## 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

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

Joy Christian wrote:
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.

Hmm... Yeah, even with more iterations, the scalars will just be cancelling out. So what is wrong? How do we reduce this part further?

(- A2B2A1B1 + A2B1A1B2 + A1B2A2B1 - A1B1A2B2)
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### Re: Joy's derivation of the 2*sqrt(2) bound

FrediFizzx wrote:
Joy Christian wrote:
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.

Hmm... Yeah, even with more iterations, the scalars will just be cancelling out. So what is wrong? How do we reduce this part further?

(- A2B2A1B1 + A2B1A1B2 + A1B2A2B1 - A1B1A2B2)

May be there are some sign errors in this part (- A2B2A1B1 + A2B1A1B2 + A1B2A2B1 - A1B1A2B2). I will re-derive the whole thing to see what is going wrong.
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### Re: Joy's derivation of the 2*sqrt(2) bound

Joy Christian wrote:
FrediFizzx wrote:Hmm... Yeah, even with more iterations, the scalars will just be cancelling out. So what is wrong? How do we reduce this part further?

(- A2B2A1B1 + A2B1A1B2 + A1B2A2B1 - A1B1A2B2)

May be there are some sign errors in this part (- A2B2A1B1 + A2B1A1B2 + A1B2A2B1 - A1B1A2B2). I will re-derive the whole thing to see what is going wrong.

Ok, good. Could the fact that the vectors a1, a2, b1 and b2 are 3D have anything to do with it? Should they maybe be 2D? Doesn't seem like it would matter though. Anyways, I am out for the night.
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### Re: Joy's derivation of the 2*sqrt(2) bound

Found the error. The correct expression is 4 - A2A1B2B1 + A2A1B1B2 + A1A2B2B1 - A1A2B1B2 (note the orders).
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### Re: Joy's derivation of the 2*sqrt(2) bound

Joy Christian wrote:Found the error. The correct expression is 4 - A2A1B2B1 + A2A1B1B2 + A1A2B2B1 - A1A2B1B2 (note the orders).

OK, that works in GAViewer to give a scalar and a bivector.
Code: Select all
>> -(A2 A1 B2 B1)+(A2 A1 B1 B2)+(A1 A2 B2 B1)-(A1 A2 B1 B2)ans = 2.57 + -0.66*e2^e3 + 1.20*e3^e1 + 1.81*e1^e2

But those terms are not in the original list that you gave for the square of the integrand. Can you give the expansion again? The error must be in what you originally gave for that. Thanks.
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### Re: Joy's derivation of the 2*sqrt(2) bound

FrediFizzx wrote:
Joy Christian wrote:Found the error. The correct expression is 4 - A2A1B2B1 + A2A1B1B2 + A1A2B2B1 - A1A2B1B2 (note the orders).

OK, that works in GAViewer to give a scalar and a bivector.
Code: Select all
>> -(A2 A1 B2 B1)+(A2 A1 B1 B2)+(A1 A2 B2 B1)-(A1 A2 B1 B2)ans = 2.57 + -0.66*e2^e3 + 1.20*e3^e1 + 1.81*e1^e2

But those terms are not in the original list that you gave for the square of the integrand. Can you give the expansion again? The error must be in what you originally gave for that. Thanks.

The original expansion I made is fine. Below I reproduce it again. What we have to remember is that A's and B's commute. This allows a lot of terms to cancel out or reduce to 1 or -1. What is left is 4 - A2A1B2B1 + A2A1B1B2 + A1A2B2B1 - A1A2B1B2 = 4 + [ A1, A2 ] [ B2, B1 ], as in Eq. (23) of my first paper I linked above (of 2007).

Next, the bivector term reduces to zero in the large N limit, as in the Albert Jan Wonnink's program. So, for your example, we get $\sqrt{4 + 2.57} = 2.563$.

