
#55
Feb2107, 03:31 PM

P: 161

(Figure 1) D(a) < w(s) [Source] w'(s') > D'(b') Two objects flyapart [w with property s (a unitvector); w' with property s' (a unitvector)] to respectively interact with detectors D (oriented a, an arbitrary unit vector) and D' (oriented b', an arbitrary unit vector). The detectors D (D') respectively project s (s') onto the axis of detectororientation a (b'). Let w and w' be created in a state such that (1) s + s' = 0; say, zero total angular momentum. That is: (2) s' = s. Then the lefthand result is a.s and the righthand result is s'.b'; each a dotproduct. To derive the related correlation, we require (using a recognised notation http://en.wikipedia.org/wiki/Column_vector ), with < ... > denoting an average: (3) <(a.s) (s'.b')> (4) =  <(a.s) (s.b')> (5) =  <[(ax ay az) (sx, sy, sz)] [(sx sy sz) (bx', by', bz')]> (6) =  (ax ay az) <(sx, sy, sz) (sx sy sz)> (bx', by', bz') (7) =  (ax ay az) <s.s> (bx', by', bz') (8) =  (ax ay az) <1> (bx', by', bz') (9) =  a.b' (10) =  cos (a, b'). Let s and s' be classical angularmomenta. Then (to the extent that we meet all the Belltheorem criteria) the result is a wholly classical refutation of Bell's theorem. [It is Bell's constrained realism that we reject; thereby maintaining the commonsense locality clearly evident above.) E and OE! QED? Critical comments most welcome, (though I'll be away for a day or so),wm 



#56
Feb2107, 03:57 PM

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1. Do you disagree that in a classical world, if the results of two measurements exhibit a 100% correlation, this must be either because one measurement determined the other, or because the results of both measurements were determined by some other event or events? 2. Do you disagree if you have a 100% probability that two measurements using the same settings always give the same results as one another on repeated trials, and the two measurements have a spacelike separation and we assume no possibility of FTL signals, then the only way to explain the perfect 100% correlation in a classical world is to assume that on each trial the results were predetermined by some event or events (presumably the emission of both signals/objects from a common origin at the source) which lie in the past light cone of both measurementevents? (if you disagree, can you suggest an alternative explanation?) 3. Do you disagree that if there was any random element to the results of either experimenter's measurement on a given trial where they both use the same settings (and I'm only talking about randomness in the outcome an experimenter will get if we know both his detector setting and the precise state of the signal/object they're measuringthe original event or events which determined the state of both object/signals at the source may still have a random element), then the probability they both get the same answer could not be 100%? If you agree that the answers must have been predetermined on the trials where they both picked the same detector setting, then if we also add the assumption that this predetermining event or events could not in any way be affected by information about what detector settings each experimenter will choose (note that this condition can be assured in a classical universe obeying locality if each experimenter chooses their setting randomly shortly before the measurement, so that a signal moving at the speed of light would not have time to travel from the event of an experimenter picking a setting to the event of the other experimenter making their measurement), then this means the answers must have been predetermined on *every* trial, even the ones where they pick different settings. 



#57
Feb2107, 03:59 PM

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So the correlation is not found by taking the expectation of the product of their expectation values, but rather by taking the product of the outcomes (the +1 or 1 for each), and weighting that with the relative probabilities for this to happen ASSUMING that, whatever probability distribution is given on the Aside (as a function of the local setting and the local unit vector) for the +1 and the 1, it is INDEPENDENT of the probability distribution on the Bside (as a function of the local setting and the local unit vector there). 



