OK Corral: Local versus non-local QM

[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:

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

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 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:

$$\frac{1}{2\pi} \left( \begin{array}{cc} \pi & 0 \\ 0 & \pi \end{array} \right)$$

or:

$$\left( \begin{array}{cc} \frac{1}{2} & 0 \\ 0 & \frac{1}{2} \end{array} \right)$$

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).

 

[JesseM]

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'.

But the derivation I linked to plus the additional comments I made does show that, just using some different notation. s.a*s'.b' in Bell's notation just means the product of the two measurement results (one using angle a and the other using angle b), where each measurement yields either spin-up (+1) or spin-down (-1), so on a given trial the product will be +1 if both are measured to have the same spin on their respective measurement axes, and -1 if they are measured to have opposite spin on their respective measurement axes. What the linked page shows is that on a given trial, the probability that s.a*s'.b' is +1 will be sin^2((a-b)/2), and the probability that s.a*s'.b' is -1 will be cos^2((a-b)/2). And by the definition of "expectation value", <s.a*s'.b'> must be:

(+1)*Probability(s.a*s'.b' = +1 on each trial) + (-1)*Probability(s.a*s'.b' = -1 on each trial)

As I showed in my comments, when you work this out using the above probabilities, you conclude that <s.a*s'.b'> will be equal to -cos(a - b).

 

[vanesch]

For those who are interested, I did explicitly the calculation of Bell's expression.

The expansion over the singlet state is a bit clumsy:

the singlet state is 1/sqrt(2) ( |+> |-> - |->|+>)
|+> is the (1,0) element, and |-> is the (0,1) element in the hilbert space.

So we expand the expectation value:

< singlet | O1 O2 | singlet >as:

( < + | < - | - < - | < + |) O1 O2 ( |+> |-> - |->|+>)

= < + | O1 | +> < - |O1 |-> + < - | O1 | - > < + | O2 | + > - < + |O1 |-> < - | O2 | + > - < -| O1 | + > < + | O2 | - >EDIT: where I forgot the front factor of 1/2, because of the double presence of the square root.
(but in the notebook, it is ok).

See attachment.

 

[DrChinese]

1. 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? 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. 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.

3. 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?

1. Sure, we are talking about the situation where QM makes a prediction. In this case, the prediction is not that the Inequality is violated, it is the "cos theta" relationship. That the Inequality is violated is applicable only when local realism is also present. There is no A, B and C in QM of course, only A and B.

2. As I pointed out, such an attempt will not work using the path described. The logic statement I showed was equivalent to Bell's Theorem is:

IF Inequality=fails AND Local Realism=demonstrated, THEN QM=Limited Validity

So all wm would be doing with his classical experiment is proving it is a classical experiment where QM doesn't apply or is wrong. I don't see it as disproving Bell's Theorem. I guess when wm or someone actually conceptualizes and executes such an experiment, we'll have more to discuss. Right now, I would place it up there with theories of perpetual motion machines.

 

[vanesch]

For those who are interested, I did explicitly the calculation of Bell's expression.

BTW, it occurred to me that the way Bell writes his stuff, and the mistake wm made, is a nice illustration of how quantum theory can get around doing "local" things in a way that a classical view cannot.

wm made the calculation of the correlation, thinking he was doing a kind of classical calculation, where the "sign" of (s.a) determined the outcome at Alice, and the sign of (s.b) determined the outcome at Bob. The outcomes were supposed to be +1 or -1. So the true correlation would in fact be:

< sign(a.s) . sign(b.s) >, and not < (a.s) (b.s) >

However, by some mathematical coincidence, if s is a uniformly distributed unit vector in R^3, these two expressions come out the same.

As JesseM and I demonstrated, however, they do not equate -(a.b), but rather -(a.b)/2 or -(a.b)/3, depending on whether one considers them in 2 or in 3 dimensions.

Nevertheless, the thing is that the ACTUAL RESULT OF MEASUREMENT, if it is truly "sign(a.s)" (hence, a numerical value of +1 or -1 for each trial) is then indeed "locally produced" (because only depending upon a and s)).

As we see, however, the correlation then comes out to be -(a.b)/2, which doesn't violate the Bell inequalities - as expected.

Now, quantum theory does, apparently, the same thing. So why can't we say that in its "inner workings", instead of transporting a unit vector s, this funny 3-some of Pauli matrices is transported ?

The reason is that in the expression < (sigma.a) (sigma.b) >, we write down a *quantum-mechanical* expectation value of an OPERATOR. We do not write the STATISTICAL expectation value of A PRODUCT OF TWO RESULTS.

In other words, the outcome at Bob was NOT (sigma.b) ! It was ONE OF ITS EIGENVALUES (which happens to be +1 or -1). As such, we cannot really say that "we transport the outcome at Bob, which is (sigma.b), to Alice, where her outcome is (sigma.a), and multiply the two together".

If that were true, indeed, this would have been a local mechanism. But the result at Alice is NOT (sigma.a), and the result at Bob is NOT (sigma.b). The results are of the kind +/- 1...

Or are they ?

