A real EPR-Bohm scenario

[Joy Christian]

***
Hi Fred,

Set f = 0 in the above R code to see what happens. We already know what happens from the original simulation, but try it out anyways, just for fun.

Then set f = 0.7 to see what happens. The exact setting should be f = 1, but that reduces the number of events to 0. So to generate a plot we have to set f = 0.7.

This proves that the new version is exactly the same simulation as the original one linked above. It is just written slightly differently. It shows that the g-function encodes the topology of the 3-space, with f = 0 being that of R^3 and f = 0.7 being stronger-than-S^3. We know all this, but just reconfirming for the new version.

***

Already tried all that. I haven't verified that the A and B single averages will be close to zero though I don't see a problem there just looking at the data.
.

Just checked the following, and they check out as expected:

corrs[i] = sum(A)/length(A) # Verifies < A > = 0 for all vectors a

corrs[j] = sum(B)/length(B) # Verifies < B > = 0 for all vectors b

***

[FrediFizzx]

Just checked the following, and they check out as expected:

corrs[i] = sum(A)/length(A) # Verifies < A > = 0 for all vectors a

corrs[j] = sum(B)/length(B) # Verifies < B > = 0 for all vectors b

***

Great. Now there is absolutely no doubt that the model gives full prediction capability event by event. We can predict if we get up or down at each detection station for each event knowing just lambda, e, a and b. It is freakin' fabulous!

[Joy Christian]

Just checked the following, and they check out as expected:

corrs[i] = sum(A)/length(A) # Verifies < A > = 0 for all vectors a

corrs[j] = sum(B)/length(B) # Verifies < B > = 0 for all vectors b

***

Great. Now there is absolutely no doubt that the model gives full prediction capability event by event. We can predict if we get up or down at each detection station for each event knowing just lambda, e, a and b. It is freakin' fabulous!

Given the hidden variables or shared randomness (e, s), we have a prescription for definite results A(a; e, s) = +/-1 and B(b; e, s) = +/-1 observed by Alice and Bob.

Given N particle pairs observed by Alice and Bob within the 3-sphere, the correlations among their results A and B work out to be E(a, b) = -a.b. End of the story.

***

[thray]

Nice!

[Joy Christian]

***
The measurement functions of Alice and Bob are now very easy to understand:



and

,

where is the x-axis and the pair is the initial state of the spin system.

***

[FrediFizzx]

***
The measurement functions of Alice and Bob are now very easy to understand:



and

,

where is the x-axis and the pair is the initial state of the spin system.

***

Oh, something new already. What is the x-axis? Propagation axis? Or a lab frame axis?

BTW, your last post on RW to Jay is really good.
.

[Joy Christian]

***
The measurement functions of Alice and Bob are now very easy to understand:



and

,

where is the x-axis and the pair is the initial state of the spin system.

***

Oh, something new already. What is the x-axis? Propagation axis?

BTW, your last post on RW to Jay is really good.

.

Thanks. I thought Jay has been misled enough on RW, so it was time to put an end to all that nonsense about Bell inequalities.

x-axis is the constant x-axis of the lab coordinates. The propagation axis is the z-axis.

cos(Angles) in the simulation are nothing but the cosines of the angles of a and b with respect to the x-axis of the lab.

***

[FrediFizzx]

Oh, something new already. What is the x-axis? Propagation axis?

BTW, your last post on RW to Jay is really good.

.

Thanks. I thought Jay has been misled enough on RW, so it was time to put an end to all that nonsense about Bell inequalities.

x-axis is the constant x-axis of the lab coordinates. The propagation axis is the z-axis.

cos(Angles) in the simulation are nothing but the cosines of the angles of a and b with respect to the x-axis of the lab.

***

Yeah, I thought so. I added the lab frame to my post after you already grabbed it for a quote. Still waking up here.

You know that I was trying to get the extra +/- 1 toggle in there for quite some time now. Crazy that it was already in there but hidden. It just made physical sense that it had to be there. At least to me.

Now to figure out how to get it all to work with the GAViewer program event by event. Perhaps it is possible now.

