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
 A: 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,
$$
\langle\theta|p_\theta\rangle=\sqrt{\frac{R}{2\pi\hbar}}e^{ip_\theta\theta R/\hbar}
$$ 
