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...)


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.


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