$\renewcommand{\Re}{\operatorname{Re}}$ $\renewcommand{\Im}{\operatorname{Im}}$ $\newcommand{\erf}{\operatorname{erf}}$ $\newcommand{\dag}{\dagger}$ $\newcommand{\const}{\mathrm{const}}$
By methods of Real Analysis one can costruct all types of spectra. However, it is much more interesting to see which kinds of spectra can appear "naturally'', for operators arising in applicationsm rather than artificial constructs.
Example 1: Dirichlet Laplacian
Example 2: Neumann Laplacian
Example 3: Laplacian on the sphere
Example 4: Schrödinger operator with growing potential
Example 5: Free particle Schrödinger operator
$L=-\frac{1}{2}\Delta$ in $\mathbb{R}^d$ has continuous spectrum $[0,+\infty)$.
Example 6: Schrödinger operator with decaying potential
Operator (\ref{eqn-13.5.1}) with potential $V(x)\to 0$ as $|x|\to \infty$ has a continuous spectrum $[0,+\infty)$ but it can have a finite or infinite number of negative eigenvalues $E_n<0$ of finite multiplicity; these eigenvalues can accummulate only to $-0$.
Example 7: Schrödinger operator with a constant magnetic field
Example 8: Schrödinger operator with a constant magnetic field. II
Example 9: Free particle Dirac operator
Let
\begin{equation}
L=\sum _{j=1}^3 \gamma^j (-i\partial_{x_j}) + \gamma^0 m, \qquad m>0
\label{eqn-13.5.5}
\end{equation}
(where $\gamma^j$ are Dirac matrices) has a continuous spectrum $(-\infty,-m]\cup [m,\infty)$.
Perturbing it by a potential $V(x)$, $V(x)\to 0$ as $|x|\to \infty$ \begin{equation} L=\sum _{j=1}^3 \gamma^j (-i\partial_{x_j}) + m\gamma^0 +V(x) I, \qquad m>0 \label{eqn-13.5.6} \end{equation} can add a finite or infinite number of eigenvalues in spectral gap $(-m,m)$. They can accumulate only to the borders of the spectral gap.
Remark 1
We can consider Dirac operator in 2D and 3D with a constant magneric field.
In 2D and with $m=0$ we need only three matrices, not four, and insted of Dirac matrices we can take Pauli matrices.
Example 10: Periodic Schrödinger operator
Consider Schröodinger operator (\ref{eqn-13.5.1}) with periodic potential in $\mathbb{R}^d$: $V(x+\boldsymbol{a})=V(x)$ for all $\boldsymbol{a}\in \Gamma$ where $\Gamma$ is a lattice of periods, see Definition 4.B.1) Then $L$ has a band spectrum.
Namely on the elementary cell (see Definition 4.B.3) $\Omega$ consider operator $L(\boldsymbol{k})$ where $\boldsymbol{k}\in \Omega^*$ is a quasimomentum; $L(\boldsymbol{k})$ is given by the same formula as $L$ but is defined on functions which are quasiperiodic with quasimomentum $\boldsymbol{k}$. Its spectrum is discrete: $\sigma (L(\boldsymbol{k}))=\{E_n (\boldsymbol{k}): n=1,2,\ldots\}$.
Then spectrum $\sigma (L)$ consists of spectral bands \begin{gather} \sigma_n:= [\min _{\boldsymbol{k}\in \Omega^*} E_n(\boldsymbol{k}) ,\max _{k\in \Omega^*} E_n(\boldsymbol{k})],\qquad \sigma(L) =\bigcup_{n=1}^\infty \sigma_n; \label{eqn-13.5.7} \end{gather} these spectral bands can overlap.
On can prove that $E_n (\boldsymbol{k})$ really depend on $\boldsymbol{k}$ and are not taking the same value on some set of non--zero measure (another notion from Real Analysis) which implies that the spectrum $\sigma(L)$ is continuos.
Example 11: Periodic Schrödinger operator. II
In the space $\ell^2(\mathbb{Z})$ (which is the space of sequences $u_n$, $n=\ldots, -2,-1,0, 1,2,\ldots$ such that $\|u\|^2:= \sum_{n=-\infty} ^{\infty}|u_n|^2<\infty$) consider almost Mathieu operator (which appears in the study of quantum Hall effect) \begin{equation} (Lu)_n =u_{n+1}+u_{n-1}+2\lambda \cos (2\pi (\theta +n\alpha)) \label{eqn-13.5.9} \end{equation} with $|\lambda|\le 1$. Assume that $\alpha$ is a Diophantine number (which means it is an irrational number which cannot be approximated well by rational numbers; almost all irrational numbers (including all algebraic like $\sqrt{2}$) are Diophantine).
