# Stone–von Neumann theorem?

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?

• I believe you are looking for the so-called Zak transform — see e.g. Cartier, P., Quantum mechanical commutation relations and theta functions, Proc. Sympos. Pure Math. 9, 361-383 (1966). ZBL0178.28401. Jan 16 '19 at 8:14
• Stone–von Neumann, not Stone von Neumann :-) Jan 16 '19 at 10:16
• @FrancoisZiegler I cannot find thhe paper online. However the link you give does not really seem to give the right transform, as it is periodic in one variable, no? Jan 16 '19 at 13:54
• @SerkanSüner I was just trying to give one good reference. You can read about the transform in many other places, e.g. Folland’s book (1989, §1.10) or the original papers quoted there: Weil (1964), Zak (1968), Brezin (1970), etc. Jan 17 '19 at 19:50

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.
• mhmm, let's see $(T_1(t)T_2(s)u)(n_1,n_2) = e^{-itn_2} (T_2(s) u )(n_1-1,n_2) = e^{-itn_2}e^{is(n_1-1)} u(n_1-1,n_2-1)$ whereas the other side gives $(T_2(s)T_1(t)u)(n_1,n_2) = e^{isn_1} (T_1(t) u )(n_1,n_2-1) = e^{isn_1}e^{-it(n_2-1)} u(n_1-1,n_2-1).$ So sorry, how do we get the $e^{-its}$ factor needed? It seems we rather get a sum in the exponent. Jan 16 '19 at 13:50