Pictures Paths Particles Processes

198 November 7, 2013
with constants c 1,2 to be determined. This is simple, since we can study the
annihilation of the charged scalar-antiscalar pair into an off-shell photon : under
the handlebar operation, the amplitude becomes
p
p 2
1
k = iQ√¯h (c 2 p 2 µ − c 1 p 1 µ ) k µ
= iQ √¯h (c 2 p 2 µ − c 1 p 1 µ ) ( p 1µ + p 2µ
)
= iQ√¯h (c 2 − c 1 )s . (7.94)
2
We see that c 1 = c 2 is required, and therefore the first Feynman rule for scalar
electrodynamics (sQED) reads
q
p
µ Q
sQED
↔ i (p + q)µ
¯h
vertex
sQED Feynman rules, version 7.1 (7.95)
Let us now consider the more complicated process of annihilation into two onshell
photons. With the above vertex two diagrams are involved :
p
1
p
2
k
The amplitude is then given, with m indicating the scalar’s mass, by
1
k 2
M = −i¯hQ 2 (p 1 + (p 1 − k)) · ɛ 1 ((p 1 − k) + (−p 2 )) · ɛ 2
(p 1 − k 1 ) 2 − m 2 + (k 1 ↔ k 2 )
(
= −2i¯hQ 2 (p1 · ɛ 1 )(p 2 · ɛ 2 )
+ (p )
1 · ɛ 2 )(p 2 · ɛ 1 )
(7.96)
(p 1 · k 1 ) (p 2 · k 1 )
The test of current conservation now fails, since
p
M⌋ ɛ1→k 1
= −2i¯hQ 2 ((p 2 · ɛ 2 ) + (p 1 · ɛ 2 )) = −2i¯hQ 2 (k 1 · ɛ 2 ) . (7.97)
The solution is to introduce a four-point vertex into the Feynman rules 23 :
23 For reasons lost in the mists of time, such a vertex is called a sea-gull vertex, although to
me it does not look very gully nor even particularly birdy.
1
p
2
k
2
k
1