OK Corral: Local versus non-local QM

[wm]

I don't think it is a technical term. Look at my post #42. The "common sense realism" is the idea that these correlations between dice throws of Alice and Bob "just cannot be" if there is no common causal origin.
It comes down to the idea that, when correlations occur where they are not a priori expected for some logical reason, there must be some causal link (by common origin, or by influence), and that this causal link must be found in some ontologically existing mechanism.
The reason to adhere to it, is that it is our main (and only) technique of inference and empirical enquiry.

Indeed, if you throw a switch, and a light goes on, and you throw it again, and the light goes out, and you do this 20 times, you expect there to be some kind of ontological mechanism to exist which explains this. It doesn't necessarily mean that the switch causes the light to go on. A technician might observe you through a camera, and steer the light as you throw a non-connected switch. Or worse, he might switch on and off a light, and "send you some brain waves" which make you throw the switch. Or even better, when you walked into the room, a scanner might have found out the exact state of your brain, and calculated at what moments you will throw the switch. One extravagant explanation even worse than the other. But you prefer that, over: well, correlations happen. There's no relationship. This desire of explaining correlations is, I think, what is meant by "common sense realism": there must be something real, which is the mechanism which explains the correlations.

(Emphasis added.)

YES! And going one step deeper: We expect there to be something real which accounts for the correlata.

Such is enlightened human nature, perhaps?

Recalling that the preparation of the highly-correlated singlet state (with its spherical symmetry), produces one of the most highly correlated states (= highly correlated correlata) that we know of.

wm

 

[wm]

No, not in the least. Bell's realism is well-defined, and it is reasonable to reject it as an assumption (while keeping locality as an assumption). That does NOT mean it is a poor definition - it is a good one and that is a big part of why Bell's Theorem is so strong!

A lot of people reject realism as an assumption, and I am hardly the first. Don't confuse the "assumption" with the "definition of the assumption" - they are entirely different.

Sorry Doc, but I'm lost and confused again. Beyond belief!

<<<Here is my assumption: LEFT BLANK.

Please, dear Professor, Do not confuse my assumption with the definition of my assumption.

Ah (light dawns): perhaps you DrC are relying on non-local transmission of my assumption.>>>

My problem! But to say ''a lot of people reject realism'' without in any way qualifying the realism of which you speak ... well that continues to be beyond me.

For now, I think it best that I find my old maths ... and maybe become (with hard study) a mathematician.

Believing, as I do, that: Maths is the best logic; and I've much to learn = comprehend.

Respectfully, wm

 

[DrChinese]

DrC,

1. Interesting: I see no mention of perturbation in your citation?

But it is specifically addressed in my definition of ''common-sense local realism'': <link removed> Would you (Einstein, Bell) disapprove? (PS: AND NB: I'm sure it can be improved!)

2. To quote my erstwhile friend: ''If you deny Bell realism (as I DrChinese am prone to do), none of that matters.''
Sorry; English is not my first language. Your proneness (= lying horizontal?) I misunderstood.

3. Non-local impact? What would that be? Peres and many others poo-poo any physicality with non-locality (I seem to recall). I am with them re physicality, since (as I before wrote to you?): How can an abstract non-physical wave not in space-time influence a concrete object in space-time?

1. The issue of perturbation by the measurement apparatus can be looked at any way you want to. It really does not affect the mathematical description of realism, which is:

1 >= P(A, B, C) >= 0

The reason is that A, B and C individually are elements of reality, because they can be predicted in advance in a Bell test. The issue is not whether the measurement somehow distorts the results, it is whether these elements of reality (EPR) exist simultaneously independently of the ACT of observation.

2. There are 2 primary assumptions in Bell's Theorem, and Bell test results lead us to deny at least one of them. You are fully justified in rejecting either one, it is simply a matter of personal preference. My preference happens to be to accept locality and deny realism, but the reverse is an acceptable position. The unjustified conclusion is to reject neither.

3. Non-locality is an issue of vital relevance. It is one of the 2 primary assumptions in Bell's Theorem, and cannot be dismissed lightly.

 

[DrChinese]

My problem! But to say ''a lot of people reject realism'' without in any way qualifying the realism of which you speak ... well that continues to be beyond me.

Again, the definition is as follows:

1 >= P(A, B, C) >= 0

This is what Bell realism is, and it remains an accepted standard definition.

