PS10-solutions - Phys 253a 1 Problem Set 10 Solutions...

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