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Understanding Bell's inequality

  1. Feb 5, 2016 #1
    I am not sure if I am allowed to ask this, but here's my shot:

    I find all the explanations of Bell's theorem summed up here, very different in interpretation and also (for me) incomprehensible. I have these simple questions:

    How does the Bell inequality, stated as N(A, not B) + N(B, not C) ≥ N(A, not C), relate to the position of the detectors? Do A, B and C represent binary values or invertable continuous values? (What do they mean??) What is the most comprehensible source discussing Bell's inequality making use of this formula? (I fully understand the formula itself, but not its significance)

    Put differently: in a Venn-diagram, what do A, B and C represent?

    Thanks.
     
    Last edited: Feb 5, 2016
  2. jcsd
  3. Feb 5, 2016 #2

    mathman

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    In a Venn diagram A, B, C are events. Fundamentally Bell's inequality is a mathematical theorem.
     
  4. Feb 5, 2016 #3
    Such as measuring a value?
     
  5. Feb 5, 2016 #4
    The areas in a Venn diagram made up of three areas:A, B and C which overlap to some degree.
     
  6. Feb 5, 2016 #5

    Mentz114

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    Here is yet another simple derivation of the correlation limit (attached). Using Venn diagram logic. I believe there is a simpler arithmetic version but you're probably fed up with these proofs !
     

    Attached Files:

  7. Feb 5, 2016 #6

    Strilanc

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    A simple way to think of the Bell inequalities is as a game.

    Two separated players will each flip a coin. Then, based on the coin flip, they each pick "yes" or "no". Their goal is for the choices to disagree when and only when the coins both land heads. They want to satisfy ##x \land y = a \oplus b##, where ##x## and ##y## are the respective random coin flips while ##a## and ##b## are their respective choices.

    Classically, the best strategies for this game win 75% of the time. Doing better requires cheating (e.g. communicating during the game, rigging the coin flips so they're not random, etc). This isn't too hard to intuit: ##a## can depend on ##x## but not ##y##; similarly ##b## can depend on ##y## but not ##x##. So really you're trying to pick ##f## and ##g## that maximize how often ##x \land y = f(x) \oplus g(y)## is satisfied. Probabilistic strategies are just weighted averages of the deterministic strategies, and the deterministic strategies all win 25% or 75% of the time, so classical strategies are stuck in the [0.25, 0.75] range.

    But you can use quantum-entanglement-based strategies to push your win rate past 85%.
     
  8. Feb 5, 2016 #7

    DrChinese

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    "A" would be the + outcomes of a polarization (spin) measurement at some angle a. "Not A" would be the - outcomes (note that + and - are arbitrary labels, you could also label it 1 and 0, or H and V).

    So "B" would be the + outcomes of a polarization (spin) measurement at some angle b. And so on.

    With an entangled pair, you can only measure 2 of the 3 angles at a time. So you would need to run a suitable number of trials of the 3 pairs: N(A, not B), N(B, not C) and N(A, not C). That would give you averages that can be plugged in.

    For a classical example: You have a room with 100 people in it. The A group are men, Not A are women. The B group are parents, Not B are people with no children. C are right handed, Not C are left handed. So we substitute into the formula N(A, not B) + N(B, not C) ≥ N(A, not C) and get:

    [1] N(men, not parents) + N(parents, left handed) >= N(men, lefthanded)

    The classical version will always be true, no matter how you select the people to go into the room. That is because the above inequality, with suitable rearrangement, becomes:

    [2] N(men, not parents, right handed) + N(women, parents, left handed) >= 0

    I can explain how I got from [1] to [2] if you like.
     
