Schwartz/distribution spaces and Hilbert Fock conventions

For $f\in H$, put $f_m=f{\mathbf{1}}_{[-m,m]}$. Then $f_m\in D(Q)$ because $\mathrm{\mid }x\mathrm{\mid }\le m$ on its support, and $f_m\to f$ in $L^2$ by dominated convergence applied to $\mathrm{\mid }{f}_m-f{\mathrm{\mid }}^2\le \mathrm{\mid }f{\mathrm{\mid }}^2$; hence $D(Q)$ is dense. For $f,g\in D(Q)$, $xf\bar{g}$ is integrable by Cauchy–Schwarz, and real-valuedness of $x$ gives $⟨Qf,g⟩=⟨f,Qg⟩$.

Equip $\mathcal{G}=\{f:{\int }_{\mathbb{R}}(1+x^2)\mathrm{\mid }f(x){\mathrm{\mid }}^{2}dx<\mathrm{\infty }\}$ with the explicit weighted inner product $⟨f,g{⟩}_{\mathcal{G}}:={\int }_{\mathbb{R}}(1+x^2)f(x)\overline{g(x)}dx$. Multiplication by $\sqrt{1+x^2}$ is an onto isometry from $\mathcal{G}$ to $L^2(\mathbb{R})$, with inverse multiplication by $(1+x^2{)}^{-1\mathrm{/}2}$, so $\mathcal{G}$ is Hilbert. Its norm equals $\mathrm{\parallel }f{\mathrm{\parallel }}_Q$ on $D(Q)$. Suppose $f\in \mathcal{G}$ is orthogonal to every $g_n$ in this weighted inner product. Since the $H_n$ have successive degrees and nonzero leading coefficients [F2], their finite span is all polynomials. Therefore $h(x)=(1+x^2)f(x)g_0(x)$ lies in $L^1$ by Cauchy–Schwarz and $\int h(x)x^{j}dx=0$ for every $j\ge 0$.

Let $g\in D(Q^{\ast })$ and write $k=Q^{\ast }g$. For each $m$, the vector $f_m=(xg-k){\mathbf{1}}_{[-m,m]}$ belongs to $D(Q)$ since $x$ is bounded there and $g,k\in L^2$. The adjoint identity for $f_m$ yields $0=⟨Q{f}_m,g⟩-⟨f_m,k⟩={\int }_{-m}^m\mathrm{\mid }xg-k{\mathrm{\mid }}^{2}dx$. Thus $xg=k$ almost everywhere on every $[-m,m]$, hence on $\mathbb{R}$; since $k\in L^2$, $g\in D(Q)$ and $Q^{\ast }g=Qg$. Conversely, symmetry in step 1.1 gives $D(Q)\subseteq D(Q^{\ast })$ and $Q^{\ast }f=Qf$. Therefore $Q=Q^{\ast }$.

Fix $\xi \in \mathbb{R}$ and let $S_N(x)={\sum }_{j=0}^N(-2\pi i\xi x{)}^j\mathrm{/}j!$. By [F4], $S_N(x)\to e^{-2\pi i\xi x}$ and $\mathrm{\mid }{S}_N(x)\mathrm{\mid }\le e^{2\pi \mathrm{\mid }\xi \mathrm{\mid }\mathrm{\mid }x\mathrm{\mid }}$. The majorant $\mathrm{\mid }h(x)\mathrm{\mid }{e}^{2\pi \mathrm{\mid }\xi \mathrm{\mid }\mathrm{\mid }x\mathrm{\mid }}$ is integrable: Cauchy–Schwarz bounds its integral by $\mathrm{\parallel }f{\mathrm{\parallel }}_{\mathcal{G}}(\int (1+x^2)g_0(x{)}^2{e}^{4\pi \mathrm{\mid }\xi \mathrm{\mid }\mathrm{\mid }x\mathrm{\mid }}dx{)}^{1\mathrm{/}2}$, and the latter integral is finite because completing the square gives $-x^2+4\pi \mathrm{\mid }\xi \mathrm{\mid }\mathrm{\mid }x\mathrm{\mid }=-(\mathrm{\mid }x\mathrm{\mid }-2\pi \mathrm{\mid }\xi \mathrm{\mid }{)}^2+(2\pi \mathrm{\mid }\xi \mathrm{\mid }{)}^2$, so [F5] applies. Dominated convergence and the vanishing moments from step 1.2 imply $\widehat{h}(\xi )={\lim }_N\int h{S}_N=0$. Since $\xi$ was arbitrary, $\widehat{h}$ vanishes identically.

