quantum mechanics - Properties of spectrum of a self-adjoint operator on a separable Hilbert space

So, if I understand it correctly, the spectrum of a self-adjoint operator on a Hilbert space $H$ consists of two parts: $ \newcommand{\ket}[1]{\,\lvert{#1}\rangle} \newcommand{\op}[1]{\hat{#1}} $

• point spectrum which is a set of eigenvalues $a$ that satisfy $\op{A} \ket{a} = a \ket{a}$ for a vector $\ket{a} \in H$;


• continuous spectrum consisting of values $a$ that, roughly speaking, satisfy the same equation as eigenvalues, but for a vector $\ket{a}$ which do not belong to the Hilbert space, but rather is a part of the so-called rigged Hilbert space.

Collectively values $a$ that satisfy $\op{A} \ket{a} = a \ket{a}$ for a vector $\ket{a}$ either from a Hilbert space or the rigged Hilbert space are called approximate eigenvalues while the vectors themselves are referred to as approximate eigenvectors.

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Now, in the limiting case when a self-adjoint operator on a Hilbert space has only point spectrum, i.e. a spectrum consisting of eigenvalues only, a set of corresponding eigenvectors is a basis which by dimension theorem is countable if a Hilbert space is separable. Consequently, the set of eigenvalues of such operator is also countable. And here comes my first question: is the same true even when continuous spectrum is not empty, i.e. is the set of eigenvalues of a self-adjoint operator on a separable Hilbert space always countable, regardless or not does the operator has approximate eigenvalues that are not eigenvalues? Or, in other words, is the point spectrum of a self-adjoint operator on a separable Hilbert space always discrete regardless of the presence of continuous spectrum?

My second question is (in some sense) the exact opposite of the first one. Is continuous spectrum of a self-adjoint operator on a separable Hilbert space always continuous? where the later continuity is understood is as the opposite to discreteness, i.e. as uncountability of the set of approximate eigenvalues that are not eigenvalues.

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These might appear as purely mathematical questions, but I'm primarily interested in the physical implications. For instance, I would like to know is the spectrum of a self-adjoint operator (that represents an observable) purely discrete only when it is limited to point spectrum? And is it purely continuous (as the opposite to discrete) only when it is limited to continuous spectrum?

Answer

(1) Yes, the point spectrum is countable in your hypotheses: otherwise the operator would have an uncountable set of pairwise orthogonal vectors since eigenvectors of a self-adjoint operator with different eigenvalues are orthogonal. This is impossible because, in every Hilbert space, every set of (normalized) orthogonal vectors can be completed into a Hilbert basis by Zorn's lemma and every Hilbert basis is countable if the space is separable.

(2) No, it is not necessarily true that the continuous spectrum is uncountable. You may have one only point in the continuous spectrum for instance. This is the case for the spectrum of the self-adjoint operator $1/H$, where $H$ is the Hamiltonian of the harmonic oscillator. The only point of the continuous spectrum is $0$.

COMMENT. However I do not believe that discrete is a really appropriate adjective for the point spectrum. My impression is that your idea of discrete involves the fact that the points are isolated. This is not the case for the point spectrum in general also if the Hilbert space is separable. You may have a point spectrum coinciding with rational numbers, which are dense in $\mathbb R$ as well known.

Indeed, there are other decompositions of the spectrum. Within a certain approach it is defined the so-called discrete spectrum as the part of point spectrum made of isolated eigenvalues whose eigenspaces are finite-dimensional.

If the Hilbert space is not separable, it is even possible to construct a self-adjoint operator whose point spectrum is the whole $\mathbb R$.

COMMENT 2. It is not necessary to introduce the notion of rigged Hilbert space to define notions of approximated eigenvalues and eigenvectors. Given a self-adjoint operator $A:D(A)\to H$ in the Hilbert space $H$, it is possible to prove that $\lambda \in \sigma_c(A)$ if and only if $\lambda$ is not an eigenvalue (in proper sense) and, for every $\epsilon>0$ there is $\psi_\epsilon \in D(A)$ with $||\psi_\epsilon||=1$ such that $||A\psi_\epsilon - \lambda \psi_\epsilon|| < \epsilon$.