Lindblad Operators

A Lindblad operator is an operator that labels one dissipative, noisy, or monitored channel in a Markovian master equation. In the convention

$$\frac{d\rho}{dt} = - \frac{i}{\hbar}[H,\rho] + \sum_k \gamma_k \mathcal D[L_k]\rho, \qquad \gamma_k\ge0,$$

the dissipator is

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

Helpful background. The Lindblad–GKSL equation defines the generator; Kraus operators describe finite-time operations. A Lindblad operator instead belongs to a continuous-time generator, and is not generally a measured observable.

Operators, rates, and physical processes

Work with finite-dimensional operators first; oscillator operators below require suitable domains and finite moments. The table gives common choices, with the Hamiltonian and bath assumptions still needed to interpret energy transfer.

Operator	Common meaning
$\sigma_-=\lvert g\rangle\langle e\rvert$	Excited-to-ground decay
$\sigma_+=\lvert e\rangle\langle g\rvert$	Incoherent excitation or pumping
$Z$	Pure dephasing in the Z basis; energy preserving when $[H,Z]=0$
$a$, $a^\dagger$	Oscillator loss and gain; thermal equilibrium requires the appropriate paired rates
$n=a^\dagger a$	Number dephasing
$X,Y,Z$ with equal rates	Isotropic Pauli depolarization
$J_-=\sum_j\sigma_-^{(j)}$	Collective decay, including interference between emitters

Rates may be separated or absorbed:

$$\gamma\mathcal D[L]=\mathcal D[\sqrt\gamma L].$$

If $L$ is dimensionless, $\gamma$ has units of inverse time and $J=\sqrt\gamma L$ has units of inverse square-root time. Compare generators using the full product, not the printed operator alone.

What the Operator Does

The term $L\rho L^\dagger$ transfers weight through the action of $L$. The anticommutator term removes the corresponding weight needed to preserve trace. Both pieces are required.

For a pure input $\rho=\lvert\psi\rangle\langle\psi\rvert$, the positive part gives

$$L\rho L^\dagger = L\lvert\psi\rangle\langle\psi\rvert L^\dagger.$$

If $L\lvert\psi\rangle$ points toward a different state, the dissipator can move population. If $L$ is diagonal in an orthonormal basis, the dissipator tends to suppress coherences between different eigenvalues without moving populations.

The operator $L$ therefore identifies a preferred structure: a decay direction, a measured observable, a noise axis, a loss mode, or a collective channel.

Decay Operators

For a two-level atom or qubit with excited state $\lvert e\rangle$ and ground state $\lvert g\rangle$,

$$L=\sigma_-=\lvert g\rangle\langle e\rvert$$

gives amplitude damping:

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

The populations obey

$$\dot\rho_{ee} = -\Gamma\rho_{ee}, \qquad \dot\rho_{gg} = \Gamma\rho_{ee}.$$

The coherence decays at half the population-decay rate, apart from Hamiltonian phase rotation:

$$\dot\rho_{eg} = -\frac{\Gamma}{2}\rho_{eg}.$$

This is the generator version of the Amplitude-Damping Channel.

For rate conventions, finite-time solutions, and thermal variants, see Amplitude Damping Master Equation.

For a harmonic oscillator, $L=a$ describes loss of one quantum. At finite temperature, both loss and gain appear:

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

The $a$ term removes quanta; the $a^\dagger$ term adds quanta from a thermally occupied bath.

Dephasing Operators

A Hermitian Lindblad operator often represents phase noise or continuous unread monitoring of an observable. For qubit pure dephasing, one common convention is

$$L=\sigma_z, \qquad \frac{d\rho}{dt} = \frac{\Gamma_\phi}{2}\mathcal D[\sigma_z]\rho.$$

In the $\sigma_z$ basis,

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

the populations are unchanged and

$$\dot\rho_{01} = -\Gamma_\phi\rho_{01}.$$

More generally, suppose $L=L^\dagger$ and

$$L\lvert m\rangle = \ell_m\lvert m\rangle.$$

Then

$$\left(\mathcal D[L]\rho\right)_{mn} = - \frac12 (\ell_m-\ell_n)^2\rho_{mn}.$$

Coherences between states with different eigenvalues decay; coherences inside degenerate eigenspaces are protected from that dissipator. This is the operator-level reason dephasing selects a preferred basis.

For the corresponding generator as a named model, see Pure Dephasing Master Equation.

