Question

Consider 3D cube with N particles. At ground state, calculate the total energy of the system. . Calculate the energy of the ground state of the system and the maximum particle energy called the Fermi energy. Given that there are

Answer 1

Let us consider a gas of $N$ non-interacting fermions in 3D cube of volume $V=L^3$. We will use occupied numbers to describe this system. Set $r_n=\sqrt{n_x^2+n_y^2+n_z^2}$ is a radius of "filled states" sphere in $n$-space.

For the ground state we have $n_x=n_y=n_z=1$. So, the energy of the ground state is

$$E_{\operatorname{GS}}=\frac{\pi^2 \hbar^2}{2mL^2}r_n^2=\frac{3\pi^2 \hbar^2}{2mL^2}$$

The number of states within the radius is

$$N=2 \cdot \frac{1}{8} \cdot \frac{4}{3}\pi r_n^3$$

where we have added a factor of 2 because fermions have two spin states, the factor of 1/8 indicates that we are just using one eighth of the sphere in n-space because all the quantum numbers must be positive.

Then we can relate the Fermi energy to the number of particles in the cube:

$$E_F=\frac{\pi^2 \hbar^2}{2mL^2}r_n^2=\frac{\pi^2 \hbar^2}{2mL^2} \left( \frac{3N}{\pi}\right)^{2/3}=\frac{\pi^2 \hbar^2}{2m} \left( \frac{3N}{\pi V}\right)^{2/3}$$

And we can integrate to get the total energy of all the fermions:

$$E_{\operatorname{total}}=2\cdot \frac{1}{8} \int\limits_0^{r_n}4\pi r^2 \frac{r^2\pi^2 \hbar^2}{2mL^2} \, dr=\frac{\pi^3 \hbar^2}{2mL^2} \cdot \frac{r_n^5}{5}=\frac{\pi^3 \hbar^2}{10mL^2} \cdot \left( \frac{3N}{\pi}\right)^{5/3}$$