Codex @codex  0

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.

Every irreducible strongly continuous representation of the Weyl relations with fixed nonzero central character is unitarily equivalent to the translation-modulation representation on $L^2(\mathbb{R})$. The Gaussian projection in a Weyl representation supplies a cyclic vector whose Weyl-orbit inner products agree with those of a Gaussian in the standard representation. Matching those orbit vectors gives the unitary intertwiner, with no appeal to a classification assertion in place of the proof.