Joy Christian's colourful exploding balls experiment

[gill1109]

Richard, I have looked at your R program and you are using the same formula for the sawtooth curve as me i.e. -1+ 2* angle /180.

I assumed, and noted so in an earlier post, that the maximum difference was 0.707-0.5 = 0.207 at angle = 45 degrees.

But now my calculations show that the maximum is for angle = 40 degrees.

Angle xxxxx -Cosine(angle) xxxxx -1+ 2* angle /180 xxxxx Difference
40 xxxxx -0.766 xxxxx -0.556 xxxxx -0.210
45 xxxxx -0.707 xxxxx -0.5 xxxxx -0.207

I had assumed that 45 deg was chosen because it gave the maximum difference, but maybe not? I may have made a silly calculation slip up, but I have checked it several times and cannot find an error.

Ben, if you change one of the differences from 45 degrees to something else, you have to change at least one of the four angles (we have two of Alice's, two of Bob's, right?).

If you make one of the differences smaller, the other differences will get change, the final answer will be worse.

If you want to have the absolute value of all four correlations equal, then 45 degrees is the best choice.

The reason we choose 0 and 90 for Alice, and 45 and 135 for Bob, is because we want at least two settings for Alice, at least two settings for Bob. The best way to use our resources is to use just two settings for Alice, just two settings for Bob. Then it turns out that the best choice of two pairs of settings is such that the four correlations are equal in absolute value, three of one sign, one of the other sign. Then it turns out that 0 and 90 for Alice, and 45 and 135 for Bob is the best.

What does best mean? It means the best chance of getting the Nobel prize. The strongest possible statistical evidence that the correlation is the negative cosine, not the triangle wave. The most bang for your buck.

[Ben6993]

Michel, I have worked through your cases 1 and 2 and experiments I and II. I generated outcomes by tossing coins. Five times to give the averages. In case 1 the three outcomes that you note are inescapable. I.e. there is no room for experimental error.

In case 2, my <A> and <B> were both -0.2. So <A> + <B> = -0.4.
<A> + <B> must be less that 1 +1 = 2. My value of -0.4 is within experimental error, or sampling error.

For two fair coins, the expectation of <A> + <B> will tend to zero for large samples, despite the theoretical range being from -2 to +2. My sample size was only 5.

For experiment I, I am not sure what you mean by special. (Presumably not 'biased'?) My value was -0.4 and that was non-spooky sampling variability.

For experiment II, there is no experimental error in <A> and <B>.

Next, I am trying to apply that to the CHSH experiment. Correlation coefficients are always calculated on pairs of particles at A and B. A pair is two different particles so that implies your case 2 and/or experiment I. The outcomes at A and B can be identical so that does rule out case 1 and experiment II. .... Sorry, I have not had a slap-in-the-face Eureka moment yet. Maybe I just need to see more steps. Or maybe you mean that tabulated data will never beat the sawtooth curve, which I already agree with.

I do not see what is wrong with CHSH/4 as an estimate of the correlation, except that it can be run from a table of data. Any good estimate of a correlation will do fine, it does not have to use CHSH, it does not have to use a table (for me, anyway). The links you gave about the pitfalls of various sampling methods in simulations were very interesting, but there is no simulating in the macroscopic experiment so effects like that are not applicable here? CHSH seem to be being associated with trickery, but I see it simply as an estimate of a correlation coefficient. Picking any sample of angles and finding the correlations, as you suggest, would give a good test. That would simply be a more general version of what CHSH is doing, and more costly, but equivalent in principle. The issue seems really to be how to do those calculations without using a table or tables and I do not know how that can be done?

[minkwe]

So you say that E(0, 135), E(0, 45), E(90, 45), and E(90, 135) will have the quantum correlations in Joy's experiment. Richard (and I as well) says E(0, 135) will be 0.5, the three others -0.5. Where is the "debunked CHSH logic" in that?

