JesseM said: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
Sorry, I didn't realize that with the way you allow s and a to vary, a.s could be negative; but see below.wm said:Dear JesseM, this is a bit rushed, BUT:
If I anywhere have probabilities going negative, JUST SHOOT ME!
But all the Bell inequalities I know of are based on the assumption that each measurement can only have two distinct outcomes, like "spin-up" and "spin-down". So, I was assuming that when the detector projects s onto a using the dot product, the resulting value is then used as a probability to display one of the two possible results, which I assign values +1 and -1 (following the convention used in the CHSH inequality which we've discussed before). I hadn't noticed that a.s could be negative, but you could always remedy this by making the probability equal to (1/2)(a.s) + 1/2, which will be a value between 0 and 1. Obviously this would mean you'd have to modify my equations above...if (1/2)*(a.s) + 1/2 is the probability of the left detector getting result +1, and 1 - [(1/2)*(a.s) + 1/2] = 1/2 - (1/2)*(a.s) is the probability of the left detector getting result -1, and (1/2)*(s'.b') + 1/2 is the probability of the right detector getting result +1, and 1 - [(1/2)*(s'.b') + 1/2] = 1/2 - (1/2)*(s'.b') is is the probability of the right detector getting result -1, then the expectation value for the product of their two results is:wm said:a.s is a dot product that make take on values from -1 to +1. It cannot be a probability in my classical maths.
To add to the issue I brought up in my previous post about the need for only two possible outcomes, there also seems to be an error in your math here. Let's say a = 0 degrees, b' = 60 degrees, and s = 90 degrees. Since they all are of unit length, the dot product of any of these two vectors is just the cosine of the angle between them. So from (4) we have - (a.s) (s.b') = - cos(90) * cos(30) = 0. But from (9) we have have - a.b' = - cos(60) = -0.5, so (4) does not seem to be equal to (9). Steps (5) and (6) in your proof don't make sense to me--in (5), is that supposed to be two column vectors multiplied by each other? The dot product is supposed to be a row vector times a column vector, not a column vector times a column vector. If you avoid vector notation and just write out both dot products from (4) in terms of components, it seems to me (5) would be something like this:wm said: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').
vanesch said:The problem is: the outcome is not a continuous quantity! It is a discrete quantity, with PROBABILITY equal to the numbers you give, with a shift. You only give the expectation values of the outcomes, but the trick is that each individual outcome is a +1 or a -1, and not a continuous value in between both (although their expectation is of course).
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 A-side (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 B-side (as a function of the local setting and the local unit vector there).
wm said: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 high-school text on vector-analysis).
In physics it is important to understand what physical quantity the terms in an equation stand for. In Bell's paper a and b represent possible angles of the stern-gerlach device used to measure the spin of the two particles, and these measurements will always yield one of two results, which in the "Formulation" paragraph on p. 1 of the paper you refer to he labels as +1 and -1. The "expectation value" refers to the average expected value of the product of the two measurements, which would be:wm said: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'.
I don't see how the correspondence principle would imply that the expectation value for an experiment in which each experimenter can get any result between +1 and -1 on a given trial would be identical to the expectation value for an experiment in which each experimenter can only get one of two results, either +1 or -1. Is this what you're claiming here?wm said: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.
If you were indeed able to reproduce the result that the expectation value is -cos(a, b) in a purely classical experiment, where on each trial each experimenter gets either +1 or -1 and the expectation value is for the average of the products of their two answers, and your classical experiment obeyed the conditions of Bell's theorem like the source not having foreknowledge of the detector settings, then yes, this would show that QM was compatible with local hidden variables. The problem is you didn't do this--you seem to assume that each experiment can yield a continuous spectrum of values rather than just +1 or -1, and even if you make the assumption I mentioned above where the probability of getting +1 is (1/2)*(a.s) + 1/2, so that the expectation value is indeed just a.s*s'.b', there seems to be an error in your "high school math", since this is not equal to -cos(a, b).wm said:3. I hope we might agree on the following important point: Since the space-like experimental results were derived by me in terms of high-school maths, AND without any reference whatsoever to non-locality, there must be an equivalent QM derivation equally devoid of non-locality.