Then the spectrum $\sigma(L)$ is continuous (no eigenvalues!) but it is singular continuous: for any $\varepsilon>0$ it can be covered by the infinite sequence of segments of the total length $<\varepsilon$. As an example of such set see Cantor set.
This was completely investigated only in the end of the 20-th century.
Remark 2,
Complete analysis of this operator shows the difference between two notions of negligible set:
Example 12. Sobolev equation
Consider Sobolev' equation for rotating liquid \begin{gather} \Delta u_{tt}= \omega^2 u_{zz}\qquad x\in \Omega , \ -\infty< t <\infty \end{gather} in the bounded domain $\Omega$ with Dirichlet boundary condition \begin{gather} u|_{\partial \Omega}=0 \end{gather} with $\omega>0$ which can be rewitten as \begin{gather} u_{tt}=-(-\Delta_D)^{-1} u_{zz} \end{gather} with self-adjoint bounded operator $L:= -(-\Delta_D)^{-1} \partial_z^2$ in $H^1_0(\Omega)=\{u\colon |\nabla u| <\infty,\ u|_{\partial \Omega}=0\}$.
Separating $t$ from spatial variables $(x,y,z)$, $u(x,y,z,t)=e^{i\tau t}v(x,y,z)$ we get \begin{align} &\Delta' v +(1-\tau^{-2}\omega^2)v_{zz}=0,&& \Delta' :=\partial_x^2+\partial_y^2,\label{eqn-13.5.13}\\ &v|_{\partial\Omega}=0. \end{align}
This is a Dirichlet problem in spatial variables only. Equation (\ref{eqn-13.5.13}) is elliptic as $|\tau|>\omega$ and hyperbolic as $|\tau|<\omega$ with the "time'' $z$. In the case $|\tau|\ge \omega$ there is only trivial solution.
Not so as $|\tau|<\omega$. Consider rectangle box $\Omega={(x,y,z) \colon 0<x <\pi /a, 0<y<\pi /b, 0<z<\pi /c}$; then we can separate variables $(x,y,z)$: \begin{gather} v(x,y,z)= \sin (m x/a)\sin (ny/b)\sin (kz/c)\label{eqn-13.5.15}\\ \tau^2 = F(m,n,k):= \frac{ \omega^2 (m^2 a^2 + n^2 b^2)}{m^2a^2+b^2n^2 +c^2k^2}. \label{eqn-13.5.16} \end{gather}
Then spectrum of operator $L$ is $\sigma(L)=\tau^2$ with $\sigma_p (L)$ consisting of eigenvalues of infinite multiplicity given by (\ref{eqn-13.5.16}) with $(m,n,k)\in \mathbb{Z}^3_+$. everywhere dense on this interval and with $\sigma_{ac}(L)=\sigma_{sc}(L)= \emptyset$.
Indeed (\ref{eqn-13.5.15}) form a complete orthogonal system and if $\tau^2 =F(m,n,k)$ then $\tau^2 =F(lm,ln,lk)$ for any $l\in \mathbb{Z}_+$.
Example 13. The same is true for 2D-version of Sibolev equation as $\Omega=\{(x,z) \colon 0<x <\pi /a, 0<z<\pi /c\}$ is a rectangular box.
Example 14. Consider the same 2D problem but in the disk $\Omega =\(x,z)\colon x^2+y^2<1\}$. We claim that the spectrum is still $\tau^2\le \omega^2$ but it is continuous withot any eigenvalues.
To prove this and discuss the difference with the rectangular observe that we get a Dirichlet problem for wave equation \begin{align} &v_{xx}-k^2u_{zz}=0, && k^2:= \omega^2\tau^{-2}-1\\ &v|_{\partial \Omega}=0. \end{align} Then $v(x,z)=\phi (kx +z) +\psi kx-z)$ wth $\phi (.)$ and $\psi (.) $ constant along lines $kx+z =\const$ and $kx-z =\const$ respectively. Consider brocken lines consisting of these segments with ``reflections'' at the boundary.
One can prove that in the disk the spectrum is absolutely continues. \end{enumerate}
Remark. We have seen in different example amazing fact: geometry of the domain is very important for spectum of operator.