Is Quantum Theory's Assumptions More Flawed Than Bell's?

[Hydr0matic]

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...

 

[Jonathan Scott]

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.

 

[Hydr0matic]

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?

 

[DrChinese]

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.

 

[JesseM]

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).

 

[Hydr0matic]

Thanks for the answers, and sorry for my delay..

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"?

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.

... 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.

 

[JesseM]

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!

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.

 

[Hydr0matic]

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.

 

[Doc Al]

Your polarizers are all 90.0° apart! Not a particularly interesting case.

 

[Hydr0matic]

Your polarizers are all 90.0° apart! Not a particularly interesting case.

The polarizers are aligned. The angle values is the lightwave polarization.

 

[Doc Al]

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.

 

[Hydr0matic]

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.

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.

 

[Doc Al]

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!

 

[JesseM]

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.

 

[Hydr0matic]

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.

 

[JesseM]

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.

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?

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?

 

[Hydr0matic]

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?

 

[Hydr0matic]

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

 

[DrChinese]

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.

 

[DrChinese]

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

 

[Hydr0matic]

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).

 

[JesseM]

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.

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.

 

[JesseM]

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.

 

[DrChinese]

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. :)

 

[Hydr0matic]

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.

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 =)

 

[Hydr0matic]

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.

 

[Hydr0matic]

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.

 

[JesseM]

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.

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!

 

[Hydr0matic]

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.

 

[JesseM]

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!

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.

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.

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.

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.

 

[DrChinese]

...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.

 

[Hydr0matic]

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.

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.

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.

You could avoid this problem by sticking to Mermin's example

You're right, let's do that.

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.

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.

 

[Hydr0matic]

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.

 

[JesseM]

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.

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.

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.

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.

 

[Hydr0matic]

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° <-

 

[JesseM]

With this setup, but negated result condition (RR or GG), I get:

What do you mean by this? Are you assuming that the wave sent to Alice has the same angle as the wave sent to Bob, rather than the assumption in your earlier experiment that the two waves were offset by 90 degrees? That's what you really should do if you want to satisfy my condition (1) above and ensure that whenever they choose the same angle, then regardless of what angle they both choose, they'll always get the same result (As you say below, I think it also works if the two waves are offset by 90 degrees and the detector threshold is cos^2(45), but it's a lot easier to understand conceptually if we just assume the waves have the same angle).

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.

The reason this works is that if they both choose some angle theta, and the angle of Alice's wave is phi while the angle of Bob's wave is phi + 90 (in degrees rather than radians, of course), then at Alice's detector we have the intensity reduced by cos^2(theta - phi) and at Bob's we have it reduced by cos^2([theta - phi] - 90), which is just equal to sin^2(theta - phi). Since the relation sin^2(omega) + cos^2(omega) = 1 holds for arbitrary values of omega, it must hold for omega = (theta - phi), so it must be true that if cos^2(theta - phi) is larger than 0.5 then sin^2(theta - phi) is smaller than 0.5, and vice versa...and of course, cos^2(45) is exactly 0.5! The only problematic case would be if theta - phi = 45, but you could solve this by saying that Alice's detector goes off if the wave's intensity coming out of the polarizer is greater than or equal to 0.5 its intensity before hitting the polarizer, while Bob's detector goes off if the wave's intensity coming out of the polarizer is greater than 0.5 its intensity before hitting the polarizer. This would ensure that they always got opposite results on every trial where they picked the same detector angle regardless of the angle of the two waves (which must be offset by 90 degrees).

However, it really is a lot simpler conceptually to assume both waves have the same angle, and Alice and Bob always get the same result on every trial where they pick the same detector angle. Unless you really prefer the idea of a 90-degree offset between the two waves.

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)

2109 out of 10000 what, exactly?

-------- 65.58% Non-detections in subset (RR)
-------- 32.41% Total Non-detections (RR)

Your notation is still confusing to me, I thought RR meant that they both got a red light, i.e. neither detector went off...wouldn't that mean you have 100% non-detections in this subset? What's the difference between the percentage of "Non-detections" and the percentage of "Total Non-detections"? And why are you looking at subsets based on whether the detector went off, rather than subsets based on what angle was chosen by Alice and Bob? What I was asking for earlier was to look at the results both in the subset of trials where they chose the same angle, and the results in the subset of trials where they chose different angles.

-------- Angle distribution: 11:1114, 12:1109, 13:1155, 21:1158, 22:1099, 23:1075, 31:1071, 32:1086, 33:1133,

What do these numbers represent? What is the significance of the numbers 11, 12, 13, 21, 22, 23, 31, 32, and 33, and the significance of the numbers following each one?

edit: never mind, it just occurred to me that you are probably using 1,2,3 to represent the three possible angles rather than A,B,C, so you're just counting the number of trials with each possible combination of angles.

(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° <-

OK, by my count you have 68 trials here, with 22 of these being ones where Alice and Bob chose the same setting (which is very close to 1/3 of the trials, so that's about what we should expect if they are choosing randomly). If we remove the trials where they chose the same setting and look only at the subset where they chose different settings, we have:

(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: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: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:G, C:R) (Alice:C, Bob:B) Result:(RG)Diff Wave:330.9473623306877°
(A:G, B:R, C:R) (Alice:C, Bob:B) Result:(RR)Same Wave:14.23186093208101° <-
(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: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: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: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: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: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:G, B:R, C:G) (Alice:B, Bob:B) Result:(RR)Same Wave:22.312254673238424°
(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° <-

Of these 46, there were 16 trials where they got the same result--that's about 0.35 of all the trials in this subset, which is greater than 1/3. Of course you can only be confident about whether or not a given data set is violating a Bell inequality when you look at a much larger number of trials, but from the data you've given so far the indication is that it isn't. So what I would suggest is doing something like the above for a larger run, this time having two counters that ignore trials such as (Alice:A, Bob:A) where they both chose the same setting, and on trials where they didn't choose the same setting the counter for "diff detector settings" is increased by one (so you're counting the total number of trials where Alice and Bob chose different settings), and the counter for "diff detector settings, same result" is increased by one only if the result was "Same" rather than "Diff" (counting the total number of trials where Alice and Bob chose different settings AND got the same result). After a large number of trials you can compare the two counters--if your program is working correctly, you should find that the value of the second counter is 1/3 or more the value of the first counter.

Meanwhile, I would really appreciate it if you would look over my proof of the Bell inequality, and tell me if you can see any reason why it shouldn't apply to your program if it's working correctly.

 

[Hydr0matic]

First of all, after your last post I realized I'd made quite a big mistake. Instead of counting the trials with results equal or opposite in a subset of angles same or different, I counted trials where both conditions matched in the total set.

To your questions...
Yes, RR means red red, no detection at alice or bob.
The total non-detections are removed since it wasn't relevant.
I replaced the 11, 12.. with AA, AB.. respectively.

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.

I have looked through your proof carefully and there is indeed no reason why it shouldn't apply. I've also tested it, and I got the following results:

With photon polarization Same and cut-off cos(20), when Alice and Bob chose same angle:

-------- RESULT: 3364/3364 (1)
-------- Non-detections (RR):2618/3364(77.82%)
-------- Angle distribution: BB:1185, AB:1044, AC:1140, BC:1105, AA:1096, CB:1104, CC:1083, BA:1150, CA:1093,
-------- Angles: A:0° B:120° C:240°
-------- Pairs in subset: 3364/10000

ResultCounts
Alice: (same as Bob)
RRR:3290
RGR:2255
GRR:2220
RRG:2235

When they chose the same angle I get a 100% correlation regardless of the cut-off value. The cut-off only affects the number of non-detections.

