OK Corral: Local versus non-local QM

[DrChinese]

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.

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.

 

[JesseM]

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.

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?

 

[wm]

Quick reply; more later

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, 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]

Common-sense local realism!

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]

Once more to BELLIAN realism!

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]

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.

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.

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 $$\theta$$ between 0 and 2pi should be:

$$- \frac{1}{2\pi} \int_{0}^{2\pi} cos(\theta - a)*cos(\theta - b) \, d\theta$$

Are you claiming that this would work out to $$- cos (a - b)$$? 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]

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 modeled 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]

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]

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.

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.

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]

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]

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).

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.

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".

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]

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.

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]

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.

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.

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.

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.

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.

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]

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]

Time to answer

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.

wm

 

[Doc Al]

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]

Can you up-date me?

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]

Seeking to locate my error

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 anyone else so far for that matter).

Can you help me, please?

Thanks, wm

 

[Doc Al]

Start with this one:

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]

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.

(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.

 

[wm]

?

Start with this one:

Originally Posted by JesseM
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'). (Emphasis added.)



DocAl, Please, and with great respect: I can't find where that is said by me ... I don't find it in the original maths; and I haven't so far found it in a later post??

Do you have a source?

PS: Early on, JesseM had some confused views on my experiment and its maths so I wonder if the error might be his?

Also: If your question implies other errors, I'd be happy to correct or comment, as appropriate. Like you, I'd like to move ahead.

wm

 

[vanesch]

DocAl, Please, and with great respect: I can't find where that is said by me ... I don't find it in the original maths; and I haven't so far found it in a later post??

Do you have a source?

In your post:

https://www.physicsforums.com/showpost.php?p=1252012&postcount=55

how do you go from (4) to (7) ?

The intermediate notation is confusing, and in as much as it is interpreted by several of us here, wrong, because it seems indeed that you use the "rule" that:

(a.s)(b.s) = (s.s)(a.b)

which is not correct.

Maybe you don't use that rule, but then the notation in (5) and (6) is completely unintelligible, and the transition from (4) to (7) ununderstandable.

EDIT:
to show that this rule is not correct, it is sufficient to have a counter example, and JesseM gave you one, to which you replied that he failed to understand that s was a random vector. But that doesn't show in the notation, because in (7), we still have the expectation symbols there, which make one think that the notation inside applies for each possible unit vector s individually.

 

[Doc Al]

DocAl, Please, and with great respect: I can't find where that is said by me ... I don't find it in the original maths; and I haven't so far found it in a later post??

This is getting tedious. JesseM has picked apart your math (in your post #55) line by line. Either address his concerns or it's time to shut this thread down.

 

[JesseM]

I think I understand wm's mistake now. He is not using either the dot product or arithmetical multiplication, but rather matrix multiplication (which I'll represent using the symbol x); and he didn't mention it until now, but he is using the convention that (sx, sy, sz) with commas represents a column vector and (sx sy sz) without commas represents a row vector. So his mistake is in going from 6 to 7, where he treats a column vector x row vector, namely (sx, sy, sz) x (sx sy sz), as equal to the dot product of (sx, sy, sz) with itself; in fact, in matrix multiplication a 3-component column vector x a 3-component row vector gives a 3 x 3 matrix, not a scalar like the dot product. Unlike in arithmetical multiplication, in matrix multiplication order is critical.

 

[wm]

Matrix average vs dot.product average?

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: 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. 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.

Jesse, you may be at the nub of the problem.

But from my readings, I thought that I was using a well-accepted compact notation. One that would not lead us astray!? One that I would like to stay with (for compactness).

And just to be sure, that is why I referenced the wiki page in support.

Now I have to ask: Why would you want to go down the MATRIX path when the s.s' dot product does the job?

Or have I taken an invalid route? I did it off the top of my head, and have done it often, BUT I would want to be correct mathematically.

PLEASE: This would be a point where I would appreciate some hand-holding and the detailed explanations that you are good at.

I guess the question is: What is <s.s'>? Is it other than unity?

Please be expansive here, for maybe this is where I need to apologise and bail-out?

