A classical challenge to Bell's Theorem?

[DrChinese]

BTW I do not doubt that Kristel and Hans might have spent a lot of their valuable time with you. Although I do doubt that, that time was spent on, let-alone winning, the "DrC challenge."

No, I did not write to them and ask them to take the DrChinese challenge. And there is no real winning or losing of the challenge anyway. The point of the exercise is to force everyone to a point where we strip away the words and focus on key elements that we can agree on.

For example, any model that cannot make counterfactual predictions should not, in my opinion, be called realistic. But if YOU define it as realistic, but we agree there are no counterfactual predictions, then we have accomplished something. Then it is up to each person to label as they see fit.

I was interested for the purpose of understanding if or how any algorithm could even begin to accomplish violation of a Bell inequality.

 

[Gordon Watson]

GW Statement 1: I derive the results for both W (the classical OP experiment) and Y (the well-known Aspect (2004) experiment) in a classical way.

Delta Kilo Response 1: No, you don't. You did not provide classical derivation for Y. Instead you just "borrowed" the result from Aspect paper. Aspect makes it very clear eq(6) was derived using QM rather than classical model.

..
NB: I have edited the quotations from "past-posts" here for clarity in this post. No meanings have been changed; no new data added. GW
..

Dear Delta Kilo, there was no "borrowing" from Aspect (2004). I simply "took it" as an example. It was the ideal example because it's on-line AND because of the very point that you make: "Aspect makes it very clear eq(6) was derived using QM." Moreover, my classical analysis pre-dates the original Aspect (2000).

Further: There is no implication anywhere in my writings that Aspect or Bell used a classical model to derive the QM result.

Further: Contrary to your bald claim, "No, you don't," I DO derive the results for both W (the classical OP experiment) and Y (the well-known Aspect (2004) experiment) in a classical way!

This was explained in my reference to Malus, which I believe is central to addressing some of your concerns: https://www.physicsforums.com/showpost.php?p=3879566&postcount=112

Also see posts leading to, and including: https://www.physicsforums.com/showpost.php?p=3874480&postcount=102

With more to follow on the above matters, and as required,

GW

GW Statement 2: Analytically, via my way: Going the whole-way (100%, say, with Y) is as easy as going half-way (50%, with W).

Delta Kilo Response 2: No, it isn't. There is a big difference: one satisfies Bell's inequality, another violates it.

Dear Delta Kilo, you again make a bald statement that is contrary to facts: "No, it isn't."

Since I did the ANALYSES, I have some right to assess which was the easiest ANALYSIS for me. I assess Y to be ANALYTICALLY easier than W: because W requires some integration whereas the Y result falls out without such (i.e., from observation)!

For example (using Bill's generalised short-cut: https://www.physicsforums.com/showpost.php?p=3878616&postcount=110; recalling that V is the general conditional for the style of experiments that we are analysing):

$E(AB)_{Aspect(2004)} = E(AB)_Y = \int d\lambda \rho (\lambda ) (AB)_Y = \int d\lambda \rho (\lambda )A(\textbf{a}, \lambda )B(\textbf{b}, \lambda )_Y$

$= \int d\lambda \rho (\lambda )[ 2 \cdot P(B^+|Y,\,A^+) - 1] = \int d\lambda \rho (\lambda)[2⋅cos^2(\textbf{a}, \textbf{b}) - 1]$

$= cos[2(\textbf{a}, \textbf{b})] = QED!$


This result may be compared to the classical OP experiment W, Y's classical equivalent (as previously defined) with some integration not included to facilitate comparisons:

$E(AB)_{OP} =(AB)_W = \int d\phi \rho (\phi ) (AB)_W = \int d\phi \rho (\phi )A(\textbf{a}, \phi )B(\textbf{b}, \phi )_W$

$=\int d\phi \rho (\phi )[ 2 \cdot P(B^+|W,\,A^+) - 1] = \int d\phi \rho (\phi)[(1/2)⋅cos[2(\textbf{a}, \textbf{b})] + 1 - 1]$

$= (1/2)⋅cos[2(\textbf{a}, \textbf{b})] = QED!$
...


$E(AB)_{Bell (1964)} = E(AB)_Z = \int d\lambda \rho (\lambda ) (AB)_Z = \int d\lambda \rho (\lambda )A(\textbf{a}, \lambda )B(\textbf{b}, \lambda' )_Z = - \int d\lambda \rho (\lambda )A(\textbf{a}, \lambda )B(\textbf{b}, \lambda )_Z$

$= -\int d\lambda \rho (\lambda )[ 2 \cdot P(B^+|Z,\,A^+) - 1] = -\int d\lambda \rho (\lambda)[2⋅cos^2[(\textbf{a}, \textbf{b})/2] - 1]$

$= -cos(\textbf{a}, \textbf{b}) = - \textbf{a.b} = QED!$

PS: Since, in Z, λ' = -λ (from λ + λ' = 0; the pair-wise conservation of angular momentum). That is: The minus-sign is physically significant in the spin-half EPRB case (s = 1/2) because λ and λ' are pair-wise detectably anti-parallel. This is not significant in the spin-one Aspect case (s = 1) because λ and λ' are pair-wise indetectably anti-parallel.

These differences are made clear when the classical challenge in the OP is addressed. For it is then seen that, for ALL the subject experiments W, X, Y and Z, Bell's A and B contain cos[2s(a, x)] or cos[2s(b, x)], respectively: where x is the relevant hidden-variable $\phi$ or λ, and s is the intrinsic spin.

