Finite-mode quadratures and symplectic data

On $L^2({\mathbb{R}}^r)$, multiplication by real $x_j$ is symmetric. If $g$ lies in its adjoint domain with adjoint value $h$, testing against compactly supported functions on each box where $x_j$ is bounded gives $h=x_{j}g$ almost everywhere there; the boxes exhaust ${\mathbb{R}}^r$, so $x_{j}g\in L^2$ and the adjoint domain is exactly $D(M_{x_j})$. Thus $M_{x_j}$ is self-adjoint. Let $Dg(k):=(2\pi {)}^{-r\mathrm{/}2}g(k\mathrm{/}(2\pi ))$. The $L^1$ Fourier definition in [F2] gives, for $f\in \mathcal{S}$, ${\mathcal{F}}_{\mathrm{a}\mathrm{n}\mathrm{g}}f=D{\mathcal{F}}_{2\pi }f$ by substitution in the defining integrals. For every $g\in L^2$, [F2] and the change-of-variables formula in [F2] give $\mathrm{\parallel }Dg{\mathrm{\parallel }}_2^2=(2\pi {)}^{-r}\int \mathrm{\mid }g(k\mathrm{/}(2\pi )){\mathrm{\mid }}^{2}dk=\int \mathrm{\mid }g(\xi ){\mathrm{\mid }}^{2}d\xi ,$ so $D$ is an isometry; its inverse is $D^{-1}h(\xi )=(2\pi {)}^{r\mathrm{/}2}h(2\pi \xi )$, so it is unitary. For Schwartz seminorms $p_{\alpha ,\beta }(g)={\sup }_k\mathrm{\mid }{k}^{\alpha }{\mathrm{\partial }}^{\beta }g(k)\mathrm{\mid }$, $p_{\alpha ,\beta }(Dg)=(2\pi {)}^{\mathrm{\mid }\alpha \mathrm{\mid }-\mathrm{\mid }\beta \mathrm{\mid }-r\mathrm{/}2}{p}_{\alpha ,\beta }(g),$ and the corresponding inverse formula has finite reciprocal factors. Hence $D$ and $D^{-1}$ are continuous on Schwartz space. Together with [F2] this proves that ${\mathcal{F}}_{\mathrm{a}\mathrm{n}\mathrm{g}}$ is unitary on $L^2$ and a topological automorphism of Schwartz space. Conjugating $M_{k_j}$ by it defines the self-adjoint operator $-i{\mathrm{\partial }}_{x_j}$; on Schwartz functions the kernel formula and integration by parts give ${\mathcal{F}}_{\mathrm{a}\mathrm{n}\mathrm{g}}(-i{\mathrm{\partial }}_{x_j}f)=k_j{\mathcal{F}}_{\mathrm{a}\mathrm{n}\mathrm{g}}f$. Schwartz functions are a core for multiplication: truncate by smooth compact cutoffs and then mollify on the resulting bounded support; dominated convergence controls $f$ and $x_{j}f$, and boundedness of $x_j$ on that support controls the mollification error. Fourier invariance of Schwartz space gives the same core assertion for the derivative. Conjugation by $\mathcal{U}$ therefore defines the stated self-adjoint quadratures and common core.

On that core, [F1] gives $Q_j=\sqrt{\mathrm{\hslash }\mathrm{/}2}(a_j+a_j^{\ast })$ and $P_j=-i\sqrt{\mathrm{\hslash }\mathrm{/}2}(a_j-a_j^{\ast })$. The creation and annihilation CCR in [F1] imply the displayed quadrature commutators by bilinearity; all products are defined on the finite Hermite span. Direct multiplication of the formula for $W$ gives the phase $e^{-i\mathrm{\hslash }(q\cdot p^{\mathrm{\prime }}-p\cdot q^{\mathrm{\prime }})\mathrm{/}2}$, which is $e^{-i\mathrm{\hslash }{z}^{T}J{z}^{\mathrm{\prime }}\mathrm{/}2}$ in interleaved notation. Its strong continuity and irreducibility follow from [F4] after the parameter-preserving change $(q_0,p_0)=(-\sqrt{\mathrm{\hslash }}p,q\mathrm{/}\sqrt{\mathrm{\hslash }})$ in the root Schrödinger formula. The real symplectic form is unchanged by that map and the central parameter is $\mathrm{\hslash }\mathrm{\ne }0$. Applying [F5] proves the representation-equivalence clause with precisely its hypotheses.

The positive trace-class spectral expansion in [F6] has weights summing to one. For any eigenvalue $w>0$, changing the orthonormal basis of its eigenspace changes the sum of squared norms $\sum \mathrm{\parallel }{R}_j\psi {\mathrm{\parallel }}^2$ by a unitary finite-dimensional change of basis when the eigenspace is finite-dimensional; positive trace-class compactness makes each nonzero eigenspace finite dimensional. The same basis invariance holds for the sesquilinear cross sum by polarization, and the zero eigenspace contributes nothing. Cauchy–Schwarz then gives absolute convergence of each first-moment series and of each cross-moment series under the stated second-moment bound. Thus the definitions of $d,V$ are finite and basis independent; taking the real part makes $V_{jk}=V_{kj}\in \mathbb{R}$. Finally, every finite-dimensional cutoff of Fock space has finite matrix dimension, and for square matrices $A,B$, $\mathrm{tr}(AB-BA)=0$ by expanding the diagonal entries, whereas the identity has positive trace. Hence a finite cutoff cannot satisfy the exact CCR $[Q_j,P_j]=i\mathrm{\hslash }I$.