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Closure relation in infinite dimensions

  1. Apr 28, 2015 #1
    The closure relation in infinite dimension is : ∫|x><x|dx =I (identity operator),but if we apply the limit definition of the integral the result is not logic or intuitive.
    The limit definition of the integral is a∫b f(x)dx=lim(n-->∞) [i=1]∑[i=∞]f(ci)Δxi, where Δxi=(b-a)/n (n--.>∞) and ci=a+((b-a)/n)Δxi.
    Apply the limit definition of the integral in the case of the closure relation above, it seems not logic and intuitive that
    ∫|x><x|dx=I(identity operator). ????
    so how it comes ∫|x><x|dx=I(identity operator). I should take it for granted or there is a prove for it ?
     
    Last edited: Apr 28, 2015
  2. jcsd
  3. Apr 28, 2015 #2

    bhobba

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    You need to study Rigged Hilbert Spaces:
    http://physics.lamar.edu/rafa/webdis.pdf [Broken]

    See page 104 on the Gelfand-Maurin Theorem.

    Be warned - its what mathematicians call non trivial - meaning its hard.

    Intuitively however its along these lines. Consider position as a very large number of discreet values xi. ∑|xi><xi| = 1. Define xi' as xi/√Δx. So it becomes Σ|xi'><xi'| Δx. Take the limit as Δx goes to zero and you get ∫ |x><x| dx = 1. Of course |x> then has infinite length so doesn't belong to a Hilbert space - to get around that you need Rigged Hilbert Spaces.

    Thanks
    Bill
     
    Last edited by a moderator: May 7, 2017
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