You're looking for a derivation of why quantum mechanics predicts that the expectation value is -a.b? Why would this be useful, since here we are just trying to figure out whether this expectation value can be reproduced in a classical experiment?wm said: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)?
Yes, please state whatever theorems from your vector textbook you are making use of in your proof. But in the meantime, could you please check the math on my example of a = 0 degrees, b' = 60 degrees, and s = 90 degrees? Do you disagree that in this case, - a.s*s.b' = - cos(90)*cos(30) = - (0)*(0.866) = 0, while - cos(a, b') = - cos(60) = -0.5? If you agree with my math on this example, then it seems clear there must be an error in your proof somewhere, unless I misunderstood what you claimed to have proved.wm said: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 high-school text on vector-analysis).
Have you ever studied the basics of QM? Derivations of probabilities and expectation values have nothing to do with one's interpretation, they basically just involve finding state vector representing the quantum state of the system, expanding it into a weighted sum of eigenvectors of the operator representing the variable you want to measure (energy, for example), and then the square of the complex amplitude for a given eigenvector represents the probability that you'll get a given value when you measure that variable (the value corresponding to a particular eigenvector is just the eigenvalue of that vector). And of course, once you know the probability for each possible value, the expectation value is just the sum of each value weighted by its probability. If you're not familiar with the general way probabilities and expectation values are derived in QM, then a specific derivation of the expectation value for the spins of two entangled electrons won't make much sense to you. And like I said, the derivation itself would have nothing to say about locality or nonlocality, it's just when you apply Bell's theorem to the predictions of QM that you see they are not compatible with local hidden variables.wm said: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?
When you say wm is not satisfying the requirement of realism, are you referring to the idea that wm's example involves negative probabilities? If you look at wm's post #65, I don't think that's the problem--while it's true that the dot product of the vector s sent from the source and the vector a representing the experimenters measurement setting can be negative, I don't think wm intended for this dot product to be a probability in the first place. Rather, it seems to me that wm has just failed to realize that Bell was assuming that each experiment could only yield two discrete results, spin-up or spin-down; wm is instead imagining an experiment where each experiment can yield a continuous infinity of results between -1 and 1, and the "expectation value" he's calculating is for the product of the real number that each experimenter gets as a result.DrChinese said:will repeat what I have stated previously: there is a mathematical requirement that you are skipping entirely, and that is the requirement of realism.
JesseM said:When you say wm is not satisfying the requirement of realism, are you referring to the idea that wm's example involves negative probabilities? ...
But why do you think wm's example violates this assumption of realism? In his example, if one experimenter can pick for his measurement setting a one of three possible angles a = A, a = B, or a = C, then if you know the angle of the vector S sent out by the source, you can determine in advance the value of A.S if the experimenter picks angle A, and the value of B.S if the experimenter picks B, and the value of C.S if the experimenter picks C. Of course these values may be any real number between -1 and 1 depending on the angles involved, whereas in the experiments Bell is dealing with, "local realism" means that if you know the hidden state of the particle, then you can determine in advance whether each of the three detector angles will yield either the result -1 or +1, with no other possibilities.DrChinese said:It is the assumption - requirement - that there simultaneously be an A, B and C to discuss.
JesseM said:But why do you think wm's example violates this assumption of realism? In his example, if one experimenter can pick for his measurement setting a one of three possible angles a = A, a = B, or a = C, then if you know the angle of the vector S sent out by the source, you can determine in advance the value of A.S if the experimenter picks angle A, and the value of B.S if the experimenter picks B, and the value of C.S if the experimenter picks C. Of course these values may be any real number between -1 and 1 depending on the angles involved, whereas in the experiments Bell is dealing with, "local realism" means that if you know the hidden state of the particle, then you can determine in advance whether each of the three detector angles will yield either the result -1 or +1, with no other possibilities.
