OK Corral: Local versus non-local QM

[wm]

?

Start with this one:

Originally Posted by JesseM
It seems like you got the rules for the dot product confused with the rules for multiplication, you can't say that (a.s)*(s.b') is equivalent to (s.s)*(a.b'). (Emphasis added.)



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

Do you have a source?

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

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

wm

 

[vanesch]

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

Do you have a source?

In your post:

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

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

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

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

which is not correct.

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

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

 

[Doc Al]

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

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

 

[JesseM]

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

 

[wm]

Matrix average vs dot.product average?

To add to that, I think it would help a lot if you would address my previous request to make explicit where you are using the dot product and where you are using multiplication: And if you want to use matrix multiplication as opposed to ordinary arithmetical multiplication, you could use the symbol x to denote that. If you do, perhaps you could also label each vector as either a row vector or a column vector like (ax, ay, az)_c or (ax, ay, az)_r. Wait, so are you saying that commas vs. no commas denotes column vectors vs. row vectors? In this case (sx, sy, sz)x(sx sy sz) will not be a single number, but rather a 3x3 matrix.

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

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

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

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

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

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

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

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

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

 

[JesseM]

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

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

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

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

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

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

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

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

 

[DrChinese]

1. I disagree that "Bell's theorem" primarily revolves around picking specific angles, if that's what you mean by "That 'exercise for the reader' IS Bell's Theorem". The proof involves finding an inequality that should hold for arbitrary angles under local realism;

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

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

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

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

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

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

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

 

[wm]

Sos! Sos! Sos!

In your post:

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

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

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

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

which is not correct.

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

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

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

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

Since s' = -s we have:

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

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

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

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

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

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

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

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


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

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

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

Thanks, wm

 

[JesseM]

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

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

 

[DrChinese]

Just a quick note that not all of the above cases are always valid even in a classical correlation scenario. (e.g [2] and [7] can be illegal simultaneous events)

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

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

 

[JesseM]

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

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

Since s' = -s we have:

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

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

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

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

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

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

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

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

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

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

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

 

[DrChinese]

1. Bell provides a general proof that a certain inequality can never be violated under local realism, a statement of the form "for all experiments obeying local realism and satisfying certain conditions, this inequality will be satisfied".

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

3. And that's what wm tried to do--find a single example of a local realist experiment which would satisfy Bell's conditions and yet violate an inequality.

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

2. I agree totally.

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

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

 

[JesseM]

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

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

I guess you could say that if one agrees with Bell's theorem, then local realism makes the prediction that "for all experiments satisfying X conditions, inequality Y will be satisfied". And in this case, QM can give an example of the form "here's an experiment satisfying X conditions which violates inequality Y", thus proving QM is incompatible with local realism. But the problem here is that wm believes there's a flaw in Bell's theorem, so he does not agree that local realism makes the prediction "for experiments satisfying X conditions, inequality Y will be satisfied" in the first place; he's trying to disprove Bell's theorem by showing that local realism can also give an example of the form "here's an experiment satisfying X conditions which violates inequality Y".

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

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

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

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

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

 

[JesseM]

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

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

3. And that's what wm tried to do--find a single example of a local realist experiment which would satisfy Bell's conditions and yet violate an inequality.

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

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

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

 

[wm]

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

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

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

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

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

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

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

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

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

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

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

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

Thanks, as always, wm

 

[wm]

Correct format for derivation?

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

<SNIP>

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

Thanks, as always, wm

Jesse, what abt I do something like this? (ROUGHLY)

- <a.s * s.b'>

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

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

...

...

= - (ax ay az) [100, 010, 001] (bx, by, bz)

= - a.b'

Would that matrix representation (and the one *) be OK?

wm

 

[JesseM]

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

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

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

Actually, one minor physical issue occurred to me--you have the vectors as 3-vectors, but if you want to mimic the type of spin measurements made in QM, they should really be 2-vectors. This is because, when you measure spin using Stern-Gerlach magnets, the long axis of the magnet has to be alligned parallel to the particle's path, so you just have the freedom to rotate the magnets around this axis at any angle (this is why in discussions of Bell's theorem people often talk about each experimenter choosing 'an angle'--if they had 3 degrees of freedom, they would each have to select 2 distinct angles instead).

So, I'd amend your (5) to look like this:

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

Of course since the two quantities in brackets give scalars, strictly speaking the middle x could be replaced by a *, but leaving it as matrix multiplication makes it easier to go to step (6):

(6) = - (ax ay)x<(sx, sy)x(sx sy)>x(bx', by')

That's the last step in your proof I'd agree with.

