Gaussian covariance and uncertainty

Write $X_j=R_j-d_{j}I$ and $G_{jk}={\sum }_n{w}_n⟨X_j{\psi }_n,X_k{\psi }_n⟩$. The series converge absolutely by Cauchy–Schwarz and the second-moment bounds. On compactly supported smooth functions, integration by parts gives $⟨R_{j}f,R_{k}f⟩-⟨R_{k}f,R_{j}f⟩=-i\mathrm{\hslash }{J}_{jk}\mathrm{\parallel }f{\mathrm{\parallel }}^2\mathrm{.}$ For the same-mode $Q_j,P_j$ pair, this is $i\mathrm{\hslash }\int x_j{\mathrm{\partial }}_j\mathrm{\mid }f{\mathrm{\mid }}^2=-i\mathrm{\hslash }\mathrm{\parallel }f{\mathrm{\parallel }}^2$ after the vanishing boundary term; distinct coordinate pairs and equal-type pairs give zero. The joint graph-norm density in [F2] extends the identity to each ${\psi }_n$ in the common domain. Centering by the real scalars $d_j$ does not change the antisymmetric difference. Summing with weights yields $G-G^T=-i\mathrm{\hslash }J$, while $\mathrm{Re}G=V$ by definition; hence $G=V-i\mathrm{\hslash }J\mathrm{/}2$. For any $c\in {\mathbb{C}}^{2r}$, ${\sum }_n{w}_n{\parallel {\sum }_j{c}_j{X}_j{\psi }_n\parallel }^2={\bar{c}}^T{G}^{T}c\ge 0.$ Since $G^T=V+i\mathrm{\hslash }J\mathrm{/}2$, the uncertainty matrix is positive semidefinite.

Now suppose the target $V+i\mathrm{\hslash }J\mathrm{/}2$ is positive semidefinite. For real $x$, $x^{T}Vx=x^{\ast }(V+i\mathrm{\hslash }J\mathrm{/}2)x\ge 0$, so $V$ is nonnegative. If a nonzero real $x$ had $x^{T}Vx=0$, put $H=V+i\mathrm{\hslash }J\mathrm{/}2$. For every complex $y$ and real $t$, positivity gives $(x+ty{)}^{\ast }H(x+ty)=2t\mathrm{Re}(y^{\ast }Hx)+t^2{y}^{\ast }Hy\ge 0$. Allowing both signs of $t$ forces $\mathrm{Re}(y^{\ast }Hx)=0$; replacing $y$ by $iy$ forces the imaginary part to vanish as well. This holds for every $y$, so take $y=Hx$ to obtain $\mathrm{\parallel }Hx{\mathrm{\parallel }}^2=0$ and hence $Hx=0$. Its real and imaginary parts give $Vx=0$ and $Jx=0$, contradicting invertibility of $J$. Thus $V$ is positive definite. Apply [F5] to obtain $V=SD{S}^T$ with $D=\mathrm{diag}({\nu }_1,{\nu }_1,\dots ,{\nu }_r,{\nu }_r)$ and ${\nu }_j\ge \mathrm{\hslash }\mathrm{/}2$.

For one mode and $\bar{n}\ge 0$, set $\eta =\frac{\bar{n}}{1+\bar{n}},{\rho }_{\bar{n}}=(1-\eta ){\sum }_{m=0}^{\mathrm{\infty }}{\eta }^m\mathrm{\mid }m⟩⟨m\mathrm{\mid },$ with ${\rho }_0=\mathrm{\mid }0⟩⟨0\mathrm{\mid }$. The geometric series makes this a positive trace-class operator of trace one. For $\bar{n}>0$, define $\mathrm{\mid }\alpha ⟩=e^{-\mathrm{\mid }\alpha {\mathrm{\mid }}^2\mathrm{/}2}{\sum }_{m=0}^{\mathrm{\infty }}\frac{{\alpha }^m}{\sqrt{m!}}\mathrm{\mid }m⟩\mathrm{.}$ Its norm is one by the exponential series. The vector ${\psi }_{\alpha }:=W(q_0,p_0)g_0$, with $\alpha =\sqrt{\mathrm{\hslash }\mathrm{/}2}(-p_0+i{q}_0)$, is a scalar phase times $\mathrm{\mid }\alpha ⟩$: the differential formula gives $a\psi =\alpha \psi$, and comparing coefficients in the complete number basis gives the displayed series up to one constant phase. In Schrödinger coordinates ${\psi }_{\alpha }(x)={\pi }^{-1\mathrm{/}4}{e}^{i\sqrt{\mathrm{\hslash }}{q}_0(x+\sqrt{\mathrm{\hslash }}{p}_0\mathrm{/}2)}{e}^{-(x+\sqrt{\mathrm{\hslash }}{p}_0{)}^2\mathrm{/}2}\mathrm{.