I Looking For Help Understanding Bell's Theorem, Hidden Variables & QM

[PeterDonis]

Would that be the case even if the photons were always identical except in cases [1] & [8]?

What does this even mean?

 

[PeterDonis]

The table of possible values would be different, since there are two photons, not one. There would be 64 possible cases, not 8.

Apparently you didn't grasp what this means. The table is not QM; it is a proposed local hidden variable model in which each photon carries with it a set of predetermined results for all three of the possible measurements, A, B, and C. The 8 rows in the table for the one-photon case are the 8 possible sets of predetermined results. For two photons, there are 64 possible sets of predetermined results (8 for each photon, so 8 x 8 = 64 total).

 

[Nugatory]

Sorry, just going back to this. In tests of Bell's inequality, where entangled pairs of photons are used, would it be be possible that in cases [1] & [8] the two photons would have the following HVs:

Photon 1: A+B+C+
Photon 2: A-B-C-

or vice versa?

Possible in some hypothetical hidden variable theory? Yes, but do remember that we can't measure these hypothetical hidden variables (if we could they wouldn't be "hidden"). All we have is that the measurements that we can make, and these say that:
1) When the detectors at both sides are set at the same angle we get the opposite results (+ at one detector, - at the other) every single time, probability 100%, no exceptions.
2) When the detectors are set at different angles, we get opposite results with probability $\cos^2\theta$, where $\theta$ is the angle between the detector settings (you will see different formulations of this depending on whether we're working with photons polarized on the same or different axes or spin-1/2 particles in the singlet state, and whether the entanglement is such that opposite results are found on the same axis or perpendicular axes).

So if these A+B+C+/A-B-C- pairs are happening, then other pairs must also generated more or less often so that after we've measured many pairs randomly distributed across all the possible configurations of the hidden variables we end up with results agreeing with #1 and #2 above. Bell’s theorem says that you won’t be able to construct such a distribution.

You may find thus old Scientific American article helpful: https://static.scientificamerican.com/sciam/assets/media/pdf/197911_0158.pdf as it directly addresses the behavior of hypothetical hidden variables of the form $A\pm{B}\pm{C}\pm$.

 

[Lynch101]

Apparently you didn't grasp what this means. The table is not QM; it is a proposed local hidden variable model in which each photon carries with it a set of predetermined results for all three of the possible measurements, A, B, and C. The 8 rows in the table for the one-photon case are the 8 possible sets of predetermined results. For two photons, there are 64 possible sets of predetermined results (8 for each photon, so 8 x 8 = 64 total).

I know it's a proposed local HV model, but to test it in the lab pairs of photons are required, no?

If both photons always have the same HVs there will only be 8 cases, not 64.

I was wondering if those two "special" cases - where both photons have the "pass all 3 filters" HVs or the "do not pass all 3" HVs - could instead have one photon with the "pass all 3 filters" HVs while the other one would have the "do not pass all 3".

 

[Nugatory]

If both photons always have the same HVs there will only be 8 cases, not 64.

The hidden variables carry enough information to calculate the outcome for each of six (two sides, three possible measurements at each side) possible +/- measurements. That is, the hidden variables carry enough information to answer 64 questions of the form "If this photon arrives at detector X when it is set to angle Y will the result be Z?" where X is left or right, Y is one of the three angles and Z is plus or minus. Whether the hidden variables are the same at both sides (that is, the answer to any two questions that differ only in the choice of X will be the same) doesn't reduce the number of questions, it just changes how we've encoded the hidden variables that we use to calculate the answer.

 

[PeterDonis]

If both photons always have the same HVs

Why would you think that?

 

[Lynch101]

Possible in some hypothetical hidden variable theory? Yes, but do remember that we can't measure these hypothetical hidden variables (if we could they wouldn't be "hidden"). All we have is that the measurements that we can make, and these say that:
1) When the detectors at both sides are set at the same angle we get the opposite results (+ at one detector, - at the other) every single time, probability 100%, no exceptions.

Ah yes. I was confusing myself on this. I had it in mind that it was the opposite case with photons, that when measured at the same angle they would be perfectly correlated (as opposed to anti-correlated). I was confusing that with the idea that once a photon passes through a filter at a given polarisation, it remains polarised at that angle.

