Lindblad Theorem

The Lindblad theorem characterizes the generators of Markovian completely positive trace-preserving quantum dynamics. In finite dimensions, it says that a norm-continuous quantum dynamical semigroup is CPTP if and only if its generator has Gorini–Kossakowski–Sudarshan–Lindblad form.

The theorem is structural. It does not derive the generator of a particular atom, circuit, molecule, detector, or reservoir. It tells you what form the generator must have once the semigroup, complete-positivity, and trace-preservation assumptions are imposed.

For the equation as a working master equation, see Lindblad–GKSL Equation. For how to interpret the operators, see Lindblad Operators.

The Setting

Let $\mathcal H$ be finite-dimensional and let $\mathcal T(\mathcal H)$ be the vector space of trace-class operators, which is just the full matrix algebra in finite dimension.

A quantum dynamical semigroup is a family of linear maps

$$\{\Phi_t:t\ge0\}$$

such that

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

The family is Markovian in the time-homogeneous semigroup sense: evolution from $s$ to $s+t$ is the same map $\Phi_t$, independent of the earlier history.

The physical requirements are:

• $\Phi_t$ is trace preserving for every $t\ge0$;
• $\Phi_t$ is completely positive for every $t\ge0$;
• $\Phi_t$ is continuous at $t=0$ in the operator norm, so a bounded generator exists.

The generator is

$$\mathcal L = \lim_{t\downarrow0} \frac{\Phi_t-\mathrm{id}}{t},$$

and the semigroup is

$$\Phi_t=e^{t\mathcal L}.$$

Finite-Dimensional Theorem

In finite dimensions, a linear map $\mathcal L$ generates a norm-continuous CPTP semigroup $\Phi_t=e^{t\mathcal L}$ if and only if there exist a Hermitian operator $H=H^\dagger$ and operators $L_j$ such that

$$\mathcal L(\rho) = - \frac{i}{\hbar}[H,\rho] + \sum_j \left( L_j\rho L_j^\dagger - \frac12 \{L_j^\dagger L_j,\rho\} \right).$$

Equivalently, one may write $L_j=\sqrt{\gamma_j}V_j$ with $\gamma_j\ge0$:

$$\mathcal L(\rho) = - \frac{i}{\hbar}[H,\rho] + \sum_j \gamma_j \left( V_j\rho V_j^\dagger - \frac12 \{V_j^\dagger V_j,\rho\} \right).$$

The first direction says: if the generator has this form, then $e^{t\mathcal L}$ is a CPTP semigroup.

The converse says: if $e^{t\mathcal L}$ is a CPTP semigroup under the assumptions above, then $\mathcal L$ can be written in this form.

Heisenberg-Picture Form

The adjoint generator acts on observables. With the trace pairing

$$\operatorname{Tr}[A\,\mathcal L(\rho)] = \operatorname{Tr}[\mathcal L^\dagger(A)\rho],$$

the Heisenberg-picture form is

$$\mathcal L^\dagger(A) = \frac{i}{\hbar}[H,A] + \sum_j \left( L_j^\dagger A L_j - \frac12 \{L_j^\dagger L_j,A\} \right).$$

Trace preservation of $\mathcal L$ is equivalent to

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

Thus the observable identity stays fixed:

$$\Phi_t^\dagger(I)=I.$$

What Each Assumption Does

Semigroup property

The semigroup condition

$$\Phi_{t+s}=\Phi_t\Phi_s$$

imposes time homogeneity and memorylessness at the reduced level. It rules out explicit memory kernels and most time-dependent driving protocols.

Complete positivity

Complete positivity means

$$(\Phi_t\otimes\mathrm{id}_R)(X)\ge0 \quad \text{whenever} \quad X\ge0$$

for every finite reference system $R$. It is stronger than positivity on the system alone and is essential for consistency when the system may be entangled with a spectator. See Completely Positive Maps.

Trace preservation

Trace preservation keeps total probability normalized:

$$\operatorname{Tr}\Phi_t(\rho)=\operatorname{Tr}\rho.$$

At the generator level this becomes

$$\operatorname{Tr}\mathcal L(\rho)=0.$$

The anticommutator term in the Lindblad form is what makes each dissipator trace preserving.

Continuity

Continuity at $t=0$ ensures that an infinitesimal generator exists. In finite dimensions this is usually technically mild. In infinite dimensions it becomes one of the places where unbounded generators and domains enter.

