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A time-independent Lindblad equation becomes an affine linear equation for the Generalized Bloch representation. If $\mathcal{L}(I)={\sum }_d[V_d,V_d^{†}]$, then $A_{mn}=\mathrm{Tr}({\sigma }_m\mathcal{L}({\sigma }_n))$ and $c_m=\mathrm{Tr}({\sigma }_m\mathcal{L}(I))/N$ for traceless basis indices. The offset vanishes exactly when the dynamics are unital. Equilibria solve $A{s}_{\ast }=-c$, and deviations evolve as $s(t)-s_{\ast }=e^{At}(s(0)-s_{\ast })$.

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 Lindbladian $\mathcal{L}$ is the linear superoperator on density operators defined by the right-hand side of a Lindblad equation, so that $\dot{\rho }=\mathcal{L}(\rho )$.

Excited-state decay is suppressed in the ideal dark-state limit. The dark state has zero amplitude on $|2⟩$, so a spontaneous-emission jump of the form $\sqrt{\Gamma }|j⟩⟨2|$ annihilates it. There is consequently no excited-state level population to decay in perfect adiabatic following. Finite speed introduces bright-state admixture and a small excited level population; its integrated loss is of order $\Gamma \int P_2(t)dt$. Thus robustness is conditional on good adiabaticity, not immunity to arbitrarily large decay at fixed pulse strength and duration. In a strongly overdamped regime the relevant relaxation gap can also limit following.

Moderate envelope perturbations are tolerated because this transfer is fixed by the endpoints of the dark-state path, not by an accurately calibrated resonant pulse area. Common amplitude changes do not change the angle, although reducing them too far closes the useful gap. Independent smooth changes still work if they preserve the counter-intuitive order, the endpoint ratios, resonance and a sufficiently large gap relative to $|\dot{\theta }|$. A wrong sequence, missing overlap, rapid fluctuations or unstable relative optical phase can spoil the transfer.

Ground-state decoherence is not suppressed by the dark-state mechanism. During the overlap, the dark orthogonal projector has quantum coherence ${\rho }_{13}=-\sin \theta \cos \theta$. Dephasing of the $1$–$3$ transition removes precisely this relative-phase information, even if $P_2=0$ initially. For example, with $L=\sqrt{\gamma /2}(|1⟩⟨1|-|3⟩⟨3|)$ in a Lindblad equation, its dissipator gives ${\dot{\rho }}_{13}{|}_{\mathrm{deph}}=-\gamma {\rho }_{13}$. It therefore drives a nontrivial dark superposition away from its pure orthogonal projector. The transfer should be slow enough for adiabaticity but fast compared with the relevant ground-state quantum coherence lifetime. Ideal protection from excited-state spontaneous emission does not protect this ground-state quantum coherence.

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.

A quantum optimal control problem specifies a terminal task, admissible fields and resource costs. Pure-state transfer can minimize $1-|⟨{\psi }_d|\psi (T)⟩{|}^2$; mixed-state transfer on an accessible spectral orbit can minimize $\Vert \rho (T)-{\rho }_d{\Vert }_{\mathrm{HS}}^2/2$; preparing an observable value can minimize $-\mathrm{Tr}(O\rho (T))$. A gate task can minimize $1-|\mathrm{Tr}(U_d^{†}U(T)){|}^2/n^2$ for an $n$-dimensional Hilbert space, which is insensitive to an overall phase. These different terminal functions encode different tasks even when the dynamics are the same. Fluence penalties, amplitude bounds, bandwidth limits, or the total duration can be added to express experimental requirements.

For example, minimizesubject to $\dot{\rho }={\mathcal{L}}_f(\rho )$ and a specified $\rho (0)$. For a closed system, ${\mathcal{L}}_f(\rho )=-i[H_0+{\sum }_m{f}_m{H}_m,\rho ]$; for an open system add the known Lindblad equation dissipator. Introducing a Hermitian costate $\Lambda$ and taking the Hilbert-Schmidt inner product gives the forward and backward equations and functional gradientFor Hamiltonian controls the second term is $\mathrm{Tr}(\Lambda [-i{H}_m,\rho ])$. The adjoint is defined by the Hilbert-Schmidt inner product; in particular, the closed-system costate equation is $\dot{\Lambda }=-i[H,\Lambda ]$, integrated backward from its terminal condition. An unconstrained interior optimum has zero gradient; amplitude constraints replace this by the corresponding constrained first-order conditions.

