QFT PS4 Solutions: Free Quantum Field Theory (22/10/18 ...€¦ · QFT PS4 Solutions: Free Quantum...

QFT PS4 Solutions: Free Quantum Field Theory (22/10/18) 1

Solutions 4: Free Quantum Field Theory

1. Heisenberg picture free real scalar field

We have

φ(t,x) =

∫d3p

(2π)31√2ωp

(ape−iωpt+ip·x + a†pe

iωpt−ip·x)

(1)

(i) By taking an explicit hermitian conjugation, we find our result that φ† = φ. You need to note that

all the parameters are real: t,x,p are obviously real by definition and if m2 is real and positive

semi-definite then ωp is real for all values of p. Also (â†)† = â is needed.

(ii) Since π = φ̇ = ∂tφ classically, we try this on the operator (1) to find

π(t,x) = −i∫

d3p

(2π)3

√ωp2

(ape−iωpt+ip·x − a†peiωpt−ip·x

)(2)

(iii) In this question as in many others it is best to leave the expression in terms of commutators. That

is exploit

[A+B,C +D] = [A,C] + [A,D] + [B,C] + [B,D] (3)

as much as possible. Note that here we do not have a sum over two terms A and B but a sum over

an infinite number, an integral, but the principle is the same.

The second way to simplify notation is to write e−iωpt+ip·x = e−ipx so that we choose p0 = +ωp.

The simplest way however is to write Âp = âpe−iωpt+ip·x and Â†p = â

†pe−iωpt+ip·x. You can quickly

check these satisfy the same commutation relations as âp and â†p. In fact Âp = e

iHtâpe−iHt = âp(t)

for a free field, i.e. we have just applied time evolution to the free field Heisenberg picture operators.

We have the usual commutation relations for the annihilation and creation operators with contin-

uous labels (here the three-momenta) 1[ap, a

†q

]= (2π)3δ3(p− q) , [ap, aq] =

[a†p, a

†q

]= 0 . (4)

The first field commutation relation is then

[φ(t,x), π(t,y)] =

∫d3p

(2π)31√2ωp

∫d3q

(2π)3(−i)

√ωq2[

(ape−iωpt+ip·x + a†pe

iωpt−ip·x), (aqe−iωqt+iq·y − a†qeiωqt−iq·y)

](5)

= (−i)∫

d3p

(2π)3d3q

(2π)3

√ωp4ωq(

−e−i(ωp−ωq)t+ip·x−iq·y[ap, a

†q

]+ ei(ωp−ωq)t−ip·x+iq·y

[a†p, aq

])(6)

= (−i)∫

d3p

(2π)3d3q

(2π)3

√ωp4ωq

(−e−i(ωp−ωq)t+ip·x−iq·y(2π)3δ3(p− q)

+ei(ωp−ωq)t−ip·x+iq·y(2π)3(−δ3(p− q)))

(7)

= i

∫d3p

(2π)31

2

(e+ip·(x−y) + e−ip·(x−y)

)(8)

=i

22δ3(x− y) (9)

1Note that this gives the annihilation and creation operators units of energy−3 (in natural units). The usual[ai, a

†j

]= iδij

is for the case of discrete labels.


Thus we find the canonical equal-time commutation relation

[φ(t,x), π(t,y)] = iδ3(x− y) . (10)

For the commutators [φ̂(t,x), φ̂(t,y)

]= [π̂(t,x), π̂(t,y)] = 0 . (11)

we see that they are of similar form[φ̂(t = 0,x), φ̂(t = 0,y)

]=

[ ∫ d3p(2π)3

1√2ωp

(âpe+ip·x + â†pe

−ip·x),∫d3q

(2π)31√2ωq

(âqe+iq·y + â†qe

−iq·y)]

(12)

[π̂(t = 0,x), π̂(t = 0,y)] =[ ∫ d3p

(2π)3(−i)

√ωp2

(âpe+ip·x − â†pe−ip·x),∫

d3q

(2π)3(−i)

√ωq2

(âqe+iq·y − â†qe−iq·y)

](13)

Thus all we need to show that

Cs =

∫d3pd3q

(ωpωq)s/2

[(âpe

+ip·x + sâ†pe−ip·x), (âqe

−iq·y + sâ†qe+iq·y)

], (14)

is zero where s = ±1 as the field (momenta) commutator is proportional to C+ (C−). Using (3)and (4) we find exactly as before that

Cs =

∫d3pd3q

(ωpωq)s/2

([âp, âq] e

+ip·x+iq·y + s[âp, â

†q

]e+ip·x−iq·y

+s[â†p, âq

]e−ip·x+iq·y +

[â†p, â

†q

]e−ip·x−iq·y

)(15)

=

∫d3pd3q

(ωp)s

(sδ(p− q)e+ip·(x−y) − sδ(p− q)e−ip·(x−y)

)(16)

= s

∫d3p

(ωp)s

(e+ip·(x−y) − e−ip·(x−y)

)(17)

If you now change integration variable in the second term to p′ = −p you will find it is of exactlythe same form as the first term and these cancel giving zero as required.

Aside - Trying to be Careful

This is really rubbish isn’t it! The integrals here look divergent as they scale as energy (=momen-

tum=mass) to the power (3− s). The oscillatory factor just means we are adding +∞ and −∞ aswe integrate out at high energy scales, at large |p|. The whole thing looks (and is) badly defined.QFT is not very reasonable about divergences.

Riemann-Lebesgue lemma2 is the real answer here but unlikely to be helpful to Imperial Physics

students who aren’t given the relevant mathematical tools. If we set up the problem more carefully

we could apply this lemma and you’d be happy. In practice I haven’t set up the maths carefully. I

think you will find most QFT text books are the same. We really should define everything carefully

to make sure that such integrals are always defined properly.

2See https://bit.ly/2KOsAkA.


