# Quantum mechanic newbie: why complex amplitudes, why Hilbert space? [duplicate]

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I'm just starting learning quantum mechanics by myself (2 "lectures" so far) and I was wondering

• why we need to define quantum states in a complex vector space rater than a real one?

• Also I was wondering why this vector space has to be a Hilbert space (rather than a pre-Hilbert space)? when do we need the property that the vector space is complete (i.e. every Cauchy sequence converge)?

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## marked as duplicate by Qmechanic♦Mar 28 at 20:20

This question has been asked before and already has an answer. If those answers do not fully address your question, please ask a new question.

The two questions are possible duplicates of physics.stackexchange.com/q/8062/2451 and physics.stackexchange.com/q/20822/2451 respectively. –  Qmechanic Aug 6 '12 at 2:57
The first is a duplicate, the second is new. –  Ron Maimon Aug 6 '12 at 8:47

## 1 Answer

Regarding your second question, the requirement that the inner product space be complete is imposed to have available a number of important theorems, among which the spectral theorem is particularly important.

After a bit of time, though, you might notice that there's a couple of fishy things in that Hilbert space axiom. For one, we put a lot of stock in position and momentum eigenstates, i.e. delta-function and plane-wave wavefunctions, which are strictly speaking not inside the Hilbert space. On the other hand, the normal Hilbert spaces have a number of wavefunctions, such as $$\psi(x)=\frac{1}{\sqrt{\pi}}\frac{1}{\sqrt{1+x^2}}$$ which violate physical intuition in one way or another (this one has infinite position dispersion). The resolution is an amendment to the Hilbert space axiom in terms of rigged Hilbert spaces.

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There is another fishy thing remedied by rigged Hilbert space: Many relevant operators (e.g., position and momentum operators) only map part of the Hilbert space into itself. –  Arnold Neumaier Aug 11 '12 at 12:22