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?


1 Answer 1


Obviously yes, because it is separable. Other than that -- this is, regarding how to find that separable decomposition -- it depends which description of the channel you are given.

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).

  • $\begingroup$ 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? $\endgroup$ Apr 24, 2019 at 0:09
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    $\begingroup$ Yes. Of course, if you want the full state afterwards (your $\sigma_{AB}$) you shouldn't trace B. $\endgroup$ Apr 24, 2019 at 18:35

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