### Exercise 3.

A bipartite quantum state \(\rho_{AB}\) is said to be *separable* if it can be written as a convex combination of product states, i.e., if there exists an ensemble \(\{p_i, \sigma_A^{(i)} \otimes \tau_B^{(i)}\}\), with \(\sigma_A^{(i)} \in \mathcal{B}(\mathcal{H}_A)\) and \(\tau_B^{(i)}\in \mathcal{B}(\mathcal{H}_B)\), such that \[
\rho_{AB} = \sum_i p_i \sigma_A^{(i)} \otimes \tau_B^{(i)}.
\] This allows us to extend the definition of entanglement to mixed states: *a mixed state is entangled if it is not separable.*

Show that if \(\rho_{AB}\) is separable then \(I_c^{A>B} \le 0\).

What implication does this have on the conditional entropy \(S(A|B)\)?

Since any separable state can be written as a convex combination of product pure states, any separable state \(\rho_{AB}\) may be written as \(\rho_{AB} = \sum_i p_i \psi_A^{(i)} \otimes \phi_B^{(i)}\) for some pure states \(\psi_A^{(i)}\) and \(\phi_B^{(i)}\). Next, because the quantum conditional entropy is concave^{1}, the coherent information is convex. Thus \[
I_c^{A>B}(\rho) \leq \sum_i p_i I_c^{A>B} ( \psi_A^{(i)} \otimes \phi_B^{(i)}) = \sum_i p_i [ S(\phi_B^{(i)}) - S(\psi_A^{(i)} \otimes \phi_B^{(i)})].
\] But each entropy is the entropy of a pure state and thus vanishes. Thus, \(I_c^{A>B}(\rho) \leq 0\). Thus conditional entropy is therefore non-negative on separable states.

Exercise 9 of Example Sheet 3↩