Answer to Question #91046 in Quantum Mechanics for Jian

Question

Consider a three-dimensional cubical well with N particles. Given that all particles occupy the ground state, calculate the total energy of the system. Now assume that each energy level can hold no more than two particles. 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

The total energy of the system in the first situation will be (for equal $L$ and all $n=1$):

$E_{tot}=N\cdot\frac{h^2}{8m}\Big(\frac{n_x^2}{L_x^2}+\frac{n_y^2}{L_y^2}+\frac{n_z^2}{L_z^2}\Big)=\frac{3Nh^2}{8mL^2}.$

By analogy, for maximum two particles at each level the energy is

$E_{g.s.}=\frac{3h^2}{4mL^2}.$

The Fermi energy for two particles ($V=L^3,\space n=N/2$) at each level is

$E_F=\frac{h^2}{8\pi^2 m}\cdot\Big(3\pi^2\frac{n}{V}\Big)^{2/3}=\frac{h^2}{8mL}\cdot\Big(\frac{3N}{2\pi^2}\Big)^{2/3}.$