Dismiss Notice
Join Physics Forums Today!
The friendliest, high quality science and math community on the planet! Everyone who loves science is here!

Who would be interested in an EPR thought experiment?

  1. Aug 17, 2015 #1
    If a thought experiment with hidden variables gave the same results as EPR experiments, would that be a publishable (even if minor) contribution?
     
  2. jcsd
  3. Aug 17, 2015 #2

    DrChinese

    User Avatar
    Science Advisor
    Gold Member

    It is generally accepted that there can be nonlocal hidden variables. So if your idea is along that lines... no one will really be interested.

    If your idea is for local hidden variables, you will run into Bell's Theorem which says that is not possible. Bell supercedes EPR. I would not even consider putting forth such an idea until you thoroughly and expertly understand Bell. I can give you some links if that helps.
     
  4. Aug 17, 2015 #3

    Nugatory

    User Avatar

    Staff: Mentor

    By "EPR experiments", do you mean the experiments that confirmed that quantum mechanics does violate the Bell inequalities?

    A thought experiment with local realistic hidden variables that violates Bell's inequality would be a publishable and major contribution - but only if it can survive peer review, a necessary precondition for publishing anything.

    The point of the peer review is to weed out the papers that are based on errors, faulty premises, or faulty logic. If a "thought experiment with hidden variables" matches the quantum mechanical prediction, the odds that it is not based on errors, faulty premises, or faulty logic are very close to zero (because this would imply an undiscovered flaw in the proof of Bell's Theorem) so the odds of such a paper actually making it through peer review are also very close to zero.
     
  5. Aug 17, 2015 #4
    I was thinking of the common 3 detector experiment where the detectors results have 0.25 agreement when they are set at two different settings (AB, AC, BC) and 1.0 agreement when they are set at the same (AA, BB, CC).

    I can think of a simple way to generate that result using ordinary objects. I am confident of those probabilities, although I don't think that is any solid basis to argue for or against hidden variables.

    I would not mind reading further, although I have a good folder of papers, and I have the google.

    I like this summary:
    http://drchinese.com/David/Bell_Theorem_Easy_Math.htm

    "if you randomly select any of the 3 pairs ([AB], [BC] or [AC]) enough times then you would expect to get matching results (++ or --) AT LEAST a third of the time. "

    Obviously the QM results disagree. But if I had physical measurements of real objects (say mass, for example) that also gave a matching result 0.25 of the time, would that matter?

    And I thoroughly agree that putting forth any idea without thorough understanding is a bad idea. I was considering whether I even should take that next step of reaching absolute certainty of the probabilities. I tend to think I'm more likely just looking at some detail that was long ago put to rest. But if the probabilities I give are correct is that just unimportant, or something that would need some further explanation?

    EDIT: I apologize for not connecting your name to the link ... I really did not see your name. Your link is well written and I should have seen your name.
     
  6. Aug 17, 2015 #5

    Nugatory

    User Avatar

    Staff: Mentor

    Depends on how interesting you find the problem.... You have an error somewhere, either in your calculations or in your understanding of what the exact challenge is. Finding the error could be a fair amount of fun, and educational too.
     
  7. Aug 17, 2015 #6

    Nugatory

    User Avatar

    Staff: Mentor

    Strictly speaking, you mean "non-parallel" not orthogonal, right?
     
  8. Aug 17, 2015 #7

    atyy

    User Avatar
    Science Advisor

    I don't think there is any generally accepted solution to the problem of hidden variables for the standard model, or for an arbitrary relativistic quantum theory, so those would still be research problems. There are proposals, like http://arxiv.org/abs/1205.1992, but I don't think there is consensus on these proposals at this time. Even "more mainstream" problems like a lattice standard model are still unsolved https://www.physicsforums.com/threads/status-of-lattice-standard-model.823860/, so there are interesting problems open.
     
  9. Aug 17, 2015 #8
    I think I see math errors in my spreadsheet. It was an old one and it was the reminder from several links that the odds should be 0.33 not 0.25 that reminded me of looking at it as a probability puzzle a long time ago. I had several odds generating sets and I can easily get higher than 0.33, but I think the lower bound is 0.33, on closer examination. I think that I have one set with split probabilities ... so that the odds are lower for some pairs but not the others. So for example, having p(AB) =0.25 is offset by p(BC)=0.41 ... still an average of 0.33.

    I would have to say I am always puzzled by what would happen if you had 3 entangled particles and sent them to all 3 detectors. It would seem impossible for p(AB)=p(AC)=p(BC) to all still be 0.25. The requirement of the reduction of the result to 2 outcomes means that the DATA has to be in the form written in the table. If A is 1 and B is 0, then C must agree with A or B. It seems like a 3rd entangled particle leads to a problem with the probabilities. I think that is what I generated as a puzzle result.

    It IS a fun puzzle. I would doubt that I see a "classical" system that generates the probabilities also. And it was fun to go look at spreadsheets and check the numbers.
     