The quick and dirty way here is to remember that these integrals here are part of an operator

expression coming from commutators even if it is a unit operator. To evaluate these expressions the

operator must act on something. So you should let your expression act on a dummy function f(p).

Now this dummy function has to be of the right type, some well behaved function. Basically that

has to be something that falls off for large |p| “nicely”. The functions needed will always respectour space-time symmetries, so here will always be functions of |p|. A suitable example might besomething that falls off as exp(−p.p/(2σ2)) where you can choose sigma to be as big as you like(take it to infinity only after everything else is done). Now your integrals are of the right form for

Riemann-Lebesgue lemma to apply. Alternatively, as each integral is now well behaved and finite,

you can start to manipulate them. In our case we would need f(p) = f(−p) for our odd/evenarguments to work but the space-time symmetries guarantee that any function we have in practice

will have that symmetry.

To be more precise we should set everything up carefully. This is what the Axiomatic QFT3 is very

careful about.

(iv) Given ωp =√p2 +m2 then ωp = ω−p and hence∫

d3xφ(t = 0,x)eiq·x =

∫d3x

d3p

(2π)31√2ωp

(ape

ip·x + a†pe−ip·x

)eiq·x (18)

=

∫d3x

d3p

(2π)31√2ωp

(ape

i(p+q)·x + a†pei(q−p)·x

)(19)

=

∫d3p

(2π)31√2ωp

(ap(2π)

3δ(3)(q + p) + a†p(2π)3δ(3)(q− p)

)(20)

=1√2ωq

(a−q + a

†q

)(21)

∫d3xπ(t = 0,x)eiq·x = −i

∫d3x

d3p

(2π)3

√ωp2

(ape

ip·x − a†pe−ip·x)

eiq·x (22)

= −i∫

d3xd3p

(2π)3

√ωp2

(ape

i(p+q)·x − a†pei(q−p)·x)

(23)

= −i∫

d3p

(2π)3

√ωp2

(ap(2π)

3δ(3)(q + p)− a†p(2π)3δ(3)(q− p))

(24)

= −i√ωq2

(a−q − a†q

)(25)

Thus solving for a−q and a†q gives

aq =

∫d3x e−iq·x

(√ωq2φ(0,x) + i

1√2ωq

π(0,x)

)(26)

a†q =

∫d3x eiq·x

(√ωq2φ(0,x)− i 1√

2ωqπ(0,x)

)(27)

(28)

3See https://bit.ly/2wrg1ID.


Hence since [φ(x), φ(y)] = [π(x), π(y)] = 0[ap, a

†q

]=

∫d3x e−ip·xd3y eiq·y (29)

×

[√ωp2φ(0,x) + i

1√2ωp

π(0,x),

√ωq2φ(0,y)− i 1√

2ωqπ(0,y)

](30)

=

∫d3x d3y e−ip·x+iq·y

(−i√

ωp4ωq

[φ(0,x), π(0,y)] + i

√ωq4ωp

[π(0,x), φ(0,y)]

)(31)

=

∫d3x d3y e−ip·x+iq·y

(√ωp4ωq

δ(3)(x− y) +√

ωq4ωp

δ(3)(y − x))

(32)

=

∫d3x ei(q−p)·x

(√ωp4ωq

+

√ωq4ωp

)(33)

= (2π)3δ(3)(p− q) (34)

(v) Taking the derivative we have

∇φ(x) =∫

d3p

(2π)3ip√2ωp

(ape

ip·x − a†pe−ip·x)

(35)

so4

P = −∫

d3xπ(x)∇φ(x) (36)

= −∫

d3xd3p

(2π)3d3q

(2π)3

√ωp2

(ape

ip·x − a†pe−ip·x) q√

2ωq

(aqe

iq·x − a†qe−iq·x)

(37)

= −∫

d3xd3p

(2π)3d3q

(2π)3

√ωpq

2√ωq

(38)

×{

ei(p+q)·xapaq − ei(p−q)·xapa†q − e−i(p−q)·xa†paq + e−i(p+q)·xa†pa†q}

(39)

= −∫

d3p

(2π)3d3q

(2π)3

√ωpq

2√ωq

(40)

×{

(2π)3δ(3)(p + q)(apaq + a

†pa†q

)− (2π)3δ(3)(p− q)

(apa

†q + a

†paq

)}(41)

= −∫

d3p

(2π)3p

2

(−apa−p − a†pa

†−p − apa†p − a†pap

)(42)

Now p(apa−p + a

†pa†−p

)is an odd function under p→ −p and so integrates to zero. Thus

p =

∫d3p

(2π)3p

2

(apa

†p + a

†pap

)(43)

=

∫d3p

(2π)3p

2

(2a†pap + (2π)

3δ(3)(0))

(44)

=

∫d3p

(2π)3p a†pap (45)

since again pδ(3)(0) is an odd function (albeit it poorly defined!).

4NOTE THIS IS FOR t = 0 should add in more exponentials for non-tirvial times!


The interpretation is that a†pap d3p is the operator giving the number of quanta in a small volume

d3p centred at momentum p. This will indeed contribute p to the total momenta. We see the same

type of term for the energy, the Hamiltonian operator H, but we get a factor of ωp not p in that

case.

(vi) From (2) we have∫d3x Π2 =

∫d3x − i

∫d3p

(2π)3

√ωp2


).− i

∫d3q

(2π)3

√ωq2

(aqe−iωqt+iq·x − a†qeiωqt−iq·x

)(46)

= −∫d3x

∫d3p

(2π)3

∫d3q

(2π)3

√ωpωq

4


)(aqe−iωqt+iq·x − a†qeiωqt−iq·x

)(47)

Using that ∫d3xeip·x = (2π)3δ(p) (48)

we apply the∫d3x and use the resulting delta function of δ3(p ± q) to eliminate the q integral.