2) When the detectors are set at different angles, we get opposite results with probability $\cos^2\theta$, where $\theta$ is the angle between the detector settings (you will see different formulations of this depending on whether we're working with photons polarized on the same or different axes or spin-1/2 particles in the singlet state, and whether the entanglement is such that opposite results are found on the same axis or perpendicular axes).

That part I remember.

So if these A+B+C+/A-B-C- pairs are happening, then other pairs must also generated more or less often so that after we've measured many pairs randomly distributed across all the possible configurations of the hidden variables we end up with results agreeing with #1 and #2 above. Bell’s theorem says that you won’t be able to construct such a distribution.

You may find thus old Scientific American article helpful: https://static.scientificamerican.com/sciam/assets/media/pdf/197911_0158.pdf as it directly addresses the behavior of hypothetical hidden variables of the form $A\pm{B}\pm{C}\pm$.

Thanks for the explanation and the article. Both have been very helpful.

 

[Lynch101]

The hidden variables carry enough information to calculate the outcome for each of six (two sides, three possible measurements at each side) possible +/- measurements. That is, the hidden variables carry enough information to answer 64 questions of the form "If this photon arrives at detector X when it is set to angle Y will the result be Z?" where X is left or right, Y is one of the three angles and Z is plus or minus. Whether the hidden variables are the same at both sides (that is, the answer to any two questions that differ only in the choice of X will be the same) doesn't reduce the number of questions, it just changes how we've encoded the hidden variables that we use to calculate the answer.

Apologies, I misread what PD meant by that, I thought he meant 64 combinations of particle pairs, as opposed to possible outcomes.

I was ignoring the 3 combinations where the filter settings were the same - is that the case when it comes to Bell tests? Also with my confusion about both photons having the same HVs, I was ignoring the opposite filter pairings i.e. I was only looking at AB and not BA (because I thought it wouldn't make a difference to the probability).

I don't suppose you can see at a glance, where I've gone wrong with the table below?

It only has 48 possible outcomes (excluding combinations where both filters are the same e.g. AA). It also has a 50% match rate, so I must be doing something wrong

Including the cases where both filters are the same would bring it down to 33.33% but it would have 62 possible outcomes, instead of 64.

 

[Lynch101]

Why would you think that?

The standard reason, I was confusing myself!

I had it in mind that there was perfect correlations with photons when the filters were set to the same orientation on both sides of the experiment. I think I had conflated that with the idea that once a photon passes through a filter at a given polarisation, it remains polarised at that angle.

 

[Nugatory]

I had it in mind that there was perfect correlations with photons when the filters were set to the same orientation on both sides of the experiment.

There could be, it depends on how we generate our entangled pairs (google for "spontaneous parametric down-conversion" ). Some processes will produce two entangled photons that will both pass filters oriented in the same direction; others will produce entangled photons that will both pass filters perpendicular to one another.

And do note that I worded it as "will both pass filters", not "are both polarized in the same/orthogonal direction". The difference matters.

 

[Lynch101]

There could be, it depends on how we generate our entangled pairs (google for "spontaneous parametric down-conversion" ). Some processes will produce two entangled photons that will both pass filters oriented in the same direction; others will produce entangled photons that will both pass filters perpendicular to one another.

And do note that I worded it as "will both pass filters", not "are both polarized in the same/orthogonal direction". The difference matters.

Ah! I've probably been piecing together different things I've read in a completely garbled manner.

Thanks for the clarification and for the suggestion. I'll have read up about spontaneous parametric down-conversion.

Would the individual photon pairs I posted in the most recent attempt at a table represent a hypothetically possible HV "model"?

I know it wouldn't match the observed correlations but in terms of adhering to passing and not passing filters; or is it way off?

 

[Lynch101]

Sorry, I know ye must be sick of seeing my confused tables at this stage, but I'm hoping it makes it easier to see the mistakes I'm making, compared to my attempts to put it in writing.

In this table I was assuming photon pairs which are anti-correlated at each angle. I weighted cases [1] & [8] such that they occur much more frequently than the others - as opposed to less frequently in my previous thinking. For ease of editing and reading, I put cases [1] & [8] together on the table.

