A classical challenge to Bell's Theorem?

[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 $$A(a,x)=\pm 1$$ $$B(b,x)=\pm 1$$ and$$E(A,B)= \int AB p(x) dx$$ 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:

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

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

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:

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

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

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

 

[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.

 

[harrylin]

I thought I made it clear [..]

Ah yes, in fact you did make it clear, thanks - I just hadn't looked at that part, as it appeared to refer to the line before.

 

[billschnieder]

Is it really?

Consider $\int_0^1 \operatorname{sign}(y-x) dx$, where $0 \le y \le 1$

I've tried this several times and I do not get your results. I get every time:

$\int_0^1 \operatorname{sign}(y-x) dx = sign(y-1)$

Please explain:
$\int_0^y \operatorname{sign}(y-x) dx = \int_0^y (+1) dx$
and
$\int_y^1 \operatorname{sign}(y-x)dx = \int_y^1 (-1) dx$

EDIT:
Now I see, for y < x, sign(x-y) = +1 and for y > x, sign(x-y) = -1.

It would appear therefore that sign(y-1) = 2y -1, which is not obvious !? Something seems to be off somewhere.

 

[mattt]

I've tried this several times and I do not get your results. I get every time:

$\int_0^1 \operatorname{sign}(y-x) dx = sign(y-1)$

Can't you see that THAT result is impossible? Do you understand what $$\int_0^1 sign(y-x)dx$$ means?

 

[mattt]

$$\int_0^1sign(y-x)dx$$ takes ALL the values of the interval $$[-1,1]$$ depending on the value of $$y\in[0,1]$$ whereas $$sign(y-1)$$ only takes discrete values.

 

[Gordon Watson]

I apologise for my recent interrupted communications here.

I am currently Out-Back Down-Under in Australia.

My attempts to minimise frustrations by answering all questions promptly has back-fired due to technical difficulties with my Net access.

I will keep trying but it may be a week to 10 days before I'm back to a normal Net connection.

GW

 

[billschnieder]

I thought I made it clear: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.

I should apologize to Delta Kilo for prematurely judging his integral when in fact he was correct and I was wrong.
I guess it is up to Gordon now to say what next, as it appears his challenge has been solved. One other thing to note is that even the sign function has three values (-1,0,+1).

 

[Gordon Watson]

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

Q1: 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.

Q2: So I'm wondering if that's what the classical challenge to Bell's Theorem is eventually going to be getting at.

Q3: If not, then what, exactly, is the classical challenge to Bell's theorem that the title refers to?

Tom, apologies for my last aborted attempt to get a comprehensive answer posted here. My current Outback Downunder Net connection is unreliable and crashes without warning. I will attempt short sharp replies until I get back to my office.

I've edited your post by clearly identifying the three (3) questions that I'll be addressing.

...

Tom, Q1: 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.


GW, A1: Yes; there is one such peer-reviewed paper that I know of but its discussion under this thread may way-lay us.

I am not aware of any proof that it cannot be done. My only personal requirement is that the starting point for any such demonstration must be Einstein-locality. I then take the view that any "problems" with such a theory will be found in any other assumptions: which will generally, if not necessarily, relate to misleading (defective, confused) concepts of realism.

I personally do not endorse EPR elements of physical reality. In my view, confusion arises in interpreting the nature of the "correspondence" that they introduce in their definition: in conjunction with the timing of their "then there exists".

More soon.
..

 

[Gordon Watson]

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

Q1: 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.

Q2: So I'm wondering if that's what the classical challenge to Bell's Theorem is eventually going to be getting at.

Q3: If not, then what, exactly, is the classical challenge to Bell's theorem that the title refers to?

Tom, Q2: So I'm wondering if that's what the classical challenge to Bell's Theorem is eventually going to be getting at?

GW, A2: In my view, Einstein-locality is properly represented in Bell's (1964) A and B functions. So the first part of the classical challenge was to apply Bell's (1964) protocol to the (classical) Einstein-local experiment in the OP. I was keen to see how physicists etc., would deliver physically significant functions satisfying Bell's A and B which would deliver the correct E(AB).

