- #1

nonequilibrium

- 1,439

- 2

As you all know, if a crystal contains a phonon of frequency [itex]\omega[/itex], it means that the Fourier decomposition of the physical vibration of that crystal lattice contains a plane wave of frequency [itex]\omega[/itex], and the more phonons of a given frequency, the bigger the amplitude of that component. Hence, the phonon is a nifty quasi-particle.

But it also seems obvious that the phonon cannot be localized. What would the position of a phonon mean, after all? If a phonon is an indication of the presence of a plane wave (of a certain frequency), then since plane waves are not localized, one cannot sensibly associate a place to the phonon. Sure, a lot of plane waves may make a localized

*wave*, but that's hardly relevant: it's not as if one associates one phonon to that localized wave; no, that localized wave is rather a whole bunch of delocalized phonons, exactly like a localized wave is a whole bunch of delocalized plane waves. (I'm saying this to show that it's not at all analogous to for example a psi-plane wave of an electron: in that case the electron is also not localized, but if we have localized wave packet, we can speak about the position of the electron; this reasoning is obviously

**ir**relevant for phonons.)

But I'm reading Kittel and where it talks about thermal conduction (p121, chapter 5, edition 8), it first mentions a formula of kinetic gas theory (a formula for the thermal conductivity [itex]K=\frac{1}{3}C v l[/itex] with C the heat capacity of the gas, v the average particle velocity and l the mean free path of a particle between collisions). And he then, without justification, applies this formula to phonons. This of course implies that he views phonons as localized particles, using a kind of kinetic theory of phonons. How can this make

*any*sense, in the light of what I said above?