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A Lindblad dissipator is the non-Hamiltonian generator associated with a Lindblad operator $V$. It preserves the trace because cyclicity gives $\mathrm{Tr}(V\rho V^{†})=\mathrm{Tr}(V^{†}V\rho )$, cancelling the two anticommutator terms. It preserves Hermiticity and satisfies $\mathcal{D}[V](I)=[V,V^{†}]$; it need not preserve the identity or the purity of a density operator.

Design Lindblad operators and a Quantum Hamiltonian so that desired density operators are stationary and unwanted perturbations decay. A dissipative interaction can change the spectrum and purity of a density operator, reaching targets unavailable to purely coherent Hamiltonian engineering. The engineered stationary state must be physical, and uniqueness alone must be distinguished from an explicit proof of asymptotic convergence.

Take the driving and feedback strengths to be real, so the feedback Quantum Hamiltonian is Hermitian. The lowering operator satisfies ${\sigma }^2=({\sigma }^{†}{)}^2=0$. Using the supplied convention for ${\sigma }_y$,The additional Quantum Hamiltonian term is consequentlyWith no drift Quantum Hamiltonian, the total Quantum Hamiltonian is just $\alpha {\sigma }_y$. The feedback-modified Lindblad operator isSubstitution into the measurement-feedback equation givesThus the feedback affects the Lindblad dissipator even though its extra coherent Quantum Hamiltonian contribution cancels. This is a concrete mechanism of Markovian reservoir engineering, rather than merely a classical control force added to an unchanged dissipative equation.

Write $\rho ={\sum }_n{r}_n{\sigma }_n$ in the Generalized Bloch representation. Linearity of the Lindbladian gives ${\dot{r}}_m={\sum }_n\mathrm{Tr}({\sigma }_m\mathcal{L}({\sigma }_n))r_n$. For the Hamiltonian term, cyclicity yieldsFor one Lindblad operator $V_d$, the jump term becomes $\mathrm{Tr}({\sigma }_m{V}_d{\sigma }_n{V}_d^{†})=\mathrm{Tr}(V_d^{†}{\sigma }_m{V}_d{\sigma }_n)$, and the two remaining terms giveThusEvery coefficient is real because $\mathcal{L}({\sigma }_n)$ is Hermitian. In addition, $L_{mn}=-L_{nm}$, although a dissipative $D^{(d)}$ need not be symmetric.

The Hamiltonian commutator has zero trace, and each dissipator obeys $\mathrm{Tr}(V\rho V^{†})=\mathrm{Tr}(V^{†}V\rho )$, so $\mathrm{Tr}\mathcal{L}(\rho )=0$. Consequently $\boxed{{\dot{r}}_{N^2}=0,r_{N^2}=1/\sqrt{N}}$ for normalized density operators. Orthogonality to $I/\sqrt{N}$ makes every other ${\sigma }_k$ traceless. Define $s=(r_1,\dots ,r_{N^2-1}{)}^T$. With $M=L+{\sum }_d{D}^{(d)}$, trace preservation gives the block formHere the indices of $A$ range from $1$ to $N^2-1$. Since $\mathcal{L}(I)={\sum }_d(V_d{V}_d^{†}-V_d^{†}{V}_d)$, the offset can also be written $c_m=N^{-1}\mathrm{Tr}({\sigma }_m{\sum }_d[V_d,V_d^{†}])$. This is the Affine Bloch equation; its offset is zero for unital dynamics, which preserve the identity.

The generator of every finite-dimensional, time-homogeneous Markovian quantum channel has Lindblad equation formHere $H=H^{†}$. The Lindblad operators $L_{\alpha }$ may be arbitrary bounded operators; positivity of their rates is already incorporated by rescaling them. They need not be Hermitian, traceless, normalized, or mutually orthogonal. Their description is nonunique: unitary mixing of the $L_{\alpha }$, and shifts by scalar multiples of the identity accompanied by a compensating change of $H$, leave the same Lindbladian.

An observed dissipative channel combined with direct feedback can create a modified Lindblad operator and a compensating Quantum Hamiltonian. Choosing those parameters can make a prescribed pure state stationary and globally attractive. At exceptional parameter values, the dissipator can instead leave an entire family of states stationary, so algebraic target stationarity does not establish stabilization. Checking the Affine Bloch equation and its transverse eigenvalues distinguishes these possibilities.