This EPRB-Bell (1964) result may be compared to classical experiment X, Z's classical equivalent (as previously defined) with some integration not included to facilitate comparisons:

$E(AB)_X = \int d\phi \rho (\phi ) (AB)_X = \int d\phi \rho (\phi )A(\textbf{a}, \phi )B(\textbf{b}, \phi' )_W = - \int d\phi \rho (\phi )A(\textbf{a}, \phi )B(\textbf{b}, \phi )_W$

$= - \int d\phi \rho (\phi )[ 2 \cdot P(B^+|W,\,A^+) - 1] = - \int d\phi \rho (\phi)[(1/2)⋅cos(\textbf{a}, \textbf{b}) + 1 - 1]$

$= - (1/2)⋅cos(\textbf{a}, \textbf{b}) = - \textbf{a.b}/2 = QED!$
...

Note in passing:

1. Einstein-locality is maintained through every step of the analysis.

2. The simple classical sources in W and X, delivering particles with pair-wise minimal correlations (with 1 part common orientation over 2-space) deliver one-half the correlation of the quantum sources in Y and Z. Yet these latter sources deliver particles which are pair-wise MUCH MORE highly correlated (over an infinity of common orientations in 3-space).
...

With more to follow, and as required,

GW
..

 

[lugita15]

De Raedt et al created a computer simulation which violates a Bell Inequality (winning the DrC challenge in the process) but still failing to violate Bell's Theorem (since it no longer matches the predictions of QM).

DrChinese, this seems to be interpreting Bell's theorem too broadly. Bell's theorem says nothing about whether local hidden variable theories can reproduce, say, the energy spectrum of the hydrogen atom; it only discusses whether theories can reproduce the specific correlations QM predicts for entangled particles. If a theory were to break the Bell inequality fair and square, Bell's theorem would put no further barriers to such a theory matching any other predictions of QM. So in judging a "challenge" to Bell's theorem, it seems to me that we should only focus on whether and how the model violates the Bell inequality. And in the case of de Raedt, all you need to say is that it exploits one of the experimental loopholes of currently practical Bell tests, and is thus not a valid counterexample to Bell's theorem, which after all is a rigorously proven theoretical result.

 

[billschnieder]

If you look at the simulation, you can vary the size of the window. This is only the outcomes that are "visible". Since the model is realistic, we can also display the full universe (which of course never matches the QM expectation, respecting Bell).

This makes no sense. There is no such thing as full universe. Only the outcomes matter for Bell or QM.

 

[lugita15]

This makes no sense. There is no such thing as full universe. Only the outcomes matter for Bell or QM.

The "full universe" issue you're talking about concerns the existence of counterfactual outcomes. But the "full universe" issue that DrChinese is discussing in regard to de Raedt's model is that it exploits the fair sampling loophole: the model only reproduces the predictions of QM if we take a small coincidental detection window, but if we had better experiments that would detect ALL entangled pairs emitted by the source, then de Raedt's model would be in stark disagreement with the predictions of QM.

 

[gill1109]

De Raedt et al. use the coincidence loophole. With a small coincidence window, many particles are not matched with a partner.

It is a cute trick. Each pair of particles agrees in advance what pair of settings it wants to see. If either sees the "wrong" setting, it arranges that its detection time is a little earlier or a little later than "normal". If both see the wrong setting the time interval between their arrival times is lenthened to something larger than the coincidence window. This way you can bias the correlations just how you like.

The important efficiciency parameter in experiments where the measurement times are co-determined by the particles is the chance that a particle detected in one wing of the experiment has a detected partner on the other side, ie the chance that a detected particle s part of a detected pair. If this chance is smaller than 100% you can violate CHSH. The smaller the chance, ie the more unpaired events, the bigger a deviation from the CHSH bound of 2 can be manufacturds. At about 95% you can get to 2 sqrt 2 in CHSH. In principle, you can recover the singlet correlations in this way.

The arrival times are correlated with the local hidden variables. The experimenter creates non-local correlation by selecting pairs on the basis of the arrival times.

arXiv:quant-ph/0312035
Bell's inequality and the coincidence-time loophole
Jan-Ake Larsson, Richard Gill
Europhysics Letters, vol 67, pp. 707-713 (2004)

 

[gill1109]

PS to Bill Schnieder: Christian's model comes in several variants. In all of them, the actual outcomes are perfectly anti-correlated whatever the settings. He claims to get -a.b only after dividing the covariance of -1 by two bivector standard deviations. (Already quite an absurd idea). And then he makes a simple sign mistake while doing this calculation, probably confused by his own ambiguous notation. In some versions of his model, he fixes the mistake by changing the postulates. The new postulates are mutually incompatible. ie the model is empty, the axioms cannot be satisfied. These errors have been known by him for four years or more yet he continues trying to hide them. No single paper by him on his model has been published in a peer reviewed journal.

Kracklauer uses the detection loophole and/or the coincidence loophole, just like de Raedt. Accardi uses the detection loophole. Hess and Phillip's very elaborate model was built around a little but well hidden mathematical error, dropping a crucial index in going from one formula to another.

And so on, and so on.

 

[harrylin]

You do not know that so why do you state it as though you do? [..]

I did not realize that until recently; the latest discussions on the different forums clarified this for me beyond reasonable doubt, in part thanks to Christian himself. However, all that is off topic here (you can send me a PM).

 

[Delta Kilo]

$E(AB)_{Aspect(2004)} = E(AB)_Y = \int d\lambda \rho (\lambda ) (AB)_Y = \int d\lambda \rho (\lambda )A(\textbf{a}, \lambda )B(\textbf{b}, \lambda )_Y$

$= \int d\lambda \rho (\lambda )[ 2 \cdot P(B^+|Y,\,A^+) - 1] = \int d\lambda \rho (\lambda)[2⋅cos^2(\textbf{a}, \textbf{b}) - 1]$

$= cos[2(\textbf{a}, \textbf{b})] = QED!$

GW,

Sorry but this is not a derivation, this is wishful thinking. While individual bits and pieces may look sort of all right, the steps do not follow one from another at all.

