Joy Christian's colourful exploding balls experiment

[Mikko]

Gravity is actually a problem because the balls would fall towards the ground in a parabolic motion when they fly from the centre of the "explosion" towards the laser screens. Although the spins will not be affected significantly, they would be tilted differently towards the ground for each run.

This and other relativistic effects are too small to have any effect on observed spins.

[Joy Christian]

Gravity is actually a problem because the balls would fall towards the ground in a parabolic motion when they fly from the centre of the "explosion" towards the laser screens. Although the spins will not be affected significantly, they would be tilted differently towards the ground for each run.

This and other relativistic effects are too small to have any effect on observed spins.

I hope you are right. Do you have any substantive arguments, or references, to support this?

[gill1109]

Sorry, the bet is called off, the experiment is meaningless.

Fred Diether and Michel Fodje have convinced m that the only bound one can place in experiments is 4.

Aspect, Weihs and all those others might as well have stayed at home since they just observed E11 - E12 - E21 - E22 between -4and 4, which was a forgone conclusion.

Joy's experiment will do the same.

What's the point?

[Joy Christian]

Sorry, the bet is called off, the experiment is meaningless.

Fred Diether and Michel Fodje have convinced m that the only bound one can place in experiments is 4.

Aspect, Weihs and all those others might as well have stayed at home since they just observed E11 - E12 - E21 - E22 between -4and 4, which was a forgone conclusion.

Joy's experiment will do the same.

What's the point?

My proposed experiment is not about any bounds. It is about detecting the effect of torsion in our physical space on the correlation function E(a, b).

[Heinera]

.... about a page or two ago, Richard displayed code:

## Fourth pair of measurement directions

Alpha <- 90 * pi / 180
Beta <- 135 * pi / 180
A <- sign(cos(AliceTheta - Alpha))
B <- - sign(cos(BobTheta - Beta))
E22 <- mean(A * B)

CHSH <- E12 - E11 - E21 - E22

CHSH
if (CHSH > 2.4) print("Congratulations, Joy") else
print("Congratulations, Richard")

If I understand Richard's four sets of results: the cosine for the angle between Alice's and Bob's angles give either 0.7071 or -0.7071 {as cos 45 = 1/sqrt 2}.

cos 45 is 0.7071, and cos 135 is -0.7071. For E12, the relative angle is 135, for the three others, it is 45. Since the quantum correlations are -cos(relative angle), we then get E11 = -0.7071, E12 = 0.7071, E21 = -0.7071, and E22 = -0.7071, for the quantum case.

These four sets of results give only four places on the sawtooth & cosine graph. I suspect that Richard is less likely to win with four settings than if the settings are spread out across the whole 360 degrees, but he will know better.

He knows that even with only four correlations his probability of winning is 1, since it is not very hard to prove that E12 - E11 - E21 - E22 can never exceed 2 in his program, no matter what vectors the input files contain.

[Joy Christian]

He knows that even with only four correlations his probability of winning is 1, since it is not very hard to prove that E12 - E11 - E21 - E22 can never exceed 2 in his program, no matter what vectors the input files contain.

The probability of Richard Gill's winning the bet is 0. He has just realized this. That is why he wants to call off the bet:

Sorry, the bet is called off, the experiment is meaningless.

Fred Diether and Michel Fodje have convinced me that the only bound one can place in experiments is 4.

Aspect, Weihs and all those others might as well have stayed at home since they just observed E11 - E12 - E21 - E22 between -4and 4, which was a forgone conclusion.

Joy's experiment will do the same.

[Ben6993]

Heinera wrote:

He knows that even with only four correlations his probability of winning is 1, since it is not very hard to prove that E12 - E11 - E21 - E22 can never exceed 2 in his program, no matter what vectors the input files contain.

Are you sure? (I was agreeing with you until I made a more detailed check. All these double negatives get slippery ... ) Richard's values for the cosine curve would be as you stated with E12 being 0.7071 and the other three values being -0.7071. Despite rearranging the natural order of the four terms, Richard does reverse the sign of the three negative values in his CHSH so the expected value for the cosine curve is 4 x 0.7071 = 2.8 by my calculation. Which is fine by me. Is that right?

As I understand it, an overall correlation could easily be used instead of the CHSH statistic. Richard's battery of R programs all seem to give an overall correlation, and that can be done for the aggregate of the four sets of data. It does not require a wide range of a and b settings to obtain an overall correlation. The 45 degree angle gives the maximum deviation between sawtooth (0.5) and cosine curve (0.7071) points and so could be the most efficient, and simplest, place to test.