Would those photon pairings be possible or am I (again) overlooking something*?

*This is more of a rhetorical question.

 

[Lynch101]

I still can't see where I'm going wrong in the above, with photons and filters, but I tried it out with electron spins and I think it's a bit clearer to me now - assuming the below is right.

D = Different outcomes e.g. Up/Down.

 

[PeterDonis]

I weighted cases [1] & [8] such that they occur much more frequently than the others

Why would you do that?

 

[PeterDonis]

I tried it out with electron spins

Which is exactly the same in terms of the scenario under discussion--in both cases you are looking at measurements on two qubits. Your two tables are identical except for you changing some letters around and putting more copies of rows [1] and [8] in your photon table in post #62. But if doing that is not justified with electron spins (which you appear to think it isn't since you didn't do it in post #63), then it's not justified with photon polarizations either.

 

[Lynch101]

Why would you do that?

Just to see if it were, in any way, possible to reproduce the predictions of quantum theory using hidden variables. It could also have prompted a clarification from someone that such a thing is ruled out in principle, which would help to give me a better understanding.

 

[Lynch101]

Which is exactly the same in terms of the scenario under discussion--in both cases you are looking at measurements on two qubits. Your two tables are identical except for you changing some letters around and putting more copies of rows [1] and [8] in your photon table in post #62. But if doing that is not justified with electron spins (which you appear to think it isn't since you didn't do it in post #63), then it's not justified with photon polarizations either.

The changes in letter are to represent the differences in measurement performed (U and D represent spin Up and Down).

Isn't there a slight difference in insofar as with the electron spins, there is always something detected by the measurement device, whereas with the photons, not every photon passes through the filter to the detector.

I didn't weight the cases [1] and [8] in this table because I could see it wouldn't make a difference to the outcomes, whereas it would appear to make a difference in the case of the photon polarisations. In post #36 DrChinese said we are free to change the weighting of the cases, which would suggest that it isn't ruled out in principle.

 

[PeterDonis]

I didn't weight the cases [1] and [8] in this table because I could see it wouldn't make a difference to the outcomes, whereas it would appear to make a difference in the case of the photon polarisations.

Go back and look at your tables again. As I said before, they are identical except for changes in labeling. So if weighting can make a difference for the photon table, it can also make a difference for the electron table.

 

[PeterDonis]

In post #36 DrChinese said we are free to change the weighting of the cases

Yes, and doing so will change the predicted average outcome. It has to: that's just math.

What changing the weighting won't do is give you an average that violates the Bell inequalities or their equivalents; that's impossible on any weighting. In the original table that @DrChinese was commenting on in post #36, every line in the table would contribute at least 0.333 to the average, so it is mathematically impossible to get an average less than 0.333 no matter how you weight the lines. And 0.333 does not violate the relevant inequality for that case. But the QM prediction for that case is 0.25, which does violate the inequality.

In your tables in posts #62 and #63, you've reversed the signs, so to speak, from the original table that @DrChinese was commenting on, so 0.25 or less does not violate the relevant inequality; so the fact that you can get averages of 0.25 or less by changing the weightings, while true, is pointless. You need to compute the correct inequality for that case and then look at what range of averages you are able to obtain by varying the weightings in your table; then you will see that no matter how you vary the weightings, you cannot get an average that violates the correct inequality for that case. And if you compute the correct QM pprediction for that case, you will see that it does violate the inequality.

 

[Lynch101]

Go back and look at your tables again. As I said before, they are identical except for changes in labeling. So if weighting can make a difference for the photon table, it can also make a difference for the electron table.

Is there not a difference in what is being measured and the specific measurement outcomes?

To clarify, am I incorrect in understanding that in tests of Bell's inequality, involving photon pairs where

when the detectors at both sides are set at the same angle we get the opposite results (+ at one detector, - at the other) every single time, probability 100%, no exceptions.

the ["minus"] at a detector means that the detector does not register a photon, because the photon does not pass through a given filter to the detector?

Whereas, in tests involving electron spins, when the stern gerlach magnets are aligned at the same angle, the outcomes are always opposite, but, unlike tests involving photons, the detector always registers a measurement - either "up" or "down" (allowing for detector inefficiencies)?

