8
$\begingroup$

I was reading through this paper on thermal transport in crystals, and saw that a primary function of their interest was $n_{\mu}(\mathbf{x},t)$, which they state is the phonon excitation/occupation number at position $\mathbf{x}$ and time $t$, where $\mu=(\mathbf{k},s)$ is the index of the phonon ($\mathbf{k}$=wavenumber, $s$=polarization).

How can you have a phonon occupation number at a particular location and time? I thought phonons were lattice excitations that necessarily extended across the entire crystal, i.e. they were excitations of a sort of "phonon field". I know phonons are bosons, and that for a crystal in thermal equilibrium with its environment the phonon occupation number will be given by standard Bose-Einstein occupation number factor:

$$\bar{n} (\epsilon)=\frac{1}{e^{(\epsilon-\mu)/T}-1}$$

where $\epsilon$ is the energy of the phonon, $\mu$ is the chemical potential, and $T$ is the temperature.

I understand that this is for out of equilibrium dynamics, and that typically one describes the relaxation of the spacial distribution (density) of constituent particles. Quantum-mechanically, even "classical" particles extend across the entire containing volume just as phonons do (through their wavefunction), but in the macroscopic limit we can faithfully assign a position and momentum to each of them without much loss of information. However, particles are collections/superpositions of harmonic modes of the system (think particle in a box), conspiring in such a way to give a classically definite position and momentum. Phonons, on the other hand, are the harmonic excitations themselves.

So,... how?

$\endgroup$
3
$\begingroup$

How can you give a phonon a location? The answer is: "The same way you do for an electron."

A phonon is a collective excitation of all the ions. What we typically call an "electron" in a solid isn't really a single electron. It's a collective excitation of all the electrons (and ions, for that matter) in the solid. There's a reason an "electron" in a solid has an effective mass different than the electron rest mass. So, an "electron" and a phonon are on the same theoretical footing.

Ok, so how do you give an electron a location? After all, in an ideal crystal, its wave function also extends thru the entire crystal.

You mention the "macroscopic limit", but that doesn't solve anything. There is no macroscopic limit of an electron. An electron is a point particle.

You mention building up a superposition of modes, but that's not really the ticket either. An electron's dispersion relation is nonlinear, so any superposition with something approaching a definite position will fall apart. After all, you can do a double slit experiment with an electron in vacuum.

The answer is that you can only really talk about location in the rough sense of "localization", and that requires an imperfect crystal. Usually, this is thought of as electrons (or phonons) scattering from impurities and the likes. Scattering causes localization. That's why you can talk about electron location in incoherent transport (like the Drude model) -- where there's a lot of scattering. For coherent transport (without scattering), you can't really talk about electron location.

That's not really an exact answer, but it should give you some terms to google.

Another way to think of it is this: in a perfect crystal, the electron (or phonon) wave functions are eigenstates of the momentum operator; i.e. they have a definite momentum. By the uncertainty principle, that means you can know nothing about their location. However, if you add imperfections to the crystal, the wavefunctions are no longer eigenstates of the momentum operator, so you don't know the momentum exactly. That opens the door for you to have some information about their location. In general, adding more and more imperfections means you know less and less about their momentum, which means that you can know more and more about their position. Consider the extreme case of an impurity that creates a potential well so deep that it traps an electron. Now you're in the opposite limit of where we started: we know the position very well but the momentum not at all. In short, impurities (and "scattering") can give collective excitations a position.

PS. Don't think too hard about relaxons. It's just not worth it.

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

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