  9. Feb 5, 2016 #8
    That may be helpful.
     
  10. Feb 5, 2016 #9
    I have made a Venn-diagram for the equation:

    N(A, not B) + N(B, not C) N(A, not C)

    Bell.png
     
  11. Feb 5, 2016 #10

    DrChinese

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    For a classical example: You have a room with 100 people in it. The A group are men, Not A are women. The B group are parents, Not B are people with no children. C are right handed, Not C are left handed. So we substitute into the formula N(A, not B) + N(B, not C) ≥ N(A, not C) and get:

    [1] N(men, not parents) + N(parents, left handed) >= N(men, lefthanded)

    The classical version will always be true, no matter how you select the people to go into the room. That is because the above inequality, with suitable rearrangement, becomes:

    [2] N(men, not parents, right handed) + N(women, parents, left handed) >= 0

    How did I get from [1] to [2]?

    Using ordinary classical assumptions, I expand as follows:

    [3]
    N(men, not parents) = N(men, not parents, left handed) + N(men, not parents, right handed)
    N(parents, left handed) = N(men, parents, left handed) + N(women, parents, left handed)
    N(men, left handed) = N(men, parents, left handed) + N(men, not parents, left handed)

    Make sense? Using those identities, I substitute into the original [1] and get:

    [4]
    N(men, not parents, left handed) + N(men, not parents, right handed) + N(men, parents, left handed) + N(women, parents, left handed)
    >=
    N(men, parents, left handed) + N(men, not parents, left handed)

    I notice that the 2 right hand side terms both also appear on the left hand side as well. So I eliminate them from both sides. This gives me:

    [5] N(men, not parents, right handed) + N(women, parents, left handed) >= 0

    QED. Now, keep in mind that the classical assumption I made in [3] is subject to attack in the quantum world. Precisely because classical logic fails there, and the inequality is not valid in actual experiments.
     
  12. Feb 5, 2016 #11
    Does the + vs the - mean pass vs not-pass the filter? (in case of polarisation)

    And how does it apply to electron spin?
     
  13. Feb 5, 2016 #12

    DrChinese

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    The result formula below can be evaluated for angle settings instead of attributes of people.

    [5] N(men, not parents, right handed) + N(women, parents, left handed) >= 0

    If you pick almost any 3 angles randomly, there is an inequality similar to the above that would always be TRUE for a classical world; but is FALSE in an actual Bell test. Essentially:

    Classical (the sum of expectation values of any 2 permutations must be greater than or equal to zero) :
    [6] N(A, ~B, C) + N(~A, B, ~C) >=0

    Quantum (yields a negative probability):
    [7] N(A, ~B, C) + N(~A, B, ~C) < 0

    Most people don't refer to Bell's Inequality in this fashion, but it is just as correct. The CHSH inequality is usually used for Bell tests instead.
     
  14. Feb 5, 2016 #13
    How?
     
  15. Feb 5, 2016 #14

    DrChinese

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    Actual Bell tests usually use photons going to polarizing beam splitters, which have 2 outputs. They can be labeled any way you like. Not really sure how you would do it with electrons to be honest.
     
  16. Feb 5, 2016 #15
    I don't know what 'A' and 'not A' means in terms of an experiment. Can anyone tell me?

    For instance: is 'A' the set of photons passing the filter at setting A, and 'not A' the set of photons not passing the filter at setting A?
     
  17. Feb 5, 2016 #16

    DrChinese

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    I can give you one, it's a bit difficult to explain the mapping. Pick angles to be 0, 120 and 240 degrees for A, B and C. Label the outputs of the beam splitters as + and -. Then adapting to the original [1] we get:

    [8] N(A+, B+) + N(B-, C-) >= N(A+, C-)

    The quantum expectation value is for a match (++ or --) is cos^2(theta) where theta is the angle difference. Assuming the ++ case and the -- cases are equally likely, each would be half the total. For our 3 angles, the theta is 120 degrees for any pair. Cos^2(120 degress) is .25. The quantum expectation value is for a mismatch (-+ or +-) is sin^2(theta) which is .75. This evaluates to the quantum expectation values as follows :

    [9] .125 + 125 >= .375

    And the inequality is violated. I know this last step was a big jump, sorry bout that. Keep in mind that there is no quantum expectation value for something like:

    [10] N(A+, B+, C+).

