Finite positive quadratic canonical operators and diagonalization data

Assume the Axiom of Choice and Countable Choice. Fix an integer $r\ge 1$, an action scale $\mathrm{\hslash }>0$ in joule-seconds, and the finite-mode Fock-to-Schrödinger unitary $\mathcal{U}:{\mathcal{F}}_s({\mathbb{C}}^r)\to L^2({\mathbb{R}}^r)$. In Schrödinger coordinates, let $\mathcal{S}({\mathbb{R}}^r)$ be the Schwartz space $\{f\in C^{\mathrm{\infty }}({\mathbb{R}}^r):{\sup }_{x\in {\mathbb{R}}^r}\mathrm{\mid }{x}^{\alpha }{\mathrm{\partial }}^{\gamma }f(x)\mathrm{\mid }<\mathrm{\infty }\text{ for every }\alpha ,\gamma \in {\mathbb{N}}_0^r\}\mathrm{.}$ Here ${\mathbb{N}}_0$ denotes the nonnegative integers, $x^{\alpha }={\prod }_j{x}_j^{{\alpha }_j}$, and ${\mathrm{\partial }}^{\gamma }={\prod }_j{\mathrm{\partial }}_{x_j}^{{\gamma }_j}$. Use the dense domain $\mathcal{D}:={\mathcal{U}}^{-1}\mathcal{S}({\mathbb{R}}^r)$ and define on it the operators $Q_j$ and $P_j$ by transporting multiplication by $\sqrt{\mathrm{\hslash }}{x}_j$ and differentiation $-i\sqrt{\mathrm{\hslash }}{\mathrm{\partial }}_{x_j}$ through $\mathcal{U}$. Thus, for $f\in \mathcal{S}$, $(Q_j{\mathcal{U}}^{-1}f)={\mathcal{U}}^{-1}(\sqrt{\mathrm{\hslash }}{x}_{j}f)$ and $(P_j{\mathcal{U}}^{-1}f)={\mathcal{U}}^{-1}(-i\sqrt{\mathrm{\hslash }}{\mathrm{\partial }}_{x_j}f)$. Use the commutator convention $[A,B]=AB-BA$ on common invariant domains. These quadratures satisfy $[Q_j,P_k]=i\mathrm{\hslash }{\delta }_{jk}I$ on $\mathcal{D}$ and have units $\sqrt{\mathrm{j}\mathrm{o}\mathrm{u}\mathrm{l}\mathrm{e}\cdot \mathrm{s}\mathrm{e}\mathrm{c}\mathrm{o}\mathrm{n}\mathrm{d}}$. Each product Hermite function is a polynomial times a Gaussian; all its derivatives and polynomial multiples remain a polynomial times that Gaussian and are bounded on ${\mathbb{R}}^r$. Thus the complete product Hermite family lies in $\mathcal{S}$, so $\mathcal{D}$ is dense. The finite-mode Fock-to-Schrödinger supplier establishes completeness and the quadrature actions used here. Reorder the quadrature column to grouped order $R_g:=(Q_1,\dots ,Q_r,P_1,\dots ,P_r{)}^T$.

A finite positive quadratic canonical operator is specified by a real symmetric matrix $G\in M_{2r}(\mathbb{R})$ with $x^{T}Gx>0$ for every $x\mathrm{\ne }0$, and a real constant $c$. The coefficient $G$ has units of inverse seconds and $c$ has units of joules. On $\mathcal{D}$ define the symmetrically ordered operator $H_{G,c}^{(0)}:=\frac{1}{4}{\sum }_{a,b=1}^{2r}{G}_{ab}(R_{g,a}{R}_{g,b}+R_{g,b}{R}_{g,a})+cI\mathrm{.}$ The real multiplication and derivative operators preserve Schwartz space, so this finite sum maps $\mathcal{D}$ into ${\mathcal{F}}_s({\mathbb{C}}^r)$. Integration by parts makes each $Q_j,P_j$ symmetric on $\mathcal{D}$. For $f,g\in \mathcal{D}$, symmetry and invariance of $\mathcal{D}$ give $⟨R_{g,a}{R}_{g,b}f,g⟩=⟨R_{g,b}f,R_{g,a}g⟩=⟨f,R_{g,b}{R}_{g,a}g⟩$; the same identity with $a,b$ interchanged proves that each symmetrized summand is symmetric. Since $G_{ab}=G_{ba}\in \mathbb{R}$, the finite sum is symmetric. Its quadratic coefficients are strictly positive in the stated matrix sense. The statement does not yet assert that this symmetric operator is self-adjoint.

In grouped coordinates set $J_g:=\left(\begin{array}{cc}0& I_r\\ -I_r& 0\end{array}\right),D_{\omega }:=\mathrm{diag}({\omega }_1,\dots ,{\omega }_r,{\omega }_1,\dots ,{\omega }_r)\mathrm{.}$ A finite diagonalization certificate for $G$ consists of matrices $U,V\in M_r(\mathbb{C})$ and positive angular frequencies ${\omega }_j>0$ in inverse seconds such that, with ${}^{\ast }$ denoting conjugate transpose and ${}^T$ transpose without conjugation, $U{U}^{\ast }-V{V}^{\ast }=I_r,U{V}^T=V{U}^T,$ $S_g:=\left(\begin{array}{cc}\mathrm{Re}(U+V)& -\mathrm{Im}(U-V)\\ \mathrm{Im}(U+V)& \mathrm{Re}(U-V)\end{array}\right),S_g^T{J}_g{S}_g=J_g,S_g^{T}G{S}_g=D_{\omega }\mathrm{.}$ This certificate is explicit finite-dimensional data. No claim is made here that every positive $G$ has such a certificate; the theorem on this page uses it as a stated hypothesis. Each term in $H_{G,c}^{(0)}$ has units of joules because $Q_j$ and $P_j$ have units $\sqrt{\mathrm{j}\mathrm{o}\mathrm{u}\mathrm{l}\mathrm{e}\cdot \mathrm{s}\mathrm{e}\mathrm{c}\mathrm{o}\mathrm{n}\mathrm{d}}$.

The Axiom of Choice

The Axiom of Countable Choice (${\mathrm{A}\mathrm{C}}_{\omega }$)

Finite-mode quadratures and symplectic data

Finite-mode Fock to Schrödinger identification

Hermite recurrence