# A classical challenge to Bell's Theorem?

by Gordon Watson
Tags: bell, challenge, classical, theorem
P: 375
This post moved from "Nick Herbert's proof?"

at the request of the OP.

 Quote by gill1109 When I said randomness I did not refer to unpredictable (experimental) phenomena. When you toss a coin, the result depends deterministically on the initial conditions. That is familiar everyday randomness which is merely practical unpredictability. QM on the other hand says that nature is intrinsically random. There is no hidden layer "explaining" what actually will happen. The randomness is spontaneous. Inexplicable. Without antecedent. Effects without a cause.
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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 Bell-test set-up and the CHSH inequality.

We replace the quantum-entanglement-producing 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 linear-polarization; though each day the common pulse polarization-orientation is different .

Let x denote any variable of your choosing. Then (as in a standard Bell-analysis) 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 $\leq$ 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.

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.
P: 375
Moved from "Nick Herbert's proof?" http://www.physicsforums.com/showthread.php?t=589134

 Quote by gill1109 GW: if you don't tell me the functions A(a,x), B(b,x) and rho(x) I obviously cannot calculate E(A(a)B(b)). However I can tell you that A(a1)B(b1)-A(a1)B(b2)-A(a2)B(b2)-A(a2)B(b1) (a function now only of x) can only take the values +2, 0 and -2. One way to see that is to note that the product of the four terms A(a1)B(b1), A(a1)B(b2), A(a2)B(b2), A(a2)B(b1) is +1, hence an even number of these four terms is equal to +1 and an even number is equal to -1. Therefore if A(a1)B(b1)=+1, then A(a1)B(b2)+A(a2)B(b2)+A(a2)B(b1) = +3 or -1; if A(a1)B(b1)=-1, then A(a1)B(b2)+A(a2)B(b2)+A(a2)B(b1) = +1 or -3. Now just check the 2x2 combinations. Hence the average of A(a1)B(b1)-A(a1)B(b2)-A(a2)B(b2)-A(a2)B(b1) (averaged over x) cannot exceed 2, either.
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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?
P: 375
Moved from "Nick Herbert's proof?" http://www.physicsforums.com/showthread.php?t=589134

 Quote by gill1109 Effects without a cause (not just in Herbert's experiment, but in QM in general): I mean that quantum events in general have no cause. By quantum events I mean results of measuring pure states which are not certain, but for which QM can only tell us the probability.
Thanks. Very puzzling, as worded, so I need to think about it.

To be clear: When the Green light (say) blinks on a detector in a Bell-test, are you saying that this event has, ultimately (when analyzed), no upstream-cause (e.g., in ordinary 3-space)?

Also: To "measure a pure state" is to perturb it, right? A complicating factor?

P: 724
A classical challenge to Bell's Theorem?

 if you don't tell me the functions A(a,x), B(b,x) and rho(x) I obviously cannot calculate E(A(a)B(b)). However I can tell you that A(a1)B(b1)-A(a1)B(b2)-A(a2)B(b2)-A(a2)B(b1) (a function now only of x) can only take the values +2, 0 and -2. One way to see that is to note that the product of the four terms A(a1)B(b1), A(a1)B(b2), A(a2)B(b2), A(a2)B(b1) is +1, hence an even number of these four terms is equal to +1 and an even number is equal to -1. Therefore if A(a1)B(b1)=+1, then A(a1)B(b2)+A(a2)B(b2)+A(a2)B(b1) = +3 or -1; if A(a1)B(b1)=-1, then A(a1)B(b2)+A(a2)B(b2)+A(a2)B(b1) = +1 or -3. Now just check the 2x2 combinations. Hence the average of A(a1)B(b1)-A(a1)B(b2)-A(a2)B(b2)-A(a2)B(b1) (averaged over x) cannot exceed 2, either.
Let us change the original scenario a little as follows: We remove the source altogether and replace it with a mystical being who governs a mystical parameter (x) which combines with their chosen angles to produce a +/-1 result. Each day over many years, he instantly decides what parameter (x) is, the instant before Alice and Bob make their measurements, whoever does it first. The only condition being that the same (x) parameter is governing both experiments.

Then (as in a standard Bell-analysis) 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 non-local. What gives?
P: 375
 Quote by billschnieder Let us change the original scenario a little as follows: We remove the source altogether and replace it with a mystical being who governs a mystical parameter (x) which combines with their chosen angles to produce a +/-1 result. Each day over many years, he instantly decides what parameter (x) is, the instant before Alice and Bob make their measurements, whoever does it first. The only condition being that the same (x) parameter is governing both experiments. Then (as in a standard Bell-analysis) 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 non-local. What gives?
Bill, you talkin' to me? (In that you cite gill1109.)

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.
P: 724
 Quote by Gordon Watson Bill, you talkin' to me? (In that you cite gill1109.) 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.
Sorry for hijacking your thread Gordon, I was just responding to the portion by gill1109. To answer your questions, and more on topic.

- 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.
 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.
 P: 141 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 time-space 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.
 P: 141 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.
P: 375
 Quote by billschnieder Sorry for hijacking your thread Gordon, I was just responding to the portion by gill1109. To answer your questions, and more on topic. - 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 maximum attainable is of course +2 as gill1109 calculated.
Bill, I see no hijacking; so, for me, no problem at all. Then, leaving your 'aside' aside for the moment:

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?
P: 375
 Quote by ThomasT 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.
Dear ThomasT, the OP is intended to be as 'subtle and as simple' as it gets! (Perhaps it fails?)

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?)
 Sci Advisor PF Gold P: 5,441 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(A-B)) 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?
P: 375
 Quote by DrChinese 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(A-B)) 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?
Thanks DrC, nice reply; I much appreciate your having a go. We seem to be in agreement thus far.

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.
 P: 141 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.
PF Gold
P: 5,441
 Quote by Gordon Watson Thanks DrC, nice reply; I much appreciate your having a go. We seem to be in agreement thus far. 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.
1. Does that mean I don't get a 100?

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...]
P: 724
 Quote by DrChinese 1. Does that mean I don't get a 100? 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...]
1. I think Gordon wants to see your equations for A(a,x) and B(b,x) for the scenario he described and show how you arrived at E(AB) = .25 + .5(cos^2(a-b)) from those equations.
3. (a, b, c, d) are the 4 angles for the CHSH experiment I understand.
 P: 375 This is getting to be quite a love-in; for which, Many thanks! TomT ever-friendly 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!
P: 375
 Quote by gill1109 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.
Thank you. Yes; agreed; for the classical example given in the OP. That is: The OP's classical example remains consistent (under any setting) with the classical (traditional) CHSH.

Indeed, would you agree that no experiment (real or imagined) has ever contradicted a mathematical truism? In fact Feynman's defective analysis of the double-slit experiment ("no one understands") in part arises from his belief in what is NOT (in general) a mathematical truism:

P(x|X) + P(x|Y) = P(x|Z) (!?)

So I would welcome your having a go at the essential challenge in the OP. That is, use Bell's widely accepted local-realistic protocol to analyse what is clearly a local-realistic 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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