This gives us∫d3x Π2 = −

∫d3p

(2π)3ωp2

(apa−pe

−2iωpt − apa†p − a†pap + a†pa†−pe

+2iωpt)

(49)

From (2) we have∫d3x (∇φ)2 =

∫d3x

∫d3p

(2π)31√2ωp

(ipape

−iωpt+ip·x − ipa†peiωpt−ip·x)

.

∫d3q

(2π)31√2ωq

(iqaqe

−iωqt+iq·x − iqa†qeiωqt−iq·x)

(50)

=

∫d3x

∫d3p

(2π)3

∫d3q

(2π)3p.q√4ωpωq

(iape

−iωpt+ip·x − ia†peiωpt−ip·x)

.(iaqe

−iωqt+iq·x − ia†qeiωqt−iq·x)

(51)

Again we apply the∫d3x and use the resulting delta function of δ3(p±q) to eliminate the q integral

to find ∫d3x (∇φ)2 =

∫d3p

(2π)3p2

2ωp

(apa−pe

−2iωpt + apa†p + a

†pap + a

†pa†−pe

2iωpt)

(52)

Finally, in the same manner we find that∫d3x (φ)2 =

∫d3p

(2π)31

2ωp

(apa−pe

−2iωpt + apa†p + a

†pap + a

†pa†−pe

2iωpt)

(53)


Putting this together we find

H =1

2

∫d3x Π2 + (∇φ)2 +m2φ2 (54)

=1

2

∫d3p

(2π)3ω2p + p

2 +m2

2ωp

(apa

†p + a

†pap

)1

2

∫d3p

(2π)3−ω2p + p2 +m2

2ωp

(apa−pe

−2iωpt + a†pa†−pe

2iωpt)

(55)

=1

2

∫d3p

(2π)3ωp

(apa

†p + a

†pap

)=

∫d3p

(2π)3ωp

(a†pap +

1

2(2π)3δ3(0)

)(56)

with ωp = +√

p2 +m2

(vii)

[H, a†kak] =

[∫d3p

(2π)3ωp

(a†pap +

1

2

), a†kak

](57)

=

[∫d3p

(2π)3ωpa

†pap, a

†kak

](58)

=

∫d3p

(2π)3ωp

[a†pap, a

†kak

]= 0 (59)

Note that this result is not true for any interesting i.e. interacting theory. The interactions mix the

modes of different momenta leading to a lack of conservation of particle number. Only a continuous

symmetry can guarantee conserved numbers and those are usually linked to total numbers of various

particles, not the individual quanta of one particle at one momentum.

2. Time evolution of annihilation operator

(i) From (56) and using the commutation relation[ap, a

†q

]= (2π)3δ(p− q) , [ap, aq] =

[a†p, a

†q

]= 0 , (60)

we have that

[H, ap] =

∫d3q

(2π)3

[ωqa

†qaq +

1

2, ap

](61)

=

∫d3q

(2π)3ωq

(a†qaqap − apa†qaq

)(62)

=

∫d3q

(2π)3ωq

(a†qaqap − (a†qaqap + (2π)3δ(p− q)a−p)

)(63)

= −∫

d3qωqδ(p− q))ap = −ωpap (64)

as required.

(ii) We want to prove that

(itH)nap = ap [it(H − ωp)]n (65)


using [H, ap] = −ωpap. First note that trivially the relation holds for n = 1. Now if we assume therelation holds for n = m then

(itH)m+1ap = (itH)(itH)map (66)

= itHap [it(H − ωp)]m (67)= it(apH − ωpap) [it(H − ωp)]m (68)= ap [it(H − ωp)]m+1 . (69)

So the relation holds for n = m+ 1 if it holds for n = m. Thus, by induction, it holds for all n ≥ 0.

(iii) Start by expanding the first exponential

eiHtape−iHt =

( ∞∑n=0

1

n!(iHt)n

)ape−iHt (70)

=

(∑n

1

n!(iHt)nap

)e−iHt (71)

=

(∑n

1

n!ap [it(H − ωp)]n

)e−iHt (72)

= ap

(∑n

1

n![it(H − ωp)]n

)e−iHt (73)

= apeit(H−ωp)e−iHt (74)

= apeit(H−ωp−H) (75)

(76)

Thus we see that5

eiHtape−iHt = ape

−itωp (77)

where we have used the obvious facts that [H,H] = 0 and [ωp, H] = 0.

The corresponding equation for a†p follows from hermitian conjugation(eiHtape

−iHt)† = (ape−itωp)† (78)⇒

(e−iHt

)†(ap)

† (eiHt)† = (e−itωp)† (ap)† (79)⇒ e+iHta†pe−iHt = e+itωpa†p (80)⇒ eiHta†pe−iHt = a†pe+iωpt . (81)

3. Delta Functions

The Dirac delta function is defined through∫ +∞−∞

dx δ(x− x0)f(x) = f(x0) . (82)

You should always start from this equation when using a delta function.

5Another neat way to prove this by taking the derivative of the equation with respect to t and solving the resulting

operator valued differential equation.


(i) Consider

J =

∫dy δ(g(y))f(y) . (83)

Assume that the zero’s of g are at Z = {y0|g(y0) = 0} and are widely spaced. Then the only placeswhere the integral (83) has a non-zero contribution is in the region of one of these zeros as we need

the argument of the delta function to be zero from (82). So we can write J as

J =∑y0∈Z

∫ y0+�y0−�

dy δ(g(y))f(y) . (84)

So consider one of these zeros, say y0, and expand the function g around this zero to find that

g(y) = g(y0) + (y − y0)g′(y0) +O((y − y0)2). By definition G(y0) = 0 so we have for small � that

J =∑y0∈Z

∫ y0+�y0−�

dy δ((y − y0)g′(y0))f(y) . (85)

Now change variable to x = (y − y0)g′(y0) to match the form given in the definition of the deltafunction (82). The change of variables is gives us

J =∑y0∈Z

∫ η−η

dx

g′(y0)δ(x)f(y) , η = g′(y0)� . (86)

Now in order to apply the formula (82) we note that we must be running from below the zero of

the argument of the delta function, from below x0 in (82), to above it. For the case of g′(y0) > 0

there is no problem as η is positive and we have the same form as (82). The range of integration

can be extended to infinity without any problem. When g′(y0) < 0 however, η is negative and we

are running past the zero in the wrong direction. However easy to switch direction but we get an

overall minus sign in this case. This then cancels the negative sign of g′(y0) in the denominator.

