= Solution
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 $k=\sqrt{2mE}/\hbar$ and $\kappa=\sqrt{2m(V_0-E)}/\hbar$. The hard-wall condition and decay select
$$
\chi(x)=\begin{cases}0,&x\leq0,\\C\sin(kx),&0<x<a,\\C\sin(ka)e^{-\kappa(x-a)},&x\geq a.\end{cases}
$$
<Continuity> 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 $k\cos(ka)=-\kappa\sin(ka)$, giving
$$
\boxed{E=\frac{\hbar^2k^2}{2m},\qquad\tan(ka)=-\frac{k}{\sqrt{2mV_0/\hbar^2-k^2}}}.
$$
The 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.
\Image[/past-exam-of-the-mathematics-course-of-the-university-of-cambridge/2002/ib/paper-1-half-line-ground-state.png]
{title=Nodeless <bound state> in a finite well with a hard wall}
To see the precise existence condition, set $z=ka$ and $K=a\sqrt{2mV_0}/\hbar$. 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 $(\sin z-z\cos z)/\sin^2z>0$. Thus the bound ground state exists exactly when $a\sqrt{2mV_0}/\hbar>\pi/2$, or $V_0>\pi^2\hbar^2/(8ma^2)$, 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.
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