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What does this example say about the applicability of Bell's inequalities?

  1. May 16, 2012 #1
    It is commonly believed that Bell's inequalities are a theoretical derivation of a condition that must be satisfied by locally causal theories. Therefore, it is often concluded that violation of these inequalities by experiments provides very strong evidence (if not conclusive proof, the only doubt being due to imperfect experiments) that reality is non-local.

    However, I present a macroscopic example using coins where, Bell's theorem is always violated in the experiments. I would like to stimulate discussion of

    1) why this specific example exhibits such behaviour?
    2) where is the non-locality, or non-reality, or any other spooky conclusion?
    3) what does this mean for the popular belief that Bell's inequalities are applicable to the EPRB experimental results?

    Here it goes:

    We have three coins labelled "a", "b", "c", if we toss all three a very large number of times, it follows that the inequality |ab + ac| - bc <= 1 will never be violated for any individual case and therefore for averages |<ab> + <ac>| - <bc> <= 1 will also never be violated (where heads = +1 & tails = -1).

    Proof: For three coins with outcomes ±1, there are 8 possibilities. the LHS for each possibility is always <=1 as illustrated below:
    a,b,c = (+1,+1,+1): |(+1) + (+1)| - (+1) <= 1, obeyed
    a,b,c = (+1,+1,-1): |(+1) + (-1)| - (-1) <= 1, obeyed
    a,b,c = (+1,-1,+1): |(-1) + (+1)| - (-1) <= 1, obeyed
    a,b,c = (+1,-1,-1): |(-1) + (-1)| - (+1) <= 1, obeyed
    a,b,c = (-1,+1,+1): |(-1) + (-1)| - (+1) <= 1, obeyed
    a,b,c = (-1,+1,-1): |(-1) + (+1)| - (-1) <= 1, obeyed
    a,b,c = (-1,-1,+1): |(+1) + (-1)| - (-1) <= 1, obeyed
    a,b,c = (-1,-1,-1): |(+1) + (+1)| - (+1) <= 1, obeyed

    The Bell inequality Challenge:
    Find a locally realistic situation which violates the above inequality. According to Bell and proponents, this is impossible to do.

    Experimental Violation:
    We have 3 coins labeled "a","b","c", inside a special box. A button on the box releases only two of the coins at random when pressed. The coins must be returned to the box before the next press, therefore only two of the three coins can ever be outside of the box at the same time.

    Since we can not toss all three at once, and since the inequality only contains averages of pairs of outcomes, we assume the inability to measure all three simultaneously is inconsequential. We decide to perform the experiment by tossing just the pairs a very large number of times and group the results into 3 runs for the pairs (a,b), (a,c), (b,c) tosses. Using the pairs, we calculate <ab>, <ac> and <bc>. Even though the data for each pair appears random, we find that from our data <ab> = -1, <ac> = -1 and <bc> = -1, which violates the inequality when substituted into the LHS. (i.e 3 <= 1 according to the inequality). This was supposed to be impossible according to the challenge!!! Does this mean "local causality" or realism is false?

    The Explanation:
    Consider the following: Each coin has a programmable bias which can be changed by the box just before it is released but not after. The box has an internal clock which keeps track of the time (t) in seconds. The above scenario [<ab> = -1, <ac> = -1 and <bc> = -1 ] can then easily be realized if the special box operates as follows:

    Every time a button is pressed, calculate calculate sin(t) where t is the time read off the internal clock. If sin(t) > 0, program coin "a" to be biased for heads (+1) and coin "c" to be biased for tails (-1). Then randomly pick two of the three coins. If coin "b" is one of the picks, program coin "b" to be biased for tails (-1) if the other pick is coin "a", otherwise program coin "b" to be biased for heads (-1). If sin(t) <= 0 reverse all the signs.

    a) The box will always produce <ab> = -1, <ac> = -1 and <bc> = -1, no matter how many times the coins are tossed, (1 or 50 billion).
    b) The results will be random
    c) The inequality will always be violated no matter how many times the coins are tossed (1 or 50 billion)
    d) The box and coins operate in a completely locally causal manner.
    e) The result of each toss is non-contextual and predetermined from the moment the coins are produced
    f) There are no loopholes in the experiment
    f) There is no spooky business happening
    g) YET Bell's inequality is violated.

