Spectral Measures: Analytic Elements Given a Hilbert space $\mathcal{H}$.
Consider a Hamiltonian:
$$H:\mathcal{D}(H)\to\mathcal{H}:\quad H=H^*$$
Denote the convergence radius by:
$$\frac{1}{\rho(\varphi)}:=\limsup_{k\to\infty}\sqrt[k]{\frac{1}{k!}\|H^k\varphi\|}$$
Introduce analytic elements:
$$\varphi_\omega\in\mathcal{C}^\omega(H):=\{\varphi\in\mathcal{D}(H):\rho(\varphi)>0\}$$
Then they belong to the domain:
$$|z|<\rho(\varphi_\omega):\quad\varphi_\omega\in\mathcal{D}(e^{izH})$$
Especially they expand as series:
$$|z|<\rho(\varphi_\omega):\quad e^{izH}\varphi_\omega=\sum_{k=0}^\infty\frac{1}{k!}(iz)^kH^k\varphi_\omega$$
This gives rise to an analytic form:
$$|\Im{z}|<\rho(\varphi_\omega):\quad F(z):=\langle e^{izH}\varphi_\omega,\psi\rangle$$
How to prove this from scratch?
 A: Meanwhile I found an answer...
Denote the probability measures by:
$$E:\mathcal{B}(\mathbb{R})\to\mathcal{B}(\mathcal{H}):\quad\nu_\varphi(A):=\|E(A)\varphi\|^2$$
They belong to the domain as:
$$\|e^{izH}\varphi\|=\int|e^{iz\lambda}|^2\mathrm{d}\nu_\varphi(\lambda)\leq2\sum_{k=0}^\infty\frac{1}{k!^2}|z|^{2k}\int|\lambda|^{2k}\mathrm{d}\nu_\varphi(\lambda)=2\sum_{k=0}^\infty\frac{1}{k!^2}|z|^{2k}\|H^k\varphi\|^2<\infty$$
So they expand as series since:
$$\|e^{izH}\varphi-\sum_{k=0}^K\frac{1}{k!}z^kH^k\varphi\|^2=\int|e^{iz\lambda}-\sum_{k=0}^K\frac{1}{k!}z^k\lambda^k|^2\mathrm{d}\nu_\varphi(\lambda)\\\leq2\sum_{k=K+1}^\infty\frac{1}{k!^2}|z|^{2k}\int|\lambda|^{2k}\mathrm{d}\nu_\varphi(\lambda)=2\sum_{k=K+1}^\infty\frac{1}{k!^2}|z|^{2k}\|H^k\varphi\|^2\to0$$
Fix a point on the real line:
$$t_0\in\mathbb{R}:\quad|z-t_0|<\rho(\varphi_\omega)$$
So they give rise to an analytic form:
$$\langle e^{izH}\varphi_\omega,\psi\rangle=\langle e^{i(z-t_0)H}\varphi_\omega,e^{-it_0H}\psi\rangle=\sum_{k=0}^\infty\frac{1}{k!}i^k(z-t_0)^k\langle H^k\varphi_\omega,e^{-it_0H}\psi\rangle$$
That finishes the proof!
