For a bipartite quantum system evolving under some master equation, is the time derivative of the reduced density matrix equal to the partial trace of the time derivative of the matrix?
In other words, is the following true:
$\dot{\rho}_{A} = Tr_B(\dot{\rho}_{AB})$
(Where $\rho_A = Tr_B(\rho_{AB})$)
If not, is there some other simple method to find $\dot{\rho}_{A}$ from $\dot{\rho}_{AB}$?
By definition,
$$\rho_A = \mathrm{Tr}_B (\rho_{AB}) = \sum_B \langle B| \rho_{AB} |B \rangle$$
Differentiating both sides
$$\dot{\rho}_A = \sum_B \langle B| \dot{\rho}_{AB} |B \rangle = \mathrm{Tr}_B (\dot{\rho}_{AB})$$
because the basis are time-independent. I have checked the literature I do not know any practical case where time-dependent basis are used to compute averages; but, you ask above, in a comment, about using a time-dependent basis. Let us see what happen if you chose a time-dependent basis $| B(t) \rangle$. In this case the above equation is augmented, a priori, by a term
$$\sum_B \langle \dot{B} | \rho_{AB} |B(t) \rangle + \sum_B \langle B(t)| \rho_{AB} | \dot{B} \rangle$$
but using $|\dot{B} \rangle = (H/i\hbar) | B(t) \rangle$ and the conjugate
$$-\sum_B \langle B(t) | (H/i\hbar) \rho_{AB} |B(t) \rangle + \sum_B \langle B(t)| \rho_{AB} (H/i\hbar) | B(t) \rangle = 0 \>\>\>\>\>\>\>\> (1)$$
where, in the last step, I have used the cyclic invariance of the trace.
Therefore the equation $\dot{\rho}_A = \mathrm{Tr}_B (\dot{\rho}_{AB})$ is valid for both time-dependent and time-independent basis.
EDIT: In response to mistaken comments I am adding some extra details. The left-hand side of equation (1) can be written as
$$-\mathrm{Tr}_B \{(H/i\hbar)\rho_{AB}\} + \mathrm{Tr}_B \{\rho_{AB}(H/i\hbar)\}$$
Using now the cyclic invariance of the trace $\mathrm{Tr}_B \{XY\} = \mathrm{Tr}_B \{YX\}$ for $X = (H/i\hbar)$ and $Y = \rho_{AB}$, we obtain
$$-\mathrm{Tr}_B \{\rho_{AB} (H/i\hbar)\} + \mathrm{Tr}_B \{\rho_{AB}(H/i\hbar)\} = 0$$
explaining the zero in equation (1).
Yes, as long as $B$ doesn't depend on time.