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

The 2-matrix in the center actually does not work out to be a unit matrix. One thing to note is that if s is an individual unit vector, while it's true that (sx sy)x(sx, sy), i.e. the dot product of s with itself, is always 1, it's not true that the 2x2 matrix (sx, sy)x(sx sy) is always a unit matrix; for example, if sx = 0.5 and sy = 0.866, the matrix works out to be:

(0.5)*(0.5) (0.5)*(0.866)
(0.866)*(0.5) (0.866)*(0.866)

or

0.25 0.433
0.433 0.75

However, you'd probably point out that we are interested in the average expectation value of this matrix when s is allowed to take any angle from 0 to 2pi. We know that if the angle of s is \theta, then s_x = cos(\theta ) and s_y = sin(\theta ). So, the matrix would be:

\left( \begin{array}{cc} cos^2(\theta ) &amp; cos(\theta )*sin(\theta ) \\<br /> sin(\theta )*cos(\theta ) &amp; sin^2(\theta ) \end{array} \right)

For each of these four components, to find the expectation value we must integrate them from 0 to 2pi, then multiply the result by (1/2pi)...see the end of my previous post for an explanation of why the expectation value of a function based on an arbitrary angle would be calculated in this way.

Using the integrator, we have:

\int cos^2(\theta ) \, d\theta = (1/2)*(\theta + cos(\theta )*sin(\theta ))
\int sin(\theta )*cos(\theta ) \, d\theta = (-1/2)*cos^2(\theta )
\int sin^2(\theta ) \, d\theta = (1/2)*(\theta - cos(\theta )*sin(\theta ))

So, taking each function f(\theta ) and plugging in the limits of integration f(2pi) - f(0), the expectation value for the matrix is:

\frac{1}{2\pi} \left( \begin{array}{cc} \pi &amp; 0 \\<br /> 0 &amp; \pi \end{array} \right)

or:

\left( \begin{array}{cc} \frac{1}{2} &amp; 0 \\<br /> 0 &amp; \frac{1}{2} \end{array} \right)

So, it looks like this will end up just being another way of proving that - <a.s*s.b'> is equal to -(1/2)*cos(a - b), which I had proved earlier by just doing one big integral.

 

[wm]

Getting there

Actually, one minor physical issue occurred to me--you have the vectors as 3-vectors, but if you want to mimic the type of spin measurements made in QM, they should really be 2-vectors. This is because, when you measure spin usingStern–Gerlach_experiment[/URL], the long axis of the magnet has to be alligned parallel to the particle's path, so you just have the freedom to rotate the magnets around this axis at any angle (this is why in discussions of Bell's theorem people often talk about each experimenter choosing 'an angle'--if they had 3 degrees of freedom, they would each have to select 2 distinct angles instead).

So, I'd amend your (5) to look like this:

(5) = - <[(ax ay)[b]x[/b](sx, sy)][b]x[/b][(sx sy)[b]x[/b](bx', by')]>

Of course since the two quantities in brackets give scalars, strictly speaking the middle [b]x[/b] could be replaced by a *, but leaving it as matrix multiplication makes it easier to go to step (6):

(6) = - (ax ay)[b]x[/b]<(sx, sy)[b]x[/b](sx sy)>[b]x[/b](bx', by')

That's the last step in your proof I'd agree with. The 2-matrix in the center actually does not work out to be a unit matrix. One thing to note is that if [b]s[/b] is an individual unit vector, while it's true that (sx sy)[b]x[/b](sx, sy), i.e. the dot product of [b]s[/b] with itself, is always 1, it's [i]not[/i] true that the 2x2 matrix (sx, sy)[b]x[/b](sx sy) is always a unit matrix; for example, if sx = 0.5 and sy = 0.866, the matrix works out to be:

(0.5)*(0.5) (0.5)*(0.866)
(0.866)*(0.5) (0.866)*(0.866)

or

0.25 0.433
0.433 0.75

However, you'd probably point out that we are interested in the average expectation value of this matrix when [b]s[/b] is allowed to take any angle from 0 to 2pi. We know that if the angle of [b]s[/b] is $$\theta$$, then $$s_x = cos(\theta )$$ and $$s_y = sin(\theta )$$. So, the matrix would be:

$$cos^2(\theta ) \,\,\,\, cos(\theta )*sin(\theta )$$
$$sin(\theta )*cos(\theta ) \,\,\,\, sin^2(\theta )$$

$$\left( \begin{array}{cc} 1- cos^2(\theta ) & cos(\theta )*sin(\theta ) \\ -sin(\theta )*cos(\theta ) & sin^2(\theta ) \end{array} \right)$$

For each of these four components, to find the expectation value we must integrate them from 0 to 2pi, then multiply the result by (1/2pi)...see the end of my previous post for an explanation of why the expectation value of a function based on an arbitrary angle would be calculated in this way.

