

(4.94)

Equivalently and more simply,∑s

u`(~p, s)um(~p, s) =[ 1

2p0(−i pcγc +m)

]`m∑

s

v`(~p, s)vm(~p, s) =[ 1

2p0(−i pcγc −m)

]`m.

(4.95)

So∑s,s′

vγau uγbv =∑

s′,j,`,m,n

vj(p′, s′)γajù`(p, s) um(p, s)γbmnvn(p′, s′)

=1

2p0

∑s′,j,`,m,n

vj(p′, s′)γaj`(−i/p+m)`mγ

bmnvn(p′, s′) (4.96)

=1

2p0

∑s′,j,`,m,n

vn(p′, s′)vj(p′, s′)γaj`(−i/p+m)`mγ

bmn

=1

2p02p′0

∑sj,`,m,n

(−i /p′ −m)njγaj`(−i/p+m)`mγ

bmn

=1

2p02p′0Tr[(−i /p′ −me)γ

a(−i/p+me)γb].

The muon part of the term (4.91) is

uγav (uγbv)∗ = uγav vγbu = vγbu uγav. (4.97)

So the sum over all spins gives for the muons∑t,t′

uγav vγbu =1

2k02k′0Tr[(−i /k′ −mµ)γb(−i/k +mµ)γa]. (4.98)

Using the formula (4.86) for the trace of 4 gammas, we evaluate the trace

for the electrons in stages:

Tr[(−i /p′ −me)γa(−i/p+me)γ

b] = − Tr(/p′γa/pγ

b)−m2

eTr(γaγb

). (4.99)


The 4-gamma term is

Tr[(−i /p′)γa(−i/p)γb] = − Tr[p′cγcγapdγ

dγb]

= − p′cpdTr[γcγaγdγb]

= − 4p′cpd(ηcaηdb − ηcdηab + ηcbηad

)= − 4

(p′apb − p′p ηab + p′bpc

). (4.100)

The 2-gamma term is

−m2eTr(γaγb

)= − 4m2

eηab. (4.101)

So their sum is

Tr[(−i /p′−me)γa(−i/p+me)γ

b] = −4[p′apb+p′bpa+ηab(m2

e−p′p)]. (4.102)

The similar term for muons is

Tr[(−i /k′−mµ)γa(−i/k+mµ)γb] = −4[k′akb+k′bka+ηab(m

2µ−k′k)

]. (4.103)

Their product is


b] Tr[(−i /k′ −mµ)γa(−i/k +mµ)γb] (4.104)

= 16[p′apb + p′bpa + ηab(m2

e − p′p)][k′akb + k′bka + ηab(m

2µ − k′k)

]= 32

[(p′k′)(pk) + (p′k)(pk′) +m2

µ(pp′) +m2e(k′k)].

So apart from the delta function and the (2π)’s, the squared amplitude

summed over final spins and averaged over initial spins is

1

4

∑s,s′

|M |2 =e4

2(2π)6p0p′0k0k′0(p′k′)(pk) + (p′k)(pk′) +m2

µ(pp′) +m2e(k′k)

(p+ p′)4

(4.105)

where (p+ p′)4 = [(p+ p′)2]2.

In the rest frame of a collider, q2 = (p+ p′)2 = 4E2 = 4p02. The cosine of

the scattering angle θ is cos θ = ~p · ~k/(|~p||~k|).Since me ≈ mµ/200, I will neglect me in what follows. One then gets

p · p′ = −2E2, p · k = E(|~k| cos θ−E), and p · k′ = p′ · k = −E(|~k| cos θ+E).


Thus

1

4

∑s,s′

|M |2 =e4[E2(E − k cos θ)2 + E2(E + k cos θ)2 + 2m2

µE2]

2(2π)6E416E4

=e4[(E − k cos θ)2 + (E + k cos θ)2 + 2m2

µ

]32(2π)6E6

=e4[2E2 + 2k2 cos2 θ + 2m2

µ

]32(2π)6E6

=e4[E2 + k2 cos2 θ +m2

µ

]16(2π)6E6

(4.106)

The squared amplitude is

|A|2 = (2π)2|δ4(p+ p′ − k − k′)|2 1

4

∑ss′

|M |2 (4.107)

=V T

(2π)2δ4(p+ p′ − k − k′) 1

4

∑ss′

|M |2. (4.108)

since

δ4(0) =

∫d4x

(2π)4=

V T

(2π)4(4.109)

in which V is the volume of the universe and T is its infinite time. We

also need to switch from delta-function normalization of states to unit nor-

malization. The relation is that between a continuum delta function and a

Kronecker delta

δ3(~p′ − ~p) =V

(2π)3δpp′ . (4.110)

Since there are two particles in the initial and final states, the probability is

P =[(2π)3

V

]4|A|2Nf (4.111)

where Nf is the number of final states. For a range of momenta d3kd3k′, the

number of final states is

Nf =[ V

(2π)3

]2d3kd3k′. (4.112)


So the rate P/T is

R =[(2π)3

V

]4 |A|2T

Nf =[(2π)3

V

]4 |A|2T

[ V

(2π)3

]2d3kd3k′

=[(2π)3

V

]4 |A|2T

Nf =[(2π)3

V

]2 |A|2T

d3kd3k′

=(2π)4

Vδ4(p+ p′ − k − k′) 1

4

∑ss′

|M |2d3kd3k′

(4.113)

The flux of incoming particles is u/V where u = |~v1 − ~v2 is the relative

velocity, which with c = 1 for massless electrons is u = 2. So the differential

cross-section is

dσ =1

2(2π)4δ4(p+ p′ − k − k′) 1

4

∑ss′

|M |2d3kd3k′. (4.114)

Integrating over d3k′ sets ~k′ = −~k, and so

dσ =1

2(2π)4δ(p0 + p′0 − 2k0)

1

4

∑ss′

|M |2d3k

=1

2(2π)4δ

(p0 + p′0 − 2

√k2 +m2

µ

) 1

4

∑ss′

|M |2k2dkdΩ

(4.115)

where k = |~k|. The derivative of the delta function is k2p0/k02, so

dσ =1

2(2π)4 1

4

∑ss′

|M |2k2dΩk02

2kp0= 2π4 1

4

∑ss′

|M |2kk02

p0dΩ

= 4π4 1

4

∑ss′

|M |2kk02

p0dΩ.

(4.116)


So then

dσ = 4π4e4[E2 + k2 cos2 θ +m2

µ

]4(2π)6E6

kk02

p0dΩ

=e4[E2 + k2 cos2 θ +m2

µ

]4(2π)2E5

k dΩ

=e4

4(2π)2E3

[1 +

k2

E2cos2 θ +

m2µ

E2

]k dΩ

=e4

4(2π)2E3

[1 +

E2 −m2µ

E2cos2 θ +

m2µ

E2

]k dΩ (4.117)

=e4

4(2π)2E2

√1−

m2µ

E2

[1 +

m2µ

E2+

(1−

m2µ

E2

)cos2 θ

]dΩ

=α2

16E2

√1−

m2µ

E2

[1 +

m2µ

E2+

(1−

m2µ

E2

)cos2 θ

]dΩ.

The total cross-section is the inetgral over solid angle

σ =α2

16E2

√1−

m2µ

E2

[(1 +

m2µ

E2

)4π +

(1−

m2µ

E2

)∫cos2 θ dΩ

]

=α2

16E2

√1−

m2µ

E2

[(1 +

m2µ

E2

)4π +

(1−

m2µ

E2

)∫cos2 θ 2πd cos θ

]

=πα2

4E2

√1−

m2µ

E2

[(1 +

m2µ

E2

)+

1

3

(1−

m2µ

E2

)]

=πα2

3E2

√1−

m2µ

E2

(1 +

1

2

m2µ

E2

)(4.118)

where E is the energy of each electron in the lab frame of the collider. The

energy of the collider is Ecm = 2E. At very high energies, the x-section is

σ = πα2/(3E2).

4.10 Electron-muon scattering 111

4.10 Electron-muon scattering

For the process e−(p1) +µ−(p2)→ e−(p′1) +µ−(p′2), the Feynman rules give

A = − i e2

(2π)2δ4(p1 + p2 − p′1 − p′2)

u(p′1, s′)γau(p1, s) u(p′2, t

′)γau(p2, t)

(p1 − p′1)2

= − 2πi δ4(p1 + p2 − p′1 − p′2)M. (4.119)

The product of the traces over q2 = (p1 − p′1)2 is

Tr[(−i /p′1 +me)γa(−i /p1 +me)γ

b] Tr[(−i /p′2 +mµ)γa(−i /p2 +mµ)γb]

(p1 − p′1)2.

(4.120)

For e+ − e− → µ+ + µ− we got


b] Tr[(−i /k′ −mµ)γa(−i/k +mµ)γb]

(p+ p′)2. (4.121)

If in this amplitude for e+ − e− → µ+ + µ− we make these replacements

p→ p1, p′ → −p′1, k → p′2, and k′ → −p2, (4.122)

it becomes

Tr[(i/p′1−me)γ

a(−i/p′1 +me)γb]Tr[(i/p2

−mµ)γa(−i/p′2 +mµ)γb]/(p1 + p′2)2

=Tr[(−i/p′1 +me)γ

a(−i/p′1 +me)γb] Tr[(−i/p2

+mµ)γa(−i/p′2 +mµ)γb]

(p1 − p′1)2

(4.123)

which is exactly the same as the amplitude (4.120) for e−(p1) + µ−(p2) →e−(p′1) +µ−(p′2) scattering. So apart from the delta function and the (2π)’s,

the squared amplitude summed over final spins and averaged over initial

spins is the one for e+ − e− → µ+ + µ− but with the substitutions (4.122).

That is, for e−(p1) + µ−(p2) → e−(p′1) + µ−(p′2) scattering, the summed

amplitudes are

1

4

∑s,s′,t,t′

|M |2 =e4[(p′1p2)(p1p

′2) + (p′1p

′2)(p1p2)−m2

µ(p1p′1)−m2

e(p2p′2)]

2(2π)6p01p′01 p

02p′02 (p1 − p′1)4

.

(4.124)

The equality of the two amplitudes (4.120) and (4.121) under the corre-

spondence (4.122) is an example of crossing symmetry. More generally,

the scattering amplitude for a process in which an incoming particle of 4-

momentum p that interacts with other particles is equal to the scattering


amplitude for the process in which the antiparticle of the incoming particle

leaves with 4-momentum −p after interacting with the same other particles

S(−p, nc; · · · | · · · ) = S(· · · |p, n; · · · ) (4.125)

in which type nc is labels the antiparticle of type n. Since −p0 + p0 = 0,

this is not the amplitude for a physical process in which each particle has its

appropriate energy, p0i =

√m2i + ~p2

i . This symmetry caused huge excitement

on the West Coast during the 1960s.

Let k = |~p′1| = p′01 be energy or equivalently the modulus (magnitude) of

the 3-momentum of the (massless) final electron. Let E = p02 be the energy

of the initial muon. Then the differential x-section in the lab frame of the

collider is

dσ

dΩ=

α2

2k2(E + k)2(1− cos θ)2

[(E + k)2 + (E + k cos θ)2 −m2

µ(1− cos θ)]

(4.126)

in which the cosine of the scattering angle of the electron is

cos θ =~p1 · ~p1

|~p1||~p1|. (4.127)

The total x-section σ is infinite.

4.11 One-loop QED


L = −1

4F abB FBab − ψB [γa (∂a + ieBA

aB) +mB]ψB (4.128)

in which the subscript B means that the values of these “bare” terms include

infinite quantities that may be used to cancel the infinities that arise in

perturbation theory. The actual physical terms are defined as

ψ ≡ Z−1/22 ψB (4.129)

Aa ≡ Z−1/23 AaB (4.130)

e ≡ Z1/23 eB (4.131)

m ≡ mB + δm (4.132)

so that in particular eAa = eBAaB. In these terms, the action density is

L = L0 + L1 + L2 (4.133)


Vacuum polarization

Figure 4.3 One-loop Feynman diagram for vacuum polarization (arbitrarycolors).

where L0 is the physical or naive free action density

L0 = − 1

4F abFab − ψ (γa∂

a +m)ψ, (4.134)

L1 is the interaction

L1 = − ieAaψγaψ, (4.135)

and L2 is a sum of infinite “counterterms”

L2 = − 1

4(Z3 − 1)F abFab − (Z2 − 1)ψ (γa∂

a +m)ψ

+ Z2 δm ψψ − ie(Z2 − 1)Aaψγaψ. (4.136)

Vacuum polarization is a one-loop correction to the photon propagator. Its

Feynman diagram is in figure 4.3. Let’s use L1 to compute it. The amplitude

S is

S = 〈q′, s′|T[ei

∫L1 d4x

]|q, s〉 = 〈q, s|T

[ee

∫Aa(x)ψ(x)γaψ(x) d4x

]|q, s〉.

(4.137)


The one-loop correction is due to the order-e2 term

S2 =1

2〈q′, s′|T

[e2

∫Aa(x)ψ(x)γaψ(x) d4x

∫Ab(y)ψ(y)γbψ(y) d4y

]|q, s〉.

(4.138)

We choose to absorb the photon at y and cancel the factor of one-half. The

electromagnetic field is

Ab(y) =

∫d3p√

(2π)32p0

∑s′

[eip·yeb(p, s

′)a(p, s′) + e−ip·ye∗b(p, s′)a†(p, s′)

].

(4.139)

So S2 is

S2 =e2

(2π)32q0〈0|T

[ ∫e−iq

′·xe∗a(q′, s′)ψ(x)γaψ(x) d4x

×∫eiq·yeb(q, s)ψ(y)γbψ(y) d4y

]|0〉. (4.140)

The term L1 is normally ordered (it should be surrounded by colons), so

fermions emitted by ψ(y) and ψ(y) must be absorbed by ψ(x) and ψ(x).

Thus S2 is

S2 =e2e∗a(q

′, s′)eb(q, s)

(2π)32q0

∫e−iq

′·xeiq·y〈0|T [ψα(x)γaαβψδ(y)]|0〉

× 〈0|T [ψβ(x)ψγ(y)γbγδ]|0〉 d4x d4y. (4.141)

Moving ψδ(y) to the left of ψα(x) adds a minus sign, so

S2 = − e2e∗a(q′, s′)eb(q, s)

(2π)32q0

∫e−iq

′·xeiq·y〈0|T [ψδ(y)ψα(x)γaαβ]|0〉

× 〈0|T [ψβ(x)ψγ(y)γbγδ]|0〉 d4x d4y. (4.142)

This minus sign is an instance of Feynman’s rule that closed fermion loops

make minus signs. In the common notation ψ = ψ†β, his propagator is

〈0|Tψ`(x)ψm(y)

|0〉 ≡ − i∆`m(x− y)β

= − i∫

d4q

(2π)4

(− i/q +m)`m


= − i∫

d4q

(2π)4

( 1

i/q +m− iε

)`meiq·(x−y).

(4.143)


Putting all this together, we get

S2 =e2e∗a(q

′, s′)eb(q, s)

(2π)32q0

∫e−iq

′·xeiq·y∫

d4k

(2π)4

(− i/k +m)δαk2 +m2 − iε

eik·(y−x)γaαβ

×∫

d4p

(2π)4

(− i/p+m)βγ

p2 +m2 − iεeip·(x−y)γbγδ d

4x d4y. (4.144)

Dirac’s delta functions simplify this 16-dimensional integral to

S2 =e2e∗a(q

′, s′)eb(q, s)δ4(q′ − q)

(2π)32q0

∫d4p

[− i(/p− /q) +m

]δα

(p− q)2 +m2 − iε

× γaαβ(− i/p+m)βγ

p2 +m2 − iεγbγδ. (4.145)

The product of matrices is a trace

S2 =e2e∗a(q

′, s′)eb(q, s)δ4(q′ − q)

(2π)32q0

∫d4p

Tr[− i(/p− /q) +m

]γa(− i/p+m)γb

[(p− q)2 +m2 − iε](p2 +m2 − iε)

.

(4.146)

Equivalently, since TrAB = TrBA, we have

S2 =e2e∗a(q

′, s′)eb(q, s)δ4(q′ − q)

(2π)32q0

∫d4p

Tr

(− i/p+m)γb[− i(/p− /q) +m

]γa

(p2 +m2 − iε)[(p− q)2 +m2 − iε].

(4.147)

or interchanging a and b

S2 =e2ea(q, s)e

∗b(q′, s′)δ4(q′ − q)

(2π)32q0

∫d4p

Tr

(− i/p+m)γa[− i(/p− /q) +m

]γb

(p2 +m2 − iε)[(p− q)2 +m2 − iε](4.148)

SW extracts from this the definition

Π∗ab(q) =−ie2

(2π)4

∫d4p

Tr

(− i/p+m)γa[− i(/p− /q) +m

]γb

(p2 +m2 − iε)[(p− q)2 +m2 − iε]. (4.149)

We use the Feynman trick

1

AB=

∫ 1

0

dx

[(1− x)A+ xB]2(4.150)


to write

1

(p2 +m2 − iε)[(p− q)2 +m2 − iε]=

∫ 1

0

dx(p2 +m2 − iε)(1− x)] + [(p− q)2 +m2 − iε]x

2

=

∫ 1

0

dx[p2 +m2 − iε− 2p · q x+ q2 x

]2(4.151)

=

∫ 1

0

dx[(p− xq)2 +m2 − iε+ x(1− x) q2

]2 .The next step is to make the integral finite somehow. One can use the

counterterms, one can regard quantum field theory as merely a way to in-

terpolate between values taken from experiment, or one can impose dimen-

sional regularization as we’ll see momentarily. If we assume that we have

made the integral finite one way or another, then we can shift p to p+ x q.

After this shift, the quantity (4.152) becomes

Π∗ab(q) =−ie2

(2π)4

∫ 1

0dx

∫d4p

Tr

(−i(/p+ x/q) +m)γa[− i(/p− /q(1− x)) +m

]γb[

p2 +m2 − iε+ x(1− x) q2]2 .

(4.152)

The trace identities of section 4.8 give

Tr

(−i(/p+ x/q) +m)γa[− i(/p− /q(1− x)) +m

]γb

= 4[− (p+ xq)a(p− (1− x)q)b + (p+ xq) · (p− (1− x)q) ηab

− (p+ xq)b(p− (1− x)q)a +m2ηab]. (4.153)

The variable p0 is being integrated along the real axis of the complex p0

plane from −∞ to +∞. We now imagine that we have made the p-integral

sufficiently finite that we can rotate the p0 contour by 90 degrees so that

it runs up the imaginary axis in the complex p0 plane from p0 = −i∞to p0 = i∞. This is a Wick rotation. We set p0 = ip4 and integrate over

p4 = −ip0, which is real, from −∞ to∞. We also set q0 = iq4 and q4 = −iq0.

So now

p · q = ~p · ~q − p0q0 = ~p · ~q + p4q4. (4.154)

All the Lorentz signs are now positive and

p0p0η00 = ip4ip4(−1) = p4p4 = p4p4δ44 (4.155)

as well as

q0q0η00 = iq4iq4(−1) = q4q4 = q4q4δ44. (4.156)


The quantity q2 is real whether q0 is real or imaginary; it is positive when

q0 < ~q2 whether or not q0 is imaginary.

We now can drop iε and express the quantity

Π∗ab(q) = Π∗ab(~q, q0) = Π∗ab(~q, iq4) (4.157)

either as

Π∗ab(q) =−4ie2

(2π)4

∫ 1

0dx

∫d4p

1[p2 +m2 − iε+ x(1− x) q2

]2×[− (p+ xq)a(p− (1− x)q)b + (p+ xq) · (p− (1− x)q) ηab

− (p+ xq)b(p− (1− x)q)a +m2ηab]

(4.158)

or with p4 = −ip0 real and d4pe = dp1dp2dp3dp4 as

Π∗ab(q) =4e2

(2π)4

∫ 1

0dx

∫d4pe

1[p2 +m2 + x(1− x) q2

]2×[− (p+ xq)a(p− (1− x)q)b + (p+ xq) · (p− (1− x)q) δab

− (p+ xq)b(p− (1− x)q)a +m2δab]

(4.159)

in which all Lorentz signs are +1 and q4 = −iq0 is imaginary if q is real.

The next step is to say how we regularize the integral over d4p. Since the

1970s, the method of choice has been dimensional regularization because

it preserves Lorentz and gauge invariance. In dimensional regularization,

the p integration is taken to be over an arbitrary number d of spacetime

dimensions.

The denominator of the integral (4.159) is a function of p2 which is taken

to be the square of the radius of a sphere in d dimensions. Terms in the

numerator that are odd powers of pa vanish. We set

papp =p2 δab

d. (4.160)

The area of a sphere of unit radius in d dimensions is

Ad =2πd/2

Γ(d/2)(4.161)

as we can verify by computing an integral of the gaussian exp(−x2) in d


dimensions using both rectangular and spherical coordinates.∫e−x

2ddx =

(∫ ∞−∞

e−x2dx

)d= πd/2

= Ad

∫ ∞0rd−1e−r

2dr = Ad

1

2Γ(d/2).

(4.162)

The integral (4.159) converges as long as d is complex or a fraction or

anything but an even integer. Nonetheless, the divergencies in that integral

reappear as powers of 1/(d−4) as d→ 4. We’ll use the counterterms (4.136)

in L2 to cancel them.

The area of a sphere of radius k =√p2 in d dimensions is Ad k

d−1. So

using the formulas (4.160 and 4.161) of dimensional regularization, we can

write the integral (4.159) as

Π∗ab(q) =4e2Ad(2π)4

∫ 1

0dx

∫ ∞0kd−1dk

1[k2 +m2 + x(1− x) q2

]2 (4.163)

×[− 2

k2δab

d+ 2qaqbx(1− x) + [k2 − x(1− x)q2]δab +m2δab

].

The needed integrals are∫ ∞0

kd−1

(k2 + c2)2dk = 1

2 (c2)d/2−2 Γ(d/2) Γ(2− d/2) (4.164)∫ ∞0

kd+1

(k2 + c2)2dk = 1

2 (c2)d/2−1 Γ(1 + d/2) Γ(1− d/2). (4.165)

Using these integrals, we get


∫ 1

0dx

[(1− 2

d

)δab(c2)d/2−1 Γ(1 + d/2) Γ(1− d/2)

(4.166)

+[(2qaqb − q2δab)x(1− x) +m2δab

](c2)d/2−2Γ(d/2)Γ(2− d/2)

]in which

c2 = m2 + x(1− x) q2. (4.167)

The identity

Γ(z + 1) = z Γ(z) (4.168)

implies that

Γ(2− d/2) = (1− d/2) Γ(1− d/2), (4.169)


and that

Γ(1 + d/2) = (d/2) Γ(d/2), (4.170)

so

Γ(1 + d/2) Γ(1− d/2) = (2/d− 1)−1Γ(d/2) Γ(2− d/2). (4.171)

So we can combine terms in the integral (4.172)


Γ(d/2)Γ(2− d/2)

∫ 1

0dx

[− δabc2 (4.172)

+[(2qaqb − q2δab)x(1− x) +m2δab

]](c2)d/2−2.

Using the value (4.167) of c2, we have

−δabc2 +(2qaqb−q2δab)x(1−x)+m2δab = 2(qaqb−q2δab)x(1−x). (4.173)

Thus we find


Γ(d/2)Γ(2− d/2)(qaqb − q2δab)

×∫ 1

0dxx(1− x)[m2 + x(1− x) q2]d/2−2.

(4.174)

Current conservation and gauge invariance imply that

qa Π∗ab(q) = 0. (4.175)

One of the advantages of dimensional regularization is that it preserves the

vanishing of this quantity

qa Π∗ab(q) ∝ qa(qaqb − q2δab) = q2qb − q2qb = 0. (4.176)

This works because current conservation and the dimension of spacetime

have nothing to do with each other:

The vanishing of qa Π∗ab(q) follows from the conservation

∂aJa(x) = ∂a[ψ(x)γaψ(x)] = 0 (4.177)

of the current

Ja(x) = ψ(x)γaψ(x) (4.178)

and from the neutrality of the current which is due to the fact that ψ lowers

the charge while ψ raises it.


Apart from a constant f , the quantity Π∗ab(q) is by (4.142)

Π∗ab(q) = f

∫d4xd4yeiq(y−x)〈0|T [Ja(x)Jb(y)]|0〉. (4.179)

So

qaΠ∗ab(q) = f

∫d4xd4y qa e

iq(y−x)〈0|T [Ja(x)Jb(y)]|0〉

= f

∫d4xd4y eiq(y−x)(i∂a) 〈0|T [Ja(x)Jb(y)]|0〉

= f

∫d4xd4y eiq(y−x)〈0|T [i∂aJ

a(x)Jb(y)]|0〉

+ f

∫d4xd4y i δ(x0 − y0) eiq(y−x)〈0|[J0(x), Jb(y)]|0〉

= f

∫d4xd4y i δ(x0 − y0) eiq(y−x)〈0|[J0(x), Jb(y)]|0〉

(4.180)

since the current is conserved. But the equal-time commutator of the charge

J0(x) with any operator is proportional to the charge of that operator be-

cause J0(x) generates the U(1) rotation of the symmetry that conserves

charge. So since currents are neutral, the commutator vanishes

δ(x0 − y0) eiq(y−x)[J0(x), Jb(y)] = 0, (4.181)

and we have

qaΠ∗ab(q) = 0. (4.182)

The gamma function Γ(2 − d/2) in Π∗ab(q) (4.174) blows up as d → 4.

So we bring in the contribution to Π∗ab(q) arising from the counterterm

−(Z3 − 1)FabFab. This quadratic form in derivatives of Aa and Ab adds to

Π∗ab(q) the term

Π∗ab(q)Z3 = − (Z3 − 1)(q2δab − qaqb) (4.183)

in euclidian space. So to order e2, Π∗ab(q) is

Π∗ab(q) = (q2δab − qaqb)π(q2) (4.184)

where

π(q) = − 4e2Ad(2π)4

Γ(d/2)Γ(2− d/2)

∫ 1

0dxx(1− x)[m2 + x(1− x) q2]d/2−2

− (Z3 − 1).

(4.185)

So we can use Z3 to cancel the poles in the gamma functions.


The quantity Π∗ab(q) is a correction to the photon propagator. And Π∗ab(0)

is a correction to the mass of the photon which we want to remain at zero.

So we set

Π∗ab(0) = 0. (4.186)

This means that set set

Z3 = 1− 4e2Ad(2π)4

Γ(d/2)Γ(2− d/2) (m2)d/2−2

∫ 1

0dxx(1− x), (4.187)

so to order e2

π(q) = − 4e2Ad(2π)4

Γ(d/2)Γ(2− d/2)

∫ 1

0dxx(1− x)

×[[m2 + x(1− x) q2]d/2−2 − (m2)d/2−2

].

(4.188)

As we let d→ 4, the tricky term is

[m2 + x(1− x) q2]d/2−2 − (m2)d/2−2 = exp[(d/2− 2) ln[m2 + x(1− x) q2]

]− exp

[(d/2− 2) ln(m2)]

]= (d/2− 2) ln[1 + x(1− x) q2/m2].

(4.189)

The final result is

π(q2) =e2

2π2

∫ 1

0x(1− x) ln

[1 + x(1− x) q2/m2

]dx. (4.190)

The scattering of two particles of charges e1 and e2 that exchange mo-

mentum q = p1 − p′1 = p′2 − p2 is

S =−ie1e2

4π2δ4(p′1 + p′2 − p1 − p2)u′1γ

au1 u′2γbu2

×[ηabq2

+[q2ηab − qaqb]π(q2)

(q2)2

].

(4.191)

The spinors obey the Dirac equation in momentum space (1.264)

(i pcγc +m)u(p, s) = 0 (4.192)

so

p1au′1γau1 = imu′1γ

au1. (4.193)

The adjoint of Dirac’s equation is

u†(p, s)(−i pcγ†c +m) = 0. (4.194)


The spatial gammas are hermitian while γ0 is antihermitian, so

iu†(p, s)(−i pcγ†c +m)γ0 = iu†(p, s)γ0(i pcγc +m) = u(p, s)(i pcγc +m) = 0.

Thus

p′1au′1γau1 = −imu′1γau1. (4.195)

Subtracting, we get

qau′1γau1 = (p′1a − p1a)u

′1γau1 = (−im− (−im))u′1γ

au1 = 0. (4.196)

So the amplitude (4.191) simplifies to

S =−ie1e2

4π2q2[1 + π(q2)]δ4(p′1 + p′2 − p1 − p2)u′1γ

au1 u′2γau2. (4.197)

For small ~p1 and ~p2,

u′1γ0u1 = iu′†1 (γ0)2u1 = −iu′†1 u1 ≈ −iδs′1s1 and iu′†1 (γ0)2u1 ≈ −iδs′2s2

while

u′1~γu1 = 0 and u′2~γu2 = 0.

So the amplitude is approximately

S =−ie1e2

4π2~q2[1 + π(~q2)]δ4(p′1 + p′2 − p1 − p2) δs′1s1δs′2s2 . (4.198)

The lowest, Born, approximation to the nonrelativistic scattering due to

a spin-independent potential V (|~x1 − ~x2|) which is

SB = −2πiδ(E′1 + E′2 − E1 − E2)TB (4.199)

where

TB =δs′1s1δs′2s2

(2π)6

∫d3x1d

3x2V (|~x1 − ~x2|)e−i~p′1·~x1e−i~p

′2·~x2ei~p1·~x1ei~p2·~x2 .

We set ~r = ~x1 − ~x2 and get

SB =−i4π2

δ4(p′1 + p′2 − p1 − p2)δs′1s1δs′2s2

∫V (r)e−i~q·~r d3r. (4.200)

Comparing this formula with (4.198), we have∫V (r)e−i~q·~r d3r = e1e2

1 + π(~q2)

~q2

or

V (r) =e1e2

(2π)3

∫ei~q·~r

[1 + π(~q2)

~q2

]. (4.201)


This is the potential that the extended charge distribution

c(~r) = δ3(~r) +1

2(2π)3

∫π(~q2) ei~q·~r d3q, (4.202)

which is normalized to unity∫c(~r) d3r = 1 +

π(0)

2= 1, (4.203)

would produce if separated from itself by ~r

V (r) = e1e2

∫d3x

∫d3y

c(~x)c(~y)

4π|~x− ~y + ~r|. (4.204)

Apart from a delta function, which we may write as

δ3(~r) =1

(2π)3

∫ ∞0

eiqr cos θ 2πq2dqd cos θ

=1

(2π)2

∫ ∞0

(eiqr − eiqr

ir

)qdq,

(4.205)

the isotropic charge distribution c′(r) = c(r)− δ3(~r) is

c′(r) =1

2(2π)3

∫ ∞0

π(q2) eiqr cos θ 2πq2dqd cos θ

=1

(2π)2r

∫π(q2) sin(qr) q dq

=e2

8π4r

∫ ∞0

qdq

∫ 1

0dxx(1− x) ln

[1 +

x(1− x) q2

m2

]sin(qr)

=−ie2

16π4r

∫ ∞−∞

eiqrqdq

∫ 1

0dxx(1− x) ln

[1 +

x(1− x) q2

m2

].

(4.206)

Vacuum polarization (4.190) shifts the energy of an atomic state with

wave function ψ(r) by

∆E =

∫∆V (r)|ψ(r)|2 d3r (4.207)

where (SW 11.2.28)

∆V (r) = ∆V (r) =e1e2

(2π)3

∫d3qeiq·r

π(q2)

q2. (4.208)

For hydrogen-like electrons in states with ` = 0 and wave function at r = 0

ψ(0) =2√4π

(Zαm

n

)3/2

(4.209)


the shift is (SW 11.2.42)

∆E = = −4Z4α5m

15πn3(4.210)

and is called the Uehling effect. In the 2s state of hydrogen it is −1.122×10−7

eV.

4.12 Magnetic moment of the electron

The one-loop correction to the interaction of an electron of momentum p with

a photon of momentum p′ − p (SW’s figure 11.4(d)) is the logarithmically

divergent integral (SW 11.3.1)

Γa(p′, p) =

∫d4k[eγb(2π)4

][ −i(2π)4

−i(/p′ − /k) +m

(p′ − k)2 +m2 − iε

]γa (4.211)

×[ −i

(2π)4

−i(/p− /k) +m

(p− k)2 +m2 − iε

][eγb(2π)4

][ −i(2π)4

1

k2 − iε

].

The first step is to use Feynman’s trick

1

ABC= 2

∫ 1

0dx

∫ x

0dy

1[Ay +B(x− y) + C(1− x)

]3 (4.212)

to write the three denominators as

1

(p′ − k)2 +m2 − iε1

(p− k)2 +m2 − iε1

k2 − iε= 2

∫ 1

0dx

∫ x

0dy (4.213)

× 1[[(p′ − k)2 +m2 − iε]y + [(p− k)2 +m2 − iε](x− y) + (k2 − iε)(1− x)

]3

= 2

∫ 1

0dx

∫ x

0dy

1[[k − p′y − p(x− y)]2 +m2x2 + q2y(x− y)− iε

]3 .

We now assume that we have made the integral sufficiently finite that we

can shift k to k+p′y+p(x−y). The integral (4.211) for Γ(p′, p) then becomes

SW’s (11.3.4)

Γa(p′, p) =2ie2

(2π)4

∫ 1

0dx

∫ x

0dy

∫d4k

[γb[− i[/p′(1− y)− /k − /p(x− y)] +m

]× γa

[− i[/p(1− x+ y)− /k − /p′y] +m

]γb

]/[

k2 +m2x2 + q2y(x− y)− iε]3. (4.214)


The next step is a Wick rotation that sets k0 = ik4 and makes the denomi-

nator a function of k2 = k2 + (k4)2. The O(4) symmetry of the denominator

lets us drop terms in the numerator that are odd under reflections. Also

since Γ(n) = (n− 1)!, our formula (4.161) for the area of the unit sphere in

d dimensions gives for d = 4

A4 =2π4/2

Γ(4/2)= 2π2. (4.215)

The integral (4.214) in terms of d4k = id4ke = 2iπ2k3dk now is

Γa(p′, p) =−4π2e2

(2π)4

∫ 1

0dx

∫ x

0dy

∫ ∞0

k3dk

[− k2γbγcγaγcγb

4

+ γb[− i[/p′(1− y)− /p(x− y)] +m

]γa

×[− i[/p(1− x+ y)− /p′y] +m

]γb

]/[

k2 +m2x2 + q2y(x− y)]3. (4.216)

Dirac’s spinors obey his equation (1.264) in momentum space

u′(i /p′ +m) = 0 and (i/p+m)u = 0. (4.217)

So we may (eventually) write the vertex correction (4.216) as

u′Γa(p′, p)u =−4π2e2

(2π)4

∫ 1

0dx

∫ x

0dy

∫ ∞0

k3dk

× u′[γa[− k2 + 2m2(x2 − 4x+ 2) + 2q2[y(x− y) + 1− x]

]+ 4imp′a(y − x+ xy) + 4impa(x2 − xy − y)

]u/[

k2 +m2x2 + q2y(x− y)]3. (4.218)

Under the reflection y → x− y, the factor y − x+ xy, which multiplies p′a,

becomes x − y − x + x(x − y) = x2 − xy − y, which multiplies pa. This

reflection maps the lower half of the triangle (x, y)|0 ≤ x ≤ 1; 0 ≤ y ≤ xinto its upper half. The jacobian is unity. So we can replace each factor by

the average of the two factors

− 1

2x(1− x) =

1

2(y − x+ xy + x2 − xy − y). (4.219)


The vertex (4.218 then reduces to

u′Γa(p′, p)u =−4π2e2

(2π)4

∫ 1

0dx

∫ x

0dy

∫ ∞0

k3dk (4.220)

× u′[γa[− k2 + 2m2(x2 − 4x+ 2) + 2q2[y(x− y) + 1− x]

]+−2im(p′a + pa)x(1− x)

]u

/[k2 +m2x2 + q2y(x− y)

]3.

This term is a correction to a term that involves 〈p′, s′|Aa(x)Ja(x)|p, s〉and we ignore the electromagnetic field Aa(x) except for its momentum. So

we are dealing with a matrix element of a current Ja(x)

〈p′, s′|Ja(x)|p, s〉 = 〈p′, s′|e−iP ·xJa(0)eiP ·x|p, s〉 = ei(p−p′)·x〈p′, s′|Ja(0)|p, s〉.

