# Different parts of spectrum appearing in the spectral theorem in terms of generalized eigenvectors

Question: How to exactly relate both expansions quoted below: Can one be "transformed" into the other? What is the interplay between the various parts of the spectrum appearing?

In Ref. 1 it is stated that

The most general form of the spectral theorem for an operator $$A$$ representing a physical observable is $$[\ldots]$$ : $$\phi = \sum |a_i)(a_i|\phi\rangle + \int \mathrm da \, |a\rangle \langle a|\phi\rangle\quad , \tag{4.4g}$$ where the sum is over the discrete spectrum and the integral is over the (absolutely) continuous spectrum of $$A$$. It can happen that some or all values $$a_i$$ appearing in the sum also appear in the integral. Then they are called discrete eigenvalues in the continuous spectrum. If this happens for $$a_k$$ then $$|a_k)$$ is still orthogonal to all $$|a\rangle$$ including $$|a_k\rangle$$ $$[\ldots]$$.

Likewise, Ref 2. gives the resolution of the identity in terms of generalized eigenvectors as:

$$[\ldots]$$ the question naturally arises whether for every self-adjoint operator $$A$$ and for every $$|\psi\rangle$$ we can always write $$|\psi\rangle = \sum\limits_{\lambda_n\in \sigma_p(A)} \sum\limits_{\alpha\in I(\lambda_n)} |\lambda_n;\alpha\rangle\langle \lambda_n;\alpha|\psi\rangle + \int_{\sigma_c(A)}\mathrm d\lambda \, \sum\limits_{\alpha\in I(\lambda)}|\lambda;\alpha\rangle\langle \lambda;\alpha|\psi\rangle \tag{2.43} \quad .$$ $$[\ldots]$$ The answer to this question is affirmative [if we interpret the integration of $$(2.43)$$ as a direct integral] for any self-adjoint operator which does not have a continuous singular spectrum. This last condition allows us to write $$\mathrm d\lambda$$ in $$(2.43)$$, instead of having to write another, more general measure.

To elaborate:

I am confused by all the different parts of the spectrum involved in both expansions. I know that the spectrum of a self-adjoint operator $$A$$ is the union of the point and continuous spectrum: $$\sigma(A)= \sigma_p(A) \, \cup\, \sigma_c(A)$$, where these sets are disjoint. Thus, in the expansion of Ref. 2, I think there should be no point of the spectrum which appears both in the sum and in the integral - and embedded eigenvalues are taken care of already by "putting them" in the sum.$$^\ddagger$$

In Ref.1 the author somehow splits the spectrum in the discrete $$\sigma_d(A)\subset \sigma_p(A)$$ and absolutely continuous $$\sigma_{ac}(A)$$ spectrum. Is this really correct? As far as I understand, the discrete spectrum consists of eigenvalues which are also isolated. So there cannot be any point from $$\sigma(A)$$ in its neighborhood and the integration thus would be somehow meaningless, no? Perhaps the author means the pure point spectrum $$\sigma_{pp}(A) = \bar \sigma_p(A)$$ instead of $$\sigma_d(A)$$? In any case, that a point appears in both the sum and the integral can only happen if both parts are not disjoint.

But do these points belong to the point or continuous spectrum then? This is indeed very confusing, since both parts of the spectrum are disjoint (as stated above) and thus, a point in the spectrum either corresponds to a genuine eigenvector (and therefore is an eigenvalue) or to a generalized eigenvector - how can both hold true at the same time?

Summarizing: I have the feeling that both expressions must be equivalent and hence it should be possible to transform one into the other. But I don't really understand the relations between all the various parts of the spectrum enough to clearly see how both expressions are related. In particular, I don't understand how a single point appears with both an eigenvector and a generalized vector in the first expansion.

References:

1. Arno Bohm. Quantum Mechanics: Foundations and Applications. Springer, third edition. Chapter I.4, page 12.

2. Quantum Mechanics I. Galindo and Pascual. Springer. Chapter 2, page 48.

$$^\ddagger$$ See also this related question. The answer there works with the notion of spectral measures instead of generalized eigenvectors, tho.

The two types of spectra, $$\sigma_c$$ and $$\sigma_p$$, are disjont; so that the first statement you quoted is mathematically speaking wrong: $$\sigma_c(A) \cap \sigma_p(A)=\emptyset$$. But the book seems to be not very mathematically minded so a bit vague statements are definitely tolerable! Furthermore, one should take the fact into account that each element of the continuous spectrum has zero (spectral) measure (see (3) in the proposition below).

The second decomposition you quote is correct. However the discrete spectrum is made of isolated points whose eignespace is finite-dimensional. This second type of decomposition refers to the decomposition of the spectrum into discrete (the one said above) and and essential spectrum (its complement). A third decomposition you mention is the one into absolutely continuous spectrum and singular spectrum.

Notice that isolated points of the spectrum are always part of the point spectrum, but the vice versa is false. I do not understand well your remark about issues with integration for points in the discrete spectrum: see the last item of the following proposition where $${\cal B}(\mathbb{R})$$ is the $$\sigma$$-algebra of Borel sets in $$\mathbb{R}$$.

