The fixed-mode isometry of Fock space and the transfer of number and ladder operators

[The powers of $\phi$ are orthonormal sector vectors] For $n\ge 0$ put ${\mathrm{\Phi }}_n:={\phi }^{\otimes n}$, with ${\mathrm{\Phi }}_0:=1$. By [F2] with $u=\phi$ and $\mathrm{\parallel }\phi \mathrm{\parallel }=1$ one has $\mathrm{\parallel }{\mathrm{\Phi }}_n\mathrm{\parallel }=1$ for every $n$, $P_{+}^{(n)}{\mathrm{\Phi }}_n={\mathrm{\Phi }}_n$, and $⟨{\mathrm{\Phi }}_n,{\mathrm{\Phi }}_n⟩={\prod }_{i=1}^n⟨\phi ,\phi ⟩=1$; for $m\mathrm{\ne }n$ the pairing $⟨{\mathrm{\Phi }}_m,{\mathrm{\Phi }}_n⟩$ vanishes because ${\mathrm{\Phi }}_m$ is supported in the direct-sum coordinate $m$ and ${\mathrm{\Phi }}_n$ in coordinate $n$, and the inner product of $F(H)$ is the sum of the coordinate pairings by [F1]. Hence $({\mathrm{\Phi }}_n{)}_{n\in \mathbb{N}}$ is an orthonormal family with ${\mathrm{\Phi }}_n\in \mathrm{ran}{P}_{+}^{(n)}$ for every $n$.

[The isometry] Define $U_{\phi }$ on the finite linear span of $(e_n)$ by $U_{\phi }{e}_n={\mathrm{\Phi }}_n$ and extend by linearity; by step 1.1, $\mathrm{\parallel }{U}_{\phi }\psi {\mathrm{\parallel }}^2={\sum }_n\mathrm{\mid }{\psi }_n{\mathrm{\mid }}^2=\mathrm{\parallel }\psi {\mathrm{\parallel }}^2$ for every finite combination, so $U_{\phi }$ is isometric on the dense finite-particle subspace $F_0(\mathbb{C})$ of Hilbert direct-sum construction clause (c) and extends uniquely to an isometry $U_{\phi }:F(\mathbb{C})\to F(H)$ with the same norm identity for all $\psi \in F(\mathbb{C})$ (A bounded linear operator between normed spaces, Parseval equivalences for an orthonormal family). Its range is the closed linear span of $({\mathrm{\Phi }}_n)$, since an isometry maps the closed span of an orthonormal family onto the closed span of its image and the image family is orthonormal; this proves (a).

[Ladder transfer on the exact domains] Write $\psi ={\sum }_n{\psi }_n{e}_n$. By [F4] and [F5], $\psi \in D(A_{+}(1))$ exactly when ${\sum }_n(n+1)\mathrm{\mid }{\psi }_n{\mathrm{\mid }}^2<\mathrm{\infty }$, while $\psi \in D(A_{-}(1))$ exactly when ${\sum }_n(n+1)\mathrm{\mid }{\psi }_{n+1}{\mathrm{\mid }}^2={\sum }_{m\ge 1}m\mathrm{\mid }{\psi }_m{\mathrm{\mid }}^2<\mathrm{\infty }$. For the unit mode $\phi$, [F2] gives $P_{+}^{(n+1)}(\phi \otimes {\psi }_n{\mathrm{\Phi }}_n)={\psi }_n{\mathrm{\Phi }}_{n+1}$ and [F4] gives ${\mathrm{\Psi }}_{\phi }^{(n)}({\psi }_{n+1}{\mathrm{\Phi }}_{n+1})={\psi }_{n+1}{\mathrm{\Phi }}_n$. Therefore the two exact domain sums for $U_{\phi }\psi$ equal the respective scalar sums, without any extra weighted output assumption. On these domains, $(A_{+}(\phi )U_{\phi }\psi {)}_n=\sqrt{n}{\psi }_{n-1}{\mathrm{\Phi }}_n=(U_{\phi }{A}_{+}(1)\psi {)}_n$ for $n\ge 1$, with zero vacuum coordinate, and $(A_{-}(\phi )U_{\phi }\psi {)}_n=\sqrt{n+1}{\psi }_{n+1}{\mathrm{\Phi }}_n=(U_{\phi }{A}_{-}(1)\psi {)}_n$. This proves (c), including its coherent-vector specialization by [F5].

