Weyl's criterion says that for a self-adjoint operator $H$, $\lambda\in\sigma(H)$ if and only if there exists a sequence $(\phi_n)_{n=1}^{\infty}$ so that $\vert \vert \phi_n \vert \vert=1$ and $$\lim_{n\rightarrow\infty} \vert \vert (H-\lambda)\phi_n \vert \vert =0$$
Moreover, it characterizes the essential spectrum by stating that $\lambda\in \sigma_{ess}(H)$ if and only if the Weyl sequence above can be chosen to be orthogonal.
I was wondering if there is a version of Weyl's criterion which allows one to distinguish the absolutely continuous part of the spectrum within $\sigma_{ess}(H)$. Meaning, can you possibly add an additional requirement on the Weyl sequence (maybe even just a sufficient condition) which ensures that $\lambda\in \sigma_{ac}(H)$?.
My motivation is I have two operators $H_1,H_2$ which are "similar enough" to have "similar" Weyl sequences. Given $\lambda\in\sigma_{ac}(H_1)$, I want to show that $\lambda\in\sigma_{ac}(H_2)$ as well. Due to nice properties of my operators, I can show that I can use the Weyl sequences of $H_1$ to construct "similar" Weyl sequences for $H_2$, which shows that $\lambda\in\sigma_{ess}(H_2)$. But naively it is possible that $\lambda$ is in fact in the singular continuous spectrum of $H_2$ or something, and the existence of an orthogonal Weyl sequence doesn't tell me anything about this. If there is an additional requirement I can put on the Weyl sequence in order to ensure that it in fact corresponds to AC spectrum, that'd solve my problem, but I am not familiar with such a criterion. Does anyone here know of something like this?
Thanks in advance.