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Suppose Alice shares $m$ pairs of maximally entangled states with Bob and $n$ pairs of maximally entangled states with Charlie. It is clear that by measuring their states, Alice can generate correlations with Bob and Alice can also generate correlations with Charlie. If Alice does entanglement swapping, she can also make Bob and Charlie share correlations.

Is there a way for Alice, Bob and Charlie to jointly share tripartite correlations without communication? The goal is for them to all have outcome "0" or all have outcome "1".

EDIT: Just want to emphasize that communication is forbidden. Otherwise, the problem is trivial since the classical outcome "0" or "1" can be copied and redistributed.

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  • $\begingroup$ To be clear, so the question is essentially whether, given a pair of (maximally) entangled states $\psi_{AC_1}$ and $\psi_{BC_2}$, it is possible to act locally on $C_1 C_2$ and obtain tripartite entanglement of $ABC$ (and the analogous situation with more than two pairs of qubits)? $\endgroup$
    – glS
    Apr 10, 2020 at 15:59
  • $\begingroup$ @glS yes almost. I only care about tripartite classical correlations in the end. It doesn't have to be tripartite entanglement. $\endgroup$ Apr 10, 2020 at 16:00

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No, it is not possible without communication.

To see why, consider B and C, and just ignore A -- since they cannot communicate, for anything B and C can do A's presence is irrelevant.

Then, B and C's measurement outcomes are completely uncorrelated (since they don't share any entanglement, or correlations, just each of them holds a maximally mixed state). Thus, they will see no correlations whatsoever, neither classical or quantum.


More technically, you can just write the reduced density operator of B+C jointly, which is the maximally mixed state (obtained by tracing A), which gives completely uncorrelated (and random) measurement outcomes.


Finally, note that with communication, A+B+C can prepare any tripartite state they want by using teleportation (A prepares the whole state and teleports B's and C's share to them.)

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