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

Joy Christian wrote:The original expansion I made is fine. Below I reproduce it again. What we have to remember is that A's and B's commute. This allows a lot of terms to cancel out or reduce to 1 or -1. What is left is 4 - A2A1B2B1 + A2A1B1B2 + A1A2B2B1 - A1A2B1B2 = 4 + [ A1, A2 ] [ B2, B1 ], as in Eq. (23) of my first paper I linked above (of 2007).

Joy Christian wrote: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.

Well........... now I am confused. You say that the A's and B's commute but then you say "the bivectors do not commute". Do you mean the A's commute with the A's and B's with the B's?

BTW, the scalar result of 2.57 in GAviewer does match 4(a1 x a2).(b2 x b1). So we know that part is correct.
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### Re: Joy's derivation of the 2*sqrt(2) bound

FrediFizzx wrote:Well........... now I am confused. You say that the A's and B's commute but then you say "the bivectors do not commute". Do you mean the A's commute with the A's and B's with the B's?

BTW, the scalar result of 2.57 in GAviewer does match 4(a1 x a2).(b2 x b1). So we know that part is correct.

Sorry about the confusion. A's and B's do not commute among themselves --- e.g., [A1, A2] $\not=$ 0. But A's commute with B's, because A and B are space-like separated.
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### Re: Joy's derivation of the 2*sqrt(2) bound

Joy Christian wrote:
FrediFizzx wrote:Well........... now I am confused. You say that the A's and B's commute but then you say "the bivectors do not commute". Do you mean the A's commute with the A's and B's with the B's?

BTW, the scalar result of 2.57 in GAviewer does match 4(a1 x a2).(b2 x b1). So we know that part is correct.

Sorry about the confusion. A's and B's do not commute among themselves --- e.g., [A1, A2] $\not=$ 0. But A's commute with B's, because A and B are space-like separated.

So the first original term -A2B2A1B1 after reduction can be rewritten as -A2A1B2B1? Is that correct?
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### Re: Joy's derivation of the 2*sqrt(2) bound

FrediFizzx wrote:
Joy Christian wrote:
FrediFizzx wrote:Well........... now I am confused. You say that the A's and B's commute but then you say "the bivectors do not commute". Do you mean the A's commute with the A's and B's with the B's?

BTW, the scalar result of 2.57 in GAviewer does match 4(a1 x a2).(b2 x b1). So we know that part is correct.

Sorry about the confusion. A's and B's do not commute among themselves --- e.g., [A1, A2] $\not=$ 0. But A's commute with B's, because A and B are space-like separated.

So the first original term -A2B2A1B1 after reduction can be rewritten as -A2A1B2B1? Is that correct?

Yes, that is correct.
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### Re: Joy's derivation of the 2*sqrt(2) bound

Joy Christian wrote:
FrediFizzx wrote:So the first original term -A2B2A1B1 after reduction can be rewritten as -A2A1B2B1? Is that correct?

Yes, that is correct.

Ok good. I think I have it all straight now and we are ready to analyze for dependency/independency. I will put it all together in a couple of hours.
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### Re: Joy's derivation of the 2*sqrt(2) bound

So taking eq. (18) in this paper and squaring the integrand results in the following formula. We have relabelled the A's and B's so easier to check with GAViewer and perhaps easier to follow.
Step 1:
(A1B1 + A1B2 + A2B1 - A2B2) (A1B1 + A1B2 + A2B1 - A2B2) =
A1B1A1B1 + A1B2A1B1 + A2B1A1B1 - A2B2A1B1 + A1B1A1B2 + A1B2A1B2 + A2B1A1B2 - A2B2A1B2 + A1B1A2B1 + A1B2A2B1 + A2B1A2B1 - A2B2A2B1 - A1B1A2B2 - A1B2A2B2 - A2B1A2B2 + A2B2A2B2

And it is obvious that the expansion is correct. Now, the terms in the result can be collected as follows but keeping the A orders and B orders the same.
Step 2:
A1A1B1B1 + A1A1B2B1 + A2A1B1B1 - A2A1B2B1 + A1A1B1B2 + A1A1B2B2 + A2A1B1B2 - A2A1B2B2 + A1A2B1B1 + A1A2B2B1 + A2A2B1B1 - A2A2B2B1 - A1A2B1B2 - A1A2B2B2 - A2A2B1B2 + A2A2B2B2

