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

Bell inequality

  1. Dec 12, 2014 #1
    I went through a paper last week about the Bell inequality and how it is incompatible with QM. Something along the lines of probability in classical regards being 1/3 but in quantum mechanics it is 1/4. It went into some basic principles of how this is determined through quantum entanglement to be 2 separate variables measured on 2 entangled particles. OK then. I file it away in the pile labeled "not fully understood". But then something doesn't sit right... wouldn't 2 variables with 2 possible values be 1/4 chance of any of the 4 possible "states" occurring? That's just classical probability.
  2. jcsd
  3. Dec 12, 2014 #2

    Stephen Tashi

    User Avatar
    Science Advisor

    You'll have to ask a specific question if you want a specific answer.

    Why would 2 variables with two possible values have a 1/4 probability of each combination of values? Is the probability that a randomly selected person is male and not retired equal to 1/4 ?
  4. Dec 12, 2014 #3
    I was trying to explain it in a simple boolean manner, 0 or 1, true or false, heads or tails fits best... 2 coins with 2 possible states. 25% both heads 25% heads/tails 25% tails/heads 25% both tails.
  5. Dec 12, 2014 #4

    Stephen Tashi

    User Avatar
    Science Advisor

    That's clear. Now what is the question? Are you asking why ordinary probability theory gives an answer of 1/3 in the quantum entanglement problem? To answer that I, myself, would need to hear a statement of the entanglement problem you read about. Perhaps someone else knows the problem just from the mention of the word "entaglement".
  6. Dec 12, 2014 #5
    I think I need to be a bit more organised... I read it from a link in a post here several days ago and now I can't find it. :-(
    Aha! Browser history...
    I'm going to read through it again hoping to catch what I missed or realize where I misunderstood.
  7. Dec 12, 2014 #6
    I just absorbed another version found in another thread here: http://www.felderbooks.com/papers/bell.html
    It makes the same case of what should be 5/9 chance turns out 50/50.

    But just before reading that version I had the "aha!" moment and realised this is because entanglement gives 100% (or close to, ideally) chance of pairs being opposite spin in any 1 of the 3 axes. We can only measure 1. So then we double up pairs, and we measure 2 axes on 1 of each pair. Then when we think we can predict what the measurements of the other pair will be it turns out random?
  8. Dec 13, 2014 #7
    I'm still not sure if I explained that well enough or correctly, or if the original paper by Bell? would provide any additional insight. All I am trying to do is relate what is expected locally (meaning classically as in all 3 axes in 1 particle should always be the opposite spin in all 3 axes of entangled pairs) to what truly occurs non-locally (the measurements in the other 2 axes is affected in the pair by measuring the first axis on the first particle). This is truly the gist of the inequality, is it not?
  9. Dec 13, 2014 #8


    User Avatar

    Staff: Mentor

    Another very good resource for your level of understanding is this article.
  10. Dec 13, 2014 #9
    I haven't finished reading it yet, but can I assume there is verification that we know for sure a singlet state that we've partnered that we can measure vectors A,B and verify we know what C is or is that inferred from Bell's Theorem exclusively?
  11. Dec 13, 2014 #10


    User Avatar
    Science Advisor

    I don't think anyone understands your question or what article you are referring to, since there are now 3 references in the thread.
  12. Dec 13, 2014 #11
    Forget it I'm good thanks for the links everyone.
  13. Dec 13, 2014 #12
    Detection-Loophole-Free Test of Quantum Nonlocality, and Applications, arXiv:1306.5772)

    Here's a recent proof...
Share this great discussion with others via Reddit, Google+, Twitter, or Facebook