    How come?
  2. jcsd
  3. May 16, 2012 #2


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    Oh, I am going to be SOOOOOO sorry that I stepped into this..... But, altogether against my better judgement.... Here goes.....

    With every button press, the box is producing a pair of coins, one biased for heads and one biased for tails. How is that any different than me randomly leaving one glove out of a pair behind when I leave home, and then "discovering" that there is a perfect anti-correlation between the handedness of the glove I have with me and the one that I left at home?
  4. May 16, 2012 #3
    I guess the difference between the thought experiment and your explanation is that the experiment states there are pre-determined results, but bell's inequality is still violated.
  5. May 16, 2012 #4
    Hi Nugatory,
    I'm not sure what you are asking but the difference is that the perfect anti-correlation is between the two coins actually produced and the person doing the experiment does not know anything other than what they actually measure and therefore can not use such hidden knowledge to calculate correlations.
  6. May 16, 2012 #5
    You basically have taken up David Mermin’s challenge (Is the moon there when nobody looks? Reality and the quantum theory) to find an instruction set to account for the experimental results. Everyone recognizes the obvious instruction set, that is, the source guarantees the correlation (or anti-correlation) when choosing the same coin, orientation angle, etc. You now offer a second instruction set in the form of sin(t) when choosing different coins to demonstrate that Bell’s inequality can be violated for a locally realistic experiment when selecting two of the three coins. Sin(t) is significant in that it makes the connection to the EPRB experiments.

    Why wouldn’t the instructions sets come in pairs? This seems necessary to explain the random results.
  7. May 16, 2012 #6
    Haha, same sentiments here.

    So here's my take:
    Essentially there is nothing much different from your instruction set and an entangled pair. The reason why your experiment violates Bell's inequalities, I believe is because it, like the case of EPR pairs, does not exist in a definite state until you press the button (ie making the measurement).

    One could see the EPR pairs as such: (consider the (|01> + |10>)/sqrt(2) case).
    So, I have a source producing them. When I make a measurement on the first qubit, the following "program", similar to yours, may run:
    Pick a number t. If sin(t) > 0, program qubit 1 to be |0> and qubit 2 to be |1>. If sin(t)<0, program qubit 1 to be |1> and qubit 2 to be |0>. (sin(t) is positive 50% of the time)
    Why this violates Bell's inequalities is because the qubits do not exist in a definite state until I perform my measurement.

    So, because your coins don't have a definite state until you press the button, they violate Bell's inequalities. But there is nothing strange or spooky there, because you have a machine that performs the program. But for two qubits separated over large distances? What sort of "program" and "system" can connect them? - That is where the strangeness and nonlocality comes in.
  8. May 16, 2012 #7
    Not sure what you mean by definite state in the above. But each outcome of the toss is definitely predetermined. Even though you do not know the outcome of the toss until you toss the coins. Comparing to EPRB, the pressing the button would be equivalent to a source releasing two anti-correlated particles, or is that considered a "measurement" in EPRB too? And the measurement would be the tossing of the coins.

    The coins in the above example exist in a definite state from the "source" (box) prior to being measured even though the outcome is unknown by the experimenter until after the measurement (toss).

    Replace the box and button with the particle source in EPRB. The measurement then is only the tossing, not the pressing of the button. The box which produces the coins, has a definite non-spooky behaviour, each coin has a definite non-spooky behaviour. The behaviours pre-exist any measurement. Everything is local and realistic. If you like, we can separate the coins and send them off at the speed of light to remote locations where they are tossed.