Then since a bivector squared is -1 we further reduce to;
Step 3:
1 - B2B1 - A2A1 - A2A1B2B1 - B1B2 + 1 + A2A1B1B2 + A2A1 - A1A2 + A1A2B2B1 +1 + B2B1 - A1A2B1B2 + A1A2 + B1B2 +1

Which further reduces to;
Step 4:
4 - A2A1B2B1 + A2A1B1B2 + A1A2B2B1 - A1A2B1B2

We saw earlier that this is the result we want and is verified by GAViewer to be correct and after taking the square root and applying many iterations gives us $2 \sqrt{2}$. Now the original dependency between expectation terms is shown by this;

A1B1 + A1B2 + A2B1 - A2B2

So if an independency is creeping in here, I would say that it has to be in step 2. Otherwise this is a mystery as to how (A1B1 + A1B2 + A2B1 - A2B2) with dependent expectation terms can return $2 \sqrt{2}$ since it is mathematically proven that it can't be larger than 2. However, I think Joy claims it is because the original math proof doesn't use bivectors and GA. Joy, please correct me if I am wrong about that.

PS: The reason I think it has to be in Step 2 because we saw earlier without the rearrangement, the result came out to be 2 since the scalar cancelled out.
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### Re: Joy's derivation of the 2*sqrt(2) bound

FrediFizzx wrote:Now the original dependency between expectation terms is shown by this;

A1B1 + A1B2 + A2B1 - A2B2

So if an independency is creeping in here, I would say that it has to be in step 2. Otherwise this is a mystery as to how (A1B1 + A1B2 + A2B1 - A2B2) with dependent expectation terms can return $2 \sqrt{2}$ since it is mathematically proven that it can't be larger than 2. However, I think Joy claims it is because the original math proof doesn't use bivectors and GA. Joy, please correct me if I am wrong about that.

I think the best way to see the difference between the (in)dependent scalar valued functions and the bivector valued functions is to write your Step 4 as follows:

F = 4 - A2A1B2B1 + A2A1B1B2 + A1A2B2B1 - A1A2B1B2 = 4 + [ A1, A2 ] [ B2, B1 ].

Now if A1 and A2 (and likewise B1 and B2) are scalars, then the two commutators vanish, and we get F = 4, so that S = $\sqrt{F}$ = 2.

But if A1 and A2 (and likewise B1 and B2) are bivectors, then the two commutators do not vanish, each giving a maximum value of 2, so that F = 4 + 4 and S = $2\sqrt{2}$.
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### Re: Joy's derivation of the 2*sqrt(2) bound

Joy Christian wrote:
FrediFizzx wrote:Now the original dependency between expectation terms is shown by this;

A1B1 + A1B2 + A2B1 - A2B2

So if an independency is creeping in here, I would say that it has to be in step 2. Otherwise this is a mystery as to how (A1B1 + A1B2 + A2B1 - A2B2) with dependent expectation terms can return $2 \sqrt{2}$ since it is mathematically proven that it can't be larger than 2. However, I think Joy claims it is because the original math proof doesn't use bivectors and GA. Joy, please correct me if I am wrong about that.

I think the best way to see the difference between the (in)dependent scalar valued functions and the bivector valued functions is to write your Step 4 as follows:

F = 4 - A2A1B2B1 + A2A1B1B2 + A1A2B2B1 - A1A2B1B2 = 4 + [ A1, A2 ] [ B2, B1 ].

Now if A1 and A2 (and likewise B1 and B2) are scalars, then the two commutators vanish, and we get F = 4, so that S = $\sqrt{F}$ = 2.

But if A1 and A2 (and likewise B1 and B2) are bivectors, then the two commutators do not vanish, each giving a maximum value of 2, so that F = 4 + 4 and S = $2\sqrt{2}$.

