# Closure relation in infinite dimensions

1. Apr 28, 2015

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. Apr 28, 2015

### bhobba

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