A classical challenge to Bell's Theorem?

[billschnieder]

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.

I'm not postulating anything.

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?

V is the experimental *conditions* under which A and B are obtained. P(AB|V) is thus appropriate and in that expression V is the probability space.

And ρ(λ) is defined over V?.

For the sake of illustration, assume the λ is discrete for our experiment V, ie

$E(AB)_V = \sum_{\lambda} (A^{\lambda}_a\cdot B^{\lambda}_b)P(\lambda|V), \;\;\;\; \lambda \in [\lambda_1, \lambda_2, \lambda_3, \lambda_4]$
$= A^{\lambda_1}_a\cdot B^{\lambda_1}_b\cdot P(\lambda_1|V) + A^{\lambda_2}_a\cdot B^{\lambda_2}_b\cdot P(\lambda_2|V) + A^{\lambda_3}_a\cdot B^{\lambda_3}_b\cdot P(\lambda_3|V) + A^{\lambda_4}_a\cdot B^{\lambda_4}_b\cdot P(\lambda_4|V)$
This expression is the same as Bell's (2), for the discrete lambda case, with "a" and "b" added for ease of comparison. Within the sum, $A^{\lambda}_a$ is a function of lambda, but once we expand it out, we just have outcomes.

And the most important, where did the settings a and b go?

See above.

Can we please get the notation straight, so that E(a,b) is a function of a and b as it should be?

It does not matter. My notation is much better than yours because it makes it clear we are calculating the expectation value for the product A*B where A and B are outcomes. Besides, "a" and "b" are not variables, they are fixed settings for a given experiment V.

Where did I say anything at all about P(AB|X) = P(A|X)P(B|X)?

You do not say that but that is what you mean. You want the probability to be factorable. All the math you are doing is just a convoluted way of saying the same thing so I'm just being blunt.

 

[DrChinese]

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

In the computer simulation, I have the complete information on all 10,000 raw trials (or whatever number you pick). That way I can see why any unmatched pairs ended up as unmatched. I can also see whether they do or do not contribute to violating the inequality. And I can see if they follow the QM expectation.

So I have the full universe of the trials, and am not limited as in actual experiments. This is what the de Raedt team calls an event by event simulation. This is what I call a realistic demonstration.

 

[DrChinese]

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.

Sure, I don't consider it an actual counterexample to Bell. To win my challenge, you have to provide a realistic dataset and it must be "local" (separable in this case) and it must violate an Inequality. They met those criteria.

But as you say, they loosely exploit the coincidence time loophole. I say loosely because there is no physical mapping between what really occurs and the simulation itself. In other words, the algorithm is ad hoc tuned to achieve this result and there is no idea that the underlying reality works that way.

 

[billschnieder]

In the computer simulation, I have the complete information on all 10,000 raw trials (or whatever number you pick). That way I can see why any unmatched pairs ended up as unmatched. I can also see whether they do or do not contribute to violating the inequality. And I can see if they follow the QM expectation.

So I have the full universe of the trials, and am not limited as in actual experiments. This is what the de Raedt team calls an event by event simulation. This is what I call a realistic demonstration.

Please could we continue this discussion over in the relevant thread? Thanks!

 

[billschnieder]

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.

Huh? Can you or can you not give an answer to the following question:
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?

What has small data set got to do with it? [++, -+, + -, - +, + +, - -, - +] IS the population. Use your law of large numbers to randomly pick from that.

 

[billschnieder]

Continuing ...
$E(AB)_V = \sum_{\lambda} (A^{\lambda}_a\cdot B^{\lambda}_b)P(\lambda|V), \;\;\;\; \lambda \in [\lambda_1, \lambda_2, \lambda_3, \lambda_4]$
$= A^{\lambda_1}_a\cdot B^{\lambda_1}_b\cdot P(\lambda_1|V) + A^{\lambda_2}_a\cdot B^{\lambda_2}_b\cdot P(\lambda_2|V) + A^{\lambda_3}_a\cdot B^{\lambda_3}_b\cdot P(\lambda_3|V) + A^{\lambda_4}_a\cdot B^{\lambda_4}_b\cdot P(\lambda_4|V)$
If you think this expression does not correctly represent Bell's locality condition, then surely, you must think the same about Bell's equation (2):

$E(AB) = \int_{\Lambda} (A^{\lambda}_a \cdot B^{\lambda}_b) \;\rho (\lambda )d\lambda = \int_{\Lambda} A(\textbf{a}, \lambda ) \cdot B(\textbf{b}, \lambda )\;\rho (\lambda )d\lambda$

The ONLY difference between the two is that in the former λ is discrete but in the latter it is continuous?

Now let us say $A^{\lambda_1}_a = A(a,\lambda_1) = +1 \;,\; B^{\lambda_1}_b = B(b,\lambda_1) = +1 \;,\; A^{\lambda_2}_a = A(a,\lambda_2) = +1 \;,\; B^{\lambda_2}_b = B(b,\lambda_2) = -1 \;,\;$
$A^{\lambda_3}_a = A(a,\lambda_3) = -1 \;,\; B^{\lambda_3}_b = B(b,\lambda_3) = +1 \;,\; A^{\lambda_4}_a = A(a,\lambda_4) = -1 \;,\; B^{\lambda_4}_b = B(b,\lambda_4) = -1$

It therefore follows that
$P(\lambda_1|V) = P(A^+_aB^+_b|V) \;,\; P(\lambda_2|V) = P(A^+_aB^-_b|V) \;,\; P(\lambda_3|V) = P(A^-_aB^+_b|V) \;,\; P(\lambda_4|V) = P(A^-_aB^-_b|V) \;,\;$
And it is immediately obvious that

