Ground state of a finite square well ($k\tan (ka/2)=\kappa$)

For $V(x)=0$ on $0<x<a$ and $V(x)=V_w>0$ outside, the lowest bound state is even about $a/2$, has no nodes, and has $0<E<V_w$. Write $k=\sqrt{2mE}/\hslash$ and $\kappa =\sqrt{2m(V_w-E)}/\hslash$. The interior wavefunction is proportional to $\cos [k(x-a/2)]$; the exterior tails decay exponentially. Wavefunction matching at a finite potential step gives the displayed condition. Its lowest root has $0<ka<\pi$, so $E<{\pi }^2{\hslash }^2/(2m{a}^2)$, the infinite square well ground state energy. As $V_w$ increases, the penetration length $1/\kappa$ tends to zero and that limiting energy is approached from below.

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 $2a{V}_0$.