(4.221)

Since the current Ja(x) is conserved

0 = ∂aJa(x) (4.222)

it follows that

0 = 〈p′, s′|∂aJa(x)|p, s〉 = ∂a〈p′, s′|Ja(x)|p, s〉

= ∂aei(p−p′)·x〈p′, s′|Ja(0)|p, s〉 = i(p− p′)a〈p′, s′|Ja(0)|p, s〉. (4.223)

Does the amplitude (4.220) obey this rule of current conservation? To check

this, we note that since Dirac’s spinors obey his equation (4.217 or 4.237),

we see that the γa term does

i(p− p′)au′γau = u′(i/p− i /p′)u = u′(m−m)u = 0. (4.224)

So does the p′ − p term:

(p− p′)au′(p′a − pa)u = u′(−m2 +m2)u = 0. (4.225)

If we integrate the charge density J0(x) over all space, we get the charge

operator Q

Q =

∫J0(x, t) d3x. (4.226)

Because the current is conserved (4.222), the charge operator Q has a van-

ishing time derivative

∂Q

∂t=

∫∂0J

0(x, t) d3x = −∫∇ · J(x, t) d3x

= −∫J(x, t) · dσ = 0 (4.227)


as long as the integral over the surface σ vanishes. (Massless scalar fields

could pose a problem here.) The matrix element of the charge operator is

an integral of the matrix element of the charge density, which is the time

component of the current (4.221),

〈p′, s′|Q|p, s〉 = 〈p′, s′|∫J0(x, t) d3x|p, s〉

=

∫ei(p−p

′)·xd3x〈p′, s′|J0(0) d3x|p, s〉

= (2π)3δ3(p− p′)〈p′, s′|J0(0) d3x|p, s〉. (4.228)

The state |p, s〉 is a state of one electron (of momentum p and spin s in the

z direction), so it has charge −e. Therefore

Q|p, s〉 = −e|p, s〉, (4.229)

and so

〈p′s′|Q|p, s〉 = − e〈p′s′|p, s〉 = −eδ3(p− p′)δss′ . (4.230)

Comparing this equation with the matrix element (4.228) of the charge

operator, we find that

〈p′, s′|J0(0) d3x|p, s〉 = − e

(2π)3δss′ . (4.231)

SW writes the matrix element of a current Ja of a particle of charge −eas (10.6.9)

〈p′, s′|Ja(0)|p, s〉 = − ie

(2π)3u(p′, s′)Γa(p′, p)u(p, s). (4.232)

Lorentz invariance requires the last factor here to be

u(p′, s′)Γa(p′, p)u(p, s) = u(p′, s′)[γaF (q2)− i

2m(p+ p′)aG(q2)

+(p− p′)a

2mH(q2)

]u(p, s). (4.233)

The hermiticity of the current implies that F,G,&H are real functions of

q2 = (p′ − p)2. Charge conservation (4.223) requires that H(q2) = 0. So the

vertex correction (4.220) is the sum of two form factors F (q2) and G(q2)

u′Γa(p′, p)u = u′[γaF (q2)− i

2m(p+ p′)aG(q2)

]u. (4.234)


Letting p′ → p in this equation and in the matrix element (4.232), we find

〈p, s′|Ja(0)|p, s〉 = − ie

(2π)3u(p, s′)Γa(p, p)u(p, s)

= − ie

(2π)3u(p, s′)

[γaF (0)− i

mpaG(0)

]u(p, s).

(4.235)

Since Dirac’s gamma matrices are defined by the anticommutation relation

γa, γb = 2ηab, it follows that

γa, i/p+m = γa, ipbγb +m = 2ipbηab + 2mγa = 2ipa + 2mγa. (4.236)

Dirac’s spinors obey his equation

u′(i /p′ +m) = 0 and (i/p+m)u = 0. (4.237)

So between spinors of the same momentum, the anticommutation relation

(4.236) gives

u(p, s′)γa, i/p+mu(p, s) = 0 = u(p, s′)(2ipa + 2mγa)u(p, s)

or

u(p, s′)γau(p, s) = − ipa

mu(p, s′)u(p, s). (4.238)

Using the formulas (1.273) for the spinors

u(p, s) =m− i/p√

2p0(p0 +m)u(0, s)

v(p, s) =m+ i/p√

2p0(p0 +m)v(0, s),

(4.239)

we can compute the inner product

u(p, s′)u(p, s) = iu†(p, s′)γ0u(p, s)

=i

2p0(p0 +m)u†(0, s′)(m+ ipaγ

a†)γ0(m− i/p)u(0, s)

=i

2p0(p0 +m)u†(0, s′)γ0(m− ipaγa)(m− i/p)u(0, s)

=1

2p0(p0 +m)u(0, s′)(m− i/p)(m− i/p)u(0, s). (4.240)

The formula (4.238) for the terms linear in gamma tells us that

u(0, s′)paγau(0, s) = −ipaδa0 u(0, s′)u(0, s) = ip0u(0, s′)u(0, s). (4.241)


Also the anticommutation relation γa, γb = 2ηab tells us that

/p/p = papbγaγb =

1

2papbγa, γb = papbη

ab = −m2. (4.242)

So we find that the inner product is

u(p, s′)u(p, s) =u(0, s′)u(0, s)

2p0(p0 +m)(2m2 + 2mp0) =

m

p0u(0, s′)u(0, s). (4.243)

The zero-momentum spinors are

u(~0,m = 12) =

1√2

1

0

1

0

and u(~0,m = −12) =

1√2

0

1

0

1

,

v(~0,m = 12) =

1√2

0

1

0

−1

and v(~0,m = −12) =

1√2

−1

0

1

0

.(4.244)

So the inner product (4.243) is

u(p, s′)u(p, s) =m

p0δss′ . (4.245)

Incidentally, SW’s pseudounitarity condition (1.209, SW’s 5.4.32)

βD†(Λ)β = D−1(Λ) (4.246)

leads to a much shorter derivation of this (4.245) result. Using the formulas

(4.238 & 4.245), we can write the expression (4.235) for a matrix element of

Ja(0) as

〈p, s′|Ja(0)|p, s〉 = − ie

(2π)3u(p, s′)

[− ipa

mF (0)− i

mpaG(0)

]u(p, s)

= − epa

(2π)3mu(p, s′)

[F (0) +G(0)

]u(p, s)

= − e

(2π)3

pa

p0δss′[F (0) +G(0)

]. (4.247)

Setting a = 0 and using the expression (4.231) for the matrix element of the

charge density

〈p′, s′|J0(0) d3x|p, s〉 = − e

(2π)3δss′ , (4.248)

we get the normalization condition

F (0) +G(0) = 1 (4.249)


for these quantities when computed with counterterms.

We can use Dirac’s equation (4.237) and the anticommutation relation

(4.236) to get

u′[i(p+ p′)a + 2mγa]u = u′[i(p+ p′)a − i /p′γa − iγa/p]u= u′

[12 iγ

a, /p′+ 12 iγ

a, /p − i /p′γa − iγa/p]u

= u′(

12 i[γ

a, γb](p′ − p)b)u. (4.250)

We now can write the coefficient of the first term in the vertex (4.234)

u′Γa(p′, p)u = u′[γaF (q2)− i

2m(p+ p′)aG(q2)

]u (4.251)

as

u′Γa(p′, p)u = − i

2mu′[(p+ p′)a

(F (q2) +G(q2)

)− 1

2[γa, γb](p′ − p)bF (q2)

]u. (4.252)

The generators of rotations and boosts in the Dirac representation of the

Lorentz group are (1.175) for spatial values of i and j

J ij = − i

4[γi, γj ] = − i

4

[−i(

0 σi−σi 0

),−i

(0 σj−σj 0

)]=

1

2εijk

(σk 0

0 σk

) (4.253)

and

J i0 = − i

4[γi, γ0] = − i

4

[−i(

0 σi−σi 0

),−i

(0 1

1 0

)]=i

2

(σi 0

0 −σi

).

(4.254)

So their p = p′ = 0 matrix elements between the spinors (4.244) are

u(0, s′)[γi, γj ]u(0, s) = = 4iεijk

(J

(1/2)k

)s′s

= 2iεijkσks′s

u(0, s′)[γi, γ0]u(0, s) = 0.(4.255)

So for small momenta and to lowest order in p and p′, the vertex (4.252) is

u(p′, s′)Γ(p′, p)u(p, s) ≈ 1

m

[(p− p′)× J (1/2)

k,s′s

]F (0)

=1

2m

[(p− p′)× σs′s

]F (0).

(4.256)


In weak, static fields, the interaction hamiltonian is

H ′ = −∫J(x) ·A(x) d3x. (4.257)

So the interaction is

〈p′, s′|H ′|p, s〉 =ieF (0)

(2π)32m

∫ei(p−p

′)·xA(x) · (p− p′)× σs′s d3x

=eF (0)

(2π)32m

∫ei(p−p

′)·xB(x) · σs′s d3x (4.258)

where B(x) = ∇ × A(x) is the magnetic field. So for a slowly varying

magnetic field,

〈p′, s′|H ′|p, s〉 =eF (0)

2mB · σs′s δ3(p− p′). (4.259)

So the magnetic moment is

µ = − e

2mF (0). (4.260)

The form factor G(q2) is

G(q2) = − 4π2e2

(2π)4

∫ 1

0dx

∫ x

0dy

∫ ∞0

k3dk4m2x(1− x)[

k2 +m2x2 + q2y(x− y)]3 .

(4.261)

The k integral is finite:

G(q2) = − 4π2e2

(2π)4

∫ 1

0dx

∫ x

0dy

4m2x(1− x)

4[m2x2 + q2y(x− y)

] (4.262)

= − e2m2

4π2

∫ 1

0dx

∫ x

0dy

x(1− x)

m2x2 + q2y(x− y). (4.263)

The magnetic moment of the electron to order e3 is

µ = − e

2m

(1−G(0)

)(4.264)


where

G(0) = − e2m2

4π2

∫ 1

0dx

∫ x

0dy

x(1− x)

m2x2

= − e2m2

4π2

∫ 1

0dx

∫ x

0dy

(1− x)

m2x

= − e2m2

4π2

∫ 1

0dx

(1− x)

m2= − e2

4π2

∫ 1

0dx (1− x)

= − e2

8π2. (4.265)

So the absolute value of the magnetic moment of the electron to order e3 is

µ =e

2m

(1 +

e2

8π2

)=

e

2m

(1 +

α

2π

)= 1.001161

e

2m(4.266)

first calculated by Julian Schwinger in 1948. The 2018 experimental value

by Gabrielse et al. (PRA 83, 052122 (2011)) is

µexp = (1.00115965218073± 0.00000000000028)e

2m(4.267)

in which 73 could be 73 ± 28. Kinoshita and others have done QED calcu-

lations to 5 loops. Their best estimate (as of 2012) including all standard-

model effects is (arXiv:1205.5368)

µth = 1.00115965218178e

2m. (4.268)

The standard model value of g − 2 agrees with experiment to 11 decimal

digits

µSM = µexp = 1.00115965218e

2m. (4.269)

Most of the theoretical uncertainty is due hadronic effects and to uncertainty

in the experimental value of the fine-structure constant α.

4.13 Charge form factor of electron

If we add in the contribution due to vacuum polaritzation, then the charge

form factor of electron is

F (q2) = π(q2) +4π2e2

(2π)4

∫ 1

0dx

∫ x

0dy

∫ ∞0

k3dk

× k2 − 2m2(x2 − 4x+ 2)− 2q2[y(x− y) + 1− x][k2 +m2x2 + q2y(x− y)

]3 (4.270)

4.13 Charge form factor of electron 133

which diverges logarithmically. Let’s take the value of this quantity at q2 = 0

from experiment. Then the charge form factor is

F (q2) = F (0) +4π2e2

(2π)4

∫ 1

0dx

∫ x

0dy

∫ ∞0

k3dk

×

[k2 − 2m2(x2 − 4x+ 2)− 2q2[y(x− y) + 1− x][

k2 +m2x2 + q2y(x− y)]3

− k2 − 2m2(x2 − 4x+ 2)[k2 +m2x2

]3

]. (4.271)

The k integral gives

F (q2) = F (0) +4π2e2

(2π)4

∫ 1

0dx

∫ x

0dy (4.272)

−[q2(− x3 + 2y2 − 2xy(1 + 2y) + x2(1 + 4y)

)−(m2x2 + q2x2(x− y)y

)ln(m2x2)

+ x2[m2x2 + q2(x− y)y] ln[m2x2 + q2(x− y)y]

]/[

2x2(m2x2 + q2(x− y)y)]

(4.273)

5

Nonabelian gauge theory

5.1 Yang and Mills invent local nonabelian symmetry

The action of a Yang-Mills theory is unchanged when a spacetime-dependent

unitary matrix U(x) = exp(−ita θa(x)) maps a vector ψ(x) of matter fields

to ψ′i(x) = Uij(x)ψj(x). The symmetry ψ†(x)U †(x)U(x)ψ(x) = ψ†(x)ψ(x)

is obvious, but how can kinetic terms like ∂iψ† ∂iψ be made invariant?

Yang and Mills introduced matrices Ai = taAai of gauge fields in which

the hermitian matrices ta are the generators of the Lie algebra of a compact

gauge group and so obey the commutation relation

[ta, tb] = ifabctc (5.1)

in which the fabc are real, totally antisymmetric structure constants. More

importantly, they replaced ordinary derivatives ∂i by covariant deriva-

tives

Diαβ = ∂iδαβ +Aiαβ ≡ ∂iδαβ + taαβ Aai (5.2)

and required that covariant derivatives of fields transform like fields so that

(Dψ)′ = UDψ or

D′ψ′ =(∂i +A′i

)Uψ =

(∂iU + U∂i +A′iU

)ψ = U (∂i +Ai)ψ. (5.3)

The nonabelian gauge field transforms as

A′i(x) = U(x)Ai(x)U †(x)− (∂iU(x))U †(x). (5.4)

In full detail, this is

A′iαδ(x) = Uαβ(x)Aiβγ(x)(U †)γδ(x)− (∂iUαβ(x)) (U †)βδ(x). (5.5)

The nonabelian Faraday tensor is

Fik = [Di, Dk] = ∂iAk − ∂kAi + [Ai, Ak] (5.6)

5.2 SU(3) 135

in which both F and A are matrices in the Lie algebra. The nonabelian

Faraday tensor transforms as

F ′ik(x) = U(x)FikU−1(x) = U(x)[Di, Dk]U

−1(x). (5.7)

The trace Tr[FikFik] of the Lorentz-invariant product FikF

ik is both

Lorentz invariant and gauge invariant. The nonabelian generalization of the

abelian action density

L = − 1

4FjkF

jk − ψ[γk(∂k + ieAk) +m]ψ (5.8)

is

L = − 1

4Tr[FjkF

jk]− ψ[γk(∂k + ieAk) +m]ψ. (5.9)

In more detail, the Fermi action is

ψ[γk(∂k+ieAk)+m]ψ = ψ`α[γk``′(∂kδαβ+ietaαβA

ak)+mδ``′δαβ

]ψ`′β. (5.10)

Use of matrix notation and of summation conventions is necessary.

5.2 SU(3)

The gauge group of quantum chromodynamics is SU(3) which has eight

generators. The Gell-Mann matrices are

λ1 =

0 1 0

1 0 0

0 0 0

, λ2 =

0 −i 0

i 0 0

0 0 0

, λ3 =

1 0 0

0 −1 0

0 0 0

,

λ4 =

0 0 1

0 0 0

1 0 0

, λ5 =

0 0 −i0 0 0

i 0 0

, λ6 =

0 0 0

0 0 1

0 1 0

,

λ7 =

0 0 0

0 0 −i0 i 0

, and λ8 =1√3

1 0 0

0 1 0

0 0 −2

. (5.11)

The generators ta of the 3 × 3 defining representation of SU(3) are these

Gell-Mann matrices divided by 2

ta = λa/2 (5.12)

(Murray Gell-Mann, 1929–).

136 Nonabelian gauge theory

The eight generators ta are orthogonal with k = 1/2

Tr (tatb) =1

2δab (5.13)

and satisfy the commutation relation

[ta, tb] = ifabc tc. (5.14)

The trace formula

f cab = − ik

Tr(

[ta, tb] t†c

). (5.15)

gives us the SU(3) structure constants as

fabc = − 2iTr ([ta, tb]tc) . (5.16)

They are real and totally antisymmetric with f123 = 1, f458 = f678 =√

3/2,

and f147 = −f156 = f246 = f257 = f345 = −f367 = 1/2.

While no two generators of SU(2) commute, two generators of SU(3) do.

In the representation (5.11,5.12), t3 and t8 are diagonal and so commute

[t3, t8] = 0. (5.17)

They generate the Cartan subalgebra of SU(3). The generators defined

by Eqs.(5.12 & 5.11) give us the 3× 3 representation

D(α) = exp (iαata) (5.18)

in which the sum a = 1, 2, . . . 8 is over the eight generators ta. This repre-

sentation acts on complex 3-vectors and is called the 3.

Note that if

D(α1)D(α2) = D(α3) (5.19)

then the complex conjugates of these matrices obey the same multiplication

rule

D∗(α1)D∗(α2) = D∗(α3) (5.20)

and so form another representation of SU(3). It turns out that (unlike in

SU(2)) this representation is inequivalent to the 3; it is the 3.

There are three quarks with masses less than about 100 MeV/c2—the

u, d, and s quarks. The other three quarks c, b, and t are more massive;

mc = 1.28 GeV, mb = 4.18 GeV, and mt = 173.1 GeV. Nobody knows

why. Gell-Mann and Zweig suggested that the low-energy strong interac-

tions were approximately invariant under unitary transformations of the

three light quarks, which they represented by a 3, and of the three light

antiquarks, which they represented by a 3. They imagined that the eight

5.2 SU(3) 137

light pseudoscalar mesons, that is, the three pions π−, π0, π+, the neutral

η, and the four kaons K0, K+, K−K0, were composed of a quark and an

antiquark. So they should transform as the tensor product

3⊗ 3 = 8⊕ 1. (5.21)

They put the eight pseudoscalar mesons into an 8.

They imagined that the eight light baryons — the two nucleons N and P ,

the three sigmas Σ−, Σ0, Σ+, the neutral lambda Λ, and the two cascades

Ξ− and Ξ0 were each made of three quarks. They should transform as the

tensor product

3⊗ 3⊗ 3 = 10⊕ 8⊕ 8⊕ 1. (5.22)

They put the eight light baryons into one of these 8’s. When they were writ-

ing these papers, there were nine spin-3/2 resonances with masses somewhat

heavier than 1200 MeV/c2 — four ∆’s, three Σ∗’s, and two Ξ∗’s. They put

these into the 10 and predicted the tenth and its mass. In 1964, a tenth

spin-3/2 resonance, the Ω−, was found with a mass close to their prediction

of 1680 MeV/c2, and by 1973 an MIT-SLAC team had discovered quarks

inside protons and neutrons. (George Zweig, 1937–)

For a given quark, say the up quark, the action of quantum chromody-

namics is

L = − 1

4Tr[FjkF

jk]− ψ[γk(∂k + ieAk) +m]ψ (5.23)

in which the index c on ψc takes on the values 1, 2, 3 which we may think of

as red, green, and blue. The Faraday tensor is a 3× 3 matrix

Fik = [Di, Dk] = ∂iAk − ∂kAi + [Ai, Ak]. (5.24)

The trace term in the QCD action (5.23) is

Tr[FjkF

jk]

= Tr[(∂iAk−∂kAi+[Ai, Ak]

)(∂iAk−∂kAi+[Ai, Ak]

)](5.25)

Quarks and gluons are confined in color-singlet particles. This effect is

robust and mysterious.

6

Standard model

6.1 Quantum chromodynamics

If to a pure SU(3) gauge theory we add massless quarks in the fundamental

or defining representation, then we get a theory of the strong interactions

called quantum chromodynamics or QCD. Thus, let ψ be a complex

3-vector of Dirac fields

ψ =

ψrψgψb

(6.1)

(so 12 fields in all). This complex 12-vector could represent u or “up” quarks.

In terms of the covariant derivative

Dµ = I∂µ +Aµ(x) = I∂µ + ig8∑b=1

tbAbµ(x), (6.2)

the action density is

L =1

2g2Tr [FµνF

µν ]− ψ (γµDµ +m)ψ (6.3)

where

Fµν = [Dµ, Dν ]. (6.4)

Nonperturbative effects are supposed to “confine” the quarks and massless

gluons. There are 6 known “flavors” of quarks—u, d, c, s, t, b.

6.2 Abelian Higgs mechanism

Suppose φ = (φ1 + iφ2)/√

2 is a complex scalar field with action density

L = − ∂aφ∗∂aφ+ µ2φ∗φ− λ(φ∗φ)2. (6.5)

6.2 Abelian Higgs mechanism 139

The minimum of the potential V = − µ2φ∗φ + λ(φ∗φ)2 is given by the

equation

0 =∂V

∂|φ|2= −µ2 + 2λ|φ|2. (6.6)

It is a circle with

|φ|2 =µ2

2λ≡ v2

2. (6.7)

As is customary, we pick one asymmetrical minimum at φ = µ/√

2λ, so the

mean values in the vacuum are

〈0|φ1|0〉 = v =

√µ2

λand 〈0|φ2|0〉 = 0. (6.8)

The U(1) symmetry has been spontaneously broken. So

φ =v√2

+φ1√

2+ i

φ2√2

(6.9)

where now φ1 is the departure of the real part of the field from its mean

value v. So the potential V is

V = − µ2φ∗φ+ λ(φ∗φ)2

= − 1

2µ2[(v + φ1)2 + (φ2)2

]+λ

4

[(v + φ1)2 + (φ2)2

]2= − µ4

4λ+ µ2φ2

1 + . . .

(6.10)

where the dots denote cubic and quartic in φ1 and φ2. The quadratic part

of the action determines the properties of the particles of the physical fields.

The cubic and higher-order terms tell us how the particles interact with each

other.

The first part of the last homework problem is to find the masses of the

particles of this theory.

Scalar particles of zero mass that arise from spontaneous symmetry break-

ing are called Goldstone bosons.

Now we give the theory a local U(1) symmetry by adding an abelian gauge

field to the action density

L = − (Daφ)∗Daφ+ µ2φ∗φ− λ(φ∗φ)2 − 1

4FabF

ab (6.11)

in which

Da = ∂a + ieAa and Fab = [Da, Db]. (6.12)

140 Standard model

Once again the field assumes a mean value in the vacuum, let us say (6.9).

But we write the field φ as

φ =1√2

(v + φ1 + iφ2). (6.13)

The term −(Daφ)∗Daφ then is

−(Daφ)∗Daφ = − (∂aφ∗ − ieAaφ∗)(∂aφ+ ieAaφ)

= − 1

2[∂aφ1 − i∂aφ2 − ieAa(v + φ1 − iφ2)]

× [(∂aφ1 + i∂aφ2 + ieAa(v + φ1 + iφ2)]

= − 1

2(∂aφ1 − eAaφ2)(∂aφ1 − eAaφ2)

− 1

2(∂aφ2 + eAa(v + φ1))(∂aφ2 + eAa(v + φ1)).

(6.14)

So we set

Ba = Aa +1

ev∂aφ2 or Aa = Ba −

1

ev∂aφ2 (6.15)

and find that

−(Daφ)∗Daφ = − 1

2(∂aφ1 − eBaφ2 +

φ2

v∂aφ2)(∂aφ1 − eBaφ2 +

φ2

v∂aφ2)

− 1

2(evBa + eBaφ1 −

φ1

v∂aφ2)(evBa + eBaφ1 −

φ1

v∂aφ2)

− 1

2

(∂aφ1∂aφ1 + e2v2BaBa

)+ . . . (6.16)

in which the dots denote cubic and quartic terms.

The second part of the last homework problem is to find the masses of

the particles of this theory.

6.3 Higgs’s mechanism

The local gauge group of the Glashow-Salam-Weinberg electroweak theory

is SU(2)` × U(1). What’s weird is that the SU(2)` symmetry applies only

to the left-handed quarks and leptons and to the Higgs boson, a complex

doublet (or 2-vector) of scalar fields H.

Let’s leave out the fermions for the moment, and focus just on the Higgs

and the gauge fields. The gauge transformation is

H ′(x) = U(x)H(x)

A′µ(x) = U(x)Aµ(x)U †(x) + (∂µU(x))U †(x)(6.17)


in which the 2× 2 unitary matrix U(x) is

U(x) = exp

[i gσa

2αa(x) + i g′

Y I

2β(x)

]. (6.18)

The generators here are the 3 Pauli matrices and the matrix I/2, where I

is the 2× 2 identity matrix.

The action density of the theory (without the fermions) is

L = − (DµH)†DµH +1

4kTr(FµνF

µν) +m2|H|2 − λ|H|4 (6.19)

in which the covariant derivative for the Higgs doublet is

DµH =

(I∂µ + i g

σa

2Aaµ + i g′

Y I

2Bµ

)H. (6.20)

The potential energy of the Higgs field is

V = −m2|H|2 + λ|H|4. (6.21)

Its minimum is where

0 =∂V

∂|H|2= 2λ|H|2 −m2. (6.22)

So

|H| = m√2λ

=v√2. (6.23)

By making an SU(2)` × U(1) gauge transformation, we can transform this

mean value to

H0 = 〈0|H(x)|0〉 =1√2

(0

v

). (6.24)

This transformation is called “going to unitary gauge.”

In this gauge, the Higgs potential is

V (v) = − 1

2m2v2 +

1

4λv4, (6.25)

and its second derivative is

V ′′(v) = 3λv2 −m2 = 2m2 = m2H . (6.26)

The mass of the Higgs then is

mH =√

2m =√

2λ v. (6.27)

Experiments at LEP2 and at the LHC have revealed value of v to be

v = 246.22 GeV. (6.28)

142 Standard model

In 2012, the Higgs boson was discovered at the LHC. Its mass has been

measured to be

mH = 125.18± 0.16 GeV. (6.29)

The self coupling λ therefore is

λ =m2H

2 v2=

1

2

(125.18

246.22

)2

= 0.12924. (6.30)

After spontaneous symmetry breaking, mass terms for the gauge bosons

emerge from the kinetic action of the Higgs doublet − (DµH)†DµH. Since

the generators of a compact group like SU(2)` × U(1) are hermitian, the

part of the kinetic action that contains the mass terms is

Lm = −H†(−i g σ

a

2Aaµ − i g′

Y I

2Bµ

)(i gσa

2Aaµ + i g′

Y I

2Bµ

)H.

(6.31)

In unitary gauge (6.24), these mass terms are

Lm = − 1

2(0, v)

(gσa

2Aaµ + g′

I

2Bµ

)(gσa

2Aµa + g′

I

2Bµ

)(0

v

)= − 1

8(0, v)

(gA3

µ + g′Bµ g(A1µ − iA2

µ)

g(A1µ + iA2

µ) −gA3µ + g′Bµ

)×(gAµ3 + g′Bµ g(Aµ1 − iA

µ2 )

g(Aµ1 + iAµ2 ) −gAµ3 + g′Bµ

)(0

v

)= − v2

8

(g(A1

µ + iA2µ), −gA3

µ + g′Bµ)( g(Aµ1 − iA

µ2 )

−gAµ3 + g′Bµ

)= − v2

8

[g2(A1µA

µ1 +A2

µAµ2

)+(−gA3

µ + g′Bµ) (−gAµ3 + g′Bµ

)].

(6.32)

The normalized complex, charged gauge bosons are

W±µ =1√2

(A1µ ∓ iA2

µ

)(6.33)

and the normalized neutral ones are

Zµ =1√

g2 + g′2

(gA3

µ − g′Bµ)

(6.34)

and the photon, which is the normalized linear combination orthogonal to

Zµ

Aµ =1√

g2 + g′2

(g′A3

µ + g Bµ). (6.35)


It remains massless. In terms of these properly normalized fields, the mass

terms are

LM = − g2v2

4W−µ W

+µ − (g2 + g′2)v2

8ZµZ

µ. (6.36)

So the W+ and the W− get the same mass

MW = gv

2= 80.370± 0.012 GeV/c2. (6.37)

while the Z (also called the Z0) has mass

MZ =√g2 + g′2

v

2= 91.1876± 0.0021 GeV/c2. (6.38)

The accuracy and precision of the mass of the Z boson is an order of mag-

nitude higher than that of the W boson because of measurements made in

the Large Electron Collider.

It is the photon. The fine-structure constant is

α =e2

4π~c= 1/137.035 999 139(31) ≈ 1/137.036. (6.39)

Why do three gauge bosons become massive? Because there are three

Goldstone bosons corresponding to three ways of moving 〈0|H|0〉 without

changing the Higgs potential. Why does one gauge boson stay massless?

Because one linear combination of the generators of SUL(2) ⊗ U(1) maps

〈0|H|0〉 to zero.

In terms of these mass eigenstates, the original gauge bosons are

A1µ =

1√2

(W+µ +W−µ

)A2µ =

1

i√

2

(W−µ −W+

µ

)A3µ =

1√g2 + g′2

(g′Aµ + gZµ

)Bµ =

1√g2 + g′2

(gAµ − g′Zµ

).

(6.40)

The 4× 4 matrices

P` =1

2

(1 + γ5

)=

(1 0

0 0

)and Pr =

1

2

(1− γ5

)=

(0 0

0 1

)(6.41)

project out the upper and lower two components of a 4-component Dirac

field. The SU(2)` gauge fields ~Aµ interact only with the upper two com-

ponents ψ` = P`ψ, while the U(1)Y gauge field Bµ interacts with all four

components of a Dirac field.

144 Standard model

Thus the covariant derivative for a fermion of coupling g′ to the U(1)

gauge field Bµ with hypercharge Y and coupling g to the SU(2)` gauge

fields is

Dµ = I∂µ + i gσa

2Aaµ P` + i g′ Y I Bµ

= I∂µ + i g

[σ1

2

1√2

(W+µ +W−µ

)+σ2

2

1

i√

2

(W−µ −W+

µ

)(6.42)

+σ3

2

1√g2 + g′2

(g′Aµ + gZµ

)]P` + i g′ Y

I√g2 + g′2

(gAµ − g′Zµ

).

The SU(2)` generators are

~T =1

2~σ, T± =

1

2(σ1 ± iσ2) and T 3 =

σ3

2. (6.43)

In terms of them, the covariant derivative is

Dµ = I∂µ + ig√2

(W+µ T

+ +W−µ T−)P` + i

1√g2 + g′2

Zµ(g2T 3P` − g′2Y I

)+ i

gg′√g2 + g′2

Aµ(T 3P` + Y I

)(6.44)

Equivalently, the left and right covariant derivatives are

Dµ` =[I∂µ + i

g√2

(W+µ T

+ +W−µ T−)+ i

1√g2 + g′2

Zµ(g2T 3 − g′2Y I

)+ i

gg′√g2 + g′2

Aµ(T 3 + Y I

) ]P` (6.45)

and

Dµr =[I∂µ − i

g′2Y I√g2 + g′2

Zµ + igg′Y I√g2 + g′2

Aµ

]Pr. (6.46)

The interaction strength or coupling constant of the photon Aµ is

e =gg′√g2 + g′2

> 0. (6.47)

The charge operator is

Q = T 3` + Y I. (6.48)

When acting on the doublet

E` =

(νee

)(6.49)


to which we assign Y = −1/2, the charge Q gives 0 as the charge of the

neutrino and − 1 as the charge of the electron. The photon-lepton term

then is

gg′√g2 + g′2

Aµ(T 3 + Y I

)E` = eAµ

(T 3 + Y I

)E`

= eAµ

(0 0

0 −1

)(νee

)= −eAµ e

(6.50)

in which the first e is the absolute value of the charge (6.47) of the electron

and the second is the field of the electron.

The weak mixing angle θw is defined by(Z

A

)=

(cos θw − sin θwsin θw cos θw

)(A3

B

). (6.51)

Equations (6.34 and 6.35) identify these trigonometric values as

cos θw =g√

g2 + g′2and sin θw =

g′√g2 + g′2

. (6.52)

Since the charge is Q = T 3 + Y I, the hypercharge Y I = Q − T 3, and so

fields couple to the Z as

1√g2 + g′2

Zµ[g2T 3 − g′2Y

]=

1√g2 + g′2

Zµ[(g2 + g′2)T 3 − g′2Q

](6.53)

and to the photon as

gg′√g2 + g′2

Aµ(T 3 + Y

)=

gg′√g2 + g′2

AµQ. (6.54)

We also have

g2 + g′2√g2 + g′2

=√g2 + g′2 =

g

cos θw(6.55)

and

g′2√g2 + g′2

=√g2 + g′2

g′2

g2 + g′2=

g

cos θwsin2 θw. (6.56)

So the coupling to the Z is

1√g2 + g′2

Zµ[(g2 + g′2)T 3 − g′2Q

]=

g

cos θwZµ(T 3 − sin2 θwQ

)(6.57)

and, if we set

e = g sin θw, (6.58)

146 Standard model

then the coupling to the photon A is

gg′√g2 + g′2

AµQ = g sin θw AµQ = eAµQ. (6.59)

In these terms, the covariant derivative (6.44) is

Dµ = I∂µ + ig√2

(W+µ T

+ +W−µ T−)P` + i

1√g2 + g′2

Zµ(g2T 3P` − g′2Y

)+ i

gg′√g2 + g′2

Aµ(T 3P` + Y

)= I∂µ + i

g√2

(W+µ T

+ +W−µ T−)P` + i

g

cos θwZµ(T 3P` − sin2 θwQ

)+ ieAµQ

(6.60)

in which the generators T± and T 3 are those of the representation to which

the fields they act on belong. When acting on left-handed fermions, they

are half the Pauli matrices, T = 12σ. When acting on right-handed fermions,

they are zero, T = 0, and so the explicit appearance of P` is unnecessary.

Since g = e/ sin θw, the couplings involve one new parameter θw.

Our mass formulas (6.37 and 6.38) for the W and the Z show that their

masses are related by

MW = gv

2=

g√g2 + g′2

√g2 + g′2

v

2= cos θw

√g2 + g′2

v

2= cos θwMZ .

(6.61)

Experiments have determined the masses and shown that at the mass of the

Z and in the MS convention

sin2 θw = 0.23122(4) or θw = 0.50163 (6.62)

and that

v = 246.22 GeV. (6.63)

6.4 Quark and lepton interactions

The right-handed fermions ur, dr, er, and νe,r are singlets under SUL(2)⊗UY (1). So they have T 3 = 0. The definition (6.48) of the charge Q

Q = T 3 + Y I (6.64)

then implies that

Yr = Qr. (6.65)

6.4 Quark and lepton interactions 147

That is, Yνe,r = 0, Yer = −1, Yur = 2/3, and Ydr = −1/3.

The left-handed fermions are in doublets

E` =

(νee−

)and Q` =

(u

d

)(6.66)

with T 3 = ±1/2. So the choices YE = −1/2 and YQ = 1/6 and the definition

(6.48) of the charge Q give the right charges:

QE` = Q

(νee−

)=

(0

−e−)

and QQ` = Q

(u

d

)=

(2u/3

−d/3

). (6.67)

Fermion-gauge-boson interactions are due to the covariant derivative (6.60)

acting on either the left- or right-handed fields. On right-handed fermions,

the covariant derivative is just

Drµ = I∂µ + i

g

cos θwZµ(− sin2 θwQ

)+ ieAµQ

= I∂µ − igsin2 θwcos θw

ZµQ+ ieAµQ

= I∂µ − iesin θwcos θw

ZµQ+ ieAµQ = I∂µ − ie tan θw ZµQ+ ieAµQ.

(6.68)

So the covariant derivative of a neutral right-handed fermion is just the

ordinary derivative.

On left-handed fermions, the covariant derivative is

D`µ = I∂µ + i

g√2

(W+µ T

+ +W−µ T−)+ i

g

cos θwZµ(T 3 − sin2 θwQ

)+ ieAµQ

= I∂µ + ie√

2 sin θw

(W+µ T

+ +W−µ T−)+ i

e

cos θwZµ

(T 3

sin θw− sin θwQ

)+ ieAµQ.

(6.69)

For the first family or generation of quarks and leptons, the kinetic

action density is

Lk = − E` /DÈ` − Er /D

rEr −Q` /D

`Q` −Qr /D

rQr (6.70)

in which /D ≡ γµDµ. The 4 × 4 matrix γ5 = γ5 plays the role of a fifth

(spatial) gamma matrix γ4 = γ5 in 5-dimensional space-time in the sense

that the anticommutator

γa, γb = 2ηab (6.71)

148 Standard model

in which η is the 5×5 diagonal matrix with η00 = −1 and ηaa = 1 for a = 1,

2, 3, 4. In Weinberg’s notation, γ5 is

γ5 =

(1 0

0 −1

). (6.72)

The combinations

P` =1

2(1 + γ5) and Pr =

1

2(1− γ5) (6.73)

are projection operators onto the left- and right-handed fields. That is,

P`Q = Q` and P`Q` = Q` (6.74)

with a similar equation for Pr. We can write Lk as

Lk = − E /D`PÈ − E /D

rPrE −Q /D

`P`Q−Q /D

rPrQ

= −[E /D

`PÈ + E /D

rPrE +Q /D

`P`Q+Q /D

rPrQ

].

(6.75)

Homework set 4, problem 1: Show that

E` /DÈ` = [1

2(1 + γ5)E]†iγ0 /D` 1

2(1 + γ5)E = E /D` 1

2(1 + γ5)E. (6.76)

Recall that in Weinberg’s notation

ψ = ψ†iγ0 = ψ†β = ψ†(

0 I

I 0

)(6.77)

in which I is the 2× 2 identity matrix.

6.5 Quark and Lepton Masses

The Higgs mechanism also gives masses to the fermions, but somewhat ar-

bitrarily. Dirac’s action density (4.56) has as its mass term

−mψψ = − imψ†γ0ψ = −imψ†γ0(P` + Pr)ψ = −imψ†γ0(P 2` + P 2

r )ψ.

