Vacuum emission Dyson convergence

[Free flow and vertex bounds] Diagonalizing the finite Hermitian $H_a$ and using [F3], [F4] gives $U_0(t)=e^{it{H}_a\mathrm{/}\mathrm{\hslash }}\otimes e^{it{H}_f\mathrm{/}\mathrm{\hslash }}$. Thus $U_0(t)$ preserves each ${\mathcal{K}}_m$, commutes with $F$ on $\mathcal{G}$, and is isometric and strongly continuous for the graph norm: both $U_0(t)x$ and $F{U}_0(t)x=U_0(t)Fx$ are continuous. On ${\mathcal{K}}_m$, the finite-sector extension of $B$ from [F2] satisfies $\mathrm{\parallel }Bx\mathrm{\parallel }\le \mathrm{\mid }{\eta }_0\mathrm{\mid }\gamma \sqrt{m+1}\mathrm{\parallel }x\mathrm{\parallel }$ and takes values in ${\mathcal{K}}_{m+1}$. Its free conjugate therefore satisfies $\mathrm{\parallel }V(t)x\mathrm{\parallel }\le \mathrm{\mid }{\eta }_0\mathrm{\mid }\gamma \sqrt{m+1}\mathrm{\parallel }x\mathrm{\parallel }$. The factor $\mathrm{\mid }{\eta }_0\mathrm{\mid }$ is included exactly once in this bound; $\gamma$ and $C$ contain no coupling.

[Cutoff commutator and graph-domain extension] For $g$ supported in $\{0<\epsilon \le \mathrm{\Lambda }\}$, $\epsilon g\in \mathfrak{h}$ and $\mathrm{\parallel }\epsilon g\mathrm{\parallel }\le \mathrm{\Lambda }\mathrm{\parallel }g\mathrm{\parallel }$. On symmetric elementary tensors with factors in $D(\epsilon )$, the sector multiplier and creation formula give $[H_f,A_{+}(g)]=A_{+}(\epsilon g)$. For contraction the removed factor contributes $-⟨\epsilon u,g⟩=-⟨u,\epsilon g⟩$, so $[H_f,A_{-}(g)]=-A_{-}(\epsilon g)$. These identities extend to every finite-photon graph vector, with no claim that all of $F_0$ lies in $D(H_f)$. Indeed, in sector $n$ first cut each mode to $\epsilon \le R$; dominated convergence in the Euclidean representation [F3] gives convergence of both the vector and its energy multiplier. On that cutoff subspace, energy is bounded by $nR$, so finite sums of elementary cutoff tensors, symmetrized, approximate in graph norm by tensor density. This supplies a graph core in each ${\mathcal{G}}_m$. The field and its candidate commutator are bounded on ${\mathcal{K}}_m$ by [F2], with respective bounds $2\sqrt{m+1}\mathrm{\parallel }g\mathrm{\parallel }$ and $2\mathrm{\Lambda }\sqrt{m+1}\mathrm{\parallel }g\mathrm{\parallel }$. Applying the core identity to a graph-approximating sequence makes the output and its energy image converge; closedness of $H_f$ proves graph-domain preservation and the identity on the limit. Hence for $x\in {\mathcal{G}}_m$, $Bx\in {\mathcal{G}}_{m+1}$ and $FBx=BFx+Q_{B}x,Q_B:={\eta }_0{\sum }_j{D}_j\otimes (A_{+}(\epsilon g_j)-A_{-}(\epsilon g_j)),\mathrm{\parallel }{Q}_{B}x\mathrm{\parallel }\le \mathrm{\mid }{\eta }_0\mathrm{\mid }\gamma \mathrm{\Lambda }\sqrt{m+1}\mathrm{\parallel }x\mathrm{\parallel }\mathrm{.}$