  10. Aug 17, 2015 #9

    DrChinese

    User Avatar
    Science Advisor
    Gold Member

    Yup... :) glad you made sense of what I meant...

    -DrC
     
  11. Aug 17, 2015 #10
     
  12. Aug 17, 2015 #11

    Nugatory

    User Avatar

    Staff: Mentor

    You can google for "three-particle entanglement"; unfortunately the analysis of these states is appreciably harder and doesn't lend itself to the "...with easy math" approach that works so well with two particle states.
     
  13. Aug 17, 2015 #12

    morrobay

    User Avatar
    Gold Member

    Rather that implying an undiscovered flaw in the proof of Bell's Theorem. Why not a more complete understanding of hidden variable physical mechanisms that could produce results that match quantum mechanical predictions ?
    A local realistic or non realistic hidden variable theory that matches QM predictions.
     
  14. Aug 17, 2015 #13

    Nugatory

    User Avatar

    Staff: Mentor

    Is it possible that there exists a right triangle whose hypotenuse squared is not equal to the sum of the squares of the sides AND that there is no flaw in the proof of the Pythagorean theorem, which says that no such right triangle can exist?

    No, of course not. If there is no such flaw, then there can be no such right triangle.
     
  15. Aug 17, 2015 #14

    bhobba

    User Avatar
    Science Advisor
    Gold Member

    If QM is correct then a local realistic theory is ruled out. But of course you can have a theory that approximates QM to a high degree of accuracy but is local and realistic, and such have been proposed eg:
    http://arxiv.org/pdf/quant-ph/9508021.pdf

    As for a non realistic hidden variable theory Feynman's sum over histories formulation is technically just that. Its not realistic that objects take all paths at once.

    I always have a bit of a smile with comments like the above. Physicists are very creative in what they come up with, and all that saying things like - why not a more complete understanding of whatever - suggests is a lack of knowledge of the literature.

    Thanks
    Bill
     
  16. Aug 17, 2015 #15
    I agree that the math precludes the hidden variable solution. But I don't have any problem with the sum over histories being also a description. Since one of the conclusive things we end up agreeing on is precisely that the path is fundamentally unknown, not just "hidden" and depending on hidden variables, the possibility of every path seems not ruled out at all.

    Thanks, I think. Not easy math answers from anyone on these boards is a daunting prospect ...

    I will probably never give up playing with odds tables, even though I agree 100% with the hidden variables math. I just have played dice games throughout my life and somehow find it fun to calculate 3 detectors with rules for 6-20-sided dice and that sort of nonsense. I completely find the math conclusive, but then I will start wondering what would happen if I added on more dice and one more set of arbitrary rules ... there is always going to be a mistake, but there is still something fun about it. As I said, a legacy of too many dice games, I guess. I always enjoyed preparing and calculating all the odds. The odds I recalled were incorrect ... if I set the odds for p(AB) and p(AC) at 0.25, the odds of p(BC) go up. And a simple math error was hidden in that calculation also. It didn't take long to find.

    I'll have to have a go at the 3rd entangled particle. I don't think it is necessary to rule out hidden variables, but it sounds interesting.
     
  17. Aug 17, 2015 #16

    Nugatory

    User Avatar

    Staff: Mentor

    Before you do that, you'll want to understand the physics behind DrChinese's "...with simple math" model. Neither the quantum mechanical prediction about the probability of getting spin-up being equal to the square of the cosine nor the rule that measurements on the same axis must always yield opposite results were plucked out of thin air - they come from quantum mechanical analysis of a two-particle spin-entangled system. You'll need analogous rules for the three-particle system, and you won't be able to develop them unless you understand the underlying physics some.
     
  18. Aug 18, 2015 #17
    Just to nit-pick, unless my brain has lost a few more billion neurons in vital places, which is quite possible, that would go to zero at 90 degrees - correct for polarization, not correct for electron spins. And I think you are talking about such particles because you say
    Electron spin follows the rule: the square of the cosine of half the angle. 90 degrees then gives you a probability of cos2(90/2) = (√2/2)2 = 1/2 which is what you want. :biggrin:
     
    Last edited: Aug 18, 2015
  19. Aug 18, 2015 #18

    morrobay

    User Avatar
    Gold Member

    If θ is angle between a and b for two spin 1/2 particles then the probability for double detection,
    +1 , +1 or -1,-1 =
    1/2 sin2 θ/2 For opposite results : 1/2 cos2 θ/2
    See paqe 29 http://arxiv.org/pdf/quant-ph/0209123v2.pdf
     
  20. Aug 19, 2015 #19
    For each of the two cases, +,- and -, +.
    What I said.
     
Know someone interested in this topic? Share this thread via Reddit, Google+, Twitter, or Facebook




Similar Discussions: Who would be interested in an EPR thought experiment?
  1. EPR thought experiment (Replies: 17)

  2. EPR thought experiment. (Replies: 13)

Loading...