Exploring Bell's Assumptions and Quantum Theory

In summary: So the probability of passing through increases by a factor of 2 when you measure the polarization of one of these photons in addition to the cos(45) percent chance of a single photon passing through.
  • #1
Hydr0matic
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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:
Now, the assumptions are:
A: Bell's assumptions (locality, non-superdeterminism, objective single outcomes)
B: quantum theory

from A, Bell derives his inequalities (mathematically), and from B, one derives violations of those inequalities. Hence a contradiction.

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 separate 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 separate 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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  • #2
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.
 
  • #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 separate 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?
 
  • #4
Hydr0matic said:
My inquery was about the nature of polarization, and whether or not the ciriteria in Bell's inequalities of three separate 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?

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.
 
  • #5
Hydr0matic said:
When testing Bell one measures separate 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 separate 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...
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).
 
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  • #6
Thanks for the answers, and sorry for my delay..

DrChinese said:
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.
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" 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"?

JesseM said:
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.
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.


JesseM said:
... conclusions which are violated in QM. Note that in local realism, ...
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.
 
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  • #7
Hydr0matic said:
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.
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!
Hydr0matic said:
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?
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.
 
  • #8
JesseM said:
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!
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.
 
  • #9
Your polarizers are all 90.0° apart! Not a particularly interesting case.
 
  • #10
Doc Al said:
Your polarizers are all 90.0° apart! Not a particularly interesting case.
The polarizers are aligned. The angle values is the lightwave polarization.
 
  • #11
Hydr0matic said:
The polarizers are aligned. The angle values is the lightwave polarization.
Oops--I misread your post. I'll rephrase my response:
Your polarizers are aligned! Not a particularly interesting case.
 
  • #12
JesseM said:
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.

Doc Al said:
Your polarizers are aligned! Not a particularly interesting case.
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.
 
  • #13
Hydr0matic said:
1 & 2 give same results.
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 Theorem with Easy Math
[/URL].
But as you can see there's a fundamental difference.
Indeed!
 
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  • #14
Hydr0matic said:
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.
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.
 
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  • #15
Doc Al said:
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.
I've put together a non-trivial javascript application based on my idea:
http://pf.soderholmdesign.com/bell.html

From Mermin's paper:
quantum theory predicts (and experiment confirms) that the photons will be found to have the same polarizations (lights flashing the same colors in the analogous gedanken experiment) if they are measured along the same direction –feature number 1
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.

But if the polarizations are measured at 120° angles, then theory predicts (and experiment confirms) that they will be the same only a quarter of the time [ ¼ = cos2 (120°)]
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###
So if the difference between the two detectors is 0, the probability of getting opposite answers is cos^2 (0) = 1
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.


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.
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.
 
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  • #16
Hydr0matic said:
I've put together a non-trivial javascript application based on my idea:
http://pf.soderholmdesign.com/bell.html

From Mermin's paper:
quantum theory predicts (and experiment confirms) that the photons will be found to have the same polarizations (lights flashing the same colors in the analogous gedanken experiment) if they are measured along the same direction –feature number 1
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.
Hydr0matic said:
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.
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?
Hyr0matic said:
Now we set Alice != Bob and RR/GG. The generated result with 50,000 pairs is:
-------- 11246/50000 (0.22492)
...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?
 
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  • #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?
 
  • #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, 33:5557Setup:
Alice != Bob
RR/GG
cos(40°)
Photon:Same
0/120/240
 
  • #19
JesseM said:
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.

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.
 
  • #20
Hydr0matic said:
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?

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
 
  • #21
DrChinese said:
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
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).
 
  • #22
Hydr0matic said:
The R is red (no detection) and G is green. Used in Mermin's paper.
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.
Hydr0matic said:
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.
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 anyone 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.
 
  • #23
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.
 
  • #24
Hydr0matic said:
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).

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. :)
 
  • #25
JesseM said:
Right, but what does RG/GR mean? 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.
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.

JesseM said:
Look over this proof and see if you can find any reason it wouldn't apply to your example;
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 =)
 
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  • #26
So I guess I am asking: is your routines supposed to emulate the quantum value or the classical (local realistic) value?

None of the above. My polarizers aren't governed by chance, so Bell does not apply.
 
