Citing Wikipedia (https://en.wikipedia.org/wiki/Lindbladian), the Lindblad equation takes the following form: \begin{align} {\displaystyle {\dot {\rho }}=-{\frac {\mathrm {i} }{\hbar }}[H,\rho ]+\sum _{n,m=1}^{N^{2}-1}h_{n,m}\left(L_{n}\,\rho \,L_{m}^{\dagger }-{\frac {1}{2}}\left(\rho \,L_{m}^{\dagger }\,L_{n}+L_{m}^{\dagger }\,L_{n}\,\rho \right)\right)}. \end{align} Here, the $L_n$ are said to form a basis of the operators acting on the system Hilbert-space, and $H$ is the operator acting solely on the system. In wikipedia it is stated that the general form can always be brought to the "diagonal form", if we diagonalize the matrix $h_n,m$: \begin{align} {\displaystyle {\dot {\rho }}=-{i \over \hbar }[H,\rho ]+\sum _{i}^{}\gamma _{i}\left(L_{i}\rho L_{i}^{\dagger }-{\frac {1}{2}}\left\{L_{i}^{\dagger }L_{i},\rho \right\}\right)}. \end{align} Every proof I have seen for the diagonal form requires that the $L_i$ are raising and lowering operators from and to energy eigenstates: \begin{align} L_i = |a \rangle \langle b | \end{align} With $|a \rangle$ and $| b \rangle$ being eigenstates of $H$, or linear combinations of such operators with the same energy differences between state $a$ and state $b$. One example is the the often used Adam-Steck Script (https://atomoptics.uoregon.edu/~dsteck/teaching/quantum-optics/ page 144) or the Heinz-Peter Breuer Book on open quantum Systems (equation 3.208)), where such a derivation is carried out.

When one wants to describe resonance fluorescence in a 2-level system however, one usually assumes an equation like the one given in the steck script for the master equation of the system:

\begin{align} \partial_t \rho(t) = -\frac{i}{\hbar} [H_S,\rho(t)] +\gamma \{ S \rho(t)S^{\dagger} - \frac{1}{2}[S^{\dagger} S \rho(t) + \rho(t) S^{\dagger} S] \} \end{align} With $\gamma$ being a decay constant, and $S$ being the lowering operator from $|1\rangle$ to $|0\rangle$. But when describing resonance fluorescence, and the system is driven by an oscillating field, $|0\rangle$ and $|1\rangle$ are not eigenstates anymore, which means the derivation given in the steck script can't be used here. If one insisted of still using the lowering operator $S = |0\rangle \langle 1|$, then there should be additional terms. Appearing. Why isn't that the case?

  • $\begingroup$ What is $\hat{H}_S$ in your last equation (I mean, mathematically) ? Also, note that $\hat{L}_i$ are ladder operators, not projectors. $\endgroup$
    – Abezhiko
    Aug 29, 2023 at 12:03
  • $\begingroup$ @Abezhiko I edited accordingly, you are right, projection operators is the wrong term here. $H_S$ is the hamilton operator acting solely on the system - not the environment - , in the example it acts on the 2-level system. When describing resonance-fluorescence, $H_S$ contains off diagonal terms that describe the action of an oscillating field, exciting the system from the ground state to the excited state, and vice versa. $\endgroup$ Sep 5, 2023 at 22:36

1 Answer 1


The Lindblad operators can be anything in principle, the point of the Lindblad equation is to be a completely positive trace preserving map.

In physics the most usual case is that you model a thermal environment acting on a system using a Lindblad equation. In that case, under some conditions (RWA, ergoticity, etc... (see Breuer chapter 3)) and the steady state of the Lindblad equation will be a thermal state.

That's not the only relevant physical case though, because sometimes the environment is not in a thermal state. The most relevant situation here is when the environment is in a thermal coherent state, which happens if there is a laser in the electromagnetic field for example, which is the case in your reference. In that case the environment is NOT a thermal environment, and there no a priori reason to expect the steady state of your Lindblad equation to give a thermal state.

Usually, if you have a system Hamiltonian $\sigma_z$, a laser term would give a contribution which is either $\sigma_x$ or $\sigma_y$ or a combination of both, while keeping the dissipative part the same.

  • $\begingroup$ Your last paragraph is at the core of my question. Why is the dissipative part kept the same? The requirements are not fulfilled anymore. Don't be confused, I did edit the question quite a lot, because I wasn't very good in expressing my problem in the first place. $\endgroup$ Sep 5, 2023 at 22:40
  • $\begingroup$ I'm interested in answering your question, but I want to make sure I understand it first. Your question seems to have two parts: part one: In order to diagonalize the dissipator matrix $h_{n,m}$ do I need $L_i$ to be eigenoperotors of the system (the answer is definitely no); part two: how would a case where the dissipator doesn't thermalize the system be derived microscopically, for example with a laser term in the Hamiltonian. Would answering those 2 answer you question, or am I missing something? $\endgroup$
    – peep
    Sep 6, 2023 at 6:25
  • $\begingroup$ I don't exactly understand part 2: The system in question is a 2 level system coupled to the EM field via dipole interaction, for simplicity in vacuum. That means the EM field (= dissipator ?) is in a thermal state. Additionally it's driven by a laser. Described classically, the laser terms only appear in H of the system, and make the ladder operators of the system non-eigen-operators. Is that equivalent with saying that the system doesn't thermalize? $\endgroup$ Sep 6, 2023 at 8:32
  • $\begingroup$ Regarding part 1: Derivations of the diagonal form use the assumption that the $L_i$ are eigen-operators, switching from the $ L_i$ to some linear combination thereof, say $l_i = \sum_j m_{ij} L_j$, will make the lindblad equation non- diagonal, no? Such a non-diagonal form hen should also govern the mentioned 2-level system. $\endgroup$ Sep 6, 2023 at 8:52

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