}$ Set $y=x+\sqrt{\mathrm{\hslash }}{p}_0+\sqrt{\mathrm{\hslash }}p\mathrm{/}2$. In $⟨W(q,p){\psi }_{\alpha },{\psi }_{\alpha }⟩$, the Gaussian factor is $e^{-y^2-\mathrm{\hslash }{p}^2\mathrm{/}4}$ and the phase is $\sqrt{\mathrm{\hslash }}qy-\mathrm{\hslash }q{p}_0+\mathrm{\hslash }{q}_{0}p$. For ${\alpha }_0>0$ and real $\beta$, [F9] gives ${\int }_{\mathbb{R}}{e}^{-{\alpha }_0{y}^2+i\beta y}dy=\sqrt{\frac{\pi }{{\alpha }_0}}{e}^{-{\beta }^2\mathrm{/}(4{\alpha }_0)}$ by the explicit substitutions $t={\alpha }_0\mathrm{/}\pi$ and $\xi =-\beta \mathrm{/}(2\pi )$ in the supplier's $2\pi$-frequency formula. Taking ${\alpha }_0=1$ and $\beta =\sqrt{\mathrm{\hslash }}q$ gives the normalized factor ${\pi }^{-1\mathrm{/}2}\int e^{-y^2+i\sqrt{\mathrm{\hslash }}qy}dy=e^{-\mathrm{\hslash }{q}^2\mathrm{/}4}$. Therefore, for $u=\mathrm{Re}\alpha$ and $v=\mathrm{Im}\alpha$, $⟨W(q,p){\psi }_{\alpha },{\psi }_{\alpha }⟩=\exp \text{\hspace{0.17em}⁣}(i\sqrt{2\mathrm{\hslash }}(qu+pv)-\frac{\mathrm{\hslash }}{4}(q^2+p^2))\mathrm{.}$ The phase is the mean $d=(\sqrt{2\mathrm{\hslash }}u,\sqrt{2\mathrm{\hslash }}v)$. The geometric weights have mean occupation ${\sum }_{m}m(1-\eta ){\eta }^m=\eta \mathrm{/}(1-\eta )=\bar{n}$. For a number vector, $Q\mathrm{\mid }m⟩=\sqrt{\mathrm{\hslash }\mathrm{/}2}(\sqrt{m+1}\mathrm{\mid }m+1⟩+\sqrt{m}\mathrm{\mid }m-1⟩),$ and $P\mathrm{\mid }m⟩=-i\sqrt{\mathrm{\hslash }\mathrm{/}2}(\sqrt{m}\mathrm{\mid }m-1⟩-\sqrt{m+1}\mathrm{\mid }m+1⟩)$ by [F8]. Orthogonality of number vectors gives zero means and $⟨m,Q^{2}m⟩=⟨m,P^{2}m⟩=\mathrm{\hslash }(m+1\mathrm{/}2)$. Also $QP+PQ=-i\mathrm{\hslash }(a^2-(a^{\ast }{)}^2)$, whose diagonal matrix entries vanish because $a^2$ and $(a^{\ast }{)}^2$ change number by two. Thus the diagonal state has zero symmetrized cross covariance. Summing against the geometric weights gives $⟨Q^2⟩=⟨P^2⟩=\frac{\mathrm{\hslash }}{2}(2\bar{n}+1)=\nu \mathrm{.}$ Thus its quadrature second moments are finite and its covariance is $\nu I_2$. The probability density $(\pi \bar{n}{)}^{-1}{e}^{-(u^2+v^2)\mathrm{/}\bar{n}}$ on the complex plane is normalized because [F9] at zero frequency gives ${\int }_{\mathbb{R}}{e}^{-u^2\mathrm{/}\bar{n}}du=\sqrt{\pi \bar{n}}$ in each coordinate, and [F12] factors the absolutely integrable product integral. By [F12] and two applications of [F9] with ${\alpha }_0=1\mathrm{/}\bar{n}$, ${\beta }_u=\sqrt{2\mathrm{\hslash }}q$ and ${\beta }_v=\sqrt{2\mathrm{\hslash }}p$, its characteristic factor is ${\int }_{\mathbb{C}}\frac{e^{-\mathrm{\mid }\alpha {\mathrm{\mid }}^2\mathrm{/}\bar{n}}}{\pi \bar{n}}{e}^{i\sqrt{2\mathrm{\hslash }}(qu+pv)}dudv=e^{-\mathrm{\hslash }\bar{n}(q^2+p^2)\mathrm{/}2}\mathrm{.}$ To identify the mixture with ${\rho }_{\bar{n}}$, the $(m,n)$ matrix entry of $\mathrm{\mid }\alpha ⟩⟨\alpha \mathrm{\mid }$ is $e^{-\mathrm{\mid }\alpha {\mathrm{\mid }}^2}{\alpha }^m{\bar{\alpha }}^n\mathrm{/}\sqrt{m!n!