A classical challenge to Bell's Theorem?

[DrChinese]

So what are the FUNCTIONS A(a,x) and B(a,x), please clearly state those functions. Remember they can only have values ±1. cos^2(x-a) can not be a valid function according to Bell. A and B must be step functions according to Bell so you have to be integrating the product of two step functions rather than the harmonic ones you have.

There is nothing wrong with a function like A(a,x)=cos^2(x-a) when it produces a +1 outcome or a -1 outcome. zonde assumes everyone knows this, we are looking for +1/+1 and -1/-1 permutations.

Again, this seems to be a pointless exercise. I have yet to see the slightest indication that this is anything more than a wild goose chase. Gordon, is there a point in here? If you and bill want to just a batch of baseless claims (which is all that has happened to date), we may as well stop here.

I will say again: this example follows Bell and comes to a completely expected conclusion. Here is Gordon's classical algorithm which respects Bell.

 

[Gordon Watson]

There is nothing wrong with a function like A(a,x)=cos^2(x-a) when it produces a +1 outcome or a -1 outcome. zonde assumes everyone knows this, we are looking for +1/+1 and -1/-1 permutations.

Again, this seems to be a pointless exercise. I have yet to see the slightest indication that this is anything more than a wild goose chase. Gordon, is there a point in here? If you and bill want to just a batch of baseless claims (which is all that has happened to date), we may as well stop here.

I will say again: this example follows Bell and comes to a completely expected conclusion. Here is Gordon's classical algorithm which respects Bell.

..

Uh? HERE is what? I don't see it? Please repost. Or have you done it non-locally?


MOST PEOPLE who RESPECT BELL (or at least understand his maths -- which I understand is not a strength of yours; SEE BELOW) would by now have gotten the point of the challenge!

Bill Schnieder and zonde seem to get it; or be having a go.

Note that I am (via the OP) interested in how Bell's supporters tackle the challenge.

I am not so interested in them jumping on the bandwagon of my approach.

Please read my posts again. And ask clearer questions. I'm not into gaming with Bell.

ONE POINT (for you) IS to start where BELL STARTS!

PS: And, since you did not think of doing it my way, make sure you come with your own way. BUT: For sure, your A(a,x)=cos^2(x-a) is just plain silly!

..

 

[Gordon Watson]

So what are the FUNCTIONS A(a,x) and B(a,x), please clearly state those functions. Remember they can only have values ±1. cos^2(x-a) can not be a valid function according to Bell. A and B must be step functions according to Bell so you have to be integrating the product of two step functions rather than the harmonic ones you have.


Another gem. But something tells me, "don't go there". This is Gordon's thread so

Bill, you seem to understand the challenge very clearly. So PLEASE go for it here, especially while I'm away.

So: A BIG Thank-You from me.

PS: This --- "integrating the product of two step functions" -- is the hard way to go!

So look for a simpler solution that most everyone here might understand.

But you're on the ball, IMHOWMSBW!

Thanks again,

Gordon
..

..

 

[Gordon Watson]

Gordon: the point is that whatever the functions A, B and whatever the angles and whatever the probability distribution of the hidden variables, CHSH will be satisfied.

..
gill1109,

With respect: That is no point of the OP, as far I see it?

Here's an interesting real point: I had hoped (for sure), that you would have understood the challenge. For I was certainly sure that you would have seen how to proceed.

From the clues given, do you still not see it, via my way?

BUT the point was to see it via your way. Once I post my way here, I've lost the chance to see how Bell's supporters INDEPENDENTLY answered the "challenge" .. for which there are already too many clues here.

So, for me now (and maybe you could help): Where might I best discuss the consequences of this challenge on the Net?

Thanks, as always,

Gordon
..

 

[Gordon Watson]

Thanks zonde! Very helpful.

Is there a sin*sin component necessary too in addition to the cos*cos one above? That's what I was starting with...

DrC, trying to be helpful: START ANEW!

Start with BELL!

Gordon

 

[DrChinese]

MOST PEOPLE who RESPECT BELL (or at least understand his maths -- which I understand is not a strength of yours; SEE BELOW) would by now have gotten the point of the challenge!...

I certainly don't need to defend my understanding of Bell or the math behind it. You are free to think as you like.

