 Finite-dimensional Lindbladians have stationary states

For any initial density operator, average its time-independent Lindblad equation trajectory as ${\bar{\rho }}_T=T^{-1}{\int }_0^T\rho (t)dt$. These averages stay in the compact convex set of density operators. Along a convergent subsequence, $\mathcal{L}({\bar{\rho }}_T)=(\rho (T)-\rho (0))/T\to 0$, so the limit is stationary. Thus a finite-dimensional physical Affine Bloch equation has at least one physical equilibrium, even though an arbitrary affine real equation can have none.