When they chose different angles, I get:

-------- RESULT: 3744/6636 (0.5641952983725136)
-------- Non-detections (RR):3744/6636 (56.42%)
-------- Angle distribution: BC:1110, AA:1077, CA:1092, BB:1149, AB:1036, CB:1115, CC:1138, BA:1179, AC:1104,
-------- Angles: A:0° B:120° C:240°
-------- Pairs in subset: 6636/10000

ResultCounts
Alice: (same as Bob)
GRR:2213
RRG:2187
RGR:2200
RRR:3400

Now, clearly this is NOT a violation of Bell's inequality. HOWEVER, as you can see, 100% of the result are non-detections. In fact, unless the cut-off value is less than cos(30), ALL correlations are non-detections. Now, I've read a couple of times that one of the main objections to the Bell tests is the low detector efficiency. So I would assume that in real tests, detectors cut-off value would range within cos(0) and cos(30). In which case, whenever they checked results at different angles, they were all non-detections.
I would imagine that this is the reason why they are using sources that emit photons with opposite polarization instead of the same.

However, it really is a lot simpler conceptually to assume both waves have the same angle, and Alice and Bob always get the same result on every trial where they pick the same detector angle. Unless you really prefer the idea of a 90-degree offset between the two waves.

Judging from the results above, it would probably be more appropriate to use sources with opposite polarization.

So, if we change the polarization in my application to Opposite, and do the above tests again. First counting opposite results when Alice and Bob chose the same angle:

cut-off°
10 --- RESULT: 728/3295 (0.220) --- Non-detections (RR):2567/3295 (728+2567=3295)
20 --- RESULT: 1487/3367 (0.441) --- Non-detections (RR):1880/3367 (1487+1880=3367)
30 --- RESULT: 2214/3310 (0.668) --- Non-detections (RR):1096/3310 (2214+1096=3310)
40 --- RESULT: 2926/3288 (0.889) --- Non-detections (RR):362/3288 (2926+362=3288)
45 --- RESULT: 3269/3269 (1.000) --- Non-detections (RR):0/3269
50 --- RESULT: 2991/3389 (0.882) --- Non-detections (RR):0/3389 (2991=3389-398) All RRs turn into GGs (>45)
60 --- RESULT: 2200/3338 (0.659) --- Non-detections (RR):0/3338 (2200=3338-1138)

Analysing these results we see that condition 1 is only truly fulfilled with cut-off value cos(45). However, when the cut-off angle is less than 45° all non-opposite results are non-detections. So, whenever Alice or Bob detects something, the other one always get opposite result.

Counting opposite results when Alice and Bob chose different angles:

cut-off°
5 ----- RESULT: 759/6712 (0.113) --- Non-detections (RR):5953/6712 (759+5953=6712)
10 --- RESULT: 1450/6710 (0.216) --- Non-detections (RR):5260/6710 (1450+5260=6710)
15 --- RESULT: 2264/6676 (0.339) --- Non-detections (RR):4412/6676 (2264+4412=6676)
20 --- RESULT: 2287/6664 (0.343)---- Non-detections (RR):3984/6664 (2287+3984=6664-393) (393 GGs)
30 --- RESULT: 2199/6732 (0.326) --- Non-detections (RR):3412/6732 (2199+3412=6732-1121)
40 --- RESULT: 2184/6669 (0.327) --- Non-detections (RR):2619/6669 (2184+2619=6669-1866)
45 --- RESULT: 2273/6720 (0.338) --- Non-detections (RR):2277/6720 (2273+2277=6720-2170)
50 --- RESULT: 2250/6637 (0.339) --- Non-detections (RR):1803/6637 (2250+1803=6637-2584)
60 --- RESULT: 2241/6646 (0.337) --- Non-detections (RR):1077/6646 (2241+1077=6646-3328)
70 --- RESULT: 2284/6675 (0.342) --- Non-detections (RR):385/6675 (2284+385=6675-4006)
75 --- RESULT: 2217/6519 (0.340) --- Non-detections (RR):0/6519 (2217+0=6675-4458)
80 --- RESULT: 1487/6745 (0.220) --- Non-detections (RR):0/6745 (1487+0=6745-5258)
85 --- RESULT: 743/6661 (0.111) --- Non-detections (RR):0/6661 (743+0=6661-5918)

First of all, notice that cut-offs 5,10,80 & 85 give results that violate Bell's inequality(*). More tests reveal that results from cut-off values higher than cos(15) or lower than cos(75) violate Bell's(*).
Second, notice that for all other cut-off values, the result is approximately 1/3. So the classical lightwave model predicts that, when Alice and Bob measure different angles of pairs with opposite polarization, they will get opposite results with probability 1/3 OR LESS, regardless of cut-off value.

* Since it's only cut-off cos(45) that truly satisfies condition 1, one cannot say that cut-offs at 5,10,80 & 85 violate Bell's inequality.

Examining cut-off values with other ABC angles, for example 0°,45° and 90°, one finds that the boundry for violation of Bell's* is changed to 22.5°. i.e the boundry is half of the difference between the measured angles. (Or equal to, if you consider the whole range -22.5° <> 22.5°)

Here is a complete result with 100 trials from the above test at cut-off cos(10):
Lines marked with <- at the end are part of the subset.

-------- RESULT: 1552/6613 (0.23468924845002267)
-------- Non-detections (RR):5061/6613(76.53%)
-------- Angle distribution: AA:1138, CA:1102, AB:1118, BA:1102, AC:1058, CC:1119, CB:1092, BC:1141, BB:1130,
-------- Angles: A:0° B:120° C:240°
-------- Pairs in subset: 6613/10000

