Question

8. Two identical particles of mass m interact with an external potential which is harmonic, but they do not interact with each other. The Hamiltonian of the system is À = pi + P + mw€ (zł + x2). H = 2m + 2m + 2 What are the energy levels of the system? Given that the particles are bosons, determine the partition function. Repeat the calculation given that the two particles are fermions.

Isn't it distinguishable? Because they do not interact with each other.
But the answer of boson and fermion are different each other..
How can I get the answer?

Answer 1

The fact that the particles are not interacting means that the potential between them is not to be considered. For example, while calculating the atomic energy levels, we consider that the interaction between the electrons is negligible and we only consider the interaction of electrons with the nucleus.

Similarly, here, we consider only the interaction of the particles with the potential. From the equation of Hamiltonian of the system, we can say that the potential is harmonic oscillator potential.

So, energy of particle in state n1 is$e_1 = \hbar \omega(n_1+1/2)$

So, Total energy of two particles is

$\epsilon_{n1,n2} =e_1+e_2 = \hbar \omega(n_1+1/2) + \hbar \omega(n_1+1/2) = \hbar \omega(n_1+n_2 +1)$

Where n1 and n2 are integers.

The partition function of a single particle in quantum harmonic oscillator is given by

$Z = \frac{e^{-\hbar \omega/2K_bT}}{(1-e^{-\hbar \omega/K_bT)}}$

The probability of a particle being in n th state is

$p(n) = \frac{1}{Z}*e^{-(n+1/2)\hbar \omega/K_bT} = e^{-n\hbar \omega/K_bT} *(1-e^{-\hbar \omega/K_bT})$

$p(n1,n2) = p(n1)*p(n2) = e^{-(2n_1)\hbar \omega/K_bT} *(1-e^{-\hbar \omega/K_bT})(1-e^{-2\hbar \omega/K_bT})$

(Since here, the second state is dependent on whether the first state is filled)

$Z= \frac{1}{p(n1,n2)}*e^{-(n_1+n_2+1)\hbar \omega/K_bT}$

a) For bosons for ground state, n1,n2 = 0

So,

$Z= \frac{e^{-\hbar \omega/K_bT}}{(1-e^{-\hbar \omega/K_bT})(1-e^{-2\hbar \omega/K_bT})}$

b) For fermions, due to exclusion principle,

n1 =0 and n2 =1

$Z= \frac{e^{-2\hbar \omega/K_bT}}{(1-e^{-\hbar \omega/K_bT})(1-e^{-2\hbar \omega/K_bT})}$