Using [url=http://integrals.wolfram.com/index.jsp]the integrator[/url], we have:

$$\int cos^2(\theta ) \, d\theta = (1/2)*(\theta + cos(\theta )*sin(\theta ))$$
$$\int sin(\theta )*cos(\theta ) \, d\theta = (-1/2)*cos^2(\theta )$$
$$\int sin^2(\theta ) \, d\theta = (1/2)*(\theta - cos(\theta )*sin(\theta ))$$

So, taking each function $$f(\theta )$$ and plugging in the limits of integration f(2pi) - f(0), and then multiplying the result by the (1/2pi) which was outside the integral, we find that the average expected value for the four components of the matrix works out to:

1/2 0
0 1/2

So, it looks like this will end up just being another way of proving that - <[b]a.s[/b]*[b]s.b'[/b]> is equal to -(1/2)*cos(a - b), which I had proved earlier by just doing one big integral.[/QUOTE]

Jesse, I think we're getting there.

BUT: Why do you say theta takes on values O = 2pi? It seems to me that if you want to use angles (I prefer unit vectors) then you need to integrate over 4pi ([U]which would give you the missing factor of 2 that you're looking for[/U])?

The 2pi divisor in you calculation would remain unchanged because the [U]angular differences[/U] can only range thereover.

This would then give the same result as that matrix that you gave earlier: for averaging inside the matrix by observation only, you produce exactly the unit matrix.

Do you see that (by observation without any maths; maybe except in your head) your matrix reduces to the unit-matrix (1, [I]U[/I], [I]I[/I], [I]E[/I])? I prefer 1 (but it is part of my problem here): Should I write it as [I]U[/I] in QM? What is best?

Thus it seems I was mathematically wrong in representing the unit-matrix as a plain 1 (= [B]s.s[/B]) and so just writing <[B]s.s[/B]> = <1>, etc. Actually I was more reading the equations physically and thinking my short-cut was Ok, given that the spaces were defined by the opening gambit.

I know you seem to say that mine/yours is not the unit matrix, but would you reconsider it and let me know. I cannot see why it is not; for you half make the case yourself above. PERHAPS you maybe overlooking that the ensemble-average is over an infinity of exemplars: so, on averaging,

(1) sisj = sij = dij (Kronecker)?

(In my opinion, all elements are averaging zero, except on the main diagonal which all average 1).

PS: Regarding the S-G orientation, and given I'm considering idealised experiments, and given the spherical symmetry of the classical state that I'm using: I find it easier to remain general with unit-vectors; for when we get it right there will be no error, even if the experiments today cannot match it.

Of course, practically/loosely, running along the flight-axis a little longer gives time for the spins to stabilise and form the two-peaked S-G output. But once we get our present maths correct, dichotomic outputs like S-G fall naturally from my equations via a simple realistic re-definition of what constitutes an experimental outcome. But this is getting ahead of where we are.

PS: Once I get your reply I think it's time to re-present my experiment and maths. To tidy lots of ends up? Do you agree?

Thanks, [B]wm[/B]

 

[JesseM]

BUT: Why do you say theta takes on values O = 2pi? It seems to me that if you want to use angles (I prefer unit vectors)

How would you propose to integrate over all possible unit vectors without having an angle in the integral? I assume you want the probability distribution to have the property that the probability of getting a unit vector whose angle is between 0 and 20 is the same as the probability of getting a unit vector whose angle is between 20 and 40, and so forth, so the probability of an angle between A and A+C is always the same as the probability of getting one between B and B+C.

Also, note that for every possible choice of angle, the components sx and sy of the unit vector with that angle are uniquely determined (ie sx = cos(angle) and sy = sin(angle)); and every possible unit vector corresponds to an angle in this way, the mapping between distinct angles and distinct unit vectors is one-to-one.

then you need to integrate over 4pi (which would give you the missing factor of 2 that you're looking for)?

