# Dimensionality in quantum mechanics [closed]

I was reading Introduction to quantum mechanics by David J. Griffiths and came across following paragraph:

" 3. The eigenvectors of a hermitian transformation span the space.

As we have seen, this is equivalent to the statement that any hermitian matrix can be diagonalized. This rather technical fact is, in a sense, the mathematical support on which much of a quantum mechanics leans. It turns out to be a thinner reed then one might have hoped, because the proof does not carry over to infinite-dimensional spaces."

My thoughts:

If much of a quantum mechanics leans on it, but the proof does not carry over to infinite-dimensional spaces, then hermitian transformations with infinite dimensionality are spurious.

But there is infinite set of separable solutions for e.g. particle in a box. So Hamiltionan for that system has spectrum with infinite number of eigenvectors and is of infinite dimensionality.

If we can't prove that this infinite set of eigenvectors span the space then how can we use completness all the time?

Am I missing something here? Any missconceptions?

I'd appriciate any help.

• I'm voting to close this question as off-topic because it is so much better suited for Physics.SE. – Wildcat Jun 16 '17 at 12:29
• You may want to read about the spectral theorem. You need more assumptions for infinite-dimensional spaces (compactness). – Felipe S. S. Schneider Jun 16 '17 at 13:12
• The real issue is not whether this belongs on Physics, but instead I would suggest that the Math SEs are better. Beyond that, however, is to note that this really is one of those points where a math proof may not be relevant to physics or chemistry. It is clear that in real physical problems that only a few symmetries tend to occur, and that a readily identifiable set of eigenfunctions (sin/cos, Bessel, Legendre, ...) are known for those. – Jon Custer Jun 16 '17 at 15:09
• @FelipeSchneider what kinds of assumptions? I'll check spectral theorem. – voldermot Jun 16 '17 at 19:23
• @f13 I think the closest you may get to the finite-dimensional case is to ask for compact self-adjoint operators. Those are diagonalisable (in the sense that there is an orthonormal basis consisting of eigenvectors of it). Kreyszig is a good read on this (specially the last two chapters). – Felipe S. S. Schneider Jun 16 '17 at 21:21