Angle operator $\hat{\phi}$ doesn't exist when doing quantum mechanics on the circle $S^1$?

Question

$\newcommand{\ket}[1]{|#1\rangle}$When doing quantum mechanics on the circle $S^1$, it is well documented (yet seemingly controversial) that a self-adjoint "angle/position operator" $\hat{\phi}$ acting like $\hat{\phi} \ket{\phi} = \phi \ket{\phi}$ does not exist.

See:

• Why doesn't the phase operator exist?
• Why doesn't the uncertainty principle contradict the existence of definite-angular momentum states?
• Are angles ($\theta$ and $\phi$) in spherical coordinates treated as operators in quantum mechanics?

That being said, I still see many people in recent papers using this approach. Am I missing something or is this the case of physicists' being defiant/uninformed/disregarding rigor? Perhaps there is a way to construct a non self-adjoint "angle-type" operator $\hat{\phi}$ (by taking limits or something etc) and this is what they mean?

An alternative approach instead quantizes via the two self-adjoint operators $\hat{c} = \widehat{\cos{\phi}}$ and $\hat{s} = \widehat{\sin{\phi}}$. Is this the only proper approach when doing quantum mechanics on the circle?

Answer 1

The rigorous way is to use $e^{i\phi}=c+is$ and the conjugate angular momentum operators, as the Hilbert space is spanned by single-valued functions on the circle. But physicists often get away with using $\phi$ but then remembering that physical results have to be $2\pi$ periodic in $\phi$.