[Conjugated commutator and graph continuity] Since $F$ commutes with $U_0(t)$, step 1.2 gives on ${\mathcal{G}}_m$ $FV(t)x=V(t)Fx+Q(t)x,Q(t):=U_0(t)Q_B{U}_0(-t),$ with $\mathrm{\parallel }Q(t)x\mathrm{\parallel }\le \mathrm{\mid }{\eta }_0\mathrm{\mid }\gamma \mathrm{\Lambda }\sqrt{m+1}\mathrm{\parallel }x\mathrm{\parallel }$. The conjugations in $Q(t)$ are retained. Equivalently, the sector formulas give $D_j(t)=e^{it{H}_a\mathrm{/}\mathrm{\hslash }}{D}_j{e}^{-it{H}_a\mathrm{/}\mathrm{\hslash }}$ and $g_j(t)=e^{it\epsilon \mathrm{/}\mathrm{\hslash }}{g}_j$, with $V(t)={\eta }_0{\sum }_j{D}_j(t)\otimes \varphi (g_j(t)),Q(t)={\eta }_0{\sum }_j{D}_j(t)\otimes (A_{+}(\epsilon g_j(t))-A_{-}(\epsilon g_j(t)))\mathrm{.}$ The same positive phase occurs in the annihilation vector argument because that argument is conjugate-linear, as in [F2]. The resulting graph bound is $\mathrm{\parallel }V(t)x{\mathrm{\parallel }}_{\mathcal{G}}\le \mathrm{\mid }{\eta }_0\mathrm{\mid }\gamma \sqrt{m+1}(\mathrm{\parallel }Fx\mathrm{\parallel }+(1+\mathrm{\Lambda })\mathrm{\parallel }x\mathrm{\parallel })\mathrm{.}$ Thus $V(t):{\mathcal{G}}_m\to {\mathcal{G}}_{m+1}$ is uniformly bounded in $t$ for these graph norms. Strong graph continuity of $U_0$, the fixed graph-bounded $B$ between these spaces, and the expression $U_0(t)B{U}_0(-t)x$ show that $t\mapsto V(t)x$ is graph continuous. On the full $\mathcal{G}$, [F1] and the graph isometry of $U_0$ additionally give $\mathrm{\parallel }V(t)x\mathrm{\parallel }\le K\mathrm{\parallel }x{\mathrm{\parallel }}_{\mathcal{G}}$ for one finite $K$ independent of $t$, and strong continuity as a map $\mathcal{G}\to \mathcal{K}$. No global map $V(t):\mathcal{G}\to \mathcal{G}$ is asserted.

[Chronological products and their graph bounds] Define $P_n(t_1,\dots ,t_n):=V(t_n)\cdots V(t_1){\mathrm{\Psi }}_0$, with $P_0={\mathrm{\Psi }}_0$. At the $k$-th vertex, the earlier product is in ${\mathcal{G}}_{k-1}$, so steps 1.1 and 2.1 give a factor $\mathrm{\mid }{\eta }_0\mathrm{\mid }\gamma \sqrt{k}$ in its Hilbert bound. Consequently $P_n\in {\mathcal{G}}_n$ and $\mathrm{\parallel }{P}_n\mathrm{\parallel }\le (\mathrm{\mid }{\eta }_0\mathrm{\mid }\gamma {)}^n\sqrt{n!}$. The graph commutator gives recursively $\mathrm{\parallel }F{P}_k\mathrm{\parallel }\le \mathrm{\mid }{\eta }_0\mathrm{\mid }\gamma \sqrt{k}(\mathrm{\parallel }F{P}_{k-1}\mathrm{\parallel }+\mathrm{\Lambda }\mathrm{\parallel }{P}_{k-1}\mathrm{\parallel })$; since $F{\mathrm{\Psi }}_0=0$, induction yields $\mathrm{\parallel }F{P}_n\mathrm{\parallel }\le n\mathrm{\Lambda }(\mathrm{\mid }{\eta }_0\mathrm{\mid }\gamma {)}^n\sqrt{n!}$. Each product is jointly continuous in its time variables for the graph norm: at each stage use strong graph continuity of the next $V$ on a fixed vector and its uniform graph bound to control the changing previous vector. For $n=0$ the Hilbert bound is $1$ and the energy bound is $0$, including when $\mathrm{\mid }{\eta }_0\mathrm{\mid }\gamma =0$.

[Graph Bochner terms and the genuine recurrence] The graph space $\mathcal{G}$ is Banach: a graph-Cauchy sequence has limits of both the vectors and their $F$ images in the Hilbert space, and closedness of $F$ identifies them as a graph limit. Each ${\mathcal{G}}_n$ is closed in it. Step 3.1 and [F5] therefore make every iterated chronological integral a Bochner integral in ${\mathcal{G}}_n$. Its graph and Hilbert norm bounds follow by integrating the preceding product bounds over the simplex; the volume is $t^n\mathrm{/}n!$, obtained recursively from ${\int }_0^t{s}^{n-1}\mathrm{/}(n-1)!ds=t^n\mathrm{/}n!$. Each factor of the integral contributes $1\mathrm{/}\mathrm{\hslash }$, giving $\mathrm{\parallel }{\mathrm{\Psi }}_n(t)\mathrm{\parallel }\le \frac{(\mathrm{\mid }{\eta }_0\mathrm{\mid }Ct{)}^n}{\sqrt{n!}},\mathrm{\parallel }F{\mathrm{\Psi }}_n(t)\mathrm{\parallel }\le \frac{n\mathrm{\Lambda }(\mathrm{\mid }{\eta }_0\mathrm{\mid }Ct{)}^n}{\sqrt{n!}}\mathrm{.}$ For $n=0$ these bounds mean $1$ and $0$. Integrating the latest vertex last gives the actual left-vertex recurrence ${\mathrm{\Psi }}_{n+1}(t)=\frac{1}{i\mathrm{\hslash }}{\int }_0^{t}V(s){\mathrm{\Psi }}_n(s)ds\mathrm{.}$ For fixed $s$, move $V(s)$ through the inner integral as a bounded linear map ${\mathcal{G}}_n\to {\mathcal{G}}_{n+1}$ by [F5]; the earlier vertices have times below $s$, with $V(s)$ on the left. This also proves graph continuity of each term, and proves (a) and graph-domain membership in (b).

