Answer to Question #126117 in Quantum Mechanics for christopher seebaran

Question

Consider a one-dimensional infinite square 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. Find an expression for the density of states for the infinite square well.

Answer 1

Total energy of the system for N at the ground level $(n=1)$ is

$$E_{tot}=N.\frac{n^2h^2}{8mL^2}=\frac{Nh^2}{8mL^2}$$

The ground is now occupied by 2 particles, hence according to the previous expression;

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

Fermi energy for the 2 electrons for each state, $(n=N/2)$

$$E_F=\frac{h^2}{8\pi^2m}.(\frac{n\pi}{L})^2=\frac{h^2}{8m}.(\frac{N}{2L})^2.$$

Expression for the density of states for the infinite square well:

the density of states in k-space is;

$$n(k')=\frac{dN(k')}{dk'}=\frac{k'L^2}{(2\pi)}.$$

$$E=\frac{k^2h^2}{(2M}. \frac{DE}{dk}=\frac{kh^2}{M}.$$

$$n(E)=\frac{dN}{dE}=\frac{DN}{dk}.\frac{dk}{dE}=\frac{ML^2}{(2\pi h^2)}$$