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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.

For a bound state, $0<E<U_0$: the Hamiltonian operator is nonnegative, and an energy at or above the exterior potential does not give a nonzero normalizable wavefunction. The time-independent Schrödinger equation and the hard wall at zero therefore giveContinuity of the wavefunction and its derivative at the finite step gives $k\cot (ka)=-\kappa$. Put $z=ka$ and $z_0=a\sqrt{2m{U}_0}/\hslash$. The matching equation becomesIt requires $\cot z<0$, so solutions lie in the intervals $((j+1/2)\pi ,(j+1)\pi )$ for $j=0,1,\dots$. Squaring, with that sign restriction retained, gives $z_0=z/|\sin z|$. On each such interval this function increases strictly from $(j+1/2)\pi$ to infinity: its derivative is $(|\sin z|-z\mathrm{sgn}(\sin z)\cos z)/{\sin }^{2}z>0$, because the second term in the numerator is positive there. Each interval therefore contributes exactly one bound state if and only if $z_0>(j+1/2)\pi$. These are the bound-state thresholds for a square well with one hard wall.

Exactly one normalizable bound state exists if and only ifThe printed strict upper inequality misses the threshold case. At the upper endpoint, the prospective second state has $E=U_0$ and $\kappa =0$. Its exterior solution is constant or linear, hence is not square-integrable unless zero; the zero exterior solution and matching would force the entire solution to vanish. Thus there is still exactly one bound state at that endpoint. The strict upper inequality is valid if threshold values are additionally excluded, but that exclusion is not part of the stated assumptions.