Register to reply

OK Corral: Local versus non-local QM

by wm
Tags: corral, local, nonlocal, versus
Share this thread:
wm
#73
Feb23-07, 04:20 AM
P: 161
Quote Quote by JesseM View Post
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:

[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').
JesseM, I appreciate both your sticking-with-me and your going-after-me BUT it seems your maths is a bit rusty! As far as I can see, there is no error in my maths; and there is more to come.

Thanks also for the cross-references that you give me.

Now, to the maths. Does this help you:

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.

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 unit-vectors and two random (but opposite) unit-vectors (of infinite variety).

5. I'm pretty sure that I did not mix the rules?

6. I'm pretty sure that the maths is spot-on. 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 correct-step by correct-step.

Thanks again, in haste for now, wm
wm
#74
Feb23-07, 04:36 AM
P: 161
Quote Quote by JesseM View Post
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.
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). Rather, I use it (the most general common-sense local realism) IN THAT I allow specifically the measurement outcome to be a consequential perturbation of the particle-detector 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
wm
#75
Feb23-07, 04:43 AM
P: 161
Quote Quote by DrChinese View Post
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.
Limited comment: On the point of which Bellian assumption to reject, I'd like to suggest that if you studied the implication of Bell's maneuver in his unnumbered equations THEN you might just find it easy to reject Bellian realism and retain locality.

Quick suggestion only, wm PS: It works for me!
JesseM
#76
Feb23-07, 08:24 AM
Sci Advisor
P: 8,470
Quote Quote by wm View Post
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.
Yes, that's why I talked about angles rather than vectors, no other feature of the vector is relevant here.
Quote Quote by wm
3. (5) looks OK to me. I don't see two column vectors multiplied.
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.
Quote Quote by wm
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).
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 [tex]\theta[/tex] between 0 and 2pi should be:

[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(a-b), 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.
DrChinese
#77
Feb23-07, 09:30 AM
Sci Advisor
PF Gold
DrChinese's Avatar
P: 5,385
Quote Quote by wm View Post
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).
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). These 8 permutations cannot be modelled with probabilities in the range 0 to 1 inclusive (0% to 100%) at certain angles (such as Bell's example A=0, B=45, C=90 degrees for spin 1/2 particles, or Mermin's example A=0, B=120, C=240 for spin 1 particles). No matter how you try, you will never be able to fill out the following table with values (for the ?) that match experiment (the ratio of any 2 columns must match the QM prediction) AND are non-negative:

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 counter-example 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 non-negative.

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. 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.
DrChinese
#78
Feb23-07, 09:42 AM
Sci Advisor
PF Gold
DrChinese's Avatar
P: 5,385
Quote Quote by JesseM View Post
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, ...
JesseM,

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
JesseM
#79
Feb23-07, 09:47 AM
Sci Advisor
P: 8,470
Quote Quote by DrChinese View Post
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).
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.
Quote Quote by DrChinese View Post
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.
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.
Quote Quote by DrChinese
Don't let the derivation wm presents fool you, because it does not disprove Bell in any way.
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
#80
Feb23-07, 10:45 AM
Sci Advisor
PF Gold
DrChinese's Avatar
P: 5,385
Quote Quote by JesseM View Post
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.
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.
JesseM
#81
Feb23-07, 11:30 AM
Sci Advisor
P: 8,470
Quote Quote by DrChinese View Post
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 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.

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 choose--for 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).
Quote Quote by DrChinese
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.
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.
Quote Quote by DrChinese
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.
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".
Quote Quote by DrChinese
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.
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.
enotstrebor
#82
Feb23-07, 11:31 AM
P: 122
Quote Quote by DrChinese View Post

Case A B C %
----- -- -- -- -----
[1] + + + ?
[2] + + - ?
[3] + - + ?
[4] + - + ?
[5] - + + ?
[6] - + - ?
[7] - - + ?
[8] - - - ?

.
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)

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.

Quote Quote by DrChinese View Post
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.
What follows is speculative but raises the question of assumptions.

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 bi-vectored 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 bi-vectored 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 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.

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 non-locality (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".
JesseM
#83
Feb23-07, 11:50 AM
Sci Advisor
P: 8,470
Quote Quote by enotstrebor View Post
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)
Bell's theorem allows for arbitrary probabilities to be assigned to each possible "hidden" state, including a probability of zero.
Quote Quote by enotstrebor
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.
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.
Quote Quote by enotstrebor
One of the implicit assumptions in Bell type inequalities is that "we" truly understand the physics rather than only understanding the mathematics.
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.
Quote Quote by enostrebor
Fundamentally, Bell's inequality and associate interpretation of EPR assumes no hidden physics.
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.
Quote Quote by enostrebor
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.
If you're implying each particle could be a classical quadrupole, then no, this could not possibly explain quantum experiments which violate Bell inequalities.
Quote Quote by enostrebor
Finally, it has been said that a mathematical model is correct if it produces the correct experimental result. Bell's inequalities do not.
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.
JesseM
#84
Feb23-07, 01:43 PM
Sci Advisor
P: 8,470
Quote Quote by enotstrebor View Post
But a lack of understanding of the physics is exactly what Feynman meant when he said "No one really understands QM".
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.
wm
#85
Feb23-07, 02:06 PM
P: 161
Quote Quote by DrChinese View Post
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.
I am happy to address any question in this thread (in that the thread was initiated by me to question some prominent views which I cannot comprehend -- having struggled hard to do so).

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 tied-up 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 high-school 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 counter-examples in my small area of interest were long-held and wrong (as shown pre-eminently by John Bell).

2. that John Bell himself was not happy with his theorem and had not given up on finding a simple constructive counter-example (as I read him).

3. I am not a John Bell, but his simple approach has motivated me to have-a-go; notwithstanding that many others have had-a-go and failed.

4. So as soon as there is some general agreement that my high-school 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: common-sense 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 verbally-minded, the simple acid test that I expect to meet is that my views will be consistent with high-school math and logic.

7. Please note that other interpretations of QM support locality; and I support locality.

Sincerely, wm
Doc Al
#86
Feb23-07, 02:17 PM
Mentor
Doc Al's Avatar
P: 41,475
Quote Quote by wm View Post
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.
So when do you plan to address your math error that JesseM has taken pains to point out?
wm
#87
Feb23-07, 02:23 PM
P: 161
Quote Quote by JesseM View Post
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).
Jesse, As I wrote: I will add the up/down pieces as soon as we are agreed that the maths to-date is OK.

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 one-way 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
wm
#88
Feb23-07, 02:27 PM
P: 161
Quote Quote by Doc Al View Post
So when do you plan to address your math error that JesseM has taken pains to point out?
Doc Al; I am apparently blind to my math error (which happens) but I sincerely am not sure what error is being identified by Jesse on this occasion (or any one else so far for that matter).

Can you help me, please?

Thanks, wm
Doc Al
#89
Feb23-07, 02:31 PM
Mentor
Doc Al's Avatar
P: 41,475
Start with this one:
Quote Quote by JesseM View Post
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
#90
Feb23-07, 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:
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.
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.
Quote Quote by wm
(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.)
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.


Register to reply

Related Discussions
Is BM Bohmian Local actually Local Quantum Physics 17
Are so-called non-local interactions really non-local? Quantum Physics 3
Local dialect you never knew was local. General Discussion 52
A proof that all signal local theories have local interpretations. Quantum Physics 3