 

[Lynch101]

Thank you PD, although I don't quite follow your reasoning, these are the kinds of insights I was hoping to get, to help me get a better understanding.

Yes, and doing so will change the predicted average outcome. It has to: that's just math.

I understand that, which is why I was playing around with the weighting to see if it would change the predicted average outcome to match the predictions of quantum theory.

What changing the weighting won't do is give you an average that violates the Bell inequalities or their equivalents; that's impossible on any weighting. In the original table that @DrChinese was commenting on in post #36, every line in the table would contribute at least 0.333 to the average, so it is mathematically impossible to get an average less than 0.333 no matter how you weight the lines. And 0.333 does not violate the relevant inequality for that case. But the QM prediction for that case is 0.25, which does violate the inequality.

In your tables in posts #62 and #63, you've reversed the signs, so to speak, from the original table

I can see that every line in DrC's table would contribute at least 0.333. However, I was under the (incorrect?) impression that DrC's table represents the case of a single photon passing through consecutive filters, while (some) tests of Bell's Inequality involve photon pairs, each passing through a single misaligned filter, such that:

when the detectors at both sides are set at the same angle we get the opposite results (+ at one detector, - at the other) every single time, probability 100%, no exceptions.

This led me to attempt to make a table representing that, by reversing the signs.

If the photon pairings form cases [1] and [8] are possible pairings, then those lines would contribute less than 0.333.

I'm not saying they are possible. I was just filling in values which would satisfy the criterion that

when the detectors at both sides are set at the same angle we get the opposite results (+ at one detector, - at the other) every single time, probability 100%, no exceptions.

I was hoping that, if they're not actually possible, then someone might point that out and I would thereby get a better understanding. In much the same way you were able to see at a glance, that I was inadvertently invoking superdeterminism in a previous post.

that @DrChinese was commenting on, so 0.25 or less does not violate the relevant inequality; so the fact that you can get averages of 0.25 or less by changing the weightings, while true, is pointless. You need to compute the correct inequality for that case and then look at what range of averages you are able to obtain by varying the weightings in your table; then you will see that no matter how you vary the weightings, you cannot get an average that violates the correct inequality for that case. And if you compute the correct QM pprediction for that case, you will see that it does violate the inequality.

Am I incorrect in my understanding that the inequality tells us what the (minimum) predicted average outcome would be?

Can this not also be read from a table representing the sample space* for statistically independent events?*I'm using "sample space" here because the table represents a finite set, not the abstract infinite ensemble.

 

[Nugatory]

the ["minus"] at a detector means that the detector does not register a photon, because the photon does not pass through a given filter to the detector?

An experiment set up this way is of limited usefulness, because every stray photon coming through from outside and blundering into a detector will register as a +/- pair at the current setting.
More often a two-channel polarizer is used, so the two outcomes are is/isn’t deflected and we need a four photodetectors total.

 

[Lynch101]

An experiment set up this way is of limited usefulness, because every stray photon coming through from outside and blundering into a detector will register as a +/- pair at the current setting.
More often a two-channel polarizer is used, so the two outcomes are is/isn’t deflected and we need a four photodetectors total.

Thank you Nugatory, this is the kind of insight I was hoping for, to help me better understand what I was missing.

 

[Nugatory]

I can see that every line in DrC's table would contribute at least 0.333. However, I was under the (incorrect?) impression that DrC's table represents the case of a single photon passing through consecutive filters, while (some) tests of Bell's Inequality involve photon pairs, each passing through a single misaligned filter, such that:

Although @DrChinese does not go into this detail, the two formulations are equivalent.

Say that we are working with pairs entangled such that measurements on the same axis will always yield opposite results. The logic of any hidden variable theory (implicit in the original EPR paper) says that measuring the right-hand particle at angle B with a - result is effectively a measurement of the left-hand particle at that angle with a + result. Thus, we can treat one measurement of each particle as two measurements of the same particle, with one result obtained directly and the other by inference. DrC's table uses the latter format, with the advantage that we have to juggle only three quantities, not six.