But in wm's notation a and b don't represent 2 particular angles, they're just variables representing arbitrary angles chosen by the left and right detectors. You could easily say that the left detector has a choice between 3 possible angles a=A, a=B, and a=C, and likewise that the right detector has a choice between the same 3 possible angles b=A, b=B, and b=C. When you talk about Bell saying that A and B are not enough, but that you also need to consider C, don't A, B, and C represent three particular detector settings that each experimenter can choose between?DrChinese said: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.
JesseM said:To add to the issue I brought up in my previous post about the need for only two possible outcomes, there also seems to be an error in your math here. Let's say a = 0 degrees, b' = 60 degrees, and s = 90 degrees. Since they all are of unit length, the dot product of any of these two vectors is just the cosine of the angle between them. So from (4) we have - (a.s) (s.b') = - cos(90) * cos(30) = 0. But from (9) we have have - a.b' = - cos(60) = -0.5, so (4) does not seem to be equal to (9). Steps (5) and (6) in your proof don't make sense to me--in (5), is that supposed to be two column vectors multiplied by each other? The dot product is supposed to be a row vector times a column vector, not a column vector times a column vector. If you avoid vector notation and just write out both dot products from (4) in terms of components, it seems to me (5) would be something like this:
- (a_x * s_x + a_y * s_y + a_z * s_z)*(b'_x*s_x + b'_y*s_y + b'_z*s_z)
But this is not equal to -(a_x * b'_x + a_y * b'_y + a_z * b'_z), even if you stipulate that (s_x * s_x + s_y * s_y + s_z * s_z) = 1. 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').
JesseM said:But why do you think wm's example violates this assumption of realism? In his example, if one experimenter can pick for his measurement setting a one of three possible angles a = A, a = B, or a = C, then if you know the angle of the vector S sent out by the source, you can determine in advance the value of A.S if the experimenter picks angle A, and the value of B.S if the experimenter picks B, and the value of C.S if the experimenter picks C. Of course these values may be any real number between -1 and 1 depending on the angles involved, whereas in the experiments Bell is dealing with, "local realism" means that if you know the hidden state of the particle, then you can determine in advance whether each of the three detector angles will yield either the result -1 or +1, with no other possibilities.
DrChinese said:It is not that he violates it, it is that he has not included it. You can see that his "formula" is simply the early part of Bell's paper. So nothing has happened! It is as if he showed 1+1=2 and says that proves classical reality. He is trying to assert that classical local hidden variables is equivalent to the predictions of QM, which we already know is completely wrong.
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.
Yes, that's why I talked about angles rather than vectors, no other feature of the vector is relevant here.wm said:1. Note first that a, b', s and s' are unit-vectors, NOT angles.
2. And Yes; the dot product between the unit-vectors is the cosine of the angular difference.
Then your notation is unclear to me. What exactly do (ax ay az), (sx, sy, sz), (sx sy sz), and (bx', by', bz') represent, if not 4 column vectors? Of course, it's all right if they are, as long as you understand that the dot product is not equal to the product (in terms of matrix multiplication) of two column vectors, it's equal to the product of a row vector and a column vector...could you rewrite your proof so that you always include both the symbol you've been using for a dot product (.) and the symbol for multiplication (*), to distinguish between them? For instance, I assume that in (5) when you writewm said:3. (5) looks OK to me. I don't see two column vectors multiplied.
OK, but I don't see that you really took that into account in your proof either. If the two angles of the detectors are, in radians, a and b, then the expectation value when you allow s to take any angle \theta between 0 and 2pi should be:wm said: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 unit-vectors and two random (but opposite) unit-vectors (of infinite variety).
wm said:JesseM, DrC says/claims that I violate realism (UNQUALIFIED)::: DESPITE THEIR BEING MULTITUDINOUS VARIETIES and me repeatedly requesting that he be specific).
As far as I am aware, I DO NOT violate the realism specifically defined by me. (I'll post it later).