Fourier uniqueness [F3] gives $h=0$ almost everywhere. Since $1+x^2>0$ and $g_0(x)>0$ for every $x$, this implies $f=0$ in $\mathcal{G}$. Thus the orthogonal complement of the Hermite span in $\mathcal{G}$ is zero; [F6] makes that span dense in $\mathcal{G}$. Because the $\mathcal{G}$ norm is the graph norm of $Q$, the Hermite span is a graph core, proving the statement.

Measurable representatives may be chosen for $f,f^{\mathrm{\prime }},g,g^{\mathrm{\prime }}$. Replacing a representative on a null set changes its product only on a product-null set: cover the other sigma-finite factor by countably many finite-measure sets and apply the rectangle formula [F1] on each. Thus the product rule is well defined on $L^2$ classes. By [F7], $f\overline{f^{\mathrm{\prime }}}$ and $g\overline{g^{\mathrm{\prime }}}$ are integrable; Fubini [F2] gives the product norm $\mathrm{\parallel }f(x)g(y){\mathrm{\parallel }}_2^2=\mathrm{\parallel }f{\mathrm{\parallel }}_2^2\mathrm{\parallel }g{\mathrm{\parallel }}_2^2$, so each elementary product belongs to the target space. It also gives $⟨f(x)g(y),f^{\mathrm{\prime }}(x)g^{\mathrm{\prime }}(y){⟩}_{L^2(X\times Y)}=⟨f,f^{\mathrm{\prime }}{⟩}_{H_X}⟨g,g^{\mathrm{\prime }}{⟩}_{H_Y}\mathrm{.}$ By sesquilinearity, the rule ${\sum }_j{\alpha }_j{f}_j\otimes g_j\mapsto {\sum }_j{\alpha }_j{f}_j(x)g_j(y)$ preserves inner products on finite sums. If two tensor presentations define the same algebraic tensor, their difference has zero tensor norm by [F5], hence its product function has zero $L^2$ norm; the rule is well-defined on the algebraic tensor product and extends uniquely as an isometry $T$ to the completion.

Choose sigma-finite exhaustions $X_k\uparrow X$ and $Y_k\uparrow Y$ with finite measure, and put $Z_k=X_k\times Y_k$. By the rectangle formula [F1], $(\mu \times \nu )(Z_k)<\mathrm{\infty }$, and $Z_k\uparrow X\times Y$. For fixed $k$, let ${\mathcal{C}}_k$ be the family of product-measurable $E\subseteq Z_k$ such that ${\mathbf{1}}_E$ belongs to the closed range of $T$. It contains $Z_k$, since ${\mathbf{1}}_{Z_k}=T({\mathbf{1}}_{X_k}\otimes {\mathbf{1}}_{Y_k})$. If $E\subseteq F$ are in ${\mathcal{C}}_k$, then ${\mathbf{1}}_{F\setminus E}={\mathbf{1}}_F-{\mathbf{1}}_E$ is in that closed range. If $E_j\uparrow E$ in $Z_k$, then $\mathrm{\parallel }{\mathbf{1}}_E-{\mathbf{1}}_{E_j}{\mathrm{\parallel }}_2^2=(\mu \times \nu )(E\setminus E_j)\to 0$ by [F6], so $E\in {\mathcal{C}}_k$. Thus ${\mathcal{C}}_k$ is a lambda-system on $Z_k$. For every measurable rectangle $A\times B$, its restriction to $Z_k$ is $(A\cap X_k)\times (B\cap Y_k)$, whose indicator is the image of ${\mathbf{1}}_{A\cap X_k}\otimes {\mathbf{1}}_{B\cap Y_k}$; these restricted rectangles are a generating pi-system by [F4]. Dynkin's theorem [F4] gives ${\mathbf{1}}_E\in \overline{\mathrm{ran}T}$ for every product-measurable $E\subseteq Z_k$.

