Quantum Field Theory
– HW 2 Solution Key
HW 2 Solution Key
1. In class, we showed that the general solution to the KleinGordon equation is:
φ
(
x
) =
i
d
3
p
(2
π
)
3
1
r
2
E
v
p
p
a
v
p
e
i
(
v
p
·
vx

E
v
p
t
)
+
a
*
v
p
e

i
(
v
p
·

E
v
p
t
)
P
(a) What is the energy for this Feld? ±or both this and for subsequent parts, I expect you to reduce
the equation to a single 3dimensional integral over
v
p
or
v
q
or similar.
Sol.
Based on the expression above, we get:
˙
φ
(
x
) =
−
i
i
d
3
p
(2
π
)
3
E
p
r
2
E
v
p
p
a
v
p
e
i
(
v
p
·

E
v
p
t
)
−
a
*
v
p
e

i
(
v
p
·

E
v
p
t
)
P
∂
i
φ
(
x
) =
i
i
d
3
p
(2
π
)
3
p
i
r
2
E
v
p
p
a
v
p
e
i
(
v
p
·

E
v
p
t
)
−
a
*
v
p
e

i
(
v
p
·

E
v
p
t
)
P
both of which we’ll need for the ensuing calculations.
The energy density can be found by multiplying out the
T
00
component of the stress energy
tensor:
ρ
(
x
) =
1
2
˙
φ
2
+
1
2
(
∇
φ
)
2
+
1
2
m
2
φ
2
Multiplying (and collecting) the whole shebang is going to yield a total of 12 terms:
ρ
(
x
)
=
1
2
i i
d
3
pd
3
q
(2
π
)
6
1
r
4
E
p
E
q
×
b
−
E
p
E
q
p
a
p
a
q
e

i
(
p
μ
+
q
μ
)
x
μ
−
a
p
a
*
q
e

i
(
p
μ

q
μ
)
x
μ
−
a
*
p
a
*
q
e
i
(
p
μ

q
μ
)
x
μ
+
a
*
p
a
*
q
e
i
(
p
μ

q
μ
)
x
μ
P
−
v
p
·
v
q
p
a
p
a
q
e

i
(
p
μ
+
q
μ
)
x
μ
−
a
p
a
*
q
e

i
(
p
μ

q
μ
)
x
μ
−
a
*
p
a
*
q
e
i
(
p
μ

q
μ
)
x
μ
+
a
*
p
a
*
q
e
i
(
p
μ

q
μ
)
x
μ
P
+
m
2
p
a
p
a
q
e

i
(
p
μ
+
q
μ
)
x
μ
+
a
p
a
*
q
e

i
(
p
μ

q
μ
)
x
μ
+
a
*
p
a
*
q
e
i
(
p
μ

q
μ
)
x
μ
+
a
*
p
a
*
q
e
i
(
p
μ

q
μ
)
x
μ
PB
These are ugly, but fortunately, they can be grouped. ±irst, note that we are actually computing:
E
=
i
d
3
xρ
(
x
)
Now, let’s consider the
a
p
a
q
terms:
RHS
=
1
2
i i i
d
3
x
d
3
pd
3
q
(2
π
)
6
1
r
4
E
p
E
q
(
−
E
p
E
q
−
v
p
·
v
q
+
m
2
)
a
p
a
q
e

i
(
p
μ
+
q
μ
)
x
μ
=
1
2
i i
d
3
pd
3
q
(2
π
)
3
1
r
4
E
p
E
q
δ
(
v
p
+
v
q
)
(
−
E
p
E
q
−
v
p
·
v
q
+
m
2
)
a
p
a
q
e

i
(
p
0
+
q
0
)
t
=
1
2
i
d
3
p
(2
π
)
3
1
R
4
E
2
p
(
−
E
2
p
+

v
p

2
+
m
2
)
a
v
p
a

v
p
e

i
(
p
0
+
q
0
)
t
=
0
This vanishes since the term in the parentheses is necessarily zero by our dispersion relation:
E
2
p
=

v
p

2
+
m
2
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View Full DocumentThe
a
*
p
a
*
q
term vanishes identically – and necessarily – since the two terms are complex conjugates
to one another. This leaves only the
aa
*
terms. I’m only going to write out the terms with the
negative exponents, and leave
C.C.
to correspond to the complex conjugate.
E
=
1
2
i i i
d
3
x
d
3
pd
3
q
(2
π
)
6
1
r
4
E
p
E
q
(
E
p
E
q
+
v
p
·
v
q
+
m
2
)
a
p
a
*
q
e

i
(
p
μ

q
μ
)
x
μ
+
C.C.
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 Spring '09
 Energy, Work, Quantum Field Theory, AP, Trigraph, Commutator, Canonical commutation relation, ap aq

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