  • #27
What you call "the classical case" with hidden variables and local realism, is actually NOT completely classical. Because the interpretation of what happens at the polarizers is based on photons and QM. Not lightwaves and determinism.
 
  • #28
Hydr0matic said:
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 think you're completely missing the point here, nothing in the proof I gave said anything about there being any chance elements when they hit the polarizers, in fact I (and Bell) explicitly assumed that the two waves were created with states that predetermined what results they'd give when they hit a polarizer at any possible angle; that's the whole point of Bell's theorem, it shows that if you assume such predetermined answers (as a way of explaining the observation that they always give identical results with the same polarizer setting), then the answers when the experimenters choose different settings will obey various Bell inequalities, like the inequality that says they will give the same answer with different polarizer settings on at least 1/3 of the trials.
Hydr0matic said:
My polarizers aren't governed by chance, so Bell does not apply.
Again, Bell assumes in the proof of the Bell inequalities that there are variables associated with the waves (hidden or otherwise) that completely determine whether or not they will pass through a polarizer at any given angle; if you think Bell was assuming there was a chance element in whether a given wave passes through a given polarizer, you've badly misunderstood the whole argument!
 
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  • #29
I have understood Bell's hidden variables. But there's a difference between having predetermined outcomes of a random process and having predetermened outcomes of correlated values. Let me try an example of my point with correlated properties.

We have a set of people on which we are measuring 3 properties - skin color, hair color and eye color. Each property has possible values Light (L) and Dark (D). So possible combinations are accordingly:

SHE (Skin, Hair, Eyes)
-------
LLL
LLD
LDL
LDD
DLL
DLD
DDL
DDDAnd each of these are seemingly equally probable.
When we start to measure we find that, for some reason, Bell's inequality is violated. When Alice measures an L color, Bob always measures L too. Same for D respectively.

Now, we can either interpret this as:

1. When Alice measures Skin color L, the persons hair and eyes "instantly collapse" into the same color. Bell's inequaility is violated, and our understanding about how skin, hair and eyes work is inaccurate.

2. All SHE combinations are NOT equally probable. In this case, there are actually only two possible combinations - LLL and DDD. The properties Skin, Hair and Eyes are CORRELATED, and only dependant on ONE property - Rase. Hence, Bell does not apply.

Ofcourse, this example is extreme correlation, but it's hard to think of normal properties correlated like polarization :/Let's take a concrete example with photons and polarization.
Let's say we have a source that emits photons with opposite polarization (90° diff), and a specific pair is emitted with polarization 0° and 90°. Now, for this pair Alice choose angle 0° and Bob choose 120°.
In QM & Local Realism, the probability of detection at Alice is cos^2(0-0) = 1, and the probability of detection at Bob is cos^2(120-90)=0.75.

In my case, if the detectors have a cut-off value of cos(20°), then the probability of detection at Alice is 1 (cos(0)>cos(20)), and the probability of detection at Bob is 0 (cos(30)<cos(20)).

If you apply this scenario in my application (with cut-off cos(20°)), I get the following results:

Subset when Alice & Bob get opposite result:
-------- 2198/10000 (0.2198)
-------- 0% Non-detections in subset (RR)
-------- 58.66% Total Non-detections (RR)
-------- Angle distribution: 11:1137, 12:1133, 13:1174, 21:1042, 22:1088, 23:1056, 31:1104, 32:1148, 33:1118,

Subset when Alice & Bob get same results: (different run than above)
-------- 4444/10000 (0.4444)
-------- 90.62% Non-detections in subset (RR) <-- *
-------- 58.67% Total Non-detections (RR)
-------- Angle distribution: 11:1090, 12:1156, 13:1097, 21:1111, 22:1142, 23:1095, 31:1121, 32:1097, 33:1091,

Notice that here *, Bell predicts no difference between nr of RR and GG. it should be 50/50. But due two the high cut-off value you get 90% non-detections.
 
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  • #30
Hydr0matic said:
I have understood Bell's hidden variables. But there's a difference between having predetermined outcomes of a random process and having predetermened outcomes of correlated values. Let me try an example of my point with correlated properties.

