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

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.

• In Le Bellacs Statistical Physics some of this is discussed if I remember correctly fulviofrisone.com/attachments/article/413/…
– user224659
Oct 17, 2020 at 10:07
• @TheoreticalMinimum Thank you for recommending a nice textbook!
– asdf
Oct 17, 2020 at 23:14
• I found it nice at first as well. But after reading through it I was also disappointed at a few places. It feels a bit unfinished from times to times. But it was better than the other undergraduate texts I read on statistical physics and helped me with my oral exam a lot.
– user224659
Oct 17, 2020 at 23:42

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 $$\vec S_1\cdot\vec S_2$$.