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

  • $\begingroup$ 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. $\endgroup$ – 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, $$ \langle\theta|p_\theta\rangle=\sqrt{\frac{R}{2\pi\hbar}}e^{ip_\theta\theta R/\hbar} $$

| cite | improve this answer | |
  • $\begingroup$ 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 $[p_\theta, \theta]=-i \hbar$ $\endgroup$ – user7757 Jun 11 '14 at 6:08
  • $\begingroup$ @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? $\endgroup$ – Kyle Kanos Jun 11 '14 at 17:37
  • $\begingroup$ 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 $\endgroup$ – user7757 Jun 12 '14 at 9:13
  • $\begingroup$ @ramanujan_dirac: I had found that paper back in March as well, but please note that that momentum operator is for 3D spherical coordinates and not 2D polar coordinates that OP seems to be describing. $\endgroup$ – Kyle Kanos Jun 12 '14 at 13:22

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.