Quantum Well Energies

Writing the 1D SE:

$\large{-\alpha \frac{d^2}{dx^2} \Psi = E \, \Psi \\ \frac{d^2}{dx^2} \Psi = -\frac{E}{\alpha} \, \Psi}$

The general solution should be:

$\large{\Psi = A \, e^{j \, \sqrt{E/\alpha} \, x} + B \, e^{-j \, \sqrt{E/\alpha} \, x}}$

Assuming that the infinite potentials outside the well impose the conditions that $\Psi(0) = 0$ and $\Psi(\ell) = 0$ , we have the following.

$\large{0 = A + B \implies B = -A \\ 0 = e^{j \, \sqrt{E/\alpha} \, \ell} - \, e^{-j \, \sqrt{E/\alpha} \, \ell} \\ e^{j \, \sqrt{E/\alpha} \, \ell} = e^{-j \, \sqrt{E/\alpha} \, \ell} }$

By inspection, the last equation is only satisfied when:

$\large{\sqrt{\frac{E}{\alpha}} \, \ell = \pm \, n \, \pi}$

In the above equation, $n$ is an integer. The corresponding energy expression is:

$\large{E = \frac{n^2 \, \pi^2}{\ell^2} \alpha}$

Thus, the boundary conditions on $\Psi$ , combined with the Schrodinger equation, give rise to discrete energy levels in quantum mechanics.

In this case, $\ell = 3$ , so the difference between the two lowest energy levels is:

$\large{E_2 - E_1 = \frac{\pi^2}{9} (2^2 - 1^2) \alpha = \frac{1}{3} \pi^2 \, \alpha}$

Writing $E = \alpha k^2$ we obtain the differential equation $\psi''(x) + k^2\psi(x) \; = \; 0$ together with the boundary conditions $\psi(0) = \psi(3) =0$ . Thus we deduce that $\psi(x) =A\sin kx$ where $\sin3k = 0$ . Thus we deduce that $k = \tfrac13n\pi$ where $n \in \mathbf{Z}$ . We cannot have $n=0$ , since that would make $\psi \equiv 0$ . Negative values of $n$ just return the same wave functions as positive ones. Thus we can assume that $k = \tfrac13\pi n$ for $n \in \mathbb{N}$ , and we obtain energy levels $E_n \; = \; \tfrac19 n^2 \pi^2 \alpha$ for $n \in \mathbb{N}$ , and so $E_2 - E_1 = \tfrac13\pi^2\alpha$ , making the answer $\boxed{\tfrac13}$ .