Two-level Gibbs state has the physical-time KMS boundary factor

Let $\mathcal{H}={\mathbb{C}}^2$ with orthonormal basis $\mathrm{\mid }0⟩,\mathrm{\mid }1⟩$, and let $H=\epsilon \mathrm{\mid }1⟩⟨1\mathrm{\mid }$ for an energy $\epsilon >0$ in joules. For $\beta >0$ in J${}^{-1}$, the finite Gibbs state has

$Z=1+e^{-\beta \epsilon },p_0=Z^{-1},p_1=e^{-\beta \epsilon }{Z}^{-1}\mathrm{.}$

Define $S_{+}=\mathrm{\mid }1⟩⟨0\mathrm{\mid }$ and $S_{-}=\mathrm{\mid }0⟩⟨1\mathrm{\mid }$. With ${\alpha }_t(X)=e^{itH\mathrm{/}\mathrm{\hslash }}X{e}^{-itH\mathrm{/}\mathrm{\hslash }}$, the correlation for $A=S_{-}$ and $B=S_{+}$ is

$F_{S_{-},S_{+}}(z)=p_0{e}^{i\epsilon z\mathrm{/}\mathrm{\hslash }}\mathrm{.}$

At the upper strip edge $z=t+i\beta \mathrm{\hslash }$, the exponential contributes $e^{-\beta \epsilon }$, so the boundary value is $p_1{e}^{i\epsilon t\mathrm{/}\mathrm{\hslash }}$, exactly the reversed-order Gibbs expectation. The full two-level Gibbs state is KMS for every matrix pair by Finite KMS condition exactly determines the Gibbs density; this example checks the physical-time sign and strip-width convention on a nontrivial pair. No measurement or material parameter is being reported.

For the reversed pair $A=S_{+}$, $B=S_{-}$, the lower value is $p_1{e}^{-i\epsilon t\mathrm{/}\mathrm{\hslash }}$ and the upper value is $p_0{e}^{-i\epsilon t\mathrm{/}\mathrm{\hslash }}$; the same Boltzmann factor relation holds with the energy difference reversed. All quantities here are model parameters or exact finite-matrix expressions, not empirical measurements.