Let $\mathbf{A}$ be a possibly-unbounded, linear self-adjoint operator on an infinte-dimensional, complex separable Hilbert space $\mathcal{H}$, and suppose we know the matrix elements $\langle j|\mathbf{A}|k\rangle$ of $\mathbf{A}$ on a basis $\{|n\rangle\}_{n\in\mathbb{N}}$.

Are there theorems helping in the explicit calculation of the spectrum of $\mathbf{A}$ using the aforementioned matrix elements? If not, are there theorems giving necessary and/or sufficient conditions for the existence of the continuous spectrum of $\mathbf{A}$?

Thanks in andvance.


Suppose $\mathcal{H}=L^{2}[0,\pi]$. Suppose $\mathcal{M}$ is a countably dense subset of compactly supported $C^{\infty}$ functions on $(0,\pi)$. Apply Gram-Schmidt in order to obtain an orthonormal basis $\{ f_{k} \}_{k=1}^{\infty}$ of smooth compactly supported functions on $(0,\pi)$. The following operators $L_{\alpha,\beta}$ are selfadjoint $$ L_{\alpha,\beta} = -\frac{d^{2}}{dx^{2}}, $$ where $\mathcal{D}(L_{\alpha,\beta})$ consists of all twice absolutely continuous functions $f\in\mathcal{H}$ for which $f''\in\mathcal{H}$ and $$ \cos\alpha f(0)+\sin\alpha f'(0) = 0,\\ \cos\beta f(\pi)+\sin\beta f'(\pi)= 0. $$ (Here $\alpha,\beta$ are real angles.) All of these operators are selfadjoint, and they all contain $\{ f_{k} \}$ in their domains. And they all agree on these elements, i.e., $L_{\alpha,\beta}f_{k}=L_{\alpha',\beta'}f_{k}$. These operators have discrete spectrum, and the spectrum is very different as you allow $\alpha$, $\beta$ to vary. For example, the conditions $$ f(0) = 0,\;\; f(\pi)=0 $$ lead to eigenfunctions $\sin(nx)$ with spectrum $\{ 1,2^{2},3^{2},\cdots\}$. The conditions $$ f'(0)=0,\;\; f(\pi)=0 $$ lead to eigenfunctions $\cos((n+1/2)x)$ with spectrum $\{ (1/2)^{2},(3/2)^{2},(5/2)^{2},\cdots\}$. Every point of $[0,\infty)$ is in the spectrum of one of the $L_{\alpha,\beta}$.

I'm not aware of anything that will help much with the continuous spectrum.

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  • $\begingroup$ Thank You for the reply. However, I do not understand how it is related to my question. Could You please help me in understandig it? $\endgroup$ – Ittiolo Jun 29 '15 at 8:28
  • $\begingroup$ @Ilcapitano : I've given you a family of operators $\{ L_{\alpha,\beta} \}$ for which $(L_{\alpha,\beta}f_j,f_k)=(L_{\alpha',\beta'}f_j,f_k)$ for all $k,j$ and all $\alpha,\alpha',\beta,\beta'$, where $\{ f_k\}$ is an orthonormal basis. And the eigenvalues of $L_{\alpha,\beta}$ vary drastically with $\alpha$,$\beta$. So you cannot determine the spectrum from the matrix $m_{j,k}=(Lf_j,f_k)$. There is a fundamental problem in trying to determine the characteristics of an unbounded selfadjoint operator from such a matrix because the operator is not unquely determined by the matrix. $\endgroup$ – COVID-20 Jun 29 '15 at 15:03
  • $\begingroup$ Ok, It is clear to me now, Thank You again. $\endgroup$ – Ittiolo Jun 29 '15 at 16:30
  • $\begingroup$ @Ilcapitano : You're welcome. This is a strange pathology that was discovered by John von Neumann, and is one of the main reasons that matrices were abandoned for general operators of Quantum. $\endgroup$ – COVID-20 Jun 29 '15 at 16:37

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