$E(AB)_Y = \int d\lambda \rho (\lambda ) (AB)_Y = \int d\lambda \rho (\lambda )A(\textbf{a}, \lambda )B(\textbf{b}, \lambda )_Y$

That's Bell's LR condition all right. Fair enough.

$= \int d\lambda \rho (\lambda )[ 2 \cdot P(B^+|Y,\,A^+) - 1]$

What the heck is that? Why are you integrating over λ a probability which is not a function of λ, what is that supposed to mean? (Thankfully there is no harm done, but it is just pointless and conceptually wrong thing to do, as P(B|A) is itself an implicit integral over all possible λ. I assume you are integrating the whole thing and not just multiplying ∫ρ(λ)dλ by the expression in the square brackets, which would be equally silly) Where did Y come from?

But crucially ... how did you get here form there?

I mean I sort of know where you got this expression from, don't bother telling me, but... you didn't actually derive it from the previous line of your 'proof', did you? I mean, you did not actually start with the first expression and somehow transmogrify it into the second one by applying strict rules of math, did you?

You do understand that there is no connection whatsoever between what you just did and the previous line, don't you? That the two expressions were written under different set of assumptions, specifically that the second line describes all possible set-ups where certain symmetry conditions are met while the first one only applies to LR ones? That it is incorrect in general to put equal sign between them? That there are plenty of counter-examples where $2 \cdot P(B^+|Y,\,A^+) - 1$ cannot be factorized into $A(\textbf{a}, \lambda )B(\textbf{b}, \lambda) \rho(\lambda)$? etc. etc.

$= \int d\lambda \rho (\lambda)[2⋅cos^2(\textbf{a}, \textbf{b}) - 1]$

Now, this $cos^2$ was taken from Aspect paper, where it was obtained as a prediction of QM. This happens to be exactly the case I mentioned earlier which cannot be factorized into $A(\textbf{a}, \lambda )B(\textbf{b}, \lambda) \rho(\lambda)$. And so your 'derivation' instead of being 'incorrect in general' becomes just 'wrong' .

 

[billschnieder]

While individual bits and pieces may look sort of all right, the steps do not follow one from another at all.

But crucially ... how did you get here form there?

From what I understand, V represents the experimental conditions.
$E(AB)_V = \int_{\Lambda} (A^\lambda \cdot B^\lambda)_V \, \rho (\lambda )_V \, d\lambda$
Instead of integrating over continuous variables λ, we can instead do a summation over discrete outcomes
$E(AB)_V = \sum_{ij} (A^i\cdot B^j)P(A^iB^j|V), \;\;\;\; ij \in [++, +-, -+, --]$
Note that the only difference between the discrete case and above case is that $A^\lambda$ is a function of λ, and $A^i$ is an outcome. But in both equations, we have the separable product $A \cdot B$ satisfying the locality condition.

$= P(A^+B^+|V) - P(A^+B^-|V) - P(A^-B^+|V) + P(A^-B^-|V)$
$= P(A^+|V)P(B^+|V,\,A^+) - P(A^+|V)P(B^-|V,\,A^+) - P(A^-|V)P(B^+|V,\,A^-) + P(A^-|V)P(B^+|V,\,A^-)$
$= P(A^+|V)\left [P(B^+|V,\,A^+) - P(B^-|V,\,A^+) \right ]- P(A^-|V) \left [P(B^+|V,\,A^-) - P(B^-|V,\,A^-) \right ]$

Since:
$P(A^+|V) = P(A^-|V) = \frac{1}{2}$
and
$P(B^+|W,\,{A^+}) + P(B^-|W,\,{A^+}) = 1$
$P(B^+|W,\,{A^-}) + P(B^-|W,\,{A^-}) = 1$
Therefore substituting above, we get
$E(AB)_V = 2 \cdot P(B^+|V,\,A^+) - 1$
$E(AB)_V = -2 \cdot P(B^+|V,\,A^-) + 1$
All that remains now is to use Malus law to get $P(B^+|V,\,A^+)$ (or $P(B^+|V,\,A^-)$ for the case where A(·) = - B(·)) according to the experimental conditions V.

 

[Gordon Watson]

GW,

Sorry but this is not a derivation, this is wishful thinking. While individual bits and pieces may look sort of all right, the steps do not follow one from another at all.
That's Bell's LR condition all right. Fair enough.What the heck is that? Why are you integrating over λ a probability which is not a function of λ, what is that supposed to mean? (Thankfully there is no harm done, but it is just pointless and conceptually wrong thing to do, as P(B|A) is itself an implicit integral over all possible λ. I assume you are integrating the whole thing and not just multiplying ∫ρ(λ)dλ by the expression in the square brackets, which would be equally silly) Where did Y come from?

But crucially ... how did you get here FROM there? [GW EDIT]

I mean I sort of know where you got this expression from, don't bother telling me, but... you didn't actually derive it from the previous line of your 'proof', did you? I mean, you did not actually start with the first expression and somehow transmogrify it into the second one by applying strict rules of math, did you?

