Bound-state thresholds for a square well with one hard wall (${\mathcal{U}}_j={\hslash }^2{\pi }^2(j+1/2{)}^2/(2m{a}^2)$)

For a zero-potential interval of width $a$, an infinite wall at its left end, and constant exterior potential $U_0$ to its right, the $j$th bound state branch exists precisely when $U_0>{\mathcal{U}}_j$, for $j=0,1,\dots$. Matching the wavefunction and its derivative gives $z\cot z=-\sqrt{z_0^2-z^2}$ with $z_0=a\sqrt{2m{U}_0}/\hslash$. At equality the new threshold solution is not normalizable, so exactly one bound state corresponds to $U_0\in ({\mathcal{U}}_0,{\mathcal{U}}_1]$.