In the classical experiment, particles whose <impact parameter> lies between $b$ and $b+db$ occupy an incident annulus of area
$$
d\sigma=2\pi b\,|db|.
$$
For a <spherically symmetric potential>, azimuthal symmetry scatters this annulus into the solid-angle strip
$$
d\Omega=2\pi\sin\theta\,|d\theta|.
$$
Thus the <central-potential classical differential cross-section> is
$$
\boxed{\frac{d\sigma}{d\Omega}
=\frac{b}{\sin\theta}
\left|\frac{db}{d\theta}\right|.}
$$
If the relation between $b$ and $\theta$ has several branches, one sums this expression over every branch. The corresponding diagram consists of an incoming line offset by $b$ from the centre, an outgoing asymptote making angle $\theta$ with the incident direction, and the annulus and solid-angle strip just described.

For quantum scattering by a <central potential>, the far-field <wavefunction> is
$$
\psi(\mathbf r)\sim e^{ikz}
+f(\theta)\frac{e^{ikr}}r,
\qquad r\to\infty,
$$
where $f(\theta)$ is the <scattering amplitude>. The incident plane wave has <probability current>
$$
\mathbf J_{\rm in}=\frac{\hbar k}{m}\,\widehat{\mathbf z}.
$$
To leading order in $1/r$, the scattered spherical wave has radial current
$$
\mathbf J_{\rm sc}
=\frac{\hbar k}{m}\frac{|f(\theta)|^2}{r^2}
\,\widehat{\mathbf r}.
$$
The scattered rate through area $r^2d\Omega$ divided by the incident flux is therefore the <quantum differential cross-section from a scattering amplitude>:
$$
\boxed{\frac{d\sigma}{d\Omega}=|f(\theta)|^2.}
$$

Solution

Past exam of the mathematics course of the University of Cambridge / 2019 / ii / Paper 4 / 33B / a / Solution

If
$$
\psi\sim e^{ikz}+f(\theta)\frac{e^{ikr}}r,
$$
then the ratio of scattered radial <probability current> through $r^2d\Omega$ to the incident probability flux is
$$
\frac{d\sigma}{d\Omega}=|f(\theta)|^2.
$$

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