You do understand that there is no connection whatsoever between what you just did and the previous line, don't you? That the two expressions were written under different set of assumptions, specifically that the second line describes all possible set-ups where certain symmetry conditions are met while the first one only applies to LR ones? That it is incorrect in general to put equal sign between them? That there are plenty of counter-examples where $2 \cdot P(B^+|Y,\,A^+) - 1$ cannot be factorized into $A(\textbf{a}, \lambda )B(\textbf{b}, \lambda) \rho(\lambda)$? etc. etc.Now, this $cos^2$ was taken from Aspect paper, where it was obtained as a prediction of QM. This happens to be exactly the case I mentioned earlier which cannot be factorized into $A(\textbf{a}, \lambda )B(\textbf{b}, \lambda) \rho(\lambda)$. And so your 'derivation' instead of being 'incorrect in general' becomes just 'wrong' .[GW emphasis added]

Delta Kilo, thanks for continuing to engage with the maths here and their physical significance. I am keen to learn of errors and will not hide from them. However, you seem to be a bit time-pressured (like me) at the moment: You appear to have missed the links that I put in my post (specifically for you):

1. The following link was to remind you of "the Malus Method." We use "the Malus Method" -- just like the famous Malus of old -- in that we study the results of experiments and draw conclusions in the form of equations. NB: "Malus' Law" (as such) is strictly limited to W (s = 1): We can generalise over s = 1/2 and s = 1; AND TO MULTI-PARTICLE EXPERIMENTS which were unknown to him!

LINK: "This was explained [edit: POORLY] in my reference to Malus, which I believe is central to addressing some of your concerns: https://www.physicsforums.com/showpost.php?p=3879566&postcount=112

2. LINK: Also see posts leading to, and including: https://www.physicsforums.com/showpost.php?p=3874480&postcount=102

3. This next link was meant to prevent your: What the heck is that? It's what I term "Bill's generalised short-cut."

LINK: "For example (using Bill's generalised short-cut: https://www.physicsforums.com/showpost.php?p=3878616&postcount=110 recalling that V is the general conditional for the style of experiments that we are analysing):"

SO would you mind reviewing your position, please, then commenting on what does not flow correctly for you? I will then provide a detailed mathematical analysis, with no short-cuts.

I am confident that the derivations are sound; every relevant element of the physical reality included in the equations: their physical significance at one with Einstein-locality.

Thanks again, GW

PS: Bill's neat post (above) contains some TYPOS that he might like to fix. So I suggest you hold off on commenting there until they're fixed. Thanks.

 

[Delta Kilo]

From what I understand, V represents the experimental conditions.
$E(AB)_V = \int_{\Lambda} (A^\lambda \cdot B^\lambda)_V \, \rho (\lambda )_V \, d\lambda$
Instead of integrating over continuous variables λ, we can instead do a summation over discrete outcomes
$E(AB)_V = \sum_{ij} (A^i\cdot B^j)P(A^iB^j|V), \;\;\;\; ij \in [++, +-, -+, --]$
Note that the only difference between the discrete case and above case is that $A^\lambda$ is a function of λ, and $A^i$ is an outcome.

Yes, sure we can. But why don't we actually try to prove it instead of just saying it?

To make the transition explicit we can split $\Lambda$ into non-overlapping subsets corresponding to distinct outcomes:
$\Lambda = \bigcup_{ij} \Lambda_{ij|V}: \forall \lambda \in \Lambda_{ij|V} \Rightarrow A(\lambda,V)=A_i, B(\lambda,V)=B_j$,
and then show that the integral can be rewritten as a sum of integrals over these subsets:
$\int_{\Lambda} A(\lambda,V) B(\lambda,V) \rho (\lambda) d\lambda =$
$= \sum_{ij}\int_{\Lambda_{ij|V}} A(\lambda,V) B(\lambda,V) \rho (\lambda) d\lambda =$
$= \sum_{ij}A_i B_j \int_{\Lambda_{ij|V}} \rho (\lambda) d\lambda =$
$= \sum_{ij} A_i B_j P(A_iB_j|V)$
where
$P(A_i,B_j|V) = \int_{\Lambda_{ij|V}} \rho(\lambda) d\lambda$

The probability integral can be rewritten as
$P(A_i,B_j|V) = \int_{\Lambda_{ij|V}} \rho(\lambda) d\lambda = \int_\Lambda \textbf{1}_{\Lambda_{ij|V}}(\lambda) \rho(\lambda) d\lambda = \int_\Lambda \delta(A(\lambda,V)-A_i) \delta(B(\lambda,V)-B_j) \rho(\lambda) d\lambda$
(1_X(x)= {1 when x∈X, else 0} is a set membership indicator function , δ is a Kronecker delta)

Note, this expression for the probability P(Ai,Bi|V), is part of the derivation. You cannot simply disregard it later on. Your expression for E(AB) as a sum of probabilities is only valid if the probability can be expressed in this particular form.

But in both equations, we have the separable product $A \cdot B$ satisfying the locality condition.

No, we don't. We don't have separate local settings a and b. Instead we have global V which presumably includes a and b along with any other settings and both A and B depend on the entire V.

Now, if we repeat the derivation above but starting from proper LR condition
$E(a,b)=\int A(a,\lambda) B(b,\lambda) \rho(\lambda) d \lambda$
We will eventually get the same expression for E(a,b)
$E(a,b)=\sum_{ij} A_i B_j P(A_i,B_j|a,b)$
but the expression for the probability will look slightly different:
$P(A_i,B_j|a,b) = \int_\Lambda \delta(A(a,\lambda)-A_i) \delta(B(b,\lambda)-B_j) \rho(\lambda) d\lambda$
The difference is of course that A does not depend on b and vice versa.

Now, QM predicts $P(A=1,B=1|a,b) = \frac{1}{2}cos^2(a-b)$.

I can easily concoct the ingredients to make the probability to come out right in the general case:
$P(A=1,B=1|V) =\int_\Lambda \delta(A(\lambda,V)-1) \delta(B(\lambda,V)-1) \rho(\lambda) d\lambda$

For example:
$V = a-b$
$\lambda \in [-1..1], \rho(\lambda)=\frac{1}{2}$
$A(\lambda,V) = sign(\lambda)$
$B((\lambda,V) = sign(\lambda) sign(cos^2(V) - |\lambda|)$

If I now substitute all that into the integral and take it, the answer is going to be
$P(A=1,B=1|a,b)=\frac{1}{2}cos^2(a-b)$

However, you will notice that B depends on V which includes both a and b.

