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Multiple entangled particles and no-communication

  1. Apr 11, 2014 #1
    Does the quantum no-communication theorem apply to multiple particle entanglement as well? To illustrate the idea, suppose we had particles A, B and C with entangled spin. We measure A and B locally using, say, detector angle 0 for particle A and angle 45 for particle B. Particle C is measured elsewhere with unknown detector angle.

    Could the correlation between measured spins of A and B somehow depend on the detector angle used for C? With only two entangled particles, it's quite easy to accept that we cannot tell anything about distant detector angle or correlation between results just by looking local measured results (which seem completely random). This is not as clear with three particles, because we have two results available locally and we might have more information contained in these.

    Does anyone know some experiment that could be applicable here, or some other way to solve this? Thanks.
     
  2. jcsd
  3. Apr 11, 2014 #2
    There's a recent paper here, which you might find useful:
    http://arxiv.org/abs/1203.6315

    An older submission to Nature here:
    http://www.nature.com/nature/journal/v403/n6769/abs/403515a0.html

    The wikipedia page here might be a good start if you don't want to dive right in:
    http://en.wikipedia.org/wiki/Greenberger–Horne–Zeilinger_state
     
    Last edited: Apr 11, 2014
  4. Apr 12, 2014 #3
    Thanks craigi, I guess multiple entanglement is so new area in research that these questions remain open for some time.
     
  5. Apr 13, 2014 #4

    Strilanc

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    If what I understand from quantum computing is representative of quantum physics, then no.

    The basic reason that there can't be communication, even with entanglement and many people present, is that the operations each side can apply commute. That is to say, if you isolate Alice and Bob then it doesn't matter if Alice runs her part of the computation first or last or even interleaved with Bob's part of the computation. So, if Bob learns in the middle of his computation that he needs Alice to apply a different operation than expected... it's already too late. The situation is equivalent to her having already finished her computation.

    This independence doesn't change when you introduce more people. The matrix multiplications will still commute. For Alice and Bob it's [itex]A \otimes B = (A \otimes I) \cdot (I \otimes B) = (I \otimes B) \cdot (A \otimes I)[/itex], where [itex]\otimes[/itex] is a tensor product and [itex]\cdot[/itex] is just matrix multiplication. For Alice, Bob, and Charlie it's the same basic thing: [itex]A \otimes B \otimes C = (A \otimes I \otimes I) \cdot (I \otimes I \otimes C) \cdot (I \otimes B \otimes I) = ...[/itex].

    As an exercise / example, try dragging some gates onto the circuit in this simple simulator (I apologize that it's not very clear). Make it as confusing as you want. Then, while watching the "Current outputs", drag a gate back and forth. The outputs don't change unless you re-order gates that depend on the same wires. That's because operations on different wires ("isolated" operations) commute.

    It's still possible to do some interesting kinds of coordination (e.g. the Bell inequalities), but communication is not possible. Bob can't ever, in the middle of the experiment, indicate to Alice that she needs to change what she was planning to do already.
     
  6. Apr 14, 2014 #5
    Good point, quantum correlation doesn't seem to be sensitive to the order of measurements. We could even bring in relativity and arrange so that it's observer-dependent which order the measurements are made, and the correlation would still be the same.

    This could be something that helps to solve this question, in some cases at least. Suppose that Alice and Bob (at the same location) can determine, using measurement data, what angle Charlie (at different location) has put his detector. Alice and Bob could check the data and send a signal (at light speed) containing this info to Charlie - but what if Charlie has not made his measurement yet? This would be impossible situation: Charlie cannot get info about his detector angle before he has made his own measurement.

    So at least in the cases where Alice and Bob make their measurements before Charlie with good margin, there cannot be such correlation that tells anything about Charlie's detector angle. The situation could be different, if Alice and Bob don't have enough time to transmit the data before Charlie makes his measurement. Need to think about this a bit more, thanks for your reply.
     
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