The Stone-von-Neumann theorem states (in rough terms) that a pair of operators $\left(\hat{Q}, \hat{P}\right)$ satisfying the exponentiated canonical commutation relation $e^{is\hat{Q}}e^{it\hat{P}} = e^{-ist}e^{it\hat{P}}e^{is\hat{Q}}$ acting irreducibly are unitarily equivalent to the standard position and momentum operators $\hat{q}, \hat{p}$ acting on $L^2(\mathbf{R})$.
I would like to figure how to extend this result. Precisely, given 2 operators $\hat{A}, \hat{B}$ satisfying the commutation relations $\left[\hat{B}, \hat{A}\right] = if(\hat{A})$ (for some Borel function $f$) acting irreducibly, can one say anything about how $\left(\hat{A}, \hat{B}\right)$ relates to $\left(\hat{q}, \frac{1}{2}\left(f(\hat{q})\hat{p} + \hat{p}f(\hat{q})\right)\right)$ (which indeed satisfy the latter commutation relation)?
Loosely speaking, the Stone-von-Neumann theorem corresponds to $f = 1$ (this is not rigorously true as it only applies to operators satisfying the exponentiated canonical commutation relation, which is stronger than the basic one). To start (hopefully) simple, I would now like to look at $f(x) = x$. I have tried to see whether the proof of the Stone-von-Neumann may be adapted there, following the derivation given in Hall, Quantum Theory for Mathematicians, paragraph 14.3, but this does not seem promising. I anticipate though that this case may still be simple enough as the relation $\left[\hat{A}, \hat{B}\right] = i\hat{A}$ gives you a nice $2$-dimensional Lie algebra structure; this would no longer be the case for general $f$, e.g $f(x) = x^2$.
Any idea how to tackle the problem?
P.S.: Also, do not hesitate to tell if you think the question would the better suited for Math StackExchange.
EDIT: This paper may be relevant to the discussion in the comments.