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Basic Bell

  1. Mar 4, 2008 #1
    The other thread here made me read up a bit on Bell, EPR, Aspect's and so on..

    An important part is ofcourse the assumptions discussed intensively in the other thread. vanesch expressed my view about this:
    Why isn't it obvious here that the mistaken assumption is within quantum theory? Is the theory so solid and consistent that core scientific principles like logic, locality and objectivity is up for grabs? What am I missing here?

    When testing Bell one measures seperate properties A, B and C, in which case the inequality most hold true. Personally I consider polarization a classical phenomenon, so I don't really see how measuring the polarization of a lightwave at different angles could be considered three seperate properties? Clearly, if a photon has polarization α and it goes through a polarizer at angle α-45, it's new amplitude will be cos(45), and if property A = α, property B = A * cos(45). Which is not seperate.

    So my point here is, couldn't it be the quantum description of polarization that is the inaccurate assumption? As I understand it, it's based on chance, right? A photon in the above setup has a cos(45) percent chance of passing through? Correct me if I'm wrong...
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  3. Mar 4, 2008 #2

    Jonathan Scott

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    Whether quantum theory is correct or not, experiments by Aspect and others have confirmed that the experimental results (involving large enough numbers of results to be statistically very significant) are consistent with the predictions of quantum theory and violate Bell's inequality.

    The earlier experiments had some loopholes, for example relating to low detection rates and the possibility of light-speed signals, so the violation of Bell's inequality was not directly observed but rather inferred via a theoretical model. However, later experiments involved such tricks as detectors which were being dynamically switched between different orientations so fast that light-speed signals could not explain the results. There is still some debate about whether all possible experimental loopholes have been closed, but it now seems more plausible that quantum theory is correct, however weird it may seem, than the alternative that nature is somehow capable of exploiting different loopholes in different experiments to simulate the quantum result in every case.
  4. Mar 4, 2008 #3
    I'm not debating the experimental results, loopholes or claiming any silly realist theory about light-speed signals being transferred.

    My inquery was about the nature of polarization, and whether or not the ciriteria in Bell's inequalities of three seperate properties A, B & C are actually met.

    Am I correct in saying that, in quantum theory, polarization is governed by chance, and in classical theory, it is not?
  5. Mar 5, 2008 #4


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    There is no actual evidence that a particle - such as a photon - has definite values for specific properties independent of the act of observation. So no, I would say that a particle does NOT have a definite polarization at angles A, B and C independently of a measurement. But it is possible to assume the counter-position if you are willing to also assume superluminal causation.

    As to chance: there is no evidence that there is a specific underlying "cause" of polarization (and thus quantum theory is already complete). In classical theories, there is a specific cause but we don't know enough about the initial conditions to know what it is.
  6. Mar 5, 2008 #5


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    Each individual photon has a cos^2 (45) percent chance of passing through (this is just like Malus' law except with the intensity of a classical wave passing through a polarizer replaced by the probability of a quantum particle passing through the polarizer), but remember that you're measuring the respective polarizations of a pair of entangled photons with opposite polarization. Because their polarizations are opposite, then if the two polarizers are set to the same angle, the photons are guaranteed to yield opposite results with probability 1--if the first one goes through the second doesn't, and vice versa. Bell's proof starts from the assumption that the results when the angle is the same are perfectly correlated in this way, and then shows that, under local realism, this would imply certain conclusions about the probabilities when different angles are chosen for the two photons, conclusions which are violated in QM. Note that in local realism, the only way to explain this perfect correlation when the angle is the same would be to assume that the two photons were created in such a way that they had predetermined answers to whether they would go through a polarizer at a given angle--if there is any randomness, it can only be at the moment of their creation when these predetermined answers are assigned (if there was any randomness when they actually reached the polarizer, then under local realism there'd be no way to explain why they always give opposite results).
    Last edited: Mar 6, 2008
  7. Mar 7, 2008 #6
    Thanks for the answers, and sorry for my delay..

    Hmm, ok. I thought polarization of light was one of those simple classical phenomenas that was well understood. Check
    http://en.wikipedia.org/wiki/Polarization" [Broken] for example. I feel stupid saying this but isn't the polarization of the emitted wave just parallell to the oscillation of the charge? Or what did you mean by "specific cause"?

