 Solution

The stationary Schrödinger equation is $-{\hslash }^2{\psi }^{\prime \prime }/(2m)+V\psi =E\psi$. A nonzero bound-state function is square-integrable on the real line and can be scaled to ${\int }_{\mathbb{R}}|\psi {|}^{2}dx=1$. Odd parity means $\psi (-x)=-\psi (x)$.

Set $k=\sqrt{2m(V_0+E)}/\hslash$ and $\kappa =\sqrt{-2mE}/\hslash$, both positive. Oddness and decay giveContinuity at $a$ gives $B=A\sin (ka)$, and derivative continuity gives $-\kappa B=Ak\cos (ka)$. A nonzero solution cannot have $\sin (ka)=0$, since that would force both the value and derivative there to vanish. Dividing therefore yields $\kappa =-k\cot (ka)$. With $\eta =\kappa a$ and $\xi =ka$,The matching at $-a$ follows automatically by odd parity.

In the required sketch, the second equation is a first-quadrant circle arc of radius $R=\sqrt{2m{V}_0{a}^2}/\hslash$. The curve $-\xi \cot \xi$ starts with limiting value $-1$ at zero, rises to zero at $\pi /2$, and then rises to $+\infty$ as $\xi \uparrow \pi$. Thus a positive-$\eta$ intersection in this range must have $\xi >\pi /2$. The accompanying plot shows the circle below threshold and one above it.

Consequently $R<\pi /2$ rules out all odd states: the first positive branch already requires that radius, and later branches require still larger $\xi$. This is precisely $\boxed{2m{V}_0<(\pi \hslash /(2a){)}^2}$. At equality the intersection has $\eta =0$ and would not decay at infinity, so it too is not a normalizable bound state. An odd state first appears only for a strict increase beyond threshold.