# Entanglement breaking quantum channels

An entanglement breaking quantum channel is defined as one where $$\sigma_{AB}=(\Phi_A\otimes I_B)(\rho_{AB})$$ is separable, even for entangled inputs $$\rho_{AB}$$. Of course, if the input $$\rho_{AB}$$ is already separable, then we have $$\rho_{AB} = \sum_k \lambda_k \rho_A^k\otimes \rho_B^k$$. Then,

$$\sigma_{AB} = (\Phi\otimes I)\rho_{AB} = \sum_k\Phi(\rho^k_A)\otimes \rho^k_B$$

One can see that $$\sigma_{AB}$$ is indeed separable.

My question is: If $$\rho_{AB}$$ is entangled and given that $$\Phi$$ is entanglement breaking, can one write down the output state in a manifestly separable form?

An canonical form for entanglement-breaking channels is the "measure and prepare" form $$\Phi(\rho) = \sum_k \sigma_k \mathrm{tr}(\rho P_k)\ ,$$ with $$\sigma_k$$ density matrices, and $$\{P_k\}_k$$ a POVM. In that case, it is indeed straightforward to write down such a separable decomposition (which should be obvious to find, so I won't explicitly spell it out).
• Thanks for answering. To check if I have understood correctly - the POVM is a local measurment in the Hilbert space $A$ only, correct? So Alice and Bob hold an entangled pair and if Alice wants to apply an entanglement breaking channel, she performs a measurement on her half of the state. She then outputs a state $\sigma_k$ with probability $tr(\rho_A P_k)$, where $\rho_A = tr_B(\rho_{AB})$. Is this correct? Apr 24 '19 at 0:09
• Yes. Of course, if you want the full state afterwards (your $\sigma_{AB}$) you shouldn't trace B. Apr 24 '19 at 18:35