Well, we COULD say, if we wanted to, that the outcome at Alice is not +1 or -1, but (sigma.a). And we COULD say that the outcome at Bob is (sigma.b). But that's a funny situation! It would mean that Alice didn't, after all, get a genuine numerical result such as -1 or +1, but rather a mathematical operator over hilbert space. If that were true, then wm's reasoning would be correct in a way. We take the result at Alice (again: it is not -1 or +1, but an operator over hilbert space!), which is determined purely by what happens at Alice, and similarly at Bob's, and at the point of their meeting, they multiply their outcomes (which, again, are not -1 or +1, but are now operators over hilbert space) and hurray, we get the right correlations.

But what could that possibly mean, that Alice didn't get -1 or +1 at a trial, but each time an operator ? Well, it means that Alice got BOTH results. It means that Alice and Bob now have a quantum-mechanical description, and that they are in a superposition of having -1 and +1 (the operator contains both eigenvalues). This is exactly the MWI view on things, and it illustrates how in MWI, there is indeed no problem with locality. But the price to pay is rather high: you cannot say anymore that Alice got a measurement result which was each time -1 or +1 !

Now, independently of interpretation, the reason why the quantum formalism can make predictions which defy classical theories is that in the formalism of quantum theory, there is a difference between the mathematical representation of a measurement (which is a hermitean operator), and actual individual results of a measurement (which are eigenvalues of that hermitean operator). In a classical theory, the representation of a measurement is necessarily its outcome.

 

[wm]

For those who are interested, I did explicitly the calculation of Bell's expression.

The expansion over the singlet state is a bit clumsy:

the singlet state is 1/sqrt(2) ( |+> |-> - |->|+>)
|+> is the (1,0) element, and |-> is the (0,1) element in the hilbert space.

So we expand the expectation value:

< singlet | O1 O2 | singlet >


as:

( < + | < - | - < - | < + |) O1 O2 ( |+> |-> - |->|+>)

= < + | O1 | +> < - |O1 |-> + < - | O1 | - > < + | O2 | + > - < + |O1 |-> < - | O2 | + > - < -| O1 | + > < + | O2 | - >


EDIT: where I forgot the front factor of 1/2, because of the double presence of the square root.
(but in the notebook, it is ok).

See attachment.

Hi vanesch

I see only 5 viewers so far of your welcome appended note. I wonder if others too are having trouble accessing it?

Can you tell me its format please; or where we might get the right software to read it?

I cannot open it.

Thanks, wm

 

[vanesch]

Hi vanesch

I see only 5 viewers so far of your welcome appended note. I wonder if others too are having trouble accessing it?

Can you tell me its format please; or where we might get the right software to read it?

I cannot open it.

Thanks, wm

Ah, sorry. It is a mathematica notebook. You can freely download a reader for them on the wolfram website http://www.wolfram.com

I guess this is the path: http://www.wolfram.com/products/mathreader/

 

[wm]

Many thanks!

Ah, sorry. It is a mathematica notebook. You can freely download a reader for them on the wolfram website http://www.wolfram.com

I guess this is the path: http://www.wolfram.com/products/mathreader/

:!) :!) Thanks; got it; great. :!) :!)

But there are some odd looking symbols: like

o/oo. = i?

And it starts:

pauli1 = ::0, 1<, :1,0<<. = Pauli matrix in mathematica notation?

(I can work them out if you are too busy -- but what about all the new readers that will soon come to it -- see below.)

Sorry also I question but I need to be certain that there is no implication of non-locality whatsoever in the derivation. (That's why I've been requesting such a derivation here for so long.)

Because:

1. I believe that QM, correctly understood, can derive most of its results locally; and I'm pretty sure you've done that.

2. I'd like to put such LOCAL maths into my own high-school framework and understanding.

[ Being a good girl ''I don't do non-local''. ]

3. Most important of all: I'd like to comment on locality in QM by saying:

The EPR-Bohm correlation can be derived wholly locally using the prescription in Bell (1964, equation (3)); refer vanesch (Physics Forums) https://www.physicsforums.com/showpost.php?p=1255185&postcount=122

Would you agree with this statement?

4. AND SO, finally: Do you need to spend a little time to polish up your notebook page? BECAUSE I think you will find it becoming very popular. It is a very nice result. Especially from my point of view as a localist: IN FACT, I suggest it would be beaucoup worth the trouble to post it in LaTeX. (Or maybe I could learn my LaTeX on it; with some help from JesseM :!) to whom I owe much.)

Thanks, wm

 

[vanesch]

Thanks; got it; great. :!)

But there are some odd looking symbols: like

o/oo

And it starts:

pauli1 = ::0, 1<,1 :1,0<< ?

That's very strange. I re-downloaded the notebook and it looks ok for me. (ok, I don't use the MathReader, but mathematica 4.1, but at least, the file is not corrupt)

I ask about them because I need to be sure that there is no implication of non-locality whatsoever in the derivation. (That's why I've been requesting such a derivation here for so long.)

Because:

1. I believe that QM, correctly understood, can derive most of its results locally; and I'm pretty sure you've done that.

2. I'd like to put such LOCAL maths into my own high-school framework and understanding. (''I don't do non-local''. )

3. Most important of all: I'd like to comment on locality in QM by saying:

The EPR-Bohm correlation can be derived wholly locally using the prescription in Bell (1964, equation (3)); refer vanesch (Physics Forums) https://www.physicsforums.com/showpost.php?p=1255185&postcount=122

Would you agree with this statement?

4. AND SO, finally: Do you need to spend a little time to polish up your notebook page? BECAUSE I think you will find it becoming very popular.

