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B Preserving local realism in the EPR experiment

  1. Dec 26, 2017 #1
    What is wrong with the idea that the spooky correlation in the EPR experiment is simply the result of the initial difference in rotation between the two polarizers in this experiment? So if you rotate one of the polarizers relative to the other polarizer, that initial act of rotation is what causes the cosine squared correlation. In other words, what we think of a N degree rotation of a polarizer or any object for that manner, does not actually rotate the internals of the object in a naive linear fashion, but with some sort of relativistic time dilation effect corresponding to the mathematics and result of the experiment.

    I certainly find this explanation more plausible than others that I have read about and so what is wrong with it?
     
    Last edited: Dec 26, 2017
  2. jcsd
  3. Dec 26, 2017 #2

    nomadreid

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  4. Dec 26, 2017 #3
    Thanks for the response. I did an initial read of the link you posted, but either I don't think I am asking the same question or I am having trouble correlating the answer to my question.

    Let me clarify my experience with EPR, I have done many computer simulations in my naive attempt to simulate reality and get the cosine squared behavior that quantum mechanics predicts. I understand pretty well what a failed result looks like. I understand that I have to run the simulation for any combination of polarizer angles. I also know how to cheat in the simulation and get the simulation to be the same as quantum mechanics by choosing a random vector for one of the polarizers, but for the other polarizer, choosing a rotation offset relative to the first polarizer vector that is cosine squared of the difference between the angles between the polarizers. This is obviously considered cheating because one polarizer can't know about the other polarizer. However it is exactly this cheating scenario I am considering in this post. I attached a snippet of code, showing the basic algorithm used and I could probably provide the entirety of the code if it is really necessary, but assuming it is accepted that this cheat works, than it is unnecessary.

    Let me attempt to emphasize and clarify what I am asking. I am asking specifically about that the act of rotation of the polarizer, before the experiment is even run or at least before an entangled photon used in the experiment even makes it to the polarizer.

    So here is how I am thinking about rotating a polarizer relative to another polarizer. Please tell me where my thinking is wrong or implausible. It takes a force to rotate an object. That force acts at the speed of light and the state inside the object are subject to internal relativistic speeds. If you were to rotate an object by say 1 degree relative to another object of the same state and orientation, you would never be able to make the states between those objects differ by 1% because the state change caused by the force has to propagate and that propagation is a never ending change subject to relativistic time dilation. So what appears to be a 1% rotation at a macroscopic level does not equate to a 1% change in state. Is this a plausible explation for EPR? Where is my thinking misguided?

    Please don't moderate this post for espousing theories. I am trying to stay clear of saying anything misleading, but it is hard to get help without explaining my thought process.


    Code (Text):
    void RunMethod32(out bool leftUp, out bool rightUp)
        {
            float dotProduct;
            float probability;
            Quaternion rotation;
            Vector3 v;

            leftUp = rightUp = false;

            v = this.GetRandomVector(true);

            // Determine Left side
            dotProduct = Vector3.Dot(v, leftParticle.spinDirection);
            probability = Mathf.Pow(dotProduct, 2f);
            if (probability > 0.5f) {
                leftUp = true;
            }

            // Determine right side
            float rotationAngle = Mathf.Pow(Mathf.Sin((rightPolarizer.angle - leftPolarizer.angle) * Mathf.Deg2Rad), 2f) * Mathf.PI / 2f * Mathf.Rad2Deg;
            rotation = Quaternion.AngleAxis(rotationAngle, rightParticle.direction);  // angle is in degrees
            Vector3 newVector = new Vector3();
            newVector = rotation * v;
            newVector.Normalize();

            dotProduct = Vector3.Dot(newVector, rightParticle.spinDirection);
            probability = Mathf.Pow(dotProduct, 2);
            if (probability > 0.5f) {
                rightUp = true;
            }
        }
     
  5. Dec 26, 2017 #4

    PeterDonis

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    Because the difference in angle between the polarizers, if there is one, doesn't have to be an "initial" difference. The direction in which each polarizer points can be chosen just before each measurement is made, in such a way that the events at which the polarizer directions are chosen are spacelike separated--i.e., there is not enough time for a light signal to travel from one to the other. That means no information about either polarizer's setting can get to the other measurement in time to affect its outcome.

    But the polarizers don't have to be rotated in this sense. There are plenty of ways to construct a polarizer whose direction of measurement can be changed without applying a force or rotating anything.

    Anyway, focusing on the internal details of how the polarizers are set is a red herring. See my comment above.
     
  6. Dec 26, 2017 #5

    PeterDonis

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    Quite apart from the specific question in this thread, this is simply false in general. I strongly advise taking some time to learn proper relativistic kinematics from a textbook.
     
