This is basically the same question I made on physics stack exchange The spectrum of the Hamiltonian in quantum mechanics, but I got no answers so far and decided to move it to mathoverflow with some minor modifications.
The original question goes like this. 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 $\psi$ is a scattering state for $H$ 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?
Add: I want to add just one small point to the original question. Wikipedia says that absolutely continuous spectrum physically corresponds to free states whilst pure point spectrum corresponds to bound states. This would partly answer my question, but it does not seem to be in agreement with the above discussion often found in physics textbooks, since a scattering or free state is also an eigenvector of $H$, so its associated eigenvectors should be in $\sigma_{pp}(H)$ as far as I understand. In addition, even if this was the case and I am just misunderstanding something, I still do not understand why one needs these two kinds of characterization of the spectrum, since the $\sigma_{d}(H)$ and $\sigma_{ess}(H)$ seem enough to address the question of whether a state is bound or not, and how states evolve in time.