$E(AB)_V = \sum_{\lambda} (A^{\lambda}_a\cdot B^{\lambda}_b)P(\lambda|V), \;\;\;\; \lambda \in [\lambda_1, \lambda_2, \lambda_3, \lambda_4]$
$= \sum_{ij} (A^i_a\cdot B^j_b)P(ij|V), \;\;\;\; ij \in [++, +-, -+, --]$
$= A^+_a\cdot B^+_b\cdot P(A^+_aB^+_b|V) + A^+_a\cdot B^-_b\cdot P(A^+_aB^-_b|V) + A^-_a\cdot B^+_b\cdot P(A^-_aB^+_b|V) + A^-_a\cdot B^-_b\cdot P(A^-_aB^-_b|V)$

So again, this expression obtains directly from Bell's equation (2) without any postulates. We have separable products of outcomes with the outcome at each station depending ONLY on the relevant lambda and the setting completely satisfying Bell's locality condition. You can not argue that it violates locality without rejecting Bell's original equation (2) on the same grounds. The rest of the analysis as laid out in post #185 follows.

Now if you want to argue that Malus law is non-local, that is a different issue, but there is nothing non-local in the above analysis.

 

[harrylin]

Huh? [..] [++, -+, + -, - +, + +, - -, - +] IS the population. Use your law of large numbers to randomly pick from that.

That doesn't make any sense to me at all, and that was the reason for my first comment. As such a bug usually takes at least 20 or 30 posts to solve (if at all), I told you that I will not hijack GW's thread for it, and also where it should be discussed if you want to discuss it.

 

[billschnieder]

That doesn't make any sense to me at all; and as such a bug usually takes at least 20 or 30 posts to solve (if at all), I told you that I will not hijack this thread for it, and also where it should be discussed if you want to discuss it. No more of that in GW's thread.

I agree that student t-tests and law of large numbers are off-topic but I did not bring those in here, you did. I asked a question which continues to be relevant for this thread, and the question was meant to illustrate the point that locality does not mean factorability of a probability.

 

[Delta Kilo]

I'm not postulating anything.

Bill, you cannot have it both ways. Either the two expressions for E(AB) are equivalent or not. If they are not, then there is no proof and there is nothing to discuss further. But if they are equivalent, as you yourself said, then the probability integral follows immediately from this.

V is the experimental *conditions* under which A and B are obtained. P(AB|V) is thus appropriate and in that expression V is the probability space.

You are confusing different terms here.V in P(AB|V) is a condition. It is not a probability space. In a typical Bell setup V would be experimental conditions, including angles a and b, while probability space for λ would include all possible states of th emitted photons.

My notation is much better than yours because it makes it clear we are calculating the expectation value for the product A*B where A and B are outcomes. Besides, "a" and "b" are not variables, they are fixed settings for a given experiment V.

By ignoring a and b, your notation completely misses the point. Do I have to remind you that the whole thing revolves about E(AB) being a function of a and b: some such functions are allowed by Bell's inequality while some others are not. If you do not keep track of a and b, you will not be able to apply bell's inequality to it and the whole thing does not make sense.

You do not say that but that is what you mean. You want the probability to be factorable. All the math you are doing is just a convoluted way of saying the same thing so I'm just being blunt.

Please, Bill, don't put words in my mouth, I mean exactly what I say. If you can't see the difference between two equations, I'm sorry, but it's your problem.

 

[Delta Kilo]

Continuing ...
$E(AB)_V = \sum_{\lambda} (A^{\lambda}_a\cdot B^{\lambda}_b)P(\lambda|V), \;\;\;\; \lambda \in [\lambda_1, \lambda_2, \lambda_3, \lambda_4]$
$= A^{\lambda_1}_a\cdot B^{\lambda_1}_b\cdot P(\lambda_1|V) + A^{\lambda_2}_a\cdot B^{\lambda_2}_b\cdot P(\lambda_2|V) + A^{\lambda_3}_a\cdot B^{\lambda_3}_b\cdot P(\lambda_3|V) + A^{\lambda_4}_a\cdot B^{\lambda_4}_b\cdot P(\lambda_4|V)$

It's OK as long as settings a and b are separate from V. If V is allowed to depend on a or b, then changing a may affect outcome B and therefore Bell's locality condition is not satisfied.

Now let us say $A^{\lambda_1}_a = A(a,\lambda_1) = +1 \;,\; B^{\lambda_1}_b = B(b,\lambda_1) = +1 \;,\; A^{\lambda_2}_a = A(a,\lambda_2) = +1 \;,\; B^{\lambda_2}_b = B(b,\lambda_2) = -1 \;,\; A^{\lambda_3}_a = A(a,\lambda_3) = -1 \;,\; B^{\lambda_3}_b = B(b,\lambda_3) = +1 \;,\; A^{\lambda_4}_a = A(a,\lambda_4) = -1 \;,\; B^{\lambda_4}_b = B(b,\lambda_4) = -1 \;,\;$

Well, the values you assign to A and B do not depend on the settings a and b. In fact, changing a and/or b would have no effect on the outcome whatsoever.

$E(AB)_V = \sum_{\lambda} (A^{\lambda}_a\cdot B^{\lambda}_b)P(\lambda|V), \;\;\;\; \lambda \in [\lambda_1, \lambda_2, \lambda_3, \lambda_4]$
$= \sum_{ij} (A^i_a\cdot B^j_b)P(ij|V), \;\;\;\; ij \in [++, +-, -+, --]$
$= A^+_a\cdot B^+_b\cdot P(A^+_aB^+_b|V) + A^+_a\cdot B^-_b\cdot P(A^+_aB^-_b|V) + A^-_a\cdot B^+_b\cdot P(A^-_aB^+_b|V) + A^-_a\cdot B^-_b\cdot P(A^-_aB^-_b|V)$

Well, since A and B do now depend on the actual values of a and b, the resulting E(AB) is a constant with respect to a and b, and therefore trivially satisfies Bell's inequality. But it is not very interesting, it is?

