# What's phonon mean free path

This is probably a naive question but still.

Phonons are quasiparticles that emerge when we quantize motion of a lattice. In this sense, they have no location in space, they are just energy quanta of certain vibration modes. Then, why do people speak about phonon mean free path as though phonon is a real particle moving in space?

• Because the vibrations (and thermal transport) are real, can be described as phonons, those phonons do scatter, and the scattering can be observed experimentally? – Jon Custer Sep 25 '15 at 20:36
• Phonons do scatter, but does it mean we can think of a moving particle? To me, scattering means disappearance of a quantum in one mode, and appearance in another. Still, none of them really move in space. Am I wrong? – Kirill Moskovtsev Sep 28 '15 at 15:23
• But phonons are quanta. Raman scattering makes and consumes them (Stokes vs anti-Stokes) for example. – Jon Custer Sep 28 '15 at 16:16

For each phonon state (without any interactions between phonons) one may assign some wave vector $\vec{q}$, mode index $j$ and specific energy $\hbar \omega_{j}(\vec{q})$. Furthermore, for this wave vector value and mode there exist some displacement pattern of atoms in the lattice determined by the eigenvector $\vec{e}(\vec{q}j)$. The group velocities $\vec{v}_{j}(\vec{q})$ of phonos can be obtained from the gradient of $\vec{v}_{j}(\vec{q}) = \nabla_{\vec{q}} \omega_{j}(\vec{q})$ and it is the velocity of this phonon "wave packet" (or displacement pattern) in the lattice.
Sometimes one describes the interaction of phonons by using, for example, time-dependent perturbation theory. One assumes that these harmonic phonon states are stationary and interactions occur between these harmonic phonon states. For instance, in 3-phonon process two phonons may annihilate and third is created or vice versa (energy and crystal-momentum conservation is assumed to be valid). One may thus calculate, for example, the transition rate (one per time unit) due to 3-phonon processes by using Fermi-Golden rule (or some other perturbative technique). The relaxation time $\tau_{j}(\vec{q})$ for each phonon state $\vec{q}j$ is sometimes defined as an inverse of this transition rate.
The mean free path due to 3-phonon processes is defined as $v_{j}(\vec{q}) \tau_{j}(\vec{q})$, which is the mean distance for phonon to travel until it interacts. Other interactions may also have some effect on mean free path. If the interactions are sufficiently strong this elementary excitation phonon picture starts to become inaccurate and the group velocities etc. are rather ill-defined.