Notes on Weinberg’s Quantum Theory of Fields …...Notes on Weinberg’s Quantum Theory of Fields Volume I: Fundamentals Jimmy Qin Summer 2019 A theorist today is hardly considered

Notes on Weinberg’s Quantum Theory of FieldsVolume I: Fundamentals

Jimmy Qin

Summer 2019

A theorist today is hardly considered respectable if he or she has not introduced at least one newparticle for which there is no experimental evidence.

It is with Isaac Newton that the modern dream of a final theory really begins.

I managed to get a quick PhD, though when I got it I knew almost nothing about physics. But Idid learn one big thing: that no one knows everything, and you don’t have to.

These are notes on Weinberg’s text The Quantum Theory of Fields. I read it after reading throughalmost all of Schwartz’ Quantum Field Theory. Volume I is on theoretical foundations, QED, andloop calculations; volume II is on RG, non-Abelian gauge theory, and symmetry breaking; volumeIII is on supersymmetry. Besides being the definitive textbook for physicists and mathematicianson quantum field theory, Weinberg’s text is also a (perhaps the) cornerstone of QFT studies inthe philosophy of science.

These notes are written in question-answer form instead of the usual narrative form. This isbecause Weinberg is a hard book, and I think it is best to digest it in small bites. Also includedare some general questions and answers which are not in Weinberg but are good to know, andmy solutions to some of the interesting problems. In these notes, I will try to understand the bigpicture and the physics behind the ideas, so many of these questions will be elementary. I will tryto understand the material with as little math as possible. My emphasis will be on giving physicalarguments and intuition rather than rigor.

Question 1. Why is relativistic quantum field theory so hard?

Answer 1. Relativistic quantum field theory, the subject of this three-volume series, is verydifficult. In Novum Organon, the philosopher Francis Bacon postulated that science proceeds bydeveloping axioms. Experiment leads to axioms, and the results of those axioms are benchmarkedagainst further experiment.

• Classical mechanics has three axioms: Newton’s laws.

• Classical electrodynamics has four axioms: Maxwell’s equations.

• Special relativity has two axioms: Einstein’s postulates.

1

Jimmy Qin Notes on Weinberg’s QFT

• Nonrelativistic quantum mechanics has five or so axioms: Born interpretation of matrixelements, Schrodinger equation, etc.

• General relativity has one axiom: Einstein’s equation.

Actually, all of the above subjects have more axioms implicitly assumed. To solve a problem inthe above subjects, you just apply the axioms. However, you would be hard-pressed to state the“axioms of quantum field theory.” Attempts at formulating such an axiomatic development (i.e.by Wightman) have fallen short of encompassing all of quantum field theory. Besides, they arevery far removed from the physics.

While it is true that the axioms of special relativity and quantum mechanics are imported intoquantum field theory, we are missing some other axioms which would tell us how to constructthe theory. For example, the idea of wavefunction must be modified from quantum mechanics toquantum field theory, and there is no axiom for that. There are only very general principles, likeenergy conservation and cluster decomposition. Showing that these principles in fact constrainthe space of reasonable theories to QFT as we know it today is the thesis of Weinberg’s book.

Contents

1 History 5

2 Relativistic Quantum Mechanics 5

2.1 Wigner’s theorem . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 5

2.2 Poincare algebra . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 6

2.3 Wigner little group . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 8

2.4 Discrete symmetries . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 13

2.5 Problems . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 14

3 Scattering Theory 16

3.1 S-matrix . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 16

3.2 Symmetries of the S-matrix . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 18

3.3 Rates and cross-sections . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 20

3.4 Implications of unitarity . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 24

3.5 Resonances . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 26

3.6 Problems . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 28

4 The Cluster Decomposition Principle 30

4.1 Operator decomposition in a†q and aq . . . . . . . . . . . . . . . . . . . . . . . . . 30

4.2 Factorization of the S-matrix . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 33

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4.3 Which Hamiltonians satisfy cluster decomposition? . . . . . . . . . . . . . . . . . 35

4.4 Problems . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 36

5 Quantum Fields and Antiparticles 37

5.1 General free fields . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 38

5.2 Causal scalar fields . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 40

5.3 Causal vector fields . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 41

5.4 Spinor representation of Lorentz group . . . . . . . . . . . . . . . . . . . . . . . . 43

5.5 Causal Dirac field . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 45

5.6 General irreps of the Lorentz group . . . . . . . . . . . . . . . . . . . . . . . . . . 48

5.7 CPT theorem; spin-statistics theorem . . . . . . . . . . . . . . . . . . . . . . . . . 50

5.8 Massless fields . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 50

5.9 Problems . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 52

6 Feynman Rules 53

6.1 Dyson’s derivation in position space . . . . . . . . . . . . . . . . . . . . . . . . . . 53

6.2 Propagators . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 55

6.3 Problems . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 57

7 The Canonical Formalism 59

7.1 Canonical variables . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 59

7.2 Lagrangian formalism . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 61

7.3 Global symmetries and Noether theorem . . . . . . . . . . . . . . . . . . . . . . . 64

7.4 Lorentz invariance of L implies Lorentz invariance of S-matrix . . . . . . . . . . . 65

7.5 Examples of canonical quantization and transition to interaction picture . . . . . . 66

7.6 Problems . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 67

8 Electrodynamics 68

8.1 Gauge invariance . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 69

8.2 Difficulties with quantizing the photon field . . . . . . . . . . . . . . . . . . . . . 71

8.3 QED in the interaction picture; photon propagator . . . . . . . . . . . . . . . . . 72

8.4 p-form gauge fields . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 74

8.5 Problems . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 75

9 Path Integrals 75

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9.1 Hamiltonian version of path-integral formula; S-matrix . . . . . . . . . . . . . . . 75

9.2 Lagrangian version of path-integral formula . . . . . . . . . . . . . . . . . . . . . 76

9.3 Path-integral derivation of Feynman rules . . . . . . . . . . . . . . . . . . . . . . 79

9.4 Path-integral formulation of QED . . . . . . . . . . . . . . . . . . . . . . . . . . . 80

9.5 Problems . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 81

10 Non-perturbative methods 83

10.1 Symmetries . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 84

10.2 Polology . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 85

10.3 Field and mass renormalization . . . . . . . . . . . . . . . . . . . . . . . . . . . . 88

10.4 Renormalized charge; Ward-Takahashi identity . . . . . . . . . . . . . . . . . . . . 91

10.5 Gauge invariance of the S-matrix . . . . . . . . . . . . . . . . . . . . . . . . . . . 94

10.6 More on the LSZ reduction formula . . . . . . . . . . . . . . . . . . . . . . . . . . 96

11 One-loop radiative corrections in QED 97

11.1 Counterterms . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 97

11.2 Vacuum polarization . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 98

11.3 Interlude: Form factors . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 101

11.4 Anomalous magnetic moment . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 103

11.5 Electron propagator . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 105

11.6 Problems . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 106

12 General renormalization theory 106

12.1 Degrees of divergence . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 107

12.2 Cancellation of divergences . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 108

12.3 Is renormalizability necessary? . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 109

12.4 Problems . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 111

13 Infrared effects 112

13.1 Soft photon amplitudes . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 112

13.2 Virtual soft photons . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 114

13.3 Cancellation of divergences . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 116

4


1 History

2 Relativistic Quantum Mechanics

Question 2. What is the difference between relativistic quantum mechanics and quantum fieldtheory?

Answer 2. Relativistic quantum mechanics is still quantum mechanics, in the sense that nothinghas been second-quantized yet. It is the study of how wavefunctions Ψ behave under relativis-tic transforms, such as Lorentz boosts and rotations, and how they behave under more general(anti)unitaries, such as PCT . Quantum field theory must be consistent with relativistic quantummechanics, but now the wavefunctions are second-quantized, and particle number is no longerconserved.

Just like special relativity is the study of how objects appear to observers in different frames,relativistic quantum mechanics is the study of how wavefunctions appear to observers in differentframes. In fact, the normalization N(p) of a wavefunction under Lorentz transform (see subsectionon Wigner little group) can be interepreted as a length-contraction correction to the normalizationof the wavefunction.

2.1 Wigner’s theorem

Question 3. How is relativistic quantum mechanics consistent between observers?

Answer 3. The physically distinguishable quantities in a Hilbert space are called rays: equiva-lence classes of vectors (wavefunctions) that differ by only a phase. If there are two observers Aand B, RQM is consistent if

|〈RA|RAn 〉|2 = |〈RB|RB

n 〉|2,

where R denotes a ray and the superscript denotes which observer is used. This is sloppy notationand I really can only take the inner product of wavefunctions which belong to the rays indicatedabove. Wigner proved that any transformation between A and B observers,

U(RA) = RB,

must be either unitary or antiunitary. The easiest way to think about this is that only thesetransformations conserve probability. A unitary U and antiunitary A are linear and antilinear,respectively, and satisfy

〈Φ|U †Ψ〉 = 〈UΦ|Ψ〉, 〈Φ|A†Ψ〉 = 〈Ψ|UΦ〉.

Question 4. Prove Wigner’s unitary-antiunitary theorem.

Answer 4. The idea of Wigner’s theorem (conservation of probability) is extremely simple. How-ever, there are some awkward steps involved in going between rays and state vectors.

Let observer A be denoted by no prime and observer B be denoted by a prime, ′, and call T thetransformation between A’s rays and B’s rays. We demand

|〈Ψ1|Ψ2〉|2 = |〈Ψ′1|Ψ′2〉|2 for all Ψ1 ∈ R1,Ψ2 ∈ R2,Ψ′1 ∈ TR1,Ψ

′2 ∈ TR2.

5


Suppose |Ψk〉 is an orthonormal basis of state vectors in A’s frame. If the postulates of quantummechanics are valid in B’s frame and the assumption above holds, then |Ψ′k〉 is an orthonormalbasis of state vectors in B’s frame.

Now we must specify how T , which acts on rays, can turn into an action U on state vectors |Ψ〉. Let|Ψ′k〉 = U |Ψk〉. From the above, we find that for general expansions Ψ =

∑k CkΨk,Ψ

′k =

∑k C′kΨ′k,

we have (for example, |〈 1√2(Ψk + Ψl)|Ψ〉|2 = |〈 1√

2(Ψ′k + Ψ′l)|Ψ′〉|2)

|Ck|2 = |C ′k|2, |Ck + Cl|2 = |C ′k + C ′l |2, |Ck + Cl + Cm|2 = |C ′k + C ′l + C ′m|2, · · · for all l,m, n · · ·

The first two equalities give

Re(CkCl

) = Re(C ′kC ′l

), Im(CkCl

) = ±Im(C ′kC ′l

) =⇒ CkCl

=C ′kC ′l

orCkCl

= (C ′kC ′l

)∗.

The first case is unitary and the second is antiunitary. It is not hard to show that the operator

U cannot be a hybrid of unitary CkCl

=C′kC′l

and antiunitary CkCl

= (C′kC′l

)∗ parts; this would violate

|Ck+Cl+Cm|2 = |C ′k+C ′l +C′m|2. Combining this with |Ck|2 = |C ′k|2 gives C ′k = ζCk or C ′k = ζC∗k ,

where ζ is a phase factor independent of k. The phase ζ can be eliminated by rotating each statevector Ψk by ζ. Effectively, ζ is the extra freedom afforded to rays over state vectors.

Finally, we can show U is unitary in the first case and antiunitary in the second case by showing,for example, 〈UΨ|UΦ〉 = 〈Ψ|Φ〉 or 〈Φ|Ψ〉.

Unsurprisingly, this logic can be generalized. See https://arxiv.org/abs/1810.10111.

2.2 Poincare algebra

Doing manipulations on the Poincare algebra is good practice for any kind of algebra, so I willstudy this example closely.

Question 5. Why is the generator ωµν antisymmetric?

Answer 5. For any Lie group with transformations Λµν that preserves a metric, we must have

gµσ = ΛµνΛσρgνρ.

Substituting the infinitesimal transformations Λµν = δµν + ωµν gives ωµν + ωνµ = 0.

This holds for SO(n) and SO(n,m), generally. It holds whenever distances are preserved. Another,equivalent way to think about it is that

v′µ = gµσΛσνvν =⇒ v′µv′µ = gµσgµπΛπρΛ

σνvρvν .

For distances to be preserved, we would like

gµσgµπΛπρΛ

σν = ΛµρΛν

µ = δρν .

Expanding Λµρ = δµρ + ωµρ around the identity gives the antisymmetry relation.

Yet another way to think of it is: let Q(t) be an orthogonal matrix function, QTQ = I, andsuppose Q(t = 0) = I. Evaluating the derivative at t = 0, we find

QTQ+QT Q = 0 =⇒ QT = −Q =⇒ ωT = −ω,

where ω is a generator of the Lie algebra.

6

https://arxiv.org/abs/1810.10111


Question 6. Describe the difference between Lorentz transformation (or Poincare algebra) andthe unitary that implements the transformation on a quantum field.

Answer 6. Infinitesimally, the difference is

Λµν = δµν + ωµν versus U(1 + ω) = 1 +

i

2ωρσJ

ρσ.

Λ acts on x or p and U acts on |Ψ〉. Generally, there exists a unitary U(Λ, a) that describes howthe wavefunction changes under the composition of a Poincare transformation Λ and a translationa. Here are some distinctions:

• ωµν is a matrix acting on 4-vectors (in this sense, it is a matrix). However, Jµν is the operatoracting on wavefunctions Ψ. This is because U acts on a Fock space, so it manifestly mustbe an operator. Λ does not act on the Fock space.

• ωµν is a scalar (well, a matrix made of scalars) and describes the magnitude of rotations andboosts. It is the generalization to 4 × 4 matrices of the 3 × 3 SO(3) algebra. Jµν is not ascalar. It is an operator, though its eigenvalues have physical meaning (for example, angularmomentum or energy).

• However, both ω and J are antisymmetric.

• ω doesn’t care about what kind of particle we are acting on, but J does. For example, theSO(1,3) algebra described by ω describes rotations, and

U ∼ e−iSθ,

where S is the spin. So clearly, the spin is encoded in J . Even though different particles aredescribed by different kinds of J , the J ’s for any particle always satisfy the same commutationrelations, as described below.

Question 7. Construct the infinitesimal unitary operator that transforms wavefunctions underthe Poincare group. Derive the algebra for the generators of this unitary.

Answer 7. Lorentz boosts are characterized by the “angles” ωµν of the rotation matrix and trans-lations are merely characterized by ερ, the shift. To leading order, the change of a wavefunctionunder Lorentz boost and translation must be linear in the angles and displacements,

U(1 + ω, ε) = 1 +i

2ωρσJ

ρσ − iερP ρ + · · · .

If U is unitary, then Jρσ and P ρ are Hermitian. By the composition rule U(Λ2, a2)U(Λ1, a1) =U(Λ1Λ2,Λ2a1 + a2), we find that

U(Λ, a)U(1 + ω, ε)U−1(Λ, a) = U(Λ(1 + ω)Λ−1,Λε− ΛωΛ−1a).

Taking Λ, a to themselves be infinitesimal gives the Lie algebra of the Poincare group. This isjust like how commutators pop up in response functions, χ ∼ [A(t), A(0)], because there is ane−iHtA(0)eiHt structure, which becomes a commutator.

Conclusion: if you want to find the algebra of a Lie group and know the composition of the un-derlying spatial transformations (for example, R12 = R1R2 for rotations), use U,U−1 as bread tomake a sandwich with U as the filling. This will generate a commutator in the case where both Uand U are infinitesimal. An important point is that the structure of the transformations on theFock space (Lie algebra of the Poincare group) depends intimately on the structure of the trans-formations on the underlying physical space (compositions of Lorentz boosts and translations).

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2.3 Wigner little group

Question 8. Why is Wigner little group useful? Why do we need it?

Answer 8. Typically it is most convenient for us to describe particles with momentum (and spin)eigenstates, as in Ψpσ. These are eigenvectors of the momentum operator. The problem is thatΨpσ is not an eigenvector of the Lorentz group (boosts and rotations).

The Wigner little group preserves momentum but not spin projection, although it does preservetotal spin of a particle. It will turn out to be useful later.

Question 9. What is Wigner’s classification?

Answer 9. Wigner’s classification uses two Casimirs (P 2 and W 2, where P µ is the 4-momentumand W µ is the Pauli-Lubjanski pseudovector) to classify the irreducible unitary representations ofthe Lorentz group. These represent particles (or one-particle states) and are infinite-dimensional.See https://en.wikipedia.org/wiki/PauliLubanski_pseudovector.

However, later we will embed these infinite-dimensional unitary representations into finite-dimensionalnonunitary representations.

Question 10. How do one-particle states Ψpσ transform under the Lorentz group?

Answer 10. Recall that the Lorentz group is basically a glorified rotation group. What we arelooking for here are essentially glorified Clebsch-Gordan coefficients 〈Λp, σ′|pσ〉,

U(Λ)|pσ〉 =∑σ′

〈Λp, σ′|pσ〉|Λp, σ′〉.

So generally, states labeled with momentum and spin will go to a certain momentum, but theirspin will split. The magnitude of the spin vector, however, is preserved.

Question 11. What is the method of induced representations?

Answer 11. The Clebsch-Gordan coefficients from above are different for each representation ofthe Poincare group. The claim of induced representations is: the Clebsch-Gordan coefficients forthe entire Lorentz group is uniquely determined by the representation of the little group.

This is not hard to show for the example of the Lorentz group. We would like to find the Clebsch-Gordan coefficients Cσ′σ(Λ, p) in the Lorentz-boost formula

U(Λ)Ψpσ =∑σ′

Cσ′σ(Λ, p)ΨΛp,σ′ .

Choose the reference momentum kµ, and define the states Ψpσ by

Ψpσ = N(p)U(L(p))Ψkσ.

Here, N(p) is a normalization factor and Lµνkν = pµ. It’s not obvious why the spin is conserved in

this definition; it doesn’t make physical sense to me. In other words, it doesn’t look like a physicallyreasonable definition.Let W be a Lorentz transformation in the little group, W µ

ν kν = kµ. Then

U(W )Ψkσ =∑σ′

Dσ′σ(W )Ψkσ′ .

8

https://en.wikipedia.org/wiki/Pauli–Lubanski_pseudovector


WritingU(Λ)Ψpσ = N(p)U(L(Λp))U(L−1(Λp)ΛL(p))Ψkσ

and setting W (Λ, p) = L−1(Λp)ΛL(p) gives

U(Λ)Ψpσ =N(p)

N(Λp)

∑σ′

Dσσ′(W (Λ, p))ΨΛp,σ′ .

Therefore, Cσ′σ(Λ, p) =N(p)

N(Λp)Dσ′σ′(W (Λ, p)) . W (Λ, p) is called the Wigner rotation.

How can we think about this? Consider the little group of a massive particle, SO(3). This isa subgroup of the full inhomogeneous Lorentz group, SO(1, 3)+; in particular, SO(3) encodesthe behavior of the wavefunction under rotations. The claim of induced representations is thatif I tell you how the wavefunction behaves under rotations, you will also know how it behavesunder boosts, and under combinations of boosts and rotations. This makes sense if you think ofboth the little-group transformations and the Lorentz transformations of the spin-j particle Ψ as(2j + 1) × (2j + 1) matrices, and the wavefunction as Ψp = (Ψp,−j,Ψp,−j+1, · · · ,Ψp,j)

T . Clearly,different spins have differently-sized matrices. You can also think about it like this: the Lorentzgroup is split into two parts:

Lorentz group = little group⊗ pure Lorentz boosts from kµ.

The latter is already specified by the definition Ψpσ = N(p)U(L(p))Ψkσ.

Question 12. In spirit, how is the above transformation law different for massless particles?

Answer 12. Massless particles travel at the speed of light, so we cannot ever “pass” them. There-fore, their helicity, σ, (eigenvalue of operator J3) is conserved. As we show below, the eigenvaluesa, b of the generators A,B of the ISO(2) little group must be equal to zero. The Wigner rotationtherefore only cares about the operator J3, and we can write

U(Λ)Ψpσ =

√(Λp)0

p0eiσθ(Λ,p)ΨΛp,σ.

Here,W (Λ, p) = L−1(Λp)ΛL(p) = S(α(Λ, p)β(Λ, p))R(θ(Λ, p)).

Question 13. How does Ψpσ = N(p)U(L(p))Ψkσ encode length contraction?

Answer 13. The normalization is

N(p) =

√k0

p0=

√1

γ.

This makes sense if you think of N(p)2d3~p as the number of particles. Because d3~pE

is Lorentz-invariant, we demand that N(p)2E be Lorentz invariant. E ∼ γ =⇒ N(p)2 ∼ γ−1.

This is essentially a length contraction argument except we are contracting the “length” of the~p-vector instead of the ~x-vector.

9


Question 14. We know that Λp = ΛL(p)k = L(Λp)k. So, is it true that L(Λp) = ΛL(p)?

Answer 14. Nope. We have to choose a standard Lorentz transformation convention, which isthe same for both L(p) and L(Λp). This is because the matrix problem

Λk = p

does not return a unique solution for Λ. Examples of standard Lorentz transformations are givenby eqs. (2.5.24) and (2.5.44) for massive and massless particles, respectively.

Question 15. The little groups of massive and massless particles are SO(3) and ISO(2), respec-tively. How can we think about them?

Answer 15. In fact, SO(3) and ISO(2) are nearly the same group. They both have three gen-erators. As in Wigner’s original paper, a transform to lightcone coordinates shows that ISO(2)leaves 3 degrees of freedom empty, even though the reference momentum (k, 0, 0, k) has only twononzero entries.

A nice way to think about them is to analogize from the regular rotation group. If we start withSO(3) and break the symmetry along z = (0, 0, 1), we are left with two degrees of freedom. If westart with SO(3) and break the symmetry along (1/

√2, 1/√

2, 0), we are still left with two degreesof freedom, even though (1/

√2, 1/√

2, 0) has only one nonzero entry.

Question 16. How is the little group of massless particle related to gauge invariance?

Answer 16. Weinberg (p. 71) shows that the unitary which encodes the little group transformson wavefunctions Ψ corresponding to massless particles is

U(W (θ, α, β)) = 1 + iαA+ iβB + iθJ3.

Here, A = −J13 + J10, B = −J23 + J20, and J3 = J12. Because [A,B] = 0, the wavefunction canbe diagonalized in terms of eigenvalues of A,B. This turns out to give the U(1) gauge symmetryof QED. “If we find one set of nonzero eigenvalues of A,B, then we find a whole continuum.”Because [J3, A] = iB and [J3, B] = −iA, the operator J3 rotates the wavefunction in the ABplane. We find that

AΨθkab = (a cos θ − b sin θ)Ψθ

kab, BΨθkab = (a sin θ + b cos θ)Ψθ

kab, where Ψθkab = U−1(R(θ))Ψkab.

There is no such continuum of eigenstates observed in nature. So, we require the eigenvalues tobe a = b = 0. The state is merely labeled by its J3 eigenvalue, which we call σ, or the helicity.

Question 17. How does the massless particle little group show that the helicity of the photonmust be σ = ±1?

Answer 17. Main point: since ISO(2) is a non-compact Lie group, the only finite dimensionalunitary representation of ISO(2) is one-dimensional. So each photon can exist only in one helicityeigenstate. Two photons of σ = 1 and σ = −1 can actually be thought of as two particles of“different species,” which are related only by P symmetry, or T symmetry. (However, massiveparticles of given helicity, such as a massive Weyl spinor, cannot be regarded as particles of‘different species” because a boost past the massive spinor changes its helicity.)

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This derivation is based on https://www.physicsforums.com/threads/reduced-density-matrices-and-lorentz-transformation.

315387/#post-2223048.

This is mostly mathematical, but hopefully we can also extract some physical intuition. Let’s firststudy why the Pauli-Lubjanski pseudovector is relevant for labeling the irreps of the Lorentz group.The punchline is that the Pauli-Lubjanski psuedovector arises from the stabilizer, or Wigner littlegroup.

Let the Lorentz transform on 4-vectors (such as momenta) be q′µ = Λµνq

ν , where Λµν = δµν + ωµν .

The Wigner little group is built from transforms such that

ωµν qν = 0 =⇒ ωµν = εµνπηq

πnη,

where nη is an arbitrary 4-vector. A typical element of the little group for momentum qµ cantherefore be written as

Uq(n, a) = eiaµPµ− i

2εµνπηqπnηJµν .

This motivates the introduction of the Pauli-Lubjanski pseudovector, defined as

W µ =1

2εµνπηJνπPη.

It satisfies the properties

0 = WµPµ, 0 = [Wµ, Pν ], [M

µν ,W η] = iηµηW ν − iηνηW µ, [W µ,W ν ] = iεµνηπWηPπ.

Thus, elements of the little group for momentum qµ can be written in terms of the Pauli-Lubjanskipseudovector as

Uq(n, a) = eiaµqµ

einµWµ

.

Obviously, this follows because (1) P µ acts like qµ on eigenstates |q〉 (2) W µ and P ν commute, soBaker-Campbell-Hausdorff formula implies we can split the exponential. This suggests that theinhomogeneous Lorentz algebra can be characterized by these two parts of the unitary:

• The first (inhomogeneous) part is the Casimir P 2. This describes the eiaµqµ

part of theunitary. It is not really part of the homogeneous Lorentz group, because homogeneousLorentz group does not include translations.

• The second (homogeneous) part is the Casimir W 2. This describes the einµWµ

part of theunitary. In fact, the relation [W µ,W ν ] = iεµνηπWηPπ = iεµνηπWηqπ for fixed momentum qµ

implies that the components of the Pauli-Lubjanski pseudovector form a Lie algebra whenrestricted to fixed momentum. This will turn out to be the Lie algebra of the Wigner littlegroup.

Because einµWµ

simply rotates between different states with momentum qµ, the spin properties ofthe particles are determined by the Pauli-Lubjanski pseudovector. Because we do not see particlesof infinite spin, we demand on physical grounds to study only the finite-dimensional representationsof the Wigner little group. Because the rest of the inhomogeneous Lorentz group is labeled by P 2,we will study the cases P 2 = m2 > 0 and also P 2 = 0. You could raise the question of why we canjust choose to study the special momentum q1 = (m,~0) rather than q2 = (γm, ~p). The answer isthat the representation of the Lorentz group is labeled by the Casimirs P 2 and W 2. Because q1

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https://www.physicsforums.com/threads/reduced-density-matrices-and-lorentz-transformation.315387/#post-2223048

https://www.physicsforums.com/threads/reduced-density-matrices-and-lorentz-transformation.315387/#post-2223048


and q2 have the same value of P 2, they give the same representation of the Lorentz group. So, wemay as well go with the simplest choice. Another way to see this is to note that substituting

W µ =1

2εµνπηJνπPη into [W µ,W ν ] = iεµνηπWηPπ

gives a factor of P 2 on both sides of the Lie bracket equation.

Now we proceed as before.

• Massive little group: Choose qµ = (m,~0). We find that

W0 = 0,Wi = mJi =⇒ [W i,W j] = iεijk0Wkq0 = imεijkWk.

This is the familiar SU(2) or SO(3) Lie algebra. It is compact, so there are nontrivialfinite-dimensional representations.

• Massless little group: Choose qµ = (E, 0, 0, E). The Pauli-Lubjanski pseudovector turns outto be

W µ = (J12, J32 + J02, J31 + J01, J12) = (J12, E1, E2, J12).

Therefore, the Casimir is given by

W µWµ = E21 + E2

2 .

So mathematically, P 2 = 0 does not imply W 2 = 0. However, we will still want W 2 = 0for physical reasons: if W 2 > 0, then the representation of ISO(2) is infinite-dimensional,and the particles are interpreted to have infinite spin. (Why is the representation infinite-dimensional for W 2 = E2

1 + E22 > 0? Note that we can choose E1 and E2 from a circle

of radius√W 2. Therefore, there is a continuum of eigenstates and the irrep is infinite-

dimensional. The irreps for SO(3) are finite-dimensional because they are labeled only byJ2 and not J2

1 + J22 or something.) Therefore, in the massless case we have

Uq(a, n) = eiaµqµ

eiθJ12 ,

and for particles with finite spin, ISO(2) is purposely “broken by hand” to U(1). If we labelthe eigenvalue

J12|q, h〉 = h|q, h〉,we find that h = 0,±1

2,±1, · · · by the θ → θ + 2π invariance. h is called the helicity

quantum number.

In fact, the relationW µ = hP µ

is true in all reference frames, even though we derived it for a special choice of momentum.This is because both sides are (pseudo)vectors which transform the same way under Lorentzgroup.

Conclusion: it is perhaps dangerous to think of the photon as having “only two helicity eigen-states,” because the helicity eigenstate σ = 1 is from a completely different irrep than σ = −1. Abetter way to say it is: the physical requirement that particles have finite spin leads us to labelmassless particles by the helicity quantum numbers h = 0,±1

2,±1, · · · . Because the helicities

h = ±σ are related by P or T reversal, we have to group them together when doing physics. Theyare not, however, considered to be “different states of the same particle,” but rather “differentparticles which are intimately related.”

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2.4 Discrete symmetries

Question 18. Describe the four parts of the Lorentz group.

Answer 18. The proper orthochronous Lorentz group is that subgroup which has det Λ = 1and Λ0

0 ≥ 1. (The choices are det Λ = ±1 and (Λ00)2 = 1+Λ0

iΛ0i .) We denote the parity and time

reversal operators on points in spacetime by P,T , respectively. We denote the induced parityand time reversal operators on wavefunctions by P,T, respectively. In other words,

P = U(P, 0),T = U(T , 0).

The other three parts of the Lorentz group are obtained by acting on the proper orthochronousLorentz group with P,T ,PT .

Question 19. What does it mean, mathematically, for a wavefunction to be invariant under parityor time symmetries?

Answer 19. To a good approximation (except for parity violation and time-reversal violations,which occur only at high energy scale), the regular composition rules for Poincare transforms arealso satisfied for these discrete symmetry transforms:

PU(Λ, a)P−1 = U(PΛP−1,Pa),TU(Λ, a)T−1 = U(T ΛT −1,T a).

Just as before, we can find the transformation rules for Jµν , P µ under the parity and time reversalsymmetries by taking Λ, a to be infinitesimal. We decide that P must be unitary and T must beantiunitary by observing how the energy operator H := P 0 transforms,

PiHP−1 = iH,TiHT−1 = −iH,

and demanding there be no states of negative energy.

However, energy is preserved under both P and T,

PHP−1 = H,THT−1 = H,

and the distinction is just the unitary-antiunitary difference.

Question 20. What do P and T do to massive particle wavefunctions Ψpσ?

Answer 20. There is a nice way to do this, which uses the fact that 0 is its own negative. Letk be the reference momentum, ~PΨkσ = 0. 0 is invariant under P and P doesn’t change angularmomentum, so

PΨkσ = ηΨkσ,

where η is a σ-independent phase (you can show this, for example, by hopping up and down σwith the J± operator), or by using the same phase argument in the proof of Wigner’s unitary-antiunitary theorem, since we know P is unitary. Because PL(p)P−1 = L(Pp), we find

PΨpσ = PN(p)U(L(p))Ψkσ = ηΨPp,σ.

This is modified for T, which flips angular momentum. In this case, T is antiunitary and the phaseis no longer σ-independent. The result follows from the surprising fact that T L(p)T −1 = L(Pp)and is

TΨpσ = ζ(−1)j−σΨPp,−σ.

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Question 21. How to understand PL(~β)P−1 = L(P~β) and T L(~β)T −1 = L(P~β)?

Answer 21. I introduced the vector β so it’s easier to think about the Lorentz boost along β.As shown in a previous question, energy H is preserved under both parity and time. Rather thanmultiplying these relations out, you can think of them as follows. Introduce a four-vector (E, ~q)on which the Lorentz transforms will act. Note that

L(P~β)(E, ~q) = L(−~β)(E, ~q).

We havePL(~β)P−1(E, ~q) = PL(~β)(E,−~q).

This means that P will reverse the sign of every unpaired vector in the result L(~β)(E,−~q). That

will merely give L(−~β)(E, ~q). Similarly,

T L(~β)T −1(E, ~q) = T L(~β)(E,−~q).

2.5 Problems

Question 22. Suppose that observer O sees a W boson (spin-1 and m > 0) with momentum ~pin the y-direction and spin z-component σ. A second observer O′ moves relative to the first withvelocity ~v in the z-direction. How does O′ describe the W state?

Answer 22. This is just like in classical mechanics, where a special relativity problem asks youto find what person A reads on person B’s clock, or how long person B’s hot dog is. Except a lotharder now, because there is not just one state - the answer is a superposition of different states.

We will use

U(Λ)Ψpσ =

√(Λp)0

p0

∑σ′

Dσσ′(W (Λ, p))ΨΛp,σ′

and also the standard boost L(p) given in (2.5.24), along with W (Λ, p) = L−1(Λp)ΛL(p). Theexplicit form of the boost along z gives

pµ = (0, |~p|, 0,√~p2 +m2), (Λp)µ = (0, |~p|,−γβ

√~p2 +m2, γ

√~p2 +m2).

Above, β := |~v|/c and γ := (1 − β)−1/2. The rest of the solution is very long and probablynot worth it. Here is an outline: substitute pµ and (Λp)µ into (2.5.24) to find L(p) and L(Λp).Λ, the Lorentz boost, is already known. Then use Mathematica or something to find the inverseL(Λp)−1, and multiply to find W (Λ, p). Finally, substitute this into the explicit form (2.5.20-22) of

D(j)σ′σ(W (Λ, p)), which gives the Clebsch-Gordan coefficients. The spin-projection is still considered

to be along the z-axis, but now there are multiple spin-projections that could be observed.

Question 23. Suppose that observer O sees a photon with momentum ~p in the y-direction andpolarization vector in the z-direction. A second observer O′ moves relative to the first with velocity~v in the z-direction. How does O′ describe the same photon?

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Answer 23. This is like the previous problem but easier, because we know the helicity is conserved.So, O′ sees the photon having spin-1 in the z-direction.

Since we chose the helicity in the z-direction, we must choose the reference momentum k to be inthe z-direction,

k = (E, 0, 0, E).

