Past exam of the mathematics course of the University of Cambridge / 2002 / ib / Paper 1 / 9D (Quantum Mechanics)

A normalizable bound state has $0<E<V_0$: the potential is nonnegative, while exponential decay is needed on the infinite right-hand region. Set $k=\sqrt{2mE}/\hslash$ and $\kappa =\sqrt{2m(V_0-E)}/\hslash$. The hard-wall condition and decay selectContinuity at $a$ has already been used. Since the finite step contains no delta interaction, integrating the Schrödinger equation across it also requires derivative continuity. Thus $k\cos (ka)=-\kappa \sin (ka)$, givingThe ground-state solution uses the first branch $\pi /2<ka<\pi$ and has no internal node. It starts at zero, rises to one maximum inside the well, decreases to a positive value at $a$, and joins smoothly onto a decaying exponential tail.

To see the precise existence condition, set $z=ka$ and $K=a\sqrt{2m{V}_0}/\hslash$. The first-branch matching equation gives $K=z/\sin z$ on $\pi /2<z<\pi$. This increases strictly from $\pi /2$ to infinity, since its derivative is $(\sin z-z\cos z)/{\sin }^{2}z>0$. Thus the bound ground state exists exactly when $a\sqrt{2m{V}_0}/\hslash >\pi /2$, or $V_0>{\pi }^2{\hslash }^2/(8m{a}^2)$, as in the bound-state thresholds for a square well with one hard wall. At equality $\kappa =0$ and the exterior solution is not square-integrable. Below it there is no normalizable energy eigenfunction of the requested ground-state type. The sketch assumes the binding condition and uses one representative solution, not a specified numerical well depth.