## Quaternion simulation no hidden variable

Foundations of physics and/or philosophy of physics, and in particular, posts on unresolved or controversial issues

### Re: Quaternion simulation no hidden variable

FrediFizzx wrote:
gill1109 wrote:I know a theorem from machine learning and AI (a distributed computing no-go theorem) which tells me not to try...

Suit yourself. I have given all the info necessary for anyone to do a sim on a different program.

a = RandomInteger[{1, 360}]; Random angle in one degree increments
b = RandomInteger[{1, 360}];
A = -Sign[Cos[a Degree]];
B = Sign[Cos[b Degree]];

If QM could do event by event outcome prediction is this perhaps the prediction it would give?

Fred, your model says that the outcomes A and B at detectors 1 and 2 are completely separate, deterministic, functions of the settings given to each detector. The expectation of the product of the outcomes, i.e., the thing which physicsts call the correlation, is -sign(cos(alpha)) * sign(cos(beta)), where I have represented the settings as angles measured in radians. The correlation can only be -1 or +1 and it depends on alpha - beta. It is not difficult to figure out what it is! No point in doing a simulation.
gill1109
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### Re: Quaternion simulation no hidden variable

gill1109" with more boring worthless commentary wrote:
FrediFizzx wrote:
gill1109 wrote:I know a theorem from machine learning and AI (a distributed computing no-go theorem) which tells me not to try...

Suit yourself. I have given all the info necessary for anyone to do a sim on a different program.

a = RandomInteger[{1, 360}]; Random angle in one degree increments
b = RandomInteger[{1, 360}];
A = -Sign[Cos[a Degree]];
B = Sign[Cos[b Degree]];

If QM could do event by event outcome prediction is this perhaps the prediction it would give?

Fred, your model says that the outcomes A and B at detectors 1 and 2 are completely separate, deterministic, functions of the settings given to each detector. The expectation of the product of the outcomes, i.e., the thing which physicsts call the correlation, is -sign(cos(alpha)) * sign(cos(beta)), where I have represented the settings as angles measured in radians. The correlation can only be -1 or +1 and it depends on alpha - beta. It is not difficult to figure out what it is! No point in doing a simulation.

You said, "I think we need to see the entire code." Why? If you don't believe the plot above then do the freakin' code yourself in R.
.
FrediFizzx
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### Re: Quaternion simulation no hidden variable

FrediFizzx wrote:
gill1109" with more boring worthless commentary wrote:
FrediFizzx wrote:
gill1109 wrote:I know a theorem from machine learning and AI (a distributed computing no-go theorem) which tells me not to try...

Suit yourself. I have given all the info necessary for anyone to do a sim on a different program.

a = RandomInteger[{1, 360}]; Random angle in one degree increments
b = RandomInteger[{1, 360}];
A = -Sign[Cos[a Degree]];
B = Sign[Cos[b Degree]];

If QM could do event by event outcome prediction is this perhaps the prediction it would give?

Fred, your model says that the outcomes A and B at detectors 1 and 2 are completely separate, deterministic, functions of the settings given to each detector. The expectation of the product of the outcomes, i.e., the thing which physicsts call the correlation, is -sign(cos(alpha)) * sign(cos(beta)), where I have represented the settings as angles measured in radians. The correlation can only be -1 or +1 and it depends on alpha - beta. It is not difficult to figure out what it is! No point in doing a simulation.

You said, "I think we need to see the entire code." Why? If you don't believe the plot above then do the freakin' code yourself in R.
.

I did.
gill1109
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### Re: Quaternion simulation no hidden variable

I finally figured out how to get the x-axis relabelled correctly and with the gridlines at 90 degree intervals.

Now it is easier to see that there is a curious change every 90 degrees. Of course the change at 180 is understandable as perfect correlation when a = -b. Buy why at 90 and 270?
.
FrediFizzx
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### Re: Quaternion simulation no hidden variable

FrediFizzx wrote:I finally figured out how to get the x-axis relabelled correctly and with the gridlines at 90 degree intervals.

