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Variation on Quantum cross covariance and CHSH

  1. Nov 23, 2014 #1
    I tried another approach to the problem of covariance like in Bell's theorem :


    from the definition ##Cov(A,B)=\langle\Psi|A\otimes B|\Psi\rangle-\langle\Psi|A\otimes 1|\Psi\rangle\langle\Psi|1\otimes B|\Psi\rangle## (##A=diag(1,-1)=B##)

    we can see that this 'average' is in fact a quadratic form and a quartic form in the wave-function. To put all on equal footage we can write the first term as ##\langle\Psi|\langle\Psi|A\otimes B|\Psi\rangle|\Psi\rangle##

    then it's all a quartic form. Thus we could find a kind of extended measurement operator acting on ##|\Psi\rangle|\Psi\rangle## :

    ##\hat{Cov}(A,B)=A\otimes B\otimes 1_4-A\otimes 1_4\otimes B##

    hence a 16x16 symmetric matrix and thus an observable. We have obviously

    ##\langle\Psi|\langle\Psi\hat{Cov}(A,B)|\Psi\rangle|\Psi\rangle=Cov(A,B)##

    But the eigenvalues of this operator are 2,-2 and 0. and in this configuration we get <Cov>=-2/2+0/2=-1 as in the usual calculation.

    1) Since A and B can have values only 1,-1 the 0 eigenvalue is obvious, but how is 2 possible.

    2) Is it possible to prove that the average of Cov is in [-1,1] since we work only with a product of psi with itself ?

    3) the CHSH operator being Cov(A,B-Cov(A,B')+Cov(A',B)+Cov(A',B'), I could find the eigenvalues with a trial version of mathematica to be 4,-4 and 0.

    Thus we can see that contrary to a global hidden variable model the CHSH values 2 or -2 never appear.

    We can see here that CHSH can get only integer value as opposed to CHSH without subtracting the product of the averages which is 2Sqrt(2).

    4) How to find the maximum value of CHSH when averaged, is it 2Sqrt(2) as when one does not make this construction ?

    Does this construction makes any sense, and how is the cloning of the wavefunction to be interpreted, or is this just more similar to a crackpottery ?
     
  2. jcsd
  3. Nov 28, 2014 #2
    Thanks for the post! This is an automated courtesy bump. Sorry you aren't generating responses at the moment. Do you have any further information, come to any new conclusions or is it possible to reword the post?
     
  4. Nov 29, 2014 #3
    What i could add is that if every term in Chsh were independent and generated by global variables we would have the possible values 6,8 but quantum mechanics restrict it to 4.
     
  5. Nov 29, 2014 #4
    How do you get 6,8 from a linear combination 4 terms,each of which has extremes of [-1,1]?

    Maybe you are asking why/how QM restricts it to 2√2 when independent global variables will give 4 (as you found in your Mathematica calculation)?
     
    Last edited: Nov 29, 2014
  6. Nov 29, 2014 #5
    We have 8 terms or 4 times c=<ab>-<a><b>. at measurement c can get qm value 2,-2,0, so 4x2=8
     
  7. Nov 29, 2014 #6
    In other word the exact covariance c=-1 is in fact an average of -2 and 0
     
  8. Dec 2, 2014 #7
    This is a mathematical artifact but it has the advantage to give integer eigenvalues for Chsh.

    Admitting that we can only have global hidden variables then chsh should take values 0,2,4 but there is no isomorphism between operator sum and eigenvalue sum.

    If we compute this eay we have eigenvalues for chsh that are 0,4 which seems more intuitive relatively to experiments and moreover it is testable if 2 appear in an experiment.
     
  9. Dec 2, 2014 #8

    atyy

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    The maximum value of CHSH attainable in quantum mechanics is ##2 \sqrt{2}## and is called Tsirelson's bound: http://en.wikipedia.org/wiki/Tsirelson's_bound.

    I believe the derivation of Tsirelson's bound holds for finite dimensional Hilbert spaces. The problem remains open for infinite dimensional Hilbert spaces: http://www.math.tau.ac.il/~tsirel/Research/bellopalg/main.html.

    Under a less restrictive assumption called no signalling, the maximum value of CHSH attainable is 4: http://arxiv.org/abs/quant-ph/9508009.
     
    Last edited: Dec 2, 2014
  10. Dec 3, 2014 #9
    What is remarkable in this calculation of the eigenvalues of chsh is that they are independent of the measurement settings contrary to qm.

    This last fact is a bit disturbing since how by adding 1 and -1 four times we could get 2sqrt(2) ?
     
