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If $A$ is self-adjoint (not necessarily bounded), then is $\langle \psi |A \psi \rangle \in \sigma(A)$ where $\psi$ is a normalized vector in the domain of $A$? This seems to be true, but I can't think of a relatively easy way to prove it.

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  • $\begingroup$ What is $\psi$? In any case, I don't think this is true for arbitrary $\psi$ in your Hilbert space, even with very nice operators $A$. In fact, perhaps you want to ask a more specific or constrained version of the question? $\endgroup$ Jun 18, 2020 at 19:05
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    $\begingroup$ Why should it be true? For matrices this would imply that $\langle \psi | A \psi \rangle $ has only finitely many different values. $\endgroup$ Jun 18, 2020 at 19:06
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    $\begingroup$ You may be thinking of the Courant min/max principle (en.wikipedia.org/wiki/Courant_minimax_principle), which relates eigenvalues to mins/maxes of expressions like the one you're thinking of. $\endgroup$
    – Chris
    Jun 18, 2020 at 19:19
  • $\begingroup$ Indeed, as @Chris comments, for bounded, self-adjoint $A$, the sup of such values over $|\psi|\le 1$ is $\pm$ the largest eigenvalue of $A$. This is also sometimes known as Rayleigh-Ritz theorem/principle, especially in finite dimensions. $\endgroup$ Jun 18, 2020 at 19:53
  • $\begingroup$ $\langle A\psi,\psi\rangle$ is in the convex hull of the spectrum of $A$ if $\psi$ is a unit vector. That's all you can say. $\endgroup$ Jun 20, 2020 at 1:09

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If $A=A^*$ has unit eigenvectors $\psi_1$,$\psi_2$ with corresponding eigenvalues $\lambda_1\ne\lambda_2$, then $\psi_1\perp\psi_2$ and the following is a unit vector $$ \psi(t) = t\psi_1+\sqrt{1-t^2}\psi_2, \;\;\; 0 \le t \le 1. $$ Furthermore, $$ \langle A\psi(dt),\psi(t)\rangle=\langle \lambda_1t\psi_1+\lambda_2\sqrt{1-t^2}\psi_2,t\psi_1+\sqrt{1-t^2}\psi_2\rangle \\ = \lambda_1t^2+\lambda_2(1-t^2), \;\; 0 \le t \le 1. $$ It follows that $\langle A\psi,\psi\rangle$ includes every value in $[\lambda_1,\lambda_2]$, even when there are no elements of the spectrum between $\lambda_1$ and $\lambda_2$.

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