Thanks, wm

Unfortunately, the derivation in QM is local, or non-local, at one's interpretation. As I said, in order to be able to consider it "local", one needs to make the hypothesis that there is no genuine unique measurement result at Alice and Bob, for each particle pair. Only then is one allowed to say that the result is an operator (and not simply a real value). And in that case, one can apply the kind of reasoning you wanted to apply with the unit vector s.

The STANDARD way of looking upon things in quantum mechanics, is non-local, or undefined. Using the projection, which affects, when Alice measures, also the state at Bob, is obviously non-local as a "calculation".

Now, depending on whether one assigns any "reality" to the wavefunction, this either means 1) (wavefunction is real) that the projection is an "action-at-a-distance" or 2) (wavefunction is not real) that this is an abstract calculational procedure which has nothing to do with any local or non-local mechanism.

 

[JesseM]

Unfortunately, the derivation in QM is local, or non-local, at one's interpretation. As I said, in order to be able to consider it "local", one needs to make the hypothesis that there is no genuine unique measurement result at Alice and Bob, for each particle pair. Only then is one allowed to say that the result is an operator (and not simply a real value). And in that case, one can apply the kind of reasoning you wanted to apply with the unit vector s.

It's probably worth expanding on this to make sure wm understands the sense in which QM can be "local" according to mainstream physics. I'm sure vanesch would agree that Bell's theorem rules out conventional local realism in which each measurement yields a unique result; but there is a loophole in which you can regain locality if you accept something like the many-worlds interpretation in which there is no "collapse of the wavefunction" on measurement, instead each spin measurement simply results in a superposition of states which includes both a state where the experimenter saw a result of spin-up and a state where the experimenter saw a result of spin-down. And the key to preserving locality is that the universe doesn't have to decide how to link the versions of experimenter #1 over here with the versions of experimenter #2 over there until there has been time for a signal moving at the speed of light to pass between them.

On a previous thread I gave a simple picture which attempts to show conceptually how you can preserve locality as long as you imagine each experimenter splitting into multiple "copies" with each measurement (although this picture should be taken with a grain of salt since there are problems with using a simple frequentist notion of counting 'copies' to derive subjective probabilities of seeing different results in the many-worlds interpretation). Recall that one of the Bell inequalities says that if Alice and Bob always get opposite spins + and - when they measure along the same axis, then when they measure along different axes, conventional single-universe local realism implies the probability of getting opposite results should be greater than or equal to 1/3. But here's my conceptual picture showing how if you accept they each split into multiple copies with each measurement, you can explain how they'll get opposite results on different axes on less than 1/3 of trials:

say Bob and Alice are each receiving one of an entangled pair of photons, and their decisions about which spin axis to measure are totally deterministic, so the only "splitting" necessary is in the different possible results of their measurements. Label the three spin axes a, b, and c. If they always find opposite spins when they both measure their photons along the same axis, a local hidden-variables theory would say that if they choose different axes, the probability they get opposite spins must be at least 1/3 (assuming there's no correlation between their choice of which axes to measure and the states of the photons before they make the measurement). The actual probability of opposite spins along different axes depends on the difference in their detector angles, but all that's important is that it's less than 1/3, so for the sake of the argument let's say that when Alice chooses axis c and Bob chooses axis a, they only get opposite results 1/4 of the time, a violation of Bell's inequality.

So now suppose that when Bob makes a measurement on axis a in one location and Alice makes a measurement on axis c in another, each splits into 8 parallel versions, with 4 measuring spin + and 4 measuring spin -. Label the 8 Bobs like this:

Bob 1: a+
Bob 2: a+
Bob 3: a+
Bob 4: a+
Bob 5: a-
Bob 6: a-
Bob 7: a-
Bob 8: a-

Similarly, label the 8 Alices like this:

Alice 1: c+
Alice 2: c+
Alice 3: c+
Alice 4: c+
Alice 5: c-
Alice 6: c-
Alice 7: c-
Alice 8: c-

Note that the decision of how they split is based only on the assumption that each has a 50% chance of getting + and a 50% chance of getting - on whatever axis they choose, no knowledge about what the other one was doing was needed. And again, only when a signal traveling at the speed of light or slower passes from one to the other does the universe need to decide which Alice shares the same world with which Bob...when that happens, they can be matched up like this:

Alice 1 (c+) <--> Bob 1 (a+)
Alice 2 (c+) <--> Bob 2 (a+)
Alice 3 (c+) <--> Bob 3 (a+)
Alice 4 (c+) <--> Bob 5 (a-)
Alice 5 (c-) <--> Bob 4 (a+)
Alice 6 (c-) <--> Bob 6 (a-)
Alice 7 (c-) <--> Bob 7 (a-)
Alice 8 (c-) <--> Bob 8 (a-)

This insures that each one has a 3/4 chance of finding out the other got the same spin, and a 1/4 chance that the other got the opposite spin. If Bob and Alice were two A.I.'s running on classical computers in realtime, you could simulate Bob on one computer and Alice on another, make copies of each according to purely local rules whenever each measured a quantum particle, and then use this type of matching rule to decide which of the signals from the various copies of Alice will be passed on to which copy of Bob, and you wouldn't have to make that decision until the information from the computer simulating Alice was actually transmitted to the computer simulating Bob. So using purely local rules you could insure that, after many trials like this, a randomly-selected copy of A.I. Bob or A.I. Alice would record the same type of statistics that's seen in the Aspect experiment, including the violation of Bell's inequality.

