Density of States at the Fermi Energy

[nboogerz]

The density of states at the fermi energy is given by

D(E_F)=(3/2)n/E_F

I understand the density of states is the number of states per energy per unity volume, accounting for n/E_F. I don't understand how the 3/2 multiplying factor accounts for the volume?

 

[PhysTech]

Dimensionally you are correct. But in this case, unfortunately, you have to perform the detailed calculus steps in order to get that factor. First let us determine the expression for $n$. In $\bf k$-space you need to count the total number of occupied states. This can be computed as seen in the steps below \begin{eqnarray} n&amp;=&amp;2\int_{{\rm FS}}\frac{d^{3}\mathbf{k}}{(2\pi)^{3}} \\<br /> &amp;=&amp; \frac{2}{(2\pi)^{3}}\int_{0}^{k_{F}}dk\int_{0}^{2 \pi}d\phi\int_{0}^{\pi}d\theta\left(k^{2}<br /> \sin(\theta)\right) \\<br /> &amp;=&amp; \frac{2}{(2\pi)^{3}}\left(\int_{0}^{k_{F}}dk\, k^{2}\right)<br /> \left(\int_{0}^{2\pi}d\phi\right)<br /> \left(\int_{0}^{\pi}d\theta\,\sin(\theta)\right) \\<br /> &amp;=&amp; \frac{2}{2\pi^{2}}\int_{0}^{k_{F}}dk\, k^{2} \\<br /> &amp;=&amp; \frac{k_{F}^{3}}{3\pi^{2}}<br /> \end{eqnarray} where $\int_{{\rm FS}}$ is an integral from the origin till the (spherical) Fermi Surface (FS). The $k^{2} \sin(\theta)$ in the second step is simply the Jacobian in spherical coordinates. Now, $n$ is the total number of available (and filled) states for $k\le k_{F}$. The total number of states available up to some arbitrary $k$ is simply N(k)=\frac{k^{3}}{3\pi^{2}} The density of states (for the isotropic case) is given by \begin{eqnarray} D(E) &amp;=&amp; \frac{dN(E)}{dE}\\<br /> &amp;=&amp; \frac{dN(k)}{dk}\left(\frac{dE}{dk}\right)^{-1} \end{eqnarray} For a parabolic dispersion we have E=\frac{\hbar^{2}k^{2}}{2m^{*}} Therefore, at $k=k_F$ we have \begin{eqnarray} D(E_{F}) &amp;=&amp; D(E(k_{F}))\\<br /> &amp;=&amp; \frac{m^{*}k_{F}}{\hbar^{2}\pi^{2}}\\<br /> &amp;=&amp; \frac{k_{F}^{3}}{\pi^{2}}\left(\frac{\hbar^{2}k_{F}^{2}}{m^{*}}\right)^{-1}\\<br /> &amp;=&amp; \frac{3}{2}\left(\frac{k_{F}^{3}}{3\pi^{2}}\right)<br /> \left(\frac{\hbar^{2}k_{F}^{2}}{2m^{*}}\right)^{-1} \end{eqnarray} From the above expressions you can make the appropriate substitutions D(E_{F}) = \frac{3}{2}nE_{F}^{-1}