Source: Cambridge Quantum Mechanics, Example Sheet 1, Question 7; accompanies David Tong’s notes.
Consider a square well potential with $U(x) = -U$ for $|x| < a$
and $U(x) = 0$ otherwise ($U$ is a positive constant). Show that there are no
bound states (normalisable energy eigenfunctions) which satisfy
$\chi(-x) = -\chi(x)$ (i.e.\ which have odd parity) if
\[a^2 U < \frac{(\pi\hbar)^2}{8m}\,.
\tag{1.1}\]
For the state to be bounded, we require $-U<E<0$. Solving the TISE, we find
\[\chi(x) = \begin{cases}
Ae^{\kappa x} & x\leq-a,\\
C\sin(kx)+ D\cos(kx) & |x|<a,\\
Ee^{-\kappa x} & x\geq a,
\end{cases}\tag{1.2}\]
with
\[\kappa = \frac{\sqrt{-2mE}}{\hbar} \qquad \text{and} \qquad k = \frac{\sqrt{2m(U+E)}}{\hbar}.\tag{1.3}\]
For the wavefunction to be odd, we require $-A=E$ and $D=0$. Imposing the boundary conditions, we find
\[\frac{\tan ka}{ka} = -\frac{1}{\kappa},\tag{1.4}\]
which is an implicit transcendental equation for $E$. We can rewrite the equation as
\[k\cot ka = -\kappa(k),\tag{1.5}\]
where
\[\kappa^2 + k^2 = \frac{2mU}{\hbar^2}.\tag{1.6}\]
Figure 1. Graphical solution of the odd-parity bound-state equation $k\cot ka = -\kappa(k)$, where $\kappa = \sqrt{2mU/\hbar^2 - k^2}$. The dashed semicircle ($U < U_c$) does not intersect $k\cot ka$ — no odd bound state exists. The solid semicircle ($U > U_c$) intersects the first branch (dot) — one odd bound state exists. The critical value is $U_c = \pi^2\hbar^2/8ma^2$, at which the circle just reaches $k = \pi/2a$ with $\kappa = 0$.
Since $\cot$ has a greater magnitude of gradient than a circle; for a solution to exist, we require the radius of the circle to be greater than $ka=\pi/2$. That is,
\[\frac{\sqrt{2mU}}{\hbar} > \frac{\pi}{2a} \implies Ua^2 > \frac{\pi^2\hbar^2}{8m}.\tag{1.7}\]