Physics 253a
1
Problem Set 4 Solutions
October 7, 2010
1.
(a) The tree level diagram has a single vertex
φ
φ
φ
ig
Recall that the LSZ formula amputates the external legs, so we just have
i
M
=
ig
.
It turns out this process is actually kinematically disallowed. Imagine boosting
to the initial
φ
particle’s rest frame. In this frame it’s clear that there’s no way
for a particle of mass
m
to decay to two particles of mass
m
, since that would
violate conservation of energy. Mathematically, this fact comes in when we add
back the momentum conserving delta function
δ
4
(
p
1

p
2

p
3
), which simply
vanishes when all three momenta are on shell. For now, where just interested in
playing around with Feynman rules, so we needn’t worry.
(b) More than one graph is possible here, but we’ll find out later that only one
of them contributes to the scattering amplitude (again, it’s related to the LSZ
formula):
p
1
k
+
p
1
p
2
k
+
p
3
p
3
k
φ
φ
φ
Here
k
is an undetermined momentum that runs around the loop.
Note that
momentum is conserved at the upperright vertex because
p
1

p
2
=
p
3
. Once
again, LSZ says we can ignore the external legs, so the Feynman rules give
i
M
=
Z
d
4
k
(2
π
)
4
(
ig
)
3
i
k
2

m
2
i
(
k
+
p
1
)
2

m
2
i
(
k
+
p
3
)
2

m
2
Note that the symmetry factor is 1.
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Physics 253a
2
(c) Let’s label our graph like this
x
1
x
3
x
2
y
1
y
2
y
3
This graph stands for the set of Wick contractions of
h
0

T
φ
(
x
1
)
φ
(
x
2
)
φ
(
x
3
)
1
3!
ig
3!
Z
d
4
y
1
φ
(
y
1
)
3
ig
3!
Z
d
4
y
2
φ
(
y
2
)
3
ig
3!
Z
d
4
y
3
φ
(
y
3
)
3

0
i
such that each external point is contracted into exactly one interaction point.
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 SCHWARTZ
 Physics, Momentum, Richard Feynman, k2, British K class submarine, Feynman diagram

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