JesseM said: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, ...
wm does not specifically name 3 angles A, B, and C, but I think this was "left as an exercise for the reader" as it were. If he was correct that he had a classical example mirroring the conditions of the quantum experiment, and the expectation value for the product of the two results for any two angles X and Y was -cos(X-Y), then this is identical to the quantum expectation value, so it should be trivial to pick three specific angles A, B, C which would violate an inequality stated in terms of expectation values like the CHSH inequality (although in the specific case of the CHSH inequality you only need 2 possible angles for each detector), since we know all these inequalities can be violated in QM. Of course, as I've said before, his classical example does not mirror the conditions of the quantum experiment since he has more than two possible results. Also, my calculations above suggest the expectation value in his classical experiment would actually be -(1/2)cos(X-Y).DrChinese said:I will repeat for the Nth time: it is not that you violate the realism requirement, the problem is that you IGNORE it. It is quite specific as I have said previously: you must consider A, B and C and that leads to 8 permutations (2^3).
Bell didn't present a counterexample to local realism, he presented a general proof that local realism could never work (although I suppose you could say he did this by picking an example of a quantum experiment which could never be replicated in a universe obeying local realism). So if wm was able to come up with a classical example which replicated all Bell's conditions and also violated an inequality, this would be a counterexample to Bell's proof, just like if you tried to give a proof that there was no prime number larger than 13, I could present 17 as a counterexample.DrChinese said: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.
Of course I agree with this, but for different reasons (again, because he does not replicate the conditions of the experiment were each experimenter can only get one of two possible results, either spin-up or spin-down, and also because his proof of the expectation value seems to be incorrect).DrChinese said:Don't let the derivation wm presents fool you, because it does not disprove Bell in any way.
JesseM said:1. wm does not specifically name 3 angles A, B, and C, but I think this was "left as an exercise for the reader" as it were.
2. Bell didn't present a counterexample to local realism, he presented a general proof that local realism could never work (although I suppose you could say he did this by picking an example of a quantum experiment which could never be replicated in a universe obeying local realism).
3. So if wm was able to come up with a classical example which replicated all Bell's conditions and also violated an inequality, this would be a counterexample to Bell's proof, just like if you tried to give a proof that there as no prime number larger than 13, I could present 17 as a counterexample.
I think you're confusing the issue by using A and B to represent both specific angles and general variables representing arbitrary angles chosen by each detector. It would be simpler if you said that a was an arbitrary angle chosen by the left detector, b an arbitrary angle chosen by the right one, then you could have A, B, and C be specific choices of angles for either detector.DrChinese said:1. That "exercise for the reader" IS Bell's Theorem. wm is asserting that A and B work, therefore it works in all situations.
I don't understand what you mean by "counter-example" here. Bell provides a general proof that a certain inequality can never be violated under local realism, a statement of the form "for all experiments obeying local realism and satisfying certain conditions, this inequality will be satisfied". Logically, any statement of the form "for all X, Y is true" can be disproved with a single counterexample of the form "there exists on X such that Y is false". And that's what wm tried to do--find a single example of a local realist experiment which would satisfy Bell's conditions and yet violate an inequality. But he did it incorrectly, because he didn't satisfy the conditions, and his math for the expectation value was wrong anyway, with the correct expectation value I don't think you could violate any inequality using his experiment.DrChinese said:3. He doesn't consider the A/B/C condition. It is not possible to provide a counter-example to Bell, because Bell is itself a counter-example. The only way to disprove Bell would be to show that the counter-example is flawed.
But I disagree, wm came along and tried to show a classical example that would satisfy Bell's conditions and yet give an expectation value which, with the correct choice of angles, could violate an inequality (like my choice of angles for the CHSH inequality above). If he had actually satisfied Bell's conditions and if his calculation of the expectation value were correct, this would disprove Bell's theorem; but of course he didn't do this, and since I can follow Bell's theorem and see that it is logically airtight, I am totally confident he'll never be able to do this, just like I'm confident no one will find a counterexample to the statement "there are no even prime numbers larger than 2".DrChinese said: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.
