Why does the "entanglement entropy" provide a necessary and sufficient condition for bipartite entanglement?

Question

The usual argument I’ve seen is that, well, if you have some separable (unentangled) pure state $|\psi\rangle= |\psi_1\rangle \otimes |\psi_2\rangle $ then $\text{Tr}_2 (|\psi\rangle \langle \psi|) = |\psi_1\rangle \langle \psi_1 |$ and therefore the entropy of entanglement $S_E$ between subsystem $1$ and $2$ is $0$ as $$ S_E(\rho, 1|2)= S_N(\rho_1), $$ where $S_N$ is the Von Neumann entropy and $\rho_1$ is the reduced density matrix describing subsystem $1$. There is no doubt about the above.

Now, why is this a reversible statement? Why if $S_E(\rho, 1|2) \neq 0$ then subsystem $1$ is entangled with subsystem $2$.

Answer 1

Consider a bipartite system and let $\rho$ denote a pure state density operator with reduced density matrices $\rho_1$ and $\rho_2$.

OP proved that if $\rho$ is a product state, then $S_{\mathrm E} (\rho) = 0$. By modus tollens it follows that if $S_{\mathrm E}(\rho) \neq 0$ for a bipartite pure $\rho$, then $\rho$ cannot be a product state (and thus both subsystems are entangled).

However, we can prove an even stronger result:

Note that $\rho_1$ and $\rho_2$ are pure if and only if $\rho$ is a product state, which can be shown using the Schmidt decomposition. Moreover, the von Neumann entropy of a density matrix is zero if and only if it is pure.

Consequently, the entropy of entanglement of a pure bipartite state $\rho$ is zero if and only if it is a product state:

$$S_{\mathrm E} (\rho) = 0 \Longleftrightarrow \rho = \rho_1 \otimes \rho_2 $$