Built on Facts

Quantum mechanics is difficult. “Yeah, yeah,” you say, “tell me something new”. But it’s not only a bit tricky to learn. Mathematically there’s simply not a way to get exact solutions to all of the problems we face. They just don’t exist in a nice short closed form. But we’d like solutions to those kinds of problems even if we have to use an approximation. One of the most effective tools in our mathematical kit for handling those problems is perturbation theory.

Let’s say we have a problem we don’t know how to do. There’s a decent chance it’s similar to a problem we do know how to do. Maybe it’s just the easy problem but with an additional small force which changes things just a little. That little extra bit can be thought of as a perturbation, and we can get good answers by modifying our original solution to the easy problem by adding in the result of the perturbation.

This isn’t just a quantum thing either. Perturbation theory has a long history in classical mechanics, where it first distinguished itself as a great tool for calculating the orbits of planets. These days though, it’s one of the main tools in the arsenal of the quantum physicist. It’s not a tool that can do everything - many solutions can’t be written as a perturbation in any order. Further, physicists who can develop non-perturbative solutions to difficult questions are often on the track toward real breakthroughs.

Back when I was an undergrad we took the GRE physics exam. To study for it, they gave us a book with a sample exam. One of the problems is a simple first-order perturbation to the quantum harmonic oscillator. Shall we try it? Problem and solution are below.

For this solution I’ll assume you’ve had a little introduction to perturbation theory, so we’ll go through the solution in detail but we won’t derive perturbation theory from scratch. That’s what textbooks are for! Here’s our problem:

The raising and lowering operators for the quantum harmonic oscillator satisfy

and

for energy eigenstates |n> with energy E_n. Which of the following gives the first-order shift in the n = 2 energy level due to the perturbation

where V is a constant?

The problem then gives five possible solutions, but I’ll not print those. We don’t need the help! But first a few notes. We know the energy levels for the harmonic oscillator. But with this added small potential, that energy level will be shifted slightly. How big is the shift? The problem specifies a first-order perturbation and there’s no degeneracies (which wouldn’t really matter in first-order anyway), so the energy shift is given by

The perturbation itself looks a little weird, since it’s given in terms of the raising and lowering operators instead of anything normal-sounding like x or p. But if you’re reading this far you probably know that you can write x and p in terms of those operators. The particular potential here is proportional to x². But that’s just lagniappe, you don’t need that information to solve the problem. Here’s what we have for our perturbation with n = 2.

Expand that sucker out, and pull out the V since it’s just a constant which specifies the strength and units of the perturbation.

This is nice and linear, so each term can be handled separately.

Now we operate from the left using the rules for raising and lowering operators, conveniently given in the statement of the problem. First operate using the rightmost operator, and then operate on the result with the remaining operator. Apologies for all the detail for those who know how to do this already, but I’d have liked the detail when I was learning it! Anyway, after doing the first part we have

Now pull the various roots outside the brackets because they’re just constants, and operate with the remaining operators.

Pull those roots out and we get

But the states are orthogonal. That means <n|m> = 1 if n = m, but <n|m> = 0 otherwise. That means the final answer after we get rid of the 0 terms and combine the roots is

Which is our final answer. Notice that only the terms with one raising and one lowering operator end up giving a nonzero contribution. In general if we have a perturbation like xⁿ only the terms with an equal number of raising and lowering operators will give a contribution, because otherwise n and m are not equal. Yes, it seemed like a lot of work if it’s your first time looking at it. But once you get used to it, that problem will take about 45 seconds. Don’t worry about getting too used to it though. There’s a whole universe of much more hideously difficult perturbation problems out there! But we’ll come to them when we come to them. Hope this helps!