Number operator domain

[Coordinate adjoint, vacuum and positivity] The coordinate Hilbert-space construction used in [F1] is valid under the present hypotheses: the sector spaces are complete closed tensor subspaces, the pairing converges by Cauchy-Schwarz, and a Cauchy sequence has coordinate limits. For each finite coordinate sum, the Cauchy norm bound passes to the limit; taking the supremum over those finite sums proves that the limit is square summable and that convergence holds in the full sum. Finite-coordinate truncations give density of $F_0$. Thus these Hilbert and density facts do not require invoking the Axiom of Choice premise of the general operator-sum theorem. [F1, F2, algebra] Every vector of $F_0$ lies in $D(N)$, so $N$ is densely defined. For $\psi ,\phi \in D(N)$, the sequence form of Cauchy-Schwarz in [F2] makes the coordinate pairing series absolutely convergent and gives $⟨N\psi ,\phi ⟩={\sum }_{n}n⟨{\psi }_n,{\phi }_n⟩=⟨\psi ,N\phi ⟩$; hence $N\subseteq N^{\ast }$. If $\phi \in D(N^{\ast })$ and $\eta =N^{\ast }\phi$, test the adjoint identity with every vector ${\iota }_n(u)$ in a single sector. It gives $⟨nu,{\phi }_n⟩=⟨u,{\eta }_n⟩$ for all $u$, so ${\eta }_n=n{\phi }_n$ for every $n$, including ${\eta }_0=0$. Consequently ${\sum }_n{n}^2\mathrm{\parallel }{\phi }_n{\mathrm{\parallel }}^2=\mathrm{\parallel }\eta {\mathrm{\parallel }}^2<\mathrm{\infty }$, proving $\phi \in D(N)$ and $N^{\ast }\phi =N\phi$. Thus $N=N^{\ast }$ on exactly the displayed domain. For $\psi \in D(N)$, $⟨N\psi ,\psi ⟩={\sum }_{n}n\mathrm{\parallel }{\psi }_n{\mathrm{\parallel }}^2\ge 0$, and the vacuum satisfies $N\mathrm{\Omega }=0$. The finite-coordinate truncations ${\psi }^{(M)}$ lie in $F_0$ and satisfy ${\psi }^{(M)}\to \psi$ and $N{\psi }^{(M)}\to N\psi$ by convergence of ${\sum }_n{n}^2\mathrm{\parallel }{\psi }_n{\mathrm{\parallel }}^2$, so $F_0$ is a graph core. This retains all conclusions of (a) under its original Countable Choice hypothesis.

[Commutation on the core] Let $f\in H$ and $\psi \in F_0$; all vectors below lie in $F_0$ by [F3]. Using the sector formulas of [F3], for $n\ge 1$ the $n$-th coordinate of $S_{+}(f)N\psi$ is $\sqrt{n}{P}_{+}^{(n)}(f\otimes (n-1){\psi }_{n-1})=(n-1)(S_{+}(f)\psi {)}_n$, while the $n$-th coordinate of $N{S}_{+}(f)\psi$ is $n(S_{+}(f)\psi {)}_n$; both coordinates vanish for $n=0$. Hence $N{S}_{+}(f)\psi -S_{+}(f)N\psi =S_{+}(f)\psi$. Similarly, the $n$-th coordinate of $S_{-}(f)N\psi$ is $\sqrt{n+1}{\mathrm{\Psi }}_f^{(n)}((n+1){\psi }_{n+1})=(n+1)(S_{-}(f)\psi {)}_n$, and that of $N{S}_{-}(f)\psi$ is $n(S_{-}(f)\psi {)}_n$; their difference is $-(S_{-}(f)\psi {)}_n$ for every $n$, proving the second identity.

