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QFT PS4 Solutions: Free Quantum Field Theory (9/5/18) 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†peiωpt−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 (a†)† = a is needed.
(ii) Since π = φ = ∂tφ classically, we try this on the operator (1) to find
π(t,x) = −i∫
d3p
(2π)3
√ωp
2
(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 Ap = ape−iωpt+ip·x and A†p = a†pe
−iωpt+ip·x. You can quickly
check these satisfy the same commutation relations as ap and a†p. In fact Ap = eiHtape−iHt = ap(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)
√ωq
2[(ape
−iωpt+ip·x + a†peiωpt−ip·x), (aqe
−iωqt+iq·y − a†qeiωqt−iq·y)]
(5)
= (−i)∫
d3p
(2π)3d3q
(2π)3
√ωp
4ωq(−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
√ωp
4ωq
(−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.
QFT PS4 Solutions: Free Quantum Field Theory (9/5/18) 2
Thus we find the canonical equal-time commutation relation
[φ(t,x), π(t,y)] = iδ3(x− y) . (10)
For the commutators [φ(t,x), φ(t,y)
]= [π(t,x), π(t,y)] = 0 . (11)
we see that they are of similar form[φ(t = 0,x), φ(t = 0,y)
]=
[ ∫ d3p
(2π)31√2ωp
(ape+ip·x + a†pe−ip·x),∫d3q
(2π)31√2ωq
(aqe+iq·y + a†qe−iq·y)]
(12)
[π(t = 0,x), π(t = 0,y)] =[ ∫ d3p
(2π)3(−i)
√ωp
2(ape+ip·x − a†pe−ip·x),∫
d3q
(2π)3(−i)
√ωq
2(aqe+iq·y − a†qe−iq·y)
](13)
Thus all we need to show that
Cs =
∫d3pd3q
(ωpωq)s/2
[(ape+ip·x + sa†pe−ip·x), (aqe−iq·y + sa†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
([ap, aq] e+ip·x+iq·y + s
[ap, a
†q
]e+ip·x−iq·y
+s[a†p, aq
]e−ip·x+iq·y +
[a†p, a
†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 exactly
the 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 −∞ as
we 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.
QFT PS4 Solutions: Free Quantum Field Theory (9/5/18) 3
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 respect
our space-time symmetries, so here will always be functions of |p|. A suitable example might be
something 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/even
arguments 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
(apeip·x + a†pe−ip·x
)eiq·x (18)
=
∫d3x
d3p
(2π)31√2ωp
(apei(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
√ωp
2
(apeip·x − a†pe−ip·x
)eiq·x (22)
= −i∫
d3xd3p
(2π)3
√ωp
2
(apei(p+q)·x − a†pei(q−p)·x
)(23)
= −i∫
d3p
(2π)3
√ωp
2
(ap(2π)3δ(3)(q + p)− a†p(2π)3δ(3)(q− p)
)(24)
= −i√ωq
2
(a−q − a†q
)(25)
Thus solving for a−q and a†q gives
aq =
∫d3x e−iq·x
(√ωq
2φ(0,x) + i
1√2ωq
π(0,x)
)(26)
a†q =
∫d3x eiq·x
(√ωq
2φ(0,x)− i 1√
2ωqπ(0,x)
)(27)
(28)
3See https://bit.ly/2wrg1ID.
QFT PS4 Solutions: Free Quantum Field Theory (9/5/18) 4
Hence since [φ(x), φ(y)] = [π(x), π(y)] = 0[ap, a
†q
]=
∫d3x e−ip·xd3y eiq·y (29)
×
[√ωp
2φ(0,x) + i
1√2ωp
π(0,x),
√ωq
2φ(0,y)− i 1√
2ωqπ(0,y)
](30)
=
∫d3x d3y e−ip·x+iq·y
(−i√
ωp
4ωq[φ(0,x), π(0,y)] + i
√ωq
4ωp[π(0,x), φ(0,y)]
)(31)
=
∫d3x d3y e−ip·x+iq·y
(√ωp
4ωqδ(3)(x− y) +
√ωq
4ωpδ(3)(y − x)
)(32)
=
∫d3x ei(q−p)·x
(√ωp
4ωq+
√ωq
4ωp
)(33)
= (2π)3δ(3)(p− q) (34)
(v) Taking the derivative we have
∇φ(x) =
∫d3p
(2π)3ip√2ωp
(apeip·x − a†pe−ip·x
)(35)
so4
P = −∫
d3xπ(x)∇φ(x) (36)
= −∫
d3xd3p
(2π)3d3q
(2π)3
√ωp
2
(apeip·x − a†pe−ip·x
) q√2ωq
(aqeiq·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!
