 Solution

Normalize the incident plane wave to unit amplitude. The scattering asymptotics are $\psi (\mathbf{r})\sim e^{ikz}+f(\theta )e^{ikr}/r$; the outgoing coefficient $f$ is the scattering amplitude. The general axisymmetric free solution for $r>0$ isusing Spherical Bessel functions, spherical Neumann functions and Legendre polynomials.

For a real short-range central potential, the exterior radial solution in each partial wave is proportional to $\cos {\delta }_{\ell }{j}_{\ell }-\sin {\delta }_{\ell }{n}_{\ell }$, whose large-$r$ form is $\sin (kr-\ell \pi /2+{\delta }_{\ell })/(kr)$. This defines the scattering phase shift. Multiplying this combination by $(2\ell +1)i^{\ell }{e}^{i{\delta }_{\ell }}$ matches the incoming spherical component of the incident plane wave: the factor $e^{i{\delta }_{\ell }}$ cancels the shifted incoming phase. The outgoing coefficient instead acquires $e^{2i{\delta }_{\ell }}$. Subtracting the plane wave's outgoing component therefore givesFor the barrier's $s$-wave, the regular reduced radial function inside is $u(r)=A\sinh (\kappa r)$, with $\kappa =\sqrt{{\gamma }^2-k^2}$. Outside take $u(r)=B\sin (kr+{\delta }_0)$. Continuity of $u$ and its derivative at $a$ equates logarithmic derivatives:Use the branch with ${\delta }_0\to 0$ as $k\to 0$. The matching equation gives ${\delta }_0=k(\tanh (\gamma a)/\gamma -a)+o(k)$, so the $s$-wave scattering amplitude tends toIts negative is the repulsive spherical barrier scattering length, positive for this repulsive barrier.