Imperial College London MSci EXAMINATION May 2009 …...Quantum Field Theory May 2009 ANSWERS May 4, 2009 Imperial College London MSci EXAMINATION May 2009 This paper is also taken

Quantum Field Theory May 2009 ANSWERS May 4, 2009

Imperial College London

MSci EXAMINATION May 2009

This paper is also taken for the relevant Examination for the Associateship

QUANTUM FIELD THEORY

For 4th-Year Physics Students

Thursday, 21st May 2009: 10:00 to 12:00

Answer THREE questions.

Marks shown on this paper are indicative of those the Examiners anticipate assigning.

General Instructions

Complete the front cover of each of the THREE answer books provided.

If an electronic calculator is used, write its serial number at the top of the front cover of

each answer book.

USE ONE ANSWER BOOK FOR EACH QUESTION.

Enter the number of each question attempted in the box on the front cover of its

corresponding answer book.

Hand in THREE answer books even if they have not all been used.

You are reminded that Examiners attach great importance to legibility, accuracy

and clarity of expression.

c© Imperial College London 2009

xxxx/y/zzzANSWERS 1 Turn over for questions


1. Consider the classical real scalar field φ(x) with Lagrangian density

L = 12∂µφ∂

µφ− V (φ).

You are given that the classical Poisson bracket satisfies{φ(t, xxx), π(t, yyy)

}PB

= −{π(t, yyy), φ(t, xxx)

}PB

= δ(3)(xxx − yyy),

while{φ(t, xxx), φ(t, yyy)

}PB

= 0 and{AB,C

}PB

= A{B,C

}PB

+ B{A,C

}PB

.

(i) What is the definition of the momentum density π(x) conjugate to φ(x)? What

is it equal to in this case?

Rewrite L in terms of φ and ∇∇∇φ and hence show that the Hamiltonian H is

given by

H =

∫d3x

(12π2 + 1

2∇∇∇φ · ∇∇∇φ+ V (φ)

).

What is V (φ) for a free scalar field of mass m? What is the minimum value of

H in this case? [6 marks]

ANSWER: By definition, π = ∂L/∂φ = φ. Here L = 12 φ2 − 12∇∇∇φ · ∇∇∇φ − V (φ).

Hence

H =

∫d3x

(πφ− L

)=

∫d3x

(φ2 − 12 φ

2 + 12∇∇∇φ · ∇∇∇φ+ V (φ)

)=

∫d3x

(12π2 + 1

2∇∇∇φ · ∇∇∇φ+ V (φ)),

as required. For a free scalar field of mass m, V (φ) = 12m2φ2. Since H is then a sum

of squares, H ≥ 0, with the minimum realised by φ = 0.

(ii) Define

T µν = ∂µφ∂νφ− ηµν(12∂λφ∂

λφ− V (φ)),

Mµνρ = T µνxρ − T µρxν.

Show that ∂µTµν = 0 if φ satisfies the equation of motion ∂µ∂

µφ+ V ′(φ) = 0

(where V ′(φ) = dV/dφ) and hence that ∂µMµνρ = 0. [5 marks]

ANSWER: By the chain rule, ∂µV (φ) = (∂µφ)V ′(φ) where V ′(φ) = dV (φ)/dφ,

hence

∂µTµν = ∂µ(∂µφ∂νφ)− 12η

µν∂µ(∂λφ∂λφ) + ηµν∂µV (φ)

= (∂µ∂µφ)∂νφ+ (∂µφ)(∂µ∂

νφ)− (∂λφ)(∂ν∂λφ) + (∂νφ)V ′(φ)

= (∂νφ)(∂µ∂

µφ+ V ′(φ))

= 0.

xxxx/y/zzz ANSWERS 2


We have ∂λxσ = δλ

σ hence,

∂µMµνρ = ∂µ (Tµνxρ − Tµρxν)

= (∂µTµν)xρ + Tµν(∂µx

ρ)− (∂µTµρ)xν − Tµρ(∂µx

ν)

= Tµνδµρ − Tµρδµν

= T ρν − T νρ

= 0,

since by definition Tµν = T νµ.

(iii) Consider the integrals

Qµ =

∫d3xT 0µ(t, xxx),

Qµν =

∫d3xM0µν(t, xxx).

How do they depend on t? What does Qµ represent physically? What about

Qi j , where i , j = 1, 2, 3?

