Faithful scalar-kernel detailed balance gives a relaxation gap

If $d=1$, then ${\mathcal{M}}_0=\{0\}$ and the only density matrix is $\sigma$, so all assertions are immediate. For the remainder assume $d>1$. By [F1], self-adjointness and $D(I)=0$ give $\mathrm{Tr}(\sigma D(A))=⟨D(A),I{⟩}_{\sigma }=⟨A,D(I){⟩}_{\sigma }=0$. For the Hamiltonian part, cyclicity [F6] and $[H_{\mathrm{s}},\sigma ]=0$ give $\mathrm{Tr}(\sigma K(A))=0$. Since $G$ is the trace dual of $\mathcal{L}$, differentiating $\mathrm{Tr}(\sigma e^{tG}A)$ now shows it is constant. Applying the same calculation with $K=0$ shows $\mathrm{Tr}(\sigma e^{tD}A)=\mathrm{Tr}(\sigma A)$. Thus $\sigma$ is stationary for both dual semigroups.

Let ${\mathcal{L}}_D(X)={\sum }_j(L_{j}X{L}_j^{\ast }-\frac{1}{2}\{L_j^{\ast }{L}_j,X\})$, the Schrödinger GKSL generator with zero Hamiltonian and the same jump matrices. The finite trace calculation in [F1], [F6] shows its trace dual is $D$. Applying [F2] to ${\mathcal{L}}_D$ and [F3] to its channel semigroup makes $e^{tD}$ unital and completely positive. The Schwarz inequality [F4] and stationarity from step 1.1 give

Differentiating at $t=0$ yields $\mathrm{Re}⟨D(A),A{⟩}_{\sigma }\le 0$; self-adjointness in [F1] makes this quantity real.

The subspace ${\mathcal{M}}_0$ is the orthogonal complement of $\mathbb{C}I$, and [F1] gives $D({\mathcal{M}}_0)\subseteq {\mathcal{M}}_0$. The finite-dimensional spectral theorem [F5] applies to the self-adjoint endomorphism $D$ on $(M_d,⟨\cdot ,\cdot {⟩}_{\sigma })$. Step 2.1 makes all eigenvalues nonpositive; $\ker D=\mathbb{C}I$ excludes zero on ${\mathcal{M}}_0$. Because $d>1$, ${\mathcal{M}}_0\mathrm{\ne }\{0\}$, so its finite nonempty list of eigenvalues has maximum $-\gamma <0$, and $⟨D(A),A{⟩}_{\sigma }\le -\gamma \mathrm{\parallel }A{\mathrm{\parallel }}_{\sigma }^2$.

By [F6] and $[H_{\mathrm{s}},\sigma ]=0$, $K$ is anti-self-adjoint for $⟨\cdot ,\cdot {⟩}_{\sigma }$. Both $K$ and $D$ preserve ${\mathcal{M}}_0$: for $D$ this follows from step 1.1, and for $K$ from the same trace-cyclicity calculation. For $A_t=e^{tG}A$ with centered $A$, differentiating gives

Thus $e^{2\gamma t}\mathrm{\parallel }{A}_t{\mathrm{\parallel }}_{\sigma }^2$ is nonincreasing, proving the asserted decay. Cauchy–Schwarz [F5] then bounds each centered correlation by $e^{-\gamma t}\mathrm{\parallel }A{\mathrm{\parallel }}_{\sigma }\mathrm{\parallel }B{\mathrm{\parallel }}_{\sigma }$, an integrable function.

The decay in step 4.1 makes $Y:={\int }_0^{\mathrm{\infty }}{e}^{tG}Adt$ converge in the finite-dimensional space ${\mathcal{M}}_0$. Differentiating $e^{tG}A$ and integrating gives $-GY=A$. If $GZ=0$ for centered $Z$, then step 4.1 gives $\mathrm{\parallel }Z{\mathrm{\parallel }}_{\sigma }=\mathrm{\parallel }{e}^{tG}Z{\mathrm{\parallel }}_{\sigma }\le e^{-\gamma t}\mathrm{\parallel }Z{\mathrm{\parallel }}_{\sigma }$, so $Z=0$; hence $-G{\mathrm{\mid }}_{{\mathcal{M}}_0}$ is invertible and $Y=(-G{\mathrm{\mid }}_{{\mathcal{M}}_0}{)}^{-1}A$.

Let $Y$ be as in step 5.1. Then $\mathrm{\Gamma }(A)=\mathrm{Re}⟨Y,-G(Y){⟩}_{\sigma }$. The anti-self-adjoint $K$ contributes zero real part, while step 3.1 gives $⟨Y,-D(Y){⟩}_{\sigma }\ge \gamma \mathrm{\parallel }Y{\mathrm{\parallel }}_{\sigma }^2\ge 0$. Therefore $\mathrm{\Gamma }(A)\ge 0$.

Let $X=\rho -\sigma =X^{\ast }$, and take any $A$ with $\mathrm{\parallel }A{\mathrm{\parallel }}_{\mathrm{\infty }}\le 1$. Set $A_0=A-\mathrm{Tr}(\sigma A)I$. Stationarity and trace duality give $\mathrm{Tr}(({\rho }_t-\sigma )A)=\mathrm{Tr}(X{e}^{tG}{A}_0)$. By cyclicity [F6], the trace-norm dual formula [F4] applies to this product order. Put $C=e^{tG}{A}_0$. Since $\mathrm{Tr}(XC)=\mathrm{Tr}(CX)=\mathrm{Tr}((C{\sigma }^{1\mathrm{/}2})({\sigma }^{-1\mathrm{/}2}X))$, the Hilbert–Schmidt trace inequality $\mathrm{\mid }\mathrm{Tr}(UV)\mathrm{\mid }\le \mathrm{\parallel }U{\mathrm{\parallel }}_2\mathrm{\parallel }V{\mathrm{\parallel }}_2$ yields

The identity $\mathrm{\parallel }{A}_0{\mathrm{\parallel }}_{\sigma }^2=\mathrm{Tr}(\sigma A^{\ast }A)-\mathrm{\mid }\mathrm{Tr}(\sigma A){\mathrm{\mid }}^2\le \mathrm{\parallel }A{\mathrm{\parallel }}_{\mathrm{\infty }}^2\le 1$ gives a bound independent of $A$. Taking the supremum in [F4] proves the stated trace-norm relaxation estimate.