I cannot help you further beyond telling you that a classical model will obey Bell inequalities, as I have already shown you numerous times in this thread. You have not really provided a challenge, because you do not assert that your model will provide results consistent with the predictions of quantum mechanics. That would be a necessary part of any challenge.

This is the end of my involvement in this thread as a poster. Please do not make any subsequent statements which are inconsistent with generally accepted physics, or I will report you as I will continue to monitor the thread and your posts.

Gordon, I have tried to be patient. But honestly, your recent comments have gotten unusually rude for you. (Probably due to billschnieder's involvement, as rude is his norm - and that is a kind assessment.) It is always interesting to see folks justifying their rudeness by the supposed "correctness" of their viewpoint. Well, that is sadly par for the course in all types of discourse; but I would hope that intelligent people would somehow be clever enough to make their points without being so snippy.

 

[zonde]

MOST PEOPLE who RESPECT BELL (or at least understand his maths -- which I understand is not a strength of yours; SEE BELOW) would by now have gotten the point of the challenge!

Bill Schnieder and zonde seem to get it; or be having a go.

Sorry Gordon but I don't get the point of your challenge.
I just saw DrChinese posting this formula P(ab)=.25+.5(cos^2(a-b)) and Bill asking where did he get this formula.
As I knew what chain of reasoning leads to this formula I posted it.

And I am not sure I understand (can check correctness of) math of Bell theorem so I find Nick Herbert's type of proof much more convincing.

 

[Mark M]

Gordon, I don't see what you're trying to prove here. I'm assuming you're asking for a derivation of the CHSH inequality with your given conditions. If so, I'll give it a go:

Starting with A(a,x)=\pm 1 B(b,x)=\pm 1 andE(A,B)= \int AB p(x) dx We can write E(A,B)= \int A(a,x)B(b,x)p(x)dx Since a, a', b, b' are settings for the detector we show that E(a,b)-E(a,{b}')=\int [A(a,x)B(b,x)-A(a,x)B({b}'x)]p(x)dx =\int A(a,x)B(b,x)[1 \pm A({a}'x)B({b}'x)] p(x)dx - \int A(a,x)B({b}',x)[1 \pm A({a}',x)B({b}',x)]p(x)dx Using the first inequality, we know that the quantities [1 \pm A({a}',x)B({b}',x)]p(x) and [1 \pm A({a}',x)B({b},x)]p(x) are non-negative. Also, we will use the triangle inequality on each side \left | E(a,b)-E(a,{b}') \right |\leq \int [1 \pm A({a}',x)B({b}',x)]p(x)dx]+\int [1 \pm A(a,x)B(b,x)]p(x)dx
Since \int p(x)=1 we can simplify to \left \lfloor E(a,b)-E(a,{b}') \right \rfloor\leq 2\pm[E({a}',{b}')+E({a}',b)] Which includes the CHSH inequality, with a maximum value of 2. As per usual.

QED

 

[billschnieder]

Gordon, I don't see what you're trying to prove here.

Gordon is challenging us to derive the classical result P(ab)=¼+½(cos²(a-b)) for the experiment he proposed, by starting where Bell started. With two separable functions A(a,x) and B(b,x) defined with a codomain ±1.

Zonde has provided a derivation of the above classical result by starting from the two functions:

A(a,x) = cos²(x−a)
B(b,x) = cos²(x−a)

However, this deviates from Bell because Bell insisted that A(a,x) and B(b,x) can only have values ±1, so the two functions must obey that if they are to follow Bell. In Bell's original paper, he suggested A(a,x) = sign(a · x), and B(a,x) = - sign(b · x) where a,b,x are vectors. Those functions do satisfy the A(a,x) = ±1. So the challenge is to use functions of that type or any other type which has ONLY values ±1 and derive the well known classical result for the experiment described in the OP.

 

[Gordon Watson]

I certainly don't need to defend my understanding of Bell or the math behind it. You are free to think as you like.

I cannot help you further beyond telling you that a classical model will obey Bell inequalities, as I have already shown you numerous times in this thread. You have not really provided a challenge, because you do not assert that your model will provide results consistent with the predictions of quantum mechanics. That would be a necessary part of any challenge.

This is the end of my involvement in this thread as a poster. Please do not make any subsequent statements which are inconsistent with generally accepted physics, or I will report you as I will continue to monitor the thread and your posts.