Write down the components of Qµ and show explicitly that one is related to

H. [4 marks]

ANSWER: They are Noether charges, independent of t since the corresponding

currents Tµν and Mµνρ are conserved. Physically Qµ correspond to the total conserved

four-momentum Pµ. The three charges J i = 12εi jkQjk are the total angular momentum

of the field. Explicitly we have

T 00 = ∂0φ∂0φ− η00(12∂λφ∂

λφ− V (φ))

= φ2 − L,T 0i = ∂0φ∂ iφ− η0i

(12∂λφ∂

λφ− V (φ))

= −π∇∇∇iφ.

Hence Qµ = (P 0, PPP ) where

PPP = −∫

d3xπ∇∇∇φ,

and P 0 = H.

(iv) Show that{Qµ, φ(x)

}PB

= −∂µφ(x) and comment on the result.

Comment briefly on what you expect for the Poisson brackets between different

components of Qµ and Qµν. [5 marks]

ANSWER: The only non-zero contributions to{T 0µ(t, yyy), φ(t, xxx)

}PB

are{T 00(t, yyy), φ(t, xxx)

}PB

= 12

{π2(t, yyy), φ(t, xxx)

}PB

= π(t, yyy){π(t, yyy), φ(t, xxx)

}PB

= −π(t, yyy)δ(3)(yyy − xxx) = −∂0φ(t, yyy)δ(3)(yyy − xxx),{T 0i(t, yyy), φ(t, xxx)

}PB

= −{π(t, yyy)∇∇∇iφ(t, yyy), φ(t, xxx)

}PB

= −∇iφ(t, yyy){π(t, yyy), φ(t, xxx)

}PB

= ∇iφ(t, yyy)δ(3)(yyy − xxx) = −∂ iφ(t, yyy)δ(3)(yyy − xxx).

xxxx/y/zzz ANSWERS 3[This question continues overleaf . . . ]


Since Qµ are independent of time we can choose to evaluate them at t = x0 so that{Qµ, φ(x)

}PB

=

∫d3y{T 0µ(t, yyy), φ(t, xxx)

}PB

= −∫

d3y∂µφ(t, yyy)δ(3)(yyy − xxx)

= −∂µφ(x).

This reflects the fact that Qµ are the generators of the translation symmetry

group. Similarly Qµν are the generators of the Lorentz symmetry group. Collectively

{Qµ, Qµν} form a representation of the Poincare group under the Poisson bracket.

[Total 20 marks]



2. Consider a free real scalar field φ(x) with conjugate momentum density π(x) = φ(x).

Define the operator

appp =

∫d3x

e−ippp·xxx√2Eppp

[Epppφ(0, xxx) + iπ(0, xxx)] ,

where Eppp =√|ppp|2 +m2.

(i) The equal-time commutation relations (ETCRs) state that[φ(t, xxx), π(t, yyy)

]= iδ(3)(xxx − yyy).

Are the fields φ(x) and π(x) in the Schrodinger or Heisenberg picture? Why

is this picture more natural in a relativistic theory?

What are the ETCRs for[φ(t, xxx), φ(t, yyy)

]and

[π(t, xxx), π(t, yyy)

]? [4 marks]

ANSWER: The fields operators are in the Heisenberg picture since they ex-

plicitly depend on time. This means they are function of all four coordinates

xµ. This is the natural picture in a relativistic theory, since one can then write

relativistically invariant expressions. The ETCRs for the remaining commutators are[φ(t, xxx), φ(t, yyy)

]=[π(t, xxx), π(t, yyy)

]= 0.

(ii) Derive an expression for a†ppp in terms of φ(0, xxx) and π(0, xxx). Using the ETCRs

show that [appp, a

†qqq

]= (2π)3δ(3)(ppp − qqq),[

appp, aqqq]

= 0,[a†ppp, a

†qqq

]= 0.

[6 marks]

ANSWER: By definition, since φ and hence π are Hermitian,

a†ppp =

∫d3x

(e−ippp·xxx)∗√2Eppp

[Epppφ(0, xxx) + iπ(0, xxx)]†

=

∫d3x

eippp·xxx√2Eppp

[Epppφ(0, xxx)− iπ(0, xxx)] .



Hence[appp, a

†qqq

]=

∫d3x d3y


eiqqq·yyy√2Eqqq

[Epppφ(0, xxx) + iπ(0, xxx), Eqqqφ(0, yyy)− iπ(0, yyy)

]=

∫d3x d3y

e−ippp·xxx+iqqq·yyy√4EpppEqqq

(iEqqq[π(0, xxx), φ(0, yyy)

]− iEppp

[φ(0, xxx), π(0, yyy)

])=

∫d3x d3y

e−ippp·xxx+iqqq·yyy√4EpppEqqq

(−i)(Eppp + Eqqq)iδ(3)(xxx − yyy)

=

∫d3x

Eppp + Eqqq√4EpppEqqq

e−i(ppp−qqq)·xxx

= (2π)3δ(3)(ppp − qqq).