Thus we have that

J =∑y0∈Z

∫ η′−η′

dx

|g′(y0|)δ(x)f(y) , η′ = |g′(y0)|� . (87)

Now we apply (82) to find that

J =

∫dy δ(g(y))f(y) =

∑y0∈Z

f(y0)

|g′(y0)|, (88)

(ii) By inspection

I =

∫d4k

(2π)4δ(k2 −m2)f(k2, p2, (k − p)2) (89)

must be Lorentz invariant if k and p are four vectors as the expression is made up of Lorentz scalars.

The arguments k2, p2, (k−p)2 and m2 are all Lorentz scalars. The measure, d4k is Lorentz invariantbecause any boost to new variables k′µ = Λµνkν produces a Jacobian factor in the transformation

|Λ| but this is 1 by definition of the Lorentz transformations. Thus all elements of this integral areinvariant.

There is only one variable p in this problem so we can only be a function of the only remaining

variable p (and of course it can depend on m or other constant parameters). The only invariant we

can build out of this is p2 so the result must be a function of p2.


The result

I =∑

k0=±ω

∫d3k

2ωf(m2, p2, (k − p)2) , ω = |

√k2 −m2| . (90)

follows from (88). Consider the k0 integration where we have that g(k0) = (k0)2−ω2. This has two

zeros at k0 = ±ω. We find that g′(k0) = 2k0 so at the zeros we have g′(k0) = 2|k0| = 2ω.

Since I is Lorentz invariant as a whole, and all the other terms in the form (90) are Lorentz invariant,

we deduce that d3k/(2ω) is also Lorentz invariant. You could also prove this directly by changing

variables to a boosted frame k′µ = Λµνkν .

(iii) Consider the integral

K =

∫ +∞−∞

dp0 f(p0)

(i

p0 − ω + i�− ip0 − ω − i�

)(91)

The poles and contours used in the two terms are shown in figure 1.

For the first term we can distort the integration path so that near the pole at p0 − ω + i� we runalong a semi circle centred on ω of radius η > 0 running above the axis (positive imaginary part),

see figure 2. We will assume η is infinitesimal so that we do not encounter any poles in the function

f . We run along the real axis for the rest of the path. We can now take the limit � → 0 to placethe pole on the real axis as for η > 0 the path does not run through the pole.

K+ =

∫C+

dp0 f(p0)i

p0 − ω(92)

For the second term we move the contour of integration so near its pole at p0 − ω − i� so that nowthis path runs along the real axis except for a semi-circle centred on ω of radius η > 0 but this time

the semi-circle is below the axis (negative imaginary part), see figure 2.

K− = −∫C−

dp0 f(p0)i

p0 − ω(93)

If we reverse the direction of integration in this second case we absorb the overall minus sign. Since

the whole result is K = K+ + K− when we put the two together, the contributions coming from

two integrations along the real axes now cancel. The only part remaining is an integration around

a small circle of radius η centred on the pole at k0 = ω, see figure 3. Note we are running round

this pole in a negative sense so that the residue there tells us this integral is equal to −2πi timesthe residue. The residue is simply if(p0 = ω) with the the factor of i coming from the numerator.

Thus we find that we have

K = 2πf(p0 = ω) (94)

However using the definition of the Dirac delta function (82) we also have that this may be written

as

K = 2π

∫ +∞−∞

dp0f(p0)δ(p0 − ω) (95)

Now comparing (91) and (95) we can identify that

2πδ(p0 − ω) =i

p0 − ω + i�− ip0 − ω − i�

(96)


w-ie w-ie

Figure 1: Figure showing two contours used for two terms in the delta function representation K. Here

the poles are placed off the real axis and contours run along the real line.

w

C-

C+

w

-C-

C+

Figure 2: Figure showing two contours used for two terms in the delta function representation, K+ of

(92) uses C+ shown on the left while on the right is C− used by K− of (93). The poles of the integrand

are now on the real axis but the contours follow semicircles above or below the real line to avoid the the

poles.

w

Figure 3: Absorbing the minus sign of the second term of (91) by reversing the direction of the countour

C−, the only non-zero contribution now comes from a small circle around the pole running in the negative

direction..


4. The Advanced Propagator

We have

DA(x) = −θ(−x0) 〈0|[φ(x), φ(0)]|0〉 (97)

(i) We start by evaluating DA. We have

[φ(x), φ(0)] =

∫d3p

(2π)3d3q

(2π)31√

4ωpωq

([ap, a

†q

]e−ip·x +

[a†p, aq

]eip·x

)∣∣∣p0=ωp

(98)

=

∫d3p

(2π)3d3q

(2π)31√

4ωpωq(2π)3δ(3)(p− q)

(e−ip·x − eip·x

)∣∣p0=ωp

(99)

=

∫d3p

(2π)31

2ωp


)∣∣p0=ωp

(100)

Hence since 〈0|0〉 = 1 we have6

DA(x) = −θ(−x0) 〈0|[φ(x), φ(0)]|0〉 (101)

= −θ(−x0)∫

d3p

(2π)31

2ωp


)∣∣p0=ωp

(102)

= −θ(−x0)∫

d3p

(2π)31

2ωp

(e−iωpx

0eip·x − eiωpx0e−ip·x

)(103)

