Phys 253a
1
Problem Set 10 Solutions
December 2, 2010
1.
(a) The phase space integral for 1
→
2 is
1
2
m
φ
Z
d
3
p
(2
π
)
3
2
E
p
d
4
q
(2
π
)
3
E
q
(2
π
)
4
δ
4
(
P

p

q
)
This is easiest to evaluate in the center of mass frame,
P
= (
m
φ
,
0
). If we first do
the integral over
q
, the spatial part of the delta function enforces
0

p

q
= 0,
so
q
=

p
and
E
q
=
E
p
. We get
1
2(2
π
)
2
m
φ
Z
d
3
p
4
E
2
p
δ
(
m
φ

2
p
m
2
e
+
p
2
)
This is spherically symmetric in
p
, so we can switch to spherical coordinates
4
π
8(2
π
)
2
m
φ
Z
p
2
dp
p
2
+
m
2
e
δ
(
m
φ

2
p
m
2
e
+
p
2
)
=
1
8
πm
φ
p
2
p
2
+
m
2
e
p
m
2
e
+
p
2
2
p
√
m
2
e
+
p
2
=
1
2
m
φ
=
1
16
πm
φ
s
1

4
m
2
e
m
φ
(b)
i. The amplitude is
i
M
=
ig
S
u
(
p
)
v
(
q
), which satisfies
X
spins
M
2
=
g
2
S
Tr[(
6
p
+
m
e
)(
6
q

m
e
)]
=
4
g
2
S
(
p
·
q

m
2
e
)
=
2
g
2
S
(
m
2
φ

4
m
2
e
)
Γ
=
g
2
S
m
φ
8
π
(1

4
x
2
)
3
/
2
ii. The amplitude is
i
M
=

g
P
u
(
p
)
γ
5
v
(
q
), so we get
X
spins
M
2
=

g
2
S
Tr[
γ
5
(
6
p
+
m
e
)
γ
5
(
6
q

m
e
)]
=

4
g
2
S
(

p
·
q

m
2
e
)
=
2
g
2
S
m
2
φ
Γ
=
g
2
S
m
φ
8
π
p
1

4
x
2
Phys 253a
2
iii. The amplitude is
i
M
=
ig
V
μ
u
(
p
)
γ
μ
v
(
q
). We get
1
3
X
spins
M
2
=
1
3
g
2
V
Tr[(
6
p
+
m
e
)
γ
μ
(
6
q

m
e
)
γ
ν
]
X
i
i
μ
i
ν
=
4
g
2
V
3
(
p
μ
q
ν
+
p
ν
q
μ

(
p
·
q
+
m
2
e
)
η
μν
)

η
μν
+
P
μ
P
ν
m
2
φ
!
=
4
g
2
V
3

2
p
·
q
+ 2
(
p
·
P
)(
q
·
P
)
m
2
φ
+ 3(
p
·
q
+
m
2
e
)
!
=
4
g
2
V
3
(2
m
2
e
+
m
2
φ
)
Γ
=
g
2
V
m
φ
12
π
(1 + 2
x
2
)
p
1

4
x
2
iv. The amplitude is

g
A μ
u
(
p
)
γ
μ
γ
5
v
(
q
), so we have
1
3
X
spins
M
2
=
1
3
g
2
A
Tr[(
6
p
+
m
e
)
γ
μ
γ
5
(
6
q

m
e
)
γ
ν
γ
5
]
X
i
i
μ
i
ν
=
4
g
2
A
3
((
m
2
e

p
·
q
)
η
μν
+
p
μ
q
ν
+
q
μ
p
ν
)

η
μν
+
P
μ
P
ν
m
2
φ
!
=
4
g
2
A
3
(
m
2
φ

4
m
2
e
)
Γ
=
g
2
A
m
φ
12
π
(1

4
x
2
)
3
/
2
v. The tricky part of this one is to write down the polarization sum for an
antisymmetric tensor.
We’ll assume it’s transverse to pick out the spin 1
part. Then we know the polarization sum
Σ
μνρσ
=
X
i
i
μν
i
*
ρσ
must satisfy
P
μ
Σ
μνρσ
= 0 (and similarly if we dot
p
into any index), along
with Σ
μνρσ
=

Σ
νμρσ
= Σ
ρσμν
. The only objects we have to build Σ with
are
η
μν
and
P
μ
, and the only tensors we can build out of these with the
correct index symmetries are
η
μρ
η
νσ

η
μσ
η
νρ
and
P
μ
P
ρ
η
νσ

P
μ
P
σ
η
νρ
+
P
ν
P
σ
η
μρ

P
ν
P
ρ
η
μσ
Phys 253a
3
The transverseness condition implies that we should take the combination
X
i
i
μν
i
ρσ
=
η
μρ
η
νσ

η
μσ
η
νρ

1
m
2
φ
(
P
μ
P
ρ
η
νσ

P
μ
P
σ
η
νρ
+
P
ν
P
σ
η
μρ

P
ν
P
ρ
η
μσ
)
(Actually, this only fixes the spin sum up to an overall scale.
We must
additionally check a special case, say ...) Alternatively, we can write this as
X
i
i
μν
i
*
ρσ
=
C
μρ
C
νσ

C
μσ
C
νρ
Where
C
μρ
:=
η
μρ

P
μ
P
ρ
m
2
φ
is manifestly transverse. The amplitude is
i
M
=

g
T
μν
u
(
p
)
σ
μν
v
(
q
). We
should be a little careful about complex conjugation:
(
uσ
μν
v
)
*
=
1
2
(
u
†
γ
0
i
[
γ
μ
, γ
ν
]
v
)
*
=

1
2
v
†
i
[
γ
ν
†
, γ
μ
†
]
γ
0
u
=

1
2
vi
[
γ
ν
, γ
μ
]
u
=
vσ
μν
u
Thus,
X
spins
M
2
=
g
2
T
Tr[
σ
μν
(
6
q

m
e
)
σ
ρσ
(
6
p
+
m
e
)]
X
i
i
μν
i
*
ρσ
=

g
2
T
4
Tr[[
γ
μ
, γ
ν
](
6
q

m
e
)[
γ
ρ
, γ
σ
](
6
p
+
m
e
)]Σ
μρνσ
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 SCHWARTZ
 Center Of Mass, Mass, Quantum Field Theory, zeropoint energy, Vacuum energy, Casimir effect