# Continuous spectrum and weak solution of eigenvalue equation

• poliopeti
In summary, on a densely defined closed operator A on the Hilbert space H, one can construct a topological dual D*(A) with respect to a finer topology than H. Approximate eigenvalues of A can be found if z is in the continuous spectrum of A.

#### 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

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

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

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

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

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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?

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 !

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

Daniel.

## 1. What is a continuous spectrum?

A continuous spectrum is a set of eigenvalues in an eigenvalue equation that forms a continuous interval on the real number line. This means that the eigenvalues can take on any value within this interval, rather than being discrete or distinct values.

## 2. How is a continuous spectrum different from a discrete spectrum?

In a discrete spectrum, the eigenvalues are distinct and can only take on specific values. This results in a set of isolated points on the real number line, rather than a continuous interval.

## 3. What is a weak solution of an eigenvalue equation?

A weak solution is a solution to an eigenvalue equation that does not satisfy the necessary conditions for a classical solution. This means that the solution may not be continuous or differentiable, but still satisfies the eigenvalue equation in a generalized sense.

## 4. How do continuous spectrum and weak solutions relate to each other?

Weak solutions can arise from eigenvalue equations with a continuous spectrum, as the solutions may not be classical but still satisfy the equation. In general, continuous spectrum and weak solutions are more common in non-self-adjoint or non-compact operators.

## 5. What are some applications of studying continuous spectrum and weak solutions?

Continuous spectrum and weak solutions have applications in many areas of mathematics and physics, such as quantum mechanics, differential equations, and functional analysis. They also have practical applications in engineering, particularly in the study of vibrations and oscillations.