## Joy's derivation of the 2*sqrt(2) bound

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

### Joy's derivation of the 2*sqrt(2) bound

Joy Christian wrote:
Heinera wrote:All local models wil have a bound of 2.

Not all local models. Here is a local model that has bound of $2\sqrt{2}$ : http://arxiv.org/pdf/1501.03393.pdf [see Eq. (26)].

I think that this derivation is worthy of a topic of its own. There must be some independency of the expectation terms from Bell-CHSH. In eq. (18) are the $A_{a}(\lambda^{k})$'s the same in both the first and second expectation terms?
FrediFizzx
Independent Physics Researcher

Posts: 2905
Joined: Tue Mar 19, 2013 7:12 pm
Location: N. California, USA

### Re: Joy's derivation of the 2*sqrt(2) bound

FrediFizzx wrote:In eq. (18) are the $A_{a}(\lambda^{k})$'s the same in both the first and second expectation terms?

Yes, they are the same. They are, however, not scalar numbers. They are bivectors. They have both direction as well as magnitude. They are also non-commuting numbers. Moreover, they are standardized variables, or standard scores, not scalar numbers, or raw scores like A = +/-1. They should be properly understood as

"A_a = +/-1 about the direction a."

Thus we are playing a completely different ball game. The manipulation of these bivectors requires the rules of geometric algebra. So this is nothing like anything being played in the Bell saga, either before or after my work. The corresponding raw scores are nevertheless scalar numbers, A = +/-1, as in Eqs. (7) and (8).

I think Michel-type analysis is not appropriate for my Eq. (18). But my derivation does expose a deeply unphysical nature of Bell's story for the EPR=B correlation.
Joy Christian
Research Physicist

Posts: 2793
Joined: Wed Feb 05, 2014 4:49 am
Location: Oxford, United Kingdom

### Re: Joy's derivation of the 2*sqrt(2) bound

Joy Christian wrote:
FrediFizzx wrote:In eq. (18) are the $A_{a}(\lambda^{k})$'s the same in both the first and second expectation terms?

Yes, they are the same. They are, however, not scalar numbers. They are bivectors. They have both direction as well as magnitude. They are also non-commuting numbers. Moreover, they are standardized variables, or standard scores, not scalar numbers, or raw scores like A = +/-1. They should be properly understood as

"A_a = +/-1 about the direction a."

Thus we are playing a completely different ball game. The manipulation of these bivectors requires the rules of geometric algebra. So this is nothing like anything being played in the Bell saga, either before or after my work. The corresponding raw scores are nevertheless scalar numbers, A = +/-1, as in Eqs. (7) and (8).

I think Michel-type analysis is not appropriate for my Eq. (18). But my derivation does expose a deeply unphysical nature of Bell's story for the EPR=B correlation.

Yes, it is a "different ball game". But it is easy to prove that for scalar A's and B's (+/- 1's), that it is mathematically impossible to violate the bound of 2 for Bell-CHSH since the expectation terms are dependent. And we have seen that with independent expectation terms the bound becomes 4. So it might be natural to assume that there is some independency coming in somewhere for your derivation. If that is the case, it might be worthwhile to investigate this further. We are sure that bound of $2\sqrt{2}$ is correct.
FrediFizzx
Independent Physics Researcher

Posts: 2905
Joined: Tue Mar 19, 2013 7:12 pm
Location: N. California, USA

### Re: Joy's derivation of the 2*sqrt(2) bound

I suspect that the independency creeps in during the squaring of the integrand of eq. (18) and subsequent reduction to the RHS of eq. (20). I guess we need to go thru that in more detail. I just did a simple kind of expression expansion in Mathcad; not sure if it is on the right track to get to the RHS of eq. (20).

$(A1B1 + A1B2 + A2B1 - A2B2)^2$ =

$A1^2 B1^2 + 2 A1^2 B1 B2 + A1^2 B2^2 + 2 A1 A2 B1^2 - 2 A1 A2 B2^2 + A2^2 B1^2 - 2 A2^2 B1 B2 + A2^2 B2^2$

$A1 = A_{a}(\lambda^{k})$, etc.

