# Why in 2-level system can a diagonal Lindblad equation be used to describe decay, when $|0\rangle$ and $|1\rangle$ aren't energy eigenstates anymore?

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?

• What is $\hat{H}_S$ in your last equation (I mean, mathematically) ? Also, note that $\hat{L}_i$ are ladder operators, not projectors. Aug 29, 2023 at 12:03
• @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. Sep 5, 2023 at 22:36

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.

• 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. Sep 5, 2023 at 22:40
• 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?
– peep
Sep 6, 2023 at 6:25
• 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? Sep 6, 2023 at 8:32
• 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. Sep 6, 2023 at 8:52