 

[wm]

Preliminary classical maths for EPR-Bohm correlations

Sorry Doc, but I'm lost and confused again. Beyond belief!

<<<Here is my assumption: LEFT BLANK.

Please, dear Professor, Do not confuse my assumption with the definition of my assumption.

Ah (light dawns): perhaps you DrC are relying on non-local transmission of my assumption.>>>

My problem! But to say ''a lot of people reject realism'' without in any way qualifying the realism of which you speak ... well that continues to be beyond me.

For now, I think it best that I find my old maths ... and maybe become (with hard study) a mathematician.

Believing, as I do, that: Maths is the best logic; and I've much to learn = comprehend.

Respectfully, [B]wm

This is a preliminary draft from wm, for critical comment, please. It responds to various requests for a classical derivation of the EPR-Bohm correlations which would nullify Bell's theorem. It's off the top of my head; and a more complex denouement might be required (and can be provided) to satisfy mathematical rigour:

(Figure 1) D(a) -<- w(s) [Source] w'(s') ->- D'(b')Two objects fly-apart [w with property s (a unit-vector); w' with property s' (a unit-vector)] to respectively interact with detectors D (oriented a, an arbitrary unit vector) and D' (oriented b', an arbitrary unit vector). The detectors D (D') respectively project s (s') onto the axis of detector-orientation a (b').

Let w and w' be created in a state such that

(1) s + s' = 0; say, zero total angular momentum. That is:

(2) s' = -s.

Then the left-hand result is a.s and the right-hand result is s'.b'; each a dot-product.

To derive the related correlation, we require (using a recognised notation http://en.wikipedia.org/wiki/Column_vector ), with < ... > denoting an average:

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

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

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

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

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

(8) = - (ax ay az) <1> (bx', by', bz')

(9) = - a.b'

(10) = - cos (a, b').

Let s and s' be classical angular-momenta. Then (to the extent that we meet all the Bell-theorem criteria) the result is a wholly classical refutation of Bell's theorem. [It is Bell's constrained realism that we reject; thereby maintaining the common-sense locality clearly evident above.)

E and OE! QED?

Critical comments most welcome, (though I'll be away for a day or so),wm

 

[JesseM]

1. It seems to me that some of your parenthetic comments (''otherwise I don't see any way of explaining how both experimenters always get the same answer when they make the same measurement'') would be helped by your doing the maths in detail so that you understand every step.

I don't agree this would help, there is very little in the way of "math" here, and trying to write this in a more formal way would provide no conceptual illumination. But maybe it would help if I spelled out the reasoning and assumptions a little more clearly.

1. Do you disagree that in a classical world, if the results of two measurements exhibit a 100% correlation, this must be either because one measurement determined the other, or because the results of both measurements were determined by some other event or events?

2. Do you disagree if you have a 100% probability that two measurements using the same settings always give the same results as one another on repeated trials, and the two measurements have a spacelike separation and we assume no possibility of FTL signals, then the only way to explain the perfect 100% correlation in a classical world is to assume that on each trial the results were predetermined by some event or events (presumably the emission of both signals/objects from a common origin at the source) which lie in the past light cone of both measurement-events? (if you disagree, can you suggest an alternative explanation?)

3. Do you disagree that if there was any random element to the results of either experimenter's measurement on a given trial where they both use the same settings (and I'm only talking about randomness in the outcome an experimenter will get if we know both his detector setting and the precise state of the signal/object they're measuring--the original event or events which determined the state of both object/signals at the source may still have a random element), then the probability they both get the same answer could not be 100%?

If you agree that the answers must have been predetermined on the trials where they both picked the same detector setting, then if we also add the assumption that this predetermining event or events could not in any way be affected by information about what detector settings each experimenter will choose (note that this condition can be assured in a classical universe obeying locality if each experimenter chooses their setting randomly shortly before the measurement, so that a signal moving at the speed of light would not have time to travel from the event of an experimenter picking a setting to the event of the other experimenter making their measurement), then this means the answers must have been predetermined on *every* trial, even the ones where they pick different settings.

2. This is not to say that my maths will always be correct; nor that your words are unintelligible. But more and more I find that those who offer ''almost non-intelligible maths'' (or none at all) are those whose words I struggle most to comprehend.

If you have difficulty comprehending anything, could you try to explain what point is confusing you?

3. As for determinism: I am most certainly that way inclined! Take any anti-parallel detector settings in EPRB and the detectors punch out identical (++) XOR (--) results till kingdom come.

