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

$$\newcommand{\ket}{|#1\rangle}$$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:

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?

• Could you give an example of a recent paper using such an operator, just for context? Dec 6, 2021 at 16:35

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 $$\phi$$.
• Are you saying physicists' define "$\hat{\phi} = -i \log( \hat{c} + i \hat{s})$" and then because the log has different branch cuts they have to choose one before this is well defined? Dec 7, 2021 at 18:37
• I think that's right, it's often convenient to be able to directly work with $\phi$, but then it has to be restricted to $[0,2\pi)$, like a branch of the log as you said, and also one has to "remember" somehow that $\phi$ is only defined up to $2\pi$. Dec 7, 2021 at 19:42