# Physical (Quantum Mechanical) Significance of completeness of Hilbert Spaces.

I'm not sure if the question is very 'mathematical',but I'm asking any way. I have a basic knowledge of quantum mechanics and I'm studying Hilbert spaces.

I was wondering what is the physical significance of the completeness of a Hilbert Space wrt QM. i.e. why is the completeness required?

How do mathematical properties of Hilbert spaces or Banach spaces play a role in Quantum Mechanics?

• One important point is that you can have a countable orthonormal basis. Otherwise, you are stuck with overcountable Hamel bases. – Raskolnikov Apr 25 '12 at 11:42
• Intersting question. I suspect Hilbert spaces are the setting of choice because of mathematical convenience, but I'd like to know more. – Olivier Bégassat Apr 25 '12 at 12:08
• Every non complete space with a inner product is a subspace of a complete one which is it closure, So if you are in a space where all the Cawchy sequence don't converges you may assume that you are in the closure. About the Hilbert basis the pre-Hilbert space has to be separable! Of course I am talking in mathematician view point! – checkmath Apr 25 '12 at 12:13

## 2 Answers

There's an MO post with some good thoughts on the role of completeness from a mathematical point of view. (I especially like Andrew Stacey's idea of "going to a party hosted by the Square Integrables in their posh mansion, but spending the whole time hanging out with the Schwartz family" :-)

With a view to quantum mechanics, note that the Riesz representation theorem (which establishes the bra-ket correspondence) depends on completeness. You can trace the role of completeness in this proof, which uses the orthogonal decomposition theorem, whose proof in turn uses completeness.

The history section of the Wikipedia article on Hilbert spaces also has some relevant information.

completeness in hilbert space means that the metric associated with inner product converges,let draw a 1/x as a function on a graph,you will see its area converges to zero. as x approaches to infinity 1/x goes to zero