= First odd bound-state threshold of a square well
{title2=$V_{\rm odd}=\pi^2\hbar^2/(2ma^2)$}
For a symmetric attractive well of width $a$ and depth $V_0$, the first odd <bound state> obeys $-k\cot(ka/2)=q>0$, where $q$ is the inverse tail <length>. Its lowest branch has $\pi<ka<2\pi$. Since $k^2<2mV_0/\hbar^2$, existence requires $V_0>V_{\rm odd}$; it is also sufficient by continuity of the matching equation on that branch. At equality the exterior solution has zero decay rate and is not square-integrable. Thus a well at or below this threshold has only its even <ground state>.
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