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?

  • $\begingroup$ 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. $\endgroup$ Jan 16 '19 at 8:14
  • $\begingroup$ Stone–von Neumann, not Stone von Neumann :-) $\endgroup$ Jan 16 '19 at 10:16
  • $\begingroup$ @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? $\endgroup$ Jan 16 '19 at 13:54
  • $\begingroup$ @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. $\endgroup$ 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.

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

  • 1
    $\begingroup$ 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. $\endgroup$ Jan 16 '19 at 13:50
  • $\begingroup$ You're right, I was remembering something different... $\endgroup$
    – Nik Weaver
    Jan 16 '19 at 13:56

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