    Just to understand what you mean by definite state, do you think dynamic systems have a definite state?
  9. May 16, 2012 #8
    Sin(t) does two things:
    - is responsible for the random results.
    - produces P(+1) = P(-1) = 0.5

    Since the experimenters do not know anything about the how fast the internal clock of the box is or even that there is an internal clock, they have no way of selecting sin(t)>0 or sin(t) <=0. Thus they get random results which are symmetric around zero.

    You are right, both of these features are very reminiscent of the EPRB experiments (on purpose).
  10. May 17, 2012 #9
    No. The releasing of two anti-correlated particles is not equivalent to the button pressing in your experiment. In your experiment, even before pressing the button, the outcome is already known to be anti-correlated, just that coin A, for instance can be either +1 or -1. So the source releasing the particles is just "setting up". While particle 1 has 50% probability of being spin up or spin down, it is neither spin up nor spin down until you make a measurement!

    Pressing the button in your case IS a measurement, because it assigns a value to coin A. So before you toss, you have already "disturbed the system".

    Going back to the EPR pairs, when we measure (for example with the polarising axis oriented in between the two polarisation modes), the photon actually experiences two things:
    First, the photon is forced into being either horizontally or vertically polarised. ("Button pressing")
    Second, the photon passes through the polariser with a probability given by Malus' Law. ("Tossing")
    Last edited: May 17, 2012
  11. May 17, 2012 #10
    how do we get ab (or ac, bc) = -1?

    not sure if I understand your experiement or nomenclature. however from what I understand-
    half of the time ab will be plus 1 and half of the time it will be -1. the average of ab would then be zero.
  12. May 17, 2012 #11
    Bill, what is the main assumption of Bell's theorem? Let me remind you, and I quote:
    Now, what have we here:
    Can you spot the difference?

    The whole point of Bell setup is to have two spatially separated detectors where settings can be changed independently and at will and settings of one detector do not influence the results of another. Your model does not have anything like that. You cannot change settings at all, let alone independently and spatially separated.

    Yes, the implications of Bell's theorem and it's experimental violations are profound, but the formulation and the proof are straightforward and uncontroversial. I mean, I could understand if it was your first post on the subject, but just how many times have we been through that A(a,λ) B(b,λ) stuff already? If you would bother to try to apply it to your setup, it would be immediately obvious that it does not fit, that in your case A in fact depends on b. I'm sorry to say but all this example shows is billschnieder is either trolling hard or is descending into crackpottery.
  13. May 17, 2012 #12
    Just like for the EPR case in which we know due to conservation of momentum that the particles are anti-correlated. What is the point?

    But that is just an assumption you are making. The difference between the real EPR experiment and the coin case above is that I have explained to you the detailed mechanism, which experimenters in the EPR case have no clue about. All they know is that they toss and get either a +1 and a -1. You can't argue that pressing the button in one case is different from "setting-up" in another based on having full information in one case and zero information in another.

    Had I known that you would get hung up on the "pressing". I would have removed it completely. It is inconsequential. I could have said the box decides by itself at what time it releases the pair of coins without any pressing. How then would you argue that the system is "disturbed"?
  14. May 17, 2012 #13
    That is not correct. You missed this part:

    The predetermined outcome for "a" is always opposite to that of "b". when the "ab" pair is produces Same for "ac" and same for "bc". So the products "a*b", "a*c" and "b*c" are always -1, therefore their averages after many tosses are always -1, not zero.

    EDIT: There was a typo in the original where I said "heads (-1)" instead of heads "(+1)". That may have confused the issue.
  15. May 17, 2012 #14
    Despite the typo in the OP which used the phrase "Bell's theorem" in place of "Bell's inequalities", this is not a discussion of Bell's theorem but Bell's inequalities. The point is not to claim that the EPR exeriment is the same as the coin-toss experiment. So don't tell me that the example fails because "coins are not photons" or any similar claims etc.