It turns out to be is impossible to find such A and B for the LR case:
$P(A_i,B_j|a,b) = \int_\Lambda \delta(A(a,\lambda)-A_i) \delta(B(b,\lambda)-B_j) \rho(\lambda) d\lambda$

 

[billschnieder]

Yes, sure we can. But why don't we actually try to prove it instead of just saying it?

There is nothing to prove here. The two expressions are equivalent!

In probability theory, the expected value (or expectation, or mathematical expectation, or mean, or the first moment) of a random variable is the weighted average of all possible values that this random variable can take on. The weights used in computing this average correspond to the probabilities in case of a discrete random variable, or densities in case of a continuous random variable. From a rigorous theoretical standpoint, the expected value is the integral of the random variable with respect to its probability measure.

 

[billschnieder]

It turns out to be is impossible to find such A and B for the LR case:
$P(A_i,B_j|a,b) = \int_\Lambda \delta(A(a,\lambda)-A_i) \delta(B(b,\lambda)-B_j) \rho(\lambda) d\lambda$

Your argument is essentially that $P(A^+B^+|V)= P(A^+|V)P(B^+|V,\,A^+)$ violates locality, but this is false. You want $P(A^+B^+|V)= P(A^+|V)P(B^+|V)$, but even a simple thought experiment will show how wrong this is. Consider for example the simple experiment (T) which involves coin tosses by two people (heads = +, tails = -)

A B
+ +
- +
+ -
- +
+ +
- -
- +

What is $P(A^+B^+|T)$??
what is $P(A^+|T)$?
what is $P(B^+|T)$?

Still think $P(A^+B^+|V)= P(A^+|V)P(B^+|V,\,A^+)$ violates locality? Still think the probability must be separable if LR is true? Or do you think the coin toss example is non-local?

Note that $P(B^+|V,\,A^+)$ simply means for a given experiment V, we find all the pairs for which A+ was observed on one side, and calculate the fraction of B+ within the same set of pairs, that was observed on the other side. It does not mean there is a non-local influence between A+ and B+.

I should also correct a typo in the previous post where I used W instead of V.

 

[billschnieder]

We don't have separate local settings a and b. Instead we have global V which presumably includes a and b along with any other settings and both A and B depend on the entire V.

This is an interesting statement because it applies to EVERY experiment that has ever been performed on this issue. So you are essentially criticizing Aspect et al for treating their data in a non-local way.

Which of these expressions do you think experimentalists use to calculate $P(A^+B^+|V)$
$P(A^+B^+|V)= P(A^+|V)P(B^+|V,\,A^+)$
OR
$P(A^+B^+|V)= P(A^+|V)P(B^+|V)$?

Certainly they can only use one of the above, not knowing anything about λ.

 

[harrylin]

[..] $P(B^+|V,\,A^+)$ simply means for a given experiment V, we find all the pairs for which A+ was observed on one side [..].

Here you refer to the statistical analysis of a small sample. However, P usually stands for probability. As in your coin toss example, the probability of head (fair coin) P=0.5 even if you throw for example heads twice.

 

[billschnieder]

Here you refer to the statistical analysis of a small sample. However, P usually stands for probability. As in your coin toss example, the probability of head (fair coin) P=0.5 even if you throw for example heads twice.

I did not say the coin was fair and what if the experimnent T can never be repeated because the coins were so fragile that they could only be tossed 7 times. That is why the calculation has to be conditioned on the specific experimental conditions T whose outcomes you have.

T = "two coins tossed 7 times by two people A and B giving outcomes [++, -+, + -, - +, + +, - -, - +]"
Calculate $P(A^+B^+|T)$.
then calculate $P(A^+|T)P(B^+|T,A^+)$ and $P(A^+|T)P(B^+|T)$ and verify that it is not possible to reason correctly in such a simple experiment if you adopt to so called "locality condition" being suggested.

BTW, are you implying that what I'm asking is not a probability?

 

[Gordon Watson]

Yes, sure we can. But why don't we actually try to prove it instead of just saying it?

I totally agree! But, with respect, because I much appreciate your engagement with the issues here: You (Delta Kilo), often breach of your own suggestion. Too often (and beyond my humble opinion) your strong opinions are NOT supported by strong analysis (as has been shown on occasion). A recent example is addressed below.

I will be attempting to address my own short-comings too! I am currently under extreme time-pressures. My goal is to keep any consequences out of my posts from here on in: with apologies for recent indiscretions!

NB: I have modified the conditionals in what follows: From V to W°Y (W XOR Y) so that we are addressing specific experiments (W°Y) and not a general one (V). I'd like to see this notation used here from now on; or let W and Y be used alone, as the argument requires.

I'll also stick to λ and λ' as the hidden-variables in all cases; trusting we all agree that some pair-wise function F_W°Y(λ, λ') = 0 is implied in all our studies: λ' being the HV sent to Bob.

Instead of integrating over continuous variables λ, we can instead do a summation over discrete outcomes
$E(AB)_{W°Y} = \sum_{ij} (A^i\cdot B^j)P(A^iB^j|W°Y), \;\;\;\; ij \in [++, +-, -+, --]$
Note that the only difference between the discrete case and above case is that $A^\lambda$ is a function of λ, and $A^i$ is an outcome. But in both equations, we have the separable product $A \cdot B$ satisfying the locality condition.