Why do you say that? You're aware that 2pi in radians is equivalent to 360 degrees, right? And just like 380 degrees is exactly the same angle as 20 degrees, so 3pi degrees is exactly the same angle as pi degrees (and the vectors for each would have all the same components, so they'd actually be the same vector). If you integrate over 2pi, you're integrating over every possible distinct angle that s can take, and thus every possible distinct unit vector s.

Do you see that (by observation without any maths; maybe except in your head) your matrix reduces to the unit-matrix (1, U, I, E)?

No, it ends up being (1/2) times the unit matrix.

If you don't trust in integrals, we could try doing a numerical approximation, i.e. pick a significant number of evenly-spaced angles between 0 and 360 for s, find a.s*s.b for each angle, add them all together, and divide by the total number of angles to find the average. If we use a fairly large number of angles--say, 36 (angles at 10-degree intervals) or 72 (angles at 5-degree intervals) then this should be a pretty good approximation for the case where the angle can vary continuously, and then we can check whether the result is close to my predicted value of (1/2)cos(a - b) or close to your original predicted value of cos(a - b). Would you like me to take a shot at this and see the result?

PERHAPS you maybe overlooking that the ensemble-average is over an infinity of exemplars

Of course I'm not overlooking it, that's why I did an integral rather than a sum. And like I said, there's a one-to-one relationship between the set of all possible unit vectors and the set of all possible angles, that's why I integrated over every possible distinct angle.

so, on averaging,

(1) sisj = sij = dij (Kronecker)?

I assume sij represents the component in the ith row and jth column of the matrix, while si represents the ith element of the column vector, and sj represents the jth element of the row vector? If so, why do you think that "averaging" would give the Kronecker delta? This is simply wrong, I've already shown that when you average over all possible unit vectors s, with components s1 = cos(angle) and s2 = sin(angle), the result ends up being (1/2)*dij.

PS: Regarding the S-G orientation, and given I'm considering idealised experiments, and given the spherical symmetry of the classical state that I'm using: I find it easier to remain general with unit-vectors; for when we get it right there will be no error, even if the experiments today cannot match it.

I'm using unit vectors too, just unit 2-vectors rather than unit 3-vectors. If you use 3-vectors the correspondence between the classical case of projecting angles and the quantum case of measuring spins becomes murkier; what's more, all the integrals I've presented would have to be different, they'd have to be double integrals where you integrate over two different angles \theta and \phi, not single integrals where you integrate over one angle.

 

[wm]

sx*sx = 1 (me) or 1/2 (you)?

How would you propose to integrate over all possible unit vectors without having an angle in the integral? I assume you want the probability distribution to have the property that the probability of getting a unit vector whose angle is between 0 and 20 is the same as the probability of getting a unit vector whose angle is between 20 and 40, and so forth, so the probability of an angle between A and A+C is always the same as the probability of getting one between B and B+C.

Also, note that for every possible choice of angle, the components sx and sy of the unit vector with that angle are uniquely determined (ie sx = cos(angle) and sy = sin(angle)); and every possible unit vector corresponds to an angle in this way, the mapping between distinct angles and distinct unit vectors is one-to-one. Why do you say that? You're aware that 2pi in radians is equivalent to 360 degrees, right? And just like 380 degrees is exactly the same angle as 20 degrees, so 3pi degrees is exactly the same angle as pi degrees (and the vectors for each would have all the same components, so they'd actually be the same vector). If you integrate over 2pi, you're integrating over every possible distinct angle that s can take, and thus every possible distinct unit vector s. No, it ends up being (1/2) times the unit matrix.

If you don't trust in integrals, we could try doing a numerical approximation, i.e. pick a significant number of evenly-spaced angles between 0 and 360 for s, find a.s*s.b for each angle, add them all together, and divide by the total number of angles to find the average. If we use a fairly large number of angles--say, 36 (angles at 10-degree intervals) or 72 (angles at 5-degree intervals) then this should be a pretty good approximation for the case where the angle can vary continuously, and then we can check whether the result is close to my predicted value of (1/2)cos(a - b) or close to your original predicted value of cos(a - b). Would you like me to take a shot at this and see the result? Of course I'm not overlooking it, that's why I did an integral rather than a sum. And like I said, there's a one-to-one relationship between the set of all possible unit vectors and the set of all possible angles, that's why I integrated over every possible distinct angle. I assume sij represents the component in the ith row and jth column of the matrix, while si represents the ith element of the column vector, and sj represents the jth element of the row vector? If so, why do you think that "averaging" would give the Kronecker delta? This is simply wrong, I've already shown that when you average over all possible unit vectors s, with components s1 = cos(angle) and s2 = sin(angle), the result ends up being (1/2)*dij. I'm using unit vectors too, just unit 2-vectors rather than unit 3-vectors. If you use 3-vectors the correspondence between the classical case of projecting angles and the quantum case of measuring spins becomes murkier; what's more, all the integrals I've presented would have to be different, they'd have to be double integrals where you integrate over two different angles \theta and \phi, not single integrals where you integrate over one angle.

