= Golden-rule relaxation of a spin coupled to an oscillator bath
{title2=$\Gamma_{\downarrow\uparrow}=\Gamma_0(n_\Omega+1),\quad\Gamma_{\uparrow\downarrow}=\Gamma_0n_\Omega$}
For transverse coupling with equal mode amplitudes $\lambda$, <Fermi's golden rule> and the oscillator <creation operator> and <annihilation operator> give
$$
\Gamma_{\downarrow\uparrow}=\Gamma_0(n_\Omega+1),\qquad
\Gamma_{\uparrow\downarrow}=\Gamma_0n_\Omega,\qquad
\Gamma_0=\frac{2\pi\lambda^2}{\hbar^2}g(\Omega),
$$
where the first index is the final spin and $g$ is the <density of states> per unit angular frequency. Decay creates a quantum and carries the factor $n_\Omega+1$; excitation annihilates one and carries $n_\Omega$. For nonuniform resonant occupations, $n_\Omega$ is their density-weighted mean.
With a thermal <Bose-Einstein distribution>, $n_\Omega/(n_\Omega+1)=e^{-\beta\hbar\Omega}$. The two-state <master equation> therefore has the <Boltzmann distribution> as its unique stationary populations when $\Gamma_0>0$. The population deviation relaxes at rate $\Gamma_0(2n_\Omega+1)$. If there is no resonant coupling, both rates vanish and stationarity does not determine the populations.
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