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If $A$ is an unbounded, self-adjoint operator on a separable Hilbert space with spectral family $E_t$, then $\exp(iA)$, defined by functional calculus with function $f(t)=exp(it)$, is an unitary operator. My question is, what about the spectral family of $exp(iA)$? Is again the unbounded $E_t$? Must not be a bounded one? How to calculate it?

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  • $\begingroup$ I guess $E_t = 1_{(-\infty,t]}(A)$? Are you familiar with the spectral measure induced by a spectral family, $E(B)=1_B(A)$ for any Borel set $B\subseteq \mathbb{C}$? $\endgroup$ May 24, 2017 at 18:50
  • $\begingroup$ Yes $E_t=E (]-\infty,t])$. I familiar with spectral measure $\endgroup$
    – dioxoid
    May 24, 2017 at 19:10

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For a closed, densely defined, normal operator $T$ in a Hilbert space $\mathcal{H}$, denote by $E^T(B)=1_B(T)$ the spectral measure induced by $T$, and if $B\subseteq \mathbb{C}$ is a Borel set and $x\in\mathcal{H}$, let $E^T_x(B):=\langle x, E^T(B) x\rangle$. Recall that $E^T$ is uniquely determined by the family of complex measures $(E^T_x)_{x\in\mathcal{H}}$. Now, if $f:\mathbb{C}\rightarrow \mathbb{C}$ is any measurable function, then $$ E_x^{f(T)}(B) = \langle x,1_B(f(T))x\rangle = \langle x,1_{f^{-1}(B)}(T)x\rangle = E_x^T(f^{-1}(B)). $$ It follows that $E^{f(T)}(B)=E^T(f^{-1}(B))$.

If $A$ is self-adjoint and $U=e^{iA}$, then the spectral measure of $U$ is supported in the unit circle $S\subseteq \mathbb{C}$, and is explicitly given by the formula $E^U(B)=E^A(f^{-1}(B))$, with $f(t)=e^{it}$.

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  • $\begingroup$ So if $T=A=\int_RtdE_t$ is self adjoint and $U=e^{iA}=\int_Re^{it}dE_t$, then if B is a Borel subset of the circle T, $E^U(B)=E^A(-i\ln(B))$ and $U=\int_T\lambda dE_{-i\ln\lambda}$ ? $\endgroup$
    – dioxoid
    May 24, 2017 at 22:26
  • $\begingroup$ With a appropriate definition of the logarithm, $E^U(B)=E^A(-i\ln(B)))$, and $U=\int_S \lambda \, dE^U(\lambda) = \int_{\mathbb{R}} e^{i\lambda} \, dE^A_\lambda$ $\endgroup$ May 24, 2017 at 22:30
  • $\begingroup$ Ok, many thanks $\endgroup$
    – dioxoid
    May 24, 2017 at 22:33

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