Jesse,

1. I integrate over 4pi steradians (SOLID ANGLES, NOT PLANAR), which gives my integral the missing 2; also in the 3x3 matrix ... -> I.

2. ... which thus agrees with my evaluation of the 3x3 matrix that you generated.

3. That is: <sx*sx> = <sy*sy> = <sz*sz> = 1. All other elements average 0; the 4pi steradians being in play throughout.

4. The 3x3 unit-matrix I results.

5. Thus my -a.b' result stands OK (it seems to me).

Do we differ? And this might save me taking up your offer above, which looks a bit tedious to me.

PS: -1 </= si </= 1. -1 </= sj </= 1.

<si*si> = <sj*sj> = <{1, ..., 0} + {0, ..., 1}> = 1/2 + 1/2 = 1;

<si*sj> = <{-1, ..., 0, ..., 1}> = 0;

where { ... } indicates the infinite set of values each expression may take.

Gruss (in appreciation), wm

 

[vanesch]

3. That is: <sx*sx> = <sy*sy> = <sz*sz> = 1. All other elements average 0; the 4pi steradians being in play throughout.

No, this is the mistake.

If s is a random unit vector on a sphere, then < sx*sx > works out to be 1/3.

If it is on a circle, then it works out to be 1/2.

You can easily see this:

the COMPONENTS of a unit vector can never be larger than 1, right ? They vary between -1 and 1. So their square varies between 0 and 1, right ?

Now, for something that varies between 0 and 1, to have an AVERAGE value of 1, that means that it must ALWAYS be 1.

So sx^2 must always be 1, so sx must be -1 or 1.
same for sy.

So this means that you only consider unit vectors of the form (1,1), (1,-1), (-1,1) and (-1,-1). But these are not unit vectors !

But that means the average of the square of their components cannot be equal to 1.

Now, how to find quickly the right value ?

We know that s.s = 1.

So sx.sx + sy.sy = 1.

Take the expectation value of this equation, then we find:

< sx.sx > + < sy.sy > = 1

If we assume, by symmetry, that < sx.sx > = < sy.sy >, then they are equal to 1/2.

EDIT: btw, that expression that:

< sx.sx > + < sy.sy > = 1

by itself already proves that < sx.sx > can only be 1 if < sy.sy > = 0.

 

[JesseM]

Jesse,

1. I integrate over 4pi steradians (SOLID ANGLES, NOT PLANAR), which gives my integral the missing 2; also in the 3x3 matrix ... -> I.

So you're insisting on using 3-vectors rather than 2-vectors? Do you understand that the integral in 3 dimensions would have to be completely different than any of the integrals I've presented, that it would have to be a double integral over two separate angles \theta and \phi, with \theta varying from 0 to 2pi and \phi varying from 0 to pi, and using the area element dS = sin(\phi ) d\phi d\theta? Have you actually evaluated this integral for each component of the 3x3 matrix produced from (sx, sy, sz)x(sx sy sz)? If not, I have no basis for trusting your intuition that it would work out to the unit matrix, since this intuition was wrong in the case of 2-vectors. And actually evaluating this integral would be more work for me...can you first tell me whether you agree that, in the case of 2-vectors, the quantity <(sx, sy)x(sx sy)> will not work out to the unit matrix, but instead to 1/2 times the unit matrix? If you don't believe my math in the simpler 2D case, it seems to me you're unlikely to believe it in the 3D case if it doesn't work out to the unit matrix, since the integral involved is a lot more complicated in 3D. And, like I said, if my integrals in the 2D case aren't convincing to you, it might help if we actually computed a numerical approximation using a large number of evenly-spaced angles from 0 to 2pi.

edit: didn't see vanesch's argument when writing this, but it looks good to me, at least for the diagonal components of the matrix...you'd need a different argument to show the off-diagonal components are 0.

