vanesch said:
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.