I Hidden phase in polarization tests of Bell's inequality?

[Nugatory]

I'm curious though, how do you create three angle correlations with pairs of polarized photons?

We position the detectors in different configurations while we run a large number of entangled pairs through our experiment. Eventually we end up with statistically significant results for the AB correlation, the BC correlation, and the AC correlation.

You may think you see a flaw in this approach: if particles with a particular value of the hypothetical hidden variable are hypothetically more likely to be detected in one of the three configurations than the other two the correlations will be not be calculated from equivalent subsets of the pairs. This is the well-known “fair sampling loophole”, and more sophisticated experiments have been able to close it.

 

[DrChinese]

1) i did not know this. I thought entangled photon pairs were created on demand. I'm not exactly sure how this affects my idea though, since you can freely change what the blue area stands for.

3c) the rule is if the vector falls into the colored area, the detector blips. A photon vector is just some undetermined property of the photon that is constant from it's departure from the source (not necessarily polarization, i just haven't thought about it), and is the same for both entangled partners.

3d) this is an issue with the circle representation, because the area overlap scales linearly with the angle rotation, which obviously means that what you said is true. Further below I've shown that you can get non-linear overlaps with angles, if your shape is not an exact circle. I haven't figured out the exact shape that would reproduce a cos^2 dependence, which is a valid objection and another reason why I posted this, but I don't think is impossible.

... Overall, I liked how PeterDonis put it. I still have to fully comprehend what it means to have factorizable probabilities in this context, but I'll dig deeper into it. ... i need to find an algorithm that can find a shape to scale the overlapping area correctly, with a cosine^2 dependence.

4c) Aren't those photons still entangled though? Regardless of the origin of the entanglement, the vector being the same would be a requisite for them to be entangled. Whatever the process that created the entanglement, also "aligned" the vectors.

1) Your model can be adjusted for this, yes. Just wanted to make sure you understood how PDC works. The ratio of photons that enter the crystal to the number that are converted to entangled pairs is on the order of 100 million to 1. It's completely random.

3c and 3d) That's what I thought. Yours is the one of the most common initial models to consider. But unfortunately it cannot be "fixed" to present the quantum expectation value (cos^2). A tweak to the model to make it appear to work at one angle simply breaks the model elsewhere.

Please be aware that for a number of years, it was assumed that a model might be developed that would work (as you have attempted). Of course, that was before Bell's 1964 paper proved conclusively that no such model was possible. Once you add in the Bell's requirement that there can be no coordination or communication between the L and the R detector settings for the separate outcomes - a requirement sometimes referred to as "realism" or "non-contextuality" - then everything falls apart.

4c) These photons are entangled after they are created, and never have interacted. The act of entangling them is done remotely (i.e. at a distance in spacetime) in a process called entanglement swapping (which is a form of quantum teleportation). They can't have the same initial phase vector, even accidently or coincidentally. Of course, the details of such swapping is pretty complex and there is no point of analyzing how swapping works until you understand why Bell prevents the development of a local realistic model of entanglement.

The existence of even one experiment demonstrating entanglement of photons that have never existed in a common region of spacetime should be enough to exclude all local realistic models of entanglement. It's not even clear a nonlocal realistic model is feasible at this point.

 

[PeroK]

I'm basing all this on popular explanations i found online, and basically all of them used explanations with two angles.

Bell's inequality uses three angles. See, for example, Modern QM by JJ Sakurai. By using three different angles, it becomes impossible for classical probabilities to achieve the experimental results.

I think I understand what you are doing now. By using two angles, you are trying to show that hidden variables can reproduce the $\cos^2$ distribution. I can see how that might also be impossible, but it's not the basis of Bell's inequality and Theorem.

I'm curious though, how do you create three angle correlations with pairs of polarized photons?

You have two detectors that can each be set at three different angles.

 

[PeroK]

I'm curious though, how do you create three angle correlations with pairs of polarized photons?

