We haven’t done an actual straight-up physics problem in a while, much less one above the level of undergraduate freshman physics. There’s a reason: it’s roughly as niche as it’s possible for an internet post to be. But on the other hand, surely someone ought to do it every once in a while. So here we go with a roughly 3rd-year undergrad physics major problem in quantum mechanics. If you’ve not studied it before, fear not! As always I’ll try to convey the overall path without relying on the details of the math.

*Consider a particle in a one-dimensional square well (0 < x < L), potential zero in the box and infinite outside. We know its eigenfunctions and energy levels. Now perturb the box with a quadratic potential such that inside the box the potential is now V = αx ^{2} for some small α. Find the first-order correction to the energy levels.*

Often in physics there’s problems that you can solve exactly. Even more often there’s problems you can’t. Before pitching up your hands in frustration, there’s plenty of things to try in order to make some progress on those problems that can’t be solved exactly. One of those ways is to take a problem you can solve, and modify it a bit so that it approximates the situation you can’t solve. And one way to do that is with a technique called perturbation theory. You take a known solution, perturb it to look more like the unknown solution, and use the resulting approximation as your answer.

Introductory quantum mechanics courses will derive the methods of doing perturbation theory, so we’ll just quote the results. To first order (informally, an initial “pretty good” approximation) the change in the energy levels of a system will be:

Where the psi are the wavefunctions of the particle in the unperturbed box. I’ve subscripted the n so that we can keep track of which state is changing. In the case of the particle in a box, that little bit of compact Dirac notation means to take the state, write it twice, put the potential in the middle, and integrate. Our life is simpler in this case because the wavefunction is purely real* and the potential operator is just a function of x. That gives us this:

You can integrate that mess by hand, but I’m going to let Mathematica do it:

That can be greatly simplified by noting that the sine of 2*pi*n is zero for all integers n. The cosine of 2*pi*n is 1 for all integers n. After a little more algebra, that gives:

Which is the solution to our problem. It’s got a few interesting features. First, it bumps all energy levels up by a constant factor of L squared times alpha. Second, it then moves the energy levels back down a bit in proportion to 1/n^2. This makes some sense. The higher energy levels should still “look like” square well levels in the sense that their spacing is not asymptotically affected. The change in the lower levels is more direct, and their spacing is more altered – in other words, you would see it if you were looking at the system with a spectrograph.

So it’s not the most enthralling thing in the world, but if I can help out an occasional confused undergrad then it will be my official good deed for the day.

*Or at least the spatial part of the separated wavefunction is. There’s also the time-dependent factor of e^{-iEt/h}, but it’s not relevant here.