But I wasn't just asking if you were "inclined" this way, I was asking if you agreed it would be impossible to explain the results any other way in a classical universe obeying locality where there is no way for one measurement to causally affect the other. If you disagree or are not sure, please go through my various questions about the need for determinism above to see if you disagree with any individually.

4. You have completely missed the question that I await an answer on. Let me answer it now, without the external input that I was hoping for: It is my view that Bell (dissatisfied with his own theorem) was open to any hidden-variable theory; local or non-local. However, in my view, non-local hidden-variables are so trivial as to be unworthy of the great man. For (it seems to me) one postulates that a measurement reveals a non-local hidden-variable in one wing of the experiment AND THEN that revelation is non-locally transmitted to the other wing. UGH!

You say "you have completely missed the question that I await an answer on", but you didn't actually say what this question was. And again, I am not really interested in historical questions about Bell's opinions; I am interested in trying to show that it is possible to prove that quantum results absolutely rule out local hidden variables, which you disagreed with earlier when I asked you about it.

5. Please recall that my earlier CLASSICAL experiment classically refuted Bellian Inequalities, not Bell's theorem.

OK, but why do you think this is relevant? I'm not aware of any physicist in history who denied that it's trivial to violate the inequalities classically if you are allowed to violate the conditions of Bell's theorem; on the other thread I showed you a very simple way of doing this in a question-and-answer game where I get to hear both questions before answering "yes" or "no".

PS: Is there a simple spot on PF where I can pick up on LaTeX? (My search revealed too much.) Though I'll probably post in a simpler but wholly adequate fashion.

Yes, see the sticky thread Introducing LaTeX Math Typesetting at the top of the "Math & Science Tutorials" forum.

7. So (to be clear): That earlier classical experiment of mine was directed at Bellian Inequalities only. The next is also wholly classical, but seeks to meet the more general Bellian conditions and establish the EPRB correlation -a.b' (per OP) in response to our discussion.

When you say "the more general Bellian conditions", do you mean the conditions of Bell's theorem, including the one I mentioned that there can be no correlation between the state of the signals/objects emitted by the source and the experimenters' choice of detector settings on each trial? If so, please present it--I'm quite confident you are either missing one of the conditions of Bell's theorem, or that your example does not actually violate any of the inequalities.

 

[vanesch]

Then the left-hand result is a.s and the right-hand result is s'.b'; each a dot-product.

The problem is: the outcome is not a continuous quantity! It is a discrete quantity, with PROBABILITY equal to the numbers you give, with a shift. You only give the expectation values of the outcomes, but the trick is that each individual outcome is a +1 or a -1, and not a continuous value in between both (although their expectation is of course).

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

 

[DrChinese]

Let s and s' be classical angular-momenta. Then (to the extent that we meet all the Bell-theorem criteria) the result is a wholly classical refutation of Bell's theorem.

wm,

1. This is basicly akin to saying "let's assume Bell's Theorem is wrong", which is hand-waving. You have to provide us something that yields results consistent with QM AND is local AND meets the realism requirement. You can't just say you have accomplished this because s and s' are classical.

2. Specifically, what are the expected probabilities for the 8 permutations:

A+B+C+
A+B+C-
...
A-B-C-

You will find that you cannot fill in such a table with non-negative numbers and still match QM. In other words, you have ignored the realism requirement entirely.

-DrC

 

[JesseM]

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

(a.s)*(s'.b') + (1 - a.s)*(1 - s'.b') - (a.s)*(1 - s'.b') - (1 - a.s)*(s'.b')

or

4*(a.s)*(s'.b') - 2*(a.s + s'.b') + 1

 

[wm]

The problem is: the outcome is not a continuous quantity! It is a discrete quantity, with PROBABILITY equal to the numbers you give, with a shift. You only give the expectation values of the outcomes, but the trick is that each individual outcome is a +1 or a -1, and not a continuous value in between both (although their expectation is of course).

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

Thanks for this comment.

1. Could you let me see how you would do the QM derivation, please?

2. Here's what I was thinking with my classical maths: In the double-peaked output from an S-G magnet, we allocate +1 xor -1 in accord with the direction of the output. Say: +1 = UP; -1 = DOWN.

That is, we do not allocate a different number to those particles which arrive (say) at the bottom of the UP distribution as opposed to those which have emerged at the up-side of the UP distribution. All the UPs get +1, etc.

