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CHSH and the triangle inequality

  1. May 4, 2012 #1
    Hello everybody,

    I've been trying to understand the CHSH proof as it is listed on Wikipedia:
    http://en.wikipedia.org/wiki/CHSH_inequality

    I got to this without any problem:
    [itex]E(a, b) - E(a, b^\prime) = \int \underline {A}(a, \lambda)\underline {B}(b, \lambda)[1 \pm \underline {A}(a^\prime, \lambda)\underline {B}(b^\prime, \lambda)]\rho(\lambda)d\lambda - \int \underline {A}(a, \lambda)\underline {B}(b^\prime, \lambda)[1 \pm \underline {A}(a^\prime, \lambda)\underline {B}(b, \lambda)]\rho(\lambda)d\lambda[/itex]

    However, now it mentions two things to get to the next step:
    - The fact that [itex][1 \pm \underline {A}(a^\prime, \lambda)\underline {B}(b^\prime, \lambda)]\rho(\lambda)[/itex] and [itex][1 \pm \underline {A}(a^\prime, \lambda)\underline {B}(b, \lambda)]\rho(\lambda)[/itex] are non-negative (easy enough to see).
    - The triangle inequality "to both sides" (how?)

    And the next equation is:
    [itex]|E(a, b) - E(a, b^\prime)| \leq \int [1 \pm \underline {A}(a^\prime, \lambda)\underline {B}(b^\prime, \lambda)]\rho(\lambda)d\lambda + \int [1 \pm \underline {A}(a^\prime, \lambda)\underline {B}(b, \lambda)]\rho(\lambda)d\lambda[/itex]

    I don't understand how it gets to this last equation from the one before. Could somebody please explain?


    Thanks in advance,
    Gespex
     
    Last edited: May 4, 2012
  2. jcsd
  3. May 5, 2012 #2
    So you have:
    [itex]E(a, b) - E(a, b^\prime) = \int \underline {A}(a, \lambda)\underline {B}(b, \lambda)[1 \pm \underline {A}(a^\prime, \lambda)\underline {B}(b^\prime, \lambda)]\rho(\lambda)d\lambda - \int \underline {A}(a, \lambda)\underline {B}(b^\prime, \lambda)[1 \pm \underline {A}(a^\prime, \lambda)\underline {B}(b, \lambda)]\rho(\lambda)d\lambda[/itex]

    Taking absolute value of both sides
    [itex]|E(a, b) - E(a, b^\prime)| = |\int \underline {A}(a, \lambda)\underline {B}(b, \lambda)[1 \pm \underline {A}(a^\prime, \lambda)\underline {B}(b^\prime, \lambda)]\rho(\lambda)d\lambda - \int \underline {A}(a, \lambda)\underline {B}(b^\prime, \lambda)[1 \pm \underline {A}(a^\prime, \lambda)\underline {B}(b, \lambda)]\rho(\lambda)d\lambda|[/itex]

    Using [itex]|A \pm B| \le |A| + |B|[/itex]:

    [itex]|E(a, b) - E(a, b^\prime)| \le |\int \underline {A}(a, \lambda)\underline {B}(b, \lambda)[1 \pm \underline {A}(a^\prime, \lambda)\underline {B}(b^\prime, \lambda)]\rho(\lambda)d\lambda| + |\int \underline {A}(a, \lambda)\underline {B}(b^\prime, \lambda)[1 \pm \underline {A}(a^\prime, \lambda)\underline {B}(b, \lambda)]\rho(\lambda)d\lambda|[/itex]

    Then using [itex]| \int f(x) dx | \le \int |f(x)| dx[/itex], (which, if you think of it, is just a variation of the above):

    [itex]|E(a, b) - E(a, b^\prime)| \le \int |\underline {A}(a, \lambda)\underline {B}(b, \lambda)[1 \pm \underline {A}(a^\prime, \lambda)\underline {B}(b^\prime, \lambda)]\rho(\lambda) | d\lambda + \int |\underline {A}(a, \lambda)\underline {B}(b^\prime, \lambda)[1 \pm \underline {A}(a^\prime, \lambda)\underline {B}(b, \lambda)]\rho(\lambda)|d\lambda[/itex]

    And since [itex]|\underline {A}(a, \lambda)\underline {B}(b, \lambda)|=1[/itex], and both [itex]1 \pm \underline {A}(a^\prime, \lambda)\underline {B}(b^\prime, \lambda) \ge 0[/itex] and [itex]\rho(\lambda) \ge 0[/itex], we get:

    [itex]|E(a, b) - E(a, b^\prime)| \leq \int [1 \pm \underline {A}(a^\prime, \lambda)\underline {B}(b^\prime, \lambda)]\rho(\lambda)d\lambda + \int [1 \pm \underline {A}(a^\prime, \lambda)\underline {B}(b, \lambda)]\rho(\lambda)d\lambda[/itex]
     
  4. May 5, 2012 #3
    Thanks a lot! The proof's easy enough to follow like that. But this statement seems to me to contradict Wikipedia (though it's probably just my ignorance):
    [itex]|\underline {A}(a, \lambda)\underline {B}(b, \lambda)|=1[/itex]

    As Wikipedia says "Since the possible values of A and B are −1, 0 and +1". So couldn't it also be that [itex]|\underline {A}(a, \lambda)\underline {B}(b, \lambda)|=0[/itex]?

    It also says "where [itex]\underline {A}[/itex] and [itex]\underline {B}[/itex] are the average values of the outcomes", so not just -1, 0 or 1, but any value from -1 to 1, which would mean only that: [itex]0 \le |\underline {A}(a, \lambda)\underline {B}(b, \lambda)|\le 1[/itex]

    Or am I missing something here?
     
  5. May 5, 2012 #4
    Oh, I'm sorry, you are right of course, it should have been [itex] |\underline {A}(a, \lambda)\underline {B}(b, \lambda)| \le 1[/itex]. It does not change anything though.
     
  6. May 5, 2012 #5
    Oops, yeah, good point. Thanks!
     
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