Now it is easier to see that there is a curious change every 90 degrees. Of course the change at 180 is understandable as perfect correlation when a = -b. Buy why at 90 and 270?

This is really strange. In fact, impossible. There is a 60-degree range of angles (difference between their settings) at which Alice and Bob's outcomes are equal and opposite. I suspect a programming error, sorry! No LHV can generate this correlation function.
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### Re: Quaternion simulation no hidden variable

There is no error in the programming so not impossible. Easy to see what is going on with these functions.

a = RandomInteger[{1, 360}]; Random angle in one degree increments
b = RandomInteger[{1, 360}];
A = -Sign[Cos[a Degree]];
B = Sign[Cos[b Degree]];

A will be -1 from 0 to 90; +1 from 90 to 270; and -1 from 270 to 360.
B will be just the opposite; +1 from 0 to 90; -1 from 90 to 270; and +1 from 270 to 360.

So that explains the changes at 90 and 270.
.
FrediFizzx
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### Re: Quaternion simulation no hidden variable

FrediFizzx wrote:There is no error in the programming so not impossible. Easy to see what is going on with these functions.

a = RandomInteger[{1, 360}]; Random angle in one degree increments
b = RandomInteger[{1, 360}];
A = -Sign[Cos[a Degree]];
B = Sign[Cos[b Degree]];

A will be -1 from 0 to 90; +1 from 90 to 270; and -1 from 270 to 360.
B will be just the opposite; +1 from 0 to 90; -1 from 90 to 270; and +1 from 270 to 360.

So that explains the changes at 90 and 270.
.

Exactly. Now shift B relative to A by 1, 2, 3 ... . The length of the initial segment where (A, B) = (-1, +1) decreases by 1 with each increment of d by 1. A new segment where (A, B) = (-1, -1) starts to grow by 1 with each step. Similarly the next two segments of together of length 90. Similar things happen at the next 90 positions in A. And so on, round the other side of the circle. This should give the triangle wave. The zig-zag curve.
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### Re: Quaternion simulation no hidden variable

More boring freakin' nonsense. Do the simulation in R and if it doesn't match my plot I will tell you where your mistake is.
.
FrediFizzx
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### Re: Quaternion simulation no hidden variable

FrediFizzx wrote:Do the simulation in R and if it doesn't match my plot I will tell you where your mistake is.

I did do it in R, and got the result which I expected.

Have a nice weekend! Keep safe.
gill1109
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### Re: Quaternion simulation no hidden variable

gill1109 wrote:
FrediFizzx wrote:Do the simulation in R and if it doesn't match my plot I will tell you where your mistake is.

I did do it in R, and got the result which I expected.

Have a nice weekend! Keep safe.

No, my code was wrong. I tried again and I got what you got! OK. So what?

Look at the model: Alice chooses a setting a, sees outcome A = +/-1 (deterministic function of a)
Bob chooses a setting b, sees outcome B = +/-1 (deterministic function of b).
The mean of AB = +/-1 for given (a, b) is a deterministic function of (a, b). Plot it as a function of b - a and you see the curve you drew. This curve is pretty but not particularly interesting.

To investigate Bell’s inequality you must pick two values of a, two values of b, and look at the four correlations AB. Each is equal to +/-1. You’ll see that any one minus the sum of the other three lies between -2 and +2. That’s because for four numbers +/-1 denoted A1, A2, B1, B2 it is always true that A1B1 - (A1B2 + A2B2 + A2B2) equals +/-2.

Code: Select all
`a <- 1:360b <- 1:360AB <- outer(-sign(cos(pi * a / 180)), sign(cos(pi * b / 180)), "*")d <- outer(a, b, "-")ABvec <- as.vector(AB)dvec <- as.vector(d)out <- aggregate(x = ABvec, by = list(dvec), FUN = mean)dvals <- out[ , 1]corrs <- out[ , 2]plot(dvals, corrs, type = "l")lines(dvals, -cos(pi * dvals/180), col = "magenta")`

You can see the plot in higher resolution here
https://gill1109.com/2021/01/09/r-stuff/