Well, in what sense is this a counter-example to local realism, as opposed to a general proof that local realism cannot replicate quantum predictions? Again, when I use the word counter-example, I'm thinking of disproving a statement of the form "for all X, Y is true" by coming up with an example of the form "there exists a particular X such that Y is false". I guess you could say that if one agrees with Bell's theorem, then local realism makes the prediction that "for all experiments satisfying X conditions, inequality Y will be satisfied". And in this case, QM can give an example of the form "here's an experiment satisfying X conditions which violates inequality Y", thus proving QM is incompatible with local realism. But the problem here is that wm believes there's a flaw in Bell's theorem, so he does not agree that local realism makes the prediction "for experiments satisfying X conditions, inequality Y will be satisfied" in the first place; he's trying to disprove Bell's theorem by showing that local realism can also give an example of the form "here's an experiment satisfying X conditions which violates inequality Y". As a general approach to disproving Bell's theorem this makes sense, it's just that he thinks he's found such an example but he actually hasn't, because his example does not actually satisfy the X conditions of Bell's theorem (specifically the one about each experiment yielding one of two possible answers), and also his math for the expectation value is wrong, with the correct expectation value I'm not sure he could violate any Bellian inequality even if you ignore the first issue.DrChinese said:A review of that shows it as a fine counter-example to the original theory (local realism). And guess what? wm now must prove this wrong too, because it too is a counter-example to be contended with.
DrChinese said:Case A B C %
----- -- -- -- -----
[1] + + + ?
[2] + + - ?
[3] + - + ?
[4] + - + ?
[5] - + + ?
[6] - + - ?
[7] - - + ?
[8] - - - ?
.
DrChinese said:So in conclusion: looking at examples which support your local realism hypothesis is a waste of time, since you must address Bell's counter-example and it is impossible to refute that.
Bell's theorem allows for arbitrary probabilities to be assigned to each possible "hidden" state, including a probability of zero.enotstrebor said:Just a quick note that not all of the above cases are always valid even in a classical correlation scenario. (e.g [2] and [7] can be illegal simultaneous events)
You can't give a classical example which satisfies all the conditions laid out in Bell's theorem (for example, the hidden states sent out by the source must be uncorrelated to the choice of detector settings, which in a classical universe can be ensured by having the experimenters choose their detector settings too late for one's measurement-event to lie in the future light cone of the other's choice-event) and still violates an inequality. If you think you have one, please present it.enotstrebor said: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.
Not really, Bell's theorem is just trying to rule out a certain set of assumptions about "the physics" of the situation, assumptions which go by the name of local realism; but it isn't giving any opinion on what should be put in their place once you've ruled them out.enotstrebor said:One of the implicit assumptions in Bell type inequalities is that "we" truly understand the physics rather than only understanding the mathematics.
No, it simply rules out assumptions about "hidden physics" which fall into the category of local realism. You are free to believe in nonlocal hidden physics as in Bohm's interpretation of QM, for example.enostrebor said:Fundamentally, Bell's inequality and associate interpretation of EPR assumes no hidden physics.
If you're implying each particle could be a classical quadrupole, then no, this could not possibly explain quantum experiments which violate Bell inequalities.enostrebor said: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 Stern-Gerlach 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 Stern-Gerlach experimental results actually make physics sense.
You're missing the point, Bell's theorem is a sort of proof-by-contradiction; Bell showed logically that if the laws of physics respect local realism, then certain inequalities must be satisfied in certain types of experiments; since we can see the inequalities are actually violated in these types of experiments, this proves that the laws of physics do not respect local realism. It isn't supposed to tell you anything else about how the laws of physics do work.enostrebor said:Finally, it has been said that a mathematical model is correct if it produces the correct experimental result. Bell's inequalities do not.
Incidentally, Feynman actually derived his own version of a proof that local hidden variables could not reproduce the results of measurement of entangled particles, in his classic lecture Simulating Physics With Computers where he also first brought up the idea of a "quantum computer"--see sections 5-6 on pages 6-8 of the PDF (which is on pages 476-480 of the book that the pdf is scanned from). Interestingly, Feynman does not mention Bell in this lecture, and a comment here by physicist Michael Nielsen says "If I recall correctly, in a 1986 or 1987 festschrift paper for David Bohm (proceedings edited by Basil Hiley), Feynman comes pretty close to saying that he discovered Bell’s theorem before Bell." Another physicist disagrees with his recollection of the paper, so someone would have to check it to see what Feynman actually says.enotstrebor said:But a lack of understanding of the physics is exactly what Feynman meant when he said "No one really understands QM".
