= Hyperbolic-tangent S-wave exterior solution
{c}
For the exterior wave
$$
\psi(r)=\frac{e^{-ikr}}r
+\frac{k+i\lambda\tanh(\lambda r)}{k-i\lambda}
\frac{e^{ikr}}r,
$$
the partial-wave scattering matrix is
$$
S_0(k)=-\frac{k+i\lambda}{k-i\lambda}
=\frac{\lambda-ik}{\lambda+ik}.
$$
Thus $\tan\delta_0=-k/\lambda$, the scattering length is $1/\lambda$, and the pole at $k=i\lambda$ represents a bound state of energy $-\hbar^2\lambda^2/(2m)$ whose exterior wavefunction is proportional to $\operatorname{sech}(\lambda r)/r$.
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