Proof Sketch

The full proof belongs in a rigorous treatment of quantum dynamical semigroups. The finite-dimensional strategy can be summarized as follows.

First, choose an orthonormal operator basis consisting of the identity and traceless operators:

$$F_0=\frac{I}{\sqrt d}, \qquad \operatorname{Tr}(F_a^\dagger F_b)=\delta_{ab}, \qquad a,b\ge1.$$

Trace preservation and Hermiticity preservation constrain the generator. The Hamiltonian part is the anti-Hermitian commutator part. The remaining dissipative part can be written using a coefficient matrix $c_{ab}$:

$$\mathcal L_{\mathrm{diss}}(\rho) = \sum_{a,b\ge1} c_{ab} \left( F_a\rho F_b^\dagger - \frac12 \{F_b^\dagger F_a,\rho\} \right).$$

Complete positivity of the semigroup forces the Kossakowski matrix $c=(c_{ab})$ to be positive semidefinite.

Because $c\ge0$, it can be diagonalized:

$$c_{ab} = \sum_j \lambda_j u_{ja}u_{jb}^*, \qquad \lambda_j\ge0.$$

Defining

$$L_j = \sqrt{\lambda_j} \sum_a u_{ja}F_a$$

puts the dissipative part in Lindblad form. Conversely, if $\mathcal L$ has Lindblad form, its short-time map has Kraus form to first order, and the semigroup generated by $\mathcal L$ is CPTP for all $t\ge0$.

This sketch hides the technical work in showing exactly how complete positivity at infinitesimal times constrains $c_{ab}$, but it captures the theorem’s structure: positivity of a coefficient matrix becomes a sum of dissipators.

What the Theorem Guarantees

If the assumptions hold, the theorem guarantees:

• trace preservation for all $t\ge0$;
• complete positivity for all $t\ge0$;
• a time-independent generator;
• a representation in terms of a Hamiltonian and Lindblad operators;
• a legitimate Markovian quantum dynamical semigroup.

It does not guarantee:

• that a microscopic weak-coupling derivation is valid;
• that the bath is thermal;
• that the listed Lindblad operators are unique;
• that a jump trajectory is physically observed;
• that a time-dependent or non-Markovian equation is covered;
• that unbounded operators are harmless.

This distinction is central in applications. A generator can be mathematically valid while still being the wrong model for a given physical regime.

Relation to Finite-Time Channels

Every $\Phi_t$ in a Lindblad semigroup is a quantum channel and therefore has a Kraus representation. The converse is not true: an arbitrary channel $\Phi$ need not sit inside a continuous semigroup

$$\Phi=\Phi_{t_*}=e^{t_*\mathcal L}$$

with a time-independent Lindblad generator.

The theorem is about continuous-time semigroups, not about isolated finite-time noise maps. Embedding a channel into a Markovian semigroup is a separate problem.

Time-Dependent Generators

Many useful master equations have

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

with explicitly time-dependent rates and operators. If $\mathcal L_t$ has Lindblad form with nonnegative rates for every time, then the evolution is CP-divisible under suitable regularity assumptions. But it is not a time-homogeneous semigroup unless $\mathcal L_t$ is constant.

If some instantaneous rates become negative, the equation is outside the direct Lindblad semigroup theorem. The finite-time map may still be physical, but complete positivity must be checked separately. This is common in time-convolutionless master equations.

Infinite-Dimensional Warnings

Many physically important systems have infinite-dimensional Hilbert spaces: oscillators, fields, particles on a line, Brownian motion, and scattering continua. In these settings, formal Lindblad-looking equations can involve unbounded $H$ or $L_j$.

Then several finite-dimensional shortcuts fail:

• the generator may be unbounded and defined only on a dense domain;
• the exponential $e^{t\mathcal L}$ may require semigroup theory rather than a matrix exponential;
• trace preservation can fail if domains and boundary terms are mishandled;
• complete positivity of a formal differential expression may not imply a well-defined CPTP semigroup;
• different operator topologies lead to different continuity assumptions.

The bounded-generator Lindblad theorem is still the right reference point, but infinite-dimensional master equations require domain and closability checks that are not visible from the compact formula alone.