PS: -1 </= si </= 1. -1 </= sj </= 1.

You didn't really tell me what si and sj represent...again, are they components of column vector and row vector, with i ranging from 1-3 (in the case of a 3-vector, it'd be 1-2 for a 2-vector) and j ranging from 1-3 as well? If so, I agree that si and sj would both be between -1 and 1. However, if the vector s is equally likely to be any angle then its components are not equally likely to take any value between -1 and 1.

<si*si> = <sj*sj> = <{1, ..., 0} + {0, ..., 1}> = 1/2 + 1/2 = 1;

I don't understand the reason why you're taking a sum of two values between 0 and 1 there, when si*si is just the square of a single number between -1 and 1, which will be a single number between 0 and 1. In any case, if each angle is equally likely, you can't assume that si is equally likely to take any value between -1 and 1, nor can you assume that si*si is equally likely to take any value between 0 and 1, if you were making either of those assumptions.

 

[vanesch]

Have you actually evaluated this integral for each component of the 3x3 matrix produced from (sx, sy, sz)x(sx sy sz)? If not, I have no basis for trusting your intuition that it would work out to the unit matrix, since this intuition was wrong in the case of 2-vectors.

For uniform distributions, unit vectors in N dimensions work out to have an expectation value of < si.si > = 1/N, simply because:

< s1.s1 > + < s2.s2 > + ... + < sn.sn > = < s.s > = 1

By symmetry (uniform distribution), all the terms are equal and hence equal to 1/N.

So in 3 dimensions, that would be 1/3.

 

[vanesch]

edit: didn't see vanesch's argument when writing this, but it looks good to me, at least for the diagonal components of the matrix...you'd need a different argument to show the off-diagonal components are 0.

That's not hard either, in the case of a uniform distribution.

Consider, say, < sx.sy > = C. C is a number that is a property of the uniform distribution of vectors s. It shouldn't depend on my specific choice of coordinate axes (the description of a uniform distribution should be identical in all orthogonal coordinate axes). C is "the expectation value of the product of the first and the second coordinate".

Let me flip the y-axis, and keep the x and z. This changes everywhere sy into -sy, while keeping sx and sz.

If, in this new system, we calculate "the expectation value of the product of the first and the second coordinate", then we find
C' = < sx.(-sy) > - < sx.sy >= - C.

But we agreed that C was a number that shouldn't depend on a precise choice of axes. So we must have C = - C. Hence, C = 0.

So < sx.sy > = 0

By symmetry (flip the axes!), in general < si.sj > = 0 if i not equal to j.

 

[DrChinese]

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

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

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

I think we have pretty well zeroed in on the issues, thanks. Again, referencing Bell itself :

1. The Inequality occurs when you map local hidden variable functions into QM expectation values. Thus Bell proves: IF QM=correct predictions (3,7) AND local realism=assumed (2,14), THEN Inequality must be true (15). From the contranegative, if the Inequality is demonstrated to be false by counter-example, then either "QM=correct predictions" is false OR "local realism=assumed" is false. Bell provides such counter-example in (22). Hopefully there is no dispute about this so far.


2. Yes, I realize wm does not accept Bell's proof as valid. Specifically, he denies "IF QM=correct predictions (3,7) AND local realism=assumed (2,14), THEN Inequality must be true (15)". OK, perhaps Bell's proof itself is wrong or somehow flawed. You can't demonstrate this by any type of counter-example, you must show that there is a mistake in the proof itself. If so, which step is it?

Showing a classical setup that violates the Inequality - as a way to invalidate the proof - does nothing, because you need a QM expectation value to compare it to. How can you have a QM expectation value if it is a classical setup? These are mutually exclusive by definition! Note that since Bell's proof is equivalent to the following:

IF the Inequality is violated, THEN QM=limited validity OR local realism=bad assumption.

and therefore also:

IF the Inequality is violated AND local realism=demonstrated, THEN QM=limited validity.

This is what you would have, i.e. QM is of limited validity and doesn't apply. And it wouldn't, because it is a classical experiment. And yet the proof is still standing, intact!

 

[JesseM]

1. The Inequality occurs when you map local hidden variable functions into QM expectation values. Thus Bell proves: IF QM=correct predictions (3,7) AND local realism=assumed (2,14), THEN Inequality must be true (15).