 

[Nugatory]

Am I incorrect in my understanding that the inequality tells us what the (minimum) predicted average outcome would be?

The inequality can be presented in various ways; you may find the wikipedia article on the CHSH inequality helpful. However, when we're considering the simple cases of this thread the Scientific American article I linked above may be most helpful: The probability of getting (for example) an A+B- result must be less than or equal to the sum of the probabilities of getting an A+C+ result and of getting a B-C- result. Phrased this way, the inequality is true almost by inspection: every A+B- case is either A+B-C+ (a subset of A+C+) or A+B-C- (a subset of B-C-).

Yet the quantum mechanical $\cos^2\theta$ results will violate this inequality.

 

[Lynch101]

An experiment set up this way is of limited usefulness, because every stray photon coming through from outside and blundering into a detector will register as a +/- pair at the current setting.
More often a two-channel polarizer is used, so the two outcomes are is/isn’t deflected and we need a four photodetectors total.

Do I have it correct that (some) experiments involving photons count the average number of outcomes which are the same, while electron spin experiments count the average number that are different?

 

[Nugatory]

Do I have it correct that (some) experiments involving photons count the average number of outcomes which are the same, while electron spin experiments count the average number that are different?

No. Look at the CHSH inequality I linked to above, and especially what goes into the definition of the various $E$ quantities.

 

[Nugatory]

@Lynch101 ch101 you might be interested in seeing what the results of a simulation using your hidden variable assignments from your post #62 above (I used a simulation program I wrote many years ago) compared with simulated results using teh quantum-mechanical $cos^2$ rule.

/usr/lib/jvm/java-1.11.0-openjdk-amd64/bin/java -javaagent:/snap/intellij-idea-community/467/lib/idea_rt.jar=43735:/snap/intellij-idea-community/467/bin -Dfile.encoding=UTF-8 -classpath /home/redacted/IdeaProjects/Bell/out/production/Bell Bell 100000 0 120 240
100000 pairs at angles 0, 120, 240
Classical results:
0:0: ++: 0       +-: 5747    -+: 5549    --: 0         matches:0       total:11296    0.0%
0:1: ++: 1422    +-: 4221    -+: 4153    --: 1359      matches:2781    total:11155    24.9%
0:2: ++: 1457    +-: 4103    -+: 4204    --: 1354      matches:2811    total:11118    25.3%
1:0: ++: 1352    +-: 4076    -+: 4012    --: 1459      matches:2811    total:10899    25.8%
1:1: ++: 0       +-: 5476    -+: 5401    --: 0         matches:0       total:10877    0.0%
1:2: ++: 1369    +-: 4194    -+: 4110    --: 1392      matches:2761    total:11065    25.0%
2:0: ++: 1397    +-: 4283    -+: 4207    --: 1428      matches:2825    total:11315    25.0%
2:1: ++: 1424    +-: 4201    -+: 4166    --: 1410      matches:2834    total:11201    25.3%
2:2: ++: 0       +-: 5397    -+: 5677    --: 0         matches:0       total:11074    0.0%
a+b+:8297 b+c+:8395 a+c+:8386
a+b-:2881 b+c-:2779 a+c-:2885
a-b+:2711 b-c+:2816 a-c+:2751
a-b-:8165 b-c-:8276 a-c-:8411
a+b- <= a+c+ + b-c-, 2881 <= 8386 + 8276, true (-82.7%)
a+c- <= a+b+ + b-c-, 2885 <= 8297 + 8276, true (-82.6%)
b+c- <= a+b+ + a-c-, 2779 <= 8297 + 8411, true (-83.4%)
Quantum results:
0:0: ++: 0       +-: 5712    -+: 5584    --: 0         matches:0       total:11296    0.0%
0:1: ++: 4159    +-: 1451    -+: 1414    --: 4131      matches:8290    total:11155    74.3%
0:2: ++: 4298    +-: 1360    -+: 1367    --: 4093      matches:8391    total:11118    75.5%
1:0: ++: 4046    +-: 1374    -+: 1402    --: 4077      matches:8123    total:10899    74.5%
1:1: ++: 0       +-: 5406    -+: 5471    --: 0         matches:0       total:10877    0.0%
1:2: ++: 4143    +-: 1351    -+: 1424    --: 4147      matches:8290    total:11065    74.9%
2:0: ++: 4304    +-: 1377    -+: 1446    --: 4188      matches:8492    total:11315    75.1%
2:1: ++: 4163    +-: 1381    -+: 1385    --: 4272      matches:8435    total:11201    75.3%
2:2: ++: 0       +-: 5559    -+: 5515    --: 0         matches:0       total:11074    0.0%
a+b+:2825 b+c+:2732 a+c+:2737
a+b-:8236 b+c-:8415 a+c-:8486
a-b+:8177 b-c+:8310 a-c+:8397
a-b-:2816 b-c-:2809 a-c-:2813
a+b- <= a+c+ + b-c-, 8236 <= 2737 + 2809, false (48.5%)
a+c- <= a+b+ + b-c-, 8486 <= 2825 + 2809, false (50.6%)
b+c- <= a+b+ + a-c-, 8415 <= 2825 + 2813, false (49.3%)

