Steady States and Relaxation

A steady state of a time-independent Markovian master equation is a density operator that the generator leaves unchanged. If

$$\dot\rho = \mathcal L(\rho),$$

then a steady state satisfies

$$\mathcal L(\rho_{\mathrm{ss}})=0, \qquad \operatorname{Tr}\rho_{\mathrm{ss}}=1, \qquad \rho_{\mathrm{ss}}\ge0.$$

Steady states are fixed points of the finite-time evolution:

$$\Phi_t(\rho_{\mathrm{ss}}) = e^{t\mathcal L}\rho_{\mathrm{ss}} = \rho_{\mathrm{ss}}.$$

The main questions are not only “what is the fixed point?” but also:

• Is it unique?
• Which initial states approach it?
• How fast do deviations decay?
• Are there conserved quantities or dark subspaces?
• Is the steady state thermal, nonequilibrium, or engineered?
• Are there long-lived metastable modes before the final steady state is reached?

This page is the conceptual hub for those questions. Relaxation and Thermalization owns the distinct closed-system problem, where the global state evolves unitarily and only restricted probes or subsystems can approach thermal predictions. The numerical workflow for finite Markovian generators is in Solving Lindblad Equations.

Liouvillian Kernel

The linear map $\mathcal L$ is often called the Liouvillian. Its kernel is

$$\ker\mathcal L = \{X:\mathcal L(X)=0\}.$$

Physical steady states are the positive, trace-one elements of this kernel. The kernel may also contain non-Hermitian or nonpositive operators; those are useful for linear algebra, but they are not states by themselves.

In finite dimensions, a time-independent completely positive trace-preserving semigroup has at least one stationary state. A practical way to see this is to average a trajectory over a long time:

$$\bar\rho_T = \frac{1}{T} \int_0^T dt\,\Phi_t(\rho_0).$$

The state space is compact, and accumulation points of such averages are stationary under mild finite-dimensional assumptions.

Trace preservation is a statement about the adjoint generator:

$$\mathcal L^\dagger(I)=0.$$

Thus the identity is always a left zero mode. Right zero modes are stationary operators. Left zero modes beyond the identity are conserved observables or sector labels, and they often explain why the steady state is not unique.

Relaxation Modes

If $\mathcal L$ is diagonalizable, one can expand operators in right eigenoperators:

$$\mathcal L(R_a)=\lambda_a R_a.$$

Then

$$e^{t\mathcal L}R_a = e^{\lambda_a t}R_a.$$

For a stable finite-dimensional CPTP semigroup, nonstationary modes have

$$\operatorname{Re}\lambda_a\le0.$$

The real part gives decay or growth. Physical stability rules out positive real parts. The imaginary part gives oscillation frequency:

$$e^{\lambda_a t} = e^{\operatorname{Re}\lambda_a t} e^{i\,\operatorname{Im}\lambda_a t}.$$

The zero eigenvalues describe stationary operators. Eigenvalues with real parts close to zero describe slow relaxation.

Many Liouvillians are nonnormal, so right eigenoperators need not be orthogonal and may be ill conditioned. Spectra are still informative, but numerical conclusions should be checked with residuals, positivity, and direct time evolution.

Liouvillian Gap

When the steady state is unique and all other modes decay, the Liouvillian gap is commonly defined as

$$\Delta_{\mathcal L} = \min_{\lambda_a\ne0} \left[ - \operatorname{Re}\lambda_a \right],$$

where the minimum is taken over nonstationary modes. The associated relaxation time scale is roughly

$$\tau_{\mathrm{rel}} \sim \frac{1}{\Delta_{\mathcal L}}.$$

This statement is only a first estimate. Prefactors, nonnormality, initial-state overlap with slow modes, and time-dependent observables can matter. Still, the gap is the main first diagnostic for how quickly a Markovian system forgets its initial condition.

If $\Delta_{\mathcal L}=0$ because there are multiple stationary sectors or purely imaginary modes, the system does not relax to a single time-independent state from all initial conditions.