QFT PS4 Solutions: Free Quantum Field Theory (9/5/18) 5
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
√ωp
2
(ape−iωpt+ip·x − a†peiωpt−ip·x
).− i
∫d3q
(2π)3
√ωq
2
(aqe−iωqt+iq·x − a†qeiωqt−iq·x
)(46)
= −∫d3x
∫d3p
(2π)3
∫d3q
(2π)3
√ωpωq
4
(ape−iωpt+ip·x − a†peiωpt−ip·x
)(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)3ωp
2
(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
†−pe2iωpt
)(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
†−pe2iωpt
)(53)
QFT PS4 Solutions: Free Quantum Field Theory (9/5/18) 6
Putting this together we find
H =1
2
∫d3x Π2 + (∇φ)2 +m2φ2 (54)
=1
2
∫d3p
(2π)3ω2p + p2 +m2
2ωp
(apa
†p + a†pap
)1
2
∫d3p
(2π)3−ω2
p + p2 +m2
2ωp
(apa−pe−2iωpt + a†pa
†−pe2iωpt
)(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)
QFT PS4 Solutions: Free Quantum Field Theory (9/5/18) 7
using [H, ap] = −ωpap. First note that trivially the relation holds for n = 1. Now if we assume the
relation 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. The Advanced Propagator
We have
DA(x) = −θ(−x0) 〈0|[φ(x), φ(0)]|0〉 (82)
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.
QFT PS4 Solutions: Free Quantum Field Theory (9/5/18) 8
(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
(83)
=
∫d3p
(2π)3d3q
(2π)31√
4ωpωq(2π)3δ(3)(p− q)
(e−ip·x − eip·x
)∣∣p0=ωp
(84)
=
∫d3p
(2π)31
2ωp
(e−ip·x − eip·x
)∣∣p0=ωp
(85)
Hence since 〈0|0〉 = 1 we have6
DA(x) = −θ(−x0) 〈0|[φ(x), φ(0)]|0〉 (86)
= −θ(−x0)∫
d3p
(2π)31
2ωp
(e−ip·x − eip·x
)∣∣p0=ωp
(87)
= −θ(−x0)∫
d3p
(2π)31
2ωp
(e−iωpx0eip·x − eiωpx0e−ip·x
)(88)
= −θ(−x0)∫
d3p
(2π)31
2ωp
(e−iωpx0eip·x − eiωpx0eip·x
)p↔ −p in second term(89)
= −θ(−x0)∫
d3p
(2π)3
(1
2ωpe−ip·x
∣∣∣∣p0=ωp
+1
(−2ωp)e−ip·x
∣∣∣∣p0=−ωp
)(90)
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 ε 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 ε→ 0+ limit.
First approach
Consider
I1(x) =
∫d4p
(2π)4i
(p0 − iε)2 − p2 −m2e−ip·x (91)
= −∫
d3p
(2π)3
∫ ∞−∞
dp0
2πi
1
(p0 − ωp − iε)(p0 + ωp − iε)e−ip·x (92)
where, as usual, ωp =√p2 +m2. The p0 integration is along the real axis with poles in the
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.
