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For small fixed positive $E$, the stated approximate curve has maxima atIts formal minima are zero at $\kappa a=n\pi$. The peak height is the $S$-wave unitarity limit, and the peaks narrow as $E/V_0$ decreases.

To identify the bound-state threshold, let the bound-state energy be $-\epsilon$ with $\epsilon >0$. The reduced radial wave is proportional to $\sin (Kr)$ inside and $e^{-\chi r}$ outside, with $K=\sqrt{2m(V_0-\epsilon )}/\hslash$ and $\chi =\sqrt{2m\epsilon }/\hslash$. Matching gives $K\cot (Ka)=-\chi$. As a new bound state emerges at $\epsilon \downarrow 0$,Thus the limiting peak locations coincide with the appearance of successive $S$-wave bound states. At the exact threshold the zero-energy radial solution is not square-integrable; a normalizable bound state appears just beyond it. Finite-energy maxima shift slightly from the zero-energy positions.

The formal minima require care because the assumption used in part d fails there. Retaining $ka$ in the exact matching condition gives the spherical square-well scattering lengthaway from a threshold. At $\kappa a=n\pi$, $a_s=a$, so the true low-energy S-wave scattering contribution is $4\pi a^2$, not zero. Genuine leading-order zeros of that contribution instead satisfy $\tan q=q$, for example $q\simeq 4.493409$ and $7.725252$. The original sketch compares the exact finite-energy $S$ wave with the requested approximation, making this nonuniformity visible. Dominance by the $S$ wave must be checked separately near a higher-partial-wave threshold. The scattering-length divergence and its bound-state interpretation are also derived in institutional lecture notes.^[ref]

For binding energy $\epsilon >0$, match the inner reduced radial wave $\sin (Kr)$ to an outer decaying exponential. The resulting equation is $K\cot (Ka)=-\chi$, with $K^2=2m(V_0-\epsilon )/{\hslash }^2$ and ${\chi }^2=2m\epsilon /{\hslash }^2$. A bound level reaches zero energy when $\kappa a=(n+1/2)\pi$. These are divergences of the spherical square-well scattering length. The exact zero-energy threshold wave is not square-integrable; a normalizable bound state appears on the more attractive side.