* Math: Density of states

A math theorem tells that for a function $f(x)$,

$$\lim_{L\to \infty}\frac{1}{L}\sum_{n=0}^{\infty}f(n/L)=\int_0^{\infty}f(x)dx.$$ (8)

Assume standing waves in a box of size $L^3$ with wave vectors

$${\bf k}=(\pi n_x/L,\pi n_y/L,\pi n_z/L).$$

Assume we wish to calculate a function $f({\bf k})=f(k)$ that only depends on the modulus of ${\bf k}$. We use

$$\lim_{L\to \infty}\frac{1}{L^3}\sum_{n_x,n_y,n_z=0}^{\infty}f\left(\pi n_x/L,\pi n_y/L,\pi n_z/L)\right) = \frac{1}{\pi^3}\int_0^{\infty}f({\bf k})dk_xdk_ydk_z,$$ (9)

i.e.

$$\lim_{L\to \infty}\frac{1}{L^3}\sum_{k_x,k_y,k_z>0}f(k) = \frac{1}{\pi^3} \int_0^{\infty}f(k)dk_xdk_ydk_z =\frac{1}{\pi^3 2^3}\int_{-\infty}^{\infty}f(k)dk_xdk_ydk_z$$

$$= \frac{1}{(2\pi)^3}\int_{0}^{\infty}dk 4\pi k^2f(k) =\int_0^{\infty}dk\frac{k^2}{2\pi^2}f(k)$$

$$= \int_0^{\infty}d\omega\frac{\omega^2}{2\pi^2c^3}f(\omega) =:\int_0^{\infty}d\omega\rho(\omega)f(\omega)$$

$$\rho(\omega) := \frac{\omega^2}{2\pi^2c^3}.$$ (10)

Here, $\rho(\omega)$ is the density of states if there was only one polarization direction per possible `state' (wave vector ${\bf k}$). If there are two polarization directions, like for unpolarized light, the density of states is twice as large, i.e. $\rho(\omega):=\frac{\omega^2}{\pi^2c^3}$.