Hopefully Jay will get off the "obfuscation trail" they are trying to get him to follow. Yes, had he spent the time learning your model better would have been good.
.

[FrediFizzx]

Strange. I already had GAViewer working event by event in a CHSH type of setup with this code.

Code: Select all

//Adaptation of Albert Jan Wonnink's original code
//http://challengingbell.blogspot.com/2015/03/numerical-validation-of-vanishing-of.html

function getRandomLambda()
{
   if( rand()>0.5) {return 1;} else {return -1;}
}
function getRandomUnitVector() //uniform random unit vector:
   //http://mathworld.wolfram.com/SpherePointPicking.html
{
   //v=randGaussStd()*e1+randGaussStd()*e2+randGaussStd()*e3;
   v=randGaussStd()*e1+randGaussStd()*e2+0.000*e3;

   return normalize(v);
}
batch test()
{
   set_window_title("Test of Joy Christian's CHSH derivation");
   N=5000; //number of iterations (trials)
   I=e1^e2^e3;
      s=0;
   a1=sin(0)*e1 + cos(0)*e2 + 0.000*e3;
   b1=sin(pi/4)*e1 + cos(pi/4)*e2 + 0.000*e3;
      a2=sin(pi/2)*e1 + cos(pi/2)*e2 + 0.000*e3;
   b2=sin(3*pi/4)*e1 + cos(3*pi/4)*e2 + 0.000*e3;


      for(nn=0;nn<N;nn=nn+1) //perform the experiment N times
   {
            lambda=getRandomLambda(); //lambda is a fair coin,
                    //resulting in +1 or -1
            e=getRandomUnitVector();
      mu=lambda * I;  //calculate the lambda dependent mu
            if((asin(e/e1)-asin(a1/e1))>pi/3.71) {C1=I.a1;} else {C1=(-I.a1);}
            if((asin(-e/e1)-asin(b1/e1))>pi/3.71) {D1=(-I.b1);} else {D1=I.b1;}
            if((asin(e/e1)-asin(a2/e1))>pi/3.71) {C2=I.a2;} else {C2=(-I.a2);}
            if((asin(-e/e1)-asin(b2/e1))>pi/3.71) {D2=(-I.b2);} else {D2=I.b2;}
            E1=mu.a1;  //E = {a_k B_k(L)}
            F1=mu.b1;  //F = {b_j B_j(L)}
            A1=C1 E1;  //eq. (1) of arXiv:1103.1879, A(a, L) = {-a_j B_j}{a_k B_k(L)}
            B1=F1 D1;  //eq. (2) of arXiv:1103.1879, B(b, L) = {b_j B_j(L)}{b_k B_k}
            E2=mu.a2;  //E = {a_k B_k(L)}
            F2=mu.b2;  //F = {b_j B_j(L)}
            A2=C2 E2;  //eq. (1) of arXiv:1103.1879, A(a, L) = {-a_j B_j}{a_k B_k(L)}
            B2=F2 D2;  //eq. (2) of arXiv:1103.1879, B(b, L) = {b_j B_j(L)}{b_k B_k}
      q=0;
            if(lambda==1) {q=((-C1) (A1 B1) (-D1))-((-C1) (A1 B2) (-D2))+((-C2) (A2 B1) (-D1))+((-C2) (A2 B2) (-D2));}
            else {q=((-D1) (B1 A1) (-C1))-((-D2) (B2 A1) (-C1))+((-D1) (B1 A2) (-C2))+((-D2) (B2 A2) (-C2));}
            s=s+q;
            print(A1);
            print(A2);
            print(B1);
            print(B2);
      }
      Joy_CHSH=abs(s/N);
   print(Joy_CHSH, "f");
      prompt();
}

Not exactly sure why pi/3.71 yet in the if else lines???? And the result is,

Joy_CHSH = 2.828427

[Joy Christian]

Strange. I already had GAViewer working event by event in a CHSH type of setup with this code.