[Uniform graph convergence] Fix $T<\mathrm{\infty }$ and put $a:=\mathrm{\mid }{\eta }_0\mathrm{\mid }CT$. The graph majorant is ${\sum }_{n\ge 0}(1+n\mathrm{\Lambda })a^n\mathrm{/}\sqrt{n!}$. For $a>0$ the ratio of consecutive scalar terms tends to zero: it is $a(1+(n+1)\mathrm{\Lambda })\mathrm{/}((1+n\mathrm{\Lambda })\sqrt{n+1})$. It is eventually at most $1\mathrm{/}2$, proving convergence by a geometric tail comparison. If $a=0$, every positive-order term is zero and only the zeroth term remains. Thus the series converges uniformly for $0\le t\le T$ in the Banach graph space, and its sum $\mathrm{\Psi }(t)$ is graph continuous and belongs to $D(H)=\mathcal{G}$.

[Integral equation and Hilbert derivative] The uniform bound $\mathrm{\parallel }V(s)x\mathrm{\parallel }\le K\mathrm{\parallel }x{\mathrm{\parallel }}_{\mathcal{G}}$ from step 2.1 makes $V(s){\sum }_{n\le M}{\mathrm{\Psi }}_n(s)$ converge uniformly in Hilbert norm on $[0,T]$ to $V(s)\mathrm{\Psi }(s)$. This limiting curve is continuous by graph continuity of $\mathrm{\Psi }$, strong continuity of $V:\mathcal{G}\to \mathcal{K}$, and the same uniform bound. Sum the recurrences of step 4.1 and use the Bochner norm inequality [F5] to pass to the limit in their Hilbert integrals. The result is $\mathrm{\Psi }(t)={\mathrm{\Psi }}_0+(i\mathrm{\hslash }{)}^{-1}{\int }_0^{t}V(s)\mathrm{\Psi }(s)ds$. The Banach-valued fundamental theorem now gives $\mathrm{\Psi }\in C^1([0,T];\mathcal{K})$, with the right derivative at zero, and $i\mathrm{\hslash }{\mathrm{\Psi }}^{\mathrm{\prime }}(t)=V(t)\mathrm{\Psi }(t)$, $\mathrm{\Psi }(0)={\mathrm{\Psi }}_0$. Along with step 5.1 this proves (b); graph continuity is used for domain control, while the differential equation is a Hilbert-space derivative.

[Identification with the self-adjoint dynamics] Put $Y(t):=U_0(-t)\mathrm{\Psi }(t)$. It is graph continuous and belongs to $D(H)$. The difference quotient splits into the difference of the free group on the fixed domain vector $\mathrm{\Psi }(t)$ and the free group times the difference quotient of $\mathrm{\Psi }$, so [F4] and step 6.1 justify the product derivative in Hilbert norm. Thus $i\mathrm{\hslash }{Y}^{\mathrm{\prime }}=H_{0}Y+U_0(-t)V(t)\mathrm{\Psi }=H_{0}Y+BY=HY$ and $Y(0)={\mathrm{\Psi }}_0$. By [F1], [F4], the Stone curve $e^{-itH\mathrm{/}\mathrm{\hslash }}{\mathrm{\Psi }}_0$ is another strong domain-valued solution because ${\mathrm{\Psi }}_0\in \mathcal{G}=D(H)$. For their difference $w$, symmetry of $H$ gives $\frac{d}{dt}\mathrm{\parallel }w(t){\mathrm{\parallel }}^2=2\mathrm{Re}⟨w^{\mathrm{\prime }}(t),w(t)⟩=2\mathrm{Re}(-\frac{i}{\mathrm{\hslash }}⟨Hw(t),w(t)⟩)=0.$ Both initial values agree, so $w=0$, proving (c) for every finite $t\ge 0$.

[Tail, vacuum and zero branches] By step 7.1 and unitarity, the truncation error at time $t$ equals in norm the interaction-picture tail, so it is at most ${\sum }_{n\ge N}(\mathrm{\mid }{\eta }_0\mathrm{\mid }Ct{)}^n\mathrm{/}\sqrt{n!}$; for $0\le t\le T$ this is at most the stated $T$ tail, proving (d). The vacuum zeroth term has $F{\mathrm{\Psi }}_0=0$ and norm one. At $t=0$ all positive-order integrals vanish. If ${\eta }_0=0$, or if $\gamma =0$ (all contributing $D_j$ or $g_j$ zero), then $B=V=0$, all positive-order terms vanish, the interaction-picture sum is exactly ${\mathrm{\Psi }}_0$, and the tail is zero for every $N\ge 1$. The argument supplies finite-time convergence and the actual self-adjoint evolution, with the original scope limitations retained.