How to generate correlations of ANY strength: Linear to Box

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

How to generate correlations of ANY strength: Linear to Box

Postby Joy Christian » Sat Mar 08, 2014 11:42 pm

As we had been discussing, here is the most accurate simulation of my 3-sphere model for the EPR-Bohm correlation: http://rpubs.com/jjc/13965.

The theoretical description and conceptual understanding of the 3-sphere model can be found on my blog: http://libertesphilosophica.info/blog/.

The initial function (or the cumulative distribution function, if you like), as defined in eq. (7) of this document, has been taken now to be

.

Not surprisingly, it turns out that the strength of the correlation generated by this function can be changed simply by changing the geometric constant .

For example, try replacing the geometric constant under the square-root in the above simulation with the following options:

Linear: Box.

This further supports my view that quantum correlations are purely geometrical and topological effects. They have nothing to do with non-locality or non-reality.
Last edited by Joy Christian on Sun Mar 09, 2014 1:39 am, edited 2 times in total.
Joy Christian
Research Physicist
 
Posts: 2793
Joined: Wed Feb 05, 2014 4:49 am
Location: Oxford, United Kingdom

Re: How to generate correlations of ANY strength: Linear to

Postby gill1109 » Sun Mar 09, 2014 1:20 am

Nice!

There are two pi's in those formulas, which one are you changing?

Terminology. In probability theory, the cumulative distribution function F of a random variable X is the function x mapsto Prob(X <= x). I don't think that your f should be called a cumulative distribution function. Your theta_0 is drawn at random from the Uniform([0, k_1]) distribution. So f(theta_0) = - 1 + 2 /sqrt(1 + 3 theta_0 /k_2) is another random variable with its own cumulative distribution function.

Here k_1 and k_2 were originally both equal to pi but you are now proposing to let one of them be different?
gill1109
Mathematical Statistician
 
Posts: 2812
Joined: Tue Feb 04, 2014 10:39 pm
Location: Leiden

Re: How to generate correlations of ANY strength: Linear to

Postby Joy Christian » Sun Mar 09, 2014 1:33 am

gill1109 wrote:Nice!

There are two pi's in those formulas, which one are you changing?

Terminology. In probability theory, the cumulative distribution function F of a random variable X is the function x mapsto Prob(X <= x). I don't think that your f should be called a cumulative distribution function. Your theta_0 is drawn at random from the Uniform([0, k_1]) distribution. So f(theta_0) = - 1 + 2 /sqrt(1 + 3 theta_0 /k_2) is another random variable with its own cumulative distribution function.

Here k_1 and k_2 were originally both equal to pi but you are now proposing to let one of them be different?



I am changing the under the square-root; so I am changing your k_2 (in other words, I am changing the ratio "circumference over diameter").

Good point about the terminology.

On a different note, I have been trying to re-derive the above function from Pearle's "probability density" (22) but with no avail. His eq. (22) is not even normalized to 1, and I don't understand how it follows from his eq. (12) as he claims. What is more, I don't understand how you derived the square-root in the above function. The expression I get from integrating (22) from 0 to r is quite different. I understand Pearle's logic, but I am having great deal of difficulty following his mathematics.
Joy Christian
Research Physicist
 
Posts: 2793
Joined: Wed Feb 05, 2014 4:49 am
Location: Oxford, United Kingdom

Re: How to generate correlations of ANY strength: Linear to

Postby gill1109 » Sun Mar 09, 2014 3:25 am

There is strange terminology and a lot of misprints in Pearle.

In formula (22), " 4 / 3 " has to be replaced by " 2 / 3 ".

Despite the remark made just before equation (22), rho(r) is *not* a probability density.

From formula (1) we see that 4 pi r^2 rho(r) is a probability density.

So I found that 4 pi r^2 rho(r) = 4 / 3 . 2 . sin(pi r /2) / (1 + cos pi r / 2) ^ 3

and even I can integrate that from zero up to r getting the cumulative distribution function

4/3 . ( -1 / 4 + 1 / (1 + cos pi r / 2) ^2 )

The model tells us to pick a direction e uniformly at random on S^2, and compare the absolute values of the inner product a^T e and b^T e each with cos(pi r/ 2), generating outcomes sign(a^T e) and sign(b^T e) only if the relevant absolute inner product is larger.