Here's is Bell's inequality. We intend to measure spin or polarization at each of the two detectors for (the same) three different angles. First, we have the distibution of hidden variables that I posted above. Whatever the overall distribution of variables, it must boil down to some distibution for any three given angles. I'll write these as probabilities this time:
$$p_1: (+, +, +) \ \ \ (-,-,-)$$$$p_2: (+, +, -) \ \ \ (-,-,+)$$$$p_3: (+, -, +) \ \ \ (-,+,-)$$$$p_4: (+, -, -) \ \ \ (-,+,+)$$$$p_5: (-, +, +) \ \ \ (+,-,-)$$$$p_6: (-, +, -) \ \ \ (+,-,+)$$$$p_7: (-, -, +) \ \ \ (+,+,-)$$$$p_8: (-, -, -) \ \ \ (+,+,+)$$Let's label the angles as $a, b, c$ in the order given. We now do a large number of experiments and vary the detectors, so that we have a large number of experiments where the two detectors are set at different angles: detector 1 at angle $a$ with dector 2 at angle $b$; detector 1 at angle $a$ with detector 2 at angle $c$; detector 1 at angle $b$, with detector 2 at angle $c$; and, vice versa. We now look at some cases:
$$p(a+, b+) = p_3+p_4$$This is the case where we have the first detector set to angle $a$ and get $+$ and also have the second detector set to angle $b$ and get $+$ as well.

Now, this is Bell's idea using the third angle, we also have
$$p(a+, c+) = p_2 + p_4$$$$p(c+, b+) = p_3 + p_7$$This gives us the inequality:
$$p(a+, b+) \le p(a+,c+) + p(c+, b+)$$This is because the left-hand side is just $p_3 + p_4$ and the right-hand side is $p_3 + p_4 + p_2 + p_7$.

The final thing is to show that QM predicts a violation of this inequality in some cases. We can let $a = 0, b= 2\theta$ and $c = \theta$. So, the third angle bisects the other two. Now, QM predicts:
$$p(a+, b+) = \sin^2 \theta $$$$p(a+, c+) = p(c+, b+) = \sin^2 \frac \theta 2$$And, Bell's inequality only holds if we have:
$$\sin^2 \theta \le 2\sin^2 \frac \theta 2$$You can actually see now where the issue of non-linearity of the $\sin^2$ distribution is critical. Although, note that it took the introduction of a third, intermediate angle to force the contradiction.

In any case, this equaility simply does not hold. In fact, if we use the trig identities $2\sin^2 \frac \theta 2 = 1 - \cos \theta$ and $\sin^2 \theta = 1 - \cos^2 \theta$, then Bell's inequality becomes:
$$\cos \theta < cos^2 \theta$$Which fails for any angle $0 < \theta < \frac \pi 2$.

This mathematics proves that your supposed elliptical distribution (or any distribution) cannot reproduce the QM probability distribution in all cases. All classical distributions must obey Bell's inequality - which QM violates.

In any case, that is Bell's inequality and why QM violates it.

 

[Leureka]

Here's is Bell's inequality. We intend to measure spin or polarization at each of the two detectors for (the same) three different angles. First, we have the distibution of hidden variables that I posted above. Whatever the overall distribution of variables, it must boil down to some distibution for any three given angles. I'll write these as probabilities this time:
$$p_1: (+, +, +) \ \ \ (-,-,-)$$$$p_2: (+, +, -) \ \ \ (-,-,+)$$$$p_3: (+, -, +) \ \ \ (-,+,-)$$$$p_4: (+, -, -) \ \ \ (-,+,+)$$$$p_5: (-, +, +) \ \ \ (+,-,-)$$$$p_6: (-, +, -) \ \ \ (+,-,+)$$$$p_7: (-, -, +) \ \ \ (+,+,-)$$$$p_8: (-, -, -) \ \ \ (+,+,+)$$Let's label the angles as $a, b, c$ in the order given. We now do a large number of experiments and vary the detectors, so that we have a large number of experiments where the two detectors are set at different angles: detector 1 at angle $a$ with dector 2 at angle $b$; detector 1 at angle $a$ with detector 2 at angle $c$; detector 1 at angle $b$, with detector 2 at angle $c$; and, vice versa. We now look at some cases:
$$p(a+, b+) = p_3+p_4$$This is the case where we have the first detector set to angle $a$ and get $+$ and also have the second detector set to angle $b$ and get $+$ as well.

