Fock Space: NESS Given the CAR-algebra with Hamiltonian dynamics:
$$\tau^t[a^\#(\eta)]=a^\#(e^{itH}\eta)\quad(H:\mathcal{D}\to\mathcal{H})$$
(Caution that the Hamiltonian is usually unbounded.)
Consider a KMS-state:
$$F(t):=\omega(A\tau^t[B]):\quad F(t+i\beta)=\omega(\tau^t[B]A)$$
Regard its two-point functions:
$$\left|\omega(a^*(\zeta)a(\eta))\right|\leq\|\zeta\|\cdot\|\eta\|\quad(\omega(a^*(\eta)a(\eta))\geq0\}$$
So it has a density by Lax-Milgram:
$$0\leq T\leq1:\quad\omega(a^*(\zeta)a(\eta))=\langle\eta,T\zeta\rangle$$
It allows an analytic extension by:
$$\omega(a^*(\zeta)\tau^t[a(\eta)])=\omega(a^*(\zeta)a(e^{itH}\eta))=\langle e^{itH}\eta,T\zeta\rangle$$
(Caution of antilinearity!!)
So for entire elements it follows:
$$\langle e^{\beta H}\eta,T\zeta\rangle=\omega(a(\eta)a^*(\zeta))=\omega(\langle\eta,\zeta\rangle1-a^*(\zeta)a(\eta))=\langle\eta,\zeta\rangle\cdot1-\langle\eta,T\zeta\rangle\quad(\eta\in\mathcal{C}^\omega)$$
Thus on entire elements it holds:
$$T(1+e^{\beta H})\eta=\eta\quad(\eta\in\mathcal{C}^\omega)$$
Whence one obtains the statistics:
$$T=\frac{1}{1+e^{\beta H}}$$
(Note that the entire elements were dense!)

But why did the entire elements remain entire:
  $$\zeta\in\mathcal{C}^\omega\implies\frac{1}{1+e^{\beta H}}\zeta\in\mathcal{C}^\omega$$

 A: It's convenient to use the "multiplication operator" form of the Spectral Theorem, in which the Hilbert space is $L^2(M, \mu)$ and $H$ corresponds to multiplication by a real-valued function $h$.  The entire vectors correspond
to functions $v$ such that $e^{sh} v \in L^2(M,\mu)$ for all real $s$.
$(1 + e^{\beta H})^{-1}$ corresponds to multiplication by the bounded function $1/(1 + e^{\beta h})$ (and thus is a bounded operator).  If $w = (1 + e^{\beta h})^{-1} v$ where $v \in C^\omega$, 
then $w \in C^\omega$ as well since
$e^{sh} /(1 + e^{\beta h}) \le e^{(s-\beta)h}$. This says $(1 + e^{\beta H})^{-1} \mathcal C^\omega \subseteq \mathcal C^\omega$.  The reverse implication is also true, so in fact $(1 + e^{\beta H})^{-1} \mathcal C^\omega = \mathcal C^\omega$.
A: First note that the inverse has full domain:
$$\int_\mathbb{R}\left|\frac{1}{1+e^{-\beta\lambda}}\right|^2\mathrm{d}\nu_\zeta(\lambda)\leq\int_\mathbb{R}1\mathrm{d}\nu_\zeta(\lambda)=\|\zeta\|^2<\infty$$
So one has a common domain:
$$\mathcal{D}\left(H^k\circ\frac{1}{1+e^{-\beta H}}\right)=\mathcal{D}\left(\frac{H^k}{1+e^{-\beta H}}\right)\cap\mathcal{D}\left(\frac{1}{1+e^{-\beta H}}\right)=\mathcal{D}\left(\frac{H^k}{1+e^{-\beta H}}\right)$$
Therefore the formal check is rigorous:
$$\left\|H^k\left(\frac{1}{1+e^{-\beta H}}\zeta\right)\right\|^2=\left\|\frac{H^k}{1+e^{-\beta H}}\eta\right\|^2=\int_\mathbb{R}\left|\frac{\lambda^k}{1+e^{-\beta\lambda}}\right|^2\mathrm{d}\nu_\zeta(\lambda)\leq\int_\mathbb{R}|\lambda|^{2k}\mathrm{d}\nu_\zeta(\lambda)$$
Concluding that entire elements remain entire.