(6.78)

Since γ0, γ5 = 0, this mass term is

−mψψ = − imψ†Prγ0P`ψ − imψ†P`γ

0Prψ

= − im (Prψ)†γ0P`ψ − im (P`ψ)†γ0Prψ

= − imψ†rγ0ψ` − imψ†`γ

0ψr = −mψrψ` −mψ`ψr.

(6.79)

Incidentally, because the fields ψ` and ψr are independent, we can redefine

them as

ψ′` = eiθ ψ` and ψ′r = eiφ ψr (6.80)


at will. Such a redefinition changes the mass term to

−m′ ψrψ` −m′∗ ψ`ψr = −mei(θ−φ) ψrψ` −me−i(θ−φ) ψ`ψr. (6.81)

So the phase of a Dirac mass term has no significance.

The definition (6.77) of ψ shows that the Dirac mass term is

−mψψ = −mψ†βψ = −m(ψ†` , ψ

†r

)(0 I

I 0

)(ψ`ψr

)= −m

(ψ†`ψr + ψ†rψ`

).

(6.82)

These mass terms are invariant under the Lorentz transformations

ψ′` = exp(−z · σ)ψ` and ψ′r = exp(z∗ · σ)ψr(6.83)

because

ψ′†` ψ′r = ψ†` exp(−z∗ · σ) exp(z∗ · σ)ψr = ψ†` ψr

ψ′†r ψ′` = ψ†r exp(z · σ) exp(−z · σ)ψ` = ψ†r ψ`.

(6.84)

They are not invariant under rigid, let alone local, SU(2)` transforma-

tions. But we can make them invariant by using the Higgs field H(x). For

instance, the quantity Q†`Hdr is invariant under local SU(2)` transforma-

tions. In unitary gauge, its mean value in the vacuum is

〈0|Q†`Hdr|0〉 =1√2v 〈0|d†`dr|0〉. (6.85)

So the term

− cdQ†`Hdr − c∗d d†rH†Q` (6.86)

is invariant, and in the vacuum it is

〈0| − cdQ†`Hdr − c∗d d†rH†Q`|0〉 =

1√2v 〈0| − cd d†`dr − c

∗d d†rd`|0〉 (6.87)

which gives to the d quark the mass

md =|cd|√

2v. (6.88)

Note that we must add one new parameter cd to get one new mass md. This

parameter cd is complex in general, but the mass md depends only upon the

absolute value and not upon its phase of cd.

Similarly, the term

− ceE†`Her − c∗e e†rH†E` (6.89)

150 Standard model

is invariant, and in the vacuum it is

〈0| − ceE†`Her − c∗e e†rH†Q`|0〉 =

1√2v 〈0| − ce e†èr − c

∗e e†re`|0〉 (6.90)

which gives to the electron the mass

me =|ce|√

2v. (6.91)

Again, we must add one new (complex) parameter ce to get one new mass

me.

The mass of the up quark requires a new trick. The Higgs field H trans-

forms under SU(2)` × U(1) as

H ′(x) = exp

[i gσa

2αa(x) + i g′

I

2β(x)

]H(x). (6.92)

If for clarity, we leave aside the U(1) part for the moment, then the Higgs

field H transforms under SU(2)` as

H ′(x) = exp

[i gσa

2αa(x)

]H(x). (6.93)

Let us use H∗ to be the complex column vector whose components are H†1and H†2 . How does σ2H

∗ transform under SU(2)`? Suppressing our explicit

mention of the space-time dependence and using the asterisk to mean hermi-

tian conjugation when applied to operators but complex conjugation when

applied to matrices and vectors, we have, since σ2 is imaginary with σ22 = I

while σ1 and σ3 are real,

(σ2H∗)′ = σ2

[exp

(i gσa

2αa)H

]∗= σ2 exp

(−i g σ

∗a

2αa)H∗

= σ2 exp

(−i g σ

∗a

2αa)σ2σ2H

∗ = exp

(i gσa

2αa)σ2H

∗.

(6.94)

Thus, the term

− cuQ†`σ2H∗ur − c∗uu†rHTσ2Q` (6.95)

is invariant under SU(2)`. In the vacuum of the unitary gauge, it is

〈0| − cuQ†`σ2H∗ur − c∗uu†rHTσ2Q`|0〉 =

1√2v〈0|icuu†ùr − ic

∗uu†ru`|0〉

=1√2v〈0|icuu†ùr − ic

∗uu†ru`|0〉

(6.96)


which gives the up quark the mass

mu =|cu|√

2v. (6.97)

But there are three families of generations of quarks and leptons on which

the gauge fields act simply:

F1 =

u

d

νee

′

, F2 =

c

s

νµµ

′

, and F3 =

t

b

νττ

′

. (6.98)

The quark and lepton flavor families are

Q′1 =

(u

d

)′, Q′2 =

(c

s

)′, and Q′3 =

(t

b

)′;

E′1 =

(νee

)′, E′2 =

(νµµ

)′, and E′3 =

(νττ

)′.

(6.99)

These are called the flavor eigenstates or more properly flavor eigenfields,

designated here with primes. They are the ones on which the W± act simply.

The weak interactions useW−µ T− to map the flavor up fields u′1 = u′, u′2 = c′,

u′3 = t′ into the flavor down fields d′1 = d′, d′2 = s′, d′3 = b′, and W+µ T

+ to

map the flavor down fields d′i into the flavor up fields u′i.

The action density

3∑i,j=1

−cdij Q′†ìHd′rj − c∗dij d

′†rjH

†Q′ì (6.100)

gives for the d′, s′, and b′ quarks the mixed mass terms

v√2

3∑i,j=1

−cdij d′†ìd′rj − c∗dij d

′†rjd′ì =

v√2

3∑i,j=1

−cdij d′†ìd′rj − c∗dij d

′†rjd′ì.

(6.101)

The 3× 3 mass matrix Md with entries

[Md]ij =v√2cdij (6.102)

need have no special properties. It need not be hermitian because for each i

and j, the term (6.101) is hermitian. But every 3× 3 complex matrix has a

singular-value decomposition

Md = LdΣdR†d (6.103)

152 Standard model

in which Ld and Rd are 3 × 3 unitary matrices, and Σd is a 3 × 3 diagonal

matrix with nonincreasing positive singular values on its main diagonal.

The singular value decomposition works for any N ×M (real or) complex

matrix. Every complex M × N rectangular matrix A is the product of an

M × M unitary matrix U , an M × N rectangular matrix Σ that is zero

except on its main diagonal which consists of its nonnegative singular values

Sk, and an N ×N unitary matrix V †

A = U ΣV †. (6.104)

This singular-value decomposition is a key theorem of matrix algebra. One

can use the Matlab command “[U,S,V] = svd(A)” to perform the svd A =

USV †.

The singular values of Σd are the masses mb, ms, and md:

Σd =

mb 0 0

0 ms 0

0 0 md

. (6.105)

So the mass eigenfields of the left and right down-quark fields are

dri = R†dijd′rj and d†ì = d′†`jLdji or dì = L†dijd`j . (6.106)

The inverse relations are

d′ri = Rdijdrj and d′†ì = d†`jL†dji or d′ì = Ldijd`j (6.107)

or in matrix notation

d′r = Rd dr, d′†r = d†r R†d, d′†` = d†`L

†d, and d′` = Ld d` (6.108)

in which

d` =

bsd

`

(6.109)

are the down-quark fields of definite masses.

Similarly, the up quark action density

3∑i,j=1

−cuijQ′†ìσ2H∗u′rj − c∗uiju

′†rjH

Tσ2Q′ì (6.110)

gives for the three known families the mixed mass terms

iv√2

3∑i,j=1

cuiju′†ì u′rj − c∗uiju

′†rj u′ì =

iv√2

3∑i,j=1

cuiju′†ì u′rj − c∗uiju

′†rj u′ì. (6.111)

6.6 CKM matrix 153

The 3× 3 mass matrix Mu with entries

[Mu]ij =iv√

2cuij (6.112)

need have no special properties. It need not be hermitian because for each

i and j, the term (6.111) is hermitian. But every 3× 3 complex matrix Mu

has a singular value decomposition

Mu = LuΣuR†u (6.113)

in which Lu and Ru are 3× 3 unitary matrices, and Σu is a 3× 3 diagonal

matrix with nonincreasing positive singular values on its main diagonal.

These singular values are the masses mt, mc, and mu:

Σu =

mt 0 0

0 mc 0

0 0 mu

. (6.114)

So the mass eigenfields of the left and right up-quark fields are

uri = R†uiju′rj and u†ì = u′†`jLuji or uì = L†uiju

′`j . (6.115)

The inverse relations are

u′ri = Ruijurj and u′†ì = u†`jL†uji or u′ì = Luiju`j (6.116)

or in matrix notation

u′r = Ru ur, u′†r = u†r R†u u′†` = u†`L

†u, and u′` = Luu` (6.117)

in which

u` =

tcu

`

(6.118)

are the up-quark fields of definite masses.

6.6 CKM matrix

We will use the labels u, c, t and d, s, b for the states that are eigenstates of

the quadratic part of the hamiltonian after the Higgs mechanism has given a

mean value to the real part of the neutral Higgs boson in the unitary gauge.

The u, c, t quarks have the same charge 2e/3 > 0 and the same T 3 = 1/2, so

they all have the same electroweak interactions. Similarly, the d, s, b quarks

have the same charge −e/3 < 0 and the same T 3 = −1/2, so they also all

have the same electroweak interactions.

154 Standard model

The right-handed covariant derivative (6.68)

Drµ = I∂µ − ie tan θw ZµQ+ ieAµQ (6.119)

just sends the fields of these mass eigenstates into themselves multiplied by

their charge and either a Z or a photon. That is,

u′†r Drµu′r = u†rR

†uD

rµRuur = u†rD

rµur

d′†r Drµd′r = d†rR

†dD

rµRddr = d†rD

rµdr

(6.120)

In these terms, the interactions of the Z and the photon with the right-

handed fields are diagonal both in mass and in flavor.

But the left-handed covariant derivative (6.69) is

D`µ = I∂µ + i

e√2 sin θw

(W+µ T

+ +W−µ T−)

+ie

cos θwZµ

(T 3

sin θw− sin θwQ

)+ ieAµQ.

(6.121)

So we have(u′†` d′†`

)D`µ

(u′`d′`

)=(u†`L

†u d†`L

†d

)D`µ

(Lu u`Ld d`

). (6.122)

Some of the unitary matrices just give unity, L†uLu = I and L†dLd = I like

R†uRu = I and R†dRd = I in the right-handed covariant derivatives (6.120).

Thus the interactions of the Z and the photon with the both the right-

handed fields and with the left-handed fields are diagonal both in mass and

in flavor. The Z and the photon do not mediate top-to-charm or charm-to-

up or µ− → e− + γ decays. Also, the Higgs mass terms are diagonal, so the

neutral Higgs boson can’t mediate such processes. Thus, in the standard

model, there are no flavor-changing neutral-currents.

The only changes are in the nonzero parts of T± which become

T+ckm =

(0 L†uLd0 0

)≡(

0 V

0 0

)and T−ckm =

(0 0

L†dLu 0

)≡(

0 0

V † 0

)(6.123)

in which the unitary matrix V = L†uLd is the CKM matrix (Nicola Cabibbo,

Makoto Kobayashi, and Toshihide Maskawa). The left-handed covariant

derivative on the mass eigenfields then is

D`µ = I∂µ + i

e√2 sin θw

(W+µ T

+ckm +W−µ T

−ckm

)+i

e

cos θwZµ

(T 3

sin θw− sin θwQ

)+ ieAµQ.

(6.124)

6.6 CKM matrix 155

It has a second part that acts more or less like the right-handed covariant

derivative, but the first part uses W−µ T− to map the up fields u, c, t into

linear combinations of the down fields d, s, b and W+µ T

+ to map the down

fields into linear combinations of the up fields. The W±µ terms are sensitive

to the CKM matrix V = L†uLd. We write them suggestively as

(u c t d s b

)†( 0 VW+µ

V †W−µ 0

)

u

c

t

d

s

b

(6.125)

=

u c t V

dsb

†( 0 W+µ

W−µ 0

)

u

c

t

V

dsb

.

(6.126)

By choosing the phases of the six quark fields, that is, u(x) → eiθuu(x)

. . . b(x) → eiθbb(x), one may make the CKM matrix L†uLd real apart from

a single phase. The existence of that phase probably is the cause of most

of the breakdown of CP invariance that Fitch and Cronin and others have

observed since 1964. The magnitudes of the elements of the CKM matrix V

are

V =

|Vud| |Vus| |Vub||Vcd| |Vcs| |Vcb||Vtd| |Vts| |Vtb|

=

0.97427 0.22536 0.00355

0.22522 0.97343 0.0414

0.00886 0.0405 0.99914

. (6.127)

Although there is only one phase exp(iδ) in the CKM matrix V , the exper-

imental constraints on this phase often are expressed in terms of the angles

α, β, and γ defined as

α = arg [−VtdV ∗tb/ (VudV∗ub)]

β = arg [−VcdV ∗cb/ (VtdV∗tb)] (6.128)

γ = arg [−VudV ∗ub/ (VcdV∗cb)] .

If V is unitary, then α + β + γ = 180. From B → ππ, ρπ, and ρρ decays,

the limits on the angle α are roughly

α = (85.4± 4). (6.129)

156 Standard model

From B± → DK± decays, the limits on the angle γ are roughly

γ = (68.0± 8). (6.130)

So the angle β is about 26.6.

One of the quark-Higgs interactions is

−cdijQ′†ìHd′rj = −

√2

vQ′†`Mdd

′rH = −

√2

vQ′†` LdΣdR

†dd′rH

= −√

2

vQ†`ΣddrH = −

√2

vQ†`

(0

(v + h)/√

2

)Σddr

= − d†`

(1 +

h

v

)Σddr = −mdid

†ì

(1 +

h

v

)dri.

(6.131)

A similar term describes the coupling of the up quarks to the Higgs

−muiu†ì

(1 +

h

v

)uri. (6.132)

Thus, the rate of quark-antiquark to Higgs is proportional to the mass of

the quark in the standard model.

6.7 Lepton Masses

We can treat the leptons just like the quarks. The up leptons are the flavor

neutrinos ν ′e, ν′µ, and ν ′τ , and the down leptons are the flavor charged leptons

e′, µ′, and τ ′. The action density

3∑i,j=1

−ceij E′†ìHe′rj − c∗eij e

′†rjH

†E′ì (6.133)

gives for the e′, µ′, and τ ′ the mixed mass terms

v√2

3∑i,j=1

−ceij e′†ìe′rj − c∗eij e

′†rje′ì. (6.134)

The 3× 3 mass matrix Me with entries

[Me]ij =v√2ceij (6.135)


Me = LeΣeR†e (6.136)

in which Le and Re are 3 × 3 unitary matrices, and Σe is a 3 × 3 diagonal

matrix with nonincreasing positive singular values mτ , mµ, and me on its

main diagonal.


6.8 Before and after symmetry breaking

Before spontaneous symmetry breaking, all the fields of the standard model

are massless, and the local symmetry under SU(2)` ⊗ U(1) is exact. Under

these gauge transformations, the left-handed electron and neutrino fields are

rotated among themselves. If e′` is a linear combination of itself and of ν ′e`,

then these two fields, e′` and ν ′e`, must be of the same kind. The left-handed

electron field is a Dirac field. Thus, the left-handed neutrino field also must

be a Dirac field. This makes sense because before symmetry breaking, all

the fields are massless, and so there is no problem combining two Majorana

fields of the same mass, namely zero, into one Dirac field. Thus, there are

three flavor Dirac neutrino fields ν ′e`, ν′µ`, and ν ′τ`.

A massless left-handed neutrino field ν ′` satisfies the two-component Dirac

equation

(∂0I −∇ · σ) ν ′`(x) = 0 (6.137)

which in momentum space is

(E + p · σ) ν ′`(p) = 0. (6.138)

Since the angular momentum is J = σ/2, and E = |p|, we have

p · J ν ′`(p) = −1

2ν ′`(p). (6.139)

The left-handed neutrino field ν ′` annihilates neutrinos of negative helicity

and creates antineutrinos of positive helicity.

Since the neutrinos are massive, there may be right-handed neutrino fields.

As for the up quarks, we can use them to make an action density

3∑i,j=1

−cνijE′†ìσ2H∗ν ′rj − c∗νijν

′†rjH

Tσ2E′ì (6.140)

that is invariant under SU(2)` ⊗ U(1) and that gives for the neutrinos the

mixed mass terms

3∑i,j=1

iv√2

(cνijν′†ìν′rj − c∗νijν

′†rjν′ì). (6.141)

The 3× 3 mass matrix Mν with entries

[Mν ]ij =iv√

2cνij (6.142)


Mν = LνΣνR†ν (6.143)

158 Standard model

in which Lν and Rν are 3× 3 unitary matrices, and Σν is a 3× 3 diagonal

matrix with nonincreasing positive singular values mντ , mνµ , and mνe on its

main diagonal (here, I have assumed that the neutrino masses mimic those

of the charged leptons and quarks, rising with family number). The neutrino

CKM matrix then would be L†ν Le, but since we are accustomed to treating

the charged leptons as flavor and mass eigenfields, we apply the neutrino

CKM matrix to the neutrinos rather than to the charged leptons. Thus the

neutrino CKM matrix is

Vν = L†e Lν . (6.144)

By choosing the phases of the six lepton fields, we can make the neutrino

CKM matrix real except for CP -breaking phases. If the neutrinos are Dirac

fields, then there is one such phase; if not, there are three.

So far, I have assumed that the mass terms for the neutrinos are the usual

Dirac mass terms. However, the right-handed Majorana neutrino fields ν ′rare not affected by the SU(2)` ⊗ U(1).

Note that a gauge transformation between e and νe rotates the operators

a(p, s, e) and a(p, s, νe) into each other. This rotation makes sense only when

the two particles have the same mass. In the standard model, such a gauge

transformation makes sense only before symmetry breaking when all the

particles are massless. Moreover, only when the particles are massless can

one say that they are left- or right-handed. While the particles are massless,

the operator a(p,−) annihilates a particle of negative helicity and occurs

only in a left-handed field, while the operator a(p,+) annihilates a particle

of positive helicity and occurs only in a right-handed field. But when the

particles are massive, the operator a(p, 12) annihilates a particle that is spin

up and occurs in both left-handed and right-handed fields. So a symmetry

transformation that acted on the operator a(p, 12) would change both left-

handed and right-handed fields.

The left-handed fields of the neutrino and electron are

νe,`(x) =

∫u(p,−)

a1(p,−, νe) + ia2(p,−, νe)√2

eipx (6.145)

+v(p,+)a†1(p,+, νe) + ia†2(p,+, νe)√

2e−ipx

d3p

(2π)3/2(6.146)

e`(x) =

∫u(p,−)

a1(p,−, e) + ia2(p,−, e)√2

eipx (6.147)

+v(p,+)a†1(p,+, e) + ia†2(p,+, e)√

2e−ipx

d3p

(2π)3/2(6.148)

where (p,−, νe) means momentum p, spin down, and electron flavor, and


(p,+, νe) means momentum p, spin up, and electron flavor. These fields sat-

isfy equations like (6.137–6.139) apart from their interactions with other

fields. Since a gauge transformation maps the fields νe,`(x) and e`(x) into

each other, we know that when all the fields are massless, before symmetry

breaking, there are (for each momentum) at least two neutrino and antineu-

trino states

1√2

[a†1(p,−, νe)− ia†2(p,−, νe)

]|0〉 (6.149)

1√2

[a†1(p,+, νe) + ia†2(p,+, νe)

]|0〉 (6.150)

for each of the three flavors, f = e, µ, τ . So there are at least six neutrino

(and antineutrino) states.

The right-handed electron field exists and interacts with gauge bosons and

other fields. So there are 12 electron states a†i (p,±, ef )|0〉 for i = 1 and 2 and

for the three flavors, f = e, µ, τ . We don’t know yet whether a right-handed

neutrino field exists or interacts with other fields. So there may be only 6

neutrino states or as many as 12.

Neutrino oscillations tell us that neutrinos have masses. If there are 12

neutrino states, then there can be three massive Dirac neutrinos analagous

to the e, µ, and τ or six massive Majorana neutrinos or some intermediate

combination. If there are only 6, then there can be 3 massive Majorana

neutrinos.

The Majorana mass terms for the right-handed neutrino fields are

6∑ij=1

1

2

[imijν

′Tri σ2 ν

′rj + (imijν

′Tri σ2 ν

′rj)†]. (6.151)

They are Lorentz invariant because under the Lorentz transformations (6.83)

ν ′′Trσ2ν′′r = ν ′Tr exp(z∗ · σT)σ2 exp(z∗ · σ) ν ′r

= ν ′Tr σ2 exp(−z∗ · σ) exp(z∗ · σ) ν ′r = ν ′Tr σ2 ν′r.

(6.152)

The Majorana mass terms (6.151) are unrelated to the scale v of the Higgs

field’s mean value. One can show that the complex matrix mij is symmetric.

One then must combine the mass matrix in (6.151) with the mass matrix

Mν in (6.142). The resulting mass matrix will have a singular-value decom-

position with six singular values that would be the masses of the “physical”

neutrinos. If these six masses are equal in pairs, then the three pairs would

form three Dirac neutrinos.

Whether or not there are right-handed neutrinos, we can make Majorana

mass terms like ν`Tσ2ν`, which are Lorentz invariant but not invariant under

160 Standard model

SU`(2) or UY (1). We can make them gauge invariant by using a triplet~φ = σiφi of Higgs fields that transforms as ~φ′ · ~σ = g(~φ · ~σ)g† for g ∈ SU`(2)

and that carries a value of Y = −1. Then if σ2 has Lorentz indices and σ′2has SU`(2) indices, the term

E`Tσ2σ

′2(~φ · ~σ)E` (6.153)

is both Lorentz invariant and gauge invariant. If the potential V (~φ) has

minima at ~φ 6= 0, then this term violates lepton number and gives a Ma-

jorana mass to the neutrino. Another possibility is to say that at higher

energies a theory with new fields of very high mass Λ plays a role, and that

when one path-integrates over these heavy fields, one is left with an effective,

nonrenormalizable term in the action

− (H†E`)2

Λ(6.154)

which gives a Majorana mass to ν`.

Models with both right-handed and left-handed neutrinos are easier to

think about, but only experiments can tell us whether right-handed neutri-

nos exist.

What is known experimentally is that there are at least three masses that

satisfy

∣∣∆m221

∣∣ ≡ ∣∣m22 −m2

1

∣∣ = (7.53± 0.18)× 10−5 eV2∣∣∆m232

∣∣ ≡ ∣∣m23 −m2

2

∣∣ = (2.44± 0.06)× 10−3 eV2 normal mass hierarchy∣∣∆m232

∣∣ ≡ ∣∣m23 −m2

2

∣∣ = (2.52± 0.07)× 10−3 eV2 inverted mass hierarchy.

(6.155)

If the neutrinos are Dirac particles, then they have a CKM matrix like

that of the quarks with one CP -violating phase. But whereas one chooses

to make the mass and flavor eigenfields the same for the up quarks u, c, t,

for the leptons one makes the mass and flavor eigenfields the same for the

down or charged leptons e, µ, τ . So the neutrino CKM matrix actually is

V = L†eLν . If they are three Majorana particles, then their CKM matrix has

two extra CP -violating phases α12 and α31. A common convention for the

6.9 The Seesaw Mechanism 161

neutrino CKM matrix is

V =

1 0 0

0 cos θ23 sin θ23

0 − sin θ23 cos θ23

cos θ13 0 sin θ13e−iδ

0 1 0

− sin θ13eiδ 0 cos θ13

(6.156)

×

cos θ12 sin θ12 0

− sin θ12 cos θ12 0

0 0 1

1 0 0

0 eiα12/2 0

0 0 eiα31/2

. (6.157)

This convention without the last 3 × 3 matrix also is used for the quark

CKM matrix. The current estimates are

sin2(2θ12) = 0.846± 0.021 (6.158)

sin2(2θ23) = 0.999+0.001

−0.018normal mass hierarchy (6.159)

sin2(2θ23) = 1.000+0.000

−0.017inverted mass hierarchy (6.160)

sin2(2θ13) = 0.093± 0.008. (6.161)

Two of these are big angles: 2θ12 ≈ 2θ23 = π/2±nπ. In the normal hierarchy,

the lightest neutrino is about 2/3 electron, 1/6 muon, and 1/6 tau; the very

slightly heavier neutrino is about 1/3 electron, 1/3 muon, and 1/3 tau; and

the much heavier heavier neutrino is about 1/6 electron, 5/12 muon, and

5/12 tau.

6.9 The Seesaw Mechanism

Why are the neutrino masses so light? Suppose we wish to find the eigen-

values of the real, symmetric mass matrix

M =

(0 m

m M

)(6.162)

in which m is an ordinary mass and M is a huge mass. The eigenvalues µ of

this hermitian mass matrix satisfy det (M− µI) = µ(µ−M)−m2 = 0 with

solutions µ± =(M ±

√M2 + 4m2

)/2. The larger mass µ+ ≈M+m2/M is

approximately the huge mass M and the smaller mass µ− ≈ −m2/M is tiny.

The physical mass of a fermion is the absolute value of its mass parameter,

here m2/M .

The product of the two eigenvalues is the constant µ+µ− = detM = −m2

so as µ− goes down, µ+ must go up. In 1975, Gell-Mann, Ramond, Slansky,

and Jerry Stephenson invented this “seesaw” mechanism as an explanation

162 Standard model

of why neutrinos have such small masses, less than 1 eV/c2. If mc2 = 10

MeV, and µ−c2 ≈ 0.01 eV, which is a plausible light-neutrino mass, then

the rest energy of the huge mass would be Mc2 = 107 GeV. This huge mass

would be one of the six neutrino masses and would point at new physics,

beyond the standard model. Yet the small masses of the neutrinos may be

related to the weakness of their interactions.

Before leaving the subject of fermion masses, let’s look more closely at

Dirac and Majorana mass terms. A Dirac field is a linear combination of

two Majorana fields of the same mass

ψ =1√2

(L+ i`

R+ ir

)(6.163)

in which L and ` are two-component left-handed spinors, and R and r are

two-component right-handed spinors. The Dirac mass term

mψψ = imψ†γ0ψ = mψ†(

0 I

I 0

)ψ

= m1

2

(L† − i`†, R† − ir†

)(0 I

I 0

)(L+ i`

R+ ir

)= m

1

2

(L† − i`†, R† − ir†

)(R+ ir

L+ i`

)= m

1

2

[(L† − i`†

)(R+ ir) +

(R† − ir†

)(L+ i`)

](6.164)

= m1

2

(R† − ir†

)(L+ i`) + h.c.,

in which h.c. means hermitian conjugate, gives mass m to the particle and

antiparticle of the Dirac field ψ.

We may set

R = iσ2 L∗ ⇐⇒ L = −iσ2R

∗ (6.165)

r = iσ2 `∗ ⇐⇒ ` = −iσ2 r

∗ (6.166)

or equivalently (R1

R2

)=

(L†2−L†1

)⇐⇒

(L1

L2

)=

(−R†2R†1

)(6.167)

(r1

r2

)=

(`†2−`†1

)⇐⇒

(`1`2

)=

(−r†2r†1

)(6.168)

which are the Majorana conditions (1.252 & 1.253). Since R† = −iLTσ2, we

6.10 Neutrino Oscillations 163

can write the Dirac mass term (6.164) in terms of left-handed fields as

mψψ =1

2m(−i LT − `T

)σ2 (L+ i`) + h.c. (6.169)

=1

2m(LT − i`T

)(−i σ2) (L+ i`) + h.c. (6.170)

=1

2m(L1 − i`1, L2 − i`2

)(0 −1

1 0

)(L1 + i`1L2 + i`2

)+ h.c. (6.171)

=1

2m(L1 − i`1, L2 − i`2

)(−L2 − i`2L1 + i`1

)+ h.c. (6.172)

=1

2m (L1 − i`1) (−L2 − i`2) + (L2 − i`2) (L1 + i`1) + h.c. (6.173)

The fermion fields anticommute, so the Dirac mass term is

mψψ =1

2m (−2L1L2 − 2`1`2) + h.c. = −m (L1L2 + `1`2) + h.c., (6.174)

and it says that the fields L and ` have the same mass m, as they must if

they are to form a Dirac field.

Since L† = iRT σ2, we also can write the Dirac mass term in terms of the

right-handed fields as

mψψ =1

2m(RT − irT

)iσ2 (R+ ir) + h.c. (6.175)

=1

2m(R1 − ir1, R2 − ir2

)( 0 1

−1 0

)(R1 + ir1

R2 + ir2

)+ h.c. (6.176)

= m (R1R2 + r1r2) + h.c. (6.177)

So the fields R and r have the same mass m, as they must if they are to

form a Dirac field.

The Majorana mass term for a right-handed field r of mass m evidently

is

mr1 r2 + h.c. (6.178)

6.10 Neutrino Oscillations

The phase difference ∆φ between two highly relativistic neutrinos of mo-

mentum p going a distance L in a time t ≈ L varies with their masses m1

and m2 as

∆φ = t∆E =LE

p∆E =

LE

p

(√p2 +m2

1 −√p2 +m2

2

)(6.179)

164 Standard model

in natural units. We can approximate this phase by using the first two terms

of the binomial expansion of the square roots with y = 1 and x = m2i /p

2

∆φ = LE

(√1 +m2

1/p2 −

√1 +m2

2/p2

)≈ LE∆m2

p2≈ L∆m2

E(6.180)

or in ordinary units ∆φ ≈ L∆m2c3/(~E).

7

Path integrals

7.1 Path integrals and Richard Feynman

Since Richard Feynman invented them over 70 years ago, path integrals

have been used with increasing frequency in high-energy and condensed-

matter physics, in optics and biophysics, and even in finance. Feynman used

them to express the amplitude for a process as a sum of all the ways the

process could occur each weighted by an exponential of its classical action

exp(iS/~). Others have used them to compute partition functions and to

study the QCD vacuum. (Richard Feynman, 1918–1988)

7.2 Gaussian integrals and Trotter’s formula

Path integrals are based upon the gaussian integral (??) which holds for

real a 6= 0 and real b ∫ ∞−∞

eiax2+2ibx dx =

√iπ

ae−ib

2/a (7.1)

and upon the gaussian integral (??)∫ ∞−∞

e−ax2+2ibx dx =

√π

ae−b

2/a (7.2)

which holds both for Re a > 0 and also for Re a = 0 with b real and Im a 6= 0.

The extension of the integral formula (7.1) to any n × n real symmetric

nonsingular matrix sjk and any real vector cj is (exercises ?? & ??)∫ ∞−∞

eisjkxjxk+2icjxj dx1 . . . dxn =

√(iπ)n

det se−icj(s

−1)jkck (7.3)

166 Path integrals

in which det a is the determinant of the matrix a, a−1 is its inverse, and

sums over the repeated indices j and k from 1 to n are understood. One

may similarly extend the gaussian integral (7.2) to any positive symmetric

n× n matrix sjk and any vector cj (exercises ?? & ??)∫ ∞−∞

e−sjkxjxk+2icjxj dx1 . . . dxn =

√πn

det se−cj(s

−1)jkck . (7.4)

Path integrals also are based upon Trotter’s product formula (Trotter,

1959; Kato, 1978)

ea+b = limn→∞

(ea/n eb/n

)n(7.5)

both sides of which are symmetrically ordered and obviously equal when

ab = ba.

Separating a given hamiltonian H = K + V into a kinetic part K and a

potential part V , we can use Trotter’s formula to write the time-evolution

operator e−itH/~ as

e−it(K+V )/~ = limn→∞

(e−itK/(n~) e−itV/(n~)

)n(7.6)

and the Boltzmann operator e−βH as

e−β(K+V ) = limn→∞

(e−βK/n e−βV/n

)n. (7.7)

7.3 Path integrals in quantum mechanics

Path integrals can represent matrix elements of the time-evolution oper-

ator exp(− i(tb − ta)H/~) in which H is the hamiltonian. For a particle of

mass m moving nonrelativistically in one dimension in a potential V (q), the

hamiltonian is

H =p2

2m+ V (q). (7.8)

The position and momentum operators q and p obey the commutation rela-

tion [q, p] = i~. Their eigenstates |q′〉 and |p′〉 have eigenvalues q′ and p′ for

all real numbers q′ and p′

q |q′〉 = q′ |q′〉 and p |p′〉 = p′ |p′〉. (7.9)

These eigenstates are complete. Their outer products |q′〉〈q′| and |p′〉〈p′|


provide expansions for the identity operator I and have inner products (??)

that are phases

I =

∫ ∞−∞|q′〉〈q′| dq′ =

∫ ∞−∞|p′〉〈p′| dp′ and 〈q′|p′〉 =

eiq′p′/~

√2π~

. (7.10)

Setting ε = (tb− ta)/n and writing the hamiltonian (7.8) over ~ as H/~ =

p2/(2m~) + V/~ = k + v, we can write Trotter’s formula (7.6) for the time-

evolution operator as the limit as n→∞ of n factors of e−iεke−iεv

e−i(tb−ta)(k+v) = e−iεk e−iεv e−iεk e−iεv · · · e−iεk e−iεv e−iεk e−iεv. (7.11)

The advantage of using Trotter’s formula is that we now can evaluate the

matrix element 〈q1|e−iεk e−iεv|qa〉 between eigenstates |qa〉 and |q1〉 of the

position operator q by inserting the momentum-state expansion (7.10) of

the identity operator I between these two exponentials

〈q1|e−iεk e−iεv|qa〉 = 〈q1| e−iεp2/(2m~)

∫ ∞−∞|p′〉〈p′| dp′ e−iε V (q)/~|qa〉 (7.12)

and using the eigenvalue formulas (7.9)

〈q1|e−iεk e−iεv|qa〉 =

∫ ∞−∞e−iεp

′2/(2m~) 〈q1|p′〉 e−iεV (qa)/~ 〈p′|qa〉 dp′. (7.13)

Now using the formula (7.10) for the inner product 〈q1|p′〉 and the complex

conjugate of that formula for 〈p′|qa〉, we get

〈q1|e−iεk e−iεv|qa〉 = e−iε V (qa)/~∫ ∞−∞

e−iεp′2/(2m~) ei(q1−qa)p′/~ dp

′

2π~. (7.14)

In this integral, the momenta that are important are very high, being of

order√m~/ε which diverges as ε→ 0; nonetheless, the integral converges.

If we adopt the suggestive notation q1 − qa = ε qa and use the gaussian

integral (7.1) with a = −ε/(2m~), x = p, and b = εq/(2~)∫ ∞−∞

exp

(− i ε p2

2m~+ i ε

q p

~

)dp

2π~=

√m

2πiε~exp

(iε

~mq2

2

), (7.15)

then we find

〈q1|e−iεk e−iεv|qa〉 =1

2π~e−ε V (qa)

∫ ∞−∞

exp

(− i εp

′2

2m~+ i

ε qa p′

~

)dp′

=( m

2πi~ε

)1/2exp

[iε

~

(mq2

a

2− V (qa)

)]. (7.16)

The dependence of the amplitude 〈q1|e−iεk e−iεv|qa〉 upon q1 is hidden in the

formula qa = (q1 − qa)/ε.

168 Path integrals

The next step is to use the position-state expansion (7.10) of the identity

operator to link two of these matrix elements together

〈q2|(e−iεk e−iεv

)2|qa〉 =

∫ ∞−∞〈q2| e−iεk e−iεv |q1〉〈q1| e−iεk e−iεv |qa〉 dq1

(7.17)

=m

2πi~ε

∫ ∞−∞

exp

[iε

~

(mq2

1

2− V (q1) +

mq2a

2− V (qa)

)]dq1

where now q1 = (q2 − q1)/ε.

By stitching together n = (tb − ta)/ε time intervals each of length ε and

letting n→∞, we get

〈qb|e−niεH/~|qa〉 =

∫〈qb|e−iεk e−iεv|qn−1〉 · · · 〈q1|e−iεk e−iεv|qa〉 dqn−1 · · · dq1

=( m

2πi~ε

)n/2∫exp

[iε

~

n−1∑j=0

mq2j

2− V (qj)

]dqn−1 · · · dq1

=( m

2πi~ε

)n/2∫exp

i ε~

n−1∑j=0

Lj

dqn−1 · · · dq1 (7.18)

in which Lj = mq2j /2− V (qj) is the lagrangian of the jth interval, and the

qj integrals run from −∞ to ∞. In the limit ε → 0 with nε = (tb − ta)/ε,this multiple integral is an integral over all paths q(t) that go from qa, ta to

qb, tb

〈qb|e−i (tb−ta)H/~|qa〉 =

∫eiS[q]/~Dq (7.19)

in which each path is weighted by the phase of its classical action

S[q] =

∫ tb

ta

L(q, q) dt =

∫ tb

ta

(mq(t)2

2− V (q(t))

)dt (7.20)

in units of ~ and Dq = (mn/(2πi~(tb − ta)))n/2 dqn−1 . . . dq1.