DrChinese said:1. That "exercise for the reader" IS Bell's Theorem. wm is asserting that A and B work, therefore it works in all situations. That is roughly like saying all prime numbers are even (because you only looked at cases that agree with your hypothesis). Make no mistake: wm is simply advocating traditional local realism. I went to his web page to make sure, and yup, there it is as big as day. He calls it "common sense realism" but it is local hidden variables with no new anything. He is simply acting as if Bell's Theorem is not valid.
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 counter-example to Bell, because Bell is itself a counter-example. The only way to disprove Bell would be to show that the counter-example 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 counter-examples. So Mermin can construct a very simple counter-examples 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 counter-example to the original theory (local realism). And guess what? wm now must prove this wrong too, because it too is a counter-example 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 counter-example to local realism. And guess what? wm must prove this wrong too.
So my point is simple: there is no such thing as a counter-example to a counter-example, the counter-example must actually be proven wrong. And in this case, we now have multiple counter-examples to consider. So the burden of (dis)proof has grown exponentially larger.
So when do you plan to address your math error that JesseM has taken pains to point out?wm said:6. For those like me, that are not verbally-minded, the simple acid test that I expect to meet is that my views will be consistent with high-school math and logic.
JesseM said:wm does not specifically name 3 angles A, B, and C, but I think this was "left as an exercise for the reader" as it were. If he was correct that he had a classical example mirroring the conditions of the quantum experiment, and the expectation value for the product of the two results for any two angles X and Y was -cos(X-Y), then this is identical to the quantum expectation value, so it should be trivial to pick three specific angles A, B, C which would violate an inequality stated in terms of expectation values like the CHSH inequality (although in the specific case of the CHSH inequality you only need 2 possible angles for each detector), since we know all these inequalities can be violated in QM. Of course, as I've said before, his classical example does not mirror the conditions of the quantum experiment since he has more than two possible results. Also, my calculations above suggest the expectation value in his classical experiment would actually be -(1/2)cos(X-Y).
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. Bell didn't present a counterexample to local realism, he presented a general proof that local realism could never work (although I suppose you could say he did this by picking an example of a quantum experiment which could never be replicated in a universe obeying local realism). So if wm was able to come up with a classical example which replicated all Bell's conditions and also violated an inequality, this would be a counterexample to Bell's proof, just like if you tried to give a proof that there was no prime number larger than 13, I could present 17 as a counterexample. Of course I agree with this, but for different reasons (again, because he does not replicate the conditions of the experiment were each experimenter can only get one of two possible results, either spin-up or spin-down, and also because his proof of the expectation value seems to be incorrect).
Doc Al said:So when do you plan to address your math error that JesseM has taken pains to point out?
JesseM said: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').
And if you want to use matrix multiplication as opposed to ordinary arithmetical multiplication, you could use the symbol x to denote that. If you do, perhaps you could also label each vector as either a row vector or a column vector like (ax, ay, az)_c or (ax, ay, az)_r.Then your notation is unclear to me. What exactly do (ax ay az), (sx, sy, sz), (sx sy sz), and (bx', by', bz') represent, if not 4 column vectors? Of course, it's all right if they are, as long as you understand that the dot product is not equal to the product (in terms of matrix multiplication) of two column vectors, it's equal to the product of a row vector and a column vector...could you rewrite your proof so that you always include both the symbol you've been using for a dot product (.) and the symbol for multiplication (*), to distinguish between them? For instance, I assume that in (5) when you write
(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.
Wait, so are you saying that commas vs. no commas denotes column vectors vs. row vectors? In this case (sx, sy, sz)x(sx sy sz) will not be a single number, but rather a 3x3 matrix.wm said:(The column vectors include the commas! as I recall ... but its the commas that differentiate one-way or the other in accord with HS maths.)