Filling the Box with Fermions

If we fill a cold box with $N$ 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 $N$ 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 $E_F$ are filled, and all the states with energies larger than $E_F$ are empty. (Non zero temperature will put some particles in excited states, but, the idea of the Fermi energy is still valid.)

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

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

The number of states within the radius is

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

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 $n$-space. The factor of ${1\over 8}$ indicates that we are just using one eighth of the sphere in $n$-space because all the quantum numbers must be positive.

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

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

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

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

where the last step shows how the total energy depends on the number of particles per unit volume $n$. It makes sense that this energy is proportional to the volume.

The step in which $E_F$ and $E_{tot}$ is related to $N$ is often useful.

$$E_F={\pi^2\hbar^2\over 2m}\left({3N\over\pi L^3}\right)^{2\over 3}$$

$$E_{\mathrm tot}={\pi^3\hbar^2\over 10mL^2}\left({3N\over\pi}\right)^{5\over 3}$$