= Z 1 Z 2 e^22 π
−i∆[
−
e−((^1) a+i∆)r
1
a+i∆
+
e−((^1) a−i∆)r
1
a−i∆
]∞
0
= Z 1 Z 2 e^22 π
−i∆[
1
1
a+i∆−
1
1
a−i∆]
= Z 1 Z 2 e^22 π
−i∆[ 1
a−i∆−1
a−i∆
1
a^2 + ∆
2]
= Z 1 Z 2 e^22 π
−i∆[
− 2 i∆
1
a^2 + ∆2]
=
4 πZ 1 Z 2 e^2
1
a^2 + ∆
2Since∆ =~ ~kf−~ki, we have ∆^2 =k^2 f+ki^2 − 2 kfkicosθ. For elastic scattering, ∆^2 = 2k^2 (1−cosθ).
The differential cross section is
dσ
dΩ=
μ^2
4 π^2 ̄h^4∣
∣
∣V ̃(∆)~
∣
∣
∣
2=
μ^2
4 π^2 ̄h^4∣
∣
∣∣^4 πZ^1 Z^2 e2
1
a^2 + 2k(^2) (1−cosθ)
∣
∣
∣∣
2=
∣
∣
∣
∣
∣
μZ 1 Z 2 e^2
̄h^2
2 a^2 +p(^2) (1−cosθ)
∣
∣
∣
∣
∣
2=
∣
∣
∣
∣
∣
Z 1 Z 2 e^2
̄h^2
2 μa^2 + 4Esin(^2) θ
2
∣
∣
∣
∣
∣
2In the last step we have used the non-relativistic formula for energy and 1−cosθ=^12 sin^2 θ 2.
The screened Coulomb potential gives a finite total cross section. It corresponds well with the
experiment Rutherford did in whichαparticles were scattered from atoms in a foil. If we scatter
from a bare charge where there is no screening, we can take the limitin whicha→∞.
∣
∣
∣
∣
∣Z 1 Z 2 e^2
4 Esin^2 θ 2∣
∣
∣
∣
∣
2The total cross section diverges in due to the region around zero scattering angle.
30.2 Scattering from a Hard Sphere
Assume a low energy beam is incident upon a small, hard sphere of radiusr 0. We will assume that
̄hkr 0 < ̄hso that only theℓ= 0 partial wave is significantly affected by the sphere. As with the
particle in a box, the boundary condition on a hard surface is that the wavefunction is zero. Outside
the sphere, the potential is zero and the wavefunction solution willhave reached its form for large
r. So we set
(
e−i(kr^0 −ℓπ/2)−e^2 iδℓ(k)ei(kr^0 −ℓπ/2)
)
=
(
e−ikr^0 −e^2 iδ^0 (k)eikr^0)
= 0
e^2 iδ^0 (k)=e−^2 ikr^0