3
$\begingroup$

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.

$\endgroup$
2
$\begingroup$

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.

| cite | improve this answer | |
$\endgroup$
  • $\begingroup$ 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

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.