# 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$) would be better. • I've now modified my original answer to fit with your modified question. – DisintegratingByParts Jul 9 '15 at 19:11 ## 2 Answers Take any orthonormal basis$\{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$L^{2}$. • 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. – SepulzioNori Jul 9 '15 at 18:59 • @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. – DisintegratingByParts Jul 9 '15 at 19:02 • 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. – SepulzioNori Jul 9 '15 at 21:20 • 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. – DisintegratingByParts Jul 9 '15 at 22:40 • 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? – DisintegratingByParts Jul 9 '15 at 22:46 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 • Thank You. I do not understand the construction, can You please be more specific? For example, what are$1_{A}(\lambda)$and$\delta_{\lambda}\$? – SepulzioNori Jul 10 '15 at 8:08
• You're welcome. :) – C-Star-W-Star Jul 10 '15 at 13:23
• I included another link: There the construction is given in full detail. – C-Star-W-Star Jul 10 '15 at 13:23
• They are both the characteristic function. (I shortened it though.) – C-Star-W-Star Jul 10 '15 at 13:24