  7. Dec 27, 2017 #6

    DrChinese

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    OK, this code might handle cases of so-called perfect correlations properly, hard to know completely what you are doing without comments.

    But it is not local. For local emulation, you cannot reference the left angle setting when calculating the right hand result, and vice versa.

    Please note that the EPR thought experiment and Bell tests are different. It is not possible to write a computer program that reproduces the results of Bell tests.
     
  8. Dec 27, 2017 #7

    vanhees71

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    Of course you can write a computer Monte-Carlo program with precisely the probabilities given by QT, or did I understand something wrong?
     
  9. Dec 27, 2017 #8

    stevendaryl

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    The issue for local realism would be to have three computers connected in a network. Call them A, B, C. Computer C can send messages to A or B, but receives messages from neither. Computers A and B send no messages, but receive messages from C. The programs running on A and B have settings, where a setting amounts to a specification of a direction in space (we can specify this by a pair of real numbers [itex]\theta, \phi[/itex], with [itex]0 \leq \theta \leq \pi[/itex] and [itex]0 \leq \phi \lt 2\pi[/itex]).

    We have a number of rounds, where each round consists of the following steps:
    1. C generates a pair of messages, [itex]m_a[/itex] and [itex]m_b[/itex], using whatever means (random or not). (These represent the states of the electron and positron, respectively)
    2. C sends [itex]m_a[/itex] to computer A, and sends [itex]m_b[/itex] to computer B.
    3. After the message [itex]m_a[/itex] is sent, but before it is processed by A, a user chooses a setting for that computer: [itex]\theta_a, \phi_a[/itex]. (These represent the Stern-Gerlach spin measurement orientation)
    4. Then, computer A runs an algorithm [itex]F_A(m_a, \theta_a, \phi_a)[/itex] to determine a result, [itex]R_a[/itex], either +1 or -1. (This represents the process whereby the particle interacts with the Stern-Gerlach device to produce an outcome of spin-up or spin-down)
    5. Similarly, a setting is chosen for computer B, and it computes a result [itex]F_B(m_b, \theta_b, \phi_b)[/itex] (The result, [itex]R_b[/itex] is again, either +1 or -1)
    This simulation would count as a local realistic simulation of EPR if, regardless of how the settings are chosen at step 3 (except that there would need to be enough variety to get good sampling), for any fixed [itex]\theta_a, \theta_b, \phi_a, \phi_b[/itex], we would have:

    [itex]\langle R_a R_b \rangle = -cos(\psi)[/itex]

    where [itex]\psi[/itex] is the angle between the direction specified for computers A and B and where [itex]\langle \rangle[/itex] is the average value, over many runs.

    Bell's theorem implies that no matter what algorithms are being used by computers A, B or C, you can't reproduce the predictions of QM. There are various loopholes or ways to cheat, though, including:
    1. If A and B are allowed to communicate.
    2. If the settings are chosen predictably (and the pattern is used in the design of the program running on C)
    3. If the settings are chosen before C sends its messages, and C is told the choices.
     
  10. Dec 27, 2017 #9

    vanhees71

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    Hm, this I don't understand, because I could program the computer with a Monte-Carlo algorithm that precisely simulates the probabilities predicted by QT. Then I should be able to use the so produced simulated "data" as I would use the data of a true experiment with photons (or whatever is used to realize the Bell states). I can as well as with the true experimental data choose the partial ensemble to postselect whatever I like also with the simulated "fake data", or am I again missing your point?

    As I understand it the point is that with a local deterministic theory I don't get the statistics leading to the violation of Bell's inequality, which is predicted by QT. If I precisely use the probabilities to Monte-Carlo simulate the quantum statistics, I should find the violation of Bell's inequality. Of course, this proves nothing, because I program the computer such as to get the result predicted by QT.
     
  11. Dec 27, 2017 #10

    DrChinese

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    Not for entangled cases. You don't even need to write code, because you can't even hand pick values that work like QT.

    Of course, that assumes you are writing something where the observer's choice of measurement settings are not known in advance.
     
  12. Dec 27, 2017 #11

    Demystifier

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    Of course it's possible. The catch is that spacelike separated (i.e. ##(\Delta x)^2-c^2(\Delta t)^2>0)## events in the real world are represented by computer memory states which may not be spacelike separated in the computer. But if you want computer to simulate QM in real time and space, then it's impossible.
     
    Last edited: Dec 27, 2017
  13. Dec 27, 2017 #12

    stevendaryl

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    But there are three different "actors" involved in EPR: the two detectors and the source of correlated twin pairs. If you simulate all three on the same computer, then of course, you can reproduce the predictions of QM. But if the detectors and the source are simulated on different computers, then you can't. Not without either communication among the computers or "superdeterminism" (the settings of the detectors is fixed ahead of time and known to the source).
     