Let $E$ be any completed-product-measurable set with finite product measure. By the completion definition [F1], it differs from a product-measurable set $E_0$ by a subset of a product-null set, so ${\mathbf{1}}_E={\mathbf{1}}_{E_0}$ as $L^2$ classes. Since $Z_k\uparrow X\times Y$, [F6] gives $(\mu \times \nu )(E_0\setminus Z_k)\to 0$, so ${\mathbf{1}}_{E_0}={\lim }_k{\mathbf{1}}_{E_0\cap Z_k}$ in $L^2$; step 2.1 puts each truncated indicator in the closed range of $T$. Every finite-measure simple function for the completed measure is a finite linear combination of indicators of completed-measurable finite-measure sets; replacing each set by its product-measurable representative gives the same $L^2$ class. Such simple functions are dense by [F3]. Consequently the range of $T$ is dense in the full target $L^2$ space. Since the range of an isometry from a complete space is closed, $T$ is onto and therefore unitary. Its uniqueness follows because elementary tensors span a dense subspace.

Each sector $H_s^{(n)}$ or $H_a^{(n)}$ is a closed subspace of its Hilbert tensor power, being the range of the corresponding orthogonal projection [F2]. The square-summable direct sum of these sectors is complete: for a Cauchy sequence, each coordinate converges in its complete sector; finite partial sums of squared coordinate norms are bounded by the common sequence bound, so the coordinate limits are square-summable; taking coordinate limits in the Cauchy estimate over each finite set of coordinates and then the supremum proves convergence in the direct-sum norm. Truncating a square-summable sequence after sector $N$ converges in norm, so ${\mathcal{D}}_{\mathrm{f}\mathrm{i}\mathrm{n}}$ is dense.

On sector $n$, creation and contraction have norms at most $\sqrt{n+1}\mathrm{\parallel }h\mathrm{\parallel }$ and $\sqrt{n}\mathrm{\parallel }h\mathrm{\parallel }$, respectively: the contraction has norm at most $\mathrm{\parallel }h\mathrm{\parallel }$ on its first factor by Cauchy–Schwarz and extends to the tensor product with the same bound [F4]; each symmetrizer or antisymmetrizer has norm at most one [F2]. For $\mathrm{\Phi }\in H_s^{(n)}$ and $\mathrm{\Psi }\in H_s^{(n+1)}$, use $P_{s,n+1}\mathrm{\Psi }=\mathrm{\Psi }$ and self-adjointness of $P_{s,n+1}$ to compute $⟨a_n^{†}(h)\mathrm{\Phi },\mathrm{\Psi }{⟩}_2=\sqrt{n+1}⟨h\otimes \mathrm{\Phi },\mathrm{\Psi }{⟩}_2=\sqrt{n+1}⟨\mathrm{\Phi },(C_h\otimes I^{\otimes n})\mathrm{\Psi }{⟩}_2=⟨\mathrm{\Phi },a_{n+1}(h)\mathrm{\Psi }{⟩}_2$; the tensor product pairing gives the middle equality because $C_h(v)=⟨h,v{⟩}_2$. The same calculation with $P_{a,n+1}$ proves the fermionic relation.

