B Preserving local realism in the EPR experiment

[kurt101]

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?

 

[nomadreid]

Your question is handled here: https://physics.stackexchange.com/q...-bells-experiment-be-a-linear-function-of-ang

 

[kurt101]

Your question is handled here: https://physics.stackexchange.com/q...-bells-experiment-be-a-linear-function-of-ang

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.

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;
        }
    }

 

[PeterDonis]

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?

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.

It takes a force to rotate an object.

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.

 

[PeterDonis]

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.

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.

 

[DrChinese]

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;
        }
    }

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.

 

[vanhees71]

Of course you can write a computer Monte-Carlo program with precisely the probabilities given by QT, or did I understand something wrong?

 

[stevendaryl]

Of course you can write a computer Monte-Carlo program with precisely the probabilities given by QT, or did I understand something wrong?

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 \theta, \phi, with 0 \leq \theta \leq \pi and 0 \leq \phi \lt 2\pi).

We have a number of rounds, where each round consists of the following steps:

1. C generates a pair of messages, m_a and m_b, using whatever means (random or not). (These represent the states of the electron and positron, respectively)
   
2. C sends m_a to computer A, and sends m_b to computer B.
3. After the message m_a is sent, but before it is processed by A, a user chooses a setting for that computer: \theta_a, \phi_a. (These represent the Stern-Gerlach spin measurement orientation)
   
4. Then, computer A runs an algorithm F_A(m_a, \theta_a, \phi_a) to determine a result, R_a, 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 F_B(m_b, \theta_b, \phi_b) (The result, R_b 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 \theta_a, \theta_b, \phi_a, \phi_b, we would have:

\langle R_a R_b \rangle = -cos(\psi)

where \psi is the angle between the direction specified for computers A and B and where \langle \rangle 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.

 

[vanhees71]

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.

 

[DrChinese]

Of course you can write a computer Monte-Carlo program with precisely the probabilities given by QT, or did I understand something wrong?

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.

 

[Demystifier]

It is not possible to write a computer program that reproduces the results of Bell tests.

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.

 

[stevendaryl]

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.

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).

 

[stevendaryl]

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.

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.

 

[kurt101]

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.

 

[DrChinese]

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.

That rabbit hole contains a lot of rabbits. 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.

 

[Mentz114]

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.

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.

 

[vanhees71]

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).

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).

 

[vanhees71]

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.

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).

 

[kurt101]

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.

 

[DrChinese]

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, ...

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.

 

[DrChinese]

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. ...

You know it doesn't vary with time already. Otherwise there would not be perfect correlations unless the timing of observations was synchronized.

 

[entropy1]

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.

 

[zonde]

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?

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?

 

[zonde]

The probabilities are given by QT, and this I simulate with a Monte-Carlo code.

QT gives the probabilities using non-local description. And of course you can reproduce QT predictions with non-local model.

 

[DrChinese]

... which can easily be circumvented in computer code...

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.

 

[entropy1]

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.

Yes, I mean that it is necessary to cheat to reproduce the experiments.

 

[stevendaryl]

All I need to do is to use the probabilities given by QT.

I was using the spin-1/2 case, but we can use the photon case if you like. So you have Bob in one room with his computer, Alice in another room with her computer, and Charlie in a third room with his computer. Charlie sends two messages, one to Alice and one to Bob, representing a twin pair of entangled photons leaving from some common source. Alice and Bob each independently choose a filter orientation. Alice's algorithm takes the message from Charlie and Alice's orientation choice and returns a number +1 or -1 (to represent passing the filter or being blocked by the filter). Bob's algorithm takes the message from Charlie and Bob's orientation choice and returns a number +1 or -1.

So what do you mean by "use the probabilities given by QT"? QT predicts joint probabilities:

P(A = B|o_A, o_B) = cos^2(\psi) (where \psi is the angle between Alice's and Bob's orientations)

where A is Alice's result, o_A is Alice's orientation choice, B is Bob's result, o_B is Bob's orientation choice.

But Alice's computer doesn't have access to Bob's orientation, and Bob's computer doesn't have access to Alice's orientation. So how can you implement that joint probability?

 

[vanhees71]

Of course, I have to simulate the full thing, i.e., I have to use the quantum probabilities predicted from the entangled state of the full system (whether you use photons, spin-1/2 or whatever else in this gedanken experiment doesn't matter, but I guess it's technically most simple with photons since you can most easily create large ensembles of Bell states of polarization-entangled photon pairs, which I think are available already in undergrad labs at the universities; I guess it's also possible with ultracold neutrons, but that's already more difficult). To get these probabilities of course you have to take into account also the full experimental setup, including the specific (relative) orientation of A's and B's polarizers etc. etc.

Of course, if I create independent random squences for the two single photons, I'll never get the quantum correlations out, but that's also as in a real experiment: In order to have entanglement you have to create the corresponding state and ensure that it's not destroyed by any influence from "the environment" before you perform the actual measurement, i.e., of course sending simply two streams of unpolarized photons around, you'll never find correlations as described by entanglement!