QFT PS4 Solutions: Free Quantum Field Theory (9/5/18) 9
integrand as shown here
-wp+ie +w
p+ie
(93)
The dp0 integrand
f(p0,p) =1
(p0 − ωp − iε)(p0 + ωp − iε)e−ip·x (94)
has simple poles at
p0 = ±ωp + iε. (95)
Near these poles the integrand looks like f ≈ R±/(p0 ∓ ωp + iε) with residues R± given by
R± = ± 1
2ωpe−ip·x
∣∣p0=±ωp+iε
(96)
The idea is that we think of our expression for the advanced propagator in (90) as being of the form
DA(x) = −θ(−x0)∫
d3p
(2π)3(R+ +R−) . (97)
In order for this to match I1(x) of (92) 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−
)(98)
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∞ . (99)
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
QFT PS4 Solutions: Free Quantum Field Theory (9/5/18) 10
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 . (100)
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 x0 < 0 . (101)
Putting the two cases together gives us the desired result∫C
dp0
2πif(p0,p) = θ(−x0)
(R+ +R−
). (102)
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 (103)
= −∫
d3p
(2π)3
∫C
dp0
2πi
1
(p0 − ωp)(p0 + ωp)e−ip·x (104)
where, as usual, ωp =√p2 +m2. The dp0 integrand
f(p0,p) =1
(p0 − ωp)(p0 + ωp)e−ip·x (105)
QFT PS4 Solutions: Free Quantum Field Theory (9/5/18) 11
has simple poles at
p0 = ±ωp (106)
with residues f ≈ R±/(p0 ∓ ωp) near p0 = ±ωp given by
R± = ± 1
2ωpe−ip·x
∣∣p0=±ωp
(107)
Observing that we can rewrite
DA(x) = −θ(−x0)∫
d3p
(2π)3(R+ +R−) (108)
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−
)(109)
Consider the following contour
C
p−Ep E
(110)
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 (111)
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− (112)
so that indeed ∫C
dp0
2πif(p0,p) = θ(−x0)
(R+ +R−
)(113)
QFT PS4 Solutions: Free Quantum Field Theory (9/5/18) 12
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
∣∣ ≤ eIm p0x0
|p0|2 − ω2p
(114)
where the last step comes from writing |p0|2 = |(p0)2 − ω2p + ω2
p| ≤ |(p0)2 − ω2p| + |ω2
p| using the
triangle 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
(115)
we have∣∣∣∣∫C−
dp01
(p0)2 − ω2p
e−ip0x0∣∣∣∣ ≤ ∫
C−dp0
∣∣∣∣ 1
(p0)2 − ω2p
e−ip0x0∣∣∣∣ (116)
≤∫C−
dp0eIm p0x0
|p0|2 − ω2p
(117)
≤∫C−
dp01
|p0|2 − ω2p
(118)
=π|p0|
|p0|2 − ω2p
(integrate on C− with dp0 = |p0|dθ) (119)
→ 0 as p0 →∞ (120)
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 (121)
= −∫C
d4p
(2π)4ie−ip·x (122)
= −iδ(4)(x) (123)
(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.)
4. Charge of a complex scalar field
We have by definition
Q = i
∫d3x
(Φ†Π† −ΠΦ
)(124)
QFT PS4 Solutions: Free Quantum Field Theory (9/5/18) 13
(i) We have
i
∫d3xΠ(x)Φ(x) =
∫d3x
d3p
(2π)3d3q
(2π)3
√ωp
4ωq
(cpeip·x − b†pe−ip·x
)(bqeiq·x + c†qe−iq·x
)(125)
=
∫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
}(126)
=
∫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
)}(127)
=1
2
∫d3p
(2π)3
(cpb−p − c†pb
†−p + cpc
†p − b†pbp
)(128)
Now
i
∫d3xΦ†(x)Π†(x) = −
(i
∫d3xΠ(x)Φ(x)
)†(129)
=1
2
∫d3p
(2π)3
(c−pbp − c†−pb†p − cpc†p + b†pbp
)(130)
and hence
Q =1
2
∫d3p
(2π)3
(c−pbp − cpb−p + c†pb
†−p − c
†−pb†p + 2b†pbp − 2cpc
†p
)(131)
=
∫d3p
(2π)3
(b†pbp − cpc†p
)(132)
=
∫d3p
(2π)3
(b†pbp − c†pcp − (2π)3δ(3)(0)
)(133)
=
∫d3p
(2π)3
(b†pbp − c†pcp
)+ infinite constant (134)
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) (135)
Π(x) =
∫d3p
(2Π)3(−i)
√ωp
2(cpe
ip·x − b†pe−ip·x) (136)
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) , (137)
along with the other zero commutators of the ETCR:
0 = [Φ(x),Φ(y)]x0=y0 =[Φ(x),Φ†(y)
]x0=y0
(138)
= [Π(x),Π(y)]x0=y0 =[Π(x),Π†(y)
]x0=y0
(139)
=[Φ(x),Π†(y)
]x0=y0
= 0 , (140)
QFT PS4 Solutions: Free Quantum Field Theory (9/5/18) 14
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)
](141)
= −i∫d3y [Φ(y)Π(y),Φ(x)] (142)
= −i∫d3y
(−iΦ(y)δ3(x− y)
)(143)
= −Φ(x) (144)
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
)(145)
= −∫
d3p
(2π)3d3q
(2π)31√2ωq
(2π)3δ(3)(p− q)(bpeiq·x + b†pe−iq·x
)(146)
= −∫
d3p
(2π)31√2ωp
(bpeip·x + b†pe−ip·x
)(147)
= −Φ(x) (148)
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 (149)[
b†pbp, b†q
]= 0 (150)[
c†pcp, b†q
]= b†p
[cp, b
†q
]= (2π)3δ(3)(p− q)b†p (151)[
c†pcp, bq
]= 0 (152)
(iii) We have that [Q, η(x)] = qη(x) for some operator η with c-number charge q. Consider
η → η′ = eiθQηe−iθQ (153)
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) (154)
= ieiθQ.Qη.e−iθQ − ieiθQ.ηQ.(e−iθQ) (155)
= ieiθQ.[Q, η].e−iθQ = ieiθQ.qQ.e−iθQ = iqeiθQQ.e−iθQ (156)
⇒ dη′
dθ= iqQ (157)
QFT PS4 Solutions: Free Quantum Field Theory (9/5/18) 15
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.
