Past exam of the mathematics course of the University of Cambridge / 2007 / iii / Paper 10 / 3 / b / Solution

Let $\tilde{W}$ be the given regular irreducible family. Write $a=(x,y)$ and $a\wedge b=x_1{y}_2-y_1{x}_2$. Define the strongly integrated operatorThe Gaussian is integrable and the operators are unitary, so the integral defines a bounded operator on each vector. The Weyl relation and $\tilde{W}(a{)}^{\ast }=\tilde{W}(-a)$ give $P^{\ast }=P$. For its square, the coefficient at $\tilde{W}(c)$ isby the two-dimensional Gaussian integral. Thus $P^2=P$. The same Gaussian integration, with an inserted Weyl operator, givesThis is the Gaussian projection in a Weyl representation.

It is nonzero. Otherwise all its conjugates would vanish; the Weyl relation makes their matrix coefficients the Fourier transforms of $e^{-|a{|}^2/2}(\tilde{W}(a)\xi ,\eta )$, with the dual variable $2b\wedge a$. Uniqueness of the Fourier transform of an integrable function would make each such function zero. Strong continuity then makes its value at zero vanish, contradicting $(\xi ,\xi )>0$ for a nonzero vector.

Choose a unit vector $v\in PH$. Irreducibility makes the span of $\tilde{W}(a)v$ dense. If $w\in PH$ is perpendicular to $v$, the boxed identity makes it perpendicular to this entire span, so $w=0$. Hence $P$ has rank one, and its identity gives $(\tilde{W}(a)v,v)=e^{-|a{|}^2/2}$.

In the concrete representation of part (a), the Gaussian $v_0(t)=(2/\pi {)}^{1/4}{e}^{-t^2}$ has precisely this expectation. With the linear-in-first inner-product convention, both orbit families have Gram matrixTherefore the map $\sum c_{j}W(a_j)v_0\mapsto \sum c_j\tilde{W}(a_j)v$ is well-defined and isometric. Its source and target spans are dense, so it extends to a unitary operator intertwiner. This proves the unitary equivalence, the Stone-von Neumann theorem in the normalization of this question, rather than invoking that theorem as the proof.