Similarly

[appp, aqqq

]=

∫d3x d3y


e−iqqq·yyy√2Eqqq

[Epppφ(t, xxx) + iπ(t, xxx), Eqqqφ(t, yyy) + iπ(t, yyy)

]=

∫d3x d3y

e−ippp·xxx−iqqq·yyy√4EpppEqqq

(iEqqq[π(t, xxx), φ(t, yyy)

]+ iEppp

[φ(t, xxx), π(t, yyy)

])=

∫d3x d3y

e−ippp·xxx−iqqq·yyy√4EpppEqqq

i(Eppp − Eqqq)iδ(3)(xxx − yyy)

=

∫d3x

Eqqq − Eppp√4EpppEqqq

e−i(ppp+qqq)·xxx

= 0,

since E−ppp = Eppp. We also have[a†ppp, a†qqq

]= −

[appp, aqqq

]†= 0.

(iii) Given

φ(x) =

∫d3p

(2π)31√2Eppp

(apppe−ip·x + a†pppeip·x

),

where p0 = Eppp, use the results from part ii to show that at unequal times

[φ(x), φ(y)

]=

∫d3p

(2π)31

2Eppp

(e−ip·(x−y) − eip·(x−y)

).

What properties must this commutator have if the field theory is to respect

microcausality? [5 marks]



ANSWER: Substituting we have[φ(x), φ(y)

]=

∫d3q

(2π)3d3p

(2π)31√

2Eqqq√

2Eppp

[(aqqqe−iq·x + a

†qqqeiq·x

),(apppe−ip·y + a

†pppeip·y

)]=

∫d3q

(2π)3d3p

(2π)31√

4EqqqEppp

([aqqq, a

†ppp

]e−iq·xeip·y +

[a†qqq, appp

]eiq·xe−ip·y

)=

∫d3q

(2π)3d3p

(2π)31√

4EqqqEppp

((2π)3δ(3)(qqq − ppp)e−iq·x+ip·y

− (2π)3δ(3)(qqq − ppp)e−iq·x+ip·y)

=

∫d3q

(2π)31

2Eqqq

(e−iq·(x−y) − eiq·(x−y)

).

Microcausality implies that operators commute if they are space-like separated. In

particular this requires[φ(x), φ(y)

]= 0 if (x − y)2 < 0.

(iv) You are given that δ(x2 − a2) = (1/2a) [δ(x − a)− δ(x + a)]. Show that the

expression for[φ(x), φ(y)

]given in part iii can be rewritten as

[φ(x), φ(y)

]=

∫d4p

(2π)3δ(p2 −m2)e−ip·(x−y),

where d4p = d3p dp0. Comment on the Lorentz transformation properties of

this expression, and give a brief argument that the ETCRs take the same form

in all inertial frames.

What is the significance of this result? [5 marks]

ANSWER: We have p2 −m2 = (p0)2 − E2ppp. Using the given relation we then have∫d4p

(2π)3δ(p2 −m2)e−ip·(x−y)

=

∫d3p

(2π)31

2Eppp

(e−ip·(x−y)

∣∣∣p0=Eppp

− e−ip·(x−y)∣∣∣p0=−Eppp

)=

∫d3p

(2π)31

2Eppp

(e−iEppp·(x

0−y0)+ippp·(xxx−yyy) − eiEppp·(x0−y0)+ippp·(xxx−yyy)

)=

∫d3p

(2π)31

2Eppp

(e−ip·(x−y) − eip·(x−y)

),

where we have changed variables ppp → −ppp in the second term. Since the new expression

is manifestly Lorentz covariant, this shows that[φ(x), φ(y)

]is Lorentz covariant.



Thus, evaluating in an arbitrary inertial frame,[φ(t, xxx), φ(t, yyy)

]=[φ(x), φ(y)

]x0=y0=0

=

∫d3p

(2π)31

2Eppp

(eippp·(xxx−yyy) − e−ippp·(xxx−yyy)

)= 0 (odd integrand),[

φ(t, xxx), π(t, yyy)]

= ∂y0[φ(x), φ(y)

]x0=y0=0

=

∫d3p

(2π)3i

2

(eippp·(xxx−yyy) + e−ippp·(xxx−yyy)

)= iδ(3)(xxx − yyy),[

π(t, xxx), π(t, yyy)]

= ∂x0∂y0[φ(x), φ(y)

]x0=y0=0

=

∫d3p

(2π)3Eppp2

(eippp·(xxx−yyy) − e−ippp·(xxx−yyy)

)= 0 (odd integrand).

