A classical challenge to Bell's Theorem?

[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.
..

 

[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.

$E(AB)_V = \int dx \, \rho (x)AB$
$= \int dx \, \rho (x) \left [<br /> P({A^+}{B^+}|V) <br /> - P({A^+}{B^-}|V) <br /> - P({A^-}{B^+}|V) <br /> + P({A^-}{B^-}|V)<br /> \right ]$
$= \int dx \, \rho (x) \left [<br /> P(A^+|V)P(B^+|V,\,A^+) <br /> - P(A^+|V)P(B^-|V,\,A^+)<br /> - P(A^-|V)P(B^+|V,\,A^-)<br /> + P(A^-|V)P(B^+|V,\,A^-)<br /> \right ]$
$= \int dx \, \rho (x) \left [ <br /> P(A^+|V)\left [P(B^+|V,\,A^+) - P(B^-|V,\,A^+) \right ] <br /> - P(A^-|V)\left [P(B^+|V,\,A^-) - P(B^-|V,\,A^-) \right ]<br /> \right ]$
$= \int dx \, \rho (x) \frac{1}{2}\left [ <br /> P(B^+|V,\,A^+) - P(B^-|V,\,A^+) - P(B^+|V,\,A^-) + P(B^-|V,\,A^-) \right ]$
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:

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

 

[DrChinese]

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.

...and thus leaving the (future) relative angle settings of the observation devices as the only relevant quantities in the outcomes. We live in an observer dependent universe.

 

[Gordon Watson]

...and thus leaving the (future) relative angle settings of the observation devices as the only relevant quantities in the outcomes. We live in an observer dependent universe.

..
DrC, I would welcome and appreciate your expanding on the point that you seek to make. It is not at all clear to me. Thanks. GW
..

GENERAL NOTE TO READERS (in passing):

Where I write cos2(a, b) I mean cos[2(a, b)] and NOT cos²( a, b).​

..

 

[Gordon Watson]

I can verify that this is correct as follows:

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

Many thanks, Bill, much appreciated: with a small point for the future.

Note that a and b are detector orientations: often defined as unit-vectors in a 2-space orthogonal to the line-of-flight; or in 3-space.

The latter is important in considering the spherically-symmetric singlet-state.

So, in Bell-studies, the most general way to represent the angle between these vectors (in the argument of a trig-function) is (a, b).

PS: My apologies if this sounds like a lecture, rather than an explanation of what I do!

 

[billschnieder]

Many thanks, Bill, much appreciated: with a small point for the future.

Note that a and b are detector orientations: often defined as unit-vectors in a 2-space orthogonal to the line-of-flight; or in 3-space.

The latter is important in considering the spherically-symmetric singlet-state.

So, in Bell-studies, the most general way to represent the angle between these vectors (in the argument of a trig-function) is (a, b).

That's right.

 

[billschnieder]

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).

It appears you have classically reproduced the QM result E(AB) for the Aspect experiment. However, I'm not sure how you obtained $P(B^+|Y, A^+) = \cos^2(a,b)$

In other words, why is $P(B^+|Y, A^+) = \cos^2(a,b)$ for Aspect 2004 (Y), different from $P(B^+|W, A^+) = \frac{1}{2}cos^2(a,b) + \frac{1}{4}$ for the the classical case in the OP (W)? Thanks to Delta Kilo, we do have a locally causal derivation of the W case. Do you have a derivation of the Aspect case that is locally causal? Is it a straightforward application of Malus?

Also since
$P(B^+|W,\,{A^+}) + P(B^-|W,\,{A^+}) = 1, \; \; <br /> P(B^+|W,\,{A^-}) + P(B^-|W,\,{A^-}) = 1$
and $P(B^+|W,\,{A^+}) = P(B^-|W,\,{A^-})$

Your condition V-4 can be reduced to:

$E(AB)_V= \int dx \, \rho (x) \left [ 2 \cdot P(B^+|V,\,A^+) - 1 \right ]$

 

[DrChinese]

..
DrC, I would welcome and appreciate your expanding on the point that you seek to make. It is not at all clear to me. Thanks. GW

Why sure...

As you point out, all of the input parameters (initial conditions) essentially cancel out. That leaves the output parameters (essentially the observation conditions) as the only remaining variables in the equation, those being A and B in the cos^2() coincidence function. It is those settings and those alone that determine coincidence. Bell showed us that each A, B pairing is unique in the sense that others cannot simultaneously exist (at least in some combinations). Ergo our final results are uniquely dependent on the observer.

