Filling the Box with Fermions

If we fill a cold box with \bgroup\color{black}$N$\egroup fermions, they will all go into different low-energy states. In fact, if the temperature is low enough, they will go into the lowest energy \bgroup\color{black}$N$\egroup states.

If we fill up all the states up to some energy, that energy is called the Fermi energy. All the states with energies lower than \bgroup\color{black}$E_F$\egroup are filled, and all the states with energies larger than \bgroup\color{black}$E_F$\egroup are empty. (Non zero temperature will put some particles in excited states, but, the idea of the Fermi energy is still valid.)

\begin{displaymath}\bgroup\color{black} {\pi^2\hbar^2\over 2mL^2}(n_x^2+n_y^2+n_z^2)={\pi^2\hbar^2\over 2mL^2}r_n^2 <E_F \egroup\end{displaymath}

Since the energy goes like \bgroup\color{black}$n_x^2+n_y^2+n_z^2$\egroup, it makes sense to define a radius \bgroup\color{black}$r_n$\egroup in \bgroup\color{black}$n$\egroup-space out to which the states are filled.


The number of states within the radius is

\begin{displaymath}\bgroup\color{black} N=(2)_{spin}{1\over 8}{4\over 3}\pi r_n^3 \egroup\end{displaymath}

where we have added a factor of 2 because fermions have two spin states. This is an approximate counting of the number of states based on the volume of a sphere in \bgroup\color{black}$n$\egroup-space. The factor of \bgroup\color{black}${1\over 8}$\egroup indicates that we are just using one eighth of the sphere in \bgroup\color{black}$n$\egroup-space because all the quantum numbers must be positive.

We can now relate the Fermi energy to the number of particles in the box.

\begin{displaymath}\bgroup\color{black} E_F={\pi^2\hbar^2\over 2mL^2}r_n^2
...i^2\hbar^2\over 2m}\left({3n\over\pi}\right)^{2\over 3} \egroup\end{displaymath}

We can also integrate to get the total energy of all the fermions.

\begin{displaymath}\bgroup\color{black} E_{\mathrm tot}=2{1\over 8}\int\limits_0...
...hbar^2\over 10m}\left({3n\over\pi}\right)^{5\over 3}L^3 \egroup\end{displaymath}

where the last step shows how the total energy depends on the number of particles per unit volume \bgroup\color{black}$n$\egroup. It makes sense that this energy is proportional to the volume.

The step in which \bgroup\color{black}$E_F$\egroup and \bgroup\color{black}$E_{tot}$\egroup is related to \bgroup\color{black}$N$\egroup is often useful.

\bgroup\color{black}$\displaystyle E_F={\pi^2\hbar^2\over 2m}\left({3N\over\pi L^3}\right)^{2\over 3} $\egroup
\bgroup\color{black}$\displaystyle E_{\mathrm tot}={\pi^3\hbar^2\over 10mL^2}\left({3N\over\pi}\right)^{5\over 3} $\egroup

Jim Branson 2013-04-22