I say for four correlations each measured on a disjoins sets of particles, each a fair sample of the population E(0, 135), E(0, 45), E(90, 45), and E(90, 135) will each agree with QM. I say for correlations all determined from the same set of particles E(0, 135), E(0, 45), E(90, 45), and E(90, 135) will not all agree with QM. You and Richard say it doesn't matter whether we use a single set of multiple disjoint sets. Your argument is debunked here: viewtopic.php?f=6&t=44

The criterion that the average should be compared to 0.6 is just a way to still settle the bet if the correlations should turn out to be somewhere in between 0.5 and 0.7071.

Astounding that you are still unable to see that the results of the bet depend on whether we are measuring on a single set of particles or different sets of particles. Now I know you still disagree with me that there is a difference between single set and disjoint sets. But assume for a moment that there is a difference. Will you then recognize that if I am right, the way the experiment is performed will determine the outcome of the bet? I'm not talking about a silly addition procedure of results you already have. I'm talking about how you design the experiment and which correlations you calculate in the first place. I'm talking about how you obtain the numbers that you can then add and average. It seems to me anyone genuinely interested in the bet should want to know whether the way the experiment is performed can influence the outcome or not.

I understand very well that the answer to the question I asked you is "yes". I hoped you understood it as well.

Too bad you do not want to understand the issue. If you change your mind, you can find a more detailed explanation here: viewtopic.php?f=6&t=44

[minkwe]

Michel, I have worked through your cases 1 and 2 and experiments I and II. I generated outcomes by tossing coins. Five times to give the averages. In case 1 the three outcomes that you note are inescapable. I.e. there is no room for experimental error.

In case 2, my <A> and <B> were both -0.2. So <A> + <B> = -0.4.
<A> + <B> must be less that 1 +1 = 2. My value of -0.4 is within experimental error, or sampling error.

For two fair coins, the expectation of <A> + <B> will tend to zero for large samples, despite the theoretical range being from -2 to +2. My sample size was only 5.

Very good. Did you assume a specific kind of local realistic coin in your case 2? Specifically, did you assume that the probability distribution of the coins for H, T was (0.5, 0.5). If you did, I suggest you try a different probability distribution, like the one I gave p(H, T) = (0.625, 0.375). Using that distribution, repeat both experiments and see what results you get.

For experiment I, I am not sure what you mean by special. (Presumably not 'biased'?) My value was -0.4 and that was non-spooky sampling variability.

For experiment II, there is no experimental error in <A> and <B>.

By special, I simply mean a specific type of coin. Case 1 and 2 are theoretical limits on ALL types of local-realistic coins with any probability distribution from p(H,T) = (0.0, 1.0) all the way to p(H,T) = (1.0, 0.0). My special coins are a specific type of local-realistic coin with a specific distribution p(H, T) = (0.625, 0.375).

Next, I am trying to apply that to the CHSH experiment. Correlation coefficients are always calculated on pairs of particles at A and B. A pair is two different particles so that implies your case 2 and/or experiment I. The outcomes at A and B can be identical so that does rule out case 1 and experiment II. .... Sorry, I have not had a slap-in-the-face Eureka moment yet.

See viewtopic.php?f=6&t=44. Tabulated data can beat the sawtooth curve. Isn't the data from experiment tabulated. we just have 4 different tables not one.

I do not see what is wrong with CHSH/4 as an estimate of the correlation, except that it can be run from a table of data.

Do the experiment I suggested and maybe this will become clear.

Any good estimate of a correlation will do fine, it does not have to use CHSH, it does not have to use a table (for me, anyway). The links you gave about the pitfalls of various sampling methods in simulations were very interesting, but there is no simulating in the macroscopic experiment so effects like that are not applicable here?

That is why I came up with the coin example. It is macroscopic and illustrates the issue clearly.

CHSH seem to be being associated with trickery, but I see it simply as an estimate of a correlation coefficient.

