Past exam of the short options paper shared by Prelims and Parts A and B of the physics course of the University of Oxford / 2012 / S18 / 2

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.

The spherical potential has a rotationally symmetric Fourier transform, so choose the polar axis of the integration along $\mathbf{q}$. With $q=|{\mathbf{k}}_f-{\mathbf{k}}_i|=2k\sin (\theta /2)$,The first Born approximation therefore givesThere is no azimuthal dependence. At $q=0$, the bracket is $(q{r}_0{)}^3/3+O(q^5)$, so $f_k(0)=-2m{V}_0{r}_0^3/(3{\hslash }^2)$ is finite, not zero. The approximation assumes scattering weak enough for the incident plane wave to replace the distorted wave inside the interaction integral.

Apart from the finite forward limit, amplitude zeros obey $\sin x-x\cos x=0$, or $\tan x=x$, where $x=q{r}_0$. The first positive root is $x_1\simeq 4.49341$. Since the accessible range is $0\le x\le 2k{r}_0$,Thus there are no zeros for $k<k_c$; the first enters at the largest momentum transfer, $\theta =\pi$, when $k=k_c$.

For larger roots, $x_n$ lies just below $a_n=\pi (n+\frac{1}{2})$, $n=1,2,\dots$. Expanding $\tan (a_n-\epsilon )\simeq 1/\epsilon$ gives $x_n=a_n-1/a_n+O(a_n^{-3})$. Keeping its leading term yields the requested diffraction estimateExactly, replace the numerator $\pi (n+\frac{1}{2})$ by the corresponding root $x_n$. The displayed approximation is a large-$q{r}_0$ root approximation: taking $k$ large at fixed low order does not make $x_1$ equal to $3\pi /2$. For example, the exact first angle at high $k$ is approximately $4.49341/(k{r}_0)$, not $4.71239/(k{r}_0)$.