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A classical challenge to Bell's Theorem?

by Gordon Watson
Tags: bell, challenge, classical, theorem
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harrylin
#217
Apr30-12, 11:01 AM
P: 3,187
Quote Quote by billschnieder View Post
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
#218
Apr30-12, 11:07 AM
P: 683
Quote Quote by harrylin View Post
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
#219
Apr30-12, 11:29 AM
P: 273
Quote Quote by billschnieder View Post
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.

Quote Quote by billschnieder View Post
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.

Quote Quote by billschnieder View Post
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.

Quote Quote by billschnieder View Post
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
#220
Apr30-12, 11:55 AM
P: 273
Quote Quote by billschnieder View Post
Continuing ...
[itex]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][/itex]
[itex]= 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)
[/itex]
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.

Quote Quote by billschnieder View Post
Now let us say [itex]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 \;,\;[/itex]
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.

Quote Quote by billschnieder View Post
[itex]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][/itex]
[itex]= \sum_{ij} (A^i_a\cdot B^j_b)P(ij|V), \;\;\;\; ij \in [++, +-, -+, --][/itex]
[itex]= 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)[/itex]
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
#221
Apr30-12, 12:05 PM
P: 683
Quote Quote by Delta Kilo View Post
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 [itex]E(A_aB_b)_V[/itex] which was implicit in the derivation. All that matters is that we calculate the expectation values of the type [itex]E(A_aB_b)_V[/itex] 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
#222
Apr30-12, 01:03 PM
P: 683
Quote Quote by Delta Kilo View Post
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
#223
Apr30-12, 02:08 PM
P: 683
Therefore adding the "a" and "b" labels? We have:

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

and
Quote Quote by Gordon Watson View Post
All that remains now is to use Malus' Method to get [itex]P(B^+|Q,\,A^+)[/itex] (or [itex]P(B^+|Q,\,A^-)[/itex] 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 [itex]P(B^+_b|Q,\,A^+_a)[/itex] or [itex]P(B^+_b|Q,\,A^-_a)[/itex] is implied in the above quote.
billschnieder
#224
Apr30-12, 02:28 PM
P: 683
Quote Quote by Gordon Watson View Post
[itex]A({\textbf{a}}, \lambda) = \int d\lambda\delta (\lambda - {\textbf{a}}^+\bigoplus {\textbf{a}}^-) cos[2s({\textbf{a}}, \lambda)] = \pm 1.

[/itex]

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

Where [itex]\bigoplus[/itex] = 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:
[itex]A(a, \lambda) = \int d\lambda\; \delta (\lambda - a) \cos[2s\cdot(a,\lambda)] = \pm 1[/itex]
where (a,λ) is the angle between the vectors a and λ. What does XOR add to the whole thing? Do you mean [itex] a \in [a^+, a^-][/itex]?

Am I understanding you correctly?
Gordon Watson
#225
Apr30-12, 04:39 PM
P: 375
Quote Quote by billschnieder View Post
Gordon, I wouldn't blame DK for misunderstanding the expression because it is unclear. The way I understand it is like:
[itex]A(a, \lambda) = \int d\lambda\; \delta (\lambda - a) \cos[2s\cdot(a,\lambda)] = \pm 1[/itex]
where (a,λ) is the angle between the vectors a and λ. What does XOR add to the whole thing? Do you mean [itex] a \in [a^+, a^-][/itex]?

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: δaλ → a+a-. ALT: δaλ → {a+, a-} for W, X, Y, Z. With apologies for the hurried short-hand.
.......
EDIT:

Bill, et al., to avoid the XOR: [itex]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.[/itex]

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 ([itex]\delta_{\textbf{a}} \lambda[/itex]) is a process that transforms ([itex]\rightarrow[/itex]) an element of a set ([itex]\lambda \in \Lambda[/itex]) into exactly one element of a dichotomic set [itex](\left \{{\textbf{a}}^+ , {\textbf{a}}^- \right \})[/itex].
.......

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
#226
Apr30-12, 07:36 PM
P: 273
Quote Quote by billschnieder View Post
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: [itex]A^{\lambda_1}_a = A(a,\lambda_1) = +1[/itex], 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.

Quote Quote by billschnieder View Post
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.

Quote Quote by billschnieder View Post
A and B already take different values for different values of a and b!
Not according to the formulas you wrote.

Quote Quote by billschnieder View Post
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={aj}, b={bk}
[itex]A^{\lambda_i}_{a_j}=A(a_j,\lambda_i)=\pm 1[/itex] 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
#227
Apr30-12, 08:00 PM
P: 683
Quote Quote by Delta Kilo View Post
You assign (potentially different) values for A and B for each distinct value λi, but not for each distinct a and b: [itex]A^{\lambda_1}_a = A(a,\lambda_1) = +1[/itex], 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:

[itex]A(a,\lambda_1) = +1 \;,\;
A(a,\lambda_2) = +1 \;,\;
A(a,\lambda_3) = -1 \;,\;
A(a,\lambda_4) = -1 \;,\;
[/itex]
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 in to 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:

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

I do not see a legitimate criticism here. What you are saying does not make any sense.
Delta Kilo
#228
Apr30-12, 08:19 PM
P: 273
Quote Quote by Gordon Watson View Post
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
#229
Apr30-12, 08:39 PM
P: 683
Quote Quote by Delta Kilo View Post
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
#230
Apr30-12, 09:03 PM
P: 273
Quote Quote by billschnieder View Post
if f(λ) = cos(λ), then ∫f(λ)dλ = sin(λ) a function of λ, contrary to your ridicule above.
[img]facepalm.jpg[/img]
I'm outta here. Good luck with your quest.
Gordon Watson
#231
Apr30-12, 11:20 PM
P: 375
Quote Quote by Delta Kilo View Post
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 http://www.physicsforums.com/showpos...&postcount=225

GW
lugita15
#232
Apr30-12, 11:39 PM
P: 1,583
Quote Quote by billschnieder View Post
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
#233
May1-12, 03:34 AM
P: 375
Quote Quote by Gordon Watson View Post
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 [itex]P(B^+|Q,\,A^+)[/itex] (or [itex]P(B^+|Q,\,A^-)[/itex] 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
Quote Quote by harrylin View Post
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 -- δaλ, δbλ' -- 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.

Quote Quote by harrylin View Post
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.)

Quote Quote by harrylin View Post
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
#234
May1-12, 08:51 AM
P: 683
Quote Quote by lugita15 View Post
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 . :-?


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