Density of States at the Fermi Energy

by nboogerz
Tags: fermi energy
nboogerz is offline
Mar2-13, 03:53 PM
P: 4
The density of states at the fermi energy is given by


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?
Phys.Org News Partner Physics news on
Vacuum ultraviolet lamp of the future created in Japan
Understanding the energy and charge transfer of ions passing through membranes
High-temperature plasmonics eyed for solar, computer innovation
PhysTech is online now
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}^{\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]

Register to reply

Related Discussions
density of states, fermi energy Advanced Physics Homework 1
Fermi energy (density of electron) Quantum Physics 1
Fermi energy, electron density, conduction electrons of Ru Atomic, Solid State, Comp. Physics 0
energy density of surface states Quantum Physics 0
Fermi's Golden Rule and Density of States Quantum Physics 8