1
$\begingroup$

Let $T$ be an unbounded self-adjoint operator on a Hilbert space and let $E(\lambda )$ be the associated spectral measure and $R(\lambda ) = (T-\lambda )^{-1}$ the resolvent. By Stone's theorem we have $$ (1) \qquad \frac{dE}{d\lambda } = \frac{1}{2\pi i} (R(\lambda + i0) - R(\lambda - i0)), $$ where $R(\lambda + i0) = \lim _{\varepsilon \searrow 0}R(\lambda + i\varepsilon )$ etc.

Assume I also know an operator $U_{\lambda }$ such that $$ U_{\lambda }TU_{\lambda }^{-1} = \lambda $$ is multiplication by $\lambda $.

Can I somehow combine these identities to express the right hand side of (1) in terms of $U_{\lambda }$?

$\endgroup$
3
  • $\begingroup$ I don't get the question, if $U_\lambda$ is invertible then $T =\lambda$. $\endgroup$
    – Marc Palm
    Commented Dec 5, 2013 at 13:29
  • $\begingroup$ $U_\lambda $ does not commute with $\lambda $. Compare with $\mathcal{F}(-\Delta )\mathcal{F}^{-1}=\xi ^2$ where $\mathcal{F}$ is the Fourier transform. $\endgroup$
    – flavio
    Commented Dec 5, 2013 at 13:42
  • $\begingroup$ Ah you mean multiplication operator by a function $\lambda \mapsto \lambda$, sorry;) But, why not $U$ but $U_\lambda$? Also you need a more general function or not? f(\lambda)? $\endgroup$
    – Marc Palm
    Commented Dec 5, 2013 at 13:47

1 Answer 1

1
$\begingroup$

So you mean $UTU^{-1} = M_x$? Then yes, of course. First you have $UR(\lambda)U^{-1} = (M_x - \lambda)^{-1} = M_{(x-\lambda)^{-1}}$. So $U(dE/d\lambda)U^{-1} = \lim (1/2\pi i)(M_{(x - \lambda - i\epsilon)^{-1}} - M_{(x - \lambda + i\epsilon)^{-1}})$. Or if you like, $dE/d\lambda = U^{-1}({\rm right\, side})U$.

$\endgroup$
5
  • $\begingroup$ Thanks for your answer. However, I would like to get rid of $E$ completely and express the RHS in my equation (1) only in terms of $U$. Also, $U$ depends on $\lambda $ and $UTU^{-1}=M_\lambda $. $\endgroup$
    – flavio
    Commented Dec 6, 2013 at 9:23
  • $\begingroup$ If $U$ depends on $\lambda$ then what is $\lambda$ exactly? @Nik Weaver: The formula given by OP for the spectral density seems to be correct, since it is just the usual inversion formula for the Stieltjes (or Cauchy) transform of a measure. $\endgroup$ Commented Dec 6, 2013 at 9:50
  • $\begingroup$ @MateuszWasilewski: $\lambda $ is any real number. $\endgroup$
    – flavio
    Commented Dec 6, 2013 at 10:09
  • $\begingroup$ @Mateusz: you're right, I've edited to correct my answer. $\endgroup$
    – Nik Weaver
    Commented Dec 6, 2013 at 17:33
  • $\begingroup$ @flavio: you seem very confused. I'd suggest you find someone knowledgeable to talk to about this in person. $\endgroup$
    – Nik Weaver
    Commented Dec 6, 2013 at 17:38

Your Answer

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

Not the answer you're looking for? Browse other questions tagged or ask your own question.