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?


1 Answer 1


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.

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

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