 Solution

Far from a short-range potential, the particle is locally free. The imposed incoming boundary condition is a unit-amplitude plane wave travelling along $+z$, while an outgoing free spherical wave has radial dependence $e^{ikr}/r$. The scattering coefficient can depend on direction, givingFor the Schrödinger probability current, $\mathbf{j}=(\hslash /m)\mathrm{Im}({\psi }^{\ast }\nabla \psi )$, the incident flux is $j_{\mathrm{in}}=\hslash k/m$. The outgoing scattered radial flux, separated from the incident beam, is $j_{\mathrm{sc},\mathrm{r}}\sim (\hslash k/m)|f_k{|}^2/r^2$. Thus the outgoing rate through $r^{2}d\Omega$, divided by incident flux, givesThis elastic-scattering result uses the same incoming and outgoing wavenumber and the usual outgoing-wave boundary condition.