If we multiply the path-integral (7.19) for 〈qb|e−i (tb−ta)H/~|qa〉 from the

left by |qb〉 and from the right by 〈qa| and integrate over qa and qb as in

the resolution (7.10) of the identity operator, then we can write the time-

evolution operator as an integral over all paths from ta to tb

e−i (tb−ta)H/~ =

∫|qb〉 eiS[q]/~ 〈qa|Dq dqa dqb (7.21)

with Dq = (mn/(2πi~(tb − ta)))n/2 dqn−1 . . . dq1 and S[q] the action (7.20).


The path integral for a particle moving in three-dimensional space is

〈qb|e−i(tb−ta)H/~|qa〉 =

∫exp

[i

~

∫ tb

ta

12 m q

2(t)− V (q(t)) dt

]Dq (7.22)

where Dq = (mn/(2πi~(tb − ta)))3n/2 dqn−1 . . . dq1.

Let us first consider macroscopic processes whose actions are large com-

pared to ~. Apart from the factor Dq, the amplitude (7.22) is a sum of

phases eiS[q]/~ one for each path from qa, ta to qb, tb. When is this ampli-

tude big? When is it small? Suppose there is a path qc(t) from qa, ta to qb, tbthat obeys the classical equation of motion (??–??)

δS[qc]

δqjc= mqjc + V ′(qc) = 0. (7.23)

Its action may be minimal. It certainly is stationary: a path qc(t)+δq(t) that

differs from qc(t) by a small detour δq(t) has an action S[qc+δq] that differs

from S[qc] only by terms of second order and higher in δq. Thus a classical

path has infinitely many neighboring paths whose actions differ only by inte-

grals of (δq)n, n ≥ 2, and so have the same action to within a small fraction

of ~. These paths add with nearly the same phase to the path integral (7.22)

and so make a huge contribution to the amplitude 〈qb|e−i(tb−ta)H/~|qa〉. But

if no classical path goes from qa, ta to qb, tb, then the nonclassical, nonsta-

tionary paths that go from qa, ta to qb, tb have actions that differ from each

other by large multiples of ~. These amplitudes cancel each other, and their

sum, which is amplitude for going from qa, ta to qb, tb, is small. Thus the

path-integral formula for an amplitude in quantum mechanics ex-

plains why macroscopic processes are described by the principle

of stationary action (section ??).

What about microscopic processes whose actions are tiny compared to ~?

The path integral (7.22) gives large amplitudes for all microscopic processes.

On very small scales, anything can happen that doesn’t break a conservation

law.

The path integral for two or more particles q = q1, . . . , qk interacting

with a potential V (q) is


∫eiS[q]/~Dq (7.24)

where

S[q] =

∫ tb

ta

[m1q

21(t)

2+ · · ·+ mkq

2k(t)

2− V (q(t))

]dt (7.25)

and Dq = Dq1 · · ·Dqk.

170 Path integrals

Example 7.1 (A free particle) For a free particle, the potential is zero,

and the path integral (7.19, 7.20) is the ε→ 0, n→∞ limit of

〈qb|e−i tH/~|qa〉 =( m

2πi~ε

)n/2(7.26)

×∫

exp

[imε

2~

((qb − qn−1)2

ε2+ · · ·+ (q1 − qa)2

ε2

)]dqn−1 · · · dq1.

The q1 integral is by the gaussian formula (7.1)

m

2πi~ε

∫eim[(q2−q1)2+(q1−qa)2]/(2~ε)dq1 =

√m

2πi~2εeim(q2−qa)2/(2~2ε). (7.27)

The q2 integral is (exercise ??)

m

2√

2πi~ε

∫eim[(q3−q2)2+(q2−qa)2/2]/(2~ε)dq2 =

√m

2πi~3εeim(q3−qa)2/(2~3ε).

(7.28)

Doing all n− 1 integrals (7.26) in this way and setting nε = tb − ta, we get

〈qb| e−i (tb−ta)H/~ |qa〉 =

√m

2πi~nεexp

[im(qb − qa)2

2~nε

]=

√m

2πi~(tb − ta)exp

[im(qb − qa)2

2~(tb − ta)

].

(7.29)

The path integral (7.26) is perfectly convergent even though the velocities

qj = (qj+1−qj)/ε that are important are very high, being of order√~/(mε).

It is easier to compute this amplitude (7.29) by using the outer products

(7.10) (exercise ??).

In three dimensions, the amplitude to go from qa, ta to qb, tb is

〈qt|e−i(tb−ta)H/~|q0〉 =

(m

2πi~(tb − ta)

)3/2

exp

[im(qb − qa)2

2~(tb − ta)

]. (7.30)

7.4 Path integrals for quadratic actions

If a path q(t) = qc(t) + x(t) differs from a classical path qc(t) by a detour

x(t) that vanishes at the endpoints x(ta) = 0 = x(tb) so that both paths

go from qa, ta to qb, tb, then the difference S[qc + x] − S[qc] in their actions

vanishes to first order in the detour x(t) (section ??). Thus the actions of


the two paths differ by a time integral of quadratic and higher powers of the

detour x(t)

S[qc + x] =

∫ tb

ta

12mq(t)

2 − V (q(t)) dt

=

∫ tb

ta

12m (qc(t) + x(t))2 − V (qc(t) + x(t)) dt

=

∫ tb

ta

[m

2q2c +mqcx+

m

2x2 − V (qc)− V ′(qc)x−

V ′′(qc)

2x2

−V′′′(qc)

6x3 − V ′′′′(qc)

24x4 − . . .

]dt (7.31)

=

∫ tb

ta

[m2q2c − V (qc)

]dt+

∫ tb

ta

[m

2x2 − V ′′(qc)

2x2

−V′′′(qc)

6x3 − V ′′′′(qc)

24x4 − . . .

]dt

= S[qc] + ∆S[qc, x]

in which S[qc] is the action of the classical path, and the detour x(t) is a

loop that goes from x(ta) = 0 to x(tb) = 0.

If the potential V (q) is quadratic in the position q, then the third V ′′′

and higher derivatives of the potential vanish, and the second derivative is

a constant V ′′(qc(t)) = V ′′. In this quadratic case, the correction ∆S[qc, x]

depends only on the time interval tb − ta and on ~, m, and V ′′

∆S[qc, x] = ∆S[x] =

∫ tb

ta

[12mx2(t)− 1

2V′′ x2(t)

]dt. (7.32)

It is independent of the classical path.

Thus for quadratic actions, the path integral (7.19) is an exponential of the

action S[qc] of the classical path multiplied by a function f(tb− ta, ~,m, V ′′)of the time interval tb − ta and of ~, m, and V ′′


∫eiS[q]/~Dq =

∫ei(S[qc]+∆S[x])/~Dq

= eiS[qc]/~∫ei∆S[x])/~Dx

= f(tb − ta, ~,m, V ′′) eiS[qc]/~.

(7.33)

The function f = f(tb − ta, ~,m, V ′′) is the limit as n → ∞ of the (n − 1)-

172 Path integrals

dimensional integral

f =

[mn

2πi~(tb − ta)

]n/2∫ei∆S[x])/~dxn−1 . . . dx1 (7.34)

where

∆S[x] =tb − tan

n∑j=1

1

2m

(xj − xj−1)2

[(tb − ta)/n]2− 1

2V ′′ x2

j (7.35)

and xn = 0 = x0.

More generally, the path integral for any quadratic action of the form

S[q] =

∫ tb

ta

u q2(t) + v q(t)q(t) + w q2(t) + s(t) q(t) + j(t) q(t) dt (7.36)

is (exercise ??)

〈qb|e−i(tb−ta)H/~|qa〉 = f(ta, tb, ~, u, v, w) eiS[qc]/~. (7.37)

The dependence of the amplitude upon s(t) and j(t) is contained in the

classical action S[qc].

These formulas (7.33–7.37) may be generalized to any number of particles

with coordinates q = q1, . . . , qk moving nonrelativistically in a space of

multiple dimensions as long as the action is quadratic in the q’s and their

velocities q. The amplitude is then an exponential of the action S[qc] of

the classical path multiplied by a function f(ta, tb, ~, . . . ) that is independent

of the classical path qc

〈qb|e−i(tb−ta)H/~|qa〉 = f(ta, tb, ~, . . . ) eiS[qc]/~. (7.38)

Example 7.2 (A free particle) The classical path of a free particle going

from qa at time ta to qb at time tb is

qc(t) = qa +t− tatb − ta

(qb − qa). (7.39)

Its action is

S[qc] =

∫ tb

ta

1

2m q2

c dt =m(qb − qa)2

2(tb − ta)(7.40)

and for this case our quadratic-potential formula (7.38) is

〈qb|e−i(tb−ta)H/~|qa〉 = f(tb − ta, ~,m) exp

[im(qb − qa)2

2~(tb − ta)

](7.41)

which agrees with our explicit calculation (7.30) when f(tb − ta, ~,m) =

[m/(2πi~(tb − ta))]3/2.


Example 7.3 (Bohm-Aharonov effect) From the formula (??) for the ac-

tion of a relativistic particle of mass m and charge e, it follows (exercise ??)

that the action a nonrelativistic particle in an electromagnetic field with no

scalar potential is

S =

∫ tb

ta

[12mq

2 + eA · q]dt =

∫ qb

qa

[12mq + eA

]· dq . (7.42)

Since this action is quadratic in q, the amplitude for a particle to go from

qa at ta to qb at tb is an exponential of the classical action

〈qb|e−i(tb−ta)H/~|qa〉 = f(tb − ta, ~,m, e) eiS[qc]/~ (7.43)

multiplied by a function f(tb−ta, ~,m, e) that is independent of the path qc.

A beam of such particles goes horizontally past but not through a vertical

pipe in which a vertical magnetic field is confined. The particles can go

both ways around the pipe of cross-sectional area S but do not enter it. The

difference in the phases of the amplitudes for the two paths is a loop integral

from the source to the detector around the pipe and back to the source∮ [mq

2+ eA

]· dq~

=

∮mq · dq

2~+e

~

∫SB · dS =

∮mq · dq

2~+eΦ

~(7.44)

in which Φ is the magnetic flux through the cylinder.

Example 7.4 (Harmonic oscillator) The action

S =

∫ tb

ta

1

2mq2(t)− 1

2mω2q2(t) dt (7.45)

of a harmonic oscillator is quadratic in q and q. So apart from a factor f ,

its path integral (7.33–7.35) is an exponential

〈qb|e−i(tb−ta)H/~|qa〉 = f eiS[qc]/~ (7.46)

of the action S[qc] (exercise ??)

S[qc] =mω

[(q2a + q2

b

)cos(ω(tb − ta))− 2qaqb

]2 sin(ω(tb − ta))

(7.47)

of the classical path

qc(t) = qa cosω(t− ta) +qb − qa cosω(tb − ta)

sinω(tb − ta)sinω(t− ta) (7.48)

that runs from qa, ta to qb, tb and obeys the classical equation of motion

mqc(t) = − ω2qc(t).

The factor f is a function f(tb−ta, ~,m,mω2) of the time interval and the

174 Path integrals

parameters of the oscillator. It is the n→∞ limit of the (n−1)-dimensional

integral (7.34)

f =

[mn

2πi~(tb − ta)

]n/2 ∫ei∆S[x])/~ dxn−1 . . . dx1 (7.49)

over all loops that run from 0 to 0 in time tb − ta in which the quadratic

correction to the classical action is (7.35)

∆S[x] =tb − tan

n∑j=1

1

2m

(xj − xj−1)2

[(tb − ta)/n]2− 1

2mω2 x2

j , (7.50)

and xn = 0 = x0.

Setting tb− ta = T , we use the many-variable imaginary gaussian integral

(7.3) to write f as

f =[ mn

2πi~T

]n/2∫eiajkxjxkdxn−1 . . . dx1 =

[ mn

2πi~T

]n/2√(iπ)n−1

det a(7.51)

in which the quadratic form ajkxjxk is

nm

~T

n∑j=1

[− xjxj−1 +

1

2(x2j + x2

j−1)− (ωT )2

2n2x2j

](7.52)

which has no linear term because x0 = xn = 0.

The (n− 1)-dimensional square matrix a is a tridiagonal Toeplitz matrix

a =nm

2~T

y −1 0 0 · · ·−1 y −1 0 · · ·0 −1 y −1 · · ·0 0 −1 y · · ·...

......

. . .. . .

. (7.53)

Apart from the factor nm/(2~T ), the matrix a = (nm/(2~T ))Cn−1(y) is a

tridiagonal matrix Cn−1(y) whose off-diagonal elements are −1 and whose

diagonal elements are

y = 2− (ωT )2

n2. (7.54)

Their determinants |Cn(y)| = detCn(y) obey (exercise ??) the recursion

relation

|Cn+1(y)| = y |Cn(y)| − |Cn−1(y)| (7.55)


and have the initial values |C1(y)| = y and |C2(y)| = y2 − 1. The trigono-

metric functions Un(y) = sin[(n+ 1)θ]/ sin θ with y = 2 cos θ obey the same

recursion relation and have the same initial values (exercise ??), so

|Cn(y)| = sin(n+ 1)θ

sin θ. (7.56)

Since for large n

θ = arccos(y/2) = arccos

(1− ω2t2

2n2

)≈ ωT

n, (7.57)

the determinant of the matrix a is

det a =( nm

2~T

)n−1|Cn−1(y)| =

( nm2~T

)n−1 sinnθ

sin θ

≈( nm

2~T

)n−1 sin(ωT )

sin(ωT/n)≈( nm

2~T

)n−1 n sinωT

ωT.

(7.58)

Thus the factor f is

f =[ mn

2πi~T

]n/2√(iπ)n−1

det a=[ mn

2πi~T

]n/2√(2πi~Tnm

)n−1 ωT

n sinωT

=

√mω

2πi~ sinωT. (7.59)

The amplitude (7.46) is then an exponential of the action S[qc] (7.47) of the

classical path (7.48) multiplied by this factor f


√mω

2πi~ sinω(tb − ta)(7.60)

× exp

i

~mω

[(q2a + q2

b

)cos(ω(tb − ta))− 2qaqb

]2 sin(ω(tb − ta))

.

As these examples (7.2 & 7.4) suggest, path integrals are as mathemati-

cally well defined as ordinary integrals.

7.5 Path integrals in statistical mechanics

At the imaginary time t = −i~β = −i~/(kT ), the time-evolution oper-

ator e−itH/~ becomes the Boltzmann operator e−βH whose trace is the

176 Path integrals

partition function Z(β) at inverse energy β = 1/(kT )

Z(β) = Tr(e−βH

)=∑n

〈n|e−βH |n〉 (7.61)

in which the states |n〉 form a complete orthonormal set, k = 8.617 × 10−5

eV/K is Boltzmann’s constant, and T is the absolute temperature. Partition

functions are important in statistical mechanics and quantum field theory.

Since the Boltzmann operator e−βH is the time-evolution operator e−itH/~

at the imaginary time t = − i~β, we can write it as a path integral by

imitating the derivation of the preceding section (7.3). We will use the same

hamiltonian H = p2/(2m) + V (q) and the operators q and p which have

complete sets of eigenstates (7.9) that satisfy (7.10).

Changing our definitions of ε, k, and v to ε = β/n, k = βp2/(2m), and

v = βV (q), we can write Trotter’s formula (7.7) for the Boltzmann operator

as the n→∞ limit of n factors of e−εk e−εv

e−βH = e−εk e−εv e−εk e−εv · · · e−εk e−εv e−εk e−εv. (7.62)

To evaluate the matrix element 〈q1|e−εk e−εv|qa〉, we insert the identity op-

erator 〈q1|e−εk I e−εv|qa〉 as an integral (7.10) over outer products |p′〉〈p′| of

momentum eigenstates and use the inner products 〈q1|p′〉 = eiq1p′/~/√

2π~and 〈p′|qa〉 = e−iqap

′/~/√

2π~

〈q1|e−εk e−εv|qa〉 =

∫ ∞−∞〈q1|e−εp

2/(2m)|p′〉〈p′|e−εV (q)|qa〉 dp′

= e−ε V (qa)

∫ ∞−∞

e−εp′2/(2m) eip

′(q1−qa)/~ dp′

2π~. (7.63)

If we adopt the suggestive notation q1 − qa = ε~ qa and use the gaussian

integral (7.2) with a = ε/(2m), x = p, and b = εq/2∫ ∞−∞

exp

(− ε p

2

2m+ iε q p

)dp

2π~=

√m

2πε~2exp

(− ε mq

2

2

), (7.64)

then we find

〈q1|e−εk e−εv|qa〉 = e−ε V (qa)

∫ ∞−∞

exp

(−ε p

′2

2m+ i ε p′ qa

)dp′

2π~

=( m

2π~2ε

)1/2exp

[− ε(m q2

a

2+ V (qa)

)](7.65)

in which q1 is hidden in the formula q1 − qa = ~ ε qa.


The next step is to link two of these matrix elements together

〈q2|(e−εk e−εv

)2|qa〉 =

∫ ∞−∞〈q2|e−εk e−εv|q1〉〈q1|e−εk e−εv|qa〉dq1 (7.66)

=m

2π~2ε

∫ ∞−∞

exp

− ε[m q2

1

2+ V (q1) +

m q20

2+ V (qa)

]dq1.

Passing from 2 to n and suppressing some integral signs, we get

〈qb|e−nεH |qa〉 =∫∫∫∞−∞〈qb|e

−εk e−εv|qn−1〉 · · · 〈q1|e−εk e−εv|qa〉 dqn−1 . . . dq1

=(

m2π~2ε

)n/2∫∫∫∞−∞exp

[− ε∑n−1

j=0

(m q2

j

2 + V (qj)

)]dqn−1 . . . dq1.

Setting du = ~ε = ~β/n and taking the limit n → ∞, we find that the

matrix element 〈qb|e−βH |qa〉 is the path integral

〈qb|e−βH |qa〉 =

∫e−Se[q]/~Dq (7.67)

in which each path is weighted by its euclidian action

Se[q] =

∫ ~β

0

mq2(u)

2+ V (q(u)) du, (7.68)

q is the derivative of the coordinate q(u) with respect to euclidian time

u = ~β, and Dq ≡ (nm/2π ~2β)n/2 dqn−1 . . . dq1.

A derivation identical to the one that led from (7.62) to (7.68) leads in a

more elaborate notation to

〈qb|e−(βb−βa)H |qa〉 =

∫e−Se[q]/~Dq (7.69)

in which each path is weighted by its euclidian action

Se[q] =

∫ ~βb

~βa

mq2(u)

2+ V (q(u)) du, (7.70)

and q and Dq are the same as in (7.68).

If we multiply the path integral (7.69) from the left by |qb〉 and from the

right by 〈qa| and integrate over qa and qb as in the resolution (7.10) of the

identity operator, then we can write the Boltzmann operator as an integral

over all paths from ta to tb

e−(βb−βa)H =

∫|qb〉 e−Se[q]/~ 〈qa|Dq dqa dqb (7.71)

with Dq = (mn/(2πi~(tb− ta)))n/2 dqn−1 . . . dq1 and Se[q] the action (7.70).

178 Path integrals

To get the partition function Z(β), we set qb = qa ≡ qn and integrate over

all n q’s letting n→∞

Z(β) = Tr e−βH =

∫〈qn|e−βH |qn〉 dqn

=

∫exp

[−1

~

∫ ~β

0

mq2(u)

2+ V (q(u)) du

]Dq

(7.72)

where Dq ≡ (nm/2π ~2 β)n/2 dqn . . . dq1. We sum over all loops q(u) that go

from q(0) = qn at euclidian time 0 to q(~β) = qn at euclidian time ~β.

In the low-temperature limit, T → 0 and β →∞, the Boltzmann operator

exp(−βH) projects out the ground state |E0〉 of the system

limβ→∞

e−βH = limβ→∞

∑n

e−βEn |En〉〈En| = e−βE0 |E0〉〈E0|. (7.73)

The maximum-entropy density operator (section ??, example ??) is the

Boltzmann operator e−βH divided by its trace Z(β)

ρ =e−βH

Tr(e−βH)=e−βH

Z(β). (7.74)

Its matrix elements are matrix elements of Boltzmann operator (7.68) di-

vided by the partition function (7.72)

〈qb|ρ|qa〉 =〈qb|e−βH |qa〉

Z(β). (7.75)

In three dimensions with q(u) = dq(u)/du, the qa, qb matrix element of

the Boltzmann operator is the analog of equation (7.68) (exercise ??)


∫exp

[− 1

~

∫ ~β

0

mq2(u)

2+ V (q(u)) du

]Dq (7.76)

where Dq ≡ (nm/2π ~2β)3n/2 dqn−1 . . . dq1, and the partition function is

the integral over all loops that go from q0 = qn to qn in time ~β

Z(β) =

∫exp

[−1

~

∫ ~β

0

mq2(u)

2+ V (q(u)) du

]Dq (7.77)

where now Dq ≡ (nm/2π ~2 β)3n/2 dqn . . . dq1.

Because the Boltzmann operator e−βH is the time-evolution operator

e−itH/~ at the imaginary time t = − iu = −i~β = −i~/(kT ), the path

integrals of statistical mechanics are called euclidian path integrals.


Example 7.5 (Density operator for a free particle) For a free particle, the

matrix element of the Boltzmann operator e−βH is the n = β/ε→∞ limit

of the integral

〈qb|e−βH |qa〉 =( m

2π~2ε

)n/2(7.78)

×∫

exp

[− m(qb − qn−1)2

2~2ε· · · − (q1 − qa)2

2~2ε

]dqn−1 · · · dq1.

(7.79)

The formula (7.2) gives for the q1 integral( m

2π~2ε

)1/2∫e−[m(q2−q1)2+m(q1−qa)2]/(2~2ε)dq1 =

e−m(q2−qa)2/(2~22ε)

√2

. (7.80)

The q2 integral is (exercise ??)( m

4π~2ε

)1/2∫e−m(q3−q2)2/(2~2ε)−m(q2−qa)2]/(4~2ε)dq2 =

e−m(q3−qa)2/(2~23ε)

√3

.

(7.81)

All n− 1 integrations give


√m

2π~2ε

e−m(qb−qa)2/(2~2nε)

√n

=

√m

2π~2βe−m(qb−qa)2/(2~2β).

(7.82)

The partition function is the integral of this matrix element over qa = qb

Z(β) =

(m

2π~2β

)1/2 ∫dqa =

(m

2π~2β

)1/2

L (7.83)

where L is the (infinite) 1-dimensional volume of the system. The qb, qamatrix element of the maximum-entropy density operator is

〈qb|ρ|qa〉 =e−m(qb−qa)2/(2~2β)

L. (7.84)

In 3 dimensions, equations (7.82 & 7.83) are


(mkT

2π~2

)3/2

e−m(qb−qa)2/(2~2β) and Z(β) =

(mkT

2π~2

)3/2

L3.

(7.85)

Example 7.6 (Partition function at high temperatures) At high temper-

atures, the time ~β = ~/(kT ) is very short, and the density operator (7.85)

180 Path integrals

for a free particle shows that free paths are damped and limited to distances

of order ~/√mkT . We thus can approximate the path integral (7.77) for

the partition function by replacing the potential V (q(u)) by V (qn) and then

using the free-particle matrix element (7.85)

Z(β) ≈∫e−βV (qn) exp

[−1

~

∫ ~β

0

mq2(u)

2du

]Dq (7.86)

=

∫e−βV (qn)〈qn|e−βH |qn〉 dqn =

(mkT

2π~2

)3/2 ∫e−βV (qn) dqn.

7.6 Boltzmann path integrals for quadratic actions

Apart from the factor Dq ≡ (nm/2π ~2β)n/2 dqn−1 . . . dq1, the euclidian

path integral


∫exp

[−1

~

∫ ~β

0

mq2(u)

2+ V (q(u)) du

]Dq (7.87)

is a sum of positive terms e−Se[q]/~ one for each path from qa, 0 to qb, β. If a

path from qa, 0 to qb, β obeys the classical euclidian equation of motion

md2qcedu2

= m qce = V ′(qce) (7.88)

then its euclidian action

Se[q] =

∫ ~β

0

m q2(u)

2+ V (u) du (7.89)

is stationary and may be minimal. So we can approximate the euclidian

action Se[qce + x] as we approximated the action S[qc + x] in section 7.4.

The euclidian action Se[qce+x] of an arbitrary path from qa, 0 to qb, β is the

stationary euclidian action Se[qce] plus a u-integral of quadratic and higher

powers of the detour x which goes from x(0) = 0 to x(~β) = 0

Se[qce + x] =

∫ ~β

0

[m2q2ce + V (qce)

]du+

∫ ~β

0

[m

2x2 +

V ′′(qce)

2x2

+V ′′′(qce)

6x3 +

V ′′′′(qce)

24x4 + . . .

]du

= Se[qce] + ∆Se[qce, x], (7.90)

7.6 Boltzmann path integrals for quadratic actions 181

and the path integral for the matrix element 〈qb|e−βH |qa〉 is

〈qb|e−βH |qa〉 = e−Se[qce]/~∫e−∆Se[qce,x]/~Dx (7.91)

as n→∞ where Dx = (nm/2π ~2β)n/2 dqn−1 . . . dq1 in the limit n→∞.

If the action is quadratic in q and q, then the integral ∆Se[qce, x] over the

detour x is a gaussian path integral that is independent of the path qce and

so is a function f only of the parameters β, m, ~, and V ′′

〈qb|e−βH |qa〉 = e−Se[qce]/~∫e−∆Se[x]/~Dx

= f(β, ~,m, V ′′) e−Se[qce]/~(7.92)

where

f(β, ~,m, V ′′) =

[mn

2π~2β

]n/2∫e−∆Se[x]/~dxn−1 . . . dx1,

∆Se[x] =~βn

n∑j=1

m

2~2

(xj − xj−1)2

(β/n)2+

1

2V ′′ x2

j , (7.93)

and xn = 0 = x0.

Example 7.7 (Density operator for the harmonic oscillator) The path

qce(β) that satisfies the classical euclidian equation of motion (7.88)

qce(u) =d2qce(u)

du2= ω2qce(u) (7.94)

and goes from qa, 0 to qb, ~β is

qce(u) =sinh(ωu) qb + sinh[ω(~β − u)] qa

sinh(~ωβ). (7.95)

Its euclidian action is (exercise ??)

Se[qce] =

∫ ~β

0

mq2ce(u)

2+mω2q2

ce(u))

2du

=mω

2~ sinh(~ωβ)

[cosh(~ωβ)(q2

a + q2b )− 2qaqb

].

(7.96)

Since V ′′ = mω2, our formulas (7.92 & 7.93) for quadratic actions give as

the matrix element

〈qb|e−βH |qa〉 = f(β, ~,m,mω2) e−Se[qce]/~ (7.97)

182 Path integrals

in which

f(β, ~,m,mω2) =

[mn

2π~2β

]n/2∫e−∆Se[x]/~) dxn−1 . . . dx1,

∆Se[x] =~βn

n∑j=1

m

2~2

(xj − xj−1)2

(β/n)2+mω2 x2

j

2,

(7.98)

and xn = 0 = x0. We can do this integral by using the formula (7.4) for a

many variable real gaussian integral

f =[ mn

2π~2B

]n/2∫e−ajkxjxkdxn−1 . . . dx1 =

[ mn

2π~2B

]n/2√(π)n−1

det a(7.99)

in which the positive quadratic form ajkxjxk is

nm

2~2B

n∑j=1

[− 2xjxj−1 + x2

j + x2j−1 +

(~ωB)2

n2x2j

](7.100)

which has no linear term because x0 = xn = 0.

The matrix a is (nm/(2~2B))Cn−1(y) in which Cn−1(y) is a square, tridi-

agonal, (n− 1)-dimensional matrix whose off-diagonal elements are −1 and

whose diagonal elements are y = 2 + (~ωβ)2/n2. The determinants |Cn(y)|obey the recursion relation |Cn+1(y)| = y |Cn(y)| − |Cn−1(y)| and have the

initial values C1(y) = y and C2(y) = y2 − 1. So do the hyperbolic functions

sinh(n+1)θ/ sinh θ with y = 2 cosh θ. So we set Cn(y) = sinh(n+1)θ/ sinh θ

with θ = arccosh(y/2). We then get as the matrix element (7.97)


√mω

2π~ sinh(~ωβ)exp

[−mω[cosh(~ωβ)(q2

a + q2b )− 2qaqb]

2~ sinh(~ωβ)

].

(7.101)

The partition function is the integral over qa of this matrix element for

qb = qa

Z(β) =

√mω

2π~ sinh(~ωβ)

∫exp

[− mω[cosh(~ωβ)− 1]q2

a

~ sinh(~ωβ)

]dqa

=1√

2[cosh(~ωβ)− 1]. (7.102)

The matrix elements of the maximum-entropy density operator (7.74) are

〈qb|ρ|qa〉 =〈qb|e−βH |qa〉

Z(β)(7.103)

=

√mω[cosh(~ωβ)− 1]

π~ sinh(~ωβ)exp

[−mω[cosh(~ωβ)(q2

a + q2b )− 2qaqb]

2~ sinh(~ωβ)

]

7.7 Mean values of time-ordered products 183

which reveals the ground-state wave functions

limβ→∞

〈qb|ρ|qa〉 = 〈qb|0〉〈0|qa〉 =

√mω

π~e−mω(q2

a+q2b )/(2~). (7.104)

The partition function gives us the ground-state energy

limβ→∞

Z(β) = limβ→∞

1√2[cosh(~ωβ)− 1]

= e−βE0 = e−β~ω/2. (7.105)

7.7 Mean values of time-ordered products

In the Heisenberg picture, the position operator at time t is

q(t) = eitH/~q e−itH/~ (7.106)

in which q = q(0) is the position operator at time t = 0 or equivalently

the position operator in the Schrodinger picture. The position operator q at

the imaginary time t = −iu = −i~β = −i~/(kT ) is the euclidian position

operator

qe(u) = qe(~β) = euH/~q e−uH/~. (7.107)

The time-ordered product of two position operators is

T [q(t1)q(t2)] =

q(t1) q(t2) if t1 ≥ t2q(t2) q(t1) if t2 ≥ t1

= q(t>) q(t<) (7.108)

in which t> is the later and t< the earlier of the two times t1 and t2. Similarly,

the time-ordered product of two euclidian position operators at euclidian

times u1 = ~β1 and u2 = ~β2 is

T [qe(u1)qe(u2)] =

qe(u1) qe(u2) if u1 ≥ u2

qe(u2) qe(u1) if u2 ≥ u1

= qe(u>) qe(u<). (7.109)

The matrix element of the time-ordered product (7.108) of two position

operators and two exponentials e−itH/~ between states |a〉 and |b〉 is

〈b|e−itH/~T [q(t1)q(t2)]e−itH/~|a〉 = 〈b|e−itH/~q(t>)q(t<)e−itH/~|a〉 (7.110)

= 〈b|e−i(t−t> )H/~q e−i(t>−t< )H/~q e−i(t+t< )H/~|a〉.

184 Path integrals

We use the path-integral formula (7.21) for each of the exponentials on the

right-hand side of this equation and find (exercise ??)

〈b|e−itH/~T [q(t1)q(t2)]e−itH/~|a〉 =

∫〈b|qb〉q(t1)q(t2)eiS[q]/~〈qa|a〉Dq

(7.111)

in which the integral is over all paths that run from −t to t. This equation

simplifies if the states |a〉 and |b〉 are eigenstates of H with eigenvalues Emand En

e−it(En+Em)/~〈n|T [q(t1)q(t2)]|m〉 =

∫〈n|qb〉q(t1)q(t2)eiS[q]/~〈qa|m〉Dq.

(7.112)

By setting n = m and omitting the time-ordered product, we get

e−2itEn/~ =

∫〈n|qb〉eiS[q]/~〈qa|n〉Dq. (7.113)

The ratio of (7.112) with n = m to (7.113) is

〈n|T [q(t1)q(t2)]|n〉 =

∫〈n|qb〉q(t1)q(t2)eiS[q]/~〈qa|n〉Dq∫

〈n|qb〉eiS[q]/~〈qa|n〉Dq(7.114)

in which the integrations are over all paths that go from −t ≤ t< to t ≥ t> .

The mean value of the time-ordered product of k position operators is

〈n|T [q(t1) · · · q(tk)]|n〉 =

∫〈n|qb〉q(t1) · · · q(tk)eiS[q]/~〈qa|n〉Dq∫

〈n|qb〉eiS[q]/~〈qa|n〉Dq(7.115)

in which the integrations are over all paths that go from some time before

t1, . . . , tk to some time them.

We may perform the same operations on the euclidian position operators

by replacing t by −iu = −i~β. A matrix element of the euclidian time-

ordered product (7.109) between two states is

〈b|e−uH/~T [qe(u1)qe(u2)]e−uH/~|a〉 = 〈b|e−uH/~qe(u>)qe(u<)e−uH/~|a〉(7.116)

= 〈b|e−(u−u> )H/~q e−(u>−u< )H/~q e−(u+u< )H/~|a〉.

As u→∞, the exponential e−uH/~ projects (7.73) states in onto the ground

state |0〉 which is an eigenstate of H with energy E0. So we replace the


arbitrary states in (7.116) with the ground state and use the path-integral

formula (7.71 for the last three exponentials of (7.116)

e−2uE0/~〈0|T [qe(u1)qe(u2)]|0〉 =

∫〈0|qb〉q(u1)q(u2)e−Se[q]/~〈qa|0〉Dq.

(7.117)

The same equation without the time-ordered product is

e−2uE0/~〈0|0〉 = e−2uE0/~ =

∫〈0|qb〉e−Se[q]/~〈qa|0〉Dq. (7.118)

The ratio of the last two equations is

〈0|T [qe(u1)qe(u2)]|0〉 =

∫〈0|qb〉q(u1)q(u2)e−Se[q]/~〈qa|0〉Dq∫

〈0|qb〉e−Se[q]/~〈qa|0〉Dq(7.119)

in which the integration is over all paths from u = −∞ to u =∞. The mean

value in the ground state of the time-ordered product of k euclidian position

operators is

〈0|T [qe(u1) · · · qe(uk)]|0〉 =

∫〈0|qb〉 q(u1) · · · q(uk) e−Se[q]/~〈qa|0〉Dq∫

〈0|qb〉e−Se[q]/~〈qa|0〉Dq.

(7.120)

7.8 Quantum field theory

Quantum mechanics imposes upon n coordinates qi and conjugate momenta

pk the equal-time commutation relations

[qi, pk] = i ~ δi,k and [qi, qk] = [pi, pk] = 0. (7.121)

In the theory of a single spinless quantum field, a coordinate qx ≡ φ(x)

and a conjugate momentum px ≡ π(x) are associated with each point x of

space. The operators φ(x) and π(x) obey the commutation relations

[φ(x), π(x′)] = i ~ δ(x− x′)[φ(x), φ(x′)] = [π(x), π(x′)] = 0

(7.122)

inherited from quantum mechanics.

To make path integrals, we will replace space by a 3-dimensional lattice of

186 Path integrals

points x = a(i, j, k) = (ai, aj, ak) and eventually let the distance a between

adjacent points go to zero. On this lattice and at equal times t = 0, the field

operators obey discrete forms of the commutation relations (7.122)

[φ(a(i, j, k)), π(a(`,m, n))] = i~a3δi,` δj,m δk,n

[φ(a(i, j, k)), φ(a(`,m, n))] = [π(a(i, j, k)), π(a(`,m, n))] = 0.(7.123)

The vanishing commutators imply that the field and the momenta have

“simultaneous” eigenvalues

φ(a(i, j, k))|φ′〉 = φ′(a(i, j, k))|φ′〉 and π(a(i, j, k))|π′〉 = π′(a(i, j, k))|π′〉(7.124)

for all lattice points a(i, j, k). Their inner products are

〈φ′|π′〉 =∏i,j,k

√a3

2π~eia

3φ′(a(i,j,k))π′(a(i,j,k))/~. (7.125)

These states are complete∫|φ′〉〈φ′|

∏i,j,k

dφ′(a(i, j, k)) = I =

∫|π′〉〈π′|

∏i,j,k

dπ′(a(i, j, k)) (7.126)

and orthonormal

〈φ′|φ′′〉 =∏i,j,k

δ(φ′(a(i, j, k))− φ′′(a(i, j, k))) (7.127)

with a similar equation for 〈π′|π′′〉.The hamiltonian for a free field of mass m is

H = 12

∫π2 + c2(∇φ)2 +

m2c4

~2φ2 d3x =

a3

2

∑v

π2v + c2(∇φv)2 +

m2c4

~2φ2v

(7.128)

where v = a(i, j, k), πv = π(a(i, j, k)), φv = φ(a(i, j, k)), and the square of

the lattice gradient (∇φv)2 is[(φ(a(i+ 1, j, k))− φ(a(i, j, k)))2 + (φ(a(i, j + 1, k))− φ(a(i, j, k)))2

+ (φ(a(i, j, k + 1))− φ(a(i, j, k)))2]/a2. (7.129)

Other fields or terms, such as c3φ4/~, can be added to this hamiltonian.