I'm thinking not; but lots of helpful notes will help me to decide. Thanks, wm

 

[JesseM]

Jesse, you may be at the nub of the problem.

But from my readings, I thought that I was using a well-accepted compact notation. One that would not lead us astray!? One that I would like to stay with (for compactness).

I hadn't seen it before, so it would have helped if you mentioned it instead of just linking to the wikipedia page which had a bunch of other information on it as well. But now that I understand it's fine.

Now I have to ask: Why would you want to go down the MATRIX path when the s.s' dot product does the job?

Or have I taken an invalid route? I did it off the top of my head, and have done it often, BUT I would want to be correct mathematically.

Yes, you made an invalid step. The only sense in which it makes sense the row vector (ax ay az) multiplied by the column vector (sx, sy, sz) is the same as the dot product of a.s is when you are using matrix multiplication--if you don't want to be using matrix multiplication, then you can't replace a.s with (ax ay az)(sx, sy, sz) and have it make sense. But if you're using matrix multiplication, a column vector (sx, sy, sz) times a row vector (sx sy sz) is not the dot product s.s, instead it is a 3 x 3 matrix. This is because when you multiply two matrices (and a vector is a type of matrix) A and B to get a product C, the rule is that the dot product of the first row of A and the first column of B gives the number in the first row, first column of C; the dot product of the first row of A and the second column of B gives the number in the first row, second column of C; and so forth. There's a little tutorial which might help understand how it works in the second section of this page. So if you apply these rules when A is a row vector with 3 components and B is a column vector with 3 components, then A only has one row and B only has one column, meaning that C is a 1 x 1 matrix (a scalar). But if A is a column vector with 3 components and B is a row vector with 3 components, A has 3 rows and B has 3 columns, so there are 9 possible combinations when you take the dot product of a row of A and a column of B; in this case C is a 3x3 matrix with 9 components.

I guess the question is: What is <s.s'>? Is it other than unity?

The error is in your previous step, where you substituted <s.s> for <(sx, sy, sz)(sx sy sz)>. They are not equivalent.

 

[DrChinese]

1. 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;

Well, I both agree strongly and disagree strongly. You are quite right that Bell says: IF local realism is to be accepted, THEN the Inequality must hold for ALL arbitrary settings. Logic (contranegative) dictates that this proposition is equivalent to: IF the Inequality does NOT hold for ALL arbitrary settings, THEN local realism is NOT to be accepted. Bell did not spell out this part of the argument, it must be inferred.

But Bell NEVER asserts that the Inequality FAILS for ALL possible angle settings... and it doesn't! In his (22) he says that for ac=cos(90 degrees), ab=bc=cos(45 degrees), the Inequality is violated and therefore "the quantum mechanical expectation value cannot be represented, either accurately or arbitrarily closely, in the form (2)." He has provided one specific counter-example, and that is all he needs to do to prove the contranegative above. Of course, with the formula in hand you can create other settings that will also violate the inequality.

This would certainly explain why you and I see things differently. You believe wm is providing the counter-example, while I see Bell as providing the counter-example. If you are correct, then why does Bell essentially start with wm's results (3) and (7) and then go on to say that "In this simple case there is no difficulty in the view that the result of every measurement is determined by the value of an extra variable..." and the like.

On the other hand, shortly after (22) he make the generalization (V.) that for systems where there are MORE than 2 assumed hidden variables ("dimensionality greater than two", i.e. more than just A and B), there will always be a case in which QM is incompatible with "seperable predetermination" for the "two dimensional subspaces" in which observations can actually be performed.

Please note that I keep quoting and referencing Bell's original paper extensively. Not once have either you disputed anything with a similar reference. I think if you look back at the paper and see the context, you may look at my argument in a little different light.

I realize that the language of Bell does not always map directly onto what we often take for granted as being Bell's Theorem. I do consider Bell precise, and the paper is (in my opinion) a masterpiece - especially considering how much ground had to be broken to get to the final result. But the paper fully stands 40+ years later.

I must say that I have gained much out of this exchange, because I always convert the argument from spin 1/2 to spin 1 particles when I am thinking about the matter. So I have had to re-read it a bit more closely to keep in the discussion. :)

 

[wm]

Sos! Sos! Sos!

