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

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 $$2$$.

• This is a question of logic: if A always implies B then the falseness of B always implies the falseness of A. Aug 9, 2021 at 19:38

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