Now if you allow A and B to take different values depending on the values of a and b respectively, things will become slightly more interesting.

 

[billschnieder]

Bill, you cannot have it both ways. Either the two expressions for E(AB) are equivalent or not.

See post #216 for the proof that the two are equivalent.

By ignoring a and b, your notation completely misses the point. Do I have to remind you that the whole thing revolves about E(AB) being a function of a and b. If you do not keep track of a and b, you will not be able to apply bell's inequality to it and the whole thing does not make sense.

No it does not miss any point. You are making a big deal about nothing. If you like, I can write it as $E(A_aB_b)_V$ which was implicit in the derivation. All that matters is that we calculate the expectation values of the type $E(A_aB_b)_V$ following Bell's local-realistic program. If those expectation values violate Bell's inequalities, then the problem is with the inequalities not local realism.

 

[billschnieder]

It's OK as long as settings a and b are separate from V. If V is allowed to depend on a or b, then changing a may affect outcome B and therefore Bell's locality condition is not satisfied.

Same thing applies to Bell's ρ(λ). I'm not doing anything different here.

Well, since A and B do now depend on the actual values of a and b, the resulting E(AB) is a constant with respect to a and b, and therefore trivially satisfies Bell's inequality.

Of course, E(AB) IS a constant for fixed settings "a" and "b". Of course, if you allow "a" and "b" to vary, the E(AB) will vary as well. Also it is not true that it "trivially satisfies Bell's inequality", where did you see that?

Now if you allow A and B to take different values depending on the values of a and b respectively, things will become slightly more interesting.

A and B already take different values for different values of a and b! But as you well know, each time you calculate E(AB), you are calculating for a given fixed pair of settings "a" and "b". It makes no sense in that case to allow "a" and "b" to vary. Here we are deriving an expression for a given pair of settings, without regard for what value "a" actually has or what value "b" actually has. All we are saying is that only one pair of values is operational in the derivation, even though the result is general for any other pair of values you may like to choose. So contrary to your claim, the outcomes do depend on the the setting in the same way as in Bell's equation.

 

[billschnieder]

Therefore adding the "a" and "b" labels? We have:

$E(A_aB_b)_V = 2 \cdot P(B^+_b|V,\,A^+_a) - 1$
$E(A_aB_b)_V = -2 \cdot P(B^+_b|V,\,A^-_a) + 1$

and

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.

Of course it should be understood that $P(B^+_b|Q,\,A^+_a)$ or $P(B^+_b|Q,\,A^-_a)$ is implied in the above quote.

 

[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; s = intrinsic spin.

Gordon, I wouldn't blame DK for misunderstanding the expression because it is unclear. The way I understand it is like:
$A(a, \lambda) = \int d\lambda\; \delta (\lambda - a) \cos[2s\cdot(a,\lambda)] = \pm 1$
where (a,λ) is the angle between the vectors a and λ. What does XOR add to the whole thing? Do you mean $a \in [a^+, a^-]$?

Am I understanding you correctly?

 

[Gordon Watson]

Gordon, I wouldn't blame DK for misunderstanding the expression because it is unclear. The way I understand it is like:
$A(a, \lambda) = \int d\lambda\; \delta (\lambda - a) \cos[2s\cdot(a,\lambda)] = \pm 1$
where (a,λ) is the angle between the vectors a and λ. What does XOR add to the whole thing? Do you mean $a \in [a^+, a^-]$?

Am I understanding you correctly?

Quick response, so you can get back to me while I'm away for a few hours (if you need to); with apologies to DK:

1. You should always feel free to pick me up on my formatting. I thought it would be clear that [2s(a, λ)] was the argument of a trig function; (a, λ) is exactly as you say.

2. So your CENTRAL expansion is correct.

3. BUT your RHS ±1 is incorrect. You only have only delivered +1. That's where XOR comes in: to deliver the PLUS XOR the MINUS one.

4. a is a, the orientation of the principal axis of Alice's device.

5. The set {a⁺, a^-} represents the orientations that λ may be transformed to via the particle/device interaction: δ_aλ → a⁺°a^-. ALT: δ_aλ → {a⁺, a^-} for W, X, Y, Z. With apologies for the hurried short-hand.
...
EDIT:

Bill, et al., to avoid the XOR: $A({\textbf{a}}, \lambda) = \int d\lambda\; (\delta_{\textbf{a}} \lambda \rightarrow \left \{{\textbf{a}}^+ , {\textbf{a}}^- \right \}) \cos[2s\cdot({\textbf{a}}, \lambda)] = \pm 1.$

Though the insistence on a (a-BOLD) throughout does nothing to match the prettiness in Bill's offering!?

Apart from that deficiency, an advantage of this format is this (imho):

1. A function is a process that transforms an element of a set into exactly one element of another set.

2. The Alice-device/hidden-variable interaction ($\delta_{\textbf{a}} \lambda$) is a process that transforms ($\rightarrow$) an element of a set ($\lambda \in \Lambda$) into exactly one element of a dichotomic set $(\left \{{\textbf{a}}^+ , {\textbf{a}}^- \right \})$.
...

PS: It was a mistake for me to introduce this maths distraction as a point of interest for DK, etc. The maths here goes through on the basis that Bell's A and B are sound representations of Einstein-locality AND we accept that such functions exist. For me, DK's important contribution is an adequate approximation to Bell's A and B: it was on that basis that we are where we are now. We can come back to these deltas later, if need be.

