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.

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.

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?


2 Answers 2


(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 $||A\psi_\epsilon - \lambda \psi_\epsilon|| < \epsilon$.

  • $\begingroup$ Looks like I have a wrong notion of countability in my mind... $\endgroup$
    – Wildcat
    Jun 20, 2016 at 11:04
  • $\begingroup$ A set $S$ is said to be countable if there is an injective map $f : S \to \mathbb N$... $\endgroup$ Jun 20, 2016 at 11:41
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    $\begingroup$ Indeed the second statement is false rigorously speaking. It is more or less correct if the Hilbert space is separable... $\endgroup$ Jun 23, 2016 at 14:49
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    $\begingroup$ I think an explicit example is helpful where on the one hand it is evident that the statement is false, on the other hand it is also clear that the example has no much physical sense. Consider, in $H = L^2(\mathbb R)$ (thus separable), a Hilbert basis $\{\psi_k\}_{k \in {\mathbb Q} \cap [0,1]}$, $\mathbb Q$ being the set of rational numbers and define the bounded self-adjoint operator $A := \sum_{k \in {\mathbb Q} \cap [0,1]} k |\psi_k\rangle \langle \psi_k|$. It is an easy exercise to prove that $D(A)=H$, $\sigma(A)= [0,1]$, but $ \sigma_p(A)= {\mathbb Q} \cap [0,1]$. $\endgroup$ Jun 23, 2016 at 15:42
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    $\begingroup$ You see that the spectrum fills the complete interval $[0,1]$ but not every point of the spectrum is an improper eigenvalue, the points in $\mathbb Q$ are proper eigenvalues whose eigenvectors are just the (normalizable) vectors $\psi_k$. The irrational points of $[0,1]$ are instead elements of the continuous spectrum, but there is no associated non-normalizable eigenfunctions associated with each of them. $\endgroup$ Jun 23, 2016 at 15:44

First, let's answer the questions precisely as you worded them:

The point spectrum is always discrete in the sense that it consists of at most countably many points.

This is true by proving the following results: a) the space spanned by all eigenvectors is a closed subspace of the Hilbert space, hence we have an orthonormal system of eigenvectors, b) two eigenvectors of two distinct eigenvalues are always orthogonal, and c) an orthonormal system of a separable Hilbert space is countable. The last is implied by the fact that separable Hilbert spaces allow for countable Schauder bases (orthonormal bases) and the fact that two bases must have the same cardinality.

However, note that there is a sense in which the eigenvalues are not necessarily discrete in a different sense of the word: The closure of the set of eigenvalues can be bigger. As an example, consider a Hilbert space $\mathcal{H}$ with orthonormal basis $|\psi_n\rangle$ and define the operator $K$ as $K:|\psi_n\rangle\mapsto 1/n|\psi_n\rangle$. Eigenvalues are $1/n$ which accumulate at $0$, which itself is not an eigenvalue.

On the other hand, the continuous spectrum can consist of just one point.

As an example, consider the same operator as above. Since the spectrum is always closed, $0$ is inside the spectrum, but it is not an eigenvalue. You can easily show that it is the only point of the spectrum which is not an eigenvalue (the reason is that $K$ is compact and for compact operators, the eigenvalues are dense in the spectrum). Hence $0$ is the continuous spectrum of the operator.

[One could argue that this part of the spectrum is not acutally what you mean by "continuous spectrum" from your definition - but I'm going to go with the usual text book definition of continuous spectrum here, which also implies the dichotomy of point spectrum vs. continuous spectrum.]

But now let me tell you that your first paragraph is problematic for many reasons and introduce a lot more stuff.

First of all, there is something called a discrete spectrum, which however is not equivalent to the point spectrum of the operator.

Definition: Let $A$ be a self-adjoint operator on a separable Hilbert space. The discrete spectrum consists of all isolated eigenvalues, i.e. eigenvalues $\lambda$ with finite multiplicity such that for some $\varepsilon>0$, we have $\sigma(A)\cap (\lambda-\varepsilon,\lambda+\varepsilon)=\{\lambda\}$. The complement of the discrete spectrum is called the essential spectrum.