We have a set of people on which we are measuring 3 properties - skin color, hair color and eye color. Each property has possible values Light (L) and Dark (D). So possible combinations are accordingly:

SHE (Skin, Hair, Eyes)
-------
LLL
LLD
LDL
LDD
DLL
DLD
DDL
DDD


And each of these are seemingly equally probable.
When we start to measure we find that, for some reason, Bell's inequality is violated. When Alice measures an L color, Bob always measures L too. Same for D respectively.
How would that be a violation of Bell's inequality?? Bell assumes that when both experimenters pick the same property (say, skin color) they will always get the same result. In this case, the Bell inequality I described says that when they pick different properties to measure, they must get the same result at least 1/3 of the time. There is nothing in the inequality that says they can't get the same result 100% of the time when they measure different properties!
Hydr0matic said:
Now, we can either interpret this as:

1. When Alice measures Skin color L, the persons hair and eyes "instantly collapse" into the same color. Bell's inequaility is violated, and our understanding about how skin, hair and eyes work is inaccurate.

2. All SHE combinations are NOT equally probable. In this case, there are actually only two possible combinations - LLL and DDD. The properties Skin, Hair and Eyes are CORRELATED, and only dependant on ONE property - Rase. Hence, Bell does not apply.

Ofcourse, this example is extreme correlation, but it's hard to think of normal properties correlated like polarization :/
Please read over my proof again carefully:

'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.'

I have made no assumption about the ratio of "homogeneous" states to "inhomogeneous" states, I have only pointed out that for a homogeneous state, when they pick different detector settings they are guaranteed to get the same result with probability 1, and for an inhomogeneous state, when they pick different detector settings they have a 1/3 chance of getting the same result. Therefore, whatever the mix of states emitted by the source, the probability of getting the same result with different detector settings must be somewhere in the range 1/3 - 1. It is quite possible that the source emits only homogeneous states, in which case they are guaranteed to get the same result on every trial. But provided we assume that on every trial the source sends them objects with identical states (whether homogeneous or inhomogeneous), it is not possible for the probability of getting the same result when they choose different detector settings to be less than 1/3; that's all the Bell inequality here is saying.
Hydr0matic said:
Let's take a concrete example with photons and polarization.
Let's say we have a source that emits photons with opposite polarization (90° diff), and a specific pair is emitted with polarization 0° and 90°. Now, for this pair Alice choose angle 0° and Bob choose 120°.
I thought we were going to stick to Mermin's example where they would get the same result if they picked the same angle, not opposite results? With opposite results my proof above would have to be slightly modified, but you'd end up with the conclusion that under local realism, if Alice and Bob choose all detector settings with equal frequency, then on the subset of trials where they choose different detector settings, they must get opposite results at least 1/3 of the time.
Hydr0matic said:
In QM & Local Realism, the probability of detection at Alice is cos^2(0-0) = 1, and the probability of detection at Bob is cos^2(120-90)=0.75.
In QM if the two photons are entangled, you cannot view the results at Alice and the results at Bob as independent; if the difference between their two detector settings is 120, then the probability of their getting opposite results here must be cos^2(120) = 0.25.

And what do you mean by "local realism"? Didn't I just tell you that Bell did not assume any random element when the wave hits the detector? So why are you talking as if local realism gives a "probability of detection" for Bob? If the 90 degree angle of the wave is supposed to represent the "hidden variables" associated with it, then this angle should guarantee with probability 1 what result Bob gets when he sets his detector to 120, according to Bell.
Hydr0matic said:
In my case, if the detectors have a cut-off value of cos(20°), then the probability of detection at Alice is 1 (cos(0)>cos(20)), and the probability of detection at Bob is 0 (cos(30)<cos(20)).
But if we repeat this experiment over many trials, with the waves always at 0 and 90 but with Alice and Bob picking their settings randomly, the problem is that this will not even satisfy the basic assumption made in deriving the Bell inequality, which is that when they choose the same detector setting, there is a perfect correlation between their results (they always get the same result, or they always get opposite results). Look at what happens:

If she sets detector to 0: probability of detection = 1 (cos^2(0) > cos^2(20))
If she sets detector to 120: probability of detection = 0 (cos^2(120) < cos^2(20))
If she sets detector to 240: probability of detection = 0 (cos^2(240) < cos^2(20))

And if Bob's wave is at 90, then:

If he sets detector to 0: probability of detection = 0 (cos^2(90 - 0) < cos^2(20))
If he sets detector to 120: probability of detection = 0 (cos^2(120 - 90) < cos^2(20))
If he sets detector to 240: probability of detection = 0 (cos^2(240 - 90) < cos^2(20))

So if they both set their detectors to 0 they'll get opposite results, but if they both set their detectors to 120 or 240 they'll get the same results. You could avoid this problem by sticking to Mermin's example where they are always supposed to get the same result with the same detector setting, and assuming they both get waves polarized at the same angle (say, 0). Then as long as both Alice and Bob have the same threshold for detection, they're guaranteed to get the same result on any trial where they pick the same detector setting. In this case, if Alice and Bob choose their settings on each trial randomly, and you look at the subset of trials where they chose different settings, you should find that they got the same answer at least 1/3 of the time; if you don't then there is presumably an error in your program.
 
  • #31
Hydr0matic said:
...Now, we can either interpret this as:

1. When Alice measures Skin color L, the persons hair and eyes "instantly collapse" into the same color. Bell's inequaility is violated, and our understanding about how skin, hair and eyes work is inaccurate.

2. All SHE combinations are NOT equally probable. In this case, there are actually only two possible combinations - LLL and DDD. The properties Skin, Hair and Eyes are CORRELATED, and only dependant on ONE property - Rase. Hence, Bell does not apply.

Ofcourse, this example is extreme correlation, but it's hard to think of normal properties correlated like polarization :/


Let's take a concrete example with photons and polarization.
Let's say we have a source that emits photons with opposite polarization (90° diff), and a specific pair is emitted with polarization 0° and 90°. Now, for this pair Alice choose angle 0° and Bob choose 120°.
In QM & Local Realism, the probability of detection at Alice is cos^2(0-0) = 1, and the probability of detection at Bob is cos^2(120-90)=0.75.

In my case, if the detectors have a cut-off value of cos(20°), ...

What is a "cut-off value"? There is no experimental quantity or parameter I am aware of that corresponds to this. So I assume it is part of a rule for your routine. If so, what is its relevence to the discussion?

Also, it is not correct that the underlying properties in your "classical" example need to be equally probable. The only question is whether you can come up with an example that mimics the results of actual polarization experiments. Basically, the only thing relevant for that topic is the value of the angle between 2 polarization settings and whether there was a Y or N result (sometimes notated as H or V instead).

With 3 possible settings (assuming realism), there are 8 outcome permutations (2^3). When those 3 settings are mutually 120 degrees apart, you CANNOT generate a set of outcomes that match QM using any algorithm. You can only match QM with an algorithm for for 2 possible settings. And that algorithm must converge on the value of cos^2.
 
  • #32
JesseM said:
How would that be a violation of Bell's inequality?? Bell assumes that when both experimenters pick the same property (say, skin color) they will always get the same result.
Sorry, was writing fast during work. Was assuming they measured different properties.

JesseM said:
In QM if the two photons are entangled, you cannot view the results at Alice and the results at Bob as independent; if the difference between their two detector settings is 120, then the probability of their getting opposite results here must be cos^2(120) = 0.25.
If Alice has probability 1 and Bob has 0.75, then there is a 0.25 probability of opposite results.


JesseM said:
And what do you mean by "local realism"? Didn't I just tell you that Bell did not assume any random element when the wave hits the detector? So why are you talking as if local realism gives a "probability of detection" for Bob?
Bell didn't assume the randomness at the polarizers no, he just moved it to the source. I.e. the "randomness of the polarizer" is created at the source and carried as hidden variables.

JesseM said:
You could avoid this problem by sticking to Mermin's example
You're right, let's do that.

JesseM said:
where they are always supposed to get the same result with the same detector setting, and assuming they both get waves polarized at the same angle (say, 0). Then as long as both Alice and Bob have the same threshold for detection, they're guaranteed to get the same result on any trial where they pick the same detector setting.
I do. Set Alice=Bob and subset(RG or GR) (opposite results) and you get 0. Which means 100% are the same.

JesseM said:
In this case, if Alice and Bob choose their settings on each trial randomly, and you look at the subset of trials where they chose different settings, you should find that they got the same answer at least 1/3 of the time; if you don't then there is presumably an error in your program.
I don't. I get ≈0.22 ... I'm going to add a angle result counter to my application (RRR:n1, RRG:n2, RGR:n3 ...), so maybe we can see what's wrong... Give me half an hour tops.
 
