Past exam of the mathematics course of the University of Cambridge / 2020 / ii / Paper 3 / 34C (Applications of Quantum Mechanics)

For a real short-range spherically symmetric potential, the stationary wavefunction must solve the Schrodinger equation, be regular at the origin whenever the potential is finite there, and satisfy the scattering boundary conditionHere $e^{ikz}$ is the incident plane wave, the second term is an outgoing spherical wave, and its coefficient $f(\theta )$ is the scattering amplitude.

Comparing the stated asymptotic expression with the partial-wave expansion of a scattering amplitude giveswhere $P_l$ is a Legendre polynomial. The outgoing-to-incoming coefficient in channel $l$ isPartial-wave unitarity requiresor equivalentlyThe real scattering phase shift ${\delta }_l$ is defined modulo $\pi$ by

For S-wave scattering, write the radial wavefunction as $R_0(r)=u(r)/r$. In the attractive shell $a<r\le 2a$, the hard-core boundary condition $u(a)=0$ givesOutside the potential, chooseContinuity of the logarithmic derivative at $r=2a$ yieldsPut $C=\kappa \cot (\kappa a)$. The tangent addition formula givesand solving for $\tan {\delta }_0$ gives

LetUsing $\sin (2ka)=2ka+O(k^3)$ and $\cos (2ka)=1+O(k^2)$ in the phase-shift formula givesBy the definition of the scattering length from a partial-wave S-matrix,Thus $a_s$ diverges exactly whenAt each such value, an S-wave bound state lies at the continuum threshold: the exterior zero-energy radial solution becomes constant, its logarithmic derivative vanishes, and the system has a zero-energy scattering resonance. As the well deepens through the critical value, a new bound state appears and the scattering length passes through infinity and changes sign.