Thus although we chose a particular inertial frame to define the ETCRs, the resulting

quantum theory is independent of the original choice of frame, as required for a

relativistic theory.

[Total 20 marks]



3. This question is about the free classical Dirac field ψ(x). The Lagrangian is given

by

L = iψγµ∂µψ −mψψ,where the gamma matrices γµ satisfy

{γµ, γν

}= 2ηµν1 and ψ = ψ†γ0 .

(i) The field ψ(x) is a spinor. How many components does it have? After quan-

tization what is the spin of the corresponding single-particle states?

Treating ψ(x) and ψ(x) as independent fields, show that the Euler–Lagrange

equation for ψ(x) is the Dirac equation

iγµ∂µψ −mψ = 0.

[4 marks]

ANSWER: A spinor field has four components. After quantization, the corre-

sponding single particle states are spin one-half. The Euler–Lagrange equations read

∂µ(∂L/∂(∂µψ)) = ∂L/∂ψ. Since ∂L/∂(∂µψ) = 0 we have

∂L∂ψ

= iγµ∂µψ −mψ = 0,

as required.

(ii) Using (γµ)† = γ0γµγ0 show that if ψ(x) satisfies the Dirac equation then

i∂µψ(x)γµ +mψ(x) = 0.

Hence show that the current jµ = ψγµψ is conserved, that is ∂µjµ = 0, if ψ

satisfies the Dirac equation. [4 marks]

ANSWER: Taking the conjugate of the Dirac equation

0 = − (iγµ∂µψ −mψ)† γ0

= i∂µψ†(γµ)†γ0 +mψ†γ0

= i∂µψ†γ0γµγ0γ0 +mψ†γ0

= i∂µψγµ +mψ.

Hence

∂µjµ = ∂µ

(ψγµψ

)= (∂µψ)γµψ + ψγµ(∂µψ)

= imψψ − imψψ

= 0,

as required.



(iii) Show that the complex conjugate of L is given by

L∗ = L − i∂µ(ψγµψ

).

Consider the real Lagrangian

L′ = 12

i[ψγµ(∂µψ)− (∂µψ)γµψ

]−mψψ.

Treating ψ(x) and ψ(x) as independent fields, show that the corresponding

Euler–Lagrange equation for ψ(x) is again the Dirac equation. [5 marks]

ANSWER: We have (γ0)† = γ0γ0γ0 = γ0, so that

L∗ =(

iψγµ∂µψ −mψψ)∗

= −i(ψ†γ0γµ∂µψ

)† − (mψ†γ0ψ)†= −i(∂µψ

†)(γµ)†γ0ψ −mψ†γ0ψ= −i(∂µψ

†)γ0γµγ0γ0ψ −mψ†γ0ψ= −i(∂µψ)γµψ −mψψ= L − i

[(∂µψ)γµψ + ψγµ(∂µψ)

]= L − i∂µ(ψγµψ).

For L′ we have

∂L′

∂(∂µψ)= −12 iγµψ,

∂L′

∂ψ= 12 iγµ∂µψ −mψ,

so that the Euler–Lagrange equation reads

∂µ(−12 iγµψ

)− 12 iγµ∂µψ +mψ = −iγµ∂µψ +mψ = 0,

as required.

(iv) The matrix γ5 = iγ0γ1γ2γ3 has the property that{γ5, γµ

}= 0. Show that

the “axial vector” current jµ5 = ψγµγ5ψ satisfies

∂µjµ5 = 2imψγ5ψ.

What does this result imply about the symmetries of the Dirac equation when

m = 0?

Identify the corresponding infinitesimal transformation of ψ and demonstrate

directly whether or not L is invariant under this transformation. [7 marks]

ANSWER: We have

∂µjµ5 = ∂µ

(ψγµγ5ψ

)= (∂µψ)γµγ5ψ + ψγµγ5(∂µψ)

= (∂µψ)γµψ − ψγ5γµ(∂µψ)

= imψγ5ψ − imψγ5ψ

= 2imψγ5ψ.



Thus if m = 0 the axial current is conserved and we have an additional symmetry of

the Dirac equation (by Noether’s theorem). The corresponding infinitesimal transfor-

mation is

δψ = iαγ5ψ,

δψ = (iαγ5ψ)† γ0 = −iαψ†γ†5γ0 = iαψ†γ0γ5γ

0γ0 = iαψγ5,

since

γ†5 = (iγ0γ1γ2γ3)† = −iγ0(γ3γ2γ1γ0)γ0 = −iγ0(γ0γ1γ2γ3)γ0 = −γ0γ5γ0.

