Finite-mode Fock, Gaussian and Bogoliubov foundations

For fixed $n$, the symmetric tensors are the range of the averaging projection on $({\mathbb{C}}^r{)}^{\otimes n}$. The elementary tensors in the basis $e_1,\dots ,e_r$ span the tensor power, so their averages span that range. Two such averages with different occupation vectors are orthogonal, because at least one tensor slot has orthogonal basis vectors in every paired term. For occupation vector $N$ the squared norm of the averaged tensor is ${\prod }_j{n}_j!\mathrm{/}n!$ by counting the ${\prod }_j{n}_j!$ permutations that preserve the ordered occupation word among $n!$ permutations. Multiplying by $\sqrt{n!\mathrm{/}{\prod }_j{n}_j!}$ normalizes it. The CCR in [F2], applied successively to the vacuum, show that this normalized vector is exactly $\mathrm{\mid }N⟩$; therefore the vectors $\mathrm{\mid }N⟩$ are an orthonormal basis of each sector. The sector direct sum in [F1] makes their union a complete orthonormal family of Fock space.

The target family $(g_N)$ is complete orthonormal by [F3]. Hence the linear map $\mathrm{\mid }N⟩\mapsto g_N$ preserves norms on finite linear combinations, extends by completion to an isometry on Fock space, and is surjective because its range contains the complete family $(g_N)$ and is closed. Uniqueness follows from density of the finite span.

On a basis vector $\mathrm{\mid }N⟩$, the ladder formulas [F2] and Hermite recurrences [F3] give $\mathcal{U}{a}_j{\mathcal{U}}^{-1}{g}_N=\sqrt{n_j}{g}_{N-e_j}=\frac{x_j+{\mathrm{\partial }}_{x_j}}{\sqrt{2}}{g}_N,$ and $\mathcal{U}{a}_j^{\ast }{\mathcal{U}}^{-1}{g}_N=\sqrt{n_j+1}{g}_{N+e_j}=\frac{x_j-{\mathrm{\partial }}_{x_j}}{\sqrt{2}}{g}_N\mathrm{.}$ Linearity proves these identities on the finite span. Adding and subtracting the two identities gives the displayed $Q_j,P_j$ formulas there. The formulas are only asserted on this core, where all terms are defined.

Each listed matrix is symplectic by direct multiplication with $J_0$: $U_X^T{J}_0{U}_X=J_0$ iff $X=X^T$, $L_Y^T{J}_0{L}_Y=J_0$ iff $Y=Y^T$, and $D_A^T{J}_0{D}_A=J_0$. A quarter-turn on a subset is itself a block diagonal product of two-dimensional matrices $\left(\begin{array}{cc}0& -1\\ 1& 0\end{array}\right)$ and identities, so is symplectic; on one selected coordinate it equals the product $U_{-1}{L}_1{U}_{-1}$ in that coordinate. Thus all proposed generators are symplectic.

First suppose the upper-left block $A$ is invertible. The block equations in [F1] imply $A^{-1}B=(A^{-1}B{)}^T$ and $C{A}^{-1}=(C{A}^{-1}{)}^T$ by multiplying $A{B}^T=B{A}^T$ and $A^{T}C=C^{T}A$ by the relevant inverses. They also give $D=A^{-T}+C{A}^{-1}B$ from $A^{T}D-C^{T}B=I$. Therefore direct block multiplication yields $S=L_{C{A}^{-1}}{D}_A{U}_{A^{-1}B},$ which is the required factorization in this case.

For arbitrary $A$, let $k=\mathrm{rank}A$. Choose invertible real matrices $L,R$ with $LAR=\mathrm{diag}(I_k,0)$, using row and column reduction, and replace $S$ by $S^{\mathrm{\prime }}=D_{L}S{D}_R$. This replacement is symplectic. Write its upper-right block as $B^{\mathrm{\prime }}$ and partition rows and columns after the first $k$. The identity $A^{\mathrm{\prime }}{B}^{\mathrm{\prime }T}=B^{\mathrm{\prime }}{A}^{\mathrm{\prime }T}$ in [F1], with $A^{\mathrm{\prime }}=\mathrm{diag}(I_k,0)$, forces the lower-left rectangular block of $B^{\mathrm{\prime }}$ to vanish. Since the first $r$ rows of $S^{\mathrm{\prime }}$ are independent, the bottom $r-k$ rows of $B^{\mathrm{\prime }}$ restricted to its last $r-k$ columns form an invertible matrix $B_{22}$. Right-multiply $S^{\mathrm{\prime }}$ by the inverse partial Fourier rotation $F$ on those last coordinates, $(q_j,p_j)\mapsto (p_j,-q_j)$. It is a product of the inverse listed generators. The resulting upper-left block is block upper triangular with diagonal blocks $I_k$ and $-B_{22}$, hence is invertible. This argument also covers $k=0$ and $k=r$ (with the empty block omitted).

