= Solution
Use $a|n\rangle=\sqrt n|n-1\rangle$, $a^\dagger|n\rangle=\sqrt{n+1}|n+1\rangle$, $\sigma_z|e\rangle=|e\rangle$ and $\sigma_z|g\rangle=-|g\rangle$. Only one interaction term acts on each of the two <quantum states>. Hence
$$
\boxed{\mathcal H|e,n\rangle=\hbar\left(n\omega+\frac{\omega_0}{2}\right)|e,n\rangle
+\hbar g\sqrt{n+1}|g,n+1\rangle,}
$$
$$
\boxed{\mathcal H|g,n+1\rangle=\hbar\left((n+1)\omega-\frac{\omega_0}{2}\right)|g,n+1\rangle
+\hbar g\sqrt{n+1}|e,n\rangle.}
$$
The interaction exchanges one atomic excitation with one field quantum. The square-root factor is the photon-number enhancement of the transition <matrix> element.
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