Hence

δL = i[δψγµ∂ψ + ψγµ∂µ(δψ)

]−m

[δψψ + ψδψ

]= (iα)i

[ψγ5γ

µ∂µψ + ψγµγ5∂µψ]− (iα)m

[ψγ5ψ + ψγ5ψ

]= −2iαmψγ5ψ.

Thus we have δL = 0 only if m = 0 as expected.

[Total 20 marks]

xxxx/y/zzz ANSWERS 11 Please turn over


4. Consider a free Dirac field ψ(x) given by

ψ(x) =

∫d3p

(2π)31√2Eppp

2∑s=1

[aspppu

s(p)e−ip·x + bs†ppp vs(p)eip·x

],

where p0 = Eppp =√|ppp|2 +m2. The only non-vanishing anti-commutation relations

are {arppp, a

s†qqq

}={brppp, b

s†qqq

}= (2π)3δ3(ppp − qqq)δrs .

The Hamiltonian H and the operator Q are given by

H =

∫d3p

(2π)3

∑s=1,2

Eppp(as†ppp a

sppp + bs†ppp b

sppp

),

Q =

∫d3p

(2π)3

∑s=1,2

(as†ppp a

sppp − bs†ppp bsppp

).

(i) Define the vacuum state |0〉 and the single particle states |ppp, s,−〉 and |ppp, s,+〉.What do the labels ppp, s and ± denote?

Define a two-particle state, and show that the particles are fermions.

[6 marks]

ANSWER: By definition

asppp|0〉 = bsppp|0〉 = 0, |ppp, s,−〉 =√

2Epppas†ppp |0〉, |ppp, s,+〉 =

√2Epppb

s†ppp |0〉.

The label ppp denotes the three-momentum of the state, s = 1, 2 denotes the two spin

states, and ± denote the charge of the state, + for a positron and − for an electron.

The two electron state is

|ppp, r,−; qqq, s,−〉 =√

2Eppp√

2Eqqqar†ppp a

s†qqq |0〉 = −

√2Eppp

√2Eqqqa

s†qqq a

r†ppp |0〉

= −|qqq, s,−; ppp, r,−〉,

demonstrating that the particles obey fermi statistics.

(ii) Show that

ar†qqq arqqq|ppp, s,−〉 = (2π)3δ(3)(ppp − qqq)δrs |qqq, r,−〉,

br†qqq brqqq|ppp, s,+〉 = (2π)3δ(3)(ppp − qqq)δrs |qqq, r,+〉.

Hence show that |0〉, |ppp, s,−〉 and |ppp, s,+〉 are eigenstates of H and Q and

give the eigenvalues. [6 marks]



ANSWER: Using the properties of the ground state we have

ar†qqq a

rqqq|ppp, s,−〉 =

√2Epppa

r†qqq a

rqqqas†qqq |0〉

=√

2Epppar†qqq

{arqqq, a

s†qqq

}|0〉

= (2π)3δ(3)(ppp − qqq)δrs√

2Epppar†qqq |0〉,

= (2π)3δ(3)(ppp − qqq)δrs |qqq, r,−〉.

Similarly

br†qqq b

rqqq|ppp, s,+〉 =

√2Epppb

r†qqq b

rqqqbs†qqq |0〉

=√

2Epppbr†qqq

{brqqq, b

s†qqq

}|0〉

= (2π)3δ(3)(ppp − qqq)δrs√

2Epppbr†qqq |0〉,

= (2π)3δ(3)(ppp − qqq)δrs |qqq, r,+〉.

Since arqqq|0〉 = brqqq|0〉 = 0 we have

H|0〉 = Q|0〉 = 0,

while using the above results we have

H|ppp, s,±〉 =

∫d3q

(2π)3

∑r=1,2

Eqqq(2π)3δ(3)(ppp − qqq)δrs |qqq, r,±〉 = Eppp|ppp, s,±〉,

Q|ppp, s,±〉 = ±∫

d3q

(2π)3

∑r=1,2

(2π)3δ(3)(ppp − qqq)δrs |qqq, r,±〉 = ±|ppp, s,±〉,

where we have used the factor that the commutators between asppp (or as†ppp ) and bsppp (or

bs†ppp ) vanish.

(iii) Using the results of part ii, write down an expression, similar to those for H

and Q, for an operator Sz with eigenvalues

Sz |0〉 = 0, Sz |ppp, 1,±〉 = ±12|ppp, 1,±〉, Sz |ppp, 2,±〉 = ∓1

2|ppp, 2,±〉.

What does this operator measure? [3 marks]

ANSWER: Consider the operator

Sz =1

2

∫d3p

(2π)3

(a1†ppp a1ppp − a

2†ppp a2ppp − b

1†ppp b1ppp + b

2†ppp b2ppp

).

