 Solution

A travelling wave, for example $A\cos (kx-\omega t)$, has moving constant-phase surfaces. A standing wave, for example $2A\sin (kx)\cos (\omega t)$, has stationary nodes and can be formed from two equal counterpropagating waves.

For the half-line finite well with $E>V_0$, define $k=\sqrt{2mE}/\hslash$ and $q=\sqrt{2m(E-V_0)}/\hslash$. The hard wall requires $\psi =0$ for $x<0$ and $\psi (0)=0$, so the general allowed forms areAt the finite step both $\psi$ and ${\psi }^{\prime }$ are continuous:There is no requirement to match the derivative to zero across the infinite wall.

For a bound state $0<E<V_0$, put $\kappa =\sqrt{2m(V_0-E)}/\hslash >0$. Normalizability excludes the growing exponential, leaving $A\sin (kx)$ inside and $D{e}^{-\kappa (x-a)}$ outside. Matching their logarithmic derivatives givesThe exterior expression is equivalent to $C{e}^{-\kappa x}$ by rescaling its constant. Negative cotangent first occurs for $\pi /2<ka<\pi$; all higher allowed intervals also have $ka>\pi /2$. Thus ${\lambda }_1=2\pi /k<4a$. Since $k<\sqrt{2m{V}_0}/\hslash$, a normalizable bound state requiresThis is also sufficient: on the lowest negative-cotangent branch, the two matching sides cross whenever $a\sqrt{2m{V}_0}/\hslash >\pi /2$. The printed non-strict inequality includes the threshold limit. At equality $E=V_0$ and the exterior is not square-integrable, so it is not itself a bound state.

For the helium mass and energy scale given, the limiting width isActual binding requires a width just larger than this threshold in the stated ideal potential.