The maximal bosonic direct-sum maps are closed. For example, if ${\mathrm{\Phi }}^{(j)}\to \mathrm{\Phi }$ and $a(h){\mathrm{\Phi }}^{(j)}\to \mathrm{\Theta }$ in Fock norm, boundedness of each sector map gives coordinatewise ${\mathrm{\Theta }}_{n-1}=a_n(h){\mathrm{\Phi }}_n$; since $\mathrm{\Theta }$ is square-summable, the defining sum for $D(a(h))$ is finite and $a(h)\mathrm{\Phi }=\mathrm{\Theta }$. The same coordinate argument applies to $a^{†}(h)$. Truncating a vector and its image after sector $N$ shows ${\mathcal{D}}_{\mathrm{f}\mathrm{i}\mathrm{n}}$ is a graph core for each. For the adjoint, testing against vectors supported in sector $n+1$ forces the adjoint output coordinate in sector $n+1$ to be $a_n^{†}(h){\mathrm{\Phi }}_n$; square-summability of these coordinates is exactly the displayed domain of $a^{†}(h)$. Conversely, if that sum is finite, the sector pairing of step 1.2 and Cauchy–Schwarz in the direct sum give the adjoint identity for every vector in $D(a(h))$. Thus $a(h{)}^{\ast }=a^{†}(h)$. Testing $a^{†}(h{)}^{\ast }$ against input sector $n$ similarly forces its output coordinate to be $a_{n+1}(h){\mathrm{\Phi }}_{n+1}$; square-summability and the same pairing give exactly $D(a(h))$ and the converse inclusion. Hence the maximal maps are mutually adjoint.

For $n\ge 1$ and a symmetric elementary tensor $v_1\vee \cdots \vee v_n:=P_{s,n}(v_1\otimes \cdots \otimes v_n)$, contraction and creation give $a(h)(v_1\vee \cdots \vee v_n)=n^{-1\mathrm{/}2}{\sum }_{j=1}^n⟨h,v_j{⟩}_2{v}_1\vee \cdots \widehat{v_j}\cdots \vee v_n$ and $a^{†}(k)(v_1\vee \cdots \vee v_n)=\sqrt{n+1}{P}_{s,n+1}(k\otimes v_1\otimes \cdots \otimes v_n)$, by expanding the permutation averages [F2]. In the composition $a(h)a^{†}(k)$, the contraction term for the newly inserted $k$ is $⟨h,k{⟩}_2(v_1\vee \cdots \vee v_n)$; each of the other $n$ contraction terms is exactly the corresponding term of $a^{†}(k)a(h)$. Hence their difference is $⟨h,k{⟩}_{2}I$. On sector zero, $a(h)\mathrm{\Omega }=0$, $a^{†}(k)\mathrm{\Omega }=k$, and $a_1(h)k=⟨h,k{⟩}_2$ give the same commutator directly. Symmetry of $P_{s,n}$ makes two creations commute; two contractions commute because contracting distinct tensor slots in either order gives the same sum. Finite elementary symmetric tensors are dense in each sector and the sector maps are bounded there, so the identities hold on all of ${\mathcal{D}}_{\mathrm{f}\mathrm{i}\mathrm{n}}$.

For a unit vector $h$, put $K=h^{\perp }$. In the $n$th antisymmetric sector, every $v\in H$ decomposes as $v=⟨h,v{⟩}_{2}h+(v-⟨h,v{⟩}_{2}h)$ with the second term in $K$, so $H=\mathbb{C}h\oplus K$. Tensors split orthogonally into those with no $h$ factor and those with exactly one: expand each tensor factor in $\mathbb{C}h\oplus K$; terms with two $h$ factors vanish under antisymmetrization, and different placements of the single $h$ are orthogonal. The map $\sqrt{n+1}{P}_{a,n+1}(h\otimes \cdot )$ is an isometry from the no-$h$ part of sector $n$ onto the one-$h$ part of sector $n+1$, since the $n+1$ placements are orthogonal and each has coefficient $1\mathrm{/}(n+1)$ in the antisymmetrizer. It vanishes on the one-$h$ part. Its adjoint $c(h)$ reverses this isometry and vanishes on the no-$h$ part. Thus $c(h)c^{†}(h)+c^{†}(h)c(h)=I$ and both maps have norm one. Scaling gives $\{c(h),c^{†}(h)\}=\mathrm{\parallel }h{\mathrm{\parallel }}^{2}I$ and norms $\mathrm{\parallel }h\mathrm{\parallel }$ for $h\mathrm{\ne }0$; for $h=0$ these statements hold because both operators vanish. Antisymmetry gives $c^{†}(h{)}^2=0$; taking adjoints gives $c(h{)}^2=0$. Polarizing the diagonal anticommutator at $h+k$ and $h+ik$, using that $c(\cdot )$ is conjugate-linear and $c^{†}(\cdot )$ is linear, yields $\{c(h),c^{†}(k)\}=⟨h,k{⟩}_{2}I$: if $q(u)=\{c(u),c^{†}(u)\}$, then the sesquilinear polarization identity is $B(h,k)=\frac{1}{4}(q(h+k)-q(h-k)-iq(h+ik)+iq(h-ik))$, where $B(h,k)=\{c(h),c^{†}(k)\}$; substituting $q(u)=\mathrm{\parallel }u{\mathrm{\parallel }}^{2}I$ gives the claimed scalar. Polarizing $c(u{)}^2=0$ and $c^{†}(u{)}^2=0$ gives the two same-type anticommutators. All fermionic maps are bounded, so these identities hold on the whole Fock space.

