Total ground state energy of N fermions in a 3D box

[kamaln]

Hello everybody, I just have no idea how to start this problem so i was hoping you guys would point me in the right direction and then i'll be able to go on by myself


the problem asks to show that the total ground state energy of N fermions in a three dimensional box is given by E total = 3/5*N*(E fermi)



some relevant equations to this are:
D(E)dE=2(1/8)4pi*r^2 dr

N=[V(2m(E fermi))^(3/2)] / [3*(hbar^3)pi^2]

E fermi = [(hbar^2)/2m]*[(3pi^2*N)/V]^2/3

I don't have any attempt at a solution because i don't know how to begin proving that E total = 3/5*N*(E fermi). Thanks in advance for any help that I get.

 

[jbsteele]

The first step is to define the total ground state energy. The total ground state energy of N fermions in a three-dimensional box is the sum of the single-particle energies of each fermion, given by: E_total = ∑_i^N E_i where E_i is the single-particle energy of the ith fermion. To calculate E_i, first use the equation D(E)dE=2(1/8)4πr2 dr to find the density of single-particle states with energy between E and E + dE. This gives:D(E) = [(2m/hbar^2)*(E/π)]^(3/2) Then, use this to calculate the Fermi energy E_Fermi, given by the equation: E_Fermi = [(hbar^2)/2m]*[(3π^2*N)/V]^2/3 Finally, use the equation for the density of single-particle states to calculate the single-particle energy of each fermion: E_i = (hbar^2/2m)*[(3π^2*i)/V]^2/3 where i is the index of the fermion, i.e. the ith fermion has an energy E_i. Now that the single-particle energies have been found, the total ground state energy can be calculated by summing these energies: E_total = ∑_i^N E_i = ∑_i^N (hbar^2/2m)*[(3π^2*i)/V]^2/3 After simplifying, this equation can be written as: E_total = (N/V)*(hbar^2/2m)*[(3π^2*N)/V]^2/3 which, after substituting the expression for the Fermi energy, reduces to: