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.


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.

  • $\begingroup$ I've now modified my original answer to fit with your modified question. $\endgroup$ – COVID-20 Jul 9 '15 at 19:11

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 $L^{2}$.

| cite | improve this answer | |
  • $\begingroup$ 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. $\endgroup$ – Ittiolo 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. $\endgroup$ – COVID-20 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. $\endgroup$ – Ittiolo 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. $\endgroup$ – COVID-20 Jul 9 '15 at 22:40
  • 1
    $\begingroup$ 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? $\endgroup$ – COVID-20 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

| cite | improve this answer | |
  • $\begingroup$ Thank You. I do not understand the construction, can You please be more specific? For example, what are $1_{A}(\lambda)$ and $\delta_{\lambda}$? $\endgroup$ – Ittiolo 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

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.