Unique Versus Multiple Steady States

A unique steady state means there is exactly one density operator $\rho_{\mathrm{ss}}$ satisfying $\mathcal L(\rho_{\mathrm{ss}})=0$. It does not automatically mean every convergence estimate is fast, but it does mean the long-time state is independent of the initial state within the modeled state space.

Multiple steady states can arise from:

• conserved quantities;
• disconnected classical sectors;
• decoherence-free subspaces;
• symmetry-protected dark subspaces;
• absorbing states in a rate equation;
• missing weak processes that would connect sectors in a more complete model.

For a Pauli Rate Equation, this is graph theory: disconnected communicating classes or absorbing states give multiple stationary distributions. For a quantum Liouvillian, the same idea appears as a nontrivial zero-mode structure of $\mathcal L$ and $\mathcal L^\dagger$.

In experiments and simulations, an apparent family of steady states can also be a metastable manifold: the system looks stationary for a long time, but an extremely slow mode eventually selects a unique state.

Thermal and Nonequilibrium Steady States

A thermal steady state has the Gibbs form

$$\rho_\beta = \frac{e^{-\beta H_S}} {\operatorname{Tr}(e^{-\beta H_S})}$$

or the appropriate renormalized equilibrium state for the approximation being used. A weak-coupling equilibrium bath usually leads to this state only when the rates satisfy Detailed Balance.

Stationarity alone is weaker:

$$\mathcal L(\rho_\beta)=0$$

does not by itself prove detailed balance. A generator can have a stationary distribution while supporting irreversible cycles, currents, or coherences. Detailed balance rules out equilibrium currents by relating forward and reverse processes.

With multiple baths, drives, measurement feedback, biased reservoirs, or engineered dissipation, the steady state is generally a nonequilibrium steady state. It can be time independent while carrying heat, particle, photon, or information currents.

Dark States and Attractors

A pure state $\lvert\psi\rangle$ is a simple dark steady state if

$$L_k\lvert\psi\rangle=0 \quad \text{for all }k, \qquad H\lvert\psi\rangle=E\lvert\psi\rangle.$$

Then

$$\rho_\psi = \lvert\psi\rangle\langle\psi\rvert$$

satisfies $\mathcal L(\rho_\psi)=0$.

This condition is sufficient, not necessary. Lindblad operators are not unique, and pure stationary states can sometimes be written in forms where the jumps do not literally annihilate the vector. The invariant object is the full Liouvillian.

A dark state is also not necessarily an attractor. If several states are dark, the generator may protect a subspace rather than prepare one unique state. Reservoir Engineering is largely the art of making the desired state or subspace stationary while ensuring all unwanted components are drained toward it; Dissipative State Preparation gives the focused preparation checklist.

Metastability

Metastability means the system first approaches a long-lived low-dimensional manifold and only later relaxes to the final steady state. Spectrally, it appears when a few eigenvalues lie very close to zero while the rest are separated by a larger decay scale.

Schematically, suppose

$$0 = \lambda_0, \qquad |\operatorname{Re}\lambda_1|, \ldots, |\operatorname{Re}\lambda_m| \ll |\operatorname{Re}\lambda_{m+1}|.$$

Then after the fast modes die out, the state is well approximated by a combination of the stationary state and the $m$ slow modes. On intermediate times it can look as if there are several steady states, even if the exact steady state is unique.

Metastability is common near phase transitions, in weakly broken symmetries, in systems with rare switching events, and in engineered reservoirs with small leakage out of a protected sector.

Examples

Pure dephasing

For qubit pure dephasing in the $\sigma_z$ basis,

$$\dot\rho_{00}=0, \qquad \dot\rho_{11}=0, \qquad \dot\rho_{01}=-\Gamma_\phi\rho_{01}$$

in the rotating frame. Every diagonal state is stationary. The steady-state manifold is not a single point:

$$\rho_{\mathrm{ss}} = \begin{pmatrix} p&0\\ 0&1-p \end{pmatrix}, \qquad 0\le p\le1.$$

Pure dephasing destroys coherence but does not choose a thermal population. See Pure Dephasing Master Equation.