So far Delta Kilo's submission is the only candidate aiming to answer this first part. And though I only need one example (from a physicist) to proceed with the classical analysis, I was hoping for more. However, with Delta Kilo's example, I have no need to produce my own: in the hope that other submissions might still come through, especially from physicists who believe in non-locality.

The crucial point is that we accept that there ARE physically significant functions satisfying Bell's A and B; functions that satisfy physicists who believe in non-locality. We do not want to get the end of the analysis and then start arguing about that point!

Tom, Q3: If not, then what, exactly, is the classical challenge to Bell's theorem that the title refers to?[/QUOTE]

Tom, A3: The final component of the challenge is to locate any error in my classical reasoning as it applies to "Herbert's proof" where I reason from the classical OP to experiments such as Herbert's, though I prefer to do that via well-known experiments with entangled particles; e.g., EPRB (Bell 1964) or Aspect (2000). That is the source of the question-mark in the title.

 

[Gordon Watson]

I should apologize to Delta Kilo for prematurely judging his integral when in fact he was correct and I was wrong.
I guess it is up to Gordon now to say what next, as it appears his challenge has been solved. One other thing to note is that even the sign function has three values (-1,0,+1).

Thanks for acknowledging Delta Kilo's effort, which I thought was pretty good! It is certainly the best so far! And it serves as a reasonable basis from which we can move ahead.

However, the sign-function, as you rightly say, can take zero (0) as a value: when the goal is A = ±1 only; B = ±1 only.

I was hoping that between the two of you (as reasonable physicists), we might get something beyond any shadow-of-a-doubt.

The important point is that neither of you are saying that A and B cannot exist.

And I know that such a function does exist, BUT I'm hoping no one puts my particular function here until more attempts come in.

I had hoped DrC and ttn would have made submissions. I do STILL want to see how physicists answer that part of the challenge!***

For I do NOT want to arrive at the end of my analysis: to find that the existence of a valid Bellian A and B is argued or denied, especially by those who believe in non-locality.

I hope to move to a more stable Net connection soon (as I begin the return trip to my office) so that I can map out the maths of my analysis of the OP via Bell's protocol. That's what is next (from my point of view), and that is what I'll deliver, DV.

***PS: I had hoped that many valid submissions would pour in while I was away from my office. My apologies for any consequent delay and confusion caused by my so being. (I had thought that Bell's supporters would be keen to show (as I know) that there is NO defect in Bell's A and B formulation of Einstein-locality. BUT it does take a little thought, and some classical know-how, I guess.)
..

 

[Gordon Watson]

Continuing:

1. We accept (for now; see next, #2) that there are physically significant functions that meet Bell's (1964) requirements: A(a, x) = ±1, B(b, x) = ±1, x being substituted for Bell's λ.

2. Hoping for such functions to be submitted (in the context of the OP) by physicists (and others), but especially by non-localists, we hold back on offering our own such functions so as not to influence (and thereby reduce), further possible offerings. [PS: Our offerings are now known to many so we again request, for the given reason, that they not be posted here just yet.] NB: We are able to proceed satisfactorily without any of them: despite our continuing interest in all of them.

3. To facilitate analysis and critique, we designate A = +1 by A+ when convenient; etc; and show Alice's outputs thus {1} or {-1}; Bob's outputs similarly (but in bold) thus {1} or {-1}. This format is designed to facilitate tracking of the respective Alice/Bob outputs: which are totally consistent with Einstein-locality, being generated by particle-device interactions that are space-like separated.

4. Our analysis will be wholly CLASSICAL throughout as we seek to learn of any errors (including typos) or confusions: and correct them. P denotes a classical probability; NO quantum-logic, negative-probabilities, etc., are involved here.