Numerically, represent the fields by piecewise constant amplitudes $f_{m,j}$ on slices, or by a finite basis $f_m(t)={\sum }_j{a}_{mj}{b}_j(t)$ such as splines or Fourier modes. For piecewise constant closed-system dynamics, $U_j=e^{-i{H}_j\Delta t}$ and $U(T)=U_K\cdots U_1$. Forward products and backward costates provide all slice derivatives in one pair of propagations. In gradient ascent pulse engineering, update these amplitudes to improve the chosen objective, with projection onto amplitude bounds if needed. The exact slice derivative is the derivative of the matrix exponentialReplacing it by $-i\Delta t{H}_m{U}_j$ is only a short-slice approximation when $H_j$ and $H_m$ do not commute. Other gradient descent or nonlinear optimization methods can optimize the same finite parameter set; direct-adjoint looping avoids a separate dynamical simulation for each control coefficient.

A pulse is accepted after propagation with the actual constrained waveform confirms its terminal error and resource use. Sampling model uncertainties in the objective can favor robust pulses. Nonconvexity means a numerical stationary point need not be a global optimum, so initial guesses, convergence tolerances and a sufficiently resolved time grid matter.

A steady state is a time-independent density operator satisfying $\mathcal{L}({\rho }_{\ast })=0$, equivalently a constant physical Bloch vector $s_{\ast }$ satisfying $\boxed{A{s}_{\ast }=-c}$. Let $K=N^2-1$. By the rank criterion for a linear system, this equation has a solution exactly whenWhen it is consistent, every algebraic solution has the form $s_0+v$ with $v\in \ker A$, so the solution affine space has dimension $K-\mathrm{rank}A$ by the rank-nullity theorem. Physical steady states are its intersection with the positive density operator body. The algebraic equilibrium is unique exactly when $\mathrm{rank}A=K$, in which case $\boxed{s_{\ast }=-A^{-1}c}$.

An arbitrary affine differential equation need not have an equilibrium: $A=0$, $c\ne 0$ is inconsistent. For the finite-dimensional, time-independent Lindblad equation under discussion, however, a physical steady state always exists. Indeed, the averages ${\bar{\rho }}_T=T^{-1}{\int }_0^T\rho (t)dt$ remain positive with trace one. Compactness supplies a convergent subsequence as $T\to \infty$, andThe limit is a physical steady state, proving that Finite-dimensional Lindbladians have stationary states and that the rank consistency condition is automatically satisfied for this physical generator.

In this setting, full rank also characterizes uniqueness among physical steady states. To justify the converse when $A$ is singular, a nonzero $v\in \ker A$ gives a fixed traceless Hermitian operator $X={\sum }_k{v}_k{\sigma }_k$. Write its positive-negative decomposition $X=X_{+}-X_{-}$. Time-average the evolutions of the two positive operators along a common convergent subsequence; the limits $Y_{\pm }$ are positive stationary operators and satisfy $X=Y_{+}-Y_{-}$. Their traces are equal because $\mathrm{Tr}X=0$. If there were only one stationary density operator, both $Y_{\pm }$ would be this matrix times the same trace, forcing $X=0$, a contradiction. Hence $\boxed{\text{unique physical steady state}⟺\mathrm{rank}A=N^2-1}$.

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.

A quantum channel is a linear completely positive trace-preserving map. Its Kraus representation isKraus operators are not unique: two representations of the same channel are related by an isometry on the Kraus index, and by a unitary when both lists are minimal and equally long. A differentiable Markovian family of infinitesimal channels gives the Lindblad equation

Quantum control chooses externally adjustable interactions to steer a quantum state or implement a desired unitary operator. Its achievable transformations depend on the Hamiltonian or Lindblad equation, the available control inputs, and physical constraints such as bandwidth and amplitude. Quantum optimal control optimizes performance within those constraints.

Quantum optimal control minimizes a state-transfer, observable, or gate error together with resource costs, subject to a controlled Schrödinger equation or Lindblad equation. Forward propagation of the state and backward propagation of a costate compute the gradient efficiently. The optimum depends on the admissible time-dependent fields and laboratory constraints.

A quantum-trajectory unravelling represents a Lindblad equation as an ensemble of stochastic pure-state histories. In one jump unravelling, a state evolves between jumps under the non-Hermitian effective Hamiltonian $H_{\mathrm{eff}}=H-\frac{i}{2}{\sum }_{\alpha }{L}_{\alpha }^{†}{L}_{\alpha }$, while a jump of type $\alpha$ sends it to a normalized multiple of $L_{\alpha }|\psi ⟩$.

An arbitrary real affine equation has steady states exactly when $\mathrm{rank}A=\mathrm{rank}[A\mid -c]$. Its solution set is $s_{\ast }+\ker A$. For a finite-dimensional time-independent Lindblad equation, physical stationary states always exist by time averaging trajectories. If $A$ is invertible the steady state is unique. Conversely, a nonzero traceless Hermitian stationary direction can be decomposed into the difference of two equally weighted density operators; time averaging their trajectories yields two distinct stationary density operators. Thus singularity of the physical Bloch matrix also obstructs physical uniqueness.