Using the probability integral transform one sees that the probability distribution is easier described by saying that the (1 + cos pi r /2) ^ 2 is uniformly distributed on the interval from 1 to 4

How to do this? Take U ~ Unif(0, 1) and invert

U = 4/3 ( -1/4 + (1 + cos pi r / 2) ^2 )

to find

(1 + cos pi r / 2) ^2 = 4 / (3U + 1) = 4 / V where V ~ Unif(1, 4)

The trick here is the following result: if X is a random variable with continuous cumulative distribution function F, then F(X) has the uniform([0, 1]) distribution. So if we can invert the equation U = F(X), ie we can then write X = F^{-1}(U) where U ~ Unif(0,1), and it follows that X has the cumulative distribution function F.
gill1109
Mathematical Statistician
 
Posts: 2812
Joined: Tue Feb 04, 2014 10:39 pm
Location: Leiden

Re: How to generate correlations of ANY strength: Linear to

Postby gill1109 » Sun Mar 09, 2014 5:05 am

PS Florin Moldoveanu wrote me the following (note: I disagree with his correction of equation 22)

I double checked Pearl's computations, and I found many problems. All checks out until formula 16 (including 16 and including appendix A and B) Then it goes downhill. (17) has a typo, the correct formula is:

mu(x) = - 1/(1-x^2)^{1/2} d/dx [(1-x^2)^{1/2} / 4x d/dx g(x) x^2]

Equation (19) is messed up big time, 1/2 g(x) is the sum of the 2 integrals over [(1-x^2)^{1/2} + x] and not the first integral over (1-x^2) + the second integral over x^2. As a consequence Eq. 23 is wrong as well, but 23 does follow from the incorrect 19. Section V of the paper is based on (23) and is invalid.

I could not obtain Eq. 20 (I got a similar slightly more complex one-I am still working on it and maybe it will simplify to 20)

Eq. 21 has a typo and should read:

h(x) = 1/4 C pi 1/(1+x)^3

Eq. 22 has a typo and should read:

rho(r) r^2 = 1/3 sin (1/2 pi r) / (1+cos(1/2 pi r))^3

It was a major pain to straighten out the math on page 1421 because it is a sloppy paper and there are many false trails one need to explore to understand the intention. On Eq. 20 I got a very similar expression but I am lacking the overall 1/1-x^2 factor outside the square brackets, and in the second integral I have an additional term of (1-z^2)^{1/2}. The first integral in the square bracket is equal with g(0)/2.
gill1109
Mathematical Statistician
 
Posts: 2812
Joined: Tue Feb 04, 2014 10:39 pm
Location: Leiden

Re: How to generate correlations of ANY strength: Linear to

Postby Joy Christian » Sun Mar 09, 2014 5:35 am

gill1109 wrote:So I found that 4 pi r^2 rho(r) = 4 / 3 . 2 . sin(pi r /2) / (1 + cos pi r / 2) ^ 3


This probability density is still not normalized to 1 for r in [0, 1]. It normalizes to 2/pi. It appears that the correct normalizing constant on the r.h.s. of (22) is 4pi/3.
Joy Christian
Research Physicist
 
Posts: 2793
Joined: Wed Feb 05, 2014 4:49 am
Location: Oxford, United Kingdom

Re: How to generate correlations of ANY strength: Linear to

Postby gill1109 » Sun Mar 09, 2014 5:54 am

You may well be right! I make algebraic errors all the time (algebraically challenged ;) ?). I always have to check my final result against a computer simulation ...

Joy Christian wrote:
gill1109 wrote:So I found that 4 pi r^2 rho(r) = 4 / 3 . 2 . sin(pi r /2) / (1 + cos pi r / 2) ^ 3


This probability density is still not normalized to 1 for r in [0, 1]. It normalizes to 2/pi. It appears that the correct normalizing constant on the r.h.s. of (22) is 4pi/3.


PS: yes, dammit, I forgot the factor coming from the transformation from [0, 1] to [0, pi/2]. We differed precisely by a factor pi/2, right?
gill1109
Mathematical Statistician
 
Posts: 2812
Joined: Tue Feb 04, 2014 10:39 pm
Location: Leiden

Re: How to generate correlations of ANY strength: Linear to

Postby gill1109 » Sun Mar 09, 2014 6:13 am

Code: Select all
r <- seq(from=1, to = 199, by = 2)/200 ## 100 midpoints of 100 intervals of width 1/100 between 0 and 1
length(r)

pr <- (4 * pi / 3) * sin(pi * r /2) / ((1 + cos(pi * r / 2)) ^ 3)  ## A probability density?
plot(r, pr, type = "l")

sum(pr)*(1/100)    ## Yes!
gill1109
Mathematical Statistician
 
Posts: 2812
Joined: Tue Feb 04, 2014 10:39 pm
Location: Leiden

Re: How to generate correlations of ANY strength: Linear to

Postby FrediFizzx » Sun Mar 09, 2014 3:16 pm

...
Good news, Joy. The new function seems to work well in the MinkweReed Mathematica simulation also.