 

[stevendaryl]

Of course, if I create independent random squences for the two single photons, I'll never get the quantum correlations out,

In the setup I described, the source machine, C, creates pairs of messages (representing photons). They aren't independent, they're as correlated as the machine wants to make them. But the detection processes, at machines A and B, are independent.

 

[vanhees71]

Sure, that's as in the real experiment. So I still don't get what you are after in making a computer simulation. If it's simulates the probabilities predicted by QT by input of the code it will produce precisely what QT predicts (provided there's no bug in the code). Isn't this just tautological?

 

[stevendaryl]

Sure, that's as in the real experiment. So I still don't get what you are after in making a computer simulation. If it's simulates the probabilities predicted by QT by input of the code it will produce precisely what QT predicts (provided there's no bug in the code). Isn't this just tautological?

It's not tautological, because it's false. You can't simulate QT this way. That's an immediate consequence of Bell's theorem.

 

[vanhees71]

Ok, I'm not understanding this claim. Let's leave it at that.

 

[entropy1]

Sure, that's as in the real experiment. So I still don't get what you are after in making a computer simulation. If it's simulates the probabilities predicted by QT by input of the code it will produce precisely what QT predicts (provided there's no bug in the code). Isn't this just tautological?

If you manage to correctly simulate the experiment without reference to the relative angle of the detectors, I don't know how you do it!

 

[vanhees71]

Of course, you have to precisly simulate the probabilities given for the assumed experimental setup. Of course, I have to use the relative angle of the detectors to calculate these probabilities, and thus this information goes into the code.

 

[Mentz114]

If you manage to correctly simulate the experiment without reference to the relative angle of the detectors, I don't know how you do it!

Isn't it sufficient to project the second measured photon ( say B) into the angle of A's polarizer if A gets a hit ?
If A's detector does not click ( no transmission) what can we say about B's alignment ?
Also, I'm not sure if the pair are produced with a random alignment or some prepared value.

I'm referring to what goes in the code, of course.

 

[entropy1]

Isn't it sufficient to project the second measured photon ( say B) into the angle of A's polarizer if A gets a hit ?

If A gets a hit, we know its angle. But how does B know if A got a hit?

 

[Mentz114]

If A gets a hit, we know its angle. But how does B know if A got a hit?

B does not need to know that. The paired photon may have changed state after the first measurement. That is what entanglement is supposed to mean.

In the simulation that is the only thing that differs from a random setup.

 

[vanhees71]

Well, in the usual setup the measurement of B doesn't do anything on A. To ensure this you make the measurement acts spacelike separated. That's what's so mind boggling about these stronger-than-classically-possible correlations described by an almost trivial math. As easy is the math as mind boggling are the implications, particularly if you stick to traditional collapse assumptions of the early days! The worrying immediately stops if you simply accept that nature is inherently probabilistic and that there is a very successful formalism called QT that tells you probabilities and only probabilities.

 

[stevendaryl]

Of course, you have to precisly simulate the probabilities given for the assumed experimental setup. Of course, I have to use the relative angle of the detectors to calculate these probabilities, and thus this information goes into the code.

But the relative angle is not known until the last minute. The situation is the following:

You have three devices: C, which is a source of message pairs, simulating photon pair production, and A and B, which simulate the measurement events.

• Every "round", C sends out a pair of messages, m_A to A and m_B to B.
• After the messages are sent, but before they are read, settings for A and B are chosen, independently. The settings are two angles, \theta_A and \theta_B.
• Device A determines an output, R_A(\theta_A), which is either +1 or -1, based on the message received from C and the setting \theta_A.
• Similarly, device B determines an output, R_B(\theta_B) based on its message and setting.
• Over many, many rounds, we can gather statistics for the correlation: \langle R_A(\theta_A) R_B(\theta_B) \rangle as a function of the pair of settings, \theta_A, \theta_B.

Bell's inequality implies that |\langle R_A(\theta_A) R_B(\theta_B) \rangle| \leq 2, no matter what algorithms are used by A, B, and C, as long as

1. There are no communications among A, B, C other than those specified.
2. The settings \theta_A and \theta_B for each round are unpredictable by C.

On the other hand, if instead of C sending messages, it generates a pair of entangled photons, and sends one to A and one to B, then you can violate the inequality. (Inside A and B, instead of a computer algorithm, you have polarizing filters and photon detectors, and each sends out +1 if the photon passes through the filter at the orientation specified by \theta_A or \theta_B.

 

[entropy1]

Because, in a computer, this would have to be a HV-setup, it would not reproduce the desired result.

 

[Mentz114]

Well, in the usual setup the measurement of B doesn't do anything on A. To ensure this you make the measurement acts spacelike separated. That's what's so mind boggling about these stronger-than-classically-possible correlations described by an almost trivial math. As easy is the math as mind boggling are the implications, particularly if you stick to traditional collapse assumptions of the early days! The worrying immediately stops if you simply accept that nature is inherently probabilistic and that there is a very successful formalism called QT that tells you probabilities and only probabilities.

