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A classical challenge to Bell's Theorem?

 
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Apr30-12, 01:03 PM   #222
 

A classical challenge to Bell's Theorem?


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.
Apr30-12, 02:08 PM   #223
 
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 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.
Apr30-12, 02:28 PM   #224
 
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?
Apr30-12, 04:39 PM   #225
 
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.
Apr30-12, 07:36 PM   #226
 
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 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 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 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.
Apr30-12, 08:00 PM   #227
 
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.
Apr30-12, 08:19 PM   #228
 
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.
Apr30-12, 08:39 PM   #229
 
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.
Apr30-12, 09:03 PM   #230
 
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.
Apr30-12, 11:20 PM   #231
 
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
Apr30-12, 11:39 PM   #232
 
Blog Entries: 1
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.
May1-12, 03:34 AM   #233
 
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 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 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 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
May1-12, 08:51 AM   #234
 
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 . :-?
May1-12, 02:43 PM   #235
 
Quote by billschnieder View Post
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!
May1-12, 03:22 PM   #236
 
Quote by Gordon Watson View Post
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
Quote by Gordon Watson View Post
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.
May1-12, 03:51 PM   #237
 
Quote by billschnieder View Post
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
May1-12, 05:20 PM   #238
 
Quote by Gordon Watson View Post
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/su...e/einstein.htm
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