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EPR paradox and time of collapse

  1. Dec 28, 2015 #1
    According to the EPR-paradox, if we have a pair of two entangled spin-1/2 fermions A and B and measure z-component of A, B collapses immediately as well(i'm using these letters for both particles and their observers). The 'canonical' solution is then to state that it is not possible to transfer information by measuring a particle and hence this does not violate the axioms of special relativity. I have two problems/questions with this.

    1)I am not really convinced that measuring A won't transfer any information to B: it transfers the time of measuring. In a (e.g. double-slit-like) interference experiment, it would be possible to test whether B has already been collapsed at time t' or not. And with a sufficient number of entangled pairs, A could then even send any message to B by using a language such as Morse code.

    2)Quantum mechanics has been succesfully unified with special relativity by now into relativistic quantum field theories. Now, one of the main aspects of special relativity is the absence of an absolute notion of simultanity. So if A and B collapse together, in which reference frame is that when A and B have a nonzero relative velocity? Maybe the time-coordinate in the FLRW metric of spacetime?

    I'm pretty sure I'm not the first one to which these questions occur. I would expect there can be a rigorous argument why my reasoning in 1) is wrong, I don't think 2) is within reach of experiment, but maybe there are theoretical arguments why one of the frames is the important one?
     
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  3. Dec 28, 2015 #2

    DrChinese

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    1) This is incorrect. There is no experiment which can determine the time entanglement ceased.

    2) The "time of collapse" (if there even is such a thing) as mentioned cannot be determined, and there is no difference in the observable outcome based on reference frames.
     
  4. Dec 28, 2015 #3
    Thanks for your answer, but can you elaborate a bit why this is true?

    For example, if a B-particle moves trough a magnetic field in z-direction and there are as well two horizontal slits at different z-values and the particles are unobserved, B will go trough both of the slits such that enough B's create an interference pattern on a screen behind the slits. When an observation of A already made B collapse to a specific z-component of the spin, all interference is lost for B, it would seem to me? I know this would destroy the particles at the end, but that doesn't seem to be a problem to me.
     
    Last edited: Dec 28, 2015
  5. Dec 28, 2015 #4

    zonde

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    As far as QM is valid there is no experiment that gives different results based on time of measurement on the other side. That conclusion comes from no-communication theorem. It explores exactly that question from perspective of QM formalism.
     
  6. Dec 28, 2015 #5

    DrChinese

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    Good thinking! And fairly easy to test this.

    However, entangled particles do not behave this way. You can see this in an enlightening article by Anton Zeilinger, p. 290, Figure 2.

    Experiment and the foundations of quantum physics
     
  7. Dec 28, 2015 #6
    thanks, very interesting!
     
  8. Dec 31, 2015 #7

    jfizzix

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    Another way of looking at it:
    If all you're doing is looking at a single particle,,,
    then there is nothing you can do to tell if that particle is half of an entangled pair or not.

    In order to see the entanglement, you would have to get extra outside information, which would have to be transmitted to you.
     
  9. Jan 3, 2016 #8
    I agree on that one, but if A and B both have 10 000 particles entangled with one-to-one correspondance, B can perform a double slit-experiment on his particles and according to DrChinese's link, interference for B will be lost because of the in-principle possibility for A to perform measurements on her particles, regardless of whether he actually does or not.

    Seems to confirm that collapse of A cannot be detected by B. But it raises a new question(on which I don't see an answer in the text right away): If A destroys her particles, according to the text, B observes observes interference again. Doesn't this 'inverse-collapse' instantly transmit information from A to B so that B knows what A did?
     
  10. Jan 3, 2016 #9

    DrChinese

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    Yours is very good thinking on this! This is a somewhat complicated issue. jfizzix is absolutely correct, but then so is my reference. So how to resolve?

    The short version: Entangled photons are not coherent. That is also why they do not self interfere in a double slit apparatus (pardon me if this is not the best description of this, someone may be able to say it better than I). If you make them coherent so they self interfere - say by diffracting through a pinhole - they are no longer entangled (because you measured position in the process of creating coherence) and you are back where you started.
     
  11. Jan 4, 2016 #10
    Thanks for answering but it still sounds a bit vague to me. Maybe coupling of spin and slit(position) by a magnetic field as I described before would spoil the proper symmetrization of the many-particle wavefunction where e.g. if A and B each have one particle the spatial part is symmetric and the spin-part antisymmetric?
     
  12. Jan 4, 2016 #11

    zonde

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  13. Jan 12, 2016 #12

    Don't you think it is likely that every particle is entangled with some other (usually unknown) particle(s)?
     
  14. Jan 14, 2016 #13

    jfizzix

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    I think it is likely that most particles share a minute amount of entanglement with one another, and it may well be possible that every particle in the universe is entangled with every other particle in the universe, however infinitesimal this connection may be.

    However...

    The only things we can make a decent claim to know are what we can measure, and there is simply no way in the current understanding of physics to tell whether a single particle is entangled with something else just by looking at that single particle.

    If you're interested in a more technical explanation, I believe this is known as the no-signaling theorem.
     
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