
#73
Feb2307, 04:20 AM

P: 161

Thanks also for the crossreferences that you give me. Now, to the maths. Does this help you: 1. Note first that a, b', s and s' are unitvectors, NOT angles. 2. And Yes; the dot product between the unitvectors is the cosine of the angular difference. 3. (5) looks OK to me. I don't see two column vectors multiplied. 4. Note that in your equational comparison, you appear to have overlooked the fact that one expression is an ensemble average, initially over two fixed unitvectors and two random (but opposite) unitvectors (of infinite variety). 5. I'm pretty sure that I did not mix the rules? 6. I'm pretty sure that the maths is spoton. But please have a look at the above comments and let me know. I've more to come but I would like to take it correctstep by correctstep. Thanks again, in haste for now, wm 



#74
Feb2307, 04:36 AM

P: 161

As far as I am aware, I DO NOT violate the realism specifically defined by me. (I'll post it later). Rather, I use it (the most general commonsense local realism) IN THAT I allow specifically the measurement outcome to be a consequential perturbation of the particledetector interaction. The s and s' are real (and random); the a and b' are arbitrary (as they should be). The consequential projection is real. QED it seems to me? For now, I'll leave it to you two to come to some agreement. Regards, wm 



#75
Feb2307, 04:43 AM

P: 161

Quick suggestion only, wm PS: It works for me! 



#76
Feb2307, 08:24 AM

Sci Advisor
P: 8,470

(5) =  <[(ax ay az) (sx, sy, sz)] [(sx sy sz) (bx', by', bz')]> presumably this would be (5) =  <[(ax ay az).(sx, sy, sz)]*[(sx sy sz).(bx', by', bz')]> correct? But then when you write for (6) (6) =  (ax ay az) <(sx, sy, sz) (sx sy sz)> (bx', by', bz') Would this be (6) =  (ax ay az).<(sx, sy, sz).(sx sy sz)>.(bx', by', bz') or (6) =  (ax ay az)*<(sx, sy, sz).(sx sy sz)>*(bx', by', bz') Or what? Neither makes sense to me. Again, it really seems to me you are mixing up the rules for the dot product and ordinary multiplication here. [tex] \frac{1}{2\pi} \int_{0}^{2\pi} cos(\theta  a)*cos(\theta  b) \, d\theta[/tex] Are you claiming that this would work out to [tex] cos (a  b)[/tex]? If you enter Cos[x  a] Cos[x  b] into the integrator, you get something fairly complicated: (x*Cos[a  b])/2 + ((Cos[2*x]*Sin[a + b])/2 + (Cos[a + b]*Sin[2*x])/2)/2 Calculating (expressionabove(x=2pi)  expressionabove(x=0)) gives: (2pi*Cos[a  b])/2 + ((Sin[a + b])/2)/2  ((Sin[a + b])/2)/2 Huh, so the integral actually does work out to pi*Cos[a  b], and if you multiply by the factor of 1/2pi outside the integral it comes out to 1/2 * Cos[a  b], only off from what you had by a factor of 1/2. Someone should check my math, and I still don't think your own proof makes sense, but this would at least suggest your end result is almost right. Even so, I still don't see this as a counterexample to what Bell proved about the impossibility of local hidden variables explaining quantum results, since Bell was assuming in his 1964 paper that each experimenter only sees +1 or 1 on each trial, and that when they choose the same angle they get opposite results (do you disagree that he was making these assumptions?) On the other hand, you seem to assume each experimenter can get a continuous result between 1 and +1, and even if you adopt my suggestion of having the number a.b be the basis for a probability (1/2)(a.b)  (1/2) of getting +1, you still would not ensure that the experimenters always get opposite results when they pick the same angle (anyway, if I'm right about the extra factor of 1/2 in front of the cos(ab), then you're not duplicating the quantum expectation value exactly, so you may not even violate the inequality in the 1964 paper in the first place, I'd have to check that). Most of the Bell inequalities I know of depend on the assumption that experimenters get opposite (or identical) results on the same measurement setting, this is the key reason for the conclusion of determinism which I talked about in an earlier post...the only exception I know of is the CHSH inequality, but I don't think you'll find 4 angles a, b, a' and b' such that (1/2)cos(a, b) − (1/2)cos(a, b′) + (1/2)cos(a′, b) + (1/2)cos(a′, b′) is not between 2 and 2, which is what you'd need to violate that inequality. 