  • #33
DrChinese said:
What is a "cut-off value"? There is no experimental quantity or parameter I am aware of that corresponds to this. So I assume it is part of a rule for your routine. If so, what is its relevence to the discussion?
The cut-off is the sensability of the detector. If the cut-off value is cos(20°), any photon that diverges more than 20° relative the polarizer will not be detected.

In my setup, whether or not a photon is detected is not a matter of probability. If the cut-off value is cos(20), then all photons with relative angle > 20 will not be detected. All others will.

The only random element in the setup is the polarization of the emitted photons.
 
  • #34
Hydr0matic said:
Sorry, was writing fast during work. Was assuming they measured different properties.
I understood that. You snipped the next part where I said "There is nothing in the inequality that says they can't get the same result 100% of the time when they measure different properties!" You seem to think that it violates the inequality for them to consistently get the same result when they measure different properties, but that's not a violation at all. The only violation is when both of the following are true:

1. When they pick the same property to measure, they get the same (or opposite) results consistently

2. When they pick different properties to measure, they get the same (or opposite results) less than 1/3 of the time.
Hydr0matic said:
If Alice has probability 1 and Bob has 0.75, then there is a 0.25 probability of opposite results.
But as I already pointed out, if Alice receives a wave polarized at 0 and Bob receives one polarized at 90, and their three detector settings are 0, 120, and 240, and their detector's threshold is cos^2(20), it is not guaranteed that if they choose the same detector setting, they'll get opposite results. In particular, if they both choose 120 or they both choose 240, then neither of their detectors will go off. So, this example doesn't satisfy condition (1) above.
Hydr0matic said:
You're right, let's do that.

I do. Set Alice=Bob and subset(RG or GR) (opposite results) and you get 0. Which means 100% are the same.
And this is for the case where you assume both waves are polarized at the same angle rather than 0 and 90 as in your earlier example, right? If both waves are polarized at the same angle, I agree they will always get the same result on any trial where they both pick the same angle.
Hydr0matic said:
I don't. I get ≈0.22 ... I'm going to add a angle result counter to my application (RRR:n1, RRG:n2, RGR:n3 ...), so maybe we can see what's wrong... Give me half an hour tops.
Sounds good, take your time.
 
  • #35
JesseM said:
But as I already pointed out, if Alice receives a wave polarized at 0 and Bob receives one polarized at 90, and their three detector settings are 0, 120, and 240, and their detector's threshold is cos^2(20), it is not guaranteed that if they choose the same detector setting, they'll get opposite results. In particular, if they both choose 120 or they both choose 240, then neither of their detectors will go off. So, this example doesn't satisfy condition (1) above.
With this setup, but negated result condition (RR or GG), I get:

With cut-off cos(20):
-------- 1782/10000 (0.1782) (82.18% opposite)
-------- 100% Non-detections in subset (RR)
-------- 59.42% Total Non-detections (RR)
-------- Angles: A:0° B:120° C:240°

With cut-off cos(40):
-------- 336/10000 (0.0336) (96.64% opposite)
-------- 100% Non-detections in subset (RR)
-------- 29.49% Total Non-detections (RR)
-------- Angles: A:0° B:120° C:240°

With cut-off cos(45):
-------- 0/10000 (0) (100% opposite)
-------- NaN% Non-detections in subset (RR)
-------- 21.73% Total Non-detections (RR)
-------- Angles: A:0° B:120° C:240°

Note that, with cut-off > cos(45) all non-opposites are non-detections (RR). With cut-off < cos(45), all non-opposites turn into all-detections (GG). Apparently, cos(45) is the boundry where condition 1 is satisfied.

I added a count when you choose "Photon polarization:Same". When Alice!=Bob, subset(RR or GG) and cut-off cos(40), I get:

-------- 2109/10000 (0.2109)
-------- 65.58% Non-detections in subset (RR)
-------- 32.41% Total Non-detections (RR)
-------- Angle distribution: 11:1114, 12:1109, 13:1155, 21:1158, 22:1099, 23:1075, 31:1071, 32:1086, 33:1133,
-------- Angles: A:0° B:120° C:240°

ResultCounts:
RRG:2207
RGR:2263
RGG:1111
GRG:1168
GRR:2170
GGR:1081

Apparently, RRR and GGG are not possible results with this setup. And detections at 2 angles seem to be about half as seldom as detections at one angle.



