# Stationary state of Lindblad equation

Is it true that a generic operator that is annihilated by the Lindblad superoperator (with both Hamiltonian and dissipative parts of the dynamics) has to be annihilated separately by both the Hamiltonian dynamics and the dissipative dynamics, or are there cases where it is only the sum of the two terms that vanishes, but not each of them separately?

• They don't have to be annihilated by both individually. You can always absorb some terms from one into the other (they are not unique) so this can't be required. Commented Mar 29 at 3:41

To give an explicit example, consider $$H=\sigma_x$$, $$V=|0\rangle\langle 1|$$, and $$\mathcal L:=-i[H,\cdot]+V(\cdot)V^\dagger-\frac12V^\dagger V(\cdot)-(\cdot)\frac12V^\dagger V\,.$$ The (in fact unique) steady state of the corresponding dynamics then is $$\sigma=\begin{pmatrix}\frac59&\frac{2i}9\\-\frac{2i}9&\frac49\end{pmatrix}$$ (so $$\mathcal L(\sigma)=0$$), but $$-i[H,\sigma]=\begin{pmatrix} -\frac{4}{9} & \frac{i}{9} \\ -\frac{i}{9} & \frac{4}{9} \end{pmatrix}\neq 0\tag1$$ (and $$V\sigma V^\dagger-\frac12V^\dagger V\sigma-\sigma \frac12V^\dagger V\neq 0$$ as it is of course the negative of (1)).