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Suppose $A$ is an unbounded self-adjoint operator in a Hilbert space $H$ with discrete spectrum $$\lambda_0 < \lambda_1 < \cdots$$ bounded below with lowest eigenvalue $\lambda_0$, lowest eigenstate $\psi_0$, and spectral gap $\lambda_1 - \lambda_0 > 0$.

Let $B$ be a bounded self-adjoint operator in $H$ with absolutely continuous spectrum.

Are there any conditions on $A, B$ which imply that the spectral measure $d\mu$ of $O=A+B$ at $\psi_0$ is bounded, i.e. $d\mu$ compact support?

Recall that the spectral measure of a self-adjoint operator $O$ in a Hilbert space $(H, \langle \cdot, \cdot \rangle)$ at a normalized state $\psi \in H$, $||\psi||=1$ is the probability measure $d\mu$ on $\mathbb{R}$ determined by $O$ and $\psi$ implicitly by

$$\int_{- \infty}^{+\infty} e^{\textbf{i} t \lambda} d \mu(\lambda) = \langle \psi , e^{\textbf{i} t O} \psi \rangle $$ for all $t \in \mathbb{R}$.

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  • $\begingroup$ "the spectral measure $d\mu$ of $O=A+B$ at $\psi_0$ is bounded, i.e. $d\mu$ compact support" Do you mean that $O$ does not have a (discrete) lowest eigenvector? $\endgroup$ – Keith McClary Dec 11 '18 at 4:01
  • $\begingroup$ No. My question very well could apply to O=A+B with discrete spectrum and lowest eigenvalue with a gap. Suppose this case applies, so O=A+B has discrete spectrum. A priori the spectral measure of O at $psi_0$ is supported on the discrete spectrum of O which is usually infinite. Then in this case I’m asking what assumptions guarantee this support is on finitely many eigenvalues of O. $\endgroup$ – Swallow Tail Dec 11 '18 at 19:21
  • $\begingroup$ I think I understand the question. I can't think of an example. Do you have any? $\endgroup$ – Keith McClary Dec 13 '18 at 17:59

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