 

[Delta Kilo]

Of course, E(AB) IS a constant for fixed settings "a" and "b". Of course, if you allow "a" and "b" to vary, the E(AB) will vary as well.

You assign (potentially different) values for A and B for each distinct value λi, but not for each distinct a and b: $A^{\lambda_1}_a = A(a,\lambda_1) = +1$, etc. Here it gives the same outcome for all possible a and b, and not just for a given fixed settings a and b. If you meant something different, then fix your math.

Also it is not true that it "trivially satisfies Bell's inequality", where did you see that?

If I substitute different values of a and b into your formulas I will get the same output E(a,b)=E(b,c)=E(a,c). Of course Bell's inequality is going to be satisfied.

A and B already take different values for different values of a and b!

Not according to the formulas you wrote.

But as you well know, each time you calculate E(AB), you are calculating for a given fixed pair of settings "a" and "b". It makes no sense in that case to allow "a" and "b" to vary. Here we are deriving an expression for a given pair of settings, without regard for what value "a" actually has or what value "b" actually has. All we are saying is that only one pair of values is operational in the derivation, even though the result is general for any other pair of values you may like to choose. So contrary to your claim, the outcomes do depend on the the setting in the same way as in Bell's equation.

I guess, what you are saying is you first consider some fixed setting of a and b and assign values to A and B for each λi. Then you repeat the entire process for different settings a and b etc.

But it is not that simple: if you change the value of a and leave b intact then you are allowed to assign different values to A(a,λi), but you must keep the same values for B(b,λi) and P(λi|V), because these do not depend on a. Same for B and b.

Even better way of doing that would be to define a={a_j}, b={b_k}
$A^{\lambda_i}_{a_j}=A(a_j,\lambda_i)=\pm 1$ and similarly for B.
Then your result for E(AB) will be local realistic. Unfortunately you will not be able to reproduce QM predictions with it.

 

[billschnieder]

You assign (potentially different) values for A and B for each distinct value λi, but not for each distinct a and b: $A^{\lambda_1}_a = A(a,\lambda_1) = +1$, etc. Here it gives the same outcome for all possible a and b, and not just for a given fixed settings a and b. If you meant something different, then fix your math.

No I don't, what are you talking about? Couldn't you see that:

$A(a,\lambda_1) = +1 \;,\; A(a,\lambda_2) = +1 \;,\; A(a,\lambda_3) = -1 \;,\; A(a,\lambda_4) = -1 \;,\;$
The outcome of the function depends on BOTH "a" and "λ" ! The way you should read this is: λ1 is the hidden variable which together with the setting "a" results in a +1 outcome. Change the *value* of "a" and of course you will need a different *value* for λ1 to obtain the result +1. This is not different for Bell's case. The only outcomes possible are ±1. For any given *value* of "a", λ1 represents the corresponding *value* of λ that together give you a +1 outcomes. Don't you see that?

If I substitute different values of a and b into your formulas I will get the same output E(a,b)=E(b,c)=E(a,c). Of course Bell's inequality is going to be satisfied.

Huh? This is FALSE. This is like saying if you substituted different values of θ into cos(θ) you get the same output. We are deriving the general form. Simply changing the symbols does not affect the general mathematical form of the result, however changing the *values* of the symbols, changes the *value* of the result. The result in this case is:

$E(A_aB_b)_V = 2 \cdot P(B^+_b|V,\,A^+_a) - 1$
Different *values* for "a" and "b" give you different *values* for $P(B^+_b|V,\,A^+_a)$ and therefore a different value for $E(A_aB_b)_V$

I do not see a legitimate criticism here. What you are saying does not make any sense.

 

[Delta Kilo]

It was a mistake for me to introduce this maths distraction

Sure, never let the facts ruin a good story.

Sorry, I can't even begin to discuss your 'equation', it makes me cringe.

Did you finally figure out why ∫f(λ)dλ is not a function of λ? I suggest you do that before embarking on a quest to resolve nature's deepest mysteries.

 

[billschnieder]

Did you finally figure out why ∫f(λ)dλ is not a function of λ?

if f(λ) = cos(λ), then ∫f(λ)dλ = sin(λ) a function of λ, contrary to your ridicule above.

 

[Delta Kilo]

if f(λ) = cos(λ), then ∫f(λ)dλ = sin(λ) a function of λ, contrary to your ridicule above.

facepalm.jpg
I'm outta here. Good luck with your quest.

 

[Gordon Watson]

Sure, never let the facts ruin a good story.

Sorry, I can't even begin to discuss your 'equation', it makes me cringe.

Did you finally figure out why ∫f(λ)dλ is not a function of λ? I suggest you do that before embarking on a quest to resolve nature's deepest mysteries.

DK, no hard feelings on my part.

Just be sure to recall that your "having-a-go" got us started on this path. That's something! (Maybe one for the grandkids? )

PS: I thought you and Bill were doing a good job.

So come back soon ... now that you're a little wiser (as am I), thanks to Bill.

PS: I've improved that cringe-making expression; see https://www.physicsforums.com/showpost.php?p=3891633&postcount=225

GW

 

[lugita15]

if f(λ) = cos(λ), then ∫f(λ)dλ = sin(λ) a function of λ, contrary to your ridicule above.

That's an indefinite integral AKA an antiderivative. What's being discussed in this thread is a definite integral, and the definite integral of f(λ) with respect to λ is obviously independent of λ.

 

[Gordon Watson]

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

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.

Harald, thanks for your plain speaking: note that what follows is to be understood as IMHO.