    The coin toss example above is self contained and parallels can be drawn. The inequalities are demonstrated to be valid in the theoretical treatment, yet violated in the experiment just like with Bell. The theoretical proof assumes all three outcomes from three coins are present at the same time, just like Bell case. The experimenters are limitted to measuring only two of the three outcomes at a time, just like in the Bell case. The experiment violates the inequalities, just like in the Bell case. The inequalities in both are Bell's inequalities. The calculation of expectation values in the experiments are done in the same way. In both cases, the outcomes are +1 or -1.

    So the question again is this. How come the inequalities which were supposed to be valid for the "coin-toss" experiment as demonstrated earlier in the OP, get violated by the experiment? It is clear that the experiment is local and realistic.

    Are you suggesting that if Alice and Bob in the real EPR experiment were given a fixed value of "a", "b", "c" and not allowed to change them, the inequality will never be violated? Or are you suggesting that if Alice and Bob were just 1m apart and not spacially separated in the real EPR experiment, the inequality will not be violated? Hopefully you now see (by answering these questions) that these issues you raise are irrelevant for the main point here. Now rather than throw accusations around, it could be usefuly if you calmed down and explain why you think the proven inequalities which appear valid from the theoretical proof in the OP, does not apply to the experiment described in the OP.
    Last edited: May 17, 2012
  16. May 17, 2012 #15
    Bill, you asked when comparing the theoretical and the experimental,

    Why the violation?

    1) The machine selects two of the coins at any one time instead of three. This has been discussed before.

    2) The fact that the results for <ab>,<bc>, and <ac> = -1. Not sure of the deeper meaning of this fact. Surely, you had a reason for choosing this value as the outcome.
    Last edited: May 17, 2012
  17. May 17, 2012 #16
    It's not that the Bell inequality will never be violated in such a case, it's that the inequality won't be derivable in the first place. It's only when you assume that the particle behavior is independent of the settings of distant polarizers that you are able to derive the inequality.
  18. May 17, 2012 #17
    ...<groan>... it's the same bloody thing! The inequality is the main result of the theorem! It is only valid as long as the assumptions of the theorem are satisfied. In your case the vital assumption is not satisfied, so the inequality is inapplicable.

    1. Do you agree that your setup does not satisfy Bell's "vital assumption", as quoted earlier?
    2. Do you agree that the derivation of Bell's inequality requires "viltal assumption" to be valid?
    3. Can you now make a great leap of logic and put 1. and 2. together and agree that Bells inequalities are inapplicable to your setup?
  19. May 18, 2012 #18


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    This algorithm, by your own admission, produces -1 (perfect correlation) for every trial. No reason to put the other stuff in. You may as well say that a perfectly correlated pair comes out every time with a random orientation of +1-1 or -1+1.

    So then the question becomes: does a perfectly correlated outcome violate a Bell Inequality? That is after all the criteria for the EPR state (elements of reality). Clearly the answer should be NO if this is a good analogy, because this case was considered specifically by Bell. He considered this as being simple.

    But here the answer is actually YES (as you say), because we have not specified any angle settings. You get the same results at all angle settings, including 0 and 90 (which should be anti-correlated relative to each other).

    So we conclude the following:

    a) This analogy does not mimic a Bell setup as it produces predictions which are counter to any experiment. Ergo, it predicts a different result than QM, which matches experiment. Therefore the main result of Bell, stated below stands:

    No physical theory of local Hidden Variables can ever reproduce all of the predictions of Quantum Mechanics.

    b) It is not truly local, as the results are observer dependent in most cases (whenever b is one of the 3 selected, which occurs about 2/3 of the time). In other words, the outcome b is predicated on whether the observer gets a or c for the other coin. This is a contextual setup, which essentially will always violate the locality/separability condition. Note that b appears to be undefined when a and c are selected, though you could fix that and would need to if you claim to have a realistic example.

    This analogy doesn't pass the test, no matter what you say about your Bell Inequality being violated.
  20. May 18, 2012 #19
    No it is not. Bell's theorem relates the inequalities to QM. The inequalities themselves do not need QM to be derived. Get it? Just because the inequalities are relevant for the theorem does not mean the inequalities by themselves are not worthy of independent discussion without being side-tracked by reference to the theorem.

