Intuition behind spectrum of an operator

Question

I am trying to understand better the concept of the spectrum of an operator, and it gives me some troubles, I would like to know if there is an intuitive idea behind this, for example, a part of the spectrum are the eigenvalues (I think this is the point spectrum), and they have a very strong "geometric" interpretation.

But, what happens with the elements in the continuous spectrum or the residual spectrum? And we can also define something like the spectral radius ($r_\sigma (T)=\sup \{|\lambda|\colon \lambda \in \sigma (T) \}$), all this stuff can be somehow interpreted? Or how should I approach them to understand them in a better way?

Also I am not sure how to "think" the spectrum when I see linear transformations related to matrices. If you can help me to see this, it would be very helpful.

Thanks.

Answer 1

This is not a complete answer, but since you are looking for intuition, it may be helpful.

Some background: Historically, the spectrum originated in quantum mechanics, and in particular in the energy states of atoms. In this context, there is an operator $H$ (typically on a space like $L^2(\mathbb{R})$) whose spectrum describes the possible states of the atom. There are two kinds of states: "bound states" are those where all the particles of the atom stay close together (they are "bound"), whereas "scattering states" are those where the particles fly apart. The only states which can stay constant in time are bound states; the scattering states necessarily entail changes to the atom in time.

Mathematically, the bound states correspond to eigenvalues of $H$ and the scattering states correspond to the rest of the spectrum. The bound states have corresponding eigenvectors in $L^2$, which typically look something like $e^{-kx^2}$, and the atom is allowed to be in such a state. On the other hand, the scattering states have pseudo-eigenvectors which are not in $L^2$ and typically look something like $e^{ikx}$. Though the scattering states do not live in $L^2$, it is possible to form combinations of them which do live in $L^2$, e.g. by a Fourier transform. So the atom cannot live in a single scattering state, but it can live in a combination of them, which can be arbitrarily close (in a suitable sense) to a pure scattering state.

Here are some concrete examples:

• Consider the "multiply by x" operator $\hat{x}$ on $L^2(\mathbb{R})$. For any $f\in L^2(\mathbb{R})$ and any $x_0\in\mathbb{R}$ we have $\hat{x}f (x_0) = x_0 f(x_0)$. This operator has no eigenvalues, but its spectrum is the entire real line. Intuitively, a delta function $\delta(x-x_0)$ is a pseudo-eigenvector to $\hat{x}$ with pseudo-eigenvalue $x_0$, and morally we have $\hat{x} \delta(x-x_0) = x_0 \delta(x-x_0)$.
• Similarly, the Laplacian $\nabla^2$ has pseudo-eigenvectors $e^{ikx}$ for any $k\in \mathbb{R}$, corresponding to pseudo-eigenvalues $-k^2$, because $\nabla^2 e^{ikx} = -k^2 e^{ikx}$. So the entire non-positive real line is in the spectrum of $\nabla^2$.

In both of these examples, the problem is that the "eigenvectors" are not in the domain of the operator--they live in some generalized space (either distributions in the first case, or non-square-integrable functions in the second). I'm not sure if this is always the case, but it is certainly typical that the part of the spectrum of an operator which is not eigenvalues corresponds to generalized eigenvectors which live in some larger space.

So summarily, my intuition for the part of the spectrum which is not proper eigenvalues is that it corresponds to some pseudo-eigenvector which does not properly live in the domain of your operator, but is still somehow "there" as some kind of complicated limiting object, e.g. a distribution or non-normalizable function.