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A classical challenge to Bell's Theorem? 
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#1
Mar2512, 04:11 PM

P: 375

This post moved from "Nick Herbert's proof?"
http://www.physicsforums.com/showthread.php?t=589134 at the request of the OP. Effects without a cause in Herbert's experiment? I presume that you believe that some quantum events have no cause; not classical effects? So I would welcome any and all comments and calculations on the following scenario, based on a typical Belltest setup and the CHSH inequality. We replace the quantumentanglementproducing source with a classical source which sends a short pulse of light to Alice and Bob each day (over many years), each pulse correlated by having the same linearpolarization; though each day the common pulse polarizationorientation is different . Let x denote any variable of your choosing. Then (as in a standard Bellanalysis) Alice's results are represented by (1) A(a, x) = ±1 where a is any analyzer orientation of her choosing; Bob's by (2) B(b, x) = ±1 where b is any analyzer orientation of his choosing; (3) 0 [itex]\leq[/itex] ρ(x); (4) ∫ρ(x) dx = 1. Please (after Bell, showing all your workings), calculate the expectation: (5) E(AB) = ∫AB ρ(x) dx. Please provide the maximum value achievable for the CHSH inequality under these conditions. With thanks in advance, GW EDIT added with move: I'd like to understand how physicists and mathematicians deal with the above wholly classical setting in the context set by Bell (1964) when arriving at his theorem. Thanks. 


#2
Mar2512, 04:24 PM

P: 375

Moved from "Nick Herbert's proof?" http://www.physicsforums.com/showthread.php?t=589134
Thanks for this comprehensive reply. It is appreciated. BUT: Given the wholly classical setting in my example, and the use of Bell's (1964) formulation, I thought there was enough info there for physicists and mathematicians to proceed? Or (at least), explain why they cannot? 


#3
Mar2512, 04:35 PM

P: 375

Moved from "Nick Herbert's proof?" http://www.physicsforums.com/showthread.php?t=589134
To be clear: When the Green light (say) blinks on a detector in a Belltest, are you saying that this event has, ultimately (when analyzed), no upstreamcause (e.g., in ordinary 3space)? Also: To "measure a pure state" is to perturb it, right? A complicating factor? 


#4
Mar2512, 06:59 PM

P: 678

A classical challenge to Bell's Theorem?
Then (as in a standard Bellanalysis) Alice's results are represented by (1) A(a, x) = ±1 where a is any analyzer orientation of her choosing; Bob's by (2) B(b, x) = ±1 where b is any analyzer orientation of his choosing; (3) 0 ≤ ρ(x); (4) ∫ρ(x) dx = 1. I wonder what the CHSH inequality will look like. I can bet it will be identical to the one derived by gill1109 above, even though the scenario is manifestly nonlocal. What gives? 


#5
Mar2512, 07:33 PM

P: 375

1. Not sure about your mystical being? Purpose =? (Is something more needed to clarify the OP?) 2. The CHSH inequality formulation will be the same, imho: since the experimental outcomes are ±1, no matter the settings a, b, etc. 3. To clarify the OP (if that's your issue): Having derived the expectation E(AB) for the classical setting  from your functions for A and B = ±1  what then is the related maximum value that that classical setting might yield for the CHSH inequality? That is: What a, b, c, d settings yield the maximum value in the CHSH formula, and what is that maximum? 4. Is it gill1109's +2? 5. Did you mean to say "the scenario is manifestly LOCAL"??? 6. So  addressing your "What gives"  just give me your answers to the OP  or tell me why you can't. Especially as it seems that Bell might think you can; the given situation being wholly classical and involving no more than Bell's proposed (1964, etc.) analytical formulation. 


