Does $S(\rho_{AB})=S(\rho_A)+S(\rho_B)$ imply that $\rho_{AB}=\rho_A\otimes\rho_B$? Let $\rho_{AB}$ $\in$ $\mathbb{H_A} \otimes \mathbb{H_B}$ be a density operator with reduced density operators $\rho_A = tr_B[\rho_{AB}]$ and $\rho_B = tr_A[\rho_{AB}]$. I assume that the von Neumann Entropy of $\rho_A$ is the same as the von Neumann Entropy of $\rho_B$, $S(\rho_A) = S(\rho_B)$. Now, if I have an expression like this $$S(\rho_{AB})=S(\rho_A)+S(\rho_B)$$
does this mean that $\rho_{AB}$ has to be the tensor product of its marginals, so $\rho_{AB}=\rho_A \otimes \rho_B$?
The reason I think so is, since I have the same amount of information on both sides of the expression written above, and the mutual information between subsystems $A$ and $B$, $I(A:B)=S(\rho_{AB}||\rho_A \otimes \rho_B)=tr[\rho_{AB}(\log (\rho_{AB})-\log(\rho_a \otimes\rho_B))]$, is zero iff $\rho_{AB}=\rho_A \otimes \rho_B$.
 A: Yes. In more detail, for any density operator $\rho_{AB}$, one has that 
$$
S(\rho_{AB}) \leq S(\rho_A)+ S(\rho_B)
$$
with equality if and only if $\rho_{AB} = \rho_A\otimes\rho_B$. This follows from the usual proof for subadditivity and applying Klein's inequality, which states that 
$$
S(\rho\Vert\sigma)\geq0
$$
for all density operators $\rho$ and $\sigma$, with equality if and only if $\rho=\sigma$ and where $S(\rho\Vert\sigma)$ is the relative entropy defined as
$$
S(\rho\Vert\sigma) = \mathrm{Tr}(\rho\log\rho)-\mathrm{Tr}(\rho\log\sigma).
$$
Now, setting $\rho=\rho_{AB}$ and $\sigma=\rho_A\otimes\rho_B$, we have
\begin{align*}
S(\rho_{AB}) =-\mathrm{Tr}(\rho\log\rho) 
& \leq -\mathrm{Tr}(\rho\log\sigma)\\
&=-\mathrm{Tr}\bigl(\rho_{AB}\bigl((\log\rho_A)\otimes I_B+I_A\otimes(\log\rho_B)\bigr)\bigr)\\
&=-\mathrm{Tr}(\rho_{A}\log\rho_A)-\mathrm{Tr}(\rho_B\log\rho_B))\\
&=S(\rho_A)+S(\rho_B),
\end{align*}
where the inequality in the first line follows from Klein's inequality, and equality holds if and only if $\rho=\sigma$.