Code: Select all

//Adaptation of Albert Jan Wonnink's original code
//http://challengingbell.blogspot.com/2015/03/numerical-validation-of-vanishing-of.html

function getRandomLambda()
{
   if( rand()>0.5) {return 1;} else {return -1;}
}
function getRandomUnitVector() //uniform random unit vector:
   //http://mathworld.wolfram.com/SpherePointPicking.html
{
   //v=randGaussStd()*e1+randGaussStd()*e2+randGaussStd()*e3;
   v=randGaussStd()*e1+randGaussStd()*e2+0.000*e3;

   return normalize(v);
}
batch test()
{
   set_window_title("Test of Joy Christian's CHSH derivation");
   N=5000; //number of iterations (trials)
   I=e1^e2^e3;
      s=0;
   a1=sin(0)*e1 + cos(0)*e2 + 0.000*e3;
   b1=sin(pi/4)*e1 + cos(pi/4)*e2 + 0.000*e3;
      a2=sin(pi/2)*e1 + cos(pi/2)*e2 + 0.000*e3;
   b2=sin(3*pi/4)*e1 + cos(3*pi/4)*e2 + 0.000*e3;


      for(nn=0;nn<N;nn=nn+1) //perform the experiment N times
   {
            lambda=getRandomLambda(); //lambda is a fair coin,
                    //resulting in +1 or -1
            e=getRandomUnitVector();
      mu=lambda * I;  //calculate the lambda dependent mu
            if((asin(e/e1)-asin(a1/e1))>pi/3.71) {C1=I.a1;} else {C1=(-I.a1);}
            if((asin(-e/e1)-asin(b1/e1))>pi/3.71) {D1=(-I.b1);} else {D1=I.b1;}
            if((asin(e/e1)-asin(a2/e1))>pi/3.71) {C2=I.a2;} else {C2=(-I.a2);}
            if((asin(-e/e1)-asin(b2/e1))>pi/3.71) {D2=(-I.b2);} else {D2=I.b2;}
            E1=mu.a1;  //E = {a_k B_k(L)}
            F1=mu.b1;  //F = {b_j B_j(L)}
            A1=C1 E1;  //eq. (1) of arXiv:1103.1879, A(a, L) = {-a_j B_j}{a_k B_k(L)}
            B1=F1 D1;  //eq. (2) of arXiv:1103.1879, B(b, L) = {b_j B_j(L)}{b_k B_k}
            E2=mu.a2;  //E = {a_k B_k(L)}
            F2=mu.b2;  //F = {b_j B_j(L)}
            A2=C2 E2;  //eq. (1) of arXiv:1103.1879, A(a, L) = {-a_j B_j}{a_k B_k(L)}
            B2=F2 D2;  //eq. (2) of arXiv:1103.1879, B(b, L) = {b_j B_j(L)}{b_k B_k}
      q=0;
            if(lambda==1) {q=((-C1) (A1 B1) (-D1))-((-C1) (A1 B2) (-D2))+((-C2) (A2 B1) (-D1))+((-C2) (A2 B2) (-D2));}
            else {q=((-D1) (B1 A1) (-C1))-((-D2) (B2 A1) (-C1))+((-D1) (B1 A2) (-C2))+((-D2) (B2 A2) (-C2));}
            s=s+q;
            print(A1);
            print(A2);
            print(B1);
            print(B2);
      }
      Joy_CHSH=abs(s/N);
   print(Joy_CHSH, "f");
      prompt();
}

Not exactly sure why pi/3.71 yet in the if else lines???? And the result is,

Joy_CHSH = 2.828427

Isn't this the same as this one: viewtopic.php?f=6&t=200&p=5549#p5628?

***

[FrediFizzx]

Strange. I already had GAViewer working event by event in a CHSH type of setup with this code.

