Perturbed Infinite Well

Let's take a look at a problem from quantum mechanics. Suppose we have an infinite square well of length L that contains a particle. Inside the well, the particle is subject to a potential given by

$V(x) = 2V_0 (\cos(\frac{2\pi x}{L}) - 1) (O < x < L)$

And suppose we wish to calculate the associated spectrum of such a particle. Our first instinct might be to try to calculate this analytically, but you'll quickly realise that the differential equation that we end up with her is somewhat non-trivial. This problem can be tackled using time independent perturbation theory, however. First we need the zero order basis functions for this system. Fortunately, this is very easy to calculate. If we use the time-independent Schrodinger equation for the flat square well, we find that:

$\big(-\frac{\hbar^2}{2m}\frac{d^2}{dx^2} - E\big)\psi = 0$

The solutions to this equation are sine functions, specifically, we find that

$\psi^{0}_n = \sqrt{\frac{2}{L}}\sin(\frac{n\pi x}{L})$
$E^{(0)}_n = \frac{n^2 \hbar^2 \pi^2}{2mL^2}$

These solutions are valid for all $n \ge 1$, naturally, and describe sine waves bound within the box. With our basis set in hand, we can now proceed to the perturbation theory. The first order correction to the energy is given by

$E_n = E^{(0)}_n + E^{(1)}_n$

Where

$E^{(1)}_n = \big<\psi^{(0)}_n|V|\psi^{(0)}_n\big>$

You can the derivation for this result here, if you like.  This is what we'll want to work out. Let's expand this whole mess.

$E^{(1)}_n = \frac{4V_0}{L}\int_0^{L}\sin(\frac{n\pi x}{L})^2 (\cos(\frac{2\pi x}{L}) - 1) dx$

We can write $\sin(\frac{n\pi x}{L})^2 = \frac{1 - \cos(\frac{2n\pi x}{L})}{2}$ on our way to solving this integral. This gives us the following

$E^{(1)}_n = \frac{4V_0}{L}\int_0^{L} \frac{1 - \cos(\frac{2n\pi x}{L})}{2} (\cos(\frac{2\pi x}{L}) - 1) dx$

This gives us four terms to evaluate.

$E^{(1)}_n = \frac{4V_0}{L} \int_0^{L}\big( \cos(\frac{2\pi x}{L}) + \frac{1}{2}\cos(\frac{2n\pi x}{L}) - \frac{1}{2} - \frac{1}{2}\cos(\frac{2n\pi x}{L})\cos(\frac{2\pi x}{L})\big) dx$

The first three terms are trivial. If you recall your Fourier theory, you'll remember also that integrals in the form of the fourth term are also exactly zero unless $n = 1$, hence this collapses to:

$E^{(1)}_n = -2V_0(1 + \delta_{n1})$

Where $\delta_{n1}$ is the Kronecker delta. Each level experiences an energy shift equal to $-2V_0$, except, curiously, the first level, which experiences a shift twice as large.