I don't know if this is addressed to me but - I'm not worried nor ever have been by any of the great 'problems' in QT !

Like you I believe if we can calculate probabilities we can simulate any experiment.

Entanglement changes the probabilities a lot and it is easy enought to show this.

My first run of 400 produced these quartets (count, coincidences, $\chi^2$ contribution, expected value = count/2 )

A₁B₁₌( 105, 48, 0.38, 52.5)
A₁B₂₌( 103, 25, 13,2, 51.5)
A₂B₁₌( 100, 77, 14.5, 50)
A₂B₂₌( 92, 56, 2.17, 46)

The sum of the $\chi^2$ terms is 30.5 which is in the very small percentile of the $\chi^2$ distribution with 3 dof.

Clearly allowing the first projection to set both photons has skewed the probabilities drastically compared to the unentangled set up.

 

[vanhees71]

But the relative angle is not known until the last minute. The situation is the following:

View attachment 217546

You have three devices: C, which is a source of message pairs, simulating photon pair production, and A and B, which simulate the measurement events.

• Every "round", C sends out a pair of messages, m_A to A and m_B to B.
• After the messages are sent, but before they are read, settings for A and B are chosen, independently. The settings are two angles, \theta_A and \theta_B.
• Device A determines an output, R_A(\theta_A), which is either +1 or -1, based on the message received from C and the setting \theta_A.
• Similarly, device B determines an output, R_B(\theta_B) based on its message and setting.
• Over many, many rounds, we can gather statistics for the correlation: \langle R_A(\theta_A) R_B(\theta_B) \rangle as a function of the pair of settings, \theta_A, \theta_B.

Bell's inequality implies that |\langle R_A(\theta_A) R_B(\theta_B) \rangle| \leq 2, no matter what algorithms are used by A, B, and C, as long as

1. There are no communications among A, B, C other than those specified.
2. The settings \theta_A and \theta_B for each round are unpredictable by C.

On the other hand, if instead of C sending messages, it generates a pair of entangled photons, and sends one to A and one to B, then you can violate the inequality. (Inside A and B, instead of a computer algorithm, you have polarizing filters and photon detectors, and each sends out +1 if the photon passes through the filter at the orientation specified by \theta_A or \theta_B.

Sure, I'm aware of this, but no matter when you choose the orientation (even in the last femtosecond before the envisaged photon hits the detector) you know the probabilities you have to simulate, when using this setup to calculate the corresponding probabilities (or expectation values). The probabilities are known of course; it's what's predicted by QT, and that's what's simulated with a correspondingly programmed MC simulator.

 

[stevendaryl]

Sure, I'm aware of this, but no matter when you choose the orientation (even in the last femtosecond before the envisaged photon hits the detector) you know the probabilities you have to simulate,

I'm not exactly sure what your point is. Yes, we know the probabilities for joint detection. The question is whether you can simulate the detection process in the way that I have sketched out and get those probabilities. The answer is "no".

When you choose the orientation makes a difference, because if the orientation was planned in advance, then device C could use that information to reproduce the predictions of QM.

 

[stevendaryl]

I don't really know what you (@vanhees71) mean by a "Monte Carlo" simulation for this experiment, but the only way you can reproduce the predictions of quantum mechanics for this case is if the settings for the detectors are known by the simulation code. In other words, by cheating (according to the rules laid out).

 

[Mentz114]

I don't really know what you (@vanhees71) mean by a "Monte Carlo" simulation for this experiment, but the only way you can reproduce the predictions of quantum mechanics for this case is if the settings for the detectors are known by the simulation code. In other words, by cheating (according to the rules laid out).

https://en.wikipedia.org/wiki/Monte_Carlo_method

 

[stevendaryl]

https://en.wikipedia.org/wiki/Monte_Carlo_method

But in a Monte Carlo simulation, the inputs are generated, as well as the outputs, which means that the inputs (the detector settings, in this case) are known in advance.

 

[DrChinese]

Of course, I have to use the relative angle of the detectors to calculate these probabilities, and thus this information goes into the code.

As mentioned in post #25: 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". That is to say that using the relative angle in the code to generate the answer is a cheat. If you don't cheat, you can't even hand pick data sets that match the probability predictions of QM.

So hopefully we are all in agreement on this point, which was what was in error in the OP's original code - which made use of the relative angle.

 

[forcefield]

As mentioned in post #25: 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". That is to say that using the relative angle in the code to generate the answer is a cheat. If you don't cheat, you can't even hand pick data sets that match the probability predictions of QM.

So hopefully we are all in agreement on this point, which was what was in error in the OP's original code - which made use of the relative angle.

Isn't it enough if "Charlie" (or "C") knows both measurement settings as said by @stevendaryl in post #43 ?

 

[entropy1]

Isn't it enough if "Charlie" (or "C") knows both measurement settings as said by @stevendaryl in post #43 ?

The measurement settings are not fixed in principle; they can be changed at the last moment.

 

[forcefield]

The measurement settings are not fixed in principle; they can be changed at the last moment.

I know that - are you saying that I can't simulate entanglement without changing measurement settings at the last moment ?