If you refer to the "Malus Method" as the main issue, keep in mind that its use is still limited (here) to classical analysis with a focus on ontology (i.e., the nature of λ, the HVs; the nature of particle/device interactions -- δ_aλ, δ_bλ' -- from a classical point of view). In that way it differs from some "QM Methods". And in that way it DOES provide "a mechanism": for the method itself was prompted by the search for the "underlying" mechanics; and it would not be up for discussion if nothing of interest had been found: the interesting point in the OP being that of finding functions satisfying Bell's A and B.

In brief, the mechanics goes thus: The HV-carrying particles, their HVs pair-wise correlated by recognised mechanisms, separate and fly to Alice and Bob. Interaction with the respective devices leads to a local transformation of each HV, most clearly seen in W where photons (initially pair-wise linearly-polarised identically) are transformed into pairs with different linear-polarisations. (Representing a fact accepted early in the foundations of QM: a "measurement" perturbs the measured object.) ... ... ...

Since the classical analysis is straight-forward, and Einstein-local (but see below), I suggest you study it and then see how it applies to your interest in Herbert's Paradox and its mechanics.

If you ensure that every step in your classical analysis satisfies Einstein-locality, the accompanying part of the analysis MUST relate to determining the distribution of the Einstein-local outcomes. That brings in probability theory ("maths is the best logic") to derive the frequencies that will be found experimentally. And, classically, you need to clearly distinguish between causal independence and logical dependence.

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.

Malus' famous Law is strictly limited to W. To move beyond that we move to Malus' Method: doing what we expect he would have done classically if he (like us) was confronted with data from multi-particle (Alice and Bob, EPRB-style experiments; GHSZ, GHZ, CRB, etc.) experiments. (NB: Malus' Law makes interesting reading in the QM context of particles being detected one-at-a-time; perhaps trickier than your comment suggests, in my view.)

PS. Your "Note in passing" that "Einstein-locality [EL, per GW] 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".

My "note in passing" could equally have been "NB" or "friendly reminder to the diligent reader" -- it was (IMHO) incidental to the discussion in that EL is not the main point to be proved. Rather the main point , it seems to me, is to shoot-down the classical analysis if it fails to be totally faithful to EL. For if EL is breached, anywhere in the classical analysis, then that analysis would be next to worthless.

So you should check to see how EL is dealt with (once and for all, at the start of the analysis), and then ensure that the remaining classical maths is focussed on determining the frequencies of the various outcomes that will be found experimentally: with no unintended disruption or fiddling-with EL; nor cheating.

As to what Bell proved, it is my opinion that he proved that EPR elements of physical reality are untenable. (A conclusion I support.) So, imho, it is possible to see EL maintained in Bell's work, and popular ideas about reality condemned.

Do you wonder then: Where does the classical analysis here depart from Bell's analysis?

You will see that nowhere here, classically, do we address a third device, at orientation c, in the same context as discussing an experiment with Alice (device-orientation a) and Bob (b).** That move by Bell, it seems to me, confirms his focus on EPR elements of reality. For, otherwise, he needs must recognise that a measurement locally perturbs the measured system ... and until that perturbation, EPR elements of reality (generally) do not exist (IMHO). Or, to put it another way: the move to c follows from an acceptance of EPR's epr; though there may be other views of reality that also permit it ... remembering that Bell's theorem is not a property of quantum theory (Peres 1995, 162), so it is not unreasonable to examine the extent to which it is NOT a property of classical theory.

PS: Discussion of this line would be best in a new thread, it seems to me. (The focus here should be on finding errors in the classical approach.)

** That is: The classical analysis ranges over (a, b), (b, c), (a, c); reflecting all possible real experiments, but no impossible ones. Also: The HVs are classically sourced from infinite sets so that (here, in this case) no two pairs of particles are the same (P = 0).

With thanks again,

GW

 

[billschnieder]

That's an indefinite integral AKA an antiderivative. What's being discussed in this thread is a definite integral, and the definite integral of f(λ) with respect to λ is obviously independent of λ.

Oh, I did not know there is a type of integral ("anti-derivative") of f(λ) that results in a function of λ, thank you, and that Gordon was not allowed to use any such integral in this thread . :-?

 

[Gordon Watson]

Oh, I did not know there is a type of integral ("anti-derivative") of f(λ) that results in a function of λ, thank you, and that Gordon was not allowed to use any such integral in this thread . :-?

Bill, I'm no saint, and a recent visit to FQXi shows me that there are "debaters" worse than you and I, or DrC and ttn. BUT I think that many at PF need to get the "SOL" out of their systems. (L= liver.) I already retracted earlier SOLs (of mine) here, and am still working on it. How about you join me? Your stuff is too good to get you suspended, or seen in the wrong light. Though I am not yet sure that you and I are on the same page re our world-views and the nature of reality? (I'm trying to use maths to eliminate the many words that are spoken on the subject.)

If I had the time I'd start a thread: "Bell's theorem and Einstein-locality: Bill and Gordon in concert?"

For we seem to have similar detractors!?

So, for now, without getting off thread: Do you have any reservations about Einstein-locality?

Thanks, from a reforming

GW

EDIT: PS: Thanks lugita15, I thought your comment was fair, good and helpful! THANKS!

 

[billschnieder]

Bill, I'm no saint, and a recent visit to FQXi shows me that there are "debaters" worse than you and I, or DrC and ttn. BUT I think that many at PF need to get the "SOL" out of their systems. (L= liver.)

Point taken, SOLs are out.

If I had the time I'd start a thread: "Bell's theorem and Einstein-locality: Bill and Gordon in concert?"

Oh no please don't. Such a thread will be closed faster non-local causes propagate . Arrgghhh I did it again, more SOLs.

So, for now, without getting off thread: Do you have any reservations about Einstein-locality?

No. I do not share your view here, especially the one expressed in the following quote

As to what Bell proved, it is my opinion that he proved that EPR elements of physical reality are untenable. (A conclusion I support.)

