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I Computing CHSH violation bound

  1. Dec 21, 2017 #1
    It seems that the upper bound of the CHSH inequality is ##2\sqrt{2}##
    How is it analytically derived?
     
  2. jcsd
  3. Dec 21, 2017 #2

    stevendaryl

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    That's not an upper bound of the CHSH inequality. The upper bound is 2. The fact that actual correlations are [itex]2\sqrt{2}[/itex] shows that the inequality is violated by QM.
     
  4. Dec 22, 2017 #3
    I'm sorry I did not express it correctly. I was talking about the quantum mechanical bound not the hidden variable bound.
    How is ##2\sqrt{2}## analytically calculated?
     
  5. Dec 23, 2017 #4

    stevendaryl

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    I don't know how to prove it for an arbitrary QM system, but in the particular case of anti-correlated spin-1/2 particles, I can prove it.

    In that case, we have:

    [itex]E(a,b) = - cos(b-a)[/itex]

    So the quantity of interest is:
    [itex]C(a,b,a',b') = E(a,b) + E(a,b') + E(a', b') - E(a',b)[/itex]
    [itex]= -cos(b-a) -cos(b'-a) -cos(b'-a') + cos(b-a') [/itex]

    If we let [itex]\alpha = b-a, \beta = b'-a, \gamma = b'-a'[/itex], then we have:

    [itex]C(\alpha, \beta, \gamma) = -cos(\alpha) - cos(\beta) -cos(\gamma) + cos(\alpha + \gamma - \beta)[/itex]

    For a minimum or maximum, the partial derivatives with respect to [itex]\alpha, \beta, \gamma[/itex] must all be zero. This implies:
    1. [itex]sin(\alpha) - sin(\alpha + \gamma - \beta) = 0[/itex]
    2. [itex]sin(\beta) + sin(\alpha + \gamma - \beta) = 0[/itex]
    3. [itex]sin(\gamma) - sin(\alpha + \gamma - \beta) = 0[/itex]
    We can use some trigonometry:
    • If [itex]sin(A) = sin(B)[/itex], then either [itex]A=B[/itex] or [itex]B = \pi - A[/itex]
    • If [itex]sin(A) = -sin(B)[/itex], then either [itex]A=-B[/itex] or [itex]B = \pi + A[/itex]
    So we have 8 possibilities:
    1. [itex]\alpha = \alpha + \gamma - \beta[/itex] and [itex]\beta = -(\alpha+ \gamma - \beta)[/itex] and [itex]\gamma = \alpha + \gamma - \beta[/itex]
    2. [itex]\alpha = \alpha + \gamma - \beta[/itex] and [itex]\beta = -(\alpha+ \gamma - \beta)[/itex] and [itex]\gamma = \pi -(\alpha + \gamma - \beta)[/itex]
    3. [itex]\alpha = \alpha + \gamma - \beta[/itex] and [itex]\beta = \pi + (\alpha+ \gamma - \beta)[/itex] and [itex]\gamma = \alpha + \gamma - \beta[/itex]
    4. [itex]\alpha = \alpha + \gamma - \beta[/itex] and [itex]\beta = \pi + (\alpha+ \gamma - \beta)[/itex] and [itex]\gamma = \pi - (\alpha + \gamma - \beta)[/itex]
    5. [itex]\alpha = \pi - (\alpha + \gamma - \beta)[/itex] and [itex]\beta = -(\alpha+ \gamma - \beta)[/itex] and [itex]\gamma = \alpha + \gamma - \beta[/itex]
    6. [itex]\alpha = \pi - (\alpha + \gamma - \beta)[/itex] and [itex]\beta = -(\alpha+ \gamma - \beta)[/itex] and [itex]\gamma = \pi - (\alpha + \gamma - \beta)[/itex]
    7. [itex]\alpha = \pi -(\alpha + \gamma - \beta)[/itex] and [itex]\beta = \pi + (\alpha+ \gamma - \beta)[/itex] and [itex]\gamma = \alpha + \gamma - \beta[/itex]
    8. [itex]\alpha = \pi - (\alpha + \gamma - \beta)[/itex] and [itex]\beta = \pi + (\alpha+ \gamma - \beta)[/itex] and [itex]\gamma = \pi - (\alpha + \gamma - \beta)[/itex]
    These can be solved to yield the possibilities:

    1. [itex]\alpha = \beta = \gamma = 0[/itex]
    2. Impossible
    3. Impossible
    4. [itex]\alpha = 0, \beta = \gamma = \pi[/itex]
    5. Impossible
    6. [itex]\alpha = \gamma = 0, \beta = -\pi[/itex]
    7. [itex]\alpha = \beta = \pi, \gamma = 0[/itex]
    8. [itex]\alpha = \gamma = \frac{3\pi}{4}, \beta = \frac{5\pi}{4}[/itex]
    To find out whether these are minima or maxima, we plug them back into the expression for [itex]C(\alpha, \beta, \gamma)[/itex]
    1. [itex]C(\alpha, \beta, \gamma) = -1-1-1+1 = -2[/itex]
    2. Impossible
    3. Impossible
    4. [itex]C(\alpha, \beta, \gamma) = -1+1+1+1 = +2[/itex]
    5. Impossible
    6. [itex]C(\alpha, \beta, \gamma) = -1+1-1-1 = -2[/itex]
    7. [itex]C(\alpha, \beta, \gamma) = +1+1-1+1 = +2[/itex]
    8. [itex]C(\alpha, \beta, \gamma) =+\sqrt{2}/2 +\sqrt{2}/2 +\sqrt{2}/2+\sqrt{2}/2 = 2\sqrt{2}[/itex]
    So, [itex]2\sqrt{2}[/itex] is the biggest value that you can get with anti-correlated spin-1/2 particles. It's not at all obvious to me why that is the best you can possibly do with any QM system.
     
  6. Dec 23, 2017 #5
    Excellent! Thank you very much.
    There is one more question however, since 2 and -2 occur it seems that ##-2\sqrt{2}## should also appear.
     
  7. Dec 24, 2017 #6

    stevendaryl

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    Now, that makes me think I must have done something wrong. Let me recheck it.
     
  8. Dec 24, 2017 #7

    stevendaryl

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    Just trying some values, I found that there is a minimum at [itex]\alpha = \gamma = \frac{\pi}{4}, \beta = \frac{7\pi}{4}[/itex], and that gives:

    [itex]C(\alpha, \beta, \gamma) = -2 \sqrt{2}[/itex]

    So, it looks I should have used a more general condition:
    • If [itex]sin(A) = sin(B)[/itex], then either [itex]B = A + 2n\pi[/itex] or [itex]B = (2n+1)\pi - A[/itex]
    • If [itex]sin(A) = -sin(B)[/itex], then either [itex]B = A + (2n+1)\pi[/itex] or [itex]B = 2n\pi - A[/itex]
     
  9. Dec 24, 2017 #8
    Yes, and it is only a minor detail. I seems reasonable to accept only solutions in the range ##[0,2\pi]## so maybe instead of ##B=\pi-A## putting ##B=\pi\pm A## will suffice
     
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