PS8_253A_solutions.pdf - Problem Set 8 Solutions Physics...

Problem Set 8 Solutions
Physics 253A
Fall 2018
1(a) The tree level diagrams for Moller scattering e - e - → e - e - are
i M =
e - ( p
2
)
e - ( p
1
)
e - ( p
4
)
e - ( p
3
)
- (1)
Note the relative sign, since the u -channel graph can be obtained from the t -channel graph by swapping the final state
fermions. In Feynman gauge,
i M = i M
t
- i M
u
(2)
= ( - ie )
2
¯ u
3
γ
μ
u
1
- ig
μν
( p
1
- p
3
) 2
¯ u
4
γ
ν
u
2
- ( - ie )
2
¯ u
3
γ
μ
u
2
- ig
μν
( p
1
- p
4
) 2
¯ u
4
γ
ν
u
1
, (3)
so that
M = e
2
(¯ u
3
γ
μ
u
1
) (¯ u
4
γ
μ
u
2
)
t
-
(¯ u
3
γ
μ
u
2
) (¯ u
4
γ
μ
u
1
)
u
. (4)
The complex conjugate of M is
M
*
= e
2
(¯ u
2
γ μ u
4
) (¯ u
1
γ
μ
u
3
)
t
-
(¯ u
1
γ μ u
4
) (¯ u
2
γ
μ
u
3
)
u
. (5)
The matrix element squared is thus
|M|
2
= |M
t
|
2
+ |M
u
|
2
- M
t
M
*
u
- M
u
M
*
t
(6)
= e
4
(¯ u
3
γ
μ
u
1
) (¯ u
4
γ
μ
u
2
) (¯ u
2
γ
ν
u
4
) (¯ u
1
γ
ν
u
3
)
t 2
+
(¯ u
3
γ
μ
u
2
) (¯ u
4
γ
μ
u
1
) (¯ u
1
γ
ν
u
4
) (¯ u
2
γ
ν
u
3
)
u 2
(7)
-
(¯ u
3
γ
μ
u
1
) (¯ u
4
γ
μ
u
2
) (¯ u
1
γ
ν
u
4
) (¯ u
2
γ
ν
u
3
)
t u
-
(¯ u
3
γ
μ
u
2
) (¯ u
4
γ
μ
u
1
) (¯ u
2
γ
ν
u
4
) (¯ u
1
γ
ν
u
3
)
t u
. (8)
Note that |M
u
|
2
is related to |M
t
|
2
by 1 ↔ 2, and that M
u
M
*
t
is related to M
t
M
*
u
by the same transformation.
So we need to compute 2 of these 4 terms and can recycle our results for the remaining 2.
Next we compute the spin sums using the identity
∑
s
[ u
s
( p )]
a
[¯ u
s
( p )]
b
= (
/
p + m )
ab
. Beginning with the 1 /t
2
terms,
|M
t
| 2 ≡
1
4
X
spins
|M
t
|
2
=
e 4
4 t 2
X
spins
(¯ u
3
)
a
( γ
μ
)
ab
( u
1
)
b
(¯ u
4
)
c
( γ
μ
)
cd
( u
2
)
d
(¯ u
2
)
e
( γ
ν
)
ef
( u
4
)
f
(¯ u
1
)
g
( γ
ν
)
gh
( u
3
)
h
(9)
=
e
4
4 t 2
X
spins
[( u
3
)
h
(¯ u
3
)
a
( γ
μ
)
ab
( u
1
)
b
(¯ u
1
)
g
( γ
ν
)
gh
] [( u
4
)
f
(¯ u
4
)
c
( γ
μ
)
cd
( u
2
)
d
(¯ u
2
)
e
( γ
ν
)
ef
] (10)
=
e
4
4 t 2
[(
/
p
3
+ m )
ha
( γ
μ
)
ab
(
/
p
1
+ m )
bg
( γ
ν
)
gh
] [(
/
p
4
+ m )
fc
( γ
μ
)
cd
(
/
p
2
+ m )
de
( γ
ν
)
ef
] (11)
=
e 4
4 t 2
Tr [(
/
p
3
+ m ) γ
μ
(
/
p
1
+ m ) γ
ν
] Tr [(
/
p
4
+ m ) γ
μ
(
/
p
2
+ m ) γ
ν
] (12)
1

=
e
4
4 t 2
h
Tr (
/
p
3
γ
μ
/
p
1
γ
ν
) + m
2
Tr ( γ
μ
γ
ν
)
i h
Tr (
/
p
4
γ
μ
/
p
2
γ
ν
) + m
2
Tr ( γ
μ
γ
ν
)
i
, (13)
where, to get to the last line, we used the fact that the trace of an odd number of gamma matrices vanishes. While
it looks like it took many steps to reach the last line, with some experience one will be skip directly from line (7) to
line (12). We can now evaluate these traces using the formulas
Tr ( γ
μ
γ
ν
) = 4 g
μν
, (14)
Tr ( γ
μ
γ
ν
γ
ρ
γ
σ
) = 4 ( g
μν
g
ρσ
- g
μρ
g
νσ
+ g
μσ
g
νρ
) . (15)
This leads to
|M
t
| 2 =
4 e
4
t 2
(
p
μ
3
p
1 ν
- p
1
· p
3
δ
μ
ν
+ p
3 ν
p
μ
1
+ 4 m
2
δ
μ
ν
) (
p
4 μ
p
ν
2
- p
2
· p
4
δ
ν
μ
+ p
ν
4
p
2 μ
+ 4 m
2
δ
ν
μ
)
(16)
=
8 e 4
t 2
( p
1
· p
2
)
2
+ ( p
1
· p
4
)
2
- 2 m
2
p
1
· p
3
+ 2 m
4
(17)
=
2 e
4
t 2
(
s
2
+ u
2
+ 8 m
2
t - 8 m
4
)
. (18)
To reach the second line above, we used relations like p
1
· p
2
= p
3
· p
4
which follows from ( p
1
+ p
2
)
2
= ( p
3
+ p
4
)
2
. To
get to the last line, we used
s = ( p
1
+ p
2
)
2
= 2 m
2
+ 2 p
1
· p
2
, (19)
t = ( p
1
- p
3
)
2
= 2 m
2
- 2 p
1
· p
3
, (20)
u = ( p
1
- p
4
)
2
= 2 m
2
- 2 p
1
· p
4
. (21)
Having computed |M
t
| 2 we can write down |M
u
| 2 by sending 1 ↔ 2, or equivalently t ↔ u :
|M
u
| 2 =
2 e
4
u 2
( s
2
+ t
2
+ 8 m
2
u - 8 m
4
) . (22)
Moving on to the next term in |M|
2
, it follows from equation (8) that
M
t
M *
u
≡
1
4
X
spins
M
t
M
*
u
= -
e 4
4 t u
Tr
h
(
/
p
3
+ m ) γ
μ
(
/
p
1
+ m ) γ
ν
(
/
p
4
+ m ) γ
μ
(
/
p
2
+ m ) γ
ν
i
(23)
= -
e
4
4 t u