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 $(2)$?