Notice, I did not sample settings of Alice and Bob at random. No need to do that, since the measurement functions are two deterministic functions of the relevant setting. It is a local, deterministic model, and hence it is a local realistic model. There is no violation of any Bell inequality... nothing weird. No connection at all between the two measurement stations!
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### Re: Quaternion simulation no hidden variable

I finally got around to cleaning up the Mathematica code,

EPRsims/abalone.pdf

However, there is a problem with the sign function as A and B are zero too many times so about 10 percent of events are dropped. So, we changed to these A and B functions,

If[Cos[(a) Degree] < 0, A = -1, A = 1];
If[Cos[(b) Degree] < 0, B = 1, B = -1];

Which avoids that problem.

EPRsims/abalone2.pdf

And we still have the same result.

.
FrediFizzx
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### Re: Quaternion simulation no hidden variable

FrediFizzx wrote:I finally got around to cleaning up the Mathematica code,

EPRsims/abalone.pdf

However, there is a problem with the sign function as A and B are zero too many times so about 10 percent of events are dropped. So, we changed to these A and B functions,

If[Cos[(a) Degree] < 0, A = -1, A = 1];
If[Cos[(b) Degree] < 0, B = 1, B = -1];

Which avoids that problem.

EPRsims/abalone2.pdf

And we still have the same result.

Yes, your result is right. Good to see the code. So ... what does it tell you?

Here’s my code.
https://gill1109.com/2021/01/09/r-stuff/

Alice chooses a setting a, sees outcome A = +/-1 (deterministic function of a)
Bob chooses a setting b, sees outcome B = +/-1 (deterministic function of b).
The mean of AB, identically equal to +/-1 for given (a, b), is a deterministic function of (a, b). The mean of AB given a - b, when you pick a and b independently and uniformly at random, is something else. It's what you see in the plot. But alas, this beautiful plot tells us nothing.

To investigate Bell’s inequality you must pick two values of a, two values of b, and look at the four correlations AB. Each is equal to +/-1. You’ll see that any one, minus the sum of the other three, lies between -2 and +2. That’s because for four numbers +/-1 denoted A1, A2, B1, B2 it is always true that A1B1 - (A1B2 + A2B2 + A2B1) equals +/-2.

Bell's theorem is vindicated yet again, as of course it has to be. There is no counterexample to a true mathematical theorem. You can call it junk physics if you like. Bell does not claim that it is physics. In fact: Bell would conclude that it is *not* true physics. Bell would conclude that the true physical world is not like this, since stringent experiment has shown that the real world does not conform to local realism. Bell already showed that QM did not conform to local realism.
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### Re: Quaternion simulation no hidden variable

CHSH = 2.53315
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FrediFizzx
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### Re: Quaternion simulation no hidden variable

FrediFizzx wrote:CHSH = 2.53315

Which two settings do you let Alice use, and which two settings do you let Bob use?

I find CHSH equal to -2 for some choices, +2 for others, since *all* correlations rho(alpha, beta) = +/- 1

Am I right in thinking that *you* compute each of your four correlations rho(alpha, beta) by *assuming* that rho(alpha, beta) = f(alpha - beta) for some periodic function f, with period 2 pi?

If you want to do that, you must first experimentally verify your assumption. You don’t. It doesn’t hold.

It is true that this is a serious mistake which quite a few experimenters have made in the past.
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### Re: Quaternion simulation no hidden variable

When there is enough non-linearity the bounds of CHSH can be exceeded by optimum selection of angles just like they do for experiments. So here we have another example that shoots the heck out of Bell's junk theory and probably Gill's theory also.
.
FrediFizzx
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### Re: Quaternion simulation no hidden variable

gill1109 wrote:
FrediFizzx wrote:CHSH = 2.53315

Which two settings do you let Alice use, and which two settings do you let Bob use?

I find CHSH equal to -2 for some choices, +2 for others, since *all* correlations rho(alpha, beta) = +/- 1

Am I right in thinking that *you* compute each of your four correlations rho(alpha, beta) by *assuming* that rho(alpha, beta) = f(alpha - beta) for some periodic function f, with period 2 pi?