∗5. Time evolution and propagators of a complex scalar field
(i) Consider two sets of annihilation and creation operators a†1, a†2, a1 and a2 obeying[
aip, a†jq
]= (2π)3δ3((p− q))δij , [aip, ajq] =
[a†ip, a
†jq
]= 0 , i, j = 1, 2 . (158)
We define
bp =1√2
(a1p + ia2p) , cp =1√2
(a1p − ia2p) . (159)
so that
b†p =1√2
(a†1p − ia
†2p
), c†p =
1√2
(a†1p + ia†2p
). (160)
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] (161)
which you should prove if this is not familiar.
The commutators between a b or c annihilation and a b or c creation operator are all of the form
1
2
[(a1p + ispa2p) ,
(a†1q + isqa
†2q
)](162)
where sp, sq = ±1. Expanding we have that
1
2
[(a1p + ispa2p) ,
(a†1q + isqa
†2q
)](163)
=1
2
[a1p, a
†1q
]+ isq
1
2
[a1p, a
†2q
]+ isp
1
2
[a2p, a
†1q
]− spsq
1
2
[a2p, a
†2q
](164)
=1
2δ3(p− q) + 0 + 0− spsq
1
2δ3(p− q) (165)
=1
2(1− spsq)δ3(p− q) . (166)
So we find
[bp, c†q] = 0 sp = +1, sq = +1 (167)
[bp, b†q] = δ3(p− q) sp = +1, sq = −1 (168)
[cp, c†q] = δ3(p− q) sp = −1, sq = +1 (169)
[cp, b†q] = 0 sp = −1, sq = −1 (170)
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[(a1p + ispa2p) , (a1q + isqa2q)] (171)
QFT PS4 Solutions: Free Quantum Field Theory (9/5/18) 16
where sp, sq = ±1. Since annihilation operators commute with each other even if they are of the
same 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 b† or c† creation operators are also zero.
(ii) Given the question below for Q we can save time by considering b†pbp + sc†pcp where s = ±1. We
then have that
b†pbp + sc†pcp =1
2
(a†1p − ia
†2p
)(a1p + ia2p)
+s
2
(a†1p + ia†2p
)(a1p − ia2p) (172)
=1
2
(a†1pa1p + ia†1pa2p − ia
†2pa1p + a†2pa2p
)+s
2
(a†1pa1p − ia
†1pa2p + ia†2pa1p + a†2pa2p
)(173)
=1
2
((1 + s)
(a†1pa1p + a†2pa2p
)+ i(1− s)
(a†1pa2p − a
†2pa1p
))(174)
When s = +1 we have the desired result that
b†pbp + c†pcp = a†1pa1p + a†2pa2p . (175)
From this we see that (here Zp is the zero point energy for mode p)
H =∑i=1,2
∫d3p ωp
(a†ipaip + Zp
)(176)
=
∫d3p ωp
(b†pbp + c†pcp + 2Zp
)(177)
(iii) With the Hamiltonian given by (177), to show that
ΦH(t,x) =
∫d3p
(2π)31√2ωp
(bpe−ipx + c†pe
+ipx) (178)
given the form at t− 0 (where px ≡ pµxµ and p0 = +ωp) we can use the same method as used for
the real field. That is since like all operators7
ΦH(t,x) = exp{iHt}ΦH(t = 0,x) exp{−iHt} (179)
all we need to do is look at the behaviour of
exp{iHt}bp exp{−iHt} and exp{iHt}c†p exp{−iHt} . (180)
As the c and b operators commute, the only part that matter that matters is the b†b term in the
first case and the c†c term in the second case. More formally we can commute one part of the
Hamitonian through the bp or the c†p operator to be cancelled. The problem reduces to that of a
single type of operator, i.e. we are left with
exp{iHbt}bp exp{−iHbt} and exp{iHct}c†p exp{−iHct} , (181)
7As this is a free Hamiltonian it is the same in Heisenberg and Scroodinger pictures and so it needs no subscript.