This is because of how helicity is defined. To find the Wigner rotation, use the explicit form of Λto compute that

p = (|~p|, 0, |~p|, 0),Λp = (γ|~p|, 0, |~p|,−γβ|~p|).Note that generally, E 6= |~p|. For example, photons can be red-shifted.

Now, use eq. (2.5.44) in Weinberg.

• For p, we have

u =|~p|E, p = (0, 1, 0).

Inserting these into (2.5.45) and using the canonical 2D rotation between y and z, gives (inWeinberg notation, where time is the last entry in a 4-vector):

L(p) =

1 0 0 0

0 0 u2+12u

u2−12u

0 −1 0 0

0 0 u2−1u2

u2+12u

• For Λp, we have

u =γ|~p|E

, p = (0, γ−1,−β).

The corresponding L(Λp) turns out to be a generalization of the above,

L(Λp) =

1 0 0 0

0 −β u2+12u

γ u2−12u

γ

0 −γ −u2+12u

β −u2−12u

β

0 0 u2−1u2

u2+12u

You can use mathematica to compute the inverse, L−1(Λp). The amazing (disappointing?) resultis that

W (Λ, p) = L−1(Λp)ΛL(p) =

(1 A

AT B

).

I used block-diagonal notation. According to (2.5.32), this means that θ = 0, even though α, β 6= 0.So, there is no phase picked up, and

U(Λ)Ψpσ =√γΨΛp,σ .

Why is there no relative phase θ? I guess it is because θ is the angle connected to rotations aboutthe z-axis. However, in constructing the states Ψpσ and ΨΛp,σ, we never had to rotate about thez axis, only around the x-axis. So there is no relative phase picked up, although there may be arelative phase if we chose a different momentum.

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3 Scattering Theory

3.1 S-matrix

Question 24. Mathematically, what is the difference between an “in” and “out” state?

Answer 24. The sign of ±iε on the denominator. The “in” state is called Ψ+ and the “out” stateis called Ψ−. More on this later.

Question 25. What are some properties of “in” and “out” states?

Answer 25. “In” and “out” states are taking to be non-interacting, in the sense that the wave-function is factorizable into the product of wavefunctions describing the individual particles. Thecorresponding Lorentz transformation law is

U(Λ, a)Ψα = e−iaµ(pµ1 +pµ2 +··· )

√(Λp1)0(Λp2)0 · · ·

p01p

02 · · ·

∑σ′1σ′2···

D(j1)

σ1σ′1(W (Λ, p1))D

(j1)

σ1σ′1(W (Λ, p1)) · · ·Ψα′ ,

where α = p1σ1; p2σ2; · · · , α′ = Λp1σ1; Λp2σ2; · · ·

Some more properties:

1. “In” and “out” states are Heisenberg-picture states. They describe the whole history of asystem of particles, and hence do not evolve in time by themselves.

2. “In” and “out” states Ψ±α could be called a “smooth adiabatic evolution” of the non-interacting states Φ±α , through the unitary Ω(τ) = ei

∫dτHe−i

∫dτH0 .

Ψ±α = Ω(∓∞)Φα.

3. For a given set of particles and momenta, the “in” and “out” states are the only possiblestates. This is because there are only two ways to shift the contour, ±iε. More physically,we need |τ | → ∞, and there are two possibilities: τ → −∞ and τ → ∞. Strictly speakingthis is a little misleading, because in a scattering process the “in” and “out” states are notin the same state α, but the point remains.

Question 26. What is the formal definition of “in” and “out” state?

Answer 26. Let H = H0 + V , and let these have eigenfunctions

H0Φα = EαΦα,HΨα = EαΨα.

Let g(α) ∈ C be a smooth “envelope” function that tells us which states are in our state vector.Define the states

Ψ±g =

∫dαe−iEαtg(α)Ψ±α ,Φg =

∫dαe−iEαtg(α)Φα.

IfΨ±g → Φg for t→ ±∞

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for every choice of g(α), then we say Ψ±α is an “in” or “out” state. The introduction of the arbitraryenvelope g(α) is basically to make sure the phases and magnitudes of Ψ±α and Φα match up, andmatch up between different α’s as well. (For example, we can always multiply an eigenvector byan arbitrary complex number, and it is still a good eigenvector. This construction removes thatfreedom.)

The eigenvalue equation(H0 − Eα)Ψ±α = −VΨ±α

is not exactly invertible unless we perturb the contour. We can also add a solution of the homo-geneous equation, (H0 − Eα)Φα = 0. The result is

Ψ±α = Φα + (Eα −H0 ± iε)−1VΨ±α ,

which gives the Lippmann-Schwinger equation

Ψ±α = Φα +

∫dβ

T±βαΦβ

Eα − Eβ ± iε, where T±βα = 〈Φβ,VΨ±α 〉.

Question 27. What is the S-matrix, and what is the S-operator? How to interpret them?

Answer 27. The S-matrix is the overlap between the “in” state Ψ+α and the “out” state Ψ−β ,

Sβα = 〈Ψ−β |Ψ+α 〉.

The rate of reaction is proportional to |Sβα−δ(α−β)|2, since Sβα = δ(α−β) in the noninteractingcase. Sβα is a unitary matrix,

∫dβSγβS

∗αβ = δ(γ − α).

The S-operator is constructed to give the same matrix elements for non-interacting states as theS-matrix does for interacting states. It is S in the expression

Sβα := 〈Φβ|SΦα〉.

Substituting Ψ±α = Ω(∓∞)Φα gives

S = Ω(∞)†Ω(−∞) = U(∞,−∞).

It may seem funny that the definition Sβα = 〈Ψ−β |Ψ+α 〉 of S-matrix has no time-evolution inside.

This makes sense if we think about it in Heisenberg picture, where the wavefunctions are constantin time. In a sense, the time-evolution is already inside Ψ+

α , and we just want to see in which“out” state it will end up in. Another way to think about it is that the effect of the potential,V , is already encoded in Ψ+

α and Ψ−β . Therefore, the S-matrix already accounts for the effect ofthe scattering potential on incoming particles. However, Φα and Φβ do not feel the scatteringpotential, which is why we have to insert the S-operator into the matrix element by hand.

Question 28. Prove the following important decomposition of the S-matrix and interpret it phys-ically,

Sβα = δ(β − α)− 2πiδ(Eα − Eβ)T+βα, where T+

βα = 〈Φβ|VΨ+α 〉.

Answer 28. Before the math, let’s think about it:

• Sβα ∝ δ(Eα − Eβ). This means that energy is conserved.

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• The reason Sβα could possibly be different from δ(β−α) is that we are comparing apples tooranges: the “in” state Ψ+

α and the “out” state Ψ−β are in different bases of the same Hilbertspace. In order to compute the S-matrix element, we must switch to a common basis. Inthe calculation below, I will express both of these in the Φα basis.

• Because Tβα ∝ V , Sβα = δ(β − α) in the case that V = 0.

• How to interpret the scattering term? Weinberg shows (pg 115) that

Ψ+α (t)→ e−iEαt(Φα − 2πi

∫dβδ(Eα − Eβ)T+

βαΦβ).

So, it’s fair to say that T+βα encodes the part of the “in” α wavefunction that doesn’t end

up in the “out” α wavefunction after time-evolution. We know this because the “out”wavefunction Ψ−β exactly matches up with the non-interacting wavefunction Φβ at t → ∞.In some sense,

Ψ+α = Φα +

∫dγ

T+γαΦγ

Eα − Eγ + iε

already has the structure of the S-matrix above. Even though every state in this decomposi-tion is at the same energy (and hence evolves at the same rate), Ψ+

α simply contains differentstates than Ψ−α does. Ultimately, we can trace this back to the fact that Ω(∞) 6= Ω(−∞).

I will give a different proof from Weinberg. Use

Ψ+α = Φα +

∫dγ

T+γαΦγ

Eα − Eγ + iε,Ψ−β = Φβ +

∫dη

T−ηβΦη

Eβ − Eη − iε.

Just smash them together and use the property that 〈Φβ,Φγ〉 = δβγ, and that (E ± iε)−1 =1EP ∓ iπδ(E). The result is

〈Ψ−β |Ψ+α 〉 = δαβ − iπδ(Eα − Eβ)T+

βα + iπδ(Eα − Eβ)(T−αβ)∗,

and we can argue T+βα = −(T−βα)∗ by sending both Ψ+

α and Ψ−β to t =∞ or t = −∞. If this soundsweird, it happens to be exactly what Weinberg does on page 114...

This gives the result. Weinberg does it by evolving Ψ+α to t = ∞ and expressing Ψ+

α (t → ∞) ina basis of the “out” states. This is why he ended up with T+

βα. You could just as well evolve Ψ−βbackwards to t = −∞ and express it in a basis of the “in” states. Then you would get (T−βα)∗. Insome sense, what I did above is exactly intermediate between these two extremes.

3.2 Symmetries of the S-matrix

Question 29. What is meant by “Lorentz invariance of the S-matrix,” and when does it hold?

Answer 29. A theory is Lorentz invariant if there exists a unitary U(Λ, a) acting on both the “in”and “out” states as

Sβα = 〈Ψ−β |Ψ+α 〉 = 〈U(Λ, a)Ψ−β |U(Λ, a)Ψ+

α 〉.However, such a unitary U(Λ, a) exists only for certain Hamiltonians. The problem is that “in”and “out” states are differently defined, so we cannot be sure that U(Λ, a) acts in the same way on

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both of them. We can solve this problem if we express both the “in” and “out” states in the samebasis, so there is no ambiguity. We use the non-interacting basis of wavefunctions, Φα, and letU0(Λ, a) induce the corresponding transformation on the non-interacting states Φ. Invariance issatisfied if

Sβα = 〈Φβ|SΦα〉 = 〈U0(Λ, a)Φβ|SU0(Λ, a)Φα〉 =⇒ [U0(Λ, a), S] = 0.

U0 is the unitary that implements the Lorentz group transforms under the non-interacting Hamil-tonian, H0. All of its generators must commute with the S-operator,

[H0, S] = [~P0, S] = [ ~J0, S] = [ ~K0, S] = 0.

Weinberg (3.3.19) shows that an interacting theory is also Lorentz-invariant if

[V , ~P0] = [V , ~J0] = 0.

There is a discussion in the book of which of the above conditions are more easily satisfied than theothers. It is very subtle so I will not include it here. The main takeaway is that Lorentz invarianceis a property of a theory, not something that comes automatically by itself for any theory.

Question 30. How does the S-matrix encode conservation of 4-momentum?

Answer 30. Use Sβα = 〈U(Λ, a)Ψ−β |U(Λ, a)Ψ+α 〉 and

U(Λ, a)Ψα = e−iaµ(pµ1 +pµ2 +··· )

√(Λp1)0(Λp2)0 · · ·

p01p

02 · · ·

∑σ′1σ′2···

D(j1)

σ1σ′1(W (Λ, p1))D

(j1)

σ1σ′1(W (Λ, p1)) · · ·Ψα′ .

The result is thatSβα ∝ eiaµΛµν (pν1+pν2+···−p′ν1 −p′ν2 −··· ).

But Sβα has no aµ-dependence. Therefore,∑i

pνi =∑j

p′νj .

We typically writeSβα = δ(β − α)− 2πiδ(4)(pβ − pα)Mβα.

Physically, the way to say this is to argue that Sβα is a translationally-invariant scalar and combinewith S ∼ eipx.

Question 31. What are “internal symmetries of the S-matrix?”

Answer 31. It may be true that the wavefunctions Ψ+α are not invariant under a symmetry, but

the S-matrix is. The spirit is similar to the treatment of Lorentz invariance. Basically, if thereis an internal symmetry T , we would like there to be a unitary U(T ) which acts the same wayon both “in” and “out” states. Similarly to the derivation above, we introduce U0(T ) to act onnon-interacting states, and U(T ) is a symmetry of the S-matrix if

[U0(T ), S] = 0 =⇒ [U0(T ),H0] = [U0(T ),V ] = 0.

Because of the decomposition of “in” and “out” states into superpositions of non-interactingstates, we may take U(T ) = U0(T ). This is not true for the Lorentz symmetry, because the boost

generators ~K have to be modified for the interacting case. The argument is a little subtle (pg119).

Here are some examples of internal symmetries.

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• Charge conservation: Let U(T (θ))ΨQ = eiQθΨQ. T (θ) is the U(1) symmetry of, say,Maxwell’s electrodynamics, and Q is the charge of the wavefunction ΨQ. When this isimplemented on an S-matrix that is invariant under Ψ→ U(T (θ))ΨQ, the conclusion is that

q1 + q2 + · · · = q′1 + q′2 + · · · ,

so charge is conserved. This is exactly the same argument we used to show that momentumis conserved, above.

• Time reversal: Time reversal is antiunitary, so we have to switch the order of entries in theinner product which defines the S-matrix. More physically, time reversal switches “in” stateswith “out” states :

〈Ψ−β |Ψ+α 〉 = 〈TΨ+

α |TΨ−β 〉 =⇒ Sβα = ST α,T β.

We can show that the time reversal T0 in T0Φα = ΦT α does, in fact, obey the commutations[T,H0] = [T,V ] = 0. For example, starting from Ψ±α = Ω(∓∞)Φα and applying T from theleft, we get

Ψ∓T α = TΩ(∓∞)Φα = [T,Ω(∓∞)]Φα + Ω(∓∞)(TΦα) = [T,Ω(∓∞)]Φα + Ψ∓T α.

Therefore, [T,Ω(∓∞)] = 0. Another way to think of it is to use the fact that T is antiunitary,so

TΩ(−∞)Φα = Ω(∞)ΦT α, or in other words TΨ+α = Ψ−T α.

• Parity: We can show [P,H0] = [P,V ] = 0 merely by looking at explicit forms of H0,V andseeing if they are invariant under ~x→ −~x.

• CPT invariance: Because CPT is antiunitary, the same arguments as for T-invariance ofthe S-matrix lead to

Sβα = SC PT α,C PT β.

This gives some meaning to “antiparticles are particles moving backwards in time,” becausethe order of α and β had to be switched (by antiunitarity). In particular,

Mαα = MC PT α,C PT α.

By the optical theorem (see next section), the total reaction rate from an initial state of par-ticles is the same as the total reaction rate from the same initial state, but with antiparticlesof reversed spin. For the case of one particle, this means that unstable particles have exactlythe same lifetime as their antiparticles.

3.3 Rates and cross-sections

This section is more illuminating than Schwartz, but the treatments are a bit different in inter-pretation, so it would be nice to read both together.

Question 32. The “master formula” of relativistic scattering theory is

dΓ(α→ [β, β + dβ]) =(2π)3Nα−2

V Nα−1δ(4)(pβ − pα)|Mβα|2dβ.

I will not include the derivation because Weinberg’s is exceedingly easy to follow. Interpret thisformula and explain what everything means.

20


Answer 32. Here are some insights into the above formula.

• We use Γ, a reaction rate, because the production of particles in the range [β, β + dβ] isexpected to be linear with time.

• The above formula only works when we ignore the δβα part of the S-matrix. In other words,it accounts only for the cases where a nontrivial reaction occurred. That is why we got|Mβα|2.

• The factor of V Nα−1 on the denominator makes sense: If we only have Nα = 1 particleparticipating in the reaction, then it shouldn’t matter how large the volume of the containeris. For Nα > 1, we add an extra V −1 for every reactant particle, because they have to bumpinto each other in the container.

• Why the 3Nα − 2 factor? This is from the normalization of the wavefunctions to haveprobability unity of any given particle being in the box, and also the time normalization.Essentially, we start with (2π)3(Nα−1), and then we multiply by (2π)2 from the definition ofMβα in terms of Sβα, and then divide by 2π because of the time delta-function. Not veryilluminating.

• How to interpret the phase-space factor δ(4)(pβ−pα)dβ? Schwartz gives a slightly differentformula for the scattering rate, because of his normalizations of the wavefunctions, and callshis phase-space factor the Lorentz-invariant phase space factor, dΠLIPS. Weinberg’s factor, ofcourse, is not Lorentz invariant because of length contraction and stuff, but he does mentiondΠLIPS. See pg. 67 and 138.

A good way to think about this is:∫

d3~p2E

is the manifestly Lorentz-invariant condition thatthe external particle be on mass-shell,∫

d4pδ(p2 −m2) =

∫d3~p

2E.

Question 33. How can we get a scattering cross-section σ from the scattering rate Γ?

Answer 33. Typically, people can’t smash more than two particles together at once. Thus, thescattering cross-section is defined as the rate per flux,

dσ(α→ [β, β + dβ]) =dΓ(α→ [β, β + dβ])

Φα

,

where Φα = uα/V is the flux. Here, uα is the relative velocity of the two projectiles. In generalframes, it turns out to be

uα =1

E1E2

√(p1 · p2)2 − (m1m2)2.

This is a definition so that σ is a Lorentz-invariant scalar. See Weinberg pg. 138.

Question 34. Describe old-fashioned perturbation theory.

Answer 34. We want to calculate T+βα, and to do so we can use Lippmann-Schwinger equation.

Multiplying on the left with 〈Φβ|V and defining Vβα = 〈Φβ|V|Φα〉 gives

T+βα = Vβα +

∫dγ

VβγVγαEα − Eγ + iε

+

∫dγdγ′

VβγVγγ′Vγ′α(Eα − Eγ + iε)(Eα − E ′γ + iε)

+ · · · .

21


There is already a δ(Eβ −Eα) enforced in the definition of Sβα. This is the same as second-orderperturbation theory in nonrelativistic quantum mechanics. In that case, you just take α = β andthat gives you the shift in energy to second order in V .

Question 35. Explain why time-dependent perturbation theory is said to be manifestly Lorentz-invariant. Why is old-fashioned perturbation theory not manifestly Lorentz-invariant?

Answer 35. Time-dependent perturbation theory is said to be manifestly Lorentz-invariant be-cause the S-operator can be written

S = 1 +∞∑n=1

(−i)n

n!

∫d4x1 · · · d4xnTU (x1) · · ·U (xn),

where U (xi) is the interaction density, V (ti) =∫d3~xiU (xi), and we assume U (xi) transforms as a

Lorentz scalar. The measure d4x is Lorentz-invariant due to the cancellation of length-contractionand time-dilation. Additionally, we demand that the time-ordering above is Lorentz-invariant.

• If x1− x2 is timelike, the time-ordering between x1 and x2 cannot be changed, i.e. if t1 > t2in one frame, then t1 > t2 in all frames. There is nothing to worry about here.

• If x1 − x2 is spacelike, the time-ordering between x1 and x2 can be flipped by appropriateLorentz boost. This destroys Lorentz invariance unless

[U (x),U (x′)] = 0 for x− x′ spacelike.

This is required because the time-ordering operator T commutes things past each other ifthey are not in the right time-order (we are not even in second quantization yet!), and gen-erally this introduces nontrivial commutators. We need them to vanish for the T -operationto be harmless.

The above is a rather strong condition, although the true condition for Lorentz invariance is alittle weaker. Old-fashioned perturbation theory is said to not be Lorentz invariant because theintegration

∫dγ looks like

∫d3~p1d

3~p2 · · · . Even if you wanted to slip out of this trap by writing,for example,

S ⊃∫

(d3~p1d

3~p2 · · ·E1E2 · · ·

)VβγVγαE1E2 · · ·Eα − Eγ + iε

,

it wouldn’t work because generally Eα and Eγ contain particles moving at different velocities, soEα and Eγ don’t transform in the same way under a boost, so the denominator Eα−Eγ + iε doesnot transform as a single Lorentz scalar.

Question 36. If someone asks me why people use time-ordered perturbation theory instead ofold-fashioned perturbation theory, what should I say?

Answer 36. There are a few very powerful reasons to use the time-ordered perturbation theoryinstead of old-fashioned perturbation theory.

• Time-ordered perturbation theory is manifestly Lorentz-invariant, but OFPT is not.

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• OFPT has lots of annoying denominators (for example, think about second-order perturba-tion theory from nonrelativistic QM). Time-ordered perturbation theory doesn’t have anydenominators in the series expansion, at least before you insert the propagators, which havethe denominators.

• Time-ordered perturbation theory gives you an extraordinarily powerful theorem – the Gell-Mann - Low theorem – that is hard to show in OFPT.

However, old-fashioned perturbation theory can also be useful.

• If we want only the leading-order term in a process, often OFPT or Fermi Golden Rule ismuch faster than using the Feynman rules. For example, if you’re doing QED to tree-level,OFPT is way easier, since the virtual photon has only an undetermined energy, and it’s easyto do an energy integral. One of the reasons for this is that you don’t have to rememberany Feynman rules; you just use second-quantization formalism. However, once you start tohave loops, it becomes easier to do the standard integrals from propagators.

• There is no time-ordering necessary in OFPT. You do, however, have to account for thefact that virtual particles can be either advanced or retarded, which leads to twice as manyintermediate states (and hence intermediate energies) as you’d expect. (See Schwartz QFTsection 4.1 for an example of this. He derives static Coulomb interaction from second-orderperturbation theory on the electron-phonon vertex.)

• OFPT can give you effective Hamiltonians, as I just mentioned. Eugene Demler likes doingthis; it’s essentially how he taught us Anderson Poor Man’s Scaling. I don’t see how time-ordered perturbation theory can give you an effective Hamiltonian because it seems to be onlyuseful for finding S-matrix elements. Of course, you can get the nonperturbative effectiveHamiltonian from path integral methods.

Finally, some good things to remember:

• A Feynman diagram has perfectly good meaning in both the time-ordered perturbationtheory and OFPT.

• Wick’s theorem holds in both formalisms.

• The idea of “field” as a distinct object is not really necessary in OFPT (or in many-bodyphysics). As in many-body physics, there is not a notion of a “field” φ(x) in the same senseas we use it in relativistic QFT. I mean this in the sense that in many-body physics, we take

ψ(x) =∑k

eikxak

but in relativistic QFT, our field is different. We have to take

φ(x) =

∫k

1√2ωk

(akeikx + a†ke

−ikx).

This is because the “field” in many-body physics does not have to transform as an irrep underthe Lorentz group. In my opinion, you can do everything in many-body physics without the

23


idea of “field,” if you know how to work with second quantization. From this viewpoint,ψ(x) is merely the Fourier transform of an annihilation operator, but φ(x) is a true field inthe sense that it is a nontrivial combination of creation and annihilation operators, whichare necessary to preserve causality (see the chapter on constructing fields).

• The bare Green functions are not the same in both formalisms. OFPT has the repeatedstructure 〈α|V|β〉, which involves doing lots of contractions. However, these contractions areexpectations or number operators with time-evolution. This is how we use Green functionsin many-body physics, and the “Green function,” which is really a contraction of creationand annihilation operators, literally doesn’t care about the Hamiltonian or Lagrangian weare using. (See the problem in section 4.4 of these notes for an example.)

Actually, I lied. It cares a little bit; usually the Green function is kind of like

Gk(t, t′) = in(k)e−iωkt =⇒ Gk(ω) =n(k)

ω − ωk

.

Here, n(k) is the distribution function, which is another way of saying number operator,which shows we are still in the second-quantization regime with no need for “fields,” per se.Certainly this is different from the Green functions in relativistic QFT.

In contrast, in time-ordered perturbation theory, things arise from a Lagrangian. The Greenfunctions, or propagators, are really nontrivial and involve spin sums and weird things likethat, which care a lot about what Lagrangian they come from.

• In OFPT the “fields” have no time-dependence, only position- or momentum-dependence. Intime-ordered PT, the fields have time-dependence. You can see the difference, for example,between the many-body theory

H =∑k

c†kk2

2mck +

∑kk′q

V (q)c†kc†k′ck+qck′−q

and the relativistic theoryL = ψ(t,x)(i/∂ −m)ψ(t,x).

• One final note on this subtle topic: In OFPT or many-body physics, the “fields” we workwith are always pure creation or annihilation operators, either in x-space or in k-space. Inrelativistic QFT, the fields are nontrivial combinations of a† and a.

3.4 Implications of unitarity

This section proves a surprising amount of interesting results. For example, Weinberg derivesLiouville’s theorem and Boltzmann’s H-theorem from unitarity, S†S = SS† = 1, a result due toC.N. Yang. Amazing.

Question 37. What is “forward scattering?”

Answer 37. This is something that confused me for a long time, but now it’s pretty obvious.Look at the decomposition

Sβα = δβα − 2πiδ(pβ − pα)Mβα.

The forward scattering term is the Mαα term. This makes so much sense. Essentially, the pertur-bation V not only scatters to different states; it can also send you back to the original state.

24


Question 38. Describe the optical theorem. What does it have to do with unitarity?

Answer 38. Weinberg derives the following optical theorem,

Im Mαα = −uασα16π3

, or Im f(α→ α) =kσα4π

.

Here, σα is the total cross-section

σα =

∫dβdσ(α→ [β, β + dβ])

dβ.

A consequence of the optical theorem is that the solid angle ∆Ω in which the scattering amplitudef(α→ β) is close to its maximum value f(α→ α) shrinks with energy,

∆Ω ∼ 1

k2σα.

The takeaway is that higher-energy beams give better-resolved diffraction peaks. Girma Hailutaught me this in nonrelativistic QM, so there is nothing new here and I will not include thederivation. A good question is “what does this have to do with unitarity?” The answer is thatthe derivation hinges on the relation

δ(γ − α) =

∫dβS∗βγSβα.

Question 39. Summarize the conclusions of Weinberg’s partial wave expansion.

Answer 39. Weinberg expands the wavefunction Φ~pEN , where N refers to all discrete indices, inspherical harmonics, just like in nonrelativistic quantum mechanics.

The result is that generally, there is no notion of the phase shift because different particles areproduced. However, in the case that the outgoing particles must of the same species as theincoming particles, then the S-matrix is diagonal in N , and there is a notion of the phase shift.This holds, for example, at very low energies (but still relativistic velocities) such that the boundstate which must form for a non-diagonal outgoing state is unattainable. Then,

Sjl′s′n′,lsn(E) = e2iδjlsn(E)δl′lδs′sδn′n,

because the S-matrix must be unitary. Here, N = jlsn and δjlsn(E) is the energy-dependent phaseshift. The total cross-section turns out to be related to

σ ∝∑jls

(2j + 1) sin2 δjlsn(E),

which is the same as in nonrelativistic QM. Conclusion: the partial wave expansion applies even atrelativistic velocities. The interpretation of partial wave expansion in terms of phase shifts applieseven at relativistic velocities, if the S-matrix is guaranteed to be diagonal. This is all a result ofthe unitarity of the S-matrix.

25


3.5 Resonances

This is an interesting section with a high density of important points.

Question 40. What is a resonance? What are some mechanisms for the formation of a resonance?

Answer 40. A resonant state or bound state is an intermediate state containing a singleunstable particle, R, which eventually decays into the particles observed as the final state. If R islong-lived, then the cross-section exhibits a peak, known as a resonance, around

ECM ≈ ER,

where ECM is the scattering energy in the CM frame and ER is the rest energy of the unstableparticle, R. The particle is said to be long-lived if the decay rate is much less than the characteristicfrequency of the particle, i.e. if

Γdecay ER/~.

There are different ways a resonance can come about.

• Strong and weak interactions: Perhaps the Hamiltonian has form

H = H0 + Vstrong + Vweak.

Both V-terms are interaction terms. The external particles asymptotically tend to Φ-eigenstates of H0, but the long-lived intermediate bound state, R, is an eigenstate ofH0 + Vstrong. The Vweak perturbation allows R to decay into the outgoing particles.

An example is the decay of pions, https://en.wikipedia.org/wiki/Pion#Charged_pion_decays. It is an eigenstate of H0 = Hkin and Vstrong = QED but not an eigenstate of theweak force, which is interpreted as Vweak.

• A bound state can be long-lived because of a potential barrier that makes escape of con-stituent particles difficult. For example, α-decay of heavy nuclei has to quantum-tunnelthrough the strong potential barrier due to the attractive strong force.

• Some reaction require statistically unlikely circumstances to take place. For example, if anexcited state of a heavy nucleus decays only when its energy is concentrated on a singleneutron, then there will be a long lifetime because this is unlikely.

Question 41. What does a resonance look like in the scattering amplitude T+βα?

Answer 41. Regardless of the microscopic process, we can physically demand the following con-ditions for a resonance:

• The bound state R must decay exponentially with time, P(R) ∼ e−Γt.

• There must be a “jump” (i.e. a smoothed-out divergence) in the scattering amplitude atEα = ER. Physically, this means the energy of the incoming particles Eα should match theenergy of the bound state ER near the resonance.

26

https://en.wikipedia.org/wiki/Pion#Charged_pion_decays

https://en.wikipedia.org/wiki/Pion#Charged_pion_decays


Starting with the Lippmann-Schwinger equation for “in” wavefunction with envelope function g(α)

Ψ+g (t) = Φg(t) +

∫dαdβ

e−iEαtg(α)T+βαΦβ

Eα − Eβ + iε,

we see that the integral

I +β =

∫dαe−iEαtg(α)T+

βα

Eα − Eβ + iε

produces a resonance ifT+βα ∼ (Eα − ER + iΓ/2)−1 + const.

The interpretation is that a certain part of the input wavefunction becomes the resonance ast → ∞, since not all of the wavefunction goes into the Eα = ER − iΓ/2 singularity. A goodquestion concerns the two-singularity structure in the above expression for I +

β . Is the overallδ(Eα −Eβ) energy conservation enforced even when the complex integration uses the Eα = ER −iΓ/2 singularity? It is. Weinberg (pg 115) gives

Ψ+g (t)→ Φ+

g (t)− 2πi

∫dβe−iEβtΦβ

∫dαδ(Eα − Eβ)g(α)T+

βα as t→∞.

The scattering part can be rewritten as∫dβΦβ

∫dαδ(Eα − Eβ)g(α)e−iEαtT+

βα.

The singularity in T+βα enforces Eα = ER and the delta function simultaneously enforces Eα = Eβ.

As you can see, not all of the wavefunction evolves into the resonance. The majority of thewavefunction never decays at all.

Question 42. Describe how the resonance shows up in the cross-section, σ(E).

Answer 42. Unsurprisingly, there is a Breit-Wigner peak, just like in nonrelativistic QM. Thescattering amplitude in the CM frame takes the form

SN ′N(E) = (S0)N ′N +RN ′N

E − ER + iΓ/2.

The important result (3.8.18) for decays into two-body states is

σ(n→ n′;E) ∝ 1

k2× ΓnΓn′

(E − ER)2 + Γ2/4,

where Γn = Γ∑

ls |ulsn|2. n refers to a two-body channel, so the above refers to scattering fromone two-body channel to another. This is the classical Lorentzian form that everybody knows.

Question 43. Explain the bound on maxσ(E) for a resonance.

Answer 43. Refer to the cross-section above and note that Γn ≤ Γ. We therefore have

maxE

σ ∼ (2π

k)2 = λ2,

27


which means that cross-sections at a single resonance are roughly bounded by a square wavelength.This applies in classical physics (for example, the resonance of a receiving antenna) and in quantumphysics as well. You can imagine this as follows: suppose a photon is impinging on an electronwhich lives in a stationary atom. Not only does the photon have to hit the electron with the rightenergy to excite it to a higher-energy state, it must also hit the electron in the right phase of its“cycle.” In other words, the electron must happen to be hit in some sweet spot range ∆t of e−iωt

oscillations. So,ω∆t ∼ 1 =⇒ b = c∆t ∼ cω−1 ∼ λ,

where b is the impact parameter. So σ ∼ b2 ∼ λ2.

3.6 Problems

Question 44. Consider a theory with a separable interaction; that is,

〈Φβ|V |Φα〉 = guβu∗α,

where g ∈ R is a coupling constant and uα ∈ C such that∑

α |uα|2 = 1. Use the Lippmann-Schwinger equation (3.1.16) to find explicit solutions for the “in” and “out” states and the S-matrix.

Answer 44. The solution is

Ψ±α = Φα +

∫dβ

T±βαΦβ

Eα − Eβ ± iε,

where

T±βα = guβu∗α + g2

∫dγ

uβu∗γuγu

∗α

Eα − Eγ ± iε+ g3

∫dγdγ′

uβu∗γuγu

∗γ′uγ′u

∗α

(Eα − Eγ ± iε)(Eα − Eγ′ ± iε)+ · · · .

Not sure anybody would say this is “explicit.” I can make it look nicer by introducing a param-

eter t± =∫dγ

u∗γuγEα−Eγ±iε . Here, Eα must be the initial energy of the reactants α; see https://en.

wikipedia.org/wiki/Perturbation_theory_(quantum_mechanics)#Second-order_and_higher_

corrections. So the expansion of the transfer matrix can be written

T±βα = guβu∗α(1 + gt± + g2t±

2+ · · · ),

and folding up the geometric series gives

T±βα =guβu

∗α

1− gt±.

Substituting into Lippmann-Schwinger equation gives an expression for Ψ± in terms of Φ, whichis assumed to be known, like plane waves. Substituting into the decomposition of the S-matrix interms of the scattering matrix T+

βα gives the S-matrix.

Question 45. Express the differential cross-section dσdΩ

for two-body scattering in the lab frame,in which one of the particles is intially at rest, in terms of kinematic variables and the matrixelement Mβα.

28

https://en.wikipedia.org/wiki/Perturbation_theory_(quantum_mechanics)#Second-order_and_higher_corrections




Answer 45. According to (3.4.15), the cross-section is defined

dσ(α→ β) = (2π)4u−1α |Mβα|2δ(4)(pβ − pα)dβ.

Here, uα is the velocity of the incoming particle in the lab frame. Let’s assume 2→ 2 scattering,so there are two outgoing particles. Therefore,

δ(4)(pβ − pα)dβ = δ(3)(~p′1 + ~p′2 − ~p1)δ(E ′1 + E ′2 − E)d3~p′1d3~p′2.

Integrating away the momentum-conserving δ-function with d3~p′1 gives

δ(4)(pβ − pα)dβ = δ(√

(~p1 − ~p′2)2 +m′21 +√~p′22 +m′22 − E)p′22 dp

′2dΩ.

Using δ(f(x)) = δ(x− x0)/|f ′(x0)| gives

δ(4)(pβ − pα)dβ = (|~p′2| − |~p1| cos θ√(~p1 − ~p′2)2 +m′21

+|~p′2|√

~p′22 +m′22)|~p′2|2dΩ.

