# How can I derive the Phonon Density of Normal Modes in two and three dimensions?

The density of normal modes in 1D has already been discussed in another post (LINK)

In essence, the formula is:

$$D(\omega) = \frac{1}{2\pi}\int\mathrm{d}q\,\delta(\omega-\omega(q))$$

resulting in a phonon level density: $$D(\omega)=\frac{1}{\pi}\frac{dk}{d \omega}$$

I want to derive this phonon level density for the 2-D and the 3-D case aswell. I have already looked at this powerpoint (LINK) and I have gone through the relevant pages in Ashcroft/Mermin but I am very confused.

What formula do I use for the two and three dimensional case?

There are some that have $$L$$ as a factor in front of the integral. Others have $$\frac{1}{V}$$ or $$\frac{1}{(2 \pi)^d}$$. Some are written as a sum, others as an integral. Some have vectors in the integral, others don't. Can someone maybe show me how the two or three dimensional phonon-level density/density of normal modes is derived and what formula I should use for this?

Edit 1: I found a formula for the 2D case but I am not sure how it was derived: $$D(\omega)=\frac{kL^2}{2\pi}\frac{dk}{d \omega}$$

In the two dimensional case the dispersion relation is: $$\omega(k_x,k_y)=\sqrt{\frac{2K}{M}(2-2\cos{k_x a}-2\cos{k_y a})}$$

Taking the derivative (I hope I am differentiating correctly):

$$\frac{d \omega }{d \vec{k}}=\frac{2a (\sin{k_x a}+ \sin{k_y a})}{2\sqrt{\frac{2K}{M}(2-2\cos{k_x a}-2\cos{k_y a})} }$$

This result correctly reproduces the Van-Hove singularity. The dispersion relation reaches a maximum at the edge of the first Brillouin-Zone ($$\pm \frac{\pi}{a})$$, which leads to $$\frac{d \omega}{d\vec{k}}=0$$ at $$\frac{\pi}{a} \implies D(\omega)=\frac{kL^2}{2\pi}\frac{1}{0} \longrightarrow$$ singularity

The one thing that still makes no sense to me is how I can go from the general equation for the density of modes (phonon-level density) taken from Ashcroft/Mermin p.465:

$$D(\omega)=\sum_s \int \frac{dS}{(2\pi)^3} \frac{1}{\lvert \vec{\nabla} \omega_s(\vec{k})\rvert}$$

to the 2D density: $$D(\omega)=\frac{kL^2}{2\pi} \frac{dk}{d \omega}$$

Any ideas?

One way to do this is to start by solving for the allowed $\mathbf{k}$ values for an $L \ \times \ L \ \times L$ crystal using the usual procedure found in any book on solid state physics (if you're unfamiliar, see A & M pg. 430-432). The allowed $| \mathbf{k} |$ values are integer multiples of $\frac{2 \pi}{L}$.

The corresponding density of k-states is then the reciprocal, $\left( \frac{L}{2\pi} \right) ^d$. In 3 dimensions the (differential) number of states within a thin shell of radius from $k \rightarrow k + dk$ is

$$dN = (number \ of \ states \ per \ unit \ k \ volume) \times (differential \ shell \ volume)$$

$$= \left( \frac{L}{2\pi} \right) ^3 \left( 4 \pi k^2 dk \right)$$

Note that the second term is just $dS$ in k-space. The DOS is

$$D(\omega) = \frac{dN}{d\omega} = \frac{dN}{dk}\frac{dk}{d\omega} = \left( \frac{L}{2\pi} \right) ^3 \left( 4 \pi k^2 \right) \frac{dk}{d\omega} = \left( \frac{L}{2\pi} \right) ^3 S \frac{dk}{d\omega} = \frac{k^2 L ^3}{2\pi^2}\frac{dk}{d\omega}$$

where $S$ is the surface area of the sphere. Note that this is equivalent to

$$L^3\int \frac{dS}{(2\pi)^3}\frac{dk}{d\omega}$$

which is the result given in A & M with a factor of $L^3$ out front (the DOS is sometimes given per unit volume, which would divide out the $L^3$).

The subscript $s$ in the A & M DOS is just the phonon branch (solution for $\omega$). You sum over $s$ to get the contribution to the DOS of the acoustic and optical phonon branches. For a single branch, you can drop the sum.

• That is a great explanation. It makes total sense now. Thank you very much!
– qmd
Jun 20, 2016 at 14:37