Can we proceed from the above?
FrediFizzx
Independent Physics Researcher

Posts: 2905
Joined: Tue Mar 19, 2013 7:12 pm
Location: N. California, USA

### Re: Joy's derivation of the 2*sqrt(2) bound

FrediFizzx wrote:I suspect that the independency creeps in during the squaring of the integrand of eq. (18) and subsequent reduction to the RHS of eq. (20). I guess we need to go thru that in more detail. I just did a simple kind of expression expansion in Mathcad; not sure if it is on the right track to get to the RHS of eq. (20).

$(A1B1 + A1B2 + A2B1 - A2B2)^2$ =

$A1^2 B1^2 + 2 A1^2 B1 B2 + A1^2 B2^2 + 2 A1 A2 B1^2 - 2 A1 A2 B2^2 + A2^2 B1^2 - 2 A2^2 B1 B2 + A2^2 B2^2$

$A1 = A_{a}(\lambda^{k})$, etc.

Can we proceed from the above?

No. This does not seem right. You have to worry about the non-commutativity of all the bivectors involved. Try the following in the GA viewer instead:

$(A1B1 + A1B2 + A2B1 - A2B2) (A1B1 + A1B2 + A2B1 - A2B2) = \,?$ ,

where all the products are geometric products. I have done this by hand many times in the past. It is a good idea to check my calculation in the GA viewer in any case.
Joy Christian
Research Physicist

Posts: 2793
Joined: Wed Feb 05, 2014 4:49 am
Location: Oxford, United Kingdom

### Re: Joy's derivation of the 2*sqrt(2) bound

Joy Christian wrote:
FrediFizzx wrote:I suspect that the independency creeps in during the squaring of the integrand of eq. (18) and subsequent reduction to the RHS of eq. (20). I guess we need to go thru that in more detail. I just did a simple kind of expression expansion in Mathcad; not sure if it is on the right track to get to the RHS of eq. (20).

$(A1B1 + A1B2 + A2B1 - A2B2)^2$ =

$A1^2 B1^2 + 2 A1^2 B1 B2 + A1^2 B2^2 + 2 A1 A2 B1^2 - 2 A1 A2 B2^2 + A2^2 B1^2 - 2 A2^2 B1 B2 + A2^2 B2^2$

$A1 = A_{a}(\lambda^{k})$, etc.

Can we proceed from the above?

No. This does not seem right. You have to worry about the non-commutativity of all the bivectors involved. Try the following in the GA viewer instead:

$(A1B1 + A1B2 + A2B1 - A2B2) (A1B1 + A1B2 + A2B1 - A2B2) = \,?$ ,

where all the products are geometric products. I have done this by hand many times in the past. It is a good idea to check my calculation in the GA viewer in any case.

Yes, I must be missing some terms since what I have above reduces simply to 4 which gives us 2 after taking the square root. I will see if I can figure out how to do it in GAViewer.
FrediFizzx
Independent Physics Researcher

Posts: 2905
Joined: Tue Mar 19, 2013 7:12 pm
Location: N. California, USA

### Re: Joy's derivation of the 2*sqrt(2) bound

So in GAViewer I just pick some random angles.

Code: Select all
>> a1ans = 0.62*e1 + 0.67*e2 + 0.40*e3>> a2ans = 0.18*e1 + 0.71*e2 + 0.69*e3>> b1ans = -0.15*e1 + 0.58*e2 + -0.80*e3>> b2ans = -0.73*e1 + -0.02*e2 + 0.69*e3