Here, |~p′2| is taken to be that magnitude which satisfies the overall δ-function. The expression isugly but it can be solved for. Finally,

dσ

dΩ(α→ β) = (2π)4u−1

α |Mβα|2(|~p′2| − |~p1| cos θ√(~p1 − ~p′2)2 +m′21

+|~p′2|√

~p′22 +m′22)|~p′2|2.

Note that there is a cos θ factor, so this is manifestly not spherically symmetric. That makessense, because uα picks out a preferred direction.

Question 46. Derive the time-dependent perturbation expansion (3.5.8) directly from the expan-sion (3.5.3) of old-fashioned perturbation theory.

Answer 46. (3.5.3) is

T+βα = Vβα +

∫dγ

VβγVγαEα − Eγ + iε

+

∫dγdγ′

VβγVγγ′Vγ′α(Eα − Eγ + iε)(Eα − E ′γ + iε)

+ · · · .

(3.5.8) is

S = 1− i∫ ∞−∞

dt1V (t1) + (−i)2

∫ ∞−∞

dt1

∫ t1

−∞dt2V (t1)V (t2) + · · · .

The first thing to note is that T+βα is a scalar and S is an operator (obviously, it’s called the

S-operator). It’s easiest to go backwards. Recall that

Sβα = 〈Φβ|S|Φα〉 = δβα − 2πiδ(Eα − Eβ)T+βα

= δβα − i∫ ∞−∞

dt1〈Φβ|V (t1)|Φα〉+ (−i)2

∫dt1dt2

∫dγ〈Φα|V (t1)|Φγ〉〈Φγ|V (t2)|Φβ〉+ · · · .

Everything here is in Heisenberg picture, so V (t1) = eiH0tV e−iH0t. This implies

−i∫ ∞−∞

dt1〈Φβ|V (t1)|Φα〉 = −i∫ ∞−∞

dt1ei(Eβ−Eα)tVβα = −2πiδ(Eβ − Eα)Vβα.

29


The next term is

−∫dγ

∫ ∞−∞

dt1ei(Eα−Eγ)t

∫ t1

−∞dt2e

i(Eγ−Eβ)t2VαγVγβ.

Inserting the right convergence factor, we find∫ t1

−∞dt2e

i(Eγ−Eβ)t2eεt2 =ei(Eγ−Eβ)t1

i(Eγ − Eβ − iε)

and then

−∫dγ

VαγVγβi(Eγ − Eβ − iε)

∫ ∞−∞

dt1ei(Eα−Eγ)t1ei(Eγ−Eβ)t1 = −2πiδ(Eα − Eβ)

∫dγ

VαγVγβEα − Eγ + iε

,

as desired. So, the two expansions are the same, order-by-order.

4 The Cluster Decomposition Principle

Question 47. Heuristically, what is the cluster decomposition principle and how is it encodedin creation and annihilation operators?

Answer 47. The cluster decomposition principle is the following:

Physical processes occuring at large separation are independent.

It turns out that if the Hamiltonian is expressed as a sum of products of creation and annihilationoperators, then the S-matrix will automatically satisfy the cluster decomposition principle. This isalso why creation and annihilation operators are used in nonrelativistic statistical QM. Ultimately,it is why we use creation and annihilation operators to introduce the idea that particle number isnot necessarily conserved.

4.1 Operator decomposition in a†q and aq

Question 48. How are many-particle states normalized?

Answer 48. We would like to normalize many-particle states in preparation for the machinery ofraising and lowering operators. If one-particle states, described by q (which includes all quantumnumbers), are normalized as

〈Φq|Φq′〉 = δ(q − q′),

then to preserve the Bose-Einstein and Fermi-Dirac statistics, many-particle states must be nor-malized as

〈Φq′1q′2···q′N |Φq1q2···qM 〉 = δNM

∑P

(−1)FP∏i

δ(qi − q′i).

Here, P is a permutation of the particles, FP is the fermionic index and tracks how many anti-symmetric switches we made, and N,M are the numbers of particles in each state. Usually, onlyone of the terms in this sum can be nonzero at a time. However, we must include all the possible

30


permutations because switching, for example, q′1 with q′2 may also give something that is nonzero.For example,

〈Φq′1q′2|Φq1q2〉 = δ(q′1 − q1)δ(q′2 − q2)± δ(q′2 − q1)δ(q′1 − q2).

This still isn’t clear to me. Think about it more

Question 49. Describe the action of creation and annihilation operators.

Answer 49. Creation operators add a particle to the wavefunction; annihilation operators takeaway a particle. However, because of the (anti)symmetrization, if there are identical particles, wedon’t know which one to take away, so we must account for all possibilities! So, the definitions are

a†qΦq1···qN = Φqq1···qN and aqΦq1···qN =N∑r=1

(±1)r+1δ(q − qr)Φq1···qr−1qr+1···qN .

The ± is for creation/annihilation operators which are bosonic and fermionic, respectively. Formore detail, see my notes on “2nd Quantization.” Both my notes and Weinberg’s book follow thelogic that we should only introduce canonical commutators after we define the action of a†, a onwavefunctions. Weinberg’s method gives an interesting derivation of the canonical commutator:

aq′a†qΦq1···qN = δ(q′ − q)Φq1···qN +

N∑r=1

(±1)r+2δ(q′ − qr)Φqq1···qr−1qr+1···qN ,

a†qaq′Φq1···qN =N∑r=1

(±1)r+1δ(q′ − qr)Φqq1···qr−1qr+1···qN .

Question 50. To set up discussion of cluster decomposition and Hamiltonians, prove the followingtheorem:

Any operator O can be expressed as a sum of products of creation and annihilation operators,

O =∞∑N=0

∞∑M=0

∫dN~q′

∫dM~qa†q′1

· · · a†q′NaqM · · · aq1CNM(~q′, ~q).

Here, ~q = (q1, · · · qM) and ~q′ = (q′1, · · · , q′N).

Answer 50. First, let’s think about what this means. We would like to reproduce the matrixelements of O between different wavefunctions, for example 〈Φq′1···q′M |O|Φq1···qN 〉. What the abovetheorem means is that, because |Φq′1···q′M 〉 and |Φq1···qN 〉 differ only by their particle content, andthe matrix element itself is a scalar, we can reproduce the matrix elements of any O by engineeringa combination of particle-content-changes (i.e. creation and annihilation operators), along withsome scalar coefficients CNM .

The proof follows by induction. Letting |Φ0〉 be the vacuum, we have C00 = 〈Φ0|O|Φ0〉. Nowfor the inductive step. Suppose we have successfully replicated all matrix elements of O forN < L,M ≤ K or N ≤ L,M < K. We would like to show that we can replicate the matrixelement for N = L,M = K.

Pick two vectors, |Φp1···pK 〉 and 〈Φp′1···p′L|, and compute the matrix element 〈Φp′1···p′L|O|Φp1···pK 〉.If N > L or M > K, the matrix element 〈Φp′1···p′L|a

†q′1· · · a†q′NaqM · · · aq1|Φp1···pK 〉 is equal to zero

because either the bra or ket is annihilated. There are three possibilities

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• If N ≤ L and M < K, we know CNM already.

• Same for N < L and M ≤ K.

• If N = L and M = K, then the evaluation of the above gives

〈Φp′1···p′L|O|Φp1···pK 〉 ⊃∑P,P ′

CLK(P ′p′1, · · · ,P ′p′L,Pp1, · · · ,PpK).

This is a little different from Weinberg, who got just L!K!CLK(q′1 · · · q′L, q1 · · · qK). I think there is no reason that CLK for particle interchanges

should be equal to each other; the only restriction is on their sum, via the matrix element of O.

This follows, for example, by looking at the matrix element∫d2~q′d3~qCNM(~q′, ~q)〈Φp′1p

′2|a†q′1a

†q′2aq3aq2aq1|Φp1p2p3〉.

We find that

aq3aq2aq1|Φp1p2p3〉 =3∑

r123=1

(±1)r1+r2+r3+3δ(q1 − pr1)δ(q2 − pr2)δ(q3 − pr3)|Φ0〉

= −3∑

r123=1

δ(q1 − pr1)δ(q2 − pr2)δ(q3 − pr3)|Φ0〉

because r1 + r2 + r3 = 1 + 2 + 3 = 6, in some order. There will also be an overall (−1) from〈Φp′1p

′2|a†q′1a

†q′2

, so there is an overall positive sign.

Conclusion: there is enough freedom to choose the matrix elements CLK(~q′, ~q) to match the matrixelements of O. According to my logic, the construction is not unique.

Question 51. Discuss the implications of the above theorem for various operators.

Answer 51. We consider several examples.

• Additive operators: For example, the free-particle Hamiltonian is additive because

H0|Φq1···qN 〉 = (E1 + · · ·+ EN)|Φq1···qN 〉.

Such an operator can be written using only the N = M = 1 term. Moreover, this is theonly way to write H0 (because H0 conserves particle number, N = M . Separability impliesN = M = 1). Thus, the free-particle Hamiltonian can always be expressed as

H0 =

∫dqE(q)a†qaq.

Momentum and z-projection of the spin are also additive.

• Symmetry operators: The unitaries that implement Λµν ,P,C,T can be expressed in terms

of creation and annihilation operators. The interpretation of such a unitary constructedin creation and annihilation operators is, for example for C: annihilate all particles in thewavefunction and then create everything again, but with the opposite charge (and perhaps

32


a phase). In fact, the behavior of wavefunctions Ψ or Φ under symmetry operations meansthat creation and annihilation operators must obey very similar rules.

For example,

U(Λ, a)Ψpσ = e−i(Λp)·a

√(Λp)0

p0

∑σ′

D(j)σ′σ(W (Λ, p))ΨΛp,σ′

and the construction of one-particle states, Ψpσ = a†pσΨ0, implies

U(Λ, a)a†pσU(Λ, a)−1 = e−i(Λp)·a

√(Λp)0

p0

∑σ′

D(j)σ′σ(W (Λ, p))a†Λp,σ′ .

We have to include a U−1 because of how we create many-particle states. Think about it!(The vacuum is invariant, U |Ψ0〉 = |Ψ0〉.)The operators C,P,T act on creation and annihilation operators in similar ways to howthey act on wavefunctions.

4.2 Factorization of the S-matrix

Question 52. State the cluster decomposition principle in mathematical language.

Answer 52. Let there be N scattering processes occuring simultaneously but at large separationsfrom each other. Then the S-matrix element for the meta-process that includes all smaller processesshould factorize,

Sβ1+···+βN ,α1+···+αN → Sβ1α1 · · ·SβNαN as |~ri − ~rj| → ∞.

Above, αi and βi are multi-particle states. All particles in a single state, say αi, are localized aroundthe position ~ri in 3-space. Factorization of the S-matrix ensures a corresponding factorization oftransition probabilities, so experimental results will be uncorrelated.

Question 53. State the cluster decomposition principle in terms of the connected S-matrix,SCβα.

Answer 53. Let α and β be states containing particles. The cluster decomposition principle isequivalent to the following:

SCβα vanishes if any particle in α + β is far away from the other particles in α + β.

To summarize, cluster decomposition says that S-matrix should factorize, and that the connectedS-matrix should vanish. (Under the large-separation condition, of course.)

Question 54. Define the connected S-matrix.

Answer 54. Take the total states α and β and partition their particle content into an equalnumber of clusters, where order of particles within a cluster does not matter:

α = α1 + · · ·+ αn and β = β1 + · · ·+ βn.

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There are many different ways to partition the particles. In the above, we partitioned both statesinto n clusters. It is also possible to have n = 1, which is the trivial partition, and it is not possibleto have clusters which are empty. (To see why, suppose that αi is non-empty but βi is empty. Thescattering process αi → βi automatically violates energy conservation, since the vacuum has noenergy. It also violates other things; for example, often [H, J2] = 0.) We define SCγσ to satisfy

Sβα =∑

all partitions

(−1)FSCβ1α1,

where F is the fermionic index due to the process of partitioning. The connected S-matrix isconstrained to be unique; it is a single matrix regardless of what the overall process Sβα it happensto be participating in.

For example, we can use small particle numbers to determine the connected S-matrices iteratively:

Sq′q = SCq′q = δ(q′ − q),

Sq′1q′2,q1q2 = SCq′1q′2,q1q2 + SCq′1q1SCq′2q2± SCq′1q2S

Cq′2q1

, etc.

Question 55. What is the physical interpretation of the connected S-matrix?

Answer 55. To get some intuition, let’s look at the expression for SCq′1q′2,q1q2,

SCq′1q′2,q1q2 = Sq′1q′2,q1q2 − (SCq′1q1SCq′2q2± SCq′1q2S

Cq′2q1

).

What this means is: take the entire S-matrix and subtract away the processes that are products oftwo 1 → 1 processes. Therefore, SCq′1q′2,q1q2

includes only scattering processes in which particles q′1and q′2 nontrivially interact with each other. In general, SCq′1···q′n,q1···qm

means the particles q′1 · · · q′nall have to be entangled somehow, and that it is impossible to factorize the scattering processrepresented by SCq′1···q′n,q1···qm

into products of smaller scattering processes. To be entangled with

someone else, you have to be close to them! Hence, the connected S-matrix vanishes if everyoneisn’t “in the same room,” so to speak.

That explains why α = q1 · · · qn must all be close together. Why must the particles in β = q′1 · · · q′mbe both close together and close to cluster α? If everyone went to a party in a single room, it isimpossible for one of them to apparate and re-appear 500 miles away from the party in an instant.So, all of the partygoers are still relatively localized after they walk out of the party, and all of αmust be close to all of β. (This implicitly assumes that the scattering process α→ β is relativelyfast, so particles have not had time to travel r →∞ distance away from the scattering center.)

Question 56. Show with an example that the vanishing of the connected S-matrix for largeseparations is equivalent to the factorization of the original S-matrix.

Answer 56. This is a nice example and comes from Weinberg, but I did a simpler case. Supposethe scattering process is 123 → 1′2′3′. Further, assume that 1, 2′ are close, and 231′3′ are close,but 1, 2′ are far away from 231′3′. For simplicity, let all particles be bosonic.

We can decompose the S-matrix into connected S-matrices,

S1′2′3′,123 = Sc1′2′3′,123 + Sc1′2′,12Sc3′3 + Sc1′2′,13S

c3′2 · · ·+ Sc1′3S

c2′2S

c3′1.

Most of these terms get killed. The only terms that don’t get killed are

S1′2′3′,123 = Sc1′3′,23Sc2′1 + Sc2′1S

c1′2S

c3′3 + Sc2′1S

c1′3S

c3′2 = S2′1S1′3′,23.

That is the result of the cluster decomposition principle for regular S-matrices.

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Question 57. How is vanishing of the connected S-matrix for large spatial separations encodedin its momentum-space form?

Answer 57. The Fourier transform from ~k-space to ~x-space is

SC~x′,~x =

∫~p′,~p

ei~p′1·~x′1 · · · ei~p′N ·~x′N e−i~p1·~x1e−i~pM ·~xMSC~p′,~p.

Physically, we demand that SC~x′,~x be invariant under any global translation,

~x′, ~x→ ~x′ + ~a, ~x+ ~a.

This means the overall phase factors from the exponential is not allowed to change, and hencethere is a constraint

p′1 + · · ·+ p′N − p1 − · · · − pM = 0.

Therefore, the vanishing of the connected S-matrix for large spatial separations is encoded in therequirement that

SC~p′,~p ∝ δ(3)(∑

~p′ −∑

~p).

In fact, the above requirement is also enforced if

SC~p′,~p ∝ δ(3)(∑

~p′1 −∑

~p1)δ(3)(∑

~p′2 −∑

~p2),

where the particle content of both incoming and outgoing states has been partitioned into twoparts. But, the two-δ-function structure is not allowed by cluster decomposition, because it implieswe could move cluster 1 very far away from cluster 2 and still have a nonvanishing matrix element.

Conclusion: Cluster decomposition implies SC~p′,~p has a single momentum-conserving delta function,δ(∑~p′ −

∑~p).

4.3 Which Hamiltonians satisfy cluster decomposition?

Question 58. Which Hamiltonians satisfy the cluster decomposition principle?

Answer 58. It is always possible to write the Hamiltonian in raising and lowering operators,

H =∞∑N=0

∞∑M=0

∫dN~q′

∫dM~qa†q′1

· · · a†q′NaqM · · · aq1hNM(~q′, ~q).

The claim is that

H yields an S-matrix that satisfies cluster decomposition if (and perhaps only if?) the coefficientfunction hNM(~q′, ~q) contains only a single momentum-conserving δ-function, δ(

∑~q′ −

∑~q).

This condition is automatically true for the free Hamiltonian H0, so to be true for the totalHamiltonian H it must also be true for the interaction, V . The proof is a little involved (pg.183), but I can give the main part of the argument. In the previous section, we found that clusterdecomposition implies that

SC~p′,~p ∝ δ(3)(∑

~p′ −∑

~p),

35


with no other smaller momentum-conserving δ-functions. Also, recall that

Sβα =∞∑n=0

(−i)n

n!

∫ ∞−∞

dt1 · · · dtn〈Φβ|TV(t1) · · · V(tn)|Φα〉.

Consider only the terms in the calculation of the connected S-matrix,

Sβα =∞∑n=0

(−i)n

n!

∫ ∞−∞

dt1 · · · dtn〈Φβ|TV(t1) · · · V(tn)|Φα〉C .

This connected matrix element is represented graphically by a Feynman diagram in which everypoint is connected to every other by some path on the graph, and each vertex carries some δ-functions inside. The question is, “how many momenta do we need to fix?”

Let the connected graph have V vertices, I internal lines, and L loops; assume for simplicity thateach vertex carries exactly one δ-function, as required by the theorem. There are V δ-functions,of which I − L are used to fix internal momenta. So, we have V − I + L δ-functions left to fixexternal momenta. For a connected graph, Euler’s theorem gives V − I + L = 1. This is exactlythe overall δ-function in SC~q′,~q′ ⊃ δ(

∑~q′ −

∑~q).

Each term in V physically must have at least one δ-function. What this theorem says is that ifany term in V had more than one δ-function, we would overspecify the momenta of the externalparticles, in violation of the cluster decomposition principle.

4.4 Problems

Question 59. Consider an interaction

V = g

∫~p1~p2~p3~p4

δ(3)(~p1 + ~p2 − ~p3 − ~p4)× a†~p1a†~p2a~p3a~p4 .

Here, a~p annihilates a spinless boson of mass M > 0. Use perturbation theory to calculatethe S-matrix element for scattering of these particles in the CM frame, to O(g). What is thecorresponding differential cross-section?

Answer 59. According to (3.4.30), the differential cross-section for 2→ 2 scattering in CM frameis

dσ(α→ β)

dΩ=

(2π)4k′E ′1E′2E1E2

E2k|Mβα|2

where k := |~p1| = |~p2|, k′ := |~p′1| = |~p′2|.

First, we will calculate the S-matrix element. The nontrivial part of Sβα is

T+βα = 〈Φβ|VΨ+

α 〉 ≈ 〈Φβ|V|Φα〉 to O(g).

Using Φα = a†~p1a†~p2|0〉, Φβ = a†~p′1

a†~p′2|0〉 gives

Sβα ≈ g

∫1234

δ(~ktot)〈0|a~p2a~p1a†~k1a†~k2a~k3a~k4a†~p′1a†~p′2|0〉.

36


There are three corresponding diagrams,

The result is

Sβα/g = 4(δ(~p1 + ~p2 − ~p′1 − ~p′2) + 4δ(~p1 − ~p′1)δ(~p2 − ~p′2) + 2δ(~p1 − ~p′1)δ(~p2 − ~p′2)(

∫d3~k

(2π)3)).

We can see the first term is connected (because there is only an overall 3-momentum conservingδ-function) and the second and third terms are not connected. In fact, the third term does nothave any scattering of the original particles at all, so there is a divergent integral. We will justignore it. The first term is real scattering that can change directions and stuff; the second termis forward scattering which contributes to scattering but preserves the directions of the originalparticles.

If we are only looking for scattering not in the forward direction, then the first term is the onlyrelevant one. Using Sβα = δβα − 2πiδ(pβ − pα)Mβα and

Sβα = δ(β − α)− 2πiδ(Eα − Eβ)T+βα, where T+

βα = 〈Φβ|VΨ+α 〉

and ignoring the trivial part gives Mβα = 4g. Finally,

dσ(α→ β)

dΩ= 16g2 (2π)4k′E ′1E

′2E1E2

E2k

Question 60. A coherent state Φλ is defined to be an eigenstate of the annihilation operatorsaq with eigenvalues λ(q). Construct such a state as a superposition of the multi-particle statesΦq1···qN .

Answer 60. The action of an annihilation operator is

aqΦq1···qN =N∑r=1

(±1)r+1δ(q − qr)Φq1···qr−1qr+1···qN .

Just like in the path integral formulation of, for example, superconductivity, we have

aq|q〉 = λ(q)|q〉, where |q〉 = eλ(q)a†q |0〉.

For bosons, at least, the coherent state can thus be defined

Φλ = (∏i

eλ(qi)a†qi )|0〉.

This is because bosonic operators commute.

5 Quantum Fields and Antiparticles

This is a long chapter. Besides constructing fields, it also covers spin-statistics theorem, antipar-ticles, and CPT theorem. There is also a lot of interesting generality on how to construct fields oflarger representations of the Lorentz group.

37


5.1 General free fields

Question 61. What is the motivation for introducing fields?

Answer 61. We saw in the previous chapter that the foolproof way to construct a Hamiltonianwhich satisfies cluster decomposition is to write it in terms of creation and annihilation operators(with a single 3-momentum-conserving δ-function). We also saw in chapter 3 that we can guaranteethis Hamiltonian is Lorentz-invariant if the interaction satisfies

V (t) =

∫d3~xU (~x, t), where [U (x),U (x′)] = 0 for (x− x′)2 ≥ 0.

Therefore, we should construct U out of creation and annihilation operators. It turns out that toguarantee such a U is a good Lorentz scalar, we should build U out of fields, which are constructedsuch that under Lorentz transform, the transformation of the fields are position-independent:

U0(Λ, a)ψ±l (x)U−10 (Λ, a) =

∑l

Dll(Λ−1)ψ±

l(Λx+ a).

This is because adding or subtracting creation and annihilation operators, for example a†~p1+ a†~p2

,

doesn’t work right away since a†~p1and a†~p2

have different transformation rules under a Lorentz

transform. We need to patch this problem up by tying a† and a to some coefficients vl and ulwhich make the combinations ul(x, ~pσn)a(~pσn) and ul(x, ~pσn)a(~pσn) transform in the same way,regardless of ~pσn. As you can imagine, the number of spin indices l tells us the spin of the particle.The matrix Dll tells us what representation we are working in. Although the representationtechnically doesn’t have to be irreducible, it is easiest to work with irreps. (For example: masslessspin-1, massive spin-1/2, etc.)

Then, the interaction density U can always be written

U (x) =∑NM

∑l′1···l′N

∑l1···lM

gl′1···l′N l1···lMψ−l′1

(x) · · ·ψ−l′N (x)ψ+l1

(x) · · ·ψ+lM

(x),

which is a scalar if the coefficients gl′1···l′N l1···lM are Lorentz covariant with the representations ofthe fields,

gl′1···l′N l1···lM =∑l′1···l′N

∑l1···lM

Dl′1 l′1(Λ−1) · · ·Dl′N l

′N

(Λ−1)Dl1 l1(Λ−1) · · ·DlM l′M

(Λ−1).

Basically, this means that under a Lorentz transform, the changes in ψ± are absorbed into thecoupling constant g, so that the overall interaction U is invariant - a Lorentz scalar.

To summarize: “The cluster decomposition principle together with Lorentz invariance thus makesit natural that the interaction density should be constructed out of the annihilation and creationfields.”

Question 62. How can we construct a field?

Answer 62. The annihilation (+) and creation (−) fields are taken to be

ψ+(x) =∑σn

∫d3~pul(x; ~pσn)a(~pσn) and ψ−(x) =

∑σn

∫d3~pvl(x; ~pσn)a†(~pσn).

38


(It is a little weird to denote annihilation by (+).) From the previous chapter, we already knowthe transformation rules of the creation and annihilation operators,

U(Λ, a)a†pσU(Λ, a)−1 = e−i(Λp)·a

√(Λp)0

p0

∑σ′

D(j)σ′σ(W (Λ, p))a†Λp,σ′ .

The idea is to tweak the transformation of ul and vl such that the fields ψ±(x) transform in theway we would like (see previous question).

The result (Weinberg pg 195-6) turns out to be uniquely expressible in terms of the standardmomentum, just like in Wigner little group. For massive particles (where the standard momentum

is ~k = 0), the result is

ul(x; ~pσn) =1

(2π)3/2eipxul(~pσn), vl(x; ~pσn) =

1

(2π)3/2e−ipxvl(~pσn).

If p = L(p)k, where k is the standard momentum, we find

ul(~pσn) =

√k0

p0

∑l

Dll(L(p))ul(~kσn), vl(~pσn) =

√k0

p0

∑l

Dll(L(p))vl(~kσn).

We recognize the factor√

k0

p0 as the normalization 1√2E~p

in Schwartz or Peskin and Schroeder.

Here, it is interpreted as a natural way to cancel the factor√

(Λp)0

p0 in the transform of the creation

or annihilation operator.

Question 63. How does the above encode the Klein-Gordon equation?

Answer 63. Klein-Gordon equation can be interpreted as a consequence of translational invari-ance, from

ul(x; ~pσn) =1

(2π)3/2eipxul(~pσn), vl(x; ~pσn) =

1

(2π)3/2e−ipxvl(~pσn).

Applying = ∂µ∂µ to ψ±l gives the Klein-Gordon equation,

(−m2)ψ±l (x) = 0.

This follows because m2 = pµpµ. Another way to think about it is that the Klein-Gordon equation

is simply a reflection of the fact that a spatial field is the Fourier transform of a momentum-spacefield. For example,

f(~x) =

∫~p

ei~p·~xf(~p) =⇒ −∇2f(~x) =

∫~p

~p2ei~p·~xf(~p)

would satisfy a Klein-Gordon-esque equation if ~p2 = const.

For a photon, ψ = 0. This returns the E = c|~p| relation.

Question 64. As of now, why is the construction

U (x) =∑NM

∑l′1···l′N

∑l1···lM

gl′1···l′N l1···lMψ−l′1

(x) · · ·ψ−l′N (x)ψ+l1

(x) · · ·ψ+lM

(x),

with correctly-transforming gl′1···l′N l1···lM and (uv)l(~pσn), still not good enough?

39


Answer 64. We haven’t considered the condition [U (x),U (x′)] = 0 for (x − x′)2 ≥ 0. Thisis a problem because [ψ+

l (x), ψ−l′ (y)]± 6= 0. (Here, ± means the [, ] symbol is interpreted as ananticommutator for fermions. I use + to denote commutator and − to denote anticommutator,but the convention in Weinberg is the other way.)

We will solve this problem by constructing the interaction out of the linear combinations ψl(x) =κlψ

+l (x) + λlψ

−l (x), where the coefficients κ, λ are chosen such that the causality condition

[ψl(x), ψl′(y)]± = [ψl(x), ψ†l′(y)]± = 0 for (x− y)2 ≥ 0

is satisfied. This is a “causality” condition because it means that if x − y is spacelike, no signalfrom x can affect measurements at y.

Question 65. How does conservation of charge imply the existence of antiparticles?

Answer 65. This is an interesting argument. If we believe that charge is conserved, we must have

[Q,U ] = 0.

If we construct U out of ψl(x) fields (see previous question), we must also have [Q, ψl(x)] =−qlψl(x) for some ql. Why? This means that each field carries a well-defined charge. It guarantees,for example, that the state |lm〉 = ψl(x)ψm(y)|Φ0〉 has a definite charge,

Q|lm〉 = Qψl(x)ψm(y)|Φ0〉 = −(ql + qm)ψl(x)ψm(y)|Φ0〉.

In particular, both the creation and annihilation parts of ψl(x) must carry the same charge. Ifparticles of charge ql are created by the creation sector, then particles of charge −ql must beannihilated in the other sector. This is the reason for antiparticles. We will use the label n forthe antiparticle of n.

In particular, U is then “charge-neutral” if the sum of charges of the fields is zero. This is easyto guarantee if we use Hermitian conjugates. For example,

U ⊃ ψ†ψ.

This is not always the case; for example, if a field is charge-neutral, then we need only one copyof it and do not require a Hermitian conjugate. This is like the QED coupling or the Yukawacoupling.

In retrospect, I feel this argument is not very strong because there were a lot of assumptionshidden in the requirement that we should have a field ψl(x) = κlψ

+l (x)+λlψ

−l (x) in the first place.

But Weinberg seems to have no problem with it.

5.2 Causal scalar fields

Now we get to our first example of a real, physical field. A scalar field is one that transforms withthe scalar representation of the Lorentz group, D(Λ) = 1. Because this is the trivial representation,

any scalar field must have spin zero. Adjusting the normalization factor√

k0

p0 (i.e. through the

constants in ψl(x) = κlψ+l (x) + λlψ

−l (x)) gives

φ+(x) =

∫d3~p

(2π)3/2

1√2p0

a~peipx and φ−(x) = (φ+(x))†.

40


Question 66. How can we enforce the causality condition and charge conservation with scalarfields?

Answer 66. Recall that the causality condition is

[φ(x), φ(y)]± = [φ(x), φ†(y)]± = 0 for (x− y)2 ≥ 0.

The point is to choose the right coefficients in ψl(x) = κlψ+l (x) + λlψ

−l (x) such that this is true.

We can compute

[φ+(x), φ−(y)]± = ∆+(x− y) :=1

(2π)3

∫d3~p

2p0eipx =

m

4π2√

(x− y)2K1(m

√(x− y)2).

This is nonzero. However, it is even in xµ − yµ for (x− y)2 > 0. Therefore,

[φ(x), φ(y)]± = κλ(1∓ 1)∆+(x− y), [φ(x), φ†(y)]± = (|κ|2 ∓ |λ|2)∆+(x− y) for (x− y)2 ≥ 0.

These can vanish only if the particle is a boson, and if |κ| = |λ|.

We can choose any relative phase between κ and λ, but once we choose it for some φ(x), we mustmake it the same for every φ(x′), where x 6= x′. We will choose the simplest convention κ = λ = 1.Also, we need φ(x) to be a field of a single charge, so we will introduce a new field φc+, φc− whichhas the opposite charge. The constructed field is therefore

φ(x) = φ+(x) + φc−(x) =

∫d3~p

(2π)3/2√

2p0(a~pe

ipx + ac†~p e−ipx).

Conclusion: scalar fields must be spinless but do not have to be charge-neutral.

The corresponding expression in Schwartz’ QFT is eq. (2.78). He implicitly assumed that a†

creates a particle of zero charge.

5.3 Causal vector fields

Besides the trivial rep, the next-simplest representation of the Lorentz group is the vector repre-sentation, in which

D(Λ)µν = Λµν .

We will consider only massive fields here, like W±, Z0. The annihilation and creation parts of thevector field are

φ+µ(x) =∑σ

∫d3~p

(2π)3/2uµ(~pσ)a~pσe

ipx where uµ(~pσ) =

√m

p0L(p)µνu

ν(0, σ),

φ−µ(x) =∑σ

∫d3~p

(2π)3/2vµ(~pσ)a†~pσe

−ipx where vµ(~pσ) =

√m

p0L(p)µνv

ν(0, σ).

Question 67. Why must the particle represented by a vector field be spin-0 or spin-1?

41


Answer 67. The explanation is in Weinberg pg. 207-8. It’s not particularly illuminating; basicallyhe argues that the conditions (5.3.12-13)∑

σ

u0(0, σ)( ~J (j))2σσ = 0,

∑σ

ui(0, σ)( ~J (j))2σσ = 2ui(0, σ)

can only be satisfied in two cases: (1) u0 is nonzero and ui are all zero, i.e. spin-0 (2) u0 is zeroand all the ui are nonzero, i.e. spin-1.

Heuristically, how can we think about this? TODO

Question 68. We showed in the previous question that the spin of a particle represented by avector field must be either 0 or 1. Describe the spin-0 case.

Answer 68. The spin-0 case is just the derivative of the scalar spin-0 field,

φµ(x) = ∂µφ(x).

It is therefore easy to write down its form in terms of creation and annihilation operators.

Question 69. Similar to the previous question, but describe the spin-1 case.

Answer 69. Remember that we are working with massive particles only. Also remember thatfor j = 1, only the spatial components ui and vi are nonzero. See Weinberg pg 209-210 for thecomplete derivation; the main point is that we can choose the field to be

φ+µ(x) = φ−µ†(x) =∑σ

∫d3~p

(2π)3/2√

2p0eµ(~pσ)a(~pσ)eipx.

Here, eµ(~pσ) := Lµν (~p)eν(0σ) and we can choose

eµ(0, 0) = (0, 0, 1, 0)T , eµ(0, 1) = − 1√2

(1, i, 0, 0)T , eµ(0,−1) =1√2

(1,−i, 0, 0)T .

Involved in the choosing of these directions were the raising and lowering operators.

Now, we are going to enforce the causality condition by tuning the parameters in φµ(x) =κφ+µ(x) + λφ−µ(x). The argument is the same as it was for the scalar field.

Introducing the matrix Πµν(~p) :=∑

σ eµ(~pσ)eν∗(~pσ) = ηµν + pµpν/m2 gives

[φ+µ(x), φ−ν(y)]± =

∫d3~p

(2π)32p0eip(x−y)Πµν(~p) = (ηµν − ∂µ∂ν

m2)∆+(x− y).

Again, this is even in xµ − yµ for spacelike separations (x − y)2 ≥ 0. Therefore, the sameconstruction is valid: the spin-one particles must be bosons and we must also have |κ| = |λ|.Again, we must conserve charge, so we introduce the antiparticle field φcµ±. The complete spin-1vector field is

vµ(x) =∑σ

∫d3~p

(2π)3/2√

2p0(eµ(~pσ)a~pσe

ipx + eµ∗(~pσ)ac†~pσe−ipx).

Important point: there are three polarizations (i.e. σ = −1, 0, 1) but each polarization is a 4-vector.And we are doing a 3-dimensional integral, even though the underlying manifold is 4-dimensional.We should try not to confuse the 3 with the 4, and know what they point to each time we usethem.