[Heinera]

Heinera wrote:

He knows that even with only four correlations his probability of winning is 1, since it is not very hard to prove that E12 - E11 - E21 - E22 can never exceed 2 in his program, no matter what vectors the input files contain.

Are you sure? (I was agreeing with you until I made a more detailed check. All these double negatives get slippery ... ) Richard's values for the cosine curve would be as you stated with E12 being 0.7071 and the other three values being -0.7071. Despite rearranging the natural order of the four terms, Richard does reverse the sign of the three negative values in his CHSH so the expected value for the cosine curve is 4 x 0.7071 = 2.8 by my calculation. Which is fine by me. Is that right?

That is right. The expression could just as well be written in the natural order -E11 + E12 - E21 - E22.

As I understand it, an overall correlation could easily be used instead of the CHSH statistic. Richard's battery of R programs all seem to give an overall correlation, and that can be done for the aggregate of the four sets of data. It does not require a wide range of a and b settings to obtain an overall correlation. The 45 degree angle gives the maximum deviation between sawtooth (0.5) and cosine curve (0.7071) points and so could be the most efficient, and simplest, place to test.

Now I'm afraid you lost me here What do you mean when you say overall correlation?

[gill1109]

He knows that even with only four correlations his probability of winning is 1, since it is not very hard to prove that E12 - E11 - E21 - E22 can never exceed 2 in his program, no matter what vectors the input files contain.

The probability of Richard Gill's winning the bet is 0. He has just realized this. That is why he wants to call off the bet:

Sorry, the bet is called off, the experiment is meaningless.

Fred Diether and Michel Fodje have convinced me that the only bound one can place in experiments is 4.

Aspect, Weihs and all those others might as well have stayed at home since they just observed E11 - E12 - E21 - E22 between -4and 4, which was a forgone conclusion.

Joy's experiment will do the same.

Joy: I am not talking about our bet till Michel and Fred have gotten some sense into their heads. Please help me to get some sense into their heads. They are arguing that the only thing anyone can sensibly say is that an experimentally observed E12 - E11 - E21 - E22 is less than or equal to 4.

Their "arguments" apply equally well to the bound under local realism derived by Clauser, Horne, Shimony and Holt (2), as the bound under quantum mechanics derived by Tsirelson (2 sqrt 2).

Experiment might in principle give any result between -4 and 4, since in experiment, we (almost always) only measure each particle in one direction.

They are obsessed by the meaning of the word "bound" and seem not to realise that in experiments we have to take account of statistical error.

Anyone who thinks the Tsirelson bound is correct, and predicts that a CHSH style experiment on quantum systems will always give a result less than or equal to 2 sqrt 2, is out of their minds.

The correct prediction is that a CHSH type experiment on quantum systems will usually give a result less than or equal to 2 sqrt 2 plus some multiples of the standard error, which is of the order of 1 / sqrt N, N being the number of runs in the experiment.

A similar reasonable prediction for someone who thinks the CHSH bound is correct, is that a CHSH type experiment on classical systems will usually give a result less than or equal to 2 plus some multiples of the standard error, which is of the order of 1 / sqrt N, N being the number of runs in the experiment.

[gill1109]

PS: I kindly asked Michel (and now I also kindly ask Fred) to do me a pleasure, and try a little experiment, which I described over at http://www.sciphysicsforums.com/spfbb1/viewtopic.php?f=6&t=40 and which should not take up more than half an hour of their precious time. I am quitting this forum and quitting this bet till they have done me the pleasure of doing this experiment and reporting their findings in that new thread. There is a little piece of R code there, which actually comes straight out of the R code which I propose for use in our bet. I would like to see Python and Mathematica versions of that code, because I need to convert the bet data analysis code to a few other languages, so that everyone can agree that it's OK. So let's start with translations and tests of the small code snippet at http://www.sciphysicsforums.com/spfbb1/viewtopic.php?f=6&t=40

Do these guys want to see this experiment done? Or do they think that experiments are meaningless since with logic they already know that the only sensible bound is "4" and we don't need to do an experiment to see that that bound is satisfied?

Does physics make predictions about outcomes of experiments, yes or no?

Are all predictions from physics "certain"?

