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Difference between Bell and CHSH?

  1. Jan 22, 2016 #1
    Is it correct to say that Chsh does not need an average but as soon as AB-AB'+A'B+A'B' is 4, 0 or -4 it means it is nonlocal ?
     
  2. jcsd
  3. Jan 22, 2016 #2

    Nugatory

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    No. You have to run a large number of pairs through and compute the average. All local realistic theories predict that the expectation value of the average will be between -2 and 2; quantum mechanics predicts that in some configurations (which are the ones we choose to test) the expectation value can be as high as ##2\sqrt{2}## which is greater than 2. Basic probability theory (the law of large numbers) says that if we do a large number of trials the probability of the average deviating significantly from the expectation value becomes vanishingly small; thus if we see an average significantly greater than 2 we conclude that the behavior of the system is not governed by a local realistic theory.

    In this sense it works like the Bell inequality - it's just easier to design experiments around the CHSH inequality so it's used more often.
     
  4. Jan 22, 2016 #3
    Though i often fall upon the proof that says that if we admit a variable then this sum factorizes to A |B+B'|+A'|B-B'| when the first parenthesis is 2 the other is zero and vice versa. Hence it cannot be 0 4 -4.

    We can find this proof in wikipedia .

    This is then obviously wrong ?
     
  5. Jan 23, 2016 #4

    jfizzix

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    Mathematically, the limits of the sum are between -4 and 4, since each of the four elements are between -1 and 1. One could imagine probabability distributions that approach these upper and lower bounds, but they would not be realizeable as measurement statistics according to our current theories.

    However, if we use quantum mechanics to describe the spin-polarization measurements, we find the sum is constrained between -2.828.. and 2.828...

    If we say that the correlations are explained by local hidden variables, then the sum has to be between -2 and 2.

    So if you find the sum is larger than 2, you violate a Bell inequality.
    If you find the sum is larger than 3, then that would be totally awesome as it would be the first evidence of an experimental violation of quantum theory.
    As it stands, there have been many very careful experiments to try and violate the CHSH Bell inequality, and it seems that while the sum can be greater than 2, it never appears to be greater than 2.828...

    So nonlocality appears possible, but quantum mechanics remains an accurate description of what's going on.
     
  6. Jan 24, 2016 #5
    Yes. What I mean is that if one gets measurement results 2,2,0 for this sum of operators the average is less than 2 1.33.

    But since we got a zero result it means it cannot be local following chsh.

    However if we use a Kronecker sum in quantum mechanics we get that the singlet state is no more an eigenstate hence there is a variance. Namely quantum then delivers 2.82+\-.707
     
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