We live in an observer dependent universe. All of the major interpretations of QM agree on this point in some fashion: Copenhagen, MWI, dBB/BM, TS (Time Symmetric).

 

[Gordon Watson]

It appears you have classically reproduced the QM result E(AB) for the Aspect experiment. However, I'm not sure how you obtained $P(B^+|Y, A^+) = \cos^2(a,b)$

See below, noting that I've changed the order of your questions.

Thanks to Delta Kilo, we do have a locally causal derivation of the W case. Do you have a derivation of the Aspect case that is locally causal? Is it a straightforward application of Malus?

Thanks indeed to Delta Kilo! It being understood that the classical analysis essentially proceeds on the basis that there is at least one physically significant formulation of Bell's functions A(a, λ) = ±1 and B(b, λ) = ±1.

The derivation in both W and Y is locally causal to the same (and essential) extent that Bell's protocol (see OP) is locally causal. That is, we capture Einstein-locality (an essentially classical concept) via Bell's functions A(a, λ) = ±1 and B(b, λ) = ±1: which we are happy to restrict to classical functions, in keeping with our classical analysis.

The application of Malus is straight-forward, bearing in mind that Malus examined the results of "one-sided" experiments and gave us his famous classical Malus' Law (ML) -- see below. He did not have double-sided experiments (involving Alice and Bob), but we easily follow him by examining and generalising Alice and Bob's classical results (+1 and -1 and their correlation) just as Malus did with his own classical results.

In other words, why is $P(B^+|Y, A^+) = \cos^2(a,b)$ for Aspect 2004 (Y), different from $P(B^+|W, A^+) = \frac{1}{2}cos^2(a,b) + \frac{1}{4}$ for the the classical case in the OP (W)?

Since the sources in W and Y differ, we allow that the HVs differ too: Let ∅ be the pair-wise common HV in W (i.e., ∅ is the linear polarisation; replacing x); let λ (replacing x) be the pair-wise common HV in Y (where, following Bell, we allow that the particles are unpolarised); let s denote the relevant intrinsic spin; let δ_a∅ → a denote the interaction of a particle (its HV ∅) with a test-device oriented a such that the result is the transition ∅ → a, etc. Then, with a little study, and noting that s = 1 in W and Y:

ML: P(δ_b∅ → b|W, ∅, s) = cos²[s(b, ∅)]; etc.


ML-W: $P(B^+|W, A^+)$ =

(ML-W1) P(δ_b∅ → b|W, ∅, s, δ_a∅ → a) =

(ML-W2) [P(δ_b∅ → b, δ_a∅ → a|W, ∅, s)]/[P(δ_a∅ → a|W, ∅, s)] =

(ML-W3) 2∫d∅ ρ(∅){cos²[s(b, ∅)]}{cos²[s(a, ∅)]} =

(ML-W4) (1/2) cos²(a, b) + 1/4.


ML-Y: $P(B^+|Y, A^+)$ =

(ML-Y1) P(δ_bλ → b|Y, λ, s, δ_aλ → a) = cos²[s(b, a)].

Each Malus Law arises from the classical analysis of classical outcomes in experiments. The differing results for W and Y that you ask about arise because of the differing sources: all else being the same. (You have already checked the W result, so you should be able to discern ML-W in play. You can check Aspect 2004 to see ML-Y in play; it falls out of the experimental results, essentially by observation.)

Also since
$P(B^+|W,\,{A^+}) + P(B^-|W,\,{A^+}) = 1, \; \; <br /> P(B^+|W,\,{A^-}) + P(B^-|W,\,{A^-}) = 1$
and $P(B^+|W,\,{A^+}) = P(B^-|W,\,{A^-})$

Your condition V-4 can be reduced to:

$E(AB)_V= \int dx \, \rho (x) \left [ 2 \cdot P(B^+|V,\,A^+) - 1 \right ]$

Correct, with many similarly instructive re-workings. For example: The product AB can only take values from the set {+1, -1}, so we need only assess the probabilities for these two AB values.

However, imho, we need to maintain the expository value of the version given in (V-1) - (V-4) above (at POST #100) ... perhaps with re-formatted maths. For it is easy to lose sight of the respective Einstein-local outcomes (+1, -1) and their origin in Bell's A(a, λ) = ±1 and B(b, λ) = ±1.

Note that, for EPRB (say condition Z; s = 1/2):

$E(AB)_Z= - \int dx \, \rho (x) \left [ 2 \cdot P(B^+|Z,\,A^+) - 1 \right ]$,

since B(b, λ) = - A(b, λ) ... after Bell (1964).