The matrix $S^{\mathrm{\prime }}F$ has invertible upper-left block, so step 1.2 factors it into the stated generators. Since $S=D_L^{-1}(S^{\mathrm{\prime }}F)F^{-1}{D}_R^{-1}$ and the inverses of $D_L,D_R,F$ are of the same listed types, the same is true of $S$. This proves the factorization for every symplectic matrix.

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 $[Q_j,P_j]=i\mathrm{\hslash }I$.

The formula for $W$ is a unitary composition of translation and a modulus-one multiplier; its inverse is $W(-q,-p)$. Direct substitution gives $W(q,p)W(q^{\mathrm{\prime }},p^{\mathrm{\prime }})=e^{-i\mathrm{\hslash }(q\cdot p^{\mathrm{\prime }}-p\cdot q^{\mathrm{\prime }})\mathrm{/}2}W(q+q^{\mathrm{\prime }},p+p^{\mathrm{\prime }})\mathrm{.}$ For a real symmetric $X$, the chirp $[C_{X}f](x)=e^{i{x}^{T}Xx\mathrm{/}2}f(x)$ is unitary, and substitution in the Weyl formula gives $C_X^{\ast }W(q,p)C_X=W(q+Xp,p)$, the upper shear. For invertible real $A$, $[D_{A}f](x)=\mathrm{\mid }\det A{\mathrm{\mid }}^{1\mathrm{/}2}f(Ax)$ is unitary by [F5] applied to the diffeomorphism $x\mapsto Ax$ and the integrable function $\mathrm{\mid }f{\mathrm{\mid }}^2$; direct substitution gives $D_A^{\ast }W(q,p)D_A=W(A^{-T}q,Ap)$, an invertible coordinate change.

Let ${\mathcal{F}}_{\mathrm{a}\mathrm{n}\mathrm{g}}$ be the exact angular-frequency unitary of [F3]. Substitution into the Fourier integral on Schwartz functions gives $[{\mathcal{F}}_{\mathrm{a}\mathrm{n}\mathrm{g}}W(q,p)f](k)=e^{i\sqrt{\mathrm{\hslash }}p\cdot (k-\sqrt{\mathrm{\hslash }}q\mathrm{/}2)}[{\mathcal{F}}_{\mathrm{a}\mathrm{n}\mathrm{g}}f](k-\sqrt{\mathrm{\hslash }}q),$ which is exactly $[W(p,-q){\mathcal{F}}_{\mathrm{a}\mathrm{n}\mathrm{g}}f](k)$. Both sides are bounded operators, and Schwartz functions are dense, so this identity holds on all of $L^2$. Applying it to selected coordinates gives partial Fourier rotations. Conjugating an upper shear by the full or partial Fourier rotation gives every lower shear, since the matrix identity is $\left(\begin{array}{cc}0& -I\\ I& 0\end{array}\right)U_X\left(\begin{array}{cc}0& I\\ -I& 0\end{array}\right)=L_{-X}$. Thus each generator and inverse in [F2] has a unitary Weyl implementer.

Each generator unitary maps Schwartz space onto itself: this follows from the product rule for the smooth quadratic chirp, the chain rule for an invertible linear coordinate change, and Fourier invariance of Schwartz space. On any connected parameter neighborhood on which $A$ remains invertible, the chirp and coordinate formulas vary continuously in every Schwartz seminorm; the Fourier factor is fixed. If $G_1^{\ast }W(z)G_1=W(T_{1}z)$ and $G_2^{\ast }W(z)G_2=W(T_{2}z)$, then $(G_1{G}_2{)}^{\ast }W(z)(G_1{G}_2)=W(T_2{T}_{1}z)$. Therefore, for any finite factorization $T=T_k\cdots T_1$ supplied by [F2], multiplying the corresponding generator unitaries in the order $G_1\cdots G_k$ gives a unitary $M_T$ with $M_T^{\ast }W(z)M_T=W(Tz)$ for every label $z$. This proves existence and does not invoke a general metaplectic existence theorem.

Suppose $M_1,M_2$ implement the same $T$. Then $X=M_2{M}_1^{\ast }$ commutes with every $W(z)$ by applying their two covariance identities. Since the Weyl system is irreducible by [F1], every spectral projection of either bounded self-adjoint operator $\mathrm{Re}X=(X+X^{\ast })\mathrm{/}2$ and $\mathrm{Im}X=(X-X^{\ast })\mathrm{/}(2i)$ commutes with every $W(z)$ and has invariant range. Irreducibility forces each such projection to be $0$ or $I$. The spectral measure in [F4] then implies each self-adjoint part is scalar: if a bounded self-adjoint operator had two distinct spectral values, disjoint neighborhoods of those values would yield a nonzero proper spectral projection. Hence $X=\lambda I$ for some $\lambda \in \mathbb{C}$; since $X$ is unitary, $\mathrm{\mid }\lambda \mathrm{\mid }=1$. Thus $M_1,M_2$ differ only by a phase.

