Discussion
A particle of mass m is confined in an infinite square well with a shelf of height \(V_{0}\) extending half way across so that :
\[ V(x) = \begin{cases} V_{0} & \text{for } 0 \lt x \lt a/2 \\ 0 & \text{for } a/2 \lt x \lt a \\ \infty & \text{otherwise} \end{cases} \]
Using WKB approximation, the allowed energies \(E_{n}\) (n = 1, 2, 3,....) for this particle are, (given \(E_{n}^{0} = n^{2}\pi^{2}\hbar^{2} / 2ma^{2}\) are the energies of particle in an infinite well with no shelf) :
Answer: D
The WKB energies are \(E_{n}^{0} + \frac{V_{0}}{2}\left(1 + \frac{V_{0}}{8E_{n}^{0}}\right)\).
With infinite walls the WKB condition is \(\int_{0}^{a}p\,dx = n\pi\hbar\). The momentum is \(\sqrt{2m(E - V_{0})}\) over the shelf and \(\sqrt{2mE}\) beyond it, each for a/2. For \(E \gg V_{0}\) expand \(\sqrt{E - V_{0}}\) to second order in \(x = V_{0}/E\):
\[ \begin{aligned} \tfrac{a}{2}\sqrt{2mE}\left(2 - \tfrac{x}{2} - \tfrac{x^{2}}{8}\right) &= n\pi\hbar \\ 2mEa^{2}\left(1 - \tfrac{x}{4} - \tfrac{x^{2}}{16}\right)^{2} &= n^{2}\pi^{2}\hbar^{2} \\ E\left(1 - \tfrac{x}{2} - \tfrac{x^{2}}{16}\right) &= E_{n}^{0} \end{aligned} \]
\[ \begin{aligned} E &= E_{n}^{0} + \frac{V_{0}}{2} + \frac{V_{0}^{2}}{16E_{n}^{0}} \\ &= E_{n}^{0} + \frac{V_{0}}{2}\left(1 + \frac{V_{0}}{8E_{n}^{0}}\right) \end{aligned} \]
\(E_{n}^{0} + V_{0}/2\), the average of the shelf over the well, is only the first-order result, \(E_{n}^{0} + V_{0}\) would raise the whole floor, and the factor \(V_{0}/4E_{n}^{0}\) doubles the second-order term.
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