Lindblad–GKSL Equation

The Lindblad–GKSL equation is the standard time-local master equation for Markovian open quantum dynamics that preserves trace and complete positivity.

In one common convention,

$$\frac{d\rho}{dt} = - \frac{i}{\hbar}[H,\rho] + \sum_k\gamma_k \left( L_k\rho L_k^\dagger - \frac12 \{L_k^\dagger L_k,\rho\} \right), \qquad \gamma_k\ge0.$$

Here $H$ is Hermitian, the $L_k$ are Lindblad operators, and the rates $\gamma_k$ are nonnegative. The same equation is often written by absorbing $\sqrt{\gamma_k}$ into $L_k$.

For fixed points, gaps, and relaxation modes, see Steady States and Relaxation. For a validation-first numerical guide using matrix-vectorized Liouvillians, steady states, and spectrum checks, see Solving Lindblad Equations.

Compact Notation

Define the dissipator

$$\mathcal D[L]\rho = L\rho L^\dagger - \frac12 \{L^\dagger L,\rho\}.$$

Then the equation becomes

$$\frac{d\rho}{dt} = - \frac{i}{\hbar}[H,\rho] + \sum_k\gamma_k\mathcal D[L_k]\rho.$$

The word “dissipator” is conventional. It does not always mean energy dissipation. A term with $L=Z$ can describe pure dephasing without changing energy populations for a Hamiltonian proportional to $Z$.

Meaning of the Terms

Term	Role
$-\frac{i}{\hbar}[H,\rho]$	coherent Hamiltonian dynamics
$L_k\rho L_k^\dagger$	jump, noise, scattering, or environmental action
$-\frac12\{L_k^\dagger L_k,\rho\}$	trace-preserving normalization term
$\gamma_k$	nonnegative rate in a Markovian semigroup convention

The anticommutator term is not optional. Without it, the trace would generally grow or shrink incorrectly.

Trace Preservation

For one dissipator,

$$\operatorname{Tr}\mathcal D[L]\rho = \operatorname{Tr}(L\rho L^\dagger) - \frac12\operatorname{Tr}(L^\dagger L\rho) - \frac12\operatorname{Tr}(\rho L^\dagger L).$$

Using cyclicity of the trace,

$$\operatorname{Tr}(L\rho L^\dagger) = \operatorname{Tr}(L^\dagger L\rho) = \operatorname{Tr}(\rho L^\dagger L).$$

Therefore

$$\operatorname{Tr}\mathcal D[L]\rho=0.$$

The Hamiltonian commutator also preserves trace, so the full equation preserves $\operatorname{Tr}\rho$.

Complete Positivity Intuition

For a short time step $dt$, define

$$K_k = \sqrt{\gamma_k dt}\,L_k$$

for the jump/noise operators, and

$$K_0 = I - \frac{i}{\hbar}Hdt - \frac12 \sum_k \gamma_k L_k^\dagger L_k\,dt.$$

Then, to first order in $dt$,

$$\rho(t+dt) = K_0\rho(t)K_0^\dagger + \sum_k K_k\rho(t)K_k^\dagger + O(dt^2).$$

Expanding this expression gives the Lindblad–GKSL equation. Also,

$$K_0^\dagger K_0 + \sum_kK_k^\dagger K_k = I+O(dt^2).$$

Thus the infinitesimal evolution has the structure of a trace-preserving Kraus map to first order. The semigroup construction promotes these infinitesimal completely positive steps into a completely positive trace-preserving evolution for finite time.

This is not a proof of the theorem, but it is the right physical memory aid.

Semigroup Setting

The original finite-dimensional theorem concerns quantum dynamical semigroups:

$$\Phi_{t+s} = \Phi_t\Phi_s, \qquad \Phi_0=\mathrm{id}, \qquad t,s\ge0.$$

The semigroup property expresses time-homogeneous memoryless reduced dynamics. If

$$\rho(t)=\Phi_t(\rho(0)),$$

and $\Phi_t=e^{t\mathcal L}$ is completely positive and trace preserving, then the generator $\mathcal L$ has Lindblad–GKSL form under the Lindblad theorem assumptions.

