3

A standard case study for the continuous spectrum of linear operators in $L^2(\mathbb{R})$ involves multiplication and differentiation operators, as mentioned e.g. in

among many other places.

A feature that seems common in these examples is that while these "continuous eigenvalues" are not eigenvalues, meaning that they do not correspond to eigenvectors belonging to the Hilbert space where the operator is defined, they can still be kinda understood as generalised functions in said space.

Namely, for the multiplication operator $\hat x$ in $L^2(\mathbb{R})$, we could consider the delta functions to be the eigenvectors of the operator, on account of the fact that $$\hat x \delta(x-x_0)=x_0 \delta(x-x_0)$$ makes perfect sense in the space of distributions. Similar reasoning works for $i\frac{d}{dt}$, being the functions $e^{ikx}$ perfectly defined as distributions (with suitable test space).

Is there any generality to these facts, or are they specific to position and momentum operators in $L^2$ spaces? In other words, is there any general way to understand continuous eigenvalues as proper eigenvalues upon suitably enlarging the space on which the linear operator is defined to act? My basic reasoning for thinking this might be possible is that continuous eigenvalues $\lambda\in\sigma_c(T)$, being elements of the approximate point spectrum, satisfy $\|(T-\lambda I)f_n\|\to0$ for some sequence $(f_n)$. Of course, this sequence doesn't generally converge to an element, but if there were a way to meaningfully interpret $\lim_n f_n$ as an element of some enlarged space, then it would seem to make sense to call $\lim_n f_n$ an eigenvector of $T$.

Admittedly, it's less obvious how to do so in other cases. For example, the multiplication operator $T:\ell^2(\mathbb{N})\to\ell^2(\mathbb{N})$, $Tx\equiv (x_n/n)_{n\in\mathbb{N} }$, has $0\in\sigma_c(T)$, but I'm not sure how I'd interpret $\lim_n e_n$ as some kind of "distribution". Though perhaps a "distribution" in this example would simply mean a zero functional in $(\ell^2(\mathbb{N}))^*$? I'm not sure how to make the analogy of multiplying a function by a distribution in this case, but at the same time there might be a general enough way to do it.

glS
  • 7,963
  • You are basically describing a Rigged Hilbert Space, which generalizes eigenvectors to be dual elements of a dense subspace. I suspect that some reading on this topic will answer your question. – whpowell96 Sep 10 '24 at 17:57
  • @whpowell96 I've heard of those, but my (extremely naive) understanding of the subject is that that's a way to do this in the particular case of $L^2$ operators. At least, most places I've seen rigged Hilbert spaces discussed deal with (subspaces of) $L^2(\mathbb{R}^n)$ specifically. Here I'm wondering if there isn't a way to make sense of something similar more generally. After all, distributions are nothing but functionals on a dense subspace, as you say, so could something like that work, for example, for operators in $\ell^2$? – glS Sep 10 '24 at 18:43
  • 1
    You are correct that they originated for $L^2$ spaces but they can be defined on any Hilbert space and dense subspace – whpowell96 Sep 10 '24 at 18:45
  • @whpowell96 I'll look into it, thanks. Although, is there a reason why Hilbert spaces specifically? Is there a clear reason why it should break for continuous eigenvalues of linear operators in a Banach space instead? Also, do you perhaps have any reference to suggest? I'm only finding sources on RHSs discussing specifically $L^2(\mathbb{R}^n)$ for quantum mechanics – glS Sep 10 '24 at 18:56
  • 2
    To avoid being weighted too much by the quantum mechanics context, it might be worth googling based on the Gelfand triple nomenclature instead of rigged Hilbert spaces. – Semiclassical Sep 10 '24 at 20:12
  • Chapter 2 of Quantum Mechanics in Rigged Hilbert Space Language by de Madrid and the included references may be a good place to start. It of course has a focus on $L^2$ spaces because it is QM but it states many of the theorems in general and includes references to more detailed sources – whpowell96 Sep 10 '24 at 21:16
  • Intuitively you use the boundary conditions to figure out which generalized formal eigenvectors corresponds to points in the spectrum. – Steen82 Sep 11 '24 at 12:06
  • @Steen82 could you elaborate on that? I'm not sure what you mean. How would you apply that to a simple example, say continuous eigenvalues for shift or multiplication operators (with continuous non-point eigenvalues) in $\ell^p$? – glS Sep 11 '24 at 12:14

0 Answers0