Thus, the number (+1 xor -1) being allocated in line with the direction (UP xor DOWN) irrespective of the position in either distribution: I thought that +1 xor -1 could equally be allocated (equally arbitrarily) in accord with the sign of the dot-product in my classical example.

?

Thanks again, wm

 

[wm]

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

(a.s)*(s'.b') + (1 - a.s)*(1 - s'.b') - (a.s)*(1 - s'.b') - (1 - a.s)*(s'.b')

or

4*(a.s)*(s'.b') - 2*(a.s + s'.b') + 1

Dear JesseM, this is a bit rushed, BUT:

If I anywhere have probabilities going negative, JUST SHOOT ME!

a.s is a dot product that make take on values from -1 to +1. It cannot be a probability in my classical maths.

I haven't look at the rest of your post. I will (later) if you want me to?

wm

 

[JesseM]

Dear JesseM, this is a bit rushed, BUT:

If I anywhere have probabilities going negative, JUST SHOOT ME!

Sorry, I didn't realize that with the way you allow s and a to vary, a.s could be negative; but see below.

a.s is a dot product that make take on values from -1 to +1. It cannot be a probability in my classical maths.

But all the Bell inequalities I know of are based on the assumption that each measurement can only have two distinct outcomes, like "spin-up" and "spin-down". So, I was assuming that when the detector projects s onto a using the dot product, the resulting value is then used as a probability to display one of the two possible results, which I assign values +1 and -1 (following the convention used in the CHSH inequality which we've discussed before). I hadn't noticed that a.s could be negative, but you could always remedy this by making the probability equal to (1/2)(a.s) + 1/2, which will be a value between 0 and 1. Obviously this would mean you'd have to modify my equations above...if (1/2)*(a.s) + 1/2 is the probability of the left detector getting result +1, and 1 - [(1/2)*(a.s) + 1/2] = 1/2 - (1/2)*(a.s) is the probability of the left detector getting result -1, and (1/2)*(s'.b') + 1/2 is the probability of the right detector getting result +1, and 1 - [(1/2)*(s'.b') + 1/2] = 1/2 - (1/2)*(s'.b') is is the probability of the right detector getting result -1, then the expectation value for the product of their two results is:

[(1/2)*(a.s) + 1/2]*[(1/2)*(s'.b') + 1/2]
+ [1/2 - (1/2)*(a.s)]*[1/2 - (1/2)*(s'.b')]
- [(1/2)*(a.s) + 1/2]*[1/2 - (1/2)*(s'.b')]
- [1/2 - (1/2)*(a.s)]*[(1/2)*(s'.b') + 1/2]

Surprisingly, this all seems to simplify to an expectation value of (a.s)*(s'.b'), which is what you were calculating in the first place! Were you making this assumption about the probabilities all along, or is it just lucky? Either way, I think the math in your proposed proof that this is equal to -cos(a, b') is incorrect, see my next post for more on that point. Also note that if you make this assumption about probabilities, the results of the two detectors will not be perfectly correlated when they pick the same setting, so if you're out to challenge Bell's theorem, you can only look at inequalities like the CHSH inequality which do not make any assumption about perfect correlations with identical settings.

Alternately, you might avoid probabilities by saying that on any trial where the value of a.s was greater than or equal to -1 and smaller than 0, the experimenter will see the result spin-down (assigned a value of -1), and on any trial where the value of a.s was greater than or equal to 0 and smaller than or equal to 1, the experimenter will see the result spin-up (assigned a value of +1). Then you could say the same applies to s'.b', and calculate the expectation value of the product of their two results; but again, it would be something different than -cos (a, b'). Quickly diagramming the problem leads me to think that if s is equally likely to have any angle, then if (a, b') represents the angle between a and b' in degrees, the probability that they both get the same spin would be (a, b')/180, and the probability they get opposite spins would be [180 - (a, b')]/180, so the expectation value for the product of their results would be (a, b')/180 - [180 - (a, b')]/180, or [(a, b')/90] - 1.

Either way, I think you need to fix it so each experimenter can only get two discrete results on a given trial. If you know of any Bell inequalities that do not assume each measurement can have only one of two possible results, then please give the name of the inequality you're thinking of, or a link giving the mathematical formulation of the inequality.

