1
$\begingroup$

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?

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Browse other questions tagged or ask your own question.