42


Question 70. Describe why the photon field cannot be derived by taking a smooth limit of themassive spin-1 vector field.

Answer 70. There are many reasons why, but Weinberg (pg. 212) gives one here: suppose theinteraction density contains a current Jµ coupled to the massive vector field vµ,

U ⊃ Jµvµ.

The scattering rate is proportional to (let 〈Jµ〉 be some generic matrix element):

Γ ∼ |〈Jµ〉eµ(~pσ)|2 ∼ 〈Jµ〉〈Jν〉∗Πµν(~p),

and unfortunately Πµν blows up. However, the matrix element can remain finite only if the Wardidentity

pµJ µ = 0

holds.

5.4 Spinor representation of Lorentz group

Question 71. Explain why the Dirac formalism is important in physics. What does this have todo with the Clifford algebra?

Answer 71. We saw earlier that for an infinitesimal Lorentz transform Λµν = δµν + ωµν , where ω is

antisymmetric, that the corresponding unitary acting on wavefunctions is

D(Λ) = 1 +i

2ωµνJ µν ,

and J µν satisfies the SO(1, 3) algebra.

It turns out that if there exists matrices satisfying the Clifford algebra γµ, γν = ηµν , then wecan construct generators

J µν := − i4

[γµ, γν ]

which satisfy the SO(1, 3) algebra. This particular representation is called the spinor represen-tation of the Lorentz algebra. This is important because any irrep of the Lorentz group is eithera tensor, a spinor (as above), or a direct product of a tensor and spinor. More on this later.

Important: this spinor representation is not unitary because the boost generators are not Hermi-tian (though they are antisymmetric). Ultimately, this leads us to use

ψψ instead of ψ†ψ

as the scalar quantity, for example, in a Lagrangian.

Question 72. Describe the difference between the vector and spinor representations.

Answer 72. Any representation of the Lorentz group has the same underlying transform on space-time, Λµ

ν . However, the unitary D(Λ) that acts on wavefunctions depends on the representation.

Scalar representation : D(Λ) = 1.

43


Vector representation : D(Λ) = Λ.

Spinor representation : D(Λ) = ei2ωµνJ µν .

D is a scalar in the scalar rep, but D is a 4×4 matrix in both the vector and spinor representations.Interestingly, the γ-matrices themselves transform under the vector representation!

Conclusion: the vector and spinor representations of Lorentz group mentioned here tell us howthe wavefunctions transform under the Lorentz group.

Question 73. Explain why the γ-matrices can be interpreted as a 4-vector. How can we constructtensors (i.e. generalizations of 4-vectors) from the γ-matrices?

Answer 73. There is an identity

[J µν , γρ] = −iγµηνρ + iγνηµρ

which combined with D(Λ) = 1 + i2ωµνJ µν implies that

D(Λ)γρD(Λ)−1 = Λρσγ

σ.

So, γρ transforms as a 4-vector under Lorentz transformation.

We have already constructed the antisymmetric tensor J µν = i4[γµ, γν ] by taking commutators of

the γ-matrices. In fact, we can construct higher-order, totally-antisymmetric tensors by enlargingthe commutator, which guarantees antisymmetry:

Aρστ = γ[ργσγτ ],Pρστη = γ[ργσγτγη].

You can check that these satisfy the tensor transformation rule, i.e.

D(Λ)J ρσD−1(Λ) = ΛρµΛσ

νJ µν .

Because we live in 1 + 3 = 4 dimensions, P has the maximum number of indices of any antisym-metric tensor. Otherwise, there would be something like Q02132 that has a repeated index.

Question 74. Give convenient, explicit forms of the Dirac matrices and the corresponding gen-erators of the Lorentz group in spinor representation.

Answer 74. These are on pg. 216-17:

γ0 = −i(

0 11 0

), γj = −i

(0 σj

−σj 0

), γ5 =

(1 00 1

),

J ij =εijk2

(σk 00 σk

),J i0 =

i

2

(σi 00 −σi

).

The raising and lowering of indices proceeds as usual with the metric tensor,

γµ = ηµνγν = (−γ0, ~γ) and γ5 = γ5.

For a detailed explanation, see https://physics.stackexchange.com/questions/296772/covariant-gamma-matrices.I think that γµ = ηµνγ

ν = (−γ0, ~γ) may change depending on choice of metric, but γ5 = γ5 isindependent of East Coast/West Coast convention.

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https://physics.stackexchange.com/questions/296772/covariant-gamma-matrices


While this spinor representation of the Lorentz group is reducible (i.e. into 2-component Weylspinors), the matrices γµ are irreducible in the sense that there is no proper subspace left invariantby all the γµ. In fact, we can show the Clifford algebra must be constructed with matrices whichare 4 × 4 or larger. If the underlying space is in d = 4, there are 42 = 16 independent matricespossible. We hope to construct all matrices out of the γµ, so we take γµ to be 4× 4.

In fact, this is borne out in the df of the antisymmetric tensors 1, γµ,J µν ,Aµνρ,Pµνρη. The df inthese matrices are 1,4,6,4,1 respectively. These sum to 16. (You can think of all of these tensorsas matrices of size 4× 4).

Question 75. Outline the following nice tricks with the γ-matrices and the tensors created fromthem: (1) the parity transformation with β = iγ0 (2) restatement of tensors using the “pseu-doscalar” γ5 (3) certain transpose properties with C = γ2β.

Answer 75. It’s good to know these exist...

1. Let β := iγ0. This gives a parity transform

βγ0β−1 = γ0, β~γβ−1 = −~γ, βγ5β−1 = γ5, βJ i0β−1 = −J i0, βJ ijβ−1 = J ij.

The above transform property of γ5 (that γ5 switches sign under parity) is why γ5 is referredto as a pseudoscalar. Also useful are

βγµ†β−1 = −γµ, βJ ρσ†β−1 = J ρσ.

2. We can make the df of the tensors constructed with (enlarged) commutators of the γ-matricesvery explicit by introducing the totally antisymmetric tensor.

J ρσ =1

4ερστηγ5[γτ , γη],Aρστ = 3!iερστηγ5γη,Pρστη = 4!iερστηγ5.

This works because γ5 = −iγ0γ1γ2γ3, so the multiplication done in the commutators isalready inside γ5. (To prove these, use the anticommutation γµ, γν = 2ηµν). Neat!

3. Finally, let C := γ2β. This will be related to charge-conjugation of a Dirac spinor. We have

C γµC−1 = −γTµ ,C γ5C

−1 = γT5 ,C (γ5γµ)C −1 = −(γ5γµ)T ,CJµνC −1 = −J Tµν .

5.5 Causal Dirac field

Phew! The last section was a hard one. Now we will construct a wavefunction ψ(x), called a

Dirac spinor, that transforms under the spinor representation D(Λ) = ei2ωµνJ µν of the Lorentz

group. We will start with the usual plane-wave expansion

ψ+l (x) =

∑σ

∫d3~p

(2π)3/2ul(~pσ)a~pσe

ipx, ψc−l (x) =∑σ

∫d3~p

(2π)3/2vl(~pσ)ac†~pσe

−ipx.

The regular normalization 1√2p0

is inside the coefficient (u, v)l(~pσ). We will calculate them after

we choose the spinors (u, v)l(~kσ) at the reference momentum k = (m,~0).

45


Question 76. Outline how we can choose the spinors (u, v)l(~kσ) at the reference momentumk = (m,~0). This will give us the spinors at general momenta, (u, v)l(~pσ).

Answer 76. The derivation is difficult and subtle; it’s on page 220-22. Rather than rehashing thederivation, I will try to illuminate what the spinor representation of Lorentz group has to do withthe choice of reference spinors, (u, v)l(~kσ).

We start with the transformation of the coefficients under the Lorentz group, which is supposedto mesh with the transformations of the creation and annihilation operators:∑

σ

ul(0σ) ~J(j)σσ =

∑l

Jllul(0σ).

The main point is that we found Jll to be diagonal (see the previous section), which leads tothe idea of Weyl spinors. Enforcing the transformation condition on the spinors at the referencemomentum gives

u(0, σ) =

(u+(0, σ)u−(0, σ)

),

where u±(0, σ) =

(u 1

2±(0, σ)

u−12±(0, σ)

)is the Weyl spinor. Here,

• ± refers to the irrep of the Lorentz group (i.e. LH or RH Weyl spinor) which is connectedto the block-diagonal form of the J µν generator we saw earlier.

• m in um±(0, σ) refers to the spin of the Weyl spinor under the Wigner rotation: its eigenvalue

with respect to ~J(j)σσ . Ultimately, this came from the little-group Dσσ(W (Λ, p)).

• σ refers to the spin of this Dirac spinor in the Dirac representation: its eigenvalue withrespect to the rotational part of J µν .

As described on pg. 220, we pick the convention that ~J is aligned with the rotational part of J µν .Once we pick this convention, it means that, for example (for some undetermined constants c±):

u(0,1

2) =

c+

0c−0

because the Weyl spinors must carry a δmσ. Otherwise, part of the Weyl spinor would point inthe wrong direction and u(0, 1

2) would no longer be a eigenvector of the rotational part of J µν .

The spinors at general momenta are obtained as usual,

u(~pσ) =

√m

p0D(L(p))u(~0σ), v(~pσ) =

√m

p0D(L(p))v(~0σ).

Conclusion: the spinor representation of Lorentz group pops up in two places in our derivation of(u, v)(~pσ): (1) determination of (u, v)(~0σ) (2) transform to finite momenta, through D(L(p)).

46


Question 77. Describe how the spinors (i.e. coefficient functions) in the previous question satisfythe Dirac equation. Is this real physics or just a convention?

Answer 77. If you look at the spinors on the top of pg. 221, for example

u(0,1

2) =

c+

0c−0

,

it’s clear that u(0, 12) is not generally an eigenvector of pµγ

µ. The thing that makes the coefficientfunctions eigenvectors of the operator pµγ

µ is the convention chosen in (5.5.33).

In fact, I think the idea that the Dirac equation is a result of convention is generally true, regardlessof how you derive it (even as a “square root” of the Klein-Gordon equation). For example, seehttps://quantummechanics.ucsd.edu/ph130a/130_notes/node45.html. In these notes, thereis no arbitrariness up until φ(L) and φ(R) are introduced. Another way to think of this is that scalarsonly have two square roots (i.e. (±1)2 = 1), but matrices have many square roots. Therefore,there is no unique “square root” of the Klein-Gordon equation, which is really a matrix equation.

Question 78. Outline the construction of the Dirac field, given that we already constructed themomentum-dependent spinors, (u, v)(~pσ).

Answer 78. Again, we would like to find the coefficients in the linear combination

ψl(x) = κψ+l (x) + λψ−cl (x).

Here, l indexes each entry in a Dirac spinor. We enforce causality by requiring that [ψ(x), ψ(y)] = 0for x− y spacelike. It turns out that the causality condition implies anticommutations and hencefermionic statistics.

Question 79. Explain the vector field ended up being bosonic and the Dirac field ended up beingfermionic. Where did the difference come from?

Answer 79. Both the vector and Dirac fields are 4-component fields. However, compare thecommutations for the vector field (pg 210-211)

[vµ(x), vν†(y)]± = (|κ|2 ∓ |λ|2)(ηµν − 1

m2∂µ∂ν)∆+(x− y)

with those for the Dirac field (pg 223)

[ψl(x), ψ†l(y)]± = (|κ|2(−γµ∂µ + bum)β∆+(x− y)∓ |λ|2(−γµ∂µ + bvm)β∆+(y − x))ll.

We see that there are ∓ signs in both of them. However, ∂µ∂ν is an even function and ∂µ is anodd function. That is where the difference comes from, and we need to switch a sign for the Diracfield.

Now the question is, why does one of them have a covariant ηµν − 1m2∂

µ∂ν operator and the otherhave a Dirac operator −γµ∂µ +m? The answer is that both of these are “spin sums,” so to speak,which arise from the (anti)commutations of the total fields:

Πµν(~p) :=∑σ

eµ(~pσ)eν∗(~pσ) = ηµν +1

m2pµpν ,

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https://quantummechanics.ucsd.edu/ph130a/130_notes/node45.html


Nll(~p) :=∑σ

ul(~pσ)u∗l (~pσ) =1

2p0(−ipµγµ +m)β.

Conclusion: It seems we can determine whether a field is fermionic or bosonic merely by lookingat the numerator of the propagator (which is exactly the spin sum; see Schwartz pg. 460). If thespin sum is even, we have a boson; if it is odd, we have a fermion.

5.6 General irreps of the Lorentz group

We now study general irreps of the Lorentz group, of which the scalar, vector and Dirac represen-tations were examples. All fields can be constructed as direct sums of fields which are constructedfrom irreps.

In fact, this is not that hard. Weinberg’s rigorous derivation is equivalent to what Girma Hailutaught me in Physics 251b, i.e. that the Lorentz group is

SO+(1, 3) ∼ SU(2)× SU(2)

Z2

.

Question 80. How can we label (i.e. classify) general irreps of the proper orthochronous Lorentzgroup, SO+(1, 3)?

Answer 80. The Lorentz algebra is

[Jµν ,Jρσ] = i(Jρνησµ + Jµρηνσ − Jσνηρµ − Jµσηνρ),

and the point is to find matrices Jµν that satisfy this algebra. As we saw, it is not just the sizeof the matrix that determines the representation; for example, both the vector and Dirac-spinorrepresentations are constructed from generators which are 4× 4 matrices.

To classify general representations, we will split the six independent components of Jµν (i.e.diagonal entries are zero because Jµν is antisymmetric under µ⇔ ν) into two three vectors:

Angular momentum 3-vector : ~J = (J23,J31,J12),

Boost 3-vector : ~H = (J10,J20,J30).

There is a problem; namely, that [Ji,Hj] 6= 0. We can fix this by constructing the linearcombinations

~A =1

2( ~J + i ~H ) and ~B =

1

2( ~J − i ~H ).

This decouples the 3-vectors,

[Ai,Aj] = iεijkAk, [Bi,Bj] = iεijkBk, [Ai,Bj] = 0.

Great! The combinations ~A and ~B each satisfy an SU(2) algebra.

Therefore, the representation is labeled by (A,B), where A describes the representation of thefirst SU(2) and B describes the representation of the second SU(2). Both A and B are eitherintegers or half-integers.

48


Question 81. Discuss which generators are Hermitian and which are not, and hence which rep-resentations are unitary and which are not. Does this matter?

Answer 81. ~J , ~A , ~B are Hermitian, and ~H is anti-Hermitian. We discovered this, for example,

at the beginning of section 5.4. The fact that ~H is anti-unitary is why we had to multiply it byi when constructing ~A and ~B.

Why is ~H anti-Hermitian? This is because there is a relative (−1) in the metric, (−+ ++), anda boost operator constructed with the boost generators must preserve the invariant −dt2 + d~x2.If the metric is Euclidean, (+ + ++), then ~H would be Hermitian.

Does this matter? Weinberg pg. 231: “There is no problem in working with non-unitary repre-sentations, because the objects we are now concerned with are fields, not wavefunctions, and donot need to have a Lorentz-invariant positive norm.” This is really confusing because the Lorentzalgebra was derived on pg. 60 from unitary representations, U(Λ, a)U(1 + ω, ε)U−1(Λ, a). Andbesides, eq. (2.4.4) literally says that Jµν is Hermitian. It turns out that this only applies forthe transformations of wavefunctions, i.e. creation and annihilation operators. Because this is anexample of a valid representation, we can go backwards and find out what the algebra is. Becausethe algebra is the same for every representation, there is no logical fallacy, even if the logic is abit convoluted.

Also, Schwartz makes a big deal of this. Subsection (8.1.1) in his book is literally called “Unitarityvs. Lorentz invariance,” and he says that we would like our representations to be unitary; becausethere are no finite-dimensional unitary representations of the Lorentz group, we have to go toinfinite dimensions: hence, the Wigner little group. This makes sense because we had to introduceWigner little group when we studied relativistic QM in Weinberg Ch 2, for which we demandedthe representation be unitary.

What is the resolution? I think the resolution is on Schwartz pg. 184, “the next step towards quan-tizing a theory with spinors is to use these Lorentz group representations to generate irreducibleunitary representations of the Poincare group.” On pg. 188, he says that the coefficient functionsu(~pσ) and v(~pσ) transform in a unitary way under the Poincare group with infinite-dimensionalrepresentation.

Weinberg’s math says the same thing. According to Weinberg (5.1.6),

U0(Λ, b)ψ+l U−10 (Λ, b) =

∑l

Dll(Λ−1)ψ+

l(Λx+ b).

This turns out to give

U0(Λ, b)a~pσU−10 (Λ, b) = ei(Λp)·b

√(Λp)0

p0

∑σ

D(jn)σσ (W−1(Λ, p))a~pΛσ.

The factor√

(Λp)0

p0 is a reflection of the non-unitarity of boosts, in the sense that its magnitude is

not 1; we had to insert it in order to preserve unitarity (i.e. preserve the norm of wavefunctions).

ei(Λp)·b is fine because its magnitude is 1, and the Wigner rotation∑

σD(jn)σσ (W−1(Λ, p)) is also fine

because rotations are unitary.

Conclusion: The Lorentz-group representation of ψ(x), or a field in general, need not be unitary.To quote from earlier in the notes, “we should construct U out of creation and annihilation

49


operators. It turns out that to guarantee such a U is a good Lorentz scalar, we should build Uout of fields, which are constructed such that under Lorentz transform, the transformation of thefields are position-independent:

U0(Λ, a)ψ±l (x)U−10 (Λ, a) =

∑l

Dll(Λ−1)ψ±

l(Λx+ a).′′

If U is a Lorentz scalar, we’re good; it is irrelevant whether Dll(Λ−1) is unitary. That is why, for

example, we use ψ(x)ψ(x) instead of ψ†(x)ψ(x).

As for the transformations of a~pσ, in fact the transform U0(Λ, b)a~pσU−10 (Λ, b) listed above is unitary

but was actually fixed before the idea of field was even introduced. The point is that we need toinsert these into a field to satisfy cluster decomposition, and if we want to describe, for example,spin-3/2 particles, we need a field that has enough spin indices to accomodate the desired creationand annihilation operators.

Question 82. Why do we care about (12, 0)⊕ (0, 1

2) as a whole? Why not just reduce it to (1

2, 0)?

Answer 82. While it is true that (12, 0) ⊕ (0, 1

2) is a reducible representation of the proper or-

thochronous Lorentz group, it is not a reducible representation of the entire Lorentz group includ-ing space inversion. This is because the parity transform β gives

βA β−1 = B, βBβ−1 = A .

An irreducible (A,B) representation of the proper orthochronous Lorentz group does not providea representation of the entire Lorentz group unless A = B. For A 6= B, the solution is to take

(A,B)⊕ (B,A).

5.7 CPT theorem; spin-statistics theorem

5.8 Massless fields

Question 83. Which representations of the Lorentz group are easily adapted to massless fields?Which are more difficult?

Answer 83. The scalar and Dirac fields for m > 0 can be smoothly taken to m → 0 withoutissue. The vector field cannot. This is because the vector field has some m−1 factors which blowup (i.e. in the spin sum) but the Dirac field does not. Another way to think about it is that inthe m→ 0 vector representation, at least one of the polarization vectors blows up.

Generally, it turns out that creation and annihilation operators for massless particles of spin j ≥ 1cannot be used to construct all of the irreps (A,B) which can be constructed for massive particles.This leads to gauge invariance.

Question 84. Mathematically, why can’t some irreps of the Lorentz group be encoded in masslessfields?

Answer 84. Suppose the field lives in the (A,B) irrep of the Lorentz group; the eigenvalues of~J2 are a2, b2 and we label the corresponding coefficient function by uab(~kσ). For massive particles,

50


−A ≤ a ≤ A but for massless particles, a = ±A. (Well, we will discover this below.) We showedpreviously that the irreps of the Lorentz group are basically matrices D(Λ)ll of various size which

satisfy the right algebra. For example, (Jij)a′b′,ab = εijk[(J(A)k )a′aδb′b + (J

(B)k )b′bδa′a].

The ISO(2) little-group transformation implies that for the rotation part parametrized by θ,

σuab(~kσ) = (a+ b)uab(~kσ),−σvab(~kσ) = (a+ b)vab(~kσ).

For the translational part parametrized by α, β,

(J(A)1 − iJ (A)

2 )aa′ua′b(~kσ) = 0, (J(B)1 + iJ

(B)2 )bb′uab′(~kσ) = 0.

The above line means that the massless state is in lowest angular-momentum state for A andhighest angular-momentum state for B. Put together, this implies

a = −A, b = B, σ = B − A.

This field annihilates massless particles of helicity σ and creates massless antiparticles of helicity−σ. Because the vector representation is (1

2, 1

2), it can only describe helicity zero. The simplest

covariant massless field with helicity one is therefore (1, 0)⊕ (0, 1), which is the Fµν tensor.

Question 85. What are the transformations of the coefficient functions for a massless vector field,and how does this imply gauge-invariance?

Answer 85. Recall that the massless little group is ISO(2) and the reference momentum iskµ = (k, k, 0, 0). Because of the wavefunction transformation

U(Λ)Ψpσ =

√(Λp)0

p0eiσθ(Λ,p)ΨΛp,σ,

we know the transformation of a†~pσ and a~pσ. If the field transforms under a certain irrep Dll(Λ) ofthe Lorentz group, then we also know the transformation of the coefficient functions. If S(α, β)describes the combined rotations and boosts of the little group in the xy-plane (i.e. the “transla-tional” part of ISO(2)), then we end up with

u(~kσ) =∑l

Dll(S(α, β))ul(~kσ).

(There is also the R(θ) part of the little group, which is easy to implement.) We claim that theabove equation cannot be satisfied for general irreps of the Lorentz group.

For example, take the vector representation (12, 1

2). Let us see why the above conditions cannot

be satisfied. Now, l = µ and we can introduce the polarization vector eµ by dividing out theboost term,

uµ(~pσ) =1√2p0

eµ(~pσ).

The conditions to be satisfied are (Einstein summation convention)

eµ(~kσ)eiσθ = R(θ)µνeν(~kσ) and eµ(~kσ) = S(α, β)µνe

ν(~kσ).

Unfortunately there is no such set eµ(~kσ) which satisfies these relations. You can think aboutthis as follows: because the vector field has spin j = 1, there should be three polarization vectors.Unfortunately, the rotational part of ISO(2), which is eµ(~kσ)eiσθ = R(θ)µνe

ν(~kσ), parametrizesrotations about a single axis and hence only allows two polarization vectors. This is why one ofthe polarization vectors blows up in the m→ 0+ limit of the massive vector field.

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Question 86. How do the above relations (although they cannot all be satisfied without extraconditions) return the Coulomb gauge of the photon field?

Answer 86. eµ(~kσ)eiσθ = R(θ)µνeν(~kσ) implies eµ(~kσ) = 1√

2(0, 1,±i, 0) in (t, ~x) notation. Apply-

ing rotations and boosts gives, generally

e0(~pσ) = 0, ~p · ~e(~pσ) = 0 =⇒ a0(x) = 0, ~∇ · ~a(x) = 0.

Question 87. How can we insert massless fields into a Lagrangian?

Answer 87. There are two ways. (1) Demand gauge-invariance and ensure that the vector fieldaµ is always coupled to a current jµ such that ∂µjµ = 0. (2) construct tensor field

Fµν = ∂µaν − ∂νaµ

in the (1, 0) ⊕ (0, 1) representation. The relative minus sign between the two parts cancels theundesirable extra term in the transformations of the polarization vectors, so this field satisfies theentire ISO(2) little group. Similarly, gravitons (j = 2) can be put together in an antisymmetricway to construct the tensor field Rµνρσ, the Riemann curvature tensor.

Massless particles with j ≥ 3 would couple to conserved tensors with three or more indices, butbecause d = 1 + 3, aside from total derivatives there are none. Conclusion: “high-spin masslessparticles cannot produce long-range forces.” This logic does not carry over to high-spin massiveparticles because they are not constrained to couple to conserved tensors (i.e. no gauge-invariance).High-spin massive particles are possible (and exist); they are usually very heavy.

5.9 Problems

Question 88. Consider a free field ψµl (x) which annihilates and creates a self-charge-conjugateparticle of spin j = 3

2and mass m > 0. Show how to calculate the coefficient functions uµl (~pσ)

which multiply the annihilation operators a~pσ in this field, in such a way that the field transformsunder Lorentz transformations like a Dirac field ψl with an extra four-vector index µ. What fieldequations and algebraic and reality conditions does this field satisfy? Evaluate the matrix P µν(p),defined for p2 = −m2 by

P µνlm (p) = 2p0

∑σ

uµl (~pσ)uν∗m (~pσ).

What are the commutation relations of this field? How does the field transform under the inversionsP,C,T?

Answer 88. TODO

Question 89. Work out the transformation properties of fields of type (j, 0)⊕ (0, j) for masslessparticles of helicity ±j under the inversions P,C,T.

Answer 89. TODO

Question 90. Consider a general field ψab describing particles of spin j and mass m > 0 whichtransforms according to the (A,B) representation of the Lorentz group. Suppose it has an inter-action Hamiltonian

V =

∫d3~x(ψab(x)Jab(x) + h.c. ),

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where Jab is an external C-current. What is the asymptotic behavior of the matrix element foremitting these particles for energy E m and definite helicity? (Assume the Fourier transformof the current has values for different a, b that are of the same order of magnitude, and that donot depend strongly on E.)

Answer 90. Weinberg shows in section 5.7 of his book that you can construct a massive spin-jtheory quite easily. The point is that, as mentioned in https://physics.stackexchange.com/

questions/14932/why-do-we-not-have-spin-greater-than-2, the propagator becomes badlydivergent for high-spin massive particles. (However, the article I linked to mixes up massive andmassless arguments, so it’s not correct the whole way through.)

The heuristic result (see link) is that the propagator grows approximately as

∆ ∼ Ej−2.

This is the same as the asymptotic behavior of the matrix element. A more rigorous way is to notethat in Weinberg (5.7.33), the construction of a field (A,B) of spin j involves many derivatives,which are basically like energy.

If the propagator looks like pn, then the position-space propagator looks like x−n, so these effectsbecome increasingly short-ranged. One way to think of this is if the exponent n is very large, thenthe effect at small energies gets heavily suppressed.

(Weinberg notes in the footnote to pg. 253 that the conserved charge with the most indices in1 + 3 dimensions is the energy-momentum 4-vector, and the corresponding current with the mostindices in 1 + 3 dimensions is the stress-energy tensor. This is important for massless particlesbecause they have to be gauged, and hence coupled to a conserved current. Thus, we cannot havemeaningful massless particles with j > 2, since the stress-energy tensor has two indices. Thisis the graviton. Massive particles, the subject of this question, do not have to be coupled to aconserved current. They may just as well be coupled to a scalar current, which they are in thisquestion.)

6 Feynman Rules

Feynman does rule! Feynman, Schwinger, and Tomonaga developed in the late 1940s a methodof perturbation theory which was Lorentz invariant to all orders. In this chapter, we will deriveFeynman rules with Dyson’s method (1949). First we will derive the position-space rules; then wederive the momentum-space rules.

6.1 Dyson’s derivation in position space

Question 91. Describe Dyson’s (1949) derivation of the Feynman rules from perturbation theory.

Answer 91. The procedure is extremely similar to the first problem I solved in section 4.4 of thesenotes. The S-matrix from initial state α = ~p1σ1n1, ~p2σ2n2, · · · to final state β = ~p′1σ

′1n′1, ~p′2σ′2n′2, · · ·

is given by the Dyson series

Sβα =∞∑N=0

∫d4x1 · · · d4xN〈β|TU (x1) · · ·U (xN)|α〉,

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https://physics.stackexchange.com/questions/14932/why-do-we-not-have-spin-greater-than-2

https://physics.stackexchange.com/questions/14932/why-do-we-not-have-spin-greater-than-2


where |α〉 = a†~p1σ1n1a†~p2σ2n2

· · · |0〉, and similarly for |β〉. The interaction densities U (xi) are builtout of fields, which are built out of creation and annihilation operators just like the states |α〉 and|β〉.

We (anti)commute all annihilation operators to the RHS and all creation operators to the LHS.There may be nonzero (anti)commutators which pop up as a result. In fact, only the productof these nonzero (anti)commutators will survive, because aq|0〉 = 0 for any species q. Therefore,every operator in this Dyson series must be contracted with another in order to survive. Forexample (recall that []± stands for commutator and anticommutator, respectively):

[a~pσn, ψ†l (x)]± =

e−ipx

(2π)3/2u∗l (~pσn).

The sum of all such contractions gives the Feynman diagrams.

Question 92. List the position-space Feynman rules.

Answer 92. To calculate the contribution to the S-matrix for a given process of order Ni in eachof the interaction terms Ui(x), this is the procedure:

1. Draw all Feynman diagrams containing Ni vertices of each type i and containing a linecoming into the diagrams from the left for each particle or antiparticle in the initial state,and a line leaving the diagram from the right for each particle or antiparticle in the finalstate. Draw any number of internal lines running from one vertex to another, as required togive each vertex the proper number of attached lines.

2. The lines of any particle have an arrow pointing to the right (positive time); the lines of anyantiparticle have an arrow pointing to the left.

3. Each vertex gets a factor −igi. It is customary to redefine the coupling such that, forexample, the factor corresponding to the interaction gφ4 is g/4!.

Particles and antiparticles leaving the diagram get factors

e−ipx

(2π)3/2u∗l (~p

′σ′n′) ande−ipx

(2π)3/2vl(~p

′σ′n′), respectively.

Particles and antiparticles entering the diagram get factors

eipx

(2π)3/2ul(~pσn) and

eipx

(2π)3/2v∗l (~pσn), respectively.

A (anti)particle which doesn’t interact with anything gets a factor δ(3)(~p′ − ~p)δσ′σδn′n. Aninternal line connecting two vertices in the diagram gets a factor

−i∆lm(x− y).

4. Perform the integral∫d4x1 · · · d4xN and add up the results from each Feynman diagram.

There are sometimes extra (−1) or symmetry factors related to fermion loops and additionalscattering symmetries. The safest thing to do is just to start from Dyson’s formula and see howmany factors come out.

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6.2 Propagators

Question 93. Describe the propagator of an arbitrary field ψl(x). What does it mean?

Answer 93. Let the annihilation and creation parts of an arbitrary field ψl(x) be

ψ+(x) =∑σn

∫d3~pul(x; ~pσn)a~pσn and ψ−(x) =

∑σn

∫d3~pvl(x; ~pσn)a†~pσn.

Such a propagator arises from Dyson’s formalism when the interactions Ui are written in normalordered form, and we would like to (anti)commute ψl(x) from Ui(x) past ψ†m(y) from Uj(y).(Normal-ordering each interaction guarantees there are no contractions inside an interaction. Wechoose ψ†m(y) instead of ψm(y) because TODO). There are three cases to consider:

1. x0 > y0 and x − y is spacelike: Because time-ordering means “later on left,” and we aretrying to put creation fields on the left, it must be that (before we commute anything) theorder is ψl(x)ψ†m(y). We should therefore move the creation part of ψ†m(y) to the left of ψl(x);the nonvanishing part is

[ψ+l (x), ψ+†

m (y)]±.

2. x0 < y0 and x− y is spacelike: Same thing, except we select the creation part of ψl(x).

[ψ−†m (y), ψ−l (x)]±.

3. x − y is timelike: In previous work, we said that Lorentz invariance of the S-matrix meansthat we must have [ψl(x), ψ†m(y)] = 0 for timelike separations. We will guarantee this is truebased on the coefficients on the fields.

The result is what we will define as the propagator,

−i∆lm(x− y) := θ(x0 − y0)[ψ+l (x), ψ+†

m (y)]± ± θ(y0 − x0)[ψ−†m (y), ψ−l (x)]±.

The relative ± between the two terms arises because the T operator does not produce any com-mutators but it does produce negative signs.

The propagator ∆lm(x−y) means (for x0 ≥ y0) either a particle was created at y and destroyed at xor an antiparticle was created at x and destroyed at y. You can think of the magnitude of ∆(x−y)as how easily this process occurs. For example, ∆lm(x− y) could be like a Coulomb potential; theGreen function of the D’Alembertian operator is 1

4πrδ(t− r

c) and falls off with distance r = |~x−~y|.

Question 94. For general fields, how can we calculate the propagator?

Answer 94. This is a little involved. Here it goes...

From the previous question,

−i∆lm(x− y) := θ(x0 − y0)[ψ+l (x), ψ+†

m (y)]± ± θ(y0 − x0)[ψ−†m (y), ψ−l (x)]±

= θ(x0− y0)

∫d3~p

(2π)3

∑σ

ul(~pσn)u∗m(~pσn)eip(x−y)± θ(y0− x0)

∫d3~p

(2π)3

∑σ

v∗m(~pσn)vl(~pσn)eip(y−x).

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The spin sums are∑σ

ul(~pσn)u∗m(~pσn) =Plm(~p,

√~p2 +m2

n)

2√~p2 +m2

n

and∑σ

vl(~pσn)v∗m(~pσn) = ±Plm(−~p,−

√~p2 +m2

n)

2√~p2 +m2

n

.

Here, Plm(~p, ω) is a polynomial in ~p and ω. Apparently Plm has an general explicit form, (6.2.7).Generally, it is the numerator of the propagator, so we know what it looks like for scalars, Diracfields, and vector fields. From now on, we will simply call Plm(~p, ω) = P (p).

If we use ∆+(x) =∫

d3~p(2π)3

eipx

2p0 , we find

−i∆lm(x− y) = θ(x0 − y0)P (p)∆+(x− y) + θ(y0 − x0)P (p)∆+(y − x).

Now (6.2.10), Weinberg does something sneaky and extends the definition of P (p) to momentawhich are off-shell; he calls this P (L)(p). Substituting p = −i∂x and using derivative product rulegives

∆lm(x− y) = P (L)(−i∂x)∆F (x− y),

where −i∆F (x) := θ(x0)∆+(x) + θ(−x0)∆+(−x) is the Feynman propagator. Now we use

θ(t) = −12πi

∫e−iωtdωω+iε

, which gives

∆F (x) =

∫d4q

(2π)4

eiqx

q2 +m2 − iε=⇒ ∆lm(x− y) =

∫d4q

(2π)4

P(L)lm (q)eiqx

q2 +m2 − iε.

