Classical Spheres, Division Algebras, and the Illusion of Quantum Non-locality:

Richard Gill

Joy quoted me:

"Just fix all of Alice angles to 0 degrees, and all Bob's angles to 30 degrees ...Just calculate one correlation"

and responded:

"That is not the model. The model is S^3, which is a simply-connected topological manifold. What Fred is doing is correct. He can show the same exact cosine correlation in the 0-increment limit without the pictures. The pictures simply make things easier for everyone to understand."

The measurement settings are outside the model. They are chosen by the experimenter.

The pictures give a false impression of accuracy. Zoom in at about 30 degrees and you'll see they are systematically off target by about 0.01, and your latest "improvements" to the code aren't helping a damn.

I'm asking for one point on the curve to be calculated as precisely as possible (science), not for a propaganda poster (bluff).

Richard Gill

PS I can presently only think of two reasons why Joy and Fred don't modify the code to do what I ask:

(1) they don't know how to

(2) they do know, and have seen the result, and don't dare to show it to us.

I hope that this is just lack of imagination on my part.

I would do it myself if I had Mathematica but I don't wish to pay the gangster who sells it. I already rewrote the code in R so that everyone can do it for themselves, for free.

Richard Gill

PS this is Michel Fodje's original algorithm, ported to R, and optimized for speed and accuracy. I compute correlations *at* angles (differences between Alice and Bob's settings) which are multiples of 7.5 degrees. The plots joint the points with straight lines to make them easier to see. But it's all about the points, not the lines.

There is no binning, no unnecessary approximation, no unnecessary discretisation. The formulas I have implemented come straight out of Joy's own description of Michel's simulation.

Each point is based on a simulation of size 1 million. You can rerun the code yourself with 10 or 100 million if you like.

http://rpubs.com/gill1109/EPRB2

Joy Christian

Richard Gill,

I know you have offered money in the past as a challenge to do this or that concerning Bell's theorem. There is a decisive way to settle all controversies about the significance of Bell's theorem in the most scientific way possible. You know about my proposed experiment. There are two possible outcomes for this experiment: one predicted by you, eq.(4), and the other predicted by me, eq.(5). These simple equations are clearly explained on this page of my blog.

Let us see how strongly you believe in your claims. If they are correct, then the experiment will confirm eq.(4). But if my claims are correct, then the experiment will confirm eq.(5). It is a clear-cut experimental question. So show us, in terms of real money, how much you believe that the outcome of the experiment will be eq.(4). How much money are you willing to bet?

Joy Christian

Thomas Ray

Gee, Richard, if you believe that topological torsion isn't real and always present, your risk is zero. Joy's is the true "Randi challenge" -- an empirical demonstration that 'ghosts' are responsible for the quantum correlations we observe.

Joy Christian

Remember that you have the backing of the entire Bell establishment.

Richard Gill

I am prepared to bet up to something in the order of several thousand Euro about the outcomes of a strict-protocol Bell-CHSH-style experiment. (Such money is not peanuts for me, but I can earn the money in a few days by consulting for lawyers).

There are quite a few parameters which we need to agree on, first. It is not quite as clear-cut as you suggest.

Joy Christian

Very good. I respect that you are willing to put your money in what you believe in. The amount itself is immaterial, but it does show how much you believe in your convictions.

I agree that the terms and parameters of the experiment can be made more precise. I am open to suggestions and discussions on all matters concerning the experiment.

Richard Gill

Dear Joy

The first thing we should discuss then, is who is going to do it, when and where?

And we need to fix things like:

* number of trials (I would suggest 10 thousand or so

* who gets to choose the settings (me)

* what is the criterion for who has one (I would suggest that we draw the line exactly half way between 2 and 2 square root of 2)

* what to do with non-detections (I would suggest that they are treated as zero's and included in the correlation. ie 10 000 pairs of settings -> 10 thousand pairs of outcomes, about 2 thousand 500 of each of the four kinds.

Richard Gill

sorry for typos: "one" -> "won"

Joy Christian

(1) The first thing we should discuss, then, is who is going to do it, when and where?

I know a highly respected experimental physicist who is willing to do the experiment provided sufficient funding is available. I am not prepared to reveal the name of the person, the name of the laboratory, the name of the institution, or the name of the country at this juncture. You will have to take my word for it at this juncture. The experiment will be done as soon as funding is available---ideally through a research grant, or through private donation(s).

(2) And we need to fix things like: * number of trials (I would suggest 10 thousand or so

My suggestion would be 100 thousand or more. In my proposal I have considered one million trials.

(3) * who gets to choose the settings (me)

You are welcome to choose whatever settings you may like, and however freely you may like to choose them. It is pretty clear from my discussions of the experiment that this will not be an issue. But see number (4) below.

(4) * what is the criterion for who has one (I would suggest that we draw the line exactly half way between 2 and 2 square root of 2)

(5) * what to do with non-detections (I would suggest that they are treated as zero's and included in the correlation. ie 10 000 pairs of settings -> 10 thousand pairs of outcomes, about 2 thousand 500 of each of the four kinds.

