130_notes.dvi

per square centimeter per second to get a cross section. To do this we set the volume to beV=
(1 cm^2 )(vrel)(1 second). The relative velocity is just the momentum divided by the reduced
mass.

σ =

1

4 π^2 ̄h^3 vrel

∫

dΩfμkf

∣

∣

∣V ̃(∆)~

∣

∣

∣

2

=

1

4 π^2 ̄h^4

∫

dΩfμ^2

∣

∣

∣V ̃(∆)~

∣

∣

∣

2

dσ

dΩ

=

μ^2

4 π^2 ̄h^4

∣

∣

∣V ̃(∆)~

∣

∣

∣

2

This is a very useful formula for scattering from a weak potential or for scattering at high energy
for problems in which the cross section gets small because the Fourier Transform of the potential
diminishes for large values ofk. It is not good for scattering due to the strong interaction since cross
sections are large and do not typically decrease at high energy. Note that the matrix elements and
hence the scattering amplitudes calculated in the Born approximation are real and therefore do not
satisfy the Optical Theorem. This is a shortcoming of the approximation.

30.1 Scattering from a Screened Coulomb Potential

A standard Born approximation example is Rutherford Scattering,that is, Coulomb scattering of a
particle of chargeZ 1 ein a screened Coulomb potentialφ(r) =Zr^2 ee−r/a. The exponential represents
the screening of the nuclear charge by atomic electrons. Without screening, the total Coulomb
scattering cross section is infinite because the range of the forcein infinite.

The potential energy then is

V(r) =

Z 1 Z 2 e^2

r

e−r/a

We need to calculate its Fourier Transform.

V ̃(∆) =~ Z 1 Z 2 e^2

∫

d^3 re−i

∆~·~re−r/a

r

Since the potential has spherical symmetry, we can choose ∆ to bein thezdirection and proceed
with the integral.

V ̃(∆) =~ Z 1 Z 2 e^22 π

∫∞

0

r^2 dr

∫^1

− 1

d(cosθ)e−i∆rcosθ

e−r/a

r

= Z 1 Z 2 e^22 π

∫∞

0

r^2 dr

[

e−i∆rx

−i∆r

]x=1

x=− 1

e−r/a

r

= Z 1 Z 2 e^2

2 π

i∆

∫∞

0

dr

[

e−i∆r−ei∆r

]

e−r/a

= Z 1 Z 2 e^2

2 π

−i∆

∫∞

0

dr

[

e−(

a^1 +i∆)r

−e−(

(^1) a−i∆)r]