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

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Q:
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)
Rimini
A:

A quick search turns up a proof for finite-dimensional vector spaces:
http://www.indiana.edu/~ssiweb/C561/PDFfiles/Operators-II.pdf

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). http://www.ece.rutgers.edu/~maparker/.../Ch03S13ThmsHermOps.pdf

Aha- Here's a good reason why I couldn't remember any proof for infinite-dimensional spaces. http://faraday.uwyo.edu/~yurid/QM/Lecture%208.pdf
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)

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