ResultCounts
Alice:
RRG:1079
RRR:6759
RGR:1094
GRR:1068

Bob:
RRR:6598
RGR:1137
GRR:1140
RRG:1125


Alice: Bob:
(A:R, B:R, C:G) (A:R, B:R, C:R) (Alice:A, Bob:A) Result:(RR)Same Wave1:234.1378800941786° Wave2: 324.1378800941786°
(A:R, B:R, C:R) (A:R, B:G, C:R) (Alice:C, Bob:A) Result:(RR)Same Wave1:35.72264460666506° Wave2: 125.72264460666506° <-
(A:R, B:R, C:R) (A:R, B:R, C:R) (Alice:A, Bob:B) Result:(RR)Same Wave1:42.20000826206056° Wave2: 132.20000826206055° <-
(A:R, B:G, C:R) (A:R, B:R, C:R) (Alice:A, Bob:B) Result:(RR)Same Wave1:123.5108071256557° Wave2: 213.5108071256557° <-
(A:R, B:G, C:R) (A:R, B:R, C:R) (Alice:B, Bob:A) Result:(GR)Diff Wave1:123.5276037607258° Wave2: 213.5276037607258° <-
(A:R, B:R, C:R) (A:R, B:R, C:R) (Alice:A, Bob:A) Result:(RR)Same Wave1:160.73075126416177° Wave2: 250.73075126416177°
(A:R, B:G, C:R) (A:R, B:R, C:R) (Alice:A, Bob:C) Result:(RR)Same Wave1:118.53166896417368° Wave2: 208.53166896417366° <-
(A:R, B:R, C:R) (A:G, B:R, C:R) (Alice:C, Bob:C) Result:(RR)Same Wave1:267.1078710578241° Wave2: 357.1078710578241°
(A:R, B:R, C:R) (A:R, B:R, C:R) (Alice:B, Bob:A) Result:(RR)Same Wave1:225.4604050311087° Wave2: 315.46040503110873° <-
(A:R, B:R, C:R) (A:R, B:G, C:R) (Alice:B, Bob:A) Result:(RR)Same Wave1:37.572428656288245° Wave2: 127.57242865628825° <-
(A:R, B:G, C:R) (A:R, B:R, C:R) (Alice:C, Bob:A) Result:(RR)Same Wave1:291.2279827790366° Wave2: 21.22798277903661° <-
(A:R, B:R, C:R) (A:R, B:R, C:R) (Alice:C, Bob:C) Result:(RR)Same Wave1:71.79740547206602° Wave2: 161.79740547206603°
(A:R, B:G, C:R) (A:R, B:R, C:R) (Alice:C, Bob:B) Result:(RR)Same Wave1:293.993078538492° Wave2: 23.993078538492° <-
(A:G, B:R, C:R) (A:R, B:R, C:R) (Alice:B, Bob:C) Result:(RR)Same Wave1:3.710932337109001° Wave2: 93.710932337109° <-
(A:R, B:R, C:R) (A:R, B:R, C:R) (Alice:A, Bob:B) Result:(RR)Same Wave1:192.85550819474088° Wave2: 282.8555081947409° <-
(A:R, B:R, C:R) (A:R, B:R, C:G) (Alice:C, Bob:C) Result:(RG)Diff Wave1:155.10432541233536° Wave2: 245.10432541233536°
(A:R, B:R, C:G) (A:R, B:R, C:R) (Alice:B, Bob:C) Result:(RR)Same Wave1:53.92845630720172° Wave2: 143.9284563072017° <-
(A:R, B:R, C:R) (A:R, B:R, C:G) (Alice:A, Bob:B) Result:(RR)Same Wave1:321.7942596025368° Wave2: 51.79425960253678° <-
(A:R, B:R, C:R) (A:R, B:R, C:R) (Alice:B, Bob:A) Result:(RR)Same Wave1:340.74542582860863° Wave2: 70.74542582860863° <-
(A:R, B:R, C:R) (A:R, B:R, C:R) (Alice:B, Bob:B) Result:(RR)Same Wave1:346.45056727244116° Wave2: 76.45056727244116°
(A:R, B:G, C:R) (A:R, B:R, C:R) (Alice:B, Bob:B) Result:(GR)Diff Wave1:301.45060740427596° Wave2: 31.450607404275956°
(A:R, B:G, C:R) (A:R, B:R, C:R) (Alice:B, Bob:A) Result:(GR)Diff Wave1:292.3347606939048° Wave2: 22.334760693904798° <-
(A:G, B:R, C:R) (A:R, B:R, C:R) (Alice:A, Bob:A) Result:(GR)Diff Wave1:3.3515104072192603° Wave2: 93.35151040721927°
(A:R, B:R, C:R) (A:G, B:R, C:R) (Alice:C, Bob:A) Result:(RG)Diff Wave1:91.84621678526328° Wave2: 181.84621678526327° <-
(A:R, B:R, C:R) (A:R, B:R, C:R) (Alice:C, Bob:C) Result:(RR)Same Wave1:41.90960393497796° Wave2: 131.90960393497795°
(A:R, B:R, C:R) (A:G, B:R, C:R) (Alice:B, Bob:A) Result:(RG)Diff Wave1:269.8585535605929° Wave2: 359.8585535605929° <-
(A:R, B:R, C:R) (A:G, B:R, C:R) (Alice:A, Bob:A) Result:(RG)Diff Wave1:277.70946204703966° Wave2: 7.709462047039665°
(A:G, B:R, C:R) (A:R, B:R, C:R) (Alice:B, Bob:A) Result:(RR)Same Wave1:353.13632396738507° Wave2: 83.13632396738507° <-
(A:R, B:R, C:R) (A:R, B:R, C:R) (Alice:A, Bob:B) Result:(RR)Same Wave1:319.85682382439944° Wave2: 49.85682382439944° <-
(A:R, B:R, C:R) (A:R, B:R, C:R) (Alice:C, Bob:C) Result:(RR)Same Wave1:194.31557633767088° Wave2: 284.3155763376709°
(A:R, B:R, C:R) (A:R, B:R, C:R) (Alice:C, Bob:A) Result:(RR)Same Wave1:192.96231128480113° Wave2: 282.9623112848011° <-
(A:R, B:G, C:R) (A:R, B:R, C:R) (Alice:A, Bob:A) Result:(RR)Same Wave1:294.5512712344834° Wave2: 24.55127123448341°
(A:R, B:R, C:R) (A:R, B:R, C:R) (Alice:A, Bob:A) Result:(RR)Same Wave1:109.59643598677565° Wave2: 199.59643598677565°
(A:R, B:R, C:R) (A:R, B:R, C:R) (Alice:B, Bob:A) Result:(RR)Same Wave1:17.39269320572545° Wave2: 107.39269320572545° <-
(A:G, B:R, C:R) (A:R, B:R, C:R) (Alice:A, Bob:B) Result:(GR)Diff Wave1:359.95549344857426° Wave2: 89.95549344857426° <-
(A:R, B:R, C:R) (A:R, B:R, C:R) (Alice:C, Bob:B) Result:(RR)Same Wave1:314.0349337884509° Wave2: 44.03493378845093° <-
(A:G, B:R, C:R) (A:R, B:R, C:R) (Alice:B, Bob:A) Result:(RR)Same Wave1:181.4324188574362° Wave2: 271.43241885743623° <-
(A:R, B:R, C:R) (A:R, B:G, C:R) (Alice:B, Bob:C) Result:(RR)Same Wave1:214.97368166443803° Wave2: 304.973681664438° <-
(A:R, B:R, C:R) (A:R, B:R, C:R) (Alice:C, Bob:B) Result:(RR)Same Wave1:165.69429826353536° Wave2: 255.69429826353536° <-
(A:R, B:R, C:R) (A:R, B:R, C:G) (Alice:A, Bob:B) Result:(RR)Same Wave1:145.301710799591° Wave2: 235.301710799591° <-
(A:R, B:R, C:G) (A:R, B:R, C:R) (Alice:A, Bob:C) Result:(RR)Same Wave1:239.45457214043003° Wave2: 329.45457214043006° <-
(A:G, B:R, C:R) (A:R, B:R, C:R) (Alice:A, Bob:C) Result:(GR)Diff Wave1:352.3414943341509° Wave2: 82.34149433415092° <-
(A:R, B:R, C:R) (A:R, B:G, C:R) (Alice:A, Bob:C) Result:(RR)Same Wave1:203.99552547397013° Wave2: 293.99552547397013° <-
(A:R, B:R, C:R) (A:R, B:R, C:R) (Alice:A, Bob:B) Result:(RR)Same Wave1:348.79981944898356° Wave2: 78.79981944898356° <-
(A:R, B:R, C:R) (A:R, B:G, C:R) (Alice:A, Bob:A) Result:(RR)Same Wave1:201.12117214272135° Wave2: 291.1211721427213°
(A:R, B:R, C:R) (A:R, B:R, C:R) (Alice:B, Bob:A) Result:(RR)Same Wave1:47.82000276013689° Wave2: 137.8200027601369° <-
(A:R, B:R, C:R) (A:R, B:R, C:R) (Alice:C, Bob:A) Result:(RR)Same Wave1:19.95420944676516° Wave2: 109.95420944676516° <-
(A:R, B:G, C:R) (A:R, B:R, C:R) (Alice:C, Bob:B) Result:(RR)Same Wave1:125.7349113218338° Wave2: 215.7349113218338° <-
(A:R, B:R, C:R) (A:R, B:R, C:G) (Alice:A, Bob:B) Result:(RR)Same Wave1:337.0598713583604° Wave2: 67.0598713583604° <-
(A:R, B:R, C:R) (A:R, B:R, C:R) (Alice:A, Bob:C) Result:(RR)Same Wave1:223.4035984747488° Wave2: 313.40359847474883° <-
