# WKB approximation to find energy levels of a step potential

Suppose the following potential: $$V(x) = \begin{cases}V_0 & 0<x<\frac{a}{2} \\ 0 & \frac{a}{2}<x<a \\ \infty & \text{otherwise} \end{cases}$$ Also, assume that for every $n$, $$V_0 < \frac{n^2\pi^2\hbar^2}{2ma^2}$$ but $$V_0 \not\ll \frac{n^2\pi^2\hbar^2}{2ma^2}$$ Using the WKB approximation, find the energy levels of the potential

In my quantum theory class, we showed that under the WKB approximation $$\psi(x) \approx A\cos\phi(x) + B\sin\phi(x)$$ where $$\phi(x) = \frac{1}{\hbar}\int\limits_0^xp(x)dx,\ p(x) = \sqrt{2m(E-V(x))}$$ The given condition about $V_0$ implies something, which I have yet to find out.

If $E <V_0$, we demand nullification of the wavefunction at $0,a$, since tunneling is impossible. From this, we get that the cosine vanishes, and $$\phi(a) = \frac{1}{\hbar}\int\limits_0^ap(x)dx = \pi n$$ $$\hbar\pi n = \frac{a}{2}\sqrt{2m(E-V_0)} + \frac{a}{2}\sqrt{2mE} \Rightarrow \frac{2\hbar^2\pi^2n^2}{ma^2} = \left(\sqrt{E-V_0} + \sqrt{E}\right)^2$$ From this, $E$ can be isolated.

How can I continue for $E < V_0$? What is the condition on $V_0$ telling us?