Ask yourself, what is the main difference between the derivation of the CHSH and the experiments? It is the fact that we are unable to measure a pair of particles more than once, so we measure separate pairs and substitute the results into the CHSH expression. This is the trickery, not deliberate though. More like stumbling block.

The issue seems really to be how to do those calculations without using a table or tables and I do not know how that can be done?

The issue really is to make sure we are using a single set to compare with a single set inequality and multiple sets to compare with a multiple-set inequality. Anything else will be unfair to one of the parties of the bet. We can make any tables we like. Just the appropriate table for the appropriate purpose. 4 separate tables when comparing with QM and experiment, a single table when comparing with the CHSH. If Richard and others insist to make a single table and compare with QM and experiments, then it is rigging.

[gill1109]

Fascinating, Michel.

Variant 1: Joy's experiment generates 10,000 directions saved in two files (the second file just has the opposite directions to the first).

Variant 2: Joy's experiment generates 4 sets of 10,000 directions each saved in two files (the second file just has the opposite directions to the first).

The experimental procedure for every single run was identical (10 000 runs in variant 1, 40 000 in four groups of 10 000 in variant 2).

Michel predicts that if I compute E(0, 45) with the 10 000 runs from variant 1, I get the answer - 0.5
Michel predicts that if I compute E(0, 45) with the first 10 000 runs from variant 2, I get -0.7.

Yet the whole procedure which leads to one set of data in variant 1, is identical to the procedure which leads to four sets of data in variant 2, stopped after the first set is complete.

The calculation of E(0, 45) is the same calculation, the same formula, based on data which is prepared experimentally in identical fashion.

Of course the numbers might be slightly different but an average of 10 000 carefully performed runs should not be much different.

Suppose now an experiment of the variant 2 type is performed but stopped because of some natural disaster, after the first set of 10 000 runs are done. It is now identical in all practical respects to a completed experiment of variant 1 type.

So the intention of the experimenter has an influence on the particles???? Even though the instructions to the lab technicians were identical???? You badly need to do my little R experiment. It might teach you something by your own eyes, which is impossible to get through to you with words of formulas.

Or consider it another way round. First we do the whole experiment Joy's way, N = 10 000, computing all four correlations with the same 10 000 runs. Then Joy realizes at last that this is a bad idea, and raises some more money so that he can just pay for three more each sets of 10 000 runs. Now we have four sets. He recomputes the first correlation on the first set, and it magically jumps from - 0.5 to - 0.7? Do you really believe that?

[minkwe]

Variant 1: Joy's experiment generates 10,000 directions saved in two files (the second file just has the opposite directions to the first).
Variant 2: Joy's experiment generates 4 sets of 10,000 directions each saved in two files (the second file just has the opposite directions to the first).

The experimental procedure for every single run was identical (10 000 runs in variant 1, 40 000 in four groups of 10 000 in variant 2).

Michel predicts that if I compute E(0, 45) with the 10 000 runs from variant 1, I get the answer - 0.5
Michel predicts that if I compute E(0, 45) with the first 10 000 runs from variant 2, I get -0.7.

Yet the whole procedure which leads to one set of data in variant 1, is identical to the procedure which leads to four sets of data in variant 2, stopped after the first set is complete.

The calculation of E(0, 45) is the same calculation, the same formula, based on data which is prepared experimentally in identical fashion.

Nice try Richard. You still do not get it, even after I've dumbed it down for you many times. You are so convinced you are right, that your mind is blocked.

You say:

Michel predicts that if I compute E(0, 45) with the 10 000 runs from variant 1, I get the answer - 0.5
Michel predicts that if I compute E(0, 45) with the first 10 000 runs from variant 2, I get -0.7.

I say no such thing. Here is what I say, now pay attention carefully and note ALL the differences with your caricature above:
If you compute compute E(0, 45) AND E(0, 135), AND E(90, 45) AND E(90, 135) all from the same set of particles in variant 1, with each particle pair contributing to EACH AND EVERY ONE OF THOSE CORRELATIONS, NOT ALL of them will agree with QM. But if you calculate E(0, 45) FROM THE FIRST set in variant 2, AND E(0, 135) FROM THE SECOND set in variant 2, AND E(90, 45) FROM THE THIRD set in variant 2 AND E(90, 135) from the FOURTH set, they will ALL AGREE with QM.