Note that you wouldn't have to simulate any hidden variables in this case--you only have to decide what the spin was along the axes each one measured, you never have to decide what the spin along the other 2 unmeasured axes of each photon was.

And wm, please note that I included this loophole way back in post #133 of the other thread where I stated all the conditions which must be assumed in order to prove that quantum results are incompatible with local realism:

do you agree or disagree that if we have two experimenters with a spacelike separation who have a choice of 3 possible measurements which we label A,B,C that can each return two possible answers which we label + and - (note that these could be properties of socks, downhill skiers, whatever you like), then if they always get opposite answers when they make the same measurement on any given trial, and we try to explain this in terms of some event in both their past light cone which predetermined the answer they'd get to each possible measurement with no violations of locality allowed (and also with the assumption that their choice of what to measure is independent of what the predetermined answers are on each trial, so their measurements are not having a backwards-in-time effect on the original predetermining event, as well as the assumption that the experimenters are not splitting into multiple copies as in the many-worlds interpretation), then the following inequalities must hold:

1. Probability(Experimenter #1 measures A and gets +, Experimenter #2 measures B and gets +) plus Probability(Experimenter #1 measures B and gets +, Experimenter #2 measures C and gets +) must be greater than or equal to Probability(Experimenter #1 measures A and gets +, Experimenter #2 measures C and gets +)

2. On the trials where they make different measurements, the probability of getting opposite answers must be greater than or equal to 1/3

 

[Doc Al]

but there is a loophole in which you can regain locality if you accept something like the many-worlds interpretation in which there is no "collapse of the wavefunction" on measurement, instead each spin measurement simply results in a superposition of states which includes both a state where the experimenter saw a result of spin-up and a state where the experimenter saw a result of spin-down.

But there are folks--and I count myself among them--who think that the "many worlds" interpretation does such violence to the usual notion of reality--with its multiple copies of experimenters and experimental results--that preserving locality with such a model is at best a Pyrrhic victory. Given such a metaphysics, locality is the least of one's worries. With all due respect to vanesch (a gentlemen and scholar--unlike Deutsch, who belongs in the loony bin )--I just don't see how it solves anything.

 

[wm]

Good stuff. And worth clarifying.

BTW, it occurred to me that the way Bell writes his stuff, and the mistake wm made, is a nice illustration of how quantum theory can get around doing "local" things in a way that a classical view cannot.

(With some minor editing and emphasis throughout.)

Comment: This post is very helpful to me. Thanks.

For the moment (for reasons to later appear): Can we have a superposition?

|Y> = V|wm wrong> + W|wm right>.

wm made the calculation of the correlation, thinking he was doing a kind of classical calculation, where the "sign" of (s.a) determined the outcome at Alice, and the sign of (s.b) determined the outcome at Bob.

Comment: wm made a classical calculation, at the level of high-school maths and logic. Maybe she went astray with her short-cut involving s and s', each a unit-vector relating to angular-momentum, and thus each a unit-axial-vector = (strictly) a bi-vector. wm believes they may transform differently.

The outcomes were supposed to be +1 or -1. So the true correlation would in fact be:

< sign(a.s) . sign(b.s) >, and not < (a.s) (b.s) >

However, by some mathematical coincidence, if s is a uniformly distributed unit vector in R^3, these two expressions come out the same.

I think this maybe not a coincidence (to be developed).

As JesseM and I demonstrated, however, they do not equate -(a.b), but rather -(a.b)/2 or -(a.b)/3, depending on whether one considers them in 2 or in 3 dimensions.

Yes; treating (standard) unit-vectors, these are both well-known relations for (standard) unit-vectors. See http://mathworld.wolfram.com/Vector.html (eg; eqn (9)).

If my comments were not clear, apologies to those who thought I was ignoring these results. My focus then was on another, equally interesting, result with bi-vectors.

SO: The question still open (for me) relates to the transformation of axial-vectors (bi-vectors; relating to angular-momentum). Maybe someone knows this answer already?

I suspect that I just need to apply Pauli matrices to classical angular momentum? Maybe some has done this already too? Would this not be possible (or not be correct) for any reason?

Nevertheless, the thing is that the ACTUAL RESULT OF MEASUREMENT, if it is truly "sign(a.s)" (hence, a numerical value of +1 or -1 for each trial) is then indeed "locally produced" (because only depending upon a and s)).

As we see, however, the correlation then comes out to be -(a.b)/2, which doesn't violate the Bell inequalities - as expected.

Yes; exactly as expected because this is the correct and well-known classical transformation for standard unit-vectors.

Now, quantum theory does, apparently, the same thing. So why can't we say that in its "inner workings", instead of transporting a unit vector s, this funny 3-some of Pauli matrices is transported ?

The reason is that in the expression < (sigma.a) (sigma.b) >, we write down a *quantum-mechanical* expectation value of an OPERATOR. We do not write the STATISTICAL expectation value of A PRODUCT OF TWO RESULTS.

In other words, the outcome at Bob was NOT (sigma.b) ! It was ONE OF ITS EIGENVALUES (which happens to be +1 or -1). As such, we cannot really say that "we transport the outcome at Bob, which is (sigma.b), to Alice, where her outcome is (sigma.a), and multiply the two together".

