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B Limits of no-communication theorem

  1. Feb 12, 2016 #1

    zonde

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    I would like to post a comment for offtopic conversation in another thread.
    This is the point of no-communication theorem that measuring one particle does not change anything measurable about the other particle.
    But conclusions of no-communication theorem are limited by it's assumptions (as for any no-go theorem). To illustrate this let me describe one example.
    If you colocate Alice and Bob then by measuring both photons and comparing result you can see if entanglement is present or not. Now depending on physical model of entanglement it might be possible to break entanglement from distance and see it in local results of colocated Alice and Bob. Say if there is no direct photon-photon entanglement but instead photon-photon entanglement is realized by combination of electron-photon entanglements. Then the electron (at the source) can be distant from Alice/Bob pair. So in such a model if there is a way how to break electron-photon entanglement by having control of electron (and optical paths connecting electron and photons) it would open possibility for FTL communication.
     
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  3. Feb 12, 2016 #2

    Nugatory

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    You will have to be more specific about the communication protocol. How is the initial state prepared, which parts of that initial state are accessible to the various participants, which measurements do they perform, what calculations do they perform on these results to recover the transmitted information? Otherwise, there's nothing concrete to discuss.

    But I'm skeptical, because introducing more particles with more complicated entanglement relationships still leaves us with an initial state preparation on a Hilbert space a portion of which is accessible to colocated Alice and Bob and a portion of which is accessible to the remote owner of the electron... And that allows us to apply the no-signalling theorem.

    (Of course if the state preparation has been done by the owner of the electron, the no-signalling theorem isn't applicable and signalling is trivially possible. But in that case the state preparation is in the past light cone of the whole shebang - which is why signalling is trivially possible).
     
  4. Feb 12, 2016 #3

    Strilanc

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    Do you really think that nobody thought to check the no-communication theorem against a system with three qubits? Which assumption of the theorem do you think you're violating?

    When you run whatever test for entanglement you have in mind, on just the two photons' polarization, the test will return "no entanglement". Even if the electron hasn't been measured and the photons are still entangled into the GHZ state with the electron's spin.

    The issue is that an uninvolved entangled qubit acts like a de-facto measurement. When you consider just the state of the two photons' polarizations, i.e. when you marginalize over the electron's spin, you'll find that the marginal state is a mixed state instead of an entangled state. And since the electron isn't involved in your local testing, that marginalized classical-looking state predicts all of your observations. You need to actively involve the electron in the testing-for-entanglement if you want to see anything non-classical.
     
  5. Feb 12, 2016 #4

    zonde

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    I am not speaking about GHZ state. Of course GHZ state is covered by no-communication theorem.
    The point is different. I am trying to understand what in two photon entanglement makes exactly these two photons entangled and maintain their entanglement. Obviously there are plenty of photons around and they have noting to do with entanglement of exactly these two photons we are measuring. One thing that makes them special is the whole experimental setup i.e. source, optical paths, analyzers, detectors.

    No-communication theorem assumes that particles remain entangled i.e. entanglement visibility is not considered as measurable parameter.
     
  6. Feb 12, 2016 #5

    Strilanc

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    The no-communication theorem doesn't assume that. The theorem would be useless if it did, since we have (noisy) measurements that tell us if particles are entangled (i.e. Bell tests).

    Then what kind of entangled state are you talking about? Write down the kets and the coefficients.
     
  7. Feb 12, 2016 #6

    zonde

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    But emphasis was on measurement of entanglement visibility. Any loss of entanglement visibility is supposed to be due to imperfections of setup that result in loss of coherence between different polarization modes and individual members of ensemble i.e. local effects.

    Symmetrization of single photon states in single Fock state is always there. We do not consider two photon configurations that are not symmetrized as they are not interesting even so there are such (named distinguishable photons). And the rules how to get symmetrized states are partially know-how of experimenters.
    On the other hand the take away from Bell inequality violation for me is that under local rotation of photon polarization global symmetrization of photon states ensures that photons remain entangled. So this is the key part why entangled state can't be explained by local models.
    So if there would appear a physical model how this symmetrization actually works (with some middle party involved) it might open up a loophole in no-communication theorem by remote reduction of entanglement visibility.

    Usual photon polarization entangled state, say ##\frac{\sqrt{2}}{2}(|HH\rangle+|VV\rangle##
     
  8. Feb 12, 2016 #7

    Strilanc

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    Where's the electron's spin? That's the state of the two photons before you entangle their polarizations with the electron's spin, but I want the state of the full three-qubit system just before the third party does their thing.

    Also you should specify the observable for the measurement you plan to perform on the two photons' polarizations.
     
  9. Feb 12, 2016 #8

    zonde

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    I already said that I am not talking about GHZ state, didn't I?
    I am talking about hypothetical physical model of photon symmetrization. There is none presently. QFT takes symmetrization as granted (except the parts that are derived in spin-statistics theorem).
     
  10. Feb 12, 2016 #9

    Strilanc

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    It's very strange to me that you can't formulate your full state in terms of a superposition of kets (or a density matrix).

    Since I don't know any QFT, and your point apparently depends on QFT, I won't be able to help you.
     
  11. Feb 12, 2016 #10

    Nugatory

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    Closed. We can reopen if OP can provide a proper description of the setup he is considering.
     
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