Condition to the holomorphy of a complex function

Question

In Witten's note https://arxiv.org/abs/1803.04993, during the proof of Reeh-Schlieder theorem, he made an arguement that considering a function $$g(u)=\langle\chi|\phi(x_1)\dots e^{\mathrm{i}Hu}\phi(x_n)|\Omega\rangle\,.$$ Because the hamiltonian $H$ is bounded below by $0$, the operator $\exp(\mathrm{i}Hu)$ is holomorphic for $u$ in the upper half plane so that the function $g(u)$ is also holomorphic. Why the holomorphy of this function is connected to the boundness of $H$ and the reigon of $u$? I suppose that the upper half plane limit of $u$ is to stop the term $\exp(-Im(u)H )$ diverge to infinity, yet I am still confused by its relation to the holomorphy of the function.

Answer 1

Let $\psi, \phi$ be two Hilbert space vectors. According to the spectral theorem, a signed measure $\mu_{\psi,\phi}$ with bounded variation exists such that $$\langle \psi, \mathrm{e}^{\mathrm{i}uH} \phi\rangle = \int_{\sigma(H)} \mathrm{e}^{\mathrm{i}u\lambda} \mathrm{d} \mu_{\psi,\phi}(\lambda),$$ where $\sigma(H)$ is the spectrum of $H$. Thus, the complex derivative of $\langle \psi, \mathrm{e}^{\mathrm{i}uH} \phi\rangle$ can be computed as follows: $$\partial_u \langle \psi, \mathrm{e}^{\mathrm{i}uH} \phi\rangle = \int_{\sigma(H)} \mathrm{i} \lambda \mathrm{e}^{\mathrm{i}u\lambda} \mathrm{d} \mu_{\psi,\phi}(\lambda).$$ If the imaginary part of $u$ is strictly positive (i.e. $\Im(u) > 0)$, the integral on the r.h.s. is convergent: $$\left| \int_{\sigma(H)} \mathrm{i} \lambda \mathrm{e}^{\mathrm{i}u\lambda} \mathrm{d} \mu_{\psi,\phi}(\lambda) \right| \leq \int_{\sigma(H)} |\lambda| \mathrm{e}^{-\Im(u)\lambda} \mathrm{d} |\mu_{\psi,\phi}|(\lambda) < \infty,$$ where I used that $\sup_{\lambda \in \sigma(H)}|\lambda| \mathrm{e}^{-\Im(u)\lambda}$ is finite (this uses that the spectrum of $H$ is bounded from below and that the exponential decays faster than any polynomial). This ensures that the complex derivative of $\langle \psi, \mathrm{e}^{\mathrm{i}uH} \phi\rangle$ exists.