If that were true, indeed, this would have been a local mechanism. But the result at Alice is NOT (sigma.a), and the result at Bob is NOT (sigma.b). The results are of the kind +/- 1...

I believe it is a wholly local mechanism, relating to the conservation of angular momentum (to be developed).

Or are they ?

Yes; they certainly are. For sure! With the detectors so programmed, no one of my acquaintance has seen anything but elements from the set {+1, -1, +1', -1'}; the prime denoting Bob's results

Well, we COULD say, if we wanted to, that the outcome at Alice is not +1 or -1, but (sigma.a). And we COULD say that the outcome at Bob is (sigma.b). But that's a funny situation! It would mean that Alice didn't, after all, get a genuine numerical result such as -1 or +1, but rather a mathematical operator over hilbert space. If that were true, then wm's reasoning would be correct in a way. We take the result at Alice (again: it is not -1 or +1, but an operator over hilbert space!), which is determined purely by what happens at Alice, and similarly at Bob's, and at the point of their meeting, they multiply their outcomes (which, again, are not -1 or +1, but are now operators over hilbert space) and hurray, we get the right correlations.

vanesch, we COULD say lots of things here; BUT why would we want to say other than the truth?

1. That Alice did indeed get {+1, -1}; and while we were with her we saw these results with our own eyes and with our own friends.

2. That when we went over to Bob's lab we saw {+1', -1'}, etc.

3. And whenever we drop in at either lab (unannounced) we see the same consistent detector outputs; even when Alice and Bob were absent! Why not come with us? (For I miss your point here.)

4. Then you can join us in the correlation-checking, which we do (and can only do in this wondrously local quantum world) after we have hold of each set of related results:

Settings anti-parallel on this occasion:

Alice's paper-tape: +1, +1, -1, +1, -1, -1, ...

Bob's paper-tape: +1', +1', -1', +1', -1', -1', ...

But then; you already know the correlation from your (maybe non-local in your eyes) calculation. So all you really need is to see the tapes (and leave the correlation analysis to us). Why is not like this, please?

But what could that possibly mean, that Alice didn't get -1 or +1 at a trial, but each time an operator ? Well, it means that Alice got BOTH results. It means that Alice and Bob now have a quantum-mechanical description, and that they are in a superposition of having -1 and +1 (the operator contains both eigenvalues). This is exactly the MWI view on things, and it illustrates how in MWI, there is indeed no problem with locality. But the price to pay is rather high: you cannot say anymore that Alice got a measurement result which was each time -1 or +1 !

Seriously; save your money or give it to me.

Every Alice known to me has reported (to me, by phone or in person) the measurement results +1 and -1 only; each being a distinct printout on a permanent-record paper-tape.

Now, independently of interpretation, the reason why the quantum formalism can make predictions which defy classical theories is that in the formalism of quantum theory, there is a difference between the mathematical representation of a measurement (which is a hermitean operator), and actual individual results of a measurement (which are eigenvalues of that hermitean operator). In a classical theory, the representation of a measurement is necessarily its outcome.

I believe rather that QM is advanced probability theory and that we are quantum machines in a wholly quantum world.

That's very strange. I re-downloaded the notebook and it looks ok for me. (ok, I don't use the MathReader, but mathematica 4.1, but at least, the file is not corrupt)

OK; I'll see what I can do.

Unfortunately, the derivation in QM is local, or non-local, at one's interpretation. As I said, in order to be able to consider it "local", one needs to make the hypothesis that there is no genuine unique measurement result at Alice and Bob, for each particle pair. Only then is one allowed to say that the result is an operator (and not simply a real value). And in that case, one can apply the kind of reasoning you wanted to apply with the unit vector s.

Well I don't mind people holding non-local intepretations. BUT I'm sure glad we're living in a common-sense local and realistic quantum world.

The STANDARD way of looking upon things in quantum mechanics, is non-local, or undefined. Using the projection, which affects, when Alice measures, also the state at Bob, is obviously non-local as a "calculation".

That's fine with me: probability theory is ideally equipped to deal with non-local calculations; WHICH, incidentally, are only ever confirmed via local communications.

Now, depending on whether one assigns any "reality" to the wavefunction, this either means 1) (wavefunction is real) that the projection is an "action-at-a-distance" or 2) (wavefunction is not real) that this is an abstract calculational procedure which has nothing to do with any local or non-local mechanism.

Thank you; I'll take 2) every time. I'm not inclined to see as real abstract-elements of abstract spaces. Moreover, there is that maths theorem from 1915 that shows that any probability distribution CAN ALWAYS be represented by the absolute square of a Fourier polynomial ( = wave-function?).

With thanks again for your expansionary comments.

Regards, wm

 

[wm]

World-view?

But there are folks--and I count myself among them--who think that the "many worlds" interpretation does such violence to the usual notion of reality--with its multiple copies of experimenters and experimental results--that preserving locality with such a model is at best a Pyrrhic victory. Given such a metaphysics, locality is the least of one's worries. With all due respect to vanesch (a gentlemen and scholar--unlike Deutsch, who belongs in the loony bin )--I just don't see how it solves anything.

Doc, you seem ''sympathetic'' to the way I see the world. And the way I see MWI. (I hope so.)