By [F4], $K$ is a Hilbert space, so its symmetric tensor sectors and Fock space in the Statement are defined. Fix $k,m\ge 0$, write $n=k+m$, and let ${\psi }_m\in K_s^{(m)}$. Expanding the permutation average and using symmetry of ${\psi }_m$ gives $P_{s,n}(e^{\otimes k}\otimes {\psi }_m)={\left(\genfrac{}{}{0px}{}{n}{k}\right)}^{-1}{\sum }_{\begin{array}{c}I\subseteq \{1,\dots ,n\}\\ \mathrm{\mid }I\mathrm{\mid }=k\end{array}}{v}_I$, where $v_I$ is the tensor with $e$ in the slots in $I$ and the $m$ factors of ${\psi }_m$ in the remaining slots, symmetrized there. Indeed, for each fixed placement $I$, the $k!m!$ permutations of the equal $e$ factors and the symmetric ${\psi }_m$ give the same tensor, and $k!m!\mathrm{/}n!={\left(\genfrac{}{}{0px}{}{n}{k}\right)}^{-1}$.

Distinct placements $v_I$ are orthogonal: if $I\mathrm{\ne }{I}^{\mathrm{\prime }}$, some slot contains $e$ in one tensor and a vector of $K$ in the other, and $e\perp K$. Each $v_I$ has norm $\mathrm{\parallel }{\psi }_m\mathrm{\parallel }$, so step 1.1 gives $\mathrm{\parallel }{P}_{s,n}(e^{\otimes k}\otimes {\psi }_m){\mathrm{\parallel }}^2={\left(\genfrac{}{}{0px}{}{n}{k}\right)}^{-1}\mathrm{\parallel }{\psi }_m{\mathrm{\parallel }}^2$. The same orthogonality shows that the images for different $k$ in a fixed $n$ sector are orthogonal. Thus the factor $\sqrt{\left(\genfrac{}{}{0px}{}{n}{k}\right)}$ makes $J_{k,m}$ isometric and their sector ranges are orthogonal.

Algebraic elementary tensors in $H^{\widehat{\otimes }n}$ are dense by [F1]. Expand each factor using $H=\mathbb{C}e\oplus K$, distribute the finite tensor product, and apply $P_{s,n}$. Every resulting term has exactly $k$ factors $e$ and $m=n-k$ factors in $K$, so it lies in the range described in step 2.1. Since $P_{s,n}$ is bounded and algebraic elementary tensors are dense, their symmetrizations are dense in $H_s^{(n)}$; the range of the isometry is closed, hence it equals $H_s^{(n)}$. Taking the orthogonal direct sum over $n=k+m$ proves that $J$ is onto and unitary.

