Fock Space: NESS

Question

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: $$\zeta\in\mathcal{C}^\omega\implies\frac{1}{1+e^{\beta H}}\zeta\in\mathcal{C}^\omega$$

Answer 1

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.

Answer 2

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 $(1 + e^{\beta H})^{-1} \mathcal C^\omega = \mathcal C^\omega$.