Past exam of the mathematics course of the University of Cambridge / 2019 / ii / Paper 4 / 33B (Applications of Quantum Mechanics)

In the classical experiment, particles whose impact parameter lies between $b$ and $b+db$ occupy an incident annulus of areaFor a spherically symmetric potential, azimuthal symmetry scatters this annulus into the solid-angle stripThus the central-potential classical differential cross-section isIf 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 iswhere $f(\theta )$ is the scattering amplitude. The incident plane wave has probability currentTo leading order in $1/r$, the scattered spherical wave has radial currentThe scattered rate through area $r^{2}d\Omega$ divided by the incident flux is therefore the quantum differential cross-section from a scattering amplitude:

Write $u=kr-l\pi /2$. Sincethe outgoing-to-incoming coefficient ratio is $-S_l$. Hence the Partial-wave S-matrix element and scattering phase shift areThe scattering length from a partial-wave S-matrix is

For the stated S-wave scattering solution, $\tanh (\lambda r)\to 1$ as $r\to \infty$, soUsing the same incoming/outgoing convention,Therefore

The analytic continuation of $S_0$ has a bound-state pole of the scattering amplitude at $k=i\lambda$ in the upper half-plane. Since $E={\hslash }^2{k}^2/(2m)$, this givesTo extract the bound-state wavefunction, multiply the scattering solution by $k-i\lambda$ and take the pole limit:Thus the unnormalized exterior solution isThese data are collected in the Hyperbolic-tangent S-wave exterior solution.