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Q & A: do eigenvectors of Hermitian operators form complete basis?

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Most recent answer: 02/04/2012
Is there a math proof of the completeness of eigenfunctions for hermitian operators and what steps it takes? In which book can i find it?
- Andrea Maracci (age 24)

A quick search turns up a proof for finite-dimensional vector spaces:

I think the proof for infinite-dimensional spaces is hard, but don't remember it and can't find one yet. [see below]

Aha- Here's a proof for Hermitian operators that are bounded from below (or above).

Aha- Here's a good reason why I couldn't remember any proof for infinite-dimensional spaces.
They say "In a vector space with a finite dimension, it can be proven rigorously that eigenfunctions of a hermitian transformation span a space, so that any vector can be presented as a linear combination of the basis.
In Hilbert space such a proof exists only for several particular cases."

Mike W.

(published on 02/04/2012)

Follow-up on this answer.