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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.

The individual rates describe population relaxation: ${\gamma }_{12}$ transfers population from level one to level two, and ${\gamma }_{21}$ transfers it back. With $p_1+p_2=1$ and the conventional qubit coordinate $r_z=p_1-p_2$,Thus $T_1^{-1}={\gamma }_{12}+{\gamma }_{21}$ is the longitudinal relaxation rate. The parameter $\Gamma =T_2^{-1}$ is the transverse relaxation rate: it damps the off-diagonal coherence, or the $x,y$ components of the Bloch vector. In the usual completely positive two-level model,so $\Gamma$ includes both the coherence loss caused by population relaxation and additional pure dephasing.

There is a normalization switch in the displayed qubit Affine Bloch equation. Its constant term is appropriate to the conventional Pauli matrices and unit-radius coordinate $r_k=\mathrm{Tr}(\rho {\sigma }_k^{\mathrm{Pauli}})$. With the orthonormal generators used earlier, $s=r/\sqrt{2}$, the same physical rates would instead give a constant term $({\gamma }_{21}-{\gamma }_{12})/\sqrt{2}$. The drift matrix is unchanged. We interpret the supplied qubit equation in its conventional $r$ coordinates.

With zero controls and $g={\gamma }_{12}+{\gamma }_{21}>0$, $\Gamma >0$, the steady state is unique:The earlier normalized vector is $s_{\ast }=r_{\ast }/\sqrt{2}$. Indeed $r_x,r_y$ decay as $e^{-\Gamma t}$ and $r_z-r_{\ast ,z}$ as $e^{-gt}$.

For vanishing rates, the full equilibrium conditions of the Affine Bloch equation are $\Gamma r_x=\Gamma r_y=0$ and $g{r}_z={\gamma }_{21}-{\gamma }_{12}$, together with $\Vert r\Vert \le 1$. In particular, for nonnegative rates, $g=0$ means both population rates vanish. If then $\Gamma >0$, all diagonal density operators are stationary; if also $\Gamma =0$, every density operator is stationary. The formal case $g>0$, $\Gamma =0$ permits extra transverse equilibrium coordinates, but is excluded by the completely positive rate bound above.

Let $\mathcal{G}=-i[H,\cdot ]+{\sum }_d\mathcal{D}[V_d]$ denote the Lindbladian. Substituting $\rho ={\sum }_n{r}_n{\sigma }_n$ and projecting with the Hilbert-Schmidt inner product givesFor the Quantum Hamiltonian term, cyclicity of the trace yieldsFor each Lindblad dissipator, the same cyclic rearrangement givesThese are the required coefficient matrices, so $\dot{r}=(L+{\sum }_d{D}^{(d)})r$. They are real because both generators map Hermitian matrices to Hermitian matrices, whose coordinates are real by (a).

The trace of a commutator vanishes. AlsoThus the last row of $L+{\sum }_d{D}^{(d)}$ is zero and $\boxed{{\dot{r}}_{N^2}=0}$. It is trace preservation, not preservation of the purity of a density operator, that produces this constant coordinate.

Write $G=L+{\sum }_d{D}^{(d)}$ and $n=N^2-1$. Since $r_{N^2}=1/\sqrt{N}$, the remaining coordinates obey the Affine Bloch equationBecause $[H,I]=0$ and $\mathcal{D}[V](I)=[V,V^{†}]$, its offset can also be writtenIt vanishes for unital dynamics, but need not vanish for a general dissipative evolution.

A steady state is a time-independent physical state: its derivative vanishes. With $n=N^2-1$, the Affine Bloch equation therefore requiresAs a real linear equation, it has a solution exactly when $\mathrm{rank}A=\mathrm{rank}[A\mid -c]$. If $s_p$ is one solution, all algebraic solutions are $s_p+\ker A$, an affine set of dimension $n-\mathrm{rank}A$. Physical steady states are its intersection with the set of Bloch vectors corresponding to positive trace-one density operators. An arbitrary affine equation need not have a steady state: $A=0$, $c\ne 0$ is a counterexample.

The equation here comes from a finite-dimensional time-independent Lindblad equation, however, so it always has at least one physical steady state. To prove this, start from any density operator and average its trajectory:These averages remain positive with trace one. That set is a compact set, so some sequence has a limit ${\rho }_{\ast }$. If $\mathcal{L}$ is the time-independent generator, thenbecause density operators are bounded. Continuity gives $\mathcal{L}({\rho }_{\ast })=0$. This proves directly that Finite-dimensional Lindbladians have stationary states.

If $A$ has full rank, its algebraic solution $s_{\ast }=-A^{-1}c$ is unique and existence makes that solution physical. For these Lindblad dynamics the converse also holds, even for a boundary steady state. If $A$ is singular, choose a nonzero traceless Hermitian stationary direction $X$. The two parts of its positive-negative decomposition have the same trace $\tau >0$, so $X=\tau ({\rho }_{+}-{\rho }_{-})$ for two density operators. Time-average their trajectories along a common convergent subsequence. Their limits are stationary by the preceding argument, and their difference remains $X/\tau$, since $X$ itself is stationary. The limits are therefore distinct physical stationary states.

Thus the finite-dimensional physical Bloch equation has a unique steady state exactly when $\mathrm{rank}A=n$. These steady states of an affine Bloch equation must be distinguished from convergence to that steady state, which requires the additional spectral condition in (b). The possibly nonorthonormal coordinate basis does not affect existence or uniqueness, since it still gives an invertible parametrization of the traceless Hermitian space.

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.

A steady state is attractive here if every physical initial density operator approaches it. Subtracting its Affine Bloch equation from the trajectory equation gives $\delta \dot{s}=A\delta s$, so $\delta s(t)=e^{At}\delta s(0)$. The necessary and sufficient condition isSuch an $A$ is a Hurwitz stable matrix. To prove sufficiency without assuming diagonalizability, put it in Jordan normal form. Each Jordan block contributes $e^{\lambda t}{\sum }_{j=0}^{q-1}{t}^j{N}^j/j!$, and the negative real part makes the exponential dominate all polynomial factors. For necessity, an eigenmode with positive real part grows; an eigenmode with zero real part remains constant or oscillates, so it does not tend to zero. A nontrivial Jordan block on the imaginary axis also prevents decay.

The physical Bloch vector body has nonempty interior, so differences of physical initial vectors from $s_{\ast }$ span the full real coordinate space. Thus convergence for every physical initial state also forces $e^{At}\to 0$ on that space. This rules out hiding a nondecaying eigenmode outside the physical set. The strict spectral condition entails uniqueness and global asymptotic stability of the steady state.

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.