I agree, with no surprises. The study here of the continuous case, per Bell's (1964) protocol, shows that Bell's analysis may be reduced to the discrete case with no loss of integrity. I personally favour Bell's approach; and will probably stick with it.

No, we don't. We don't have separate local settings a and b. Instead we have global W°Y which presumably includes a and b along with any other settings and both A and B depend on the entire W°Y.

Delta Kilo, please reconsider. What you write is surely wrong?

1. a is the setting of Alice's device, b is the setting of Bob's device. The devices are cleared separated. If you mean that a and b are just arbitrary orientations in 3-space, just say so. But there are NO other settings POSSiBLE, imho! So what do you mean by "any other settings"?

2. You write: "both A and B depend on the entire W°Y." But this is surely false?

Consistent with Einstein-locality and Bell's protocol: A(a, λ) = ±1, B(b, λ') = ±1. There is no dependency of A on B or b or λ'; NOR of B on A or a or λ.

So what do you mean ... in this critical area of analysis so central to our discussions?

If I now substitute all that into the integral and take it, the answer is going to be
$P(A=1,B=1|a,b)=\frac{1}{2}cos^2(a-b)$

However, you will notice that B depends on V which includes both a and b.

It turns out to be is impossible to find such A and B for the LR case:
$P(A_i,B_j|a,b) = \int_\Lambda \delta(A(a,\lambda)-A_i) \delta(B(b,\lambda)-B_j) \rho(\lambda) d\lambda$

In relation to the challenge in the OP, here are my functions A and B for W, X, Y and Z; any new terms*** here being defined by their well-known associates; there being no suggestion here that $\lambda = {\textbf{a}}^+\bigoplus {\textbf{a}}^-$, etc., prior to the relevant Einstein-local particle-device interaction. Rather, the delta-functions put the "quantum-jumps" in the equations and not just in the talk, representing dynamical processes in dynamically defined conditions ... in line with Bell's hope (2004, p. 118):

In short-hand:

A(a, λ) = ∫dλ δ(λ - a⁺°a^-) cos[2s(a, λ) = ±1.

B(b, λ') = ∫dλ δ(λ' - b⁺°b^-) cos[2s(b, λ') = ±1.

Or better:

$A({\textbf{a}}, \lambda) = \int d\lambda\delta (\lambda - {\textbf{a}}^+\bigoplus {\textbf{a}}^-) cos[2s({\textbf{a}}, \lambda)] = \pm 1.$

$B({\textbf{b}}, \lambda') = \int d\lambda'\delta (\lambda' - {\textbf{b}}^+\bigoplus {\textbf{b}}^-) cos[2s({\textbf{b}}, \lambda')] = \pm 1.$

Where $\bigoplus$ = XOR; s = intrinsic spin.

Do you see a problem with these functions? If not, how does your integral proceed, please, with them?
...

***Edit: Footnote added:

PS:

a⁺ = a for s = 1/2 or 1; etc.

a^- = -a for s = 1/2; etc.

a^- = a ± π/2 for s = 1; etc.

The above reflect the related orthogonalities for photons and spin-half particles.

......

Thanks,

GW

 

[Gordon Watson]

...


Therefore substituting above, we get
$E(AB)_V = 2 \cdot P(B^+|V,\,A^+) - 1$
$E(AB)_V = -2 \cdot P(B^+|V,\,A^-) + 1$
All that remains now is to use Malus law to get $P(B^+|V,\,A^+)$ (or $P(B^+|V,\,A^-)$ for the case where A(·) = - B(·)) according to the experimental conditions V.

Bill, to cover ALL the specific experiments under discussion (W, X, Y, Z), I suggest it is better to state the general case:

All that remains now is to use Malus' Method to get $P(B^+|Q,\,A^+)$ (or $P(B^+|Q,\,A^-)$ according to the experimental conditions Q (be they W, X, Y or Z).

By Malus' Method I mean: Following Malus' example (ca 1808-1812, as I recall), we study the results of experiments and write equations to capture the underlying generalities.

Cheers, GW

 

[billschnieder]

$A({\textbf{a}}, \lambda) = \int d\lambda\delta (\lambda - {\textbf{a}}^+\bigoplus {\textbf{a}}^-) cos[2s({\textbf{a}}, \lambda)] = \pm 1.$

$B({\textbf{b}}, \lambda') = \int d\lambda'\delta (\lambda' - {\textbf{b}}^+\bigoplus {\textbf{b}}^-) cos[2s({\textbf{b}}, \lambda')] = \pm 1.$

Where $\bigoplus$ = XOR.

Do you see a problem with these functions? If not, how does your integral proceed, please, with them?

Thanks,

GW

I think you forgot to mention that s is the spin. Is that correct?

 

[Delta Kilo]

There is nothing to prove here. The two expressions are equivalent!

No in general they are not.

$E(a,b)=\sum_{ij} A_iB_j P(A_i,B_j|a,b)$
is applicable to any discrete random values A, B with no restrictions whatsoever, while

$E(a,b)=\int_\Lambda A(a,\lambda) B(b,\lambda) \rho(\lambda) d\lambda$
describes a particular LR setup, where A and B are both functions of random variable λ.

If you treat the two formulas above as a system of equations, and solve it with respect to P(..) you will get the answer:
$P(A_i,B_j|a,b)=\int_\Lambda \delta_{A(a,\lambda),A_i} \delta_{B(b,\lambda),B_j}\rho(\lambda) d\lambda$
In fact, given any two of the above formulas you can derive the third.

So for the first two formulas to be true simultaneously, the probability has to be representable in this particular form. Malus law cannot be represented is this form.

 

[Gordon Watson]

I think you forgot to mention that s is the spin. Is that correct?