As before Sz |0〉 = 0 and

Sz |ppp, 1,±〉 = ±1

2

∫d3q

(2π)3

2∑r=1

Eqqq(2π)3δ(3)(ppp − qqq)δr1 |qqq, r,±〉 = ±12 |ppp, 1,±〉,

Sz |ppp, 2,±〉 = ∓1

2

∫d3q

(2π)3

2∑r=1

Eqqq(2π)3δ(3)(ppp − qqq)δr2 |qqq, r,±〉 = ∓12 |ppp, 2,±〉,



as required.

(iv) One could also try to quantize the Dirac field using commutation relations.

Specifically the only non-vanishing commutators are then[arppp, a

s†qqq

]= −

[brppp, b

s†qqq

]= (2π)3δ3(ppp − qqq)δrs .

and the sign of the bs†ppp bsppp term in the Hamiltonian H changes.

If we define states in the Hilbert space in the conventional way, what are the

two problems with this quantization prescription?

How can one problem be alleviated by changing the definition of the Hilbert

space? [5 marks]

ANSWER: Consider the normalization of the putative electron states

〈ppp, r,−|qqq, s,−〉 =√

2Eppp√

2Eqqq〈0|arpppas†qqq |0〉 =

√2Eppp

√2Eqqq〈0|

[arppp, a

s†qqq

]|0〉

= −(2π)3(2Eppp)δ(3)(ppp − qqq)〈0|0〉= −(2π)3(2Eppp)δ(3)(ppp − qqq)

Thus for ppp = qqq we have a negative norm state. This corresponds to negative proba-

bilities and hence violates the standard interpretation of quantum mechanics. We also

have

H =

∫d3p

(2π)3

2∑s=1

Eppp

(as†ppp a

sppp − b

s†ppp b

sppp

).

implying that the positron states have negative energy. This is unphysical, and gives

a theory unstable to the production of the more and more positrons. One can correct

the norm problem be defining the positron states and vacuum using

bsppp = bs†ppp ,

so that[brppp, b

s†qqq

]= (2π)3δ3(ppp − qqq)δrs . However the negative energy problem still

remains since the form of H is unchanged (after normal ordering).

[Total 20 marks]



5. In both the interaction picture and the Heisenberg picture operators evolve with

time. One has

∂tOI = i[H0,I,OI

], ∂tOH = i

[HH,OH

],

where in the interaction picture, OI is the operator and H0,I is the free Hamiltonian,

while in the Heisenberg picture OH is the operator and HH is the total Hamiltonian.

(i) What are the corresponding expressions for the evolution of states |ψ(t)〉H and

|ψ(t)〉I in the Heisenberg and interaction pictures?

Comment briefly on why the interaction picture is useful in perturbation theory.

Write down the full Lagrangian density for “phi-fourth” theory and identify L0and Lint, the free and interaction Lagrangian densities, that contribute to the

free and interaction Hamiltonians respectively. [5 marks]

ANSWER: We have

∂t |ψ(t)〉I = −iHint,I |ψ(t)〉I , ∂t |ψ(t)〉H = 0.

In the perturbation expansion, one expands as a power series in Hint. In the interaction

picture, the fields still evolve with H0,I so take the same form as in the free theory.

All the dependence on Hint,I is then incorporated in the evolution of the states. In

“phi-fourth” theory we have L+ L0 + Lint where

L0 = 12∂µφ∂

µφ− 12m2φ2, Lint = − 14!λφ

4.

(ii) Show that in the two pictures ∂tHH = 0 and ∂tH0,I = 0. Hence show that

OH(t) = eiHH(t−t0)OH(t0) e−iHH(t−t0),

OI(t) = eiH0,I(t−t0)OI(t0) e−iH0,I(t−t0),

are solutions of the evolution equations for OH(t) and OI(t). [4 marks]


∂tHH = i[HH, HH

]= 0, ∂tH0,I = i

[H0,I , H0,I

]= 0.

We then have

∂tOH(t) = ∂t

(eiHH(t−t0)OH(t0) e−iHH(t−t0)

)= ∂t

(eiHH(t−t0)

)OH(t0) e−iHH(t−t0) + eiHH(t−t0)OH(t0) ∂t

(e−iHH(t−t0)

)= iHHeiHH(t−t0)OH(t0) e−iHH(t−t0) + eiHH(t−t0)OH(t0) (−iHH)e−iHH(t−t0)

= iHHOH(t)− iOH(t)HH

= i[HH,OH(t)

].



