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Below is an image of the optical density (proportional to the absorption coefficient) of KBr crystal at low temperature. Indicated at 6.6 ev and 7.7 eV are the absorption by excitons. As you can see, it has a small width. Can you explain why?

enter image description here

Source:Theory of Excitons, Robert S Knox, 1954

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up vote 1 down vote accepted

We can invert a bit the situation. You have that $\Delta E\Delta t \ge \frac{\hbar}{2}$. Thus for narrow energy spread $\Delta E$ it means that the lifetime $\Delta t$ is large.

Now the question is why $\Delta t$ is large. In this case, this must be mostly because a rather small interaction of such excitation with its environment (most probably phonons being the most relevant) letting them live long and then having narrow energy spread.

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Because Excitons are quasiparticles constitued by an electron-hole couple which fall in quasi-bound state. Therefore, while a bound state is a sharp line in the spectrum, a quasi-bound state (or a metastable state) has a very small frequency (or energy) width. This means that instead of having an infinite lifetime, it has a finite one, but very long

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This makes sense. Excitons have longer lifetimes than interband transitions. – mcodesmart Oct 24 '13 at 16:04

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