Now, this is Bell's idea using the third angle, we also have
$$p(a+, c+) = p_2 + p_4$$$$p(c+, b+) = p_3 + p_7$$This gives us the inequality:
$$p(a+, b+) \le p(a+,c+) + p(c+, b+)$$This is because the left-hand side is just $p_3 + p_4$ and the right-hand side is $p_3 + p_4 + p_2 + p_7$.

The final thing is to show that QM predicts a violation of this inequality in some cases. We can let $a = 0, b= 2\theta$ and $c = \theta$. So, the third angle bisects the other two. Now, QM predicts:
$$p(a+, b+) = \sin^2 \theta $$$$p(a+, c+) = p(c+, b+) = \sin^2 \frac \theta 2$$And, Bell's inequality only holds if we have:
$$\sin^2 \theta \le 2\sin^2 \frac \theta 2$$You can actually see now where the issue of non-linearity of the $\sin^2$ distribution is critical. Although, note that it took the introduction of a third, intermediate angle to force the contradiction.

In any case, this equaility simply does not hold. In fact, if we use the trig identities $2\sin^2 \frac \theta 2 = 1 - \cos \theta$ and $\sin^2 \theta = 1 - \cos^2 \theta$, then Bell's inequality becomes:
$$\cos \theta < cos^2 \theta$$Which fails for any angle $0 < \theta < \frac \pi 2$.

This mathematics proves that your supposed elliptical distribution (or any distribution) cannot reproduce the QM probability distribution in all cases. All classical distributions must obey Bell's inequality - which QM violates.

In any case, that is Bell's inequality and why QM violates it.

One thing that is still not clear to me: why do you group up the different measurements? (In the following D1 stands for detector 1 and D2 for detector2, while a b c are the angles)

We measure D1A and D2B, we get a probability distribution P1 (for example, getting both +);
Similarly we do D1A and D2C, getting P2.
Finally, we do D1C and D2B, getting P3.

Let's say, for example, a=0°, b=22.5°, c=45°.
We get something like this:

P1(A/B) (+ +, + -, - +, - -) = P(22.5°) (42.5%, 15%, 15%, 42.5%)

P2(A/C) (+ +, + -, - +, - -) = P(45°) (25%, 25%, 25%, 25%)

P3(B/C) (+ +, + -, - +, - -) = P(22.5°) (42.5%, 15%, 15%, 42.5%)(It looks to me like D1A/D2B should give the exact same distribution as D1B/D2A, since the detectors are identical. The same applies to P2 and P3.)

In the end, you group up all the cases: like for all the times where D1A got +, you count how many got D2B+ and D2C+. But how can you assign a probability distribution to these combinations? P1 (D1A+/D2B+) is independent from P2 (D1A+/D2C+) or P3. I don't understand how you can assign some P to (+ - -) (- + + ). It's clear you can't just add the probabilities up: for (+ - -) (- + + ), it would be 42.5% (D1A+, D2B+) + 25% (D1A+, D2C+) + 15% (D1C-, D2B+) which is 82.5% (makes no sense for only one distribution).

It's possible Im missing some probability calculus rule here, it's not my forte.

 

[Nugatory]

One thing that is still not clear to me: why do you group up the different measurements?

