= Solution
The <Pauli matrix> $\sigma_x$ flips the <spin> with unit <matrix element>. A nonzero oscillator <matrix element> changes exactly one <occupation number>, through a <creation operator> or an <annihilation operator>:
$$
\langle n_j+1|a_j^\dagger|n_j\rangle=\sqrt{n_j+1},\qquad
\langle n_j-1|a_j|n_j\rangle=\sqrt{n_j}.
$$
For a <spin> flip, the spin energy change divided by $\hbar$ is $-s_\sigma\Omega$. Thus creation and annihilation in mode $j$ have total frequency changes $\omega_j-s_\sigma\Omega$ and $-\omega_j-s_\sigma\Omega$, respectively. Distinct modes or different occupation changes lead to orthogonal final <quantum states>, so their <probabilities> add. <Fermi's golden rule> consequently gives
$$
|c_{\eta\sigma}(t)|^2=\frac{2\pi\lambda^2t}{\hbar^2}
\sum_j\left[(n_j+1)\delta(\omega_j-s_\sigma\Omega)
+n_j\delta(\omega_j+s_\sigma\Omega)\right],\quad \eta\ne\sigma.
$$
The <Dirac delta function> is even, accounting for the sign in the annihilation term. Since all $\omega_j,\Omega>0$, an initially up spin can only emit an oscillator quantum, while an initially down spin can only absorb one. Writing the oscillator <density of states> per unit angular frequency as $g(\omega)=\sum_j\delta(\omega-\omega_j)$, the continuum limit gives
$$
\boxed{|c_{\eta\sigma}(t)|^2=\frac{2\pi\lambda^2t}{\hbar^2}
\begin{cases}g(\Omega)(n_\Omega+1),&\sigma=\uparrow,\\
g(\Omega)n_\Omega,&\sigma=\downarrow,
\end{cases}\quad\eta\ne\sigma.}
$$
The $+1$ permits emission even from the oscillator vacuum; absorption requires a quantum already present. The notation $n_\Omega$ assumes a common or smoothly specified occupation at the resonant frequency. If several resonant modes have different occupations, it denotes their <density of states> weighted mean, so that $g(\Omega)n_\Omega=\sum_j n_j\delta(\Omega-\omega_j)$ after continuum smoothing. This is the <golden-rule relaxation of a spin coupled to an oscillator bath>.
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