Quantum relaxation to equilibrium?

Question

Source and context: Im reading “The Theory of Quantum Open Systems” by Breuer and Petruccione. As an application of the just derived Lindblad equation for the dynamics of the reduced density matrix $rho_S(t)$ of subsystem $S$:
$$
frac{d}{dt} rho_S(t) = -i [H_S, rho_S(t)] +mathcal{D}(rho_S(t)), tag{1}
$$
the authors “show” that for any initial state $rho_S(0)$ in contact with an stationary heat bath $rho_B = rho_{th} equiv exp (-beta H_b)/mathcal{Z}$
$$
rho_S(t) longrightarrow rho_{th} quad text{as} quad t rightarrow infty,
$$
as one naturally expects. They “show” this by proving that $rho_{th}$ is stationary, i.e. $frac{d}{dt}rho_{th}=0$ (I can follow that proof okay).

Questions:

Why do they just prove that $rho_{th}$ is stationary? (I don’t find this very impressive as $rho_{th}$ doesn’t depend on time by definition). In fact, I would have thought that according to what they showed, $rho_{th}$ is a stationary solution of subsystem $S$ as I expected $frac{d}{dt}rho_{th}=0$ to be true by definition and hence $rho_{th}$ to satisfy (1).

Wouldn’t one want to explicitly show that no matter what $rho_S(0)$ was, in the limit $trightarrow infty$ then $rho_Srightarrow rho_{th}$?

Goffredo_Gretzky · Accepted Answer

Uhm I don't exactly know which page of Breuer&Petruccione you are referring to, but maybe the following remarks can help:

In the context of GKLS (Lindblad) master equations, a stationary state is defined as an eigenvector of the Liouvillian $mathcal{L}$ with zero eigenvalue. The Liouvillian is defined as the generator of the dynamical semigroup (Eq.(3.47) of my edition of Breuer&Petruccione): $rho(t)=expmathcal{L}t[rho(0)]$. Then, given a generic (time-independent) density matrix $rho$, we say that it is a stationary state if $mathcal{L}[rho]=0$. The meaning of this can be easily deduced from the definition of Liouvillian.
The characterization of stationary states for GKLS master equation is a difficult task, with still a lot of ongoing research. There are some remarkable results (not really discussed in Breuer&Petruccione), such as:

If the Hilbert space of the system is finite, there is always at least one stationary state of the dynamics [1,2].
If the stationary state is unique (say $rho_{th}$), then the semigroup is relaxing, i.e. $rho_Srightarrow rho_{th}$ for $trightarrow infty$ and for any $rho_S$ in the state space of the system (the condition in your second point) [1].
There are some sufficient and necessary conditions about the uniqueness of stationary states. See the nice review below [3].

Under certain assumptions on the microscopic model (not always satisfied in physical systems), it can be shown that there is a unique stationary state of the Markovian master equation derived from this model, which is the Gibbs state (section 3.3.2 of your textbook).

Further references:
[1] Rivas and Huelga. "Open Quantum Systems. An Introduction.", Springer Berlin (2012)
[2] Baumgartner and Narnhofer. "Analysis of quantum semigroups with GKS–Lindblad generators: II. General." Journal of Physics A: Mathematical and Theoretical 41, 395303 (2008).
[3] Nigro. "On the uniqueness of the steady-state solution of the Lindblad–Gorini–Kossakowski–Sudarshan equation." Journal of Statistical Mechanics: Theory and Experiment 2019, 043202 (2019).

Quantum relaxation to equilibrium?

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