I am not sure what is a non-detection. As far as I can see there is no such thing as non-detection in the scheme I have proposed. So you will have to tell me what you mean by non-detections.

Joy Christian

And one more thing: There will be no Nx4 spreadsheet in my experiment. Each correlation function E(a, b) must be calculated separately, as done in eqs.(2) and (3) on this page of my blog.

Richard Gill

No, there will be no N x 4 spreadsheet

But I do insist on creating myself all N setting pairs. I'll be revealing them one by one, separately to Alice and Bob so they can use them for each of their measurements, without knowing what measurement setting the other has received.

If you have some side requirements such as, for instance, at least 24% of each of the four possible setting pairs, we can discuss that.

The four correlations are computed as the sum of the product of the outcomes divided by the number of pairs, for each of the four pairs of settings.

I understand that there are will be "non-detections" so each outcome will be +/-1.

So I come along to the site(s) with 2N settings 1, 2. N for Alice, N for Bob. At the end of the experiment we also have 2N times a +/- 1; N observed by Alice, N by Bob.

We also need to recruit a small team of referees / adjudicators, I suggest three people, I am ready to propose several names. And I understand that we both wager the same amount, X Euros (or jars of peanut butter or whatever we like to agree on).

Richard Gill

The criterion CHSH < 2 or >2 is unacceptable for many pretty obvious reasons. A pretty stupid local hidden variables model can already win half the time, by arranging that the mean value is exactly 2. The observed averages are spread on both sides of this number.

And with the half way crition, the condition that you know half the settings is unreasonable too. If you know half the setting pairs you can arrange that on that half, CHSH = 4; on the other half you get CHSH approx 2; overall you get 3.

You don't prove grand claims about the nature of physical reality in such a cheap way.

Joy Christian

And you can't fool me with any of your rigged games. The bet is off.

Richard Gill

Joy writes "And you can't fool me with any of your rigged games. The bet is off. "

What? But this game can be won in the quantum optics lab with CHSH = 2.8, approx! At least, they will probably have done it within 5 years, according to the top experts in the field. And the top experimenters are in a major competition to be there first. How am I rigging anything?

I'll fix the settings in advance and give them to our trusted adjudicators. So that it's clear that *I* am not cheating.

On my side, I am a bit worried that your secret experimentalist is actually going to use photons instead of colourful exploding balls.

Joy Christian

OK, I propose that you first prove to the adjudicators (without letting me or the experimentalists know) that QM does predict |CHSH| > 2.414 for your chosen settings. This way I can be sure that you are not cheating. In any case, if you read the last page of the attached paper you will see that settings need not be chosen before all the runs are completed. You will also see that there cannot be any non-detections in the experiment.

Name the referees or adjudicators and I will accept them or veto them.Attachment #1: 34_0806.3078v2.pdf

Richard Gill

I'm happy that the adjudicators see my settings in advance (I must trust that they will keep secret from you). I can tell you now that I am just going to pick them using the pseudo random generator in R, with a random seed which I'll dream up on the morning of the big day. So in fact I just have to send them my R code and the seed, and they have them. They can verify at any time that the numbers I bring along in a couple of envelopes (or on a couple of USB sticks) are indeed the numbers I said I'd bring.

I would propose to ask Andrei Khrennikov (Vaxjo, Sweden) and Inge Helland (Oslo) and Lucien Hardy (Perimeter Institute). Another possibilities might be Adrian Kent. But go ahead, name a few others. I have deliberately chosen people whom (a) I trust (b) I know think very differently from me about quantum foundations.

The point is that these must be three persons, whom we both trust. After all, our scientific reputation (and our money, but that is less important) will be in their hands.

Richard Gill

By the way, the settings which I deliver are binary: N times 1 or 2 for Alice, N times 1 or 2 for Bob. You have total control over which angles these "labels" should correspond to. Anything you like for Alice, and anything you like for Bob.

Richard Gill

PS the correlations are computed by splitting the data into four groups according to the setting *labels*.

The pairs are of course (1,1), (1,2) (2,1), and (2,2). A completely standard setup for a state of the art delayed choice Bell-CHSH-type experiment with event-ready detectors.

Richard Gill

You said "Remember that you have the backing of the entire Bell establishment. ". We can allow other people to put in money too. People can bet on you or bet on me. I think a huge amount of money will be put on me winning, so this money will go to your supporters when you win.

An interesting proposition?

Richard Gill

It's called "crowd-funding". Probably we can fund the experiment in this way, too.

Joy Christian

I accept Andrei Khrennikov and Lucien Hardy. And I propose Hans De Raedt, Steve Weinstein (Perimeter), and Christopher Fuchs (Perimeter) (assuming that these people will be willing to get involved). By the way, although Lucien and I are friends, I consider him to be firmly in your camp.

Joy Christian

I like your idea of "crowd-funding". That is probably the easiest way to fund this unorthodox experiment.

Richard Gill

Excellent names, all of them.

Maybe FQXI likes to help organising the "crowd-funding" part? I think this should be arranged by an independent neurtal organization of high standing.