(A:R, B:R, C:G) (Alice:B, Bob:A) Result:(RR)Same Wave:55.33394317417163° <-
(A:R, B:G, C:R) (Alice:B, Bob:C) Result:(GR)Diff Wave:280.53612197198504°
(A:R, B:G, C:R) (Alice:A, Bob:C) Result:(RR)Same Wave:124.36775910448999° <-
(A:R, B:G, C:G) (Alice:B, Bob:C) Result:(GG)Same Wave:82.01282839063036° <-
(A:R, B:R, C:G) (Alice:B, Bob:A) Result:(RR)Same Wave:59.82230096148803° <-
(A:G, B:R, C:G) (Alice:B, Bob:A) Result:(RG)Diff Wave:21.064768816942056°
(A:G, B:R, C:R) (Alice:A, Bob:C) Result:(GR)Diff Wave:196.49632211000963°
(A:G, B:R, C:R) (Alice:C, Bob:A) Result:(RG)Diff Wave:353.717587748989°
(A:G, B:R, C:G) (Alice:C, Bob:B) Result:(GR)Diff Wave:204.43728181745195°
(A:G, B:R, C:G) (Alice:C, Bob:A) Result:(GG)Same Wave:20.983012712514256° <-
(A:G, B:G, C:R) (Alice:B, Bob:C) Result:(GR)Diff Wave:142.37439956338943°
(A:G, B:R, C:R) (Alice:B, Bob:C) Result:(RR)Same Wave:178.29335014198526° <-
(A:R, B:R, C:G) (Alice:C, Bob:B) Result:(GR)Diff Wave:259.7604226951664°
(A:R, B:G, C:R) (Alice:C, Bob:C) Result:(RR)Same Wave:118.22104001867596°
(A:R, B:R, C:G) (Alice:A, Bob:A) Result:(RR)Same Wave:243.49446988242934°
(A:R, B:G, C:R) (Alice:A, Bob:B) Result:(RG)Diff Wave:309.191668357614°
(A:G, B:G, C:R) (Alice:C, Bob:A) Result:(RG)Diff Wave:323.8647271988439°
(A:G, B:R, C:R) (Alice:A, Bob:A) Result:(GG)Same Wave:13.946472427685528°
(A:R, B:R, C:G) (Alice:C, Bob:C) Result:(GG)Same Wave:234.72520468634272°
(A:R, B:G, C:R) (Alice:C, Bob:C) Result:(RR)Same Wave:313.4941304009434°
(A:G, B:R, C:G) (Alice:C, Bob:A) Result:(GG)Same Wave:31.851310639507386° <-
(A:R, B:R, C:G) (Alice:B, Bob:A) Result:(RR)Same Wave:220.09222626242854° <-
(A:R, B:R, C:G) (Alice:B, Bob:A) Result:(RR)Same Wave:246.62362276103187° <-
(A:R, B:G, C:R) (Alice:C, Bob:B) Result:(RG)Diff Wave:105.67569489556982°
(A:G, B:R, C:R) (Alice:C, Bob:C) Result:(RR)Same Wave:344.86506464034517°
(A:R, B:G, C:G) (Alice:A, Bob:A) Result:(RR)Same Wave:93.73491259844621°
(A:G, B:R, C:R) (Alice:B, Bob:B) Result:(RR)Same Wave:17.372161662275264°
(A:G, B:G, C:R) (Alice:C, Bob:B) Result:(RG)Diff Wave:330.9473623306877°
(A:G, B:R, C:G) (Alice:B, Bob:B) Result:(RR)Same Wave:33.38294181576445°
(A:G, B:R, C:R) (Alice:B, Bob:B) Result:(RR)Same Wave:162.2677212635943°
(A:G, B:R, C:R) (Alice:C, Bob:B) Result:(RR)Same Wave:14.23186093208101° <-
(A:R, B:G, C:R) (Alice:C, Bob:C) Result:(RR)Same Wave:126.43868673509705°
(A:R, B:G, C:R) (Alice:B, Bob:A) Result:(GR)Diff Wave:104.45137908158726°
(A:R, B:G, C:R) (Alice:C, Bob:B) Result:(RG)Diff Wave:119.13226576630872°
