- #101

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It isn't to me. As I said, Bell's inequalities are about a very specific situation: Alice has a device that is capable of producing a measurement result of [itex]\pm 1[/itex]. The device has (at least) two possible settings, [itex]a[/itex] or [itex]a'[/itex]. Bob similarly has a device with two possible settings, [itex]b[/itex] or [itex]b'[/itex]. For many rounds, you perform the following procedure and collect statistics:Would you say that a geometrical feature like say equality of right angles implying isotropy of the space and that figures may be moved to any location and still be congruent is a local or a nonlocal property? Does the fact that figures with arbitrary spacelike separation can be measured to be congruent if they are put together again mean that measuring one figure is affecting the measure of the other figure making it congruent with itself?

Now this particular geometrical property is commutative so it would not violate the inequalities, but a non-commutative geometrical property like euclidean rotations is not commutative and does violate the inequalities, the spatial separation(local-vs nonlocal) is not the important feature for such properties because they are pervasive(ubiquitous) and independent of spatial separation of the objects on which it is observed.

Is this more understandable?

On round number [itex]n[/itex],

- Alice chooses a setting [itex]a_n[/itex].
- She performs a measurement, and gets a result [itex]A_n[/itex]
- Bob chooses a setting [itex]b_n[/itex].
- Bob gets result [itex]B_n[/itex]

That's the context for which Bell derived his inequality. If you allow for Alice's choice [itex]a_n[/itex] to affect Bob's result, [itex]B_n[/itex], and vice-verse, then there is no reason to expect the inequality to hold. That's where locality comes in: without assuming locality, there is no reason to assume that the inequality holds. Locality in the context of Bell's theorem means a very specific thing: that Alice's choice cannot affect Bob's result, and vice-verse. So your musings about whether congruence of geometric figures is local or nonlocal don't seem to be related to Bell's notion of locality.

Bell defined a hidden variables theory to be an explanation of the correlations along the following lines:

- Each round, there is a hidden variable [itex]\lambda_n[/itex] influencing the results. The variable may take on different values on different rounds (hence the subscript).

- There are deterministic functions [itex]F_A(a, b, \lambda)[/itex] and [itex]F_B(a, b, \lambda)[/itex] such that [itex]A_n = F_A(a_n, b_n, \lambda_n)[/itex] and [itex]B_n = F_B(a_n, b_n, \lambda_n)[/itex]
- Each round, [itex]\lambda_n[/itex] is chosen randomly according to some probability distribution [itex]P(\lambda)[/itex]

[itex]A_n = F_A(a_n, \lambda_n)[/itex] and [itex]B_n = F_B(b_n, \lambda_n)[/itex]

That's the critical assumption that allows him to derive his inequality. Right off the bat, I don't see how his derivation has anything to do with whether [itex]a[/itex] and [itex]b[/itex] are described by a commutative or noncommutative algebra.

Now, what someone has shown is that QM only predicts a violation of Bell's inequality in the case where the two measurements corresponding to Alice's settings [itex]a[/itex] and [itex]a'[/itex] are described by noncommuting operators, and the two measurements corresponding to Bob's settings [itex]b[/itex] and [itex]b'[/itex] are described by noncommuting operators. But that's a fact about quantum mechanics. Bell's derivation doesn't (as far as I can see) assume anything at all about whether things commute or not. The noncommutativity is about the two choices that Alice (or Bob) might make, not about Alice's measurements versus Bob's. Alice's measurements do commute with Bob's measurements.