= Solution
Write $u=kr-l\pi/2$. Since
$$
A_l\sin u-B_l\cos u
=\frac{-iA_l-B_l}{2}e^{iu}
+\frac{iA_l-B_l}{2}e^{-iu},
$$
the outgoing-to-incoming coefficient ratio is $-S_l$. Hence the <Partial-wave S-matrix> element and <scattering phase shift> are
$$
\boxed{
S_l(k)=\frac{A_l(k)-iB_l(k)}{A_l(k)+iB_l(k)}
=e^{2i\delta_l(k)},
\qquad
\tan\delta_l(k)=-\frac{B_l(k)}{A_l(k)}.}
$$
The <scattering length from a partial-wave S-matrix> is
$$
\boxed{
a_s=-\lim_{k\to0}\frac{\tan\delta_0(k)}k
=\lim_{k\to0}\frac{B_0(k)}{kA_0(k)}.}
$$
For the stated <S-wave scattering> solution, $\tanh(\lambda r)\to1$ as $r\to\infty$, so
$$
\psi(r)\sim\frac{e^{-ikr}}r
+\frac{k+i\lambda}{k-i\lambda}\frac{e^{ikr}}r.
$$
Using the same incoming/outgoing convention,
$$
\boxed{
S_0(k)=-\frac{k+i\lambda}{k-i\lambda}
=\frac{\lambda-ik}{\lambda+ik}.}
$$
Therefore
$$
\boxed{
\delta_0(k)=-\tan^{-1}\frac{k}{\lambda}\pmod\pi,
\qquad
a_s=\frac1\lambda.}
$$
The <analytic continuation> of $S_0$ has a <bound-state pole of the scattering amplitude> at $k=i\lambda$ in the upper half-plane. Since $E=\hbar^2k^2/(2m)$, this gives
$$
\boxed{E_{\rm bound}=-\frac{\hbar^2\lambda^2}{2m}.}
$$
To extract the bound-state wavefunction, multiply the scattering solution by $k-i\lambda$ and take the pole limit:
$$
\begin{aligned}
\psi_{\rm bound}(r)
&\propto i\lambda[1+\tanh(\lambda r)]e^{-\lambda r}/r\\
&\propto \frac{\operatorname{sech}(\lambda r)}r.
\end{aligned}
$$
Thus the unnormalized exterior solution is
$$
\boxed{\psi_{\rm bound}(r)\propto
\frac{\operatorname{sech}(\lambda r)}r,\qquad r>r_0.}
$$
These data are collected in the <Hyperbolic-tangent S-wave exterior solution>.
Back to article page