QFT PS4 Solutions: Free Quantum Field Theory (9/5/18) 17
where
Hb =
∫d3p ωpb
†pbp , Hc =
∫d3p ωpc
†pcp . (182)
We have already derived these so from (77) we know that
exp{iHbt}bp exp{−iHbt} = exp{−iωpt} and exp{iHct}c†p exp{−iHct} = exp{+iωpt} .(183)
and we find the desired time evolution for the free complex scalar field in the Heisenberg picture.
(iv) (No answer for this part so far).
Consider two real8 scalar fields
φHi(t,x) =
∫d3p
(2π)31√2ωp
(aipe−iωpt+ip·x + a†ipeiωpt−ip·x
)(184)
where aip and a†ip are the operators given in (158). Show that if (??) is true then the field operators
obey
ΦH(t,x) =1√2
(φH1(t,x) + iφH2(t,x)
), Φ†H(t,x) =
1√2
(φH1(t,x)− iφH2(t,x)
). (185)
(v) The Wightman function for a complex scalar field is defined as
D(x− y) := 〈0|Φ(x)Φ†(y)| 0〉 . (186)
Substituting (178) into (186) give us
D(x− y) =
∫d3p
(2π)31√2ωp
d3q
(2π)31√2ωq〈0|(bpe−ipx + c†pe
+ipx)(b†qe+iqy + cqe
−iqy)| 0〉 (187)
=
∫d3p
(2π)31√2ωp
d3q
(2π)31√2ωq〈0|bpb†q| 0〉e−ipx+iqy (188)
as the c (c†) 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 b†q bp which again annihilates the vacuum states so does not contribute. This leaves us with
D(x− y) =
∫d3p
(2π)31
2ωpe−ip(x−y) (189)
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|Φ(y)Φ†(x)| 0〉 = D(y − x) . (190)
For the free field case in (189) we can see this explicitly.
(vi) For free complex scalar fields by substituting (178) into 〈0|Φ(x)Φ(y)| 0〉 gives us
〈0|Φ(x)Φ(y)| 0〉 =
∫d3p
(2π)31√2ωp
d3q
(2π)31√2ωq〈0|(bpe−ipx + c†pe
+ipx)(bqe−iqy + c†qe
+iqy)| 0〉(191)
=
∫d3p
(2π)31√2ωp
d3q
(2π)31√2ωq〈0|bpc†q| 0〉e−ipx+iqy (192)
as the b (c†) 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.
QFT PS4 Solutions: Free Quantum Field Theory (9/5/18) 18
(vii) The Feynman propagator for complex scalar fields is defined as
∆F(x− y) := 〈0|TΦ(x)Φ†(y)| 0〉 (193)
where T is the time ordering operator. We can express ∆F in terms of the Wightman functions for
the complex field (186)
∆F(x− y) = θ(x0 − y0)〈0|Φ(x)Φ†(y)| 0〉+ θ(y0 − x0)〈0|Φ†(y)Φ(x)| 0〉 (194)
Following the same steps as above we can show that
〈0|Φ†(y)Φ(x)| 0〉 =
∫d3p
(2π)31√2ωp
d3q
(2π)31√2ωq〈0|(b†qe+iqy + cqe
−iqy)(bpe−ipx + c†pe
+ipx)| 0〉(195)
=
∫d3p
(2π)31√2ωp
d3q
(2π)31√2ωq〈0|cq c†p| 0〉e−iqy+ipx (196)
=
∫d3p
(2π)31√2ωp
d3q
(2π)31√2ωq
(2π)3δ3(q − p)e−iqy+ipx (197)
=
∫d3p
(2π)31
2ωpe−ip(y−x) (198)
By comparing with (189) we see that
〈0|Φ†(y)Φ(x)| 0〉 = D(y − x) . (199)
This means that the Feynman propagator for complex scalar fields is
∆F(x− y) = 〈0|TΦ(x)Φ†(y)| 0〉 = θ(x0 − y0)D(x− y) + θ(y0 − x0)D(y − x) . (200)
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 + iε. (201)
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 comlex field indicates there are two independent degrees of freedom with the same mass, here
distinct particle and anti-particles.
Note that 〈0|TΦ†(y)Φ(x)| 0〉 = 〈0|TΦ(x)Φ†(y)| 0〉 if the times are unequal because the T operator
defines the order so how we write the fields under the T operator has no effect.