Then for C=((A1 B1)+(A1 B2)+(A2 B1)-(A2 B2)), I get a rotor,

ans = 0.71 + 1.76*e2^e3 + -0.38*e3^e1 + -0.63*e1^e2

Then for C*C I get another rotor,

ans = -3.16 + 2.50*e2^e3 + -0.54*e3^e1 + -0.89*e1^e2

Is this making any sense? Here is the rest of the code.
Code: Select all
function getRandomUnitVector() //uniform random unit vector:    //http://mathworld.wolfram.com/SpherePointPicking.html{   v=randGaussStd()*e1+randGaussStd()*e2+randGaussStd()*e3;   return normalize(v);}   batch test(){   set_window_title("Test of Joy Christian's arXiv:1103.1879 paper");   //N=20001; //number of iterations (trials)   I=e1^e2^e3;                         a1=getRandomUnitVector();                b1=getRandomUnitVector();                a2=getRandomUnitVector();                b2=getRandomUnitVector();                A1=I.a1;                A2=I.a2;                B1=I.b1;                B2=I.b2;                C=((A1 B1)+(A1 B2)+(A2 B1)-(A2 B2));                print(C)                print(C*C);}
FrediFizzx
Independent Physics Researcher

Posts: 2905
Joined: Tue Mar 19, 2013 7:12 pm
Location: N. California, USA

### Re: Joy's derivation of the 2*sqrt(2) bound

FrediFizzx wrote:So in GAViewer I just pick some random angles.

Code: Select all
>> a1ans = 0.62*e1 + 0.67*e2 + 0.40*e3>> a2ans = 0.18*e1 + 0.71*e2 + 0.69*e3>> b1ans = -0.15*e1 + 0.58*e2 + -0.80*e3>> b2ans = -0.73*e1 + -0.02*e2 + 0.69*e3

These are not bivectors. They are vectors. You have to use bivectors. You will still get a rotor in the end, as you are getting now. Then, when you square that rotor, you will get another rotor. Then the bivector term of that rotor will drop out if you use both the right-left perspectives, as we did in Albert Jan's program. What you should be left with is just the number 8 --- the square-root of which is $2\sqrt{2}$ .
Joy Christian
Research Physicist

Posts: 2793
Joined: Wed Feb 05, 2014 4:49 am
Location: Oxford, United Kingdom

### Re: Joy's derivation of the 2*sqrt(2) bound

Joy Christian wrote:
FrediFizzx wrote:So in GAViewer I just pick some random angles.

Code: Select all
>> a1ans = 0.62*e1 + 0.67*e2 + 0.40*e3>> a2ans = 0.18*e1 + 0.71*e2 + 0.69*e3>> b1ans = -0.15*e1 + 0.58*e2 + -0.80*e3>> b2ans = -0.73*e1 + -0.02*e2 + 0.69*e3

These are not bivectors. They are vectors. You have to use bivectors. You will still get a rotor in the end, as you are getting now. Then, when you square that rotor, you will get another rotor. Then the bivector term of that rotor will drop out if you use both the right-left perspectives, as we did in Albert Jan's program. What you should be left with is just the number 8 --- the square-root of which is $2\sqrt{2}$ .

Yes right, they are the angle vectors. But you see in the code I did A1 = I.a1, etc. The A1, A2, B1 and B2 are all bivectors. So the results are basically right. I don't think GAViewer will do an analytical type of calculation to get to the result of 8.

What I am looking for here is what the missing terms are in the expansion.
FrediFizzx
Independent Physics Researcher

Posts: 2905
Joined: Tue Mar 19, 2013 7:12 pm
Location: N. California, USA

### Re: Joy's derivation of the 2*sqrt(2) bound

FrediFizzx wrote:What I am looking for here is what the missing terms are in the expansion.

Ah... That is easy:

$(A1B1 + A1B2 + A2B1 - A2B2) (A1B1 + A1B2 + A2B1 - A2B2) = A1B1A1B1 + A1B2A1B1 + A2B1A1B1 - A2B2A1B1 + A1B1A1B2 + A1B2A1B2 + A2B1A1B2 - A2B2A1B2 + A1B1A2B1 + A1B2A2B1 + A2B1A2B1 - A2B2A2B1 - A1B1A2B2 - A1B2A2B2 - A2B1A2B2 + A2B2A2B2$

Here the orders of all the products are important, because the bivectors do not commute.
Joy Christian
Research Physicist

Posts: 2793
Joined: Wed Feb 05, 2014 4:49 am
Location: Oxford, United Kingdom

### Re: Joy's derivation of the 2*sqrt(2) bound

Joy Christian wrote:
FrediFizzx wrote:What I am looking for here is what the missing terms are in the expansion.