  14. Dec 27, 2017 #13

    stevendaryl

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    Well, the timing is not really the issue. The issue is communication. You have three actors--a source of twin-pairs, and two detectors. If the only communication allowed is one-way from the source to the detectors then there is no way to reproduce the predictions of QM. Well, there is actually the "superdeterminism" loophole--if the settings of the detectors are predictable ahead of time, you can violate Bell's inequality.
     
  15. Dec 27, 2017 #14
    I can now see my efforts are misguided. I guess I want to believe in local reality so badly that I go down the rabbit hole every so often.

    What helps me, is to think of the hypothetical situation of entangling 3 photons and sending them to 3 different polarizers. I can only make 2 of the polarizer results agree with the actual experimental result and I can't make all 3 agree, no matter what.
     
  16. Dec 27, 2017 #15

    DrChinese

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    That rabbit hole contains a lot of rabbits. :smile: You were quick to escape though. And you are correct about the 2 vs. 3 settings. It takes that 3rd to really see what's going on.
     
  17. Dec 27, 2017 #16

    Mentz114

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    You don't actually need any electronic computers. One or more people could do it with some slips of paper and maybe a coin to toss. It could make an interesting board game.
     
  18. Dec 27, 2017 #17

    vanhees71

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    Let's take the usual polarization measurement. All I do is preparing an ensemble of photons represented by the Bell state (e.g., the symmetric one ##|\Psi \rangle=1/\sqrt{2}(|HV \rangle+|VH \langle)## and then measure for each pair in the ensemble whether the one photon comes through one and the other through another polarization filter. The probabilities are given by QT, and this I simulate with a Monte-Carlo code. Then I should get all results of QT, among them the violation of Bell's inequality, which is a, seen from a purely mathematical perspective simply a feature of the probability function. I just make a measurement protocol about the simulated outcomes, and this I cannot distinguish from a measurement protocol in a real experiment. So I can simulate, of course, the measurement with a computer, but that's trivial since I precisely wrote my MC code to do so. I don't see, what this should tell me about the interpretational issues of QT at all. The only point is that according to all observations in real experiments with such prepared photon pairs agree with QT, and this I simulate is a MC code, but that has no significance (although perhaps for some pedagogical purpose to have a nice exercise in programming a MC simulation).
     
  19. Dec 27, 2017 #18

    vanhees71

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    All I need to do is to use the probabilities given by QT. Why shouldn't I be able to program this with a standard MC code? Perhaps I misunderstand which kind of code you guys have in mind, but all there is in both QT and in the real-world measurement are the probabilities (in experiment of course the relative frequencies for the outcomes in an ensemble).
     
  20. Dec 27, 2017 #19
    There is one case that I can think of where the hypothetical 3 entangled experiment does not help me.

    What if the act of entanglement is unbalanced, but complimentary? For example one of the entangled photons has a strength of 90% in one direction, but the other photon has a strength of 10% in the other direction. Where the unbalance of entanglement changes with time in either a random or predictable pattern. Note, the complimentary nature of entanglement keeps the result the same or opposite when the polarizers are aligned or 90 degrees apart.

    If entanglement was unbalanced, it would invalidate the hypothetical experiment as being meaningful, because you would never be sending the same or opposite photon to the different polarizers.

    Is it possible entanglement is unbalanced, but complimentary or does something rule out this possibility?

    Assuming this type of entanglement is possible, is this just another rabbit hole?

    Thanks.
     
  21. Dec 27, 2017 #20

    DrChinese

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    Not really sure where you are going with your comments. Bell already demonstrated that QM's predictions are incompatible with separable algorithms of any type.
     
  22. Dec 27, 2017 #21

    DrChinese

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    You know it doesn't vary with time already. Otherwise there would not be perfect correlations unless the timing of observations was synchronized.
     
    Last edited: Dec 27, 2017
  23. Dec 27, 2017 #22
    Probably superfluous, but for completeness, you can of course simulate entanglement experiments on a computer. It is only when the speed of light is taken into account, such that the measurement devices don't communicate their angles (which can easily be circumvented in computer code), that the Bell inequalities come into play.
     
  24. Dec 27, 2017 #23

    zonde

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    Yes, if you model entanglement with asymmetric hidden variables impossibility 3-way entanglement proves nothing. But have you seen this informal proof of Bell inequality?
     
  25. Dec 27, 2017 #24

    zonde

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    QT gives the probabilities using non-local description. And of course you can reproduce QT predictions with non-local model.
     
  26. Dec 27, 2017 #25

    DrChinese

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    As mentioned regarding computer simulation: you cannot simulate entanglement if Alice doesn't know Bob's choice of measurement setting, and Bob doesn't know Alice's choice of measurement setting. So yes, if you circumvent that, you cheat.
     
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