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OK Corral: Local versus non-local QM

by wm
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wm
#55
Feb21-07, 03:31 PM
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Quote Quote by wm View Post
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
#56
Feb21-07, 03:57 PM
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Quote Quote by wm View Post
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.
Quote Quote by wm
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?
Quote Quote by wm
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.
Quote Quote by wm
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.
Quote Quote by wm
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".
Quote Quote by wm
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.
Quote Quote by wm
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
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Feb21-07, 03:59 PM
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Quote Quote by wm View Post
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
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Feb21-07, 04:33 PM
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Quote Quote by wm View Post
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
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Feb21-07, 04:39 PM
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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
#60
Feb21-07, 04:47 PM
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Quote Quote by vanesch View Post
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
#61
Feb21-07, 04:55 PM
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Quote Quote by JesseM View Post
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?

Best, wm
JesseM
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Feb21-07, 05:02 PM
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Quote Quote by wm View Post
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.
Quote Quote by wm
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
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Feb21-07, 06:45 PM
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Quote Quote by wm View Post
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:

[tex]- (a_x * s_x + a_y * s_y + a_z * s_z)*(b'_x*s_x + b'_y*s_y + b'_z*s_z)[/tex]

But this is not equal to [tex]-(a_x * b'_x + a_y * b'_y + a_z * b'_z)[/tex], even if you stipulate that [tex](s_x * s_x + s_y * s_y + s_z * s_z) = 1[/tex]. It seems like you got the rules for the dot product confused with the rules for multiplication, you can't say that (a.s)*(s.b') is equivalent to (s.s)*(a.b').
wm
#64
Feb22-07, 06:50 PM
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Quote Quote by vanesch View Post
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
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Feb22-07, 09:54 PM
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Quote Quote by wm View Post
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
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Feb22-07, 10:26 PM
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Quote Quote by wm View Post
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.
Quote Quote by wm
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?
Quote Quote by wm
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).
Quote Quote by wm
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?
Quote Quote by wm
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.
Quote Quote by wm
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 [tex]sin^2 (q/2)[/tex], and the probability they get opposite results (one spin-up and one spin-down) is [tex]cos^2 (q/2)[/tex]. 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:

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

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

[tex]cos(2x) = cos^2 (x) - sin^2 (x)[/tex]

So, setting 2x = q, this becomes:

[tex]cos(q) = cos^2 (q/2) - sin^2 (q/2)[/tex]

Multiply both sides by -1 and you get:

[tex]sin^2 (q/2) - cos^2 (q/2) = - cos (q)[/tex]

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
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Feb22-07, 11:44 PM
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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
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Feb22-07, 11:58 PM
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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
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Feb23-07, 12:35 AM
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Quote Quote by JesseM View Post
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
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Feb23-07, 12:55 AM
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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
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Feb23-07, 01:08 AM
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Quote Quote by JesseM View Post
But why do you think wm's example violates this assumption of realism? In his example, if one experimenter can pick for his measurement setting a one of three possible angles a = A, a = B, or a = C, then if you know the angle of the vector S sent out by the source, you can determine in advance the value of A.S if the experimenter picks angle A, and the value of B.S if the experimenter picks B, and the value of C.S if the experimenter picks C. Of course these values may be any real number between -1 and 1 depending on the angles involved, whereas in the experiments Bell is dealing with, "local realism" means that if you know the hidden state of the particle, then you can determine in advance whether each of the three detector angles will yield either the result -1 or +1, with no other possibilities.
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
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Feb23-07, 01:20 AM
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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?


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