Amplitude damping

For zero-temperature amplitude damping,

$$\dot\rho = \Gamma\mathcal D[\sigma_-]\rho,$$

the unique steady state is

$$\rho_{\mathrm{ss}} = \lvert g\rangle\langle g\rvert.$$

The population mode decays with rate $\Gamma$, while coherences decay with rate $\Gamma/2$ before adding Hamiltonian phase rotation. Thus different observables can relax on different time scales. See Amplitude Damping Master Equation.

Two-state rate equation

For upward and downward rates,

$$\dot p_e = - \Gamma_\downarrow p_e + \Gamma_\uparrow(1-p_e),$$

the stationary population is

$$p_e^{\mathrm{ss}} = \frac{\Gamma_\uparrow} {\Gamma_\uparrow+\Gamma_\downarrow},$$

and the nonzero eigenvalue is

$$\lambda = - (\Gamma_\uparrow+\Gamma_\downarrow).$$

This is the population-sector relaxation mode of the corresponding two-level thermal master equation.

Driven two-level system

The Optical Bloch Equations have a unique steady state under ordinary radiative decay and continuous drive. The nonzero modes can be real or complex depending on drive strength and detuning. Complex modes describe damped Rabi oscillations, while their real parts determine decay envelopes.

Practical Computation

For a finite-dimensional Hilbert space, a reliable workflow is:

1. Build the Liouvillian superoperator $\mathbb L$ in a declared vectorization convention.
2. Compute the null space of $\mathbb L$.
3. Convert candidate zero modes back into matrices.
4. Impose Hermiticity, positivity, and trace normalization.
5. Count independent physical steady states.
6. Compute residuals $\lVert\mathcal L(\rho_{\mathrm{ss}})\rVert$.
7. Inspect the adjoint null space for conserved observables.
8. Compute nonzero eigenvalues and estimate the relaxation gap.
9. Compare time-domain evolution with the spectral prediction.

For nonunique steady states, do not silently normalize an arbitrary null vector and call it the steady state. The null space itself is the result.

For infinite-dimensional systems, truncation can create artificial steady states or remove real ones. Oscillator cutoffs, unbounded generators, and domain issues must be checked by convergence tests and physical limits.

Common Mistakes

Confusing stationary with attracting

$\mathcal L(\rho_\star)=0$ only says the state does not move. It does not say other states approach it.

Assuming the steady state is thermal

Thermal interpretation requires the right bath assumptions and detailed-balance relations. A driven or biased system can have a perfectly good steady state that is not Gibbs.

Ignoring conserved quantities

Extra left zero modes of $\mathcal L^\dagger$ label quantities that do not decay. They can prevent convergence to a unique state.

Calling any dark state prepared

A dark state may be protected without being globally attracting. Preparation requires that unwanted components have decay paths into the target and no stable traps elsewhere.

Reading too much from eigenvalues alone

Liouvillians can be nonnormal. Eigenvalues give essential time scales, but amplitudes, transient growth, condition numbers, and observable overlaps still matter.

Trusting a truncated oscillator steady state too quickly

If probability accumulates near the cutoff, the computed steady state may be a numerical artifact. Increase the cutoff and check observables.

Exercises

1. Fixed point from the generator. Show that if $\mathcal L(\rho_{\mathrm{ss}})=0$, then $e^{t\mathcal L}\rho_{\mathrm{ss}}=\rho_{\mathrm{ss}}$.

