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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.

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  • $\begingroup$ I've now modified my original answer to fit with your modified question. $\endgroup$ – DisintegratingByParts Jul 9 '15 at 19:11
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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 $L^{2}$.

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  • $\begingroup$ Thank You, but when I wrote: "Besides the well-known one-dimensional harmonic oscillator..." I should have added: "...and every unbounded linear self-adjoint operator with the same spectrum", which is the case you mentioned. My bad. $\endgroup$ – SepulzioNori Jul 9 '15 at 18:59
  • $\begingroup$ @Ilcapitano : The spectrum of the harmonic oscillator would be $n+\frac{1}{2}$ or some multiple of that. So they're not the same. But, you can choose any sequence of real numbers $a_1 < a_2 < a_3 < \cdots \rightarrow \infty$ and define $Lf = \sum_{n=1}^{\infty}a_n(f,e_n)e_n$. The same remarks apply. $\endgroup$ – DisintegratingByParts Jul 9 '15 at 19:02
  • $\begingroup$ Thank You. However, I was thinking of something like the operator $-\imath\frac{d}{dx}$ on $(\mathcal{L}([a\,;b])$ or the laplacian on a compact manifold. Specifically, I would like an operator the definition of which is not given by means of its action on a basis. $\endgroup$ – SepulzioNori Jul 9 '15 at 21:20
  • $\begingroup$ Every such selfadjoint operator as you describe reduces to the case I listed. That's part of the Spectral Theorem. Also, you originally wanted $L^{2}(\mathbb{R})$. The only selfadjoint version of $-i\frac{d}{dx}$ on a finite interval consists of functions with a periodic type of condition. However $-\frac{d^{2}}{dx^{2}}$ can be specified with endpoint conditions $\cos\alpha f(a)+\sin\alpha f'(a)=0$ and $\cos\beta f(b)+\sin\beta f'(b)=0$, and you can get isolated simple point spectrum that isn't evenly spaced. The eigenvalues can end up as solutions of a transcendtal $\tan x=Cx$ type equation. $\endgroup$ – DisintegratingByParts Jul 9 '15 at 22:40
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    $\begingroup$ The ordinary Legendre equation has eigenvalues $n(n+1)$. Even $-\frac{d^{2}}{dx^{2}}$ has eigenvalues $n^{2}$ when you use periodic conditions. Bessel's equation can give you eigenvalues related to the zeros of the Besell functions. Do any of these classical cases work for you? $\endgroup$ – DisintegratingByParts Jul 9 '15 at 22:46
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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

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  • $\begingroup$ Thank You. I do not understand the construction, can You please be more specific? For example, what are $1_{A}(\lambda)$ and $\delta_{\lambda}$? $\endgroup$ – SepulzioNori Jul 10 '15 at 8:08
  • $\begingroup$ You're welcome. :) $\endgroup$ – C-Star-W-Star Jul 10 '15 at 13:23
  • $\begingroup$ I included another link: There the construction is given in full detail. $\endgroup$ – C-Star-W-Star Jul 10 '15 at 13:23
  • $\begingroup$ They are both the characteristic function. (I shortened it though.) $\endgroup$ – C-Star-W-Star Jul 10 '15 at 13:24

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