Take the 2-minute tour ×
MathOverflow is a question and answer site for professional mathematicians. It's 100% free, no registration required.

I need a reference concerning a theorem that shows the following result, stated very roughly:

Given a self-adjoint differential operator densely defined on a Hilbert space, then the given Hilbert space is spanned by the eigenvectors of the operator.


1) The statement above is very rough since for example, the continuous spectrum (which does not correspond to eigenvectors in the space) has to be used.

2) The operator I am thinking about are defined on the whole line (so there is continuous spectrum).

3) I am really looking for a spectral theorem for differential operators on the whole line. One issue I have is that in most of the books, they prove such theorems for bounded and/or compact operators only.

4) Another way to phrase is to look at the Sturm-Liouville theory as stated in http://en.wikipedia.org/wiki/Sturm–Liouville_theory and be able to say something about the basis when a and b are infinite.

share|improve this question
Weidmann's book is an excellent reference on the questions you mention. books.google.hu/books/about/… –  András Bátkai Mar 26 '13 at 20:16
To be concrete let's take $\partial_x^2$ defined on $H^2(\mathbb{R}) \subset L^2(\mathbb{R})$. What do you mean by eigenvectors? The solutions to the eigenvector equation $u_{xx} = \lambda u$ don't live in $L^2$. If we think of an orthonormal system of eigenvectors as forming the columns of an orthogonal matrix which diagonalizes the operator, perhaps the appropriate generalization is to seek a unitary transformation (Fourier Transform) which conjugates the operator with multiplication (by $-k^2$). –  Aaron Hoffman Mar 26 '13 at 20:21
You are perfectly right but I am thinking in physical term when you have an observable (self-adjoint linear diff. operator) whose "eigenvectors" generate the whole physical space (i.e. $L^2$). The eigenvectors corresponding to the continuous spectrum are unphysical but they can be added continuously in the form of an integral. In your example, there is only continuous spectrum but the Fourier series is a continuous sum over the unphysical solutions of the eigenvalue problem. –  Stephane Mar 27 '13 at 0:56
add comment

3 Answers

up vote 1 down vote accepted

For differential (especially, for Sturm--Liouville) operators I would recommend Akhiezer, Glazman's "Theory of linear operators in Hilbert space" and Naimark's "Linear differential operators".

In von Neumann's classical book "Mathematical foundations of quantum mechanics" the spectral theorem is stated very roughly.

share|improve this answer
Thank you very much for the references –  Stephane Mar 26 '13 at 23:43
add comment

Reed & Simon, Methods of modern mathematical physics I: Functional analysis (Academic Press, 1980): Chapter VIII, Section 3, Theorem VIII.6 (combined with property (b) of a projection-valued measure, loc. cit.). Of course, you'd have to first prove that your differential operator is at least essentially self-adjoint.

share|improve this answer
Thank you very much for the very precise pointer –  Stephane Mar 26 '13 at 23:43
add comment

You might also be interested in a more general version of the theorem, which, while more technical, IMHO, is much more elegant.

Theorem A self-adjoint operator in a rigged Hilbert space has a complete system of generalized eigenvectors, corresponding to real generalized eigenvalues.

This is Theorem 5' in Section 4.5 of Volume IV of I. M. Gelfand's Generalized Functions (on pg. 126). They define generalized eigenvectors and eigenvalues as follows.

Let $A$ be a linear operator on a linear topological space $\Phi$. A linear functional $F$ on $\Phi$, such that $$ F(A\phi )=\lambda F(\phi ) $$ for every element $\phi$ of $\Phi$, is called a generalized eigenvector of the operator $A$, corresponding to the eigenvalue $\lambda$.

(This can be found on pg. 105 of the same text.)

The nice thing about this formulation is that (1) you don't have to worry about operators being only densely-defined (if their dense domain gives the Hilbert space the structure of a rigged Hilbert space, you can extend the operator (it must be self-adjoint) to the entire rigged Hilbert space) and (2) you don't have to formulate the theorem in terms of projection-valued measures (this is somewhat unnatural), but can formulate it in terms of honest-to-god eigenvectors and eigenvalues.

In fact, in general, I would recommend looking into the theory of rigged Hilbert spaces. According to Gelfand himself (pg. 105 of the same text):

We believe this concept [of a rigged Hilbert space] is no less (if indeed not more) important than that of a Hilbert space.

I am inclined to agree.

share|improve this answer
What does "generalized eigenvector" mean in this context? (I am only familiar with the usage from linear algebra.) In particular, what is a generalized eigenvector for the bounded, self-adjoint operator $(M f)(t) = tf(t)$ on $L^2[0,1]$? –  Yemon Choi Apr 7 '13 at 4:57
Yemon, a generalized eigenvector with eigenvalue $\lambda$ in your case is just $\delta_{\lambda}$, a tempered distribution. And this what you want to do: extend Hilbert space to some space which includes objects that you would think of as eigenvectors. Maybe a more natural example would be differentiation ($i \frac{d}{dx}$) which obviously have exponentials ("plane waves") as eigenvectors, however they are not square-integrable. –  Mateusz Wasilewski Apr 7 '13 at 9:16
I want to add that I do not agree that formulation of spectral theorem with projection-valued measures is unnatural: first, it is more geometric (does not refer to any basis; that is what I like about it even in finite-dimensional case) and it is quite well motivated that projections should correspond to (elementary) observables in quantum mechanics and spectral theorem says then that all self-adjoint operators ought to do so. –  Mateusz Wasilewski Apr 7 '13 at 9:20
@Yemon Choi I've added a verbatim copy of their definition of generalized eigenvalues and eigenvectors. –  Jonathan Gleason Apr 7 '13 at 15:26
Jonathan, thank you for clearing it up. I think it is best to always have in mind many equivalent forms of spectral theorem and I agree with you that this one is extremely valuable, because this is how physicists usually think of spectral decomposition. –  Mateusz Wasilewski Apr 7 '13 at 16:42
show 2 more comments

Your Answer


By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.