Gordon, I have tried to be patient. But honestly, your recent comments have gotten unusually rude for you. (Probably due to billschnieder's involvement, as rude is his norm - and that is a kind assessment.) It is always interesting to see folks justifying their rudeness by the supposed "correctness" of their viewpoint. Well, that is sadly par for the course in all types of discourse; but I would hope that intelligent people would somehow be clever enough to make their points without being so snippy.

..
Dear DrChinese,

So that I might apologise, I'd welcome your pointing to any rudeness on my part.

If your response has something to do with me associating you with the word silly, then I defend my position as follows:

Silly is a good Bellian word, as you no-doubt know.

I asked for a good Bellian (1964) function A(a, x) = ±1 in a specific context: the context of a simple and specific CLASSICAL experiment.

You proffered the following:

(DrC-1) A(a, x) = ±1 = cos^2 (a, x)!​

Please Note:

(a): RHS (DrC-1) can NEVER equal MINUS ONE (-1)!

(b): RHS (DrC-1) can equal +1 ONLY when a = x!

(c): HOWEVER, in the specified context, the probability that a = x (prior to the measurement interaction) is ZERO (0): P(a = x| in specifed context) = 0.


So, dear DrC: Since your proffered function cannot equal -1, and has P= 0 of equalling +1, WHEN does it EVER equal ±1?


Thus, as to 'silliness', I thought your defended mistake was worse then vN's; that's all.

Gordon

..
EDIT: My reply here was made before I saw Bill's response immediately above. To be clear, I asked for E(AB). Given the difficulties that appear to be associated with its derivation, the answer is (if my maths is correct):

(GW-1) E(AB) = (1/2) cos 2(a, b).​

..

 

[Gordon Watson]

Gordon is challenging us to derive the classical result P(ab)=¼+½(cos²(a-b)) for the experiment he proposed, by starting where Bell started. With two separable functions A(a,x) and B(b,x) defined with a codomain ±1.

Zonde has provided a derivation of the above classical result by starting from the two functions:

A(a,x) = cos²(x−a)
B(b,x) = cos²(x−a)

However, this deviates from Bell because Bell insisted that A(a,x) and B(b,x) can only have values ±1, so the two functions must obey that if they are to follow Bell. In Bell's original paper, he suggested A(a,x) = sign(a · x), and B(a,x) = - sign(b · x) where a,b,x are vectors. Those functions do satisfy the A(a,x) = ±1. So the challenge is to use functions of that type or any other type which has ONLY values ±1 and derive the well known classical result for the experiment described in the OP.

..
Dear Bill

Thanks for clarifying the OP and its challenge in my absence. I was away from the Net when a friend told me of DrC's reply (above). I am now on a slow server, attempting to correct some other wrong positions (but it is difficult).

My full participation here is still a week away. So, please, do not hesitate to add your valued comments at any time.

With thanks again,

Gordon

 

[Gordon Watson]

Sorry Gordon but I don't get the point of your challenge.
I just saw DrChinese posting this formula P(ab)=.25+.5(cos^2(a-b)) and Bill asking where did he get this formula.
As I knew what chain of reasoning leads to this formula I posted it.

And I am not sure I understand (can check correctness of) math of Bell theorem so I find Nick Herbert's type of proof much more convincing.

Dear zonde, I appreciated the fact that you engaged with the maths.

Please note that the OP requests the application of Bell's local-realistic protocol to what is clearly an Einstein-local and realistic CLASSICAL experiment. So we are not yet concerned with the maths of Bell's Theorem; the maths that arises when we study Bell's (1964), etc., inequalities.

That concern might arise when we see what is required to derive the simple classical result (if my maths is correct).

To encourage you with the classical maths in all of this, note that no Bell supporter has yet here (nor anywhere else, to my knowledge) derived the simple classical result requested in the OP ... USING BELL'S PROTOCOL.

The OP was intended to see what differing approaches emerged (differing from my own). Or, if someone said the task was IMPOSSIBLE, I was interested to learn their reasons -- for it could have indicated an error in my maths.

Please see my recent reply to DrC, and see if this clarifies the challenge for you. There have been many misleading statements here as to what the challenge is, and much running for cover. BUT there is no trick; just a very simple challenge ... especially for those who take Bell (1964), etc., seriously (as I do).