}$ by the coherent-state series. Under a planar rotation $\alpha \mapsto e^{i\theta }\alpha$, Lebesgue measure and the radial density are invariant by [F11], while ${\alpha }^m{\bar{\alpha }}^n$ gains the factor $e^{i(m-n)\theta }$; choosing $\theta$ with this factor unequal to $1$ shows that every off-diagonal entry vanishes. For $m=n$, the polar formula [F11] and $\sigma (S^1)=2\pi$ give ${\int }_{\mathbb{C}}\mathrm{\mid }\alpha {\mathrm{\mid }}^{2m}{e}^{-a\mathrm{\mid }\alpha {\mathrm{\mid }}^2}dudv=2\pi {\int }_0^{\mathrm{\infty }}{r}^{2m+1}{e}^{-a{r}^2}dr=\pi {\int }_0^{\mathrm{\infty }}{x}^m{e}^{-ax}dx,$ where $x=r^2$. For $a>0$, the last integral is $m!\mathrm{/}{a}^{m+1}$: its value at $m=0$ is $1\mathrm{/}a$, and integration by parts on $[0,R]$ followed by $R\to \mathrm{\infty }$ gives $I_m(a)=(m\mathrm{/}a)I_{m-1}(a)$; the boundary term vanishes by [F10]. Thus the $m$th diagonal entry of the mixture is $\frac{1}{\pi \bar{n}m!}{\int }_{\mathbb{C}}\mathrm{\mid }\alpha {\mathrm{\mid }}^{2m}{e}^{-(1+1\mathrm{/}\bar{n})\mathrm{\mid }\alpha {\mathrm{\mid }}^2}dudv=\frac{{\bar{n}}^m}{(1+\bar{n}{)}^{m+1}}=(1-\eta ){\eta }^m\mathrm{.}$ Let $P_K={\sum }_{m=0}^K\mathrm{\mid }m⟩⟨m\mathrm{\mid }$ and integrate the finite-rank operators $P_K\mathrm{\mid }\alpha ⟩⟨\alpha \mathrm{\mid }{P}_K$. Their diagonal entries just computed give ${\sum }_{m=0}^K(1-\eta ){\eta }^m\mathrm{\mid }m⟩⟨m\mathrm{\mid }$, which converges in trace norm to ${\rho }_{\bar{n}}$. Also $P_K\mathrm{\mid }\alpha ⟩\to \mathrm{\mid }\alpha ⟩$ in norm for each $\alpha$, and $\mathrm{\mid }⟨W(q,p)P_K\alpha ,P_K\alpha ⟩\mathrm{\mid }\le 1$. Dominated convergence [F6] therefore passes the characteristic function through the mixture. Multiplying the two scaled Gaussian Fourier factors just computed by the vacuum factor in the coherent formula gives $\mathrm{tr}({\rho }_{\bar{n}}W(q,p))=e^{-\mathrm{\hslash }(2\bar{n}+1)(q^2+p^2)\mathrm{/}4}=e^{-\nu (q^2+p^2)\mathrm{/}2},\nu =\mathrm{\hslash }(\bar{n}+1\mathrm{/}2)\mathrm{.}$ For $\bar{n}=0$, the same formula and covariance follow from [F9] with ${\alpha }_0=1$ and $\beta =\sqrt{\mathrm{\hslash }}q$ (and the $p$ coordinate).

Let $M_S$ be an implementer from [F3] and set ${\rho }_S=M_S\rho M_S^{\ast }$. The Weyl identity and cyclicity of trace against a bounded operator give ${\chi }_{{\rho }_S}(z)=\mathrm{tr}(\rho M_S^{\ast }W(z)M_S)={\chi }_{\rho }(S^{T}z)\mathrm{.}$ For a coordinate vector $e_j$, the explicit Weyl formula gives $W(t{e}_j)=e^{it{R}_j}$. Therefore $M_S^{\ast }{e}^{it{R}_j}{M}_S=W(t{S}^T{e}_j)$; its derivative on Schwartz functions at zero is $i{\sum }_k{S}_{jk}{R}_k$. By [F7], the two strongly continuous groups have unique self-adjoint generators, so $M_S^{\ast }{R}_j{M}_S=\overline{{\sum }_k{S}_{jk}{R}_k}$. The left side has Schwartz as a core because every generator unitary in [F3] preserves Schwartz. By [F2], Schwartz is dense in the joint quadrature graph norm, so every vector with all quadrature second moments lies in the domain of these finite sums and the identity acts there as written. Applying it to the eigenvectors of $\rho$ and using the absolutely convergent form sums in [F1] gives $d_S=Sd$ and $V_S=SV{S}^T$. For transformations arising from finite Bogoliubov matrices, these are exactly the core and Weyl interfaces of [F4].