Process finished with exit code 0

The first line starting 0:0 should be read as "with both detectors set to position 0/a, we had 0 ++ results, 5712 +- results, 5584 -+ results, 0 -- results, zero matches of the 11296 trials at this detector settings, 0% of the trials produced matches".
The next three lines show the number of particles that should be expected to have various properties at the first detector: for example, a+b+ is implied if we are +- in the 0:1 row or -+ in the 1:0 row).
The the next final three lines are checks to see if the inequality is valid, and as you can see it is massively violated in the quantum mechanical case.

You will also note that your hidden variable assignments don't match the quantum predictions at all. This is the issue that @PeterDonis pointed out in post #69 above - you inadvertently reversed the sense of the comparison (which is why we recommend focusing on the simplest models). You overweighted the +++/--- possibilties at opposite sides, which is (as I pointed out somewhere above) equivalent to overweighting the +++ cases in DrC' presentation, pushing the average in the wrong direction.

 

[Lynch101]

@Lynch101 ch101 you might be interested in seeing what the results of a simulation using your hidden variable assignments from your post #62 above (I used a simulation program I wrote many years ago) compared with simulated results using teh quantum-mechanical $cos^2$ rule.

/usr/lib/jvm/java-1.11.0-openjdk-amd64/bin/java -javaagent:/snap/intellij-idea-community/467/lib/idea_rt.jar=43735:/snap/intellij-idea-community/467/bin -Dfile.encoding=UTF-8 -classpath /home/redacted/IdeaProjects/Bell/out/production/Bell Bell 100000 0 120 240
100000 pairs at angles 0, 120, 240
Classical results:
0:0: ++: 0       +-: 5747    -+: 5549    --: 0         matches:0       total:11296    0.0%
0:1: ++: 1422    +-: 4221    -+: 4153    --: 1359      matches:2781    total:11155    24.9%
0:2: ++: 1457    +-: 4103    -+: 4204    --: 1354      matches:2811    total:11118    25.3%
1:0: ++: 1352    +-: 4076    -+: 4012    --: 1459      matches:2811    total:10899    25.8%
1:1: ++: 0       +-: 5476    -+: 5401    --: 0         matches:0       total:10877    0.0%
1:2: ++: 1369    +-: 4194    -+: 4110    --: 1392      matches:2761    total:11065    25.0%
2:0: ++: 1397    +-: 4283    -+: 4207    --: 1428      matches:2825    total:11315    25.0%
2:1: ++: 1424    +-: 4201    -+: 4166    --: 1410      matches:2834    total:11201    25.3%
2:2: ++: 0       +-: 5397    -+: 5677    --: 0         matches:0       total:11074    0.0%
a+b+:8297 b+c+:8395 a+c+:8386
a+b-:2881 b+c-:2779 a+c-:2885
a-b+:2711 b-c+:2816 a-c+:2751
a-b-:8165 b-c-:8276 a-c-:8411
a+b- <= a+c+ + b-c-, 2881 <= 8386 + 8276, true (-82.7%)
a+c- <= a+b+ + b-c-, 2885 <= 8297 + 8276, true (-82.6%)
b+c- <= a+b+ + a-c-, 2779 <= 8297 + 8411, true (-83.4%)
Quantum results:
0:0: ++: 0       +-: 5712    -+: 5584    --: 0         matches:0       total:11296    0.0%
0:1: ++: 4159    +-: 1451    -+: 1414    --: 4131      matches:8290    total:11155    74.3%
0:2: ++: 4298    +-: 1360    -+: 1367    --: 4093      matches:8391    total:11118    75.5%
1:0: ++: 4046    +-: 1374    -+: 1402    --: 4077      matches:8123    total:10899    74.5%
1:1: ++: 0       +-: 5406    -+: 5471    --: 0         matches:0       total:10877    0.0%
1:2: ++: 4143    +-: 1351    -+: 1424    --: 4147      matches:8290    total:11065    74.9%
2:0: ++: 4304    +-: 1377    -+: 1446    --: 4188      matches:8492    total:11315    75.1%
2:1: ++: 4163    +-: 1381    -+: 1385    --: 4272      matches:8435    total:11201    75.3%
2:2: ++: 0       +-: 5559    -+: 5515    --: 0         matches:0       total:11074    0.0%
a+b+:2825 b+c+:2732 a+c+:2737
a+b-:8236 b+c-:8415 a+c-:8486
a-b+:8177 b-c+:8310 a-c+:8397
a-b-:2816 b-c-:2809 a-c-:2813
a+b- <= a+c+ + b-c-, 8236 <= 2737 + 2809, false (48.5%)
a+c- <= a+b+ + b-c-, 8486 <= 2825 + 2809, false (50.6%)
b+c- <= a+b+ + a-c-, 8415 <= 2825 + 2813, false (49.3%)