(A:R, B:R, C:R) (A:R, B:R, C:R) (Alice:B, Bob:B) Result:(RR)Same Wave1:165.88525275241784° Wave2: 255.88525275241784°
(A:R, B:R, C:R) (A:G, B:R, C:R) (Alice:B, Bob:C) Result:(RR)Same Wave1:269.0346627988359° Wave2: 359.0346627988359° <-
(A:R, B:G, C:R) (A:R, B:R, C:R) (Alice:B, Bob:C) Result:(GR)Diff Wave1:111.42251244920784° Wave2: 201.42251244920783° <-
(A:R, B:R, C:R) (A:R, B:R, C:G) (Alice:C, Bob:B) Result:(RR)Same Wave1:157.5249478861851° Wave2: 247.5249478861851° <-
(A:R, B:R, C:R) (A:R, B:R, C:R) (Alice:B, Bob:C) Result:(RR)Same Wave1:340.1889934539382° Wave2: 70.18899345393822° <-
(A:R, B:R, C:G) (A:R, B:R, C:R) (Alice:B, Bob:C) Result:(RR)Same Wave1:248.61150797931379° Wave2: 338.6115079793138° <-
(A:G, B:R, C:R) (A:R, B:R, C:R) (Alice:A, Bob:C) Result:(GR)Diff Wave1:358.79360815426685° Wave2: 88.79360815426685° <-
(A:R, B:R, C:R) (A:R, B:R, C:R) (Alice:B, Bob:B) Result:(RR)Same Wave1:165.4625205759179° Wave2: 255.4625205759179°
(A:R, B:R, C:R) (A:R, B:R, C:R) (Alice:B, Bob:A) Result:(RR)Same Wave1:48.12877458474593° Wave2: 138.12877458474594° <-
(A:R, B:R, C:R) (A:R, B:R, C:R) (Alice:B, Bob:B) Result:(RR)Same Wave1:315.5072208466679° Wave2: 45.507220846667906°
(A:R, B:R, C:R) (A:G, B:R, C:R) (Alice:C, Bob:A) Result:(RG)Diff Wave1:263.06639095605215° Wave2: 353.06639095605215° <-
(A:R, B:R, C:R) (A:R, B:R, C:G) (Alice:C, Bob:A) Result:(RR)Same Wave1:326.99124381048114° Wave2: 56.99124381048114° <-
(A:R, B:R, C:G) (A:R, B:R, C:R) (Alice:C, Bob:C) Result:(GR)Diff Wave1:64.79753534354637° Wave2: 154.79753534354637°
(A:R, B:G, C:R) (A:R, B:R, C:R) (Alice:C, Bob:C) Result:(RR)Same Wave1:300.47969331972365° Wave2: 30.47969331972365°
(A:R, B:R, C:R) (A:R, B:R, C:R) (Alice:C, Bob:C) Result:(RR)Same Wave1:226.86417866602403° Wave2: 316.86417866602403°
(A:R, B:R, C:G) (A:R, B:R, C:R) (Alice:C, Bob:C) Result:(GR)Diff Wave1:238.73102312436706° Wave2: 328.7310231243671°
(A:R, B:R, C:R) (A:R, B:R, C:R) (Alice:B, Bob:A) Result:(RR)Same Wave1:75.58966665762914° Wave2: 165.58966665762915° <-
(A:R, B:R, C:R) (A:G, B:R, C:R) (Alice:A, Bob:B) Result:(RR)Same Wave1:277.31875612414126° Wave2: 7.3187561241412595° <-
(A:R, B:R, C:R) (A:R, B:R, C:G) (Alice:A, Bob:B) Result:(RR)Same Wave1:336.9991552178912° Wave2: 66.9991552178912° <-
(A:R, B:R, C:R) (A:R, B:R, C:R) (Alice:C, Bob:A) Result:(RR)Same Wave1:10.24960171915506° Wave2: 100.24960171915507° <-
(A:R, B:G, C:R) (A:R, B:R, C:R) (Alice:C, Bob:C) Result:(RR)Same Wave1:298.2832574224113° Wave2: 28.283257422411282°
(A:R, B:R, C:R) (A:G, B:R, C:R) (Alice:A, Bob:A) Result:(RG)Diff Wave1:99.6544700072415° Wave2: 189.6544700072415°
(A:R, B:R, C:R) (A:R, B:R, C:R) (Alice:A, Bob:A) Result:(RR)Same Wave1:222.12437116324668° Wave2: 312.1243711632467°
(A:R, B:R, C:G) (A:R, B:R, C:R) (Alice:C, Bob:B) Result:(GR)Diff Wave1:248.3276771483694° Wave2: 338.3276771483694° <-
(A:G, B:R, C:R) (A:R, B:R, C:R) (Alice:C, Bob:B) Result:(RR)Same Wave1:173.37716366170937° Wave2: 263.37716366170935° <-
(A:R, B:R, C:R) (A:R, B:R, C:R) (Alice:B, Bob:C) Result:(RR)Same Wave1:282.2478201393047° Wave2: 12.247820139304679° <-
(A:R, B:R, C:R) (A:G, B:R, C:R) (Alice:A, Bob:B) Result:(RR)Same Wave1:263.1492525837448° Wave2: 353.1492525837448° <-
(A:R, B:R, C:R) (A:R, B:G, C:R) (Alice:A, Bob:A) Result:(RR)Same Wave1:22.8958136578825° Wave2: 112.8958136578825°
(A:R, B:R, C:G) (A:R, B:R, C:R) (Alice:A, Bob:A) Result:(RR)Same Wave1:65.61605792399588° Wave2: 155.61605792399587°
(A:R, B:R, C:R) (A:R, B:R, C:R) (Alice:A, Bob:A) Result:(RR)Same Wave1:223.73056283271416° Wave2: 313.73056283271416°
(A:R, B:G, C:R) (A:R, B:R, C:R) (Alice:B, Bob:C) Result:(GR)Diff Wave1:115.15263077987316° Wave2: 205.15263077987316° <-
(A:R, B:R, C:R) (A:R, B:G, C:R) (Alice:C, Bob:B) Result:(RG)Diff Wave1:21.004246593556807° Wave2: 111.00424659355681° <-
(A:R, B:R, C:R) (A:R, B:R, C:G) (Alice:C, Bob:A) Result:(RR)Same Wave1:149.6761176325398° Wave2: 239.6761176325398° <-
(A:R, B:R, C:R) (A:R, B:R, C:G) (Alice:B, Bob:A) Result:(RR)Same Wave1:157.48701199037342° Wave2: 247.48701199037342° <-
(A:R, B:R, C:R) (A:R, B:G, C:R) (Alice:A, Bob:C) Result:(RR)Same Wave1:211.44143224704325° Wave2: 301.4414322470433° <-
(A:R, B:R, C:R) (A:R, B:R, C:G) (Alice:B, Bob:C) Result:(RG)Diff Wave1:151.7881553779931° Wave2: 241.7881553779931° <-
(A:R, B:R, C:R) (A:R, B:R, C:G) (Alice:B, Bob:A) Result:(RR)Same Wave1:336.77253467714985° Wave2: 66.77253467714985° <-
(A:R, B:R, C:R) (A:R, B:R, C:G) (Alice:C, Bob:B) Result:(RR)Same Wave1:150.84780853014178° Wave2: 240.84780853014178° <-
(A:R, B:R, C:G) (A:R, B:R, C:R) (Alice:A, Bob:B) Result:(RR)Same Wave1:66.06742015546814° Wave2: 156.06742015546814° <-
(A:R, B:G, C:R) (A:R, B:R, C:R) (Alice:B, Bob:A) Result:(GR)Diff Wave1:125.74272701948371° Wave2: 215.7427270194837° <-
(A:R, B:R, C:G) (A:R, B:R, C:R) (Alice:C, Bob:C) Result:(GR)Diff Wave1:236.48965566643° Wave2: 326.48965566643°
(A:R, B:R, C:R) (A:R, B:R, C:R) (Alice:A, Bob:B) Result:(RR)Same Wave1:109.9916145019468° Wave2: 199.9916145019468° <-
(A:R, B:R, C:R) (A:R, B:G, C:R) (Alice:B, Bob:C) Result:(RR)Same Wave1:210.52822909816123° Wave2: 300.5282290981612° <-
(A:R, B:R, C:R) (A:R, B:R, C:R) (Alice:B, Bob:B) Result:(RR)Same Wave1:315.72372067606165° Wave2: 45.723720676061646°
(A:R, B:R, C:R) (A:R, B:R, C:G) (Alice:B, Bob:B) Result:(RR)Same Wave1:339.4231793713175° Wave2: 69.42317937131747°
(A:R, B:R, C:R) (A:R, B:G, C:R) (Alice:B, Bob:B) Result:(RG)Diff Wave1:208.92772526638552° Wave2: 298.92772526638555°
(A:R, B:G, C:R) (A:R, B:R, C:R) (Alice:C, Bob:C) Result:(RR)Same Wave1:293.54305481361735° Wave2: 23.543054813617346°
(A:R, B:R, C:R) (A:G, B:R, C:R) (Alice:A, Bob:C) Result:(RR)Same Wave1:274.94079044002507° Wave2: 4.940790440025069° <-
(A:R, B:R, C:R) (A:G, B:R, C:R) (Alice:B, Bob:B) Result:(RR)Same Wave1:91.26311868162189° Wave2: 181.2631186816219°
(A:R, B:R, C:R) (A:R, B:R, C:R) (Alice:B, Bob:A) Result:(RR)Same Wave1:165.65882919660118° Wave2: 255.65882919660118° <-

