2
$\begingroup$

Kotani theory gives roughly that for ergodic operators there is a certain equivalence between absolutely continuous spectrum and an absolutely continuous density of states measure.

I would like to ask whether this is also true in the "opposite case", i.e. given a periodic Schrödinger operator $-\Delta+V$ on $\mathbb{R}^n$ (by this I mean that the potential is periodic):

Is it true that if this operator has purely absolutely continuous spectrum (which is always true in dimension $1$), then the DOS measure is absolutely continuous?- It somehow sounds very reasonable but I fail to see from what it should follow exactly?

Improve this question
$\endgroup$
4
  • $\begingroup$ I think that's clear, modulo (perhaps annoying) technical problems; it would be easier in the discrete setting probably. You can obtain the DOS by averaging spectral measures, and periodicity means that the averaging is over a compact set (it would be a finite one in the discrete case) of measures which are assumed absolutely continuous themselves. $\endgroup$
    – Christian Remling
    Jan 9, 2018 at 22:33
  • $\begingroup$ By the way, absolute continuity of the spectrum has been established under very general assumptions, see here for example: academic.oup.com/imrn/article-abstract/2001/1/1/… $\endgroup$
    – Christian Remling
    Jan 9, 2018 at 22:34
  • $\begingroup$ @ChristianRemling thank you, I understand. May I ask you a follow-up questions from what you wrote: If the potential is nice, can we assume that the DOS function has then also some regularity besides $L^1_{\text{loc}}$?-This also seems natural to me, but I am not aware of any results. And thank you for the reference, this results looks very interesting to me. $\endgroup$
    – DDriggs
    Jan 9, 2018 at 23:27
  • $\begingroup$ Yes, that definitely sounds to me like it has to be true, but of course I also have to warn you that I'm certainly no expert on higher-dimensional operators. $\endgroup$
    – Christian Remling
    Jan 10, 2018 at 0:27

0

Reset to default

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge you have read our privacy policy.