For instance, a physical theory predicts that nothing goes faster than the speed of light. Yet I can imagine an experiment which determines the speed of some particle or other, comes up with a number bigger than the "official" speed of light. On the other hand, a decent experimenter also reports how accurate he thinks his estimate is. So the experimenter is not going to go to the newspapers when his measurement outcome is within a few standard deviations of the "official" speed of light. He'll only go to the newspapers with his findings when his measurement outcome is five or more standard devations about the "official" speed of light.

Similarly the Higgs boson. They broadcast the news to the world when the deviation of the signal from what would have been expected if it hadn't existed was five standard deviations. Yet in principle any size of measurement errors is possible, and whether or not the Higgs existed, the observed signal could have been essentially of any size whatsoever.

[FrediFizzx]

CHSH seems to have some problems as applied to experiments and LHV models. Joy's experiment doesn't need to be evaluated with CHSH. It either gives E(a, b) = -a.b or it doesn't. Figure it out; it is not that hard.

[Joy Christian]

CHSH seems to have some problems as applied to experiments and LHV models. Joy's experiment doesn't need to be evaluated with CHSH. It either gives E(a, b) = -a.b or it doesn't. Figure it out; it is not that hard.

I agree with Fred. Michel also has similar views. It is easy to see why. At the end of the day we are interested in physics, not in Bell or CHSH.

The physical question is simple, as Fred so eloquently puts it: "Joy's experiment doesn't need to be evaluated with CHSH. It either gives E(a, b) = -a.b or it doesn't."

So why this obsession with CHSH? It is especially puzzling because there is no room for any mischief in my experiment, like remote parameter dependence or remote outcome dependence. We will be observing the actual spin directions, without needing to specify either a or b in advance. All the actual spins will be mapped out in advance, in an everyday, macroscopic experiment. There will be no room for non-locality or a question of non-reality. So why this bizarre obsession with CHSH?

[gill1109]

So why this obsession with CHSH? It is especially puzzling because there is no room for any mischief in my experiment, like remote parameter dependence or remote outcome dependence. We will be observing the actual spin directions, without needing to specify either a or b in advance. All the actual spins will be mapped out in advance, in an everyday, macroscopic experiment. There will be no room for non-locality or a question of non-reality. So why this bizarre obsession with CHSH?

Why indeed? Fred and Michel are totally obsessed with the word "bound".

The question is, is E(45) = -0.707 and is E(135) = +0.707 ?

Or is E(45) = -0.5 and is E(135) = +0.5 ?

In an experiment there is always experimental error. N is large but finite so there's always statistical error.

When people say "we observed a violation of CHSH" they just mean that they observed correlations much closer to +/- 0.7 than +/- 0.5, and that the statistical error is much smaller than 0.1, so that one can have confidence that this was not just due to chance.

[Joy Christian]

The physical question is simple, as Fred so eloquently puts it: "Joy's experiment doesn't need to be evaluated with CHSH. It either gives E(a, b) = -a.b or it doesn't."

[gill1109]

The physical question is simple, as Fred so eloquently puts it: "Joy's experiment doesn't need to be evaluated with CHSH. It either gives E(a, b) = -a.b or it doesn't."

Joy, I agree!!!!

Either E(45) = -0.707 and E(135) = + 0.707, or they don't.

You realise that the experimentally measured correlations will typically be off by +/- 1/sqrt N? Maybe we can't afford an incredibly huge N?

I quit the bet unless I get some help with the technical preparations. We need a Python translation, and a Mathematica translation, of the code which determines who has won. It needs to be tested and verified by independent and critical people in whom you trust.

People who are certain you are going to win. People who believe I lie and cheat. People who have some competence in programming.

I posted a small piece of that code on a separate thread. Michel keeps talking about "a silly piece of R code" and refuses to look at it. That silly piece of R code needs to be tested and verified by people who believe you are going to win the bet, by people who believe I lie and cheat and/or am spectacularly incompetent.

Otherwise, no bet. And, I suspect, no bet will mean no experiment.

[Ben6993]

This is in haste as I am setting out very soon on a day of fine art appreciation at Knowsley Hall.

I do not see what the concern is about the CHSH statistic. It seems to allow Joy to win the bet if the balls show the + or - 0.7071 correlation. The correlation could be as low as 0.61 and Joy would still win.

What I meant about preferring to see the overall correlation, rather than the CHSH statistic, is that I feel happier with a target expressed as an overall average correlation of 0.7071 (or as low as 0.61) rather than a statistic of 2.8 (or rather 2.4) just because I am happier with the target as a correlation. But I understand that Richard will have done his calculations for the bet based on the CHSH. The objective is for the balls to achieve, on average, a result nearer to 0.7071 than to 0.5. As far as I can see, the CHSH does that. What CHSH does not do is show the full cosine curve over all 360 degrees of the curve but only shows it for a few points of the curve. But if the target cannot be reached for a few points on the curve, it will not reach the target across the whole curve.

