First odd bound-state threshold of a square well ($V_{\mathrm{odd}}={\pi }^2{\hslash }^2/(2m{a}^2)$)

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 $k^2<2m{V}_0/{\hslash }^2$, existence requires $V_0>V_{\mathrm{odd}}$; 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.