 Solution

For a finite step in a one-dimensional potential and constant mass, the stationary Schrödinger equation requires continuity of the wavefunction and its first derivative. A discontinuous derivative would introduce a delta-function term in its second derivative, unsupported by a merely finite step. A discontinuous wavefunction is still more singular. A delta-function potential has a different derivative jump condition; an infinite wall instead imposes zero wavefunction and does not require matching to an exterior zero derivative.

For a bound state of the half-well, $-V_0<E<0$. The wall at zero excludes a cosine term, and square integrability excludes exponential growth outside. WriteContinuity at $a$ gives $A\sin ka=B$ and $Ak\cos ka=-qB$. HencePut $z=ka$, $w=qa$, and $\nu =a\sqrt{2m{V}_0}/\hslash$. Matching becomes $z\cot z=-w$ with $z^2+w^2={\nu }^2$, $w>0$. The negative branches of $z\cot z$ occur on $((n-1/2)\pi ,n\pi )$. There is one bound-state intersection in the $n$th interval exactly whenAt equality the new level has $q=0$, so its outside tail is constant and not normalizable: it is a threshold state, not a bound state. Thus no bound state exists for $\nu \le \pi /2$, including the strict inequality asked in the question.

Exactly two negative-energy bound states require the second threshold to have been crossed and the third not crossed:The lower endpoint is excluded and the upper endpoint included precisely because a zero-energy threshold state does not count as bound. This half-line finite-well bound-state count differs from an unrestricted symmetric one-dimensional well, which has no hard wall at its centre.