= Solution
The incoming and outgoing coefficients give the <scattering matrix>
$$
\boxed{S(k)=\frac{g(k)}{g(-k)}=\frac{(k+i\kappa)(k+i\alpha)}{(k-i\kappa)(k-i\alpha)}.}
$$
It obeys $S(-k)=S(k)^{-1}$ and $S(k^*)^*=S(k)^{-1}$, so $|S(k)|=1$ for real $k$, as required by elastic <unitarity>. With $S=e^{2i\delta}$, the <scattering phase shift> satisfies $\tan\delta=k(\kappa+\alpha)/(k^2-\kappa\alpha)$. The inverse tangent needs a continuous branch: taking $\delta(\infty)=0$ gives $\delta(0)=\pi$. Equivalently $\delta=\arctan(\kappa/k)+\arctan(\alpha/k)$ for $k>0$.
The <scattering length> follows from $k\cot\delta=-1/a_s+O(k^2)$:
$$
\boxed{a_s=\frac{\kappa+\alpha}{\kappa\alpha}.}
$$
For a putative bound state put $k=i\eta$, $\eta>0$. The incoming term grows as $e^{\eta r}$, so its coefficient must vanish. The incoming <Jost function> is $F(k)=g(-k)=(k-i\kappa)/(k+i\alpha)$; its only upper-half-plane zero is $i\kappa$. Under the regular-potential assumption this gives \b[one bound state], with $\boxed{E=-\hbar^2\kappa^2/(2m).}$ The pole at $i\alpha$ in $S$ instead comes from the outgoing numerator: it is a <redundant pole of a scattering matrix>, not a second normalizable state. The S-matrix alone, without its incoming/outgoing analytic structure, would tempt an incorrect count of two states.
For completeness, the nonsingular Eckart realization of this Jost function requires $\alpha>\kappa$, as discussed in https://dipot.ulb.ac.be/dspace/bitstream/2013/373712/4/2306.12216.pdf. The supplied positive constants alone do not establish such a realization in every ordering; a coincident pole is likewise not evidence for two states.
There is no positive-energy <scattering resonance>. Indeed $\delta'(k)=-\kappa/(k^2+\kappa^2)-\alpha/(k^2+\alpha^2)<0$, and the s-wave total <partial-wave total scattering cross-section> is $4\pi(\kappa+\alpha)^2/[(k^2+\kappa^2)(k^2+\alpha^2)]$, strictly decreasing for $k>0$. All poles are on the imaginary axis, with no oscillatory decaying resonance pole.
Back to article page