    Work through the formulas a bit and hopefully you will get the idea.
     
  18. Feb 5, 2016 #17

    Strilanc

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    The inequality experiment that Dr. Chinese is explaining (which is different than the game I described, though they're examples of the same basic thing) is explained very clearly in this youtube video.
     
  19. Feb 6, 2016 #18
    That is really helpful! Thanks! :woot:
     
    Last edited: Feb 6, 2016
  20. Feb 6, 2016 #19
    Suppose hidden variables would exist. Would that mean that there would be a direct relation between the value of the variable and the value of the measurement?

    It seems to me that the value of the measurement has to depend at least also on the factors that the beamsplitter or the polarizer represent...!
     
    Last edited: Feb 6, 2016
  21. Feb 6, 2016 #20
    I think I've got it: if the outcome of the measurement would not depend on the on the orientation of the filter, that would be just stupid. So it does. However, we can change the orientation of, say, filters FA and FB at the last moment, and still there is a correlation between the measurements. This would imply superluminal communication or something of the sort, since the measurements are now local.

    So we could assume the outcomes of the measurements do not depend of the orientations of the filters, but rather on the hidden variables in a one-to-one fashion. And that is ruled out by the Bell inequality.

    QED. :smile:
     
    Last edited: Feb 6, 2016
  22. Feb 6, 2016 #21

    Zafa Pi

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    This is the 1st time I've disagreed with DrChinese. Classical logic is the backbone of mathematics and DOES NOT FAIL! Here is a concrete example of what's happening.

    Theorem: Let Ah, At, Bh, Bt each be +1 or -1. The values may come about via some random process such as coin flipping.
    If Ah•Bh = 1 then P(At•Bt = -1) ≤ P(At•Bh = -1) + P(Ah•Bt = -1).

    Proof: P(At•Bt = -1) = P(At•Bt•Ah•Bh = -1) = P(At•Bh•Bt•Ah = -1) =
    P({At•Bh = -1 and Bt•Ah = 1} or {At•Bh = 1 and Bt•Ah = -1}) =
    P(At•Bh = -1 and Bt•Ah =1) + P(At•Bh = 1 and Bt•Ah = -1) ≤
    P(At•Bh = -1) + P(Ah•Bt = -1) QED

    The numbers Ah, At, Bh, Bt could come about as follows:
    #1 The physical set up for the Theorem:
    Alice and Bob are 2 light minutes apart, and Eve is half way between them. Alice
    has a fair coin (see probability appendix) and a device. Her device has 2 buttons
    labeled h and t, a port to receive a signal from Eve. The device also has a screen
    that will display “Eve’s signal received” when a signal from Eve is received. It will
    also display either +1 or -1 if one of the buttons is pushed. Bob has the same
    equipment and shows the same values, tho the internal workings of his device
    may be different.
    #2 The following experiment is performed:
    Eve simultaneously sends a light signal to each of Alice and Bob. When Alice’s
    device indicates Eve’s signal has been received she flips her coin. If it comes up
    heads she pushes button h, otherwise button t, and then notes what the screen
    displays. What Alice does takes less than 30 seconds. The same goes for Bob.
    #3 Notation:
    If Alice flipped a head and pushed button h, we let Ah be the value her screen
    would show. So Ah = 1 or -1 and is the result of some objective physical process.
    Similarly we let At be the value if she had flipped a tail. We let Bh and Bt be the
    analogous values for Bob.