    I don't see how this can be correct. If the polarizers are aligned at angle α, and a photon pair happens to have polarization α-45 and α+45, each photon should have a 50% chance of passing through, right? It's only the specific case when one photon equals α where the probability of opposite results is 1.

    I understand both arguments of QM and local realism with hidden variables. But I disagree with both actually, and my problem is with the chance-governed polarizers. Could someone explain to me why the Bell test experiments can't be explained simply with a classical interpretation of polarization?

    If polarizers aren't governed by chance, then correlation at different angles is expected.
    Last edited by a moderator: May 3, 2017
  8. Mar 7, 2008 #7


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    No, the photons are entangled, so you can't have a separate wavefunction for each one; you must use a combined wavefunction for the two-photon system which assigns probabilities to different combinations of outcomes. In the case where both photons have their spin measured on the same axis, this two-photon wavefunction assigns zero probability to combinations of outcomes where they fail to show opposite spins. Entanglement is weird, no doubt!
    Classical polarization gives you a deterministic prediction about the intensity of light passing through a polarizer, not a probability that a photon will make it through or not. For a single photon the quantum probability has the same value as the reduction in intensity, and for an individual member of an entangled pair this probability is unchanged, but you have the strange phenomenon of entanglement where you always find a perfect correlation between the outcomes for each member of the pair when the two experimenters set the polarizers to the same angle. Obviously there isn't really any good analogy for this sort of thing in classical physics. A local realist theory could still explain the perfect correlation in terms of each photon being assigned a predetermined answer to whether they'll pass through the polarizer at the moment they're created, but this would lead to certain predictions about the statistics you should see when you measure them with the polarizers set to different angles (the Bell inequalities), and these predictions are violated in QM. If you're not familiar with the logic, take a look at the lotto card analogy I gave in post #3 of this thread.
  9. Mar 8, 2008 #8
    Ok, let me try and give you a scenario that isn't weird :)

    The source emits 2 lightwaves with opposite polarization in opposite directions. The lightwaves are classical and deterministic, with actual values. When a lightwave hits a polarizer it always goes through, but with an amplitude that is cos(angle diff) of the emitted one. Now, the detector has a cut-off value when the post-polarizer amplitude is too low to be detected. As long as this value is higher than cos(45) (half the intensity of the lightwave), there will be a perfect correlation of opposite detections.

    Let's take an example. I know from reading about the experiments that there have been issues with the detector efficency, but for illustration purposes let's say that the detector is able to register lightwaves that only have an amplitude cos(44) of the emitted one.

    Following are pairs detected when polarizers are aligned:

    pair# | A° | B° | detection

    01. | -00,0° | 90.0° | 1,0
    02. | -22.5° | 67.5° | 1,0
    03. | -35.0° | 55.0° | 1,0 --
    04. | -45.0° | 45.0° | 0,0
    05. | -55.0° | 35.0° | 0,1 --
    06. | -67.5° | 22.5° | 0,1
    07. | -90.0° | 00.0° | 0,1

    These are emitted randomly ofcourse.

    If we instead have a detector that registers lightwaves with a post-polarizer amplitude of cos(30) or higher, detection of pairs 03 & 05 would also be 0,0 (i.e not detected).

    Classical wave theory. Aligned polarizers. Perfect correlation.
  10. Mar 8, 2008 #9

    Doc Al

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    Your polarizers are all 90.0° apart! Not a particularly interesting case.
  11. Mar 8, 2008 #10
    The polarizers are aligned. The angle values is the lightwave polarization.
  12. Mar 8, 2008 #11

    Doc Al

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    Oops--I misread your post. I'll rephrase my response:
    Your polarizers are aligned! Not a particularly interesting case.
  13. Mar 8, 2008 #12
    Hehe, Oh but I disagree. It is NOT interesting if you want a distinction between QM and hidden variables. It IS interesting if you're looking for a different interpretation. And as I see it, there's two options:

    1. Polarizers are governed by chance. Photons are entangeled. Bell's inequality is violated. Fundamental principle of locality tossed out.

    2. Polarizers are NOT governed by chance. Classic waves, no entanglement. Bell's inequalities does not apply. Locality holds.