[gill1109]

Here is what I say, now pay attention carefully and note ALL the differences with your caricature above:
If you compute compute E(0, 45) AND E(0, 135), AND E(90, 45) AND E(90, 135) all from the same set of particles in variant 1, with each particle pair contributing to EACH AND EVERY ONE OF THOSE CORRELATIONS, NOT ALL of them will agree with QM. But if you calculate E(0, 45) FROM THE FIRST set in variant 2, AND E(0, 135) FROM THE SECOND set in variant 2, AND E(90, 45) FROM THE THIRD set in variant 2 AND E(90, 135) from the FOURTH set, they will ALL AGREE with QM.

Why does it make a difference to the value of E(0, 45), whether or not I also calculate three more correlations using the same data file or different data files? The input data for variant 1 AND variant 2 equals the same computer file, containing spherical coordinates representing 10 000 directions. It's a plain computer file of 10 000 x 2 real numbers. In variant 1, the same file is taken as input for the calculation of each of the four correlations. In variant 2 we have three more files and use a different file for each correlation. But the first file is the same. All four files are prepared using identical procedure. The lab technicians who actually do the dirty work don't even know which file they are making. They just make four, in precisely the same way. After the files are made, someone gives them four names, and it doesn't matter which, it doesn't matter what the order of their names is.

You said in the first case we got +/- 0.5; while in the second case we got +/- 0.7

In the first case, I run a piece of computer code four times on the same set of data, with two parameters different, each time. So my intention for the second, third, and fourth calculation, changes the result of the first.

That is what you are saying, my poor confused friend.

Seems to me you haven't even read Joy's paper.

I do 4 times 10 000 runs. I create four files containing 10 000 x 2 numbers theta, phi.

Variant 1:
calculate E(0, 45) with input data = file 1, then
calculate E(0, 135) with input data = file 1, then
calculate E(90, 45) with input data = file 1, then
calculate E(90, 135) with input data = file 1.

Variant 2:
calculate E(0, 45) with input data = file 1, then
calculate E(0, 135) with input data = file 2, then
calculate E(90, 45) with input data = file 3, then
calculate E(90, 135) with input data = file 4.

The same deterministic computer program is used for all these four calculations. It takes as argument the name of a file, and two angles. My computer experiment which you refuse to do does exactly these two experiments. You are telling me that the first calculation of each set of 4 will give a different answer depending on what the next three calculations are. You must be out of your mind...Because you refuse to do the experiment, you are never going to find out. That's called tunnel vision. Do not go out with eyes open because you might see something which contradicts what you already "know for sure" to be true.

Variant 1 is the experiment which Joy wants to do. You recommend against it. You recommend he chooses Variant 2. Well, I am laughing all the way. One or both of you is badly confused. I win, with certainty under Variant 1; with large probability under Variant 2.

[minkwe]

Why does it make a difference to the value of E(0, 45), whether or not I also calculate the other three, on the same computer file containing coordinates representing 10 000 directions?

I'm glad you asked, because that is the root of your misunderstanding: it does not matter for E(0, 45). It matters for the other three once you have determined E(0, 45). Because they become counter-factual. They are drawn from a distribution which is opposite to what is observed, therefore they must have a probability distribution opposite to what is observed, not the same as what is observed, E(0, 45).

Because when I say the probability of obtaining one outcome X, P(X) = 0.75, the counter-factual probability of obtaining X, is not the same as P(X) because counter-factually, we are talking about what we did not get but could have gotten. It cannot have the same probability as what we actually got. It is in fact P(~X) = 0.25. Therefore the counter-factual correlations from the same set of particles are not necessarily the same as the actual correlations from different sets of particles. They will be the same only when P(X) = 0.5.