Put $J_0=\left(\begin{array}{cc}0& I\\ -I& 0\end{array}\right)$. For $f\in \mathcal{S}$, the $L^1$ definitions give the exact dilation identity ${\mathcal{F}}_{\mathrm{a}\mathrm{n}\mathrm{g}}f(k)=(2\pi {)}^{-r\mathrm{/}2}({\mathcal{F}}_{2\pi }f)(k\mathrm{/}(2\pi ))$. The change-of-variables formula with $k=2\pi \xi$ proves this dilation is unitary on $L^2$; differentiating its argument shows $p_{\alpha ,\beta }(Dg)=(2\pi {)}^{\mathrm{\mid }\alpha \mathrm{\mid }-\mathrm{\mid }\beta \mathrm{\mid }-r\mathrm{/}2}{p}_{\alpha ,\beta }(g)$, and the inverse dilation has the reciprocal seminorm factors, so it is a Schwartz automorphism. Integration by parts in the angular kernel gives ${\mathcal{F}}_{\mathrm{a}\mathrm{n}\mathrm{g}}(-i{\mathrm{\partial }}_{x_j}f)=k_j{\mathcal{F}}_{\mathrm{a}\mathrm{n}\mathrm{g}}f$. Thus the angular transform is precisely the required unitary, derivative and Schwartz-space consequence of the imported $2\pi$ transform, rather than a differently normalized Plancherel claim. Near the identity the block $A$ is invertible with positive determinant. For the factorization in [F2] define $C_{Y}f(x)=e^{i{x}^{T}Yx\mathrm{/}2}f(x)$, $E_{A}f(x)=(\det A{)}^{-1\mathrm{/}2}f(A^{-1}x)$ and $T_X={\mathcal{F}}_{\mathrm{a}\mathrm{n}\mathrm{g}}^{-1}{e}^{-i{k}^{T}Xk\mathrm{/}2}{\mathcal{F}}_{\mathrm{a}\mathrm{n}\mathrm{g}}$. Each is unitary by change of variables or [F4]. The product rule and linear change of variables show that each factor and its inverse act continuously on Schwartz space and depend smoothly on their parameters there. For example every derivative of a chirp is a polynomial times the same chirp, giving bounds by finitely many Schwartz seminorms, uniformly on compact parameter sets; differentiation in a parameter inserts only another polynomial or a first-order differential operator. The same argument applies in Fourier coordinates to $T_X$ by [F4]. Direct differentiation or substitution gives $C_Y^{\ast }{R}_g{C}_Y=L_Y{R}_g$, $E_A^{\ast }{R}_g{E}_A=D_A{R}_g$, and $T_X^{\ast }{R}_g{T}_X=U_X{R}_g$. The same substitutions in the explicit Weyl formula give covariance on all $L^2$. Thus $L(S):=C_Y{E}_A{T}_X$ is a smooth local unitary Schwartz lift of $S$, with $L(I)=I$. Shrink the neighborhood so $A$ is sufficiently close to $I$; then replacing $(Y,A,X)$ by $(\tau Y,I+\tau (A-I),\tau X)$, $0\le \tau \le 1$, gives a smooth operator path from $I$ to $L(S)$ within these formulas.

Along an absolutely continuous local parameter path the three logarithmic generators $i\mathrm{\hslash }{\dot{C}}_Y{C}_Y^{-1}$, $i\mathrm{\hslash }{\dot{E}}_A{E}_A^{-1}$ and $i\mathrm{\hslash }{\dot{T}}_X{T}_X^{-1}$ are respectively $-\mathrm{\hslash }{x}^T\dot{Y}x\mathrm{/}2$, $-i\mathrm{\hslash }((\dot{A}{A}^{-1}x)\cdot {\mathrm{\partial }}_x+\mathrm{tr}(\dot{A}{A}^{-1})\mathrm{/}2)$ and $\mathrm{\hslash }{p}^T\dot{X}p\mathrm{/}2$. These identities hold almost everywhere by finite-dimensional differentiation at the points where the parameters have derivatives. The half-trace in the middle expression is exactly the symmetric ordering of the mixed position-momentum term. Product differentiation conjugates and adds these expressions. The covariance from step 1.1 replaces each quadrature by a real linear combination, so a symmetrically ordered homogeneous quadratic stays exactly such a quadratic, with no extra scalar. If its coefficient matrix is $K_L$, differentiating the covariance and using $[R_j,R_k]=i\mathrm{\hslash }(J_0{)}_{jk}$ gives $\dot{S}=J_0{K}_{L}S$, hence $K_L=-J_0\dot{S}{S}^{-1}$. Equality of the quadratic coefficients is literal: the coefficients of second derivatives, of $x_j{\mathrm{\partial }}_k$, and of $x_j{x}_k$ determine the three real blocks, including the symmetric-order half-trace.