Code: Select all

//Adaptation of Albert Jan Wonnink's original code
//http://challengingbell.blogspot.com/2015/03/numerical-validation-of-vanishing-of.html

function getRandomLambda()
{
   if( rand()>0.5) {return 1;} else {return -1;}
}
function getRandomUnitVector() //uniform random unit vector:
   //http://mathworld.wolfram.com/SpherePointPicking.html
{
   //v=randGaussStd()*e1+randGaussStd()*e2+randGaussStd()*e3;
   v=randGaussStd()*e1+randGaussStd()*e2+0.000*e3;

   return normalize(v);
}
batch test()
{
   set_window_title("Test of Joy Christian's CHSH derivation");
   N=5000; //number of iterations (trials)
   I=e1^e2^e3;
      s=0;
   a1=sin(0)*e1 + cos(0)*e2 + 0.000*e3;
   b1=sin(pi/4)*e1 + cos(pi/4)*e2 + 0.000*e3;
      a2=sin(pi/2)*e1 + cos(pi/2)*e2 + 0.000*e3;
   b2=sin(3*pi/4)*e1 + cos(3*pi/4)*e2 + 0.000*e3;


      for(nn=0;nn<N;nn=nn+1) //perform the experiment N times
   {
            lambda=getRandomLambda(); //lambda is a fair coin,
                    //resulting in +1 or -1
            e=getRandomUnitVector();
      mu=lambda * I;  //calculate the lambda dependent mu
            if((asin(e/e1)-asin(a1/e1))>pi/3.71) {C1=I.a1;} else {C1=(-I.a1);}
            if((asin(-e/e1)-asin(b1/e1))>pi/3.71) {D1=(-I.b1);} else {D1=I.b1;}
            if((asin(e/e1)-asin(a2/e1))>pi/3.71) {C2=I.a2;} else {C2=(-I.a2);}
            if((asin(-e/e1)-asin(b2/e1))>pi/3.71) {D2=(-I.b2);} else {D2=I.b2;}
            E1=mu.a1;  //E = {a_k B_k(L)}
            F1=mu.b1;  //F = {b_j B_j(L)}
            A1=C1 E1;  //eq. (1) of arXiv:1103.1879, A(a, L) = {-a_j B_j}{a_k B_k(L)}
            B1=F1 D1;  //eq. (2) of arXiv:1103.1879, B(b, L) = {b_j B_j(L)}{b_k B_k}
            E2=mu.a2;  //E = {a_k B_k(L)}
            F2=mu.b2;  //F = {b_j B_j(L)}
            A2=C2 E2;  //eq. (1) of arXiv:1103.1879, A(a, L) = {-a_j B_j}{a_k B_k(L)}
            B2=F2 D2;  //eq. (2) of arXiv:1103.1879, B(b, L) = {b_j B_j(L)}{b_k B_k}
      q=0;
            if(lambda==1) {q=((-C1) (A1 B1) (-D1))-((-C1) (A1 B2) (-D2))+((-C2) (A2 B1) (-D1))+((-C2) (A2 B2) (-D2));}
            else {q=((-D1) (B1 A1) (-C1))-((-D2) (B2 A1) (-C1))+((-D1) (B1 A2) (-C2))+((-D2) (B2 A2) (-C2));}
            s=s+q;
            print(A1);
            print(A2);
            print(B1);
            print(B2);
      }
      Joy_CHSH=abs(s/N);
   print(Joy_CHSH, "f");
      prompt();
}

Not exactly sure why pi/3.71 yet in the if else lines???? And the result is,

Joy_CHSH = 2.828427

Isn't this the same as this one: viewtopic.php?f=6&t=200&p=5549#p5628?

***

That one doesn't have the "e" vector in it but it is the same result. Not sure what I was trying to do with this other one but possibly can tailor it to event by event.
.

[FrediFizzx]

Yeah, I am not doing the arc sin function correctly. Say we have the "e" vector as,

0.46*e1 + -0.89*e2 + -0.01*e3

So to get the arc sin it is not going to be simply e/e1. What is it going to be?
.

[FrediFizzx]

Yeah, I am not doing the arc sin function correctly. Say we have the "e" vector as,

0.46*e1 + -0.89*e2 + -0.01*e3

So to get the arc sin it is not going to be simply e/e1. What is it going to be?
.

Well, I am using a and b vectors with e3 set to zero so this can probably be simplified by having the e vector 2D as,

0.46*e1 + -0.89*e2 + 0.00*e3
.

[Joy Christian]

Yeah, I am not doing the arc sin function correctly. Say we have the "e" vector as,

0.46*e1 + -0.89*e2 + -0.01*e3

So to get the arc sin it is not going to be simply e/e1. What is it going to be?
.

