Exercise 52 gives the Boltzmann distribution for the special case of simple harmonic oscillators, expressed in terms of the constant $\mathbf{\epsilon }\mathbf{=}{\mathbf{Nh\omega }}_{\mathbf{0}}\mathbf{/}\mathbf{(}\mathbf{2}\mathbf{s}\mathbf{+}\mathbf{1}\mathbf{)}$. Exercise 53 gives the Bose-Einstein and Fermi-Dirac distributions in that case. Consider a temperature low enough that we might expect multiple particles to crowd into lower energy states:${\mathbf{k}}_{\mathbf{\text{B}}}\mathbf{T}\mathbf{=}\frac{\mathbf{1}}{\mathbf{5}}\mathbf{\epsilon }$. How many oscillators would be expected in a state of the lowest energy,$\mathbf{E}\mathbf{=}\mathbf{0}$? Consider all three-classically distinguishable. boson, and fermion oscillators - and comment on the differences.

${\mathbf{N}}_{\mathbf{\text{Boltz}}}\mathbf{\left(}\mathbf{E}\mathbf{\right)}\mathbf{=}\frac{\mathbf{E}}{{\mathbf{k}}_{\mathbf{B}}\mathbf{T}}\frac{\mathbf{1}}{{\mathbf{e}}^{\mathbf{F}\mathbf{/}{\mathbf{h}}_{\mathbf{B}}\mathbf{T}}}$

${\mathbf{N}}_{\mathbf{BE}}\mathbf{\left(}\mathbf{E}\mathbf{\right)}\mathbf{=}\frac{\mathbf{1}}{\frac{{\mathbf{e}}^{\mathbf{[}{\mathbf{K}}_{\mathbf{B}}\mathbf{T}}}{\mathbf{1}\mathbf{-}{\mathbf{e}}^{\mathbf{-}{\mathbf{EJ}}_{\mathbf{B}}\mathbf{T}}}\mathbf{-}\mathbf{1}}$

${\mathbf{N}}_{\mathbf{FD}}\mathbf{\left(}\mathbf{E}\mathbf{\right)}\mathbf{=}\frac{\mathbf{1}}{\frac{{\mathbf{e}}^{{\mathbf{\ell \pi }}_{\mathbf{B}}\mathbf{T}}}{{\mathbf{e}}^{\mathbf{1}\mathbf{/}{\mathbf{\pi }}_{\mathbf{B}}\mathbf{T}}\mathbf{-}\mathbf{1}}\mathbf{+}\mathbf{1}}$

${\mathrm{k}}_{\mathrm{B}}\mathrm{T}=\frac{1}{5}\mathrm{\xi }$

The Boltzmann distribution of equation (1) is checked first, using the energy condition of equation (5), along with $\mathrm{E}$being 0:

$\begin{array}{l}{\mathrm{N}}_{\text{Boltz}}\left(\mathrm{E}\right)=\frac{\mathrm{E}}{{\mathrm{k}}_{\mathrm{B}}\mathrm{T}}\frac{1}{{\mathrm{e}}^{\text{EIIk}{\mathrm{B}}^{\mathrm{T}}}}\\ {\mathrm{N}}_{\text{Boltz}}\left(0\right)=\frac{\mathrm{E}}{\frac{1}{5}\mathrm{E}}\frac{1}{{\mathrm{e}}^{\mathrm{\omega }0\mathrm{v}}\left(\frac{1}{5}\mathrm{E}\right)}\\ {\mathrm{N}}_{\text{Boltz}}\left(0\right)=\frac{5}{{\mathrm{e}}^0}\\ {\mathrm{N}}_{\text{Boltz}}\left(0\right)=5\end{array}$

So, the number of oscillators expected if they were classically distinguishable would be${\mathrm{N}}_{\text{Boltz}}\left(0\right)=5.$

The Bose-Einstein distribution of equation is checked with the energy condition of equation (5) along with E being 0:

$\begin{array}{c}{\mathrm{N}}_{\mathrm{BE}}\left(\mathrm{E}\right)=\frac{1}{\frac{{\mathrm{e}}^{{\mathrm{EZ}}_{\mathrm{B}}\mathrm{T}}}{1-{\mathrm{e}}^{-{\mathrm{BE}}_{\mathrm{B}}\mathrm{T}}-1}}\\ {\mathrm{N}}_{\mathrm{BE}}\left(0\right)=\frac{1}{\frac{{\mathrm{e}}^{\mathrm{\omega }\text{jx}\left(\frac{1}{5}\mathrm{E}\right)}}{1-\mathrm{E}\left(\frac{1}{5}\mathrm{E}\right)}-1}\\ =\frac{1}{\frac{{\mathrm{e}}^0}{1-{\mathrm{e}}^{-5}}-1}\\ =\frac{1}{\frac{1}{1-{\mathrm{e}}^{-5}-1}}\end{array}$

That can be rearranged some to get rid. of the complex fraction:

$\begin{array}{c}{\mathrm{N}}_{\mathrm{BE}}\left(0\right)=\frac{1}{\frac{1}{1-{\mathrm{e}}^{-5}-\frac{1-{\mathrm{e}}^{-5}}{1-{\mathrm{e}}^{-5}}}}\\ =\frac{1}{\left[\frac{1-(1-{\mathrm{e}}^{-5})}{1-{\mathrm{e}}^{-5}}\right]}=\frac{1-{\mathrm{e}}^{-5}}{{\mathrm{e}}^{-5}}\end{array}$

Solving further, we get

$\begin{array}{c}{\mathrm{N}}_{\mathrm{BE}}\left(0\right)=\frac{1}{{\mathrm{e}}^{-5}}\frac{{\mathrm{e}}^{-5}}{{\mathrm{e}}^{-5}}\\ ={\mathrm{e}}^{-5}-1\\ =147\end{array}$

So, the number of oscillators expected if they were bosons would be ${\mathrm{N}}_{\mathrm{BE}}\left(0\right)=147$.

From equation (3) the Fermi-Dirac distribution is,

${\mathrm{N}}_{\mathrm{FD}}\left(\mathrm{E}\right)=\frac{1}{\frac{{\mathrm{e}}^{{\mathrm{E\pi }}_{\mathrm{B}}\mathrm{T}}}{{\mathrm{e}}^{{\mathrm{FK}}_{\mathrm{B}}\mathrm{T}}-1}+1}$

$\begin{array}{c}{\mathrm{N}}_{\mathrm{FD}}\left(0\right)=\frac{1}{\frac{{\mathrm{e}}^{\mathrm{cov}\left(\frac{1}{5}\mathrm{E}\right)}}{{\mathrm{e}}^{\mathrm{FD}}\left(\frac{1}{5}\mathrm{E}\right)}+1}\\ =\frac{1}{\frac{{\mathrm{e}}^0}{{\mathrm{e}}^5}+1}\\ =\frac{1}{\frac{1}{{\mathrm{e}}^5}+1}\\ =\frac{1}{\frac{1}{{\mathrm{e}}^5}+\frac{{\mathrm{e}}^5}{{\mathrm{e}}^5}}\\ =\frac{1}{\left[\frac{1+{\mathrm{e}}^5}{{\mathrm{e}}^5}\right]}\\ =\frac{{\mathrm{e}}^5}{1+{\mathrm{e}}^5}\end{array}$

Therefore,

$\begin{array}{c}{\mathrm{N}}_{\mathrm{FD}}\left(0\right)=\frac{{\mathrm{e}}^5}{1+{\mathrm{e}}^5}\\ =\frac{148.4}{148.4+1}\\ =0.9933\end{array}$

So, the number of oscillators expected if they were fermions${\mathrm{N}}_{\mathrm{FD}}\left(0\right)=0.9933$ .