Suppose you have a "box" in which each particle may occupy any of $10$single-particle states. For simplicity, assume that each of these states has energy zero.

(a) What is the partition function of this system if the box contains only one particle?

(b) What is the partition function of this system if the box contains two distinguishable particles?

(c) What is the partition function if the box contains two identical bosons?

(d) What is the partition function if the box contains two identical fermions?

(e) What would be the partition function of this system according to equation $7.16$?

(f) What is the probability of finding both particles in the same single particle state, for the three cases of distinguishable particles, identical bosom, and identical fermions?

Each particle can have any $10$states each with zero energy.

So, $E\left(r_i\right)=0for1\le r_i\le 10$

Formula for particle function with box having one particle:

$Z_l=\sum _i{e}^{-\beta E\left(r_i\right)}$

where, $\beta =\frac{1}{kT}$$\beta =\frac{1}{kT}$

We substitute $E\left(r_i\right)=0$

localid="1647243851142" $$\begin{gathered} Z_l=\sum _1^{10}{e}^{\beta \left(0\right)} \\ =\sum _1^{10}1 \\ =10 \end{gathered}$$

So, Partition function of system with box having only one particle is$10$.

For two distinguishable particles partition function is:

$Z={Z_l}^2$

Substitute $Z_l=10$

role="math" localid="1647244178631" $$\begin{gathered} Z={\left(10\right)}^2 \\ =100 \end{gathered}$$

So, the partition function of the system if the box contains two distinguishable particles is$100$.

To put two identical bosons in same single particle state the possibility is $10$.

So, to put two identical bosons in different single particle state the possibility is calculated as:

$$\begin{gathered} {}_2{}^{10}C=\frac{10!}{2!8!} \\ =45 \end{gathered}$$

So, total possible states are:

$$\begin{gathered} Z=10+45 \\ =55 \end{gathered}$$

Hence, the partition function if the box contains two identical bosons is$55$.

There is no possibility to put two identical fermions in same single particle state.

For different single particle state possibilities are calculated by:

$$\begin{gathered} {}_2{}^{10}C=\frac{10!}{2!8!} \\ =45 \end{gathered}$$

So, the partition function if the box contains two identical fermions is$45$.

Partition function for $N$indistinguishable and non interacting particles is given by,

$Z=\frac{{Z_l}^N}{N!}$

Substitute $Z_l=10$and $N=2$

$$\begin{gathered} Z=\frac{{10}^2}{2!} \\ =50 \end{gathered}$$

So, the partition function of the system when box have indistinguishable particles is$50$.

For distinguishable particles number of accessible states are $Z=100$.

For both particles in same single particle state, accessible states are $10$.

So, probability of finding both particle in same single particle state is:

$$\begin{gathered} P=\frac{10}{100} \\ =10\% \end{gathered}$$

So, probability for distinguishable particles is role="math" localid="1647245922911" $10\%$.

For identical bosons number of accessible states are $Z=55$.

For both bosons in same single particle state, accessible states are $10$.

So, probability of finding both bosons in same single particle state is:

$$\begin{gathered} P=\frac{10}{55} \\ =18\% \end{gathered}$$

So, probability for identical bosons is $18\%$.

For identical fermions number of accessible states are $Z=45$.

For both fermions in same single particle state, accessible states is $0$.

So, probability of finding both fermions in same single particle state is:

$$\begin{gathered} P=\frac{0}{45} \\ =0\% \end{gathered}$$

So, probability for identical fermions is $0\%$.