Hmm, but when you say "QM=correct predictions", you're talking about some subset of its predictions rather than all possible predictions made by QM, right? After all, one of QM's predictions is that the inequality will be violated in certain experiments! Are you just talking about the prediction that whenever both experimenters measure their particles at the same angle, they always get opposite results?

From the contranegative, if the Inequality is demonstrated to be false by counter-example, then either "QM=correct predictions" is false OR "local realism=assumed" is false. Bell provides such counter-example in (22). Hopefully there is no dispute about this so far.

If my guess about what you meant in the "correct predictions" step is right, then no dispute...but if it isn't, could you clarify?

2. Yes, I realize wm does not accept Bell's proof as valid. Specifically, he denies "IF QM=correct predictions (3,7) AND local realism=assumed (2,14), THEN Inequality must be true (15)". OK, perhaps Bell's proof itself is wrong or somehow flawed. You can't demonstrate this by any type of counter-example, you must show that there is a mistake in the proof itself.

But that's my point, you can show a proof is flawed simply by presenting a counterexample in some circumstances. In general, if a proof makes a statement like "for all cases where X is true, Y is true", then producing a single case of the form "X is true, but Y is false" shows the proof must have a flaw somewhere, without identifying which step in the proof must be flawed.

In this case, if Bell proves something like "IF the entangled-particles experiment always produces opposite spins when the experimenters choose the same angle AND local realism=assumed, THEN Inequality must be true". If wm could come up with a purely classical way of duplicating all the results of the entangled-particles experiment, including both the fact that the experimenters always get opposite results when they pick the same angle, and also the fact that the inequality is FALSE when they pick certain different angles, and it was clear by construction that wm's experiment respected local realism, then this would be sufficient to show that Bell's general statement was false, so that there must be some flaw in a proof.

Showing a classical setup that violates the Inequality - as a way to invalidate the proof - does nothing, because you need a QM expectation value to compare it to. How can you have a QM expectation value if it is a classical setup?

The expectation value is simply on the spin each experimenter will find when their detector is at a given angle, which on a given trial is either spin-up (+1) or spin-down (-1). You can also look at the expectation value for the product of their two results, either the same (+1) or different (-1). Either way, you can certainly come up with a classical experiment where, on each trial, each experimenter will get either the result +1 or -1, decided based on their choice of angle combined with some classical signal or object sent from a central source, and possibly with a random element as well. If you could further set things up so that the experimenters always get opposite results when they choose the same angle, and all the conditions necessary for Bell's theorem are obeyed (like the condition that the source has no foreknowledge of what angle each experimenter will choose on a given trial), do you disagree that Bell's proof should apply in exactly the same way to this experiment, and lead you to conclude that the same inequality should be obeyed as long as the experiment does not violate local realism?

 

[wm]

EPR-Bohm correlation calculation?

That's not hard either, in the case of a uniform distribution.

Consider, say, < sx.sy > = C. C is a number that is a property of the uniform distribution of vectors s. It shouldn't depend on my specific choice of coordinate axes (the description of a uniform distribution should be identical in all orthogonal coordinate axes). C is "the expectation value of the product of the first and the second coordinate".

Let me flip the y-axis, and keep the x and z. This changes everywhere sy into -sy, while keeping sx and sz.

If, in this new system, we calculate "the expectation value of the product of the first and the second coordinate", then we find
C' = < sx.(-sy) > - < sx.sy >= - C.

But we agreed that C was a number that shouldn't depend on a precise choice of axes. So we must have C = - C. Hence, C = 0.

So < sx.sy > = 0

By symmetry (flip the axes!), in general < si.sj > = 0 if i not equal to j.

1. I hope I am wording this right, but I might be expressing a big mistake. I apologise for my lack of LaTeX (which I should fix). Perhaps if my case is hopeless that won't be necessary.2. I agree that <sisj> = 0, where si and sj are (as I understand them) the s projections on the i and j axes. But I think that <sisi> might =1; not to dispute your result for ''routine'' unit-vectors but to see if there is not a difference that we need to take into account.

3. s (and s' = -s) is a unit-vector representing angular momentum. Such vectors transform satisfactorily infinitesimally; but not in general (I believe). Does this not mean that the 1/3 factor that is correct for routine unit-vectors may be different for vectors representing angular-momentum?

4. I have in mind Bell's (1964, equation (3)); effectively:

(1) <s.a*s'.b'> = -a.b'.

