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10. Loops and Renormalization I: The ABC Theory
10.3 Dealing with the bad news: a simple example
10.3.1 Evaluating Π
[2] (q 2 )
C
We turn our attention to the actual evaluation of a one-loop amplitude, beginning with the simplest, which is −iΠ
[2] (q
2 ):
C
∫ d
4 k
i
i
−iΠ
[2] (q
2 ) = (−ig)
2
;
(10.39)
C
2
2
(2π) 4 k 2 − m + i∈ (q − k) 2 − m + i∈
A
B
in particular, we want to know the precise mathematical form of the divergence
which arises when the momentum integral in (10.39) is not cut off at an upper
limit Λ. This will necessitate the introduction of a few modest tricks from a
large armoury (mostly due to Feynman) for dealing with such integrals.
The first move in evaluating (10.39) is to ‘combine the denominators’ using
the identity (problem 10.2)
∫ 1
1
dx
=
(10.40)
AB
[(1 − x)A + xB] 2
0
(similar ‘Feynman identities’ exist for combining three or more denominator
factors). Applying (10.40) to (10.39) we obtain
∫
∫
1
d
4 k
[2]
2
−iΠ (q
2 ) = g
dx
C
(2π) 4
0
1
×
(10.41)
2
2
[(1 − x)(k 2 − m + i∈) + x((q − k) 2 − m + i∈)] 2
A
B
Collecting up terms inside the [. . .] bracket and changing the integration variable to k
′ = k − xq leads to (problem 10.3)
∫
∫
1
d
4 k
′
[2]
2
−iΠ (q
2 ) = g
dx
1
(10.42)
C
(2π) 4 (k ′2 − Δ + i∈) 2
0
where
2
2
2
Δ = −x(1 − x)q + xm B + (1 − x)m A .
(10.43)
The d
4 k
′ integral means dk
′0 d
3
k
′ , and k
′2 = (k
′0 )
2
− k
′2 .
We now perform the k
′0 integration in (10.42) for which we will need the
contour integration techniques explained in appendix F. The integral we want
to calculate is
∫
∫
∞
dk
′0
∞
dk
′0
∂
∂
=
≡
I(A)
(10.44)
[(k ′0 ) 2 − A] 2
∂A
[(k ′0 ) 2 − A]
∂A
−∞
−∞
k
′2
where A =
+ Δ − i∈. We rewrite I(A) as
∫
dz
I(A) = lim
(10.45)
R→∞ CR [z 2 − A]