The Stone–von Neumann theorem says that given two unitary groups on a Hilbert space $H$ satisfying the canonical commutation relations (CCR)
$$
U(t)V(s) = e^{-i st} V(s) U(t) \qquad \forall s, t
$$
then there is a unitary map $W:L^2(\mathbb R) \rightarrow H$ such that 
$$
W^\*U(t)W = e^{itx}, \qquad W^\*V(s)W = e^{isp}.
$$


I would like to understand now the following:


Define the following translation operators on $\ell^2(\mathbb Z^2):$
$$
(T\_1u)(n\_1,n\_2):=u(n\_1-1,n\_2),\qquad (T\_2(s)u)(n\_1,n\_2):=e^{isn\_1}u(n\_1,n\_2-1)
$$
then these two satisfy the CCR 
$$
T\_1T\_2(s) = e^{-is}T\_2(s) T\_1
$$
but only with one parameter.


I would like to ask: Is it still true that there is a unitary $W$ such that 
$$
W^\* T\_1 W = e^{ix}$$ and $$W^\*T\_2(s)W =e^{isp}?
$$ 


It seems the Stone–von Neumann theorem does not imply this since there is only one parameter. On the other hand, the example is fully explicit, so perhaps one can say something?



Stone–von Neumann theorem?
My earlier answer was incorrect. The question is ill-conceived because $T\_2$ is not a representation: it fails to satisfy $T\_2(s)T\_2(s') = T\_2(s + s')$. So of course it cannot be unitarily equivalent to a one-parameter group.


(Also, note that the Stone-von Neumann theorem only applies to irreducible representations.)


