# Exponential of unbounded self adjoint operator

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?

• 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 $B\subseteq \mathbb{C}$? – Jonas Dahlbæk May 24 '17 at 18:50
• Yes $E_t=E (]-\infty,t])$. I familiar with spectral measure – dioxoid May 24 '17 at 19:10

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 $f(t)=e^{it}$.
• 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 $U=\int_T\lambda dE_{-i\ln\lambda}$ ? – dioxoid May 24 '17 at 22:26
• With a appropriate definition of the logarithm, $E^U(B)=E^A(-i\ln(B)))$, and $U=\int_S \lambda \, dE^U(\lambda) = \int_{\mathbb{R}} e^{i\lambda} \, dE^A_\lambda$ – Jonas Dahlbæk May 24 '17 at 22:30