 Solution

In first-order time-dependent perturbation theory, the resonant time factor before taking the Fermi's golden rule limit isIts central frequency width is of order $1/t$, and its total area is $2\pi t$. To replace it by $2\pi t\delta (\nu )$, the oscillator density of states and relevant occupations must be approximately constant over that width. Thus $t\Delta \omega \gg 1$. At the same time, the perturbative flip probability must remain much smaller than one.

Define ${\Gamma }_0=2\pi {\lambda }^{2}g(\Omega )/{\hslash }^2$, so the two flip rates are ${\Gamma }_0(n_{\Omega }+1)$ and ${\Gamma }_0{n}_{\Omega }$. A sufficient common time window for both initial spins isFor a specified initial spin, the upper condition is simply its own flip rate times $t\ll 1$. Such a window requires weak enough coupling. The continuum approximation also presumes times shorter than recurrences from a resolved finite oscillator spectrum, and long enough to resolve the positive-frequency resonance from its nonresonant counterpart.