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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

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

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?

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1 Answer 1

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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.: $$\int_0^\infty \sqrt{2m\mathcal{E}^0} \,d\mathcal{E}^0 = \int_{-\infty}^\infty \sqrt{2m\mathcal{E}^0} \,d\mathcal{E}^0 \,.$$

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

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