For necessity in the converse, suppose a density operator $\rho$ has the displayed Gaussian characteristic function with parameters $d,V$, without initially assuming moment bounds. Let $E_j$ be the spectral PVM of $R_j$ and ${\mu }_j(B)=\mathrm{tr}(\rho E_j(B))$. Then $\int e^{it\lambda }d{\mu }_j(\lambda )=\chi (t{e}_j)$. With $t_n=1\mathrm{/}n$, [F13] gives $\frac{1}{2}\int {\lambda }^{2}d{\mu }_j\le {\mathrm{lim\hspace{0.17em}inf}}_{n\to \mathrm{\infty }}\int \frac{1-\cos (t_n\lambda )}{t_n^2}d{\mu }_j=\frac{d_j^2+V_{jj}}{2}<\mathrm{\infty }\mathrm{.}$ For an orthonormal density expansion with weights $w_n>0$, positivity and the definition of the nonnegative integral identify this second moment with ${\sum }_n{w}_n\int {\lambda }^{2}d{E}_{j,{\psi }_n}$. The spectral domain formula [F13] therefore places each ${\psi }_n$ in $D(R_j)$ and gives ${\sum }_n{w}_n\mathrm{\parallel }{R}_j{\psi }_n{\mathrm{\parallel }}^2<\mathrm{\infty }$. Thus the hypotheses of step 1.1 hold. To identify the resulting moment parameters with the prescribed $d,V$, fix real $z$ and let $A_z$ be the self-adjoint generator of $t\mapsto W(tz)$. For $z\mathrm{\ne }0$, extend $z$ to a symplectic basis (choose a vector with nonzero pairing, normalize it, then induct in its symplectic orthogonal complement); [F3] and the generator argument of step 2.1 show that $A_z$ has Schwartz core and equals ${\sum }_j{z}_j{R}_j$ on the joint domain. The case $z=0$ is the zero generator. The joint-domain moment bounds make ${\sum }_n{w}_n\mathrm{\parallel }{A}_z{\psi }_n{\mathrm{\parallel }}^2$ finite. In its scalar spectral probability, the bounds $\mathrm{\mid }\sin (t\lambda )\mathrm{/}t\mathrm{\mid }\le \mathrm{\mid }\lambda \mathrm{\mid }$ and $0\le (1-\cos (t\lambda ))\mathrm{/}{t}^2\le {\lambda }^2\mathrm{/}2$ permit dominated convergence. Comparing with $\chi (tz)$ gives $⟨A_z⟩=z^{T}d$ and $⟨A_z^2⟩=(z^{T}d{)}^2+z^{T}Vz$, where the second moment denotes the spectral quadratic form. These are also the directional mean and covariance forms of the now finite-moment state, so equality for every real $z$ identifies its mean and real symmetric covariance with $d,V$. Step 1.1 yields $V+i\mathrm{\hslash }J\mathrm{/}2⪰0$, proving necessity for every density operator, not just those initially assumed to have finite moments.

Put ${\bar{n}}_j={\nu }_j\mathrm{/}\mathrm{\hslash }-1\mathrm{/}2\ge 0$ and take the finite tensor product of the one-mode states ${\rho }_{{\bar{n}}_j}$. In the occupation basis it is positive trace class with trace one, and its characteristic function is the product of the one-mode functions from step 1.3, namely $\exp (-z^{T}Dz\mathrm{/}2)$. Conjugate this product state by an implementer $M_S$ from [F3]; step 2.1 gives characteristic function $\exp (-z^{T}SD{S}^{T}z\mathrm{/}2)=\exp (-z^{T}Vz\mathrm{/}2)$. Finally set $z_0=Jd\mathrm{/}\mathrm{\hslash }$ and conjugate by the Weyl unitary $W(z_0)$. The Weyl commutation relation yields $W(z_0{)}^{\ast }W(z)W(z_0)=e^{i\mathrm{\hslash }{z}_0^{T}Jz}W(z)=e^{i{z}^{T}d}W(z),$ because $d=-\mathrm{\hslash }J{z}_0$. The resulting unitary conjugate of the product thermal state is positive, trace class and trace one, and has exactly the target characteristic function. This proves sufficiency and completes the finite-mode equivalence. No infinite-mode inference is used.