Adopted finite-regulator equilibrium state

Assume Countable Choice (The Axiom of Countable Choice (${\mathrm{A}\mathrm{C}}_{\omega }$)) for the finite-regulator spectral and trace construction. For the fixed model of Adopted finite-regulator polynomial model and a declared temperature input $T>0$, use Finite-regulator temperature and inverse-energy parameter to set ${\beta }_{\mathrm{p}\mathrm{h}\mathrm{y}\mathrm{s}}=(k_{B}T{)}^{-1}$ and $\widehat{\beta }=E_{\ast }{\beta }_{\mathrm{p}\mathrm{h}\mathrm{y}\mathrm{s}}>0$. Let $h$ be the model's specified self-adjoint form realization and $H_{\mathrm{p}\mathrm{h}\mathrm{y}\mathrm{s}}=E_{\ast }h$. When the finite-mode coercivity hypotheses hold, Finite-regulator Gibbs trace and bounded expectations proves that ${\rho }_{\gamma }$ is positive trace class with finite positive trace for every dimensionless $\gamma >0$. Make its spectral construction explicit: let $({\psi }_k,{\lambda }_k{)}_{k\ge 0}$ be the complete orthonormal eigenbasis and repeated eigenvalues used in that supplier. For $f\in H_N$, define

$G_{\gamma }f:={\sum }_{k\ge 0}{e}^{-\gamma {\lambda }_k}⟨f,{\psi }_k⟩{\psi }_k\mathrm{.}$

The vector series converges in $H_N$ for every $f$, and this is the action of the supplier's operator ${\rho }_{\gamma }$, obtained there as the operator-norm limit of its finite-rank truncations. In this item the notation $e^{-\gamma h}$ means this explicitly defined operator $G_{\gamma }$. Since $H_{\mathrm{p}\mathrm{h}\mathrm{y}\mathrm{s}}{\psi }_k=E_{\ast }{\lambda }_k{\psi }_k$, define $e^{-{\beta }_{\mathrm{p}\mathrm{h}\mathrm{y}\mathrm{s}}{H}_{\mathrm{p}\mathrm{h}\mathrm{y}\mathrm{s}}}$ on $H_N$ by the corresponding series $e^{-{\beta }_{\mathrm{p}\mathrm{h}\mathrm{y}\mathrm{s}}{H}_{\mathrm{p}\mathrm{h}\mathrm{y}\mathrm{s}}}f:={\sum }_{k\ge 0}{e}^{-{\beta }_{\mathrm{p}\mathrm{h}\mathrm{y}\mathrm{s}}{E}_{\ast }{\lambda }_k}⟨f,{\psi }_k⟩{\psi }_k$. Since $\widehat{\beta }=E_{\ast }{\beta }_{\mathrm{p}\mathrm{h}\mathrm{y}\mathrm{s}}$, its coefficients equal those of $G_{\widehat{\beta }}$ term by term, and both series converge in $H_N$. Therefore, as operators on $H_N$, $e^{-{\beta }_{\mathrm{p}\mathrm{h}\mathrm{y}\mathrm{s}}{H}_{\mathrm{p}\mathrm{h}\mathrm{y}\mathrm{s}}}=G_{\widehat{\beta }}=e^{-\widehat{\beta }h}$. The Gibbs-trace supplier consequently applies to exactly the operator in the state assignment below. Adopt the normalized operator

${\rho }_T:=\frac{e^{-{\beta }_{\mathrm{p}\mathrm{h}\mathrm{y}\mathrm{s}}{H}_{\mathrm{p}\mathrm{h}\mathrm{y}\mathrm{s}}}}{Z_T}=\frac{e^{-\widehat{\beta }h}}{Z_T},Z_T:=\mathrm{Tr}(e^{-{\beta }_{\mathrm{p}\mathrm{h}\mathrm{y}\mathrm{s}}{H}_{\mathrm{p}\mathrm{h}\mathrm{y}\mathrm{s}}})=\mathrm{Tr}(e^{-\widehat{\beta }h})$

as this finite-regulator model's canonical equilibrium state. This is a physical state assignment, not a deduction that the dynamics reach equilibrium. It asserts no infinite-volume trace, no zero-temperature state, and no empirical observation.