Microscopic Origin

In weak-coupling derivations, Lindblad operators are often system coupling operators resolved into Bohr-frequency components. Start with

$$H_I = \sum_\alpha A_\alpha\otimes B_\alpha.$$

In the energy basis of $H_S$, each $A_\alpha$ is decomposed as

$$A_\alpha(\omega) = \sum_{\epsilon'-\epsilon=\hbar\omega} \Pi(\epsilon)A_\alpha\Pi(\epsilon').$$

After the Markov and secular approximation, the dissipator has the schematic form

$$\sum_{\omega,\alpha,\beta} \Gamma_{\alpha\beta}(\omega) \left( A_\beta(\omega)\rho A_\alpha^\dagger(\omega) - \frac12 \{A_\alpha^\dagger(\omega)A_\beta(\omega),\rho\} \right).$$

The positive rate matrix $\Gamma_{\alpha\beta}(\omega)$ can be diagonalized. The resulting eigenvectors define Lindblad operators as linear combinations of the $A_\alpha(\omega)$ within a frequency block.

Thus a Lindblad operator is rarely arbitrary in a microscopic model. It reflects the system operator coupled to the bath, the bath spectrum at a transition frequency, and the approximations used to remove memory.

Jump Interpretation

When an environment is monitored in a photon-counting-like way, an operator

$$J_k=\sqrt{\gamma_k}\,L_k$$

can be interpreted as a jump operator. During a short interval $dt$, the probability of a jump of type $k$ is

$$p_k = dt\, \operatorname{Tr} \left( J_k^\dagger J_k\rho \right),$$

and, for a positive-probability resolved outcome, the conditioned state after observing that jump is

$$\rho \longmapsto \frac{ J_k\rho J_k^\dagger }{ \operatorname{Tr}(J_k^\dagger J_k\rho) }.$$

If all these channels are monitored with unit efficiency and no jump is observed, the unnormalized conditional state evolves with an effective non-Hermitian Hamiltonian

$$H_{\mathrm{eff}} = H - \frac{i\hbar}{2} \sum_k J_k^\dagger J_k,$$

followed by normalization. Unobserved channels add their own dissipators to the no-detection evolution; a missed photon is not the same as no emission. Averaging the conditioned updates recovers the unconditional Lindblad equation.

The warning is important: a Lindblad operator in an equation is not automatically an actually observed jump. The jump interpretation depends on the monitoring scheme. A different measurement of the same environment can give a diffusive trajectory rather than jumps, while the unconditional master equation remains the same.

For the operational distinction between conditioned and unconditioned updates, see Quantum Instruments.

Nonuniqueness

Lindblad operators are not unique labels of physical mechanisms.

First, rates can be absorbed:

$$\gamma\,\mathcal D[L] = \mathcal D[\sqrt{\gamma}\,L].$$

Second, absorb all rates into the operators in the following formulas. This is essential when the original rates differ. The absorbed operators can be unitary-mixed. If

$$\widetilde L_a = \sum_k u_{ak}L_k$$

with $u$ unitary, then

$$\sum_a\mathcal D[\widetilde L_a] = \sum_k\mathcal D[L_k].$$

Third, shifting an operator by a scalar multiple of the identity can be compensated by changing the Hamiltonian. For one channel,

$$L\mapsto L+\alpha I$$

changes the dissipator by

$$\mathcal D[L+\alpha I]\rho-\mathcal D[L]\rho =-\frac i\hbar[H_\alpha,\rho],\qquad H_\alpha=\frac{i\hbar}{2}(\alpha^*L-\alpha L^\dagger).$$

Thus the same generator uses $L'=L+\alpha I$ and $H'=H-H_\alpha$. Both $L$ and $\alpha$ carry the absorbed-rate units here. The sign of the compensating Hamiltonian matters.

The invariant object is the full generator, not a particular list of $L_k$ written on the page.

Dark States and Pointer States

A state $\lvert\psi\rangle$ is dark with respect to a jump operator $J$ if

$$J\lvert\psi\rangle=0.$$

For amplitude damping, $\sigma_-\lvert g\rangle=0$, so the ground state is dark with respect to emission. If the Hamiltonian and all dissipators leave a dark subspace invariant, it can become a steady or Decoherence-Free Subspace. Designing dissipators so that useful states are dark and attracting is the subject of Reservoir Engineering.