5. Would you comment? And could you provide or point-me-to a step-by-step working of this relation?

Thanks, wm

 

[JesseM]

4. I have in mind Bell's (1964, equation (3)); effectively:

(1) <s.a*s'.b'> = -a.b'.

5. Would you comment? And could you provide or point-me-to a step-by-step working of this relation?

That equation is a specifically quantum-mechanical one. You do not obtain it by assuming s is an angular momentum which was pointing in some definite direction before measurement, and then getting the expectation value by averaging over all possible definite angles for the angular momentum. If you would like a quantum-mechanical derivation of the fact that the expectation value for the product of the two experimenter's spins will be the negative cosine of the angle between their detectors, i.e. -a.b = -cos(a - b), then you can look at the derivation I linked to at the end of post #66:

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.

As I said earlier in that post, though, this derivation won't be of much use to you unless you already have a basic familiarity with the way probabilities and expectation values are derived in QM.

 

[wm]

Bell's way or the highway?

That equation is a specifically quantum-mechanical one. You do not obtain it by assuming s is an angular momentum which was pointing in some definite direction before measurement, and then getting the expectation value by averaging over all possible definite angles for the angular momentum. If you would like a quantum-mechanical derivation of the fact that the expectation value for the product of the two experimenter's spins will be the negative cosine of the angle between their detectors, i.e. -a.b = -cos(a - b), then you can look at the derivation I linked to at the end of post #66: As I said earlier in that post, though, this derivation won't be of much use to you unless you already have a basic familiarity with the way probabilities and expectation values are derived in QM.

Jesse, I looked at them, and they are helpful. Thanks.

But what I am still hoping for is a derivation that starts where Bell starts <s.a*s'.b'> and ends where Bell ends -a.b'.

Note that I am not in any way disputing the result, since it can be derived in many ways. It's just that I think Bell's way is likely to be one of the cleanest (and clearest) for me.

wm

 

[vanesch]

4. I have in mind Bell's (1964, equation (3)); effectively:

(1) <s.a*s'.b'> = -a.b'.

5. Would you comment? And could you provide or point-me-to a step-by-step working of this relation?

AAAAHH ! You are quoting from the article "On the Einstein-Podolsky-Roosen paradox" Physics 1 (1964) 195-200 which is taken as the second chapter in "speakable and unspeakable..." by Bell ?

Right. I have this in front of me now, and there it is written:

< sigma_1.a sigma_2.b > = -a.b

sigma_1 and sigma_2 are (the same) 3-vectors of the 3 Pauli-matrices, but which act upon the state 1 and the state 2 respectively!

They are operators over hilbert space !

And I have to say that this notation used by Bell is extremely confusing.
Bell is calculating the operator that corresponds to "the product of the outcomes of the two measurements". This is the operator over hilbertspace which corresponds to the "measurement of the correlation". He builds that operator by taking the product of the operator "Bob's measurement" and the operator "Alice's measurement".

Now, the operator "Bob's measurement" is the "measurement along the a axis" which is nothing else but the "a-component" of the three-some of angular-momentum operators on the first particle. This threesome of operators is symbolised by sigma_1, and the "a-component" is found by doing the in-product between this 3-some of operators and the real unit vector a.

Now, for spin-1/2 particles, the angular momentum operators are the three Pauli matrices, and that is what sigma_1 stands for.

However, there's a complication: we have two particles. So, for the first particle, sigma_1 acts as a matrix, while for the second particle, it acts as a real number (it commutes).

In the same way, the operator "Alices measurement" is similar, with the dot product between the 3-some of operators (the three Pauli matrices again, but this time "acting upon the second particle only"), and the b-vector, to find the operator corresponding to Alice's measurement.

Bell then takes the product of these two operators as being the operator of the product of the outcome. This is in fact a bit sloppy, but he can get away with it, because both operators acting on different spaces, they commute.

We now have the famous observable operator, which corresponds to the measurement "the correlation between the outcomes of Alice and Bob".

He next takes the quantum-mechanical expectation value of this operator over the singlet state. That's what the brackets stand for, and what they mean is the following:

First, one let's the operator act upon the quantum state vector describing the singlet state. This gives us another state in Hilbert space.
Next, one takes the Hilbert in-product of the singlet state with the result of teh previous outcome. That's the quantum mechanical technique of finding "expectation values".

After A VERY LONG CALCULATION, this comes out to be -a.b
where this time we have the simple in-product in euclidean 3-space of two unit vectors. (at least, I take it to be the correct result, I didn't verify it).