 Solution

For unit incident plane-wave amplitude, define the scattering amplitude by $\psi \sim e^{ikz}+f(\theta )e^{ikr}/r$ at large $r$. Comparing outgoing radial flux with incident flux gives $d\sigma /d\Omega =|f(\theta ){|}^2$. Outside the potential, a real regular radial solution is proportional to $j_{\ell }(kr)\cos {\delta }_{\ell }-n_{\ell }(kr)\sin {\delta }_{\ell }$, with asymptotic phase $\sin (kr-\ell \pi /2+{\delta }_{\ell })/(kr)$. This defines the scattering phase shift modulo $\pi$.

The incident plane-wave expansion supplies coefficient $(2\ell +1)i^{\ell }$ for $j_{\ell }{P}_{\ell }$. To preserve its incoming part, replace $j_{\ell }$ by $e^{i{\delta }_{\ell }}(j_{\ell }\cos {\delta }_{\ell }-n_{\ell }\sin {\delta }_{\ell })$. Writing the asymptotic sine in exponentials shows that its incoming coefficient is unchanged and its outgoing coefficient is multiplied by $e^{2i{\delta }_{\ell }}$. The outgoing difference from the plane wave therefore gives

Put $\rho =ka$, $t=\tan {\delta }_{\ell }$, and abbreviate $j=j_{\ell }(\rho )$, $n=n_{\ell }(\rho )$. Logarithmic-derivative matching gives $a{R}_{\ell }^{\prime }(a)/R_{\ell }(a)=\rho (j^{\prime }-t{n}^{\prime })/(j-tn)$. The supplied asymptotics determine the constant Wronskian: ${\rho }^2(j{n}^{\prime }-j^{\prime }n)=1$. ConsequentlySolving for $t$ givesAt zeros of denominators it is interpreted by limits; the logarithmic-derivative definition itself presumes its stated denominators nonzero.

Near a dominant resonant partial wave, ${\sin }^2{\delta }_{\ell }\approx {\gamma }^2/[(k-k_0{)}^2+{\gamma }^2]$. Neglecting the small other partial waves yieldsThis is a resonant peak of width $2|\gamma |$ in $k$, with the partial-wave angular pattern and maximum allowed ${\sin }^2{\delta }_{\ell }=1$ at resonance. Integrating gives $\sigma \approx 4\pi (2\ell +1){\gamma }^2/[k^2((k-k_0{)}^2+{\gamma }^2)]$. A small background may matter at angular zeros of the dominant wave.