# If $A$ is self-adjoint, then is $\langle \psi |A \psi \rangle \in \sigma(A)$

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.

• 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? Commented Jun 18, 2020 at 19:05
• Why should it be true? For matrices this would imply that $\langle \psi | A \psi \rangle$ has only finitely many different values. Commented Jun 18, 2020 at 19:06
• 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. Commented Jun 18, 2020 at 19:19
• 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. Commented Jun 18, 2020 at 19:53
• $\langle A\psi,\psi\rangle$ is in the convex hull of the spectrum of $A$ if $\psi$ is a unit vector. That's all you can say. Commented Jun 20, 2020 at 1:09

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 $$\lambda_2$$.