#77
Feb2307, 09:30 AM

Sci Advisor
PF Gold
P: 5,145

Case A B C %      [1] + + + ? [2] + +  ? [3] +  + ? [4] +  + ? [5]  + + ? [6]  +  ? [7]   + ? [8]    ? It works like this: you can present a hundred derivations and examples that support classical local realism, and you will be in exactly the same spot as Einstein and Bohr were in circa 1935  a pissing match. But it only takes a single counterexample to refute any theory, and that is what Bell presented in 1965. So you can ignore the realism requirement  which I have challenged you above on  and you will learn nothing about why Bell's Theorem is so important. On the other hand, QM does not include the realism requirement. Therefore A, B and C do not need to exist simultaneously. Therefore, there are only 2^2 permutations. I can satsify the table for this quite simply (as Bell explains in the early sections of his paper): Case A B %     [1] + + QM expection value [2] +  QM expection value [3]  + QM expection value [4]   QM expection value The QM expectation value will be one thing for spin 1/2 particles, another thing for spin 1 particles, the cases will add to 100%, and all values will be nonnegative. So in conclusion: looking at examples which support your local realism hypothesis is a waste of time, since you must address Bell's counterexample and it is impossible to refute that. As a result, we conclude that either the Einstein (=Bell) locality or the Einstein (=Bell) realism assumption must be rejected; and which you choose to reject is a matter of personal interpretation. 



#78
Feb2307, 09:42 AM

Sci Advisor
PF Gold
P: 5,145

You are missing the big picture on this. It is not that wm is presenting a counterexample to Bell. wm is presenting an example of local realism, which Bell had already presented a counterexample to. You should be able to see that wm is simply replicating what Bell himself has already shown to be true: that looking at versions of local realism with A and B will work. But if you extend that same logic into a situation with A, B and C, it does NOT work. Don't let the derivation wm presents fool you, because it does not disprove Bell in any way. DrC 



#79
Feb2307, 09:47 AM

Sci Advisor
P: 8,470

In any case, I agree that wm's argument would certainly be a lot clearer if he picked some specific choices of angle for each detector, and then explained which specific Bellian inequality he thinks will be violated in his experiment with those choices of angles. 



#80
Feb2307, 10:45 AM

Sci Advisor
PF Gold
P: 5,145

2. I would definitely agree with your representation on this. 3. He doesn't consider the A/B/C condition. It is not possible to provide a counterexample to Bell, because Bell is itself a counterexample. The only way to disprove Bell would be to show that the counterexample is flawed. For example, consider the "theory" that there no primes larger than 13. Bell comes along and says, whoa! what about 17? Now wm comes along and say Bell is wrong, look at 2, 3, 5, 7, 11, 13 as my proof. No, he must show that 17 is NOT a prime to make his case. Now, because he has seen Bell's Theorem, Mermin (and many others, I just use him as an example) now knows the trick: there are certain specific situations (such as 17, 19, etc.) that are counterexamples. So Mermin can construct a very simple counterexamples to explain the situation, and that is his classic "Is the moon there when nobody looks? Reality and the quantum theory / Physics Today (April 1985) " A review of that shows it as a fine counterexample to the original theory (local realism). And guess what? wm now must prove this wrong too, because it too is a counterexample to be contended with. And what else? Now that they are armed with the "trick", these rather bright guys Greenberger, Horne and Zeilinger come up with yet another counterexample to local realism. And guess what? wm must prove this wrong too. So my point is simple: there is no such thing as a counterexample to a counterexample, the counterexample must actually be proven wrong. And in this case, we now have multiple counterexamples to consider. So the burden of (dis)proof has grown exponentially larger. 



#81
Feb2307, 11:30 AM

Sci Advisor
P: 8,470

What wm attempted to do was give a general proof that for arbitrary angles a and b, in his classical experiment the expectation value for the product of the two results would be cos(a  b). This would cover all specific angles you chould choosefor example, if a=B and b=C, then the expectation value for a large set of trials with these angles would be cos(B  C); if a=C and b=A, then the expectation value for a large set of trials with these angles would be cos(C  A); and so forth. I disagree that "Bell's theorem" primarily revolves around picking specific angles, if that's what you mean by "That 'exercise for the reader' IS Bell's Theorem". The proof involves finding an inequality that should hold for arbitrary angles under local realism; then it's just a fairly simple final step to note that the inequality can be violated using some specific angles in some specific quantum experiment, but this last step is hardly the "meat" of the theorem. For example, look at the CHSH inequality. This inequality says that if the left detector has a choice of two arbitrary angles a and a', the right detector has a choice of two arbitrary angles b and b', then the following inequality should be satisfied under local realism: 2 <= E(a, b)  E(a, b') + E(a', b) + E(a', b') <= 2 Now, suppose wm were correct that he had a classical experiment satisfying the conditions of Bell's theorem such that the expectation value E(a, b) would equal cos(a  b). In this case it we could pick some specific angles a = 0 degrees, b = 0 degrees, a' = 30 degrees and b' = 90 degrees; in this case we have E(a, b) =  cos(0) = 1, E(a, b') = cos(90) = 0, E(a', b) = cos(30) = 0.866, and E(a', b') = cos(60) = 0.5. So E(a, b)  E(a, b') + E(a', b) + E(a', b') would be equal to 1  0  0.866  0.5 = 2.366, which violates the inequality. The hard part was the proof that the expectation value was cos(a  b), just as in QM; once we have this expectation value, it's a pretty trivial exercise for the reader to find some specific angles which allow the inequality to be violated, just as in QM. Again, the problem here is that wm did not actually replicate the conditions assumed in Bell's theorem, where each measurement can only yield two possible answers rather than a continuous spectrum of answers, and also his derivation of the expectation value seems to be flawed, my math suggested the expectation value would actually be E(a, b) = (1/2)*cos(a  b). 



