Past exam of the mathematics course of the University of Cambridge / 2010 / ii / Paper 2 / 34B / Solution

The asymptotic scattering wavefunction is $\Psi \sim e^{ikz}+f(\theta )e^{ikr}/r$. For a real central potential, the scattering phase shift ${\delta }_l$ multiplies the outgoing component of each partial wave relative to the incoming one by $e^{2i{\delta }_l}$. Matching the incident-wave normalization givesSubtract the given incident plane wave expansion. Since $e^{2i{\delta }_l}-1=2i{e}^{i{\delta }_l}\sin {\delta }_l$, the scattering amplitude isOrthogonality ${\int }_{-1}^1{P}_l{P}_{j}dx=2{\delta }_{lj}/(2l+1)$ eliminates cross terms in $\sigma =\int |f{|}^{2}d\Omega$, giving $\boxed{\sigma =(4\pi /k^2)\sum _l(2l+1){\sin }^2{\delta }_l}$.

For the spherical square well, the regular reduced radial $l=0$ wave is $A\sin (\kappa r)$ inside and $B\sin (kr+{\delta }_0)$ outside, with ${\kappa }^2=k^2+{\gamma }^2$. Continuity of the wave and its derivative gives the logarithmic-derivative matching relation $\kappa \cot (\kappa a)=k\cot (ka+{\delta }_0)$, orAt zeros of the denominators the original matching relation supplies the limiting interpretation.

Under the stated deep-well approximation, put $x=ka$ and $B_0=\tan (\gamma a)/(\gamma a)$. The condition ${\delta }_0\in \pi \mathbb{Z}$ is $\tan x/x=B_0$. If $B_0<0$, its first positive solution lies in $(\pi /2,\pi )$; if $B_0=0$, it is $\pi$. For $0<B_0\le 1$, a solution lies in $(\pi ,3\pi /2)$ because there $\tan x/x$ rises continuously from zero to infinity. If $B_0>1$, the first positive solution lies in $(0,\pi /2)$, where $\tan x/x$ rises from one to infinity. The limiting pole case gives $x=\pi /2$ by the cotangent relation. Thus $\boxed{0<k_0<3\pi /(2a)}$, without calculating the integer phase index. The use of $\kappa \simeq \gamma$ presumes the located wave numbers remain small compared with $\gamma$.