Check the equations 5.74, 5.75, and 5.77 for the example in section 5.4.1

• Bosons are frequently connected with force carrier particles, whereas fermions are typically associated with matter.
• Today's particle physics, however, makes it difficult to distinguish between the two ideas.
• Under extreme conditions, weakly interacting fermions can also exhibit bosonic behavior.

Need to check if equationsandgive the same result as counting states "by hand" in example in Section 5.4.1. We calculate occupied states for distinguishable particles, fermions and bosons by formulae:

_{$$\begin{gathered} Q_d(N_{1,}{N}_{2,}{N}_{3,}....)=N!\prod _{n-1}^{\infty }\frac{{d_n}^{N_n}}{N_n!} \\ {Q}_f(N_{1,}{N}_{2,}{N}_{3,}....)=\prod _{n-1}^{\infty }\frac{d_n!}{N_n!(d_n-N_n)!} \\ {Q}_b(N_{1,}{N}_{2,}{N}_{3,}....)=\prod _{n-1}^{\infty }\frac{(N_n+d_n-1)!}{N_n!(d_n-1)!} \end{gathered}$$}

Where is number of particles in the system and is degree of degeneracy of oneparticle state. For all configurations in example and . So we have:

$$\begin{gathered} Q_d=6\prod _{n-1}^{\infty }\frac{1}{N_n!} \\ {Q}_f=\prod _{n-1}^{\infty }\frac{1}{N_n!(1-N_n)!} \\ {Q}_f=\prod _{n-1}^{\infty }\frac{N_n}{N_n!}=1 \end{gathered}$$

So, the number of occupied states

First configuration$N_{11}=3$others zero $\left\{\begin{array}{l}{Q}_d=6.\frac{1}{3!}=1\\ Q_f=0\\ Q_b=1\end{array}\right.$

Second configuration$N_5=1,N_{13}=2:$others zero$\left\{\begin{array}{l}{Q}_d=6.\frac{1}{1!}.\frac{1}{2!}=3\\ Q_f=0\\ Q_b=1\end{array}\right.$

Third configuration$N_1=2,N_{19}=1:$others zero$\left\{\begin{array}{l}{Q}_d=6.\frac{1}{1!}.\frac{1}{2!}=3\\ Q_f=0\\ Q_b=1\end{array}\right.$

Fourth configuration$N_5=N_7=N_{17}$others zerorole="math" localid="1658218029425" $\left\{\begin{array}{l}{Q}_d=6=6\\ Q_f=0\\ Q_b=1\end{array}\right.$

Using equations for number of occupied states for distinguishable particles, fermions and bosons, we calculated number of occupied states for example in Section 5.4.1.