QuantumPhysics.dvi

The fact that on the right hand side, the scattering amplitude is being evaluated for equal initial
and final momenta is responsible for the name forward scattering.

To prove this result, we start from the form of the full wave functionψin terms of the incoming
free wave functionφ, and the scattered wave functionχ, related by

ψ=φ+χ (12.58)

The probability current, integrated across a large sphereSmust vanish for this stationary problem
when we take the free partφor the full wave functionψ, leading to the following equations,
∮

S

d^2 S·Im(φ∗∇φ) =

∮

S

d^2 S·Im(ψ∗∇ψ) = 0 (12.59)

The total cross section, on the other hand, is given by the integral of the probability current for
the scattered wave functionχ, as follows,

m|v|

(2π)^3

σtot= Im

∮

S

d^2 S·χ∗∇χ (12.60)

Finally, we use the identity

Im(ψ∗∇ψ) = Im(−φ∗∇φ+ψ∗∇φ−φ∇ψ∗+χ∗∇χ) (12.61)

Putting all together, and using Stokes’ theorem, we have,

Im

∮

S

d^2 S·χ∗∇χ= Im

∫

d^3 x·(ψ∆φ∗−φ∗∆ψ) (12.62)

We now use the Schr ̈odinger equation forφandψrespectively,

∆φ = − 2 mEφ

∆ψ = − 2 mEψ+ 2mV (12.63)

The terms involvingEcancel out, and we are left with

k σtot= 2m(2π)^3 Im

∫

d^3 xφV ψ∗= 4πImf(k,k) (12.64)

The advantage of this formula is that the forward scatteringamplitude is often easier to evaluate
than the full amplitude which then requires squaring to get the cross section and integrating over
all solid angles to get the total cross section.

12.10Spherical potentials and partial wave expansion

Many potentials of interest, such as the Coulomb and Yukawa potentials, have spherical symmetry,
and only depend on a radial coordinater, but not on the spherical anglesθandφ. The spherical