 

[JesseM]

To derive the related correlation, we require (using a recognised notation http://en.wikipedia.org/wiki/Column_vector ), with < ... > denoting an average:

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

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

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

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

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

(8) = - (ax ay az) <1> (bx', by', bz')

(9) = - a.b'

(10) = - cos (a, b').

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&#039;_x*s_x + b&#039;_y*s_y + b&#039;_z*s_z)

But this is not equal to -(a_x * b&#039;_x + a_y * b&#039;_y + a_z * b&#039;_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').

 

[wm]

Some important issues continue.

The problem is: the outcome is not a continuous quantity! It is a discrete quantity, with PROBABILITY equal to the numbers you give, with a shift. You only give the expectation values of the outcomes, but the trick is that each individual outcome is a +1 or a -1, and not a continuous value in between both (although their expectation is of course).

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

Dear vanesch (and with respect: DrC and JesseM and some others as well).

This is a bit rushed as I am in a meeting BUT:

1. I'd like to point out that I began with the exact equation that Bell used [1964; equation (3)]. I get the identical result also: - a.b'.

2. NB: At the moment I have limited my derivation to that which I offered: A wholly LOCAL and CLASSICAL derivation of the EPRB correlation. That is, I have derived the limit to which your derivation must tend in accord with Bohr's Correspondence Principle.

3. I hope we might agree on the following important point: Since the space-like experimental results were derived by me in terms of high-school maths, AND without any reference whatsoever to non-locality, there must be an equivalent QM derivation equally devoid of non-locality.

4. So: May I ask you again to provide your fundamental derivation of the EPRB correlation (ie, from first principles; and preferably in the terms of the OP), beginning with Bell's equation (just as I did)?

5. I request this of you because you are a PF MENTOR and because SCIENCE ADVISER DrC has not been able to derive it and SCIENCE ADVISER JesseM is a bit confused on my mathematics (but I will sort that out soon: noting for now that there are no errors in my maths, so far as I can see from my high-school text on vector-analysis).

6. Your derivation will not be wasted as I am keen to learn. HOWEVER: If you will not be providing this important derivation; could you please point to where I might find a detailed version; preferably one that complies with your own local interpretation of QM?

7. Not to muddy the waters any further now: I respectfully suggest that there are other matters in your post which may be presented differently and more clearly. Not to be addressed now because they may be clarified when I see your EPRB derivation.

Thank you, and sincerely, wm

 

[DrChinese]

5. I request this of you because you are a PF MENTOR and because SCIENCE ADVISER DrC has not been able to derive it and SCIENCE ADVISER JesseM is a bit confused on my mathematics (but I will sort that out soon: noting for now that there are no errors in my maths, so far as I can see from my high-school text on vector-analysis).

wm,

You can pick a single item out of Bell's paper, and quote it out of context and it still won't mean anything. You might consider toning down your claims a bit until you see them all the way through.

I will repeat what I have stated previously: there is a mathematical requirement that you are skipping entirely, and that is the requirement of realism. If you ignore that, you are missing the entire point of Bell. That requirement is that there is a real probability of a specified outcome of observations at settings A, B and C which has a value between 0 and 1. You do not need the formula you are tinkering with to derive Bell's Theorem, as Mermin has shown.

 

[JesseM]

1. I'd like to point out that I began with the exact equation that Bell used [1964; equation (3)]. I get the identical result also: - a.b'.

In physics it is important to understand what physical quantity the terms in an equation stand for. In Bell's paper a and b represent possible angles of the stern-gerlach device used to measure the spin of the two particles, and these measurements will always yield one of two results, which in the "Formulation" paragraph on p. 1 of the paper you refer to he labels as +1 and -1. The "expectation value" refers to the average expected value of the product of the two measurements, which would be:

(+1)*probability(angle a yields +1, angle b yields +1) +
(+1)*probability(angle a yields -1, angle b yields -1) +
(-1)*probability(angle a yields +1, angle b yields -1) +
(-1)*probability(angle a yields -1, angle b yields +1)

This experiment is not one where the result of each measurement is an arbitrary real number between -1 and +1, and where the expectation value is the average value of the product of these two real numbers, as you seem to assume in your example. Again, Bell is assuming that each measurement always yields one of two results which are assigned values +1 and -1, so when you multiply the two values you always get the result -1 or +1 on any given trial; the expectation value refers to the average this product over many trials.