= −θ(−x0)∫

d3p

(2π)31

2ωp

(e−iωpx

0eip·x − eiωpx0eip·x

)p↔ −p in second term(104)

= −θ(−x0)∫

d3p

(2π)3

(1

2ωpe−ip·x

∣∣∣∣p0=ωp

+1

(−2ωp)e−ip·x

∣∣∣∣p0=−ωp

)(105)

We now need to introduce a p0 integration and rewrite the expression in terms of a contour inte-

gration. There are two standard ways to do this. In the first approach we shift the poles of the

integrand, introducing a small positive infinitesimal � into the integrand which is taken to zero

(from the positive side) at the end of the calculation. This is the approach used in the lectures and

it is common practice to use this notation, especially in the case of the time-ordered (Feynman)

propagator. The second approach is to make small distortions in the contour away from the real

p0 axis near the poles. This is used by Tong in his derivation of the Feynman propagator (sec.2.7.1

page 38) though Tong reverts to the first and standard notation later on (see Tong equation (3.37)).

Both methods are equivalent in the �→ 0+ limit.

First approach

Consider

I1(x) =

∫d4p

(2π)4i

(p0 − i�)2 − p2 −m2e−ip·x (106)

= −∫

d3p

(2π)3

∫ ∞−∞

dp0

2πi

1

(p0 − ωp − i�)(p0 + ωp − i�)e−ip·x (107)

6For the p ↔ −p in second term don’t forget that in changing variable in each momentum component∫ +∞−∞ dpi means

the range of integration changes from −∞ to +∞ to the other way round giving another minus sign.


where, as usual, ωp =√p2 +m2. The p0 integration is along the real axis with poles in the

integrand as shown here

-wp+ie +w

p+ie

(108)

The dp0 integrand

f(p0,p) =1

(p0 − ωp − i�)(p0 + ωp − i�)e−ip·x (109)

has simple poles at

p0 = ±ωp + i�. (110)

Near these poles the integrand looks like f ≈ R±/(p0 ∓ ωp + i�) with residues R± given by

R± = ±1

2ωpe−ip·x

∣∣p0=±ωp+i� (111)

The idea is that we think of our expression for the advanced propagator in (105) as being of the

form

DA(x) = −θ(−x0)∫

d3p

(2π)3(R+ +R−) . (112)

In order for this to match I1(x) of (107) we need to find a closed contour C such that by using the

residue theorem we can deduce that∫C

dp0

2πif(p0,p) = θ(−x0)

(R+ +R−

)(113)

If x0 > 0 then e−ip0x0 → 0 as =(p0) = −i∞. This means that an integration of this integrand f

round a large semi-circle running around the lower half plane is equal to zero∫C−

dp0

2πif(p0,p) = 0 ifx0 > 0 as =(p0)→ −i∞ . (114)

We can therefore add this integration of f around the C− semi-circle to our p0 integration along

the real axis in I1 without changing the result for I1. So we produce an expression for I1 which uses

a closed contour for the p0 integration by adding this lower semi-circle. Now no poles are enclosed


within this closed contour so the residue theorem tells us the result is zero

C+

C-

-wp+ie +w

p+ie ∫

C

dp0

2πif(p0,p) = 0 if x0 > 0 . (115)

If x0 < 0 then e−ip0x0 → 0 as =(p0) = +i∞. This means that an integration of this integrand

f round a large semi-circle running around the upper half plane will give zero. We can therefore

add this to our existing p0 integration along the real axis in I1 without changing the result. So we

produce a closed contour by adding the semi-circle above and now the residue theorem tells us that

we pick up contributions from both poles. This gives us

C+

C-

-wp+ie +w

p+ie ∫

C

dp0

2πif(p0,p) = R+ +R− if x

0 < 0 . (116)

Putting the two cases together gives us the desired result∫C

dp0

2πif(p0,p) = θ(−x0)

(R+ +R−

). (117)

Alternative approach

The second approach to these types of problem is to distort the contour away from the real p0 axis

near the poles by a small amount. So now consider

I(x) =

∫d3p

(2π)3

∫C

dp0

2π

i

p2 −m2e−ip·x (118)

= −∫

d3p

(2π)3

∫C

dp0

2πi

1

(p0 − ωp)(p0 + ωp)e−ip·x (119)

where, as usual, ωp =√p2 +m2. The dp0 integrand

f(p0,p) =1

(p0 − ωp)(p0 + ωp)e−ip·x (120)


has simple poles at

p0 = ±ωp (121)

with residues f ≈ R±/(p0 ∓ ωp) near p0 = ±ωp given by

R± = ±1

2ωpe−ip·x

∣∣p0=±ωp (122)

Observing that we can rewrite

DA(x) = −θ(−x0)∫

d3p

(2π)3(R+ +R−) (123)

in order for this to match I(x) we need to find a contour C such that∫C

dp0

2πif(p0,p) = θ(−x0)

(R+ +R−

)(124)

Consider the following contour

C

p−Ep E

(125)

If x0 > 0 then e−ip0x0 → 0 as =(p0) = −i∞ and we close the contour below and pick up no poles

C

p−Ep E

∫C

dp0

2πif(p0,p) = 0 (126)

If x0 < 0 then e−ip0x0 → 0 as =(p0) = +i∞ and we close the contour below and picking up both

poles

C

p−Ep E

∫C

dp0

2πif(p0,p) = R+ +R− (127)

so that indeed ∫C

dp0

2πif(p0,p) = θ(−x0)

(R+ +R−

)(128)


