General clarifications about the Lindblad equation

Question

I would like to ask some clarifications about the Lindblad equation:

1. Is the system guaranteed to reach a steady state after starting from a generic initial state under both unitary evolution and dissipative terms? And under solely dissipative dynamics?

2. If the system starts in a pure state, can one end up in a mixed stationary state for the whole system?

3. If I have purely dissipative dynamics, one can in some cases identify dark states as the states annihilated by the jump operators. Are those the unique stationary states or may one find other ones?

Answer 1

1. The system is not guaranteed to reach a steady state. For instance you might have a bosonic system with jump operators corresponding to pumping energy/phonons into the system where the system is unstable and there is no limit for $\langle n(t) \rangle$. The same pathology can't happen for fermionic systems whose energy is bounded above so this may be true there, I am unsure.

2. If your system starts in a pure state, the Lindblad equation will normally immediately make it a mixed state. Since many systems will reach a stationary state, this is definitely the generic outcome.

3. Maybe I am wrong about how you define dark states but I think a dark state $\rho$ is any state annihilated by the dissipation ${\cal D}[\rho_{\rm Dark}]=0$. Since the Lindblad equation with only dissipation takes the form $$\frac{d\rho}{dt}={\cal D}[\rho]$$ and a stationary state therefore satisfies ${\cal D}[\rho_{\rm ss}]=0$, this seems to be exactly the dark state condition. If you have Hamiltonian dynamics then dark states are no longer stationary states (unless they are also annihilated by the coherent Liouville-von Neumann dynamics).