# Spectral measure and Stone's theorem

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 $U_{\lambda }$?

• I don't get the question, if $U_\lambda$ is invertible then $T =\lambda$. Dec 5, 2013 at 13:29
• $U_\lambda$ does not commute with $\lambda$. Compare with $\mathcal{F}(-\Delta )\mathcal{F}^{-1}=\xi ^2$ where $\mathcal{F}$ is the Fourier transform. Dec 5, 2013 at 13:42
• 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)? Dec 5, 2013 at 13:47

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, $dE/d\lambda = U^{-1}({\rm right\, side})U$.
• 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 $UTU^{-1}=M_\lambda$. Dec 6, 2013 at 9:23
• 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. Dec 6, 2013 at 9:50
• @MateuszWasilewski: $\lambda$ is any real number. Dec 6, 2013 at 10:09