With best regards,

Gordon

 

[lugita15]

And I am not sure I understand (can check correctness of) math of Bell theorem so I find Nick Herbert's type of proof much more convincing.

That's my attitude as well. Bell's proof involves so many things, like integrating over hidden variables and factorization of conditional probabilities, that need to be studied carefully and that leave room open for confusion and debate. I think Herbert's proof "quantumtantra.com/bell2.html" gives a simple and intuitive explanation, so it's much easier to isolate points of ambiguity or disagreement.

 

[Gordon Watson]

Gordon, I don't see what you're trying to prove here. I'm assuming you're asking for a derivation of the CHSH inequality with your given conditions. If so, I'll give it a go:

Starting with A(a,x)=\pm 1 B(b,x)=\pm 1 andE(A,B)= \int AB p(x) dx We can write E(A,B)= \int A(a,x)B(b,x)p(x)dx Since a, a', b, b' are settings for the detector we show that E(a,b)-E(a,{b}')=\int [A(a,x)B(b,x)-A(a,x)B({b}'x)]p(x)dx =\int A(a,x)B(b,x)[1 \pm A({a}'x)B({b}'x)] p(x)dx - \int A(a,x)B({b}',x)[1 \pm A({a}',x)B({b}',x)]p(x)dx Using the first inequality, we know that the quantities [1 \pm A({a}',x)B({b}',x)]p(x) and [1 \pm A({a}',x)B({b},x)]p(x) are non-negative. Also, we will use the triangle inequality on each side \left | E(a,b)-E(a,{b}') \right |\leq \int [1 \pm A({a}',x)B({b}',x)]p(x)dx]+\int [1 \pm A(a,x)B(b,x)]p(x)dx
Since \int p(x)=1 we can simplify to \left \lfloor E(a,b)-E(a,{b}') \right \rfloor\leq 2\pm[E({a}',{b}')+E({a}',b)] Which includes the CHSH inequality, with a maximum value of 2. As per usual.

QED

Dear Mark M, many thanks for having a very neat go!

To understand the challenge here, please see recent posts here. I trust they are removing some confusions?

With regard to this: "I don't see what you're trying to prove here. I'm assuming you're asking for a derivation of the CHSH inequality with your given conditions. If so, I'll give it a go:"

The CHSH could be made the subject of another interesting discussion:

For those who derive it from an IDENTITY, we have the problem of an EXPERIMENT contradicting an IDENTITY!

Since I accept the experimental results; and since I do not accept that an Identity can be contradicted; well ... I hope you see my problem. Hopefully for discussion here, soon.

With thanks again,

Gordon

 

[zonde]

Gordon is challenging us to derive the classical result P(ab)=¼+½(cos²(a-b)) for the experiment he proposed, by starting where Bell started. With two separable functions A(a,x) and B(b,x) defined with a codomain ±1.

Zonde has provided a derivation of the above classical result by starting from the two functions:

A(a,x) = cos²(x−a)
B(b,x) = cos²(x−a)

However, this deviates from Bell because Bell insisted that A(a,x) and B(b,x) can only have values ±1, so the two functions must obey that if they are to follow Bell. In Bell's original paper, he suggested A(a,x) = sign(a · x), and B(a,x) = - sign(b · x) where a,b,x are vectors. Those functions do satisfy the A(a,x) = ±1. So the challenge is to use functions of that type or any other type which has ONLY values ±1 and derive the well known classical result for the experiment described in the OP.

My suggestion was that P(A=+1)=cos²(x−a) (and P(A=-1)=sin²(x−a) ).
This of course satisfies A(a,x) = ±1

 

[Gordon Watson]

My suggestion was that P(A=+1)=cos²(x−a) (and P(A=-1)=sin²(x−a) ).
This of course satisfies A(a,x) = ±1

Sorry, but I do not yet see A(a, x) = ±1.

I see probabilities.

See Bell (1964) for the Bell protocol if my representation is unclear to you.

 

[Delta Kilo]

My suggestion was that P(A=+1)=cos²(x−a) (and P(A=-1)=sin²(x−a) ).
This of course satisfies A(a,x) = ±1

Sorry, but I do not yet see A(a, x) = ±1.

I see probabilities.

See Bell (1964) for the Bell protocol if my representation is unclear to you.