Thanks Bill, now fixed. Along with a few other fixes and expansions that occurred to me on my way back to the office. Your comments on them, especially the delta-functions, would be welcome. GW

 

[Gordon Watson]

No in general they are not.

$E(a,b)=\sum_{ij} A_iB_j P(A_i,B_j|a,b)$
is applicable to any discrete random values A, B with no restrictions whatsoever, while

$E(a,b)=\int_\Lambda A(a,\lambda) B(b,\lambda) \rho(\lambda) d\lambda$
describes a particular LR setup, where A and B are both functions of random variable λ.

If you treat the two formulas above as a system of equations, and solve it with respect to P(..) you will get the answer:
$P(A_i,B_j|a,b)=\int_\Lambda \delta_{A(a,\lambda),A_i} \delta_{B(b,\lambda),B_j}\rho(\lambda) d\lambda$
In fact, given any two of the above formulas you can derive the third.

So for the first two formulas to be true simultaneously, the probability has to be representable in this particular form. Malus law cannot be represented is this form.

..
DK, might it help if you included the general conditional Q in each equation, where Q is an element of {W, X, Y, Z}? GW

EDIT:

PS: I think Bill meant "Malus' Method" -- Malus' Law being strictly limited to W? GW
..

 

[billschnieder]

No in general they are not.

Of course you understood that I did not mean generally, we are discussing Bell, and in this case the two are equivalent. The expectation value $E(AB)_V$ is defined over any probability measure of the probability space V. You can pick anyone you like and you will get the same result. ρ(λ) is just as valid a probability measure over V as ρ(ij), ij ∈ [++,+−,−+,−−]. The latter just makes it easier to compare with how E(AB) is calculated in real experiments. So you cannot think the calculation is wrong without claiming the same about how the Bell-test experimentalists analyse their data.

So for the first two formulas to be true simultaneously, the probability has to be
representable in this particular form.

I do not agree with that characterization. The chain rule of probability theory, P(AB|X) = P(A|X)P(B|X,A), is valid generally. P(AB|X) = P(A|X)P(B|X) is ONLY valid in the limited case in which P(B|X,A) = P(B|X) and even then the chain rule is still valid (ie, the two expressions give you the exact same result). There will never be a situation in which P(AB|X) = P(A|X)P(B|X) would be valid while P(AB|X) = P(A|X)P(B|X,A) is not. In any case, as demonstrated in the coin-toss example a few posts back, separability of the probability is not a requirement for locality.

It now appears you are saying Malus law is non-local. Did I understand that correctly?

 

[Delta Kilo]

Of course you understood that I did not mean generally, we are discussing Bell, and in this case the two are equivalent.

If you postulate that the two forms are equivalent, then it immediately and automatically follows from the math that the probability has this particular form I have given you.

The expectation value $E(AB)_V$ is defined over any probability measure of the probability space V. You can pick anyone you like and you will get the same result. ρ(λ) is just as valid a probability measure over V as ρ(ij), ij ∈ [++,+−,−+,−−].

I am now totally confused. What is V? Earlier it appeared to be either a label identifying particular experimental setup, or a set of parameters including settings a and b. Now you tell me it is a probability space? And ρ(λ) is defined over V?. And the most important, where did the settings a and b go? Can we please get the notation straight, so that E(a,b) is a function of a and b as it should be?

The chain rule of probability theory, P(AB|X) = P(A|X)P(B|X,A), is valid generally. P(AB|X) = P(A|X)P(B|X) is ONLY valid in the limited case in which P(B|X,A) = P(B|X) and even then the chain rule is still valid (ie, the two expressions give you the exact same result). ...

Where did I say anything at all about P(AB|X) = P(A|X)P(B|X)? I didn't, obviously, because this condition just means "for a given X, A|X and B|X are independent", which is clearly not true. Instead I said there must exist such A(a,λ), B(b,λ) and ρ(λ) so that the probability can be expressed as
$P(A_i,B_j|a,b)=\int_\Lambda \delta_{A(a,\lambda),A_i} \delta_{B(b,\lambda),B_j}\rho(\lambda) d\lambda$

or in more general form:
$P(A_i,B_j|a,b)=\int_\Lambda P(A_i|a,\lambda) P(B_j|b,\lambda)\rho(\lambda) d\lambda$
where in our particular case
$P(A_i|a,\lambda)=\delta_{A(a,\lambda),A_i}$, $P(B_i|b,\lambda)=\delta_{B(b,\lambda),B_j}$

See the difference?

It now appears you are saying Malus law is non-local. Did I understand that correctly?

That really is a matter of physical interpretation. Say you have a physical process described by a bunch of equations. Suppose you can assign regions of space-time to each physical quantity. Then if you can massage the equations in a way so that each value in the LHS in the intersections of past light-cones of all values in the RHS, then it is local-realistic, otherwise it is not.

 

[Delta Kilo]

$A({\textbf{a}}, \lambda) = \int d\lambda\delta (\lambda - {\textbf{a}}^+\bigoplus {\textbf{a}}^-) cos[2s({\textbf{a}}, \lambda)] = \pm 1.$

$B({\textbf{b}}, \lambda') = \int d\lambda'\delta (\lambda' - {\textbf{b}}^+\bigoplus {\textbf{b}}^-) cos[2s({\textbf{b}}, \lambda')] = \pm 1.$

Where $\bigoplus$ = XOR; s = intrinsic spin.

Do you see a problem with these functions?

Yes, I do see a problem. The LHS is a function of λ, while the RHS is not, as the λ has been integrated over.

 

[Gordon Watson]

Yes, I do see a problem. The LHS is a function of λ, while the RHS is not, as the λ has been integrated over.

Sorry, your short reply has me mystified:

LHS satisfies Bell's formulation.

