Golden-rule relaxation of a spin coupled to an oscillator bath (${\Gamma }_{\downarrow \uparrow }={\Gamma }_0(n_{\Omega }+1),{\Gamma }_{\uparrow \downarrow }={\Gamma }_0{n}_{\Omega }$)

For transverse coupling with equal mode amplitudes $\lambda$, Fermi's golden rule and the oscillator creation operator and annihilation operator givewhere 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, $n_{\Omega }/(n_{\Omega }+1)=e^{-\beta \hslash \Omega }$. 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.