Noncommuting entropy production and thermal secular models

By [F1], $\mathrm{spec}(H)$ is finite and the projections $P_E$ are well-defined, pairwise orthogonal, and sum to $I_d$. Therefore ${\mathrm{\Omega }}_H$ is finite and symmetric, and each $S_a(\omega )$ is a finite sum of endomorphisms of $\mathcal{H}$; when $m=0$ or a frequency block is empty, the corresponding sum is the zero matrix.

Every term in the displayed formula is a product and finite sum of matrices on $\mathcal{H}$, so $\mathcal{L}$ is a complex-linear endomorphism of $M_d(\mathbb{C})$. The Hamiltonian coefficient has units joules/(joule-seconds) = seconds${}^{-1}$; each dissipative term has units ${\gamma }_{ab}={\mathrm{s}}^{-1}$ times dimensionless matrices. Thus the formula is a well-defined finite linear map with the stated units.

For each $\omega$, positive semidefiniteness of the finite matrix $\mathrm{\Gamma }(\omega )$ gives a finite factorization ${\gamma }_{ab}(\omega )={\sum }_r\overline{c_{ra}(\omega )}{c}_{rb}(\omega )$, with $c_{ra}(\omega )$ in seconds${}^{-1\mathrm{/}2}$. Define $J_{r,\omega }:={\sum }_b{c}_{rb}(\omega )S_b(\omega )$. Expanding yields ${\sum }_r{J}_{r,\omega }X{J}_{r,\omega }^{\ast }={\sum }_{a,b}{\gamma }_{ab}(\omega )S_b(\omega )X{S}_a(\omega {)}^{\ast }$ and ${\sum }_r{J}_{r,\omega }^{\ast }{J}_{r,\omega }={\sum }_{a,b}{\gamma }_{ab}(\omega )S_a(\omega {)}^{\ast }{S}_b(\omega )$. Thus the displayed generator has the finite GKSL form of [F2] with Hamiltonian $H+H_{\mathrm{L}\mathrm{S}}$ and finite jumps $J_{r,\omega }$, so [F3] gives a finite channel semigroup; [F1] states its complete-positivity and trace-preservation content.

From the spectral definition, $[H,S_a(\omega )]=-\mathrm{\hslash }\omega S_a(\omega )$ and $S_a(\omega {)}^{\ast }=S_a(-\omega )$. Therefore $S_a(\omega )e^{-\beta H}=e^{-\beta \mathrm{\hslash }\omega }{e}^{-\beta H}{S}_a(\omega )$. Every $S_a(-\omega )S_b(\omega )$ commutes with $H$ because its two energy shifts cancel, hence it commutes with each spectral projection of $H$ and with ${\sigma }_{\beta }$. Also $[H_{\mathrm{L}\mathrm{S}},H]=0$ implies $H_{\mathrm{L}\mathrm{S}}$ preserves every eigenspace of $H$: if $Hx=Ex$, then $H(H_{\mathrm{L}\mathrm{S}}x)=H_{\mathrm{L}\mathrm{S}}(Hx)=E(H_{\mathrm{L}\mathrm{S}}x)$. Thus $H_{\mathrm{L}\mathrm{S}}$ commutes with every spectral projection of $H$ and with ${\sigma }_{\beta }$. Finally, the exponential $e^{-\beta H}$ is positive definite with positive finite trace, making ${\sigma }_{\beta }$ a faithful density.

Fix $\omega >0$ and put $q=e^{-\beta \mathrm{\hslash }\omega }$. The $\omega$ contribution is ${\mathcal{D}}_{\omega }({\sigma }_{\beta })={\sigma }_{\beta }{\sum }_{a,b}{\gamma }_{ab}(\omega )(q{S}_b(\omega )S_a(-\omega )-S_a(-\omega )S_b(\omega ))$, using $S_b(\omega ){\sigma }_{\beta }=q{\sigma }_{\beta }{S}_b(\omega )$ and commutation of the anticommutator product with ${\sigma }_{\beta }$. For the negative frequency, substitute ${\gamma }_{ab}(-\omega )=q{\gamma }_{ba}(\omega )$ and $S_a(-\omega {)}^{\ast }=S_a(\omega )$, then reindex $a,b$ to obtain ${\mathcal{D}}_{-\omega }({\sigma }_{\beta })={\sigma }_{\beta }{\sum }_{a,b}{\gamma }_{ab}(\omega )(S_a(-\omega )S_b(\omega )-q{S}_b(\omega )S_a(-\omega ))=-{\mathcal{D}}_{\omega }({\sigma }_{\beta })$. Thus every nonzero frequency pair cancels.

At $\omega =0$, the balance relation gives ${\gamma }_{ab}(0)={\gamma }_{ba}(0)$. Each $S_a(0)$ is Hermitian and commutes with $H$, hence with ${\sigma }_{\beta }$; consequently the zero-frequency contribution is ${\sigma }_{\beta }{\sum }_{a,b}{\gamma }_{ab}(0)(S_b(0)S_a(0)-S_a(0)S_b(0))=0$, after swapping $a,b$ in the first finite sum. The Hamiltonian contribution vanishes because both $H$ and $H_{\mathrm{L}\mathrm{S}}$ commute with ${\sigma }_{\beta }$. Together with step 2.1 this gives $\mathcal{L}({\sigma }_{\beta })=0$, and the exponential series then gives $T_t({\sigma }_{\beta })={\sigma }_{\beta }$ for all $t\ge 0$.