To simplify the appearance of the equations in the rest of this chapter, I

will mostly use natural units in which ~ = c = 1. To convert the value of

a physical quantity from natural units to universal units, one multiplies or

divides its natural-unit value by suitable factors of ~ and c until one gets

the right dimensions. For instance, if V = 1/m is the value of a time in


natural units, where m is a mass, then the time you want is T = ~/(mc2). If

V = 1/m is supposed to be a length, then the needed length is L = ~/(mc).We set K = a3

∑v π

2v/2 and V = (a3/2)

∑v(∇φv)2 + m2φ2

v + P (φv) in

which P (φv) represents the self-interactions of the field. With ε = (tb−ta)/n,

Trotter’s product formula (7.6) is the n→∞ limit of

e−i(tb−ta)(K+V ) =(e−i(tb−ta)K/ne−i(tb−ta)V/n

)n=(e−iεKe−iεV

)n. (7.130)

We insert I in the form (7.126) between e−iεK and e−iεV

〈φ1|e−iεK e−iεV |φa〉 = 〈φ1|e−iεK∫|π′〉〈π′|

∏v

dπ′ve−iεV |φa〉 (7.131)

and use the eigenstate formula (7.124)

〈φ1|e−iεK e−iεV |φa〉 = e−iεV (φa)

∫e−iεK(π′)〈φ1|π′〉〈π′|φa〉

∏v

dπ′v. (7.132)

The inner product formula (7.125) now gives

〈φ1|e−iεK e−iεV |φa〉 = e−iεV (φa)∏v

[∫a3dπ′v

2πea

3[−iεπ2v/2+i(φ1v−φav)π′v ]

].

(7.133)

We again adopt the suggestive notation φa = (φ1 − φa)/ε and use the gaus-

sian integral (7.1) to find

〈φ1|e−iεK e−iεV |φa〉 =∏v

[(a3

2πiε

)1/2

eiεa3[φ2

av−(∇φav)2−m2φ2av−P (φv)]/2

].

(7.134)

The product of n = (tb − ta)/ε such time intervals is

〈φb|e−i(tb−ta)H |φa〉 =∏v

[(a3n

2πi(tb − ta)

)n/2 ∫eiSvDφv

](7.135)

in which

Sv =tb − tan

a3

2

n−1∑j=0

[φ2jv − (∇φjv)2 −m2φ2

jv − P (φv)]

(7.136)

φjv = n(φj+1,v − φj,v)/(tb − ta), and Dφv = dφn−1,v · · · dφ1,v.

The amplitude 〈φb|e−i(tb−ta)H |φa〉 is the integral over all fields that go

from φa(x) at ta to φb(x) at tb each weighted by an exponential

〈φb|e−i(tb−ta)H |φa〉 =

∫eiS[φ]Dφ (7.137)

188 Path integrals

of its action

S[φ] =

∫ tb

ta

dt

∫d3x

1

2

[φ2 − (∇φ)2 −m2φ2 − P (φ)

](7.138)

in which Dφ is the n→∞ limit of the product over all spatial vertices v

Dφ =∏v

[(a3n

2πi(tb − ta)

)n/2dφn−1,v · · · dφ1,v

]. (7.139)

Equivalently, the time-evolution operator is

e−i(tb−ta)H =

∫|φb〉eiS[φ] 〈φa|DφDφaDφb (7.140)

in which DφaDφb =∏v dφa,vdφb,v is an integral over the initial and final

states.

As in quantum mechanics (section 7.4), the path integral for an action

that is quadratic in the fields is an exponential of the action of a stationary

process times a function of the times and of the other parameters in the

action

〈φb|e−i(tb−ta)H |φa〉 =

∫eiS[φ]Dφ = f(ta, tb, . . . ) e

iS[φc] (7.141)

in which S[φc] is the action of the process that goes from φ(x, ta) = φa(x)

to φ(x, tb) = φb(x) and obeys the classical equations of motion, and the

function f is a path integral over all fields that go from φ(x, ta) = 0 to

φ(x, tb) = 0.

Example 7.8 (A stationary process) The field

φ(x, t) =

∫eik·x[a(k) cosωt + b(k) sinωt] d3k (7.142)

with ω =√k2 +m2 makes the action (7.138) for P = 0 stationary because

it is a solution of the equation of motion ∇2φ − φ −m2φ = 0. In terms of

the Fourier transforms

φ(k, ta) =

∫e−ik·x φ(x, ta)

d3x

(2π)3and φ(k, tb) =

∫e−ik·x φ(x, tb)

d3x

(2π)3,

(7.143)

the solution that goes from φ(x, ta) to φ(x, tb) is

φ(x, t) =

∫eik·x

sinω(tb − t) φ(k, ta) + sinω(t− ta) φ(k, tb)

sinω(tb − ta)d3k. (7.144)


The solution that evolves from φ(x, ta) and φ(x, ta) is

φ(x, t) =

∫eik·x

[cosω(t− ta) φ(k, ta) +

sinω(t− ta)ω

˜φ(k, ta)

]d3k

(7.145)

in which the Fourier transform˜φ(k, ta) is defined as in (7.143).

Like a position operator (7.106), a field at time t is defined as

φ(x, t) = eitH/~φ(x)e−itH/~ (7.146)

in which φ(x) = φ(x, 0) is the field at time zero, which obeys the commu-

tation relations (7.122). The time-ordered product of several fields is their

product with newer (later time) fields standing to the left of older (ear-

lier time) fields as in the definition (7.108). The logic (7.110–7.115) of the

derivation of the path-formulas for time-ordered products of position opera-

tors applies directly to field operators. One finds (exercise ??) for the mean

value of the time-ordered product of two fields in an energy eigenstate |n〉

〈n|T [φ(x1)φ(x2)]|n〉 =

∫〈n|φb〉φ(x1)φ(x2)eiS[φ]/~〈φa|n〉Dφ∫

〈n|φb〉eiS[φ]/~〈φa|n〉Dφ(7.147)

in which the integrations are over all paths that go from before t1 and t2 to

after both times. The analogous result for several fields is (exercise ??)

〈n|T [φ(x1) · · ·φ(xk)]|n〉 =

∫〈n|φb〉φ(x1) · · ·φ(xk)e

iS[φ]/~〈φa|n〉Dφ∫〈n|φb〉eiS[φ]/~〈φa|n〉Dφ

(7.148)

in which the integrations are over all paths that go from before the times

t1, . . . , tk to after them.

7.9 Finite-temperature field theory

Since the Boltzmann operator e−βH = e−H/(kT ) is the time evolution

operator e−itH/~ at the imaginary time t = − i~β = −i~/(kT ), the formulas

of finite-temperature field theory are those of quantum field theory with t

replaced by −iu = −i~β = −i~/(kT ).

190 Path integrals

If as in section 7.8, we use as our hamiltonian H = K + V where K and

V are sums over all lattice vertices v = a(i, j, k) = (ai, aj, ak) of the cubes

of volume a3 times the squared momentum and potential terms

H = K + V =a3

2

∑v

π2v +

a3

2

∑v

(∇φv)2 +m2φ2v + P (φv). (7.149)

A matrix element of the first term of the Trotter product formula (7.7)



)n(7.150)

is the imaginary-time version of (7.133) with ε = ~β/n

〈φ1|e−εK e−εV |φa〉 = e−εV (φa)∏v

[∫a3dπ′v

2πea

3[−επ2v/2+i(φ1v−φav)π′v ]

].

(7.151)

Setting φav = (φ1v − φav)/ε, we find, instead of (7.134)

〈φ1|e−εK e−εV |φa〉 =∏v

[(a3

2πε

)1/2

e−εa3[φ2

av+(∇φav)2+m2φ2av+P (φv)]/2

].

(7.152)

The product of n = ~β/ε such inverse-temperature intervals is

〈φb|e−βH |φa〉 =∏v

[(a3n

2πβ

)n/2 ∫e−SevDφv

](7.153)

in which the euclidian action is

Sev =β

n

a3

2

n−1∑j=0

[φ2jv + (∇φjv)2 +m2φ2

jv + P (φv)]

(7.154)

φjv = n(φj+1,v − φj,v)/β, and Dφv = dφn−1,v · · · dφ1,v.

The amplitude 〈φb|e−(βb−βa)H |φa〉 is the integral over all fields that go

from φa(x) at βa to φb(x) at βb each weighted by an exponential

〈φb|e−(βb−βa)H |φa〉 =

∫e−Se[φ]Dφ (7.155)

of its euclidian action

Se[φ] =

∫ βb

βa

du

∫d3x

1

2

[φ2 + (∇φ)2 +m2φ2 + P (φ)

](7.156)



Dφ =∏v

[(a3n

2π(βb − βa)

)n/2dφn−1,v · · · dφ1,v

]. (7.157)

Equivalently, the Boltzmann operator is

e−(βb−βa)H =

∫|φb〉e−Se[φ] 〈φa|DφDφaDφb (7.158)


states.

The trace of the Boltzmann operator is the partition function

Z(β) = Tr(e−βH) =

∫e−Se[φ] 〈φa|φb〉DφDφaDφb =

∫e−Se[φ] DφDφa

(7.159)

which is an integral over all fields that go back to themselves in euclidian

time β.

Like a position operator (7.107), a field at an imaginary time t = −iu =

−i~β is defined as

φe(x, u) = φe(x, ~β) = euH/~φ(x) e−uH/~. (7.160)

in which φ(x) = φ(x, 0) = φe(x, 0) is the field at time zero, which obeys

the commutation relations (7.122). The euclidian-time-ordered product of

several fields is their product with newer (higher u = ~β) fields standing to

the left of older (lower u = ~β) fields as in the definition (7.109).

The euclidian path integrals for the mean values of euclidian-time-ordered-

products of fields are similar to those (7.161 & 7.148) for ordinary time-

ordered-products. The euclidian-time-ordered-product of the fields φ(xj) =

φ(xj , uj) is the path integral

〈n|T [φe(x1)φe(x2)]|n〉 =

∫〈n|φb〉φ(x1)φ(x2)e−Se[φ]/~〈φa|n〉Dφ∫

〈n|φb〉e−Se[φ]/~〈φa|n〉Dφ(7.161)

in which the integrations are over all paths that go from before u1 and u2

to after both euclidian times. The analogous result for several fields is

〈n|T [φe(x1) · · ·φe(xk)]|n〉 =

∫〈n|φb〉φ(x1) · · ·φ(xk)e

−Se[φ]/~〈φa|n〉Dφ∫〈n|φb〉e−Se[φ]/~〈φa|n〉Dφ

(7.162)

192 Path integrals


u1, . . . , uk to after them.

A distinctive feature of these formulas is that in the low-temperature

β = 1/(kT ) → ∞ limit, the Boltzmann operator is a multiple of an outer

product |0〉〈0| of the ground-state kets, e−βH → e−βE0 |0〉〈0|. In this limit,

the integrations are over all fields that run from u = −∞ to u =∞ and the

energy eigenstates are the ground state of the theory

〈0|T [φe(x1) · · ·φe(xk)]|0〉 =

∫〈0|φb〉φ(x1) · · ·φ(xk)e

−Se[q]/~〈φa|0〉Dφ∫〈0|φb〉e−Se[q]/~〈φa|0〉Dφ

.

(7.163)

Formulas like this one are used in lattice gauge theory.

7.10 Perturbation theory

Field theories with hamiltonians that are quadratic in their fields like

H0 =

∫12

[π2(x) + (∇φ(x))2 +m2φ2(x)

]d3x (7.164)

are soluble. Their fields evolve in time as

φ(x, t) = eitH0φ(x, 0)e−itH0 . (7.165)

The mean value in the ground state of H0 of a time-ordered product of these

fields is a ratio (7.148) of path integrals

〈0|T [φ(x1) . . . φ(xk)] |0〉 =

∫〈0|φb〉φ(x1) . . . φ(xn) eiS0[φ]〈φa|0〉Dφ∫

〈0|φb〉 eiS0[φ]〈φa|0〉Dφ(7.166)

in which the action S0[φ] is quadratic in the field φ

S0[φ] =

∫12

[φ2(x)− (∇φ(x))2 −m2φ2(x)

]d4x

=

∫12

[− ∂aφ(x)∂aφ(x)−m2φ2(x)

]d4x

(7.167)

and the integrations are over all fields that run from φa at a time before

the times t1, . . . , tk to φb at a time after t1, . . . , tk. The path integrals in the

ratio (7.166) are gaussian and doable.


The Fourier transforms

φ(p) =

∫e−ipxφ(x) d4x and φ(x) =

∫eipxφ(p)

d4p

(2π)4(7.168)

turn the spacetime derivatives in the action into a quadratic form

S0[φ] = −12

∫|φ(p)|2 (p2 +m2)

d4p

(2π)4(7.169)

in which p2 = p2 − p02 and φ(−p) = φ∗(p) by (??) since the field φ is real.

The initial 〈φa|0〉 and final 〈0|φb〉 wave functions produce the iε in the

Feynman propagator (2.24). Although its exact form doesn’t matter here,

the wave function 〈φ|0〉 of the ground state of H0 is the exponential (??)

〈φ|0〉 = c exp

[−1

2

∫|φ(p)|2

√p2 +m2

d3p

(2π)3

](7.170)

in which φ(p) is the spatial Fourier transform of the eigenvalue φ(x)

φ(p) =

∫e−ip·x φ(x) d3x (7.171)

and c is a normalization factor that will cancel in ratios of path integrals.

Apart from −2i ln c which we will not keep track of, the wave functions

〈φa|0〉 and 〈0|φb〉 add to the action S0[φ] the term

∆S0[φ] =i

2

∫ √p2 +m2

(|φ(p, t)|2 + |φ(p,−t)|2

) d3p

(2π)3(7.172)

in which we envision taking the limit t → ∞ with φ(x, t) = φb(x) and

φ(x,−t) = φa(x). The identity (Weinberg, 1995, pp. 386–388)

f(+∞) + f(−∞) = limε→0+

ε

∫ ∞−∞

f(t) e−ε|t| dt (7.173)

(exercise ??) allows us to write ∆S0[φ] as

∆S0[φ] = limε→0+

iε

2

∫ √p2 +m2

∫ ∞−∞|φ(p, t)|2 e−ε|t| dt d3p

(2π)3. (7.174)

To first order in ε, the change in the action is (exercise ??)

∆S0[φ] = limε→0+

iε

2

∫ √p2 +m2

∫ ∞−∞|φ(p, t)|2 dt d3p

(2π)3

= limε→0+

iε

2

∫ √p2 +m2 |φ(p)|2 d4p

(2π)4. (7.175)

194 Path integrals

Thus the modified action is

S0[φ, ε] = S0[φ] + ∆S0[φ] = − 1

2

∫|φ(p)|2

(p2 +m2 − iε

√p2 +m2

) d4p

(2π)4

= − 1

2

∫|φ(p)|2

(p2 +m2 − iε

) d4p

(2π)4(7.176)

since the square root is positive. In terms of the modified action, our formula

(7.166) for the time-ordered product is the ratio

〈0|T [φ(x1) . . . φ(xn)] |0〉 =

∫φ(x1) . . . φ(xn) eiS0[φ,ε]Dφ∫

eiS0[φ,ε]Dφ

. (7.177)

We can use this formula (7.177) to express the mean value in the vacuum

|0〉 of the time-ordered exponential of a spacetime integral of j(x)φ(x), in

which j(x) is a classical (c-number, external) current, as the ratio

Z0[j] ≡ 〈0| T

exp

[i

∫j(x)φ(x) d4x

]|0〉

=

∫exp

[i

∫j(x)φ(x) d4x

]eiS0[φ,ε]Dφ∫

eiS0[φ,ε]Dφ

.

(7.178)

Since the state |0〉 is normalized, the mean value Z0[0] is unity,

Z0[0] = 1. (7.179)

If we absorb the current into the action

S0[φ, ε, j] = S0[φ, ε] +

∫j(x)φ(x) d4x (7.180)

then in terms of the current’s Fourier transform

j(p) =

∫e−ipx j(x) d4x (7.181)

the modified action S0[φ, ε, j] is (exercise ??)

S0[φ, ε, j] = − 12

∫ [|φ(p)|2

(p2 +m2 − iε

)− j∗(p)φ(p)− φ∗(p)j(p)

] d4p

(2π)4.

(7.182)

Changing variables to

ψ(p) = φ(p)− j(p)/(p2 +m2 − iε) (7.183)


we write the action S0[φ, ε, j] as (exercise ??)

S0[φ, ε, j] = − 12

∫ [|ψ(p)|2

(p2 +m2 − iε

)− j∗(p)j(p)

(p2 +m2 − iε)

]d4p

(2π)4

= S0[ψ, ε] + 12

∫ [j∗(p)j(p)

(p2 +m2 − iε)

]d4p

(2π)4. (7.184)

And since Dφ = Dψ, our formula (7.178) gives simply (exercise ??)

Z0[j] = exp

(i

2

∫|j(p)|2

p2 +m2 − iεd4p

(2π)4

). (7.185)

Going back to position space, one finds (exercise ??)

Z0[j] = exp

[i

2

∫j(x) ∆(x− x′) j(x′) d4x d4x′

](7.186)

in which ∆(x− x′) is Feynman’s propagator (2.24)

∆(x− x′) =

∫eip(x−x

′)

p2 +m2 − iεd4p

(2π)4. (7.187)

The functional derivative (chapter ??) of Z0[j], defined by (7.178), is

1

i

δZ0[j]

δj(x)= 〈0| T

[φ(x) exp

(i

∫j(x′)φ(x′)d4x′

)]|0〉 (7.188)

while that of equation (7.186) is

1

i

δZ0[j]

δj(x)= Z0[j]

∫∆(x− x′) j(x′) d4x′. (7.189)

Thus the second functional derivative of Z0[j] evaluated at j = 0 gives

〈0| T[φ(x)φ(x′)

]|0〉 =

1

i2δ2Z0[j]

δj(x)δj(x′)

∣∣∣∣j=0

= −i∆(x− x′). (7.190)

Similarly, one may show (exercise ??) that

〈0| T [φ(x1)φ(x2)φ(x3)φ(x4)] |0〉 =1

i4δ4Z0[j]

δj(x1)δj(x2)δj(x3)δj(x4)

∣∣∣∣j=0

= −∆(x1 − x2)∆(x3 − x4)−∆(x1 − x3)∆(x2 − x4)

−∆(x1 − x4)∆(x2 − x3). (7.191)

Suppose now that we add a potential V = P (φ) to the free hamiltonian

(7.164). Scattering amplitudes are matrix elements of the time-ordered expo-

nential T exp[−i∫P (φ) d4x

](Weinberg, 1995, p. 260) Our formula (7.177)

196 Path integrals

for the mean value in the ground state |0〉 of the free hamiltonian H0 of any

time-ordered product of fields leads us to

〈0|T

exp

[−i∫P (φ) d4x

]|0〉 =

∫exp

[−i∫P (φ) d4x

]eiS0[φ,ε]Dφ∫

eiS0[φ,ε]Dφ

.

(7.192)

Using (7.190 & 7.191), we can cast this expression into the magical form

〈0|T

exp

[−i∫P (φ) d4x

]|0〉 = exp

[−i∫P

(δ

iδj(x)

)d4x

]Z0[j]

∣∣∣∣j=0

.

(7.193)

The generalization of the path-integral formula (7.177) to the ground state

|Ω〉 of an interacting theory with action S is

〈Ω|T [φ(x1) . . . φ(xn)] |Ω〉 =

∫φ(x1) . . . φ(xn) eiS[φ,ε]Dφ∫

eiS[φ,ε]Dφ

(7.194)

in which a term like iεφ2 is added to make the modified action S[φ, ε].

These are some of the techniques one uses to make states of incoming and

outgoing particles and to compute scattering amplitudes (Weinberg, 1995,

1996; Srednicki, 2007; Zee, 2010).

7.11 Application to quantum electrodynamics

In the Coulomb gauge ∇ ·A = 0, the QED hamiltonian is

H = Hm +

∫ [12π

2 + 12(∇×A)2 −A · j

]d3x+ VC (7.195)

in which Hm is the matter hamiltonian, and VC is the Coulomb term

VC =1

2

∫j0(x, t) j0(y, t)

4π|x− y| d3x d3y. (7.196)

The operators A and π are canonically conjugate, but they satisfy the

Coulomb-gauge conditions

∇ ·A = 0 and ∇ · π = 0. (7.197)

One may show (Weinberg, 1995, pp. 413–418) that in this theory, the


analog of equation (7.194) is

〈Ω|T [O1 . . .On] |Ω〉 =

∫O1 . . .On eiSC δ[∇ ·A]DADψ∫

eiSC δ[∇ ·A]DADψ

(7.198)

in which the Coulomb-gauge action is

SC =

∫12A

2 − 12(∇×A)2 +A · j + Lm d4x −

∫VC dt (7.199)

and the functional delta function

δ[∇ ·A] =∏x

δ(∇ ·A(x)) (7.200)

enforces the Coulomb-gauge condition. The term Lm is the action density

of the matter field ψ.

Tricks are available. We introduce a new field A0(x) and consider the

factor

F =

∫exp

[i

∫1

2

(∇A0+∇4−1j0

)2d4x

]DA0 (7.201)

which is just a number independent of the charge density j0 since we can

cancel the j0 term by shifting A0. By 4−1, we mean − 1/4π|x − y|. By

integrating by parts, we can write the number F as (exercise ??)

F =

∫exp

[i

∫12

(∇A0

)2 −A0j0 − 12 j

04−1j0 d4x

]DA0

=

∫exp

[i

∫12

(∇A0

)2 −A0j0 d4x+ i

∫VC dt

]DA0.

(7.202)

So when we multiply the numerator and denominator of the amplitude

(7.198) by F , the awkward Coulomb term VC cancels, and we get

〈Ω|T [O1 . . .On] |Ω〉 =

∫O1 . . .On eiS

′δ[∇ ·A]DADψ∫

eiS′δ[∇ ·A]DADψ

(7.203)

where now DA includes all four components Aµ and

S′ =

∫12 A

2 − 12 (∇×A)2 + 1

2

(∇A0

)2+A · j −A0j0 + Lm d4x. (7.204)

Since the delta-functional δ[∇ ·A] enforces the Coulomb-gauge condition,

we can add to the action S′ the term (∇ · A)A0 which is − A · ∇A0 after

198 Path integrals

we integrate by parts and drop the surface term. This extra term makes the

action gauge invariant

S =

∫12 (A−∇A0)2 − 1

2 (∇×A)2 +A · j −A0j0 + Lm d4x

=

∫− 1

4 Fab Fab+Abjb + Lm d4x.

(7.205)

Thus at this point we have

〈Ω|T [O1 . . .On] |Ω〉 =

∫O1 . . .On eiS δ[∇ ·A]DADψ∫

eiS δ[∇ ·A]DADψ

(7.206)

in which S is the gauge-invariant action (7.205), and the integral is over

all fields. The only relic of the Coulomb gauge is the gauge-fixing delta

functional δ[∇ ·A].

We now make the gauge transformation

A′b(x) = Ab(x) + ∂bΛ(x) and ψ′(x) = eiqΛ(x)ψ(x) (7.207)

in the numerator and also, using a different gauge transformation Λ′, in the

denominator of the ratio (7.206) of path integrals. Since we are integrating

over all gauge fields, these gauge transformations merely change the order

of integration in the numerator and denominator of that ratio. They are like

replacing∫∞−∞ f(x) dx by

∫∞−∞ f(y) dy. They change nothing, and so

〈Ω|T [O1 . . .On] |Ω〉 = 〈Ω|T [O1 . . .On] |Ω〉′ (7.208)

in which the prime refers to the gauge transformations (7.207) Λ and Λ′.

We’ve seen that the action S is gauge invariant. So is the measure DADψ.

We now restrict ourselves to operators O1 . . .On that are gauge invariant .

So in the right-hand side of equation (7.208), the replacement of the fields

by their gauge transforms affects only the term δ[∇ ·A] that enforces the

Coulomb-gauge condition

〈Ω|T [O1 . . .On] |Ω〉 =

∫O1 . . .On eiS δ[∇ ·A+4Λ]DADψ∫

eiS δ[∇ ·A+4Λ′]DADψ

. (7.209)

We now have two choices. If we integrate over all gauge functions Λ(x) and

Λ′(x) in both the numerator and the denominator of this ratio (7.209), then

apart from over-all constants that cancel, the mean value in the vacuum of


the time-ordered product is the ratio

〈Ω|T [O1 . . .On] |Ω〉 =

∫O1 . . .On eiS DADψ∫

eiS DADψ

(7.210)

in which we integrate over all matter fields, gauge fields, and gauges. That

is, we do not fix the gauge.

The analogous formula for the euclidian time-ordered product is

〈Ω|T [Oe,1 . . .Oe,n] |Ω〉 =

∫O1 . . .On e−Se DADψ∫

e−Se DADψ

(7.211)

in which the euclidian action Se is the spacetime integral of the energy

density. This formula is quite general; it holds in nonabelian gauge theories

and is important in lattice gauge theory.

Our second choice is to multiply the numerator and the denominator of the

ratio (7.209) by the exponential exp[−i12α∫

(4Λ)2 d4x] and then integrate

over Λ(x) in the numerator and over Λ′(x) in the denominator. This oper-

ation just multiplies the numerator and denominator by the same constant

factor, which cancels. But if before integrating over all gauge transforma-

tions, we shift Λ so that4Λ changes to4Λ−A0, then the exponential factor

is exp[−i12α∫

(A0 −4Λ)2 d4x]. Now when we integrate over Λ(x), the delta

function δ(∇ ·A+4Λ) replaces 4Λ by −∇ ·A in the inserted exponen-

tial, converting it to exp[−i12α∫

(A0 +∇ ·A)2 d4x]. This term changes the

gauge-invariant action (7.205) to the gauge-fixed action

Sα =

∫− 1

4Fab F

ab − α

2(∂bA

b)2+Abjb + Lm d4x. (7.212)

This Lorentz-invariant, gauge-fixed action is much easier to use than the

Coulomb-gauge action (7.199) with the Coulomb potential (7.196). We can

use it to compute scattering amplitudes perturbatively. The mean value of a

time-ordered product of operators in the ground state |0〉 of the free theory

is

〈0|T [O1 . . .On] |0〉 =

∫O1 . . .On eiSα DADψ∫

eiSα DADψ

. (7.213)

By following steps analogous to those the led to (7.187), one may show

200 Path integrals

(exercise ??) that in Feynman’s gauge, α = 1, the photon propagator is

〈0|T [Aµ(x)Aν(y)] |0〉 = − i4µν(x− y) = − i∫

ηµνq2 − iε

eiq·(x−y) d4q

(2π)4.

(7.214)

7.12 Fermionic path integrals

In our brief introduction (??–??) and (??–??), to Grassmann variables, we

learned that because

θ2 = 0 (7.215)

the most general function f(θ) of a single Grassmann variable θ is

f(θ) = a+ b θ. (7.216)

So a complete integral table consists of the integral of this linear function∫f(θ) dθ =

∫a+ b θ dθ = a

∫dθ + b

∫θ dθ. (7.217)

This equation has two unknowns, the integral∫dθ of unity and the integral∫

θ dθ of θ. We choose them so that the integral of f(θ + ζ)∫f(θ + ζ) dθ =

∫a+ b (θ + ζ) dθ = (a+ b ζ)

∫dθ + b

∫θ dθ (7.218)

is the same as the integral (7.217) of f(θ). Thus the integral∫dθ of unity

must vanish, while the integral∫θ dθ of θ can be any constant, which we

choose to be unity. Our complete table of integrals is then∫dθ = 0 and

∫θ dθ = 1. (7.219)

The anticommutation relations for a fermionic degree of freedom ψ are

ψ,ψ† ≡ ψ ψ† + ψ†ψ = 1 and ψ,ψ = ψ†, ψ† = 0. (7.220)

Because ψ has ψ†, it is conventional to introduce a variable θ∗ = θ† that

anti-commutes with itself and with θ

θ∗, θ∗ = θ∗, θ = θ, θ = 0. (7.221)

The logic that led to (7.219) now gives∫dθ∗ = 0 and

∫θ∗ dθ∗ = 1. (7.222)


We define the reference state |0〉 as |0〉 ≡ ψ|s〉 for a state |s〉 that is not

annihilated by ψ. Since ψ2 = 0, the operator ψ annihilates the state |0〉

ψ|0〉 = ψ2|s〉 = 0. (7.223)

The effect of the operator ψ on the state

|θ〉 = exp(ψ†θ − 1

2θ∗θ)|0〉 =

(1 + ψ†θ − 1

2θ∗θ)|0〉 (7.224)

is

ψ|θ〉 = ψ(1 + ψ†θ − 12θ∗θ)|0〉 = ψψ†θ|0〉 = (1− ψ†ψ)θ|0〉 = θ|0〉 (7.225)

while that of θ on |θ〉 is

θ|θ〉 = θ(1 + ψ†θ − 12θ∗θ)|0〉 = θ|0〉. (7.226)

The state |θ〉 therefore is an eigenstate of ψ with eigenvalue θ

ψ|θ〉 = θ|θ〉. (7.227)

The bra corresponding to the ket |ζ〉 is

〈ζ| = 〈0|(

1 + ζ∗ψ − 1

2ζ∗ζ

)(7.228)

and the inner product 〈ζ|θ〉 is (exercise ??)

〈ζ|θ〉 = 〈0|(

1 + ζ∗ψ − 1

2ζ∗ζ

)(1 + ψ†θ − 1

2θ∗θ

)|0〉

= 〈0|1 + ζ∗ψψ†θ − 1

2ζ∗ζ − 1

2θ∗θ +

1

4ζ∗ζθ∗θ|0〉

= 〈0|1 + ζ∗θ − 1

2ζ∗ζ − 1

2θ∗θ +

1

4ζ∗ζθ∗θ|0〉

= exp

[ζ∗θ − 1

2(ζ∗ζ + θ∗θ)

]. (7.229)

Example 7.9 (A gaussian integral) For any number c, we can compute

the integral of exp(c θ∗θ) by expanding the exponential∫ec θ∗θ dθ∗dθ =

∫(1 + c θ∗θ) dθ∗dθ =

∫(1− c θ θ∗) dθ∗dθ = −c. (7.230)

The identity operator for the space of states

c|0〉+ d|1〉 ≡ c|0〉+ dψ†|0〉 (7.231)

202 Path integrals

is (exercise ??) the integral

I =

∫|θ〉〈θ| dθ∗dθ = |0〉〈0|+ |1〉〈1| (7.232)

in which the differentials anti-commute with each other and with other

fermionic variables: dθ, dθ∗ = 0, dθ, θ = 0, dθ, ψ = 0, and so forth.

The case of several Grassmann variables θ1, θ2, . . . , θn and several Fermi

operators ψ1, ψ2, . . . , ψn is similar. The θk anticommute among themselves

θi, θj = θi, θ∗j = θ∗i , θ∗j = 0 (7.233)

while the ψk satisfy

ψk, ψ†` = δk` and ψk, ψl = ψ†k, ψ†` = 0. (7.234)

The reference state |0〉 is

|0〉 =

(n∏k=1

ψk

)|s〉 (7.235)

in which |s〉 is any state not annihilated by any ψk (so the resulting |0〉 isn’t

zero). The direct-product state

|θ〉 ≡ exp

(n∑k=1

ψ†kθk −1

2θ∗kθk

)|0〉 =

[n∏k=1

(1 + ψ†kθk −

1

2θ∗kθk

)]|0〉

(7.236)

is (exercise ??) a simultaneous eigenstate of each ψk

ψk|θ〉 = θk|θ〉. (7.237)

It follows that

ψ`ψk|θ〉 = ψ`θk|θ〉 = −θkψ`|θ〉 = −θkθ`|θ〉 = θ`θk|θ〉 (7.238)

and so too ψkψ`|θ〉 = θkθ`|θ〉. Since the ψ’s anticommute, their eigenvalues

must also

θ`θk|θ〉 = ψ`ψk|θ〉 = −ψkψ`|θ〉 = −θkθ`|θ〉 (7.239)

(even if the eigenvalues commuted with the ψ’s in which case we’d have

ψ`ψk|θ〉 = θkθ`|θ〉 = −ψkψ`|θ〉 = −θ`θk|θ〉).


The inner product 〈ζ|θ〉 is

〈ζ|θ〉 = 〈0|

[n∏k=1

(1 + ζ∗kψk −1

2ζ∗kζk)

][n∏`=1

(1 + ψ†`θ` −1

2θ∗` θ`)

]|0〉

= exp

[n∑k=1

ζ∗kθk −1

2(ζ∗kζk + θ∗kθk)

]= eζ

†θ−(ζ†ζ+θ†θ)/2. (7.240)

The identity operator is

I =

∫|θ〉〈θ|

n∏k=1

dθ∗kdθk. (7.241)

Example 7.10 (Gaussian Grassmann Integral) For any 2 × 2 matrix A,

we may compute the gaussian integral

g(A) =

∫e−θ

†Aθ dθ∗1dθ1dθ∗2dθ2 (7.242)

by expanding the exponential. The only terms that survive are the ones that

have exactly one of each of the four variables θ1, θ2, θ∗1, and θ∗2. Thus, the

integral is the determinant of the matrix A

g(A) =

∫1

2(θ∗kAk`θ`)

2 dθ∗1dθ1dθ∗2dθ2

=

∫(θ∗1A11θ1 θ

∗2A22θ2 + θ∗1A12θ2 θ

∗2A21θ1) dθ∗1dθ1dθ

∗2dθ2

= A11A22 −A12A21 = detA. (7.243)

The natural generalization to n dimensions

∫e−θ

†Aθn∏k=1

dθ∗kdθk = detA (7.244)

is true for any n × n matrix A. If A is invertible, then the invariance of

204 Path integrals

Grassmann integrals under translations implies that∫e−θ

†Aθ+θ†ζ+ζ†θn∏k=1

dθ∗kdθk =

∫e−θ

†A(θ+A−1ζ)+θ†ζ+ζ†(θ+A−1ζ)n∏k=1

dθ∗kdθk

=

∫e−θ

†Aθ+ζ†θ+ζ†A−1ζn∏k=1

dθ∗kdθk

=

∫e−(θ†+ζ†A−1)Aθ+ζ†θ+ζ†A−1ζ

n∏k=1

dθ∗kdθk

=

∫e−θ

†Aθ+ζ†A−1ζn∏k=1

dθ∗kdθk

= detA eζ†A−1ζ . (7.245)

The values of θ and θ† that make the argument −θ†Aθ+θ†ζ+ζ†θ of the expo-

nential stationary are θ = A−1ζ and θ† = ζ†A−1. So a gaussian Grassmann

integral is equal to its exponential evaluated at its stationary point, apart

from a prefactor involving the determinant detA. This result is a fermionic

echo of the bosonic result (??).

One may further extend these definitions to a Grassmann field χm(x) and

an associated Dirac field ψm(x). The χm(x)’s anticommute among them-

selves and with all fermionic variables at all points of spacetime

χm(x), χn(x′) = χ∗m(x), χn(x′) = χ∗m(x), χ∗n(x′) = 0 (7.246)

and the Dirac field ψm(x) obeys the equal-time anticommutation relations

ψm(x, t), ψ†n(x′, t) = δmn δ(x− x′) (n,m = 1, . . . , 4)

ψm(x, t), ψn(x′, t) = ψ†m(x, t), ψ†n(x′, t) = 0.(7.247)

As in (7.235), we use eigenstates of the field ψ at t = 0. If |0〉 is defined

in terms of a state |s〉 that is not annihilated by any ψm(x, 0) as

|0〉 =

[∏m,x

ψm(x, 0)

]|s〉 (7.248)

then (exercise ??) the state

|χ〉 = exp

(∫ ∑m

ψ†m(x, 0)χm(x)− 1

2χ∗m(x)χm(x) d3x

)|0〉

= exp

(∫ψ†χ− 1

2χ†χd3x

)|0〉 (7.249)


is an eigenstate of the operator ψm(x, 0) with eigenvalue χm(x)

ψm(x, 0)|χ〉 = χm(x)|χ〉. (7.250)

The inner product of two such states is (exercise ??)

〈χ′|χ〉 = exp

[∫χ′†χ− 1

2χ′†χ′ − 1

2χ†χ d3x

]. (7.251)

The identity operator is the integral

I =

∫|χ〉〈χ|Dχ∗Dχ (7.252)

in which

Dχ∗Dχ ≡∏m,x

dχ∗m(x)dχm(x). (7.253)

The hamiltonian for a free Dirac field ψ of mass m is the spatial integral

H0 =

∫ψ (γ · ∇+m)ψ d3x (7.254)

in which ψ ≡ iψ†γ0 and the gamma matrices (??) satisfy

γa, γb = 2 ηab (7.255)

where η is the 4× 4 diagonal matrix with entries (−1, 1, 1, 1). Since ψ|χ〉 =

χ|χ〉 and 〈χ′|ψ† = 〈χ′|χ′†, the quantity 〈χ′| exp(− iεH0)|χ〉 is by (7.251)

〈χ′|e−iεH0 |χ〉 = 〈χ′|χ〉 exp

[− iε

∫χ′ (γ · ∇+m)χ d3x

](7.256)

= exp

[∫12(χ′† − χ†)χ− 1

2χ′†(χ′ − χ)− iεχ′(γ ·∇+m)χd3x

]= exp

ε

∫ [12 χ†χ− 1

2χ′†χ− iχ′ (γ · ∇+m)χ

]d3x

in which χ′† − χ† = εχ† and χ′ − χ = εχ. Everything within the square

brackets is multiplied by ε, so we may replace χ′† by χ† and χ′ by χ so as

to write to first order in ε

〈χ′|e−iεH0 |χ〉 = exp

[ε

∫12 χ†χ− 1

2χ†χ− iχ (γ · ∇+m)χ d3x

](7.257)

in which the dependence upon χ′ is through the time derivatives.