In your post:

https://www.physicsforums.com/showpost.php?p=1252012&postcount=55

how do you go from (4) to (7) ?

The intermediate notation is confusing, and in as much as it is interpreted by several of us here, wrong, because it seems indeed that you use the "rule" that:

(a.s)(b.s) = (s.s)(a.b)

which is not correct.

Maybe you don't use that rule, but then the notation in (5) and (6) is completely unintelligible, and the transition from (4) to (7) ununderstandable.

EDIT:
to show that this rule is not correct, it is sufficient to have a counter example, and JesseM gave you one, to which you replied that he failed to understand that s was a random vector. But that doesn't show in the notation, because in (7), we still have the expectation symbols there, which make one think that the notation inside applies for each possible unit vector s individually.

I am for sure wondering what I have missed? So, from my maths post, with explanations for the moves from (4) to (7). And more to come if I'm still not clear:

(3) <(a.s) (s'.b')>

Since s' = -s we have:

(4) = - <(a.s) (s.b')>

Since we can expand a dot product using row and column vectors http://en.wikipedia.org/wiki/Column_vector we have (in an accepted compact notation):

(5) = - <[(ax ay az) (sx, sy, sz)] [(sx sy sz) (bx', by', bz')]>

Since the vectors a and b' are constant during a given experimental run, they may be removed from the ensemble average; which we do. So:

(6) = - (ax ay az) <(sx, sy, sz) (sx sy sz)> (bx', by', bz')

The ensemble average is now over a dot product between s and s. So we write:

(7) = - (ax ay az) <s.s> (bx', by', bz')

With s.s = 1 FOR ANY AND ALL s, we move to the conclusion.


PS: My maths was offered in good faith and was not intended to be a con-job: The last person I'd want to con on this subject is myself.

SO: Have I made a mistake that cannot be corrected?

If you need to use words, please be expansionary in your comments. Compact maths should be OK however.

Thanks, wm

 

[JesseM]

EDIT:
to show that this rule is not correct, it is sufficient to have a counter example, and JesseM gave you one, to which you replied that he failed to understand that s was a random vector. But that doesn't show in the notation, because in (7), we still have the expectation symbols there, which make one think that the notation inside applies for each possible unit vector s individually.

Another point on this is that although wm did use <> to indicate he was talking about expectation values, his proof didn't seem to depend in any way on how the vector s could vary; if the proof is valid for the case where s is equally likely to take any angle from 0 to 2pi, then it should also be valid for a case where s can only have a discrete number of possibilities, like if s had a 50% chance of being 90 degrees and a 50% chance of being 80 degrees. In this case, if a is 0 degrees and b' is 60 degrees, then the expectation value for - <a.s*s.b'> is just - {0.5*[cos(90)*cos(30)] + 0.5*[cos(80)*cos(20)]}, which is equal to -0.0816. But meanwhile, -cos(a - b') is equal to -cos(60), or -0.5. So unless wm claims to be making specific use of the fact that the angle of s is equally likely to take any angle, this example shows it is not true in general that - <a.s*s.b'> is equal to -cos(a - b').

 

[DrChinese]

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)

As JesseM points out, in a classical scenario ALL of the cases must be non-negative. This is the very definition of realism.

[2] and [7] are the quantum cases that are suppressed, and end up with negative probabilities for certain angle settings.

 

[JesseM]

I am for sure wondering what I have missed? So, from my maths post, with explanations for the moves from (4) to (7). And more to come if I'm still not clear:

(3) <(a.s) (s'.b')>

Since s' = -s we have:

(4) = - <(a.s) (s.b')>

Since we can expand a dot product using row and column vectors http://en.wikipedia.org/wiki/Column_vector

You can only expand a dot product in row and column vectors if you are using matrix multiplication; i.e. (ax, ay, az).(sx, sy, sz) is equal to (ax ay az)x(sx, sy, sz), where I am using x to represent matrix multiplication. So with the matrix multiplication made explicit, your step 5 is something like this:

(5) = - <(ax ay az)x(sx, sy, sz)x(sx sy sz)x(bx', by', bz')>

This is valid in itself. But your mistake is here:

(6) = - (ax ay az) <(sx, sy, sz) (sx sy sz)> (bx', by', bz')

The ensemble average is now over a dot product between s and s.