[Energy expectation] Assume now $H=\mathfrak{h}$ and $\phi$ unit with $\epsilon \phi \in \mathfrak{h}$; write $\bar{\epsilon }:=⟨\epsilon \phi ,\phi ⟩$. For fixed $n$, ${\mathrm{\Phi }}_n={\phi }^{\otimes n}\in \mathrm{ran}{P}_{+}^{(n)}$ and, by [F6], $T_n{\mathrm{\Phi }}_n={\epsilon }_n{\mathrm{\Phi }}_n$; using the Euclidean identification and the symmetry of ${\epsilon }_n$ under the permutation action, the iterated $L^2$ integral gives $⟨T_n{\mathrm{\Phi }}_n,{\mathrm{\Phi }}_n⟩={\int }_{({\mathbb{R}}^3{)}^n}({\sum }_{i=1}^n\epsilon (k_i)){\prod }_{i=1}^n\mathrm{\parallel }\phi (k_i){\mathrm{\parallel }}_{{\mathbb{C}}^3}^2{d}^3{k}_1\cdots d^3{k}_n=n\bar{\epsilon }\mathrm{\parallel }\phi {\mathrm{\parallel }}^{2(n-1)}=n\bar{\epsilon },$ each of the $n$ summands contributing the same product $\bar{\epsilon }\mathrm{\parallel }\phi {\mathrm{\parallel }}^{2(n-1)}$ by Tonelli's theorem, the product measure being $\sigma$-finite and the integrand nonnegative; this is finite because $\epsilon \phi \in L^2$ and $\phi \in L^2$ give $\int \epsilon \mathrm{\parallel }\phi {\mathrm{\parallel }}^2\le \mathrm{\parallel }\epsilon \phi \mathrm{\parallel }\mathrm{\parallel }\phi \mathrm{\parallel }<\mathrm{\infty }$. The actual domain is also justified: by the triangle inequality on the product space, $\mathrm{\parallel }{T}_n{\mathrm{\Phi }}_n\mathrm{\parallel }\le n\mathrm{\parallel }\epsilon \phi \mathrm{\parallel }$, since each factor multiplier has that norm and all other factors have norm one. Hence ${\sum }_n\mathrm{\mid }{\psi }_n{\mathrm{\mid }}^2\mathrm{\parallel }{T}_n{\mathrm{\Phi }}_n{\mathrm{\parallel }}^2\le \mathrm{\parallel }\epsilon \phi {\mathrm{\parallel }}^2{\sum }_n{n}^2\mathrm{\mid }{\psi }_n{\mathrm{\mid }}^2<\mathrm{\infty }$, so $U_{\phi }\psi \in D(H_f)$ and its finite-sector truncations converge in the graph norm: the squared graph-tail norms are the tails of these two convergent nonnegative series. Therefore for a finite combination the sector-diagonal form of $H_f$ (clause (c) of [F6]) gives $⟨H_f{U}_{\phi }\psi ,U_{\phi }\psi ⟩={\sum }_n\mathrm{\mid }{\psi }_n{\mathrm{\mid }}^{2}n\bar{\epsilon }$; the general case ${\sum }_n{n}^2\mathrm{\mid }{\psi }_n{\mathrm{\mid }}^2<\mathrm{\infty }$ follows by approximation with truncations in the graph norm of $H_f$ (both sides are continuous under this approximation, the left because $H_f$ is closed and self-adjoint, the right by dominated convergence). This proves the first two displays of (d).

[Sector and number transfer] For $\psi ={\sum }_m{\psi }_m{e}_m$ in the dense span and fixed $n$, ${\mathrm{\Pi }}_n{U}_{\phi }\psi ={\psi }_n{\mathrm{\Phi }}_n=U_{\phi }({\psi }_n{e}_n)=U_{\phi }{\mathrm{\Pi }}_n^{\mathrm{\prime }}\psi$; since both sides are bounded in $\psi$ and the span is dense, the identity extends to all $\psi \in F(\mathbb{C})$, and the norm identity is step 2.1. For $\psi \in F(\mathbb{C})$ the coefficient vector of $U_{\phi }\psi$ is again $({\psi }_n)$, so ${\sum }_n{n}^2\mathrm{\parallel }(U_{\phi }\psi {)}_n{\mathrm{\parallel }}^2={\sum }_n{n}^2\mathrm{\mid }{\psi }_n{\mathrm{\mid }}^2$: $U_{\phi }\psi \in D(N)$ exactly when $\psi \in D(N^{\mathrm{\prime }})$, and then $(N{U}_{\phi }\psi {)}_n=n{\psi }_n{\mathrm{\Phi }}_n=(U_{\phi }{N}^{\mathrm{\prime }}\psi {)}_n$ for every $n$ by [F3]. This proves (b).

[Spectral second moment and literal square domain] For $\phi \in D(\epsilon )$, expand ${\epsilon }_n^2={\sum }_i\epsilon (k_i{)}^2+{\sum }_{i\mathrm{\ne }j}\epsilon (k_i)\epsilon (k_j)$ and integrate against ${\prod }_i\mathrm{\parallel }\phi (k_i){\mathrm{\parallel }}^2$. Tonelli gives $\mathrm{\parallel }{T}_n{\mathrm{\Phi }}_n{\mathrm{\parallel }}^2=n\mathrm{\parallel }\epsilon \phi {\mathrm{\parallel }}^2+n(n-1){\bar{\epsilon }}^2$, finite by the stated domain and Cauchy-Schwarz. This is the spectral second moment and requires no $T_n^2{\mathrm{\Phi }}_n$. If $\phi \in D({\epsilon }^2)$, the inequality $({\sum }_{i=1}^n\epsilon (k_i){)}^4\le n^3{\sum }_i\epsilon (k_i{)}^4$ gives $\mathrm{\parallel }{T}_n^2{\mathrm{\Phi }}_n{\mathrm{\parallel }}^2\le n^4\mathrm{\parallel }{\epsilon }^2\phi {\mathrm{\parallel }}^2<\mathrm{\infty }$, so ${\mathrm{\Phi }}_n\in D(T_n^2)$; self-adjointness then gives $⟨T_n^2{\mathrm{\Phi }}_n,{\mathrm{\Phi }}_n⟩=\mathrm{\parallel }{T}_n{\mathrm{\Phi }}_n{\mathrm{\parallel }}^2$, exactly the original operator-square display. The vacuum has energy zero and all its moments vanish, handled separately when $n=0$.