Density of States at the Fermi Energy


by nboogerz
Tags: fermi energy
nboogerz
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#1
Mar2-13, 03:53 PM
P: 4
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?
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#2
Mar2-13, 09:45 PM
P: 24
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 [tex] \begin{eqnarray} n&=&2\int_{{\rm FS}}\frac{d^{3}\mathbf{k}}{(2\pi)^{3}} \\
&=& \frac{2}{(2\pi)^{3}}\int_{0}^{k_{F}}dk\int_{0}^{2 \pi}d\phi\int_{0}^{\pi}d\theta\left(k^{2}
\sin(\theta)\right) \\
&=& \frac{2}{(2\pi)^{3}}\left(\int_{0}^{k_{F}}dk\, k^{2}\right)
\left(\int_{0}^{2\pi}d\phi\right)
\left(\int_{0}^{\pi}d\theta\,\sin(\theta)\right) \\
&=& \frac{2}{2\pi^{2}}\int_{0}^{k_{F}}dk\, k^{2} \\
&=& \frac{k_{F}^{3}}{3\pi^{2}}
\end{eqnarray} [/tex] 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 [tex] N(k)=\frac{k^{3}}{3\pi^{2}} [/tex] The density of states (for the isotropic case) is given by [tex] \begin{eqnarray} D(E) &=& \frac{dN(E)}{dE}\\
&=& \frac{dN(k)}{dk}\left(\frac{dE}{dk}\right)^{-1} \end{eqnarray} [/tex] For a parabolic dispersion we have [tex] E=\frac{\hbar^{2}k^{2}}{2m^{*}} [/tex] Therefore, at ##k=k_F## we have [tex] \begin{eqnarray} D(E_{F}) &=& D(E(k_{F}))\\
&=& \frac{m^{*}k_{F}}{\hbar^{2}\pi^{2}}\\
&=& \frac{k_{F}^{3}}{\pi^{2}}\left(\frac{\hbar^{2}k_{F}^{2}}{m^{*}}\right)^ {-1}\\
&=& \frac{3}{2}\left(\frac{k_{F}^{3}}{3\pi^{2}}\right)
\left(\frac{\hbar^{2}k_{F}^{2}}{2m^{*}}\right)^{-1} \end{eqnarray} [/tex] From the above expressions you can make the appropriate substitutions [tex] D(E_{F}) = \frac{3}{2}nE_{F}^{-1} [/tex]


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