On a vector $e^{\otimes k}\otimes {\psi }_m$ with $n=k+m$, creation by $e$ acts by $\sqrt{n+1}{P}_{s,n+1}(e\otimes J_{k,m}(e^{\otimes k}\otimes {\psi }_m))$. Since $P_{s,n+1}$ averages over all permutations, symmetrizing the last $n$ slots first does not change the result. This equals $\sqrt{n+1}\sqrt{\left(\genfrac{}{}{0px}{}{n}{k}\right)}{P}_{s,n+1}(e^{\otimes (k+1)}\otimes {\psi }_m)$. The binomial identity $\left(\genfrac{}{}{0px}{}{n+1}{k+1}\right)=\frac{n+1}{k+1}\left(\genfrac{}{}{0px}{}{n}{k}\right)$ makes this exactly $J_{k+1,m}(\sqrt{k+1}{e}^{\otimes (k+1)}\otimes {\psi }_m)$. Thus creation becomes the stated one-mode raising map. Annihilation by $e$ is zero for $k=0$, because every first tensor factor lies in $K$. For $k\ge 1$, the placement formula of step 1.1 shows that contraction keeps exactly the $\left(\genfrac{}{}{0px}{}{n-1}{k-1}\right)$ placements with $e$ in the contracted slot. Therefore $a(e)J_{k,m}{\psi }_m=\sqrt{n}\sqrt{\left(\genfrac{}{}{0px}{}{n}{k}\right)}\frac{\left(\genfrac{}{}{0px}{}{n-1}{k-1}\right)}{\left(\genfrac{}{}{0px}{}{n}{k}\right)}{P}_{s,n-1}(e^{\otimes (k-1)}\otimes {\psi }_m)=\sqrt{k}{J}_{k-1,m}{\psi }_m$, using $\left(\genfrac{}{}{0px}{}{n}{k}\right)=\frac{n}{k}\left(\genfrac{}{}{0px}{}{n-1}{k-1}\right)$. Thus annihilation becomes the stated one-mode lowering map and leaves the $K$ factor unchanged.

For $n\ge 1$, each sector $H_s^{(n)}$ is a closed subspace of its Hilbert tensor power, because it is the range of an orthogonal projection [F3]; the zero-particle sector $H_s^{(0)}=\mathbb{C}$ is also complete. To prove that the Fock direct sum is complete for arbitrary $H$, let $({\mathrm{\Psi }}^{(k)}{)}_{k\ge 1}$ be Cauchy in its square-sum norm. Each coordinate is Cauchy in the complete sector $H_s^{(n)}$, so it has a limit ${\mathrm{\Psi }}_n$. A Cauchy sequence in a normed space is bounded (its tail lies within distance $1$ of one fixed term, and its finite initial segment is bounded); taking limits of each finite partial sum ${\sum }_{n=0}^N\mathrm{\parallel }{\mathrm{\Psi }}_n^{(k)}{\mathrm{\parallel }}^2$ shows ${\sup }_N{\sum }_{n=0}^N\mathrm{\parallel }{\mathrm{\Psi }}_n{\mathrm{\parallel }}^2<\mathrm{\infty }$, hence $\mathrm{\Psi }=({\mathrm{\Psi }}_n)$ belongs to the direct sum. Given $\epsilon >0$, choose $K$ so that $\mathrm{\parallel }{\mathrm{\Psi }}^{(k)}-{\mathrm{\Psi }}^{(\mathrm{\ell })}\mathrm{\parallel }\le \epsilon$ for $k,\mathrm{\ell }\ge K$; for fixed $k\ge K$, take $\mathrm{\ell }\to \mathrm{\infty }$ in every finite coordinate sum to obtain ${\sum }_{n=0}^N\mathrm{\parallel }{\mathrm{\Psi }}_n^{(k)}-{\mathrm{\Psi }}_n{\mathrm{\parallel }}^2\le {\epsilon }^2$ for every $N$. Taking the supremum over $N$ proves ${\mathrm{\Psi }}^{(k)}\to \mathrm{\Psi }$. Truncating the coordinate sequence proves that ${\mathcal{D}}_{\mathrm{f}\mathrm{i}\mathrm{n}}$ is dense. For $n=1$ use the identity in [F2]. For $n\ge 2$, iterate the product-tensor unitary of [F2]; induction gives a unitary from $H^{\widehat{\otimes }n}$ to the full completed-product $L^2(X^n,{\mu }_n)$. On elementary tensors the permutation $U_{\pi }$ becomes coordinate permutation, so by density the unitary intertwines $P_{s,n}$ with averaging over coordinate permutations [F3]. Its restriction identifies $H_s^{(n)}$ with the closed symmetric subspace. Each $s_n$ is Borel measurable by [A3], since it is a finite sum of measurable coordinate maps, and is invariant under all permutations. The $n=0$ sector is $\mathbb{C}$ by definition.