Many applied master equations are time-dependent generalizations:

$$\frac{d\rho}{dt} = \mathcal L_t(\rho).$$

If $\mathcal L_t$ has Lindblad form with nonnegative time-dependent rates at every time, the dynamics is CP-divisible. More general time-local equations can still be useful, but complete positivity must be checked rather than assumed.

Pure Dephasing Example

For pure dephasing in the $Z$ basis, take

$$L=Z, \qquad \gamma=\frac{\Gamma_\phi}{2}.$$

Since $Z^\dagger Z=I$,

$$\gamma\mathcal D[Z]\rho = \frac{\Gamma_\phi}{2} \left( Z\rho Z-\rho \right).$$

For

$$\rho = \begin{pmatrix} \rho_{00}&\rho_{01}\\ \rho_{10}&\rho_{11} \end{pmatrix},$$

the populations are unchanged and

$$\frac{d\rho_{01}}{dt} = -\Gamma_\phi\rho_{01}.$$

This is Markovian decoherence without energy relaxation.

See Pure Dephasing Master Equation for the continuous-time solution, $T_2$ relation, and microscopic noise interpretation.

Amplitude Damping Example

For zero-temperature relaxation of a qubit, take

$$L=\sigma_-=|0\rangle\langle1|, \qquad \gamma=\Gamma_1.$$

The master equation is

$$\frac{d\rho}{dt} = \Gamma_1\mathcal D[\sigma_-]\rho.$$

It gives

$$\frac{d\rho_{11}}{dt} = -\Gamma_1\rho_{11}, \qquad \frac{d\rho_{00}}{dt} = \Gamma_1\rho_{11},$$

and

$$\frac{d\rho_{01}}{dt} = -\frac{\Gamma_1}{2}\rho_{01}$$

apart from Hamiltonian phase rotation.

This model is dissipative: excitation leaves the qubit and enters the environment.

See Amplitude Damping Master Equation for the population solution, jump interpretation, and finite-temperature extension.

Thermal Qubit Example

At finite temperature, include both downward and upward jumps:

$$\frac{d\rho}{dt} = \Gamma_\downarrow\mathcal D[\sigma_-]\rho + \Gamma_\uparrow\mathcal D[\sigma_+]\rho.$$

For a thermal bath at inverse temperature $\beta$, detailed balance gives

$$\frac{\Gamma_\uparrow}{\Gamma_\downarrow} = e^{-\beta\hbar\omega_0}$$

under the usual weak-coupling assumptions.

The steady state is thermal in the energy basis when the rates and Hamiltonian are consistent with the bath temperature. See Thermal Master Equations for the general weak-coupling setting.

Oscillator Damping Example

For a harmonic oscillator coupled to a thermal reservoir,

$$\frac{d\rho}{dt} = \kappa(\bar n+1)\mathcal D[a]\rho + \kappa\bar n\,\mathcal D[a^\dagger]\rho.$$

The $a$ term describes loss of quanta to the bath. The $a^\dagger$ term describes thermal excitation from the bath. When $\bar n=0$, the oscillator relaxes toward the vacuum state.

Depolarizing Dynamics Example

A simple qubit depolarizing generator can be written as

$$\frac{d\rho}{dt} = \frac{\Gamma}{4} \sum_{j=x,y,z} \left( \sigma_j\rho\sigma_j-\rho \right).$$

In Bloch-vector form,

$$\rho = \frac12 \left( I+\mathbf r\cdot\boldsymbol\sigma \right),$$

this gives

$$\frac{d\mathbf r}{dt} = -\Gamma\mathbf r.$$

The state is driven isotropically toward $I/2$.

Validity and Derivation Assumptions

The Lindblad–GKSL equation is not the most general open-system equation. It is the structure associated with Markovian completely positive trace-preserving dynamics, as characterized by the Lindblad theorem.

Microscopic derivations often use approximations such as:

• initially factorized system–bath state;
• weak system–bath coupling;
• bath state stationary or near equilibrium;
• rapidly decaying bath correlations;
• Markov approximation;
• secular or rotating-wave approximation;
• coarse graining over bath memory times.