But then maybe you & I have different understandings of what "EPR elements of reality" mean.

 

[Gordon Watson]

Point taken, SOLs are out.


Oh no please don't. Such a thread will be closed faster non-local causes propagate . Arrgghhh I did it again, more SOLs.


No. I do not share your view here, especially the one expressed in the following quote


But then maybe you & I have different understandings of what "EPR elements of reality" mean.

I think this is enough for me for now, my problem is with the reading/understanding of EPR's "corresponding". But note the EDIT below [the BUT] re the way I'm reading you:


Bill = ?: 1. No. No reservations re Einstein-locality. Surely?? Else my SOL is rising from the confused (to me) reply?

Bill = ?: 2. BUT I do not share your view here, especially the one expressed in the following quote ...

OK, and fair-enough, with strong agreement that my (GW) quote (written in haste)** needs expansion and improvement ... [Bill, I'm taking "here" to mean my reply to Harald. Is there more "here" here?]

Also: SOL + smiley = Salt on the Liver = much more palatable I guess to non-vegetarians. Well done! Move to top of the reform class. (Me being a raw-food vegetarian where WTF = Where's the fruit)

**PS: Harald, Apologies; my EDIT time expired; improved expressions needed; maybe via your questioning. GW

PPS: Another vow from me: This is to be my last hurried post from the middle of a meeting. With no more from the middle of airports, etc., either! You-all'll just have to wait for MHOs where H no longer means Hurried:

GW

 

[billschnieder]

Bill: 1. No reservations re Einstein-locality. Surely?? Else my SOL is rising from the confused (to me) reply?

Bill: 2. BUT I do not share your view here, especially the one expressed in the following quote ...

And my understanding of your position is

GW-1: No reservations re Einstein-Locaity
GW-1: Reservations re EPR elements of reality


GW-1 = Bill-1 = agreement
GW-2 ?≠? Bill-2 = either disagreement or different definitions

http://www.marxists.org/reference/subject/philosophy/works/ge/einstein.htm

 

[Gordon Watson]

And my understanding of your position is

GW-1: No reservations re Einstein-Locaity
GW-1: Reservations re EPR elements of reality


GW-1 = Bill-1 = agreement
GW-2 ?≠? Bill-2 = either disagreement or different definitions

http://www.marxists.org/reference/subject/philosophy/works/ge/einstein.htm

1. Your understanding is correct. Thanks. And good!

2. To sort out our "possible" disagreement re EPR-epr may take a while. If you think it relevant here, maybe discuss it here? I think first need for me here is to re-cap with some neatly formatted equations so that critiques can focus on the essentials.

3. This is interesting, from ttn, my emphasis (and ignoring the subsequent sentence):

"Thanks to Sheldon Goldstein (private communication) for pointing out that Tim Maudlin also stresses this point in “Space-time in the quantum world,” in Bohmian Mechan- ics and Quantum Theory: An Appraisal, edited by James T. Cushing, Arthur Fine, and Sheldon Goldstein (Kluwer Academic Publishers, 1996), p. 305. Maudlin writes that “Physicists have been tremendously resistant to any claims of non-locality, mostly on the assumption (which is not a theorem) that non-locality is inconsistent with Relativity. The calculus seems to be that one ought to be willing to pay any price – even the renunciation of pretensions to accurately describe the world – to preserve the theory of Relativity. But the only possible view that would make sense of this obsessive attachment to Relativity is a thoroughly realistic one! These physicists seem to be so certain that Relativity is the last word in space-time structure that they are willing even to forego any coherent account of the entities that inhabit space-time.”" http://arxiv.org/pdf/quant-ph/0404016v2.pdf

4. Maybe I'm a thorough-going realist? BUT let's not get off thread.

 

[Gordon Watson]

And my understanding of your position is

GW-1: No reservations re Einstein-Locaity
GW-1: Reservations re EPR elements of reality


GW-1 = Bill-1 = agreement
GW-2 ?≠? Bill-2 = either disagreement or different definitions

http://www.marxists.org/reference/subject/philosophy/works/ge/einstein.htm

Please copy and paste the bits that refer to EPR-epr, or that you want discussed here because relevant to this thread. I find little to fault in Einstein's article: but EPR and their epr-definition is not so clear to me:- "corresponding" is the bug for me.

Quoting Einstein (from your cited article): "I am, in fact, firmly convinced that the essentially statistical character of contemporary quantum theory is solely to be ascribed to the fact that this [theory] operates with an incomplete description of physical systems."

 

[billschnieder]

I find little to fault in Einstein's article: but EPR and their epr-definition is not so clear to me:- "corresponding" is the bug for me.

As concerns "corresponding", here is my view, edited from a previous post in another thread:

Apparently not everyone understands correspondence is difference from equivalence ... As the following illustration demonstrates:

- Elements of reality = Wide spectrum wavelength photons from the sun
- Observation = DrC wears red goggles and looks at the sun
- Observable = Red Sun

- EPR: we can predict the observable with certainty, therefore there is an element of reality corresponding to that observable. That is, we can predict with certainty that if DrC wears red goggles and looks at the sun, he will see a red sun. Therefore there exists an element of reality (photons in the red-wavelength region) from the sun.

- EPR: Realism means the "elements of reality" ie, the red-wavelength photons, exist independently of the act of observation. Just because the "red-wavelength photons from the sun" exist independent of measurement, does not mean "red-sun" is an element of reality, it simply means "red-sun" corresponds to an element of reality which in this case is the "red-wavelength photons from the sun".