To give a bit more detail on how we close the contour, consider x0 > 0. We have (remember p0 is

complex) ∣∣∣∣ 1(p0)2 − ω2p e−ip0x0∣∣∣∣ = eIm p0x0∣∣(p0)2 − ω2p∣∣ ≤ e

Im p0x0

|p0|2 − ω2p(129)

where the last step comes from writing |p0|2 = |(p0)2 − ω2p + ω2p| ≤ |(p0)2 − ω2p| + |ω2p| using thetriangle inequality. Closing the integral below we have Im p0 ≤ 0 so that eIm p0x0 ≤ 1. Thus,evaluating first at finite |p0| we have, for the infinite semi-circular path C− on the lower half plane

-

C

(130)

we have∣∣∣∣∫C−

dp01

(p0)2 − ω2pe−ip

0x0∣∣∣∣ ≤ ∫

C−dp0

∣∣∣∣ 1(p0)2 − ω2p e−ip0x0∣∣∣∣ (131)

≤∫C−

dp0eIm p

0x0

|p0|2 − ω2p(132)

≤∫C−

dp01

|p0|2 − ω2p(133)

=π|p0|

|p0|2 − ω2p(integrate on C− with dp0 = |p0|dθ) (134)

→ 0 as p0 →∞ (135)

Hence we see that the contribution from the integration along C− is zero and we can close the

contour C along C− in order to evaluate I(x) when x0 > 0. A similar argument holds for x0 < 0.

(ii) We have, taking the derivative of e−ip·x

(∂2 +m2

)DA(x) =

∫C

d4p

(2π)4i

p2 −m2(−p2 +m2

)e−ip·x (136)

= −∫C

d4p

(2π)4ie−ip·x (137)

= −iδ(4)(x) (138)

(Note the integrand has no poles on the real axis, so none of the subtleties in the path C relevant

when defining DA(x) appear.)


∗5. Time evolution and propagators of a complex scalar field

(i) Consider two sets of annihilation and creation operators â†1, â†2, a1 and a2 obeying[

âip, â†jq

]= (2π)3δ3((p− q))δij , [âip, âjq] =

[â†ip, â

†jq

]= 0 , i, j = 1, 2 . (139)

We define

b̂p =1√2

(â1p + iâ2p) , ĉp =1√2

(â1p − iâ2p) . (140)

so that

b̂†p =1√2

(â†1p − iâ

†2p

), ĉ†p =

1√2

(â†1p + iâ

†2p

). (141)

We require several commutators to be proved. The key identity here is that for any operators

A,B,C,D we have

[A+B,C +D] = [A,C] + [A,D] + [B,C] + [B,D] (142)

which you should prove if this is not familiar.

The commutators between a b or c annihilation and a b̂ or ĉ creation operator are all of the form

1

2

[(â1p + ispâ2p) ,

(â†1q + isqâ

†2q

)](143)

where sp, sq = ±1. Expanding we have that

1

2

[(â1p + ispâ2p) ,

(â†1q + isqâ

†2q

)](144)

=1

2

[â1p, â

†1q

]+ isq

1

2

[â1p, â

†2q

]+ isp

1

2

[â2p, â

†1q

]− spsq

1

2

[â2p, â

†2q

](145)

=1

2δ3(p− q) + 0 + 0− spsq

1

2δ3(p− q) (146)

=1

2(1− spsq)δ3(p− q) . (147)

So we find

[b̂p, ĉ†q] = 0 sp = +1, sq = +1 (148)

[b̂p, b̂†q] = δ

3(p− q) sp = +1, sq = −1 (149)[ĉp, ĉ

†q] = δ

3(p− q) sp = −1, sq = +1 (150)[ĉp, b̂

†q] = 0 sp = −1, sq = −1 (151)

Note the last is the hermitian conjugate of the first so you could avoid calculating it for that reason.

The commutators between a pair of b or c annihilation operators are all of the form

1

2[(â1p + ispâ2p) , (â1q + isqâ2q)] (152)

where sp, sq = ±1. Since annihilation operators commute with each other even if they are of thesame type and at the same momentum, all these commutators are clearly zero. Taking the hermitian

conjugate, (or using a similar argument for the creation operators), we see that all the commutators

between a pair of b̂† or ĉ† creation operators are also zero.


(ii) Given the question below for Q we can save time by considering b̂†pb̂p + sĉ†pĉp where s = ±1. We

then have that

b̂†pb̂p + sĉ†pĉp =

1

2

(â†1p − iâ

†2p

)(â1p + iâ2p)

+s

2

(â†1p + iâ

†2p

)(â1p − iâ2p) (153)

=1

2

(â†1pâ1p + iâ

†1pâ2p − iâ

†2pâ1p + â

†2pâ2p

)+s

2

(â†1pâ1p − iâ

†1pâ2p + iâ

†2pâ1p + â

†2pâ2p

)(154)

=1

2

((1 + s)

(â†1pâ1p + â

†2pâ2p

)+ i(1− s)

(â†1pâ2p − â

†2pâ1p

))(155)

When s = +1 we have the desired result that

b̂†pb̂p + ĉ†pĉp = â

†1pâ1p + â

†2pâ2p . (156)

From this we see that (here Zp is the zero point energy for mode p)

Ĥ =∑i=1,2

∫d3p ωp

(â†ipâip + Zp

)(157)

=

∫d3p ωp

(b̂†pb̂p + ĉ

†pĉp + 2Zp

)(158)

(iii) With the Hamiltonian given by (158), to show that

Φ̂H(t,x) =

∫d3p

(2π)31√2ωp

(b̂pe−ipx + ĉ†pe

+ipx) (159)

given the form at t− 0 (where px ≡ pµxµ and p0 = +ωp) we can use the same method as used forthe real field. That is since like all operators7

Φ̂H(t,x) = exp{iĤt}Φ̂H(t = 0,x) exp{−iĤt} (160)

all we need to do is look at the behaviour of

exp{iĤt}b̂p exp{−iĤt} and exp{iĤt}ĉ†p exp{−iĤt} . (161)

As the c and b operators commute, the only part that matter that matters is the b̂†b̂ term in the

first case and the ĉ†c term in the second case. More formally we can commute one part of the

Hamitonian through the b̂p or the ĉ†p operator to be cancelled. The problem reduces to that of a

single type of operator, i.e. we are left with

exp{iĤbt}b̂p exp{−iĤbt} and exp{iĤct}ĉ†p exp{−iĤct} , (162)

where

Ĥb =

∫d3p ωpb̂

†pb̂p , Ĥc =

∫d3p ωpĉ

†pĉp . (163)

We have already derived these so from (77) we know that

exp{iĤbt}b̂p exp{−iĤbt} = exp{−iωpt} and exp{iĤct}ĉ†p exp{−iĤct} = exp{+iωpt} .(164)

and we find the desired time evolution for the free complex scalar field in the Heisenberg picture.

