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I A computational model of Bell correlations

  1. Nov 7, 2017 #1

    PAllen

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    I am no expert on the nuances of assumptions of various formulations of Bell theorem(s), but wonder if the following model is adequate to explain the correlations without any non-local features. If this is a known, flawed, approach, a pointer to its refutation (or an explanation) would be appreciated.

    Consider a model that any state vector implicitly includes a 'true random' seed. In Copenhagen type interpretations, when a measurement is made, this seed feeds a universal pseudo-random number generator to produce the result. Then, entangled particles simply have the feature that each one's state vector share the same 'true random seed' that is 'generated' at preparation time.

    If this works, the only causal relation is between each measurement and the preparation, and there is no influence at all between spacelike separated measurements.
     
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  3. Nov 7, 2017 #2

    PeterDonis

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    I would recommend trying to explicitly derive the correlation function between measurement results for two entangled particles as a function of the measurement settings. I think you will find that it will factorize (i.e., it will be a product of two functions, one that depends only on the settings at measurement #1, and the other that depends only on the settings at measurement #2--both will also depend on the shared seed, of course). If this is the case, then your model cannot produce correlations that violate the Bell inequalities--that's the theorem that Bell proved.
     
  4. Nov 7, 2017 #3

    DrChinese

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    Adding to PeterDonis' comment:

    There are 2 main variations to the "random" seed concept, and both fail (as you must certainly expect).

    a. There is a polarization at some random angle that both express. This fails because the resulting correlation function is separable and will NOT produce "perfect" correlations at all angles. Only entangled particles produce those correlations.

    b. There is a random (and possibly different) polarization for all possible angles. This produces perfect correlations at all angles. However: this fails because the resulting correlation function cannot match the predictions of QM at most other angles. You can hand select the results and it still won't work UNLESS you know the angles you plan to test in advance (which defeats the purpose of this exercise, as the observers Alice and Bob choose their angles freely). This is what we learned from Bell.

    It's easy to gloss over these points when you are saying the "state vector implicitly includes a 'true random' seed". So you really have to lay out some examples for yourself so you can see what is going on with the angles. I suggest working it out with 0, 120 and 240 degrees and imagine you have an entangled pair of photons that match at the same angles (perfect correlations). If you can't do that condition, you can't go to step b. And in this simple situation, when the Alice/Bob angles are different, the experimental match rate will be 25% (per QM). However, for any example where you pick values for all 3 of those angles in advance, you can't get it less than 33%.
     
    Last edited: Nov 7, 2017
  5. Nov 8, 2017 #4

    zonde

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    If you can make two copies of the seed you can make three copies of the seed just as well. Now measure one seed at zero angle, second seed at small angle to the "left" and the third seed small angle to the "right". If correlation between first and second result produce predicted value and correlation between first and third produce predicted value there is no way how correlation between second and third can produce predicted result.
     
  6. Nov 8, 2017 #5
    Nice thread, but I don't see how the OP's suggestion might further our understanding of light, or of Bell's Theorem.

    My reading of Bell is that he showed that hidden variable models (hvm) can't reproduce the predictions of quantum theory, and that even though it can be done if the two wings of the experiment are allowed to communicate ftl, Bell's Theorem doesn't imply ftl communication which is ruled out because of a lack of evidence for it, and because it's contradicted by accepted theory.

    So what we're left with is the same lack of a deep understanding of light as before Bell's Theorem and associated works.

    This is not to disparage Bell's Theorem, which is fascinating.
     
  7. Nov 11, 2017 #6

    Zafa Pi

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    The OP made a vague suggestion of how to replicate the correlations of entangled entities via local hidden variables. @PeterDonis, @DrChinese, and @zonde were attempting to help relieve him of this folly, so yes, not much new understanding there. And you're right Bell doesn't imply ftl.
    However, there are accepted interpretations that assume ftl phenomena to "explain" Bell that are compatible with standard physics.
    Yet indeed, light is a mystery, but so is gravity and why there is a universe, and ...
     
  8. Nov 11, 2017 #7
    My reasoning is (often) thus: the correlation depends on the relative angle of the both detectors, and as such must have some sort of nonlocality.
     
  9. Nov 11, 2017 #8

    vanhees71

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    Hm, I'd say light is far from being a mystery but being among the best understood phenomena (with QED). You are right, gravitation is still a mystery in the sense that there's not yet a consistent quantum theory of the gravitational interaction.
     
