# Conditions for Boundedness of Spectral Measures of Perturbations of Self-Adjoint Operators?

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 $$t \in \mathbb{R}$$.

• "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? – Keith McClary Dec 11 '18 at 4:01
• 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. – Swallow Tail Dec 11 '18 at 19:21
• I think I understand the question. I can't think of an example. Do you have any? – Keith McClary Dec 13 '18 at 17:59