#6
Mar2612, 01:26 AM

P: 678

 Without specifying the method by which the common pulse orientations are chosen, it is not possible to calculate an expectation value.Without ρ(x) we are hopeless to calculate a meaningful E(AB) even if A(a,x) and B(b,x) are clearly specified.  The maxumum attainable is of course +2 as gill1109 calculated. Aside: However, note the following extremely important point A(a1)B(b1)A(a1)B(b2)A(a2)B(b2)A(a2)B(b1) → A(a1)[B(b1)B(b2)]  A(a2)[B(b2)B(b1)] → A(a1)[B(b1)B(b2)] + A(a2)[B(b1)B(b2)] → [A(a1) + A(a2)]*[B(b1)B(b2)] > **!!! if A(a1) = A(a2) = +1, and B(b1) = B(b2) = +1 Or, A(a1) = A(a2) = 1 and B(b1) = B(b2) = 1, we obtain the maxium of 2. If A(a1) = A(a2) = ±1, OR B(b1) = B(b2) = ±1, we get a value of zero. And if A(a1) = A(a2) = 1 and B(b1) = B(b2) = +1 Or A(a1) = A(a2) = +1 and B(b1) = B(b2) = 1, we obtain the minimum of 2. This may seem like a pointless way to arrive at the same result as gill1109 except it is obvious from the emphasized expresion that the original 4 terms (A(a1)B(b1), A(a1)B(b2), A(a2)B(b2), A(a2)B(b1)) of products in the inequality originate from only 4 functions (A(a1), B(b1), A(a2), B(b2)) which must be factorizable. You can not use 4 different runs of an experiment (i, j, k, l) to obtain results from 8 functions (A(a1i), B(b1i), A(a1j), B(b2j), A(a2k), B(b2k), A(a2l), B(b1l)) and expect the inequality to work. It is a simple exercise to verify that for the case where 4 different runs of the experiment are performed, the maximum of the expression will be A(a1i)B(b1i)  A(a1j)B(b2j) A(a2k)B(b2k) A(a2l)B(b1l) <= 4 NOT 2. Some naively leave out the experiment identifyers (i,j,k,l) and fool themselves into thinking the result can be factorized. In order for the results from 4 different experiments to be factorizable the following equalities must hold A(a1i) = A(a1j) A(a2k) = A(a2l) B(b1i) = B(b1l) B(b2j) = B(b2k) Practically, this means if the experimental results consisted of a list of numbers (+1, 1) for each function and you obtained 8 columns for 4 different experiments, the data MUST be sortable such that 4 of the columns are duplicates, not only in the numbers of +1s and 1s but also in the switching pattern. Therefore it is not sufficient that A*B for one experiment gives you a certain expection value for the paired product. For the inequality to have a maximum of 2, rather than 4, the value of one pair must constrain the value of a different pair in some manner. 


#7
Mar2612, 01:57 AM

P: 1,414

I'm not saying that there's anything wrong with anything that Gordon Watson and billschnieder have said. And maybe one or both of their approaches will one day explain BI violations in a way that an ignorant layman such as myself might understand. However, currently, I don't think so. I think that a true understanding of why BI violations don't inform wrt the deep reality is more subtle, and yet simpler, than either have yet pinpointed.
Just in the humble, and perhaps quite wrong, opinion of an ignorant layperson. But, yeah, the factorizability of the entangled state would seem to be the key to it. Because this is a composite of the functions that determine individual detection. And the variable that determines individual detection isn't relevant wrt the rate of coincidental detection. Just a certain point of view. Maybe it's important, maybe not. 


#8
Mar2612, 02:30 AM

P: 72

Bill Schnieder: nothing gives. The CHSH inequality is true. Locality in all these discussions concerns only the measurement settings and the measurement outcomes. The hidden variables x or lambda may as well be known throughout the whole universe. You can think of them, if you like, as being "in the measurement apparatus" and "in the particles". But you don't have to think that way. The real assumption in deriving CHSH is the "reality" and timespace location of the outcomes of the unperformed measurements, alongside of those actually performed; and the freedom of the experimenters to choose which measurements to perform.