I don't think that quite explains why the rearrangement of A's and B's make the difference. When we had this,

F = 4 - A2B2A1B1 + A2B1A1B2 + A1B2A2B1 - A1B1A2B2 (each term is a rotor; scalar plus bivector)

The scalars cancelled out so we had F = 4 + (bivector terms) which the bivector terms cancel out with many iterations so the result is 2. Rearranging to

F = 4 - A2A1B2B1 + A2A1B1B2 + A1A2B2B1 - A1A2B1B2 (each term is a rotor; scalar plus bivector)

Gives the $2\sqrt{2}$ result.
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### Re: Joy's derivation of the 2*sqrt(2) bound

FrediFizzx wrote:
Joy Christian wrote:
FrediFizzx wrote:Now the original dependency between expectation terms is shown by this;

A1B1 + A1B2 + A2B1 - A2B2

So if an independency is creeping in here, I would say that it has to be in step 2. Otherwise this is a mystery as to how (A1B1 + A1B2 + A2B1 - A2B2) with dependent expectation terms can return $2 \sqrt{2}$ since it is mathematically proven that it can't be larger than 2. However, I think Joy claims it is because the original math proof doesn't use bivectors and GA. Joy, please correct me if I am wrong about that.

I think the best way to see the difference between the (in)dependent scalar valued functions and the bivector valued functions is to write your Step 4 as follows:

F = 4 - A2A1B2B1 + A2A1B1B2 + A1A2B2B1 - A1A2B1B2 = 4 + [ A1, A2 ] [ B2, B1 ].

Now if A1 and A2 (and likewise B1 and B2) are scalars, then the two commutators vanish, and we get F = 4, so that S = $\sqrt{F}$ = 2.

But if A1 and A2 (and likewise B1 and B2) are bivectors, then the two commutators do not vanish, each giving a maximum value of 2, so that F = 4 + 4 and S = $2\sqrt{2}$.

I don't think that quite explains why the rearrangement of A's and B's make the difference. When we had this,

F = 4 - A2B2A1B1 + A2B1A1B2 + A1B2A2B1 - A1B1A2B2 (each term is a rotor; scalar plus bivector)

The scalars cancelled out so we had F = 4 + (bivector terms) which the bivector terms cancel out with many iterations so the result is 2. Rearranging to

F = 4 - A2A1B2B1 + A2A1B1B2 + A1A2B2B1 - A1A2B1B2 (each term is a rotor; scalar plus bivector)

Gives the $2\sqrt{2}$ result.

GAviewer is probably blind to the fact that A's and B's are space-like separated and hence commute. How does it compute the product of four bivectors, say in A2B2A1B1? Does it compute A2B2 first, and then A1B1, and then (A2B2)*(A1B1)? I suspect it must be picking a wrong order somehow in the first expression of F.
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### Re: Joy's derivation of the 2*sqrt(2) bound

Joy Christian wrote:
FrediFizzx wrote:I don't think that quite explains why the rearrangement of A's and B's make the difference. When we had this,

F = 4 - A2B2A1B1 + A2B1A1B2 + A1B2A2B1 - A1B1A2B2 (each term is a rotor; scalar plus bivector)

The scalars cancelled out so we had F = 4 + (bivector terms) which the bivector terms cancel out with many iterations so the result is 2. Rearranging to

F = 4 - A2A1B2B1 + A2A1B1B2 + A1A2B2B1 - A1A2B1B2 (each term is a rotor; scalar plus bivector)

Gives the $2\sqrt{2}$ result.

GAviewer is probably blind to the fact that A's and B's are space-like separated and hence commute. How does it compute the product of four bivectors, say in A2B2A1B1? Does it compute A2B2 first, and then A1B1, and then (A2B2)*(A1B1)? I suspect it must be picking a wrong order somehow in the first expression of F.