You might try https://static.scientificamerican.com/sciam/assets/media/pdf/197911_0158.pdf which is @PeroK’s explanation presented in a more visual way. But it’s not as complicated as it looks at first glance:

When we have one detector set at angle A and the second detector set at angle B, and we’re assuming your hypothesis about a hypothetical phase angle as the hidden variable…. What is the probability that we will record + at the first detector and + at the second detector? That can only happen if the phase angle is set to produce + at the first detector at angle A, + at the second detector at angle B, and either + or - if we had instead set either detector to angle C. That’s two mutually exclusive possibilities so we add the probabilities, which is what @PeroK is doing when he writes $p(a+,b+)=p_3+p_4$.

 

[PeroK]

It's possible Im missing some probability calculus rule here, it's not my forte.

The point you may be missing is that a particle would have to have hidden variables for all possible measurement angles. If you have one angle, then a single (relevant) variable will do. But, if you have $3$ angles you must have effectively $3$ relevant hidden variables that form a sample space of $8$ possibilities. I.e. every permutation of up/down for each angle. These possibilities are mutually exclusive.

An understanding of probability theory is a prerequisite to studying QM. I can't teach you that here: you have to study that for yourself.

Post #34 in my view is elementary and a misunderstanding of it is hard to comprehend, I'm sorry to say.

PS I suspect you may also not understand the concept of correlation. This seems to be a stumbling block for many in understanding Bell's inequality. It's not only the correlation between the measurements on the two particles, but the $\cos^2$ correlation between measurement angles on the same particle, that create the classically impossible overall correlation between measurement results.

 

[Leureka]

The point you may be missing is that a particle would have to have hidden variables for all possible measurement angles. If you have one angle, then a single (relevant) variable will do. But, if you have $3$ angles you must have effectively $3$ relevant hidden variables that form a sample space of $8$ possibilities. I.e. every permutation of up/down for each angle. These possibilities are mutually exclusive.
An understanding of probability theory is a prerequisite to studying QM. I can't teach you that here: you have to study that for yourself.

Post #34 in my view is elementary and a misunderstanding of it is hard to comprehend, I'm sorry to say.

PS I suspect you may also not understand the concept of correlation. This seems to be a stumbling block for many in understanding Bell's inequality. It's not only the correlation between the measurements on the two particles, but the $\cos^2$ correlation between measurement angles on the same particle, that create the classically impossible overall correlation between measurement results.

Bear with me here. I understood your math, as you said it’s very simple. What I did not understand are your premises. I don’t see why you need a single variable for all three angles, when the most you can ever do is sample two angles. In this model the hidden variable is an area overlap which you can determine through multiple measurements of the photon, but only with always the same angle settings. Once you change your setting on one of the angles, the area overlap you are determining will be different. Since entanglement is limited to only two photons, there is no way to do this operation without losing information of the overlap on one of the angles.

To visualise what I mean, here are some other drawings I made. This time we’re going to deal with spheres, since we have an extra degree of freedom in a third angle. Note that since we’re dealing with a perfect sphere the overlaps will vary linearly with angle and so won’t reproduce a cosine distribution, but again this is a matter of choosing the correct 3d shape (like in the ellipse case).

Here, the figure is actually 3 spheres overlapped, each representing one detector setting. The coloured volume represents a “blip” when the detector interacts with the photon, and each sphere is only half coloured to represent each detector has an intrinsic 50% chance of detection. In this example, A and B are at 90 degrees, while B and C are at 45 degrees. The actual values are not important for my argument.

In this second figure, A and B are still at 90 degrees while B and C are still at 45 degrees, but the relationship between A and C changed. This is possible because for each measurement, I can only really define two sets of spheres (one for each photon-detector couple) with a third degree of freedom (left unmeasured) that has no fundamental effect on the outcome of the measurement; in fact, I can freely rotate each sphere independently on the axis defined by its partner without changing the volume overlap. If I wanted to also keep A+C at 45 degrees I would have changed the relationship between A and B, but this would not affect my measurement of A + C.

