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The incoming and outgoing coefficients give the scattering matrixIt obeys $S(-k)=S(k{)}^{-1}$ and $S(k^{\ast }{)}^{\ast }=S(k{)}^{-1}$, 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 $k\cot \delta =-1/a_s+O(k^2)$: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 one bound state, with $\boxed{E=-{\hslash }^2{\kappa }^2/(2m).}$ 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 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 ${\delta }^{\prime }(k)=-\kappa /(k^2+{\kappa }^2)-\alpha /(k^2+{\alpha }^2)<0$, 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.