
#19
Feb1013, 09:27 PM

P: 79

I do think that it's correct, as DrChinese mentioned, that not all LHV models of quantum entanglement predict a linear correlation. A linear correlation is the maximum correlation that can be gotten with a hidden variable model constrained by Bell's locality condition. There's also a minimum bound to the correlation predicted by these types of models. The QM predictions lie, mostly, outside these LHV boundaries. The LHV linear correlation produces, as far as I know, the most data points in common with QM. So, if all Bell inequalities are predicated on a linear correlation, then maybe that's why. You can see for youself that the prototype Bell LHV model, wherein the probability distribution for λ is uniform (all λ values are equally likely), will predict a linear correlation. C(a,b) = ∫ ρ(λ) A(a,λ) B(b,λ) dλ C(a,b) = ½∏ ∫ sign[cos2(aλ)] sign[cos2(bλ)] dλ = (½∏) (2∏  8ab) = 1  4θ/∏ (where θ = ab) For θ = 0 the correlation coefficient is 1, meaning either both photons are passed by their polarizers or both are absorbed. For θ = ∏/2 the correlation coefficient is 1, meaning if one photon is transmitted by its polarizer the other must be absorbed and vice versa. For θ = ∏/4 the correlation coefficient is 0, meaning there's no correlation. These three values are the only values (for θ) which match the QM prediction. For any other values of θ this LHV correlation function doesn't match QM predictions or experimental results. It's fairly easy to see that Bell inequalities must be violated by QM. The difficulty most people seem to have is identifying what, precisely, it is about Bell LHVs that causes them to be incompatible with QM. It is, as Bell himself has told us, Bell's locality condition. Even Bell's initial expression of realism in the functions (Bell's equations (1) in his paper, "On The Einstein Podolsky Rosen Paradox") that determine individual detection [A(a,λ) = ±1, B(b,λ) = ±1] is an expression of Bell locality (ie., it's made explicit that the results at A depend only on the value of λ and the setting of a, and the results at B depend only on the value of λ and the setting of b. Bell's locality condition, as embodied in the form of his equation (2) in the same paper, makes it further explicit that the individual results (already denoted as independent of each other in his equations (1)) are independent of each other in that the resulting form makes explicit the assumption that the result B doesn't depend on the setting of a and the result A doesn't depend on the setting of b. Bell's equation (2) is an almost perfect denotation of the assumption of the principle of local action. Almost, because it also encodes the assumption that the measurement outcomes, A and B, are statistically independent, which they aren't. They're statistically dependent in that prior to a detection at either end during a coincidence interval the sample space, ρ(λ), for both ends is a uniform distribution across the entire range of possible λ values. On detection at one end the sample space at the other end is changed. Which is the definition of statistical dependence. 



#20
Feb1113, 07:22 PM

Sci Advisor
P: 1,185

I recently came across a supersimple proof the CHSH form of the Bell inequality,
2 ≤ C(a,b)  C(a,d) + C(c,b) + C(c,d) ≤ +2 where C(a,b) = ∫ dλ ρ(λ) A(a,λ)B(b,λ) . Here is the proof. From the definition of C(a,b), we have C(a,b)  C(a,d) + C(c,b) + C(c,d) = ∫ dλ ρ(λ) [ A(a,λ)B(b,λ)  A(a,λ)B(d,λ) + A(a,λ)B(b,λ) + A(c,λ)B(b,λ) + A(c,λ)B(d,λ) ] . In general, A(a,λ) = ±1 and B(b,λ) = ±1. Thus, each of the four terms inside the square brackets is itself ±1 (for any value of λ). Furthermore, the product of these four terms (including the explicit minus sign as part of the 2nd term) is simply 1, because each factor of A(a,λ), etc, appears exactly twice in this product. Thus, the four terms must consist of either three +1's and one 1, or three 1's and one +1. Thus the sum of the four terms must be either +2 or 2. Hence, the integral is computing the weighted average of a set of +2's and 2's. This weighted average must then be in the range from 2 to +2 (inclusive). Hence, 2 ≤ C(a,b)  C(a,d) + C(c,b) + C(c,d) ≤ +2, which is the CHSH inequality. 



#21
Feb1113, 10:09 PM

P: 79

Just reading over the comments. Considering everything that's been said, it looks to me like spenserf should have the answers to his questions. I had skimmed over a couple of Nurgatory's comments, but he covered something that I had been unsure about along with spenserf. The nice simple proof in Avodyne's post #20 helps a lot also. Thanks.



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