This is related to a previous question of mine.
We consider a density matrix $\sigma(t)$ operating on a Hilbert space $\mathscr{H}_{s}\otimes \mathscr{H}_b$ with Hamiltonian $H = H_s \otimes \mathbb{I}_B + \mathbb{I}_s \otimes H_b + g H_{int}$. We are interested in the reduced density matrix $\rho(t) := \mathrm{Tr}_{B}[ \sigma(t) ]$.
In the Born Approximation, the evolution of the reduced density matrix is described (in the interaction-picture) by $$ \frac{d\rho_I(t)}{dt} \simeq - g^2 \int_0^t ds\ \mathrm{Tr}_b\bigg( \big[ V(t), [V(s), \rho_I(s) \otimes \varrho_{B}] \big] \bigg) $$ where $\rho_{I}(t) = e^{+ i H_s t} \rho(t) e^{- i H_s t}$, and $V(t)$ is just $H_{int}$ in the interaction-picture, and $\varrho_B$ is the initial state of the reservoir (where $\sigma(0) = \rho(0) \otimes \varrho_B$).
Suppose for concreteness that $\mathscr{H}_s$ is finite-dimensional, and the interaction is of the form $$ V(t) = \sum_{jk} A_{j} \otimes B_{k} $$ Then the above equation of motion takes the form $$ \frac{d\rho_I(t)}{dt} \simeq g^2 \sum_{jk\ell m} \int_0^t ds\ \bigg( \big[ A_{\ell}(s) \rho_I(s), A_{j}(t) \big] \langle B_{k}(t) B_{m}(s) \rangle + \big[ A_{j}(t), \rho_I(s) A_{\ell}(s) \big] \langle B_{m}(s) B_{k}(t) \rangle \bigg) $$ where we define the correlator $\langle B_{k}(t) B_{m}(s) \rangle := \mathrm{Tr}_B\left[ B_{k}(t) B_{m}(s) \varrho_B \right]$.
My Question: Suppose $\sigma(0)$ is trace 1, Hermitian and positive. This means that $\varrho(0)$ is also trace 1, Hermitian and positive. Are these properties preserved for $\varrho(t)$ with $t>0$?
I am not sure. Here are my thoughts:
Preservation of Trace: taking the trace of the above evolution equation yields $\mathrm{Tr}[\dot{\varrho}_I(t)]=0$ because the equation contains commutators of finite-dimensional matrices. As far as I can tell this is only true because of my particular choice of $H_{int}$ which is a tensor product.
Preservation of Hermicity: I would like to show explicitly that $\frac{d}{dt}\big[ \varrho(t)^\dagger - \varrho(t) \big]=0$ for the state evolved by the above approximate equation.
Preservation of Positivity: I don't know if this is true at all, or how to show it.
In my previous post, it was pointed out that $\sigma(t)$ is trace 1, Hermitian and positive. Taking the partial trace of a positive operator yields a positive operator, which also implies that the reduced density matrix is Hermitian (since bounded positive operators are self-adjoint). I may be confused, but I think this is not an answer to my question: the evolved state $\rho(t)$ from the above approximate equation of motion does not contain the full information about $\sigma(t)$. I would like to know if these properties hold true for the $\rho(t)$ predicted by the above approximation.
