# Matrix elements for the particle on a ring

$$x$$ on a ring is really the angle "$$\theta$$'' which has to be identified $$\theta \simeq \theta+2\pi$$. Consequently $$\theta$$ is not single-valued, and cannot act as an operator on the Hilbert space of periodic functions. In particular if $$\psi(\theta)$$ is periodic --- meaning that $$\psi(\theta)= \psi(\theta+2\pi)$$ --- then $$\theta\psi(\theta)$$ is not periodic. As a consequence the matrix element $$\langle n|\theta|m\rangle$$ is not definable.

I really like this question, I'm probably going to give it as an exercise sometime, once I've completely understood it. That being said, I'm still a little unsure if my answer is correct, so feel free to comment.

I assume that you have chosen the Hamiltonian to be $$\hat{H} = \frac{\hat{p}^2}{2M} = -\frac{\hbar^2}{2M}\frac{\text{d}^2}{\text{d}x^2},$$ where $$\hat{x}$$ is the "usual" position operator. I do not think this is correct. There is a hint of this even in your question, as your variable $$x$$ is not quite the position, as it goes from $$0$$ to $$2\pi$$. In fact it is the polar angle $$\varphi$$. While it might look like it should be, I think there is nevertheless a fundamental (topological?) difference between the free particle and the particle on a ring. (i.e. Even if you were to assume that $$x$$ were the position along the circumference, I do not believe it is equivalent to the one dimensional free particle.)

I believe the correct Hamiltonian for this system is $$\hat{H} = - \frac{\hbar^2}{2 M R^2} \frac{\partial^2}{\partial \varphi^2} \equiv \frac{\hat{L}_z^2}{2 M R^2}.$$

The wavefunctions therefore are $$\psi_m(x) = \frac{1}{\sqrt{2\pi}}e^{i m \varphi},$$ which are eigenfunctions of $$\hat{H}, \hat{L}_z,$$ and (in this case) $$\hat{L}^2$$. (The above state would crudely represent a particle spinning "rightward" or "leftward" if $$m$$ is positive or negative respectively.)

The sort of commutator that you want to calculate will therefore be $$[H, \hat{\varphi}],$$ where $$\hat{\varphi}$$ is some sort of "angle" operator. But such an operator is not single-valued, and as a result there are many problems with defining such operators in Quantum Mechanics, as illustrated by this fantastic and understandable paper by Levy-Leblond Who is afraid of nonhermitian operators? A quantum description of angle and phase.

Consequently, I do not think you can use any of the above relation in the case of a particle on a ring.