Of course, this is the Fourier transform of the momentum-space propagator, which is (6.3.3)

−i(2π)4

Plm(q)

q2 +m2l − iε

.

Question 95. How is the Feynman propagator, with the generic

1

k2 −m2

structure, related to the propagators we use in condensed matter field theory, which have thegeneric

1

ω − εkstructure?

Answer 95. The Feynman propagator has both retarded and advanced propagators hiding inside.This is due to how we define the fields; for example, φ(x) is a linear combination of creation andannihilation operators.

Roughly speaking, the sum of retarded and advanced propagators

1

ω − εk+

1

−ω − εk→ 1

ω2 − ε2kgives the Feynman propagator, where ε2k = k2 +m2. This is also why we only need to draw a singleFeynman diagram to represent all combinations of advanced and retarded processes. Otherwisewe would have to care about which vertices are more to the right or to the left, which direction isthe photon line going in, etc. See section 5.1 of these notes.

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6.3 Problems

Question 96. Consider the theory of a real scalar field φ with interaction (in the interaction pic-ture) V (t) = g

3!

∫d3~xφ(x)3. Calculate the connected S-matrix element for scalar-scalar scattering

to O(g2), doing all integrals. Use the results to calculate the differential cross-section for thisprocess in the CM reference frame.

Answer 96. Scalar-scalar scattering means 2→ 2 scattering. We use the Dyson formula describedin section 6.1 of these notes,

Sβα =∞∑N=0

∫d4x1 · · · d4xN〈β|TU (x1) · · ·U (xN)|α〉,

where |α〉 = a†~p1a†~p2|0〉, and similarly |β〉 = a†~p3

a†~p4|0〉.

It should be clear that when we smash the field

φ(x) =

∫k

1√2E~k

(a~keikx + a†~ke

−ikx)

into the state |α〉, for example, we will get an external factor of

1

(2π)3/2

eikx√2E~k

.

Here are the possible diagrams:

Using the position-space Feynman rules, we find that

S~p3~p4,~p1~p2 =(ig)2

(2π)6√

16E1E2E3E4

∫xy

(−i∆(x, y))(eix(p1−p2)−iy(p3−p4)+eix(p1−p3)−iy(p2−p4)+eix(p1−p4)−iy(p2−p3)).

Using the momentum-space Feynman rules, we find that S~p3~p4,~p1~p2 = −2πiδ(p0 − pf )M~p3~p4,~p1~p2 ,where

M~p3~p4,~p1~p2 =(ig)2/(2πi)

(2π)6√

16E1E2E3E4

(i

(p1 + p2)2 −m2 + iε+

i

(p1 − p3)2 −m2 + iε+

i

(p1 − p4)2 −m2 + iε

).

Introducing the Mandelstam variables s, t, u (we don’t even need to know which is which) andusing the CM differential cross-section in Weinberg (3.4.30) gives

dσ

dΩ=

g4|~k′|(2π)10E2|~k|

∣∣∣∣ 1

s−m2 + iε+

1

t−m2 + iε+

1

u−m2 + iε

∣∣∣∣2.Here, |~k| and |~k′| are the momenta of either of the scalar particles before and after the collision,and E is the total energy (which is conserved).

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Question 97. What is the contribution in Feynman diagrams from the contraction of the deriva-tive ∂µψl(x) of a Dirac field with the adjoint ψ†m(y) of the field?

Answer 97. You just take the regular propagator

∆lm(x− y) =

∫d4q

(2π)4

P(L)lm (q)eiq(x−y)

q2 +m2 − iε

and apply ∂µ. So the momentum-space contribution has an extra iqµ or something, but it looksbasically like a propagator.

Question 98. Use Gell-Mann - Low theorem to give expressions for the vevs of these Heisenberg-picture operators: Φ(x), TΦ(x)Φ(y). Use the interaction V (t) = g

3!

∫d3~xφ(x)3 and calculate the

answer to O(g) and also O(g2).

Answer 98. Take the Lagrangian density to be (in Schwartz notation with +−−− metric)

L =1

2(∂µφ)(∂µφ)− 1

2m2φ2 +

g

3!φ3.

Gell-Mann - Low theorem basically says we only need the connected diagrams, because the dis-connected ones will be divided out from the denominator. As far as I can tell, these are the onlydiagrams:

and let us do 〈Φ(x)〉 first. We expect this to be independent of x. Interpreting the diagram gives

〈Φ(x)〉 = ig

∫y

∆(x, y)∆(y, y) = 0.

This is independent of x because we are effectively integrating over all possible values of x − y.We could guess it’s if g is small enough such that there is no spontaneous symmetry breaking.The mathematical reason is that ∆(y, y) = 0 since it is neither advanced nor retarded.

The other one is just

〈Φ(x)Φ(y)〉 = ∆(x, y)− g2

∫z1z2

∆(x, z1)∆(z1, z2)2∆(z2, y).

In fact, there is another diagram at O(g2): the Hartree diagram. However, this is usually takento be zero because ∆(x, x) = 0 since it is neither retarded nor advanced. Or if you have reason tothink ∆(x, x) is not zero, you can add the Hartree contribution.

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7 The Canonical Formalism

The meaning of canonical formalism is: postulate a Lagrangian and apply the rules of canonicalquantization; i.e. with canonical (anti)commutations. This is historically how quantum fieldtheory was developed.

We saw in the previous chapter that we can calculate S-matrix elements using the Hamiltonianformalism. However, in this chapter we introduce the Lagrangian formalism. One advantageis that the Lorentz-invariance of the classical Lagrangian guarantees Lorentz-invariance of thequantum theory. Another advantage is that there will be no ad hoc non-scalar terms which mustbe added to the interaction density (for example, AµJµ for a massless photon) to compensate fornon-covariant terms in the propagators. These will pop out automatically from the Lagrangianformulation.

7.1 Canonical variables

Question 99. What is “canonical” about the fields we constructed in previous chapters, ψl(x)?

Answer 99. The free fields we constructed previously are “canonical” in the sense that theysatisfy the canonical commutations. Schematically,

[x, p] = i, [x, x] = [p, p] = 0.

We found previously that, for instance, the real scalar field satisfies

[φ(x), φ(y)] = ∆(x− y) and ∆(~x, 0) = 0, ∆(~x, 0) = −iδ3(~x).

The above properties follow from the definition ∆(x) =∫

d3~k(2π)3

eikx−e−ikx2k0 . Therefore, the field and

its time derivative (i.e. the conjugate momentum) satisfy

[φ(~x, t), φ(~y, t)] = iδ(~x− ~y).

This follows from differentiating with respect to y0.

Question 100. Describe how different kinds of fields can be interpreted as conjugate pairs ofcanonical variables.

Answer 100. The case of the real scalar field was done above. For the complex scalar field, thecommutation was [φ(x), φ†(y)] = ∆(x− y), so we have to change the canonical momentum to

(q, p) = (φ, φ†).

Or, you can decompose into real and imaginary parts and make both of them canonical fields,

(q, p) = (φ1, φ1) and (φ2, φ2)

where φ = 1√2(φ1 + iφ2).

For the spin-1 vector field vµ, it turns out that v0 can be expressed in terms of the other variables,so there are only three canonical variables, vi. (This follows from ∂µv

µ = 0, which essentially

59


removes one df. This is because spin-1 only really needs three df, while a vector field has 4 df.)We may take

(qi, pi) = (vi, vi + ∂iv0) and v0 =

∂i · pim2

.

For the Dirac field, we take(q, p) = (ψ, iψ†).

The factor of i is the factor in [x, p] = i.

Conclusion: the identification of fields as canonical variables, and their conjugates, must be donecarefully. Sometimes you must take a time-derivative (scalar and vector fields) and sometimes youdon’t have to (Dirac field). (This is because the anticommutator of ψ(x) and ψ†(x) already had aderivative, /∂.)

Question 101. In a canonical formalism, describe how a functional derivative is mapped to acommutator.

Answer 101. This is so pretty! If F [q, p] is bosonic, then it doesn’t matter whether q and p arebosonic or fermionic; we have

δF

δq= i[p, F ] and

δF

δp= i[F, q].

This is a generalization of Hamilton’s equations of motion. Beautiful!

This can be motivated by imagining F is normal ordered, with all qs on the left and all ps on theright. However, the above is a definition that resolves ambiguities when F may not be normal-ordered.

Question 102. Transform the free-field Hamiltonian from second-quantized notation into thecanonical variables.

Answer 102. Consider a real scalar field. The free-field Hamiltonian is

H =

∫~k

√~k2 +m2a†~ka~k.

Recall that the expression for the real scalar field is

φ(~x, t) =

∫~k

1√2k0

(a~keikx + a†~ke

−ikx).

Up to the zero-point energy, this turns out to be

H =1

2

∫~x

p2 + (~∇q)2 +m2q2.

(Schematically, you can see this by noting

1

2

∫~x

p2 + (~∇q)2 +m2q2 ∼ 1

2(

1

k0((k0)2 + (~k)2 +m2)) ∼ k0,

since k0 =√~k2 +m2.) I feel this is not very nice; for example, it’s not clear to me that this

expression H[p, q] is even the only one that returns H =∫~k

√~k2 +m2a†~ka~k. However, Weinberg’s

logic is that the second-quantized form of H is the one which is certain, and we are just trying tofind a way to write it with qs and ps.

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Question 103. Transform the Hamiltonian from the previous question, H[q, p], to a Lagrangian.Show it returns the usual scalar Lagrangian of relativistic QFT.

Answer 103. We use the Legendre transform

L =

∫~x

p(x)q(x)−H.

This gives

L =

∫~x

(pq − 1

2p2 − 1

2(~∇q)2 − 1

2m2q2) = −1

2

∫~x

(∂µφ∂µφ−m2φ2).

Question 104. Show that the canonical variables of a free theory can be recast into the canonicalvariables of an interacting theory via similarity transformation.

Answer 104. Let Q,P be the canonical variables in the interacting theory and q, p be the canon-ical variables in the free theory. Consider the similarity transforms which preserve the canonicalcommutators:

Q(~x, t) = eiHtq(~x, 0)e−iHt and P (~x, t) = eiHtp(~x, 0)e−iHt.

Here, H is the full Hamiltonian; these transforms commute with H, so if the interacting Hamilto-nian is expressed in the free variables as H = F [q, p], it can also be written H = F [Q,P ].

What does this mean? It means the canonical variable in the interacting Hamiltonian is a time-evolution of the canonical variable in the free Hamiltonian. First of all, when exactly we start thetime evolution is irrelevant because the start of time can always be changed by arbitrary phaseshift. (Here, the start time is t = 0. However, once we pick a start time , we must be consistentover all canonical variables.) Second, we have to evolve with the entire Hamiltonian. While itmay be true that

eiH0tqe−iH0t = q,

it is not true that we can merely evolve with the interaction V in H = H0 + V . This is becausegenerally [H0,V ] 6= 0, so the exponential may imply H0, the free Hamiltonian, has some nontrivialeffect on the “free” canonical variable.

7.2 Lagrangian formalism

It turns out that the best way to choose the Hamiltonian of a theory is to choose the Lagrangianfirst, then Legendre-transform to find the Hamiltonian. It what follows, let capital fields Ψ andΠ represent interacting (conjugate) fields; the lowercase ψ and π are the free (conjugate) fields.Careful! These fields are no longer canonical in general. Even though we can define the Hamilto-nian equations of motion, and hence describe which fields are conjugate to each other, they do notgenerally satisfy the canonical (anti)commutations. That is a product of special choices like theones in the previous section. So there is a different between having a conjugate friend and havinga conjugate friend who is also canonical.

Question 105. Describe the motivation leading to the Euler-Lagrange equations of motion.

Answer 105. From this viewpoint, the Euler-Lagrange equations are merely restatements of theequations of motion for the conjugate fields Π(~x, t). How?

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The definition and equation of motion for Π is:

Π(~x, t) =δL

δΨ(~x, t)and Π(~x, t) =

δLδΨ(~x, t)

.

(The equation of motion is a restatement of Hamilton’s equation, but with a negative sign becauseH = pq − L. The definition is motivated by, I guess, making the below action stationary?Heuristically, if ΠδΨ = δL is true, and we want

∫dtL to be stationary, then we have to absorb

this equation of motion into a total time derivative on the LHS, which means adding the termΠδΨ.) This means that ∂t(

δLδΨ(~x,t)

) = δLδΨ(~x,t)

, which is just the stationarity condition for the action

I[Ψ, Ψ] =

∫dtL[Ψ, Ψ].

So now we have an action I[Ψ, Ψ].

We expect I[Ψ, Ψ] to be a Lorentz scalar because it gives the equations of motion, which shouldbe Lorentz-covariant. Because I =

∫dtL, for Lorentz scalar we expect there to be a

∫d4x, which

implies

L =

∫d3~xL ,

where L [Ψ, ∂µΨ] is the Lagrangian density, which itself is also a Lorentz scalar (i.e. becausethe measure d4x is a Lorentz scalar). We guess it is a function of ∂µΨ because that has all theindices.

The stationarity condition of the action ∂t(δL

δΨ(~x,t)) = δL

δΨ(~x,t)can now be expressed as a stationarity

condition of L, which turns out to be the Euler-Lagrange equation

∂µ(∂L

∂(∂µΨ)) =

∂L

∂Ψ.

Conclusion: The Euler-Lagrange equation just above is a restatement of the Hamilton equationof motion for the conjugate field, under the assumption that the Lagrangian can be written as aspacetime integral of some Lagrangian density.

Question 106. Show that the Legendre transform back to the Hamiltonian eliminates the time-derivative field Ψ.

Answer 106. Consider

H =

∫~x

Π(~x, t)Ψ(~x, t)− L[Ψ(t), Ψ(t)].

By definition of Π, δHδΨ

= 0, so generally H = H[Ψ,Π]. In fact, the same equations of motion aresatisfied as in the canonical case,

Ψ =δHδΠ

, Π = −δHδΨ

.

However, as I mentioned in the introduction to this section, although Ψ and Π are conjugate, theymay not be canonical conjugates. Sometimes you can force them to be canonical by imposing the(anti)commutations. Then, the functional derivative could again be interpreted as a commutatorwith the bosonic energy H (see the previous section).

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In fact, imposing such (anti)commutations is not always possible. For example, we saw thatiψ†(x), the Hermitian conjugate of the Dirac field, is not a canonical field variable (in fact, it ismore like a conjugate variable), and thus its time-derivative does not appear on L. So fields arenot always mapped to field variables, as we will see in the next question.

Question 107. Describe the canonical quantization of a general theory.

Answer 107. As we saw in the previous question, some fields (such as the time component ofvector field or the Dirac field) appear in the Lagrangian, but without their time-derivatives. So,these fields have no canonical conjugates. Let us denote them by Cr.

The fields which do appear with their time derivatives can have canonical conjugates. Let usdenote them, and their time-derivatives, by Qn and Qn.

The canonical conjugates of the fields Qn are defined from the Lagrangian L[Q, Q, C]:

Pn(x) =δLδQn

, 0 =δLδCr

=δLδCr

.

Pn(x) =∂L∂Qn

, 0 =∂L∂Cr

=∂L∂Cr

.

The Legendre transform to the Hamiltonian involves only the P and Q variables, because the Cvariables have no conjugates:

H =

∫d3~x(PnQ

n)− L[Q, Q, C] =

∫d3~x(PnQ

n −L ).

Canonical quantization means “express everything in the canonical variables.” In other words,given the equations above, we have enough information for solve for Cr and Ql in terms of Qn andPn. Then the Hamiltonian is a functional only in the canonical variables,

H = H[Q,P ].

Question 108. For a given theory which has been canonically quantized, how do we transitionfrom Heisenberg picture to interaction picture?

Answer 108. Recall that in the previous question regarding canonical quantization, all fields Qand P were in the Heisenberg picture. We would like to switch to the interaction picture becausethat is how we compute S-matrix elements (to remember why, recall that the propagators of fieldsare calculated for H0 and the interactions are treated as perturbations. This implies the fields areevolving via H0, hence we must have been in interaction picture. Obviously, the field φ−(x) hasan e−iωt attached to it. Of course, this is post-mortem justification; the Feynman rules alreadyassume this).

To switch back to interaction picture, in which the fields evolve only with H0 instead of the fullHamiltonian H, we simply “undo” the similarity transformation described in the previous section.In fact, q = Q and p = P at t = 0, so we just rewrite

HHeisenberg[P,Q]→ Hinteracting[p, q].

The functional form is exactly the same. Then because operators evolve with H0, obviously

H0(t) = H0(t = 0) and V(t) = eiH0tV(t = 0)e−iH0t.

Unfortunately, the exact details of finding, for example, the expansion of q(x) into a†~pσ and a~pσoperators is different for every theory. See Weinberg pg. 304-5.

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7.3 Global symmetries and Noether theorem

According to Weinberg, the Lagrangian formalism “provides a natural framework for the quantum-mechanical interpretation of symmetry principles.” What he means is that Noether’s theorem,i.e. for global symmetries, is easiest to see in the Lagrangian formalism. This is because theLagrangian formalism involves a stationarity condition (small change in the field), which is easilyadapted to the global rotation of Noether’s theorem. This is described below.

Question 109. Give a short derivation of Noether’s theorem for global symmetries.

Answer 109. Suppose there is a global (ε constant) transform

Ψ(x)→ Ψ(x) + iεF (x)

which preserves the action. (Even when the action is not minimized, or equivalently when thedynamical equations are not satisfied. For example, L = ψ(/∂ −m)ψ is invariant under ψ → eiθψeven if (/∂ −m)ψ 6= 0.) Now consider the local transform (here, F is known)

Ψ(x)→ Ψ(x) + iε(x)F (x).

The action no longer vanishes; however, it can always be written in the form

δI = −∫d4xJµ(x)∂µε(x) =

∫d4x∂µJ µε(x)

which forces it to vanish when ε(x) = const. If we choose the fields to satisfy the field equations(i.e. if we choose a stationary point), then δI = 0 regardless of the local ε(x). Therefore, thecurrent J µ is conserved when the Euler-Lagrange equations are satisfied:

0 = ∂µJ µ =⇒ 0 =dF

dt, where F =

∫d3~xJ 0.

Important: this derivation only works for global symmetries. If the symmetry was not global, wewould not be able to write ∂µε(x) in the above. (Well, if the symmetry is local, then it is alsoglobal, so not such a big deal.) This derivation is the same in classical theories.

Important: the conservation of current ∂µJ µ is only true in the “ground state” of the system.You can imagine it like this: suppose I have a stretchy ribbon resting on a flat plane, enclosing afixed area A; the action is I = σL, where L is the length of the ribbon. The ground state is whenthe ribbon is a perfect circle. Even if it is not a perfect circle, the energy H is invariant undera global rotation around the z-axis by angle ε. If we allow the rotation angle ε = ε(θ), then thetotal length might change (i.e. δI 6= 0). Here, the analog of J µ is basically r(θ). We see that inthe ground state, obviously

∂θr(θ) = 0 because for a circle, r = const.

However, ∂θr(θ) 6= 0 generally for non-circular shapes. So the current is only conserved when theequations of motion hold.

From Wikipedia: “Noether’s theorem is an on-shell theorem: it relies on use of the equationsof motion - the classical path... The quantum analogs of Noether’s theorem probing off-shellquantities as well are the Ward-Takahashi identities.”

Phew! At least we cleared that up now.

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Question 110. When can we find, explicitly, the conserved charge F and the conserved currentJ µ?

Answer 110. We can always find F when the Lagrangian L (and not just its time-integral, theaction I) is invariant under the symmetry. We can always find J µ when the Lagrangian densityL is invariant under the symmetry. The derivations are easy (pg. 308-9) and the results are

F (t) = −i∫d3~x

δLδΨ

F (~x, t) and J µ = −i ∂L

∂(∂µΨ)F .

Question 111. What does it mean for the conserved charge F to generate the symmetry Ψ →Ψ + iεF?

Answer 111. Suppose the variable Ψ is, in fact, a canonical variable with conjugate Π. ThenΠ = δL

δΨ, and the formula in the previous answer becomes

F (t) = −i∫d3~xΠF (~x, t)

so we have[F (t),Ψ(~x, t)] = −F (~x, t) =⇒ e−iF εΨeiF ε = Ψ + iεF .

For example, the Hamiltonian generates translations in time. This derivation, of course, only workswhen the answer to the above question is true: when the global symmetry leaves the LagrangianL invariant, in addition to leaving the action I =

∫dtL invariant.

A nice fact to know (pg. 312-3) is that if the generators ta of the transforms of canonical fields

Qn(x)→ Qn(x) + iεa(ta)nmQ

m(x)

satsify a Lie algebra [ta, tb] = if cabtc, then so do the conserved charges, [Ta, Tb] = if cabTc.

7.4 Lorentz invariance of L implies Lorentz invariance of S-matrix

Question 112. Describe the logic of Weinberg’s argument that the Lorentz invariance of Limplies the Lorentz invariance of the S-matrix.

Answer 112. The first question in section 3.2 of these notes says that the S-matrix is Lorentz-invariant if

Sβα = 〈Φβ|SΦα〉 = 〈U0(Λ, a)Φβ|SU0(Λ, a)Φα〉 =⇒ [U0(Λ, a), S] = 0.

where U0 implements the Lorentz transforms on the wavefunctions. We already know relativisticquantum mechanics, so U0 is known (the generators are J µν

0 ). What is unknown is S = U(−∞,∞),which itself is made of generators J µν . Weinberg shows in (3.3.21) that this is satisfied if

[ ~K0,V ] = −[ ~W,H],

and matrix elements of ~W are smooth functions of the energies. Here, ~K0 generates boosts in thefree theory, ~K generates boosts in the interacting theory, and ~W = ~K − ~K0. The derivation isreally complicated.

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However, we now know that, for a particular theory L , the representation of the Lorentz groupwhich acts on the canonical variables is generated by the conserved charges J µν of Noether’stheorem, as in the previous section. If we know that L is invariant, then the explicit forms ofthe generators can be calculated, as shown in the previous section. This is the stress-energytensor. These generators are exactly the ones we are looking for, i.e. the ones we use to buildS = U(−∞,∞).

Then we just have to show that the regular commutations of the Lorentz generators are satisfied,along with the two crucial conditions above. It turns out that these are satisfied for nearly anyL which is Lorentz-invariant.

7.5 Examples of canonical quantization and transition to interactionpicture

Performing canonical quantization of a theory is kind of like solving a crossword puzzle or doingsudoku, in the sense that the problem is essentially the same every time but you can always havefun doing it.

Weinberg does three examples. The first is the scalar field with derivative coupling to a current.That is pretty easy and his derivation is clear enough. The second is the vector spin one field,which involves introducing an ad hoc condition to eliminate the scalar component. Let us workthrough the third example: the Dirac field.

Keep in mind that the original Lagrangians are for interacting theories in the Heisenberg picturewhere, for example, Ψ evolves with the interacting Hamiltonian. We want to write a decom-position H = H0 + V in the interaction picture where the fields evolve with the free Hamilto-nian, H0. Finally, we express the fields in creation and annihilation operators; we can find the(anti)commutations of the creation and annihilation operators by enforcing the (anti)commutationsof the fields themselves. For this last step, I realized that in my final project for Statistical Physicsof Fields that you always need a classical equation of motion to express fields in a† and a. This isalso true here; the classical equation of motion comes from H0 and Hamilton’s equation.

We will also get to practice this technique in the problems.

Question 113. Carry out the canonical quantization, and transition to interaction picture, of theLagrangian density

L = −Ψ(/∂ +m)Ψ−U (Ψ,Ψ).

Answer 113. First, note that Ψ has no time-derivative acting on it, so it has no conjugate variable,as we mentioned before. The canonical conjugate of Ψ is

Π =∂L

∂Ψ= −Ψγ0.

The Hamiltonian is

H =

∫d3~x(ΠΨ−L ).

We can rewrite the free part of the Lagrangian as L0 = Πγ0(/∂ +m)Ψ = ΠΨ + Πγ0(~γ · ~∇+m)Ψ,so

H = H0 + V , where

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H0 =

∫d3~xΠγ0(~γ · ~∇+m)Ψ and V =

∫d3~xU (Ψ,Ψ).

Passing to the interaction picture, in this case, means replacing all uppercase with lowercase. Theclassical equation of motion for ψ is given by Hamilton’s equation,

ψ =δH0

δπ= γ0(~γ · ~∇+m)ψ =⇒ (/∂ +m)ψ = 0.

To express the field in raising and lowering operators, we need a classical equation of motion −which we now have. We write ψ(x) as the expansion

ψ(x) =

∫~pσ

u(~pσ)eipxa~pσ + v(~pσ)e−ipxb†~pσ,

where (i/p+m)u~pσ = (−i/p+m)v~pσ = 0.

To enforce the canonical anticommutators ψα(~x, t), πβ(~y, t)(γ0)βγ = i(γ0)αγδ(~x− ~y) (see section

7.1), it turns out that we must have a~pσ, a†~p′σ′ = b~pσ, b†~p′σ′ = δ(~p− ~p′)δσ′σ. This completes thecanonical quantization; the free Hamiltonian can be expressed

H0 =

∫~pσ

p0(a†~pσa~pσ + b†~pσb~pσ),

up to an arbitrary constant.

This was pretty systematic, but why was it that [ψ(~x, t), π(~y, t)] 6= iδ(~x− ~y)? This is what I saidin the opening paragraph of my notes on section 7.2. Namely, there is a difference between havinga conjugate friend and having a conjugate friend who is also canonical.

7.6 Problems

These problems are a bit hard.

Question 114. Consider the theory of a set of real scalar fields Ψn with Lagrangian density

L = −1

2

∑mn

∂µΨn∂µΨmfnm(Ψ),

where fnm is some non-singular matrix function of the fields; this is called the nonlinear sigma-model. Carry out the canonical quantization of this theory and derive the interaction V [φ, φ] inthe interaction picture.

Answer 114. We find the conjugate momenta are

Πn =∂L

∂Ψn= −

∑m

fnm(Ψ)∂0Ψm.

This is an inspired guess, but it’s easy to check that∑nm

(f(Ψ)−1)nmΠnΠm =∑ab

f(Ψ)ab∂0Ψa∂0Ψb.

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Therefore, our Lagrangian can be written as

L = −1

2

∑nm

[(f(Ψ)−1)nmΠnΠm + fnm(Ψ)(∇Ψn) · (∇Ψm)].

Everything in this is in terms of the canonical variables (i.e. no Cr variables) so passage to theinteraction picture is accomplished by taking

Ψ→ ψ and Π→ π.

Also, I think the interaction is just what is left over from the kinetic term. So we would write(interaction picture):

L = L0 + U , where

L0 = −1

2

∑nm

[(f(ψ)−1)nmπnπm + (∇ψn) · (∇ψm)] and U = −1

2

∑nm

(fnm(ψ)− 1)(∇ψn) · (∇ψm).

Question 115. Consider the theory of a complex scalar field Φ and a real vector field V µ withcorresponding tensor Fµν = ∂µVν − ∂νVµ and covariant derivative Dµ = ∂µ − igVµ. Carry out thecanonical quantization of

L = −(DµΦ)†DµΦ− 1

4FµνF

µν − m2

2VµV

µ.

Derive the interaction in the interaction picture.

Answer 115. Just as in the photonic case, the antisymmetry of Fµν means that V 0 is not acanonical variable since its time-derivative does not appear in the Lagrangian. Therefore, thecanonical variables are V i and Φ and Φ†.

Let’s find the conjugate momenta; I will call them Wi and Π and Π†. Expanding the Lagrangiandensity gives

L = −(∂µΦ†)(∂µΦ)+ igVµ(Φ†∂µΦ−Φ∂µΦ†)+g2VµVµΦ†Φ− 1

2(∂µVν∂

µV ν−∂νVµ∂µV ν)−m2

2VµV

µ.

I find that

Π = −∂0Φ† + igV0Φ† and Wi = ∂iV0 − ∂0Vi =⇒ V 0 =1

m2~∇ · ~W.

The last comes from Euler-Lagrange equation on V 0.

Now, let’s write H =∫d3~x(ΠΦ + Π†Φ† + ~W · ~V −L ). First, we write

~V = −~∇V 0 + ~Π = ~Π− 1

m2∇(∇ · ~Π).

8 Electrodynamics

Quantum electrodynamics - particularly the theory of the photon - is really the one part of QFTthat never made any sense. So, we will study and try to make sense of it now.

Weinberg also does some tree-level scattering calculations. These are the same as in any textbookso I will not write notes on them.

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8.1 Gauge invariance

Question 116. Why do we need gauge invariance?

Answer 116. In fact, we wouldn’t need gauge invariance in electrodynamics if nature operatedaccording to different laws than it does. But it seems that electrodynamics is mediated by amassless spin-1 particle which couples to a conserved current, J µ = (ρ, ~J). (If the particle didn’tneed to couple to a conserved current, we could just use L ⊃ FµνF

µν and get out of jail free.)

Given that we need to couple a massless spin-1 particle to a conserved current, the easiest wayto do so is to embed it in a vector field Aµ, since AµJ µ is a Lorentz scalar. However, a masslessspin-1 particle cannot be embedded in a vector field without also requiring gauge invariance! Thereason for this is that the photon little group, ISO(2), and any irreducible representation of theLorentz group SO+(1, 3) are incompatible (for example, we saw in section 5.8 of these notes thatthe helicity of a massless particle in the (A,B) irrep must be σ = B−A). For example, Weinberg(5.9.31) shows that under a general Lorentz transform,

U(Λ)aµ(x)U−1(Λ) = Λµνaν(Λx) + ∂µΩ(x,Λ)

where Ω(x,Λ) is some linear combination of creation and annihilation operators.

Because the vector representation (12, 1

2) is an irrep, we cannot embed the photon in the vector

representation.

This is not true for massive particles. The reason is that the generators of the photon little groupISO(2) mix rotations and boosts, but the generators of the massive little group SO(3) consistpurely of rotations.

Question 117. Describe how gauge invariance allows us to couple the photon to an externalcurrent in a Lorentz-invariant way.

Answer 117. To ensure that the extra term in

U(Λ)aµ(x)U−1(Λ) = Λµνaν(Λx) + ∂µΩ(x,Λ)

has no effect, we require that the action be invariant under the gauge transformation

Aµ(x)→ Aµ(x) + ∂µε(x)

for any ε(x). (Here, uppercase means interacting and lowercase means free.)

Now we will couple this to a conserved current by constructing a Lagrangian that forces the currentto be conserved, so to speak. The change in the action under a gauge transform is

0 = δI =

∫d4x

δI

δAµ(x)∂µε(x) =⇒ 0 = ∂µ(

δI

δAµ(x)).

Because Noether’s theorem on continuous global symmetries implies conserved currents, we intro-duce a global U(1) symmetry into the action. Let Ψl(x) be all fields other than the photon field,and consider the U(1) symmetry

δΨl(x) = iεqlΨl(x),

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which is generated by the charge Q =∫d3~xJ 0, where ∂µJ µ = 0 by Noether’s theorem. If we

scale the U(1) charges such that

J µ =δI

δAµ(x),

then we solve two problems (introduction of a conserved current and gauge-invariance) at once.Everybody is happy!

Question 118. In the above question, we started with a global U(1) symmetry. However, QEDis based on a local U(1) symmetry. Where does this come from?

Answer 118. If we substitute J µ = δIδAµ(x)

into 0 = δI =∫d4x δI

δAµ(x)∂µε(x), we find that the

symmetry has become local, since now ε(x) doesn’t have to be a constant and the action is stillconserved.

This is because of the nontrivial identification J µ = δIδAµ(x)

. (If the U(1) symmetry was only

global, we wouldn’t call QED a gauge theory.)

Question 119. What is the logically opposite way of arriving at the AµJ µ coupling?

Answer 119. You can find this on Weinberg 342-43 and in Chapter 1 of Nakahara Geometry,Topology, and Physics. Essentially we start with a global U(1) symmetry and promote it to alocal symmetry. This necessitates introducing a covariant derivative

Dlµ = ∂µ − iqlAµ

to act on each field Ψl(x). The hypothetical mass term AµAµ, of course, is ruled out as not

gauge-invariant, so that is why (in this logic) the photon field must be massless.

Question 120. How can I think of J µ? For example, where is J µ in the usual Lagrangian forQED,

L = ψ(i /D −m)ψ?

Answer 120. Recall that J µ is a current which describes the charge of the fields which transformunder local U(1) gauge symmetry. In this Lagrangian, J µ is just what it looks like: namely,∂L∂Aµ = ψγµeψ (up to some constants). This reflects the fact that J µ comes from Noether’s

theorem applied on the charged fields; in this case, the charged field is ψ.

Question 121. How does QED return Maxwell’s equations?

Answer 121. Everyone knows this one. We take the Lagrangian

L = −1

4FµνF

µν + AµJ µ + · · · .

The classical (i.e. on-shell) field equation for the photon field is

0 = ∂µFµν + J ν .

This gives the inhomogeneous Maxwell equations ∇ · ~E = 4πρ, ∇ × ~B = 4π ~J . The other twoMaxwell equations have nothing to do with charge, so they have nothing to do with J µ. Instead,they arise from the definition of Fµν ,

0 = ∂µFνε + ∂εFµν + ∂νFεµ.

This is called the Bianchi identity.

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8.2 Difficulties with quantizing the photon field

There are multiple difficulties to surmount before we can quantize the photon field. They arerelated to gauge-invariance, if you want to think about it that way.

Question 122. Describe why A0(x), the time-component of the photon field, has no conjugatevariable. Explain how this is related to gauge invariance.

Answer 122. For L ∼ FµνFµν , we find the canonical conjugate

Πµ =∂L

∂(∂0Aµ)

gives Π0 ∼ F 00 = 0, since F µν is antisymmetric. (In fact, we studied this in Chapter 7.) This iscalled a primary constraint because it follows directly from the structure of L .

There is also a secondary constraint

∂iΠi = −∂i

∂L

∂Fi0= −∂L

∂A0

= −J 0.

This follows from TODO.