Similarly

∂tOI(t) = ∂t

(eiH0,I(t−t0)OI(t0) e−iH0,I(t−t0)

)= ∂t

(eiH0,I(t−t0)

)OI(t0) e−iH0,I(t−t0) + eiH0,I(t−t0)OI(t0) ∂t

(e−iH0,I(t−t0)

)= iH0,Ie

iH0,I(t−t0)OI(t0) e−iH0,I(t−t0) + eiH0,I(t−t0)OI(t0) (−iH0,I)e−iH0,I(t−t0)

= iH0,IOI(t)− iOI(t)H0,I= i[H0,I ,OI(t)

],

as required.

(iii) Let operators in the two pictures be related by

OI(t) = U(t, t0)OH(t)U(t, t0)−1,

where U(t0, t0) = 1.

Using the results of part ii write an expression for U(t, t0). Show that it satisfies

∂tU = −iHint,IU,

where Hint,I is the interaction Hamiltonian in the interaction picture.

[4 marks]

ANSWER: We have OH(t0) = OI(t0) and hence

U(t, t0) = eiH0,I(t−t0)e−iHH(t−t0).

Thus

∂tU = ∂t

(eiH0,I(t−t0)e−iHH(t−t0)

)= (iH0,I)eiH0,I(t−t0)e−iHH(t−t0) + eiH0,I(t−t0)(−iHH)e−iHH(t−t0)

= iH0,I

(eiH0,I(t−t0)e−iHH(t−t0)

)− i(

eiH0,I(t−t0)e−iHH(t−t0))HH

= iH0,IU − iUHH

= iH0,IU − i(UHHU−1)U

= i (H0,I −HI)U= −iHint,IU.

as required.

(iv) Define the time-ordered product

THint,I(t1)Hint,I(t2).



As an expansion in Hint,I, show explicitly that

U(t, t0) = 1− i

∫ t

t0

dt1Hint,I(t1)

+ 12

(−i)2∫ t

t0

∫ t

t0

dt1dt2THint,I(t1)Hint,I(t2) + . . .

is a solution for U(t, t0) to second order in Hint,I. [7 marks]


THint,I(t1)Hint,I(t2) = θ(t1 − t2)Hint,I(t1)Hint,I(t2) + θ(t2 − t1)Hint,I(t2)Hint,I(t1),

where θ(t) is the Heaviside step function. Now solve the equation for U = U(0) +

U(1) + . . . perturbatively in Hint,I . At each order we have

∂tU(n+1) = iHint,IU(n).

We can satisfy U(t0, t0) = 1 by taking U(0)(t0, t0) = 1 and U(n)(t0.t0) = 0 for n ≥ 1.

To zeroth-order

∂tU(0) = 0 ⇒ U(0)(t, t0) = 1,

since we have the boundary condition U(t0, t0) = 1. At first order we have

∂tU(1)(t, t0) = −iHint,I(t)U(0)(t, t0) = −iHint,I(t)

⇒ U(1)(t, t0) = −i

∫ t

t0

dt1Hint,I(t1).

At second order we have

∂tU(2) = −iHint,IU(1) = (−i)2Hint,I(t)

∫ t

t0

dt1Hint,I(t1)

⇒ U(2)(t, t0) = (−i)2∫ t

t0

dt1Hint,I(t1)

∫ t1

t0

dt2Hint,I(t2).

By definition THint,I(t1)Hint,I(t2) = THint,I(t2)Hint,I(t1). Thus∫ t

t0

dt1

∫ t

t0

dt2THint,I(t1)Hint,I(t2)

=

∫ t

t0

dt1

∫ t1

t0

dt2THint,I(t1)Hint,I(t2) +

∫ t

t0

dt2

∫ t2

t0

dt1THint,I(t1)Hint,I(t2)

= 2

∫ t

t0

dt1

∫ t1

t0

dt2THint,I(t1)Hint,I(t2) (t1 ↔ t2 in second term)

= 2

∫ t

t0

dt1

∫ t1

t0

dt2Hint,I(t1)Hint,I(t2).



Hence

U(t, t0) = 1− i

∫ t

t0

dt1Hint,I(t1)

+ 12(−i)2

∫ t

t0

∫ t

t0

dt1dt2THint,I(t1)Hint,I(t2) + . . .

as required.

[Total 20 marks]



6. This question is about Feynman diagrams in “phi-fourth” theory. Recall that in the

interaction picture the S-matrix is given by

S = T exp

(i

∫d4x : Lint(x) :

).

(i) Explain how S can be described as a perturbation expansion and write down

the first three terms in the expansion.