It's not a big deal. Just let λ={θ,ζ,χ}, θ∈[0..2∏), ζ,χ∈[0..1), A(a,λ) = sign(cos²(θ−a)-ζ), B(b,λ) = sign(cos²(θ−b)-χ)

 

[zonde]

Sorry, but I do not yet see A(a, x) = ±1.

I see probabilities.

See Bell (1964) for the Bell protocol if my representation is unclear to you.

According to what I give what other values A(a, x) can have besides ±1?

P.S. Thanks Delta Kilo for your explanation.

 

[zonde]

Is there a sin*sin component necessary too in addition to the cos*cos one above? That's what I was starting with...

Don't know if it's still interesting but anyways.

In short
P(+1)=cos^2; P(-1)=sin^2
so we have 4 combinations for A*B
cos^2*cos^2 and integral in 0-2Pi range is Pi/4+Pi/2*cos^2
sin^2*sin^2 with integral the same as above Pi/4+Pi/2*cos^2
cos^2*sin^2 with integral Pi/4+Pi/2*sin^2
sin^2*cos^2 with integral Pi/4+Pi/2*sin^2

I used this integral calculator to check what I am saying:
http://www.numberempire.com/integralcalculator.php
Just type in "cos(x-a)^2*cos(x-b)^2"
For interval 0-2Pi all terms except the one with "*x" go to zero. And then using appropriate trigonometric identity we can get more accustomed form.

 

[ThomasT]

So doesn't it all just simplifiy to the correlation function being the average (wrt a large number of pairs) of the product of the functions A and B?

C(a,b) = < A(a,λ) B(b,λ)> , then a probability distribution,

C(a,b) = ∫ ρ(λ) A(a,λ) B(b,λ) dλ, which is Bell's archetypal LR form (equation 2 in the 1964 paper), wrt which Bell's theorem proved that there's no ρ(λ) wrt which this form can produce a correlation coefficient that matches Malus Law.

So, for any value of ρ(λ), then

C(a,b) = ρ(λ) ∫ sign [cos²(a-λ)] sign [cos²(b-λ)] dλ ≠ cos²θ , where θ is the angular difference |a-b|

 

[billschnieder]

According to what I give what other values A(a, x) can have besides ±1?

Take a look at Bell's candidate function A(a,x) = sign(a · x)

This *function* maps ANY two vectors (a,x) directly to ONE outcome +1 or -1. Say we have ONE photon arriving at a detector and the "mechanism" of the detector is represented by Bell's function, the OUTCOME will be either +1 or -1 for that ONE photon. So Bell's function is correct for what he was talking about.

Now take a look at yours. You do not provide any function which can give an outcome, all you have done is say that looking at many outcomes, we get a probability that looks like this P(A=+1)=cos²(x−a). In other words, this is a statement of the probability distribution of outcomes, not a function which defines how each outcome is *generated* from (a and x).

In case this was not clear. Imagine yourself writing a simulation of the situation. With Bell's function, you can randomly generate vector pairs (a,x), one pair at a time and immediately calculate the outcomes for each pair. However in your case, you will not be able to produce a single outcome, rather you will have to generate a large number of outcomes such that the relative frequencies obeyed cos²(x−a). You provided a function for P(A), you did not provide a function for A. However, if you can specifiy the function A, and show how you obtained P(A) from A, it will move the ball a long way in the right direction.

I hope this clarifies now why your functions do not pass Bell's "sniff-test" so to speak.

 

[billschnieder]

So doesn't it all just simplifiy to the correlation function being the average (wrt a large number of pairs) of the product of the functions A and B?

C(a,b) = < A(a,λ) B(b,λ)> , then a probability distribution,

C(a,b) = ∫ ρ(λ) A(a,λ) B(b,λ) dλ, which is Bell's archetypal LR form (equation 2 in the 1964 paper), wrt which Bell's theorem proved that there's no ρ(λ) wrt which this form can produce a correlation coefficient that matches Malus Law.

So, for any value of ρ(λ), then

C(a,b) = ρ(λ) ∫ sign [cos²(a-λ)] sign [cos²(b-λ)] dλ ≠ cos²θ , where θ is the angular difference |a-b|

Hi TT,

I think sign[ cos²(a-x) ] is not a valid function since it is always +1.

