# Exercise 3

## from Example Sheet 4

### 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 concave1, 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.

1. Exercise 9 of Example Sheet 3