Wrong! Again, (sx, sy, sz)x(sx sy sz) is not a dot product with a single value, it is a 3 x 3 matrix whose 9 entries look like this:

sx*sx sx*sy sx*sz
sy*sx sy*sy sy*sz
sz*sx sz*sy sz*sz

SO: Have I made a mistake that cannot be corrected?

Yes. I showed a counterexample to your proof in post #99, and what's more, in post #76 I've given a proof that the correct value for -<(a.s)*(s.b)> would be -(1/2)cos(a - b), using the rule that if the outcome of a given experiment will be some function f(x) of a parameter x whose value is between A and B and whose probability distribution is given by p(x), then the expectation value is $$\int_{A}^{B} p(x)*f(x) \, dx$$. In this case, the outcome is a function of the angle $$\theta$$, namely $$cos(\theta - a)*cos(\theta - b)$$, and $$\theta$$ must be equally likely to take any value from 0 to 2pi, so we must use the probability distribution $$p(\theta) = 1/2\pi$$ to ensure that if we integrate the probability from 0 to 2pi, the answer is 1. Thus, the expectation value for -<(a.s)*(s.b)> must be $$- \frac{1}{2\pi} \int_{0}^{2\pi} cos(\theta - a)*cos(\theta - b) \, d\theta$$, which works out to -(1/2)cos(a - b).

 

[DrChinese]

1. 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".

2. Logically, any statement of the form "for all X, Y is true" can be disproved with a single counterexample of the form "there exists one X such that Y is false".

3. 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.

1. I agree totally, although the language should be more like: "for all theories obeying local realism and satisfying certain conditions, this inequality will be experimentally satisfied".

2. I agree totally.

3. This is the problem: Bell's Inequality was set up for the specific purpose of showing it is violated! Bell himself provides the counter-example. But look at what you have said here: it is not the counter-example to what you say in 1.

For what wm is saying to work, he needs the proposition to be: "If the Inequality is violated and the experiment obeys local realism, then Bell's argument is wrong. That is a different argument entirely, and certainly will be disputed by anyone if quantum mechanics is involved in any way, shape or form. There are people out there trying to show this, but so far no one has succeeded. I consider it akin to tilting at windmills, but to each his own. There are still people looking for perpetual motion too...

 

[JesseM]

But Bell NEVER asserts that the Inequality FAILS for ALL possible angle settings...

I never claimed it did! What comment of mine are you thinking of? In fact, I said quite the opposite, that only one example of a violation of an inequality in QM is enough to disprove local realism if you accept Bell's proof as valid, and likewise one example of a violation or an inequality in a classical scenario would be enough to show the proof is flawed (although of course I don't think such a classical example will ever be found, since I think the proof is fine). Please read this comment of mine again:

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".

This would certainly explain why you and I see things differently. You believe wm is providing the counter-example, while I see Bell as providing the counter-example.

I still don't understand what you mean when you use the word "counter-example", you aren't specifying a counterexample to what...perhaps you could restate your comments as I have done above, where someone offers a general statement of the form "for all X, Y is true" and someone offers a counterexample of the form "there exists an X such that Y is false". Again, in these terms, what I am saying is:

1. Bell's theorem claims to prove "for all experiments which are of the form X (referring to all the conditions on the experiment like the source not having foreknowledge of detector settings) AND which respect local realism, it is true that inequality Y will be satisfied." But then he shows that in QM, "there exists a quantum experiment of the form X such that inequality Y is violated." The point here is that if you accept his proof of the general statement, then a single counterexample from QM is enough to show that QM is incompatible with local realism.

2. wm does not accept Bell's proof in the first place, and he wants to show that it is flawed by demonstrating that in an ordinary classical universe, "there exists an experiment which is of form X AND which does respect local realism, yet which violates inequality Y." If he could indeed produce a single example like this, then he'd have proved there must be a flaw in Bell's proof, since the theorem asserts that all experiments of the form X which respect local realism must obey inequality Y.