Would you mind expanding on your world-view please? And pointing me to some of your papers?

I am not a physicist, but I see no reason to abandon my common-sense realism (which allows for measurement perturbation) and Einstein locality. (My post before this sets out some of my ideas.)

I do not accept what I call Bellian realism; a concept which Peres said (I believe) had nothing to do with QM. I feel that way and am truly surprised so few others see the world that way.

Thanks, wm

 

[wm]

It's probably worth expanding on this to make sure wm understands the sense in which QM can be "local" according to mainstream physics. I'm sure vanesch would agree that Bell's theorem rules out conventional local realism in which each measurement yields a unique result; but there is a loophole in which you can regain locality if you accept something like the many-worlds interpretation in which there is no "collapse of the wavefunction" on measurement, instead each spin measurement simply results in a superposition of states which includes both a state where the experimenter saw a result of spin-up and a state where the experimenter saw a result of spin-down. And the key to preserving locality is that the universe doesn't have to decide how to link the versions of experimenter #1 over here with the versions of experimenter #2 over there until there has been time for a signal moving at the speed of light to pass between them.

On a previous thread I gave a simple picture which attempts to show conceptually how you can preserve locality as long as you imagine each experimenter splitting into multiple "copies" with each measurement (although this picture should be taken with a grain of salt since there are problems with using a simple frequentist notion of counting 'copies' to derive subjective probabilities of seeing different results in the many-worlds interpretation). Recall that one of the Bell inequalities says that if Alice and Bob always get opposite spins + and - when they measure along the same axis, then when they measure along different axes, conventional single-universe local realism implies the probability of getting opposite results should be greater than or equal to 1/3. But here's my conceptual picture showing how if you accept they each split into multiple copies with each measurement, you can explain how they'll get opposite results on different axes on less than 1/3 of trials: And wm, please note that I included this loophole way back in post #133 of the other thread where I stated all the conditions which must be assumed in order to prove that quantum results are incompatible with local realism:

Jesse, it's not your fault, but I am not that good with words; I struggle.

But now that we have some maths before us; can we discuss things more in the context of specific experiments and related specific maths?

I think that it would be good to have vanesch's addendum converted out of mathematica (which I do not have) into a LaTeX post.

Is there a program that would do that? (It is only very short.)

This was something I have chased for awhile and since it is central QM, discussion of it will equally be central QM and common non-physicist mistakes.

I will still study your words, which I welcome every time. But mess-ups and time-delays are likely.

PS: To be clear, I do NOT think that QM is incompatible with common-sense local realism. NOT IN ANY WAY do I think that. Have I said that?

wm

 

[JesseM]

PS: To be clear, I do NOT think that QM is incompatible with common-sense local realism. NOT IN ANY WAY do I think that. Have I said that?

I never said that YOU think that. What I said is that mainstream physicists would all agree QM is incompatible with common-sense local realism, based on Bell's theorem (which need NOT include any assumption that measurements don't disturb the particle, if that's what you mean by 'Bellian realism') and that I am sure vanesch would agree, since when he talks about QM being 'local' he is only referring to a non-commonsense interpretation where measurements do not yield a single unique outcome. Do you think his derivation suggests the -cos(a-b) expectation value is compatible with common-sense local realism? If you do, then you have misunderstood something.

Also, do you still claim that you have a classical method of reproducing the -cos(a-b) expectation value based on sending two classical vectors s and -s to different experimenters, or have we managed to convince you that your math was incorrect in that case, and that the expectation value would be either -(1/2)*cos(a-b) or -(1/3)*cos(a-b) depending on whether s was a 2-vector or a 3-vector?

 

[Doc Al]

Doc, you seem ''sympathetic'' to the way I see the world. And the way I see MWI. (I hope so.)

Don't jump to conclusions. While I am sympathetic to your desire for a local interpretation of QM, I just don't see it as a serious possibility given Bell's theorem and current experimental results.

Would you mind expanding on your world-view please? And pointing me to some of your papers?

I certainly have no papers on this topic. One person I admire is Bell himself.

I am not a physicist, but I see no reason to abandon my common-sense realism (which allows for measurement perturbation) and Einstein locality. (My post before this sets out some of my ideas.)

I suspect that's because you don't appreciate the import of Bell's theorem.

I do not accept what I call Bellian realism; a concept which Peres said (I believe) had nothing to do with QM. I feel that way and am truly surprised so few others see the world that way.

Not sure what you mean by "Bellian realism". As I understand it, and I'm hardly an expert, is that Bell's theorem (combined with the experimental facts of QM and the reasoning of Einstein himself in EPR) leads to the conclusion that no theory satisfying Bell locality can accurately describe the world as we know it.

 

[wm]

Replying to a back-reference to another thread.

