Question

Use the WKB approximation to find the allowed energies $\left(E_n\right)$of an infinite square well with a “shelf,” of height${\mathit{V}}_{\mathbf{0}}$, extending half-way across

role="math" localid="1658403794484" $\mathbf{V}\mathbf{\left(}\mathbf{x}\mathbf{\right)}\mathbf{=}\mathbf{\{}\begin{array}{cc}{\mathbf{v}}_{\mathbf{0}\mathbf{,}}& \left(0<x<a/2\right)\\ \mathbf{0}\mathbf{,}& \left(a/2<x<a\right)\\ \mathbf{\infty }\mathbf{,}& \left(\mathrm{otherwise}\right)\end{array}$

Express your answer in terms ofrole="math" localid="1658403507865" ${\mathbf{V}}_{\mathbf{0}}\mathbf{}\mathbf{and}\mathbf{}{\mathbf{E}}_{\mathbf{n}}^{\mathbf{0}}\mathbf{\equiv }{\left(\mathrm{n\pi ħ}\right)}^{\mathbf{2}}\mathbf{/}\mathbf{2}{\mathbf{ma}}^{\mathbf{2}}$(the nth allowed energy for the infinite square well with no shelf). Assume that, but do not assume that ${\mathit{E}}_{\mathbf{1}}^{\mathbf{0}}\mathbf{>}{\mathit{V}}_{\mathbf{0}}$. Compare your result with what we got in Section 7.1.2, using first-order perturbation theory. Note that they are in agreement if either${\mathit{V}}_{\mathbf{0}}$is very small (the perturbation theory regime) or n is very large (the WKB—semi-classical—regime).

Short answer

The allowed energies of an infinite square well with a “Shelf” of height $$\begin{gathered} V_{0}is \\ {E}_n=E_n^0+\frac{V_0}{2}+\frac{V_0^2}{16E_n^0} \end{gathered}$$

Step-by-step solution

Parameters.

$\mathrm{V}\left(\mathrm{x}\right)=\{\begin{array}{cc}{\mathrm{v}}_{0,}& (0<\mathrm{x}<\mathrm{a}/2)\\ 0,& (\mathrm{a}/2<\mathrm{x}<\mathrm{a})\\ \infty ,& \left(\mathrm{otherwise}\right)\end{array}$

$E_n^0\equiv {\left(n\pi ħ\right)}^2/2m{a}^2$

Finding allowed energies (En) of an infinite square well with a “Shelf” of height V0

$$\begin{gathered} P\left(x\right)\mathrm{is}\mathrm{real},\mathrm{so} \\ {\int }_{\mathrm{x}}^0\mathrm{p}\left(\mathrm{x}\right)\mathrm{dx}=\mathrm{n\pi ħ},\mathrm{with}\mathrm{n}=1,2,3,... \\ \mathrm{and}\mathrm{p}\left(\mathrm{x}\right)=2\mathrm{m}\left[\mathrm{E}-\mathrm{V}\left(\mathrm{x}\right)\right] \\ {\int }_{\mathrm{x}}^0\mathrm{p}\left(\mathrm{x}\right)\mathrm{dx}=\mathrm{n\pi ħ} \\ {\int }_{\mathrm{x}}^{\mathrm{a}}\mathrm{p}\left(\mathrm{x}\right)\mathrm{dx}=\sqrt{2\mathrm{mE}}\left(\frac{\mathrm{a}}{2}\right)+\sqrt{2\mathrm{m}\left(\mathrm{E}-{\mathrm{V}}_0\right)}\left(\frac{\mathrm{a}}{2}\right) \\ \mathrm{Here},=\sqrt{2\mathrm{m}}\left(\frac{\mathrm{a}}{2}\right)\left(\sqrt{\mathrm{E}}+\sqrt{\mathrm{E}-{\mathrm{V}}_0}\right) \\ =\mathrm{n\pi ħ} \\ \Rightarrow \mathrm{E}+\mathrm{E}-{\mathrm{V}}_0+2\sqrt{\mathrm{E}\left(\mathrm{E}-{\mathrm{V}}_0\right)}=\frac{4}{2\mathrm{m}}{\left(\frac{\mathrm{n\pi ħ}}{\mathrm{a}}\right)}^2=4{\mathrm{E}}_{\mathrm{n},}^{0.} \\ 2\sqrt{\mathrm{E}\left(\mathrm{E}-{\mathrm{V}}_0\right)}=\left(4{\mathrm{E}}_{\mathrm{n},}^{0.}-2\mathrm{E}+{\mathrm{V}}_0\right) \\ \mathrm{Squaring}\mathrm{again}: \\ 4\mathrm{E}\left(\mathrm{E}-{\mathrm{V}}_0\right)=4{\mathrm{E}}^2-4{\mathrm{EV}}_0 \\ =16{\mathrm{E}}_{\mathrm{n},}^{0.}+4{\mathrm{E}}^2+{\mathrm{V}}_0^2-16{\mathrm{EE}}_{\mathrm{n},}^{0.}+8{\mathrm{E}}_{\mathrm{n},}^{0.}{\mathrm{V}}_0-4{\mathrm{EV}}_0 \\ \Rightarrow 16{\mathrm{EE}}_{\mathrm{n},}^{0.}=16{\mathrm{E}}_{\mathrm{n},}^{0.}+8{\mathrm{E}}_{\mathrm{n},}^{0.}{\mathrm{V}}_0-4{\mathrm{EV}}_0 \\ \Rightarrow {\mathrm{E}}_{\mathrm{n}}={\mathrm{E}}_{\mathrm{n},}^{0.}+\frac{{\mathrm{V}}_0}{2}+\frac{{\mathrm{V}}_0^2}{16{\mathrm{E}}_{\mathrm{n},}^{0.}} \end{gathered}$$

Perturbation theory gave

$E_n=E_n^0+\frac{V_0}{2}$

The extra term goes to zero for very$small{V}_{0,}or\left(\sin ce{E}_n^0~n^2\right)$,for large n.

Thus the allowed energies of an infinite square well with a “Shelf” of height $V_0$is

$E_n=E_n^0+\frac{V_0}{2}+\frac{V_0^2}{16E_n^0}$