    Eve could send each of Alice and Bob a photon from an entangled pair so that
    Ah•Bh = 1, and P(At•Bt = -1) = 3/4, while P(At•Bh = -1) + P(Ah•Bt = -1) = 1/2.
    (the state of the pair is 1/rt2(|00> +|11>), Ah and Bh measure at 0 degrees, At at 30, and Bt at -30)

    So is the Theorem false? Not at all, the logic/mathematics is impeccable and not contradicted by QM. The inequality is false however. How could this be? The hypothesis of the Theorem has all 4 numbers Ah, At, Bh, Bt existing at once and shows up in the proof. In each experiment however only two of the numbers are observed, e,g. At and Bt. The existence of the other two is inferred by counterfactual definiteness = hidden variables = realism, and it is that inference which must be questioned (assuming locality).
    MATHEMATICS PREVAILS!
     
  23. Feb 7, 2016 #22

    DrChinese

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    1. Only once? I take that as a compliment... :smile:

    2. Bell's Theorem is valid, and the inequalities are valid, but they all rest on the assumption of realism (counterfactual definiteness) as you say. That assumption, along with the assumption of locality, are subject to being attacked. Experimental results (violating the inequalities) then lead us to reject one or both of these assumptions.

    I don't think we disagree.
     
  24. Feb 7, 2016 #23

    stevendaryl

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    I've pointed this out many times when "counterfactual definiteness" is brought up. I don't think that counterfactual definiteness should be identified with realism. The way I understand how CFD comes into play in Bell's theorem is that

    realism + perfect correlations/anti-correlations implies CFD
    Because of the perfect correlations for the EPR experiments, Alice can reason along the lines of "If Bob has chosen detector setting X, then he would have gotten result Y". CFD doesn't automatically follow from realism (at least not in the way I think of it).

    To me, the sort of (local) realism that Bell assumed is best given in his essay "The Theory of Local Beables". Roughly speaking, every measurement confined to a small region of spacetime has a probability distribution for possible outcomes, and that distribution can only depend on information about conditions in the backward light-cone of the measurement. Information about distant conditions can only be relevant to the extent that those distant conditions tell you something about the backward light-cone.

    The sort of theories that Bell's notion of a local realism covers is basically any theory of the following type:
    • We can partition the universe up into small regions, and give a "state" (values of fields, positions, charges, and momenta of particles) within each region.
    • The future state of a small region can depend only on the current state of that region and neighboring regions.
    Quantum mechanics is clearly not a local realistic theory of this type. You don't need Bell's theorem to tell you that. It isn't a locally realistic theory because many-particle quantum mechanics evolves in configuration space, rather than physical 3D space. To me, Bell's theorem is really not about QM, it's about the possibility of discovering a local realistic theory that makes the same predictions as QM. (Or we can take QM completely out of the picture, and ask whether there is any local realistic theory that makes predictions consistent with the results of EPR-type experiments that have actually been performed.)

    In general, the concepts of "local" and "realism" might be too fuzzy to reason about (is MWI local? Is it realistic? What about Bohmian mechanics? What about retrocausal interpretations?). But Bell's notion of local realism seems pretty concrete, and it is apparently ruled out by experiment.
     
  25. Feb 7, 2016 #24

    DrChinese

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    I agree with this. As to whether CFD and realism are interchangeable: they are as interchangeable for purposes of our discussions. There may be nuances, but you will see them interchanged in Bell-related literature quite often. I'm not sure the nuance is really that important as to interpreting the Bell result.
     
  26. Feb 7, 2016 #25
    I recommend Richard Gill's "Statistics, Causality and Bell's Theorem": http://arxiv.org/abs/1207.5103

    His theme here is that locality together with CFD simply means that (in the context of CHSH) any theory/model that respects those two, can just as well be described by a 4xN matrix, where N is the number of runs, and for each row the first two elements are Alice's results for her two possible detector settings, and the last two elements are Bob's results for his two possible settings. If it is possible to reduce a model to such a 4xN matrix, then the upper limit of 2 for the CHSH inequality will apply when N goes to infinity, assuming the detector settings can be chosen randomly, independent of the values in the table.
     
    Last edited: Feb 7, 2016
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