This is related to gauge invariance because, given any solution Aµ(x) of the field equations, wecan always find another solution Aµ(x) + ∂µε(x) with the same value and time-derivative at t = 0but which differs from Aµ(x) at later times. The df here is one (because ε(x) comes packaged with1 df). This also suggests the solution: given that Aµ has some freedom due to gauge invariance,we can remove the freedom (i.e. make Aµ(x) a deterministic object, which makes it possible toquantize) by choosing a gauge.

Question 123. Describe with examples why choosing a gauge allows us to describe a photon4-component field with only 3 propagating degrees of freedom.

Answer 123. We found in section 7.3 of these notes that we can explicitly find the conservedcharge, J 0, in terms of the canonical variables. Well, A0 is not a canonical variable, as shown inthe previous question. The point here is that J 0 is known.

Let us first choose the Coulomb gauge, ~∇ · ~A = 0. Then

~∇ · ~A = 0 and − ∂iF i0 = J 0 =⇒ −∇2A0 = J 0

and because J 0 is known, so is A0(x).

Or we can choose Lorentz (Landau) gauge ∂µAµ = 0. The solution for A0(x) would look

different but the df are the same as Coulomb gauge, so there is also a deterministic solution for A0

in terms of all the canonical variables in the theory (even those which are for other fields besidesthe photon field, since they enter in A0 through J 0).

Question 124. Describe the quantization of QED. Specifically, outline the important points inWeinberg’s (quite difficult) derivation on pages 346-50.

Answer 124. Because of the constraints

∂iAi = 0 and ∂iΠ

i + J 0 = 0,

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the variables Ai(x) are still not free of constraints and hence there is difficulty in quantizing them,even though we have already eliminated the A0(x) component by choosing the Coulomb gauge. Itturns out that the equal-time commutators are modified to (8.3.5.):

[Ai(~x),Πj(~y)] = iδijδ(~x− ~y) + i∂j∂i(1

4π|~x− ~y|).

The derivation of this is really hard. The end result is that

H =1

2~Π2⊥ +

1

2(∇× ~A)2 − ~J · ~A+

1

2J 0A0 +HM ,

where HM is the rest of the Hamiltonian for the other fields (which may be coupled to the photonfield through the A0 term), and

~Π⊥ = ~Π− ~∇A0

is chosen because it commutes with HM , basically.

8.3 QED in the interaction picture; photon propagator

Let me simply summarize the passage to the interaction picture. Then we will study the photonpropagator in detail.

We start with the Hamiltonian from the previous question and split it into quadratic terms plusthe interaction,

H = H0 + V , where

H0 =1

2~Π2⊥ +

1

2(∇× ~A)2 +HM0 and V = − ~J · ~A+

1

2J 0A0 + VM .

In this way, we can use the quadratic free Hamiltonian to talk about a “photon propagator.” Thenwe pass to the interaction picture by applying the similarity transformation described in section7.5 of these notes. The regular Hamilton equations of motion for the conjugate pair ~a, ~π gives thewave equation

~a = 0.

Now there is a subtlety. “Since A0 is not an independent Heisenberg-picture field variable, butrather a functional of the matter fields and their canonical conjugates that vanishes in the limitof zero charges, we do not introduce any corresponding operator a0 in the interaction picture, butrather take a0 = 0.” We can use the wave equation to second-quantize the field a(x); it turnsout that the weird commutations of the previous question (in the Heisenberg picture) are satisfiedonly if the usual commutations of creation and annihilation operators in the interaction pictureare satisfied.

Conclusion: We can eliminate the time-component of the photon field in the interaction picturebut not in the Heisenberg picture. We find that quantizing the field is easy after we have gone tointeraction picture (upon reflection, in section 7.5, we also quantized after we went to interactionpicture).

Question 125. Describe the calculation of the photon propagator.

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Answer 125. First, recall from above that we could set a0 = 0 in the interaction picture. Thismeans that e0(~pσ) = 0, so there are only three basis vectors, ei(~pσ). We can normalize them suchthat

Pµν =∑σ

ei(~pσ)ej(~pσ)∗ = δij − pipj/|~p|2.

As in section 6.2 of these notes, we define the propagator

−i∆µν(x− y) = 〈T [aµ(x), aν(y)]〉

where the expectation is a vev (obviously, because we are working with free fields and there is nobackground). We find that

∆µν(x− y) =1

(2π)4

∫d4qeiq(x−y)Pµν(~q)

q2 − iε.

Question 126. Describe the subtlety in the photon propagator. What did Feynman get wrong?

Answer 126. Feynman’s pioneering paper which established the fundamentals of QED is PR 76769 (1949). Unfortunately, he got something wrong about the gauge invariance of the photonpropagator, which was noticed by Bialynicki-Birula in PR 155 1414 (1967).

This is described (well, a related phenomenon is described) on pages 354-355 of Weinberg. Thepunchline is that the photon propagator is not gauge-invariant on mass shell, but that this iscancelled by another non-gauge-invariant effect in renormalization. The easiest way to prove thisis through the path-integral formalism.

The result ((8.5.9) in Weinberg) is that we can effectively take the photon propagator to be thecovariant quantity

−i(2π)4

ηµνq2 − iε

in momentum-space. This is nice because it guarantees our matrix elements will be Lorentz scalars(which they might not be if we used a different convention for the photon propagator).

Question 127. Describe numerically the effect of adding more loops to a Feynman diagram. Inother words, how does the numerical value of the matrix element for a given process scale withthe number of loops?

Answer 127. This is from page 358. A connected diagram with V vertices, I internal lines, Eexternal lines, and L loops satisfies the graph theory relations

L = I − V + 1 and 2I + E = 3V.

There is a factor e(2π)4 from each vertex, (2π)−4 from each internal line, and a 4D momentum-space integral from each loop, which turns out to contribute π2. (This is because TODO.)

Therefore, the diagram scales like

(2π)4eE−2(e2

16π2)L = (2π)4eE−2(

α

4π)L.

The important factor is α/4π = 5.81× 10−4, which is dimensionless. Because this is small, many-loop diagrams have small contributions.

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8.4 p-form gauge fields

The strength tensor Fµν of QED can be generalized to arbitrary tensors. Let us consider anarbitrary p-form tµ1µ2···µp with exterior derivative dt. A p-form whose exterior derivative vanishesis called closed and a form which itself is an exterior derivative is called exact. Since d2 = 0, anyexact form is also closed. Poincare proved that in a simply connected region, any closed form isalso exact. (For example, the homogeneous Maxwell equations ∇× ~B = −∂t ~E,∇ · ~B = 0 tell usthat Fµν is closed. Since R4 is simply connected, it must also be exact, so there exists some Aµsuch that F = dA.)

Question 128. How can we generalize the 1-form Aµ gauge theory to a gauge theory of generalp-forms, in D dimensions?

Answer 128. Consider the generalized gauge transform

δA = dΩ =⇒ δAµ1···µp = ∂[µ1Ωµ2···µp ,

where Ω is an arbitrary (p− 1)-form. Construct the field strength tensor

F = dA

and also the Lagrangian density

L = − 1

2(p+ 1)Fµ1···µp+1F

µ1···µp+1 + J µ1···µpAµ1···µp , where ∂µ1J µ1···µp = 0.

This Noether theorem guarantees the Lagrangian (integral of the Lagrangian density) is invariantunder a gauge transform, i.e. via integration by parts. Clearly we must have p + 1 ≤ D due toantisymmetry of F .

Question 129. Prove this duality theorem:

In D dimensions, the theory of a p-form gauge field A is equivalent to the theory of a(D − p− 2)-form gauge field φ.

Answer 129. Unsurprisingly, we will use the duality transforms

F µ1···µp+1 = εµ1···µDFµp+2···µD and J µ1···µp = εµ1···µDJµp+1···µD .

The field equation and Noether conservations

∂µFµµ1···µp = −J µ1···µp and ∂µ1J µ1···µp = 0 become

dF = J and dJ = 0.

This means that J is a closed form; therefore it can be written as J = dG =⇒ dF = dG .Thus, Poincare tells us that

F = G + dφ

where φ is a (D− p− 2)-form. We still have not used one piece of information: the field equationdF = 0 =⇒ ∂µ1F

µ1···µD−p−1 = 0 thus becomes

∂µ1(dφ)µ1···µD−p−1 = ∂µ1Gµ1···µD−p−1 .

This is invariant under the new gauge transform φ→ φ+ dω. In the case where D = p− 2, dω isreplaced by a scalar (and all the terms in the Lagrangian would be constructed out of derivativesonly).

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8.5 Problems

Question 130. Calculate the differential and total cross-sections for the process

e+ + e− → µ+ + µ−

to lowest order in e. Assume that spins are not observed and use the simplest possible Lagrangian.

Answer 130. This is computed in my notes on Schwartz Quantum Field Theory. I feel thatSchwartz’ Feynman rule conventions are a little cleaner and more standard than Weinberg’s, so Iwon’t bother to do it again.

The result for the differential scattering cross-section turned out to be

(dσ

dΩ)CM =

α2

16E6

|~p||~k|

(E4 + k2p2 cos2 θ + E2(m2e +m2

µ)).

Question 131. Write a gauge-invariant Lagrangian for a charged massive vector field interactingwith the electromagnetic field.

Answer 131. The fact that the vector field is charged means that it also transforms (i.e. localphase) under the gauge transform. A possible Lagrangian density is

L =1

2(DµV

ν)†(DµVν) +1

2(DµV

µ)†(DνVν)−1

2m2(V µ)†Vµ +

1

4FµνF

µν .

9 Path Integrals

We saw in the previous chapter that the photon propagator has some gauge freedom which is“cancelled out,” so to speak, by renormalization. Instead of doing this messy stuff where wehave to choose a covariant propagator, we would like a formalism in which Lorentz covarianceis manifest in all the Feynman rules the moment we derive them. This is the path integralformalism.

Feynman rules can therefore be derived from either the canonical formalism (Schwinger’s for-malism) or from the path integral (Feynman’s formalism). However, the path integral is vastlysuperior for more complicated theories, such as non-Abelian gauge theories, in which the gaugefreedom is even harder to handle than in QED. It is also better for theories such as the nonlinearσ-model in which the Feynman rules from the canonical formalism are simply wrong! Weinbergnotes a “conservation of trouble,” in the sense that the canonical formalism makes unitarity man-ifest but not Lorentz invariance, but the path integral makes Lorentz invariance manifest butobscures unitarity.

9.1 Hamiltonian version of path-integral formula; S-matrix

The derivation of the Hamiltonian version of the path-integral formula follows from insertinglots of intermediate resolutions of the identity, exploiting the coherent states |λ〉 = eλa

†|0〉. See

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Coleman Many-Body Physics for a detailed and similar derivation in condensed matter theory.Let

F = OA(P (tA), Q(tA)),OB(P (tB), Q(tB)), · · · .

The result is (x = (t, ~x)):

〈q′t′|T [F ]|q0t0〉 =

∫ ∏x,a

dqa(x)∏x,a

dpb(x)

2πFe

i∫ t′t0dt

∑a qa(t)pa(t)−H(q,p)

.

We used the convention that all qs are to the left of all ps. You can also choose another convention.

Question 132. Why is the above form of the path integral not suitable for calculating S-matrixelements?

Answer 132. Easy: S-matrix elements are between particle states |α〉 and |β〉, not betweencoherent states (which are superpositions of an infinite number of particle states). Inserting thecompletion

〈β| =∫dqa(~x,∞)

dpb(~x,∞)

2π〈β|q′,∞〉〈q′,∞|

at t =∞ and a similar completion at t = −∞ gives

〈β|T [F ]|α〉 =

∫ ∏x,a

dqa(x)∏x,a

dpb(x)

2πFei

∫∞−∞ d4x

∑a qa(x)pa(x)−H(q,p)〈β|q′,∞〉〈q,−∞|α〉.

The big quantity is no longer inside a matrix element because we have done an unconstrainedintegral over all x = (~x, t). That way, it basically became kind of inside a trace, which means thatthe outside vectors become an identity matrix, so we can just ignore it.

In fact, it turns out that the last part, the wavefunctions

〈β|q′,∞〉〈q,−∞|α〉,

gives the iε in every propagator in the Feynman rules. The derivation is on Weinberg pg. 386-8but is quite difficult. The final result is

〈β|T [F ]|α〉 = |N |2∫ ∏

x,a

dqa(x)∏x,a

dpb(x)

2πFei

∫∞−∞ d4x

∑a qa(x)pa(x)−H(q,p)+(iε terms).

The normalization factors (such as 1detA

and such) are unimportant because they also contributeto the matrix element 〈β|α〉. Any constant factors divide out (or you can say they are cancelledout in the denominator of the Gell-Mann - Low theorem by exponentiation of the disconnecteddiagrams).

9.2 Lagrangian version of path-integral formula

The integrand in the exponentiated part of the previous formula,∑a

qa(x)pa(x)−H(q, p),

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looks quite a bit like the Lagrangian. This is not exactly true because Hamilton’s equations (ortheir equivalents, the Euler-Lagrange equations) no longer hold in the path-integral formalismwhen things are manifestly allowed to go way off-shell. Therefore, there is no “minimization” toperform and hence no true notion of a Legendre transform.

However, it turns out that we can, in fact, pass to the Lagrangian form if the Hamiltonian isquadratic in the canonical momentum variables (see pg. 389-90). This is kind of what Schwartzdoes when he derives the path integral; essentially he does something like

m

2x2 −H → L

for nonrelativistic QM and postulates that the same thing works in relativistic QFT.

Because this receives nontrivial modifications for certain kinds of important theories, such as thenonlinear sigma model, we will study it in detail here.

Question 133. Outline how, if the Hamiltonian is quadratic in the canonical momentum variables

H[Q,P ] =1

2

∑nm

∫~x~y

A~xn,~ym[Q]Pn(~x)Pm(~y) +∑n

∫~x

B~xn[Q]Pn(~x) + C[Q],

where A is real, symmetric, positive and nonsingular, we can pass from the Hamiltonian pathintegral to the Lagrangian path integral.

Answer 133. If we substitute this into the Hamiltonian version of the path integral and integrateout the momenta (which we can do because the integral is Gaussian), then we get two things:

• First, we get the factor (det(2πiA [Q]))−1/2, where A [Q] = A[Q]δ(t − t′). This will givenontrivial Feynman rules if the matrix A[Q] is a nontrivial function of Q.

• Second, we get the Lagrangian in the exponential. How? The Gaussian integral is pro-portional to the “exponential evaluated at the stationary point of its argument.” But thestationary point for the canonical momentum is merely when Hamilton’s equation

qn(~x, t) = δHpn(~x, t)

is satisfied! Therefore, we can interpret∑a

qa(x)pa(x)−H(q, p)

as a true Legendre transform after we integrate out the canonical momenta, because doingso automatically enforces Hamilton’s equation.

To find expectation values, we have to assume the O are independent of the canonical momenta.This can be achieved even if O contains time-derivatives of the canonical fields by expressing themas O[Φ(t+ dt),Φ(t)] instead of O[Π(t)]. A little weird, admittedly.

Question 134. The above only works when the Hamiltonian is quadratic in the canonical mo-menta. Why is this a reasonable assumption for many theories?

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Answer 134. Lots of theories look roughly like

H = Hkin + V(Φ),

where the kinetic term is obviously quadratic in the momenta and the interaction has no momenta-dependence at all.

Question 135. Describe the Feynman rules which could arise from the (det(2πiA [Q]))−1/2 factorafter integrating out the momenta.

Answer 135. Let us take the nonlinear sigma model

L = −1

2

∑nm

∂µΦn∂µΦm(δnm + Unm(Φ))− V (Φ).

Let Πn be the canonically conjugate momentum to Φn. It’s not hard to calculate

H =

∫d3~x(

1

2Πn(1 + U(Φ))−1

nmΠm +1

2∇Φn · ∇Φm(1 + U(Φ))nm + V (Φ)),

and obviously we findA = (1 + U(Φ))−1

nmδ(4)(x− y).

If demote the δ-function to a discrete δ-function in a lattice spacetime, where each point on thelattice has small 4-dimensional volume Ω, i.e. δ(4)(x− y)→ Ω−1δxy, then we can calculate

det A ∝ exp[− 1

Ω

∫d4xtr ln(1 + U(Φ(x))].

The result is that there is a correction of

∆L = − i2

Ω−1 tr ln(1 + U(Φ(x))

to the Lagrangian density.

I think this is so interesting. It seems that in this case, the path-integral correction to theLagrangian density is dependent in scale, because the lattice spacing depends on our energy scale.The larger the energy, the smaller the lattice spacing and the more important this correctionbecomes.

Compare this to the typical case, when we have H ⊃∫d3~x1

2Π2n and therefore

A = δ(4)(x− x′).

Replacing (1 + U)−1 with 1 gives the correction

∆L =i

2Ω−1 tr ln(1) = 0.

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9.3 Path-integral derivation of Feynman rules

Let us use the path-integral formalism to calculate quantities which look like

MlAlB ···(xAxB · · · ) =〈Ω,∞|T [ΨlA(xA)ΨlB(xB) · · · ]|Ω,−∞〉

〈Ω,∞|Ω,−∞〉.

The bottom is just a phase factor, because of slow adiabatic evolution, etc. You know the drill bynow. These vacuum expectations values are the ones we are interested in because TODO.

Question 136. Describe how to derive the Feynman rules from the path integral, given that thatthe Hamiltonian is quadratic in the canonical momenta.

Answer 136. Obviously we have

MlAlB ···(xAxB · · · ) =

∫D[ψ]ψlA(xA)ψlB(xB) · · · eiI[ψ]∫

D[ψ]eiI[ψ],

and we will split I =∫d4xL into the terms

I = I0 + I1, where

I0 =

∫d4xL0(ψ, ∂µψ) + (iε terms) and I1 =

∫d4xL1(ψ, ∂µψ).

Now, what follows is very similar to the derivation of Feynman rules in Kardar’s Statistical Physicsof Fields because Kardar also works in the path-integral formalism. To make use of Wick’s theorem(which will allow us to use simple propagators), we expand in the interaction but keep the largepart of the Lagrangian (or action) intact, because that is quadratic and hence gives a Gaussiandistribution:

eiI[ψ] = eiI0[ψ]

∞∑N=0

iN

N !(I1[ψ])N .

The question now is how to use Wick’s theorem. We assume L0 is quadratic in the fields and canbe written

I0 = −1

2

∫d4xd4x′

∑ll′

Dlx,l′x′ψl(x)ψl′(x′)

where the matrix D can contain derivatives as well, of course. For example, the simple scalarLagrangian L0 = −1

2∂µφ∂

µφ− 12m2φ2 has

Dxx′ = ∂µ∂′µδ(x− x′) +m2δ(x− x′)− iεE(~x, ~x′)δ(t− t′).

=

∫d4p

(2π)4eip(x−x

′)(p2 +m2 − iεE(~p)),

if we assume translational invariance. The iε term is a little complicated and I didn’t really botherto understand how he got it. In fact, because ε is infinitesimally small, we can just replace iεEwith iε. But anyway, you get the point. The good thing about this method is that this makes thepropagator extraordinarily easy to see. You can basically see the propagator from the previousequation already!

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Wick’s theorem gives us a product of paired expectation values, like 〈φl1(x)φl2(y)〉. We cancalculate this using Gaussian integration and the result of course is just −i∆l1l2(x, y), where

∆l1l2(x, y) = (D−1)l1x1,l2x2 .

Here, the −i came from both the eiI and the way D is defined with respect to I. Okay, from abovewe merely have

∆(x, y) =

∫d4p

(2π)4eip(x−y)(p2 +m2 − iεE(~p))−1.

The function of the iε is to make the inverse well-defined for all real values of p.

Another important example: let’s study the photon propagator. We will see in a later section thatwe can introduce another term into the Lagrangian, such that

I0 =

∫d4x(−1

4fµνf

µν − α

2(∂µa

µ)2 + (iε terms)) = −1

2

∫d4xd4x′Dµx,νx′a

µ(x)aν(x′).

Here,

Dµx,νy = (ηµν∂ρ∂′ρ−(1−α)∂µ∂ν)δ(x−y)+(iε terms) =

∫d4q

(2π)4(q2ηµν−(1−α)qµqν−iεηµν)eiq(x−y).

Now, all we have to do is invert a 4× 4 matrix. The result is

∆µx,νy =

∫d4q

(2π)4(ηµν

q2 − iε+

1− αα

qµqν

(q2 − iε)2)eiq(x−y).

This is fully covariant.

Question 137. Why are the propagators derived from path integration always manifestly covari-ant?

Answer 137. This question is not fully-formed. In fact, the propagators derived from pathintegration are manifestly covariant only if the action I0 that gave birth to them was manifestlyLorentz-invariant. This can be guaranteed for most theories of interest by integrating in or outcertain fields, for example.

A better question to ask may be why the propagators derived from the canonical formalism are notalways manifestly covariant. This is because, as we saw in section 8.3 of these notes, we derivedthe photon propagator after choosing a gauge (and quantizing the field in this gauge). When wemade the convenient choice A0(x) = 0 in the interaction picture, for example, this is completelynot Lorentz-covariant because of course Aµ doesn’t obey the regular transformation laws if A0 isalways zero.

When we derive things in the path integral, there is no explicit gauge choice to spoil Lorentzcovariance of the propagator. As shown above, we make the gauge choice after we derive a generalform of the propagator.

9.4 Path-integral formulation of QED

The point of this section is essentially to show how the term

α

2(∂µa

µ)2

80


in the Lagrangian density (see the previous section, where we found the photon propagator) arises.The logic is kind of strange, but the math is not very hard. I will just summarize the main points.

First, we start with the (not manifestly Lorentz-invariant) QED Hamiltonian in Coulomb gauge

∇ · ~A = 0 (see section 8.2 of these notes):

H =1

2~Π2⊥ +

1

2(∇× ~A)2 − ~J · ~A+

1

2J 0A0 +HM ,

where we can ignore A0 in the interaction picture.

1. First we integrate out the conjugate field Π, which is easy to do because the Hamiltonian isquadratic in Π. Roughly speaking, the result is

Z =

∫D[a, ψ] exp[i

∫d4x(

1

2( ~A)2− 1

2(∇× ~A)2+ ~J · ~A+LM)−i

∫dtVCoulomb][

∏x

δ(∇· ~A(x))].

Here, the last factor enforces the Coulomb gauge.

2. Second, we make the argument of the exponential manifestly Lorentz invariant by integratingin the A0 component of the field. Essentially how this works is that the stationary point of∫

d4x(−A0J 0 +1

2(∇A0(x))2

turns out to be exactly the Coulomb action∫dtVCoulomb, so it may be reasonable that the

Coulomb action is the low-energy limit of the action of a dynamical field A0(x). (This iswhat Girma Hailu means when he says “integrating in.”) Once this is done, we find thatthe action is manifestly Lorentz and gauge-invariant,

I[A,ψ] =

∫d4x(−1

4FµνF

µν + AµJµ + LM) + (iε terms).

3. Third, we get rid of the final product of delta functions which enforces Coulomb gauge.Think about it: this product of delta functions is equivalent to a gauge choice, so gettingrid of it is equivalent to re-introducing gauge freedom, which turns out to be equivalent tointroducing the term α

2(∂µa

µ)2 with arbitrary α into the Lagrangian density. This makesperfect sense.

The last part of this derivation is on pg. 416-417. It is kind of creative and complex.

9.5 Problems

Question 138. Consider a nonrelativistic particle of mass m in a 1D potential V (x) = m2ω2x2.

Use path-integral methods to find the probability that if the particle is at x1 at time t1, then it isbetween x and x+ dx at time t.

Answer 138. This is a well-documented example of a pain in the ass (see, for example, notes forPhysics 251b). The main difference between this Lagrangian and the Lagrangians we encounterin QFT is that here, the Lagrangian is not translationally-invariant. It can be written

L =

∫dxψ†(x, t)(−m

2∂2t −

1

2m∂2x −

1

2mω2x2)ψ(x, t).

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This is a little weird compared to the Lagrangians we’re used to from relativistic QFT (imagineif we had an interaction which looked like L ⊃ x3φ(x)4). The propagator is

〈x′t′|xt〉 =

∫ x′

x

D[x(t)]ei∫ t′t dtL[x(t)],

and we would like to know how to compute this. Since the Lagrangian is quadratic, the integral isGaussian, and up to a constant from the determinant we may set the argument of the exponentialequal to the classical value from the classical equation of motion. If the boundary conditions are(x, t)→ (x′, t′), then the classical path is known,

xcl(t′′) =

1

sinω(t′ − t)(x sin(ω(t′ − t′′))− x′ sin(ω(t′′ − t))).

Here, t′′ is an intermediate time between t and t′. Then you can obtain the classical action justby substitution into L and integrating over t′′:

Scl(x′t′, xt) =

mω

2 sin(ω(t′ − t))((x2 + x′2) cos(ω(t′ − t))− 2xx′).

This means that the propagator is

〈x′t′|xt〉 = N (t, t′)eimω

2 sin(ω(t′−t)) ((x2+x′2) cos(ω(t′−t))−2xx′),

where N (t, t′) is some normalization constant to preserve the number of particles in the wave-function. It is independent of x, x′ because the size of fluctuations are independent of x (see 251bnotes for more detail). I won’t do the math, but this can be done by assuming |xt〉 is alreadynormalized as 1 = 〈xt|xt〉 and assuming it evolves into the wavefunction |ψ(t′)〉 at time t′, where

|ψ(t′)〉 =

∫dx′〈x′t′|xt〉|x′t′〉.

Then we enforce

1 = 〈ψ(t′)|ψ(t′)〉 =

∫dx′dx′′〈x′t′|xt〉〈xt|x′′t′〉〈x′′t′|x′t′〉 ∝ N (t, t′)3.

The integrations over x′ and x′′ are both Gaussian so this is doable. Then we will know N (t, t′),at least up to a phase. Then the probability the particle is between x and x + dx at time t′ ismerely

P ((x1, t1)→ ([x, x+ dx], t)) = |〈xt|ψ(t)〉|2dx, where

〈xt|ψ(t)〉 =

∫dx2〈x2t|x1t1〉〈xt|x2t〉.

There is still an integral over x2 to do. This integration is how the different paths are of varyinglikelihood - if the standard deviation of the Gaussian integral is high, then the probability oflanding in [x, x+ dx] is also high.

Question 139. The Lagrangian density of the free spin-32

Rarita-Schwinger field, ψµ, is

L = −ψµ(/∂ +m)ψµ −1

3ψµ(γµ∂ν + γν∂µ)ψν +

1

3ψµγµ(/∂ −m)γνψν .

Use path-integral methods to find the propagator of this field.

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Answer 139. Although not mentioned in these notes, Weinberg shows in (9.5.50) that the free-particle term for fermionic path integrals is actually bilinear in the fields and their conjugatemomenta,

I0 = −∫d4xd4x′

∑ll′

Dlx,l′x′pl(x)ql′(x′)

(clearly this must be true: if it was quadratic in a Grassmann variable, the term would just vanish).It seems to be that, instead of going to the trouble of finding the conjugate momenta, we can justassume that we integrate over Dψ and Dψ. This is maybe not so rigorous, but see https:

//arxiv.org/pdf/hep-th/0404131.pdf and https://aip.scitation.org/doi/abs/10.1063/

1.5064003.

So, we just writeL = ψµΛµνψν

and try to invert Λ, which is a 4 × 4 matrix. I suppose you can do it with mathematica. Theanswer apparently (copied from second link above) is

∆µν(p) =P µν(p)

p2 +m2 − iε, where

P µν(p) = (−i/p+m)(gµν − 1

3γµγν +

2

3m2pµpν +

i

3m(γµpν − γνpµ))β.

In the massless case, https://en.wikipedia.org/wiki/RaritaSchwinger_equation says thatthe Rarita-Schwinger Lagrangian has a gauge symmetry, which would mean some kind of gaugesymmetry in the propagator, kind of like the gauge symmetry in the photon propagator. Thatwould make this propagator blow up and we would have to integrate in another field, or something.In fact, this is easy to see because there are lots of 1

mterms in the massive Rarita-Schwinger

propagator, just like in the massive spin-1 boson propagator.

10 Non-perturbative methods

We now study higher-order, or loop, processes. In this chapter we will concern ourselves withderiving general results which are valid to all orders in perturbation theory, or could be manifestlynonperturbative (the difference is that “to all orders in perturbation theory” means something istrue for all diagrams; “manifestly nonperturbative” means something can be proved true withoutdrawing any diagrams). We will not focus on regularizing integrals, because we would like toconceptually separate the physics of higher-order processes from the mathematics of making themconvergent.

We will use the following theorem extensively: The sum of all diagrams between initial state |α〉and final state |β〉 with extra vertices inserted, corresponding to operators oa(x), ob(y), etc. isgiven by

〈Ψβ|T−iOa(x),−iOb(y), · · · |Ψα〉.

There are factors of −i due to the eiL = eipq−H , for example. Often, we take the operators to bethe fields themselves. This is the perturbative expansion of e−iHint .

83

https://arxiv.org/pdf/hep-th/0404131.pdf

https://arxiv.org/pdf/hep-th/0404131.pdf

https://aip.scitation.org/doi/abs/10.1063/1.5064003

https://aip.scitation.org/doi/abs/10.1063/1.5064003

https://en.wikipedia.org/wiki/Rarita–Schwinger_equation


10.1 Symmetries

Question 140. What is meant when we say a particular matrix element, or a particular Feynmandiagram, preserves a symmetry?

Answer 140. Let us think about global symmetries first, since that allows us to use Noether’stheorem. If a theory is invariant under a global symmetry, Noether says there is a conservedcharge, call it q. The matrix element or diagram conserves the charge, or equivalently, preservesthe symmetry, if

〈Ψβ|T−iOa(x),−iOb(y), · · · |Ψα〉 vanishes unless qα − qβ = qa + qb + · · · .

Actually, this depends on the sign conventions. The above identity holds for

Q|Ψα〉 = qα and [Q,Oa(x)] = −qaOa(x).

The proof is very easy:

〈Ψβ|QT−iOa(x),−iOb(y), · · · |Ψα〉 − 〈Ψβ|T−iOa(x),−iOb(y), · · · Q|Ψα〉

= 〈Ψβ|[Q, T−iOa(x),−iOb(y), · · · ]|Ψα〉.

Now, let’s think about discrete symmetries, like PCT; we will meet Furry’s theorem later in thissection. Essentially what this means is that the diagram (or sum of diagrams) cannot contributeunless it the result is invariant under, say, C.

Finally, how about gauge symmetries? The matrix element (or sum of Feynman diagrams) ispreserved under the gauge symmetry if it doesn’t change when we shift the photon field Aµ(x)→Aµ(x) +∂µε(x). From our study of QED, we know that this is equivalent to saying that ∂µJ µ = 0,where the photon field is coupled to this current, AµJ µ. The generalization of ∂µJ µ = 0 to generalmatrix elements is what could be called the generalized Ward identity,

qµMµµ′···βα (q, q′, · · · ) = q′µ′M

µµ′···βα (q, q′, · · · ) = · · · = 0,

where

Mµµ′···βα (q, q′, · · · ) =

∫x,x′,···

e−iqxe−iq′x′ · · · 〈Ψβ|TJ µ(x),J µ′(x′), · · · |Ψα〉.

Question 141. Describe how a matrix element is invariant under translational symmetry.

Answer 141. Translational symmetry is a global symmetry, so we use the same idea as in thequestion above. The symmetry operator is Pµ and we have [Pµ,O(x)] = i∂µO(x), and we canchoose |α〉 and |β〉 to be eigenstates of the momentum. If the big matrix element is called

M = 〈Ψβ|T−iOa(x1),−iOb(x2), · · · |Ψα〉,

then we have(pβµ − pαµ)M = i(∂1µ + ∂2µ + · · · )M.

The solution isM = ei(pα−pβ)xF,

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where F depends only on the relative distances between xi and not the center coordinate x =∑i cixi,

∑i ci = 1 (any choice of c’s is fine). Taking M to Fourier space, M , gives something like

M =

∫x1x2···

e−ik1x1−ik2x2−···+i(pα−pβ)xF

which enforces conservation of momentum!

This is interpreted as: the sum of Feynman graphs conserves four-momentum, order-by-order.This is not very interesting because every vertex of a Feynman graph conserves four-momentum,so obviously the entire graph conserves four-momentum. So in this case, we can give a strongercondition: every Feynman graph conserves four-momentum, and not just their sum.

Question 142. Prove Furry’s theorem,

The sum of all Feynman graphs with an odd number of external photons (on- or off-shell) and noother external lines vanishes identically.

Describe why this theorem is true only for matrix elements, or sums of Feynman diagrams, ratherthan for individual Feynman diagrams.

Answer 142. Furry’s theorem is a consequence of C-invariance of the matrix element. The C-operator flips the charge but not the momentum or spin. If ξ is a phase, then

Ca~pσeC−1 = ξ∗a~pσ,−e and Ca~pσ,−eC

−1 = ξa~pσe.

The Lagrangian is made of Dirac spinors, not these annihilation operators. The action is

CψC−1 = −ξ∗βCψ∗.

Here, βC is a 4× 4 matrix. The important thing is that

C(ψγµψ)C−1 = −ψγµψ =⇒ CAµC−1 = −Aµ.

Otherwise the Lagrangian coupling between current and photon field would not be C-invariant.Because Aµ is odd under charge, a Feynman graph with an odd number of external photon linesand no other external lines is also odd. Therefore, it vanishes. For example, the scattering of aphoton by an external electromagnetic field receives no contributions of first order (or any oddorder) in the external field.

This does not hold for individual Feynman diagrams because vertices and propagators and thingsdo not really “conserve charge-inversion,” like they conserve momentum.

10.2 Polology

Question 143. What is polology?

Answer 143. We would like to know where the poles are in the momentum-space amplitudes ofmatrix elements. Because the poles give the biggest contributions, it may help us compute things,perform contour integrals, as well as think about the on-shell equations of motion, etc.

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In this next question, we prove the main result of this section. Denote

G(q1, · · · , qn) =

∫x1,··· ,xn

e−iq1x1···−iqnxn〈TA1(x1) · · ·An(xn)〉0,

where Ai are Heisenberg-picture operators and 〈O〉0 is the expectation value in true vacuum.Ai may be arbitrary local functions of fields and field derivatives. Contrary to the 〈Ψβ|O|Ψα〉structure we were studying in the last section, here there are no external particle fields, just thevacuum. This can be useful to us if we are working with Gell-Mann - Low theorem, for example.