    1 & 2 give same results. But as you can see there's a fundamental difference.
  14. Mar 8, 2008 #13

    Doc Al

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    Only because you chose a trivial case of aligned polarizers. Try something non-trivial, like having the polarizers set at various angles (0°, 120°, 240°), then try to explain the results using a simple model of pre-existing polarization. Read Mermin's classic article from Physics Today: http://xoomer.alice.it/baldazzi69/papers/mermin_moon.pdf" Here it is explained by our own Dr C: "[URL [Broken] Theorem with Easy Math
    Last edited by a moderator: May 3, 2017
  15. Mar 8, 2008 #14


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    It's easy to find classical situations where the two experimenters always get opposite answers when they pick the same angle, there is nothing strange about this in itself. The trouble is that in these classical situations, the Bell inequalities will not be violated--did you read my analogy with the lotto cards? Consider a situation where each experimenter is choosing between three possible angles 0°, 60°, and 120°. In quantum mechanics the probability of getting opposite answers for two entangled particles when the difference between the angles of the two detectors is theta would be cos^2 (theta). So if the difference between the two detectors is 0, the probability of getting opposite answers is cos^2 (0) = 1; if the difference between the two detectors is 60, the probability of getting opposite answers is cos^2 (60) = 0.25; and if the difference is 120, the probability of getting opposite answers is cos^2 (120) = 0.25. So if experimenters Alice and Bob perform many trials with a random choice of setting on each trial, and we look at the subset where they chose different detector settings, all of the following cases will be equally frequent:

    (Alice chose 0, Bob chose 60) = 0.25 probability of opposite results
    (Alice chose 0, Bob chose 120) = 0.25 probability of opposite results
    (Alice chose 60, Bob chose 0) = 0.25 probability of opposite results
    (Alice chose 60, Bob chose 120) = 0.25 probability of opposite results
    (Alice chose 120, Bob chose 0) = 0.25 probability of opposite results
    (Alice chose 120, Bob chose 60) = 0.25 probability of opposite results

    So, if these are equally frequent, the total probability of opposite results on the subset of trials where they chose different settings is (0.25 + 0.25 + 0.25 + 0.25 + 0.25 + 0.25)/6 = 0.25. As I explained in the lotto card analogy, if local realism is true, then one of the Bell inequalities says that if the experimenters always get opposite results when they choose the same setting, then when they choose different settings they must get opposite results at least 1/3 of the time. But here when they choose different settings they only get opposite results 1/4 of the time. There is no way you can replicate this classically.

    For example, you can look at the different cases in your example from post #8 to find what will happen if Alice and Bob are setting their polarizers to 0, 60 and 120. In your case 01, the polarization of the two light waves is 0 and 90. For the light wave that's 0 (say this is the one Alice receives), the detector will go off when the polarizer is set to 0, but not 60 or 120. For the wave that's 90 (say this is the one Bob receives), the detector will go off when the polarizer is set to 60 or 120, but not when it's set to 0. So in the subset of trials where they chose different detector settings, the cases would be:

    (Alice chose 0, Bob chose 60) = 0 probability of opposite results (both detectors go off)
    (Alice chose 0, Bob chose 120) = 0 probability of opposite results (both detectors go off)
    (Alice chose 60, Bob chose 0) = 0 probability of opposite results (neither detector goes off)
    (Alice chose 60, Bob chose 120) = 1 probability of opposite results (Alice's detector doesn't go off, Bob's does)
    (Alice chose 120, Bob chose 0) = 0 probability of opposite results (neither detector goes off)
    (Alice chose 120, Bob chose 60) = 1 probability of opposite results (Alice's detector doesn't go off, Bob's does)

    So in this case, the total probability of opposite results is (0 + 0 + 0 + 1 + 0 + 1)/6 = 1/3, which is not a violation of the Bell inequality. You can try this for other possible angles for the waves besides 0 and 90 (your cases #02 - #07) and you'll find the same thing.
    Last edited: Mar 8, 2008
  16. Mar 9, 2008 #15
    I've put together a non-trivial javascript application based on my idea:
    http://pf.soderholmdesign.com/bell.html [Broken]

    From Mermin's paper:
    So in my app, if I set Alice = Bob, RG/GR, Photon:Same & Angles:0,120,240 ... I get 0 pairs in my subset. i.e. if A & B have the same setting, they will never have opposite results.