Ah... That is easy:

$(A1B1 + A1B2 + A2B1 - A2B2) (A1B1 + A1B2 + A2B1 - A2B2) = A1B1A1B1 + A1B2A1B1 + A2B1A1B1 - A2B2A1B1 + A1B1A1B2 + A1B2A1B2 + A2B1A1B2 - A2B2A1B2 + A1B1A2B1 + A1B2A2B1 + A2B1A2B1 - A2B2A2B1 - A1B1A2B2 - A1B2A2B2 - A2B1A2B2 + A2B2A2B2$

Here the orders of all the products are important, because the bivectors do not commute.

Ok good, I have to run out for a bit but if you have time, go ahead and start on the reduction.
FrediFizzx
Independent Physics Researcher

Posts: 2905
Joined: Tue Mar 19, 2013 7:12 pm
Location: N. California, USA

### Re: Joy's derivation of the 2*sqrt(2) bound

It is much easier to do the calculation slightly differently, as I have done on the last page of my first paper on Bell: http://arxiv.org/pdf/quant-ph/0703179.pdf.
Joy Christian
Research Physicist

Posts: 2793
Joined: Wed Feb 05, 2014 4:49 am
Location: Oxford, United Kingdom

### Re: Joy's derivation of the 2*sqrt(2) bound

Joy Christian wrote:It is much easier to do the calculation slightly differently, as I have done on the last page of my first paper on Bell: http://arxiv.org/pdf/quant-ph/0703179.pdf.

Ok, if this first way doesn't show any independency, then we can look at that. So I think the expression reduces to,

4 - A2B2A1B1 + A2B1A1B2 + A1B2A2B1 - A1B1A2B2

Let me know if that is right.
FrediFizzx
Independent Physics Researcher

Posts: 2905
Joined: Tue Mar 19, 2013 7:12 pm
Location: N. California, USA

### Re: Joy's derivation of the 2*sqrt(2) bound

FrediFizzx wrote:
Joy Christian wrote:It is much easier to do the calculation slightly differently, as I have done on the last page of my first paper on Bell: http://arxiv.org/pdf/quant-ph/0703179.pdf.

Ok, if this first way doesn't show any independency, then we can look at that. So I think the expression reduces to,

4 - A2B2A1B1 + A2B1A1B2 + A1B2A2B1 - A1B1A2B2

Let me know if that is right.

Yes, this is correct.
Joy Christian
Research Physicist

Posts: 2793
Joined: Wed Feb 05, 2014 4:49 am
Location: Oxford, United Kingdom

### Re: Joy's derivation of the 2*sqrt(2) bound

Joy Christian wrote:
FrediFizzx wrote:
Joy Christian wrote:It is much easier to do the calculation slightly differently, as I have done on the last page of my first paper on Bell: http://arxiv.org/pdf/quant-ph/0703179.pdf.

Ok, if this first way doesn't show any independency, then we can look at that. So I think the expression reduces to,

4 - A2B2A1B1 + A2B1A1B2 + A1B2A2B1 - A1B1A2B2

Let me know if that is right.

Yes, this is correct.