Common Mistakes

Calling any time-local equation a Lindblad theorem case

The theorem concerns time-homogeneous semigroups with a time-independent generator. A time-local equation with time-dependent coefficients is related but not the same theorem.

Forgetting complete positivity

Trace preservation and positivity on isolated system states are not enough. The semigroup must remain positive when tensored with an arbitrary reference.

Treating negative rates as theorem-compatible

Negative rates are outside the standard semigroup theorem. They may appear in non-Markovian time-local descriptions, but the finite-time map must then be checked directly.

Inferring microscopic validity from Lindblad form

Lindblad form guarantees a legitimate Markovian semigroup. It does not prove weak coupling, fast bath decay, secularization, detailed balance, or thermalization.

Ignoring unbounded-operator domains

For oscillators and particles, writing $a$, $x$, or $p$ in a dissipator is not the end of the analysis. The generator must be defined on an appropriate operator domain and must generate a valid trace-preserving semigroup.

Exercises

Trace preservation of the Lindblad generator

Show that

$$\operatorname{Tr}\mathcal L(\rho)=0$$

for

$$\mathcal L(\rho) = - \frac{i}{\hbar}[H,\rho] + \sum_j \left( L_j\rho L_j^\dagger - \frac12\{L_j^\dagger L_j,\rho\} \right).$$

Solution

The trace of the commutator vanishes by cyclicity:

$$\operatorname{Tr}[H,\rho]=0.$$

For one dissipator,

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

Therefore

$$\operatorname{Tr} \left[ L\rho L^\dagger - \frac12\{L^\dagger L,\rho\} \right] = 0.$$

Summing over $j$ gives $\operatorname{Tr}\mathcal L(\rho)=0$.

Generator from a semigroup

Suppose $\Phi_t=e^{t\mathcal L}$ in finite dimensions. Show that

$$\mathcal L = \left. \frac{d}{dt}\Phi_t \right|_{t=0}.$$

Solution

Using the matrix exponential,

$$e^{t\mathcal L} = \mathcal I + t\mathcal L + O(t^2).$$

Hence

$$\frac{\Phi_t-\mathcal I}{t} = \mathcal L+O(t),$$

so the limit as $t\downarrow0$ is $\mathcal L$.

Time-dependent positive rates

Consider

$$\dot\rho = \gamma(t)\mathcal D[L]\rho, \qquad \gamma(t)\ge0.$$

Is this automatically a time-homogeneous semigroup?

Solution

No. Nonnegative $\gamma(t)$ supports CP-divisible time-local evolution under suitable regularity assumptions, but the maps generally depend on both start and end times:

$$\Phi(t,s) \ne \Phi(t-s,0)$$

unless $\gamma(t)$ is constant or has special time-translation-invariant structure. A time-homogeneous semigroup requires $\Phi_{t+s}=\Phi_t\Phi_s$ with the same map family indexed only by elapsed time.

Diagonalizing a Kossakowski matrix

Let $c_{ab}$ be positive semidefinite and let

$$c_{ab} = \sum_j\lambda_j u_{ja}u_{jb}^*, \qquad \lambda_j\ge0.$$

Show how the operators

$$L_j = \sqrt{\lambda_j}\sum_a u_{ja}F_a$$

turn the coefficient-matrix dissipator into a sum of Lindblad dissipators.

Solution

Substitute the definition:

$$\sum_j L_j\rho L_j^\dagger = \sum_{j,a,b} \lambda_j u_{ja}u_{jb}^* F_a\rho F_b^\dagger = \sum_{a,b} c_{ab}F_a\rho F_b^\dagger.$$

Similarly,

$$\sum_j L_j^\dagger L_j = \sum_{a,b} c_{ab}F_b^\dagger F_a.$$

The same replacement inside the anticommutator gives exactly the coefficient-matrix dissipator.

References

• 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).
• G. Lindblad, “On the generators of quantum dynamical semigroups,” Communications in Mathematical Physics 48, 119-130 (1976).
• E. B. Davies, Quantum Theory of Open Systems, Academic Press (1976).
• R. Alicki and K. Lendi, Quantum Dynamical Semigroups and Applications, Springer (1987).
• H.-P. Breuer and F. Petruccione, The Theory of Open Quantum Systems, Oxford University Press (2002).
• Á. Rivas and S. F. Huelga, Open Quantum Systems: An Introduction, Springer (2012).