On $M_d(\mathbb{C})$ use the max-entry norm $\mathrm{\parallel }X{\mathrm{\parallel }}_{\max }:={\max }_{i,j}\mathrm{\mid }{X}_{ij}\mathrm{\mid }$. The space is finite-dimensional and Banach by [F3]. In coordinates, $(\mathcal{L}X{)}_{ij}={\sum }_{k,l}{a}_{ij,kl}{X}_{kl}$ for finitely many coefficients, so $\mathrm{\parallel }\mathcal{L}X{\mathrm{\parallel }}_{\max }\le C\mathrm{\parallel }X{\mathrm{\parallel }}_{\max }$ with $C:={\max }_{i,j}{\sum }_{k,l}\mathrm{\mid }{a}_{ij,kl}\mathrm{\mid }<\mathrm{\infty }$; thus $\mathcal{L}$ is bounded. Hence [F4] gives operator-norm convergence of the exponential series, its compact-uniform convergence, and the law $T_u{T}_v=T_{u+v}$. For $0\le s\le t$, put $u=t-s$ and $v=s$. The map $T_{t-s}$ is a finite channel by the hypothesis and [F1], and fixes $\sigma$ by stationarity. Apply [A1] to $X=T_s\rho$ and $Y=\sigma$; these are positive semidefinite, so $D(T_t\rho \mathrm{\parallel }\sigma )\le D(T_s\rho \mathrm{\parallel }\sigma )$. This proves global monotonicity, including singular initial densities, because $\sigma >0$ makes every displayed relative entropy finite.

For $x>0$ and $R>0$, direct integration gives ${\int }_0^R((1+s{)}^{-1}-(x+s{)}^{-1})ds=\log x+\log (1+R)-\log (x+R)$, which tends to $\log x$ as $R\to \mathrm{\infty }$. Applying this scalar identity to each eigenvalue using [F2] yields $\log A={\int }_0^{\mathrm{\infty }}((1+s{)}^{-1}I-(A+sI{)}^{-1})ds$ for positive definite $A$. Let $u\mapsto A(u)$ be a $C^1$ path of positive-definite matrices on an open interval, or on a one-sided interval at an endpoint, around the parameter under consideration, and write $X=\dot{A}(u)$. Differentiating $(A+sI)(A+sI{)}^{-1}=I$ gives $\frac{d}{du}(A+sI{)}^{-1}=-(A+sI{)}^{-1}X(A+sI{)}^{-1}$, so $\frac{d}{du}\log A={\int }_0^{\mathrm{\infty }}(A+sI{)}^{-1}X(A+sI{)}^{-1}ds$. At a fixed parameter $u_0$, let ${\lambda }_0>0$ be the least eigenvalue of $A(u_0)$. By continuity, on a neighborhood $\mathrm{\parallel }A(u)-A(u_0)\mathrm{\parallel }<{\lambda }_0\mathrm{/}2$, so $A(u)\ge ({\lambda }_0\mathrm{/}2)I$ there; since $A^{\mathrm{\prime }}$ is locally bounded, the derivative integrand is uniformly bounded by $C({\lambda }_0\mathrm{/}2+s{)}^{-2}$ nearby for some finite $C$, an integrable function. The product rule gives $\frac{d}{du}\mathrm{Tr}(A\log A)=\mathrm{Tr}(X\log A)+{\int }_0^{\mathrm{\infty }}\mathrm{Tr}(A(A+sI{)}^{-1}X(A+sI{)}^{-1})ds$. Diagonalize $A$ with eigenvalues ${\lambda }_j>0$; the integral is ${\sum }_j{X}_{jj}{\int }_0^{\mathrm{\infty }}{\lambda }_j({\lambda }_j+s{)}^{-2}ds={\sum }_j{X}_{jj}=\mathrm{Tr}X$. Hence $\frac{d}{du}\mathrm{Tr}(A\log A)=\mathrm{Tr}(X(\log A+I))$.

For $f(t):=D(T_t\rho \mathrm{\parallel }\sigma )$, every quotient $(f(t+h)-f(t))\mathrm{/}h$ with $h>0$ is nonpositive by step 1.1. Thus $D^{+}f(t):={\mathrm{lim\hspace{0.17em}sup}}_{h\downarrow 0}(f(t+h)-f(t))\mathrm{/}h\le 0$ as an extended-real upper right Dini derivative, without inserting $\log {\rho }_t$ when ${\rho }_t$ is singular.

If ${\rho }_t>0$, continuity gives a positive-definite neighborhood (right-sided at $t=0$). By [F4], $T_t$ is differentiable in operator norm and ${\dot{\rho }}_t=\mathcal{L}({\rho }_t)$. Trace preservation implies $\mathrm{Tr}{\dot{\rho }}_t=0$, by differentiating $\mathrm{Tr}(T_t\rho )=\mathrm{Tr}\rho$. Apply step 1.2 to $A={\rho }_t$ and differentiate the linear term $-\mathrm{Tr}({\rho }_t\log \sigma )$ to obtain the displayed derivative formula. Its sign is nonpositive because the same function is nonincreasing by step 1.1.