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.

Helpful background. Completely positive maps and the Choi matrix explain positivity with a reference. The Lindblad–GKSL equation gives elementary examples of the generator.

Continuous quantum dynamical semigroups

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 has time-homogeneous reduced composition: the same map $\Phi_t$ advances any modeled state from $s$ to $s+t$. This concerns one-time reduced maps; it does not determine arbitrary multi-time experiments with interventions.

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-homogeneous composition. A generic time-dependent drive or finite-bath memory model will not satisfy it. An alternative memory-kernel representation of a semigroup can still be written, so the presence of an integral in some representation is not by itself decisive.

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. Terms involving the identity basis element combine into a commutator with a Hermitian operator. The remaining dissipative part can be written using a coefficient matrix $c_{ab}$:

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

To see where the positivity constraint enters, use the output-first Choi convention

$$J(\mathcal L)=\sum_{mn}\mathcal L(|m\rangle\langle n|)\otimes|m\rangle\langle n|.$$

The identity map has $J(\mathrm{id})=|\Omega\rangle\langle\Omega|$, with $|\Omega\rangle=\sum_m|m,m\rangle$. On the orthogonal complement of $|\Omega\rangle$, this matrix is zero. Since $J(\Phi_t)=J(\mathrm{id})+tJ(\mathcal L)+O(t^2)$ must remain positive, $QJ(\mathcal L)Q\ge0$, where $Q=I-|\Omega\rangle\langle\Omega|/d$. The traceless basis vectors span this complement; their coefficient matrix is $c$. Thus $c\ge0$. This is conditional positivity of the generator, not the false claim that $\mathcal L$ itself is a CP map.

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. Here $u_{ja}$ is component $a$ of eigenvector $j$; the index order of the displayed coefficient formula fixes the conjugation convention.

For the converse, let $R=\sum_jL_j^\dagger L_j$ and choose $h>0$ with $h\|R\|\le1$. The operators

$$K_0=e^{-iHh/\hbar}\sqrt{I-hR},\qquad K_j=\sqrt h L_j$$

satisfy $\sum_jK_j^\dagger K_j+K_0^\dagger K_0=I$ exactly. Their channel is $\Psi_h=\mathrm{id}+h\mathcal L+O(h^2)$. In finite dimension, $(\Psi_{t/n})^n\to e^{t\mathcal L}$; compositions and limits preserve complete positivity and trace. This proves sufficiency without mistaking an Euler step for an exactly physical channel.

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.

A complex coefficient matrix: worked example

For a qubit choose $F_1=X/\sqrt2$, $F_2=Y/\sqrt2$ and

$$c=\kappa\begin{pmatrix}1&i\\-i&1\end{pmatrix},\qquad\kappa>0.$$

Its eigenvalues are $2\kappa$ and zero. An eigenvector for $2\kappa$ is $(1,-i)/\sqrt2$, so one jump suffices:

$$L=\sqrt\kappa(F_1-iF_2)=\sqrt{2\kappa}\,|1\rangle\langle0|.$$

Thus the imaginary off-diagonal entries encode a directed $0\to1$ transition. Replacing $c$ by its entrywise real part would instead produce separate X and Y noise and a different stationary state. This example also explains why checking only the diagonal entries of $c$ is insufficient; positivity concerns the whole Hermitian matrix.

Mathematical validity does not establish microscopic accuracy, a bath temperature, or a measured jump record. Those require the physical derivation or preparation and monitoring model.

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.

A negative canonical rate in a traceless orthonormal basis prevents the instantaneous generator from having GKSL form. A negative coefficient in a redundant operator list need not: $2\gamma\mathcal D[Z]-\gamma\mathcal D[Z]=\gamma\mathcal D[Z]$. Even when a canonical rate is negative, maps from a fixed initial time can remain physical; check them 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.

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:

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

For a fixed nonzero dissipator, time homogeneity requires the integrated coefficient to depend only on interval length, hence $\gamma(t)$ must be constant almost everywhere. If the dissipator vanishes, its coefficient has no physical effect. 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

• 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). Publisher.
• E. B. Davies, Quantum Theory of Open Systems, Academic Press (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). DOI.
• G. Lindblad, “On the generators of quantum dynamical semigroups,” Communications in Mathematical Physics 48, 119–130 (1976). DOI.
• Á. Rivas and S. F. Huelga, Open Quantum Systems: An Introduction, Springer (2012). Author manuscript.