= Ground state of a finite square well
{title2=$k\tan(ka/2)=\kappa$}
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 $k=\sqrt{2mE}/\hbar$ and $\kappa=\sqrt{2m(V_w-E)}/\hbar$. 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 $E<\pi^2\hbar^2/(2ma^2)$, 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.
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