Phonon density of states in statistical thermodynamics

Question

Phonon density of states (DoS), $g_{(\omega_i)}$, help understand the distribution of states across frequencies of vibration. $\int g_{(\omega_i)}d\omega_i =$ the number of states between $\omega_i$ and $d\omega_i$.

DoS values aren't integers, at least none in my calculations were. Wouldn't the possible number of states in any range of frequency be an integer, or is this some quantum/wave effect?

If we think about a crystal with $N$ atoms. Each atom has its own set of frequencies, so the equations above work for each atom. Codes like Phonopy output DoS as if it represented the frequencies of vibration of the whole crystal system of all $N$ atoms. The following equations are from Dr. Brent Fultz's review of the vibrational thermodynamics of materials. $Z$ is the canonical harmonic partition function.

\begin{equation} Z_i = \frac{e^{-\beta \epsilon_i/2}}{1 - e^{-\beta \epsilon_i}}\tag{1} \end{equation}

\begin{equation} Z_N = \prod_{i}^{3N}\frac{e^{-\beta \epsilon_i/2}}{1 - e^{-\beta \epsilon_i}}\tag{2} \end{equation}

The first equation is about each of the $N$ oscillators and the second about the whole system.

I'm confused with regards to the way lattice dynamics codes like Phonopy output DoS data. Can someone help me organize my thoughts here and put the right pieces in place?

Answer 1

There are two possible things tripping you up here:

1. Phonons are collective oscillations: they involve the motion of all the atoms together. Therefore it only makes sense to talk about the phonons of the whole system, not any individual atom.
2. The density of states only makes sense as you take $N\to \infty$. For finite $N$, there is a finite/discrete number of phonon modes, and so the density of states would be a series of delta functions. As you take $N\to \infty$, the spacing between the modes goes to zero and you're left with a continuum and thus a continuous DOS.