Energy-scale independence on the block-diagonal input algebra

The orthogonal projections are Hermitian, so real ${\omega }_0,{\eta }_J$ make $H_0({\omega }_0)+D$ Hermitian, as required by [F1]. If $X\in \mathcal{A}$, then $[H_0({\omega }_0),X]=0$ because $H_0({\omega }_0)$ is scalar on each block, and $[D,X]\in \mathcal{A}$. Write $T_{\delta J}=L_{\delta }{P}_J$. For distinct $J,K\ge 1$, the ranges of $T_{\delta J}$ and $T_{\delta K}$ lie in distinct orthogonal destination blocks, so $T_{\delta J}^{†}{T}_{\delta K}=0$; terms with $J=0$ or $K=0$ vanish as well. Hence $L_{\delta }^{†}{L}_{\delta }={\sum }_J{T}_{\delta J}^{†}{T}_{\delta J}$ is block diagonal. Also $L_{\delta }X{L}_{\delta }^{†}={\sum }_J{T}_{\delta J}(P_{J}X{P}_J)T_{\delta J}^{†}$ is block diagonal. Thus the anticommutator and every other term of the generator preserve $\mathcal{A}$.

For the reference assertion, use the product inner product of [F5] on $\mathcal{H}\otimes R$ and choose ordered orthonormal bases $u_1,\dots ,u_d$ of $\mathcal{H}$ and $e_1,\dots ,e_r$ of $R$, with matrix units $U_{ij}{u}_k={\delta }_{jk}{u}_i$ and $E_{ab}{e}_c={\delta }_{bc}{e}_a$. By [F4], every operator $Y$ on $\mathcal{H}\otimes R$ has a unique expansion $Y={\sum }_{i,j,a,b}{c}_{ijab}{U}_{ij}\otimes E_{ab}$. Grouping coefficients gives $Y={\sum }_{a,b}{X}_{ab}\otimes E_{ab}$ with $X_{ab}={\sum }_{i,j}{c}_{ijab}{U}_{ij}$, uniquely. This also establishes the product-basis identification used in the Statement: the tensor products of the matrix-unit bases of the two operator spaces are sent to the basis in [F4] by the factorwise operator action. Under this identification, [F3] gives a well-defined extension on all operators, with $({\mathcal{N}}_{t,{\omega }_0}\otimes \mathrm{id})(Y)={\sum }_{a,b}{\mathcal{N}}_{t,{\omega }_0}(X_{ab})\otimes E_{ab}$. Its factorwise rule determines it uniquely, so changing the chosen orthonormal bases does not change the map.

The restricted generator on the invariant algebra $\mathcal{A}$ is exactly the same linear map for every ${\omega }_0$, since the only parameter-dependent term vanishes there. Induction on the power shows that every power of the generator has the same restriction to $\mathcal{A}$ for every ${\omega }_0$. Applying the norm-convergent exponential series of [F1] proves the claimed equality of the restricted channels at fixed $t\ge 0$.

For a density operator $\rho$, the condition $\mathrm{supp}(\rho )\subseteq V_J$ is equivalent to $\rho =P_J\rho P_J$: by [F2] and self-adjointness, containment of the support implies that $V_J^{\perp }$ is in the kernel and that the range is in $V_J$; conversely the displayed identity puts $V_J^{\perp }$ in the kernel. Thus such a $\rho$ belongs to $\mathcal{A}$. For the reference assertion, the support identity in the Statement and uniqueness of the expansion in step 1.2 imply $X_{ab}=P_J{X}_{ab}{P}_J$ for every $a,b$: multiplying ${\sum }_{a,b}{X}_{ab}\otimes E_{ab}$ on both sides by $P_J\otimes I_R$ replaces each coefficient by $P_J{X}_{ab}{P}_J$. Each extended-output coefficient ${\mathcal{N}}_{t,{\omega }_0}(X_{ab})$ is therefore independent of ${\omega }_0$ by step 2.1. By [F5] and the factorwise action in [F3], $Q_J=P_J\otimes I_R$ satisfies $Q_J^2=Q_J$ and $⟨Q_{J}v,w⟩=⟨v,Q_{J}w⟩$, first on elementary tensors and then on all vectors by sesquilinearity; its range is $V_J\otimes R$. Thus $Q_J$ is the orthogonal projection onto that subspace, and the same support-kernel argument applies to density operators on $\mathcal{H}\otimes R$.

The subsequent-recovery consequence uses only functions: let $E$ be either output operator space, $S$ a fixed set, and $Q:E\to S$ any fixed function. For any fixed-input output $Z_{{\omega }_0}\in E$ proved independent above, equality of the outputs implies equality of $Q(Z_{{\omega }_0})$. More generally put $T_{{\omega }_0}={\mathcal{N}}_{t,{\omega }_0}{\mathrm{\mid }}_{\mathcal{A}}$, a complex-linear map from $\mathcal{A}$ to itself by steps 1.1 and 2.1. Let $\mathrm{\Phi }$ be a fixed function assigning to every complex-linear map $\mathcal{A}\to \mathcal{A}$ a function $E\to S$. Since $T_{{\omega }_0}=T_{{\omega }_0\mathrm{\prime }}$, we have $\mathrm{\Phi }(T_{{\omega }_0})=\mathrm{\Phi }(T_{{\omega }_0\mathrm{\prime }})$ and therefore $\mathrm{\Phi }(T_{{\omega }_0})(Z_{{\omega }_0})=\mathrm{\Phi }(T_{{\omega }_0\mathrm{\prime }})(Z_{{\omega }_0\mathrm{\prime }})$. A fixed subsequent recovery, or one selected solely by such a fixed rule from the restricted channel, has exactly this function-composition property. No positivity or other channel property of the postprocessing function is needed for this consequence.