(A:R, B:R, C:G) (Alice:A, Bob:A) Result:(RR)Same Wave:73.66093682638146°
(A:G, B:R, C:G) (Alice:A, Bob:B) Result:(GR)Diff Wave:203.67598891160878°
(A:R, B:R, C:G) (Alice:C, Bob:A) Result:(GR)Diff Wave:236.18228462020627°
(A:R, B:R, C:G) (Alice:A, Bob:C) Result:(RG)Diff Wave:254.44418076225924°
(A:R, B:G, C:R) (Alice:C, Bob:A) Result:(RR)Same Wave:317.2556548792309° <-
(A:G, B:G, C:R) (Alice:C, Bob:C) Result:(RR)Same Wave:159.75730971236067°
(A:R, B:G, C:R) (Alice:B, Bob:C) Result:(GR)Diff Wave:135.19957046552892°
(A:G, B:R, C:G) (Alice:A, Bob:C) Result:(GG)Same Wave:38.16741779834976° <-
(A:G, B:G, C:R) (Alice:B, Bob:C) Result:(GR)Diff Wave:147.42837822243075°
(A:G, B:R, C:R) (Alice:A, Bob:C) Result:(GR)Diff Wave:197.3206120463011°
(A:G, B:G, C:R) (Alice:A, Bob:C) Result:(GR)Diff Wave:143.86133740366955°
(A:G, B:G, C:R) (Alice:C, Bob:C) Result:(RR)Same Wave:335.5625220953304°
(A:R, B:R, C:G) (Alice:C, Bob:B) Result:(GR)Diff Wave:72.24531042336667°
(A:G, B:R, C:R) (Alice:A, Bob:B) Result:(GR)Diff Wave:174.84695084948635°
(A:R, B:G, C:R) (Alice:B, Bob:B) Result:(GG)Same Wave:313.9653741499807°
(A:R, B:R, C:G) (Alice:B, Bob:B) Result:(RR)Same Wave:225.45468768130047°
(A:G, B:G, C:R) (Alice:A, Bob:A) Result:(GG)Same Wave:153.10440542797002°
(A:R, B:R, C:G) (Alice:B, Bob:A) Result:(RR)Same Wave:79.03564053287022° <-
(A:G, B:G, C:R) (Alice:A, Bob:C) Result:(GR)Diff Wave:332.278541241043°
(A:R, B:R, C:G) (Alice:B, Bob:B) Result:(RR)Same Wave:51.437506275376464°
(A:R, B:G, C:R) (Alice:B, Bob:B) Result:(GG)Same Wave:131.94170120248234°
(A:R, B:R, C:G) (Alice:C, Bob:A) Result:(GR)Diff Wave:249.933870179784°
(A:G, B:R, C:R) (Alice:B, Bob:A) Result:(RG)Diff Wave:170.61355527100176°
(A:G, B:G, C:R) (Alice:A, Bob:C) Result:(GR)Diff Wave:324.4486209365025°
(A:G, B:R, C:R) (Alice:B, Bob:C) Result:(RR)Same Wave:176.41203368780063° <-
(A:G, B:R, C:G) (Alice:C, Bob:B) Result:(GR)Diff Wave:37.198388580297745°
(A:G, B:G, C:R) (Alice:C, Bob:A) Result:(RG)Diff Wave:339.3091989363497°
(A:R, B:R, C:G) (Alice:A, Bob:A) Result:(RR)Same Wave:252.22393108586778°
(A:G, B:R, C:R) (Alice:A, Bob:A) Result:(GG)Same Wave:1.9980738584536928°
(A:G, B:R, C:G) (Alice:B, Bob:B) Result:(RR)Same Wave:22.312254673238424°
(A:R, B:R, C:G) (Alice:A, Bob:A) Result:(RR)Same Wave:67.85310289343913°
(A:R, B:G, C:R) (Alice:A, Bob:B) Result:(RG)Diff Wave:316.95728140409244°
(A:R, B:R, C:G) (Alice:C, Bob:B) Result:(GR)Diff Wave:49.25931829314021°
(A:R, B:G, C:R) (Alice:A, Bob:C) Result:(RR)Same Wave:306.7749934488405° <-
 

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