I am following along Breuer and Petruccione's book . I would like to know if the property $\rho^{\dagger} = \rho$ is preserved for evolution that is described by the Born Approximation.

For a Hilbert space $\mathscr{H}_s \otimes \mathscr{H}_{b}$ describing a system of interest and some reservoir/bath, we consider a time-independent Hamiltonian of the form $$ H = H_s \otimes \mathbb{I}_b + \mathbb{I}_s \otimes H_b + g H_{int} $$ where $g$ is some small coupling and $H_{int}$ describes some interaction between the system and the bath.

The full density matrix of the combined system and bath $\sigma(t)$ evolves via the von Neumann equation $$ \frac{d\sigma_I(t)}{dt} = - i [ V(t) , \sigma_{I}(t) ] $$ which is written in the interaction-picture, where $$ \sigma_{I}(t) := e^{+ i H_s t} \otimes e^{+ i H_b t} \sigma(t) e^{- i H_s t} \otimes e^{- i H_b t} \ \ \ \ \text{and} \ \ \ \ V(t) := e^{+ i H_s t} \otimes e^{+ i H_b t} H_{int} e^{- i H_s t} \otimes e^{- i H_b t} $$

If we define the reduced density matrix describing the system as the following partial trace $$ \rho(t) := \mathrm{Tr}_{b}[ \sigma(t) ] $$ and then $\rho_{I}(t) := e^{+ i H_s t} \rho(t) e^{- i H_s t}$, the Born Approximation says that $$ \frac{d\rho_I(t)}{dt} \simeq - g^2 \int_0^t ds\ \mathrm{Tr}_b\bigg( \big[ V(t), [V(s), \rho_I(s) \otimes \varrho_{B}] \big] \bigg) $$ where $\varrho_b$ is the initial state of the bath at $t=0$, where $\sigma(0) = \rho(0) \otimes \varrho_b$.

My Question: Suppose that $\rho(0)^{\dagger}= \rho(0)$ so that the initial density matrix is Hermitian. How can you use the above equation to show that $\rho(t)^{\dagger}= \rho(t)$ for $t>0$?

Is this possible? In the literature, I sometimes come across some statements that Lindblad equations preserve Hermicity (Lindblad equations being the above equation, after taking the Markov Approximation and then the secular/rotating-wave approximation).

Is it possible to prove this from this equation of motion? Or do we need additional assumptions? Or can this be more generally proven for $\rho$ without specifying to a particular evolution equation?

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    $\begingroup$ Positivity is preserved when taking the partial trace. A (bounded) positive operator is selfadjoint. $\endgroup$ – Valter Moretti May 31 at 5:45
  • $\begingroup$ So if $\sigma(t)$ is positive for all times $t>0$ then $\rho(t)$ is positive and the statement is trivial (so long as $\rho(t)$ is bounded)? Would this not mean you need to prove that $\rho(t)$ is bounded for $t>0$? $\endgroup$ – Greg.Paul May 31 at 15:35
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    $\begingroup$ By definition a density operator is bounded it being trace-class. $\endgroup$ – Valter Moretti May 31 at 16:09
  • $\begingroup$ If we had access to the full state $\sigma(t)$, and we took the partial trace of it then I now agree that it would have to be positive and therefore also self-adjoint. I think that my question is different than this though - for an operator $\rho(t)$ obeying the evolution equation above, and property $\rho(0)^\dagger=\rho(0)$, is the evolved state also guaranteed to be Hermitian? Unless I am missing something, I think this is a different question because the above equation is a perturbation in $H_{int}$ of the full von Neumann equation - and is therefore an approximation $\endgroup$ – Greg.Paul May 31 at 20:18
  • $\begingroup$ I think that my idea works anyway because the interaction evolutor is unitary so that it preserves boundedness and positivity. $\endgroup$ – Valter Moretti May 31 at 20:56

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