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If Alice and Bob share only classical communication resources such as noisy or perfect classical channels, is shared entanglement always equivalent to shared randomness?

In other words, must there be a quantum channel in order to exploit the power of shared entanglement? If yes, how can one prove this?

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    $\begingroup$ No, for example Bell-inequality violations are not possible with shared randomness only. $\endgroup$
    – Rammus
    Jan 21 at 7:32
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As @Rammus has mentioned in the comments one does not need a quantum channel to utilize entanglement as a resource. One can utilize quantum correlations to aptly perform tasks that are impossible to perform with classical correlations (i.e. shared randomness). For example in a nonlocal game, two players share entanglement (but don't need any kind of quantum channel) and they can use their entanglement to satisfy certain constraints that are impossible with only classical corrlations.

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Entanglement + a classical channels allows you to build a quantum channel using teleportation. Thus, adding a quantum channel does not give you additional power.

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  • $\begingroup$ excellent point! $\endgroup$
    – Condo
    Jan 22 at 17:52
  • $\begingroup$ In fact, perhaps this is an intuitive argument why QMIP=MIP*. $\endgroup$
    – Condo
    Jan 23 at 19:17
  • $\begingroup$ @Condo That's certainly the intuition how to convert a QMIP protocol into an MIP* protocol if everyone is honest. But the issue is how to make sure that nothing goes wrong is someone is dishonest. Teleportation only gives you a quantum channel if everyone is honest. $\endgroup$ Jan 23 at 19:46
  • $\begingroup$ yes, certainly there are finer points involved in the full equivalence. A great answer to the original question though! $\endgroup$
    – Condo
    Jan 24 at 21:17

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