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Consider $\hat{r} = (\hat{x}_1,\hat{p}_1,\ldots \hat{x}_n,\hat{p}_n)^\intercal$ with commutation relation \begin{equation} [\hat{r},\hat{r}^\intercal] = i\Omega. \end{equation}

I want a simple statement like:

Any two irreducible representations of the canonical commutation relations are unitarily equivalent.

so that I can prove that $\hat{r}$ and $S\hat{r}$ are unitarily equivalent when $S \in \operatorname{Sp}(2n,\mathbb{R})$. I seek out the Stone von Neumann theorem! But here I am told that

There is substantial literature on solutions to the commutation relations that do not necessarily integrate to unitary groups and for these the Stone von Neumann theorem breaks down completely...We bypass these difficulties by going to the Weyl integrated form.

which provides a reason that the theorem is always stated along these lines:

Let $(V,\Omega)$ be a finite dimensional symplectic vector space. Let $(\mathcal{H},\hat{W}(y))$ and $(\mathcal{H}',\hat{W}(y))$ be strongly continuous, irreducible, unitary representations of the Weyl relations. Then $(\mathcal{H},\hat{W}(y))$ and $(\mathcal{H}',\hat{W}(y))$ are unitarily equivalent.

Can I say something like: 'I don't want to consider the horrible cases', rigorously? Something like

Any two irreducible representations of the canonical commutation relations ''that are nice'' are unitarily equivalent

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  • $\begingroup$ Just a small comment: since Stone and von Neumann were distinct people, it's customary to place a hyphen between their names (as for Stone and Weierstrass). $\endgroup$ Feb 13 '17 at 20:10
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From a random review:

There are many theorems that answer this question, due to Rellich, Dixmier, Tillmann, Foiaş-Gehér-Sz.-Nagy, Kristensen-Mejlbo-Poulsen, and the reviewer. (...) Some of the proofs obtain the result directly, mostly via considering the operator $A=2^{−1/2}(Q+iP)$ and its adjoint $A^*$ which satisfy the formal relation $AA^*=A^*A+I$. Others deduce the Weyl-von Neumann relation. {It seems to the reviewer that none of these theorems is quite satisfactory.}

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  • $\begingroup$ Can I just say that there exists $\hat{r}$ that is a nice solution. This implies that $S\hat{r}$ is a nice solution. Therefore they are unitarily equivalent? I am looking for a rigorous wording rather than rigorous conditions but these are very helpful links all the same. $\endgroup$
    – Matta
    Feb 13 '17 at 10:38
  • $\begingroup$ @Matta Yes, you can. The "nice solution that exists" is the Schrödinger representation $W(\ell)$ of the Heisenberg group $H$ (a central extension of $\smash{\mathbf R^{2n}}$ by the circle), and what you call $S\hat r$ is the result of composing $W(\ell)$ with the automorphism of $H$ defined by $S\in\text{Sp}(2n,\mathbf R)$. Stone-von Neumann is sufficient, but not at all necessary to see that both are unitarily equivalent... $\endgroup$ Feb 13 '17 at 12:23
  • $\begingroup$ ... In fact, as the notation indicates, the construction of $W(\ell)$ depends on the choice of a lagrangian subspace $\ell\subset\smash{\mathbf R^{2n}}$ (the span of the first $n$ vectors of a symplectic basis, a.k.a. a "real polarization"), and the problem boils down to seeing that $W(\ell)$ and $W(S\ell)$ are unitarily equivalent. And for that one readily constructs an explicit intertwining operator: see e.g. Prop. 1.4.7 of Lion-Vergne (1980), available here. $\endgroup$ Feb 13 '17 at 12:23

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