Image
FrediFizzx
Independent Physics Researcher
 
Posts: 2905
Joined: Tue Mar 19, 2013 7:12 pm
Location: N. California, USA

Re: How to generate correlations of ANY strength: Linear to

Postby Joy Christian » Sun Mar 09, 2014 3:43 pm

FrediFizzx wrote:...Good news, Joy. The new function seems to work well in the MinkweReed Mathematica simulation also.


Thanks, Fred. I am not surprised. In fact I was going to ask you to check the new function in Mathematica. All previous attempts were missing the correct phase-shifts. I did manage to guess an almost correct distribution by trial and error in the previous best simulation, but even that wasn't exactly right. The above function is both analytically and numerically correct, and it has been derived rigorously. It also checks out in various consistency checks I have been performing. So good news indeed.

By the way, the slight discrepancy you see in the Mathematica version is because the Minkwe-Reed simulation is 2D, or S^1 based. It should disappear in a 3D version.
Joy Christian
Research Physicist
 
Posts: 2793
Joined: Wed Feb 05, 2014 4:49 am
Location: Oxford, United Kingdom

Re: How to generate correlations of ANY strength: Linear to

Postby FrediFizzx » Sun Mar 09, 2014 10:47 pm

Hmm.... the function works backwards to the sin^2 function. When theta is 0, the function is 1 and when theta is pi, the function is 0. What happens if we make it run from 0 to 0.5 like the original?

Ok, ran it. Ends up being half way between straight lines and the cosine curve.
FrediFizzx
Independent Physics Researcher
 
Posts: 2905
Joined: Tue Mar 19, 2013 7:12 pm
Location: N. California, USA

Re: How to generate correlations of ANY strength: Linear to

Postby FrediFizzx » Sun Mar 09, 2014 10:48 pm

Hmm.... Also strange that the topic title got truncated somehow. I tried to fix but no go.
FrediFizzx
Independent Physics Researcher
 
Posts: 2905
Joined: Tue Mar 19, 2013 7:12 pm
Location: N. California, USA

Re: How to generate correlations of ANY strength: Linear to

Postby gill1109 » Sun Mar 09, 2014 10:59 pm

This illustrates the difficulty with detection loophole models - you can get any correlations you like by just twiddling knobs, there is nothing special about the QM correlations, no Tsirelson bound CHSH <= 2 sqrt 2. You can get all the way to 4.
gill1109
Mathematical Statistician
 
Posts: 2812
Joined: Tue Feb 04, 2014 10:39 pm
Location: Leiden

Re: How to generate correlations of ANY strength: Linear to

Postby Joy Christian » Sun Mar 09, 2014 11:39 pm

gill1109 wrote:This illustrates the difficulty with detection loophole models.


My model has nothing whatsoever to do with detection loophole.
Joy Christian
Research Physicist
 
Posts: 2793
Joined: Wed Feb 05, 2014 4:49 am
Location: Oxford, United Kingdom

Re: How to generate correlations of ANY strength: Linear to

Postby FrediFizzx » Sun Mar 09, 2014 11:42 pm

Yeah, Richard still doesn't understand about states that don't exist in the first place. He stopped asking questions about that before he finished the lesson. :-(
FrediFizzx
Independent Physics Researcher
 
Posts: 2905
Joined: Tue Mar 19, 2013 7:12 pm
Location: N. California, USA

Re: How to generate correlations of ANY strength: Linear to

Postby Joy Christian » Mon Mar 10, 2014 1:12 am

A better way of stating what I have stated in the first post of this thread is that the ratio can be varied to produce correlation of any strength, from Linear to Box:


Linear: Box.


This further supports my view that quantum correlations are purely geometrical and topological effects. They have nothing to do with non-locality or non-reality.

To understand this, recall the definition of as a ratio of the circumference over the diameter of a circle. Considering that in my model is an angle between two vectors, the ratio can be thought of as representing a specific geometrical structure of the physical space. Consequently, changing the ratio within the context of my model means changing the very geometry of the physical space, say from to something else, with the larger representing the flatter geometries and the smaller representing more curvy (or tighter) geometries. It is thus not surprising that for larger the correlation tend to linear correlation, whereas for smaller the correlation tend to the strongest possible correlation, with quantum correlation emerging exactly in the "middle", corresponding precisely to the geometry.
Last edited by Joy Christian on Mon Mar 10, 2014 2:15 am, edited 1 time in total.
Joy Christian
Research Physicist
 
Posts: 2793
Joined: Wed Feb 05, 2014 4:49 am
Location: Oxford, United Kingdom

Re: How to generate correlations of ANY strength: Linear to

Postby gill1109 » Mon Mar 10, 2014 1:54 am

FrediFizzx wrote:Yeah, Richard still doesn't understand about states that don't exist in the first place. He stopped asking questions about that before he finished the lesson. :-(

Richard understands this very well. States that don't exist in the first place, depending on the measurement settings. Other people would call this a conspiracy loophole model. One discards the runs with "non existent" states, sure. The runs that remain have hidden states whose probability distribution depends on the settings.

