The rate at which a phonon with wavevector $\vec{q}$ is absorbed is given by $$\frac{1}\tau \propto n(\hbar \omega(\vec{q}))$$ This is pretty obvious to me. The more phonons there are the more often the absorption.

Meanwhile, the rate at which a phonon with wavevector $\vec{q}$ is emitted can be written as $$\frac{1}\tau \propto [1+n(\hbar \omega(\vec{q}))]$$

I would like to know the reasoning behind this. If able, I want a physical interpretation to this.

What I thought of so far:

  1. $[1+n(\hbar \omega(\vec{q}))]$ is equal to $-n(-\hbar \omega(\vec{q}))$, but I am not sure how to interpret this.
  2. Since it is an emission process, right when the emission happens the number of existing phonons with wavevector $\vec{q}$ increases by one.

1 Answer 1


If we start in a state with $n$ phonons and calculate the probability that one phonon is emitted, we typically have a matrix element like $$\langle n+1|b^\dagger|n\rangle=\sqrt{n+1}\langle n+1|n+1\rangle=\sqrt{n+1}.$$

On the other hand, if a phonon is absorbed, the matrix element will have a factor $$\langle n|b|n+1\rangle=\sqrt{n}\langle n|n\rangle=\sqrt{n}.$$

Averaging over Boltzmann distribution transforms $n$ into the phonon distribution function.

In other words, this is a consequence of Bose-Einstein statistics. The factors would be different, if we talked about the emission and absorption of electrons, which obey Fermi-Dirac statistics.

This reflects the fact that a photon can be emitted, even if there are no other photons present spontaneous emission, but it can be absorbed only when there are some photons already (otherwise, there's nothing to absorb.)

Note also that in the classical regime, where the number if phonons is very big, one can neglect $1$.


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