 

[PeroK]

I don't understand what you are doing there. The numbers are important. I think you should try to calculate to what extend your classical probability distribution can violate the Bell inequality. You need to calculate a number for the Aspect experiment.

Otherwise, all you have are ambiguous diagrams that may not achieve the miracle you claim.

 

[PeroK]

PS the Aspect experiment does sample three different angles. So, your claim that this cannot be done is false. This also suggests, I'm sorry to say, that you haven't understood the Bell test at all!

 

[Leureka]

I don't understand what you are doing there. The numbers are probably important. I think you should try to calculate to what extend your classical probability distribution can violate the Bell inequality. You need to calculate a number for the Aspect experiment.

Otherwise, all you have are ambiguous diagrams that may not achieve the miracle you claim.

I'm not trying to achieve any miracle. As I said, I'm trying to understand why some things are the way they are.

No the numbers are not important. It's a visual way to say that (in my mind) if your measurement can only output two values, there's no reason to introduce a third that can't be determined by the previous two values. If you measure at angles A and B, whatever your angle C is your measurements will not be affected. It feels wrong to assume that what you get from measuring A and C will be affected by what you had before, because each time you're sampling completely different photons.

Just take two overlapping spheres instead of three, A and B at 90 degrees. You can rotate by 360° the sphere A around B (or B around A) keeping the colored emispheres orthogonal. Any rotation by any amount always corresponds to the exact same probability of both detectors bleeping, which in turn corresponds to the overlapping volumes of the colored hemispheres (a quarter of the volume of the whole sphere). In other words, if the photon "vector" (a unit vector that starts from the centre of the sphere) lands inside this volume, both detectors will blink. But the number of photon vectors that are capable of landing in that volume is not limited to a fixed part of the sphere: any vector in the sphere can land there if you rotate sphere A and sphere B the right amount. That is why determining the correlations for A and B won't tell me anything about those for A and C or B and C: there's infinitely many different distributions of vectors also landing in C, one for each infinitesimal rotation of either A or B.

I can't provide any math because I don't know which 3D shape can reproduce a cos^2 non linear overlap dependent on angle, if that's even possible (because I'm not claiming it is, but i don't have evidence it isn't). What I already showed is that non linear overlaps ARE possible, with anything but a perfect sphere.

 

[Nugatory]

I don’t see why you need a single variable for all three angles, when the most you can ever do is sample two angles.

That is a requirement for a realistic theory, which is to say one that asserts that unmeasured quantities still have a definite value. If your hypothetical phase vector does not allow us to predict the results of the measurement on all three axes, you haven't proposed a local realistic theory as a counterexample to Bell's theorem - at best you have a more complicated model of quantum mechanics.

To see why all three angles matter for a realistic theory even when we're only measuring two, consider the experiment (which has been done!) where the measurement angles are not set until after the pair has been created and the particles are in flight. We can choose to measure on any of the three angles and the phase vector is already set (if it's not, then your theory is not local) so the phase vector must be sufficient to determine the result on any of the three axes even though we won't measure one of them.

 

[Leureka]

That is a requirement for a realistic theory, which is to say one that asserts that unmeasured quantities still have a definite value. If your hypothetical phase vector does not allow us to predict the results of the measurement on all three axes, you haven't proposed a local realistic theory as a counterexample to Bell's theorem - at best you have a more complicated model of quantum mechanics.

It might be that my model is indeed nonlocal (I just haven't got a clear answer one way or another); but the point I'm trying to get across is that if that's true, then the source of non-locality is not mysterious at all. It really boils down to how we compare measurements of a phenomenon in which both detector and photon play a role. All that is happening at the local level is asking whether the detector is able to bleep, and that will depend by both the photon vector and the local phase of the instrument. The correlations will be a natural result of the difference in phase of the two detectors with respect to the photon vector.

