 Solution

For a short-range potential, an incident plane wave of unit amplitude and wave vector $k{\mathbf{e}}_z$ has the large-distance scattering wavefunctionThe coefficient $f$ is the scattering amplitude. For a central potential it depends only on the scattering angle $\theta$. The incident probability current is $\hslash k/m$, while the scattered radial current is $(\hslash k/m)|f{|}^2/r^2$ to leading order. Their ratio givesThese are the differential scattering cross-section and total cross section for elastic scattering with the same incident and outgoing wave number.

The expression is asymptotic: it specifies the outgoing wave's leading $1/r$ term, not its finite-distance radial corrections or its form inside the interaction region. In particular, for angle-dependent $f$, the spherical term by itself has an angular-Laplacian contribution of order $r^{-3}$ to the free wave equation, which is cancelled by subleading terms in the exact solution.