Eigenvectors of self-adjoint operators as Hilbert space basis

Question

My question is strongly related to this question, but I'd like to have a few more details about the topic.

Question 1: If $\mathscr{H}$ is a Hilbert space and $T$ is a self-adjoint operator on $\mathscr{H}$ with a countable set of eigenvectors $\{\psi_{n}\}_{n\in \mathbb{N}}$, does it follow that $\{\psi_{n}\}_{n\in \mathbb{N}}$ form a complete orthonormal set (i.e. a Hilbert basis) for $\mathscr{H}$? If so, can someone please explain how does it follow from the spectral theorem?

Question 2: In the linked question, a special attention was given to operators with continuous spectrum. In this case, it seems not to be possible to form a basis for $\mathscr{H}$ composed by eigenvectors of a self-adjoint operator $T$. But one of the answers speak of a "generalized eigenfunction", which is not properly an element of a Hilbert space. How does it work? Could someone please at least sketch the ideas?

#2 Do you want a hand-waving physics description? https://physics.stackexchange.com/search?q=generalized+eigenfunction — Keith McClary, Jul 11 '21 at 19:13
#1 Can you link to a statement of the version of spectral theorem you are using? — Keith McClary, Jul 11 '21 at 19:59
Q1: If $H$ is not separable then the eigenvectors cannot be a complete orthonormal set. The closure of the span of the eigenvectors is equal to the orthogonal complement of the null-space of the operator. — daw, Jul 12 '21 at 06:01

score 3 · Answer 1 · answered Jul 12 '21 at 07:39

Let $H=L^2_{\mu}((-1,1)\cup\{1,1/2,1/3,\cdots\})$ where $\mu$ is ordinary Lebesgue measure on $(-1,0)$ and where $\mu$ has discrete atoms of unit mass on $\{ 1,1/2,1/3,1/4,\cdots \}$. The operator $A : H\rightarrow H$ defined as $(Af)(\lambda)=\lambda f(\lambda)$ is self adjoint with continuous spectrum on $(-1,0)$ and has discrete spectrum $\{1,1/2,1/3,\cdots\}$. The eigenvectors are $e_n=\chi_{{1/n}}$ for $n=1,2,3,\cdots$, but these do not form a complete basis of $H$. This type of operator comes up when studying the basic Hydrogen isotope in Physics, but the continuous spectrum is in $(0,\infty)$ instead. You have approximate eigenvectors with approximate eigenvalues $\lambda\in(-1,1)$, but no actual eigenvalues in that range.

Thanks for the answer! Very nice. But if I assume $\mathscr{H}$ is separable, do these eigenvectors form a basis? Maybe I had to harden some hypothesis? — Idontgetit, Jul 12 '21 at 11:43

Eigenvectors of self-adjoint operators as Hilbert space basis

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