# QM: How to compute position/momentum relation in polar coordinates

So if we are working in one dimensional space, we have the formula: $$\langle x|p\rangle = \frac{1}{\sqrt{2\pi\hbar}} e^{ipx/\hbar}$$

Suppose instead we are confined to a circle of radius $R$ so that the position is given by $\theta$ and the generalized momentum is $$p_\theta=-i\hbar\frac{1}{R}\frac{d}{d \theta}$$ Then what is the value of $\langle \theta|p_\theta\rangle$?

I know we can compute $\langle x|p\rangle$ by using the fact that $p$ is the generator of spatial translations. But the generator of rotations is $L_z$, and the analogous derivation becomes messty

• The operator you want to be dealing with is definitely $L_z$ and not this $p_\theta = \frac{L_z}{r}$ which is just going to give you a headache. I don't have a resource on me, but the thing you want to look up is "partial waves". Its normally used in scattering theory. It isn't neat but I don't think I have ever found an occasion when angles in QM where neat. – By Symmetry Jul 11 '14 at 10:24

I'm willing to gamble that I'm wrong on this, but it seems to me that you should be able to use the same method of solving the Cartesian $\langle x|p\rangle$ for this polar coordinate one.
In order to get $\langle x|p\rangle$, we used $$\langle x|\hat p|p\rangle=p\langle x|p\rangle=-i\hbar\frac{\partial}{\partial x}\langle x|p\rangle$$ which has a normalized solution you give. I warrant that, in a similar fashion, $$\langle\theta|\hat{p}_\theta|p_\theta\rangle=p_\theta\langle\theta|p_\theta\rangle=-\frac{i\hbar}{R}\frac{\partial}{\partial\theta}\langle\theta|p_\theta\rangle$$ Which has a similar solution: $$\langle\theta|p_\theta\rangle=Ae^{ip_\theta\theta R/\hbar}$$ Which is normalized via $$\langle\theta|\theta'\rangle=\int dp_\theta'\langle\theta|p_\theta'\rangle\langle p_\theta'|\theta'\rangle$$ $$\delta(\theta-\theta')=A^2\int dp_\theta'\exp\left[\frac{ip_\theta'\left(\theta-\theta'\right)R}{\hbar}\right]=2\pi\frac{\hbar}{R}A^2\delta\left(\theta-\theta'\right)$$ Thus, $$\langle\theta|p_\theta\rangle=\sqrt{\frac{R}{2\pi\hbar}}e^{ip_\theta\theta R/\hbar}$$
• Are you sure $\hat{p}_{\theta}=-\frac{i\bar{h}}{R}\frac{\partial}{\partial \theta}$ is the momentum operator? It doesn't satisfy the canonical commutation relation $[p_\theta, \theta]=-i \hbar$ – user7757 Jun 11 '14 at 6:08
• @ramanujan_dirac: No, I am not entirely convinced it is either. However, I was accepting the OPs premise that it is the correct way and answering the actual question: how does one compute $\langle\theta|p_\theta\rangle$ if we know $\hat{p}_\theta? – Kyle Kanos Jun 11 '14 at 17:37 • Yeah, as I suspected this is not the case, similiar to the radial momentum operator,$p_{\theta}$also undergoes a change, as the current$P_{\theta}\$ is not hermitian. Please have a look at eqn 37 this paper: scielo.org.mx/pdf/rmfe/v54n2/v54n2a8.pdf – user7757 Jun 12 '14 at 9:13