= 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, giving
$$
\psi\sim e^{ikz}+f_k(\theta,\phi)\frac{e^{ikr}}r.
$$
For the Schrödinger <probability current>, $\mathbf j=(\hbar/m){\rm Im}(\psi^*\boldsymbol\nabla\psi)$, the incident flux is $j_{\rm in}=\hbar k/m$. The outgoing scattered radial flux, separated from the incident beam, is $j_{\rm sc,r}\sim(\hbar k/m)|f_k|^2/r^2$. Thus the outgoing rate through $r^2d\Omega$, divided by incident flux, gives
$$
\boxed{\frac{d\sigma}{d\Omega}=\frac{j_{\rm sc,r}r^2}{j_{\rm in}}=|f_k(\theta,\phi)|^2.}
$$
This elastic-scattering result uses the same incoming and outgoing wavenumber and the usual outgoing-wave boundary condition.
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