Affine Bloch equation ($\dot{s}=As+c$)

A time-independent Lindblad equation becomes an affine linear equation for the Generalized Bloch representation. If $\mathcal{L}(I)={\sum }_d[V_d,V_d^{†}]$, then $A_{mn}=\mathrm{Tr}({\sigma }_m\mathcal{L}({\sigma }_n))$ and $c_m=\mathrm{Tr}({\sigma }_m\mathcal{L}(I))/N$ for traceless basis indices. The offset vanishes exactly when the dynamics are unital. Equilibria solve $A{s}_{\ast }=-c$, and deviations evolve as $s(t)-s_{\ast }=e^{At}(s(0)-s_{\ast })$.

Decay of the off-diagonal coherence of a two-level density operator. Population relaxation contributes half its total rate; additional dephasing contributes ${\gamma }_{\varphi }\ge 0$ in the usual completely positive two-level model. Thus transverse relaxation should not automatically be identified with pure dephasing alone.

For a two-level density operator, rates ${\gamma }_{12}$ from level one to two and ${\gamma }_{21}$ in the reverse direction give ${\dot{p}}_1=-{\gamma }_{12}{p}_1+{\gamma }_{21}{p}_2$. Their sum sets the longitudinal population relaxation rate, and their ratio fixes the stationary population imbalance. If both rates vanish, populations are conserved.

An arbitrary real affine equation has steady states exactly when $\mathrm{rank}A=\mathrm{rank}[A\mid -c]$. Its solution set is $s_{\ast }+\ker A$. For a finite-dimensional time-independent Lindblad equation, physical stationary states always exist by time averaging trajectories. If $A$ is invertible the steady state is unique. Conversely, a nonzero traceless Hermitian stationary direction can be decomposed into the difference of two equally weighted density operators; time averaging their trajectories yields two distinct stationary density operators. Thus singularity of the physical Bloch matrix also obstructs physical uniqueness.