Example of self-adjoint linear operator with pure point spectrum on an infinite-dimensional separable Hilbert space Let $\mathcal{H}$ be an infinite-dimensional, complex, separable Hilbert space.
Besides the well-known one-dimensional Harmonic oscillator on $\mathcal{H}=(\mathcal{L}^{2}(\mathbb{R})\,;d\mathit{l})$, does anyone know some explicit examples (domain, eigenvalues and eigenvectors) of a possibly unbounded self-adjoint linear operator having only a pure-point spectrum with no degenerancy?
References are appreciated.
Thanks in advance.
ADDENDUM
In view of TrialAndError's reply, I would like to add that an example on a functional space (e.g., $\mathcal{L}^{2}(M)$ with $M$ a measure space$) would be better.
 A: Take any orthonormal basis $\{e_{n} \}_{n=1}^{\infty}$ of a Hilbert space $H$ and define $Lf = \sum_{n=1}^{\infty}n(f,e_{n})e_{n}$ on the domain $\mathcal{D}(L)$ consisting of all $f\in H$ for which $\sum_{n}n^{2}|(f,e_n)|^{2} < \infty$. The operator $L : \mathcal{D}(L)\subset H \rightarrow H$ is a densely-defined selfadjoint linear operator that has simple eigenvalues at $n=1,2,3,4,\cdots$.
There are lots of orthonormal bases for $L^{2}$. Start with any countable dense subset and perform Gram-Schmidt on the dense subset, discarding dependent elements along the way. Then choose any sequence $\{ a_{n} \}$ of distinct real numbers for which $|a_{n}|\rightarrow \infty$. Define
$$
                  Lf = \sum_{n=1}^{\infty} a_n (f,e_n)e_n.
$$
Then $L$ is selfadjoint with spectrum consisting only of simple eigenvalues $a_n$. This works whenever $L^{2}$ is separable. If it is not separable, then $L$ cannot exist because the eigenvectors of $L$ as you have described must be a complete orthogonal basis of $L^{2}$.
A: Canonical spaces...
Given $\Lambda\subseteq\mathbb{C}$ and $\ell^{2}(\overline{\Lambda})$.
Denote for readability:
$$A\in\mathcal{B}(\mathbb{C}):\quad1_A(\lambda):=\chi_A(\lambda)$$
Construct normal operator:*
$$E(A)\varphi:=1_{\overline{\Delta}\cap A}\varphi:\quad N:=\int\lambda\mathrm{d}E(\lambda)$$
Then it has spectrum:**
$$\sigma(N)=\sigma_0(N)=\overline{\Lambda}\subseteq\mathbb{C}$$
Note that the set was arbitrary!!
*See the thread: Constructions
**See the thread: Special Spectrum
