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Suppose that $Q,P$ are self-adjoint operators which satisfy the relation $$(1) \ \ \ \ \ [Q,P]=iI$$ One can easily show that in this case $P,Q$ cannot be bounded. However one can find unbounded operators (multiplication by $x$ and $\frac1i \frac{d}{dx}$) satisfying this relation. When dealing with unbounded operators one encounters problems with the domains so in order to avoid them one proceed as follows: as any self-adjoint operator gives rises to the one parameter group of unitary operators via $P \mapsto (e^{itP})_{t \in \mathbb{R}}$ one can formulate this problem in terms of this one-parameter groups. The canonical commutation relation takes the form $$(2) \ \ \ V(s)U(t)=e^{its}U(t)V(s)$$ where $U$ corresponds to $Q$ while $V $ corresponds to $P$. One form of the Stone-von Neumann theorem states that any irreducible representation of $(2)$ (say on the space $\mathcal{H}$) is unitary equivalent to the operators of multiplication by $x$ and $\frac1i \frac{d}{dx}$ (in more details: there exists a unitary $W:L^2(\mathbb{R}) \to \mathcal{H}$ such that $W^{-1}U(t)W=e^{itQ}$ and $W^{-1}V(s)W=e^{isP}$ where $P=\frac1i \frac{d}{dx}$ and $Q=M_x$. As far as I know for the relation $(1)$ this is no longer true (probably due to issues with the domains). So I would like to clarify what is the exact relation between $(1)$ and $(2)$.

Suppose that if $U(t)$ and $V(s)$ are one parameter groups of unitaries with generators $Q$ and $P$ resp. Is it true that $P,Q$ satisfy $(1)$ if and only if $U(t)$ and $V(s)$ satisfy $(2)$?

EDIT: Let me try to give some more details how I understand my question: suppose that $P,Q$ are densely defined self-adjoint operators with the property that there is a dense linear subspace $D$ contained in the intersection $\bigcap_{n,m}dom(Q^m P^n)$. Then the implication $(1) \Rightarrow (2)$ I understand as follows: we take $P,Q$ as above satisfying $(1)$ and form $U(t)=e^{itQ}$ and $V(s)=e^{isP}$ Is it true that $U(t)$ and $V(s)$ satisfy $(2)$? Regarding the opposite implication: let $U(t),V(s)$ be a one parameter groups of unitaries (strongly continuous) and take $Q,P$ to be their infinitesimal generators. Is it true that if $U(t),V(s)$ satisfy $(2)$ then $P,Q$ satisfy $(1)$? I don't see why this question is not well defined: in particular I'm not using any notion such that ,,irreducible representation of something (possibly unbounded)''. I just wanted to clarify how one can deduce $(1)$ from $(2)$ and vice versa

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    $\begingroup$ Does this answer your question? The Stone-von Neumann theorem without exponentials? $\endgroup$ – Francois Ziegler Aug 3 '20 at 13:06
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    $\begingroup$ (1) has no immediate meaning for unbounded operators, so one has to give a precise interpretation, and one way to proceed would be (2) (until the last few lines of your post, I thought this is what you were doing). Of course, if this path is taken, then your question disappears. In any event, your question isn't precise until you clarify what (1) means for unbounded operators. $\endgroup$ – Christian Remling Aug 3 '20 at 14:10
  • $\begingroup$ To reiterate what Christian Remling said: You have to define the operator $[Q,P]$. If $P$ and $Q$ are not everywhere defined, it is not obvious what their commutator is supposed to be. What is its domain? Do you want it to be closable and densely defined? $\endgroup$ – MaoWao Aug 4 '20 at 7:44

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