PS: I appreciate your engagement with the classical maths here, and am happy to rely on you (alone, it seems) spotting any errors. DrC has a comment that I've yet to study. If he is saying that Einstein-locality holds, then he and I agree. :!) Which would be nice!

How does all this, including DrC's comment, sit with your own views?

In conclusion: For me, maths is the best logic: so I'd like to reduce any disagreements to maths. The important point here being that every relevant element of the physical reality is included in the equations.

As the Accountants say: E. and O.E. (With apologies: I'm still way from my office, on a borrowed mini-screen computer.)

 

[billschnieder]

We live in an observer dependent universe. All of the major interpretations of QM agree on this point in some fashion: Copenhagen, MWI, dBB/BM, TS (Time Symmetric).

DrC et al,
Gordon has now shown that he can obtain classically the result that agrees with QM for the Aspect experiment in a local-realistic manner, which implies a violation of Bell's inequalities and a refutation of Bell's theorem. I'm surprised this is all you have to say in return. Besides, it is not clear to me how what you say above is relevant to the issue here especially since you already agreed earlier that the OP experiment was entirely classical and local realistic and yet E(AB) contains only the angular settings. Thus, the fact that E(AB) for any experiment contains only the angular settings means squat as far as locality or realism is concerned, no?

Do you see any problem in his analysis?

 

[bohm2]

Gordon has now shown that he can obtain classically the result that agrees with QM for the Aspect experiment in a local-realistic manner, which implies a violation of Bell's inequalities and a refutation of Bell's theorem.

Given the recent PBR theorem (see 2 of many links below), isn’t non-locality implied by any “realistic” model by PBR theorem itself, irrespective of Bell’s. For instance Leifer writes:

It (PBR) provides a simple proof of many other known theorems, and it supercharges the EPR argument, converting it into a rigorous proof of nonlocality that has the same status as Bell’s theorem...Nevertheless, the PBR result now gives an arguably simpler route to the same conclusion by ruling out psi-epistemic theories, allowing us to infer nonlocality directly from EPR.

The quantum state cannot be interpreted statistically
http://lanl.arxiv.org/abs/1111.3328

Quantum Times Article on the PBR Theorem
http://mattleifer.info/2012/02/26/quantum-times-article-on-the-pbr-theorem/

 

[DrChinese]

DrC et al,
Gordon has now shown that he can obtain classically the result that agrees with QM for the Aspect experiment in a local-realistic manner, which implies a violation of Bell's inequalities and a refutation of Bell's theorem. I'm surprised this is all you have to say in return. Besides, it is not clear to me how what you say above is relevant to the issue here especially since you already agreed earlier that the OP experiment was entirely classical and local realistic and yet E(AB) contains only the angular settings. Thus, the fact that E(AB) for any experiment contains only the angular settings means squat as far as locality or realism is concerned, no?

It is really impossible at this point to comment, other that to repeat what has already been stated: his classical thought experiment yields classical probabilities (I gave those) that don't violate a Bell Inequality. Unless there is something serious offered forward, I do not plan to comment further on the example itself as I have indicated.

In cases where a Bell Inequality does not matter (as in a classical example), the observer angular settings would not imply failure of realism (observer independence).

Your pushing the idea that Gordon has disproved Bell here is particularly sad. I consider your comments either born of insincerity or ignorance or perhaps a strange form of humor, really cannot figure out what you are trying to do here.

Unless there is a new question or example put forth, I personally think this thread has reached the end of its useful life.

 

[billschnieder]

It is really impossible at this point to comment, other that to repeat what has already been stated: his classical thought experiment yields classical probabilities (I gave those) that don't violate a Bell Inequality.

Sure, we agree about that. I'm referring to his reproduction of the QM expectation value in post #102. Maybe you missed it. Check it out.

Your pushing the idea that Gordon has disproved Bell here is particularly sad.

That is strange given that all I'm asking is for someone else to verify the math in post #102 to make sure it is correct because I did not find any errors in it, and it appeared to reproduce the QM expectation classically. So check and explain why you think it is wrong *if* you think it is wrong.

Unless there is a new question or example put forth, I personally think this thread has reached the end of its useful life.

If the thread is no longer useful for you, feel free to not read it anymore. Clearly there are issues remaining to be clarified for those still participating. I assumed that if you believe Bell's theorem, you would be alarmed anytime the QM expectation values for the EPRB are reproduced in a local-realistic manner, and would seize the opportunity to point out where the error is if there was an error at all! But instead, you start dropping hints/suggestions that the thread should be closed. I wonder why ...