    I disagree that the "vital assumption" quoted earlier is relevant to this thread. Simply look at the OP and explain why the proof of the inequality stated in the first part of the OP should not apply to the second part of the OP. Everything is contained in the openning post. Not need to waste time looking elsewhere.
    Again, look at the openning post where the inequality is proven by brute force calculation of all the possibilities. Is it your claim that the inequality is different from Bell's? Just because it was proven differently? So your so called "vital assumption" may be interesting for another thread not this one.

    Again it is you who is failing to make the great leap of logic: We have the proof of the inequalities for three coins labelled "a", "b", "c", not angles. You do not argue that the inequality which is proven in the openning post is invalid. Neither do you argue that the inequality is not Bell's inequality. Then we have an example using three coins which violates the inequality just proven. The issue for discussion is: How come? Everything required for answering the question is contained in the openning post. You do not need any reference to QM predictions because there is none for the specific example with coins mentioned in the OP. We can go off trail discussing how "angles" are missing from the example, or how distant separation is missing, but nobody expects coins to be tossed at angles etc.

    So again here are the issues for this thread.
    1) Are the inequalities proven in the openning post valid for the situation described?
    2) How come it is violated by the experiment described?
    3) What does this mean about the applicability of inequalities of the type proven in the OP, to experiments of the type described in the OP?
  21. May 18, 2012 #20
    Hi DrC,
    Thanks for correcting the typo in your quote.
    Actually, it is perfect anti-correlation since the product of the two coins is always -1. Perfect correlation would be a product of +1 every time.

    If the question was changed to "does a perfectly anti-correlated outcome violate a Bell Inequality?", the is YES. To verify replace each of the expectation values with a -1 and you get 3 on the LHS.

    Again as I explained to DK, focusing on angles just muddies the water and takes us off topic. There are no angles involved in the example. "a", "b", "c" are just coins. In case you are tempted to think angles are relevant for "Bell's inequalities" as opposed to "Bell's theorem", remember than "a", "b", "c" in the mathematical derivation performed by Bell are just symbols. The derivation will proceed in exactly the same way to obtain the inequalities, no matter what those symbols represented. The algebra used to derive the inequalities makes absolutely no use of the idea that values for "a", "b", "c" can be changed, which means the inequalities are valid even if "a", "b", "c" are fixed. In our case "a", "b", "c" are just labels for three coins. The fact that the inequalities are valid even for the case in which "a", "b", "c" represent three coins is proven in the openning post. Just in case this is not clear still. Imagine that in the actual EPR experiment, the angles were fixed to just those in which a violation occured, and Alice and Bob were not allowed to change them. Would you argue in that instance that the inequality is not valid because they were not allowed to change the angles? The simple fact that a violation occured whether the angles could be changed or not is significant. That is the scenario you should be comparing this example with, if any.

    And what may those predictions of QM for a coin-toss experiment be? You are comparing apples and oranges here. No claim is made that this example reproduces QM results. The issue is the inequalities, period.
    Now that is an interesting claim. You think the example as described in the openning post is non-local? Please elaborate.

    But the coins are produced that way by the box due to conservation of it's mechanism of functioning.

    Really? At any one moment in time, each of the three coins has a definite outcome value. Prior to the tossing of the coins so it is not contextual. Unless you have a different meaning for the word.

    I did not tell you what b was in that case but of course you understand that it could be easily fixed and it will be realistic and still violate the inequality.

    So let us see:

    - Every outcome is predetermined
    - Every outcome is known for certain before the toss
    - Every coin has a definite state before the toss
    - The box and coins operate in a local and realistic manner.
    - The outcomes are non-contextual (not obeserver dependent) -- there is nothing the observers can do to change the results except refuse to toss.

    Yet the inequality is violated. Why?
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