Is the Von Neumann entropy of two initially non-interacting systems always increasing?

Question

Let $A$ and $B$ be two systems that does not interact initially ($t=0$), i.e., the density matrix of the initial total system is given by $\rho(0) = \rho_A (0) \otimes \rho_B (0)$. Suppose that interaction between the two systems is turned on after $t=0$. Then the density matrices of each system can be obtained by partial tracing: $\rho_A(t) = \text{Tr}_B \rho(t)$ and $\rho_B(t) = \text{Tr}_A \rho(t)$. I want to show that the von Neumann entropy $S(\rho_A(t)) + S(\rho_B(t))$ increases with $t$ (though I'm not sure if this is true). I tried to calculate the time derivative of $S(\rho_A(t)) + S(\rho_B(t))$ directly by using $i\hbar\dot{\rho(t)} = [H, \rho(t)]$, but it resulted in a seemingly useless messy equation. Can anyone tell me whether the statement is correct? If it is, how can I approach to prove it? I appreciate any help.

Answer 1

This is not true. The interaction $H$ (unless it has irrational eigenvalues) will have a recurrence time $T$ at which $e^{-iHT/\hbar}=\mathrm{Id}$.

At that time, the entropy will be equal to the one at $t=0$. Thus, it cannot increase all the time - unless it stays constant (which it will generally not do).

As an example, consider an initially unentangled state $\lvert\uparrow\rangle\otimes\lvert\downarrow\rangle$ and the Heisenberg interaction $\vec S_1\cdot\vec S_2$.