    Now we set Alice != Bob and RR/GG. The generated result with 50,000 pairs is:
    -------- 11246/50000 (0.22492)

    And for JesseM's example: ###OBS. Above pairs have same polarization, below have opposite###
    We set Alice=Bob, RR/GG, Photon:Opposite, Angles:0/60/120. In this case we find that, with a cos(40) detector sensibility, there is a ≈96% probability of opposite results. However, the other ≈4% are all non-detections.
    -------- 361/10000 (0.0361)
    -------- 100% Non-detections in subset (RR)

    In this case the number of non-detections is very dependent on the detector cut-off value. Lower than cos(45) actually turns non-detections to all-detections (GG). The probability of opposite answers approaches 1 at cos(45), and 2/3 at cos(0) and cos(90). Haven't thought about why.

    Set Alice != Bob, RG/GR. I get subset:
    -------- 11123/50000 (0.22246)

    Which is basically the same result as with the Mermin setup. (good sign =))

    As I interpret the result, it is a clear violation of Bell's inequalities, even bigger than with QM. Also note that, when they choose different angles, the correlation isn't that dependent on the detector cut-off value. Between cos(20) and cos(70) it stays around 0.22. With low or high detection sensability, the value goes down even more.
    Last edited by a moderator: May 3, 2017
  17. Mar 9, 2008 #16


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    Ah, I didn't realize that photon polarization worked the opposite way from the spin of particles like electrons--entangled electrons always have opposite spin when measured along the same direction.
    I don't quite follow your terminology--what does "RG/GR" mean? Are you assuming all the light waves are polarized at the same angle (say, 0 degrees) or does the source emit pairs of waves at a variety of angles? I take it you're still using the rule that, if the difference between the wave's angle and their detector's angle is theta, then their detector will go off if cos(theta) is greater than some critical value (this was cos(45) in your original example, but your applet allows you to adjust it), but it won't go off if cos(theta) is less than this value? If so, what value were you using for the numbers above? Finally, do you assume in this case that for all possible combinations of angles where Alice = Bob (i.e. [Alice=0, Bob=0], [Alice=120, Bob=120], and [Alice=240, Bob=240]), each of these combinations occurs with equal frequency?
    ...and could you answer the analogous questions here too? For example, do you assume that [Alice=0, Bob=120], [Alice=0, Bob=240], [Alice=120, Bob=0], [Alice=120, Bob=240], [Alice=240, Bob=0], and [Alice=240, Bob=120] all occur with equal frequency here?
    Last edited by a moderator: May 3, 2017
  18. Mar 9, 2008 #17
    The R is red (no detection) and G is green. Used in Mermin's paper.

    In the example from Mermin's (0,120,240), the source emits photons with random but equal polarization. In the example we talked about (0,60,120), the source emits photons with random polarization but 90° difference.

    I'm still using that rule, yes. I used cos(40) for my results.

    My functions generates pseudo-random angles for each photon and pair:

    function getRandomPhotonPair() {
    var A = Math.floor(Math.random() * 361);
    var B = opposite == 1 ? (A + 90) % 360 : A;
    return new Array(A,B);

    function getRandomPolarizerAngle() {
    var rand = Math.ceil(Math.random() * 3); <-- random angle 1, 2 or 3.
    return rand;

    If for some reason the angles aren't equally distibuted, it's due to Math.random().

    If you want I can add a counter for each angle at Alice & Bob?
  19. Mar 9, 2008 #18
    Added the count. Random function seems to work fine.

    -------- 10884/50000 (0.21768)
    -------- 66.59% Non-detections in subset (RR)
    -------- 32.77% Total Non-detections (RR)
    -------- Angle distribution: 11:5561, 12:5460, 13:5600, 21:5436, 22:5610, 23:5718, 31:5454, 32:5604, 33:5557

    Alice != Bob
  20. Mar 9, 2008 #19


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    Jesse, you really aren't wrong about that. It depends on the how the photons are generated. They can be made to be either both the same or different depending on whether they are PDC Type I or PDC Type II. So really it only matters that you say which you are assuming. Besides, you can always rotate to get the desired results, and you can also run through 1/2 wave plates etc.
  21. Mar 9, 2008 #20