 

[JesseM]

-------- RESULT: 3744/6636 (0.5641952983725136)
-------- Non-detections (RR):3744/6636 (56.42%)
-------- Angle distribution: BC:1110, AA:1077, CA:1092, BB:1149, AB:1036, CB:1115, CC:1138, BA:1179, AC:1104,
-------- Angles: A:0° B:120° C:240°
-------- Pairs in subset: 6636/10000

ResultCounts
Alice: (same as Bob)
GRR:2213
RRG:2187
RGR:2200
RRR:3400

Now, clearly this is NOT a violation of Bell's inequality. HOWEVER, as you can see, 100% of the result are non-detections.

I don't understand, what part of the listed results am I supposed to see that in? Does the 6636/10000 you got for "pairs in subset" refer to the subset of trials where Alice and Bob chose different angles? If so, then if they got RR on 3744/6636, doesn't this mean that 56.42% of the results in this subset are non-detections? Or does "non-detections" mean something different from RR? Maybe you just mean that at least one of them got R, so that on 100% of the trials in the subset where they chose different angles, their results were either RR or RG or GR, but never GG? Even if that's it, I don't understand how I could see this from what you wrote.

In fact, unless the cut-off value is less than cos(30), ALL correlations are non-detections.

You mean, in the subset of trials where they choose different angles, all correlations are non-detections? This makes sense, since if they choose different angles their two detectors are either 120 or 240 apart...if the wave's angle is between 0 and 30 away from Alice's angle, then it's got be in one of four possible ranges away from Bob's: 120-150, 120-90, 240-270, or 240-210. In the ranges 120-90 and 240-270 there's no way the cos^2 of Bob's angle could be as large as cos^2(30), and in the ranges 120-150 or 240-210, the cos^2 of Bob's angle would only reach cos^2(30) if his angle was exactly 150 or 210 away from the wave's angle...and if you're choosing the wave's angle randomly, the chances of this are infinitesimal.

Now, I've read a couple of times that one of the main objections to the Bell tests is the low detector efficiency. So I would assume that in real tests, detectors cut-off value would range within cos(0) and cos(30). In which case, whenever they checked results at different angles, they were all non-detections.

But in real tests, the detectors don't work the way they do in your classical model...if the polarization is random, each detector should ideally get R 50% of the time and G 50% of the time, but in QM the probability of the two detectors getting the same result on a given trial is simply equal to the cos^2 of the angle between the two detectors. So if Alice gets G when hers is set to 120, and Bob's is set to 240 on that trial, there should be a cos^2(240-120) = cos^2(120) = 0.25 probability that Bob also gets G.

And I think "detector efficiency" in QM doesn't work like in your model where there's a cutoff, I would guess it just uniformly drops the probability that a given photon gets through...for example if the detector efficiency were 0.2, that would mean even if their detectors were aligned, if one got G there would only have an 0.2 probability of also getting G, while if their detectors were 120 degrees apart, if one got G the other would only have an 0.25*0.2 = 0.05 probability of also getting G.

So, if we change the polarization in my application to Opposite, and do the above tests again. First counting opposite results when Alice and Bob chose the same angle:

cut-off°
10 --- RESULT: 728/3295 (0.220) --- Non-detections (RR):2567/3295 (728+2567=3295)
20 --- RESULT: 1487/3367 (0.441) --- Non-detections (RR):1880/3367 (1487+1880=3367)
30 --- RESULT: 2214/3310 (0.668) --- Non-detections (RR):1096/3310 (2214+1096=3310)
40 --- RESULT: 2926/3288 (0.889) --- Non-detections (RR):362/3288 (2926+362=3288)
45 --- RESULT: 3269/3269 (1.000) --- Non-detections (RR):0/3269
50 --- RESULT: 2991/3389 (0.882) --- Non-detections (RR):0/3389 (2991=3389-398) All RRs turn into GGs (>45)
60 --- RESULT: 2200/3338 (0.659) --- Non-detections (RR):0/3338 (2200=3338-1138)

Analysing these results we see that condition 1 is only truly fulfilled with cut-off value cos(45). However, when the cut-off angle is less than 45° all non-opposite results are non-detections. So, whenever Alice or Bob detects something, the other one always get opposite result.