5. Designating the general conditions of common Bell-tests on two-correlated particles by V, we are interested in the general application of Bell's (1964) protocol thus:

(V-1) E(AB)_V = ∫dx ρ(x) AB =

(V-2) ∫dx ρ(x) ([P(A+|V){1}][P(B+|V, A+){1} + P(B-|V, A+){-1}] + [P(A-|V){-1}][P(B+|V, A-){1} + P(B-|V, A-){-1}]) =

(V-3) ∫dx ρ(x) [P(A+|V).P(B+|V, A+) - P(A+|V).P(B-|V, A+) - P(A-|V).P(B+|V, A-) + P(A-|V).P(B-|V, A-)] =

(V-4) ∫dx ρ(x) [P(B+|V, A+) - P(B-|V, A+) - P(B+|V, A-) + P(B-|V, A-)]/2;

since x is a random variable:

(V-5) P(A+|V) = P(A-|V) = 1/2.

To be continued.
..

 

[Gordon Watson]

Continuing:

6. Designating the general conditions of the classical ("Bell-style") tests on the correlated particles in the OP by W, we are interested in the general application of Bell's (1964) protocol [from (V-1) - (V-5)] thus:

(W-1) E(AB)_W = ∫dx ρ(x) AB =

(W-2) ∫dx ρ(x) [P(B+|W, A+) - P(B-|W, A+) - P(B+|W, A-) + P(B-|W, A-)]/2 =

(W-3) ∫dx ρ(x) [(cos2(a, b)+ 2) - (- cos2(a, b)+ 2) - (- cos2(a, b)+ 2) + (cos2(a, b)+ 2)]/8 = (1/2) cos2(a, b):

QED; we have derived the correct answer for classical experiment W (from the OP): using classical principles (including the classical Malus Law), and in agreement with Bell that ∫dx ρ(x) = 1.

To be continued: but please report any errors or confusions, etc., in the interim.
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[Gordon Watson]

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?

...

Aspect (2004) -- http://arxiv.org/abs/quant-ph/0402001 -- represents an optical Bell setup. Designating its general conditions by Y, and applying Bell's protocol as part of the form that you request, we proceed from (V-4) above as follows:

(Y-1) E(AB)_{Aspect (2004)} = E(AB)_Y =

(Y-2) ∫dx ρ(x) [P(B+|Y, A+) - P(B-|Y, A+) - P(B+|Y, A-) + P(B-|Y, A-)]/2 =

(Y-3) ∫dx ρ(x) [cos²(a, b) - sin²(a, b) - sin²(a, b) + cos²(a, b)]/2 = cos2(a, b):

which is the correct result for Aspect (2004); see his equation (6).

This is not a new result, but we can now compare this quantum result (Y-3) with the classical result (W-3) above. And since the maths is straight-forward, we can focus on that maths and minimise the words, reducing the wordy discussions that seem to go nowhere.

Note that (in words) the maths here embraces causal independence (result A has no influence on B; nor vice-versa; = Einstein-locality) with logical dependence (result A logically tells us something about result B; because particles are emitted from the source pair-wise correlated).

This result needs to be critiqued with the use here of Bell's protocol and the specific identification (and tracking) of the Einstein-local outputs {1}, {-1}, etc. Recent posts at PF, such as https://www.physicsforums.com/showpost.php?p=3874539&postcount=369, show that much debate continues on the subject.

 

[Gordon Watson]

ThomasT, I believe the above goes to the heart of your observation (and concern) that the HVs delivered the random outcomes at the Alice and Bob detectors BUT seemed to disappear when it came to the correlations.

Re the outcomes: Since the HVs are generally unknown random variables (say x), and the Alice and Bob outcomes are A(a, x) and B(b, x), your point about the random outcomes is mathematically confirmed.

Re the correlations: Since the HVs are generally unknown random variables, it figures that correlations must arise from other sources (since "random" is hardly the sort of correlation we are looking for). But the A and B outcomes (from particle-device interactions), knock the HVs into shape (as it were), and so the HVs may be eliminated from the correlation functions (as we see). Thus the correlations, deriving from the distribution of the outcomes, are independent of the HVs.