Different physical derivations may justify different subsets of these assumptions. The theorem identifies the generator structure once Markovian semigroup assumptions are imposed; it does not by itself prove that a particular laboratory system satisfies those assumptions.

Nonuniqueness

The Lindblad representation is not unique.

One can unitary-mix Lindblad operators with the same rates, absorb rates into operators, shift operators by multiples of the identity with a compensating Hamiltonian change, or choose different unravelings for the same master equation. See Lindblad Operators for the physical interpretation and nonuniqueness of the operator list.

Therefore:

Lindblad operator in an equation
  does not automatically mean
actually observed jump in an experiment.

A jump interpretation requires a specified monitoring scheme. Without monitoring, the same equation describes unconditional ensemble dynamics.

Common Mistakes

• Dropping the anticommutator term.
• Calling every time-local master equation Lindblad form even when rates are negative or the generator is not CP-compatible.
• Assuming Markovian means “no environment.” It means environmental memory has been eliminated from the reduced description.
• Treating Lindblad operators as unique physical observables.
• Confusing pure dephasing with energy dissipation because both appear in dissipator notation.
• Using a Lindblad equation outside its approximation regime without checking positivity, trace preservation, and timescales.
• Assuming the steady state is thermal without detailed-balance conditions.

References

• 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).
• E. B. Davies, Quantum Theory of Open Systems, Academic Press (1976).
• H.-P. Breuer and F. Petruccione, The Theory of Open Quantum Systems, Oxford University Press (2002).
• C. W. Gardiner and P. Zoller, Quantum Noise, Springer, 3rd ed. (2004).
• H. M. Wiseman and G. J. Milburn, Quantum Measurement and Control, Cambridge University Press (2010).
• Á. Rivas and S. F. Huelga, Open Quantum Systems: An Introduction, Springer (2012).

Exercises

1. Trace preservation. Show directly that $\operatorname{Tr}\mathcal D[L]\rho=0$.

Solution

By definition,

$$\operatorname{Tr}\mathcal D[L]\rho = \operatorname{Tr}(L\rho L^\dagger) - \frac12\operatorname{Tr}(L^\dagger L\rho) - \frac12\operatorname{Tr}(\rho L^\dagger L).$$

Cyclicity gives

$$\operatorname{Tr}(L\rho L^\dagger) = \operatorname{Tr}(L^\dagger L\rho) = \operatorname{Tr}(\rho L^\dagger L).$$

The first term is canceled by the two half-terms, so the trace change is zero.

2. Short-time Kraus expansion. Use

$$K_0 = I - \frac{i}{\hbar}Hdt - \frac12\sum_k\gamma_kL_k^\dagger L_kdt, \qquad K_k=\sqrt{\gamma_kdt}\,L_k$$

to recover the Lindblad–GKSL equation to first order in $dt$.

Solution

Keeping only terms through first order,

$$K_0\rho K_0^\dagger = \rho - \frac{i}{\hbar}[H,\rho]dt - \frac12 \sum_k\gamma_k \{L_k^\dagger L_k,\rho\}dt.$$

The jump terms give

$$\sum_kK_k\rho K_k^\dagger = \sum_k\gamma_kdt\,L_k\rho L_k^\dagger.$$

Adding terms,

$$\rho(t+dt)-\rho(t) = \left[ - \frac{i}{\hbar}[H,\rho] + \sum_k\gamma_k \left( L_k\rho L_k^\dagger - \frac12\{L_k^\dagger L_k,\rho\} \right) \right]dt.$$

Divide by $dt$ and take the limit.

3. Pure dephasing rate. With $L=Z$ and $\gamma=\Gamma_\phi/2$, compute the equation of motion for $\rho_{01}$.

Solution

Since $Z\rho Z$ flips the signs of the off-diagonal entries,

$$(Z\rho Z-\rho)_{01} = -2\rho_{01}.$$

Thus

$$\frac{d\rho_{01}}{dt} = \frac{\Gamma_\phi}{2} (-2\rho_{01}) = -\Gamma_\phi\rho_{01}.$$

The coherence decays as $\rho_{01}(t)=e^{-\Gamma_\phi t}\rho_{01}(0)$ in the interaction picture.