To see why all three angles matter for a realistic theory even when we're only measuring two, consider the experiment (which has been done!) where the measurement angles are not set until after the pair has been created and the particles are in flight. We can choose to measure on any of the three angles and the phase vector is already set (if it's not, then your theory is not local) so the phase vector must be sufficient to determine the result on any of the three axes even though we won;t measure one of them.

If the measurement is a result of both photon vector AND instrument phase, whether the choice happens after or before pair creation makes no difference. The measurement will always be either a yes or no at a single detector, but the correlation will depend on both detector choices as I explained above. The photon by itself can't know whether it will pass a barrier that might or might not be there.

 

[DrChinese]

... No the numbers are not important. It's a visual way to say that (in my mind) if your measurement can only output two values, there's no reason to introduce a third that can't be determined by the previous two values. If you measure at angles A and B, whatever your angle C is your measurements will not be affected. It feels wrong to assume that what you get from measuring A and C will be affected by what you had before, because each time you're sampling completely different photons.

...I can't provide any math because I don't know which 3D shape can reproduce a cos^2 non linear overlap dependent on angle, if that's even possible (because I'm not claiming it is, but i don't have evidence it isn't). What I already showed is that non linear overlaps ARE possible, with anything but a perfect sphere.

You haven't shown anything that relates to Physics. Your model does not pass the Bell test, and not because you are having a difficult time trying to come up with overlapping shapes. It is sad that you continue to ignore Bell's Theorem, as you have been advised.

I will try to explain what others have said about Bell in terms of what you have said. According to you: "If you measure at angles A and B, whatever your angle C is your measurements will not be affected." And yet you model produces a prediction at every possible angle. A, B, C, D and so on, according to a hard and fast rule you have given us. Oh, and yes, you may plan to tweak it. But...

It's that hard and fast rule that's the problem. Because there are NO consistent models of the type you describe. (And by consistent, I mean consistent with the quantum predictions.) That's because the type of model you propose is of type "objective realistic" - meaning the experimenter's choice of angle settings in no way affects the outcomes. Which is just what you think is the case.

But that's what Bell showed is not possible. That's because quantum mechanics is what is called "subjective realistic". It is the specific 2 angles, taken as a pair, that determines the statistical results. In your model, I should be able to select as many angles as I want and get a statistical prediction for any pair. But the resulting statistics won't match QM. Yes, it looks that way to you - because you are only looking at the outcomes in front of you. Hey, Einstein thought the same exact thing! But he didn't live to read Bell.

You freely acknowledge that the match % for 0 & 22.5 degrees is 85%. You freely acknowledge that the match % for 22.5 & 45 degrees is also 85%. So how do you explain that the match % for 0 and 45 degrees is 50%? According to standard probability theory: the absolute minimum match % would actually be 85% * 85% which is about 73%. That doesn't match your model's prediction of 50%. Your model's assertion that there is a hard and fast rule - which says that the outcome is objectively realistic - leads to inconsistency.

QM says that the outcomes are only dependent on the relative difference between 2 settings. There is no meaning to the possible outcomes at other settings. This is closely related to the Uncertainty Principle, which says that spin (or polarization) has mutually non-commuting bases. Regardless, you will quickly find that the statistics don't work out.

If you like, check out a simple proof on one of my web pages. My example uses 120 degrees (match=25%) for photons instead of your 22.5 degrees (match=85%). This angle makes the problem of a model of your type more clear, and you shouldn't care what the angle is. In fact, for the angle settings of 120 degrees, you don't even need a model. You can't even hand pick them and match the quantum predictions with an objective realistic model. The closest you can get with hand picked outcomes is at least 33%.

Bell's Theorem with Easy Math

 

[Nugatory]

I should add that the impossibility of making a local realistic model work across three settings is not just a quibble - It goes directly to some of the weirdest aspects of quantum mechanics.

