So I'm reading in Ashcroft and Mermin about indirect optical transitions:


So, a photon comes in, and it only excites the electron across the indirect band gap if a phonon with the appropriate wave vector can "help" the electron by giving it momentum (they point out that the momentum of the photon is negligible). First of all, does this mean that, if the photon comes in and no phonon happens to help it, it's not absorbed, but most likely reflected? Or would something else almost certainly absorb it?

Secondly, when a phonon is there to "help" the electron, does this mean the crystal is now cooler as a result, because there are fewer/less powerful lattice vibrations (because the phonon gave up some momentum)?

The crystal has certainly gained energy from the photon, but the energy is now just in the form of... the electron having a higher energy state? (Is that actually where the energy has "gone"? Do we just say that the electron is in a higher energy state, and that's the extent of it?) As opposed to kinetic energy of the crystal, which seems like it has decreased.

If this is possible (I'm ready to be told that it's not), could it be scaled up and carefully done to cool certain semiconductors?


  • $\begingroup$ Unless the electron goes somewhere, it will give up its energy as the system relaxes back to thermal equilbrium. So in the end the temperature of your system has gone up, not down. But I don't know what the leading process for energy relaxation is an indirect gap semiconductor, maybe it depends on the details. Probably you will find the answer if you look up excitons in indirect gap semiconductors. $\endgroup$ Jul 16, 2013 at 18:39
  • $\begingroup$ What you're thinking of is very similar to (but not exactly the same as) "laser cooling." The most recent Nobel prize was awarded to David Wineland for pioneering work in laser cooling. The basic "layman" description is this: if a material has a certain energy transition from $E_1$ to $E_2$ then you shine a slightly "red-detuned" laser $\hbar\omega \lesssim E_2 - E_1$ such that $\hbar\omega+\hbar\omega_m = E_2 - E_1$ you'll absorb a photon ($\hbar\omega$) and phonon ($\hbar\omega_m$) and emit another photon at $\omega_{\rm{emit}} = (E_2 - E_1)/\hbar$. So you "ate up" a phonon permanently! $\endgroup$
    – NanoPhys
    Jul 16, 2013 at 20:25
  • $\begingroup$ @NanoPhys: That's how magneto-optical traps work, but for that to effectively cool a solid state system almost all of your excitons would have to decay to photons - is such a thing possible? $\endgroup$ Jul 17, 2013 at 15:18
  • $\begingroup$ @BebopButUnsteady: Solid-state laser-cooling, although technically challenging, is still a very promising field partly due to their abundance and increase of external quantum efficiency with decreasing temperature. From a technological standpoint we need to worry about reducing surface recombination, increasing extraction luminescence efficiency, etc. to name a few. Nevertheless, people are still working out these issues and making progress in this field. For example, check out this recent article: dx.doi.org/10.1038/nphoton.2013.87 $\endgroup$
    – NanoPhys
    Jul 18, 2013 at 3:10
  • $\begingroup$ Small clarification: by solid-state laser cooling I mean semiconductor laser cooling. I was not referring to ion-doped glasses and crystals. $\endgroup$
    – NanoPhys
    Jul 18, 2013 at 23:23

2 Answers 2


As far as I know, to overcome the wave vector difference q neccessary to absorb the photon, a phonon can either be absorbed or emitted. To avoid the second case, one would obviously have to control precisely energy and momentum of the incoming photons.

To your question regarding where the energy has gone: All the energy of a crystal can only be stored in the constituents it is made of (core ions and electrons) and in interactions between them. So yes, the electron being lifted to a higher energy state is precisely the effect of absorbing a photon, and thus the total energy of the crystal must have been raised accordingly.

I think what BebopButUnsteady already mentioned in his/her comment should make sense: Even having absorbed a phonon, the total energy of the crystal has been increased and as soon as it relaxes into thermal equilibrium, you could even end up with more phonons than before. Remember that phonons are bosons (and even quasi-particles), so they are not bound to particle number conservation, so their number and energy will just conform to boson statistics and the internal energy of the system (not remembering the details...).

  1. Indeed, in the case you described, if the phonon with the specific characteristics (dispersion relation) that would match the requirement for the energy&momentum conservation is not available, the photon will be reflected at the same frequency.

  2. When the phonon is available, and it is absorbed or emitted, the crystal energy changes. Just a small but important correction: It happens not "because the phonon gave up some momentum", but because with that momentum it gave up some energy.

The energy balance is such that the difference in the energy of two photons ($E_o - E_i$) is equal in the difference of the body (i-incoming, o-outgoing). Inside the body we can define two "channels" of the energy: the change in the energy of the electron, $\Delta E_e$, and the change in the energy due to the emission or absorption of a phonon, $\Delta E_{ph}$. All three values, in principle, can have either sign (as long as the total energy is conserved): ($E_o - E_i = \Delta E_e + \Delta E_{ph})$.

You can take a look at Fig.3 of this article: http://www-personal.umich.edu/~kaviany/researchtopics/Xulin_JHT%20Review%20Laser%20Cooling%20of%20Solids%20%282%29.pdf (Please, ignore the right portion of the picture that describes the non-radiative recombination of the electron [and the hole] - that's not directly relevant to the questions your asked.)

One of the aspects of the question you are raising is that there is a vibrational energy of the crystal lattice/ions ("ionic temperature"), and there is an energy of the electrons (an effective "electronic temperature" can be introduced). When the electrons are not at equilibrium (i.e. not occupying the lowest available states), the electronic temperature can be higher. In your case, emission of a phonon(s) lowers the crystal temperature, but raises the electronic one.

There are different things that can happen to an excited electron to equilibrate, i.e. for its temperature to become equal to that of the crystal (eventually): it can flow away (if there is another conductor attached) - that's an electric current, or it can relax to a lower energy state through some interaction with the lattice (depends on the system). The latter process would raise the lattice temperature.


Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge you have read our privacy policy.

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