Process finished with exit code 0

The first line starting 0:0 should be read as "with both detectors set to position 0/a, we had 0 ++ results, 5712 +- results, 5585 -+ results, 0 -- results, zero matches of the 11296 trials at this detector settings, 0% of the trials produced matches".
The next three lines show the number of particles that should be expected to have various properties at the first detector: for example, a+b+ is implied if we are +- in the 0:1 row or -+ in the 1:0 row).
The the next final three lines are checks to see if the inequality is valid, and as you can see it is massively violated in the quantum mechanical case.

You will also note that your hidden variable assignments don't match the quantum predictions at all. This is the issue that @PeterDonis pointed out in post #69 above - you inadvertently reversed the sense of the comparison (which is why we recommend focusing on the simplest models). You overweighted the +++/--- possibilties at opposite sides, which is (as I pointed out somewhere above) equivalent to overweighting the +++ cases in DrC' presentation, pushing the average in the wrong direction.

Thank you for taking the time to do this, I appreciate it! I'll give it a few reads to see if I can get a better grasp of it.

 

[Lynch101]

@Lynch101 ch101 you might be interested in seeing what the results of a simulation using your hidden variable assignments from your post #62 above (I used a simulation program I wrote many years ago) compared with simulated results using teh quantum-mechanical $cos^2$ rule.