Although I tailored the short proofs I gave above to a particular thought-experiment, it's quite trivial to change a few words so they cover any situation where two people can measure one of three properties and they find that whenever they measure the same property they get opposite results. If you don't see how, I can do this explicitly if you'd like. I am interested in the physics of the situation, not in playing a sort of "gotcha" game where if we can show that Bell's original proof did not cover all possible local hidden variable explanations then the whole proof is declared null and void, even if it would be trivial to modify the proof to cover the new explanations we just thought up as well. I'll try reading his paper to see what modifications, if any, would be needed to cover the case where measurement is not merely revealing preexisting spins, but in the meantime let me ask you this: do you agree or disagree that if we have two experimenters with a spacelike separation who have a choice of 3 possible measurements which we label A,B,C that can each return two possible answers which we label + and - (note that these could be properties of socks, downhill skiers, whatever you like), then if they always get opposite answers when they make the same measurement on any given trial, and we try to explain this in terms of some event in both their past light cone which predetermined the answer they'd get to each possible measurement with no violations of locality allowed (and also with the assumption that their choice of what to measure is independent of what the predetermined answers are on each trial, so their measurements are not having a backwards-in-time effect on the original predetermining event, as well as the assumption that the experimenters are not splitting into multiple copies as in the many-worlds interpretation), then the following inequalities must hold:

1. Probability(Experimenter #1 measures A and gets +, Experimenter #2 measures B and gets +) plus Probability(Experimenter #1 measures B and gets +, Experimenter #2 measures C and gets +) must be greater than or equal to Probability(Experimenter #1 measures A and gets +, Experimenter #2 measures C and gets +)

2. On the trials where they make different measurements, the probability of getting opposite answers must be greater than or equal to 1/3

Jesse,

1. Do you see here how long you have sentences?

2. Does not my classical model (of old) refute this Bellian-Inequality easily? Are you not giving conditions which my model meets?

3. Are you not saying (as I will let you):

a. That Alice may make a countable-inifinity of detector-settings, each delivering outcome of {+1, -1}.

b. That Bob may make a countable-infinity of detector-settings, each delivering outcomes of {+1', -1'}.

4. Anyway: Down-hill skiers, dirty-socks, books and the like will satisfy your inequality. More subtle, less wholly concrete objects will sink it for some detector combinations. Yes?

5. Is my conclusion not what vanesch has shown?

wm

 

[JesseM]

But there are folks--and I count myself among them--who think that the "many worlds" interpretation does such violence to the usual notion of reality--with its multiple copies of experimenters and experimental results--that preserving locality with such a model is at best a Pyrrhic victory. Given such a metaphysics, locality is the least of one's worries. With all due respect to vanesch (a gentlemen and scholar--unlike Deutsch, who belongs in the loony bin )--I just don't see how it solves anything.

Yes, I do think philosophical questions are unavoidable when discussing the many-worlds interpretation. But when discussing a possible universe where observers are constantly splitting upon measurement (which is certainly logically possible, we could simulate such a universe on a computer), I think it makes sense to relate this to the subjective probabilities experienced by each observer using the "self-sampling assumption" which the philosopher Nick Bostrom argues is implicit in all forms of anthropic reasoning. Basically, this assumption says that it makes sense in many circumstances to reason as if you were randomly sampled from the set of all observers, and to use this assumption to update your estimate of the probabilities of different events using Bayes' rule. On his website anthropic-principle.com he discusses this principle in detail, and includes many thought-experiments to show why it is plausible, such as these ones from the paper Self-Location and Observation Selection Theory:

Dungeon

The world consists of a dungeon that has one hundred cells. In each cell there is one prisoner. Ninety of the cells are painted blue on the outside and the other ten are painted red. Each prisoner is asked to guess whether he is in a blue or a red cell. (And everybody knows all this.) You find yourself in one of these cells. What color should you think it is? – Answer: Blue, with 90% probability.

In the doomsday argument FAQ he quotes a similar thought-experiment by John Leslie:

firm plan was formed to rear humans in two batches: the first batch to be of three humans of one sex, the second of five thousand of the other sex. The plan called for rearing the first batch in one century. Many centuries later, the five thousand humans of the other sex would be reared. Imagine that you learn you’re one of the humans in question. You don’t know which centuries the plan specified, but you are aware of being female. You very reasonably conclude that the large batch was to be female, almost certainly. If adopted by every human in the experiment, the policy of betting that the large batch was of the same sex as oneself would yield only three failures and five thousand successes. ... [Y]ou mustn’t say: ‘My genes are female, so I have to observe myself to be female, no matter whether the female batch was to be small or large. Hence I can have no special reason for believing it was to be large.’

If we accept this sort of reasoning as valid, then it can be applied to a situation where I am constantly being copied, since if I reason as though I am randomly sampled from the set of all copies of me, I can draw probabilistic conclusions about what I am likely to see over the result of many measurements.

 

[DrChinese]

PS: To be clear, I do NOT think that QM is incompatible with common-sense local realism. NOT IN ANY WAY do I think that. Have I said that?

You continue to ignore Bell's Theorem conveniently as if it does not exist. There is no substantive difference between your "common-sense" definition of realism (per your page) and Bell's. Even if there were, no one would care because Bell's maps to the debate of concern to Einstein, Bohr, and everyone who follows EPR's argument.

You should quit confusing people with your statements, and acknowledge as follows:

1. You believe in locality.
2. You believe in your version of realism, which is slightly different than Bell's but you are not sure how so mathematically.
3. You do not accept Bell's Theorem as valid.
4. You accept 3. as a matter of faith because you believe 1. and 2., and you think other folks should too.
5. You have no actual plan for developing your pet theory, but hope that those of us here at Physicsforums will help you.

I would like to point out that this discussion belongs in Theory Development, and not quantum physics.

 

[wm]

Don't jump to conclusions. While I am sympathetic to your desire for a local interpretation of QM, I just don't see it as a serious possibility given Bell's theorem and current experimental results.


