# Continuous spectrum and weak solution of eigenvalue equation

1. Dec 4, 2006

### poliopeti

Hi All!

Preliminaries: Let H denote the Hilbert-space, and let A be a densely defined closed operator on it, with domain $D(A) \subset H$. On D(A) one defines a finer topology than that of H such way that f_n->f in the topology on D(A) iff both f_n->f and Af_n->Af in the H-topology. Let D*(A) denote the topological dual of D(A) with respect to this topology. (This is kind of a starting point to create a Rigged Hilbert Space generated by A). Anyway, I don't want to go into too much details, one can look up A. Bohm: Quantum Theory: Foundations and Applications, Chapter 1 for further details.

Suppose that $z \in \mathbb{C}$ is in the continuous spectrum of A, ie (A-zI)^{-1}, exists, densely defined, but unbounded, ie there is a sequence f_n in D(A) (not convergent in general) such that f_n-s are normed to 1, and (A-zI)f_n->0. Then for every $\varepsilon>0$ there is an $n_\varepsilon$ such that if $n>n_\varepsilon$ we have $||(A-zI)f_n||<\varepsilon$, where ||.|| denotes the Hilbert-space norm. So we have approximate eigenvectors.

Now the question is the following. Can anyone prove - or tell me where the proof can be found - that if z is in the continuous spectrum, then the equation Af=zf has a solution in the sense of distributions (in D*(A)), ie there is some functional $T \in D*(A)$ that for every $f \in D(A)$ the equation

<T|(A-zI)f>=0

holds?

The proof should be simple, I mean I don't want to read through all the 5 volumes of Gelfand-Shilov . If someone knows the book
Kreyszig: Introductory Functional Analysis with Applications, as far as I remember he mntioned this result within, without proving it...

Cheerio,

Peter

2. Dec 4, 2006

### Hurkyl

Staff Emeritus
If you replace the $...$ constructions with [ itex ] ... [ /itex ] tags (remove the spaces), you'll invoke our LaTeX parser. (and use [ tex ] ... [ /tex ] for the $...$ construction)

Last edited: Dec 4, 2006
3. Dec 4, 2006

### Hurkyl

Staff Emeritus
It seems to me you'd want to invoke your sequence of approximate eigenvalues. e.g. something in the spirit of

$$\langle T | := \lim_{n \rightarrow \infty} \langle f_n |.$$

4. Dec 4, 2006

### poliopeti

well,

$\langle (A-zI)f_{n} \vert \phi \rangle \rightarrow 0$

is definitely true, but it does not mean that

$\langle f_{n}\vert \right.$

converges weak-* to a nonzero functional, which would require

$\forall \; \phi\in D(A)\quad \langle f_{n}\vert \phi \rangle=\langle T \vert \phi \rangle$...

Last edited: Dec 4, 2006
5. Dec 5, 2006

### Hurkyl

Staff Emeritus
You mean

$$\forall \; \phi\in D(A)\quad \lim_{n \rightarrow +\infty} \langle f_{n}\vert \phi \rangle=\langle T \vert \phi \rangle$$

don't you? Isn't that guaranteed by the topology you're using?

I think the easiest way to show $\langle T$ nonzero is to actually find a ket satisfying $\langle T | x \rangle \neq 0$. If $\langle T$ really were an eigenbra, then how would you do it?

(I don't know that you can do either. This might be the right answer, but my main point is that something in this spirit is going to be how to do this problem)

P.S. Are you assuming A to be Hermetian?

6. Dec 5, 2006

### poliopeti

Well, after some reading it turns out that the result is not trivial at all. Basically my proposition I wanted to show is called Gelfand-Maurin theorem, or Nuclear Spectral Theorem, and it states that for every number in the spectrum of A there is a generalised eigenvector, ie a distributional solution of the eigenvalue equation (of course for the point spectrum, the proper eigenvectors are also generalised eigenvectors). For further reading see

R. de la Madrid et al., Fortschr. Phys. 50, 185, (2002), quant-ph/0109154

and references therein.

So basically it is many years' work done by many people, no wonder I couldn't cope in 3 days...

Cheers,

Peter

ps.: the proof there does not use at all the sequence I've mentioned before !

Last edited: Dec 5, 2006
7. Dec 6, 2006

### dextercioby

You can also search for Rafael de la Madrid's PhD thesis which is available online. It provides a wealth of information.

Daniel.