The scattering amplitude iM is usually defined to be

〈out|iT |in〉 = (2π)4δ(4) (pout − pin) iM

where S = 1+ iT . Why is the 1 contribution not included? What are pout and

pin and what is the physical meaning of the δ-function? [4 marks]

ANSWER: In perturbation theory we treat Lint as a small quantity. Thus we can

expand S as

S = 1 + i

∫d4x : Lint(x) :

+ 12(i)2

∫ ∫d4x1d

4x2T : Lint(x1) : : Lint(x2) : + . . .

and calculate the scattering amplitude term by term (using for instance Feynman

diagrams). The 1 contribution is not included in the scattering amplitude since it

corresponds to no scattering. The quantities pin an pout are the total momentum of

the incoming and outgoing states. The δ-function implies momentum conservation.

(ii) Consider the scattering of three incoming φ particles with momenta k1, k2 and

k3 to three outgoing φ particles with momenta p1, p2 and p3.

Define the |in〉 and |out〉 states for this process.

Use the position-space Feynman rules to calculate the contribution of the fol-

lowing Feynman diagram to 〈out|iT |in〉 in terms of the propagator DF (x − y).

p1

p2

p3

k1

k2

k3

1



Given

DF (x − y) = i

∫d4p

(2π)4e−ip·(x−y)

p2 −m2 + iε

write this contribution as a function only of momenta.

Show that this agrees with the contribution to iM calculated using the

momentum-space rules. [7 marks]

ANSWER: We have

|in〉 =√

2Ekkk1√

2Ekkk2√

2Ekkk3a†kkk1a†kkk2a†kkk3|0〉.

|out〉 =√

2Eppp1√

2Eppp2√

2Eppp3a†ppp1a†ppp2a†ppp3 |0〉.

Using the Feynman rules,

contri. to 〈out|iT |in〉

= (−iλ)2∫

d4x d4y ei(p3−k2−k3)·xei(p1+p2−k1)·yDF (x − y)

= (−iλ)2∫

d4x d4yd4p

(2π)4iei(p3−k2−k3−p)·xei(p1+p2−k1+p)·y

p2 −m2 + iε

= −iλ2∫

d4p

(2π)4(2π)4δ(4)(p3 − k2 − k3 − p)(2π)4δ(4)(p1 + p2 − k1 + p)

p2 −m2 + iε

= −iλ2

p2 −m2 + iε(2π)4δ(4)(p1 + p2 + p3 − k1 − k2 − k3).

For the momentum space rules, the momentum flowing in the propagator is k2+k3−p1so

contri. to iM = (−iλ)2i

(k1 + k2 − p3)3 −m2 + iε

= −iλ2

(k1 + k2 − p3)3 −m2 + iε.

This is in agreement with the position-space calculation.

(iii) Draw a second Feynman diagram that has no loops, is not related to the

diagram in part ii by a permutation of the pi momenta or of the ki momenta,

and contributes to iM at the same order in λ.

Evaluate this diagram using the momentum-space Feynman rules. [3 marks]

ANSWER: The diagram is

p1

p2

p3

k1

k2

k3

2



The momentum flowing in the propagator is k1 + k2 + k3, so using the Feynman rules

we find

contri. to iM = (−iλ)2i

(k1 + k2 + k3)3 −m2 + iε

= −iλ2

(k1 + k2 + k3)3 −m2 + iε.

Now consider the scattering of two incoming φ particles with momenta q1 and q2to two outgoing φ particles with momenta p1 and p2.

Taking the non-relativistic limit and comparing with the Born approximation gives

iM(q1, q2, p1, p2) = −i12

[V (ppp1 − qqq1) + V (ppp1 − qqq2)

],

where V (qqq) is the Fourier transform of the classical potential V (xxx) between the two

particles.

(iv) Use the momentum-space Feynman rules to identify V (kkk) and hence calculate

the form of V (xxx) at order λ.

Draw a Feynman diagram that gives corrections to iM (and hence potentially

to V (xxx)) at order λ2. [6 marks]

ANSWER: At order λ, the only diagram is

p1

p2

k1

k2

3

This contributes

contri. to iM = −iλ.

Hence we have V (qqq) = λ. Fourier transforming gives

V (xxx) =

∫d3q

(2π)3V (qqq)e−iqqq·xxx =

∫d3q

(2π)3λe−iqqq·xxx = λδ(3)(xxx).

(This is what is known as a contact interaction.) The relevant loop Feynman diagrams

at order λ2 are



p1

p2

k1

k2

p1

p2

k1

k2

p1

p2

k1

k2

4

(which lead to a renormalization of the vertex).

[Total 20 marks]

xxxx/y/zzz ANSWERS 22 End of examination paper

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