#58
Feb2107, 04:33 PM

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1. This is basicly akin to saying "let's assume Bell's Theorem is wrong", which is handwaving. You have to provide us something that yields results consistent with QM AND is local AND meets the realism requirement. You can't just say you have accomplished this because s and s' are classical. 2. Specifically, what are the expected probabilities for the 8 permutations: A+B+C+ A+B+C ... ABC You will find that you cannot fill in such a table with nonnegative numbers and still match QM. In other words, you have ignored the realism requirement entirely. DrC 



#59
Feb2107, 04:39 PM

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Yeah, what Vanesch said. If the value a.s represents the probability of the left detector getting result +1, and (1  a.s) is the probability of the left detector a getting 1, and s'.b' is the probability of the right detector getting +1, and (1  s'.b') is the probability of the right detector getting 1, then presumably the expectation value for the correlation would be:
(a.s)*(s'.b') + (1  a.s)*(1  s'.b')  (a.s)*(1  s'.b')  (1  a.s)*(s'.b') or 4*(a.s)*(s'.b')  2*(a.s + s'.b') + 1 



#60
Feb2107, 04:47 PM

P: 161

1. Could you let me see how you would do the QM derivation, please? 2. Here's what I was thinking with my classical maths: In the doublepeaked output from an SG magnet, we allocate +1 xor 1 in accord with the direction of the output. Say: +1 = UP; 1 = DOWN. That is, we do not allocate a different number to those particles which arrive (say) at the bottom of the UP distribution as opposed to those which have emerged at the upside of the UP distribution. All the UPs get +1, etc. Thus, the number (+1 xor 1) being allocated in line with the direction (UP xor DOWN) irrespective of the position in either distribution: I thought that +1 xor 1 could equally be allocated (equally arbitrarily) in accord with the sign of the dotproduct in my classical example. ??? Thanks again, wm 



#61
Feb2107, 04:55 PM

P: 161

If I anywhere have probabilities going negative, JUST SHOOT ME! a.s is a dot product that make take on values from 1 to +1. It cannot be a probability in my classical maths. I haven't look at the rest of your post. I will (later) if you want me to? Best, wm 



#62
Feb2107, 05:02 PM

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[(1/2)*(a.s) + 1/2]*[(1/2)*(s'.b') + 1/2] + [1/2  (1/2)*(a.s)]*[1/2  (1/2)*(s'.b')]  [(1/2)*(a.s) + 1/2]*[1/2  (1/2)*(s'.b')]  [1/2  (1/2)*(a.s)]*[(1/2)*(s'.b') + 1/2] Surprisingly, this all seems to simplify to an expectation value of (a.s)*(s'.b'), which is what you were calculating in the first place! Were you making this assumption about the probabilities all along, or is it just lucky? Either way, I think the math in your proposed proof that this is equal to cos(a, b') is incorrect, see my next post for more on that point. Also note that if you make this assumption about probabilities, the results of the two detectors will not be perfectly correlated when they pick the same setting, so if you're out to challenge Bell's theorem, you can only look at inequalities like the CHSH inequality which do not make any assumption about perfect correlations with identical settings. Alternately, you might avoid probabilities by saying that on any trial where the value of a.s was greater than or equal to 1 and smaller than 0, the experimenter will see the result spindown (assigned a value of 1), and on any trial where the value of a.s was greater than or equal to 0 and smaller than or equal to 1, the experimenter will see the result spinup (assigned a value of +1). Then you could say the same applies to s'.b', and calculate the expectation value of the product of their two results; but again, it would be something different than cos (a, b'). Quickly diagramming the problem leads me to think that if s is equally likely to have any angle, then if (a, b') represents the angle between a and b' in degrees, the probability that they both get the same spin would be (a, b')/180, and the probability they get opposite spins would be [180  (a, b')]/180, so the expectation value for the product of their results would be (a, b')/180  [180  (a, b')]/180, or [(a, b')/90]  1. Either way, I think you need to fix it so each experimenter can only get two discrete results on a given trial. If you know of any Bell inequalities that do not assume each measurement can have only one of two possible results, then please give the name of the inequality you're thinking of, or a link giving the mathematical formulation of the inequality. 