For example, if you take the operator $K$ above, then the discrete spectrum is precisely the point spectrum, whereas the essential spectrum consists of $0$. On the other hand, for the identity operator $I$, clearly the point spectrum (and also the spectrum) consists of $1$, but the discrete spectrum is empty, since the eigenvalue is of infinite multiplicity.

The idea is that "discrete" actually means "all eigenvalues that are not accumulation points of eigenvalues". Clearly, the discrete spectrum consists of at most countably many values.

But that's not all. It might seem rather unfortunate to have $0$ be part of the continuous spectrum of the compact operator $K$. In particular, this makes it possible that the continuous spectrum consist of only points. At the same time, the eigenvalues of $K$ already span the whole Hilbert space $\mathcal{H}$, so there is no need to have $0$ be a real part of the spectrum. In addition, as you say, maybe you want something like "continuous spectrum is given by a continous part of the real line". To have this, you must exclude some values from the continous spectrum.

This can be done in a systematic way by definition the spectral measure of the Hilbert space operator and consider the Radon-Nikodym decomposition of the measure with respect to the Lebesgue measure into a pure point part, an absolutely continuous part and a singularly continuous part:

Theorem: Let $A$ be a self-adjoint operator on a separable Hilbert space. Let $\mu$ be the spectral measure defined for $A$, then $\mu$ can be decomposed into a pure point par $\mu_{pp}$ consisting of all eigenvalues of $A$, an absolutely continuous part $\mu_{ac}$ (which is absolutely continuous with respect to the Lebesgue measure) and a singularly continuous part $\mu_{sc}$ (the "rest").

The support of $\mu_{pp}$ (where it is nonzero) is then the eigenvalues $\sigma_{pp}(A)$, the support of $\mu_{ac}$ is called the absolutely continuous spectrum $\sigma_{ac}(A)$ and likewise we have $\sigma_{sc}$. Then

$$ \sigma(A)=\sigma_{ac}(A)\cup \sigma_{sc}(A)\cup \overline{\sigma}_{pp} $$

where the overline denotes the closure. Here is the interesting part: A measure that is absolutely continuous with respect to the Lebesgue measure cannot have countable support, as such sets have zero measure. In other words: For the absolutely (and I believe also for the singularly continuous) spectrum, the spectrum always consists of uncountably many points.

In addition, one can define a Hilbert space $\mathcal{H}_{pp}$ consisting of all eigenvectors, and Hilbert spaes $\mathcal{H}_{ac}$ and $\mathcal{H}_{sc}$ consisting of all "approximate eigenfunctions" of the absolutely and singularly continuous spectrum and those three spaces add up:

$$ \mathcal{H}=\mathcal{H}_{pp}\oplus \mathcal{H}_{ac}\oplus \mathcal{H}_{sc}$$.

In this sense, this is also the "right" decomposition.

For our operator $K$ one can then easily see that the continuous parts of the spectrum are empty, it has only pure point spectrum - exactly as we wish.

Note however that the three spectra need not be disjoint. You can have continuous spectrum and some eigenvalues lying embedded in the continuous spectrum.

Finally, let's do a little bit of physics (at least nearly). There is a beautiful theorem, which tells you that the last decomposition is the right one for physics. It is called the RAGE theorem (see for instance Theorem 5.7 in Garald Teschl's book "Mathematical Methods in Quantum Mechanics") and it basically tells you that if you consider some Hamiltonina $H$, the pure point part of the spectrum forms the bound states of the opererator in that the particles are eternally confined within some region. On the other hand, the absolutely continuous part forms the unbounded states that escape and never return (the singularly continuous part is tricky - usually you try to show it doesn't exist, but it can be interpreted as unbounded states that sort of escape at infinity, but until then return to where they started over and over again).


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