# Normalized probability distribution from the Coulomb/Rutherford scattering amplitude?

My question appears elementary, but I have been pretty vexed trying to answer it precisely. Can one use the Rutherford/Coulomb scattering amplitude to get a finite, normalized momentum-space distribution for the outgoing particle state?

Specifically, consider an incoming wave packet $\psi_{in}(\vec{k})$ which is, let's say, a Gaussian in momentum $\vec{k}$ centered around some $\vec{k}_0$. Say this is a charged particle and it scatters off a classical Coulomb potential $V(r) = \alpha/r$. Naively, one would write the outgoing wavefunction as something like

$\psi(\vec{k'}) = \psi_{in}(\vec{k'}) + i \int d^3\vec{k} \ f(\vec{k'},\vec{k}) \delta(E_{k'}-E_k) \psi(\vec{k})$

where the amplitude depends on the angle $\theta$ between $\vec{k}$ and $\vec{k'}$ as

$f(\vec{k'},\vec{k}) \sim 1/(1-\cos \theta)$.

As is well known, this thing diverges as $\vec{k} \to \vec{k'}$, so the integral blows up there, so the total cross section is divergent, and so on. That's fine. But one could ask for eg. the probability distribution

$dP(\vec{k'}) = |\psi(\vec{k'})|^2 d^3\vec{k'} \ \ \ \ (*)$

but it appears to be singular at every $\vec{k'}$ that's in the support of the incoming wavefunction, since at each such value the forward part of the amplitude diverges.

Is there a way to get around this and produce a finite distribution? What's particularly confusing is that one can actually invoke the optical theorem and show that, formally, $(*)$ integrates over all $\vec{k'}$ to unity, but on any finite region of $\vec{k'}$-space, the integral looks divergent!

(As a special case, consider the incoming packet to be a delta function $\psi_{in}(\vec{k}) = \delta(\vec{k} - \vec{k_0})$. Then the probability to scatter into any solid angle that doesn't include the forward direction $\theta = 0$ is manifestly finite, and you can use the optical theorem to write the probability for scattering into a small solid angle $\delta \Omega$ which includes the forward direction as $P = 1 - \int_{S^2 / \delta \Omega} d\Omega \frac{d \sigma}{d \Omega}$, i.e. as 1 - (the probability to scatter into any other angle). But if instead we consider an incoming wavepacket that has some finite support, this kind of trick doesn't seem to work...)

## 1 Answer

Just in case anyone was left hanging here, I think I'm satisfied for the time being by the "Coulomb wave functions". These things are exact, normalizable solutions to the Schrodinger equation with a plane-wave boundary condition. They aren't exactly of the form given above here, with a scattering amplitude in them, but you can form initial wavepackets and get well-defined answers that take forward scattering into account.