7As this is a free Hamiltonian it is the same in Heisenberg and Scroödinger pictures and so it needs no subscript.


(iv) (No answer for this part so far).

Consider two real8 scalar fields

φ̂Hi(t,x) =

∫d3p

(2π)31√2ωp

(âipe

−iωpt+ip·x + â†ipeiωpt−ip·x

)(165)

where âip and â†ip are the operators given in (139). Show that if (??) is true then the field operators

obey

Φ̂H(t,x) =1√2

(φ̂H1(t,x) + iφ̂H2(t,x)

), Φ̂†H(t,x) =

1√2

(φ̂H1(t,x)− iφ̂H2(t,x)

). (166)

(v) The Wightman function for a complex scalar field is defined as

D(x− y) := 〈0|Φ̂(x)Φ̂†(y)| 0〉 . (167)

Substituting (159) into (167) give us

D(x− y) =∫

d3p

(2π)31√2ωp

d3q

(2π)31√2ωq〈0|(b̂pe−ipx + ĉ†pe+ipx)(b̂†qe+iqy + ĉqe−iqy)| 0〉 (168)

=

∫d3p

(2π)31√2ωp

d3q

(2π)31√2ωq〈0|b̂pb̂†q| 0〉e−ipx+iqy (169)

as the c (ĉ†) operator annihilates the ket (bra) vacuum. Commuting the two b operators gives a

non-zero term containing a delta function in momentum (which will kill one integral) plus a second

term b̂†q b̂p which again annihilates the vacuum states so does not contribute. This leaves us with

D(x− y) =∫

d3p

(2π)31

2ωpe−ip(x−y) (170)

exactly we found before for real scalar fields.

Note that from the definition we have that for any type of field

(D(x− y))∗ := 〈0|Φ̂(y)Φ̂†(x)| 0〉 = D(y − x) . (171)

For the free field case in (170) we can see this explicitly.

(vi) For free complex scalar fields by substituting (159) into 〈0|Φ̂(x)Φ̂(y)| 0〉 gives us

〈0|Φ̂(x)Φ̂(y)| 0〉 =∫

d3p

(2π)31√2ωp

d3q

(2π)31√2ωq〈0|(b̂pe−ipx + ĉ†pe+ipx)(b̂qe−iqy + ĉ†qe+iqy)| 0〉(172)

=

∫d3p

(2π)31√2ωp

d3q

(2π)31√2ωq〈0|b̂pĉ†q| 0〉e−ipx+iqy (173)

as the b (ĉ†) operator annihilates the ket (bra) vacuum. However the two remaining operators

commute and again they then annihilate the vacuum so this case is zero for any time.

8These represent classical real scalar fields but their quantised versions are hermitian not real.


(vii) The Feynman propagator for complex scalar fields is defined as

∆F(x− y) := 〈0|TΦ̂(x)Φ̂†(y)| 0〉 (174)

where T is the time ordering operator. We can express ∆F in terms of the Wightman functions for

the complex field (167)

∆F(x− y) = θ(x0 − y0)〈0|Φ̂(x)Φ̂†(y)| 0〉+ θ(y0 − x0)〈0|Φ̂†(y)Φ̂(x)| 0〉 (175)

Following the same steps as above we can show that

〈0|Φ̂†(y)Φ̂(x)| 0〉 =∫

d3p

(2π)31√2ωp

d3q

(2π)31√2ωq〈0|(b̂†qe+iqy + ĉqe−iqy)(b̂pe−ipx + ĉ†pe+ipx)| 0〉(176)

=

∫d3p

(2π)31√2ωp

d3q

(2π)31√2ωq〈0|ĉq ĉ†p| 0〉e−iqy+ipx (177)

=

∫d3p

(2π)31√2ωp

d3q

(2π)31√2ωq

(2π)3δ3(q − p)e−iqy+ipx (178)

=

∫d3p

(2π)31

2ωpe−ip(y−x) (179)

By comparing with (170) we see that

〈0|Φ̂†(y)Φ̂(x)| 0〉 = D(y − x) . (180)

This means that the Feynman propagator for complex scalar fields is

∆F(x− y) = 〈0|TΦ̂(x)Φ̂†(y)| 0〉 = θ(x0 − y0)D(x− y) + θ(y0 − x0)D(y − x) . (181)

This is exactly the same form as we had for the real free field scalar propagator so the Feynman

propagator is of exactly the same form i.e. in momentum space it is equal to

∆F(p) =i

p2 −m2 + i�. (182)

Clearly these are also solutions of the Klein-Gordon equation as they were in the real free scalar

field case. Again the poles at p2 = m2 indicate that we have particle-like solutions of mass m

dominating the behaviour. The presences of two distinct types of annihliation/creation operator in

this complex field indicates there are two independent degrees of freedom with the same mass, here

distinct particle and anti-particles.

Note that 〈0|TΦ̂†(y)Φ̂(x)| 0〉 = 〈0|TΦ̂(x)Φ̂†(y)| 0〉 if the times are unequal because the T operatordefines the order so how we write the fields under the T operator has no effect.