I have not followed arguments about the target being equivalent to an average correlation of + or - 1 as I do not expect to see 0.7071 exceeded. Why worry about reaching 1 when the target is only to reach 0.60 (or CHSH = 2.4)?

[gill1109]

This is in haste as I am setting out very soon on a day of fine art appreciation at Knowsley Hall.

I do not see what the concern is about the CHSH statistic. It seems to allow Joy to win the bet if the balls show the + or - 0.7071 correlation. The correlation could be as low as 0.61 and Joy would still win.

What I meant about preferring to see the overall correlation, rather than the CHSH statistic, is that I feel happier with a target expressed as an overall average correlation of 0.7071 (or as low as 0.61) rather than a statistic of 2.8 (or rather 2.4) just because I am happier with the target as a correlation. But I understand that Richard will have done his calculations for the bet based on the CHSH. The objective is for the balls to achieve, on average, a result nearer to 0.7071 than to 0.5. As far as I can see, the CHSH does that. What CHSH does not do is show the full cosine curve over all 360 degrees of the curve but only shows it for a few points of the curve. But if the target cannot be reached for a few points on the curve, it will not reach the target across the whole curve.

I have not followed arguments about the target being equivalent to an average correlation of + or - 1 as I do not expect to see 0.7071 exceeded. Why worry about reaching 1 when the target is only to reach 0.60 (or CHSH = 2.4)?

Exactly.

Actually we should talk about a surface E(alpha, beta) not a curve E(theta). CHSH is just a clever way of looking at four, critical, points on the *surface*.

Why four? We need to vary both Alice's and Bob's settings, right? 4 = 2x2. It's the most efficient solution. The smallest experiment. The best way to allocate resources. We want a decisive experiment? If we make a bet, we need a clear, fair, unambiguous, criterion? We want to leave as little as possible to chance. CHSH is a nice example of "less is more" in science.

The sexy graphics on the cover of Nature are good for PR, good for the next round of funding. But the hard science often comes down to one number, obtained with the greatest possible care.

[Mikko]

Gravity is actually a problem because the balls would fall towards the ground in a parabolic motion when they fly from the centre of the "explosion" towards the laser screens. Although the spins will not be affected significantly, they would be tilted differently towards the ground for each run.

This and other relativistic effects are too small to have any effect on observed spins.

I hope you are right. Do you have any substantive arguments, or references, to support this?

There is a discussion of precession in MTW, but you can simply note that you can have a freely falling coordinate system that is approximately rectilinear,
orthogonal, isometric, and inertial over a region suffiient for the experiment; and its not hard to estimate the deviations.

[minkwe]

I do not see what the concern is about the CHSH statistic. It seems to allow Joy to win the bet if the balls show the + or - 0.7071 correlation. The correlation could be as low as 0.61 and Joy would still win.

Ben, the issue is actually very simple to see using the coin toss example I presented in the other thread.
1. For a single coin toss (H=+1, T=-1), where A is the result we get and B is the other result we didn't get but could have gotten A + B = 0, and for repeated tosses of a single coin <A> + <B> = 0, and E(A) + E(B) = 0. This is equivalent to the CHSH <= 2 expression. Every time we toss a coin we get one outcome but because the coins are locally realistic, we assume that the other outcome exists counter-factually. This is the usual argument used to derive the CHSH and Bell-like inequalities.
2. Since we can only get one result from a single coin, we may decide to toss two separate coins (of exactly the same type as above). We assume coins are local realistic, and it shouldn't matter if we use the counter-factual result or two actual results where A is the actual result from first coin and B is the actual result from the second coin. Since the coins are the same, we reason that the probability of H and the probability of T are the same for each coin, and it shouldn't matter if we are tossing them singly or in pairs, we naively assume that the expression <A> + <B> = 0 should still be valid for the two coins.

3. It turns out, QM predicts E(A) = E(B) = 0.25, which gives us E(A) + E(B) = 0.5. So we perform an experiment by tossing two coins, and we find that we get <A> + <B> = 0.5 consistently. Confirming the QM result and apparently violating E(A) + E(B) =0 . And we conclude naively that therefore the coins are not locally realistic.

