One of the first examples given when studying quantum mechanics is the particle on a cylinder, or particle on a ring. One finds that because of the periodic boundary conditions, the energy eigenvalues are required to be quantized. Specifically, the Schrodinger equation on a circle of radius $R$:

$$\frac{-\hbar^2}{2m}\frac{1}{R^2}\frac{\partial^2 \psi}{\partial \theta^2} = E\psi$$

has solution $\psi = \exp\left(i \theta\sqrt{\frac{2mR^2E}{\hbar^2}}\right)$. Enforcing periodic boundary conditions $\psi(\theta) = \psi(\theta + 2\pi)$ gives the requirement $$E = \frac{n^2 \hbar^2}{2mR^2}, \qquad n \in \mathbb{N}$$ That is, the energy is quantised. More relevant to this question, we find that the momentum (in the circle direction) is also quantised. The momentum in this case being the eigenvalue of the momentum operator $\hat{p} = \frac{1}{i}\frac{\partial}{\partial\theta}$: $$ \hat{p} \psi = n \psi$$

I had a discussion with a professor who claimed this was a generic feature of compact spaces - whenever the direction was compact, the momentum was quantised. I tried to see this for the case of a particle on $S^2$, but ran into some issues. In the case of $S^2$, the solutions to the Schrodinger equation are well known - they are the spherical harmonics. Specifically, we have (up to some normalisation) $$\psi = e^{im\phi}P_{l}^m(\cos\theta)$$ where $l \in \mathbb{N}$ and $m\in \mathbb{Z}$ such that $-l \leq m \leq l$. It is clear that in the case, the "$\phi$-momentum": $\hat{p}_{\phi} = \frac{1}{i}\frac{\partial}{\partial \phi}$ is quantised, but the "$\theta$-momentum": $\hat{p}_{\theta} = \frac{1}{i}\frac{\partial}{\partial \theta}$ doesn't seem to be quantised. In fact, we can compute: $$\hat{p}_{\theta}\psi = \frac{1}{i}\frac{d}{d\theta}[e^{im\phi}P_l^m(\cos\theta)] = -\frac{i}{2}e^{im\phi}\left[(l+m)(l-m+1)P_l^{m-1} - P_l^{m+1}\right]$$ where I have ommitted the $(\cos \theta)$ for notational clarity. This computation follows from properties of the $P_l^m$ functions.

The solution is not even an eigenfunction of the $\theta$-momentum operator, so I don't see how it can have quantised momentum.

I have the following two questions:

1. What's going on with this $S^2$ example?

2. Is there a slick way to see that momentum is quantised for compact spaces? Perhaps something to do with the properties of $L^2(compact)$ as opposed to $L^2(\mathbb{R}^d)$.


So momentum is associated with a translational symmetry. Angular momentum with a rotational symmetry. The way the coordinates are defined $\phi \to \phi + c$ is a rotation. But the other rotations are not $\theta \to \theta +c'$. This means you shouldn't even think to write down that $p_\theta$. It may be a good exercise to write down some other rotations.

The way you see energy quantization as the spectrum of Laplacian on compact vs noncompact. But for momentum the question does not even make sense because it might not even have a notion of momentum.

  • $\begingroup$ That professor's statement was a generic in the sense of everything a physicist is bound to use still has symmetries and are not like a high genus surface or anything ugly/beautiful like that. $\endgroup$ – AHusain Jan 24 '16 at 0:20

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