Lindblad equation

The Lindblad equation describes the generator of a time-homogeneous, completely positive, trace-preserving quantum dynamical semigroup in a finite-dimensional system. In units $\hslash =1$, the Hamiltonian $H$ supplies coherent evolution and the Lindblad operators $L_{\alpha }$ specify dissipative contributions.

A Markovian continuous-time quantum channel obeys

The Lindblad dissipator combines its gain term and two anticommutator terms so that the trace remains one. For $L=\sqrt{\gamma /2}Z$ and $H=0$, diagonal entries in the $Z$ basis stay fixed while an off-diagonal element obeys ${\dot{\rho }}_{01}=-\gamma {\rho }_{01}$: dephasing without population decay. The semigroup assumption is a specific model of environmental memory loss. ^[ref]

A finite-dimensional quantum dynamical semigroup is Davies irreducible when no proper nonzero subspace can support a state for all future times. Equivalently, if a projection $P$ satisfies ${\Phi }_t(P\rho P)=P{\Phi }_t(P\rho P)P$ for every state $\rho$ and every $t\ge 0$, then $P=0$ or $P=I$. This is a property of invariant supports, rather than merely uniqueness of a stationary density matrix.

For a two-level system with only spontaneous decay, $L=\sqrt{\gamma }|0⟩⟨1|$ and zero Quantum Hamiltonian, the ground-state line is invariant. The stationary state is unique but pure, so the dynamics is reducible. Adding a positive excitation rate $L^{\prime }=\sqrt{\kappa }|1⟩⟨0|$ removes that invariant line; the stationary populations are $\gamma /(\gamma +\kappa )$ and $\kappa /(\gamma +\kappa )$ and the dynamics is irreducible. In finite dimensions irreducibility ensures a unique faithful stationary state, which has full rank. Related notions of quantum irreducibility must be distinguished.^[ref]

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.

A Lindblad operator describes one dissipative contribution to a Markovian master equation for a density operator. In the common convention,with $\hslash =1$. Absorbing a rate into $L_{\alpha }$ gives it square-root-rate units.

For a two-level system, $L=\sqrt{\gamma }|0⟩⟨1|$ models decay from the excited state, while a diagonal operator can describe dephasing. The gain and anticommutator terms together preserve trace; retaining only the first would not define the same unconditional evolution. The operators are not unique, and a stochastic jump interpretation needs a specified monitoring of the environment. The physical object is the complete Lindbladian, including its Hamiltonian and dissipative parts.

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

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 gap is the smallest positive decay rate $-\mathrm{Re}\lambda$ among nonzero eigenvalues $\lambda$ of a relaxing Lindbladian. It controls the slowest asymptotic exponential approach to the stationary state.

For a finite-dimensional Lindbladian $\mathcal{L}$, convergence to one stationary density matrix for every initial state holds exactly when zero is a simple eigenvalue and every other eigenvalue has strictly negative real part. The decaying spectral modes then leave only the stationary component.

A scalar commutant of an operator algebra generated by $H,L_{\alpha },L_{\alpha }^{†}$ is not sufficient on its own. A useful sufficient criterion requires an additional hypothesis: a faithful stationary state, meaning a stationary density matrix that is a positive-definite matrix. Under that hypothesis, a commutant consisting only of scalar multiples of the identity operator implies uniqueness.^[ref]

For a directly checkable counterexample without a faithful stationary state, take an orthonormal basis $|0⟩,|1⟩,|e⟩$, $H=0$, andEvery density matrix supported on $\mathrm{span}\{|0⟩,|1⟩\}$ is stationary: both jump terms and the anticommutator terms vanish there. Nevertheless, the jumps, their adjoints and their products generate all nine matrix units, so their commutant is scalar. There are therefore infinitely many stationary states despite that scalar commutant. The missing faithful-state assumption changes the conclusion.

A quantum Trajectory Theory unravelling represents a Lindblad equation by an ensemble of stochastic pure states. In units $\hslash =1$, define the non-Hermitian effective HamiltonianFor a normalized state $|\psi ⟩$, channel $\alpha$ has instantaneous jump rate $r_{\alpha }=⟨\psi |L_{\alpha }^{†}{L}_{\alpha }|\psi ⟩$. During a small interval $dt$, that jump occurs with probability $r_{\alpha }dt$ and sends the state to $L_{\alpha }|\psi ⟩/\sqrt{r_{\alpha }}$. A zero rate means that channel cannot jump. Otherwise, with total no-jump probability $1-{\sum }_{\alpha }{r}_{\alpha }dt$, evolve with $I-i{H}_{\mathrm{eff}}dt$ and renormalize.

To verify the ensemble evolution, put $\rho =|\psi ⟩⟨\psi |$. The conditional normalizations cancel their probability weights, givingExpanding recovers the Lindblad equation. Different monitoring choices can give different trajectories with the same ensemble density matrix. For spontaneous emission, $L=\sqrt{\Gamma }|g⟩⟨e|$ gives jump rate $\Gamma |⟨e|\psi ⟩{|}^2$ and resets the atom to its ground state.^[ref]

A pure stationary state is dark when every jump operator acts on it as a scalar, after an allowed shift of the jump operators, and the corresponding effective Hamiltonian preserves its ray. Every trajectory that reaches such a state remains on that same ray.