#63
Feb2107, 06:45 PM

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[tex] (a_x * s_x + a_y * s_y + a_z * s_z)*(b'_x*s_x + b'_y*s_y + b'_z*s_z)[/tex] But this is not equal to [tex](a_x * b'_x + a_y * b'_y + a_z * b'_z)[/tex], even if you stipulate that [tex](s_x * s_x + s_y * s_y + s_z * s_z) = 1[/tex]. It seems like you got the rules for the dot product confused with the rules for multiplication, you can't say that (a.s)*(s.b') is equivalent to (s.s)*(a.b'). 



#64
Feb2207, 06:50 PM

P: 161

This is a bit rushed as I am in a meeting BUT: 1. I'd like to point out that I began with the exact equation that Bell used [1964; equation (3)]. I get the identical result also:  a.b'. 2. NB: At the moment I have limited my derivation to that which I offered: A wholly LOCAL and CLASSICAL derivation of the EPRB correlation. That is, I have derived the limit to which your derivation must tend in accord with Bohr's Correspondence Principle. 3. I hope we might agree on the following important point: Since the spacelike experimental results were derived by me in terms of highschool maths, AND without any reference whatsoever to nonlocality, there must be an equivalent QM derivation equally devoid of nonlocality. 4. So: May I ask you again to provide your fundamental derivation of the EPRB correlation (ie, from first principles; and preferably in the terms of the OP), beginning with Bell's equation (just as I did)? 5. I request this of you because you are a PF MENTOR and because SCIENCE ADVISER DrC has not been able to derive it and SCIENCE ADVISER JesseM is a bit confused on my mathematics (but I will sort that out soon: noting for now that there are no errors in my maths, so far as I can see from my highschool text on vectoranalysis). 6. Your derivation will not be wasted as I am keen to learn. HOWEVER: If you will not be providing this important derivation; could you please point to where I might find a detailed version; preferably one that complies with your own local interpretation of QM? 7. Not to muddy the waters any further now: I respectfully suggest that there are other matters in your post which may be presented differently and more clearly. Not to be addressed now because they may be clarified when I see your EPRB derivation. Thank you, and sincerely, wm 



#65
Feb2207, 09:54 PM

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You can pick a single item out of Bell's paper, and quote it out of context and it still won't mean anything. You might consider toning down your claims a bit until you see them all the way through. I will repeat what I have stated previously: there is a mathematical requirement that you are skipping entirely, and that is the requirement of realism. If you ignore that, you are missing the entire point of Bell. That requirement is that there is a real probability of a specified outcome of observations at settings A, B and C which has a value between 0 and 1. You do not need the formula you are tinkering with to derive Bell's Theorem, as Mermin has shown. 



#66
Feb2207, 10:26 PM

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(+1)*probability(angle a yields +1, angle b yields +1) + (+1)*probability(angle a yields 1, angle b yields 1) + (1)*probability(angle a yields +1, angle b yields 1) + (1)*probability(angle a yields 1, angle b yields +1) This experiment is not one where the result of each measurement is an arbitrary real number between 1 and +1, and where the expectation value is the average value of the product of these two real numbers, as you seem to assume in your example. Again, Bell is assuming that each measurement always yields one of two results which are assigned values +1 and 1, so when you multiply the two values you always get the result 1 or +1 on any given trial; the expectation value refers to the average this product over many trials. As I pointed out in a previous post, if you assume that each experimenter has a device which projects the vector s onto their own angle (either a or b'), like a.s, and then this continuous value is used to determine the probability (1/2)*(a.s) + 1/2 that the experimenter will get a +1 result on that trial or a 1 result, then it actually does work out that the expectation value for the product of their results will end up being a.s*s'.b' as you had in your attempted proof. But again, in this case you don't have a guarantee that when they pick the same angle they always get opposite results on a given trial, so Bell's theorem would only rule out inequalities which don't include this assumption, like the CHSH inequality. edit: by the way, if you are familiar with calculations in QM, you can look at this page for a nearly complete derivation. What they derive there is that if q represents the angle between the two detectors, then the probability that the two detectors get the same result (both spinup or both spindown) is [tex]sin^2 (q/2)[/tex], and the probability they get opposite results (one spinup and one spindown) is [tex]cos^2 (q/2)[/tex]. If we represent spinup with the value +1 and spindown with the value 1, then the product of their two results when they both got the same result is going to be +1, and the product of their results when they got different results is going to be 1. So, the expectation value for the product of their results is: [tex](+1)*sin^2 (q/2) + (1)*cos^2 (q/2) = sin^2 (q/2)  cos^2 (q/2)[/tex] Now, if you look at the page on trigonometric identities here, you find the following identity: [tex]cos(2x) = cos^2 (x)  sin^2 (x)[/tex] So, setting 2x = q, this becomes: [tex]cos(q) = cos^2 (q/2)  sin^2 (q/2)[/tex] Multiply both sides by 1 and you get: [tex]sin^2 (q/2)  cos^2 (q/2) =  cos (q)[/tex] This fills in the final steps to show that the expectation value for the product of their results will be the negative cosine of the angle between their detectors. 



