modified eigenvalues $$\lambda^\Gamma_i=\lambda_i-\Gamma.$$ For $i=1$, we have again $\lambda_1=0$ and the stationary state $R_1=\rho_{\rm ss}$, which we calculate below using an analogous method to that of Sec. \[ResetSpec\].

Next, we consider the corresponding left eigenmodes dual to the above. The adjoint equation is given by $${\mathcal{L}}_{\Gamma}^\dagger(\rho) = {\mathcal{L}}^\dagger(\rho)+\mathcal{V}_{\Gamma}^\dagger(\rho)-\Gamma\rho,$$ where $\mathcal{V}_{\Gamma}^\dagger(\rho) = \Gamma\bra{\psi}\rho\ket{\psi}I$. Note that the identity $I$ remains an eigenmode with eigenvalue 0, as expected. Inserting the original left eigenmodes, we find $${\mathcal{L}}_\Gamma^\dagger(L_k)=(\lambda^*_k-\Gamma)L_k+\Gamma\bra{\psi}L_k\ket{\psi}{I}.$$ Defining $$L^\Gamma_k=L_k+\frac{\Gamma\bra{\psi}L_k\ket{\psi}}{\lambda^*_k-\Gamma}I,$$ we see that, since the identity is annihilated by the adjoint operator, we have $${\mathcal{L}}_\Gamma^\dagger\left(L^\Gamma_k\right)=(\lambda^*_k-\Gamma)L^\Gamma_k,$$ demonstrating that the new left eigenmodes are $L^\Gamma_k$. It can be checked that $\text{Tr}({L^\Gamma_i}^\dagger R_j)=\delta_{ij}$ for $j\neq 1$ and for all $i$, as expected, since $\text{Tr}(R_j)=0$. Finally, requiring $\text{Tr}({L^\Gamma_i}^\dagger \rho^\Gamma_{ss})=\delta_{i1}$ for the new stationary state $\rho^\Gamma_{ss}$ gives the expansion coefficients of that state in the original right eigenmode basis as $$\label{SteadyStateResets}
\rho_{ss}^\Gamma=\rho_{ss}^0-\sum_{j=2}^{\text{dim}({\mathcal{H}})^2}\frac{\Gamma\bra{\psi}L^\dagger_j\ket{\psi} }{\lambda_j-\Gamma}R_j,$$ which, as can be checked, gives ${\mathcal{L}}_\Gamma\rho_{ss}^\Gamma=0$.

The practical applicability of these results depends, as in the classical case, on the system studied and whether, in particular, it has symmetries simplifying the spectral problem. There is an additional benefit in the case of closed quantum systems coming from the fact that the Lindblad equation reduces in that 