Question 144. Consider G, defined in the previous question, as a function of q2, where

q := q1 + · · · qr = −qr+1 − · · · − qn.

Here, 1 ≤ r ≤ n−1. The central result of this theorem is that G has a pole at q2 = −m2, where mis the mass of any one-particle state that has a non-vanishing matrix element between the statesA†1 · · ·A†r|Ω〉 and Ar+1 · · ·An|Ω〉. The residue at this pole is

G→ −2i√~q2 +m2

q2 +m2 − iε(2π)7δ(q1 + · · ·+ qn)

∑σ

M0|~qσ(q2 · · · qr)M~qσ|0(qr+2 · · · qn),

where

M0|~pσ(q2 · · · qr)× (2π)4δ(q1 + · · ·+ qr − p) =

∫x1···xr

e−iq1x1···−iqrxr〈Ω|TA1(x1) · · ·Ar(xr)|Ψ~pσ〉.

(We will see in the next question why the x1 and xr+1 disappeared. It has to do with absoluteversus relative coordinates.) Interpret the physical meaning of this theorem.

Answer 144. Wow, what a complicated theorem. What does it mean? Notice that with

q := q1 + · · · qr = −qr+1 − · · · − qn,

we are conceptually splitting the external fields Ai into fields coming in (1 through r) and fieldsgoing out (r + 1 through n). This holds if the times tr+1, · · · , tn > t1, · · · , tr, which is alwaysa possibility because we are integrating over all the xs and have to pass over this kind of time-ordering. What if there was a Feynman diagram such that a single line connected the lines goingin and out? Then the total matrix element would be proportional to a propagator

1

k2 +m2 − iεand it is not necessary that such a particle actually appears in the Lagrangian of the theory. It cancorrespond to a bound state for this process. The pole arises not from single Feynman diagrams,but rather from infinite sums of diagrams. This is a nonperturbative claim. Also good to knowis that the bound state does not necessarily correspond to a “single particle.” That’s not howwe should think about it. It could be like a bunch of particles having a existentially short-livedorgy or something and then leaving the party, never to see each other again. But that party itselfconstitutes the bound state, and we represent the bound state with a one-particle state. Thebound state is technically an infinite sum of diagrams made up of the fields in the Lagrangian.

Of course, there should be a bound state for every way of partitioning the external fields into “in”fields and “out” fields.

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Question 145. Outline the proof of this “polology” theorem. Where does the propagator 1k2+m2−iε

arise in the mathematics?

Answer 145. If we are to think of the first r fields coming in and the last n− r + 1 fields goingout, we need to insert a δ-function to make the time-ordering come true.

G(q1 · · · qn) ⊃∫x1···xn

e−iq1x1···−iqnxnθ(min[x01 · · ·x0

r]−min[x0r+1 · · ·x0

n])

×〈TA1(x1) · · ·Ar(xr)TAr+1(xr+1) · · ·An(x)n〉0.Because we are interested in virtual one-particle states, we can insert a not-complete basis ofone-particle states, ignoring intermediate states with two or more fields:

G(q1 · · · qn) ⊃∫x1···xn

e−iq1x1···−iqnxnθ(min[x01 · · ·x0

r]−min[x0r+1 · · ·x0

n])

×∫~pσ

〈Ω|TA1(x1) · · ·Ar(xr)|Ψ~pσ〉〈Ψ~pσ|TAr+1(xr+1) · · ·An(x)n|Ω〉.

Now some true voodoo is going to happen. We turn the Heaviside function into a function witha pole, like in regular Feynman propagator,

θ(τ) = − 1

2πi

∫ ∞−∞

e−iωτ

ω + iεdω.

To turn the 1ω

into something in terms of the fields and what we know, we switch to relativecoordinates. The overall center coordinate is arbitrarily chosen to be x1 for the incoming particlesand xr+1 for the outgoing particles. We need translational invariance for both sets of particles.Integration over the “center” coordinates x1 and xr+1 then gives the energy-conserving δ-functions

δ(√~p2 +m2 + ω − q0

1 − · · · − q0r) and δ(

√~p2 +m2 + ω + q0

r+1 + · · ·+ q0n).

As you can see here, the matrix element will be nonzero if only if momentum is conserved.However, if momentum is conserved, the matrix element is not necessarily nonzero. It means thatthe scattering process does care whether the intermediate one-particle state is Al Gore or JerryFalwell, even if they weigh the same and have the same momentum. Then we use

1

p0 −√~p2 +m2 + iε

→ −2√~p2 +m2

p2 +m2 − iε

and this gives the result.

Why is it that we cared about the one-particle intermediate states and not, say, two- or three-particle intermediate states? Apparently it is because multiple-particle intermediate states producebranch points in p, which are gentler than poles. This makes sense if you think about somethinglike

1

(p− p′)2 +m2

1

(p′)2 + (m′)2.

Here, there is certainly no obvious pole in p. There is a pole in p′, but that momentum is consideredto be related to a virtual particle only rather than related to the momenta of the incoming andoutgoing particles, p.

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In retrospect, Weinberg maybe didn’t need to phrase things in position-space. We could haveavoided writing so many

∫x1···xn integrals. The only value of the position-space integration is to

end up with momentum-conserving δ-functions.

Question 146. Explain how the “polology” of the nonrelativistic Yukawa interaction, which is anapproximation to the residual strong force between nucleons, suggested the existence of π mesons.

Answer 146. Yukawa potential in nonrelativistic theory gives a scattering like (no integrationsover energy because we assume everything is on-shell)∫

~x1~x2,~x′1~x′2

e−i(~x1·~p1+~x2·~p2)+i(~x′1·~p′1+~x′2·~p′2) e−mπ |~x1−~x2|

2π|~x1 − ~x2|δ(~x1 − ~x′1)δ(~x2 − ~x′2)

= −(2π)3δ(~p1 + ~p2 − ~p′1 − ~p′2)1

(~p1 − ~p′1)2 +m2π

.

The last term is the nonrelativistic limit of the regular QFT propagator, i.e. for low energy sothe energy term disappeared, since for nucleons N , 1

2mN(~p2

1 − (~p′1)2) is smaller than |~p1 − ~p′1| dueto the large mass. Another way to think of it is the static limit; this also happens to be how youget the Coulomb interaction 1

q2 from the more general retarded interaction 1ω2−q2 .

Based on this, Yukawa proposed the existence of a meson field with mass around 100 MeV. Theviewpoint at that time was that the Lagrangian was thus far incomplete and they were missing ameson field in the theory. The modern interpretation, based on polology, is that while a theory ofmesons is one way to understand things, that there is not an absolute need to throw in anotherfield to account for what looks like a virtual particle of mass mπ. In the modern interpretation, theπ meson, or pion, is interpreted as a bound state of two quarks or something. It does not have toappear, itself, in the fundamental Lagrangian. Of course, you can phenomenologically introducesuch a field in an effective Lagrangian. This is nice because pions turn out to be unstable with veryshort lifetime. It would not be honest to introduce them as “true fields” in the Lagrangian, becausethat would imply that you could have a free-particle state made of pions. But that free-particlestate would die out on its own, which doesn’t make sense; hence polology is a nice way to accountfor particles that mediate exchange but typically aren’t reagents or products in themselves.

10.3 Field and mass renormalization

In this section, we will study the renormalization of masses and fields. Of course, charges, couplingconstants, etc. also need to be renormalized. The main point of this section is

“The renormalization of masses and fields [and couplings] has nothing directly to do with thepresence of infinities and would be necessary even in a theory in which all momentum space

integrals were convergent.”

Question 147. What is a renormalized field?

Answer 147. A renormalized field is a field “whose propagator has the same behavior near itspole as for a free field.” The renormalized mass is “defined by the position of the pole,” and isgenerally different from the mass of the bare field.

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We will see in this section that if |q, σ〉 is the one-particle state of a field (renormalized or not), itis convenient to define a renormalized field as one whose normalization satisfies

〈Ω|Ψl|q, σ〉 =ul(qσ)

(2π)3/2.

This is always satisfied by bare fields in free theories. However, if we have interactions, then thiswill not be satisfied by the bare fields. We will find it convenient to introduce renormalized fields,which are obtained, as we know, with self-energy diagrams.

This is the point of “having the same residue.” In a sense, the residue is the number of particles car-ried by the field. We would like to have the same number of particles, but just with a different dis-persion, so we demand physically that the number of particles is equal to one. See the discussion athttps://www.physicsoverflow.org/15999/renormalization-condition-must-the-residue-the-propagator.

Question 148. How does the renormalized field appear in a simple theory? How can I see itsconnection to the 1PI graphs, or equivalently, to the self-energy diagrams?

Answer 148. Consider the Lagrangian density

L = −1

2φ(−m2)φ− V (φ).

Here, φ,m, V are all bare quantities. However, there is an interaction. Due to the interaction, φmay not be normalized according to the convention in the previous question. The solution is tointroduce a renormalized field and mass,

φ = Z−1/2φ and m2 = m2 + δm2,

such that φ is a renormalized field whose propagator has a pole at m2. Then we write

L = L0 + L1

where

L0 = −1

2φ(− m2)φ,

L1 = −1

2(Z − 1)φ(− m2)φ+

Z

2δm2φ2 − V (φ),

where V (φ) = V (√Zφ).

Question 149. How can I find Z and m for the renormalized field in the scalar theory?

Answer 149. Weinberg starts with the Lagrangian for the renormalized field to find these things.It seems a little weird to find Z and m from a theory which already contains these parameters. Iguess it works because actually m hasn’t been defined yet. In a sense, we will define it only aftersolving for Z (more on this below!).

Let us denote the self-energy, or sum of 1PI graphs, as Σ(q2) = i(2π)4Π(q2). Important point:these “1PI” graphs actually include an insertion of vertices (i.e. a single line which is not thepropagator) plus the regular loop contributions:

Π(q2) = −(Z − 1)(q2 + m2) + Zδm2 + Πloop(q2).

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https://www.physicsoverflow.org/15999/renormalization-condition-must-the-residue-the-propagator


There is no 12

in the Π(q2) because of the 12!

kind of thing on the vertices. Also, the extra (2π)4

normalization in front is to match −i(2π)−4 1q2+m2−iε , or rather, to cancel out the (2π)4 thing when

we write a geometric series (sneaky). The renormalized propagator is just

∆′(q) =1

q2 + m2 − Π(q2)− iε.

Remember that the definition of renormalized mass m is the pole of ∆′(q). If it smells like anormal propagator, it should also have a unit residue, like the bare propagator. This implies that

Π(−m2) = 0 and ∂q2Π(q2)

∣∣∣∣q2=−m2

= 0.

The solution is therefore

Z = 1 + ∂q2Πloop(q2)

∣∣∣∣q2=−m2

, Zδm2 = −Πloop(−m2).

These solutions are explicit. There is no Z- or δm2-dependence in Πloop.

For the Dirac field, as another example, the Lagrangian contains a m rather than m2. Thus,renormalization is more convenient with m = m+ δm rather than m2 = m2 + δm2.

Question 150. How is the calculation of Z and m2 above related to the things like “minimalsubtraction scheme” and whatever from Schwartz QFT ?

Answer 150. Usually, people are not so careful when setting up the idea of renormalized field.An equivalent, and quicker, way to do things is to simply calculate

Πloop(q2)

and say “okay, perhaps Πloop(q2) has some infinities, but I can subtract a linear polynomial in q2

by choosing coefficients to satisfy the condition Π(−m2) = 0.” This is equivalent to subtractingout the infinities by hand. It is the reason why Z is formally infinite.

Unfortunately, it’s not yet clear to me why the infinities should be limited to q2 terms. Hopefullywe will find out soon.

Question 151. Explain how our understanding of polology from the last section can help usunderstand the need for field renormalization. What does this have to do with the LSZ reductionformula?

Answer 151. Let us describe the idea of LSZ reduction formula in very heuristic terms. Then,we will see what it has to do with renormalized fields.

The idea of LSZ reduction formula is as follows. Consider a Lagrangian L and let Ol(x) bya Heisenberg-picture operator living in an irrep l with the same Lorentz-transform properties asthe free field ψl. Ol(x) may be totally unrelated to the fields in L .

Usually, we calculate scattering processes and matrix elements with the fields in the Lagrangian.Let me call them ψL . LSZ claims that we can develop Feynman rules to calculate the samescattering processes and matrix elements for L , except by using the Heisenberg operators Ol(x)

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which have nothing to do with L . We can think about this as follows: you have a girlfriend whowill be late, so she asked you to make lunch. However, the red line subway broke down and youwill also be late. Rather than starve and not have her matrix element, your girlfriend outsourcesall elements of the task to various sentient beings. Girlfriend’s girlfriend buys some ginger ale fromCVS and girlfriend’s other friend calls her mom who sends her dog to tell her husband to text hiscoyote to run to Felipe’s and steal some burritos, which are brought back to the dorm via carrierpigeons controlled by the CIA. In the end, she gets a lunch. It is the same scattering process asif she just calculated things with the fields in the original Lagrangian (i.e. if I bought burritos).As you can imagine, this is a very deep theorem. Perhaps it is not so surprising, since it’s kind oflike inserting lots of intermediate states in linear algebra.

The LSZ formula gives an explicit form for the “propagator” of Ol, or more precisely, the propa-gator of ψl, which has the same Lorentz properties as Ol.

Question 152. Outline the proof of the Lehmann-Symanzik-Zimmerman reduction for-mula. What an intimidating name!

Answer 152. In the notation of the previous section, suppose r = 1. This means that, when wegroup the external operators 1 through n into “in” and “out” categories, there is only one operatorin the “in” category. Let the “in” operator be Ol, where l represents an irrep of the Lorentz group,and let the “out” operators by Ai(xi). Let us write the corresponding matrix element as

Gl(q1q2 · · · ) =

∫x1x2···

e−iq1x1−iq2x2···〈Ω|TOl(x1)A2(x2) · · · |Ω〉.

Ol means that O is a Heisenberg-picture operator with the same Lorentz transform properties asa free field ψl belonging to an irrep l of the Lorentz group. It could be that Ol is some complicatedfunction of the fields, which happens to transform in the l-irrep. The polology theorem of theprevious section suggests that if there is a one-particle state |Ψq1σ〉 having nonzero matrix elementswith the “in” and “out” states, then we can write

〈Ω|Ol(0)|Ψq1σ〉 = N〈Ω|ψl|Ψq1σ〉 = (2π)−3/2Nul(q1σ),

by Lorentz-invariance. The result is

Gl(q1q2 · · · )→−2i

√q2

1 +m2

q21 +m2 − iε

∑σl′

ul(q1σ)u∗l′(q1σ)Ml′

where Ml′ is a complicated matrix element and ul′ means the same kind of operator but fora different irrep. The sum

∑σl′ ul(q1σ)u∗l′(q1σ) looks kind of like the spin sum on top of a

propagator. Neat!

10.4 Renormalized charge; Ward-Takahashi identity

As Girma Hailu likes to tell people, the electron charge is a function of energy scale. At asymptot-ically low energy it is e = 1.6× 10−19 C and at higher energy it is different. We will see how to beconsistent with defining the charge of a renormalized field. This is called renormalized charge,but the idea is broadly applicable to any kind of global symmetry, not just charge conservation.

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Recall that the Ward identity is a generalization of ∂µJ µ to multiple external currents. The ideaof renormalized charge gives us a more general and nonperturbative Ward identity, called theWard-Takahashi identity.

Question 153. Explain why the charge needs to be renormalized. Describe how to do so.

Answer 153. Let Q be the charge operator and let F (y) be a local function of the fields y. If amatrix element 〈Ω|F (y)|pσn〉 conserves charge (we saw this matrix element was important in theLSZ reduction theorem), then

Q〈Ω|F (y)|pσn〉 = 0 =⇒ qn = qF .

Of course, qF is the sum of the electric charges of the fields being multiplied together to give F (y).

So, we have charge conservation. However, charge conservation holds under a global rescaling ofall charges, qn → αqn, so it looks like we have too much freedom. To fix this, we say the “physicalelectric charges are those that determine the response of matter fields to a given renormalizedelectromagnetic field Aµ.” In other words, we demand the electron-photon coupling has the form

L ⊃ ψl(∂µ − iqlAµ)ψl.

This is convenient because the current retains the form J µ = δLδAµ

∣∣A=0

.

If the renormalized field is Aµ = 1√Z3Aµ, we should define

ql =√Z3ql.

The physical electric charge is ql, and the proportionality constant√Z3 is the same for all particles

(obviously, since it renormalizes the photon, which is not a charged particle).

Question 154. What does the above theorem tell us about the charges of different particles, atvarious energy scales?

Answer 154. The charges of the proton and electron are exactly opposite at all energy scales,because both change only by 1√

Z3, which is the same for each (because it’s for the photon!). See

also: Schwartz QFT section 19.5.

This means that there must be cancellations among the radiative corrections to the propagatorsand vertices of the charged particles. These cancellations are described by the Ward-Takahashiidentity. We derive it in the next question.

Question 155. What is the Ward-Takahashi identity? Describe its connection to the less general∂µJ µ, and explain how the more general form is related to Noether theorem for the global U(1)symmetry. Elucidate why it means there are no radiative corrections to the charge due to theelectron-photon vertices.

Answer 155. Let S ′(k) and Γµ(k, l) be the dressed electron propagator and electron-photonvertex, for kµ-momentum electron coming in and a lµ-momentum electron coming out. The q → 0limit is obviously

S ′(k)→ 1

i/k +m− iεand Γµ(k, l)→ γµ.

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The Ward-Takahashi identity is

(l − k)µΓµ(k, l) = i(S ′−1(k)− S ′−1(l)).

In the l→ k limit, this gives the original Ward identity which is just

Γµ(k, k) = −i ∂∂kµ

S ′−1(k) =⇒ Γµ(k, k) = γµ + i∂

∂kµΣ(k).

The claim is that Ward-Takahashi identity ensures the radiative corrections to the vertex functionΓµ cancel when a fermion on the mass shell interacts with an electromagnetic field with zeromomentum transfer (such as when we measure the electric charge). That is because on the mass-shell,

u′kΓµ(k, k)uk = u′kγ

µuk.

This follows by using Dirac equation for the spinors, and from ∂Σ∂/k

∣∣∣∣/k=im

= 0.

The conclusion is that only the renormalization of the photon propagator makes the charge flow.Neither the renormalization of the electron propagator or the electron-photon vertex change thecharge; their effects are “cancelled out,” so to speak, via Ward-Takahashi. In this sense, propagatorchanges by something like Σ(k) and the vertex changes by i ∂

∂kµΣ(k), and these changes turn out

to be exactly opposite.

Question 156. Derive the Ward-Takahashi identity.

Answer 156. The dressed electron-photon vertex can be written

TJ µ(x)Ψn(y)Ψm(z).

Ward-Takahashi identity follows from

∂µTJ µ(x)Ψn(y)Ψm(z) = T(∂µJ µ(x))Ψn(y)Ψm(z)+ δ(x0 − y0)T[J 0(x),Ψn(y)]Ψm(z)+ δ(x0 − z0)TΨn(y)[J 0(x), Ψm(z)].

This is because ~∇x doesn’t act on y or z, but the time-ordering means the quantity changesnontrivially when x0 = y0, for example. The terms with δ-functions are called contact termsand are described in Schwartz QFT as not contributing to the scattering amplitude, because itdoesn’t have poles. We know ∂µJ µ = 0 and also

J µ = −i∑l

∂L

∂(∂µΨl)qlΨl =⇒ [J 0(x, t),Ψl(y, t)] = −qlΨl(y, t)δ(x− y).

Then we insert the result into

− i(2π)4qS ′nn′(k)Γµn′m′(kl)S′m′m(l)δ(p+ k − l) =

∫xyz

e−i(px+ky−lz)〈Ω|TJ µ(x)Ψn(y)Ψm(z)|Ω〉.

The result is the Ward-Takahashi identity. How can we understand this? Ward-Takahashi identitymeans that the kinetic term Ψ/∂Ψ and the interaction Ψ /AΨ are renormalized in the same way, so

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they can be combined into a covariant gauge transform term, Ψ /DΨ. In this sense, the electriccharge doesn’t change because it’s kind of a ratio between /∂ and /A.

Another way to understand the derivation of Ward-Takahashi identity is as follows: note that forany operators · · · , we have

∂µ〈Ω|TJ µ(x) · · · |Ω〉 = contact terms.

Because the contact terms do not contribute to the scattering amplitude, we find that

kµMµ = 0,

where Mµ is the matrix element represented by the correlation function (or rather, the Fouriertransform of the time-ordered quantity). Actually if we include the renormalization factors ex-plicitly, it looks like

(k − l)µΓµ(k, l) = Z−12 Z1e

[S ′(k)−1 − S ′(l)−1

]and since Γµ and S ′ should be finite, this suggests Z1 = Z2, which Schwartz tried to provein his book. Here, Z1 is the renormalization factor on the electron propagator and Z2 is therenormalization factor on the electron-photon vertex. See http://www.physics.indiana.edu/

~dermisek/QFT_08/qft-II-15-1p.pdf.

10.5 Gauge invariance of the S-matrix

We claim that the S-matrix is gauge-invariant. This means the following: let a photon prop-agator a matrix element be ∆µν(q) and consider the gauge transform

∆µν(q)→ ∆µν(q) + αµqν + qµβν

or the redefinition of polarizations

εµ(kλ)→ εµ(kλ) + ckµ,

where αµ, βν , c don’t have to be constants, and don’t have to be the same for all photons in thetheory.

There will be some radiative corrections to the photon propagator. The important thing aboutgauge invariance of the S-matrix is that the dressed photon propagator is still massless. This is areally nontrivial thing, since usually the mass flows with energy scale.

Question 157. Outline a proof of the gauge-invariance of the S-matrix.

Answer 157. This follows immediately by writing

Sβα ∝∫q1q2···

∆µ1ν1(q1)∆µ2ν2(q2) · · · [eµ1(k1λ1) · · · ][Mµ1µ2···ba (momenta)]

and use the result of the Ward identity

pµMµ···ba = 0.

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http://www.physics.indiana.edu/~dermisek/QFT_08/qft-II-15-1p.pdf

http://www.physics.indiana.edu/~dermisek/QFT_08/qft-II-15-1p.pdf


This is not very clear in the book. I think the logic is kind of like this: pµMµ··· is the generalization

of Noether theorem ∂µJ µ, but it seems to hold even off-shell. That is because the gauge transformAµ → Aµ + ∂µε is supposed to leave the matrix element

Mµµ′···βα (q, q′, · · · ) =

∫x,x′,···

e−i(qx+q′x′+··· )〈β|TJ µ(x)J µ′(x′) · · · |α〉

invariant even if the photons coupling to the external currents are off-shell.

We can build the S-matrix out of these matrix elements, plus some electron propagators andelectron-photon vertices and stuff like that. Because the matrix elements are gauge-invariant evenwhen they are off-shell, so is the S-matrix.

A reflection: I think this is really interesting, since Noether’s theorem ∂µJ µ is an on-shell theorem,but this identity ∂µM

µ··· seems to hold off-shell as well, due to gauge invariance. I think this isbecause gauge invariance is the statement of a local symmetry, so we can say more powerful thingsif we have a local symmetry, compared to if we only have a global symmetry.

Here is a short derivation by Arkani-Hamed of the identity in question: http://pages.physics.cornell.edu/~ajd268/Notes/Ward_Identity.pdf. However, it is not the best because he usesan equation of motion for Aµ, which only holds on-shell. A really good resource is here, http://bolvan.ph.utexas.edu/~vadim/Classes/16f/WTI.pdf. It states on the first page that theidentities hold for “off-shell amplitudes.”

Question 158. Outline a proof of the following result:

Radiative corrections do not give the photon a mass.

Answer 158. Consider the 1PI corrections Π(q), such that

∆′µν(q) =[∆(q)−1 − Π(q)

]−1

µν.

This can be rewritten as

∆′µν(q) = ∆µν(q) + ∆µρ(q)Πρσ(q)∆′σν(q).

Gauge invariance must hold for dressed propagators as well as for bare propagators. We know thisis true because of the theorem on matrix elements, which can be thought of as bare propagatorswith lots of vertices in the middle, or dressed propagators with fewer vertices. This restrictionwith Lorentz invariance gives

qρΠρσ(q) = 0 =⇒ Πρσ(q) = (q2ηρσ − qρqσ)π(q2).

We can find the inverse:

∆′µν(q) =ηµν − ξ(q2)qµqν/q

2

(q2 − iε)(1− π(q2))where ξ(q2) = ξ(q2)(1− π(q2)) + π(q2).

This is kind of interesting. It looks like the dressed propagator differs from the bare propagatorby a momentum-dependent overall constant and a different gauge choice. Here is the crux: Π(q)is a sum of one-photon irreducible graphs, so it does not have a pole at q2 = 0. Otherwise therewould be a matrix element with a massless “bound state” which could be represented as a photon

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http://pages.physics.cornell.edu/~ajd268/Notes/Ward_Identity.pdf

http://pages.physics.cornell.edu/~ajd268/Notes/Ward_Identity.pdf

http://bolvan.ph.utexas.edu/~vadim/Classes/16f/WTI.pdf

http://bolvan.ph.utexas.edu/~vadim/Classes/16f/WTI.pdf


propagator. Thus, the pole of the dressed propagator remains at q2 = 0, since there is no otherpole at q2 = 0 to get rid of the existing one.

This isn’t particularly watertight. Physically we can argue that the remaining pole at q2 = 0implies a unique mass m = 0; however, we did not show there are no new poles. This would bereally confusing because we would not know which pole to use for the physical mass!

Question 159. Show that the renormalization factor Z3 can be related to π(q2) from the previousquestion via

Z3 = 1 + πloop(q2 = 0).

In practice, we just calculate the loop contributions and subtract a constant to make π(0) vanish.

Answer 159. This follows because ∆′ is supposed to contain all diagrams. Therefore sum aboutthe radiative corrections to ∆′, they should not change the propagator and we should get ∆′ back(kind of like a fractal?).

The gauge-invariant part of the propagator,

∆′µν(q) =ηµν

(q2 − iε)(1− π(q2))

should thus have a residue which is independent of radiative corrections. Therefore π(0) = 0.

If we write the Lagrangian as

L = −1

4F 2 + (1− Z3)

1

4F 2 + Lmatter,

then the self-energy is likeπ(q2) = 1− Z3 + πloop(q2)

because the factor 1−Z3 multiplies the q2ηρσ−qρqσ if you write the Lagrangian as L = AµXµνAν

or whatever. Setting q2 = 0 gives the desired result.

A good question is: why is the residue of the photon pole important? Another way of phrasingthis question is why the propagator should be invariant under more radiative corrections at q2 = 0(on-shell), but not necessarily for q2 > 0 (off-shell). I think this is a physical condition whichwe can understand as coming from the LSZ theorem applied to a photon propagating by itself inempty space. From Wikipedia, “[LSZ] asserts that S-matrix elements are the residues of the polesthat arise in the Fourier transform of the correlation functions as four-momenta are put on-shell.”

10.6 More on the LSZ reduction formula

This is not a section of Weinberg, but I feel that LSZ is both important and not something Iunderstand very well. So, I will study it in more depth than Weinberg gives.

I should say what I skipped in this chapter of Weinberg: Kallen-Lehmann (spectral) representation,dispersion relations. I think none of these topics are too difficult so I can always read them later.(By the way, apparently “none” is plural when referring to “not any,” so in the previous sentenceI correctly used “are” instead of “is.”) The section on form factors, which is in Weinberg Ch 10,is moved to Ch 11 of my notes. Weinberg put it in Ch 10 because it is a nonperturbative form,but I put it in Ch 11 because they are useful in renormalizing QED; either organization is fine.

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11 One-loop radiative corrections in QED

We renormalize QED to one-loop. We will perform the following calculations: (1) vacuum po-larization, which renormalizes the photon propagator (2) anomalous magnetic moment of theelectron and the α

2πcorrection (3) electron self-energy, which renormalizes the electron propaga-

tor. Before doing (2), we will take a diversion and learn about form factors, whose properties arenonperturbative.

Here are the tools we will use:

• Feynman parameters

• Wick rotation

• Dimensional regularization (due to t’Hooft and Veltman)

• Pauli-Villars regularization

I will not worry about the calculations. They are done in great detail, for example, in SchwartzQFT with possibly more organized and polished methods. I am more interested in the motivationand the exact procedures of counterterms, etc.

11.1 Counterterms

The point of counterterms is that the calculations will give infinities, but the final results are finite“if expressed in terms of the renormalized charge and mass.” A counterterm generically looks like

(Z − 1)× (some term in the renormalized fields).

The Z − 1 will be formally infinite. We will understand this better in the next sections.

Introduce the renormalized fields (the B subscript means “bare”):

ψ =1√Z2

ψB, Aµ =

1√Z3

AµB, e =√Z3eB,m = mB + δm

We must take the renormalizations of e and Aµ to be opposites because of the combinationψ(/∂ + ie /A)ψ. If Aµ got renormalized, then we must change e by the opposite factor in order forthe response of the charged particles to a renormalized electric field to be the same. This is thesame thing as something we proved in the last chapter: charge renormalization arises only fromradiative corrections to the photon propagator.

Let us rewrite the QED Lagrangian in the renormalized fields and the counterterms:

L = L0 + L1 + L2

where

L0 = −1

4F 2 − ψ(/∂ +m)ψ

L1 = −ieψ /Aψ

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L2 = −1

4(Z3 − 1)F 2 − (Z2 − 1)ψ(/∂ +m)ψ + Z2δmψψ − ie(Z2 − 1)ψ /Aψ.

Apparently, “all the terms in L2 are of O(e2) and higher, and these terms just suffice to cancelthe ultraviolet divergences in the sum of the loop graphs.” The point is that L0 + L1 is thefree field Lagrangian plus the regular electron-photon coupling, and the L2 are the counterterms.(Note: it’s not obvious that L2 should be enough to cancel all the divergences. Often, if you havea Lagrangian and want to write down the counterterms, you have to make sure you include allpossible scalars in the Lagrangian which could contribute, not just the scalars in the original bareLagrangian.) Let’s see how this plays out.

11.2 Vacuum polarization

The vacuum polarization is the virtual cloud of an electron-positron pair which is the leading-order correction to the photon propagator. It produces measuable shifts in the energy levels ofhydrogen, etc.

As in the previous chapter, define i(2π)4Π∗µν(q) as the 1PI sum with polarization indices µ andν. The complete photon propagator ∆′ is

∆′µν = ∆µπ[(1− Π∗∆)−1

]πν. (1)

Question 160. What is the leading-order contribution to the 1PI energy?

Answer 160. The leading-order correction is something like

i(2π)4Π∗µν(q) = −∫p

Tr

[−i

(2π)4

−i/p+m

p2 +m2 − iε× (2π)4eγµ × −i

(2π)4

i(/p− /q) +m

(p− q)2 +m2 − iε× (2π)4eγν

].

To preserve the Ward identity qµΠ∗µν(q) = 0, it is best to use t’Hooft and Veltman’s dimensionalregularization. See Schwartz QFT or Weinberg for more details.

Question 161. Suppose we have already calculated the value of the virtual polarization bubble.How do we get rid of the divergence?

Answer 161. The idea is that the loop diagram is not the only O(e2) contribution to the 1PIenergy; there is also the counterterm. Adding these two contributions together suggests

Π∗µν(q) = (q2ηµν − qµqν)π(q2)

where

π(q2) = −4e2Ωd

(2π)4Γ(d

2)Γ(2− d

2)[ ∫ 1

0

dxx(1− x)(m2 + q2x(1− x))d−4

2

]− (Z3 − 1).

Now we use the result from section 10.5 of these notes that the gauge-invariant part of the prop-agator,

∆′µν(q) =ηµν

(q2 − iε)(1− π(q2))

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should have a residue which is independent of radiative corrections (due to the fractal nature ofthe loop graphs). Therefore π(0) = 0. This gives us the value of Z3, at least to O(e2). Solving forZ3 and then substituting it back into the self-energy gives

π(q2) = −4e2Ωd

(2π)4Γ(d

2)Γ(2− d

2)

∫ 1

0

dxx(1− x)[(m2 + q2x(1− x))

d−42 − (m2)

d−42

].

Let us see how this form leads to the cancellation of the poles. The regularization is removed bytaking d→ 4.

• Divergence of the Γ-function: There is a divergence

Γ(2− d

2)→ 1

2− d2

− γ,

where γ is the Euler-Mascheroni constant. However, for d→ 4 we have

(m2 + q2x(1− x))d−4

2 → 1, (m2)d−4

2 → 1.

The limiting behavior of the limits just above is faster than the limiting behavior of the Γ-function, which is easy to check (or to intuit). Therefore, the divergences of the Γ-functionhave been regulated.

This is called a “cancellation of poles” because of the pole d = 4 in 12− d

2

.

• Many cancellations of finite parts: Many finite parts cancel between the loop diagram andthe counterterm, Z3 − 1. This is not really because we “wanted” them to cancel, but ratherbecause if we choose the correct value of Z3 − 1 for π(q2 = 0) = 0, then they turn out tocancel anyway. These finite parts include

– The term multiplying γ.

– The product of the pole in Γ(2− d2) and the term (4−d) lnµ, where µ is the dimensionful

generalization of the coupling constant e.

However, there is a finite part which changes with momentum q. It is the product of the pole inΓ(2 − d

2) and the linear term in the expansion of (m2 + q2x(1 − x))

d−42 and (m2)

d−42 in powers of

d− 4. This gives

π(q2) =e2

2π2

∫ 1

0

dxx(1− x) ln(1 +q2x(1− x)

m2). (Final result)

Let’s summarize this procedure. We fixed the value of Z3−1 by demanding a physically meaningfulcondition on the residue of the pole of the photon propagator. This happened to be at q2 = 0for the photon, but presumably for a massive propagator it would have been at the renormalizedmass, q2 = m2.

After fixing Z3 − 1, we found that the infinite part (Z3 − 1)∞ cancels the infinity in the loopdiagram corresponding to an incoming photon at any momentum (including off-shell momenta).Essentially, this is because a finite change in momentum of the photon doesn’t change the characterof the divergence; it does, however, change the finite part of the loop diagram.