If you have read that Scientific American article I linked earlier in the thread you will have seen that the proof of Bell’s inequality uses the same logic that I use when I assert that the number of male redheads in a room must be at least at great as not greater than the number of red-headed cigarette smokers plus the number of non-smoking men. I can verify this while measuring only two of the three properties at a time by asking all the non-smoking men to raise their hands, counting, then asking all the red-headed smokers to raise their hands, counting, then finally asking for all the red-headed men. This is basically what an experiment testing for violations of Bell’s inequality is doing, and the profoundly weird thing is that the inequality is not respected by quantum mechanics.

 

[PeroK]

No the numbers are not important.

The numbers are everything! As @DrChinese points out:

You freely acknowledge that the match % for 0 & 22.5 degrees is 85%. You freely acknowledge that the match % for 22.5 & 45 degrees is also 85%. So how do you explain that the match % for 0 and 45 degrees is 50%? According to standard probability theory: the absolute minimum match % would actually be 85% * 85% which is about 73%. That doesn't match your model's prediction of 50%.

 

[Leureka]

If you have read that Scientific American article I linked earlier in the thread you will have seen that the proof of Bell’s inequality uses the same logic that I use when I assert that the number of male redheads in a room must be at least at great as the number of red-headed cigarette smokers plus the number of non-smoking men.

3 properties:
- is male
- is redheaded
- is smoker

Male + redhead <= (redhead + smoker) + (male + nonsmoker).

Is this right? I think you meant "is less or equal to", but I might misrepresenting your example.But what happens if everytime you ask a question, the people in the room change? You might still be getting the exact same number of redheaded males, but the number of smokers is not constant. The three properties are completely independent from one another.

@DrChinese @PeroK for the same reason, why do you think AC(22.5) and BC(22.5) have anything to do with AB(45)? They are completely different measurements. This is obvious from the fact that the system is rotationally invariant, calling AC and BC with different names doesn't make them distinct.

 

[PeroK]

@DrChinese @PeroK for the same reason, why do you think AC(22.5) and BC(22.5) have anything to do with AB(45)?

In a word, because QM predicts that the results are statistically correlated. This is confirmed by experiment. That's why random hidden variables were proposed (and, are being proposed by you!!) as an alternative to QM. I suspect you don't understand the concept of correlated measurement results in the context of light polarization.

There are many experiments that prove these correlations between different measurement angles. So, when you say that "one measurement has nothing to do with another", you are right that they are different measurements, but wrong that they are uncorrelated.

It's difficult to know what you understand and what you are fundamentally misunderstanding. I suspect you are missing a lot of the fundamentals of QM, so we are just chasing shadows here.

1) Classical physics would normally predict a single (deterministic) result for measurements on a sequence (or ensemble) of identically prepared systems. There would be no statistics at all. Like, say, a cricket or baseball machine that can fire cricket balls or baseballs with the same velocity and spin every time. It's fundamental that you cannot do that in QM.

2) QM is fundamentally probabilistic. I.e. measurements on identically prepared systems produce a distribution of results. For example, if you build a machine that produces photons with a given polarization in one direction, then polarization measurements in other directions will be probabilistic.

3) If you want a classical model of polarization, then you must introduce hidden random variables to mimic the probabilistic behaviour. There are severe constraints on your variables, as described above, as they must reproduce the measurements results and correlations of QM. Not just the measurement results, as you may believe, but all the correlations as well.

And, of course, this is proved (quite simply) to be mathematically impossible.

Bell's inequality is all about correlations, correlations, correlations.

 

[berkeman]

Thread closed temporarily for Moderation...

 

[Nugatory]

The question in the original post has been answered - no, a hidden variable of the type that OP is considering cannot explain violations of Bell’s inequality - and the subsequent argument is based on OP’s misunderstanding of Bell's inequalities.

This thread will remain closed.
As with all thread closings, you can PM any mentor to ask that it be reopened to allow new and relevant content to be added.