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    I think you missed generating the third polarization. You want to generate for A, B and C and you need to generate a Y or N for each. Then you randomly pick 2 of the 3 of A, B or C and determine whether they match. You can check against this: Mermin's version simplified
  22. Mar 9, 2008 #21
    Sorry, it's just a misleading choice of variable names. The A and B in the getRandomPhotonPair() is the random polarization of the two photons. In getRandomPolarizerAngle() I generate a random number 1,2 or 3, that assigns angle 0, 120 or 240 (or which other option you've chosen).
  23. Mar 9, 2008 #22


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    Right, but what does RG/GR mean? In Mermin's paper each trial is just associated with a single red or green outcome for each experimenter--for example, 31RG would mean Alice chose setting 3 and got result R, while Bob chose setting 1 and got G. I don't understand what RG/GR means in this context, or why it's associated with Alice and Bob choosing the same detector settings while RR/GG is associated with them choosing different detector settings.
    Ok, let's stick with Mermin's example, since I'm not sure if there's any way for a source to emit entangled photons with a 90-degree difference (again, I was generalizing from the fact that entangled electrons give opposite results when measured on the same angle to the false notion that entangled photons would do the same) (edit: now I see from DrChinese's post it is possible, but let's stick with Mermin's example in any case), and cos^2 of 120 and 240 gives 0.25 just like cos^2 of 60 and 120 in my example.

    Anyway, it's really not to complicated to prove that if they always give the same result when measured at the same angle, then they must give the same result when measured on different angles at least 1/3 of the time, so I think there must be an error in your program--maybe if you add some lines to make it "show its work" in certain ways (I'll suggest ways below) we can find the error. Consider--for any angle the program chooses for the light waves on single trial, that angle alone must be enough to predetermine what answer an experimenter will get if they measure on any one of the three angles. If we label the angles by A=0, B=120 and C=240, then if the wave's angle is, say, 50, that predetermines the fact that if the experimenter chooses A they'll get R (since cos^2 (50 - 0) < cos^2 (40)), if the experimenter chooses B they'll get R (since cos^2 (120 - 50) < cos^2 (40)), and if the experimenter chooses C they'll get G (since cos^2 (240 - 50) > cos^2 (40)...though cos(240 - 50) < cos(40) since cos(240 - 50) is negative, were you calculating the threshold based on cos or cos^2? It doesn't really matter for the purposes of trying to find a counterexample to Bell's theorem, but cos^2 is more physically realistic since it appears that way in Malus' Law). The point is that on each trial, you can label the angle of the wave based on what results it would be predetermined to give for each detector setting--on one trial you might have the wave (A-R, B-R, C-G) as above, on another trial you might have (A-G, B-R, C-G), and so forth.

    Label such a predetermined state "homogeneous" if it would give the same answer for each setting, i.e. (A-G, B-G, C-G) or (A-R, B-R, C-R), and "inhomogeneous" if it would give two answers of one type and one of the other, like (A-G, B-R, C-G). For any homogeneous state, it's obvious that when Alice and Bob choose different settings, they are guaranteed to get the same result with probability 1. For an "inhomogeneous" state, there is a probability of 1/3 that Alice picks the setting that gives the "rare" answer (so if the state gives a green for A and C but a red for B, she picks B and gets red), and if we're looking only at trials where Alice and Bob pick different settings, then Bob is guaranteed with probability 1 to pick a setting that gives the "common" answer (he picks A or C which both give green), so in this case they get opposite results. There is also a probability of 2/3 that Alice picks a setting that gives the "common" answer; in this case of the two remaining settings, there is a 1/2 probability that Bob picks the one that also gives the "common" answer so they get the same result, and a 1/2 probability that Bob picks the one that gives the "rare" answer and they get opposite results. So, for an "inhomogeneous" state the probability that they get opposite results must be (probability Alice picks rare, Bob picks common) + (probability Alice picks common, Bob picks rare) = (1/3)*(1) + (2/3)*(1/2) = 1/3 + 1/3 = 2/3. Thus, for inhomogeneous states the probability they get the same result is 1/3. And we already found that for homogeneous states the probability that they get the same result was 1, so this tells us that whatever the mixture of homogenous and inhomogeneous states produced by the source, they must lead Alice and Bob to get the same result at least 1/3 of the time.