I consider your comments either born of insincerity or ignorance or perhaps a strange form of humor, really cannot figure out what you are trying to do here.

right back at ya!

 

[Gordon Watson]

It is really impossible at this point to comment, other that to repeat what has already been stated: his classical thought experiment yields classical probabilities (I gave those) that don't violate a Bell Inequality. Unless there is something serious offered forward, I do not plan to comment further on the example itself as I have indicated.

In cases where a Bell Inequality does not matter (as in a classical example), the observer angular settings would not imply failure of realism (observer independence).

Your pushing the idea that Gordon has disproved Bell here is particularly sad. I consider your comments either born of insincerity or ignorance or perhaps a strange form of humor, really cannot figure out what you are trying to do here.

Unless there is a new question or example put forth, I personally think this thread has reached the end of its useful life.

DrC, I am hoping that these questions might bring you back to this thread, for it is not yet clear (to me) the critical point that you are making:

Could you clarify your point by comparing how your criticism applies to the OP case and how it applies to Aspect 2004, please?

The similarity in the equivalent equations leads me to stick to my view that Einstein-locality rules OK. On what basis are you opposing that position, please?

 

[Gordon Watson]

Sure, we agree about that. I'm referring to his reproduction of the QM expectation value in post #102. Maybe you missed it. Check it out.That is strange given that all I'm asking is for someone else to verify the math in post #102 to make sure it is correct because I did not find any errors in it, and it appeared to reproduce the QM expectation classically. So check and explain why you think it is wrong *if* you think it is wrong.If the thread is no longer useful for you, feel free to not read it anymore. Clearly there are issues remaining to be clarified for those still participating. I assumed that if you believe Bell's theorem, you would be alarmed anytime the QM expectation values for the EPRB are reproduced in a local-realistic manner, and would seize the opportunity to point out where the error is if there was an error at all! But instead, you start dropping hints/suggestions that the thread should be closed. I wonder why ...
right back at ya!

..
Bill and DrC, I would be pleased if your ancient antagonisms do not disrupt the train of this thread.

Bill, I am still unsure of the case that you have long been making against DrC's position*** re locality and realism. So, if my stuff here helps you sharpen your view, maybe you should open another thread? Which, of course, does not mean that you desert this thread: just that you stay on focus.

DrC, please provide the comparison requested earlier. I truly want to understand (and hopefully respond to) the point that you are making. Are you just saying that our interaction with Nature influences the future trend of of events? Or just that reality is veiled from us? For I see the possibility of us being close to a serious point of agreement (along such lines). Or else I'm displaying how much I do not understand your point? Thanks in advance.

***EDIT: I am not saying that I understand DrC's position. I may be wrong in thinking that he has moved from being a non-localist (seeing some of his recent discussions with ttn); maybe he never was. Rather, I'm hoping that seeing your position will make both positions clearer to me. THAT I would welcome and appreciate. Thanks.
..

 

[DrChinese]

DrC, I am hoping that thes question might bring you back to this thread, for it is not yet clear (to me) the critical point that you are making:

Gordon, it is not appropriate to take a classical example, run a few formulas and say "Voila! Bell is wrong." I have already provided the math to show you your example is classical and does not violate Bell. I believe zonde and Mark M both showed the same thing. Sadly, billschnieder is using you in some strange way and is egging you on. I do not know his reason, but again I implore you to go back to ground zero in learning about the math of Bell.

The idea of this thread - see title - is absurd. You have never done anything so far to show otherwise despite the time I have taken to assist you. Which is why I don't think further discussion here is warranted.

And I respect de Raedt and Michielsen too much to recommend you send your classical example to them so they can analyze your ground-breaking work. So perhaps you might send it to Joy Christian instead.

 

[Gordon Watson]

Gordon, it is not appropriate to take a classical example, run a few formulas and say "Voila! Bell is wrong." I have already provided the math to show you your example is classical and does not violate Bell. I believe zonde and Mark M both showed the same thing. Sadly, billschnieder is using you in some strange way and is egging you on. I do not know his reason, but again I implore you to go back to ground zero in learning about the math of Bell.

The idea of this thread - see title - is absurd. You have never done anything so far to show otherwise despite the time I have taken to assist you. Which is why I don't think further discussion here is warranted.

And I respect de Raedt and Michielsen too much to recommend you send your classical example to them so they can analyze your ground-breaking work. So perhaps you might send it to Joy Christian instead.

..
If our posts have crossed; pleased reconsider my requests. Thanks.