Does entanglement entropy remain constant? So I am trying to figure out what the time derivative of entanglement entropy is
$$\frac{d}{dt}(-Tr[\rho^A\log_2\rho^A])$$
Where 
$$\rho^A=\sum_{aa'}|a\rangle\langle a'|\sum_bc_{ab}c_{a'b}^*$$
And 
$$|\psi\rangle = \sum_{ab} c_{ab}|a\rangle\otimes|b\rangle$$
$$H = H_A\otimes 1+1\otimes H_B$$
So I start with the TDSE to find:
$$c_{ab}(t) = c_{ab}(0)\exp(\frac{-i(E_A+E_B)t}{\hbar})$$
So now I get lost trying to find $\partial_tS$.  I know that I can rewrite the first equation as
$$-\partial_t \sum_n p_n(t)\log(p_n(t)=\sum_n\dot p_n(t)(1+\log(p_n(t))$$
Where $p_n$ are the eigenvalues of $\rho^A$.  My question is how to figure out t.hese eigenvalues to evaluate this.  I am told that this should evaluate to 0
 A: Short answer: If systems $A$ and $B$ are non-interacting, such that $H = H_A + H_B$, then 
$$
\frac{d}{dt}\left( - Tr_A\left[ \rho^A(t) \log \rho^A(t) \right] \right) = 0
$$
Here's why:
If there is no interaction, the (reduced) state of $A$ evolves under $H_A$, regardless of what happens to $B$. That is, if $\rho(0)$ is the initial state of $A$ and $B$, and $\rho(t) = e^{-i(H_A + H_B)t}\rho(0)e^{i(H_A + H_B)t}$, we have
$$
\rho^A(t) =  Tr_B \left[ \rho(t) \right] = Tr_B \left[ e^{-i(H_A + H_B)t}\rho(0)e^{i(H_A + H_B)t} \right] =\\
= e^{-iH_A t}Tr_B \left[ e^{-iH_Bt}\rho(0)e^{iH_Bt} \right] e^{iH_A t} = e^{-iH_A t}Tr_B \left[ \rho(0) \right] e^{iH_A t} = e^{-iH_A t}\rho^A(0) e^{iH_A t}
$$
and so
$$
\rho^A(t) = e^{-iH_A t}\rho^A(0) e^{iH_A t}
$$
where $\rho^A(0) = Tr_B \left[ \rho(0) \right]$. But then, it also follows that the eigenvalues of $\rho^A(t)$ never change: if $\rho^A(0) = \sum_n {|n\rangle p_n \langle n|}$, then 
$$
\rho^A(t) = \sum_n {e^{-iH_A t}|n\rangle p_n \langle n|e^{iH_A t}} = \sum_n {|n(t)\rangle p_n \langle n(t)|}
$$ 
Necessarily the entropy doesn't change either, since it is a function of the eigenvalues, but not of the eigenstates:
$$
- Tr_A\left[ \rho^A(t) \log \rho^A(t) \right] = - \sum_n {p_n \log p_n} = - Tr_A\left[ \rho^A(0) \log \rho^A(0) \right]
$$
Notice that this holds for arbitrary states of $A$ and $B$, not just for pure states $|\psi\rangle$. 
