Discreteness and gap in spectrum - is it the same?

Question

Discreteness is attributed to discontinuous evolution of eigenvalues on the spectrum of an operator. A spectral gap, is the distance between two disjoint points on the spectrum. However, as this gap can describe an interval where there are no solutions, it does not guarantee that the following eigenvalues to the right are not continuous and not discrete. Is the existence of a gap therefore sufficient for guaranteeing discreteness of the spectrum?

Answer 1

If $\mu$ is a finite, compactly-supported Lebesgue measure on $\mathbb{R}$, then the multiplication operator $M : L^2_{\mu}(\mathbb{R})\rightarrow L^2_{\mu}(\mathbb{R})$ defined by $$ (Mf)(x)=xf(x) $$ is bounded and has spectrum $\sigma(M)$ equal to the support of the measure $\mu$. Every atom of $\mu$ is an eigenvalue of $M$ of multiplicity $1$. The support of the absolutely continuous part of $\mu$ is the absolutely continuous spectrum of $M$. Etc. So it is not difficult to construct self-adjoint operators with specific types of spectra in specific places on the real line. A gap in the spectrum does not have to correspond to discrete spectrum; a gap can occur without any discrete spectrum.

Answer 2

The existence of a spectral gap does not imply that the spectrum is discrete. Example from physics: Most Hamiltonians in quantum field theory have a spectral gap but their spectrum is not discrete, e.g. the spectrum of the free Dirac Hamiltonian of mass $m>0$ is $(-\infty,-m] \cup [m,\infty)$.