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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.

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    $\begingroup$ I'm not sure why you assume that two arbitrary operators relate to $q$ and $p$ at all. If the operators in question do not have fully continuous spectrum, then certainly neither of them can be mapped to $q$, dooming your ansatz from the start (this is also part of the reason SvN needs the exponentiated relation). Worse, your 2d Lie algebra example does not have any infinite-dimensional irreducible unitary representations, hence there cannot be a SvN analogue for it. It seems to me that you need much stronger hypotheses here to say anything. $\endgroup$
    – ACuriousMind ♦
    Dec 8 '19 at 13:38
  • $\begingroup$ Thanks for your comment. Concerning the "2d Lie algebra" example you mentioned, do you mean the one involving $\hat{q}$ and $\frac{1}{2}\left(\hat{q}\hat{p} + \hat{p}\hat{q}\right)$? (In this case, $e^{it\hat{B}} = e^{\frac{it}{2}\left(\hat{q}\hat{p} + \hat{p}\hat{q}\right)}$ is a squeeze operator in the position space). Can you expand (maybe pointing to some references) on why in this case there exists no infinite-dimensional irreducible unitary representation? $\endgroup$
    – IchKenneDeinenNamen
    Dec 8 '19 at 13:50
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    $\begingroup$ I meant your algebra $[A,B] = \mathrm{i}A$ - there is only one 2d non-Abelian Lie algebra, the affine algebra in 1 dimension. All irreps of this algebra are one-dimensional, cf. math.stackexchange.com/q/791386/143136. I.e. your starting point "Two operators $A,B$ acting irreducibly with $[A,B] = \mathrm{i}f(A)$" is inconsistent for the choice $f(A) = A$ (in fact, it is inconsistent for every function of $A$ whose result is a linear combination of $A$ and $B$, because that's always the same two-dimensional algebra, just with different basis). $\endgroup$
    – ACuriousMind ♦
    Dec 8 '19 at 13:55
  • $\begingroup$ Thanks for the link! I am still confused though (sorry, my background in representation theory is a bit weak) how this applies to the operators $e^{it\hat{q}}, e^{it\left(\hat{q}\hat{p} + \hat{p}\hat{q}\right)}$ acting on $L^2(R)$. Is it that $L^2(R)$ cannot be decomposed in irreps in this case? $\endgroup$
    – IchKenneDeinenNamen
    Dec 8 '19 at 14:02
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    $\begingroup$ No, I'm saying that 1. Any irreducible action of $[A,B] = \mathrm{i}A$ is on a one-dimensional space. 2. It therefore cannot be unitarily equivalent to any pair of operators on $L^2(\mathbb{R})$, since that space is not one-dimensional but unitary equivalences preserve cardinality. $\endgroup$
    – ACuriousMind ♦
    Dec 8 '19 at 14:08
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The only theorem that could be useful, I think, is a theorem by Nelson proving what follows. Suppose we have a base for a representation of a certain Lie algebra in terms of $n$ densely defined (anti)-symmetric operators $A_j$ on a common dense domain $D$ in a Hilbert space $H$. Suppose also that $\sum_j A_j^2$ is essentially selfadjoint, then the operators $A_j$ are the generators of a unitary strongly-continuous rep of the simply connected Lie group $G$ whose Lie algebra is $g$.

Now, if the representation $U$ is irreducible and you know the classes of irreducible equivalent unitary representations of $G$, then $U$ must be unitarily equivalent to one of them. For the Weyl-Heisenberg group there is only one such representation, but this case is very peculiar. For instance you know that for $SU(2)$ there is an infinite number of those representations.

Regarding the case $[B,A] =iA$ that is the Lie algebra of the triangular subgroup of $SL(2,\mathbb{R})$. If $A^2+B^2$ is essentially selfadjoint and the rep is irreducible, your rep must be unitarily equivalent to one if that group (which are many).

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