Do you understand it now?

[gill1109]

Why does it make a difference to the value of E(0, 45), whether or not I also calculate the other three, on the same computer file containing coordinates representing 10 000 directions?

I'm glad you asked, because that is the root of your misunderstanding: it does not matter for E(0, 45). It matters for the other three once you have determined E(0, 45).

Because when I say the probability of obtaining one outcome X, P(X) = 0.75, the counter-factual probability of obtaining X, is not the same as P(X) because counter-factually, we are talking about what we did not get but could have gotten. It cannot have the same probability as what we actually got. It is in fact P(~X) = 0.25. Therefore the counter-factual correlations from the same set of particles are not necessarily the same as the actual correlations from different sets of particles. They will be the same only when P(X) = 0.5.

Gasp. How does it make a difference which order I calculate the four correlations? Do you realise that we are talking about Joy's experiment?? Have you read his experimental paper??? Carefully????

I have one file (well actually one pair of files, with equal and opposite directions) based on one experiment with one set of 10 000 exploding balls. I run the file through one program with four different parameters (the four setting pairs). The program is already written. I just run it four times. Joy approved of it. Joy approved the format of the data. Spherical coordinates. N of them. Now I repeat the complete experiment (but not the final computation) three more times (three more sets of 10 000 exploding balls) and now I have altogether four pairs of files. Exactly the same experimental procedure. There is nowhere in the experimental procedure, any mention of which settings are going to be used.

Now I have four files, prepared in idential way, without even any knowledge of what was going to be done with the results. I run them one at a time through same program... results are completey different. Even the results for the first file (the one that hasn't changed) and the first setting pair, has changed, from - 0.5 to - 0.7

Do you really believe that?

I asked you to convert the program into Python. Perhaps if you do that, you'll find out what this thread is about.

[Heinera]

Astounding that you are still unable to see that the results of the bet depend on whether we are measuring on a single set of particles or different sets of particles. Now I know you still disagree with me that there is a difference between single set and disjoint sets. But assume for a moment that there is a difference. Will you then recognize that if I am right, the way the experiment is performed will determine the outcome of the bet? I'm not talking about a silly addition procedure of results you already have. I'm talking about how you design the experiment and which correlations you calculate in the first place. I'm talking about how you obtain the numbers that you can then add and average. It seems to me anyone genuinely interested in the bet should want to know whether the way the experiment is performed can influence the outcome or not.

If you understand this so well, I am sure you can come up with a numerical demonstration. Generate two lists of vectors (10 000 elements each) that you think can achieve this. I will then run this through Richards program (the version that computes the correlations on four random disjoint subsets). If the correlations E(0, 45) , E(0, 135), E(90, 45) and E(90, 135) all come out with the QM values, I will shut up forever.

[minkwe]

If you understand this so well, I am sure you can come up with a numerical demonstration. Generate two lists of vectors (10 000 elements each) that you think can achieve this. I will then run this through Richards program (the version that computes the correlations on four random disjoint subsets). If the correlations E(0, 45) , E(0, 135), E(90, 45) and E(90, 135) all come out with the QM values, I will shut up forever.

Richard's program is a joke and a waste of time. I'm explaining to you why it is a joke and you are asking me to use it to prove my point? Seriously!? I've already produced 3 numerical demonstrations which illustrate the problem. I've dumbed it down it to a very simple numerical illustration using coins but you still do not want to see it.

[gill1109]

Give us the two lists of vectors, then, Michel. If the correlations E(0, 45) , E(0, 135), E(90, 45) and E(90, 135) all come out with the QM values, I will shut up forever, too. Goodnight all!

[Heinera]

If you understand this so well, I am sure you can come up with a numerical demonstration. Generate two lists of vectors (10 000 elements each) that you think can achieve this. I will then run this through Richards program (the version that computes the correlations on four random disjoint subsets). If the correlations E(0, 45) , E(0, 135), E(90, 45) and E(90, 135) all come out with the QM values, I will shut up forever.