Solution

Use the power series:

$$e^{t\mathcal L}\rho_{\mathrm{ss}} = \sum_{n=0}^{\infty} \frac{t^n}{n!} \mathcal L^n(\rho_{\mathrm{ss}}).$$

The $n=0$ term is $\rho_{\mathrm{ss}}$. Every term with $n\ge1$ contains at least one $\mathcal L(\rho_{\mathrm{ss}})$ and therefore vanishes. Thus

$$e^{t\mathcal L}\rho_{\mathrm{ss}} = \rho_{\mathrm{ss}}.$$

2. Pure-dephasing steady states. For a qubit with $\dot\rho_{01}=-\Gamma_\phi\rho_{01}$ and fixed populations, describe the steady-state set.

Solution

At stationarity, $\rho_{01}=0$ if $\Gamma_\phi>0$, while $\rho_{00}$ and $\rho_{11}$ are unchanged except for normalization. Therefore

$$\rho_{\mathrm{ss}} = \begin{pmatrix} p&0\\ 0&1-p \end{pmatrix}, \qquad 0\le p\le1.$$

There is a continuum of steady states, not a unique thermal state.

3. Two-state relaxation gap. The rate matrix for $p=(p_g,p_e)^{\mathsf T}$ is

$$K = \begin{pmatrix} -\Gamma_\uparrow & \Gamma_\downarrow\\ \Gamma_\uparrow & -\Gamma_\downarrow \end{pmatrix}.$$

Find its eigenvalues.

Solution

Because columns sum to zero, one eigenvalue is $0$. The trace is

$$\operatorname{Tr}K = - (\Gamma_\uparrow+\Gamma_\downarrow).$$

For a $2\times2$ matrix, the sum of eigenvalues is the trace, so the other eigenvalue is

$$\lambda_1 = - (\Gamma_\uparrow+\Gamma_\downarrow).$$

The relaxation gap is $\Gamma_\uparrow+\Gamma_\downarrow$ when the two rates connect the states.

4. Dark but not unique. Two qubits undergo collective decay with $J_-=\sigma_-^{(1)}+\sigma_-^{(2)}$. Both $\lvert gg\rangle$ and the singlet $\lvert\psi_-\rangle=(\lvert ge\rangle-\lvert eg\rangle)/\sqrt2$ are annihilated by $J_-$. Does collective decay alone uniquely prepare the singlet?

Solution

No. Both states are dark under the same jump operator:

$$J_-\lvert gg\rangle=0, \qquad J_-\lvert\psi_-\rangle=0.$$

Since the dark space is at least two-dimensional, collective decay alone does not select the singlet as a unique attractor. Additional drives, dissipators, measurement conditioning, or symmetry breaking are needed to make the target uniquely attracting.

5. Thermal versus stationary. A rate equation on three states has a stationary distribution but also a nonzero clockwise probability current around the cycle. Is that an equilibrium detailed-balance steady state?

Solution

No. Stationarity only requires the net inflow and outflow at each state to balance. Detailed balance requires each edge current to vanish pairwise:

$$W_{n\leftarrow m}p_m^{\mathrm{ss}} = W_{m\leftarrow n}p_n^{\mathrm{ss}}.$$

A circulating current violates this pairwise condition, so the state is a nonequilibrium steady state rather than an equilibrium detailed-balance state.

References

• H.-P. Breuer and F. Petruccione, The Theory of Open Quantum Systems, Oxford University Press (2002).
• R. Alicki and K. Lendi, Quantum Dynamical Semigroups and Applications, Springer (2007).
• G. Lindblad, “On the generators of quantum dynamical semigroups,” Communications in Mathematical Physics 48, 119–130 (1976).
• V. Gorini, A. Kossakowski, and E. C. G. Sudarshan, “Completely positive dynamical semigroups of N-level systems,” Journal of Mathematical Physics 17, 821–825 (1976).
• V. V. Albert and L. Jiang, “Symmetries and conserved quantities in Lindblad master equations,” Physical Review A 89, 022118 (2014).
• K. Macieszczak, M. Guță, I. Lesanovsky, and J. P. Garrahan, “Towards a theory of metastability in open quantum dynamics,” Physical Review Letters 116, 240404 (2016).