As I pointed out in a previous post, if you assume that each experimenter has a device which projects the vector s onto their own angle (either a or b'), like a.s, and then this continuous value is used to determine the probability (1/2)*(a.s) + 1/2 that the experimenter will get a +1 result on that trial or a -1 result, then it actually does work out that the expectation value for the product of their results will end up being a.s*s'.b' as you had in your attempted proof. But again, in this case you don't have a guarantee that when they pick the same angle they always get opposite results on a given trial, so Bell's theorem would only rule out inequalities which don't include this assumption, like the CHSH inequality.

2. NB: At the moment I have limited my derivation to that which I offered: A wholly LOCAL and CLASSICAL derivation of the EPRB correlation. That is, I have derived the limit to which your derivation must tend in accord with Bohr's Correspondence Principle.

I don't see how the correspondence principle would imply that the expectation value for an experiment in which each experimenter can get any result between +1 and -1 on a given trial would be identical to the expectation value for an experiment in which each experimenter can only get one of two results, either +1 or -1. Is this what you're claiming here?

3. I hope we might agree on the following important point: Since the space-like experimental results were derived by me in terms of high-school maths, AND without any reference whatsoever to non-locality, there must be an equivalent QM derivation equally devoid of non-locality.

If you were indeed able to reproduce the result that the expectation value is -cos(a, b) in a purely classical experiment, where on each trial each experimenter gets either +1 or -1 and the expectation value is for the average of the products of their two answers, and your classical experiment obeyed the conditions of Bell's theorem like the source not having foreknowledge of the detector settings, then yes, this would show that QM was compatible with local hidden variables. The problem is you didn't do this--you seem to assume that each experiment can yield a continuous spectrum of values rather than just +1 or -1, and even if you make the assumption I mentioned above where the probability of getting +1 is (1/2)*(a.s) + 1/2, so that the expectation value is indeed just a.s*s'.b', there seems to be an error in your "high school math", since this is not equal to -cos(a, b).

4. So: May I ask you again to provide your fundamental derivation of the EPRB correlation (ie, from first principles; and preferably in the terms of the OP), beginning with Bell's equation (just as I did)?

You're looking for a derivation of why quantum mechanics predicts that the expectation value is -a.b? Why would this be useful, since here we are just trying to figure out whether this expectation value can be reproduced in a classical experiment?

5. I request this of you because you are a PF MENTOR and because SCIENCE ADVISER DrC has not been able to derive it and SCIENCE ADVISER JesseM is a bit confused on my mathematics (but I will sort that out soon: noting for now that there are no errors in my maths, so far as I can see from my high-school text on vector-analysis).

Yes, please state whatever theorems from your vector textbook you are making use of in your proof. But in the meantime, could you please check the math on my example of a = 0 degrees, b' = 60 degrees, and s = 90 degrees? Do you disagree that in this case, - a.s*s.b' = - cos(90)*cos(30) = - (0)*(0.866) = 0, while - cos(a, b') = - cos(60) = -0.5? If you agree with my math on this example, then it seems clear there must be an error in your proof somewhere, unless I misunderstood what you claimed to have proved.

6. Your derivation will not be wasted as I am keen to learn. HOWEVER: If you will not be providing this important derivation; could you please point to where I might find a detailed version; preferably one that complies with your own local interpretation of QM?

Have you ever studied the basics of QM? Derivations of probabilities and expectation values have nothing to do with one's interpretation, they basically just involve finding state vector representing the quantum state of the system, expanding it into a weighted sum of eigenvectors of the operator representing the variable you want to measure (energy, for example), and then the square of the complex amplitude for a given eigenvector represents the probability that you'll get a given value when you measure that variable (the value corresponding to a particular eigenvector is just the eigenvalue of that vector). And of course, once you know the probability for each possible value, the expectation value is just the sum of each value weighted by its probability. If you're not familiar with the general way probabilities and expectation values are derived in QM, then a specific derivation of the expectation value for the spins of two entangled electrons won't make much sense to you. And like I said, the derivation itself would have nothing to say about locality or nonlocality, it's just when you apply Bell's theorem to the predictions of QM that you see they are not compatible with local hidden variables.

edit: by the way, if you are familiar with calculations in QM, you can look at this page for a nearly complete derivation. What they derive there is that if q represents the angle between the two detectors, then the probability that the two detectors get the same result (both spin-up or both spin-down) is sin^2 (q/2), and the probability they get opposite results (one spin-up and one spin-down) is cos^2 (q/2). If we represent spin-up with the value +1 and spin-down with the value -1, then the product of their two results when they both got the same result is going to be +1, and the product of their results when they got different results is going to be -1. So, the expectation value for the product of their results is:

(+1)*sin^2 (q/2) + (-1)*cos^2 (q/2) = sin^2 (q/2) - cos^2 (q/2)

Now, if you look at the page on trigonometric identities here, you find the following identity:

cos(2x) = cos^2 (x) - sin^2 (x)

So, setting 2x = q, this becomes:

cos(q) = cos^2 (q/2) - sin^2 (q/2)

Multiply both sides by -1 and you get:

sin^2 (q/2) - cos^2 (q/2) = - cos (q)

This fills in the final steps to show that the expectation value for the product of their results will be the negative cosine of the angle between their detectors.

 

[JesseM]

will repeat what I have stated previously: there is a mathematical requirement that you are skipping entirely, and that is the requirement of realism.

When you say wm is not satisfying the requirement of realism, are you referring to the idea that wm's example involves negative probabilities? If you look at wm's post #65, I don't think that's the problem--while it's true that the dot product of the vector s sent from the source and the vector a representing the experimenters measurement setting can be negative, I don't think wm intended for this dot product to be a probability in the first place. Rather, it seems to me that wm has just failed to realize that Bell was assuming that each experiment could only yield two discrete results, spin-up or spin-down; wm is instead imagining an experiment where each experiment can yield a continuous infinity of results between -1 and 1, and the "expectation value" he's calculating is for the product of the real number that each experimenter gets as a result.

 

[vanesch]

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

 

[DrChinese]

When you say wm is not satisfying the requirement of realism, are you referring to the idea that wm's example involves negative probabilities? ...

JesseM,

No, I am not referring to that.

Looking at Bell as a reference, so we're talking about the same thing:

1. wm's "- a . b" deal is the QM expectation value, referenced as (3). It really doesn't matter how you get to this, the key is that you know that it reduces to cos theta for spin 1/2 particles. Obviously, for photons it is a slightly different formula. If this was all there was to it, then Bell would have stopped after (7) or so - and wouldn't have too much.

2. Bell then goes to great pains to show that the mapping of (2) to (3), with A and B, WILL work. I.e. that a hidden A and hidden B is possible, and will get you to the QM predictions if you need it to. So we still don't have much.

3. Then around (14), Bell introduces the realism requirement: the mapping with hidden variables does not extend to an A, B AND C! He does not label it as the "realism requirement", that is something I label it as because it is present in EVERY proof of Bell's Theorem one way or another. It is the assumption - requirement - that there simultaneously be an A, B and C to discuss. If there isn't, then there is no (15) which is the Inequality.

Why do we need this assumption? Because without it we can't see the real problem that occurs when wm says:

"(4) = - <(a.s) (s.b')> becomes (9) = - a.b' "

Clearly this is the original classical hidden variable idea in disguise, which Bell says is "no difficulty". But this doesn't work if there is a c too, and Bell somehow figured that out. (Amazing accomplishment to me...)

4. In my proofs of Bell's Theorem, I always make this assumption *explicit*. I will repeat that without simultaneous A, B and C: there is no Bell's Theorem. You can reformulate the theorem in many ways, such as my negative probabilities version and my version that follows Mermin. These versions substitute easier math, or at least an easier notation for most people to follow - and is built around one specific counter-example. Bell presents a more general proof and then picks a specific example (ac=90 degrees, ab=bc=45 degrees is how I read it) to show the issues.

-DrC

 

[JesseM]

It is the assumption - requirement - that there simultaneously be an A, B and C to discuss.

But why do you think wm's example violates this assumption of realism? In his example, if one experimenter can pick for his measurement setting a one of three possible angles a = A, a = B, or a = C, then if you know the angle of the vector S sent out by the source, you can determine in advance the value of A.S if the experimenter picks angle A, and the value of B.S if the experimenter picks B, and the value of C.S if the experimenter picks C. Of course these values may be any real number between -1 and 1 depending on the angles involved, whereas in the experiments Bell is dealing with, "local realism" means that if you know the hidden state of the particle, then you can determine in advance whether each of the three detector angles will yield either the result -1 or +1, with no other possibilities.

 

[DrChinese]

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&#039;_x*s_x + b&#039;_y*s_y + b&#039;_z*s_z)

But this is not equal to -(a_x * b&#039;_x + a_y * b&#039;_y + a_z * b&#039;_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.