Looking at the EPR quote again, they say:

If, without in any way disturbing a system, we can predict with certainty (i.e., with probability equal to unity) the value of a physical quantity, then there exists an element of physical reality corresponding to this physical quantity

Note a few very important things often ignored:
1) Predict with certainty - probability 1.0 !
2) They did not say "then this physical quantity IS an element of physical reality", they say it corresponds to one.

So "spin-component along x" maybe a physical quantity which corresponds to "spin" which is an element of physical reality. If a theory can predict "spin-component along x" with CERTAINTY, then there exists an element of reality corresponding to it (ie, spin). Now you can predict spin-component along an infinite number of directions with certainty and still they will all correspond to just one element of reality, the spin.

Take three such "physical quantities" which we predict with certainty to be components along "x", "y", "z". Let us manipulate them together, do algebra and come up with some inequalities which for the purpose of this explanation, we call Bill's inequalities ,

xy + yz + xz ≥ V

Practically only one of "x", "y", "z" can be measured at a time on a particle, so therefore only the one which was measured now exists as an "actual outcome". The limitations of experimentation does not change the fact that the element of reality, "spin", exists. Bill's inequality is not testable because it is an expression involving terms which can not be simultaneously realized (actualized) in an experiment. Note that the inequality expresses a relationship between pairs of spin components of the same particle But some people, on the basis of statistics, naively think that if we average over a very large number of different particles, we can obtain the same relationship.

So if we take spin components from different particles and plug in Bill's inequalities, we find that it is violated. Does that mean elements of reality do not exist? Of course not. Now QM makes a prediction for what we will obtain if we measure "xy" on a large number of particles. Now we naively take the result from QM and plug it into the inequality and it is violated. Does that mean elements of reality do not exist? Of course not. Bill's inequality is valid so long as we understand the meaning of the terms in it. Our error is that we have now taken oranges from QM and oranges from experiments and we expect an inequality which was derived for apples to still be obeyed.

 

[Gordon Watson]

As concerns "corresponding", here is my view, edited from a previous post in another thread:



Looking at the EPR quote again, they say:


Note a few very important things often ignored:
1) Predict with certainty - probability 1.0 !
2) They did not say "then this physical quantity IS an element of physical reality", they say it corresponds to one.

So "spin-component along x" maybe a physical quantity which corresponds to "spin" which is an element of physical reality. If a theory can predict "spin-component along x" with CERTAINTY, then there exists an element of reality corresponding to it (ie, spin). Now you can predict spin-component along an infinite number of directions with certainty and still they will all correspond to just one element of reality, the spin.

Take three such "physical quantities" which we predict with certainty to be components along "x", "y", "z". Let us manipulate them together, do algebra and come up with some inequalities which for the purpose of this explanation, we call Bill's inequalities ,

xy + yz + xz ≥ V

Practically only one of "x", "y", "z" can be measured at a time on a particle, so therefore only the one which was measured now exists as an "actual outcome". The limitations of experimentation does not change the fact that the element of reality, "spin", exists. Bill's inequality is not testable because it is an expression involving terms which can not be simultaneously realized (actualized) in an experiment. Note that the inequality expresses a relationship between pairs of spin components of the same particle But some people, on the basis of statistics, naively think that if we average over a very large number of different particles, we can obtain the same relationship.

So if we take spin components from different particles and plug in Bill's inequalities, we find that it is violated. Does that mean elements of reality do not exist? Of course not. Now QM makes a prediction for what we will obtain if we measure "xy" on a large number of particles. Now we naively take the result from QM and plug it into the inequality and it is violated. Does that mean elements of reality do not exist? Of course not. Bill's inequality is valid so long as we understand the meaning of the terms in it. Our error is that we have now taken oranges from QM and oranges from experiments and we expect an inequality which was derived for apples to still be obeyed.

Bill,

Thanks for addressing a possible difference in our world views. However, it seems to me that there is so much wriggle-room in your wordage that any conclusion would be (for me) inconclusive. For example, this next bit worries me, and seems to be a stretch in the EPR context:-

"Now you can predict spin-component along an infinite number of directions with certainty and still they will all correspond to just one element of reality, the spin."

Questions: Why then bother with the prediction-without-disturbing? What relevance does "prediction" have, if things are as you say?

Possibly I'm biassed by the case that I now mount in the context of this thread:

1. Consider a number of experiments Q_i (i = 1-N), such a series forming a run: Q $\in$ {W, X, Y, Z}.

2. Here is my genuine prediction (no tricks); the primes indicating things in Bob's locale:

$P(\delta _{b}'\lambda' \rightarrow \textbf{b}^+|Q, \;experiment \; number \; i\;selected \;by \;GW) = 1.$​

3. Where, please is the EPR-epr? The one (in your words) that EPR say it corresponds to? The one (presumably) existing ["there exists"] before the selected i-th experiment is concluded?

4. Especially when I tell you that (regarding correspondence) the probability of the predicted outcome having ever occurred before (anywhere in the Cosmos) is zero:

$P(\textbf{b}^+|heretofore) = 0.$​

5. HOWEVER: If you say that, after that particular experiment is concluded (confirming the prediction), THEN there exists a new epr ... well, THEN we agree. Indeed we would agree that a new epr had been brought into EXISTENCE by Q. Then we might also agree that EPR's definition of an EPR-epr is not so good?

PS: The simplicity of my case is designed to make it easy for you to correct any defective perceptions I may have about EPR.

Thanks again,

GW

 

[billschnieder]

For example, this next bit worries me, and seems to be a stretch in the EPR context:-

"Now you can predict spin-component along an infinite number of directions with certainty and still they will all correspond to just one element of reality, the spin."

Questions: Why then bother with the prediction-without-disturbing? What relevance does "prediction" have, if things are as you say?

