Answer to Question #111831 in Classical Mechanics for Tanya Saini

Question

Q9.Consider a system of N >>1weakly interacting identical particles, with each particle
having possible non-degenerate states with energies ϵ, 2ϵ, 3ϵ,.., nϵ,... where n is a natural
number. What is the ground state energy of the system if the particles are, (a) Bosons, (b)
Fermions.

Answer 1

A system of very weakly interacting particles practically preserves the energy spectrum of single particles. Bosons have a whole spin and their Bose-Einstein statistics do not impose restrictions on the number of particles in the same quantum state. The probability amplitude of an ensemble of such particles is even with respect to permutations of all quantum numbers. Thus the ground state energy corresponds to the state when all bosons are in the lowest energy state

(1) $E_b=\epsilon\cdot N$.

Fermions have a half-integer spin and obey the Fermi-Dirac statistics in which only two particles can be in the same state one of which has spin 1/2 and the second -1/2. For such particles, the wave function is antisymmetric with respect to their permutations, so when completely identical particles are permuted (including the quantum number of spins), the probability amplitude gets the value zero. There can only be two particles at each level of the fermion ensemble so the next ones have no choice but to fill in the higher energy levels. The ground state of fermions will have the following energy.

(2)$E_f=\sum_{n=1}^{N/2}2\cdot(\epsilon n)\simeq 2\epsilon\frac{(\frac{N}{2}+1)(\frac{N}{2})}{2}\simeq\epsilon\frac{N^2}{4}$

Where 2 occurs due to a pair of electrons at the same level, and we have neglected all values of the order of one in the formula (2) compared to a very large number N>>1.

Answer: The ground state energy of the system if the particles are (a) Bosons is $\epsilon\cdot N$ , (b)

Fermions is $\epsilon\frac{N^2}{4}$ if N>>1.