Lindblad–GKSL Equation

The Lindblad–GKSL equation describes continuous quantum evolution whose finite-time maps preserve both probabilities and positivity, including when the system is entangled with an untouched reference. With time-independent coefficients it generates a quantum dynamical semigroup. Time-dependent coefficients with nonnegative rates give a broader class of CP-divisible evolutions.

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. With dimensionless $L_k$, the $\gamma_k$ have inverse-time units. The same equation is often written by absorbing $\sqrt{\gamma_k}$ into $L_k$; an absorbed jump then has units of inverse square root of time.

Helpful background. Completely positive maps explain why a reference system matters; reduced dynamics distinguishes a physical channel family from a semigroup. Numerical solutions are developed in Solving Lindblad Equations.

The dissipator and its assumptions

Work first in finite dimension, with bounded operators. For the oscillator examples below, unbounded creation and annihilation operators require a suitable generator domain; finite occupation moments are needed when deriving moment equations. A formal expression alone is not a proof of a well-defined infinite-dimensional semigroup.

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 $Z$.

A qubit: from jump rates to measured decay

Consider an undriven qubit with $H=\hbar\omega_0|e\rangle\langle e|$, $\omega_0>0$. Set $\sigma_-=|g\rangle\langle e|$, $\sigma_+=\sigma_-^\dagger$, and $Z=|g\rangle\langle g|-|e\rangle\langle e|$. In the interaction picture, downward transitions, upward transitions, and pure dephasing give

$$\dot\rho=\Gamma_\downarrow\mathcal D[\sigma_-]\rho +\Gamma_\uparrow\mathcal D[\sigma_+]\rho +\frac{\Gamma_\phi}{2}\mathcal D[Z]\rho, \qquad \Gamma_\downarrow,\Gamma_\uparrow,\Gamma_\phi\ge0.$$

Write $\rho=\begin{pmatrix}1-p&c\\c^*&p\end{pmatrix}_{g,e}$. The downward jump term is $p|g\rangle\langle g|$, and its anticommutator removes $p$ from the excited population and $c/2$ from the coherence. The upward dissipator reverses the population flow and also removes $c/2$; the Z dissipator contributes $-2c$. Therefore

$$\dot p=-\Gamma_\downarrow p+\Gamma_\uparrow(1-p), \qquad \dot c=-\left(\frac{\Gamma_\downarrow+\Gamma_\uparrow}{2}+\Gamma_\phi\right)c.$$

For $\Gamma_1=\Gamma_\downarrow+\Gamma_\uparrow>0$, subtract the stationary population $p_*=\Gamma_\uparrow/\Gamma_1$ from the first equation. Both equations now integrate directly:

$$\begin{aligned} p(t)&=p_*+[p(0)-p_*]e^{-\Gamma_1t}, &p_*&=\frac{\Gamma_\uparrow}{\Gamma_1},\\ c(t)&=c(0)e^{-\Gamma_2t}, &\Gamma_2&=\frac{\Gamma_1}{2}+\Gamma_\phi. \end{aligned}$$

Thus population measurements determine $T_1=1/\Gamma_1$, while a phase-sensitive measurement determines $T_2=1/\Gamma_2$. The laboratory-frame $\rho_{ge}$ additionally has the phase $e^{+i\omega_0t}$. Nonnegative pure dephasing requires $T_2\le2T_1$ for this model. These times refer to decay in the stated basis and protocol; a driven transient or an inhomogeneously broadened ensemble requires its own interpretation.

