Consider the Hilbert space $\mathscr{H} = L^{2}(\mathbb{R}^{d})$ and a Hamiltonian: $$H = -\frac{\hbar^{2}}{2m}\Delta + V(x)$$ for some potential function $V$. States of well-defined energy $E$ are obtained by means of the Schrödinger equation: $$H\psi(x) = -\frac{\hbar^{2}}{2m}\Delta\psi(x) + V(x)\psi(x) = E\psi(x). \tag{1}\label{1}$$ With some boundary condition on the function $\psi(x)$. In the physics literature, we often see the following characterization:
- If $E \le \lim_{|x|\to \infty}V(x)$, the solution of (\ref{1}) is called a bound state and the associate eigenvalues that admit nonzero solutions form a discrete set.
- If $E > \lim_{|x|\to \infty}V(x)$, the solution of (\ref{1}) is called a scattering state and the associate eigenvalues that admit nonzero solutions form a continuous set.
Now, define the discrete spectrum of $H$ by: $$\sigma_{d}(H) := \{\lambda \in \mathbb{R}: \mbox{$\lambda$ is an eigenvalue of $H$ with finite multiplicity}\}$$ and its essential spectrum by $\sigma_{ess}(H) := \sigma(H)\setminus \sigma_{d}(H)$, with $\sigma(H)$ denoting spectrum of $H$. It seems to me that the above two conditions can be translated as:
- A state $\psi \in L^{2}(\mathbb{R}^{d})$ is a bound state iff $H\psi(x) = E\psi(x)$, with $E \in \sigma_{d}(H)$.
- A state $\psi \in L^{2}(\mathbb{R}^{d})$ for $H$ is a scattering state iff $H\psi(x) = E\psi(x)$ with $E \in \sigma_{ess}(H)$.
Hence, it seems that the study of the discrete and essential spectrum of $H$ completely characterizes its bound and scattering states.
On the other hand, we can also define the pure point spectrum $\sigma_{pp}(H)$, the absolutely continuous spectrum $\sigma_{ac}(H)$ and the singular spectrum $\sigma_{s}(H)$ of $H$ by means of the Lebesgue Decomposition Theorem.
In mathematics books, one is usually interested in the characterization of these three spectrum. So, my question is: if the above analysis is correct, what is the point of analyzing these three other spectrum, if all relevant physical information seems to be concentrated on the discrete and essential spectrum? I mean, if I prove that $\sigma_{ac}(H)\cup \sigma_{s}(H) = \emptyset$, then the spectrum of $H$ consists only of eigenvalues and the study of the Schrödinger equation tell us all we need to know about the spectrum of $H$. That's fine to me. But otherwise, what is the physical relevance of $\sigma_{ac}(H)$ and $\sigma_{s}(H)$ when this is not the case?