    Look over this proof and see if you can find any reason it wouldn't apply to your example; if you agree it should, then you should agree there must be something wrong with the program if you got the same result less than 1/3 of the time when you look at the subset of trials where Alice and Bob chose different settings. For troubleshooting, I'd suggest having the program "show its work" in the following ways:

    1. For each trial, have the program calculate the result that the wave would give for each of the three possible detector angles A=0, B=120, C=240, and over the course of the experiment have it count how many trials fall into each possible category below, displaying the number for each at the end:

    (A-G, B-G, C-G)
    (A-G, B-G, C-R)
    (A-G, B-R, C-G)
    (A-G, B-R, C-R)
    (A-R, B-G, C-G)
    (A-R, B-G, C-R)
    (A-R, B-R, C-G)
    (A-R, B-R, C-R)

    2. For each trial, have the program take note of whether Alice and Bob chose different settings or the same setting, and have it display the total number for each at the end.

    3. For the subset of trials where they chose different settings, have the program keep a separate counter for how many trials in this subset had each of the eight possible predetermined answers above, and whether Alice and Bob got the same result (S) or different results (D) given their choice of settings, and have it display the number of trials in this subset that fall into each of these 16 categories:

    (A-G, B-G, C-G) - S
    (A-G, B-G, C-G) - D
    (A-G, B-G, C-R) - S
    (A-G, B-G, C-R) - D
    (A-G, B-R, C-G) - S
    (A-G, B-R, C-G) - D
    (A-G, B-R, C-R) - S
    (A-G, B-R, C-R) - D
    (A-R, B-G, C-G) - S
    (A-R, B-G, C-G) - D
    (A-R, B-G, C-R) - S
    (A-R, B-G, C-R) - D
    (A-R, B-R, C-G) - S
    (A-R, B-R, C-G) - D
    (A-R, B-R, C-R) - S
    (A-R, B-R, C-R) - D

    And of course, if you agree something is probably wrong with the program but don't like my troubleshooting suggestions you could just try to find the problem in your own way.
  24. Mar 9, 2008 #23


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    Actually, a possibly simpler troubleshooting method would be to run a more limited number of trials--say, 50 or 100--and have the program print out the complete information for each trial: the wave angle, the set of predetermined answers for the wave given that angle, and the detector settings chosen by Alice and Bob as well as the program's answer for whether they got the same result or different results on that trial. So, for one trial you might have something like:

    wave angle = 52; (A-R, B-R, C-G); Alice-B, Bob-C; Different

    If we then had a list of 50 or 100 trials like this, we could look and see if the program was consistently calculating a same/different answer consistent with what should be implied by the predetermined answers for that trial, as well as whether the predetermined answers were correct given the angle it chose.
  25. Mar 9, 2008 #24


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    It isn't clear that you had pre-assigned values of Y or N for each of the 3 settings PRIOR to selecting the particular A and B. You see, the classical (chance) value you get should converge on .333. On the other hand, the quantum value will converge on .25 (cos^2 120 degrees).

    So I guess I am asking: is your routines supposed to emulate the quantum value or the classical (local realistic) value?

    A. Here is how the quantum algorithm should go:

    Select a random number between 0 and 1. If it is less than .25 (cos^2 120 degrees, because that is the delta between any A, B, C), then they agree. If it is greater than .25, then they don't agree. Calculate the percentage after N trials. Obviously, it will approach .25 (total agrees divided by N).

    B. The classical one is:

    Pre-select values for A, B and C according to ANY algorithm you care to choose. Doesn't matter. Then randomly select any pair from A, B, and C. It will never be less than .333.

    C. So if you are getting .22 something, your calculations have gone awry. :)
  26. Mar 9, 2008 #25
    RG/GR means that Alice and Bob get opposite results, either RG or GR. The association isn't my choice, I just followed the example in Mermin.

    Well, that was actually my main point from the start. In a completely classical scenario, the only event governed by chance is the polarization of the emitted photon pairs. Once they hit the polarizers it's not a matter of "probability of passing through". Given a certain photon polarization, there are respective post-polarizer amplitudes at angles 0, 120 and 240. These values are correlated, so Bell's inequalities does not apply to them.

    I will read your post more closely tomorrow. Goin to work in 5 hours so need some sleep =)
    Last edited: Mar 9, 2008
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