Why can we use $$\frac{1}{V}\sum_{k \in 1BZ} \approx \int_{k\in 1BZ} \frac{dk}{(2\pi)^d}$$ where $d$ is the dimension of the lattice, for phonon related integration?

I know the above sum converted to integral is valid for free electrons in Sommerfeld model. But why can it still be used for phonons?


2 Answers 2


That formula comes from the following $$\frac{1}{V}\sum_k = \frac{1}{Na^3}\frac{(2\pi)^3}{(2\pi)^3}\sum_k = \frac{2\pi}{N_1 a}\frac{2\pi}{N_2 a}\frac{2\pi}{N_3 a}\frac{1}{(2\pi)^3}\sum_k\\ =\frac{1}{(2\pi)^3}\sum_k \Delta k_x \Delta k_y \Delta k_z $$

And now the important step

Whenever it's feasible to think that $\mathbf{N}$ is very large, practically $N\rightarrow\infty$ we can write $$\frac{1}{(2\pi)^3}\sum_k \Delta k_x \Delta k_y \Delta k_z\to\int\limits_{k\in FBZ}\frac{d^3k}{(2\pi)^3}$$

which for electrons it clearly holds, but what makes you think that it will still hold for phonons?

The mean number of phonons in a given band with wavevector $k$, is given by a Bose-Einstein distribution $$\bar{n}_{ks} = \frac{1}{e^{\beta\hbar\omega_s(k)}-1} $$ where $\omega_s(k)$ is the dispersion relation of the $s$-branch. You can easily see that, at $T=0$, the mean number of phonons is finite but not at all big, while the number of electrons still remains of the order $10^{23}$. The number of electrons in a solid will always remain big, while the number of phonons will vary depending on various physical conditions.

  • $\begingroup$ Maybe my description is not qutie clear. It's the integration related with phonon wave vector k,not with phonons. $\endgroup$
    – Yuan Fang
    Mar 7, 2020 at 15:18
  • $\begingroup$ Isn't the number of phonons equal to 0 at T=0? $\endgroup$ Mar 7, 2020 at 19:57

The logic for phonons is exactly the same as it is for electrons. The approximation relies on the fact that the crystal is very large, so the density of allowed $\vec{k}$ in reciprocal space is very high. In that limit, the sum can be approximated as an integral. (The allowed reciprocal lattice vectors are determined by the crystal lattice; it doesn't matter if you're talking about electrons or phonons.)

For reference, see the book Electrons and Phonons by J. Ziman sec 1.9 (or really any good book on solid state physics).

For the record, it has nothing to do with the Bose-Einstein distribution or the actual number of phonons in the system.


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.