 

[billschnieder]

It's not a big deal. Just let λ={θ,ζ,χ}, θ∈[0..2∏), ζ,χ∈[0..1), A(a,λ) = sign(cos²(θ−a)-ζ), B(b,λ) = sign(cos²(θ−b)-χ)

Those look like good functions but you have now split up λ into three variables. Now all you have to do is perform the integration of the product to obtain the classical result ¼+½(cos²(a-b)) which does not contain any of your 3 variables {θ,ζ,χ}.

Remember Bell's integration is: P(a,b) = ∫A(a,x)B(b,x)ρ(x)

 

[Gordon Watson]

Those look like good functions but you have now split up λ into three variables. Now all you have to do is perform the integration of the product to obtain the classical result ¼+½(cos²(a-b)) which does not contain any of your 3 variables {θ,ζ,χ}.

Remember Bell's integration is: P(a,b) = ∫A(a,x)B(b,x)ρ(x)

..

Bill, My sincere thanks for your understanding and these responses (most especially while I'm away, or on poor Net connections).

Then, to focus on the challenge here, I think it best to stay with the notation in the OP. (Or give sound reasons to change it.)

So we are seeking:

(1) E(AB) = ∫A(a,x)B(b,x)ρ(x) dx = (1/2) cos2(a, b).​

Then, since x is a random variable, it seems to me that all solutions must be based on a uniform distribution. So we have immediately:

(2) E(AB) = ρ(x) ∫A(a,x)B(b,x) dx = (1/2) cos2(a, b).​

So, by design: The options for the "non-localists" here are fairly limited.

PS: BUT they are exactly the adequate options that are available under Bell's (1964, etc.) local-realistic protocol (per OP).I trust you are seeing it that way too?

With my thanks, again,

Gordon

 

[Delta Kilo]

It's not a big deal. Just let λ={θ,ζ,χ}, θ∈[0..2∏), ζ,χ∈[0..1), A(a,λ) = sign(cos²(θ−a)-ζ), B(b,λ) = sign(cos²(θ−b)-χ)

Those look like good functions but you have now split up λ into three variables. Now all you have to do is perform the integration of the product to obtain the classical result ¼+½(cos²(a-b)) which does not contain any of your 3 variables {θ,ζ,χ}.

Remember Bell's integration is: P(a,b) = ∫A(a,x)B(b,x)ρ(x)

So what is the big deal again? According to Bell, λ can include absolutely everything except a and b, you just integrate over the whole thing.
E(a,b)=\int_\Omega A(a,\lambda) B(b,\lambda) \rho(\lambda) d\lambda = \int_0^{2\pi} \int_0^1 \int_0^1 sign( cos^2(\theta-a)-\xi) sign(cos^2(\theta-b)-\zeta) (\frac{1}{2\pi}) d\xi d\zeta d\theta = ... =cos^2 (b-a) - \frac{1}{2},
P(a,b) = \frac{E(a,b) + 1}{2}= \frac{1}{2} cos^2 (b-a) + \frac{1}{4}

 

[billschnieder]

So what is the big deal again? According to Bell, λ can include absolutely everything except a and b, you just integrate over the whole thing.
E(a,b)=\int_\Omega A(a,\lambda) B(b,\lambda) \rho(\lambda) d\lambda = \int_0^{2\pi} \int_0^1 \int_0^1 sign( cos^2(\theta-a)-\xi) sign(cos^2(\theta-b)-\zeta) (\frac{1}{2\pi}) d\xi d\zeta d\theta = ... =cos^2 (b-a) - \frac{1}{2},
P(a,b) = \frac{E(a,b) + 1}{2}= \frac{1}{2} cos^2 (b-a) + \frac{1}{4}

Your integral above is wrong! Are you trying to pull a fast one or something? Anybody else in doubt can check that:

\frac{1}{2 \pi} \int_0^{2\pi} \int_0^1 \int_0^1 \operatorname{sign} \left( \operatorname{cos}^{2} \left(a - x \right) - y \right) \operatorname{sign} <br /> \left( \operatorname{cos}^{2} \left( b - x \right) - z \right) dz dy dx<br />

= \operatorname{sign} \left( \operatorname{cos}^{2} \left( a \right) -1 \right) \operatorname{sign}\left( \operatorname{cos}^{2} \left( b \right) -1 \right)

I changed {θ,ζ,χ} to (x,y,z) to make it easier to format

 

