# Derivation of density of states (free electrons)

I am reading Condensed matter physics from M.Marder.

This is the derivation for the density of states for free electrons.

\begin{aligned} D(\mathcal{E}) &=\int[d \vec{k}] \delta\left(\mathcal{E}-\mathcal{E}_{\vec{k}}^{0}\right) \\ &=4 \pi \frac{2}{(2 \pi)^{3}} \int_{0}^{\infty} d k k^{2} \delta\left(\mathcal{E}-\mathcal{E}_{\vec{k}}^{0}\right) \\ &=\frac{1}{\pi^{2}} \int_{0}^{\infty} \frac{d \mathcal{E}^{0}}{\left|d \mathcal{E}^{0} / d k\right|} \frac{2 m \mathcal{E}^{0}}{\hbar^{2}} \delta\left(\mathcal{E}-\mathcal{E}^{0}\right) \\ &=\frac{m}{\hbar^{3} \pi^{2}} \sqrt{2 m \mathcal{E}} \end{aligned},

where

$$\int[d \vec{k}] \equiv \frac{2}{V} \sum_{\vec{k}}=\int d \vec{k} D_{\vec{k}}=\frac{2}{(2 \pi)^{3}} \int d \vec{k}$$,

The factor of 2 accounts for electron spin. V is the volume.

and $$\mathcal{E}_{\vec{k}}^{0}=\frac{\hbar^{2} k^{2}}{2 m}$$.

In step one he says that he changes the integral to polar coordinates because $$\mathcal{E}^0$$ depends upon the magnitude and not the direction of $$\mathbf{k}$$. So, where does the $$4\pi$$ come from? Shouldn't it be $$2\pi$$?

In second step he writes $$k$$ in terms of $$\mathcal{E}^0$$. In the end I will have

$$$$\frac{m}{\hbar^3 \pi^2} \int_0^\infty d \mathcal{E}^0 \sqrt{2 m \mathcal{E}^0} \delta(\mathcal{E}-\mathcal{E}^0)$$$$

I think he used the property

$$\int_{-\infty}^{+\infty} f(x) \delta(x-a) dx = f(a)$$

but the integral goes from 0 to infinity. Can you explain to me what is going on here?

## 1 Answer

In step one, the angular part of the integral is $$4\pi$$ because the $$\vec{k}$$ integral is an integral in 3 dimensions, so you're integrating over a sphere: $$\int d\vec{k} = \underbrace{\int_{0}^{2\pi} \,d\phi \int_{0}^\pi \sin\theta \,d\theta }_{=\,4\pi} \int_0^\infty k^2 \,dk \,.$$ In step two, the lower bound of the integral can be extended to $$-\infty$$, since the integrand is zero for $$\mathcal{E}^0<0$$, i.e.: $$\int_0^\infty \sqrt{2m\mathcal{E}^0} \,d\mathcal{E}^0 = \int_{-\infty}^\infty \sqrt{2m\mathcal{E}^0} \,d\mathcal{E}^0 \,.$$

• Ok, understood. Thanks!
– AA10
Sep 29, 2019 at 10:04