Counting opposite results when Alice and Bob chose different angles:

cut-off°
5 ----- RESULT: 759/6712 (0.113) --- Non-detections (RR):5953/6712 (759+5953=6712)
10 --- RESULT: 1450/6710 (0.216) --- Non-detections (RR):5260/6710 (1450+5260=6710)
15 --- RESULT: 2264/6676 (0.339) --- Non-detections (RR):4412/6676 (2264+4412=6676)
20 --- RESULT: 2287/6664 (0.343)---- Non-detections (RR):3984/6664 (2287+3984=6664-393) (393 GGs)
30 --- RESULT: 2199/6732 (0.326) --- Non-detections (RR):3412/6732 (2199+3412=6732-1121)
40 --- RESULT: 2184/6669 (0.327) --- Non-detections (RR):2619/6669 (2184+2619=6669-1866)
45 --- RESULT: 2273/6720 (0.338) --- Non-detections (RR):2277/6720 (2273+2277=6720-2170)
50 --- RESULT: 2250/6637 (0.339) --- Non-detections (RR):1803/6637 (2250+1803=6637-2584)
60 --- RESULT: 2241/6646 (0.337) --- Non-detections (RR):1077/6646 (2241+1077=6646-3328)
70 --- RESULT: 2284/6675 (0.342) --- Non-detections (RR):385/6675 (2284+385=6675-4006)
75 --- RESULT: 2217/6519 (0.340) --- Non-detections (RR):0/6519 (2217+0=6675-4458)
80 --- RESULT: 1487/6745 (0.220) --- Non-detections (RR):0/6745 (1487+0=6745-5258)
85 --- RESULT: 743/6661 (0.111) --- Non-detections (RR):0/6661 (743+0=6661-5918)

First of all, notice that cut-offs 5,10,80 & 85 give results that violate Bell's inequality(*). More tests reveal that results from cut-off values higher than cos(15) or lower than cos(75) violate Bell's(*).
Second, notice that for all other cut-off values, the result is approximately 1/3. So the classical lightwave model predicts that, when Alice and Bob measure different angles of pairs with opposite polarization, they will get opposite results with probability 1/3 OR LESS, regardless of cut-off value.

* Since it's only cut-off cos(45) that truly satisfies condition 1, one cannot say that cut-offs at 5,10,80 & 85 violate Bell's inequality.

Yes, and notice that when you make the cutoff cos^2(45) (not cos(45) as you say...I know what you meant, but the difference is important!), the fraction of trials where they get the same result is closer to 1/3 than for any other angle. You had said earlier that in the case where the two waves were 90 degrees apart, then the predetermined states of the waves were never (A:R, B:R, C:R), or (A:G, B:G, C:G)...the predetermined states were always "inhomogeneous". And as I showed in the proof, for inhomogeneous predetermined states, if Alice and Bob choose their angles randomly, then on the subset of trials where they choose different angles, the probability of getting opposite results should be exactly 1/3 (assuming that when they pick the same angle they always get opposite results). So I think if you did an even larger number of trials, the fraction in the cutoff = cos^2(45) case would tend to converge towards 0.33333...

 

[DrChinese]

I think if you did an even larger number of trials, the fraction in the cutoff = cos^2(45) case would tend to converge towards 0.33333...

That's true of course, if the underlying algorithm is correct - which I question anyway. The real issue is that there are no samples of A, B and C - regardless of how you decide to seed it - that meet the criteria that:

a) Alice and Bob always get the same result when AA, BB, or CC are selected; and
b) The sample converges on any number less than .333 when different combos are selected; and
c) Non-detections are random.

The last item is the "fair-sampling hypothesis" (JesseM is already well aware of this). This is an experiemental issue that deserves a separate discussion. Suffice it to say that no evidence of a violation of the "fair sampling hypothesis" has ever been demonstrated.

However, there are a few people who believe it that "fair sampling" does NOT occur, and that explains the overall violation of Bell's Inequality. The problem with that concept is that it points to a different (and very strange) relationship than the observed cos^2 rule. The other big problem is that as samples get relatively larger due to improvements in technology, the results converge ever more closely on the Quantum predictions.

 

[Hydr0matic]

I don't understand, what part of the listed results am I supposed to see that in? Does the 6636/10000 you got for "pairs in subset" refer to the subset of trials where Alice and Bob chose different angles? If so, then if they got RR on 3744/6636, doesn't this mean that 56.42% of the results in this subset are non-detections? Or does "non-detections" mean something different from RR? Maybe you just mean that at least one of them got R, so that on 100% of the trials in the subset where they chose different angles, their results were either RR or RG or GR, but never GG? Even if that's it, I don't understand how I could see this from what you wrote.

6636 is the subset where Alice and Bob chose different angles, yes. According to Bell, given condition 1, when we look at the results where they chose different angles, they would get the Same result at least 1/3 of the trials. So the number 3744 represents the trials (in the subset) where they got the Same result. Which were 56.42% of the trials (more than 1/3), which is not a violation.
Non-detection means RR, yes. And since we're only counting RRs and GGs (Same result), if I get 3744 RR's then ALL of the counts must be RRs.

You mean, in the subset of trials where they choose different angles, all correlations are non-detections? This makes sense, since if they choose different angles their two detectors are either 120 or 240 apart...if the wave's angle is between 0 and 30 away from Alice's angle, then it's got be in one of four possible ranges away from Bob's: 120-150, 120-90, 240-270, or 240-210. In the ranges 120-90 and 240-270 there's no way the cos^2 of Bob's angle could be as large as cos^2(30), and in the ranges 120-150 or 240-210, the cos^2 of Bob's angle would only reach cos^2(30) if his angle was exactly 150 or 210 away from the wave's angle...and if you're choosing the wave's angle randomly, the chances of this are infinitesimal.

Exactly.

But in real tests, the detectors don't work the way they do in your classical model...if the polarization is random, each detector should ideally get R 50% of the time and G 50% of the time, but in QM the probability of the two detectors getting the same result on a given trial is simply equal to the cos^2 of the angle between the two detectors. So if Alice gets G when hers is set to 120, and Bob's is set to 240 on that trial, there should be a cos^2(240-120) = cos^2(120) = 0.25 probability that Bob also gets G.

Ok, this is what bothers me. How can they assume how the polarizers work? Is there any experiment made that proves polarizers are governed by chance? Maybe I've missed it. How can physicists be more open to tossing out locality, than to reconsider assumptions about how polarizers work?

Yes, and notice that when you make the cutoff cos^2(45) (not cos(45) as you say...I know what you meant, but the difference is important!)

The choice of cos and not cos² seems more reasonable to me. Intensity is a concept relating to lots of light. I mean, what's the intensity of one photon?
If you use the absolute value of cos there's no difference anyway (cos²(a) > cos²(b) => |cos(a)| > |cos(b)|, same for =, <). But maybe I'm wrong.

You had said earlier that in the case where the two waves were 90 degrees apart, then the predetermined states of the waves were never (A:R, B:R, C:R), or (A:G, B:G, C:G)...the predetermined states were always "inhomogeneous".

For a specific case, yes. This is completely dependant on the cut-off value. A high cut-off cos(20) gives only (RRR, GRR, RGR, RRG), a cut-off cos(45) gives all states except (RRR, GGG), and cut-off cos(70) gives cos(20) inversed (GGG, RGG, GRG, GGR).

And as I showed in the proof, for inhomogeneous predetermined states, if Alice and Bob choose their angles randomly, then on the subset of trials where they choose different angles, the probability of getting opposite results should be exactly 1/3 (assuming that when they pick the same angle they always get opposite results). So I think if you did an even larger number of trials, the fraction in the cutoff = cos^2(45) case would tend to converge towards 0.33333...

You're absolutely right. From 100,000 generated pairs I got a subset of 66557, out of which 22309 were opposite results (0.33518...). It indeed converges towards 1/3.

