Finite absorbing lowering dynamics and replacement limit

If $J=0$, the space is one-dimensional, all jumps vanish and $H$ is scalar, so $\mathcal{L}=0$ and every operator already equals its trace times $\mathrm{\mid }g⟩⟨g\mathrm{\mid }$. Assume henceforth $J\ge 1$. Put $D={\sum }_{j=0}^{J}j{P}_j$ and $Q=I-P_0$. The lowering relation gives $[D,L_a]=-L_a$, including on $V_0$, while $H$ and every $L_a^{†}{L}_a$ preserve the blocks. The latter follows from orthogonality: for $u\in V_j$ and $v\in V_k$ with $j\mathrm{\ne }k$, $⟨L_{a}u,L_{a}v⟩=0$. Thus cyclicity of the finite trace gives $\mathrm{Tr}(D\mathcal{L}(X))=-\mathrm{Tr}(RX)$ for every $X$. The blockwise rate hypothesis implies $R\ge r_{\ast }Q$; also $Q\le D\le JQ$.

Let $\rho$ be a density operator and put ${\rho }_t={\mathcal{N}}_t(\rho )$ and $m(t)=\mathrm{Tr}(D{\rho }_t)$. By [F2], ${\rho }_t$ is positive and has trace one. The convergent exponential series, differentiated entrywise using [F4], gives ${\dot{\rho }}_t=\mathcal{L}({\rho }_t)$. By step 1.1, $m^{\mathrm{\prime }}(t)=-\mathrm{Tr}(R{\rho }_t)\le -r_{\ast }\mathrm{Tr}(Q{\rho }_t)\le -(r_{\ast }\mathrm{/}J)m(t)$. The trace inequalities follow by diagonalizing the positive operator ${\rho }_t$ as in [F3] and applying the quadratic-form inequalities to its eigenvectors. Let $c=r_{\ast }\mathrm{/}J>0$. The difference quotient of $t\mapsto ct$ equals $c$, and the exponential is differentiable on $\mathbb{R}$; hence the chain rule in [F4], applied on real domains, gives $(e^{ct}{)}^{\mathrm{\prime }}=c{e}^{ct}$. The product rule in [F4] now gives $(e^{ct}m(t){)}^{\mathrm{\prime }}=e^{ct}(cm(t)+m^{\mathrm{\prime }}(t))\le 0$; [F5] therefore makes $e^{ct}m(t)$ nonincreasing on every finite interval. Hence $0\le \mathrm{Tr}(Q{\rho }_t)\le m(t)\le m(0)e^{-ct}\le J{e}^{-ct}$.

Choose an orthonormal basis beginning with $g$ and continuing with vectors in the excited blocks. Write its matrix entries as $({\rho }_t{)}_{ab}$. By [F3], ${\rho }_t$ has a positive square root; applying [F6] to its images of two basis vectors gives $\mathrm{\mid }({\rho }_t{)}_{ab}{\mathrm{\mid }}^2\le ({\rho }_t{)}_{aa}({\rho }_t{)}_{bb}$. Every excited diagonal entry is between zero and $\mathrm{Tr}(Q{\rho }_t)$, and the ground entry is $1-\mathrm{Tr}(Q{\rho }_t)$. Step 2.1 and [F5] therefore show that every matrix entry of ${\rho }_t$ converges to the corresponding entry of $\mathrm{\mid }g⟩⟨g\mathrm{\mid }$, including all off-diagonal entries. In particular all excited populations vanish and the ground population tends to one.

The entrywise function in the Statement is a norm: it vanishes exactly on the zero matrix, is absolutely homogeneous, and satisfies the triangle inequality by summing $\mathrm{\mid }{X}_{kl}+Y_{kl}\mathrm{\mid }\le \mathrm{\mid }{X}_{kl}\mathrm{\mid }+\mathrm{\mid }{Y}_{kl}\mathrm{\mid }$. The rank-one operators $\mathrm{\mid }{e}_k⟩⟨e_l\mathrm{\mid }$ form a basis of the operator space. Each diagonal unit is a density operator. For $k\mathrm{\ne }l$, the polarization identity expresses $\mathrm{\mid }{e}_k⟩⟨e_l\mathrm{\mid }$ as a fixed complex linear combination of the four rank-one positive operators formed from $e_k\pm e_l$ and $e_k\pm i{e}_l$; after division by their traces these are density operators. By linearity and step 3.1, ${\mathcal{N}}_t(E_{kl})\to \mathrm{Tr}(E_{kl})\mathrm{\mid }g⟩⟨g\mathrm{\mid }$ entrywise for every matrix unit. For any linear map $T$ on this operator space, put $C_T={\max }_{k,l}\mathrm{\parallel }T(E_{kl}){\mathrm{\parallel }}_{\mathrm{e}\mathrm{n}\mathrm{t},1}$, a finite maximum. For $X={\sum }_{k,l}{x}_{kl}{E}_{kl}$, linearity and the triangle inequality give $\mathrm{\parallel }T(X){\mathrm{\parallel }}_{\mathrm{e}\mathrm{n}\mathrm{t},1}\le {\sum }_{k,l}\mathrm{\mid }{x}_{kl}\mathrm{\mid }\mathrm{\parallel }T(E_{kl}){\mathrm{\parallel }}_{\mathrm{e}\mathrm{n}\mathrm{t},1}\le C_T\mathrm{\parallel }X{\mathrm{\parallel }}_{\mathrm{e}\mathrm{n}\mathrm{t},1}$. Thus $T$ is bounded, so the operator-norm definition in [F8] applies and gives $\mathrm{\parallel }T{\mathrm{\parallel }}_{\mathrm{e}\mathrm{n}\mathrm{t},1\to \mathrm{e}\mathrm{n}\mathrm{t},1}\le C_T$. For $T={\mathcal{N}}_t-\mathcal{R}$, each matrix-unit image tends entrywise to zero, so its finite sum of absolute entries tends to zero; the maximum over the finitely many pairs also tends to zero. This proves the asserted channel convergence. The same argument applies in any fixed orthonormal basis.

For $K_k=\mathrm{\mid }g⟩⟨e_k\mathrm{\mid }$, direct expansion gives ${\sum }_k{K}_{k}X{K}_k^{†}=\mathrm{Tr}(X)\mathrm{\mid }g⟩⟨g\mathrm{\mid }$ and ${\sum }_k{K}_k^{†}{K}_k=I$. By [F7], the Choi matrix is ${\sum }_k\mathrm{\mid }{e}_k⟩⟨e_k\mathrm{\mid }\otimes \mathrm{\mid }g⟩⟨g\mathrm{\mid }=I\otimes \mathrm{\mid }g⟩⟨g\mathrm{\mid }$, whose rank is $d$; [F7] gives the minimum Kraus number. Finally, if an excited vector $v$ is annihilated by every $L_a$, then $⟨v,Rv⟩=0$, whereas step 1.1 gives $⟨v,Rv⟩\ge r_{\ast }\mathrm{\parallel }v{\mathrm{\parallel }}^2$; hence $v=0$. Together with the $J=0$ case, this proves all claims.