Richard's program is a joke and a waste of time. I'm explaining to you why it is a joke and you are asking me to use it to prove my point? Seriously!? I've already produced 3 numerical demonstrations which illustrate the problem. I've dumbed it down it to a very simple numerical illustration using coins but you still do not want to see it.

Sigh. So I'll just divide the lists into four random disjoint subsets myself, and then compute the four correlations separately, myself. Satisfied? Just give me the two lists.

[minkwe]

Gasp. How does it make a difference which order I calculate the four correlations?

Again Richard, you can calculate the correlations anyway you like, but don't be deceived that they mean the same thing no matter how you calculate them.

Richard. I toss a coin 10,000 times on a glass table with a cameras above and below the table, each recording the results of the coin. The experiment produces two files, one from each camera. You have a program which reads the file and calculates an expectation value. You think it doesn't make a difference which file you are running through your program?

Now, I toss the exact same coin 10,000 times again, same experiment as above, we get two more files. You think it doesn't make a difference which of the two files you are running through your program?

Now I chose just the files corresponding to the results from ABOVE the table, you still think they will give the same results as those corresponding to the result from UNDER the table?

[minkwe]

Give us the two lists of vectors, then, Michel. If the correlations E(0, 45) , E(0, 135), E(90, 45) and E(90, 135) all come out with the QM values, I will shut up forever, too. Goodnight all!

Using my analogy, you are asking me to give you two outcomes from the same coin which violates <A> + <B> = 0 and agrees with QM prediction for two separate coins E(A) + E(B) = 0.5
If you understood my argument, you will be asking for 4 lists of pairs of vectors which agree with QM. Or better yet, you'd have shut up already.

[minkwe]

I win, with certainty under Variant 1; with large probability under Variant 2.

Richard, if as you say the average values from Variant 1 are the same as those from Variant 2, why is there a difference in your likelihood of winning the bet between the two. Those are your own words, so please explain why according to you will certainly win it under Variant 1 but can lose it under Variant 2.

It seems you know that there is a difference. Explain the source of the difference.

[gill1109]

What Joy's experiment is about: which of these two pictures is correct?

The surfaces are theoretical correlation functions rho(a, b).

The points are theoretical correlations - four of them, according to two different theories.

Joy's experiment will measure four points, generating four observed correlations E(a, b), not shown in the images.

Will they be close to the blue or to the red points? The answer to that question will determine the outcome of the bet.




The images are made by the R script http://rpubs.com/gill1109/Wireframe
I'm investigating even better ways to visualise this. Help and advice is welcome. Maybe it's easier to do with Mathematica...

[Joy Christian]

What Joy's experiment is about: which of these two pictures is correct?

The images are made by the R script http://rpubs.com/gill1109/Wireframe
I'm investigating even better ways to visualise this. Help and advice is welcome. Maybe it's easier to do with Mathematica...

These images are very nice. You may want to use a third colour for the four points to make them visually better distinguishable from the grids.

Also, just for the record, my model makes exactly the same prediction for the four points as quantum mechanics does: http://rpubs.com/jjc/13965.

[Joy Christian]

Code: Select all

good <- abs(ca) > f & abs(cb) > f  ## Select the 'states'

Indeed: "Select the states." Not "Reject the data."

The initial or complete state of the system in the model is the pair (e, theta), NOT just the vector e by itself.

[gill1109]

Code: Select all

good <- abs(ca) > f & abs(cb) > f  ## Select the 'states'

Indeed: "Select the states." Not "Reject the data."

The initial or complete state of the system in the model is the pair (e, theta), NOT just the vector e by itself.

In the experiment, Nature selects the states. There will be no post-selection by the experimenter. No detection loophole. N runs -> N states.

4N runs -> 4N states if Joy wishes to use separate runs for the four correlations. N runs for each correlation.