Ok, this part (- A2B2A1B1 + A2B1A1B2 + A1B2A2B1 - A1B1A2B2) returns a bivector in GAViewer. Looks like the scalar cancels out in the sum of the rotors?
Code: Select all
>> -(A2 B2 A1 B1)+(A2 B1 A1 B2)+(A1 B2 A2 B1)-(A1 B1 A2 B2)ans = 0.31*e2^e3 + 0.01*e3^e1 + -0.11*e1^e2
FrediFizzx
Independent Physics Researcher

Posts: 2905
Joined: Tue Mar 19, 2013 7:12 pm
Location: N. California, USA

### Re: Joy's derivation of the 2*sqrt(2) bound

FrediFizzx wrote:Ok, this part (- A2B2A1B1 + A2B1A1B2 + A1B2A2B1 - A1B1A2B2) returns a bivector in GAViewer. Looks like the scalar cancels out in the sum of the rotors?
Code: Select all
>> -(A2 B2 A1 B1)+(A2 B1 A1 B2)+(A1 B2 A2 B1)-(A1 B1 A2 B2)ans = 0.31*e2^e3 + 0.01*e3^e1 + -0.11*e1^e2

There is something wrong. There should be a scalar, -4 (a x a') . (b' x b), in addition to the 4 and the bivector. Unless you chose your bivectors in some special way.
Joy Christian
Research Physicist

Posts: 2793
Joined: Wed Feb 05, 2014 4:49 am
Location: Oxford, United Kingdom

### Re: Joy's derivation of the 2*sqrt(2) bound

Joy Christian wrote:
FrediFizzx wrote:Ok, this part (- A2B2A1B1 + A2B1A1B2 + A1B2A2B1 - A1B1A2B2) returns a bivector in GAViewer. Looks like the scalar cancels out in the sum of the rotors?
Code: Select all
>> -(A2 B2 A1 B1)+(A2 B1 A1 B2)+(A1 B2 A2 B1)-(A1 B1 A2 B2)ans = 0.31*e2^e3 + 0.01*e3^e1 + -0.11*e1^e2

There is something wrong. There should be a scalar, -4 (a x a') . (b' x b), in addition to the 4 and the bivector. Unless you chose your bivectors in some special way.

Here is the results for each term,
Code: Select all
>> -(A2 B2 A1 B1)ans = -0.29 + 0.26*e2^e3 + 0.54*e3^e1 + 0.74*e1^e2>> A2 B1 A1 B2ans = 0.29 + 0.60*e2^e3 + -0.11*e3^e1 + -0.74*e1^e2>> A1 B2 A2 B1ans = 0.29 + -0.06*e2^e3 + -0.26*e3^e1 + -0.92*e1^e2>> -(A1 B1 A2 B2)ans = -0.29 + -0.49*e2^e3 + -0.16*e3^e1 + 0.80*e1^e2

We can see that the scalar is canceling out. Now this is only for one iteration (trial).
FrediFizzx
Independent Physics Researcher

Posts: 2905
Joined: Tue Mar 19, 2013 7:12 pm
Location: N. California, USA

### Re: Joy's derivation of the 2*sqrt(2) bound

The bivectors are just this,
Code: Select all
      A1=I.a1;      A2=I.a2;      B1=I.b1;      B2=I.b2;
FrediFizzx
Independent Physics Researcher

Posts: 2905
Joined: Tue Mar 19, 2013 7:12 pm
Location: N. California, USA

### Re: Joy's derivation of the 2*sqrt(2) bound

Let me understand your notation. A1 = A(a), B1 = B(b), A2 = A(a' ), and B2 = B(b' ). Is that correct? If so, then you have a product of four different bivectors in each line such as A2 B2 A1 B1. A geometric product of four bivectors like these should give scalar + bivector. But you are not getting that. So something is clearly wrong.
Joy Christian
Research Physicist

Posts: 2793
Joined: Wed Feb 05, 2014 4:49 am
Location: Oxford, United Kingdom

### Re: Joy's derivation of the 2*sqrt(2) bound

FrediFizzx wrote:The bivectors are just this,
Code: Select all
      A1=I.a1;      A2=I.a2;      B1=I.b1;      B2=I.b2;

Sorry, I misread your code. You do have scalars, but they are cancelling out. So you have chosen your bivectors in a very special way.
Joy Christian
Research Physicist

Posts: 2793
Joined: Wed Feb 05, 2014 4:49 am
Location: Oxford, United Kingdom

Next

### Who is online

Users browsing this forum: ahrefs [Bot] and 8 guests