Two reasons:
1) Because it should be a theory not an experiment. A theory can predict with certainty and yet experimental limitations may not allow exact measurement.
2) Completeness: a theory which by itself, i.e. without any experiment, can not predict with certainty, is not complete. If a theory says the probability is 0.98, then there is information missing from it that would have allowed it to give a CERTAIN prediction (P=1 or 0).

Possibly I'm biassed by the case that I now mount in the context of this thread

The case you mount in this thread in my view is to show that the QM result can be obtained from a locally causal theory of hidden variables, contrary to Bell.

1. Consider a number of experiments Q_i (i = 1-N), such a series forming a run: Q $\in$ {W, X, Y, Z}.

2. Here is my genuine prediction (no tricks); the primes indicating things in Bob's locale:

$P(\delta _{b}'\lambda' \rightarrow \textbf{b}^+|Q, \;experiment \; number \; i\;selected \;by \;GW) = 1.$​

3. Where, please is the EPR-epr? The one (in your words) that EPR say it corresponds to? The one (presumably) existing ["there exists"] before the selected i-th experiment is concluded?

It is easier to see if you focus on ONE photon for a moment, not a whole series. What do you think $\lambda_i$ represents? $1$ is the physical quantity which corresponds to the element of reality $\lambda_i$ which belongs to the one photon under consideration. In a complete theory, by knowing $\lambda_i$ and the mechanism that governs interactions, $\rightarrow$, you may predict with certainty that the interaction $\lambda_i \rightarrow b_i^+$ will happen resulting in the physical quantity $1$. In other words, if we can predict with certainty without in any way disturbing the system the physical quantity $1$, then the corresponding $\lambda_i$ exists.

I guess I do not quite understand your reservation to EPR's elements of reality. More later.

 

[billschnieder]

3. Where, please is the EPR-epr? The one (in your words) that EPR say it corresponds to? The one (presumably) existing ["there exists"] before the selected i-th experiment is concluded?

Imagine yourself as the i-th photon, leaving the source flying toward a device. The EPR question is, "Do you have properties $\lambda_i$ which exist as part of your "identity" which ultimately interact with Bob's device to result in a measurement outcome $+1$?"

The EPR point of view is that if a theory can predict $+1$ with certainty, then the theory is complete with respect to the hidden element of reality $\lambda_i$ which corresponds to $+1$

 

[Gordon Watson]

Imagine yourself as the i-th photon, leaving the source flying toward a device. The EPR question is, "Do you have properties $\lambda_i$ which exist as part of your "identity" which ultimately interact with Bob's device to result in a measurement outcome $+1$?"

The EPR point of view is that if a theory can predict $+1$ with certainty, then the theory is complete with respect to the hidden element of reality $\lambda_i$ which corresponds to $+1$

Bill, I'm responding to this post first, in the hope it might ease my response to #243 (see PS at foot). Or (please, Bill) it might help me avoid that need by your saying, "Ah-Ha, I see the light!" For it seems to me that this second post of yours (where it is accurate) is moving you closer to my position re EPR-eprs.

1. What you suggest above ("imagine") is exactly what I do. (Is it not?)

2. But at that level (QM-level), "reality is veiled" from us (in d'Espagnat's terminology; or h > 0).

3. But, on the classical level (where I come from), reality is unveiled.

4. So I just continue in that vein, seeking to move from the classically unveiled to unveil as much of the "veiled reality" as I can.

5. So I just use PT (here, classical Probability Theory), taking the view that a veiled reality may be represented by a probability distribution.

6. So I ("imagining, as usual") answer your good question, "Do you [dear photon i] have properties λ_i which exist as part of your "identity" which ultimately interact with Bob's device to result in a measurement outcome +1?":

Yes, of course!​

7. Then, re this from you: "The EPR point of view is that if a theory can predict +1 with certainty, then the theory is complete with respect to the hidden element of reality λ_i which corresponds to +1."

7a. I suggest that this is NOT accurate at all (see PS): and may be misleading you! See how your question (to me, as photon) differs re +1? FOR I THINK this says something quite different; or maybe I'm missing something?

7b. Can you not say: λ_i is the specific HV that delivers the output +1 during the Q_i particle/device interaction?

8. Note this re your quote in #7 above, EPR say (with no mention of "theory" but with focus on existence): "If, without any way disturbing a system, we can predict with certainty, the value of a physical quantity, then there exists an epr corresponding to this physical quantity."

9. So, is it not the case that my prediction is disturbance-free wrt the subject system?

10. So does it not follow, does it not remain the case, that the epr represented by b⁺ is brought into existence by the particle-device interaction; and not otherwise? Is it not the case that λ_i, the pre-interaction HV, is transformed (during the interaction) to become the previously-non-existent (the now-post-interaction existent) b⁺?

11. In a nutshell: Does it not remain the case that EPR's "corresponding" is just plain WRONG?

Bill, this might help you rephrase #243 and I can just reply to your next post? Or else, maybe rephrase questions for #243 with this new info in mind?

PS: See this, as part of my difficulty in responding, in #243: "1 is the physical quantity*** which corresponds to the element of reality λ_i which belongs to the one photon under consideration."

In my view, 1 represents a green-light, a printed +1, a beep, ... it represents the output of the device AS IT TOO is transformed during the interaction. Yes?

*** How about: The OUTPUT +1 "corresponds to" the element of reality b⁺! For both +1 (the device output) and b⁺ (the particle's new property) represent/denote/signal/tell-us of a photon linearly-polarised-parallel-wrt-the-principal-axis-of-the-device (s = 1) or of a spin-half particle spin-up-wrt-the-principal-axis-of-the-device (s = 1/2).

PPS: I am sure that we are NOT saying or implying the same thing; that it's NOT just semantics. Do you agree?

Thanks,

GW