Two isolated two-level sites do not relax an energy eigenstate to Gibbs form

Assume AC and Countable Choice for the canonical mode preparation. Let ${\mathcal{H}}_1={\mathcal{H}}_2={\mathbb{C}}^2$ with orthonormal basis $\mathrm{\mid }0⟩,\mathrm{\mid }1⟩$, and define the factor number maps by $N_r\mathrm{\mid }0⟩=0$, $N_r\mathrm{\mid }1⟩=\mathrm{\mid }1⟩$. The composite Hilbert space is the finite Hilbert tensor product ${\mathcal{H}}_1\widehat{\otimes }{\mathcal{H}}_2\cong {\mathbb{C}}^4$ from Hilbert tensor construction: inner product, completion and product bases, with orthonormal product basis $\mathrm{\mid }{n}_1{n}_2⟩$, $n_r\in \{0,1\}$. On this basis define the site-number operators ${\mathcal{N}}_1\mathrm{\mid }{n}_1{n}_2⟩=n_1\mathrm{\mid }{n}_1{n}_2⟩$ and ${\mathcal{N}}_2\mathrm{\mid }{n}_1{n}_2⟩=n_2\mathrm{\mid }{n}_1{n}_2⟩$. For $T>0$ kelvins set $\epsilon =k_{\mathrm{B}}T\log 2$ joules and define the noninteracting total Hamiltonian $H=\epsilon ({\mathcal{N}}_1+{\mathcal{N}}_2),H\mathrm{\mid }{n}_1{n}_2⟩=\epsilon (n_1+n_2)\mathrm{\mid }{n}_1{n}_2⟩\mathrm{.}$ The quantum Gibbs preparation at $T$ is used for the thermal comparison, and the closed-system unitary evolution is generated by $H$ with $U_t=e^{-itH\mathrm{/}\mathrm{\hslash }}$, where $t$ is in seconds and $\mathrm{\hslash }>0$ is in $\mathrm{J}\mathrm{s}$. Start the dynamics in ${\rho }_0=\mathrm{\mid }\psi ⟩⟨\psi \mathrm{\mid }$, where $\mathrm{\parallel }\psi \mathrm{\parallel }=1$ and $H\psi =\lambda \psi$ for $\lambda \in \{0,\epsilon ,2\epsilon \}$. Here $\mathrm{\mid }\psi ⟩⟨\psi \mathrm{\mid }$ means the rank-one map $v\mapsto ⟨v,\psi ⟩\psi$, with inner product linear in the first argument. This includes every unit vector in the degenerate eigenspace $\mathrm{span}\{\mathrm{\mid }01⟩,\mathrm{\mid }10⟩\}$, as well as the ground and top eigenvectors. The preparation stays constant for every time and does not converge to the Gibbs density. The choice $\psi =\mathrm{\mid }00⟩$ gives the explicit ground-state comparison below.