For Hermitian dephasing operators, an eigenstate with nonzero eigenvalue is not dark in this counting representation, but its projector is unchanged by that dissipator. A zero-eigenvalue eigenstate is dark as well. Superpositions of different eigenvalues lose phase coherence; mixtures within the eigenbasis are unchanged.

This connects Lindblad operators to pointer states and to the distinction between dephasing and dissipation.

Collective and independent emission: a worked comparison

For two emitters let $J=\sqrt\Gamma(\sigma_-^{(1)}+\sigma_-^{(2)})$. The antisymmetric single-excitation state

$$|D\rangle=\frac{|eg\rangle-|ge\rangle}{\sqrt2}$$

satisfies $J|D\rangle=0$: the two amplitudes to reach $|gg\rangle$ cancel. The symmetric state $|B\rangle=(|eg\rangle+|ge\rangle)/\sqrt2$ instead has emission rate $\langle B|J^\dagger J|B\rangle=2\Gamma$.

For two independent channels $J_1=\sqrt\Gamma\sigma_-^{(1)}$ and $J_2=\sqrt\Gamma\sigma_-^{(2)}$, probabilities add after distinguishing the channels. Both $|D\rangle$ and $|B\rangle$ emit at total rate $\Gamma$. Replacing the collective jump by two local jumps therefore destroys the dark state. Hamiltonian mixing, unequal couplings, or further noise can also remove its protection.

Exercises

Dephasing in an eigenbasis

Let $L=L^\dagger$ and $L\lvert m\rangle=\ell_m\lvert m\rangle$. Show that

$$\left(\mathcal D[L]\rho\right)_{mn} = - \frac12(\ell_m-\ell_n)^2\rho_{mn}.$$

Solution

The first term gives

$$(L\rho L)_{mn} = \ell_m\ell_n\rho_{mn}.$$

The anticommutator gives

$$\frac12(L^2\rho+\rho L^2)_{mn} = \frac12(\ell_m^2+\ell_n^2)\rho_{mn}.$$

Therefore

$$\left(\mathcal D[L]\rho\right)_{mn} = \left[ \ell_m\ell_n - \frac12(\ell_m^2+\ell_n^2) \right]\rho_{mn} = - \frac12(\ell_m-\ell_n)^2\rho_{mn}.$$

Amplitude damping populations

For $L=\sigma_-=\lvert g\rangle\langle e\rvert$, compute $\dot\rho_{ee}$ and $\dot\rho_{gg}$ in

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

Solution

Using $\sigma_+\sigma_-=\lvert e\rangle\langle e\rvert$ and $\sigma_-\rho\sigma_+=\rho_{ee}\lvert g\rangle\langle g\rvert$,

$$\dot\rho_{ee} = \Gamma \langle e|\mathcal D[\sigma_-]\rho|e\rangle = -\Gamma\rho_{ee},$$

and

$$\dot\rho_{gg} = \Gamma \langle g|\mathcal D[\sigma_-]\rho|g\rangle = \Gamma\rho_{ee}.$$

Population flows from $\lvert e\rangle$ to $\lvert g\rangle$.

Jump probability

Let $J=\sqrt{\Gamma}\sigma_-$. What is the probability of an emission jump in a short interval $dt$ for a state $\rho$?

Solution

The jump probability is

$$p = dt\,\operatorname{Tr}(J^\dagger J\rho) = \Gamma dt\, \operatorname{Tr}(\sigma_+\sigma_-\rho).$$

Since $\sigma_+\sigma_-=\lvert e\rangle\langle e\rvert$,

$$p = \Gamma dt\,\rho_{ee}.$$

Unitary mixing

Suppose $\widetilde L_a=\sum_k u_{ak}L_k$ with $u$ unitary. Show that $\sum_a\mathcal D[\widetilde L_a]\rho=\sum_k\mathcal D[L_k]\rho$.

Solution

For the positive term,

$$\sum_a \widetilde L_a\rho\widetilde L_a^\dagger = \sum_{a,k,l} u_{ak}u_{al}^* L_k\rho L_l^\dagger.$$

Unitarity gives $\sum_a u_{ak}u_{al}^*=\delta_{kl}$, so this becomes

$$\sum_k L_k\rho L_k^\dagger.$$

Similarly,

$$\sum_a \widetilde L_a^\dagger\widetilde L_a = \sum_k L_k^\dagger L_k.$$

The anticommutator parts are therefore also equal.

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