Simple question: Does my model emulate actual results or not? Not with fair sampling, subtraction of accidentals and other stuff. All the data. Because unlike the other models, this one actually predicts that there will be lots of non-detections.

So does it emulate the results or not? If not, why?

If you think there's something wrong in my application, what tests can I run to further test it?

 

[DrChinese]

Originally Posted by JesseM
But in real tests, the detectors don't work the way they do in your classical model...if the polarization is random, each detector should ideally get R 50% of the time and G 50% of the time, but in QM the probability of the two detectors getting the same result on a given trial is simply equal to the cos^2 of the angle between the two detectors. So if Alice gets G when hers is set to 120, and Bob's is set to 240 on that trial, there should be a cos^2(240-120) = cos^2(120) = 0.25 probability that Bob also gets G.

Hydromatic:
Ok, this is what bothers me. How can they assume how the polarizers work? Is there any experiment made that proves polarizers are governed by chance? Maybe I've missed it. How can physicists be more open to tossing out locality, than to reconsider assumptions about how polarizers work?

So does it emulate the results or not? If not, why?

If you think there's something wrong in my application, what tests can I run to further test it?

You are hung up on your algorithm, and it does NOT - as JesseM points out - match known experiments. The EPR Paradox STARTS with the fact when measuring the same attribute for Alice and Bob (i.e. same angle setting), that knowing the results for Alice implies what the results will be for Bob. Under local realism, the implication is that the results were predetermined. Since your formula does not do this, you are wrong before you get any further. You must change your formula so that Alice and Bob's results are perfectly correlated when measuring the same angle. Further, to match experiment, they must be R 50% of the time and G 50% of the time which I think your formula does properly.

Next comes the difficult part. You must have an algorithm that works independently on Alice and Bob. And still gives detections on both sides that ALWAYS have a detection probability not only that is independent of the other side's setting - it must also be independent of the polarization of the photon itself! This is in direct contradiction to your formula! It has been known for over 200 years (Malus, 1807) that the correct formula for detection is cos^2 theta when there is a polarizer. When it is a beamsplitter instead of a polarizer, it goes to either one or the other detectors. SO THERE IS NO CUTOFF! (sorry for shouting, but this should have been obvious to all of us earlier).

In real experiments, these elements are checked carefully when the alignments are performed to insure that the results make sense. PDC crystals actually generate pairs with known polarization when the polarization of the input pump is known. So your purported cut-off effect is actually checked during the setup, and it does not exist. A beam-splitter is used, and it is rotated to determine optimal alignments as photons pass through the entire apparati. There is no cut-off, but Malus' Law is seen in how the splitting occurs.

You also wonder if assumptions have been made about how a polarizer works? Hmmm. As mentioned, beam splitters are used which either transmit or deflect the incident beam. We are simply trying to predict results, and those predictions work according to Malus' law. And that is simply a classical requirement which happens to exactly match quantum predictions. You will want to follow that too. I personally would not characterize matching theory and experiment as "assumption". More of a requirement.

So my point is: NO, your results do not match reality. I hope you can see that your formula needs rework. Because of Bell's Theorem, you cannot have a local realistic theory that gives results identical to quantum theory. So your mission is quite impossible, since this was your original objective.

By the way, there are theorists who cling to the idea that such a formula IS possible. Emilio Santos comes to mind. He has pursued stochastic solutions (that is the kind of formula you are developing) and is convinced one is possible. Although he does get published, I can't think of anyone (outside of perhaps his co-authors) who holds out hope for such a solution in face of Bell's Theorem and subsequent tests.

 

[Hydr0matic]

DrChinese, I appreciate you joining the discussion, but I would prefer if you ask me questions - like JesseM - if there's something I'm not clear about. Shouting about "no cutoff" when you've misunderstood the meaning is obviously not very constructive.

Clarifications:

1. I am NOT trying to emulate Local Realism. My idea is nothing like Local realism. I think the idea of hidden variables is very silly. I believe that particles and waves have defined properties (polarization, position..), but the notion of predetermened outcomes of a random process, lightspeed communication between photons and so forth, is indeed very silly.

2. The cut-off value we are discussing refers to the detector, not the polarizer. The polarizers indeed follow Malus' law. The cut-off value comes into play when you replace the idea of a photon with the idea of an electromagnetic wave. Because, if you consider light as a wave, there is no limit to how small the amplitude can be. It can be infinitesimal. However, there IS a limit to what we can detect. We don't have detectors that can register infinitely weak lightwaves. There is a lightwave amplitude cut-off value, below which we cannot detect any lightwaves.

3. I don't know how your programming skills are, but it seems like you're having issues interpreting my code (which I admit is quite messy. Too lazy to fix it). My application computes random angles A,B or C for Alice and Bob independantly. It generates a photon with random polarization, then assigns either same or opposite polarization to the other photon. It then calculates - independantly for Alice and Bob - whether the relative angle between photon and polarizer is small enough to output a detectable lightwave.

You are more than welcome to continue the discussion and/or ask me questions if something needs further/better explaining...

 

[JesseM]

Sorry, forgot to answer this earlier...

But in real tests, the detectors don't work the way they do in your classical model...if the polarization is random, each detector should ideally get R 50% of the time and G 50% of the time, but in QM the probability of the two detectors getting the same result on a given trial is simply equal to the cos^2 of the angle between the two detectors. So if Alice gets G when hers is set to 120, and Bob's is set to 240 on that trial, there should be a cos^2(240-120) = cos^2(120) = 0.25 probability that Bob also gets G.

Ok, this is what bothers me. How can they assume how the polarizers work? Is there any experiment made that proves polarizers are governed by chance? Maybe I've missed it. How can physicists be more open to tossing out locality, than to reconsider assumptions about how polarizers work?

I don't understand what you mean by "how polarizers work" or "governed by chance". My statement above refers only to statistics--if we do a large number of trials, we will find that at each detector we get R on about 50% of trials and G on about 50%, but if we look at the subset of trials where the difference between detector angles was 120 degrees we'll find that only on about 25% of the trials do they get the same results (or opposite results, if we're doing an experiment where they always get opposite results when they choose the same detector angle). This is just a statement about observed results as predicted by QM, nothing about what is "really" going on with the particles or whether their behavior involves any true randomness (either at the detectors or at the moment the pair is created). The use of probability to describe the statistics just reflects our own practical lack of ability to predict the behavior in any deterministic way, it doesn't mean there couldn't be some underlying deterministic rule that explains what happens on any trial (similarly, in classical statistical mechanics you can still talk about probabilities for things like an increase or decrease in entropy, based on your ignorance of the precise microstate of a large system--the precise position and momentum of every particle making it up--even though it's assumed that given the exact microstate its behavior would be perfectly predictable). The point here is just that no local realist rule, including yours, can reproduce these statistics.

Yes, and notice that when you make the cutoff cos^2(45) (not cos(45) as you say...I know what you meant, but the difference is important!)

The choice of cos and not cos² seems more reasonable to me. Intensity is a concept relating to lots of light. I mean, what's the intensity of one photon?
If you use the absolute value of cos there's no difference anyway (cos²(a) > cos²(b) => |cos(a)| > |cos(b)|, same for =, <). But maybe I'm wrong.

But I thought you weren't talking about photons in your experiment, you were talking about a classical experiment where we sent classical waves and the detectors only went off if the intensity was above some threshold. Anyway, intensity is a concept related to the energy of light, so if we're using light of a uniform frequency like a laser, than all the photons should have the same energy, therefore intensity in a given region is directly proportional to the number of photons that hit that region over many trials (which is proportional to the probability that a single photon will hit that region in a single trial).