#9
Mar2612, 02:39 AM

P: 72

Bill Schnieder: you wanted outcomes of four experiments to match exactly (your i,j,k,l indices). The way I think about it, in one run of the experiment there are potential outcomes A1, A2, B1, B2. Alice and Bob each toss a coin to choose which outcome to actually observe (A1 or A2, B1 or B2). Then this is repeated many times. We assume that their coin tosses are independent of the physical systems generating A1, A2, B1, B2. Then the average of, say, A1 times B2 over all runs will hardly differ from the average over those runs where Alice chose "1", Bob chose "2". The averages over all runs satisfy CHSH. Hence the observed averages do too, up to statistical variation.



#10
Mar2612, 03:42 PM

P: 375

The pulse orientations are many (a very large number) and random in orientation. In my terms: 'a uniform distribution' (the same distribution that we'd expect with Bell's λ, though it's a different beast). Is that fair enough? I think that leaves you needing to specify A and B, consistent with Bell's 'analytical protocol' for the study of local realism: each equal to ± 1. Or telling me why you cannot? 


#11
Mar2612, 03:57 PM

P: 375

But: The OP is wholly classical. And you are not (entirely) an ignorant layman (being familiar with Malus, at least; as well as the point that you make right here, above). So the challenge remains. That is: Time to do some basic maths! (And cut the words?) 


#12
Mar2612, 04:23 PM

Sci Advisor
PF Gold
P: 5,299

Gordon, this is a straight classical setup, so of course the CHSH will have its traditional upper limit and no experiment will exceed it (as Richard says, within normal statistical deviation).
Specifically, the value of the function (5) E(AB) is .25 + .5(cos^2(AB)) which is the classical expectation when there is separability. To get statistics in which CHSH is violated, you must have entanglement. So really, what point are you trying to make? 


#13
Mar2612, 07:12 PM

P: 375

However, referring to the OP (and after correcting the typos in your equation above): 1. I see no equations for A and B, with each satisfying ± 1 (the boundary condition required by Bell's formulation)? 2. What maximum did you derive for the CHSH inequality under the subject conditions? 3. What were the related a, b, c and d? Completion of these tasks should bring us to the point I'm seeking to make from my wholly classical scenario. Thanks again. 


#14
Mar2712, 01:48 AM

P: 72

Gordon: the point is that whatever the functions A, B and whatever the angles and whatever the probability distribution of the hidden variables, CHSH will be satisfied.



#15
Mar2712, 08:46 AM

Sci Advisor
PF Gold
P: 5,299

2. Traditional CHSH upper limit is always 2. The candidate local realistic estimate is not a factor, as it is model dependent. 3. I don't know what you are referring to. [...gently prodding us forward as I suspect there is a point just around the corner...] 


#16
Mar2712, 09:16 AM

P: 678

3. (a, b, c, d) are the 4 angles for the CHSH experiment I understand. 


#17
Mar2712, 02:56 PM

P: 375

This is getting to be quite a lovein; for which, Many thanks!
TomT everfriendly and cautiously seeking. gill1109 helpful, balanced and conventional. DrC getting 100* for nicely trying (when he can be very)! Bill rightly helping DrC to move ahead and get a better score. DrC and I in some sort of general agreement. If only Bill would read his email and understand why I must run? Hoping to reply to all in about 12 hours; thanks again. GW */1000:)) Can do better! 


#18
Mar2812, 02:29 PM

P: 375

Indeed, would you agree that no experiment (real or imagined) has ever contradicted a mathematical truism? In fact Feynman's defective analysis of the doubleslit experiment ("no one understands") in part arises from his belief in what is NOT (in general) a mathematical truism: P(xX) + P(xY) = P(xZ) (!?) So I would welcome your having a go at the essential challenge in the OP. That is, use Bell's widely accepted localrealistic protocol to analyse what is clearly a localrealistic experiment. If you can't deliver the requisite A and B, perhaps you could explain why? At least derive the maximum value that the experiment could deliver using the traditional CHSH formula? That way I can check my own calculation, which will be brought into later discussion. Thanks. 


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