#67
Feb2207, 11:44 PM

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#68
Feb2207, 11:58 PM

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I moved the exchange between ttn and I to a new thread (what we see is bogus in MWI), because it started to hijack this one...




#69
Feb2307, 12:35 AM

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No, I am not referring to that. Looking at Bell as a reference, so we're talking about the same thing: 1. wm's " a . b" deal is the QM expectation value, referenced as (3). It really doesn't matter how you get to this, the key is that you know that it reduces to cos theta for spin 1/2 particles. Obviously, for photons it is a slightly different formula. If this was all there was to it, then Bell would have stopped after (7) or so  and wouldn't have too much. 2. Bell then goes to great pains to show that the mapping of (2) to (3), with A and B, WILL work. I.e. that a hidden A and hidden B is possible, and will get you to the QM predictions if you need it to. So we still don't have much. 3. Then around (14), Bell introduces the realism requirement: the mapping with hidden variables does not extend to an A, B AND C! He does not label it as the "realism requirement", that is something I label it as because it is present in EVERY proof of Bell's Theorem one way or another. It is the assumption  requirement  that there simultaneously be an A, B and C to discuss. If there isn't, then there is no (15) which is the Inequality. Why do we need this assumption? Because without it we can't see the real problem that occurs when wm says: "(4) =  <(a.s) (s.b')> becomes (9) =  a.b' " Clearly this is the original classical hidden variable idea in disguise, which Bell says is "no difficulty". But this doesn't work if there is a c too, and Bell somehow figured that out. (Amazing accomplishment to me...) 4. In my proofs of Bell's Theorem, I always make this assumption *explicit*. I will repeat that without simultaneous A, B and C: there is no Bell's Theorem. You can reformulate the theorem in many ways, such as my negative probabilities version and my version that follows Mermin. These versions substitute easier math, or at least an easier notation for most people to follow  and is built around one specific counterexample. Bell presents a more general proof and then picks a specific example (ac=90 degrees, ab=bc=45 degrees is how I read it) to show the issues. DrC 



#70
Feb2307, 12:55 AM

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#71
Feb2307, 01:08 AM

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In other words: his formula may have some issues with it, but no one is really doubting that Bell's (2) and (3) can be made to work together as long as you limit it to considering a and b. Bell himself says exactly that! And then he introduces c, and that leads immediately to the Inequality. So when you include the reality assumption, you get the Inequality. The Inequality is violated in nature; therefore one of the assumptions is wrong. The assumptions are locality and realism; and one of these needs to be thrown out. P.S. Besides, you can't do what you are saying about A.S, B.S and C.S.  this is precisely what Bell shows. The reason is that these 3 cannot be made to be internally consistent. I.e. the relationship A.S to B.S, A.S to C.S, and B.S to C.S won't work. 



#72
Feb2307, 01:20 AM

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