4. But if we examine the two-coin experiment very carefully, we realize that the result from one coin is mutually independent of the results from the second coin. Since H = +1 and T = -1, both A and B can independently reach a maximum of +1 for two separate coins, whereby the maximum value of A + B for a separate coins is 2 not 0. Therefore by tossing a pair of local realistic coins many times E(A) + E(B) <= 2. Which means our two coin experiment with <A> + <B> = 0.5 is fully consistent with local realistic coins. (Note the absence/irrelevance of loopholes)

5. Where was the error made? Re-evaluating the derivation of the E(A) + E(B) = 0 expression we see that the two outcomes A and B are mutually dependent. For a single coin, knowing the A outcome immediately determines the B outcome. If a coin gave H, it definitely did not give T. Whenever A = +1, B = -1 and whenever A = -1, B = +1 for a single coin. Therefore A + B = 0 is the single coin relationship that was carried over to the averages <A> + <B> =0 and expectation values E(A) + E(B) =0 . However this relationship does not apply to A and B from two separate coins because A and B can both be H or T independently of each other, impossible for a single coin.

6. It appears to be a trivial silly error but it is in fact the error in Bell's theorem. You see, QM predictions are for separate sets of particles, experimental results are from separate sets of particles, but the CHSH is derived from a single set of particles. It turns out our specific coins which gave <A> + <B> = 0.5 have a probability distribution of [H,T] of [0.625,0.375]. They produce Heads 62.5% of the time, local realistically. E(A) = E(B) = 0.625*(1) + 0.325*(-1) = 0.25, E(A) + E(B) = 0.5. Yet if we toss one of our coins on glass table and record the up-facing result on column A, and the down facing result on column B, even though <A> = 0.25, just like the result from two coins, the counter-factual result <B> will not be 0.25, it will be -0.25. so that <A> + <B> from a single coin will still never be different from 0. It is now obvious why it is wrong to substitute actual results separate systems for counter-factual results from a single system and expect to get anything meaningful.

7. So how does all of this affect Joy's experiment and the current "almost-bet"? If you perform an experiment in which you toss your single coin on a glass table and can read both results, you still will never violate E(A) + E(B) = 0, ever. Not even by experimental error, it doesn't matter what kind of coin you use. But if you perform an experiment in which you throw a pair of coins each time and read off the two actual results A and B, one from each, based on the kind of coin you use, you can violate it. But we know it is only an apparent violation, since we are using the wrong expression to compare it with. Richard has been insisting that Joy's experiment be performed by using repeated tosses of a single coin and reading from above and below the glass table, he says Joy can win if he violates E(A) + E(B) = 0 + experimental error. Joy has been insisting that he simply wants to demonstrate that his model matches QM, and since QM predicts for separate coins, we only need to do E(A) from one coin and E(B) from another coin, he says he will win if his coins produce <A> = 0.25.

[Heinera]

Heinera wrote:

He knows that even with only four correlations his probability of winning is 1, since it is not very hard to prove that E12 - E11 - E21 - E22 can never exceed 2 in his program, no matter what vectors the input files contain.

Are you sure?

In fact, the procedure to be used to analyze the experimental data is so simple that it it is a perfect Excel job. So I've written an Excel sheet that you can download here:
CHSH.xlsx
After clicking the link, just skip the pop-up that ask to sign up.

Except for the fact that I'm using Cartesian coordinates, it's the same as Richard's R code.

You will see that the sheet is pretty much self explanatory. For pedagogical purposes, it only has 10 runs of the experiment. No need to know any programming, only basic Excel. The three components of the vectors at Alice's station are in columns B,C,D, and the three components of the vectors at Bob's station are in columns F,G,H.

The computed value of the CHSH expression is in cell O19.

In this version, the list of vectors at Alice's side is just randomly generated, and Bob's vectors are just the negative of Alice's. Each time you press F9, Excel will generate a new set of random vectors.

Now, press F9 a number of times. You will see that the CHSH is exactly 2, no matter what vectors are generated. This is why Richard is guaranteed to win the bet. It has nothing to do with physics; it is a simple mathematical fact.

If you drop conservation of angular momentum and replace Bob's vectors with a completely random list (just replace Bob's 3x10 array with the same RAND() expression as in Alice's vectors), you will see that the CHSH value is now different for each press on F9, but never exceeds 2.

I also have a version with 1000 runs, but otherwise identical:
CHSH1000.xlsx
Here, you see that each of the four correlations now start converging to 0.5.