= Solution
The thermal oscillator occupation is the <Bose-Einstein distribution>. With the rate notation $\Gamma_{\eta\sigma}$ for a transition from $\sigma$ to $\eta$, the <master equation> for the up population is
$$
\frac{dp_\uparrow}{dt}=\Gamma_{\uparrow\downarrow}p_\downarrow
-\Gamma_{\downarrow\uparrow}p_\uparrow
=\Gamma_0\left[n_\Omega p_\downarrow-(n_\Omega+1)p_\uparrow\right].
$$
At <thermal equilibrium>, assuming $\Gamma_0>0$, vanishing time derivative is <detailed balance> between the two spin states. Hence
$$
\frac{p_\uparrow}{p_\downarrow}=\frac{n_\Omega}{n_\Omega+1}
=\frac{1/(e^{\beta\hbar\Omega}-1)}{e^{\beta\hbar\Omega}/(e^{\beta\hbar\Omega}-1)}
=e^{-\beta\hbar\Omega},
$$
or
$$
\boxed{p_\uparrow=e^{-\beta\hbar\Omega}p_\downarrow.}
$$
Together with normalization, this gives $p_\uparrow=1/(1+e^{\beta\hbar\Omega})$ and $p_\downarrow=e^{\beta\hbar\Omega}/(1+e^{\beta\hbar\Omega})$, the <Boltzmann distribution> for the <energy eigenvalues> $\pm\hbar\Omega/2$. At zero temperature, absorption vanishes and emission relaxes the spin into its down ground state.
The nonzero-rate condition is necessary: if $\lambda=0$ or $g(\Omega)=0$, both rates vanish, so every initial pair of populations is stationary within this approximation. In that case stationarity alone cannot imply a thermal distribution. The equilibrium conclusion therefore presumes coupling to resonant oscillator modes.
Back to article page