Joy Christian

I am not in good books of many people these days, including FQXi. Would you like to write to FQXi directorate asking them for a platform (i.e., a dedicated blog space) for this adventure? After all, it could turn out to be an experiment of the century!

Richard Gill

The adjudicators should be an odd number of person, with a chairman who will report any decisions by them to us.

You can start, by asking someone if they are willing to be chairperson. I'm sure we can quickly agree on a whole team.

Another name I would like to mention would be Gregor Weihs.

The adjudicators' job is not to lay down the rules. But to interpret the rules which we together agree on, in case of disagreement.

Richard Gill

I don't know if I am in anyone's good books or bad books - no present or past connection to FQXI at present. But I can write to someone...

Joy Christian

I veto Gregor Weihs, just as I vetoed Adrian Kent. Nothing personal. I just don't think they can be unbiased adjudicators for this particular experiment.

I will ask Steve Weinstein (the only philosopher) whether he would like to be the Chairman.

Richard Gill

I don't know Steve or his work (mea culpa) but I know people who respect him highly. Excellent start. Please go ahead and find out if he is interested. We can switch to private email if we don't want to do the next steps in the public eye.

Joy Christian

For those who may be curious, Richard Gill and I are going ahead with the bet, and discussing all further details in private until there is something significant to report.

Thomas Ray

Richard, you've spent a great many words to question others' competence in simulation and modeling. Would that you exert an equal effort to examine your own competence in analysis.

Richard Gill

Tom, I make no claim to be a universal genius.

I claim to be more competent at computational mathematics than Joy or Fred ... but that is not saying much.

Fred: please post the Mathematica code somewhere as a plain text file, so one can run it by copy-paste.

Thomas Ray

Richard, are quantum correlations caused by computational mathematics?

Richard Gill

Tom we are discussing computational mathematics. If you are not interested in that subject, please just shut up.

Robert McEachern

Richard,

"So Penrose screwed up" And so did everyone else.

"read Bell's article about Bertlemann's socks"

Read my earlier comments on Bertlemann's sock's.

Rob McEachern

Thomas Ray

" ... we are discussing computational mathematics."

You are discussing computational mathematics; we are discussing quantum correlations as locally real phenomena. The rest is details.

Thomas Ray

"Your simulation in R is inconsequential. All that is left to do is some minor improvements like increasing the degree increment resolution. As the degree increment resolution gets finer, the simulation will get closer and closer to the cosine curve."

Fred's statement is not only true, it is predicted in Joy's analytical framework, which is independent of parameters of scale. Conventional quantum theory cannot explain where quantum reality smooths out into classical functions, for the simple reason that such a domain boundary doesn't exist. Quantum correlations at every scale are a function of the topology, just as Joy has exhaustively explained.

Resolution to arbitrary accuracy in a simulation of continuous measurement functions of two random variables is quite a different animal than Richard Gill's strawman model which is dependent on measurement settings and the separations thereof.

Put that in your mathematical computation pipe and smoke it.

Richard Gill

Here is the Mathematica code for an experiment in which Alice and Bob keep their settings fixed:

spin = 1/2;

phase = 2 Pi spin;

spin2 = 2 spin;

trials = 1000000;

aliceDeg = 0;

bobDeg = 0;

aliceAngle = 0;

aliceDeg = 0;

bobAngle = Pi / 6;

bobDeg = 30;

nPP = 0;

nNN = 0;

nPN = 0;

nNP = 0;

nA = 0;

nB = 0;

test[angle_, e_, lambda_] := Module[{c, out},

c = Cos[1 (angle - e)];

If[lambda >= Abs[c], out = 0, out = Sign[c]]; out]

Do[

eVector = RandomReal[{0, 2 Pi}];

lambda = (1/2) Sin[RandomReal[{0, Pi / 2}]]^2;

eLeft = RandomReal[{0, 2 Pi}];

eRight = eLeft + 2 Pi spin;

aliceD = test[aliceAngle, eLeft, lambda];

bobD = test[bobAngle, eRight, lambda];

If[aliceD == 1, nA++];

If[bobD ==1, nB++];

If[aliceD ==1 && bobD == 1, nPP++];

If[aliceD == 1 && bobD == -1, nPN++];

If[aliceD == -1 && bobD == 1, nNP++];

If[aliceD == -1 && bobD == -1, nNN++],

{i, trials}];

corr = (nPP - nPN - nNP + nNN)/(nPP + nPN + nNP + nNN)

corr // N

-Cos[bobAngle] // N

Here is the result:

In[96]:= (nPP - nPN - nNP + nNN)/(nPP + nPN + nNP + nNN) //N

Out[96]= -0.848954

In[97]:= - Cos[bobAngle] // N

Out[97]= -0.866025

See the difference?

Richard Gill

Fred's Mathematica code, 1 million trials, Alice setting fixed at 0 degrees, Bob's at 30 degrees:

In[96]:= (nPP - nPN - nNP nNN)/(nPP nPN nNP nNN) //N

Out[96]= -0.848954

In[97]:= - Cos[bobAngle] // N

Out[97]= -0.866025

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