How to chose the collapse operators for an open quantum system?

Question

Lets assume we have an open quantum system. In the Born-Markov-Approximation, the dynamics of the density operator is described by a Lindblad-Type equation \begin{equation} \dot{\rho} = - i[H, \rho] + \sum_k \left(A_k\rho A_k^\dagger - \frac{1}{2}\{A_k^\dagger A_k, \rho \}\right). \end{equation} When describing photon losses in a cavity, the collapse operator will be the photon annihilation operator $a$. I know one derivation of this is by the Bloch-Redfield-Equation. Here, the collapse operators $A(\omega)$ are the eigenoperators of the Hamitonian, which satisfy \begin{equation} [H, A(\omega)] = -\omega A(\omega), \end{equation} weighted by the spectral function. Since $a$ satisfies the relation for the Hamiltonian $H = \omega a^\dagger a$, its an eigenoperator and thus one of the collapse operators.

Now, if we consider a more complex system, for example multiple cavities coupled with each other and maybe also some atoms, we get more complex eigenstates of the system and $a$ will no longer be a eigenoperator. Why is the photon decay still modeled by chosing $a$ as the collapse operator and not the correct eigenoperators of the Hamiltonian? Is this just an approximation? If yes, why is nobody talking about it?

Answer 1

Why is the photon decay still modeled by chosing a as the collapse operator and not the correct eigenoperators of the Hamiltonian? Is this just an approximation? If yes, why is nobody talking about it?

This is a major concern in the field of open quantum systems, not only is it widely talked about, it even have a name. This approximation is known as a "local master equation", while not using this approximation is known as "global master equation".

Unfortunately it's not straightforward to answer if this approximation is good or not, and they have problems in different situations. The local master equation does not predict a thermal steady state, and predicts a completely nonsensical steady state in some situations. For example, if you have 2 coupled harmonic oscillators (with a RWA interaction), and a bath coupled only to the first one , the steady state will be such that both harmonic oscillators have the same population. Of course if they have very different frequencies this is nowhere near the thermal state. In the case where the frequencies are similar and the coupling is weak, then the steady state will be close to the thermal one.

I can give some good references to start your research. This paper shows some advantages of the global approach, this paper shows that the local master equation can be thermodynamically consistent (which was one of the major concerns). Exact master equations will in general not be local (and usually not even Markovian). This is a big rabbit-hole in the field, therefore there will be many references with different points of view. Practically it's very hard to get the global master equation since you have to diagonalize your Hamiltonian, therefore the local one is often used out of convenience.

Answer 2

There are two reasons.

It is often the case that there is a clear energy-scale separation between the "bare" frequency of the modes and the other energies in the Hamiltonian, and so the Lindblad equation is derived considering just the "bare" modes and the rest of the terms are added at the end. For example, for the Jaynes-Cummings Hamiltonian $$ H = \omega (a^\dagger a + \frac{1}{2} \sigma^z) + g(a^\dagger \sigma^- + a \sigma^+), $$ with $\omega>> g$, the operators $a$, $a^\dagger$, $\sigma^-$ and $\sigma^+$ will well approximate the real eigenoperators. It is true that this is very often overlooked and rarely mentioned.

The second reason is that in many-body problems it might be hard or impossible to diagonalize the Hamiltonian to otbain the eigenoperators. [Recall that the eigenoperators $A(\omega)$ are transition elements between eigenstates of the Hamiltonian, $|\epsilon_n\rangle \langle \epsilon_m|$, where $\omega = \epsilon_m - \epsilon_n$.]