206 Path integrals

Putting together n = 2t/ε such matrix elements, integrating over all

intermediate-state dyadics |χ〉〈χ|, and using our formula (7.252), we find

〈χt|e−2itH0 |χ−t〉 =

∫exp

[∫12 χ†χ− 1

2χ†χ− iχ (γ ·∇+m)χd4x

]Dχ∗Dχ.

(7.258)

Integrating χ†χ by parts and dropping the surface term, we get


∫exp

[∫− χ†χ− iχ (γ ·∇+m)χd4x

]Dχ∗Dχ. (7.259)

Since − χ†χ = − iχγ0χ, the argument of the exponential is

i

∫− χγ0χ− χ (γ · ∇+m)χ d4x = i

∫− χ (γµ∂µ +m)χ d4x. (7.260)

We then have


∫exp

(i

∫L0(χ) d4x

)Dχ∗Dχ (7.261)

in which L0(χ) = − χ (γµ∂µ +m)χ is the action density (4.56) for a free

Dirac field. Thus the amplitude is a path integral with phases given by the

classical action S0[χ]


∫ei

∫L0(χ) d4xDχ∗Dχ =

∫eiS0[χ]Dχ∗Dχ (7.262)

and the integral is over all fields that go from χ(x,−t) = χ−t(x) to χ(x, t) =

χt(x). Any normalization factor will cancel in ratios of such integrals.

Since Fermi fields anticommute, their time-ordered product has an extra

minus sign

T[ψ(x1)ψ(x2)

]= θ(x0

1−x02)ψ(x1)ψ(x2)− θ(x0

2−x01)ψ(x2)ψ(x1). (7.263)

The logic behind our formulas (7.148) and (7.166) for the time-ordered prod-

uct of bosonic fields now leads to an expression for the time-ordered product

of 2n Dirac fields (with Dχ′′ and Dχ′ and so forth suppressed)

〈0|T[ψ(x1) . . . ψ(x2n)

]|0〉 =

∫〈0|χ′′〉χ(x1) . . . χ(x2n) eiS0[χ]〈χ′|0〉Dχ∗Dχ∫

〈0|χ′′〉 eiS0[χ]〈χ′|0〉Dχ∗Dχ.

(7.264)

As in (7.177), the effect of the inner products 〈0|χ′′〉 and 〈χ′|0〉 is to insert


ε-terms which modify the Dirac propagators

〈0|T[ψ(x1) . . . ψ(x2n)

]|0〉 =

∫χ(x1) . . . χ(x2n) eiS0[χ,ε]Dχ∗Dχ∫

eiS0[χ,ε]Dχ∗Dχ

. (7.265)

Imitating (7.178), we introduce a Grassmann external current ζ(x) and

define a fermionic analog of Z0[j]

Z0[ζ] ≡ 〈0| T[e∫ζψ+ψζ d4x

]|0〉 =

∫e∫ζχ+χζ d4xeiS0[χ,ε]Dχ∗Dχ∫

eiS0[χ,ε]Dχ∗Dχ

. (7.266)

Example 7.11 (Feynman’s fermion propagator) Since

i (γµ∂µ +m) ∆(x− y) ≡ i (γµ∂µ +m)

∫d4p

(2π)4eip(x−y)−i (−iγνpν +m)

p2 +m2 − iε

=

∫d4p

(2π)4eip(x−y) (iγµpµ +m)

(−iγνpν +m)

p2 +m2 − iε

=

∫d4p

(2π)4eip(x−y) p2 +m2

p2 +m2 − iε= δ4(x− y),

(7.267)

the function ∆(x− y) is the inverse of the differential operator i(γµ∂µ +m).

Thus the Grassmann identity (7.245) implies that Z0[ζ] is

〈0| T[e∫ζψ+ψζ d4x

]|0〉 =

∫e∫

[ζχ+χζ−i χ(γµ∂µ+m)χ]d4xDχ∗Dχ∫eiS0[χ,ε]Dχ∗Dχ

= exp

[∫ζ(x)∆(x− y)ζ(y) d4xd4y

].

(7.268)

Differentiating we get

〈0|T[ψ(x)ψ(y)

]|0〉 = ∆(x− y) = −i

∫d4p

(2π)4eip(x−y) −iγνpν +m

p2 +m2 − iε. (7.269)

208 Path integrals

7.13 Application to nonabelian gauge theories

The action of a generic non-abelian gauge theory is

S =

∫− 1

4FaµνFµνa − ψ (γµDµ +m)ψ d4x (7.270)

in which the Maxwell field is

Faµν ≡ ∂µAaν − ∂νAaµ + g fabcAbµAcν (7.271)

and the covariant derivative is

Dµψ ≡ ∂µψ − ig taAaµ ψ. (7.272)

Here g is a coupling constant, fabc is a structure constant (??), and ta is a

generator (??) of the Lie algebra (section ??) of the gauge group.

One may show (Weinberg, 1996, pp. 14–18) that the analog of equation

(7.206) for quantum electrodynamics is

〈Ω|T [O1 . . .On] |Ω〉 =

∫O1 . . .On eiS δ[Aa3]DADψ∫

eiS δ[Aa3]DADψ

(7.273)

in which the functional delta function

δ[Aa3] ≡∏x,b

δ(Aa3(x)) (7.274)

enforces the axial-gauge condition, and Dψ stands for Dψ∗Dψ.

Initially, physicists had trouble computing nonabelian amplitudes beyond

the lowest order of perturbation theory. Then DeWitt showed how to com-

pute to second order (DeWitt, 1967), and Faddeev and Popov, using path

integrals, showed how to compute to all orders (Faddeev and Popov, 1967).

7.14 Faddeev-Popov trick

The path-integral tricks of Faddeev and Popov are described in (Weinberg,

1996, pp. 19–27). We will use gauge-fixing functions Ga(x) to impose a gauge

condition on our non-abelian gauge fields Aaµ(x). For instance, we can use

Ga(x) = A3a(x) to impose an axial gauge or Ga(x) = i∂µA

µa(x) to impose a

Lorentz-invariant gauge.

Under an infinitesimal gauge transformation (??)

Aλaµ = Aaµ − ∂µλa − g fabcAbµ λc (7.275)

7.14 Faddeev-Popov trick 209

the gauge fields change, and so the gauge-fixing functions Gb(x), which de-

pend upon them, also change. The jacobian J of that change at λ = 0 is

J = det

(δGλa(x)

δλb(y)

)∣∣∣∣λ=0

≡ DGλ

Dλ

∣∣∣∣λ=0

(7.276)

and it typically involves the delta function δ4(x− y).

Let B[G] be any functional of the gauge-fixing functions Gb(x) such as

B[G] =∏x,a

δ(Ga(x)) =∏x,a

δ(A3a(x)) (7.277)

in an axial gauge or

B[G] = exp

[i

2

∫(Ga(x))2 d4x

]= exp

[− i

2

∫(∂µA

µa(x))2 d4x

](7.278)

in a Lorentz-invariant gauge.

We want to understand functional integrals like (7.273)

〈Ω|T [O1 . . .On] |Ω〉 =

∫O1 . . .On eiS B[G] J DADψ∫

eiS B[G] J DADψ

(7.279)

in which the operators Ok, the action functional S[A], and the differentials

DADψ (but not the gauge-fixing functional B[G] or the Jacobian J) are

gauge invariant. The axial-gauge formula (7.273) is a simple example in

which B[G] = δ[Aa3] enforces the axial-gauge condition Aa3(x) = 0 and the

determinant J = det (δab∂3δ(x− y)) is a constant that cancels.

If we translate the gauge fields by gauge transformations Λ and Λ′, then

the ratio (7.279) does not change

〈Ω|T [O1 . . .On] |Ω〉 =

∫OΛ

1 . . .OΛn e

iSΛB[GΛ] JΛDAΛDψΛ∫

eiSΛ′B[GΛ′ ] JΛ′ DAΛ′ DψΛ′

(7.280)

any more than∫f(y) dy is different from

∫f(x) dx. Since the operators Ok,

the action functional S[A], and the differentials DADψ are gauge invariant,

most of the Λ-dependence goes away

〈Ω|T [O1 . . .On] |Ω〉 =

∫O1 . . .On eiS B[GΛ] JΛDADψ∫

eiS B[GΛ′ ] JΛ′ DADψ

. (7.281)

Let Λλ be a gauge transformation Λ followed by an infinitesimal gauge

210 Path integrals

transformation λ. The jacobian JΛ is a determinant of a product of matrices

which is a product of their determinants

JΛ = det

(δGΛλ

a (x)

δλb(y)

)∣∣∣∣λ=0

= det

(∫δGΛλ

a (x)

δΛλc(z)

δΛλc(z)

δλb(y)d4z

)∣∣∣∣λ=0

= det

(δGΛλ

a (x)

δΛλc(z)

)∣∣∣∣λ=0

det

(δΛλc(z)

δλb(y)

)∣∣∣∣λ=0

= det

(δGΛ

a (x)

δΛc(z)

)det

(δΛλc(z)

δλb(y)

)∣∣∣∣λ=0

≡ DGΛ

DΛ

DΛλ

Dλ

∣∣∣∣λ=0

. (7.282)

Now we integrate over the gauge transformations Λ (and Λ′) with weight

function ρ(Λ) = (DΛλ/Dλ|λ=0)−1 and find, since the ratio (7.281) is Λ-

independent

〈Ω|T [O1 . . .On] |Ω〉 =

∫O1 . . .On eiS B[GΛ]

DGΛ

DΛDΛDADψ∫

eiS B[GΛ′ ]DGΛ′

DΛ′DΛ′DADψ

=

∫O1 . . .On eiS B[GΛ]DGΛDADψ∫

eiS B[GΛ]DGΛDADψ

=

∫O1 . . .On eiS DADψ∫

eiS DADψ

. (7.283)

Thus the mean-value in the vacuum of a time-ordered product of gauge-

invariant operators is a ratio of path integrals over all gauge fields without

any gauge fixing. No matter what gauge condition G or gauge-fixing func-

tional B[G] we use, the resulting gauge-fixed ratio (7.279) is equal to the

ratio (7.283) of path integrals over all gauge fields without any gauge fixing.

All gauge-fixed ratios (7.279) give the same time-ordered products, and so

we can use whatever gauge condition G or gauge-fixing functional B[G] is

most convenient.

The analogous formula for the euclidian time-ordered product is

〈Ω|Te [O1 . . .On] |Ω〉 =

∫O1 . . .On e−Se DADψ∫

e−Se DADψ

(7.284)

7.15 Ghosts 211

where the euclidian action Se is the spacetime integral of the energy density.

This formula is the basis for lattice gauge theory.

The path-integral formulas (7.210 & 7.211) derived for quantum electro-

dynamics therefore also apply to nonabelian gauge theories.

7.15 Ghosts

Faddeev and Popov showed how to do perturbative calculations in which

one does fix the gauge. To continue our description of their tricks, we return

to the gauge-fixed expression (7.279) for the time-ordered product

〈Ω|T [O1 . . .On] |Ω〉 =

∫O1 . . .On eiS B[G] J DADψ∫

eiS B[G] J DADψ

(7.285)

set Gb(x) = −i∂µAµb (x) and use (7.278) as the gauge-fixing functional B[G]

B[G] = exp

[i

2

∫(Ga(x))2 d4x

]= exp

[− i

2

∫(∂µA

µa(x))2 d4x

]. (7.286)

This functional adds to the action density the term −(∂µAµa)2/2 which leads

to a gauge-field propagator like the photon’s (7.214)

〈0|T[Aaµ(x)Abν(y)

]|0〉 = − iδab4µν(x− y) = − i

∫ηµνδabq2 − iε

eiq·(x−y) d4q

(2π)4.

(7.287)

What about the determinant J? Under an infinitesimal gauge transfor-

mation (7.275), the gauge field becomes

Aλaµ = Aaµ − ∂µλa − g fabcAbµ λc (7.288)

and so Gλa(x) = i∂µAλaµ(x) is

Gλa(x) = i∂µAaµ(x) + i∂µ∫

[−δac∂µ − g fabcAbµ(x)] δ4(x− y)λc(y) d4y.

(7.289)

The jacobian J then is the determinant (7.276) of the matrix(δGλa(x)

δλc(y)

)∣∣∣∣λ=0

= −iδac2 δ4(x−y)− ig fabc∂

∂xµ[Aµb (x)δ4(x− y)

](7.290)

that is

J = det

(−iδac2 δ4(x− y)− ig fabc

∂

∂xµ[Aµb (x)δ4(x− y)

]). (7.291)

212 Path integrals

But we’ve seen (7.244) that a determinant can be written as a fermionic

path integral

detA =

∫e−θ

†Aθn∏k=1

dθ∗kdθk. (7.292)

So we can write the jacobian J as

J =

∫exp

[∫iω∗a2ωa + igfabcω

∗a∂µ(Aµbωc) d

4x

]Dω∗Dω (7.293)

which contributes the terms −∂µω∗a∂µωa and

− ∂µω∗a g fabcAµbωc = ∂µω

∗a g fabcA

µcωb (7.294)

to the action density.

Thus we can do perturbation theory by using the modified action density

L′ = − 14FaµνF

µνa − 1

2 (∂µAµa)2−∂µω∗a∂µωa+∂µω

∗a g fabcA

µcωb−ψ (6D +m)ψ

(7.295)

in which 6D ≡ γµDµ = γµ(∂µ−igtaAaµ). The ghost field ω is a mathematical

device, not a physical field describing real particles, which would be spinless

fermions violating the spin-statistics theorem (example ??).

7.16 Integrating over the momenta

When a hamiltonian is quadratic in the momenta like (7.128 & 10.8), one

easily integrates over its momentum and converts it into its lagrangian. If,

however, the hamiltonian is a more complicated function of the momenta,

one usually can’t path-integrate over the momenta analytically. The parti-

tion function is then a path integral over both φ and π

Z(β) =

∫exp

∫ β

0

∫ [iφ(x)π(x)−H(φ, π)

]dtd3x

DφDπ. (7.296)

This exponential is not positive, and so is not a probability distribution for φ

and π. The Monte Carlo methods of chapter ?? are designed for probability

distributions, not for distributions that assume negative or complex values.

This is one aspect of the sign problem.

The integral over the momentum

P [φ] =

∫exp

∫ β

0

∫ [iφ(x)π(x)−H(φ, π)

]dtd3x

Dπ (7.297)

is positive, however, and is a probability distribution. So one can numerically

integrate over the momentum, make a look-up table for P [φ], and then apply

7.16 Integrating over the momenta 213

the usual Monte Carlo method to the probability functional P [φ] (Amdahl

and Cahill, 2016). Programs that do this are in the repository Path integrals

at github.com/kevinecahill.

8

Landau theory of phase transitions

8.1 First- and second-order phase transitions

In a first-order phase transition, a physical quantity, called an order

parameter, changes discontinuously. Often there is a point where the dis-

continuous phase transition becomes continuous. Such a point is called a

critical point. The continuous phase transition is called a second-order

phase transition.

Binary alloys, fluids, and helium-4 exhibit phase transitions.

The archetypical example is the ferromagnet. In an idealized ferromagnet

with spins parallel or antiparallel to a given axis, the order parameter is the

total magnetization M along this axis. At low temperatures, an external

magnetic field H will favor one spin direction. At H = 0 the states will be

in equilibrium, but small changes in H will make M change discontinuously.

There is a first-order phase transition at H = 0 at low temperatures.

At slightly higher temperatures, the spins fluctuate and the mean value

of |M | decreases. At the critical temperature Tc, tiny changes in H no

longer make M change discontinuously. At T > Tc, the phase transition is

continuous and second order.

Landau used a Gibbs free-energy potential G which obeys

∂G

∂M

∣∣∣∣T

= −H. (8.1)

He worked at T ≈ Tc and M ≈ 0 and expanded G(M) at H = 0 as

G(M) = A(T ) +B(T )M2 + C(T )M4 + . . . (8.2)

keeping only even terms because of the symmetry under M → −M . So at

8.1 First- and second-order phase transitions 215

H = 0 equation (8.1) gives

0 = −H =∂G

∂M

∣∣∣∣T

= 2B(T )M + 4C(T )M3. (8.3)

If B and C are positive, the only solution is M = 0. But if C > 0 and if for

T < Tc it happens that B < 0, then we get

2CM2 +B = 0 or M =√−B/2C. (8.4)

The approximations

B(T ) ≈ b(T − Tc) and C(T ) = c (8.5)

give

M = ±√

(b/2c)(Tc − T ) for T < Tc (8.6)

while M = 0 for T > Tc.

The minimum of the potential

G(M,H) = A(T ) +B(T )M2 + C(T )M4 −HM (8.7)

as a function of M at constant T gives equation (8.3).

One may represent the magnetization in terms of a spin density s(x) as

M =

∫s(x) d3x. (8.8)

Gibb’s free energy then becomes

G =

∫1

2(∇s)2 + b(T − Tc)s2 + cs4 −Hsd3x. (8.9)

The spin density that minimizes the free energy must make G stationary

0 =δG

δs(x)= −∇2s(x) + 2b(T − Tc)s(x) + 4cs3(x)−H(x). (8.10)

For T > Tc, the macroscopic magnetization vanishes and so the spin density

s is small. So we drop the nonlinear term and get(−∇2 + 2b(T − Tc)

)s(x) = H(x). (8.11)

The spin density D(x) generated by H(x) = H0δ3(x) is the Green’s function

of the differential equation(−∇2 + 2b(T − Tc)

)D(x) = H0 δ(x). (8.12)

216 Landau theory of phase transitions

Expressing both sides as Fourier transforms, we get for T > Tc

D(x) =

∫d3k

(2π)3

H0eik·x

k2 + 2b(T − Tc)

=H0

4πre−r/ξ

(8.13)

where the correlation length

ξ = [2b(T − Tc)]−1/2 (8.14)

which diverges as T → Tc indicating the onset of large-scale fluctuations.

Different classes of systems have different critical exponents. For exam-

ple, when T < Tc, the magnetization can go as

M ≈ (Tc − T )β (8.15)

in which instead of β = 1/2 one has β = 0.313 for some fluids and some 3D

magnets.

The susceptibility χ = (∂M/∂H)|H = 0 diverges as |T −Tc|−γ . We differ-

entiate the equation (8.3) with respect to H

0 = −H =∂G

∂M

∣∣∣∣T

= 2B(T )M + 4C(T )M3 (8.16)

getting

− 1 = 2Bχ+ 4C3M2χ. (8.17)

So

χ =−1

2B + 12CM2. (8.18)

Setting B = b(T − Tc), C = c, and M = ±√

(b/2c)(Tc − T ), we get

χ =−1

2b(T − Tc) + 6b(Tc − T )=

−1

4b(Tc − T )=

1

4b(T − Tc)(8.19)

which says γ = 1.

9

Effective field theories and gravity

9.1 Effective field theories

Suppose a field φ whose mass M is huge compared to accessible energies

interacts with a field ψ of a low-energy theory such as the standard model

Lφ = − 12∂aφ(x) ∂aφ(x)− 1

2M2φ2(x) + g ψ(x)ψ(x)φ(x). (9.1)

Compared to the mass term M2, the derivative terms ∂aφ∂aφ contribute

little to the low-energy path integral. So we drop them and represent the

effect of the heavy field φ as L0 = −12M

2φ2 +g ψψφ. Completing the square

L0 = − 12M

2(φ− g

M2ψψ)2

+g2

2M2(ψψ)2 (9.2)

and shifting φ by gψψ/M2, we evaluate the path integral over φ as∫exp

[i

∫−1

2M2φ2 +

g2

2M2(ψψ)2 d4x

]Dφ = exp

[i

∫g2

2M2(ψψ)2 d4x

]apart from a field-independent factor. The net effect of heavy field φ is thus

to add to the low-energy theory a new interaction

Lφ(ψ) =g2

2M2(ψψ)2 (9.3)

which is small because M2 is large.

If a gauge boson Ya of huge mass M interacts with a spin-one-half field ψ

as

L0 = − 12M

2YaYa + igψγaψYa (9.4)

= − 12M

2

(Ya −

ig

M2ψγaψ

)(Y a − ig

M2ψγaψ

)− g2

2M2ψγaψ ψγaψ

218 Effective field theories and gravity

then the new low-energy interaction is

SY = − g2

2M2ψγaψ ψγaψ. (9.5)

The low-energy theory of the weak interactions by Fermi and by Gell-Mann

and Feynman is an example, but with γa → γa(1 + γ5).

One way of explaining the lightness of the masses of the (known) neutri-

nos is to say that new a field of very high mass M plays a role, and that

when one path-integrates over this heavy field, one is left with an effective,

nonrenormalizable term in the action

g2

MLσ2H

∗HTσ2L (9.6)

which gives a tiny mass to ν`. Here’s how this can work: take as part of the

action density of the high-energy theory

Lψ = −ψ(/∂ +M)ψ + g ψHTσ2L+ g Lσ2H∗ψ (9.7)

where M is huge. Drop /∂ and complete the square:

Lψ0 = −M(ψ − g

MLσ2H

∗)(ψ − g

MHTσ2L

)+g2

MLσ2H

∗HTσ2L. (9.8)

The path integral over the field ψ of mass M yields a field-independent

constant and leaves in the action the term

Lν =g2

MLσ2H

∗HTσ2L = − g2

M(νh∗0 − eh−)(h+e− h0ν). (9.9)

The mean value of the Higgs field in the vacuum yields the tiny mass term

Lνm =g2v2

2Mνν. (9.10)

If the neutrino is a Dirac field, then this is a Dirac mass term. If the neutrino

is a Majorana field, then this is a Majorana mass term. In both cases, the

mass is intrinsically tiny

mν ∼g2v2

2M. (9.11)

9.2 Is gravity an effective field theory?

The action of the gravitational field is

SE =c3

16πG

∫R√g d4x =

c3

16πG

∫gik Rik

√g d4x. (9.12)

9.2 Is gravity an effective field theory? 219

The action of fields whose energy-momentum tensor is T ik interacting with

gravity is

S =c3

16πG

∫gik Rik

√g d4x+

1

2c

∫T ikgik

√g d4x. (9.13)

The action of a free scalar field of mass m in flat spacetime is

Sφf =1

2

∫ (φ2 − c2(∇φ)2 − m2c4

~2φ2

)d4x (9.14)

where d4x = cdtd3x and a dot means a time derivative. In curved spacetime

and in natural units, the action is

Sφ = − 1

2

∫ (∂iφ∂kφ gik +m2φ2

)√g d4x. (9.15)

The total action is

S =c3

16πG

∫R√g d4x− 1

2


)√g d4x. (9.16)

The change in√g ≡

√|det gik| is

δ√g = −1

2

δ det gik√g

=1

2

√g gikδgki = − 1

2

√g gik δg

ik. (9.17)

The change in the inverse of the metric is

δgik = − gijg`kδgj`. (9.18)

So when the metric changes, the action (9.15 changes by

δSφ = − 1

2

∫ [∂iφ∂kφ (

√g δgik + gik δ

√g) +m2φ2 δ

√g]d4x

= − 1

2

∫ [∂iφ∂kφ

(√g δgik + gik

12

√g gj`δg`j

)+m2φ2 1

2

√g gikδgki

]d4x

= − 1

2

∫ [∂iφ∂kφ

(δgki + gik

12 g

j`δg`j

)+m2φ2 1

2 gikδgki

]√g d4x

= − 1

2

∫ [∂iφ∂kφ+

(12∂

jφ∂`φ gj` + 12m

2φ2)gik]δgki√g d4x

=1

2

∫ [Lgik − ∂iφ∂kφ

]δgki√g d4x =

1

2

∫T ik δgki

√g d4x (9.19)

in which the energy-momentum tensor is

T ik = Lgik − ∂iφ∂kφ. (9.20)


Equivalently, in terms of the change in the inverse metric (9.18), the change

in the scalar action is

δSφ = − 1

2

∫Tik δg

ki√g d4x (9.21)

The variation of the gravitational action is

δSE =c3

16πG

∫ (Rik −

1

2gikR

)√g δgik d4x. (9.22)

The variation of the whole action must vanish

0 =c3

16πG

∫ (Rik −

1

2gikR

)√g δgik d4x− 1

2c

∫Tik δg

ki√g d4x. (9.23)

Thus Einstein’s equations are

Rik −1

2gik R =

8πG

c4Tik. (9.24)

In terms of the Planck mass mP =√~c/G, the way the world works is

Rik −1

2gik R =

8π~c3m2

P

Tik (9.25)

or in natural units

Rik −1

2gik R =

8π

m2P

Tik. (9.26)

The total action is proportional to

S =

∫R√g d4x− 8πG


)√g d4x

=

∫ [R− 8π

m2P

(∂iφ∂kφ gik +m2φ2

) ]√g d4x.

(9.27)

For a massless scalar field, this is simply

S0 =

∫ (R− 8π

m2P

∂iφ∂kφ gik

)√g d4x. (9.28)

The second term in this action S0 could arise in a path integral over a

gauge field Ai whose mass is M = mP the Planck mass∫exp

i

∫ [− 1

2m2P

(Ai −

4√π

m2P

∂iφ)(Ai − 4

√π

m2P

∂iφ)

+8π

m2P

∂iφ∂kφ gik

]√g d4x

DA (9.29)


which might arise from an interaction such as

L = − 12m

2P AiA

i + 4√πAi∂

iφ. (9.30)

Let’s try this again. The usual action is

S =

∫√g d4x

[c3

16πGR− 1

2

(φ,iφ,i +m2φ2

)](9.31)

where 1/G = M2/(~c) in which M = mP is the Planck mass. So

S =

∫√g d4x

[c2M2

16π~R− 1

2

(φ,iφ,i +m2φ2

)]. (9.32)

9.3 Quantization of Fields in Curved Space

Some good references for these ideas are:

Einstein’s Gravity in a Nutshell by Zee

Quantum Fields in Curved Space by Birrell and Davies,

Conformal Field Theory by Di Francesco, Mathieu, and Senechal

Let’s focus on scalar fields for simplicity. We usually expand a scalar field

in flat space as Fourier might have (1.56)

φ(x) =

∫d3p√

(2π)32p0


]. (9.33)

The field φ obeys the Klein-Gordon equation

(∇2 − ∂20 −m2)φ(x) ≡ (2−m2)φ(x) = 0 (9.34)

because the flat-space modes, which have p2 = −m2,

fp(x) = eip·x (9.35)

do

(∇2 − ∂20 −m2) fp(x) ≡ (2−m2) fp(x) = 0. (9.36)

In terms of these functions fp(x), the field is

φ(x) =

∫d3p√

(2π)32p0

[a(p) fp(x) + a†(p) f∗p (x)

]. (9.37)

In terms of a discrete set of modes fn, it is

φ(x) =∑n

[an(p) fn(x) + a†n f

∗n(x)

]. (9.38)


The action for a scalar field in a space described by the metric gik is

S = − 1

2

∫√g d4x

[gikφ,iφ,k +

(m2 + ξR

)φ2]

(9.39)

in which R is the scalar curvature, ξ is a constant, commas denote derivatives

as in φ,k = ∂kφ, gij is the inverse of the metric gij , and g is the absolute

value of the determinant of the metric gij . If the spacetime metric is gij , then

instead of (9.34), the field φ obeys the covariant Klein-Gordon equation

∂i

[√g gij ∂j φ(x)

]−(m2 + ξR

)√g φ(x) = 0. (9.40)

However, to simplify what follows, we will now set ξ = 0.

To quantize in the new coordinates or in the gravitational field of the

metric gij , we need solutions f ′n(x) of the curved-space equation (9.40)

∂i

[√g gij ∂j f

′n(x)

]−m2√g f ′n(x) = 0 (9.41)

which we label with primes to distinguish them from the flat-space solutions

(9.35). We use these solutions to expand the field in terms of curved-space

annihilation and creation operators, which we also label with primes

φ(x) =∑n

[a′n(p) f ′n(x) + a′†n f

′∗n (x)

]. (9.42)

The flat-space modes obey the orthonormality relations

(fp, fq) = i

∫d3x[f∗p (x) ∂tfq(x)−

(∂tf∗p (x)

)fq(x)

]= i

∫d3x[e−ipx (−iq0)eiqx − ip0e−ipx eiqx

]= 2p0(2π)3δ3(p− q),

(9.43)

(f∗p , f∗q ) = − 2p0(2π)3δ3(p− q) and (fp, f

∗q ) = 0. (9.44)

In terms of discrete modes, the flat-space scalar product is

(fn, fm) = i

∫d3x[f∗n(x) ∂tfm(x)−

(∂tf∗n(x)

)fm(x)

]= δnm (9.45)

and its orthonormality relations are

(fn, fm) = δnm, (f∗n, f∗m) = − δnm and (fn, f

∗m) = 0. (9.46)

The scalar product (9.45) is a special case of more general curved-space

scalar product

(f, g) = i

∫S

√gS d

3S va[f∗(x) ∂ag(x)−

(∂af

∗(x))g(x)

](9.47)


in which the integral is over a spacelike surface S with a future-pointing

timelike vector va, and gS is the absolute value of the spatial part of the

metric gik. This more general scalar product (9.47) is hermitian

(f, g)∗ = (g, f). (9.48)

It also satisfies the rule

(f, g)∗ = − (f∗, g∗). (9.49)

One may use Gauss’s theorem to show (Hawking and Ellis, 1973, section

2.8) that this inner product is independent of S and v. The curved-space

modes fn(x) obey orthonormality relations

(f ′n, f′m) = δnm, (f ′∗n , f

′∗m) = − δnm, and (f ′n, f

′∗m) = 0 (9.50)

like those (9.46) of the flat-space modes.

The flat-space modes fn(x) = eipnx naturally describe particles of mo-

mentum pn in flat Minkowski space. In curved space, however, there are in

general no equally natural modes. We can consider other complete sets of

modes f ′′n(x) that are solutions of the curved-space Klein-Gordon equation

(9.41) and obey the orthonormality relations (9.50). We can use any of these

complete sets of mode functions to expand a scalar field φ(x)

φ(x) =∑n

anfn(x) + a†nf∗n(x)

φ(x) =∑n

a′nf′n(x) + a′†nf

′∗n (x)

φ(x) =∑n

a′′nf′′n(x) + a′′†n f

′′∗n (x).

(9.51)

The curved-space orthonormality relations (9.50) imply that

(f`, φ) =∑n

an(f`, fn) + a†n(f`, f∗n) = a`

(f ′`, φ) =∑n

a′n(f ′`, f′n) + a′†n (f ′`, f

′∗n ) = a′`

(f ′′` , φ) =∑n

a′′n(f ′′` , f′′n) + a′′†n (f ′′` , f

′′∗n ) = a′′`

(9.52)


and that

(f∗` , φ) =∑n

an(f∗` , fn) + a†n(f∗` , f∗n) = −a†`

(f ′∗` , φ) =∑n

a′n(f ′∗` , f′n) + a′†n (f ′∗` , f

′∗n ) = −a′†`

(f ′′∗` , φ) =∑n

a′′n(f ′′∗` , f ′′n) + a′′†n (f ′′∗` , f ′′∗n ) = −a′′†` .

(9.53)

The completeness of the mode functions lets us expand them in terms of

each other. Suppressing the spacetime argument x, we have

f ′j =∑i

(αjifi + βji f

∗i

). (9.54)

The curved-space orthonormality relations (9.50) let us identify these Bo-

goliubov coefficients

(f`, f′j) =

∑i

[αji(f`, fi) + βji (f`, f

∗i )]

= αj`

(f∗` , f′j) =

∑i

[αji(f

∗` , fi) + βji (f∗` , f

∗i )]

= −βj`.(9.55)

To find the inverse relations, we use the completeness of the mode functions

f ′i to expand the fj ’s

fj =∑i

(cjif

′i + dji f

′∗i

)(9.56)

and then use the orthonormality relations (9.50) to form the inner products

(f ′`, fj) =∑i

[cji(f

′`, f′i) + dji (f ′`, f

′∗i )]

= cj`

(f ′∗` , fj) =∑i

[cji(f

′∗` , f

′i) + dji (f ′∗` , f

′∗i )]

= −dj`.(9.57)

The hermiticity (9.48) of the scalar product tells us that the cj`’s are related

to the α’s

cj` = (fj , f′`)∗ = α∗`j . (9.58)

The hermiticity (9.48) of the scalar product and the rule (9.49) show that

dj` = − (f ′∗` , fj) = − (fj , f′∗` )∗ = (f∗j , f

′`) = −β`j . (9.59)

So the inverse relation (9.56) is

fj =∑i

(α∗ijf

′i − βij f ′∗i

). (9.60)


The formulas (9.54 & 9.60) that relate the mode functions of different metrics

are known as Bogoliubov transformations.

The vacuum state for a given metric is the state that is mapped to zero

by all the annihilation operators. Our formulas (9.53) for the annihilation

and creation operators

a` = (f`, φ) and a′` = (f ′`, φ)

a†` = − (f∗` , φ) and a′†` = − (f ′∗` , φ)(9.61)

let us express the annihilation and creation operators for one metric in terms

of those for a different metric. Thus, using our formula (9.54) for the f ′’s in

terms of the f ’s, we find

a′j = (f ′j , φ) =∑i

[αji (fi, φ) + βji (f∗i , φ)

]=∑i

(αji ai − βji a†i

). (9.62)

Our formula (9.60) for the f ’s in terms of the f ′’s gives us the inverse relation

aj = (fj , φ) =∑i

[α∗ij (f ′i , φ)− βij (f ′∗i , φ)

]=∑i

(α∗ij a

′i + βij a

′†i

). (9.63)

The annihilation operators a′j define the vacuum state |0〉′ by the rules

a′j |0〉′ = 0 (9.64)

for all modes j. Thus our formula (9.63) for aj says that

aj |0〉′ =∑i

(α∗ij a

′i + βij a

′†i

)|0〉′ =

∑i

βij a′†i |0〉

′ (9.65)

The adjoint of this equation is

8〈0|a†j = 8〈0|∑i

a′iβ∗ij . (9.66)

Thus the mean value of the number operator a†jaj in the |0〉′ vacuum is

8〈0|a†jaj |0〉′ = 8〈0|

∑i

a′iβ∗ij

∑k

βkj a′†k |0〉

′. (9.67)

The commutation relations

[a′i, a′†k ] = δik (9.68)

and the definition a′j |0〉′ = 0 (9.64) of the vacuum |0〉′ imply that the average

number (9.67) of particles of mode j in the (normalized) vacuum |0〉′ is

8〈0|a†jaj |0〉′ = 8〈0|

∑ik

β∗ijβkj δik|0〉′ =∑i

β∗ijβij . (9.69)


It follows that the vacuum of one metric contains particles of the other

metric unless the Bogoliubov matrix

βj` = − (f∗` , f′j) (9.70)

vanishes. The value of βj` in the Minkowski-space scalar product (9.43) is

βj` = − (f∗` , f′j) = −i

∫d3x[f`(x) ∂tf

′j(x)−

(∂tf`(x)

)f ′j(x)

]. (9.71)

This integral for βj` is nonzero, for example, when the functions f` and f ′jhave different frequencies but are not spatially orthogonal.

An example due to Rindler. Let us consider the two metrics

ds2 = dx2 + dy2 + dz2 − dt2 = dr2 + dy2 + dz2 − r2du2 (9.72)

in which y and z are the same, r plays the role of x and u that of time. The

first metric has gik = ηik and g = |det(η)| = 1. The second metric has

gik =

−r2 0 0 0

0 1 0 0

0 0 1 0

0 0 0 1

and g = r2. (9.73)

The inverse metric is

gik =

−r−2 0 0 0

0 1 0 0

0 0 1 0

0 0 0 1

. (9.74)

The equation of motion is

∂i

[√ggik∂kf

]= m2√gf. (9.75)

So we must solve

∂u(−r−1∂uf) + ∂r(r∂rf) + ∂y(r∂yf) + ∂z(r∂zf) = m2rf. (9.76)

Now u, r, y, z are independent coordinates, so this equation of motion is

− r−1∂2uf + ∂r(r∂rf) + r∂2

yf + r∂2zf = m2rf (9.77)

or

− r−2∂2uf + r−1∂rf + ∂2

rf + ∂2yf + ∂2

zf = m2f. (9.78)

Let’s make f a plane wave in the y and z directions

f(u, r, y, z) = f(u, r)eiypy+izPz . (9.79)


If we also set

M2 = m2 + P 2y + P 2

z , (9.80)

then we must solve

− r−2∂2uf + r−1∂rf + ∂2

rf = M2f (9.81)

where f = f(u, r). We now make the Daniel-Middleton transformation,

separating the dependence of f upon r and u

f(u, r) = a(u)b(r). (9.82)

Our differential equation (9.81) reduces to

− r−2ab+ r−1ab′ + ab′′ = M2ab (9.83)

in which dots denote ∂u and primes ∂r. Dividing by a, we get

− a

a

b

r2+b′

r+ b′′ −M2b = 0. (9.84)

As a(u), we choose

a(u) = eiωu anda

a= − ω2. (9.85)

Then our equation (9.84) for b(r) is

b′′ +b′

r−(M2 − ω2

r2

)b = 0 (9.86)

or equivalently

r2b′′ + rb′ −(M2r2 − ω2

)b = 0. (9.87)

Yi’s solution is a modified Bessel function Iν(z) for imaginary ν.