Simple question: Does my model emulate actual results or not? Not with fair sampling, subtraction of accidentals and other stuff. All the data. Because unlike the other models, this one actually predicts that there will be lots of non-detections.

Keep in mind that one can do experiments where "red" does not represent a non-detection like a photon failing to make it through a polarizer, but instead red and green represent two different outcomes for a particle that is measured, like an electron either being deflected up (spin-up) or down (spin-down) by a magnetic field. In this case at least, I assume one could just throw out all the cases where one detector registered an electron but the other didn't. In the case of photons, I'm not actually sure how they make sure that they are actually measuring members of an entangled pair rather than just random background photons, and I don't know if the "red" result here would just correspond to one detector failing to go off or if it would mean something else, basically I don't know that much about the experimental side of these Bell tests. Maybe someone else can answer this question.

 

[DrChinese]

DrChinese, I appreciate you joining the discussion, but I would prefer if you ask me questions - like JesseM - if there's something I'm not clear about. Shouting about "no cutoff" when you've misunderstood the meaning is obviously not very constructive.

Clarifications:

1. I am NOT trying to emulate Local Realism. My idea is nothing like Local realism. I think the idea of hidden variables is very silly. I believe that particles and waves have defined properties (polarization, position..), but the notion of predetermened outcomes of a random process, lightspeed communication between photons and so forth, is indeed very silly.

2. The cut-off value we are discussing refers to the detector, not the polarizer. The polarizers indeed follow Malus' law. The cut-off value comes into play when you replace the idea of a photon with the idea of an electromagnetic wave. Because, if you consider light as a wave, there is no limit to how small the amplitude can be. It can be infinitesimal. However, there IS a limit to what we can detect. We don't have detectors that can register infinitely weak lightwaves. There is a lightwave amplitude cut-off value, below which we cannot detect any lightwaves.

3. I don't know how your programming skills are, but it seems like you're having issues interpreting my code (which I admit is quite messy. Too lazy to fix it). My application computes random angles A,B or C for Alice and Bob independantly. It generates a photon with random polarization, then assigns either same or opposite polarization to the other photon. It then calculates - independantly for Alice and Bob - whether the relative angle between photon and polarizer is small enough to output a detectable lightwave.

You are more than welcome to continue the discussion and/or ask me questions if something needs further/better explaining...

As said, sorry for the shouting. Also, I am a software developer (actually manager), so I think I can follow OK.

The issue is what you are saying about amplitudes and detection. This really can't apply here. Specifically, you say "whether the relative angle between photon and polarizer is small enough to output a detectable lightwave". This is demonstrably false. We know that there is no angular difference at which detection probability is affected.

You say that you believe that particles have defined properties. That is the basis for realism. You also say that communication between photons is silly, which is equivalent to saying that you believe in locality. That sounds like local realism, which is what Bell says is inconsistent with quantum mechanics.

As I said earlier, you cannot produce a formula via computer program that reproduces all of the predictions of quantum mechanics with random selection of settings for Alice and Bob. Your results will always need to depend on the relative settings for Alice and Bob, and all the other factors will essentially cancel out.

 

[Hydr0matic]

Apologies for the ridiculously late reply.

I don't understand what you mean by "how polarizers work" or "governed by chance".

This is just a statement about observed results as predicted by QM, nothing about what is "really" going on with the particles or whether their behavior involves any true randomness (either at the detectors or at the moment the pair is created). The use of probability to describe the statistics just reflects our own practical lack of ability to predict the behavior in any deterministic way, it doesn't mean there couldn't be some underlying deterministic rule that explains what happens on any trial

I'm actually not debating underlying determinism or whether there is true randomness or not. I'm saying there isn't any randomness at all. This is the key difference. In both QM and Local realism the polarizers are goverened by chance. By this I mean that, when a photon hits the polarizer there is a cos²(θ) probability that it goes through. This is not the case in classical physics. In classical physics ALL photons (lightwaves) go through, but they do so with an amplitude A = A₀ cos(θ).

Let's take an example.
Say we have a light-source that emits all photons with the same polarization. A polarizer is set with a 45° angle difference to the photon polarization. Now, in QM, all photons will hit the polarizer at a 45° difference, and therefore each photon has a cos²(45°) = 50% chance of passing through. In classical physics on the other hand, all photons (lightwaves) will pass through with an amplitude A = A₀ cos(45°) of the emitted one. IF this new amplitude A is below the detectors cut-off value, NO photons are detected.

As you can see, this is not just a matter of interpretation or underlying mechanisms. Although the theories give equivalent results for lots of light, in single-photon setups they do not.

But I thought you weren't talking about photons in your experiment, you were talking about a classical experiment where we sent classical waves and the detectors only went off if the intensity was above some threshold. Anyway, intensity is a concept related to the energy of light, so ../ /.. (which is proportional to the probability that a single photon will hit that region in a single trial).

True, my setup is completely classical. So it doesn't matter if I compare amplitude or intensity, they give the same results. To me at least, amplitude makes more sense.

basically I don't know that much about the experimental side of these Bell tests. Maybe someone else can answer this question.

Neither do I. Would appreciate that very much.

The issue is what you are saying about amplitudes and detection. This really can't apply here. Specifically, you say "whether the relative angle between photon and polarizer is small enough to output a detectable lightwave". This is demonstrably false. We know that there is no angular difference at which detection probability is affected.

Ofcourse there is. If a photon has polarization ϕ and the polarizer is set to angle θ, the probability of detection is cos²(ϕ-θ). (in QM & LR)

You say that you believe that particles have defined properties. That is the basis for realism. You also say that communication between photons is silly, which is equivalent to saying that you believe in locality. That sounds like local realism, which is what Bell says is inconsistent with quantum mechanics.

Hehe, true, I believe in realism and locality. The only difference from Local Realism, and also the key difference, is how the polarizers behave [see above]. If you remove the randomness from the polarizers and view the polarization from a classical wave perspective, I believe the experimental results will make perfect sense in a local realists world.

 

[DrChinese]

1. Ofcourse there is. If a photon has polarization ϕ and the polarizer is set to angle θ, the probability of detection is cos²(ϕ-θ). (in QM & LR)

2. Hehe, true, I believe in realism and locality. The only difference from Local Realism, and also the key difference, is how the polarizers behave [see above]. If you remove the randomness from the polarizers and view the polarization from a classical wave perspective, I believe the experimental results will make perfect sense in a local realists world.

1. The probability of passing through the polarizer is per Malus' Law. But the likelihood of being detected by the detector itself is not affected. Usually, Bell tests use polarizing beamsplitters that cause the vertical and hortizontal components to be split. There are 2 detectors, and the photon is seen at one or the other. In your formulas, it looks as if you are using something you call "cut-off" to bias your results. This is part of what I am objecting to.

But the other part is the idea that the photon had a definite polarization prior to being observed. This is one of the assumptions that Bell challenges.

2. The experimental results do NOT make sense from a classical perspective, this has already been explained many times to you and others. A classical analysis yields a different formula altogether - namely Match=(cos^2(theta)/2)+.25, if you assume Malus (i.e. the classical) as a fundamental starting point - which is what you assert. That formula has a range of .25 to .75, while the quantum formula has a range of 0 to 1. Of course, we observe the 0 to 1 range.

-DrC