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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 $U_{\lambda }$?

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  • $\begingroup$ I don't get the question, if $U_\lambda$ is invertible then $T =\lambda$. $\endgroup$
    – Marc Palm
    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
    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
    Dec 5, 2013 at 13:47

1 Answer 1

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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, $dE/d\lambda = U^{-1}({\rm right\, side})U$.

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  • $\begingroup$ 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
    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. $\endgroup$
    – Mateusz Wasilewski
    Dec 6, 2013 at 9:50
  • $\begingroup$ @MateuszWasilewski: $\lambda $ is any real number. $\endgroup$
    – flavio
    Dec 6, 2013 at 10:09
  • $\begingroup$ @Mateusz: you're right, I've edited to correct my answer. $\endgroup$
    – Nik Weaver
    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
    Dec 6, 2013 at 17:38

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