For example, take $(\Gamma_\downarrow,\Gamma_\uparrow,\Gamma_\phi)=(3,1,2)\,\mathrm{s}^{-1}$ and initially $|+\rangle=(|g\rangle+|e\rangle)/\sqrt2$. At $t_*=(\ln2)/(4\,\mathrm{s}^{-1})\simeq0.1733$ s, both decaying factors are $1/2$, so

$$\rho(t_*)=\begin{pmatrix}5/8&1/4\\1/4&3/8\end{pmatrix}_{g,e}, \qquad \operatorname{Tr}\rho(t_*)^2=\frac{21}{32}.$$

Its determinant is $11/64>0$, and its trace is one. The unique long-time state is $\operatorname{diag}(3/4,1/4)$: pure dephasing changes the approach of the coherence, not the final population. Omitting that dephasing would keep $T_1=1/4$ s but give $T_2=1/2$ s. If both transition rates vanish instead, all diagonal populations remain stationary and the formula for $p_*$ must not be used.

For a single equilibrium weak-coupling bath, detailed balance further requires $\Gamma_\uparrow/\Gamma_\downarrow=e^{-\beta\hbar\omega_0}$; the example has $\beta\hbar\omega_0=\ln3$. Positive rates alone do not establish that the actual environment is thermal. Thermal master equations derives this extra constraint. The separate amplitude-damping and pure-dephasing articles develop their microscopic and measurement interpretations.

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.

The displayed Kraus sum is exactly CP but trace preserving only through first order. The Euler update $I+dt\mathcal L$ is a different finite step and can fail positivity even for arbitrarily small $dt$. For instance, Hamiltonian evolution with $H=\hbar\Omega Z/2$ sends the initial Bloch vector $(1,0,0)$ to $(1,\Omega dt,0)$ under Euler stepping. Its length exceeds one, giving a negative eigenvalue $(1-\sqrt{1+\Omega^2dt^2})/2$. The exact rotation has length one. Use the finite propagator or a controlled positivity-preserving method when this distinction matters.

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 composition of the reduced one-time maps. It does not by itself establish that every multi-time experiment with interventions has no memory. 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:

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

For finite-dimensional, piecewise continuous bounded coefficients, if $\mathcal L_t$ has Lindblad form with nonnegative time-dependent rates, the dynamics is CP-divisible. More general time-local equations can still be useful, but complete positivity must be checked rather than assumed.

Other useful generators

Pure dephasing

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

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

With a Hamiltonian diagonal in the Z basis, this is Markovian decoherence without energy relaxation. The displayed coherence equation is in the interaction picture.

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

Oscillator damping

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

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 and monitoring

After rates are absorbed into $J_k=\sqrt{\gamma_k}L_k$, unitary mixing of the $J_k$ leaves the generator unchanged. Scalar shifts also leave it unchanged if the Hamiltonian is adjusted. These transformations are derived in Lindblad Operators.

An operator list alone does not specify a measured photon record. Jump and diffusive monitoring can produce different conditional trajectories with the same unconditional equation. State which environment measurement is performed before interpreting a term as an observed jump.

Exercises

1. Reconstructing rates. An undriven qubit is fit by the constant-rate model above, with $T_1=20\,\mu\mathrm{s}$, $T_2=30\,\mu\mathrm{s}$, and $p_*=0.10$. Find the three rates and the inferred thermal population ratio. Could the same model accommodate $T_2=50\,\mu\mathrm{s}$ while keeping $T_1$ and $p_*$ fixed?

Solution

The sum of transition rates is $\Gamma_1=T_1^{-1}=5.0\times10^4\,\mathrm{s}^{-1}$. Since $p_*=\Gamma_\uparrow/\Gamma_1$,

$$\Gamma_\uparrow=5.0\times10^3\,\mathrm{s}^{-1}, \qquad \Gamma_\downarrow=4.5\times10^4\,\mathrm{s}^{-1}.$$

The pure-dephasing rate is

$$\Gamma_\phi=\frac1{T_2}-\frac1{2T_1} =\frac1{120\,\mu\mathrm{s}}\simeq8.33\times10^3\,\mathrm{s}^{-1}.$$

The upward/downward ratio is $1/9$. If an equilibrium thermal interpretation is justified, $\beta\hbar\omega_0=\ln9$; a numerical temperature also requires the transition frequency.