  10. Nov 11, 2017 #9

    Zafa Pi

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    Can you prove ER = EPR (wormholes) is wrong?
     
  11. Nov 11, 2017 #10
    I'm not familiar with that. Is that local?

    Of course there is the option to drop realism.
     
    Last edited: Nov 11, 2017
  12. Nov 11, 2017 #11

    Zafa Pi

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    Mystery is certainly not a well defined term, but Feynman (well acquainted with QED) continually thought photons were a mystery. He claimed the double slit phenomena an ultimate mystery.

    When I mentioned that gravity was a mystery I wasn't thinking about the problems related to quantum gravity. It is more along the lines of Newton's comment that he had no explanation for how masses pulled off his laws. The notion of a gravitational field or QFT (and their wonderful predictions) may satisfy you and others, but many still feel bereft as to how those fields get created, how mediators get exchanged, how entangled photons correlate themselves, how a photon interferes with itself at the slit.

    If you think that is just the way of nature and we should just move on (a variation of shut up and calculate) well good for you. Perhaps you can console Feynman, and Einstein, in their repose. And me as well.:sorry:
     
  13. Nov 11, 2017 #12

    Zafa Pi

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    Google ER = EPR and you decide. It seems so to me.
    Indeed, dropping realism is an option, but it is hard to visualize.
     
  14. Nov 11, 2017 #13

    PeterDonis

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    ER = EPR doesn't refute Bell's Theorem. It is just a proposed "mechanism" for producing correlations that violate the Bell inequalities; it doesn't show that such correlations don't violate locality in the sense that Bell defined the term.
     
  15. Nov 11, 2017 #14

    Zafa Pi

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    Your first sentence can not be disputed, no physics can refute a mathematical theorem. I agree with the first half of your second sentence. I'm having difficulty with the second half due to several negatives, and a long dispute in another thread over Bell's sense of locality.

    In post #7 @entropy1 said there must be some sort of nonlocality. I'm not sure what he meant by nonlocality, some mean ftl phenomina (see post #5). It seems possible in my mind that the ER = EPR "mechanism" can produce the correlations that violate Bell's inequality without invoking ftl. The distant entangled particles are not distant after all.
     
  16. Nov 11, 2017 #15

    PeterDonis

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    Bell's definition was very simple: "locality" means the function describing the correlation between the two measurement results factorizes (i.e., it is the product of two functions, each of which depends only on one of the two measurement settings); "nonlocality" means it doesn't.

    Most discussions (probably including the one in the other thread you refer to) get bogged down because they don't give a precise definition to the term "locality" as Bell did.
     
  17. Nov 11, 2017 #16

    PeterDonis

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    No, it can't. ER = EPR does not create timelike or lightlike paths between the two measurement events, which is what would be needed for this.
     
  18. Nov 12, 2017 #17

    vanhees71

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    Well, at the same time, Feynman has among the best introductory chapters on QM. As Einstein said about theoretical physicsts: "Don't listen to their words but see their deeds."

    http://www.feynmanlectures.caltech.edu/III_01.html
     
  19. Nov 12, 2017 #18
    I mean the information required to constitute the correlation, when adopting realism, that is the detectors really yield the measurement result at the moment of measurement, is spacelike separated at first and can (to my understanding, yet) not be explained locally, that is, within the limits of lightspeed.

    (unless you drop realism of course)
     
  20. Nov 14, 2017 #19

    Demystifier

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    So is ER=EPR a local or a non-local theory (in the Bell sense)? I'm asking because the ER side is supposed to be described by general relativity, which is local.
     
  21. Nov 14, 2017 #20

    PeterDonis

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    Since AFAIK it makes the same predictions as standard QM, it would have to be nonlocal in the Bell sense, since its predictions violate the Bell inequalities.

    GR is local in the sense of obeying relativistic causality. But that's not the same as the definition of "local" that Bell used. QFT is also "local" in the sense of obeying relativistic causality (in QFT that translates to "field operators commute at spacelike separations"), but it makes predictions which are nonlocal in the Bell sense, since they violate the Bell inequalities. AFAIK QFT in curved spacetime is the same as QFT in flat spacetime in this respect.

    (Btw, "described by GR" is a bit of a misnomer for the "ER" side, since we are talking about quantum phenomena and GR is not a quantum theory. A better description would be that ER = EPR does QFT in a particular curved spacetime, the "Einstein-Rosen bridge" spacetime. At least that's my understanding.)
     
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