Bogoliubov’s β matrix is

βj` = − (f∗` , f′j). (9.88)

The scalar product (9.47) uses the metric of one, not both, of the solu-

tions. If we choose the flat-space metric, then we need to write the solution

f ′j(u, r, y, z) in terms of the flat-space coordinates t, x, y, z. The mean occu-

pation number (9.69) is the sum∑j

8〈0|a†jaj |0〉′ =

∑ij

|βij |2 =∑ij

|(f∗j , f ′i)|2. (9.89)


9.4 Accelerated coordinate systems

We recall the Lorentz transformations

t′ = γ(t− vx), x′ = γ(x− vt), y′ = y, and z′ = z

t = γ(t′ + vx′), x = γ(x′ + vt′), y = y′, and z = z′(9.90)

in which γ = 1/√

1− v2. The velocities (in the x direction) are

u =dx

dtand u′ =

dx′

dt′=γ(dx− vdt)γ(dt− vdx)

=(u− v)

(1− vu). (9.91)

So the accelerations (in the x direction) are

a =du

dt

a′ =du′

dt′= d

[ (u− v)

(1− vu)

]/γ(dt− vdx)

=[ du

(1− vu)+

(u− v)vdu

(1− vu)2

]/γ(dt− vdx)

=[du(1− vu)

(1− vu)2+

(u− v)vdu

(1− vu)2

]/γ(dt− vdx)

=du(1− v2)

γ(1− vu)2(dt− vdx)

=(1− v2)a

γ(1− vu)2(1− vu)

=(1− v2)a

γ(1− vu)3=

a

γ3(1− vu)3.

(9.92)

Now we let the acceleration a′ be a constant. That is, the acceleration in

the (instantaneous) rest frame of the frame moving instantaneously at u = v

in the laboratory frame is a constant, a′ = α. In this case, since u = v, we

get an equation (Rindler, 2006, sec. 3.7)

a′ = α =a

γ3(1− vu)3=

a

γ3(1− v2)3=

a

(1− v2)3/2

=1

(1− v2)3/2

du

dt=

1

(1− u2)3/2

du

dt=

d

dt

(u√

1− u2

) (9.93)

that we can integrate

u√1− u2

= α (t− t0). (9.94)


Squaring and solving for u, we find

u =dx

dt=

α(t− t0)√1 + α2(t− t0)2

(9.95)

which we can integrate to

x = α−1

∫dt

α2(t− t0)√1 + α2(t− t0)2

=

√1 + α2(t− t0)2 − 1

α+ x0. (9.96)

The proper time of the accelerating frame is τ = t′. An interval dt′ of

proper time is one at which dx′ = 0. So dt′ =√

1− v2dt or dtτ =√

1− u2dt

where u(t) is the velocity (9.95). Integrating the equation

dt′ = dτ =√

1− u2dt =

√1− α2(t− t0)2

1 + α2(t− t0)2dt =

dt√1 + α2(t− t0)2

,

(9.97)

we get

α(t′ − t′0) = α(τ − τ0) =

∫αdt√

1 + α2(t− t0)2= arcsinh(α(t− t0)). (9.98)

So

α(t− t0) = sinh(α(τ − τ0)). (9.99)

The formula (9.96) for x now gives

α(x− x0) = cosh(α(τ − τ0)). (9.100)

9.5 Scalar field in an accelerating frame

For simplicity, we’ll work with a real scalar field (1.54)

φ(x) =

∫d3p√

(2π)32p0


]. (9.101)

If the field is quantized in a box of volume V , then the expansion of the field

is

φ(x) =∑k

1√2ωkV

[a(k) eik·x + a†(k) e−ik·x

]. (9.102)

For the scalar field (9.101), the zero-temperature correlation function is


the mean value in the vacuum

〈0|φ(x, t)φ(x′, t′)|0〉 = 〈0|∫

d3p√(2π)32p0


]×∫

d3p′√(2π)32p′0

[a(p′) eip

′·x′ + a†(p′) e−ip′·x′]|0〉

= 〈0|∫

d3pd3p′

(2π)3√

2p02p′0a(p)a†(p′) eip·x−ip

′·x′ |0〉

=

∫d3pd3p′

(2π)3√

2p02p′0δ3(p− p′)eip·x−ip′·x′

=

∫d3p

(2π)32p0eip·(x−x

′). (9.103)

This integral is simpler for massless fields. Setting p = |p| and r = |x− x′|,we add a small imaginary part to the exponential and find

〈0|φ(x, t)φ(x′, t′)|0〉 =

∫d3p

(2π)32peipr cos θ−ip(t−t′−iε)

=

∫pdp dcos θ

(2π)22eipr cos θ−ip(t−t′−iε)

=

∫pdp

(2π)22

(eipr − e−ipr

ipr

)e−ip(t−t

′−iε)

=

∫ ∞0

dp

(2π)22ir

(eipr − e−ipr

)e−ip(t−t

′−iε)

=1

(2π)22ir

(− 1

i(r − (t− t′))− 1

i(r + (t− t′))

)=

1

(2π)22r

(1

r − (t− t′)+

1

r + (t− t′)

)=

1

(2π)2[(x− x′)2 − (t− t′)2]. (9.104)

This is the zero-temperature correlation function.

In the instantaneous rest frame of an accelerated observer moving in the

x direction with coordinates (9.99 & 9.100), this two-point function is

〈0|φ(x, t)φ(x′, t′)|0〉 =α2/(2π)2[

cosh(ατ)− cosh(ατ ′)]2 − [ sinh(ατ)− sinh(ατ ′)

]2= − α2

(4π)2 sinh2[α(τ − τ ′)/2](9.105)


in which the identities

cosh(ατ) cosh(ατ ′)− sinh(ατ) sinh(ατ ′) = cosh(α(τ − τ ′)) (9.106)

and

cosh(α(τ − τ ′))− 1 = 2 sinh(α(τ − τ ′)/2) (9.107)

as well as cosh2 ατ − sinh2 ατ = 1 were used.

Now let’s compute the same correlation function at a finite inverse tem-

perature β = 1/(kBT )

〈φ(0, τ)φ(0, τ ′)〉β = Tr[φ(0, 0)φ(0, t)e−βH

]/Tr(e−βH

). (9.108)

The mean value of the number operator for a given momentum k is

〈a†(k) a(k)〉β = nk = Tr[a†(k) a(k)e−βH

]/Tr(e−βH

)(9.109)

in which

H0 =∑k

ωk(a†(k) a(k) + 1

2

). (9.110)

The trace over all momenta with k′ 6= k is unity, and we are left with

〈a† a〉β = nk = Tr[a† ae−βωka

†a]/

Tr(e−βωka

†a)

= − 1

ωkTr(e−βωka†a

) ∂∂β

Tr(e−βωka

†a) (9.111)

in which a†a ≡ a†(k) a(k) and the 1/2 terms have cancelled. The trace is

Tr(e−βωka

†a)

=∑n

e−βnωk =1

1− e−βωk. (9.112)

So we have

〈a† a〉β = −(1− e−βωk

)ωk

(−ωke−βωk)(1− e−βωk

)2 =1

eβωk − 1. (9.113)

In the trace (9.108), only terms that don’t change the number of quanta

in each mode contribute. So the mean value of the correlation function at

inverse temperature β = 1/(kBT ) is

〈φ(0, τ)φ(0, τ ′)〉β =∑k

1

2kV

(e~ωk/kBT − 1

)−1eiωk(τ−τ ′)

+

[(e~ωk/kBT − 1

)−1+ 1

]e−iωk(τ−τ ′)

.

(9.114)


In the continuum limit, this τ, τ ′ correlation function is two integrals. For

massless particles, the simpler one requires some regularization because it

involves zero-point energies

Aβ =∑k

1

2kVe−ik(τ−τ ′) =

∫d3k

(2π)32ke−ik(τ−τ ′) =

∫k2dk

(2π)2ke−ik(τ−τ ′)

=

∫ ∞0

kdk

(2π)2e−ik(τ−τ ′).

(9.115)

We send τ − τ ′ → τ − τ ′ + iε and find

Aβ =

∫ ∞0

kdk

(2π)2e−ik(τ−τ ′+iε) = i

d

dτ

∫ ∞0

dk

(2π)2e−ik(τ−τ ′+iε)

= id

dτ

∫ ∞0

dk

(2π)2e−ik(τ−τ ′+iε) =

1

(2π)2

d

dτ

1

(τ − τ ′)

= − 1

(2π)2

1

(τ − τ ′)2

(9.116)

as ε→ 0.

The second integral is

Bβ =

∫ ∞0

kdk

(2π)2

(eβk − 1

)−12 cos(k(τ − τ ′)) (9.117)

which Mathematica says is

Bβ =1

(2π)2(τ − τ ′)2− csch2(π(τ − τ ′)/β)

4β2. (9.118)

Thus the finite-temperature correlation function is

〈φ(0, τ)φ(0, τ ′)〉β = Aβ +Bβ = − 1

4β2 sinh2(π(τ − τ ′)/β). (9.119)

Equating this formula to the zero–temperature correlation function (9.105)

in the accelerating frame, we find

1

4β2 sinh2(π(τ − τ ′)/β)=

α2

(4π)2 sinh2[α(τ − τ ′)/2](9.120)

which says redundantly

kBT =α

2πand πkBT =

α

2. (9.121)

So a detector accelerating uniformly with acceleration α in the vacuum feels


a nonzero temperature

T =~α

2πckB. (9.122)

This result (Davies, 1975) is equivalent to the finding (Hawking, 1974) that

a gravitational field of local acceleration g makes empty space radiate at a

temperature

T =~g

2πckB. (9.123)

Black holes are not black.

9.6 Maximally symmetric spaces

The spheres S2 and S3 and the hyperboloids H2 and H3 are maximally sym-

metric spaces. A transformation x → x′ is an isometry if g′ik(x′) = gik(x

′)

in which case the distances gik(x)dxidxk = g′ik(x′)dx′idx′k = gik(x

′)dx′idx′k

are the same. To see what this symmetry condition means, we consider the

infinitesimal transformation x′` = x` + εy`(x) under which to lowest order

gik(x′) = gik(x) + gik,`εy

` and dx′i = dxi+ εyi,jdxj . The symmetry condition

requires

gik(x)dxidxk = (gik(x) + gik,`εy`)(dxi + εyi,jdx

j)(dxk + εyk,mdxm) (9.124)

or

0 = gik,` y` + gim y

m,k + gjk y

j,i. (9.125)

The vector field yi(x) must satisfy this condition if x′i = xi + εyi(x) is to be

a symmetry of the metric gik(x). Since the covariant derivative of the metric

tensor vanishes, gik;` = 0, we may write the condition on the symmetry

vector y`(x) as

0 = yi;k + yk;i. (9.126)


In more detail, we subtract gik;`y`, which vanishes, from the condition (9.125)

that y be a Killing vector:

0 = gik,`y` + gimy

m,k + gjky

j,i − gik;`y

`

= gik,`y` + gimy

m,k + gjky

j,i − gik,`y

` + gimΓm`ky` + gkjΓ

jìy

`

= gimym,k + gjky

j,i + gimΓm`ky

` + gkjΓjìy

` (9.127)

= gim(ym,k + Γm`ky`) + gjk(y

j,i + Γjìy

`)

= gimym;k + gjky

j;i = yi;k + yk;i.

Students leery of the last step may use the vanishing of the covariant deriva-

tive of the metric tensor and the derivation property

(AB);k = A;kB +AB;k (9.128)

of covariant derivatives, to show that

0 = (gimym);k + (gjky

j);i = yi;k + yk;i. (9.129)

The symmetry vector y` is a Killing vector (Wilhelm Killing, 1847–1923).

We may use symmetry condition (9.125) or (9.126) either to find the sym-

metries of a space with a known metric or to find a metric with given sym-

metries.

Example 9.1 (Killing vectors of the sphere S2) The first Killing vector

is (yθ1, yφ1 ) = (0, 1). Since the components of y1 are constants, the symmetry

condition (9.125) says gik,φ = 0 which tells us that the metric is independent

of φ. The other two Killing vectors are (yθ2, yφ2 ) = (sinφ, cot θ cosφ) and

(yθ3, yφ3 ) = (cosφ,− cot θ sinφ). The symmetry condition (9.125) for i = k =

θ and Killing vectors y2 and y3 tell us that gθφ = 0 and that gθθ,θ = 0.

So gθθ is a constant, which we set equal to unity. Finally, the symmetry

condition (9.125) for i = k = φ and the Killing vectors y2 and y3 tell us that

gφφ,θ = 2 cot θgφφ which we integrate to gφφ = sin2 θ. The 2-dimensional

space with Killing vectors y1, y2, y3 therefore has the metric

gik =

(gθθ gθφgφθ gφφ

)=

(eθ · eθ eθ · eφeφ · eθ eφ · eφ

)=

(R2 0

0 R2 sin2 θ

)(9.130)

of the sphere S2.

Example 9.2 (Killing vectors of the hyperboloid H2) The metric

(gij) = R2

(1 0

0 sinh2 θ

). (9.131)


of the hyperboloid H2 is diagonal with gθθ = R2 and gφφ = R2 sinh2 θ.

The Killing vector (yθ1, yφ1 ) = (0, 1) satisfies the symmetry condition (9.125).

Since gθθ is independent of θ and φ, the θθ component of (9.125) implies

that yθ,θ = 0. Since gφφ = R2 sinh2 θ, the φφ component of (9.125) says

that yφ,φ = − coth θ yθ. The θφ and φθ components of (9.125) give yθ,φ =

− sinh2 θ yφ,θ. The vectors y2 = (yθ2, yφ2 ) = (sinφ, coth θ cosφ) and y3 =

(yθ3, yφ3 ) = (cosφ, − coth θ sinφ) satisfy both of these equations.

The Lie derivative Ly of a scalar field A is defined in terms of a vector

field y`(x) as LyA = yÀ,`. The Lie derivative Ly of a contravariant vector

F i is

LyF i = y`F i,` − F `yi,` = y`F i;` − F `yi;` (9.132)

in which the second equality follows from y`Γi`kFk = F `Γi`ky

k. The Lie

derivative Ly of a covariant vector Vi is

LyVi = y`Vi,` + V`y`,i = y`Vi;` + V`y

`;i. (9.133)

Similarly, the Lie derivative Ly of a rank-2 covariant tensor Tik is

LyTik = y`Tik,` + T`ky`,i + Ti`y

`,k. (9.134)

We see now that the condition (9.125) that a vector field y` be a symmetry

of a metric gjm is that its Lie derivative

Lygik = gik,` y` + gim y

m,k + gjk y

j,i = 0 (9.135)

must vanish.

A maximally symmetric space (or spacetime) in d dimensions has d trans-

lation symmetries and d(d− 1)/2 rotational symmetries which gives a total

of d(d + 1)/2 symmetries associated with d(d + 1)/2 Killing vectors. Thus

for d = 2, there is one rotation and two translations. For d = 3, there are

three rotations and three translations. For d = 4, there are six rotations and

four translations.

A maximally symmetric space has a curvature tensor that is simply related

to its metric tensor

Rijk` = c (gikgj` − gi`gjk) (9.136)

where c is a constant (Zee, 2013, IX.6). Since gkigik = gkk = d is the number

of dimensions of the space(time), the Ricci tensor and the curvature scalar

of a maximally symmetric space are

Rj` = gkiRijk` = c (d− 1) gj` and R = g`jRj` = c d(d− 1). (9.137)


9.7 Conformal algebra

An isometry requires that g′ik(x′) = gik(x

′). The looser condition

g′ik(x′) = Ω(x′) gik(x

′) with Ω(x′) > 0 (9.138)

says that two metrics are conformally related. An equivalent condition is

g′ik(x′) = x`,i′x

m,k′g`m(x) = Ω(x′)gik(x

′). (9.139)

This condition guarantees that angles do not change:

g′ikuivk√g′ijuiuj

√g′ijvivj

=gikuivk√

gijuiuj√gijvivj

. (9.140)

For the infinitesimal transformation

x′` = x` + εy`, (9.141)

we approximate the factor Ω as

Ω2(x′) ≈ 1 + εω(x′). (9.142)

The conformal condition (9.139) then says that

g`m(δì − εy`,i)(δmk − εym,k) = (1 + εω)(gik + εy`gik,`) (9.143)

or

gimym,k + g`ky

`,i + y`gik,` + ωgik = 0. (9.144)

This is Killing’s conformal condition.

Multiplication by gik and summing over i and k gives

δkmym,k + δi`y

`,i + giky`gik,` + 4ω = 0 (9.145)

or

2yk,k + y`gikgik,` + 4ω = 0. (9.146)

The function ω(x) therefore

ω = − 1

4

(2yi,i + y`gikgik,`

). (9.147)

Substituting this formula for ω into Killing’s conformal condition (9.144)

gives a condition on the metric g and on the vector y

gimym,k + g`ky

`,i + y`gik,` −

(1

2yj,j +

1

4y`gjngjn,`

)gik = 0. (9.148)


When ω = 0, Killing’s conformal condition (9.144)

gimym,k + g`ky

`,i + y`gik,` = 0 (9.149)

is the same as his isometry condition (9.125) .

Using the vanishing of the covariant derivative of the metric tensor as in

(9.127), we can write Killing’s conformal condition (9.144) more succinctly

as

yi;k + yk;i + ω gik = 0 (9.150)

or as

Lygik = −ω gik. (9.151)

9.8 Conformal algebra in flat space

To find Killing’s conformal vectors y in flat space, we set gik = ηik and use

the formula (9.147) for ω to find

ω = − 1

2yi,i (9.152)

which in d dimensions of spacetime is ω = −2yi,i/d. So in flat-space, Killing’s

conformal condition (9.144) is

yi,k + yk,i =1

2ηik y

`,` or yi,k + yk,i =

2

dηik y

`,`. (9.153)

Infinitesimal transformations

x′i = xi + ε yi (9.154)

that obey this rule generate the conformal algebra of Minkowski space.

We already know some of these vectors. The vector

y` = a` + b`kxk (9.155)

with ai and b`k constant and b antisymmetric, bik = − bki, has a vanishing

divergence y`,` = 0 and satisfies Killing’s conformal condition (9.153)

yi,k + yk,i = bik + bki =2

dηik y

`,` = 0. (9.156)

These y’s generate the translations ai and the Lorentz transformations bik.

A conformal transformation x → x′ changes the metric by no more than

an overall factor. That means that

Ω(x′) ηikdx′idx′k = ηikdx

idxk. (9.157)


So if space is just stretched by a constant factor σ, so that x′i = σxi, then

dx′i = σdxi, and the stretch condition (9.157) is satisfied with Ω(x′) = σ−2.

This kind of conformal transformation is a dilation, which some call a

dilatation. The Killing vector y is

yi = σxi (9.158)

with σ a constant. This vector obeys Killing’s flat-space conformal condition

(9.153) because y`,` = c d in d dimensions, and so

yi,k + yk,i = ηij c δjk + ηkj c δ

ji = 2ηik c =

2

dηik y

`,` . (9.159)

The inversion

x′i =r2 xi

x2(9.160)

where r is a length and x2 = xkxk also obeys Killing’s conformal condition

(9.153). The change in x′i is

dx′i =r2

x2

(δij −

2xixjx2

)dxj (9.161)

so with Ω ≡ Ω(x′)

Ωηikdx′idx′k = Ωηik

(r2

x2

)2(δij −

2xixjx2

)dxj

(δk` −

2xkx`x2

)dx`

= Ω

(r2

x2

)2 [ηj` −

2ηjkxkx`

x2− 2ηi`x

ixjx2

+4x2xjx`

(x2)2

]dxjdx`

= Ω

(r2

x2

)2

ηj` dxjdx`. (9.162)

So the stretch condition (9.157) is satisfied with

Ω(x′) =

(x2

r2

)2

. (9.163)

Since both inversions and translations obey Killing’s conformal condition

(9.153), we may combine them so as to shrink an inversion to its infinitesimal

form. We invert, translate by a tiny vector a, and re-invert and so find as


the legitimately infinitesimal form of an inversion the transformation

xi → xi

x2→ xi

x2+ ai →

(xi

x2+ ai

)/ηjk

(xj

x2+ aj

)(xk

x2+ ak

)=

(xi

x2+ ai

)/ηjk

(1

x2+

2a · xx2

+ a2

)(9.164)

= x2

(xi

x2+ ai

)/(1 + 2a · x+ a2x2

)'(xi + aix2

)(1− 2a · x)

= xi + ak(ηikx2 − 2xixk) (9.165)

which is a conformal transformation.

If we use ∂i to differentiate Killing’s conformal condition (9.153), then we

get

∂iyi,k + ∂iyk,i =2

dηik ∂

i y`,` (9.166)

or, with ∂2 = ∂i∂i and ∂ · y = ∂iyi,(

1− 2

d

)∂k ∂ · y = − ∂2yk (9.167)

which says that in two dimensions (d = 2) every Killing vector is a solution

of Laplace’s equation:

∂2yk = 0 if d = 2. (9.168)

There are infinitely many Killing vectors in two-dimensions, a fact exploited

in string theory.

Differentiating the same equation (9.167) again, we get

(d− 2) ∂i∂k ∂ · y = − d ∂2yk,i (9.169)

which says that ∂2yk,i = ∂2yi,k. Applying this symmetry and ∂2 to Killing’s

conformal condition (9.153), we get

ηik ∂2∂ · y = d ∂2yk,i. (9.170)

Comparing the right-hand sides of the last two equations (9.169 & 9.170),

we find [(d− 2) ∂i∂k + ηik∂

2]∂ · y = 0. (9.171)

Differentiating the same equation (9.167) a third time, we have

∂k (d− 2) ∂k ∂ · y = − d ∂2∂kyk (9.172)


or

(d− 1) ∂2 ∂ · y = 0. (9.173)

In the case of one dimension (d = 1), Killing’s conformal condition (9.153)

reduces to 2∂0y0 = 2∂0y0 and so places no restriction on Killing’s vector,

as reflected equations (9.167 & 9.173). Every function y0(x0) ≡ y(t) is a

conformal Killing vector. The condition (9.157) for a finite one-dimensional

conformal transformation is

dx′

dx= Ω−1(x′). (9.174)

This condition constrains the transformation x→ x′ = x′(x) only by requir-

ing that the derivative (9.174) be positive.

The vanishing (9.168) of the laplacian of the Killing vector in d = 2 di-

mensions means that every Killing vector is a solution of Laplace’s equation:

∂2yk = 0 if d = 2. (9.175)

In two-dimensional euclidian space, we let the d = 2 space be the complex

plane and let y0 and y1 be the real and imaginary parts of an analytic (or

antianalytic) function:

z = x0 + ix1 and f(z) = y0 + iy1. (9.176)

Then the Cauchy-Riemann conditions

y0,0 = y1

,1 and y0,1 = − y1

,0 (9.177)

imply that the real and imaginary parts of every analytic function f(z) are

harmonic

y0,00 + y0

,11 = 0 and y1,00 + y1

,11 = 0 (9.178)

because y0,00 = y1

10 = −y0,11 and y1

,00 = −y0,10 = −y1

,11. The real and imag-

inary parts of every antianalytic function f(z∗) also are harmonic, and

so satisfy Killing’s condition (9.175). There are infinitely many solutions

for the Killing vectors of an infinitesimal conformal transformation in two-

dimensional euclidian space.

What about finite conformal transformations in two-dimensional euclidian

space? The definition (9.157) of a conformal transformation in flat space is

η′ik = Ω(x′)ηik. (9.179)

But metrics transform like this:

η′ik =∂xj

∂x′i∂x`

∂x′kηj` and η′ik =

∂x′i

∂xj∂x′k

∂x`ηj`. (9.180)


In two-dimensional euclidian space, η is the 2×2 matrix η is the 2×2 matrix

η =

(1 0

0 1

), (9.181)

so these equations say that

Ω = Ωη00 =∂x′0

∂x0

∂x′0

∂x0η00 +

∂x′0

∂x1

∂x′0

∂x1η11 =

(∂x′0

∂x0

)2

+

(∂x′0

∂x1

)2

Ω = Ωη11 =∂x′1

∂x0

∂x′1

∂x0η00 +

∂x′1

∂x1

∂x′1

∂x1η11 =

(∂x′1

∂x0

)2

+

(∂x′1

∂x1

)2

0 = Ωη01 =∂x′0

∂x0

∂x′1

∂x0η00 +

∂x′0

∂x1

∂x′1

∂x1η11 =

∂x′0

∂x0

∂x′1

∂x0+∂x′0

∂x1

∂x′1

∂x1.

(9.182)

More succinctly, these conditions are(∂x′0

∂x0

)2

+

(∂x′0

∂x1

)2

=

(∂x′1

∂x0

)2

+

(∂x′1

∂x1

)2

∂x′0

∂x0

∂x′1

∂x0= − ∂x′0

∂x1

∂x′1

∂x1.

(9.183)

These conditions are equivalent either to

∂x′0

∂x0=∂x′1

∂x1and

∂x′1

∂x0= − ∂x′0

∂x1(9.184)

which are the Cauchy-Riemann conditions ux = vy and vx = − uy for

f = u+ iv to be an analytic function of z = x+ iy or to

∂x′0

∂x0= − ∂x′1

∂x1and

∂x′1

∂x0=∂x′0

∂x1(9.185)

which are the Cauchy-Riemann conditions ux = − vy and vx = uy for

f(z) = u+ iv to be an analytic function of z = x− iy.

9.8.1 Angles and analytic functions

An analytic function f(z) maps curves in the z plane into curves in the

f(z) plane. In general, this mapping preserves angles. To see why, we consider

the angle dθ between two tiny complex lines dz = ε exp(iθ) and dz′ =

ε exp(iθ′) that radiate from the same point z. This angle dθ = θ′ − θ is the

phase of the ratio

dz′

dz=εeiθ

′

εeiθ= ei(θ

′−θ). (9.186)


Let’s use w = ρeiφ for f(z). Then the analytic function f(z) maps dz into

dw = f(z + dz)− f(z) ≈ f ′(z) dz (9.187)

and dz′ into

dw′ = f(z + dz′)− f(z) ≈ f ′(z) dz′. (9.188)

The angle dφ = φ′ − φ between dw and dw′ is the phase of the ratio

dw′

dw=eiφ′

eiφ=f ′(z) dz′

f ′(z) dz=dz′

dz=eiθ′

eiθ= ei(θ

′−θ). (9.189)

So as long as the derivative f ′(z) does not vanish, the angle in the w-plane

is the same as the angle in the z-plane

dφ = dθ. (9.190)

Analytic functions preserve angles. They are conformal maps.

What if f ′(z) = 0? In this case, dw ≈ f ′′(z) dz2/2 and dw′ ≈ f ′′(z) dz′2/2,

and so the angle dφ = dφ′ − dφ between these two tiny complex lines is the

phase of the ratio

dw′

dw=eiφ′

eiφ=f′′(z) dz′2

f ′′(z) dz2=dz′2

dz2= e2i(θ′−θ). (9.191)

So angles are doubled, dφ = 2dθ.

In general, if the first nonzero derivative is f (n)(z), then

dw′

dw=eiφ′

eiφ=f (n)(z) dz′n

f (n)(z) dzn=dz′n

dzn= eni(θ

′−θ) (9.192)

and so dφ = ndθ. The angles increase by a factor of n.

Example 9.3 (zn) The function f(z) = zn has only one nonzero derivative

f (k)(0) = n! δnk at the origin z = 0. So at z = 0 the map z → zn scales

angles by n, dφ = ndθ, but at z 6= 0 the first derivative f (1)(z) = nzn−1 is

not equal to zero. So zn is conformal except at the origin.

Example 9.4 (Mobius transformation) The function

f(z) =az + b

cz + d(9.193)

maps (straight) lines into lines and circles and circles into circles and lines,

unless ad = bc in which case it is the constant b/d.


But if we require that f(z) be entire and invertible on the Riemann sphere

(the complex plane with the “point” z = ∞ mapped onto the north pole)

so that the transformations form the special conformal group, then the

function f(z) cannot vanish at more than one value of z or f−1(0) would

not be unique. Similarly, f(z) can’t have more than one pole. So the only

functions f(z) that are entire and invertible on the Riemann sphere are those

of the projective transformation

f(z) =az + b

cz + dwith ad− bc = 1 (9.194)

and a, b, c, and d complex, which is a special case of the fractional linear

or Mobius transformation (9.193). These matrices form the group SL(2,C)

which is isomorphic to the Lorentz group SO(3, 1). So the requirement that

the finite transformations form a group has reduced the symmetry of the

conformal group to that of the Lorentz group.

In two-dimensional Minkowski space, we must solve Killing’s equation for

the vector of an infinitesimal transformation

yk,11 = yk,00. (9.195)

So we use Dirac’s light-cone coordinates. We set

x± = t± x. (9.196)

Then

ds2 = − dt2 + dx2 = −(dt+ dx)(dt− dx) = ηikdxidxk

= η−−dx−dx− + η++dx

+dx+ + η−+dx+dx− + η+−dx

−dx+(9.197)

with η−+ = η+− = −12 and η−− = η++ = 0. So the inverse metric has

η−+ = η+− = −2. The light-cone derivatives are

∂± =1

2(∂t ± ∂x) . (9.198)

They act like this:

∂+x+ = 1 and ∂−x

− = 1

∂+x− = 0 and ∂−x

+ = 0.(9.199)

In light-cone coordinates, the equation (9.195) for the d = 2 Killing vector

is

∂−∂+yk = ∂+∂−y

k = 0. (9.200)


Any vector yk(x+) satisfies this equation because

∂−yk(x+) = y′k∂−x

+ = 0. (9.201)

Similarly, any function yk(x−) obeys the same equation. So there are two

sets of infinitely many solutions yk(x+) and yk(x−) of Killing’s equation

(9.200). We let

yk(x+, x−) = yk+(x+) + yk−(x−). (9.202)

To understand finite conformal transformations in 2-d Minkowski space,

we return to the definition (9.179)

η′ik = Ω(x′)ηik. (9.203)

of a conformal transformation in flat space and recall how (9.204) metrics

transform

η′ik =∂xj

∂x′i∂x`

∂x′kηj` and η′ik =

∂x′i

∂xj∂x′k

∂x`ηj`. (9.204)

In two-dimensional Minkowski space, η is the 2× 2 matrix

η =

(−1 0

0 1

), (9.205)

so these equations say that

− Ω = Ωη00 =∂x′0

∂x0

∂x′0

∂x0η00 +

∂x′0

∂x1

∂x′0

∂x1η11 = −

(∂x′0

∂x0

)2

+

(∂x′0

∂x1

)2

Ω = Ωη11 =∂x′1

∂x0

∂x′1

∂x0η00 +

∂x′1

∂x1

∂x′1

∂x1η11 = −

(∂x′1

∂x0

)2

+

(∂x′1

∂x1

)2

0 = Ωη01 =∂x′0

∂x0

∂x′1

∂x0η00 +

∂x′0

∂x1

∂x′1

∂x1η11 = − ∂x′0

∂x0

∂x′1

∂x0+∂x′0

∂x1

∂x′1

∂x1.

(9.206)

More succinctly, these conditions are(∂x′0

∂x0

)2

−(∂x′0

∂x1

)2

= −(∂x′1

∂x0

)2

+

(∂x′1

∂x1

)2

∂x′0

∂x0

∂x′1

∂x0=∂x′0

∂x1

∂x′1

∂x1.

(9.207)

In terms of light-cone coordinates, these conditions are

∂+x′0 ∂−x

′0 = − ∂+x′1 ∂−x

′1

∂+x′0 ∂−x

′1 = − ∂−x′0 ∂+x′1.

(9.208)


We can satisfy them by having x′k be functions either of x+ or of x−:

x′k(t, x) = xk(x+) or x′k(t, x) = xk(x−) (9.209)

but not both as would be okay (9.202) for infinitesimal transformations.

Example 9.5 (Liouville’s equation) Liouville’s equation is

φtt − φxx + 2eφ = 0 (9.210)

or

∂+ ∂−φ = − 12 e

φ. (9.211)

It has as a solution

eφ =4f ′(x+)g′(x−)

(f(x+)− g(x−))2(9.212)

in which f ′ g′ > 0.

In more than two dimensions (d ≥ 3), the vanishing (9.173) of the lapla-

cian of the divergence ∂ · y and the third-order equation (9.171) imply that

the all the second derivatives of the divergence vanish

∂i ∂k (∂ · y) = 0. (9.213)

So the divergence of Killing’s vector must be at most linear in the coordinates

∂ · y = a+ bixi for d ≥ 3. (9.214)

To see what this implies, we differentiate Killing’s conformal condition (9.153)

yi,k` + yk,i` =2

dηik(∂ · y),`, (9.215)

add to it the same equation with k and ` interchanged

yi,`k + y`,ik =2

dηi`(∂ · y),k, (9.216)

and then subtract the same equation with i, k, ` permuted

yk,ì + y`,ki =2

dηk`(∂ · y),i (9.217)

so as to get

2yi,k` =2

d

[ηik(∂ · y),` + ηi`(∂ · y),k − ηk`(∂ · y),i

]. (9.218)

So if we substitute the at-most-linear condition (9.214) into this equation

(9.218), then we find

2yi,k` =2

d[ηikb` + ηi`bk − ηk`bi] (9.219)


which says that yi can be at most quadratic in the coordinates.

yi = αi + βijxj + γijkx

jxk (9.220)

in which γijk = γikj .

Killing’s conformal condition (9.153) must hold for all x’s and is linear

in the vector y, so we can impose it successively on ai, bij , and cijk by

differentiating with respect to the variables x` and setting them equal to

zero. The vector αi is an arbitrary translation. The condition on βij is

βik + βki =2

dηik β

`` (9.221)

which says that βij is the sum

βij = bij + c ηij (9.222)

of an antisymmetric part bij = −bji that generates a Lorentz transformation

and a term proportional to ηij that generates a scale transformation. The

conformal condition (9.153) makes the quadratic term satisfy the relations

γik` + γki` = 2ηikd` (9.223)

γi`k + γìk = 2ηi`dk (9.224)

γkì + γ`ki = 2ηk`di (9.225)

in which

d` =1

dγrr`. (9.226)

Subtracting the last equation (9.225) from the sum of the first two (9.223 &

9.224), we find

γik` = ηikd` + ηi`dk − ηk`di. (9.227)

The corresponding infinitesimal transformation is

x′i = xi + 2(x · d)xi − dix2 (9.228)

which is called a special conformal transformation whose exponential

or finite form is

x′i =xi − dix2

1− 2d · x+ d2x2. (9.229)

The general Killing vector and its infinitesimal transformation are

yi = ai + bikxk + cxi + dk(η

ikx2 − 2xixk) (9.230)

and

x′i = xi + ai + bikxk + cxi + dk(η

ikx2 − 2xixk) (9.231)

9.9 Maxwell’s action is conformally invariant for d = 4 247

in which we see a translation, a Lorentz transformation, a dilation, and a

special conformal transformation.

The differential forms of the Poincare transformations and the dilation D

and the inversion k are

Pi = ∂i, Jik = (xi ∂k − xk ∂i),D = xi ∂i, and Ki = (ηik xj xj − 2xi xk) ∂k.

(9.232)

The commutators of the conformal Lie algebra are

[P i, P j ] = 0, [Ki,Kj ] = 0,

[D,P i] = − P i, [D,Jij ] = 0, [D,Ki] = Ki,

[J ij , P `] = − ηi`P j + ηj`P i, [J ij ,K`] = −ηi`Kj + ηj`Ki (9.233)

[J ij , J `m] = − ηi`J jm − ηjmJ i` + ηj`J im + ηimJ j`,

[Ki, P j ] = 2J ij + 2ηijD.

These commutation relations say that D is a Lorentz scalar, and that Ki

transforms as a vector.

There are d P ’s, dK’s, d(d− 1)/2 J ’s, and one D. That’s (d+ 2)(d+ 1)/2

generators, which is the same number of generators as SO(d + 2). In fact,

Minkowski space in d dimensions has SO(d− 1, 1) as its Lorentz group and

SO(d, 2) as its conformal algebra. The “Lorentz group” of a spacetime with

d spatial dimensions and 2 time dimensions is also SO(d, 2).

9.9 Maxwell’s action is conformally invariant for d = 4

Under a change of coordinates from x to x′, a covariant vector field Ai(x)

changes this way

A′i(x′) = xk,i′ Ak(x) =

∂xk

∂x′iAk(x). (9.234)

So under the tiny change (with a convenient minus sign)

x′ = x− y or x = x′ + y (9.235)

the change in Ai is

δAi(x) ≡ A′i(x)−Ai(x)

= A′i(x)−A′i(x′) +A′i(x′)−Ai(x)

= yj A′i,j(x′) +

(δji + yj,i′

)Aj(x)−Ai(x)

= yj A′i,j(x′) + yj,i′ Aj(x).