To begin with, your e vector is not normalized. If you want arcsin of y = sin(x), where x is the angle between e and a vectors, then arcsin(y) = x = arccos(e.a), where both e and a are unit vectors.

***

[FrediFizzx]

Yeah, I am not doing the arc sin function correctly. Say we have the "e" vector as,

0.46*e1 + -0.89*e2 + -0.01*e3

So to get the arc sin it is not going to be simply e/e1. What is it going to be?
.

To begin with, your e vector is not normalized. If you want arcsin of y = sin(x), where x is the angle between e and a vectors, then arcsin(y) = x = arccos(e.a), where both e and a are unit vectors.

***

The e vector is normalized. But duh... yeah it is just the arccos(e.a) that I want actually. Thanks.

[Joy Christian]

Given the hidden variables or shared randomness (e, s), we have a prescription for definite results A(a; e, s) = +/-1 and B(b; e, s) = +/-1 observed by Alice and Bob.

Given N particle pairs observed by Alice and Bob within the 3-sphere, the correlations among their results A and B work out to be E(a, b) = -a.b. End of the story.

***

The measurement functions of Alice and Bob are now very easy to understand:



and

,

where is the x-axis and the pair is the initial state of the spin system.

***

Donald Graft has made some silly remarks about the above simplified simulation at Retraction Watch without understanding the first thing about it. He claims that "it produces 0 outputs as well" and thus "it is just the old detection loophole" based simulation. His misunderstanding stems from not recognizing that the "initial state" or the "particle" in the simulation is represented by the pair , not just the vector ; and there is one-to-one correspondence in the model between the initial states and the measurement results . In other words, not a single "particle" goes undetected by Alice and Bob, unlike in the detection loophole model.

PS: The above feature is explicitly demonstrated in the full version of the simulation: http://rpubs.com/jjc/84238.

***

[FrediFizzx]

Well, he is not the first one that made that mistake but we already know that Graft has some serious shortcomings even understanding Bell.

Perhaps now we can figure out a version with no zeros in the A and B output streams.

[Joy Christian]

Well, he is not the first one that made that mistake but we already know that Graft has some serious shortcomings even understanding Bell.

Now it is "Mikko" who has decided to make silly remarks about my model, following Graft and HR (aka Heinera). If they refuse to abandon their flatland, R^3, then there is little we can do to help them. It is unfortunate that they --- especially the Flatlander-in-Chief Gill --- are able to influence the perception of other readers.

***

[Joy Christian]

***
Fred, I have changed some notations in the simulation slightly to make its connection with the full version more transparent: http://rpubs.com/jjc/233477.

Code: Select all

p = function(u,v){colSums(u*v)} # polarizer function of Bell

q = function(u,v,s){ifelse(abs(p(u,v)) > f, cos(Angles), 0)}

g = function(u,v,s){p(u,v)*q(u,v,s)}   # cf. http://rpubs.com/jjc/84238


***

[FrediFizzx]

***
Fred, I have changed some notations in the simulation slightly to make its connection with the full version more transparent: http://rpubs.com/jjc/233477.

Code: Select all

p = function(u,v){colSums(u*v)} # polarizer function of Bell

q = function(u,v,s){ifelse(abs(p(u,v)) > f, cos(Angles), 0)}

g = function(u,v,s){p(u,v)*q(u,v,s)}   # cf. http://rpubs.com/jjc/84238


***

I am not sure why you have the g-function as you don't call it in the simulation. I also found out that sign(p(a,e)) as you have it programmed is not quite the same as using the acos() function that I was originally using so a bit of a mystery as to why both ways work.

I'm hung up with trying to program a full arc cosine function for GAViewer. I suspect it is also the difference between the R version and Mathematica version. I need it for GAViewer so that I can get the angles properly for b - a. As you know, arc cosine only works 0 to pi and gives the wrong angles for quadrants 2, 3 and 4 based on cosine values. But knowing both cosine and sine values it is possible to have a full arc cosine function that works from 0 to 2pi correctly. I'm almost there.
.