Physics Stack Exchange is a question and answer site for active researchers, academics and students of physics. Join them; it only takes a minute:

Sign up
Here's how it works:
  1. Anybody can ask a question
  2. Anybody can answer
  3. The best answers are voted up and rise to the top

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

share|cite|improve this question
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.

share|cite|improve this answer

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

share|cite|improve this answer
This makes sense. Excitons have longer lifetimes than interband transitions. – mcodesmart Oct 24 '13 at 16:04

Your Answer


By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.