 Shallow finite-square-well energy

For a well of depth $V_0$ and full width $2a$, measured from zero exterior potential energy, even-state matching gives $q=k\tan ka$. If $a\sqrt{m{V}_0}/\hslash \ll 1$, then $ka\ll 1$ and $q\simeq 2m{V}_{0}a/{\hslash }^2$. Using $E_0=-{\hslash }^2{q}^2/(2m)$ gives the displayed leading binding energy. It agrees with the attractive delta-potential result for integrated strength $2a{V}_0$.