If you want to do that, you must first experimentally verify your assumption. You don’t. It doesn’t hold.

It is true that this is a serious mistake which quite a few experimenters have made in the past.

Since my name is on that Mathematica notebook as its author, I went back and looked at it. It has changed considerately since I translated Fodje's Python program. I took my original version (which I still have) and changed the hidden variable so that it no longer sneaks in the detection loophole, and it gives the triangle output waveform. I don't know what the curves that are being shown here represent in terms of physics.

I have been working on Gull's theorem and took a few days off to check this out. As far as Gull's theorem I have been able to simulate Gill's equation (1), the correlation function. Now I need to understand how to do the Fourier transform. Progress is slow but I want to make sure I'm correct.
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### Re: Quaternion simulation no hidden variable

jreed wrote:Since my name is on that Mathematica notebook as its author, I went back and looked at it. It has changed considerately since I translated Fodje's Python program. I took my original version (which I still have) and changed the hidden variable so that it no longer sneaks in the detection loophole, and it gives the triangle output waveform. I don't know what the curves that are being shown here represent in terms of physics. ...

Sorry, I should have added to the title that it was modified by me. But of course that should be rather obvious. Some improvements have been made to the original. This particular simulation is an attempt to see what it might take for just the detector angles along with the -1 from singlet to produce -a.b using the event be event outcomes. No hidden variable. Easy to see that something much more sophisticated than the cosine of the angles is required. I too was expecting the striaght line triangle output but got this non-linear output instead. So, this is the physics of that particular operation. Here is an updated notebook file,

EPRsims/abalone2.nb
EPRsims/abalone2.pdf

You may need to clear your browser cache for the PDF to update.
.
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### Re: Quaternion simulation no hidden variable

FrediFizzx wrote:When there is enough non-linearity the bounds of CHSH can be exceeded by optimum selection of angles just like they do for experiments. So here we have another example that shoots the heck out of Bell's junk theory and probably Gill's theory also.

Rubbish, Fred, you are not even computing the CHSH bound how real physicists compute it. I also can compute a number and get an answer bigger than two. Look: 4 * 0.7 = 2.8 ! Wow! I violated CHSH, and confirmed QM! Only a tiny bit short of 2 sqrt 2. That’s because of rounding error. Obviously, I need to get Mathematica so as to do it to four decimal places instead of 1.

Your “theory” sure is junk physics.
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### Re: Quaternion simulation no hidden variable

Last time you said I was wrong it was YOU that was wrong. Are ya sure you want to go that route again and look more foolish?
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FrediFizzx
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### Re: Quaternion simulation no hidden variable

FrediFizzx wrote:Last time you said I was wrong it was YOU that was wrong. Are ya sure you want to go that route again and look more foolish?

Well, I do make mistakes from time to time, and when I learn about them, I admit them and try to repair any damage found. Close to 70 years old, I make more mistakes now, and I'm slower to "get" things, than when I was 30.

But no matter. This is your model:

Take two large wooden disks and colour half of each black, half white. Fix to a wall. There's a pointer painted on the disks, near the edge, in the middle of the black half of the circumference. Painted on the wall, just above the top of each disk, is another pointer. Around the side of the disk, painted on the wall, equally spaced, are the numbers 1 to 360; 360 on top. Spin each disk. Wait till it stops. The pointers painted *in* the disks each point to a number alpha, beta between 1 and 360, painted on the wall. The pointers at the top of the disk painted *on the wall* point either to black, or to white, *in *the disk. That defines A = +/- 1 and B = +/- 1. You repeat this fairground game many, many times, and average A times B for each value of delta = alpha - beta.

That's the curve you drew.

Exercise: compute it analytically. Probably Mathematica can do it by computer algebra if you are clever with Mathematica! Good luck.

Hint: do the continuous case, and simplify the double integration over alpha and beta by a change of variables to alpha and delta.

Image: two fairground "wheels of fortune" which have come to rest at A = +1 (black on top), B = -1 (white on top), alpha = 330 (approx), beta = 220 (approx)
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