99


However, we found that this finite change in the loop diagram is related to the pole in Γ(2 − d2).

So, it’s not like the pole is unimportant and gets totally subtracted out. The divergent nature ofthe pole is subtracted out, but the pole gets multiplied by other things which can both (1) get ridof the divergence (2) change with photon momentum q.

To summarize, the pole is like a behind-the-scenes kind of thing; we never see a pole directly,but it is the important part of the loop diagram in the sense that only the divergence of the polecontributes to a q-dependence of the self-energy. Behind every Sultan is an evil divergent Jafar;regularization tames the out-of-control vizier running the kingdom.

Question 162. Describe the physical effects of the radiative correction to the photon propagator.

Answer 162. First of all, these radiative corrections do not give the photon a mass. We provedthis in section 10.5 of these notes, and it is also obvious here because the pole of the propagatoris still located at q2 = 0.

Heuristically, here is what happens: suppose we have an external electron Alice interacting withan external positron Bob; they interact by shooting photons at each other. The electron-positronvirtual particle pair in the photon tends to have a particular orientation relative to Alice and Bob;namely, it tends to shield Alice and Bob from each other. So the virtual positron is more likelyto be on Alice’s side, and the virtual electron is more likely to be on Bob’s side. This is the sameas if Alice and Bob were interacting with the original photon propagator, but their charges wereweaker. Indeed, this is the entire idea of the scale-dependence of the electronic charge.

It is shown in Weinberg (or, you can guess by taking the static limit of the propagators andexpanding (1) to leading order in the self-energy) that the new Coulomb potential is like

V (q) =e1e2

(2π)3

1 + π(q2)

q2.

Correspondingly, the renormalized electronic charge is

e =eB√Z3

,

where eB is the (infinite) bare electronic charge. We know from Girma Hailu that eB →∞.

(Note: based on e = eB√Z3, it might seem that there is no scale-dependence of e. It depends on

what you are talking about; if you cut off the integral∫k

at some scale Λ, then you would get ascale dependence of Z3 and hence of e as well. This is the idea of Wilson’s RG.)

Question 163. Describe how this produces measurable energy level changes in the hydrogen ormuon atomic energy levels.

Answer 163. The corrected Coulomb potential reduces to the regular Coulomb potential quitequickly,

1 + π(q2)

q2→ 1

q2for |q| < m,

where m is the electron mass (it was part of the loop calculation!). The electrons are not verymassive, so the correction is important only when |q| is large, or equivalently only when theseparation between two charged particles |r| is small.

100


If you think about the Hydrogenic wavefunctions Rnl(r), they are nonzero around the origin onlyfor l = 0. Hence, the radiative corrections to the Coulomb potential tend to break the l-degeneracyof energy levels Enlm, and especially so for l = 0 compared to l 6= 0. This is called the Uehlingeffect and the mathematical answer is given by Weinberg (11.2.42),

∆En,l=0 = −4Z4α5m

15πn3.

People saw this in experiments, Phys Rev 48 55 (1935). That seems pretty early to me.

11.3 Interlude: Form factors

Suppose we want to calculate the scattering of a particle by an external EM field to first orderin the external field, but to all orders in all other interactions, such as the virtual corrections toour particle’s propagation. We should make this general enough to handle when the external EMfield actually arises from the field of another charged particle, so the photon propagator can beoff-shell.

According to Gell-Mann - Low theorem (Weinberg section 6.4), if we introduce the external cou-plings εl into

Vε(t) = V (t) +∑l

∫x

εl(x, t)ol(x, t),

then [δrSβα

δεa(x)δεb(y) · · ·

]ε=0

= 〈β|T−iOa(x),−iOb(x), · · · |α〉

where

Oa(x, t) = Ω(t)oa(x, t)Ω−1(t) and Ω(t) = eiHte−iH0t and oa(x, t) = eiH0toa(x)e−iH0t.

Question 164. What does the above Gell-Mann - Low theorem mean?

Answer 164. This is a nonperturbative identity. The LHS is equivalent to the sum of all Feynmandiagrams connecting the external sources oa, ob, · · · . The RHS is a matrix element. What thissays is if we can know something about this matrix element, then we will also know somethingabout the nonperturbative sum of all the diagrams. This is the idea of form factors.

Question 165. How can we exploit this very powerful identity to constrain the possible matrixelements to a known form with unknown form factors?

Answer 165. Since the EM field is external, we take o = J µ(x) (so it must have been coupled toan infinitesimal εµ(x)). Our initial and final states are on-shell one-particle states (with the sameparticle). So far, our matrix element is

〈Ψp′σ′|J µ(x)|Ψpσ〉 = ei(p−p′)x〈Ψp′σ′|J µ(0)|Ψpσ〉.

By itself, translational invariance will not give us form factors. We must use two other conservationlaws: Noether theorem and Lorentz invariance. Noether theorem gives

(p′ − p)µ〈Ψp′σ′ |J µ(0)|Ψpσ〉 = 0, 〈Ψp′σ′|J 0(x = 0)|Ψpσ〉 =qδσσ′δp′,p

(2π)3,

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where q is the particle charge.

Using Lorentz invariance is harder because it involves the irrep of the particle, and hence dependson the particle’s spin. We will consider this in the following questions.

Question 166. Use Lorentz invariance to derive the form factors for a spin-zero particle.

Answer 166. This is easy because we know that |Ψp〉 is an on-shell external state,

|Ψp〉 =1

(2p0)1/2ap.

Therefore, we have

〈Ψp′σ′|J µ(x = 0)|Ψpσ〉 =qJ µ(p′, p)

(2π)3(2p′0)1/2(2p0)1/2.

J µ(p′, p) must be a 4-vector function which satisfies

(p′ − p)µJ µ(p′, p) = 0.

Therefore, J µ(p′, p) must be some linear combination of p′µ and pµ. It’s really easy to show thatsince (p′ − p)(p′ + p) = 0, we must have

J µ(p′, p) = (p′µ

+ pµ)F (k2), where k2 = (p− p′)2.

The important thing about k2 is that it depends only on pµp′µ, which is the only nontrivial constantyou can make out of p and p′. Due to the integration of scalar charge, we must have F (k2 = 0) = 1.In fact, you could write it only in terms of some F (p · p′), but this is not as pretty?

Nothing is known about F except F (k2 = 0) = 1. However, this function is not arbitrary ; it iscompletely determined by the theory. The point is that it is the only thing left to calculate.

Question 167. Use Lorentz invariance to derive the form factors for a spin-12

particle.

Answer 167. This is similar to the simpler spin-0 case. We find

〈Ψp′σ′ |J µ(x = 0)|Ψpσ〉 =iq

(2π)3up′σ′Γ

µ(p′, p)upσ.

Γµ is a 4-vector 4× 4 matrix function of p, p′, γ. What does this mean? It means for each choiceof µ, we can write Γµ as a linear combination of the 16 covariant matrices (see section 5.4 of thesenotes). Also, we have to make sure that

(p′ − p)µup′σ′Γµ(p′, p)upσ = 0.

The result is that

Γµ(p′, p) = γµF (k2)− i

2m(p′

µ+ pµ)G(k2) and F (0) +G(0) = 1.

The last equality again comes from the spatial charge condition. Often, people rewrite this as

Γµ(p′, p) = γµF1(k2) +i

2[γµ, γν ](p′ν − pν)F2(k2) and F1(0) = 1.

The relation between these two representations is

F (k2) = F1(k2) + 2mF2(k2) and G(k2) = −2mF2(k2).

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Question 168. Explain why the magnetic moment of a charged spin-12

particle is

µ =q

2mF (0),

which is µ = q2m

without radiative corrections, which is Dirac’s result from the relativistic ex-tension of Schrodinger equation. (Or is it? I thought there was a missing 1

2due to the Thomas

precession. Or is it that the Dirac form correctly includes the Thomas precession, and it was onlythe semiclassical one that didn’t?)

Answer 168. This follows from remembering that this matrix element is merely a sandwich(roughly speaking) where the pieces of bread are the one-particle states and the good stuff in themiddle is ∫

x

J µAµ.

Then, we take just the ~B-field part of Aµ and see what comes out. If Hint = −∫x

J(x) ·A(x), thematrix element is like

〈Ψp′σ′|Hint|Ψpσ〉 = −qF (0)

m(J)σσ′ ·Bδpp′ where J =

1

2σ for spin-

1

2.

This makes sense if you think of the interaction Hint as roughly proportional to a particle densityor something. So J is like a particle current from undergrad QM, and it couples to B like inclassical electrodynamics.

11.4 Anomalous magnetic moment

Now we calculate the shift in the magnetic moment of an electron. If you look at the previoussection of these notes, you can see how the magnetic moment is defined:

µ =q

2mF (0).

Here are the leading-order diagrams, to O(e2). As we will see, not all of them contribute:

We will ignore the corrections to the electron propagators; we have assumed the electrons areon-shell, and physically we want the on-shell residue to remain the same (just as for the photon),so the radiative correction is irrelevant. Anyway, we are trying to renormalize a vertex here andnot a propagator. We will not ignore the correction to the photon propagator, but it was donein the previous section and we can put it in later. Therefore, we will only calculate the last loop

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graph, which is

Γµloop(p′, p) =

∫k

(2π)4eγρ × −i(2π)4

−i(/p′ − /k) +m

(p′ − k)2 +m2 − iε× γµ

× −i(2π)4

−i(/p− /k) +m

(p− k)2 +m2 − iε× (2π)4eγρ ×

−i(2π)4

1

k2 − iε.

There are no external-particle propagators here, as expected.

Question 169. Summarize all the contributions to the vertex function Γµ.

Answer 169. There are a few contributions.

• The vanilla coupling γµ. For a meditation on interesting things that are also vanilla, seehere.

• Γµloop(p′, p), of course.

• The counterterm for the electron-photon vertex, (Z2 − 1)γµ. This cancels the divergence inΓµloop(p′, p). To fix the cancellation, we will take p′ = p so that we can use π(0) = 0 in thenext bullet point. In fact, this will be an infrared divergence rather than a UV divergence.You could guess this by looking at the powers of k in the integral for Γµloop(p′, p).

• The radiative correction to the photon propagator,

Γµphoton(p′ − p) =Πµν(q)× γνq2 − iε

, where q = p′ − p.

This has already been regularized since it includes (Z3 − 1), so it is finite.

Question 170. Describe how to get Schwinger’s famous α2π

correction to the magnetic momentof the electron.

Answer 170. Recall from the previous section that

(p′ − p)µup′σ′Γµ(p′, p)upσ = 0.

Also, the vertex is written

Γµ(p′, p) = γµF (k2)− i

2m(p′

µ+ pµ)G(k2) where k = p− p′.

In fact, it’s not physically important what Γµ(p′, p) is; the important quantity is up′σ′Γµ(p′, p)upσ.

After some brutal calculations, the result is

u′Γµloop(p′, p)u = −4π2e2

(2π)4

∫ 1

0

dx

∫ x

0

dy

∫ ∞0

κ3dκ

u′[γµ(−κ2 + 2m2(x2 − 4x+ 2) + 2q2(y(x− y) + 1− x))− 2im(p′µ + pµ)x(1− x)

]u

× (κ2 +m2x2 + q2y(x− y))−3.

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https://www.psychologytoday.com/us/blog/debunking-myths-the-mind/201009/overcoming-the-anxiety-vanilla-sex


This matches the form factor template, in the sense that there are only terms linear in pµ + p′µ,or constants.

If you add all the contributions (see previous question) together, you’ll find that F is IR-divergentand gets regulated by Z2. However, G is finite. Because F (0) + G(0) = 1 and we are interestedonly in the k = 0 case, the path of least resistance is merely to calculate G(0). It turns out that

G(0) = − e2

8π2= −0.001161.

This is Schwinger’s α2π

result, Phys Rev 73 416 (1948).

Question 171. Describe the idea of the electron’s “charge radius.” This is described in Weinbergpp. 491-493.

Answer 171. The “charge radius” is how thick the electron’s clothes are. It depends on how hotit is outside; the hotter it is, the less clothes the electron wears and the more “bare skin” you cansee.

I feel that defining a “charge radius” is really strange because everyone taught me that the cross-section of scattering off a Coulomb potential is infinite. So, is it that the electron cloud ends atthis “charge radius,” or that the electron cloud is infinitely large?

Anyway, on page 493, Weinberg defines this “radius” as having to do with the q → 0 limit of theform factor F1(q2). It turns out that

F1(q2)→ 1− a2

6q2 as q → 0

where

a2 =−e2

4π2m2

[ln(

µ2

m2) +

2

5+

3

4

]is the “charge radius.” It is positive because µ is an IR cutoff and therefore µ m. According to(11.3.27) and (11.3.28), the main contribution to this “radius” is from F (0) (of course, since wefound G(0) wasn’t divergent to one-loop). I should probably ask somebody how to think aboutthis.

11.5 Electron propagator

By now, we’re pretty good at renormalizing things and using counterterms. Let us do just onemore example. As you can imagine, the logic of renormalizing the electron propagator is the sameas that for renormalizing the photon propagator. It turns out that the loop diagram Σ∗loop hasboth a UV and an IR divergence. We will regulate the UV divergence below, and it turns out thatthe UV divergence by itself is enough to fix the counterterms. We will find out how to regulatethe IR divergence later.

Question 172. Explain how to regulate the UV divergence of the loop graph using the Pauli-Villars scheme.

Answer 172. Weinberg takes care of the divergence by using Pauli-Villars regularization on thephoton propagator,

1

k2 − iε→ lim

µ→∞

[1

k2 − iε+

1

k2 + µ2 − iε

].

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Question 173. Outline all the contributions to the electron self-energy. How do we fix thecounterterms?

Answer 173. There are the counterterms and the virtual-photon contribution, all at O(e2).Schematically, all the O(e2) contributions put together are

Σ∗(p) = −(Z2 − 1)(i/p+m) + Z2δm+ Σ∗loop(p).

Here, m is the renormalized mass. We again use the physical condition that the renormalizedpropagator ∆′(p) should have a pole at i/p = −m with residue unity. Location of pole + unitresidue is really two facts, so this gives us two equations:

δm = −Σ∗loop

∣∣i/p=−m

and Z2 − 1 = −i∂Σ∗loop

∂/p

∣∣∣∣i/p=−m

.

There are two counterterms here, but there was only one counterterm for the photon propagator.This is because the photon mass stays at zero regardless of radiative corrections. However, theelectron mass can change, so there is one additional df, and hence one additional counterterm.

11.6 Problems

Question 174. Calculate the contributions to the vacuum polarization function π(q2) and to Z3

of one-loop graphs involving a charged spinless particle of mass ms. What effect does this haveon the energy shift of the 2s state of hydrogen, if ms Zαme?

Answer 174. TODO

Question 175. Consider a neutral scalar field φ with mass mφ and self-interaction g3!φ3. To

one-loop, calculate the S-matrix element for scalar-scalar (i.e. 2→ 2) scattering.

Answer 175. TODO

12 General renormalization theory

Question 176. What was the conclusion of Dyson’s paper Phys Rev 75 486, 1736 (1949)?

Answer 176. Dyson showed that when we express all parameters of a theory in terms of “renor-malized” (i.e. physical) quantities, that the divergences cancel to all orders. Actually, this is trueonly for the renormalizable theories.

Non-renormalizable theories can have their divergences cancelled as well, but we must use an infi-nite number of counterterms. This is allowed because there are an infinite number of interactionsallowed by symmetries.

This is definitely the easiest chapter in the book. Phew!

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12.1 Degrees of divergence

We would like to estimate roughly how badly a loop graph is going to diverge. We can do this bywriting down the value of the diagram and seeing whether the integrals blow up at large or smallmomentum. There is a more systematic way to do this, however, based on graph theory.

Question 177. What is the “superficial degree of divergence?” What is so superficial about it?

Answer 177. The superficial degree of divergence, D, is the divergence with k of the integral∫dkkD−1.

It is superficial because this D is not always the true degree of divergence of the graph. This isespecially in the sense that for D = 0, you could have a log-divergent graph of a convergent graph.It depends. Or, you could be sneaky and take one loop momentum to infinity, but not the other.

Along these lines, it has been shown by Weinberg Phys Rev 118 838 (1959) that the requirementfor actual convergence of any graph is that the degree of divergence should be D < 0 for not onlythe complete graph, but also for any subintegration defined by holding any one or more linearcombinations of the loop momenta fixed.

Question 178. Explain why the superficial degree of divergence of a graph is

D = 4−∑f

Ef (sf + 1)−∑i

Ni∆i, where ∆i = 4− di −∑f

nif (sf + 1).

Answer 178. First, we introduce notation. Let there be interactions of types i, fields of type f ,and let there be nif of the field f involved in the interaction i. Let the number of derivatives inthe interaction of type i be di. The asymptotic behavior of the propagator for field f is taken tobe

∆f (k) ∼ k−2+2sf as k →∞.

For scalar fields, sf = 0. For the electron and photons, sf = 12

and sf = 0, respectively. Thedegree of divergence is

D =∑f

If (2sf + 2) +∑i

Nidi + 4[∑

f

If − (∑i

Ni) + 1].

The last term comes from the momentum δ-functions in the interactions. This turns into theanswer above.

Question 179. Describe the importance of ∆i for the interaction of type i.

Answer 179. ∆i describes the scaling of the interaction of type i and turns out to be the massdimension of the coupling constant (see next question). If we think about

D = 4−∑f

Ef (sf + 1)−∑i

Ni∆i,

it’s apparent that as we increase the number of vertices, D will decrease for ∆i > 0 and increasefor ∆i < 0. Therefore we introduce the following vocabulary:

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• An interaction with ∆i > 0 is called superrenormalizable.

• An interaction with ∆i ≥ 0 is called renormalizable.

• An interaction with ∆i < 0 is called nonrenormalizable.

A theory in which ∆i ≥ 0 for all i is termed renormalizable. For any theory, there are only afinite number of interactions i which are renormalizable. Usually we can always write all of themdown.

For intuition on this point, see http://math.ucr.edu/home/baez/renormalizability.html.

Question 180. Relate Weinberg’s notation with the sf and ∆i and whatnot to the mass dimen-sions of the fields and couplings.

Answer 180. Actually Weinberg’s method is more rigorous than the mass dimension method.But for certain things, either is fine. To get the mass dimensions, remember that S =

∫xL is

dimensionless, so [L ] = 4. Then we get the following:

• For a field f , [f ] = sf + 1.

• For an interaction g of generic form

Lint = gipdifni11 · · · fnimm ,

we find that[gi] = ∆i.

12.2 Cancellation of divergences

In this section, we will make rigorous the idea that a renormalizable theory, with ∆i ≥ 0 for alli, requires only a finite number of counterterms. We will finally explain why the countertermsare formally divergent, i.e. why (Z2 − 1) has to have an infinite part. The explanation is veryinteresting.

Question 181. Describe why a Feynman diagram with superficial degree of divergence D can bewritten as a polynomial of order D in external momenta with divergent coefficients, plus a finiteremainder.

Answer 181. Roughly speaking, the integral is like∫dkkD−1.

We can make this convergent by differentiating D+1 times with respect to any external momentum(i.e. not k). In fact, this is the idea behind “derivative regularization;” see Schwartz QFT.(Actually, there are technicalities here and this might not always work.) Then we integrate againand get a polynomial in the external momenta.

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http://math.ucr.edu/home/baez/renormalizability.html


Let us do an example to see it in action.

d2

dq2

∫ ∞0

kdk

k + q=

∫ ∞0

kdk

(k + q)2=

1

2q.

Now we integrate twice:∫q

1

2q=

1

2ln q + C and

∫q

(1

2ln q + C

)= a+ bq +

1

2q ln q.

Here, a and b are divergent constants. They are independent of both k and q!

Question 182. Explain why it is kosher to add an infinite counterterm to cancel the divergentconstants mentioned in the previous question.

Answer 182. A polynomial term in external momenta is exactly what could be produced byadding another interaction to the Lagrangian, with the right derivatives to get those momenta,etc. But how do we know this ad-hoc procedure is justified?

“All that we ever measure is the sum of the bare coupling constant and the corresponding coefficientfrom one of the divergent polynomials, so if we demand that the sum equals the (presumably finite)measured value, then the bare coupling must automatically contain an infinity that cancels theinfinity from the divergent integral over internal momenta.”

“One qualification: where the divergence occurs in a graph or subgraph with just two external lines,which appears as a radiative correction to a particle propagator, we must demand not that someeffective coupling constant equals its measured value, but rather that the complete propagator hasa pole at the same position and with the same residue as for free particles.”

The bare coupling contains the infinity. The renormalized coupling doesn’t contain the infinity;instead, the infinity goes to the counterterm.

About the second point that Weinberg makes: you can’t “measure” a propagator because, well,you need an external source interacting with a particle to measure anything, and then we wouldn’thave just a propagator anymore (for example, this is how we measured the magnetic moment ofthe electron in the previous chapter). By “free particles,” Weinberg does not mean those withbare mass. The free particles have the renormalized mass. Weinberg just calls anything with thepole structure

1

k2 −m2

a “free particle.” See section 10.3 of these notes.

The most rigorous way of eliminating superficial divergences is called the BPHZ prescription.It looks boring.

12.3 Is renormalizability necessary?

There is a lot of deep physics in this section. I will think about it carefully.

Question 183. Does the QED Lagrangian contain all Lorentz- and gauge-invariant terms?

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Answer 183. No! The termψ[γµ, γν ]ψF

µν

is both Lorentz- and gauge-invariant. However, it is not renormalizable. Experimentally, thiswould affect the magnetic moment of the electron, since it is roughly like ψψA. It seems thatnonrenormalizable terms are not allowed in QED. In fact, people historically arrived at the theoriesof weak and strong interactions by exclusing nonrenormalizable terms. But how do we know thisprocedure is valid? Weinberg’s viewpoint is that the requirement of renormalizability is not afundamental principle, but rather something which is a very good approximation at low energies.

Question 184. Explain why requiring theories to be renormalizable is a good approximation atlow energy scales.

Answer 184. This is a strange argument, but here it goes. We know from section 12.1 of thesenotes that

[gi] = ∆i.

Renormalizable theories are those for which ∆i > 0. We guess that the behavior of the couplingconstants in all theories in the standard model, etc. are basically like

gi ∼M∆i ,

where M is some very large common mass. Indeed, people think this is actually the case. Ilearned this from Girma:

Figure 1: Picture from https://www.nobelprize.org/prizes/physics/2004/

popular-information/

The more I think about it, the more outrageous this “common mass” claim becomes. Anyway,suppose ∆i < 0. We know gi ∼M∆i , and we can argue there has to be a k-dependence coming intoo, so the effect of this interaction looks like

(M

k)∆i .

k is on the order of the masses in the problem, usually, since√k2 ∼ E = γmc2. The electron

mass is 0.511 MeV. Above, it looks like M = 1015 GeV. That is a big difference and suppressesthe effect of the interaction by a lot.

For the renormalizable interactions, on the other hand, I think there is no M∆i involved in thesize of the coupling constant. See the second problem solved in this chapter.

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https://www.nobelprize.org/prizes/physics/2004/popular-information/

https://www.nobelprize.org/prizes/physics/2004/popular-information/


Question 185. Describe how nonrenormalizable interactions could be detected, even if their effectis strongly suppressed.

Answer 185. If the nonrenormalizable interaction breaks a symmetry of the low-energy theory,then maybe you could detect a symmetry breaking. Apparently people think that conservation ofbaryon and lepton number is violated by the small effects of nonrenormalizable interactions. Sincethese symmetries of the low-energy theory are not symmetries of the fundamental Lagrangian,whatever the fundamental theory is, they are termed accidental symmetries. Humorously,most of the experimentally discovered symmetries of particle physics are accidental.

Gravitation can be detected because gravitational fields add up rather than shielding each other.

From this viewpoint, the low-energy theory is an effective field theory. If we wanted to, we couldinclude some correction terms by expanding in powers of k/M .

Question 186. Explain why you can still calculate things in a nonrenormalizable field theory.

Answer 186. Suppose I have a nonrenormalizable field theory L . For simplicity, there is onlyone interaction with coupling constant g, so I can split it like

L = L0 + Lint.

If I want to calculate things to O(gn), I will have a finite number of diagrams. Therefore, I willhave a finite number of counterterms to add to the Lagrangian. This can always be done.

It’s important to note that the convergence of high-loop terms in perturbation theory dependson the size of the coupling g, and not on whether the interaction is renormalizable. Therefore,we can always calculate things to high precision even in nonrenormalizable theories. It is justthat in renormalizable theories, eventually we won’t need to add any extra counterterms. Innonrenormalizable theories, we must add some counterterms whenever we go to higher order.

12.4 Problems

Question 187. List all the renormalizable (or superrenormalizable) Lorentz-invariant terms inthe Lagrangian of a single scalar field for the following spacetime dimensionalities: 2, 3, 6.

Answer 187. This is a fun exercise. We will use ∆i = d− di −∑

f nif (sf + 1) a lot. (I replaced4→ d in the preceding equation.)

• d = 2: In d = 2, sφ + 1 = 0. Therefore any interaction with an arbitrary number of fieldsand 2 or fewer derivatives is renormalizable.

• d = 3: In d = 3, sφ + 1 = 1/2. The renormalizable interactions are: 3 fields and onederivative, three fields and no derivatives, four fields and one derivative, four fields and noderivative, five or six fields and no derivative.

• d = 6: In d = 6, sφ + 1 = 2. The only renormalizable interaction is the one with 3 fields andno derivatives.

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Question 188. Suppose that the quantum electrodynamics of electrons and photons is actuallyan effective field theory, derived by integrating out unknown particles of mass M m, where mis the electron mass. Assume gauge invariance and Lorentz invariance, but not invariance underC,P,T. What are the nonrenormalizable terms in the Lagrangian, to leading order in 1

M? Of

next to leading order?

Answer 188. The idea here is that all the terms of order O(M0) should be renormalizable. Theterms of order O(M−n), where n ≥ 1, may not. We would like to find those terms for n = 1.

For simplicity, let the particle with mass M be Ψ and the electron, of mass m, be ψ. For ideas,see Weinberg section 12.5.

I think the most general Lagrangian we can write is

L =

13 Infrared effects

Radiative corrections include both UV and IR effects. For UV effects, we saw that we needed toadd counterterms to cancel the high-energy divergences.

The same is not true for IR effects. Very generally, the divergences in IR-divergent graphs cancelwhen we add them up to get physical processes. Physically they come from photons of infinitelylarge wavelength, but not massive particles of infinitely large wavelength. This is easy to see bycomparing

igµνq2 + iε

with1

q2 +m2 + iε.

The first one diverges as q → 0; the second does not.

The main result of this chapter is the following:

The infrared divergence problem is solved with the observation that it is not really possible tomeasure the rate Γβα for a reaction α→ β involving definite numbers of photons of charged

particles, since photons of very low energy can always escape undetected. What can bemeasured is the rate Γβα(E,ET ); where no photon has energy greater than some small E and no

more than some small energy ET goes into any number of unobserved photons.

To get to this, we will have to slog through a lot of uninteresting calculations.

13.1 Soft photon amplitudes

A soft photon is one which has very low momentum, q → 0; specifically, it should momentumq Λ, where Λ is a typical energy scale of the scattering process. We would like to know theamplitude for emission of a soft photon, in the following sense:

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Figure 2: The solid lines can represent either massive particles or “hard” (i.e. q /Λ) photons.The soft photons can come off of either “reactant” α or “product” β particles.

The main result of this section is that the matrix element for emitting a single soft photon withmomentum q and polarization index µ in the process α→ β is given in the q → 0 limit by

Mµβα →Mβα

∑n

ηnenpµn

pn · q − iηnε,

where n labels a particle in α or β, ηn = ±1 for product/reactant particles, and en is the chargeof particle n. In the case of N soft photons, this is generalized to

Mµ1···µNβα (q1 · · · qN)→Mβα

N∏r=1

(∑n

ηnenpµrn

pn · qr − iηnε

).

Question 189. Describe why this is true for the case of a soft photon emitted from an electronline.

Answer 189. Compared to the usual Mβα, the emission of a photon with momentum q from anelectron with momentum p+ q (i.e. so the final electron momentum can match what is measured– momentum p), we need to multiply Mβα by

−(2π)4eγµ−i

(2π)4

−i(/p+ /q) +m

(p+ q)2 +m2 − iε.

Taking q → 0 and using

−i/p+m = 2p0∑σ′

upσ′upσ′ as well as upσγµupσ′ = −iδσσ′

pµ

p0

gives the correct result, epµ

p·q−iε . You also need to use the fact that the external particle is on-shell,

so p2 +m2 = 0.

It turns out that the general structure ηnenpµn

pn·q−iηnε holds for all different kinds of particles interactingwith the photons. You can check it, for example, with scalar QED. Generally, the numerator ofthe propagator approaches a sum of dyads of coefficient functions, which convert the new vertexmatrix into a factor proportional to pµ and a unit matrix in helicity indices.

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Question 190. Does the contribution ηnenpµn

pn·q−iηnε receive radiative corrections?

Answer 190. Nope. It is on-shell, and the residue of the mass-shell poles (or the matrix elementof the electric current between states of the same particle at equal momentum, i.e. the propagator)does not receive radiative corrections.

Question 191. Why can’t a hard photon emit a soft photon?

Answer 191. Duh. There is no photon-photon vertex in QED.

Question 192. Why don’t we count soft photons emitted from internal lines, i.e. the “circle” inthe diagram?

Answer 192. The internal lines are not on-shell, p2 +m2 6= 0. So they will not have the 1p·q -type

divergence for q → 0 that the external lines have.

Question 193. Show the following theorem:

For spin-1 massless particles, Lorentz invariance requires the conservation of whatever couplingconstant (like electric charge) governs the interaction of these particles at low energies.

Answer 193. I was very surprised by this theorem. Here is the idea. The soft photon is anoutgoing state, so to calculate numbers we need to take

Mµ(q)βα · eµ(q,±),

but eµ is not a four-vector. Under Lorentz transform, it can pick up a longitudinal component qµ.If you look back at section 8.1 of these notes,

U(Λ)aµ(x)U−1(Λ) = Λµνaν(Λx) + ∂µΩ(x,Λ).

Therefore for Lorentz invariance to hold, we must have

Mµ(q)βαqµ = 0 =⇒∑n

ηnen = 0.

Therefore, e must be conserved. More generally, any coupling constant must be conserved; we didnot have to use gauge invariance for this proof. Although I think it is equivalent to using gaugeinvariance.

13.2 Virtual soft photons

Using the result of the previous section, we calculate to all orders the effect of radiative correctionsof virtual soft photons among the charged particle lines in the process α→ β. Let the soft photonsbe defined as those for which q Λ. For now, we will also constrain q with the IR cutoff q > λ.Note the difference between Λ and λ: the first is a definition of which photons are soft; the secondis a mathematical cutoff which will be taken to 0.

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Figure 3: Orange lines are virtual soft photons.

Question 194. Let the matrix element for the process with no soft radiative corrections beMβα.What is the matrix element for the process with N soft virtual photons?

Answer 194. We use the results of the previous section. It’s not terribly easy to write down, butafter you see it it becomes obvious. The matrix element gets multiplied by

1

N !2N

[1

(2π)4

∑nm

enemηnηmJnm

]N

where Jnm = −i(pn · pm)

∫λ≤|q|≤Λ

d4q

(q2 − iε)(pn · q − iηnε)(−pm · q − iηmε).

Question 195. Describe the computation of all virtual effects.

Answer 195. We take the answer to the previous question and sum over all N , giving an expo-nential:

Mλβα =MΛ

βα exp

[1

2(2π)4

∑nm

enemηnηmJnm

].

Here Mλβα keeps all photons with momenta greater than λ and MΛ

βα keeps all photons withmomenta greater than Λ. Then you have to compute Jnm. I do not find this very pleasing. Theend result is

Γλβα =

(λ

Λ

)A(α→β)

ΓΛβα.

Here Γλβα keeps all photons with momenta greater than λ and ΓΛβα keeps all photons with momenta

greater than Λ.

Apparently, A(α→ β) (we always have A > 0) is this complicated

A(α→ β) =−1

8π2

∑nm

enemηnηmβnm

ln(1 + βnm

1− βnm), where β2

nm = 1− m2nm

2m

(pn · pm)2.

It seems that the “effect of the infrared divergences introduced by soft virtual photons whensummed to all orders is to make the rate for any charged particle process α → β vanish in theλ→ 0 limit.” That is weird. Put more and more infinities together and you end up with nothing.

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I think the ultimate idea was in the computation of

Jnm = −i(pn · pm)

∫λ≤|q|≤Λ

d4q

(q2 − iε)(pn · q − iηnε)(−pm · q − iηmε).

Obviously if you take λ = 0 initially, everything blows up. However, if you keep λ > 0 initially, thensum everything, and then take λ→ 0, then things are okay. This is like the exponential regulatorthat Girma taught me, useful for finding the Fourier transform of the Coulomb potential, forexample.

13.3 Cancellation of divergences

Our claim is that if we include real soft photons (i.e. soft external photons) along with the softvirtual photons, then we will get a cancellation of divergences. This means the scattering processwill be independent of the IR cutoff, λ.

Reference the beginning of this chapter’s notes for the notation. I am not interested in doing manyintegrals, but the final result is

Λλβα(E,ET )→ F (

E

ET;A(α→ β))

(E

Λ

)A(α→β)

ΓΛβα.

This is independent of λ.

How can we think about this? Essentially, we start with the process which only includes virtualphotons with IR cutoff λ. This process vanishes as λ → 0. However, unitarity says that if weuse λ as the IR cutoff for the virtual photons, we must also use it as the IR cutoff for the real(external) processes. This gives a divergence as λ → 0, and of course it multiplies the processwhich only includes virtual rates.

The process which includes virtual rates goes to 0. The multiplication due to real soft emissiongoes to infinity. It turns out that these effects exactly cancel and the overall rate is independentof the IR cutoff. This is the meaning of “cancellation of divergences.”

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Notes on Weinberg’s Quantum Theory of Fields …...Notes on Weinberg’s Quantum Theory of Fields Volume I: Fundamentals Jimmy Qin Summer 2019 A theorist today is hardly considered