As it turns out, the excited states of an atom are not really, strictly speaking, eigenstates. That is, they're eigenstates of the atomic hamiltonian, but they are not eigenstates of the atom-plus-EM-field hamiltonian. How do we know? Well, if you prepare such a state (excited atom, empty field) and then you leave it alone, then it will change: the atom will decay to the ground state, emitting a photon. Eigenstates of the full system never change - they're stationary states - so these excited states of the atom can't be eigenstates of the full hamiltonian.
What they are is resonances, which turn out to be fairly tricky to actually define. (For more on this, see 'Decay theory of unstable quantum systems', L. Fonda et al., Rep. Prog. Phys. 41 no. 4587 (1978).) This means is that they have a finite lifetime $\tau$, and equivalently they do not have a well-defined energy and have a finite natural linewidth $\Delta\omega$. These two are Fourier-dual versions of each other, so it is no surprise that they satisfy the relation
with the precise details depending on what exactly each symbol means.
In particular, the finite natural linewidth of the excited state means that you can excite the atom to it from the ground state even with an almost-monochromatic (i.e. finitely but very sharp) laser at a frequency that is not the centre frequency of the level. In other words, the width of the level and the natural width of the transition are exactly the same thing.