[Delta Kilo]

Your integral above is wrong! Are you trying to pull a fast one or something? Anybody else in doubt can check that:

\frac{1}{2 \pi} \int_0^{2\pi} \int_0^1 \int_0^1 \operatorname{sign} \left( \operatorname{cos}^{2} \left(a - x \right) - y \right) \operatorname{sign} <br /> \left( \operatorname{cos}^{2} \left( b - x \right) - z \right) dz dy dx<br />

= \operatorname{sign} \left( \operatorname{cos}^{2} \left( a \right) -1 \right) \operatorname{sign}\left( \operatorname{cos}^{2} \left( b \right) -1 \right)

I changed {θ,ζ,χ} to (x,y,z) to make it easier to format

Is it really?

Consider \int_0^1 \operatorname{sign}(y-x) dx, where 0 \le y \le 1
\int_0^1 \operatorname{sign}(y-x) dx = \int_0^y \operatorname{sign}(y-x) dx + \int_y^1 \operatorname{sign}(y-x) dx = \int_0^y (+1) dx + \int_y^1 (-1) dx = 2y-1

Therefore
\frac{1}{2 \pi} \int_0^{2\pi} \int_0^1 \int_0^1 \operatorname{sign} \left( \operatorname{cos}^{2} \left(a - x \right) - y \right) \operatorname{sign} <br /> \left( \operatorname{cos}^{2} \left( b - x \right) - z \right) dz dy dx<br />
= \frac{1}{2 \pi} \int_0^{2\pi} \left[ \int_0^1 \operatorname{sign} \left( \operatorname{cos}^{2} \left(a - x \right) - y \right) dy \right] \left[\int_0^1 \operatorname{sign} <br /> \left( \operatorname{cos}^{2} \left( b - x \right) - z \right) dz \right] dx
= \frac{1}{2 \pi} \int_0^{2\pi} \left( 2 \operatorname{cos}^{2} \left(a - x \right) - 1 \right) \left( 2 \operatorname{cos}^{2} \left( b - x \right) - 1 \right) dx
= \frac{1}{2 \pi} \int_0^{2\pi} \operatorname{cos} \left(2a - 2x \right) \operatorname{cos} \left(2 b - 2x \right) dx
= \frac{1}{2 \pi} \int_0^{2\pi} \frac{1}{2} \left[ \operatorname{cos} \left(2a - 2b <br /> \right) + \operatorname{cos} \left(2a + 2 b - 4x \right) \right] dx
= \frac{1}{2} \operatorname{cos} \left(2a - 2b \right)+ 0
= \operatorname{cos}^2 \left(a - b \right)-\frac{1}{2}

 

[ThomasT]

Since there seems to be some confusion about the meaning of the thread title, I have a question ... for anybody.

Can the QM prediction (for either the Stern-Gerlach setup that Bell dealt with, or an optical Bell test setup) be reproduced by the form in Bell's equation 2 in his 1964 paper?

That that can't be done has, afaik, been definitively demonstrated and is one way of stating Bell's theorem (without any inference regarding nonlocality), and rules out Bell-type LR models of quantum entanglement.

So I'm wondering if that's what the classical challenge to Bell's Theorem is eventually going to be getting at. If not, then what, exactly, is the classical challenge to Bell's theorem that the title refers to?

 

[harrylin]

= [..] \left[ \int_0^1 \operatorname{sign} \left( \operatorname{cos}^{2} \left(a - x \right) - y \right) dy \right] [..]

= [..] \left( 1 - 2 \operatorname{cos}^{2} \left(a - x \right) \right) [..]

Delta Kilo can you clarify how one can get rid of the sign like that? Thanks in advance!

 

[Delta Kilo]

Delta Kilo can you clarify how one can get rid of the sign like that? Thanks in advance!

I thought I made it clear:

Consider \int_0^1 \operatorname{sign}(y-x) dx, where 0 \le y \le 1
\int_0^1 \operatorname{sign}(y-x) dx = \int_0^y \operatorname{sign}(y-x) dx + \int_y^1 \operatorname{sign}(y-x) dx = \int_0^y (+1) dx + \int_y^1 (-1) dx = 2y-1

EDIT: Oops, just realized I had the sign wrong in the above (I had 1-2y instead of 2y-1), but they cancel each other out so the end result is the same anyway.