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As the name suggests, an entanglement breaking channel $\Phi$ is such that $(Id \otimes \Phi)[\rho]$ is always separable, even when $\rho$ is entangled. Won't such channels be useless, as they destroy entanglement? Does it make sense to say a channel that is more "entanglement breaking" is more useless?

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    $\begingroup$ Whether something is "useless" or not depends on your objective. If your goal is to prepare a separable state, then an entanglement breaking channel is useful. Moreover, entanglement breaking channel models a large class of processes wherein we measure the input state and based on the measurement outcome prepare an output state. This encompasses tasks that are part of many experiments performed in practice. $\endgroup$ Aug 25 at 18:29
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Even if such a channel is useless, it's still worthwhile to define them for things like quantum communication, to see whether or not you can use your channel to transmit quantum information. So the channels are "useful" in the sense of recognizing what type of channel you have (mathematical formalism should always make it easier to recognize when you do or do not have entanglement breaking). The usefulness of the concept is that you can find when your channels are not entanglement breaking and thus use them to transmit quantum information.

It does not seem to make sense to me to define a channel as more or less entanglement breaking, because entanglement breaking is a binary statement. You could talk about the reduction in entanglement a generic state undergoes because of a certain channel, but then you'll have to find some notion of a "generic state" because not all states will have their entanglement behave in the same way for arbitrary channels.

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