Markovian Master Equation and Uncorrelated States

[Kreizhn]

Homework Statement

In deriving the Markovian master equation for a weakly coupled system+environment scensario, we have
\frac{\partial}{\partial t} P \rho(t) = \alpha^2 \int_{t_0}^t P L(t) L(s) P \rho(s) ds
where \alpha is the coupling strenght, P\rho = \left( \mathrm{Tr} \rho \right) \otimes \rho_B for \rho_B the initial environment state, Q\rho = (\mathcal I - P ) \rho for \mathcal I the identity superoperator, and L the Liouvillian of the system satisfying L(t)\rho = -i[H_I(t), \rho(t)] where H is the interaction Hamiltonian. My goal is to show that with the knowledge, if we have an initially uncorrelated state
\rho(t_0) = \rho_A(t_0) \otimes \rho_B(t_0)
then the state remains uncorrelated for all time t \geq 0

The Attempt at a Solution

This pre-master equation comes from the fact that we've assumed the propagator G(s,t) = \mathrm{exp}\left[ \alpha \int_s^t QL(\tau) d\tau \right] is approximated by identity (which occurs in the weak coupling limit). I need to show that in this weak coupling limit, the states do not become entangled (intuitively reasonable), though I can't think of how to show that the evolution according to the above differo-integral equation keeps the states uncorrelated.

 

[Kreizhn]

Nobody has any ideas?

 

[gabbagabbahey]

Well, this is just an idea but; you might try showing that \rho(t) = \rho_A(t) \otimes \rho_B(t) is a solution, and then perhaps see if there might be some uniqueness theorem you could appeal to that guarantees it's the only solution given the initial condition \rho(t_0) = \rho_A(t_0) \otimes \rho_B(t_0)