(9.236)


To lowest order in y, this is

δAi(x) = yjAi,j(x) + yj,iAj(x). (9.237)

So the change in the Maxwell-Faraday tensor is

δFik = δ (∂iAk − ∂kAi) = ∂i (δAk)− ∂k (δAi)

= ∂i

(yjAk,j(x) + yj,k Aj(x)

)− ∂k

(yjAi,j(x) + yj,iAj(x)

)= yj∂jFik + yj,iAk,j − y

j,k Ai,j + yj,kiAj − y

j,ik Aj + yj,k ∂iAj − y

j,i ∂kAj

= yj∂jFik + yj,i Fjk + yj,k Fij . (9.238)

Thus the change in Maxwell’s action density is

δ(F ikFik

)= 2F ik δFik = 2F ik

(yj∂jFik + yj,i Fjk + yj,k Fij

)= ∂j

(yjF ikFik

)− (∂ · y)F ikFik + 2yj,i F

ikFjk + 2yj,k FikFij

= ∂j

(yjF ikFik


ikF jk + 2yj,k FikF j

i

= ∂j

(yjF ikFik


ikF jk (9.239)

= ∂j

(yjF ikFik

)− (∂ · y)F ikFik + 4yk,i F

ijF kj .

We can rewrite this as

δ(F ikFik

)= ∂j

(yjF ikFik

)+ 2F ijF kj

(yk,i + yi,k − 1

2ηik∂ · y). (9.240)

But if y is a conformal Killing vector obeying (9.153), then this change in

the Maxwell action density is

δ(F ikFik

)= ∂j

(yjF ikFik

)+ 2F ijF kj

(2

dηiky

`,` −

1

2ηik∂ · y

). (9.241)

Iff d = 4, this change is a total derivative

δ(F ikFik

)= ∂j

(yjF ikFik

)⇐⇒ d = 4. (9.242)

The action density of pure (only gauge fields), classical Yang-Mills theory

also is conformally invariant in four dimensions (d = 4). Quantum effects

introduce a mass scale, however, and break conformal invariance.

9.10 Massless scalar field theory is conformally invariant if d = 2

Under a tiny change of coordinates x = x′ + y, the change in a scalar field

φ is

δφ = yjφ,j = yjφ,j . (9.243)


So the change in the action density is

δ

(1

2φ,iφ,i

)= φ,i δφ,i = φ,i ∂iδφ = φ,i ∂i

(yjφ

,j)= φ,iφ,jyj,i + φ,iyjφ

,j,i (9.244)

=1

2φ,iφ,j (yj,i + yi,j

)+ φ,iφ

,j,i yj .

If y obeys the condition (9.153) for a conformal Killing vector, then the

change (9.244) in the action density is a total divergence in two dimensions

δ

(1

2φ,iφ,i

)=

1

2φ,iφ,j 2

dηijy

`,` + φ,iφ

,j,i yj

=1

dφ,iφ,i y

`,` + φ,iφ,i` y

` =1

dφ,iφ,i y

`,` +

1

2

(φ,iφ,i

),` y

`

= ∂`

(1

2φ,iφ,i y

`

)⇐⇒ d = 2. (9.245)

Thus the classical theory of a free massless scalar field is conformally invari-

ant in two-dimensional spacetimes.

The rest of this chapter is at best a first draft, not ready for human

consumption.

9.11 Christoffel symbols as nonabelian gauge fields

A contravariant vector V i transforms like dx′i = x′i,kdxk as

V ′i(x′) =∂x′i

∂xkV k(x) = x′i,k V

k(x) = Eik(x)V k(x). (9.246)

The 4 × 4 matrix Eik(x) = x′i,k depends upon the spacetime point x and is

a member of the huge noncompact group GL(4,R).

The insight of Yang and Mills (section 5.1) lets us define a covariant

derivative D` = ∂` +A` of a contravariant vector V i

(D`V )k = (∂`δkj +Ak`j)V

j (9.247)

that transforms as [(D`V )k

]′=∂xi

∂x′`∂x′k

∂xm(DiV )m. (9.248)


We need[(∂`δ

kj +Ak`j)V

j]′

= (∂′`δkj +A′k`j)V

′j = V ′k,`′ +A′k`jV′j

= xi,`′x′k,m(∂iδ

mn +Amin)V n = xi,`′x

′k,m(V m

,i +AminVn)

(9.249)

or

V ′k,`′ +A′k`jV′j = xi,`′x

′k,m(V m

,i +AminVn). (9.250)

Decoding the left-hand side, we get

∂`′(x′k,mV

m) +A′k`jV′j = xi,`′x

′k,m(V m

,i +AminVn) (9.251)

or

xi,`′∂i(x′k,mV

m) +A′k`jV′j = xi,`′x

′k,m(V m

,i +AminVn) (9.252)

or

xi,`′x′k,mV

m,i + xi,`′x

′k,miV

m +A′k`jV′j = xi,`′x

′k,m(V m

,i +AminVn). (9.253)

After a cancelation, we get

xn,`′x′k,mnV

m +A′k`jV′j = xi,`′x

′k,mA

minV

n (9.254)

or

A′k`jV′j = xp,`′x

′k,mA

mpnV

n − xn,`′x′k,mnV m (9.255)

which is

A′k`jx′j,mV

m = xp,`′x′k,mA

mpnV

n − xn,`′x′k,mnV m (9.256)

or after the interchange of m and n in the second term

A′k`jx′j,mV

m = xp,`′x′k,nA

npmV

m − xn,`′x′k,mnV m. (9.257)

Since the vector V m is arbitrary, we can promote this to an equation with

three free indexes

A′k`jx′j,m = xp,`′x

′k,nA

npm − xn,`′x′k,mn (9.258)

and then multiply by the inverse of x′j,m

A′k`jx′j,mx

m,i′ = xp,`′x

′k,nA

npmx

m,i′ − xn,`′x′k,mnxm,i′ (9.259)

so as to get

A′kì = xm,`′x′k,pA

pnmx

n,i′ − x

p,`′x′k,npx

n,i′ . (9.260)

Partial derivatives of the same kind commute, so x′k,np = x′k,pn. It follows


that the inhomogeneous term in the transformation law (9.260) for A is

symmetric in ` and i

− xp,`′x′k,npx

n,i′ = − xn,i′x′k,npx

p,`′ = − xp,i′x

′k,pnx

n,`′ = − xp,i′x

′k,npx

n,`′ . (9.261)

The homogeneous term xm,`′x′k,pA

pnmxn,i′ in the transformation law (9.260) for

A is symmetric in ` and i if Apnm is symmetric in n and m, Apnm = Apmn.

Thus we can use gauge fields that are symmetric in their lower indexes,

Apnm = Apmn.

The transformations are

V ′k = x′k,jVj ≡ EkjV j and V ′` = xi,`′Vi ≡ E−1T i

` Vi = E−1i`Vi (9.262)

and

A′kì = xm,`′x′k,pA

pnmx

n,i′ − x

p,`′x′k,npx

n,i′

= E−1mÈ

kpA

pnmE

−1ni − x

p,`′x′k,npx

n,i′ .

We decode the inhomogeneous term

−xp,`′x′k,npx

n,i′ = − (∂`′x

′k,n)xn,i′ = −(∂`′E

kn)xn,i′ = −(∂`′E

kn)E−1n

i

= − E−1ni(∂`′E

kn) = Ekn(∂`′E

−1ni) = x′k,n(∂`′x

n,i′)

(9.263)

So on a contravariant vector V ′ = E−1V , A′ is

A′` = E−1mÈAmE

−1 + E∂`′E−1 (9.264)

or

A′kì = E−1mÈ

kpA

pmnE

−1ni + Ekn∂`′E

−1ni . (9.265)

This gauge field is the affine connection, aka, Christoffel symbol (of the

second kind) Akì = Γkì. The usual formula for the covariant derivative of

a contravariant vector is

(D`V )k = (∂`δkj +Ak`j)V

j = (∂`δkj + Γk`j)V

j = ∂`Vk + Γk`jV

j . (9.266)

We can write the chain rule identities

∂xi

∂x′k∂x′k

∂x`= δi` =

∂x′i

∂xk∂xk

∂x′`(9.267)

in terms of the matrix Ek` = x′k,` as

∂xi

∂x′kEk` = δi` = Eik

∂xk

∂x′`(9.268)


which show that the inverse of the matrices Ek` and Eik are

E−1ik =

∂xi

∂x′k= xi,k′ and E−1k

` =∂xk

∂x′`= xk,`′ . (9.269)

The very general identity

0 = ∂iI = ∂i(C C−1) = C,iC

−1 + C C−1,i (9.270)

will soon be useful. So we can write

EmnE−1n

k,`′ = Emn ∂`′∂xn

∂x′k= Emn ∂k′

∂xn

∂x′`

= EmnE−1n

`,k′ = −Emn,k′ E−1n`.

(9.271)

A covariant vector Vj transforms like a derivative ∂j

V ′j (x′) =∂xk

∂x′jVk(x) = xk,j′ Vk(x) ≡ E−1k

j Vk. (9.272)

The Yang-Mills trick now requires that we find a gauge field B` so that the

covariant derivative of a covariant vector

(D`V )k = (∂`δjk +Bj

`k)Vj (9.273)

transforms as [(D`V )k

]′=∂xi

∂x′`∂xn

∂x′k(DiV )n. (9.274)

We need[(∂`δ

jk+B

j`k)Vj

]′= (∂′`δ

jk +B′j`k)V

′j = V ′k,`′ +B′j`kV

′j

= xi,`′xn,k′(∂iδ

pn +Bp

in)Vp = xi,`′xn,k′(Vn,i +Bp

inVp)(9.275)

or

V ′k,`′ +B′j`kV′j = xi,`′x

n,k′(Vn,i +Bp

inVp). (9.276)

Decoding the left-hand side, we get

(xn,k′ Vn),`′ +B′j`k xn,j′ Vn = xn,k′`′ Vn + xn,k′ Vn,`′ +B′j`k x

n,j′ Vn

= xn,k′`′ Vn + xn,k′ Vn,i xi,`′ +B′j`k x

n,j′ Vn.

(9.277)

So combining the last two equations, we find

xn,k′`′ Vn + xn,k′ Vn,i xi,`′ +B′j`k x

n,j′ Vn = xi,`′x

n,k′ Vn,i + xi,`′x

n,k′ B

pinVp (9.278)

or

xn,k′`′ Vn +B′j`k xn,j′ Vn = xi,`′x

n,k′ B

pinVp. (9.279)


Interchanging p and n in the third term

xn,k′`′ Vn +B′j`k xn,j′ Vn = xi,`′x

p,k′ B

nip Vn (9.280)

and then using the arbitrariness of Vn, we get

xn,k′`′ +B′j`k xn,j′ = xi,`′x

p,k′ B

nip. (9.281)

Multiplying by x′m,n gives

xn,k′`′ x′m,n +B′j`k x

n,j′ x

′m,n = xn,k′`′ x

′m,n +B′m`k = xi,`′x

p,k′ B

nip x′m,n (9.282)

or

B′m`k = xi,`′xp,k′ B

nip x′m,n − xn,k′`′ x′m,n . (9.283)

The inhomogeneous term −xn,k′`′ x′m,n is symmetric in k and `. The homoge-

neous term xi,`′xp,k′ B

nip x′m,n also is symmetric in k and `. Thus we can use

gauge fields that are symmetric in their lower indexes, Bnip = Bn

pi.

In terms of the matrices

Ek` = x′k,` and E−1k` =

∂xk

∂x′`= xk,`′ (9.284)

and in view of the identities (9.270) and (9.271), the transformation rule

(9.283) for B has the equivalent form

B′m`k = (E−1T) i` E

mnB

nipE−1p

k + Emn,`′E−1n

k. (9.285)

The transformation law for A` is

A′kì = E−1mÈ−1n

iApnmE

kp − E−1n

i∂`′Ekn. (9.286)

We write the formula (9.285) for B′ as

−B′m`k = − (E−1T) i` E

mnB

nipE−1p

k − E−1n

k Emn,`′

= − (E−1T) i` E

mnB

nipE−1p

k − E−1n

Èmn,k′

(9.287)

and do the substitutions k → i, i→ m, and m→ k

−B′kì = − E−1mÈ

knB

nmpE

−1pi − E

−1ni ∂`′E

kn. (9.288)

The inhomogeneous terms in the equations for A′ and B′ are the same. We

interchange p and n in the formula for B′

−B′kì = − E−1mÈ

kpB

pmnE

−1ni − E

−1ni ∂`′E

kn

= − E−1mÈ−1n

iBpmnE

kp − E−1n

i ∂`′Ekn.

(9.289)


Since the gauge fields are symmetric in their lower indexes, we can write

this as

−B′kì = − E−1mÈ−1n

iBpnmE

kp − E−1n

i ∂`′Ekn. (9.290)

Since the transformation laws are the same, we can identify the two gauge

fields apart from a minus sign

Akì = −Bkì = Γki`. (9.291)

The gauge fields A and B correspond to the Christoffel symbols, the Γ’s

of the usual notation,

Ai`k : D`Vi = V i

;` = V i,` + Γik`V

k = V i,` + ei · ek,`V k

Bkì : D`Vi = Vi;` = V i

,` − Γki` Vk = Vi,` − ek · ei,` Vk(9.292)

9.12 Spin connection

The Lorentz index a on a tetrad cai is subject to local Lorentz transforma-

tions. A spin-one-half field ψb also has an index b that is subject to local

Lorentz transformations. We can use the insight of Yang and Mills to in-

troduce a gauge field, called the spin connection ωabi which multiplies the

generators Jab of the Lorentz group

ωi = ωabi Jab. (9.293)

Since the generators Jab are antisymmetric Jab = −Jba, the spin connection

is also antisymmetric

ωabi = − ωbai . (9.294)

So for each spacetime index i, there are six independent ωabi ’s.

The spin connection is

ωab ` = − ckb(cak,` − Γjk` c

aj

)= caj c

kb Γjk` − c

ak,` c

kb . (9.295)

Under a general coordinate transformation and a local Lorentz transfor-

mation, the spin connection (9.295) transforms as

ω′ab ` =∂xi

∂x′`

[Lad ω

de i − (∂iL

ae)]L−1e

b. (9.296)

This example is a work in progress....

Example 9.6 (Relativistic particle) The lagrangian of a relativistic par-

ticle of mass m is L = −m√

1− q2 in units with c = 1. The hamiltonian is

9.12 Spin connection 255

H =√p2 +m2. The path integral for the first step (7.63) toward Z(β) is

〈q1|e−εH |qa〉 =

∫ ∞−∞

e−ε√p′2+m2

eip′(q1−qa)/~ dp

′

2π~(9.297)

which is a nontrivial integral. In the m→ 0 limit, it is

〈q1|e−εH |qa〉 =

∫ ∞−∞

e−ε|p′| eip

′(q1−qa)/~ dp′

2π~

= 2

∫ ∞0

e−εp′cos(p′(q1 − qa)

~

) dp′

2π~

=

∫ ∞0

e−εp′+ip′(q1−qa)/~ + e−εp−ip

′(q1−qa)/~ dp′

2π~

=( 1

ε− i(q1 − qa)/~+

1

ε+ i(q1 − qa)/~

) 1

2π~

=( 2ε

ε2 + (q1 − qa)2/~2

) 1

2π~

=1

π~ε

( 1

1 + (q1 − qa)2/(~ε)2

). (9.298)

So the partition function is n→∞ limit of the product of the n integrals

Z(β) =

(1

π~ε

)n ∫ ∞−∞

dqn1 + (qn − qn−1)2/(~ε)2

· · · dq1

1 + (q1 − qn)2/(~ε)2

in which qn = qa.

Z(β) =

(1

π~ε

)n(π~ε)n. (9.299)

In one space dimension, quantum mechanics gives the result

Z(β) =L

π~βc(9.300)

in which c has reappeared. and we must take the L→∞ limit. In two space

dimensions, it gives

Z(β) =L2

(2π~)2

∫ ∞0

e−βc√m2c2+p2

2πpdp =L2

4π~2

∫ ∞0

e−βc√m2c2+udu

=(Lmc)2

4π~2

∫ ∞0

e−βmc2√

1+vdv =(Lmc)2

4π~2

2(1 + βmc2)

(βmc2)2e−βmc

2(9.301)

=L2

2π~2

(1 + βmc2)

(βc)2e−βmc

2.

10

Field Theory on a Lattice

10.1 Scalar Fields

To represent a hermitian field ϕ(x), we put a real number ϕ(i, j, k, `) at each

vertex of the lattice and label a lattice of spacetime points x by a 4-vector

of integers as

x = a(i, j, k, `) = as (10.1)

where a is the lattice spacing and

s = (i, j, k, `) (10.2)

is a 4-vector of lattice sites. The derivative ∂iϕ(x) is approximated as

∂iϕ(x) ≈ ϕ(x+ i)− ϕ(x)

a(10.3)

in which x is the discrete position 4-vector and i is a unit 4-vector pointing

in the i direction. So the euclidian action is the sum over all lattice sites of

Se =∑x

1

2(∂iϕ(x))2 a4 +

1

2m2 ϕ2(x) a4

=∑x

1

2

(ϕ(x+ i)− ϕ(x)

)2a2 +

1

2m2 ϕ2(x) a4

(10.4)

We switch to dimensionless variables

φ = aϕ and µ = am. (10.5)

In these terms, the action density is

Se = − 1

2

∑xi

φ(x+ i)φ(x) +1

2

(8 + µ2

)φ2(x) (10.6)


or if the self interaction happens to be quartic

Se =1

2(∂iφ(x))2 a4 +

1

2m2 φ2(x) a4 +

λ

4φ4(x) a4

= − 1

2

∑xi

φ(x+ i)φ(x) +1

2

(8 + µ2

)φ2(x) +

λ

4φ4(x).

(10.7)

10.2 Finite-temperature field theory

Since the Boltzmann operator e−βH = e−H/(kT ) is the time evolution op-

erator e−itH/~ at the imaginary time t = − i~β = −i~/(kT ), the formulas

of finite-temperature field theory are those of quantum field theory with t

replaced by −iu = −i~β = −i~/(kT ).

Our hamiltonian is H = K+V where K and V are sums over the vertices

of the spatial lattice v = a(i, j, k) = (ai, aj, ak) of the cubes of volume a3

times the squared momentum and the potential energy

aH = aK + aV =a4

2

∑v

$2v +

a4

2

∑v

(∇ϕv)2 +m2ϕ2v + P (ϕv)

=1

2

∑v

π2v +

1

2

∑v

(∇φv)2 + µ2φ2v + P (φv).

(10.8)

We are assuming that the action if quadratic in the time derivatives of the

fields. A matrix element of the first term of the Trotter product formula

(7.7)



)n(10.9)

is the imaginary-time version of (7.133) with ε = ~β/n

〈φ1|e−εK e−εV |φa〉 = e−εV (φa)∏v

[∫a3dπ′v

2πea

3[−επ2v/2+i(φ1v−φav)π′v ]

].

(10.10)

Setting φav = (φ1v − φav)/ε, we find, instead of (7.134)

〈φ1|e−εK e−εV |φa〉 =∏v

[(a3

2πε

)1/2

e−εa3[φ2

av+(∇φav)2+m2φ2av+P (φv)]/2

].

(10.11)

The product of n = ~β/ε such inverse-temperature intervals is

〈φb|e−βH |φa〉 =∏v

[(a3n

2πβ

)n/2 ∫e−SevDφv

](10.12)

258 Field Theory on a Lattice

in which the euclidian action (10.4 or 10.7) is a sum over all the vertices

x = a(i, j, k, `) of the spacetime lattice

Sev =β

n

a3

2

n−1∑j=0

[φ2jv + (∇φjv)2 +m2φ2

jv + P (φv)]

(10.13)

where φjv = n(φj+1,v − φj,v)/β and Dφv = dφn−1,v · · · dφ1,v.

The amplitude 〈φb|e−(βb−βa)H |φa〉 is the integral over all fields that go

from φa(x) at βa to φb(x) at βb each weighted by an exponential

〈φb|e−(βb−βa)H |φa〉 =

∫e−Se[φ]Dφ (10.14)

of its euclidian action

Se[φ] =

∫ βb

βa

du

∫d3x

1

2

[φ2 + (∇φ)2 +m2φ2 + P (φ)

](10.15)


Dφ =∏v

[(a3n

2π(βb − βa)

)n/2dφn−1,v · · · dφ1,v

]. (10.16)

Equivalently, the Boltzmann operator is

e−(βb−βa)H =

∫|φb〉e−Se[φ] 〈φa|DφDφaDφb (10.17)


states.

The trace of the Boltzmann operator is the partition function

Z(β) = Tr(e−βH) =

∫e−Se[φ] 〈φa|φb〉DφDφaDφb =

∫e−Se[φ] DφDφa

(10.18)

which is an integral over all fields that go back to themselves in euclidian

time β. So we must require that the field ϕ is periodic in euclidian time. If

na is the extend of the lattice in euclidian time, then

ϕ(i, j, k, `+ n) = ϕ(i, j, k, `). (10.19)

As a convenience, in scalar field theoeries one usually also imposes periodic

boundary conditions in the spatial directions. One then has full periodicity:

for any four multiples of the lengths Li of the lattice in the four directions

ni = nLi, one has

ϕ(i+ n1, j + n2, k + n3, `+ n4) = ϕ(i, j, k, `). (10.20)


Like a position operator (7.107), a field at an imaginary time t = −iu =

−i~β is defined as

φe(x, u) = φe(x, ~β) = euH/~φ(x, 0) e−uH/~ (10.21)

in which φ(x) = φ(x, 0) = φe(x, 0) is the field at time zero, which obeys

the commutation relations (7.122). The euclidian-time-ordered product of

several fields is their product with newer (higher u = ~β) fields standing to

the left of older (lower u = ~β) fields as in the definition (7.109).

We set ~ = 1 to avoid clutter and irrelevant thinking. The euclidian path

integrals for the mean values of euclidian-time-ordered-products of fields are

similar to those (7.161 & 7.148) for ordinary time-ordered-products. The

euclidian-time-ordered-product of the fields φ(xj) = φ(xj , uj) is the path

integral

〈n|T [φe(x1)φe(x2)]|n〉 =

∫〈n|φb〉φ(x1)φ(x2)e−Se[φ]〈φa|n〉Dφ∫

〈n|φb〉e−Se[φ]〈φa|n〉Dφ(10.22)

in which the integrations are over all paths that go from before u1 and u2

to after both euclidian times. The analogous result for several fields is

〈n|T [φe(x1) · · ·φe(xk)]|n〉 =

∫〈n|φb〉φ(x1) · · ·φ(xk)e

−Se[φ]/~〈φa|n〉Dφ∫〈n|φb〉e−Se[φ]/~〈φa|n〉Dφ

(10.23)


u1, . . . , uk to after them.

In the low-temperature β = 1/(kT ) → ∞ limit, the Boltzmann operator

is proportional to the outer product |0〉〈0| of the ground-state kets, e−βH →e−βE0 |0〉〈0|. In this limit, the integrations are over all fields that run from

u = −∞ to u = ∞, and the only energy eigenstate |n〉 that contributes is

the ground state |0〉 of the theory

〈0|T [φe(x1) · · ·φe(xk)]|0〉 =

∫〈0|φb〉φ(x1) · · ·φ(xk)e

−Se[q]/~〈φa|0〉Dφ∫〈0|φb〉e−Se[q]/~〈φa|0〉Dφ

.

(10.24)

Formulas like this one are used in lattice gauge theory. For a free scalar field


theory, they take the form of a multiple integral over lattice sites s

Z0(β) = Tr(e−βH0) =∏s

dφse−S0e (10.25)

in which the free euclidian action (10.6) is a sum over the spacetime lattice

with nearest-neighbor coulings

S0e = − 1

2

∑xi

φ(x+ i)φ(x) +1

2

(8 + µ2

)φ2(x)

=1

2

∑n,m

φnKnmφm.(10.26)

The matrix K is n4 × n4. Its elements are

Ksu = −∑i>0

[δs+i,u + δs−i,u − 2δsu] + µ2δsu (10.27)

in which i = (1, 0, 0, 0) etc. The integral is a gaussian. If the lattice is four

dimensional with n4 sites, then Z(β) is

Z0(β) =∏s

dφs exp

−1

2

∑s,s′

φsKss′φs′

=

√((2π)n4

detK

) (10.28)

in which β = na. In the presence of a classical current J , the otherwise free

partition function becomes

Z0(β, J) =∏s

dφs exp

−1

2

∑s,s′

φsKss′φs′ +∑s

Jsφs

(10.29)

We shift the field to

φ′ = φ+K−1J (10.30)

so that

−1

2φ′Kφ′ + J · φ′ = − 1

2

(φ+ JK−1

)K(φ+K−1J

)+ J ·

(φ+K−1J

)= − 1

2φKφ+

1

2J ·K−1J.

(10.31)

10.3 The Propagator 261

and

Z0(β, J) =∏s

dφs exp

−1

2

∑s,s′

φsKss′φs′ +∑s,s′

JsK−1s,s′Js′ .

=

√((2π)n4

detK

)exp

1

2

∑s,s′

JsK−1s,s′Js′

.

(10.32)

The formula (10.29) now gives us the propagator as

〈φsφs′〉 = Z0(β, J)−1 ∂2

∂Js∂Js′Z0(β, J)

∣∣∣∣J=0

= K−1ss′ . (10.33)

To find K−1ss′ , we use∑

t

KstK−1tu = δsu =

∫ π

−π

d4k

(2π)4eik·(s−u). (10.34)

We also use this formula to write K as

Ksu =

∫ π

−π

d4k

(2π)4K(k) eik·(s−u) (10.35)

where

K(k) = 44∑j=1

sin2(kj/2) + µ. (10.36)

10.3 The Propagator

10.4 Pure Gauge Theory

The gauge-covariant derivative is defined in terms of the generators ta of a

compact Lie algebra

[ta, tb] = ifabctc (10.37)

and a gauge-field matrix Ai = igAbi tb as

Di = ∂i −Ai = ∂i − igAbi tb (10.38)

summed over all the generators, and g is a coupling constant. Since the

group is compact, we may raise and lower group indexes without worrying

about factors or minus signs.

The Faraday matrix is

Fij = [Di, Dj ] = [I∂i−Ai(x), I∂j−Aj(x)] = −∂iAj+∂jAi+[Ai, Aj ] (10.39)


in matrix notation. With more indices exposed, it is

(Fij)cd = (−∂iAj + ∂jAi + [Ai, Aj ])cd

= − igtbcd(∂iA

bj − ∂jAbi

)+(

[igtbAbi , igteAej ]

)cd.

(10.40)

Summing over repeated indices, we get

(Fij)cd = − igtbcd(∂iA

bj − ∂jAbi

)− g2AbiA

ej

([tb, te]

)cd

= − igtbcd(∂iA

bj − ∂jAbi

)− g2AbiA

ejifbef t

fcd

= − igtbcd(∂iA

bj − ∂jAbi

)− ig2AbiA

ejfbef t

fcd

= − igtbcd(∂iA

bj − ∂jAbi

)− ig2Afi A

ejffebt

bcd

= − igtbcd(∂iA

bj − ∂jAbi + gAfi A

ejffeb

)= − igtbcd

(∂iA

bj − ∂jAbi + gfbfeA

fi A

ej

)= − igtbcd F bij

(10.41)

where

F bij = ∂iAbj − ∂jAbi + gfbfeA

fi A

ej (10.42)

is the Faraday tensor.

The action density of this tensor is

LF = − 1

4F bijF

ijb . (10.43)

The trace of the square of the Faraday matrix is

Tr[FijF

ij]

= Tr[igtb F bij igt

c F ijc

]= − g2F bij F

ijc Tr(tbtc) = −g2F bij F

ijc kδbc

= − kg2 F bij Fijb .

(10.44)

So the Faraday action density is

LF = − 1

4F bijF

ijb =

1

4kg2Tr[FijF

ij]

=1

2g2Tr[FijF

ij]. (10.45)

The theory described by this action density, without scalar or spinor fields,

is called pure gauge theory.

The purpose of a gauge field is to make the gauge-invariant theories. So

if a field ψa is transformed by the group element g(x)

ψ′a(x) = g(x)abψb(x), (10.46)

10.4 Pure Gauge Theory 263

then we want the covariant derivative of the field to go as

(Diψ(x))′ =(∂i −A′i(x)

)g(x)ψ(x)

= g(x)Diψ(x) = g(x) [(∂i −Ai(x))ψ(x)] .(10.47)

So the gauge field must go as

∂ig −A′ig = − gAi (10.48)

or as

A′i(x) = g(x)Ai(x)g−1(x) + (∂ig(x))g−1(x). (10.49)

So if g(x) = exp(−iθa(x)ta), then a gauge transforms as

A′i(x) = e−iθa(x)ta Ai(x) eiθ

a(x)ta + (∂ie−iθa(x)ta) eiθ

a(x)ta . (10.50)

How does an exponential of a path-ordered, very short line integral of

gauge fields go? We will evaluate how the path-ordered exponential in which

g0 is a coupling constant

P exp(g0Ai(x)dxi

)′= P exp

([g(x)g0Ai(x)g−1(x) + (∂ig(x))g−1(x)

]dxi)

= P exp([g(x)g0Ai(x)g−1(x) + ∂i log g(x)

]dxi)(10.51)

changes under the gauge transformation (10.50) in the limitt dxi → 0. We

find

P exp(g0Ai(x)dxi

)′= P

[g(x)eg0Ai(x)dxig−1(x)

×elog g(x+dxi/2)−log g(x)+log g(x)−log g(x−dxi/2)]

= P[g(x)eg0Ai(x)dxig−1(x)

×g(x+ dxi/2)g−1(x)g(x)g−1(x− dxi/2)]

= g(x+ dxi/2)eg0Ai(x)dxig−1(x− dxi/2).

(10.52)

Putting together a chain of such infinitesimal links, we get

P exp

(∫ x

yg0Ai(x

′)dx′i)′

= g(x)P exp

(∫ x

yg0Ai(x

′)dx′i)g−1(y). (10.53)


In particular, this means that the trace of a closed loop is gauge invariant[TrP exp

(∮g0Ai(x)dxi

)]′= Trg(x)P exp

(∮g0Ai(x)dxi

)g−1(x)

= TrP exp

(∮g0Ai(x)dxi

).

(10.54)

10.5 Pure Gauge Theory on a Lattice

Wilson’s lattice gauge theory is inspired by these last two equations. Another

source of inspiration is the approximation for a loop of tiny area dx ∧ dywhich in the joint limit dx→ 0 and dy → 0 is

W = P exp

(∮g0Ai(x)dxi

)= exp

[g0

(Ay,x −Ax,y + g0[Ax, Ay]

)dxdy

]= exp

(− g0Fxy dxdy

).

(10.55)

To derive this formula, we will ignore the bare coupling constant g0 for the

moment and apply the Baker-Campbell-Hausdorf identity

eA eB = exp

(A+B +

1

2[A,B] +

1

12[A, [A,B]] +

1

12[B, [B,A]] + . . .

).

(10.56)

to the product

W = exp(Ax dx

)exp

(Ay dy +Ay,x dxdy

)× exp

(−Ax dx−Ax,y dxdy

)exp

(−Ay dy

).

(10.57)

We get

W = exp(Ax dx+Ay dy +Ay,x dxdy +

1

2[Ax dx, (Ay +Ay,x dx)dy]

)× exp

(−Ax dx−Ay dy −Ax,y dxdy +

1

2[(Ax +Ax,y dy)dx,Ay dy]

).

(10.58)

Applying again the BCH identity, we get

W = exp[(Ay,x −Ax,y + [Ax, Ay]

)dxdy

]= exp

(Fxy dxdy

), (10.59)

an identity that is the basis of lattice gauge theory.

10.5 Pure Gauge Theory on a Lattice 265

Restoring g0, we divide the trace of W by the dimension n of the matrices

ta and subtract from unity

1− 1

nTr

[P exp

(∮g0Ai(x)dxi

)]= 1− 1

nTr[

exp(g0Fxy dxdy

)]= − 1

nTr[g0Fxy dxdy +

1

2

(g0Fxy dxdy

)2].

(10.60)

The generators of SU(n), SO(n), and Sp(2n) are traceless, so the first term

vanishes, and we get

1− 1

nTr

[P exp

(∮g0Ai(x)dxi

)]= − 1

2nTr(g0Fxy dxdy

)2. (10.61)

Recalling the more explicit form (10.41) of the Faraday matrix, we have

1− 1

nTr(W ) =

g20

2nTr[taF

axy tbF

bxy

(dxdy

)2]=kg2

0

2n

(F axydxdy

)2(10.62)

in which k is the constant of the normalization Tr(tatb) = kδab. For SU(2)

with ta = σa/2 and for SU(3) with ta = λa/2, this constant is k = 1/2.

The Wilson action is a sum over all the smallest squares of the lattice,

called the plaquettes, of the quantity

S2 =n

2kg20

[1− 1

nTr(W )

]=

1

4

(F aij)2a4 (10.63)

in which a is the lattice spacing. The full Wilson action is the sum of

this quantity over all the elementary squares of the lattice and over i, j =

1, 2, 3, 4.

References

Amdahl, David, and Cahill, Kevin. 2016. Path integrals for awkward actions. arXiv,1611.06685.

Davies, P. C. W. 1975. Scalar particle production in Schwarzschild and Rindlermetrics. J. Phys., A8, 609–616.

DeWitt, Bryce S. 1967. Quantum Theory of Gravity. II. The Manifestly CovariantTheory. Phys. Rev., 162(5), 1195–1239.

Faddeev, L. D., and Popov, V. N. 1967. Feynman diagrams for the Yang-Mills field.Phys. Lett. B, 25(1), 29–30.

Hawking, S. W. 1974. Black hole explosions. Nature, 248, 30–31.Kato, T. 1978. Topics in Functional Analysis. Academic Press. Pages 185–195.Rindler, Wolfgang. 2006. Relativity: Special, General, and Cosmological. 2d edn.

Oxford University Press.Srednicki, Mark. 2007. Quantum Field Theory. Cambridge University Press.Trotter, H. F. 1959. On the product of semi-groups of operators. Proc. Ann. Math.,

10, 545–551.Weinberg, Steven. 1995. The Quantum Theory of Fields. Vol. I Foundations.

Cambridge University Press.Weinberg, Steven. 1996. The Quantum Theory of Fields. Vol. II Modern Applica-

tions. Cambridge, UK: Cambridge University Press.Zee, A. 2013. Einstein Gravity in a Nutshell. New Jersey: Princeton University

Press.Zee, Anthony. 2010. Quantum Field Theory in a Nutshell. 2d edn. Princeton

University Press.

Index

complex-variable theoryconformal mapping, 233–235

conformal algebra, 228–241conformal mapping, 233–235conjugate momenta, 81conserved quantities, 79–84Covariant derivatives

in Yang-Mills theory, 126

decuplet of baryon resonances, 129

effective field theories, 209–210energy density, 79–84

Feynman’s propagatoras a Green’s function, 57

fractional linear transformation, 234

gauge theory, 104–156gaussian integrals, 157–158Gell-Mann’s SU(3) matrices, 127Grassmann variables, 192–199groupsSU(3), 127–129SU(3) structure constants, 128and Yang-Mills gauge theory, 104–156Gell-Mann matrices, 127–129Lie groupsSU(3), 127–129SU(3) generators, 127SU(3) structure constants, 128

hamiltonian density of scalar fields, 81

Lagrange’s equationin field theory, 77–79

maximally symmetric spaces, 225–228Mobius transformation, 234

natural units, 178

octet of baryons, 129octet of pseudoscalar mesons, 128

path integrals, 157–205and gaussian integrals, 157–158and lattice gauge theories, 203

and nonabelian gauge theories, 200–204ghosts, 203–204the method of Faddeev and Popov,

200–204and perturbative field theory, 184–205and quantum electrodynamics, 188–192and the Bohm-Aharonov effect, 164–165and The principle of stationary action, 161density operator for a free particle, 170–171density operator for harmonic oscillator,

173–175euclidian, 167–170fermionic, 192–199finite temperature, 167–170for harmonic oscillator in imaginary time,

173–175for partition function, 167–170for the harmonic oscillator, 165–167in finite-temperature field theory, 181–184,

249–253in imaginary time, 167–170in quantum mechanics, 158–167in real time, 158–167Minkowski, 158–167of fields, 177–205of fields in euclidian space, 181–184, 249–253of fields in imaginary time, 181–184,

249–253of fields in real time, 177–181partition function for a free particle,

170–171principle of stationary action

in field theory, 77–79

sign problem, 204symmetries, 79–84

tensorsmaximally symmetric spaces, 225–228

time-ordered products, 175–177, 181, 183–184,251–253

268 Index

of euclidian field operators, 183–184,251–253

of field operators, 181of position operators, 175–177

Yang-Mills theory, 104–156

Documents

Physics 523 & 524: Quantum Field Theory Iquantum.phys.unm.edu/523-18/sw.pdfPhysics 523 & 524: Quantum Field Theory I Kevin Cahill Department of Physics and Astronomy ... 2.8 Feynman