6. Charge of a complex scalar field

We have by definition

Q = i

∫d3x

(Φ†Π† −ΠΦ

)(183)

(i) We have

i

∫d3xΠ(x)Φ(x) =

∫d3x

d3p

(2π)3d3q

(2π)3

√ωp4ωq

(cpe

ip·x − b†pe−ip·x)(

bqeiq·x + c†qe

−iq·x)

(184)

=

∫d3x

d3p

(2π)3d3q

(2π)3

√ωp√4ωq

×{

ei(p+q)·xcpbq + ei(p−q)·xcpc

†q − e−i(p−q)·xb†pbq − e−i(p+q)·xb†pc†q

}(185)

=

∫d3p

(2π)3d3q

(2π)3

√ωp√4ωq

×{

(2π)3δ(3)(p + q)(cpbq − c†pb†q

)+ (2π)3δ(3)(p− q)

(cpc†q − b†pbq

)}(186)

=1

2

∫d3p

(2π)3

(cpb−p − c†pb

†−p + cpc

†p − b†pbp

)(187)

Now

i

∫d3xΦ†(x)Π†(x) = −

(i

∫d3xΠ(x)Φ(x)

)†(188)

=1

2

∫d3p

(2π)3

(c−pbp − c†−pb†p − cpc†p + b†pbp

)(189)

and hence

Q =1

2

∫d3p

(2π)3

(c−pbp − cpb−p + c†pb

†−p − c

†−pb†p + 2b

†pbp − 2cpc†p

)(190)

=

∫d3p

(2π)3

(b†pbp − cpc†p

)(191)

=

∫d3p

(2π)3

(b†pbp − c†pcp − (2π)3δ(3)(0)

)(192)

=

∫d3p

(2π)3

(b†pbp − c†pcp

)+ infinite constant (193)

where we have used the fact that odd functions integrate to zero.

(ii) Substitute in the form of the fields

Φ(x) =

∫d3p

(2Π)31√2ωp

(bpeip·x + c†pe

−ip·x) (194)

Π(x) =

∫d3p

(2Π)3(−i)

√ωp2

(cpeip·x − b†pe−ip·x) (195)

and hence confirm that the ETCR (equal time commutation relations) for complex scalar fields are

[Tong (2.71), p.34]

[Φ(x),Π(y)]x0=y0 = (2π)3δ3(x− y) ,

[Φ†(x),Π†(y)

]x0=y0

= (2π)3δ3(x− y) , (196)


along with the other zero commutators of the ETCR:

0 = [Φ(x),Φ(y)]x0=y0 =[Φ(x),Φ†(y)

]x0=y0

(197)

= [Π(x),Π(y)]x0=y0 =[Π(x),Π†(y)

]x0=y0

(198)

=[Φ(x),Π†(y)

]x0=y0

= 0 , (199)

and so forth.

Note that the ETCR for the complex fields should match what we found for the real field case

exactly. These ETCR are true for ANY field. Since we are meant to treat Φ and (Φ)† as separate

fields, they should each obey exactly the same equal time commutation relations. It is trivial to

check the as we can take the hermitian conjugate of [Φ,Π] and should get the [Φ†,Π†] for free and it

should look the same. The factor of i is critical for that. This is really i~δ3(x− y) but with ~ = 1,the same factor as seen in the QM [q, p] = i~ commutator.

Then we can prove the charge-field commutator by calculating (not all operators are at equal times)

[Q,Φ(x)] = i

∫d3y

[Φ†(y)Π†(y)− Φ(y)Π(y),Φ(x)

](200)

= −i∫d3y [Φ(y)Π(y),Φ(x)] (201)

= −i∫d3y

(−iΦ(y)δ3(x− y)

)(202)

= −Φ(x) (203)

Alternatively, we can prove this as follows:

[Q,Φ(x)] =

∫d3p

(2π)3d3q

(2π)31√2ωq

([c†pbp − b†pcp, bq

]eiq·x +

[c†pbp − b†pcp, c†q

]e−iq·x

)(204)

= −∫

d3p

(2π)3d3q

(2π)31√2ωq

(2π)3δ(3)(p− q)(bpe

iq·x + b†pe−iq·x

)(205)

= −∫

d3p

(2π)31√2ωp

(bpe

ip·x + b†pe−ip·x

)(206)

= −Φ(x) (207)

where we have used [ab, c] = a [b, c] + [a, c] b and hence[b†pbp, bq

]=

[c†p, bq

]bp = −(2π)3δ(3)(p− q)bp (208)[

b†pbp, b†q

]= 0 (209)[

c†pcp, b†q

]= b†p

[cp, b

†q

]= (2π)3δ(3)(p− q)b†p (210)[

c†pcp, bq

]= 0 (211)

(iii) We have that [Q, η(x)] = qη(x) for some operator η with c-number charge q. Consider

η → η′ = eiθQηe−iθQ (212)


We proceed exactly as we did for this type of transformation for the Hamiltonian and annihilation

operator. In fact as all we used was the commutation relation, a simple substitution into the answer

given above is sufficient to convince that indeed η′ = eiθqη.

However let us illustrate this with a different proof.

dη′

dθ= iQeiθQ.η.e−iθQ +QeiθQ.η.(−iQe−iθQ) (213)

= ieiθQ.Qη.e−iθQ − ieiθQ.ηQ.(e−iθQ) (214)= ieiθQ.[Q, η].e−iθQ = ieiθQ.qQ.e−iθQ = iqeiθQQ.e−iθQ (215)

⇒ dη′

dθ= iqQ (216)

Integrating this and using the fact that if θ = 0 we have η = η′ as a boundary condition, we see

that we get our answer η′ = eiθqη. If integration of an operator equation worries you, note that

if you substitute the solution into the differential equation, the operators are not involved in the

differentiations, just the c-number parts with which you are familiar.

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