Second-quantized multipliers, joint spectral measure and exact domains

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.

Let $\mathrm{\Omega }={⨆}_{n\ge 0}{X}^n$ with the disjoint-union sigma-algebra and measure ${\mu }_{\mathrm{\Omega }}(C)={\sum }_{n\ge 0}{\mu }_n(C\cap X^n)$, using the conventions of [A2]; write $s_j$ for $s_{n,j}$ on component $n$. The Fock realization in step 1.1 is a closed subspace of $L^2(\mathrm{\Omega },{\mu }_{\mathrm{\Omega }})$ consisting of symmetric components. For $B\in \mathcal{B}({\mathbb{R}}^d)$, define $E(B)$ by multiplication by ${\mathbf{1}}_{s^{-1}(B)}$. Since $s$ is invariant under permutations, this multiplier preserves the symmetric subspace. It is an orthogonal projection because its multiplier is real and idempotent; $E(\mathrm{\varnothing })=0$, $E({\mathbb{R}}^d)=I$ and $E(B)E(C)=E(B\cap C)$ follow pointwise. For pairwise disjoint $B_k\in \mathcal{B}({\mathbb{R}}^d)$ with union $B$, the squared norm of $E(B)\mathrm{\Psi }-{\sum }_{k=0}^{N}E(B_k)\mathrm{\Psi }$ is ${\sum }_{n\ge 0}{\int }_{X^n}{\mathbf{1}}_{s_n^{-1}(B\setminus {\cup }_{k=0}^N{B}_k)}\mathrm{\mid }{\mathrm{\Psi }}_n{\mathrm{\mid }}^{2}d{\mu }_n\mathrm{.}$ The indicators tend pointwise to zero and the integrand is bounded by $\mathrm{\mid }{\mathrm{\Psi }}_n{\mathrm{\mid }}^2$, whose total integral is $\mathrm{\parallel }\mathrm{\Psi }{\mathrm{\parallel }}^2$; dominated convergence [F6] gives norm convergence. Hence $E$ is a strongly countably additive PVM on the Borel sets.

Fix a real measurable multiplier $f$ on $\mathrm{\Omega }$ that is invariant under coordinate permutations, and define $M_f\mathrm{\Psi }=f\mathrm{\Psi }$ on the symmetric domain $D(M_f)=\{\mathrm{\Psi }:f\mathrm{\Psi }\in L^2(\mathrm{\Omega })\}$. This domain is dense: the symmetric cutoffs $\mathrm{\Psi }{\mathbf{1}}_{\mathrm{\mid }f\mathrm{\mid }\le m}$ belong to it and converge in norm by dominated convergence. Reality of $f$ makes $M_f$ symmetric. If $\mathrm{\Psi }\in D(M_f^{\ast })$ and $K=M_f^{\ast }\mathrm{\Psi }$, then for each $m$ the vector $v_m=(f\mathrm{\Psi }-K){\mathbf{1}}_{\mathrm{\mid }f\mathrm{\mid }\le m}$ is symmetric and lies in $D(M_f)$, since $f$ is bounded on its support. The adjoint identity gives $0=⟨M_f{v}_m,\mathrm{\Psi }{⟩}_2-⟨v_m,K{⟩}_2={\int }_{\mathrm{\mid }f\mathrm{\mid }\le m}\mathrm{\mid }f\mathrm{\Psi }-K{\mathrm{\mid }}^{2}d{\mu }_{\mathrm{\Omega }}\mathrm{.}$ Taking the union over $m$ shows $f\mathrm{\Psi }=K$ almost everywhere, so $\mathrm{\Psi }\in D(M_f)$ and $M_f^{\ast }\mathrm{\Psi }=M_f\mathrm{\Psi }$. Conversely, for every $\mathrm{\Psi }\in D(M_f)$, reality of $f$ gives the adjoint identity against every domain vector, so $M_f^{\ast }\mathrm{\Psi }=M_f\mathrm{\Psi }$. Therefore $M_f$ is self-adjoint on its maximal multiplication domain. Apply this to $f=s_j$ to obtain each self-adjoint $H_j$, and to $f=\theta \cdot s$ to obtain $H_{\theta }$.

For each $j$, the multiplier $s_j$ defines $H_j$ with the displayed component domain because the squared norm of $H_j\mathrm{\Psi }$ is exactly ${\sum }_n\int \mathrm{\mid }{s}_{n,j}{\mathrm{\mid }}^2\mathrm{\mid }{\mathrm{\Psi }}_n{\mathrm{\mid }}^{2}d{\mu }_n$; the vacuum term is zero. For $\theta$, the same calculation gives $D(H_{\theta })=\{\mathrm{\Psi }:{\sum }_n\int \mathrm{\mid }\theta \cdot s_n{\mathrm{\mid }}^2\mathrm{\mid }{\mathrm{\Psi }}_n{\mathrm{\mid }}^{2}d{\mu }_n<\mathrm{\infty }\}$. The PVM from step 2.1 has coordinate functions $s_j$, so its indicators give the joint spectral projections for all the $H_j$ and its scalar-coordinate multipliers have these exact domains. Truncating a vector in any such domain after sector $N$ makes both its norm tail and operator-image tail tend to zero; hence ${\mathcal{D}}_{\mathrm{f}\mathrm{i}\mathrm{n}}\cap D(H_j)$ and ${\mathcal{D}}_{\mathrm{f}\mathrm{i}\mathrm{n}}\cap D(H_{\theta })$ are graph cores.

By [F6], $U_{\theta }(t)$ is unitary and $U_{\theta }(t+s)=U_{\theta }(t)U_{\theta }(s)$ on each component, including the vacuum where the phase is one. For any $\mathrm{\Psi }$, the squared norm of $U_{\theta }(t)\mathrm{\Psi }-U_{\theta }(t_0)\mathrm{\Psi }$ is the integral over $\mathrm{\Omega }$ of $\mathrm{\mid }{e}^{it\theta \cdot s}-e^{i{t}_0\theta \cdot s}{\mathrm{\mid }}^2\mathrm{\mid }\mathrm{\Psi }{\mathrm{\mid }}^2$; the integrand tends pointwise to zero and is bounded by $4\mathrm{\mid }\mathrm{\Psi }{\mathrm{\mid }}^2$, so [F6] proves strong continuity. If $\mathrm{\Psi }\in D(H_{\theta })$, the pointwise difference quotient tends to $i(\theta \cdot s)\mathrm{\Psi }$ and its squared error is bounded by $4\mathrm{\mid }\theta \cdot s{\mathrm{\mid }}^2\mathrm{\mid }\mathrm{\Psi }{\mathrm{\mid }}^2$; dominated convergence proves norm convergence. Conversely, if the difference quotient has a norm limit as $t\to 0$, take $t_m=1\mathrm{/}m$. Fatou's lemma [F6] on the disjoint union gives ${\sum }_n\int \mathrm{\mid }\theta \cdot s_n{\mathrm{\mid }}^2\mathrm{\mid }{\mathrm{\Psi }}_n{\mathrm{\mid }}^{2}d{\mu }_n\le {\mathrm{lim\hspace{0.17em}inf}}_m{\parallel \frac{U_{\theta }(t_m)\mathrm{\Psi }-\mathrm{\Psi }}{t_m}\parallel }^2<\mathrm{\infty },$ so $\mathrm{\Psi }\in D(H_{\theta })$; the forward calculation identifies the limit as $i{H}_{\theta }\mathrm{\Psi }$. This proves the exact derivative criterion and all claims.