I certainly have no papers on this topic. One person I admire is Bell himself.


I suspect that's because you don't appreciate the import of Bell's theorem.


Not sure what you mean by "Bellian realism". As I understand it, and I'm hardly an expert, is that Bell's theorem (combined with the experimental facts of QM and the reasoning of Einstein himself in EPR) leads to the conclusion that no theory satisfying Bell locality can accurately describe the world as we know it.

Thanks darn it! But while I'm here: For me I would change the last two sentences to put in my language:

"Bellian realism" is the constrained realism that we may associate with Bell (1964; for example); where A+, originally a measurement outcome, is subtly changed to a property of the particle itself; what I call the d'Espagnat move (from Sci. Am). As I understand it, and I'm hardly an expert, Bell's theorem (combined with the experimental facts of QM and the reasoning of Einstein* on locality) leads to the conclusion that no theory satisfying Bellian realism describes the world as we know it.

PS: * Einstein did not see the final draft; disagreement was later such that Einstein never again spoke to the author (Podolsky). That is why I believe many pot-shots at Einstein over EPR can be a bit misleading.

wm

 

[JesseM]

Jesse,

1. Do you see here how long you have sentences?

Well, if the long sentences are hard to follow, just ask for clarification, it's usually pretty easy to break them up into a list of distinct statements or assumptions...in the quote above, I could rewrite the assumptions like this:

do you agree or disagree that IF we have:

1. two experimenters with a spacelike separation
2. who have a choice of 3 possible measurements which we label A,B,C that can each return two possible answers which we label + and - (note that these could be properties of socks, downhill skiers, whatever you like)
3. then if they always get opposite answers when they make the same measurement on any given trial
4. and we try to explain this in terms of some event in both their past light cone which predetermined the answer they'd get to each possible measurement
5. with no violations of locality allowed
6. and also with the assumption that their choice of what to measure is independent of what the predetermined answers are on each trial, so their measurements are not having a backwards-in-time effect on the original predetermining event
7. as well as the assumption that the experimenters are not splitting into multiple copies as in the many-worlds interpretation)

THEN the following inequalities must hold:

1. Probability(Experimenter #1 measures A and gets +, Experimenter #2 measures B and gets +) plus Probability(Experimenter #1 measures B and gets +, Experimenter #2 measures C and gets +) must be greater than or equal to Probability(Experimenter #1 measures A and gets +, Experimenter #2 measures C and gets +)

2. On the trials where they make different measurements, the probability of getting opposite answers must be greater than or equal to 1/3

2. Does not my classical model (of old) refute this Bellian-Inequality easily? Are you not giving conditions which my model meets?

I would say your old model, where the source knows what detector setting Alice will use before it sends out a signal, is violating condition 6 above, "the assumption that their choice of what to measure is independent of what the predetermined answers are on each trial". In your "yoked" experiment, the angle that Alice measures the polarization is not independent of the polarization of the light sent out (and knowing both Alice's measurement angle and the polarization of the light does predetermine the result).

Of course my condition 6 also talked about a "backwards-in-time influence", which isn't true in your yoked scenario, since you assume there is enough time between Alice choosing her measurement angle and Alice actually making a measurement for a signal to have gotten back to the source and told it what the angle would be before it sent out the polarized light. I guess in condition 1 I was implicitly assuming that each experimenter's choice of detector setting was immediately before they actually made a measurement, so that there is also a spacelike separation between these two pairs of events:

1. (Alice randomly choosing her measurement angle) AND (Bob measuring the signal/object sent to him from the source)
2. (Bob randomly choosing his measurement angle) AND (Alice measuring the signal/object sent to her from the source)

So, if you add this condition explicitly to my list, then it would be impossible for the source's choice of what signals/objects to be sent out to be correlated with Alice and Bob's choice of detector settings, unless somehow the information was traveling backwards in time (or unless there was a weird cosmic conspiracy in the initial conditions of the universe which caused the source's output to be correlated with Alice and Bob's choices on each trial even though no signal could travel between them).

3. Are you not saying (as I will let you):

a. That Alice may make a countable-inifinity of detector-settings, each delivering outcome of {+1, -1}.

b. That Bob may make a countable-infinity of detector-settings, each delivering outcomes of {+1', -1'}.

Yes.

4. Anyway: Down-hill skiers, dirty-socks, books and the like will satisfy your inequality. More subtle, less wholly concrete objects will sink it for some detector combinations. Yes?

Nope, as long as you obey my conditions 1-7 above (including the clarification of what I meant by condition #1...also, note that #4 is not so much a condition as a logical conclusion necessitated by the other 6), then it is impossible for the inequalities to be violated by any experiment whatsoever. Therefore, since the inequalities are empirically violated in QM, it must be that QM violates one of the assumptions 5-7...either QM allows nonlocality, or QM allows backwards-in-time signalling, or QM allows experimenters to split into multiple copies (I suppose QM could also just violate the rules of logic, but I was assuming traditional logic must be obeyed).

5. Is my conclusion not what vanesch has shown?

Not at all--where did you get that idea? Vanesch just shows that when you use the conventional quantum rules to make predictions, you get the prediction that the expectation value is -cos(a-b). But the conventional quantum rules themselves do not say anything about locality or nonlocality, that's a matter for interpretation.