Its form is very simple, sincef(1)(k′,k) is proportional to the Fourier transform of the potential
evaluated onk′−bkwhich is proportional to themomentum transferq= ̄h(k′−k) of the process.
The first Born term for any spherically symmetric potentialU(r) may be computed as follows.
The starting point is the integral representation, which weexpress is spherical coordinates defined
with respect to the incoming momentum directionk. We find,
f(1)(q) = −1
4 π∫ 2 π0dφ∫π0dθsinθ∫∞0dr r^2 U(r)e−iqrcosθ= −1
q∫∞0dr r U(r) sin(qr) (12.48)Here, we use the notationq=|q|.
12.8.1 The case of the Coulomb potential
For the Coulomb potential, we have
U(r) =
2 m
̄h^2V(r) =−
2 mZe^2
̄h^21
r(12.49)
The first Born term is given by
f(1)(k′,k) =mZe^2
2 π ̄h^2∫
d^3 ye−i(k′−bk)·y
|y|(12.50)
It is a standard result that
∫
d^3 y
eik·y
|y|
=
4 π
k^2(12.51)
To derive it, note first that the integral depends only onk^2 in view of rotation invariance of the factor
d^3 y/|y|, and that it manifestly scales as 1/k^2. Finally the constant factor may be determined by
multiplying both sides byq^2 , using the fact thatq^2 =−∆ in the momentum space representation,
and the fact that ∆(1/|y|) = 4πδ(3)(y). Collecting all terms, we have,
f(1)(k′,k) =
2 mZe^2
̄h^2 (k′−k)^2(12.52)
If we define the angle betweenk′andkto beθ, and use the fact that|k′|=|k|, we obtain,
(k′−k)^2 = 4k^2 sin^2
θ
2(12.53)
As a result, the cross section is given by
dσ
dΩ=
∣∣
∣f(1)(k′,k)∣∣
∣2
=
m^2 Z^2 e^4
4 q^4 sin^4 θ 2