3Interacting fields

III Quantum Field Theory



3.5 Amplitudes
Now the important question is what do we do with that quantity we computed?
We first define a more useful quantity known as the amplitude:
Definition
(Amplitude)
.
The amplitude
A
f,i
of a scattering process from
|ii
to
|fi is defined by
hf|S 1 |ii = iA
f,i
(2π)
4
δ
4
(p
F
p
I
).
where
p
F
is the sum of final state 4-momenta, and
p
I
is the sum of initial
state 4-momenta. The factor of
i
sticking out is by convention, to match with
non-relativistic quantum mechanics.
We can now modify our Feynman rules accordingly to compute A
f,i
.
Draw all possible diagrams with the appropriate external legs and impose
4-momentum conservation at each vertex.
Write a factor (ig) for each vertex.
For each internal line of momentum p and mass m, put in a factor of
1
p
2
m
2
+
.
Integrate over all undetermined 4-momentum k flowing in each loop.
We again do some examples of computation.
Example
(Nucleon-antinucleon scattering)
.
Consider another tree-level process
ψ +
¯
ψ φ + φ.
We have a diagram of the form
ψ
¯
ψ
φ
φ
p
1
p
0
1
p
0
2
p
2
k
and a similar one with the two φ particles crossing.
The two diagrams then say
A = (ig)
2
1
(p
1
p
0
1
)
2
µ
2
+
1
(p
2
p
0
2
)
2
µ
2
.
Note that we dropped the
terms in the denominator because the denominator
never vanishes, and we don’t need to integrate over anything because all momenta
in the diagram are determined uniquely by momentum conservation.
Example (Meson scattering). Consider
φ + φ φ + φ.
This is a bit more tricky. There is no tree-level diagram we can draw. The best
we can get is a box diagram like this:
φ
φ
φ
φ
p
1
p
2
p
0
1
p
0
2
k
k p
0
2
k + p
0
1
p
1
k + p
0
1
We then integrate through all possible k.
This particular graph we’ve written down gives
iA = (ig)
4
Z
d
4
k
(2π)
4
i
4
(k
2
µ
2
+ )((k + p
0
1
)
2
µ
2
+ )((k + p
0
1
p
1
)
2
µ
2
+ )((k p
0
2
)
2
µ
2
+ )
.
For large
k
, this asymptotically tends to
R
d
4
k
k
8
, which fortunately means the
integral converges. However, this is usually not the case. For example, we
might have
d
4
k
k
4
, which diverges, or even
R
d
4
k
k
2
. This is when we need to do
renormalization.
Example (Feynman rules for φ
4
theory). Suppose our Lagrangian has a term
λ
4!
φ
4
.
We then have single-vertex interactions
Any such diagram contributes a factor of (
). Note that there is no
1
4!
. This
is because if we expand the term
4!
hp
0
1
, p
0
2
|:φ(x
1
)φ(x
2
)φ(x
3
)φ(x
4
): |p
1
, p
2
i,
there are 4! ways of pairing the
φ
(
x
i
) with the
p
i
, p
0
i
, so each possible interaction
is counted 4! times, and cancels that 4! factor.
For some diagrams, there are extra combinatoric factors, known as symmetry
factors, that one must take into account.
These amplitudes can be further related to experimentally measurable quan-
tities. We will not go into details here, but in The Standard Model course, we
will see that if we have a single initial state
i
and a collection of final states
{f}
,
then the decay rate of i to {f} is given by
Γ
if
=
1
2m
i
Z
|A
f,i
|
2
dρ
f
,
where
dρ
f
= (2π)
4
δ
4
(p
F
p
I
)
Y
r
d
3
p
r
(2π)
3
2p
0
r
,
and r runs over all momenta in the final states.