  11. Dec 3, 2014 #10

    DrChinese

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    Apples and oranges. One ("adding 1 and -1 four times") includes an assumption that is unwarranted , as the actual experimental max (" 2sqrt(2)") is not related to that premise.
     
    Last edited: Dec 3, 2014
  12. Dec 3, 2014 #11
    We don't add 1 and -1 four times in the CHSH inequality. We add together four quantities that each range from -1 to 1.
     
  13. Dec 4, 2014 #12
    What i mean is the result of measurement each time is 1,-1 then we add them and make an average the order of averaging and summing is not important since averaging is linear
     
  14. Dec 4, 2014 #13

    atyy

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    An equal mixture of 1 and -1 averages to 0, so that constraint is very weak. Additional assumptions are needed to get the various upper bounds of 2 (local causality), 2*sqrt(2) (quantum mechanics) or 4 (no signalling).
     
  15. Dec 4, 2014 #14
    but it is still debated if qm is nonlocal i think or should it be interpreted (like double slit) : if we would localize it would change the result ?
     
    Last edited: Dec 4, 2014
  16. Dec 5, 2014 #15

    atyy

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    There are several notions of locality.

    (1) signal locality: no classical information travels faster than light
    (2a) local causality: the probability of an outcome is independent of the measurement choice and outcome at a distant location
    (2b) local causality: a theory with classical relativistic dynamics
    (3) relativistic causality: the causes of an event are entirely in its past light cone

    The inequalities in post #13 only deal with signal locality and local causality, and there is consensus that quantum mechanics is signal local but not local causal.

    The debate as to whether quantum mechanics is "local" or not concerns the "relativistic causality" notion of locality, as discussed eg. in http://arxiv.org/abs/1311.6852.
     
  17. Dec 5, 2014 #16
    Yes, but to determine an upper bound of the CHSH you are adding extreme values 1 and -1 four times. Only the extreme values and degrees of freedom determine what you can mathematically say about the upper bound. Averages are never used to determine upper bounds. The upper bound is not the most probable value, it is the maximum value beyond which no higher value is possible within the assumptions used to derive it.

    For example, for 4 independent variables (a,b,c,d), each of which has a maximum value of 1, the upper bound of a + b + c + d = 4. (we simply add the maximum values 4 times, irrespective of what the range or average is). It is a contradiction to say that 4 independent variables each with a maximum of +1, will give an upper bound for (a + b + c + d) that is less than 4, because the only way you get an "upper bound" less than 4 is if (a,b,c,d) are not really independent or one or more of them do not have a maximum of 1.

    As concerns the CHSH, the 4 terms are not independent due to the cyclical combinations of angle settings, which we can write as S = f(a-b) + f(a-b′) + f(a′-b) - f(a′-b′)
    Substituting z = (a-b), y=(a-b'), x=(a'-b), (a'-b') = x + y - z, we end up with S = f(z) + f(y) + f(x) - f(x+y-z). which more clearly shows the nature of the dependencies. We have a sum of 3 independent terms f(z) + f(y) + f(x) with an upper bound 3, but a fourth term which depends on all the other 3. The upper bound of S is no longer that obvious. Note we aren't yet talking about the specific nature of the functions f(.), but by considering a few basic facts about the EPR experiment (specifically the fact that f(.) must be even in the EPR scenario), and doing a little algebra, it turns out the maximum value of S is obtained when x = y = -z and therefore S = 3f(x) - f(3x) The exact value of the upper bound for a given function should then just be a matter of substituting in the candidate function. For example, for f(x) = cos(x), the maximum will be at 2√2 (see http://www.wolframalpha.com/input/?i=Maximum 3cos(x) - cos(3x)) with x corresponding to the value at the "Bell-test" angles (x = y = -z = π/4, for Bell-test angles). It could be an interesting exercise to try to find an even function f() with extrema [-1, +1] for which S = 3f(x) - f(3x) = 2 for the "Bell-test" angles (ie, x = π/4)
     
  18. Dec 6, 2014 #17
    This function could be cos(x)^2 for 0 to pi/2 and -cos(x)^2 for pi/2 to pi and ee complete by symmetry and periodicity

    Though this curve is not good looking we could try curves like a(x)cos(x) with a(0)=1 and a(pi/4)=1/sqrt2 with some constraint on the second derivative to make a simple bump.

    We shall note here that qm has a max value of 2sqrt2 not only on average : the maximum eigenvalue of the chsh operator is that value. 4 obtained above is not standard qm.
     
    Last edited: Dec 6, 2014
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