Exact quantum oscillator and spin ensembles

Assume the Axiom of Countable Choice ${\mathrm{A}\mathrm{C}}_{\omega }$ (The Axiom of Countable Choice (${\mathrm{A}\mathrm{C}}_{\omega }$)) and the preparation postulate Quantum statistics and model equilibrium preparation. Let $\omega >0$ in s${}^{-1}$, $m>0$ in kg, $\mathrm{\hslash }>0$ in J s, $k_B>0$ in J K${}^{-1}$, and $T>0$ in K; set $ϵ=\mathrm{\hslash }\omega$ in joules and $\beta =(k_{B}T{)}^{-1}$ in J${}^{-1}$. Take the self-adjoint one-dimensional realization $H_{\mathrm{o}\mathrm{s}\mathrm{c}}$ of $-{\mathrm{\hslash }}^2(2m{)}^{-1}{d}^2\mathrm{/}d{x}^2+m{\omega }^2{x}^2\mathrm{/}2$ on $L^2(\mathbb{R},dx)$ with Hermite eigenvectors ${\varphi }_n$, eigenvalues $ϵ(n+1\mathrm{/}2)$, and domain $D(H_{\mathrm{o}\mathrm{s}\mathrm{c}})=\{{\sum }_n{c}_n{\varphi }_n:{\sum }_n{ϵ}^2(n+1\mathrm{/}2{)}^2\mathrm{\mid }{c}_n{\mathrm{\mid }}^2<\mathrm{\infty }\}$, as proved in Oscillator scaling and exact realization. Declare this oscillator to be a fixed-number QSM model with particle number $N=1$ and mode set $I={\mathbb{N}}_0$ in its usual order. The unitary ${\mathcal{U}}_{\mathrm{o}\mathrm{s}\mathrm{c}}:{\mathrm{\ell }}^2({\mathbb{N}}_0)\to L^2(\mathbb{R},dx)$, ${\mathcal{U}}_{\mathrm{o}\mathrm{s}\mathrm{c}}(c)={\sum }_{n\ge 0}{c}_n{\varphi }_n$, identifies its one-particle sector with this realization; the symmetric and antisymmetric one-particle sectors coincide. Adopt the canonical preparation ${\rho }_{\mathrm{o}\mathrm{s}\mathrm{c}}=Z^{-1}{e}^{-\beta H_{\mathrm{o}\mathrm{s}\mathrm{c}}}$ when its trace $Z$ is positive and finite. Define $U=\mathrm{tr}({\rho }_{\mathrm{o}\mathrm{s}\mathrm{c}}{H}_{\mathrm{o}\mathrm{s}\mathrm{c}})$ when its absolute energy moment is finite, $C_V=\mathrm{\partial }U\mathrm{/}\mathrm{\partial }T$ at fixed $m,\omega ,\mathrm{\hslash }$, and $S=-k_B\mathrm{tr}({\rho }_{\mathrm{o}\mathrm{s}\mathrm{c}}\log {\rho }_{\mathrm{o}\mathrm{s}\mathrm{c}})$ when its eigenvalue entropy sum is finite. Then $Z=e^{-\beta ϵ\mathrm{/}2}(1-e^{-\beta ϵ}{)}^{-1},U=ϵ(\frac{1}{2}+\bar{n}),\bar{n}=(e^{\beta ϵ}-1{)}^{-1},C_V=k_B(\beta ϵ{)}^2\frac{e^{\beta ϵ}}{(e^{\beta ϵ}-1{)}^2}\mathrm{.}$ Its entropy is $S=k_B[(1+\bar{n})\log (1+\bar{n})-\bar{n}\log \bar{n}]$; as $T\downarrow 0$, $U\to ϵ\mathrm{/}2$, $S,C_V\to 0$. As $\beta ϵ\downarrow 0$, $U={\beta }^{-1}+O(\beta {ϵ}^2)$.

For a separate two-level spin model, let ${\mathcal{H}}_s={\mathbb{C}}^2$ with its standard orthonormal basis, define ${\sigma }_z=\left(\begin{array}{cc}1& 0\\ 0& -1\end{array}\right)$ on its full domain $D({\sigma }_z)={\mathbb{C}}^2$, and set $H_s=-h{\sigma }_z$ for $h\in \mathbb{R}$ in joules. Declare this separate QSM model to be its fixed $N=1$ sector with two ordered modes $I_s=\{1,2\}$, identified with ${\mathrm{\ell }}^2(I_s)={\mathbb{C}}^2$ by the coordinate basis, and adopt ${\rho }_s=Z_s^{-1}{e}^{-\beta H_s}$ when $0<Z_s=\mathrm{tr}(e^{-\beta H_s})<\mathrm{\infty }$. Define $U_s,C_{V,s},S_s$ by the same mean, fixed-$h$ temperature derivative, and entropy formulas. Then $Z_s=2\cosh (\beta h)$, $U_s=-h\tanh (\beta h)$, $C_{V,s}=k_B(\beta h{)}^2{\mathrm{sech}}^2(\beta h)$, and $S_s\to 0$ as $T\downarrow 0$ when $h\mathrm{\ne }0$, whereas $S_s=k_B\log 2$ when $h=0$. These results make no generic third-law claim.