One can alternatively interpret this as a detection loophole model, if one keeps those runs, and registers the "no show" cases as missed detections.

Two closely related models for two similar but different physical phenomena.

Is the number of particle pairs which you imagine to have actually existed in your picture of physical reality the larger number which you put in at the start (e.g. 10^7) , or is the number which survive the selections (about 70%)?
gill1109
Mathematical Statistician
 
Posts: 2812
Joined: Tue Feb 04, 2014 10:39 pm
Location: Leiden

Re: How to generate correlations of ANY strength: Linear to

Postby Joy Christian » Mon Mar 10, 2014 2:22 am

My model has nothing whatsoever to do with any loopholes. A numerical simulation of an analytical model is not a model. It is just a numerical simulation of the model. Among many other places, the exact analytical model itself is described in this book-chapter: http://libertesphilosophica.info/blog/w ... hapter.pdf.

A "particle" in my model is defined by the pair , not by the vector alone.
Joy Christian
Research Physicist
 
Posts: 2793
Joined: Wed Feb 05, 2014 4:49 am
Location: Oxford, United Kingdom

Re: How to generate correlations of ANY strength: Linear to

Postby gill1109 » Mon Mar 10, 2014 7:18 am

Pearle's (1970) paper is called "Hidden-Variable Example Based upon Data Rejection". The numerical simulation of Pearle's model which we are doing can be given various physical interpretations. (1) Detection loophole model [this was Pearle's motivation]; (2) Conspiracy loophole model; (3) Joy Christian's S^3 model. At present, there does not exist a mathematical derivation by Joy Christian of the probability distribution which we are using in the Pearle model for the second component of the hidden state. Pearle himself showed that his own choice was one of many - he just picked something simple which does the trick. Joy has opportunistically adopted Pearle's choice but has not shown that it follows from his (Joy's) physical modelling.
gill1109
Mathematical Statistician
 
Posts: 2812
Joined: Tue Feb 04, 2014 10:39 pm
Location: Leiden

Re: How to generate correlations of ANY strength: Linear to

Postby Joy Christian » Mon Mar 10, 2014 9:10 am

gill1109 wrote:At present, there does not exist a mathematical derivation by Joy Christian of the probability distribution which we are using in the Pearle model for the second component of the hidden state. Pearle himself showed that his own choice was one of many - he just picked something simple which does the trick. Joy has opportunistically adopted Pearle's choice but has not shown that it follows from his (Joy's) physical modelling.


Incorrect and meaningless assertions.

Can one, or even should one, "derive" an initial state of the physical system? In Bell's local realistic framework the measurement results and depend on the measurement settings and , respectively, as well as on the unknown and uncontrollable common cause or initial state .

Here is Bell's famous paper of 1964. In the last paragraph of section 2 of this paper he writes: "... stands for any number of variables and the dependences thereon of and are unrestricted. In a complete physical theory of the type envisioned by Einstein, the hidden variables would have dynamical significance and laws of motion; our can then be thought of as initial values of these variables at some suitable instant."

In my model Bell's is taken to be the initial state and his "suitable instant" is shown in the first figure of this two-page document. More importantly, this initial state is shown to be part of the geometric structure of the 3-sphere in this derivation. It is clearly stated just after eq. (7) in this derivation that the function is an arbitrary function of . It specifies a part of the initial state of the physical system. As such, we have complete freedom to choose whatever function we like. If we choose it such that it corresponds to the quantum mechanical singlet state (which cannot be "derived" but only postulated within quantum mechanics), then that will produce the EPR-Bohm correlation. If we choose it such that it corresponds to the quantum mechanical Hardy state (which, again, cannot be "derived" but postulated within quantum mechanics), then that will produce the Hardy correlation (as reproduced in this paper). One cannot, and even should not, "derive" the initial state of the physical system. In the simulation the function thus represents an un-derived initial state of the physical system.
Joy Christian
Research Physicist
 
Posts: 2793
Joined: Wed Feb 05, 2014 4:49 am
Location: Oxford, United Kingdom

Next

Return to Sci.Physics.Foundations

Who is online

Users browsing this forum: No registered users and 81 guests

cron
CodeCogs - An Open Source Scientific Library