[Eigenvalues and spectrum] Let $n\in \mathbb{N}$ and $u\in \mathrm{ran}{P}_{+}^{(n)}$, $u\mathrm{\ne }0$. The vector ${\iota }_n(u)$ lies in $D(N)$ with $(N{\iota }_n(u){)}_m=m({\iota }_n(u){)}_m$, which is $n{\iota }_n(u)$; thus every $n\in \mathbb{N}$ is an eigenvalue when $H\mathrm{\ne }\{0\}$, since then each sector contains $e^{\otimes n}\mathrm{\ne }0$ (and the vacuum sector is $\mathbb{C}$). Now fix $\lambda \in \mathbb{C}\setminus \mathbb{N}$ and put $c_n:=(n-\lambda {)}^{-1}$. If $n>2\mathrm{\mid }\lambda \mathrm{\mid }$, then $\mathrm{\mid }n-\lambda \mathrm{\mid }\ge n-\mathrm{\mid }\lambda \mathrm{\mid }>n\mathrm{/}2$, so $\mathrm{\mid }{c}_n\mathrm{\mid }<2\mathrm{/}n\le 2$ and $n\mathrm{\mid }{c}_n\mathrm{\mid }<2$; among the finitely many remaining indices each denominator is nonzero because $\lambda \mathrm{\notin }\mathbb{N}$. Hence ${\sup }_n\mathrm{\mid }{c}_n\mathrm{\mid }<\mathrm{\infty }$ and ${\sup }_{n}n\mathrm{\mid }{c}_n\mathrm{\mid }<\mathrm{\infty }$. Define $R_{\lambda }\psi :=(c_n{\psi }_n{)}_n$. Then $\mathrm{\parallel }{R}_{\lambda }\psi {\mathrm{\parallel }}^2={\sum }_n\mathrm{\mid }{c}_n{\mathrm{\mid }}^2\mathrm{\parallel }{\psi }_n{\mathrm{\parallel }}^2\le ({\sup }_n\mathrm{\mid }{c}_n\mathrm{\mid }{)}^2\mathrm{\parallel }\psi {\mathrm{\parallel }}^2$ and ${\sum }_n{n}^2\mathrm{\parallel }{c}_n{\psi }_n{\mathrm{\parallel }}^2\le ({\sup }_{n}n\mathrm{\mid }{c}_n\mathrm{\mid }{)}^2\mathrm{\parallel }\psi {\mathrm{\parallel }}^2$, so $R_{\lambda }$ is bounded from $F(H)$ to $D(N)$ with its graph norm. Coordinatewise, $(N-\lambda I)R_{\lambda }=I$ on $F(H)$ and $R_{\lambda }(N-\lambda I)=I$ on $D(N)$; hence $R_{\lambda }$ is a bounded inverse of $N-\lambda I$. Therefore $\sigma (N)\subseteq \mathbb{N}$, and the exhibited eigenvalues give $\sigma (N)=\mathbb{N}$ for $H\mathrm{\ne }\{0\}$.

[Domain inclusion and moments] If $\psi \in D(N)$ then ${\sum }_n(n+1)\mathrm{\parallel }{\psi }_n{\mathrm{\parallel }}^2={\sum }_{n}n\mathrm{\parallel }{\psi }_n{\mathrm{\parallel }}^2+{\sum }_n\mathrm{\parallel }{\psi }_n{\mathrm{\parallel }}^2\le {\sum }_n{n}^2\mathrm{\parallel }{\psi }_n{\mathrm{\parallel }}^2+\mathrm{\parallel }\psi {\mathrm{\parallel }}^2<\mathrm{\infty }$ (using $n\le n^2$), so $D(N)\subseteq D_1$. The coordinate formula $(N\psi {)}_n=n{\psi }_n$ and orthogonality of the sectors give $\mathrm{\parallel }N\psi {\mathrm{\parallel }}^2={\sum }_n\mathrm{\parallel }n{\psi }_n{\mathrm{\parallel }}^2={\sum }_n{n}^2\mathrm{\parallel }{\psi }_n{\mathrm{\parallel }}^2$ and $⟨N\psi ,\psi ⟩={\sum }_{n}n\mathrm{\parallel }{\psi }_n{\mathrm{\parallel }}^2$, both series converging absolutely; this proves (d) and completes the proof of (a)-(d).