#82
Feb2307, 11:31 AM

P: 122

Second, one can give classical examples which also violates Bell's inequality, when one does not have the correct probability model. For example, a probability model based on behavior (making assumptions about the physics or cause of the behavior) can results in the violation of Bell's inequality. One of the implicit assumptions in Bell type inequalities is that "we" truly understand the physics rather than only understanding the mathematics. For example, the average photon behavior is determined by a single "phase" variable. There is no physics understanding, on an individual photon basis, of why some photons pass through a polarizer/analyzer and others don't. The photon's polarization properties indicates a bivectored object. What has not been considered is that the use of a single "phase" (an average interaction variable) in the present behavioral model does not fully describe the physics that is occurring at the analyzer/polarizer in deciding which photon's pass. It only gives the "on average phase based value". But if the photon is bivectored one would actually expect that the analyzer/polarizer interaction on an individual photon basis (or individual pair basis) should be determined by two "phases" or a phase (e.g. the average of the two vectors) and a second parameter (the spread of the two vectors about the phase average). If this is the case, then Bell's approach of adding hidden variables externally is asking the wrong question (?resulting in the wrong answer?). If it is the "spread" rather than the phase (average of the vectors) that is fundamental to the passage through the polarizer/analyzer then the observed probability of observing a correlated pair (correlation via the hidden phase/spread aspect) can be different than the on average single "phase" would predict. Effectively the pair passes or does't pass changing the pair probability verus the average left or average right polarizar probability. Fundamentally, Bell's inequality and associate interpretation of EPR assumes no hidden physics. A. O. Barut in a published paper essentially pointed in this direction with respect to spin 1/2 particles. It is, for example tacitly assumed that spin 1/2 particles have only a single spin up or single spin down state. In fact given SternGerlach experiments it makes more physics sense that the particle is a spin/magnetic quadrapole with two spin planes at 90 degrees (physically orthagonal rather than QM's mathematical 180 orthagonal spin planes). For such a quadrapole particle SternGerlach experimental results actually make physics sense. Finally, it has been said that a mathematical model is correct if it produces the correct experimental result. Bell's inequalities do not. So it is equally valid to assume we have the wrong mathematical model for the experimental situation (even if we do not know how this could be) as it is to assume nonlocality (even if we don't know how this could be). Maybe our understanding of the physics of particles is not as complete as our understanding of the application of the mathematical model we call QM. But a lack of understanding of the physics is exactly what Feynman meant when he said "No one really understands QM". 



#83
Feb2307, 11:50 AM

Sci Advisor
P: 8,470





#84
Feb2307, 01:43 PM

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P: 8,470





#85
Feb2307, 02:06 PM

P: 161

I accept the (exponential) burden of truth; and will get to my further questions and answers soon ( it's just that I am a bit tiedup at the moment ) because I want to learn. Especially do I want to learn why some see a small piece of the world differently ... ... when that small piece of interest to me can be built from highschool maths and logic (which is about the limit of my current questions and ability). So I'd just like to get it on the record: 1. that many prior and erroneous counterexamples in my small area of interest were longheld and wrong (as shown preeminently by John Bell). 2. that John Bell himself was not happy with his theorem and had not given up on finding a simple constructive counterexample (as I read him). 3. I am not a John Bell, but his simple approach has motivated me to haveago; notwithstanding that many others have hadago and failed. 4. So as soon as there is some general agreement that my highschool maths so far is correct, I would like to continue in that vein to mathematically answer some of the other questions here. (That is, I will move to dichotomic outcomes A = (+, ), B' = (+', '); since doubts and concerns about this issue are being expressed here.) PS: That will introduce standard probability theory (in line with Ed Jaynes' views) which is also among some questions here. 5. To differentiate my local realism from other versions falling under the same phrase, I call it CLR: commonsense local realism. I think that CLR is the way many scientists see the world (while many  but probably in the minority  think that the world cannot be seen that way). 6. For those like me, that are not verballyminded, the simple acid test that I expect to meet is that my views will be consistent with highschool math and logic. 7. Please note that other interpretations of QM support locality; and I support locality. Sincerely, wm 



#86
Feb2307, 02:17 PM

Mentor
P: 40,875





#87
Feb2307, 02:23 PM

P: 161

I sent a note re some of the maths; but I'm not sure if (having read them) you still find an error in the math? Your question about column vectors is answered in the wiki reference that was in my original post. (The column vectors include the commas! as I recall ... but its the commas that differentiate oneway or the other in accord with HS maths.) As I read my equations: My maths is nor defective on that count: so are there any other maths issues ... before we move on to ups and downs? Have I missed something which needs correction? Eh? wm 



#88
Feb2307, 02:27 PM

P: 161

Can you help me, please? Thanks, wm 



#90
Feb2307, 02:50 PM

Sci Advisor
P: 8,470

To add to that, I think it would help a lot if you would address my previous request to make explicit where you are using the dot product and where you are using multiplication:



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