BUT NB: To put it another way: The HVs can be used to derive the distributions, from which they disappear (as you should expect).

PS: THis is possibly garbled as I rush to capture what I want to say. Will save and edit later due Net problem here.

 

[Gordon Watson]

I found it easier to follow after I recast it in the following form; just makes it easier to follow without too much effort.

$E(AB)_V = \int dx \, \rho (x)AB$
$= \int dx \, \rho (x) \left [ P({A^+}{B^+}|V) - P({A^+}{B^-}|V) - P({A^-}{B^+}|V) + P({A^-}{B^-}|V) \right ]$
$= \int dx \, \rho (x) \left [ P(A^+|V)P(B^+|V,\,A^+) - P(A^+|V)P(B^-|V,\,A^+) - P(A^-|V)P(B^+|V,\,A^-) + P(A^-|V)P(B^+|V,\,A^-) \right ]$
$= \int dx \, \rho (x) \left [ P(A^+|V)\left [P(B^+|V,\,A^+) - P(B^-|V,\,A^+) \right ] - P(A^-|V)\left [P(B^+|V,\,A^-) - P(B^-|V,\,A^-) \right ] \right ]$
$= \int dx \, \rho (x) \frac{1}{2}\left [ P(B^+|V,\,A^+) - P(B^-|V,\,A^+) - P(B^+|V,\,A^-) + P(B^-|V,\,A^-) \right ]$
since for random variables
$P(A^+|V) = P(A^-|V) = \frac{1}{2}$

A kindly reader offers the above!

With thanks,
Gordon.
..

 

[billschnieder]

Continuing:

6. Designating the general conditions of the classical ("Bell-style") tests on the correlated particles in the OP by W, we are interested in the general application of Bell's (1964) protocol [from (V-1) - (V-5)] thus:

(W-1) E(AB)_W = ∫dx ρ(x) AB =

(W-2) ∫dx ρ(x) [P(B+|W, A+) - P(B-|W, A+) - P(B+|W, A-) + P(B-|W, A-)]/2 =

(W-3) ∫dx ρ(x) [(cos2(a, b)+ 2) - (- cos2(a, b)+ 2) - (- cos2(a, b)+ 2) + (cos2(a, b)+ 2)]/8 = (1/2) cos2(a, b):

QED; we have derived the correct answer for classical experiment W (from the OP): using classical principles (including the classical Malus Law), and in agreement with Bell that ∫dx ρ(x) = 1.

To be continued: but please report any errors or confusions, etc., in the interim.
..

I can verify that this is correct as follows:

$P(B^+|W,\,{A^+}) + P(B^-|W,\,{A^+}) = 1, \; \; P(B^+|W,\,{A^-}) + P(B^-|W,\,{A^-}) = 1$
$P(B^+|W,\,{A^+}) = P(_B^-|W,\,{A^-}) = \frac{1}{2}cos^2(a-b) + \frac{1}{4}$
therefore
$P(B^-|W,\,{A^+}) = P(B^+|W,\,{A^-}) = -\frac{1}{2}cos^2(a-b) + \frac{3}{4}$
from V-4:
$E(AB)_w = \int dx \, \rho (x) \frac{1}{2}\left [ P(B^+|W,\,{A^+}) - P(B^-|W,\,{A^+}) - P(B^+|W,\,{A^-}) + P(B^-|W,\,{A^-}) \right ]$
$= \int dx \, \rho (x) \frac{1}{8}\left [ \left [2cos^2(a-b) + 1 \right ] - \left [-2cos^2(a-b) + 3 \right ] - \left [-2cos^2(a-b) + 3 \right ] +\left [2cos^2(a-b) + 1 \right ] \right ]$
$= \int dx \, \rho (x)\left [ cos^2(a-b) - \frac{1}{2} \right ] = cos^2(a-b) - \frac{1}{2} = \frac{1}{2}cos(2\theta), \;\;\; for \; \theta = a - b$