Proposition. Let $$A: D(A) \to H$$ be a selfadjoint operator and let $$P: {\cal B}(\mathbb{R}) \to \mathfrak{B}(H)$$ be the projection-valued measure (also known as spectral measure) of $$A$$ so that $$A = \int_{\mathbb{R}} \lambda dP(\lambda)\:.$$ The following facts are true for $$\lambda \in \mathbb{R}$$.

(1) The support of $$P$$ (i.e., the complement of the largest open set $$A\subset \mathbb{R}$$ such that $$P_A=0$$) is $$\sigma(A)$$.

(2) $$\lambda \in \sigma_p(A)$$ if and only if $$P_{\{\lambda\}} \neq 0$$. In that case $$P_{\{\lambda\}}$$ is the orthogonal projector onto the $$\lambda$$-eigenspace of $$A$$.

(3) $$\lambda \in \sigma_c(A)$$ if and only if both conditions are true: (a) $$P_{\{\lambda\}}=0$$ and (b) $$P_{(a,b)}\neq 0$$ if $$a and $$\lambda \in (a,b)$$.

(4) If $$\lambda$$ is an isolated point of $$\sigma(A)$$ then necessarily $$\lambda \in \sigma_p(A)$$ and thus $$P_{\{\lambda \}}\neq 0$$.

(As references I could quote my books on spectral theory and QM books on spectral theory or also the more recent one.)

On account of the proposition, we can re-arrange the integral as $$A = \sum_{\lambda \in \sigma_p(A)} P_{\{\lambda\}} + \int_{\sigma_c(A)} \lambda dP(\lambda)\:.$$

• Dear Valter, as always, great answer; thank you very much. Do you know/ can you guess what the author of the first quote means however (e.g. physically)? I also took the freedom to correct two typos. Feb 20, 2023 at 10:44
• Yes it is correct, but there the issue is subtle. There the book is considering the fact if the measure used in the continuous part is proportional to the Lebesgue measure or not. Feb 20, 2023 at 11:09
• I agree with your analysis. The decomposition I used in my answer is the most natural and elementary one. However, for several specific reasons, it could be intersting/useful to study the interplay of the part of the spectral measure different from isolated points and the Lebesgue measure. Feb 20, 2023 at 11:23
• Notice that the standard point spectrum may be a dense set or even an open set. So "point spectrum" is a misleading name. THis partially justifies the further types of decompositions one encounters in the literature on the subject. Feb 20, 2023 at 11:25
• Hi @Tobias Fünke Please ask a separate question since this issue is quite complicated and I am not sure to understand well. Feb 21, 2023 at 9:17

Ah, there are many natural situations in which some large part of the discrete/eigenvalue spectrum of a (bounded, self-adjoint) Hilbert space operator lies inside the continuous spectrum. Also for unbounded self-adjoint operators.

That is, the spectral measure can be a sum of Lebesgue-type measure on some intervals, plus some Dirac measures on the discrete spectrum.

The case I'm familiar with is the Laplace-Beltrami operator (or its resolvent, if we insist on something bounded) on the modular curve, where there are infinitely-many "cuspforms" in the discrete spectrum, with eigenvalues going to $$-\infty$$, as well as continuous spectrum going from $$-1/4$$ to $$-\infty$$ (depending on one's normalization of the Laplacian). So there is no doubt that the continuous spectrum can contain (infinitely many) eigenvalues.

So-called "imbedded spectrum".

EDIT: Yes, many sources seem to define continuous spectrum in a way that makes it seem that it cannot also contain any discrete spectrum. And, yes, essentially since the Lebesgue measure of single points is $$0$$, this doesn't actually wreck anything. But it is misleading, I think.

To give a concrete example, consider the ("multiplication") operator on $$L^2(\mathbb R)$$ given by $$Tf(x)=f(x)\cdot h(x)$$, where $$h$$ is identically $$0$$ on $$(-\infty,0]$$, is identically $$1$$ on $$[0,2]$$, rises linearly to $$5$$ on $$[2,3]$$, decays linearly to $$0$$ on $$[3,4]$$, and is $$0$$ on $$[4,+\infty)$$.

The real-valued-ness means that this multiplication operator is self-adjoint.

The flat part with value $$1$$ means that any function supported on $$[0,2]$$ is an eigenvector with eigenvalue $$1$$. (Also, any function supported on $$(-\infty,0]\cup [4,+\infty)$$ is an eigenvector with eigenvalue $$0$$.) Since the graph on $$[2,4]$$ has no horizontal-flat parts, there are no other eigenvalues. But/and the fact that the range of the function restricted to $$[2,4]$$ is $$[0,5]$$ means that there are (Weyl-style) "approximate eigenvectors" for all $$\lambda\in[0,5]$$, which (small theorem) implies that the continuous spectrum is (well, to be precise, contains) $$[0,5]$$, in a natural way.

Yes, the eigenvalues $$1$$ and $$0$$ are also in $$[0,5]$$.

(So, the spectral measure is Lebesgue measure on $$[0,5]$$, plus Dirac measures at $$1$$ and $$0$$.)

• Dear Paul, thanks for your answer. Unfortunately I don't really understand it. Could you elaborate what exactly it means that the discrete spectrum lies inside the continuous spectrum? As I tried to elaborate in my post, I thought that the point and continuous spectrum are disjoint - so how is this possible? Feb 20, 2023 at 7:45