 Ground state of a finite square well

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.