Put $V(t)=J_{0}K(t)$ and $m(t)={\int }_0^t\mathrm{\parallel }V(s)\mathrm{\parallel }ds$. The iterated-integral series for the matrix equation has $n$th term bounded by $m(t{)}^n\mathrm{/}n!$: the scalar product integrand is symmetric in its $n$ variables and the ordered simplex is one of the $n!$ equal regions of their product cube, up to null diagonals. Thus the series converges uniformly, with sum bounded by $e^{m(T)}$. It satisfies $S(t)=I+{\int }_0^{t}V(s)S(s)ds$, since the series under that integral is dominated by $\mathrm{\parallel }V(s)\mathrm{\parallel }{e}^{m(T)}$. For any integrable scalar $f$, choose $M$ with ${\int }_{\mathrm{\mid }f\mathrm{\mid }>M}\mathrm{\mid }f\mathrm{\mid }<\epsilon \mathrm{/}2$; then ${\int }_E\mathrm{\mid }f\mathrm{\mid }\le M\lambda (E)+\epsilon \mathrm{/}2$, proving absolute continuity of its primitive by the finite disjoint-interval definition. Applied entrywise to the integral equation, this proves absolute continuity of $S$, and [F6] gives $\dot{S}=VS$ almost everywhere. If two solutions have the same initial value, their continuous difference obeys $\mathrm{\Delta }(t)={\int }_0^{t}V(s)\mathrm{\Delta }(s)ds$. Iterating this equation $n$ times gives $\mathrm{\parallel }\mathrm{\Delta }(t)\mathrm{\parallel }\le {\sup }_s\mathrm{\parallel }\mathrm{\Delta }(s)\mathrm{\parallel }m(T{)}^n\mathrm{/}n!$, which tends to zero. This proves uniqueness of the classical absolutely continuous solution. Product differentiation gives $(S^T{J}_{0}S{)}^{\mathrm{\prime }}=0$ almost everywhere, so [F5] and its initial value give symplecticity at every time. A finite subdivision, using uniform continuity and invertibility of $S$, makes each $S(t)S(t_j{)}^{-1}$ lie in the neighborhood of step 1.1. Set $U(t)=L(S(t)S(t_j{)}^{-1})U(t_j)$, with $U(0)=I$. To justify the Schwartz calculus, restrict each parameter path to a small compact neighborhood where the local factors are smooth. The seminorm estimates of step 1.1, also for their parameter derivatives, imply $p((L(\alpha )-L(\beta ))f)\le C_{p,q}\mathrm{\mid }\alpha -\beta \mathrm{\mid }q(f)$ there and $p({\mathrm{\partial }}_{t}L(\alpha (t))f)\le C_{p,q}\mathrm{\mid }{\alpha }^{\mathrm{\prime }}(t)\mathrm{\mid }q(f)$ almost everywhere, for some Schwartz seminorm $q$. One can obtain the first bound by integrating along the straight parameter segment inside that neighborhood; finite smaller neighborhoods suffice. Matrix inversion and the local factor parameters have bounded derivatives on those sets, and $\mathrm{\mid }{S}^{\mathrm{\prime }}\mathrm{\mid }\le e^{m(T)}\mathrm{\parallel }V\mathrm{\parallel }$ is integrable. A finite-dimensional smooth map with bounded derivative on the small convex parameter neighborhood is Lipschitz there, hence takes an absolutely continuous parameter curve to an absolutely continuous curve; at points of differentiability of that curve, its differentiability gives the chain rule directly. The same derivative estimate in each seminorm proves this assertion for the explicit Schwartz-valued factors. Products are covered by the two-variable version of those bounds. Thus these estimates prove absolute continuity in every seminorm, the chain and product rules almost everywhere, and the same assertions for the inverse factors. Step 2.1 therefore gives $i\mathrm{\hslash }\dot{U}=H_{K}U$ almost everywhere, with continuous matching at joins and covariance from step 1.1. If two such evolutions have the same initial vector, their difference $w(t)$ is absolutely continuous in $L^2$ and satisfies the same equation. Its squared norm is absolutely continuous and has derivative $2\mathrm{Re}⟨-i{H}_{K}w\mathrm{/}\mathrm{\hslash },w⟩=0$ almost everywhere, because the real symmetrically ordered expression is symmetric on Schwartz space by integration by parts. By [F5] the difference remains zero. Density and unitarity extend uniqueness to all $L^2$. Thus existence, uniqueness, inverse continuity and covariance hold for the full integrable coefficient class in the Statement.