Suppose you had three (noninteracting) particles, in thermal equilibrium in a one-dimensional harmonic oscillator potential, with a total energy$\mathit{E}\mathbf{=}\frac{\mathbf{9}}{\mathbf{2}}\mathit{h}\mathit{\omega }$ .

(a) If they are distinguishable particles (but all with the same mass), what are the possible occupation-number configurations, and how many distinct (threeparticle) states are there for each one? What is the most probable configuration? If you picked a particle at random and measured its energy, what values might you get, and what is the probability of each one? What is the most probable energy?

(b) Do the same for the case of identical fermions (ignoring spin, as we did in Section 5.4.1).

(c) Do the same for the case of identical bosons (ignoring spin).

• Fermions are typically associated with matter, whereas bosons are commonly associated with force carrier particles.
• However, in today's particle physics, the distinction between the two concepts is blurred.
• Under extreme conditions, weakly interacting fermions can also exhibit bosonic behavior.

There are three particles in a harmonic oscillator. Total energy of the system is$E_0=\frac{9h\omega }{2}$. There is need to find all possible configurations of particles if they are distinguishable, fermions or bosons. Total energy of system in harmonic potential is:

_{$$\begin{gathered} E=h\omega (n_1+n_2+n_3+\frac{3}{2}) \\ \frac{9h\omega }{2}=h\omega (n_1+n_2+n_3+\frac{3}{2}) \\ \frac{9}{2}=(n_1+n_2+n_3+\frac{3}{2}) \\ {n}_1+n_2+n_3=3 \end{gathered}$$}

Possible configurations are

$$\begin{gathered} n_1=n_2=n_3=1:(1,1,1) \\ {n}_1=0,n_2=1,n_3=2:\left(,0,1,2\right),(2,0,1),(2,1,0),(0,2,1),(1,2,0),(1,0,2) \\ {n}_1=n_2=0,n_3=3:(3,0,0),(0,3,0),(0,0,3) \end{gathered}$$

(a)

Distinguishable particles

We are interested in occupation-number configurations

(1)All three-particle in-state;$1;(0,3,0,0,0,.........)$

(2)All three-particle in a different state;$(1,1,1,0,0,.......)$

(3)Two particles in the ground state one are third;$(2,0,0,1,0,.........)$

Second configuration is the most probable one, because there are six possibilities for it to happen. First one has only one possibility, and third one three.

Now we need to find the most probable energy. Energy$E_0=\sout{h}\omega l$can be measured in two configurations: second and third. Second configuration appears 6 times in 10 (total number of possibilities) and to measure energy$E_0$we have chance of 1/3 . Third configuration appears 3 times in 10, and to measure energy we have chance of 2.1/3 . So total probability of measuring energy$E_0$is:

$$\begin{gathered} p_0=\frac{6}{10}\frac{1}{3}+\frac{3}{10}\frac{2}{3}=\frac{2}{5}{E}_0=\frac{h\omega }{2} \\ {p}_1=\frac{6}{10}\frac{1}{3}+\frac{1}{10}=\frac{3}{10}{E}_1=\frac{3h\omega }{2} \\ {p}_2=\frac{6}{10}\frac{1}{3}=\frac{1}{5}{E}_2=\frac{5h\omega }{2} \\ {p}_3=\frac{3}{10}\frac{1}{3}=\frac{1}{10}{E}_3=\frac{7h\omega }{2} \end{gathered}$$$p_0=\frac{6}{10}\frac{1}{3}$

Most probable energy is$E_0$ .

Therefore in distinguishable particles, there are three possible configurations for occupation numbers. One configuration has one state, other 6 and third 3.The most probable configuration is the second one with probability 6/10 . Most probable energy for distinguishable particles is ground state.