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Eberhard assumptions and FTL communication

  1. Dec 5, 2015 #1

    zonde

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    In this paper http://dx.doi.org/10.1103/PhysRevA.47.R747 Eberhard derives Bell type inequality from these assumptions:
    A theory is defined as being "local" if it predicts that, among these possible sequences of events [with the same number of events N], one can find four sequences (one for each setup [(α1,β1), (α2,β1), (α1,β2), (α2,β2)]) satisfying the following conditions:
    (i) The fate of photon a is independent of the value of β, i.e., is the same in an event of the sequence corresponding to setup (α1,β1) as in the event with the same event number k for (α1,β2); also same fate for a in (α2,β1) and (α2,β2); this is true for all k's for these carefully selected sequences.
    (ii) The fate of photon b is independent of the value of α, i.e., is the same in an event k of the sequences (α1,β1) and (α2,β1); also same fate for b in (α1,β2) and (α2,β2).
    (iii) Among all sets of four sequences that one has been able to find with conditions (i) and (ii) satisfied, there are some for which all averages and correlations differ from the expectation values predicted by the theory by less than, let us say, ten standard deviations.

    Now the question I am trying to answer is how we could violate condition (i) or (ii) without opening possibility of FTL communication.
    Say if we can not find two sequences (α1,β1) and (α1,β2) that are identical at Alice's end then we can communicate by making up a catalog of possible sequences that Alice can see when Bob sets his detector at β1 and another catalog of sequences that Alice can see for Bob's setting β2.
    Even if Alice sees some sequences more often for Bob's β1 and some other for Bob's β2 we still can communicate FTL.
     
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  3. Dec 5, 2015 #2

    Strilanc

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    Hide the violations in the correlations between outcomes, instead of making them apparent in the individual local outcomes, so Alice and Bob can't detect any effects until they get back together and compare notes. That will prevent them from using the violations for FTL communication, because of the note-comparing requirement.
     
  4. Dec 5, 2015 #3

    zonde

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    Yes, the question is how to do that given that hiding correlations from local outcomes seems to lead to Bell type inequality.
     
  5. Dec 5, 2015 #4

    Strilanc

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    How is that an issue? You can't use Bell inequality violations to communicate, precisely because Alice and Bob need to get back together and compare notes to detect any violations. Regardless of what Alice and Bob plan to do ahead of time, Bob's expected measurement outcomes are always 50/50 coin flips.

    If you're having trouble being convinced about that, go to this blog post. There's a few interactive widgets in it. The third one, about three quarters of the way down the page, is a "try to communicate with entanglement" game. It simulates what happens when Alice and Bob turn and measure their qubits, based on the strategy you enter.

    The goal is for Bob's move variable's value to end up equal to what the refChoice variable given to Alice started as. The table below the widget shows how well the entered strategy works. If the results statistically deviate from 50/50 within a column, you're communicating. See if what you have in mind works.
     
  6. Dec 5, 2015 #5

    zonde

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    You are missing my question, so I will restate it.

    QM predicts:
    1. Entangled particles can violate Eberhard inequality in certain measurement arrangement.
    2. Taken the same setup as in 1. for each side measurement outcomes taken separately but grouped by remote side's measurement settings are completely random within these groups.

    Provide strategy how to produce sample that satisfies 1. and 2. Intuitively there seems to be no problem with that but as strange as it seems Eberhard inequality and it's sufficient assumptions seem to suggest that this is not the case. It's hard to believe that so I am asking the question as I might be missing something.
     
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