# Why can't I add the energies in this WKB approximation example to get the allowed energies for the given potential?

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)=\begin{cases} V_0 &, \text{ if} \quad 0

This is what I did:

For the region $$0: $$\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 $$a/2: $$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 $$E_n^0 = \frac{n^2\pi ^2 \hbar ^2}{2ma^2}$$

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

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

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: $$E~\stackrel{(C)}{=}~\left(\sqrt{E_0}+\frac{V_0}{4\sqrt{E_0}}\right)^2~=~E_0+ \frac{V_0}{2}+\frac{V_0^2}{16E_0}.\tag{D}$$