 Solution

The total excitation operator $\mathcal{N}=a^{†}a+({\sigma }_z+1)/2$ commutes with the Hamiltonian. Each sector with $\mathcal{N}=n+1\ge 1$ is therefore spanned by $|e,n⟩,|g,n+1⟩$, and no term connects it to a sector of different excitation number.

In that ordered basis, with the source's Rabi detuning $\Delta =\omega -{\omega }_0$,The first term is the common sector energy; the remaining diagonal entries record the bare Rabi detuning and the off-diagonal terms exchange the excitation.

The isolated ground sector of the Jaynes-Cummings model is an important exception to a statement that the entire Hilbert space consists of such pairs: $|g,0⟩$ has no partner and is annihilated by both interaction terms. It is a one-dimensional invariant sector with energy $-\hslash {\omega }_0/2$. The full decomposition is this one-dimensional sector plus all the two-dimensional sectors above.