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Use the WKB approximation to find the allowed energies ($E_n$) of an infinite square well with a "shelf", of height $V_0$ extending half-way across:

$$V(x)=V_0 \quad , \text{if} \quad 0<x<a/2$$ $$V(x)=0 \quad , \text{if} \quad a/2<x<a$$ $$V(x)=\infty \quad , \text{otherwise}$$

This is what I did:

For the region $0<x<a/2$: $$\phi (x)=\frac{1}{\hbar}\int_0^{a/2}p(x)dx=n\pi$$ $$\frac{ap}{2}=n\pi \hbar$$

$p=\sqrt{2m(E-V_0)}$, so solving for $E$ yields:

$$E=\frac{2n^2\pi ^2 \hbar ^2}{ma^2}+V_0$$

For the region $a/2<x<a$: $$E=\frac{2n^2\pi ^2 \hbar ^2}{ma^2}$$

So then I said that we can't have 2 different allowed energies defining the entire potential, so I summed them up.

$$E_n = \frac{4n^2\pi ^2 \hbar ^2}{ma^2} + V_0$$

$$=8E_n^0 + V_0$$

where $E_n^0 = \frac{n^2\pi ^2 \hbar ^2}{2ma^2}$

...but the given answer is

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

Why isn't it correct to simply add the energies like I did?

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Hints:

  1. OP apparently thinks of the potential as two half-width infinite wells and add the two energy spectra. OP this way gets higher energy levels than the energy levels for the two individual half-width wells. This method and result are incorrect. In fact, in reality, the extra space lowers the energy levels.

  2. The important notion is the length $$\ell(V) ~=~ \frac{a}{2}\theta(V)+ \frac{a}{2}\theta(V-V_0)\tag{A}$$ of the classically accessible position region.

  3. As explained in my Phys.SE answer here, the number $n$ of bound states below energy level $E$ (in the WKB approximation) is

$$ n~\approx~ \frac{\sqrt{2m}}{h}\int_{\min(0,V_0)}^E \frac{\ell(V)~dV}{\sqrt{E-V}} ~\stackrel{(A)}{=}~\frac{\sqrt{2m}}{h}a \left(\sqrt{E}+ \sqrt{E-V_0}\right).\tag{B}$$

  1. From eq. (B) we deduce that $$ 2\sqrt{E_0}~\stackrel{(B)}{=}~\sqrt{E}+ \sqrt{E-V_0},\tag{C}$$ where $E_0$ denotes the energy levels for the system without the shelf $V_0=0$. (We have here suppressed the index $n$ from the notation.)

  2. Rearrange eq. (C) to derive the sought-for formula: $$ E~\stackrel{(C)}{=}~\left(\sqrt{E_0}+\frac{V_0}{4\sqrt{E_0}}\right)^2~=~E_0+ \frac{V_0}{2}+\frac{V_0^2}{16E_0}.\tag{D}$$

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