MIDDLE satisfies formal definition of a function.

RHS satisfies Bell's requirement.

What am I missing? Thanks.

 

[Gordon Watson]

billschnieder asked: It now appears you are saying Malus law is non-local. Did I understand that correctly?

That really is a matter of physical interpretation. Say you have a physical process described by a bunch of equations. Suppose you can assign regions of space-time to each physical quantity. Then if you can massage the equations in a way so that each value in the LHS in the intersections of past light-cones of all values in the RHS, then it is local-realistic, otherwise it is not.

DK, your position is not at all clear to me. Would you mind explaining your view in the context of W? Thanks, GW

 

[Delta Kilo]

Sorry, your short reply has me mystified:
LHS satisfies Bell's formulation.
MIDDLE satisfies formal definition of a function.
RHS satisfies Bell's requirement.
What am I missing? Thanks.

LHS says A is a function of λ, however MIDDLE is not a function of λ. Also, s appears to be a function of two arguments, but it is not defined anywhere.

 

[Gordon Watson]

LHS says A is a function of λ, however MIDDLE is not a function of λ. Also, s appears to be a function of two arguments, but it is not defined anywhere.

1. s is defined several places, including the cut that you quoted??

2. That is: s = intrinsic spin of the relevant particle.

3. Why is MIDDLE not a function of λ? In detail, please, for I'm missing your point.

Thanks, GW

 

[Delta Kilo]

1. s is defined several places, including the cut that you quoted??

2. That is: s = intrinsic spin of the relevant particle.

s appears to be a function of two arguments s=s(a,λ). Where is the definition of this function?

3. Why is MIDDLE not a function of λ? In detail, please, for I'm missing your point.

Because λ is the integration variable. This is the way integrals work. It is the same as, for example, $f(x)=\sum_{i=0}^N A_ix^i$ is not a function of i.

 

[harrylin]

I did not say the coin was fair and what if the experimnent T can never be repeated because the coins were so fragile that they could only be tossed 7 times. That is why the calculation has to be conditioned on the specific experimental conditions T whose outcomes you have.[..] BTW, are you implying that what I'm asking is not a probability?

Indeed, I merely gave my example of a special case of your example as I wrote:
"Here you refer to the statistical analysis of a small sample. However, P usually stands for probability. As in your coin toss example, the probability of head (fair coin) P=0.5 even if you throw for example heads twice."
To elaborate, you could have the following statistical sequence:
++
++
While the statistical result of a few throws was ++ for all throws, this does certainly not mean that the probability of throwing ++ is 1.
- http://en.wikipedia.org/wiki/Law_of_large_numbers
- http://en.wikipedia.org/wiki/Student's_t-distribution

 

[harrylin]

[..] All that remains now is to use Malus' Method to get $P(B^+|Q,\,A^+)$ (or $P(B^+|Q,\,A^-)$ according to the experimental conditions Q (be they W, X, Y or Z).

By Malus' Method I mean: Following Malus' example (ca 1808-1812, as I recall), we study the results of experiments and write equations to capture the underlying generalities. [..]

Likely this is indeed the main issue. For this is basically what QM did. And doing so does not provide a mechanism for how this may be possible.

The purpose of such derivations as the one you are doing, should be to determine if the same is true for a similar law about the correlation between the detections of two light rays at far away places. Merely including experimental results does not do that. Malus law for the detected light intensity of a light ray going into one direction can be easily explained with cause and effect models, but this is not done by writing down Malus law.

PS. Your "Note in passing" that "Einstein-locality is maintained through every step of the analysis", is the main point that is to be proved, as Bell claimed to have disproved it; it can't be a "note in passing".

 

[billschnieder]

Indeed, I merely gave my example of a special case of your example as I wrote:
"Here you refer to the statistical analysis of a small sample. However, P usually stands for probability. As in your coin toss example, the probability of head (fair coin) P=0.5 even if you throw for example heads twice."
To elaborate, you could have the following statistical sequence:
++
++
While the statistical result of a few throws was ++ for all throws, this does certainly not mean that the probability of throwing ++ is 1.
- http://en.wikipedia.org/wiki/Law_of_large_numbers
- http://en.wikipedia.org/wiki/Student's_t-distribution

I'm quite surprised that you too are playing these types of tricks. If I say

T = an urn with N balls, M of which are red and M-N of which are white and ask you to calculate P(Red|T), will you then tell me that because of law of large numbers ..., won't you simply use basic probability theory and say P(Red|T) = M/N

Now if I tell you that N is actually 2. will you tell me that my sample is too small? Surely you know better, that if you want to use law of large numbers you MUST consider T as the population from which you are drawing infinitely and then you can't escape my question.

So when I say:

T = "two coins tossed 7 times by two people A and B giving outcomes [++, -+, + -, - +, + +, - -, - +]"
Do you still believe that P(A^+B^+|T) is factorable into P(A^+|T) and P(B^+|T), and does that mean the situation is non-local?

 

[harrylin]

I'm quite surprised that you too are playing these types of tricks.

It looked as if you were tricking yourself (but perhaps it's just a misunderstanding of what you mean?); don't shoot the messenger!

T = an urn with N balls, M of which are red and M-N of which are white and ask you to calculate P(Red|T), will you then tell me that because of law of large numbers ..., won't you simply use basic probability theory and say P(Red|T) = M/N

Good - that was my point.

Now if I tell you that N is actually 2. will you tell me that my sample is too small? [..]

Depending on your conclusions from the outcomes - such as concluding a probability rule from a very small data set - I may warn you again not to confuse statistics with probability, as explained in the provided references.

And since this mathematical issue comes up regularly we should discuss it in the mathematics forum. Please start it as a topic there with a link from here.