You might be right.
Code: Select all
>> A2 B2 A1 B1ans = 0.29 + -0.26*e2^e3 + -0.54*e3^e1 + -0.74*e1^e2>> E=(A2 B2)E = 0.10 + -0.10*e2^e3 + 0.19*e3^e1 + 0.97*e1^e2>> F=(A1 B1)F = -0.77 + 0.39*e2^e3 + -0.44*e3^e1 + -0.25*e1^e2>> E Fans = 0.29 + -0.26*e2^e3 + -0.54*e3^e1 + -0.74*e1^e2>> E*Fans = 0.29 + -0.26*e2^e3 + -0.54*e3^e1 + -0.74*e1^e2>> E.Fans = 0.37

Because E F and E*F are the same as doing all 4 at once. If that is the case then getting a result of 2 the first way without rearrangement is just a fluke?

However we get the same kind of result when doing A2A1B2B1. And GAViewer confirms that we get 4(a1 x a2).(b2 x b1) for the scalar part of the whole sum of terms.
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### Re: Joy's derivation of the 2*sqrt(2) bound

Perhaps what Michel wrote last year will shed some light on this.
viewtopic.php?f=6&t=49&start=160&hilit=Invariance#p2578
minkwe wrote:Before I take my break however, as concerns Tsirelson's bound:

[1] S = E(a − b) + E(a − b') + E(a' − b) − E(a' − b')
Can be written as
[2] S = f (x) + f (y) + f (z) − f (w), w, x, y, z ∈ [0, 2π], and −1 ≤ f(.) ≤ 1

If where w, x, y and z were independent variables, the upper bound of S would be 4 as we've established elsewhere. But inspection of [1] shows that they are not because we can express w in terms of the other three variables w = y + z -x to get

[3] S = f (x) + f (y) + f (z) − f (y + z − x)

At the extrema of S, the partial derivatives of S are all zero:
f'(x) + f'(y + z − x) = f'(y) − f'(y + z − x) = f'(z) − f'(y + z − x) = 0, therefore y = z and f'(y) = f'(z) = -f'(x). Substituting in [3] we get

[4] Se = f(x) + 2f(y) − f(2y − x)

For even f(.), f'(x) = f'(-x) = f'(y). At the extrema therefore, y = -x and we can reduce [4] further to

[5] Se = 3f(y) - f(3y)

Therefore the maximum of S = cos(a, b) + cos(a, b') + cos(a', b) - cos(a', b') is also the maximum of the function

3cos(y) - cos(3y)

Which is 2 sqrt 2. http://www.wolframalpha.com/input/?i=Ma ... os%283x%29

Like I told you, there is nothing quantum about Tsirelson's bound, nor is any Hilbert space involved.
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### Re: Joy's derivation of the 2*sqrt(2) bound

These are all commuting numbers, so mathematically this is quite a different argument.

Is there any way to tell the GAviewer that A's and B's commute for the first F?

The only difference between the first F and the second F is the assumption [ Ai, Bj ] = 0 that is used in the second F (the first F does not use this assumption).
Last edited by Joy Christian on Fri Aug 28, 2015 11:04 am, edited 1 time in total.
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### Re: Joy's derivation of the 2*sqrt(2) bound

Joy Christian wrote:These are all commuting numbers, so mathematically this is quite a different argument.

Is there any way to tell the GAviewer that A's and B's commute for the first F?

Well, the point is that it is possible to show a violation with "dependent" terms. It is a subtle thing on exactly what the dependency is.

I don't think GAViewer will do that but I will investigate it further.
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### Re: Joy's derivation of the 2*sqrt(2) bound

FrediFizzx wrote:
Joy Christian wrote:These are all commuting numbers, so mathematically this is quite a different argument.

Is there any way to tell the GAviewer that A's and B's commute for the first F?

Well, the point is that it is possible to show a violation with "dependent" terms. It is a subtle thing on exactly what the dependency is.

Ok, puzzle is solved. You can only get a "violation" of CHSH with dependent expectation terms if the terms are counterfactual. And of course using counterfactual terms is how quantum theory shows a "violation" of CHSH. So once again Joy's model matches the QM predictions of a bound of $2 \sqrt{2}$.
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