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In laser cooling, with a model of a 2-level atomic system, spontaneous emission is stated to be dependent on the "natural line width" of the excited state of the atom. This width is defined as the inverse of the lifetime of the state. In a picture in the same article, however, this line width is represented as the width of the excited state itself (due uncertainty principle I believe). This is confusing to me, as I was persuaded that the line width was the width of the transition line from excited state to ground state.

Are these notions the same thing or in some way related?

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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 $$ \tau·\Delta\omega\approx1, $$ 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.

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If I understand your question correctly, you are confusing between two definitions of "width". One is the spread of the spectral line and the other is the lifetime of the atom in excited state. Well, these two are related(as you guessed) by uncertainty principle. $\Delta E\,\Delta t = h/4\pi$. So if the time the atom spends in excited state is less, the energy spread will be more and vice versa. So any one of them can be thought as the measure of linewidth.

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  • $\begingroup$ I see... so the more time the atom sits in the excited state, the less the energy distribution will be "wide" (I imagine a lorentzian curve) ? So when the article I'm reading states W=1/t t is referring to the time the atom sits in the excited state (after having absorbed a photon)? I find this a little tricky sorry $\endgroup$ – user94213 Sep 30 '15 at 13:51
  • $\begingroup$ Yes exactly that. But don't mind if it's not easy grasp, even Hisenberg found this uncertainty a little trickier than the other one. $\endgroup$ – Ari Sep 30 '15 at 17:06

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