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The time-evolution in quantum mechanics is given by Schrödinger's equation. For time-independent Hamiltonians, one searches for solutions of the problem $\hat{H}\psi = E\psi$.

The physicist point of view

Physicists solve $\hat{H}\psi = E\psi$, which is an eigenvalue problem, to obtain a basis of eigenvectors of $\hat{H}$. Usually, things are easier when this equation has only a discrete set of eigenvalues. Sometimes, however, this equation leads to solutions that do not belong to $\mathscr{H} = L^{2}(\mathbb{R}^{d})$, a problem associated to the existence of a continuous spectrum. These solutions are not square integrable, so they cannot be part of an eigenbasis, but it is usually said that they are still part of a complete orthonormal in a more general sense. Here, I quote Griffiths:

If the spectrum of a hermitian operator is continuous, the eigenfunctions are not normalizable, they are not in Hilbert space and they do not represent possible physical states; nevertheless, the eigenfunctions with real eigenvalues are Dirac orthonormalizable and complete (with the sum now an integral). Luckily, this is all we really require.

These solutions with continuous eigenvalues are associated to scattering states, whilst the discrete spectra determines bound states.

The mathematician point of view

Suppose I would like to attack this eigenvalue problem under a rigorous point of view. The set of eigenvalues of a self-adjoint operator $A$ on a Hilbert space $\mathscr{H}$ is defined as $\sigma_{p}(A) = \{\lambda \in \mathbb{R}: \mbox{$(A- \lambda) \psi = 0$ for some $\psi \in \mathscr{H}$}\}$. The complement of the spectrum is called the continuous spectrum, $\sigma_{c}(A) = \sigma(A)\setminus \sigma_{p}(A)$. Hence, the Schrödinger equation is, first of all, a problem of studying $\sigma_{p}(A)$.

If the spectrum of $\hat{H}$ is composed only of $\sigma_{p}(\hat{H})$, then there is a basis of eigenfunctions of $\hat{H}$ and everything is okay. However, when the Schrödinger equation has solutions which do not belong to $L^{2}(\mathbb{R}^{d})$, as before, these solutions must be part of $\sigma_{c}(A)$ instead.

The usage of the spectrum decomposition into $\sigma_{p}(\hat{H})$ and $\sigma_{c}(\mathscr{H})$ is usually limited, and one looks for other decompositions. Other characterizations include the discrete and essential spectrum, defined respectively as $\sigma_{d}(\hat{H}) = \{\lambda \in \mathbb{R}: \mbox{$\lambda$ is an eigenvalue with finite multiplicity}\}$ and $\sigma_{ess} = \sigma(\hat{H})\setminus \sigma_{d}(\hat{H})$. Another possible decomposition is $\sigma_{pp}(\hat{H})$, $\sigma_{ac}(\hat{H})$ and $\sigma_{sing}(\hat{H})$, the pure point, absolutely continuous and singular spectrum, respectively.

Once again, the existence of elements of $\sigma_{c}(\hat{H})$ indicates the existence of bound and scattering states.

Problem 1: It is natural to look for ways of characterizing bound and scattering states in terms of spectrum. However, I have seen different characterizations: some say a bound state is an eigenvector of $\hat{H}$ with eigenvalue belonging to $\sigma_{d}(\hat{H})$ and scattering states are their orthogonal elements. Some say a bound state is an element of the pure point $\mathscr{H}_{pp}$ and a scattering state is an element of $\mathscr{H}_{ac}\oplus \mathscr{H}_{sing}$. Is there any reason for different characterizations? Is there any standard one? Based on the analysis of the physics point of view, it would be natural to adopt the first one where the discrete spectrum determines the bound states. But then, why bother using the second one?

Problem 2: Can anything be rigorously proven about the "continuous spectrum" solutions of the Schrödinger equation as part of the complete orthonormal set, as claimed in physics books? In math books, I usually find analyses of spectra (things like $\sigma_{d}(\hat{H}) = [0,+\infty)$ and so on) but that's all. It is hard to even understand what one wants with each kind of spectrum decomposition: some use $\sigma_{d}(\hat{H})$ and $\sigma_{ess}(\hat{H})$, and some $\sigma_{pp}(\hat{H})$, $\sigma_{ac}(\hat{H})$ and $\sigma_{sing}(\hat{H})$. Most of the time, the discussion stops there. But, at least for practical purposes, isn't the idea of characterizing the spectrum to find a basis of eigenfunctions and/or obtaining bound and scattering states of the system under consideration?

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  • The Hamiltonian could be a complicated thing. If you are only interested in studying the continuous spectrum, it may be helpful to start by considering a simpler case. An example is the quadrature operators. They have only continuous spectra. – flippiefanus Apr 10 '23 at 04:02
  • Perhaps related: https://physics.stackexchange.com/questions/756599/what-are-the-exact-relations-between-bound-states-discrete-spectra-and-negativ – Apoorv Potnis Apr 10 '23 at 05:41
  • With no disrespect to Herr Schroedinger and a host of mathematically oriented physicists who are taking his approximation too seriously: the discrete spectrum predicted by this equation does not exist in nature. What we are actually observing when we are doing spectroscopy is a scattering problem (photons scattering on atoms) and there are no real eigenvalues in that problem at all. All spectral lines have a finite line width. What creates the discrete spectra are the mathematical choices we make to avoid having to solve the real problem. – FlatterMann Apr 10 '23 at 07:49
  • Please don't ask multiple questions at once, it makes organizing questions and answers hard. In this case, your problem 1 is a duplicate of https://physics.stackexchange.com/q/282574/50583, and Martin's answer there is excellent, discussing the difficulty in nailing down what a "bound" or "scattering state is - essentially if you insist on this dichotomy, you always get either bound states that don't really look bound or scattering states that don't really look like they're scattering. – ACuriousMind Apr 10 '23 at 10:57
  • And your problem 2 - about the existence of the "continuous eigenvectors" $\lvert x\rangle$ - is closed related to rigged Hilbert spaces (https://physics.stackexchange.com/q/43515/50583) and/or the formulation of the spectral theorem in terms of projection-valued measures, see also this Wiki article. – ACuriousMind Apr 10 '23 at 11:00

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