To be clear, I of course think that wm has failed to find an experiment of form X (ie the conditions specified in proofs of Bell's theorem) which respects local realism yet violates some Bellian inequality. I of course think Bell's theorem is solid, and that no one will find a purely classical counterexample. But that's what wm is trying to do, and all I'm saying is that it makes sense as a strategy.

 

[JesseM]

3. This is the problem: Bell's Inequality was set up for the specific purpose of showing it is violated!

I think you're missing a key part of my 3, though:

3. 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.

Bell did not provide an example of a local realist experiment that violated the inequality; rather, he provided a quantum example, and thus (if you accept his proof as valid) he showed that QM is not compatible with local realism.

For what wm is saying to work, he needs the proposition to be: "If the Inequality is violated and the experiment obeys local realism, then Bell's argument is wrong.

Exactly! That's what I think wm is trying to do. After all, his experiment is a purely classical one that does obey local realism (it just involves sending two vectors in opposite directions, and each experimenter projects the one they get onto their own vector to get a real number between -1 and +1), yet he claims that the expectation value for the product of the two experimenter's answers will be given by -cos(a - b), and I showed that this would violate the CHSH inequality for some particular choices of detector angles.

 

[wm]

You can only expand a dot product in row and column vectors if you are using matrix multiplication; i.e. (ax, ay, az).(sx, sy, sz) is equal to (ax ay az)x(sx, sy, sz), where I am using x to represent matrix multiplication. So with the matrix multiplication made explicit, your step 5 is something like this:

(5) = - <(ax ay az)x(sx, sy, sz)x(sx sy sz)x(bx', by', bz')>

This is valid in itself. But your mistake is here: Wrong! Again, (sx, sy, sz)x(sx sy sz) is not a dot product with a single value, it is a 3 x 3 matrix whose 9 entries look like this:

sx*sx sx*sy sx*sz
sy*sx sy*sy sy*sz
sz*sx sz*sy sz*sz

Yes. I showed a counterexample to your proof in post #99, and what's more, in post #76 I've given a proof that the correct value for -<(a.s)*(s.b)> would be -(1/2)cos(a - b), using the rule that if the outcome of a given experiment will be some function f(x) of a parameter x whose value is between A and B and whose probability distribution is given by p(x), then the expectation value is $$\int_{A}^{B} p(x)*f(x) \, dx$$. In this case, the outcome is a function of the angle $$\theta$$, namely $$cos(\theta - a)*cos(\theta - b)$$, and $$\theta$$ must be equally likely to take any value from 0 to 2pi, so we must use the probability distribution $$p(\theta) = 1/2\pi$$ to ensure that if we integrate the probability from 0 to 2pi, the answer is 1. Thus, the expectation value for -<(a.s)*(s.b)> must be $$- \frac{1}{2\pi} \int_{0}^{2\pi} cos(\theta - a)*cos(\theta - b) \, d\theta$$, which works out to -(1/2)cos(a - b).

Jesse, thanks for this; I like it very much. Also: Excuse my jumping in and out at the moment but I'm just grabbing bits of time hopefully to move us ahead. Which should not happen till we have agreement re my equations.

I think they are going to be just fine BUT could you tell me how you want the last correct line (whatever you deem that to be) to be written.

I ask because if I go on what's above, you look like you would like an x in three places? But would that be satisfactory?

The reason that I want this correction because all I did in my own work was to see that the matrix that resides in the middle (after all correct mathematical proceses) is just the equivalent of a unit-matrix, obtained by taking the ensemble-average inside the matrix and evaluating each element's ensemble-average.

What I'm guessing here (in a hurry) is that you would be happier to see the unit-matrix equivalent in the middle, and not my short-cut?

Sorry to be slow on the up-take; but hope you won't mind me still seeing that the fully rigorous maths will not change the answer.

As soon as I get where you want some x (and some y if necessary) I'll send in the expanded version that should then suit you; or maybe have a different error.

Thanks, as always, wm