For $T_2=50\,\mu\mathrm{s}$, the inferred $\Gamma_\phi=-5.0\times10^3\,\mathrm{s}^{-1}$ is negative. These fitted quantities cannot describe this nonnegative-rate model. That diagnoses an inconsistent model or fit; it does not by itself identify which experimental assumption fails.

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. A positive population step can fail complete positivity. Take only zero-temperature damping, $\mathcal L=\Gamma\mathcal D[|g\rangle\langle e|]$ with $\Gamma>0$, and let $x=\Gamma h$ for a finite step $h>0$. Euler stepping gives $p'=(1-x)p$ and $c'=(1-x/2)c$. For what $x$ does it preserve every diagonal state’s probabilities? Test complete positivity on $|\Omega\rangle=(|g,0\rangle+|e,1\rangle)/\sqrt2$. Then compare with the Kraus step

$$K_0=|g\rangle\langle g|+\sqrt{1-x}|e\rangle\langle e|, \qquad K_1=\sqrt{x}|g\rangle\langle e|.$$

What step restriction makes this Kraus map physical, and how would you change $x$ to obtain the exact finite-time propagator?

Solution

All diagonal probabilities remain valid for $0\le x\le1$. Nevertheless, in the basis $|g,0\rangle,|g,1\rangle,|e,0\rangle,|e,1\rangle$, Euler stepping on the first qubit gives

$$(\mathrm{id}+h\mathcal L)\otimes\mathrm{id} \bigl(|\Omega\rangle\langle\Omega|\bigr) =\frac12\begin{pmatrix} 1&0&0&1-x/2\\ 0&x&0&0\\ 0&0&0&0\\ 1-x/2&0&0&1-x \end{pmatrix}.$$

The $g0,e1$ principal block has determinant

$$\frac14\left[(1-x)-(1-x/2)^2\right]=-\frac{x^2}{16}.$$

It is negative for every $x>0$. Thus this Euler map fails complete positivity however small the positive step, even in the range where populations pass their test.

The Kraus map instead has $\sum_jK_j^\dagger K_j=I$ and is exactly CPTP for $0\le x\le1$, or $h\le1/\Gamma$. It gives the same population step but multiplies coherence by $\sqrt{1-x}$, which agrees with $1-x/2$ only to first order. After $n$ such steps, survival is $(1-\Gamma t/n)^n\to e^{-\Gamma t}$.

For the exact damping semigroup over any $h\ge0$, replace $x$ by $1-e^{-\Gamma h}$. Then survival is $e^{-\Gamma h}$ and coherence decays by $e^{-\Gamma h/2}$. The restriction $h\le1/\Gamma$ belongs to the first-order Kraus approximation, not to the exact evolution. Physicality of a discrete channel and accuracy at a chosen time step are separate checks. Preskill’s notes, §§3.4.3 and 3.5.2, develop the finite-channel and infinitesimal constructions.

References

• H.-P. Breuer and F. Petruccione, The Theory of Open Quantum Systems, Oxford University Press (2002). Publisher.
• E. B. Davies, Quantum Theory of Open Systems, Academic Press (1976).
• C. W. Gardiner and P. Zoller, Quantum Noise, 3rd ed., Springer (2004). Publisher.
• 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). DOI.
• G. Lindblad, “On the generators of quantum dynamical semigroups,” Communications in Mathematical Physics 48, 119–130 (1976). DOI.
• J. Preskill, Lecture Notes for Ph219/CS219: Quantum Information, chapter 3, §§3.4.3 and 3.5.2, California Institute of Technology (updated October 2018). Author’s notes.
• Á. Rivas and S. F. Huelga, Open Quantum Systems: An Introduction, Springer (2012). Author manuscript.
• H. M. Wiseman and G. J. Milburn, Quantum Measurement and Control, Cambridge University Press (2010). DOI.