/usr/lib/jvm/java-1.11.0-openjdk-amd64/bin/java -javaagent:/snap/intellij-idea-community/467/lib/idea_rt.jar=43735:/snap/intellij-idea-community/467/bin -Dfile.encoding=UTF-8 -classpath /home/redacted/IdeaProjects/Bell/out/production/Bell Bell 100000 0 120 240
100000 pairs at angles 0, 120, 240
Classical results:
0:0: ++: 0       +-: 5747    -+: 5549    --: 0         matches:0       total:11296    0.0%
0:1: ++: 1422    +-: 4221    -+: 4153    --: 1359      matches:2781    total:11155    24.9%
0:2: ++: 1457    +-: 4103    -+: 4204    --: 1354      matches:2811    total:11118    25.3%
1:0: ++: 1352    +-: 4076    -+: 4012    --: 1459      matches:2811    total:10899    25.8%
1:1: ++: 0       +-: 5476    -+: 5401    --: 0         matches:0       total:10877    0.0%
1:2: ++: 1369    +-: 4194    -+: 4110    --: 1392      matches:2761    total:11065    25.0%
2:0: ++: 1397    +-: 4283    -+: 4207    --: 1428      matches:2825    total:11315    25.0%
2:1: ++: 1424    +-: 4201    -+: 4166    --: 1410      matches:2834    total:11201    25.3%
2:2: ++: 0       +-: 5397    -+: 5677    --: 0         matches:0       total:11074    0.0%
a+b+:8297 b+c+:8395 a+c+:8386
a+b-:2881 b+c-:2779 a+c-:2885
a-b+:2711 b-c+:2816 a-c+:2751
a-b-:8165 b-c-:8276 a-c-:8411
a+b- <= a+c+ + b-c-, 2881 <= 8386 + 8276, true (-82.7%)
a+c- <= a+b+ + b-c-, 2885 <= 8297 + 8276, true (-82.6%)
b+c- <= a+b+ + a-c-, 2779 <= 8297 + 8411, true (-83.4%)
Quantum results:
0:0: ++: 0       +-: 5712    -+: 5584    --: 0         matches:0       total:11296    0.0%
0:1: ++: 4159    +-: 1451    -+: 1414    --: 4131      matches:8290    total:11155    74.3%
0:2: ++: 4298    +-: 1360    -+: 1367    --: 4093      matches:8391    total:11118    75.5%
1:0: ++: 4046    +-: 1374    -+: 1402    --: 4077      matches:8123    total:10899    74.5%
1:1: ++: 0       +-: 5406    -+: 5471    --: 0         matches:0       total:10877    0.0%
1:2: ++: 4143    +-: 1351    -+: 1424    --: 4147      matches:8290    total:11065    74.9%
2:0: ++: 4304    +-: 1377    -+: 1446    --: 4188      matches:8492    total:11315    75.1%
2:1: ++: 4163    +-: 1381    -+: 1385    --: 4272      matches:8435    total:11201    75.3%
2:2: ++: 0       +-: 5559    -+: 5515    --: 0         matches:0       total:11074    0.0%
a+b+:2825 b+c+:2732 a+c+:2737
a+b-:8236 b+c-:8415 a+c-:8486
a-b+:8177 b-c+:8310 a-c+:8397
a-b-:2816 b-c-:2809 a-c-:2813
a+b- <= a+c+ + b-c-, 8236 <= 2737 + 2809, false (48.5%)
a+c- <= a+b+ + b-c-, 8486 <= 2825 + 2809, false (50.6%)
b+c- <= a+b+ + a-c-, 8415 <= 2825 + 2813, false (49.3%)

Process finished with exit code 0

The first line starting 0:0 should be read as "with both detectors set to position 0/a, we had 0 ++ results, 5712 +- results, 5585 -+ results, 0 -- results, zero matches of the 11296 trials at this detector settings, 0% of the trials produced matches".
The next three lines show the number of particles that should be expected to have various properties at the first detector: for example, a+b+ is implied if we are +- in the 0:1 row or -+ in the 1:0 row).
The the next final three lines are checks to see if the inequality is valid, and as you can see it is massively violated in the quantum mechanical case.

You will also note that your hidden variable assignments don't match the quantum predictions at all. This is the issue that @PeterDonis pointed out in post #69 above - you inadvertently reversed the sense of the comparison (which is why we recommend focusing on the simplest models). You overweighted the +++/--- possibilties at opposite sides, which is (as I pointed out somewhere above) equivalent to overweighting the +++ cases in DrC' presentation, pushing the average in the wrong direction.

Just wondering Nugatory, if the simulation program you wrote is meant to roughly correspond to an experimental set-up, as represented below. I just find it easier to think things through when I have an approximation to the physical set-up.

 

[Nugatory]

Just wondering Nugatory, if the simulation program you wrote is meant to roughly correspond to an experimental set-up, as represented below. I just find it easier to think things through when I have an approximation to the physical set-up.

It is an idealized model that looks a lot like your illustration, but using idealized spin-entangled particle pairs instead of polarization-entangled photons. Instead of two-channel polarizers the detectors are idealized spin analyzers: they are randomly set at one of three angles (in these runs the possible angles were 0, 120, and 240 degrees); when a particle enters them they either signal + (a spin-up measurement at that angle) or - (a spin-down measurement).