5Quantizing the Dirac field
III Quantum Field Theory
5.3 Feynman rules
As always, the Feynman rules are much better:
(i)
An incoming fermion is given a spinor
u
r
p
, and an outgoing fermion is given
a ¯u
r
p
.
u
r
p
p
¯u
s
p
p
(ii)
For an incoming anti-fermion, we put in a
¯v
r
p
, and for an outgoing anti-
fermion we put a v
r
p
.
¯v
r
p
p
v
r
p
p
(iii) For each vertex we get a factor of (−iλ).
(iv) Each internal scalar line gets a factor of
i
p
2
− µ
2
+ iε
,
and each internal fermion we get
i(
/
p + m)
p
2
− m
2
+ iε
.
(v)
The arrows on fermion lines must flow consistently, ensuring fermion
conservation.
(vi)
We impose energy-momentum conservation at each vertex, and if we have
a loop, we integrate over all possible momentum.
(vii) We add an extra minus sign for a loop of fermions.
Note that the Feynman propagator is a 4
×
4 matrix. The indices are contracted
with each vertex, either with further propagators or external spinors.
We look at computations using Feynman rules.
Example (Nucleon scattering). For nucleon scattering, we have diagrams
p, s
q, r
p
0
, s
0
q
0
, r
0
p − p
0
p, s
q, r
p
0
, s
0
q
0
, r
0
p − q
0
In the second case, the fermions are swapped, so we get a relative minus sign.
So by the Feynman rules, this contributes an amplitude of
A = (−iλ)
2
[¯u
s
0
p
0
· u
s
p
][¯u
r
0
q
0
· u
r
q
]
(p
0
− p)
2
− µ
2
+ iε
−
[¯u
s
0
p
0
· u
r
q
][¯u
r
0
q
0
· u
s
p
]
(q
0
− p)
2
− µ
2
+ iε
!
.
Example. We look at nucleon to meson scattering
ψ
¯
ψ → φφ.
We have two diagrams
p, s
q, r
p
0
q
0
p, s
q, r
p
0
q
0
This time, we flipped two bosons, so we are not gaining a negative sign. Then
the Feynman rules give us
A = (−iλ)
2
¯v
r
q
(
/
p −
/
p
0
+ m)u
s
p
(p − p
0
)
2
− m
2
+ iε
+
¯v
r
q
(
/
p −
/
q
0
+ m)u
s
p
(p − q
0
)
2
− m
2
+ iε
.
Example. We can also do nucleon anti-nucleon scattering
ψ
¯
ψ → ψ
¯
ψ.
As before, we have two contributions
p, s
q, r
p
0
, s
0
q
0
, r
0
p − p
0
p, s
q, r
p
0
, s
0
q
0
, r
0
This time we have an amplitude of
A = (−iλ)
2
−[¯u
s
0
p
0
· u
s
p
][¯v
r
q
· v
r
0
q
0
]
(p − p
0
)
2
− µ
2
+ iε
+
[ ¯v
q
r
· u
s
p
][¯u
s
0
p
0
· v
r
0
q
0
]
(p + q)
2
− µ
2
+ iε
!
.
We have these funny signs that we have to make sure are right. We have an
initial state
|ii =
p
4E
p
E
q
b
s†
p
c
r†
q
|0i ≡ |p, s; q, ri
|fi =
p
4E
p
0
E
q
0
b
s
0
†
p
0
c
r
0
†
q
0
|0i ≡ |p
0
, s
0
; q
0
, r
0
i
To check the signs, we work through the computations, but we can ignore all
factors because we only need the final sign. Then we have
ψ ∼ b + c
†
¯
ψ ∼ b
†
+ c
So we can check
hf|:
¯
ψ(x
1
)ψ(x
1
)
¯
ψ(x
2
)ψ(x
2
): b
s†
p
c
r†
q
|0i
∼ hf|[¯v
m
k
1
ψ(x
1
)][
¯
ψ(x
2
)u
n
k
2
]c
m
k
1
b
n
k
2
b
s†
p
c
r†
q
|0i
∼ hf|[¯v
m
q
ψ(x
1
)][
¯
ψ(x
2
)u
n
p
] |0i
∼ h0|c
r
0
q
0
b
s
0
p
0
c
m†
`
1
b
n
†
`
2
[¯v
r
q
· v
m
`
1
][¯u
n
`
2
· u
s
p
] |0i
∼ −[¯v
r
q
· v
r
0
q
0
][¯u
s
0
p
0
· u
s
p
],
where we got the final sign by anti-commuting
c
m†
`
1
and
b
s
0
p
0
to make the
c
’s and
b’s get together.
We can follow a similar contraction to get a positive sign for the second
diagram.