Exercise 52 gives the Boltzmann distribution for the special case of simple harmonic oscillators, expressed in terms of the constant$\epsilon$, $\equiv N\hslash {\omega }_0/(2s+1)$and Exercise 53 gives the two quantum distributions in that case. Show that both quantum distributions converge to the Boltzmann in the limit$k_{\text{B}}T\gg$.

The quantum distributions for some energy $E$are the Bose-Einstein distribution $N_{BE}\left(E\right)$and the Fermi-Dirac distribution$N_{FD}\left(E\right)$

$N_{BE}\left(E\right)=\frac{1}{\frac{e^{\frac{E}{k_{B}T}}}{1-e^{\frac{-E}{k_{B}T}}}-1}$ ……. (1)

$N_{FD}\left(E\right)=\frac{1}{\frac{e^{\frac{E}{k_{B}T}}}{e^{\frac{E}{k_{B}T}}-1}+1}$ ……. (2)

Here,$E\to$Energy

$k_B\to$Boltzmann's constant

$T\to$Temperature

$E=\frac{N\hslash {\omega }_0}{2s+1}$ ……. (3)

Here,

$\text{N}\to$Number of oscillators

$\hslash \to$Planck's reduced constant

${\omega }_0\to$Fundamental angular frequency

$s\to$Spin

The Boltzmann distribution$N_{\text{Boltz}}\left(E\right)$is given by

$N_{\text{Boltz}}\left(E\right)=\frac{E}{k_{B}T}\frac{1}{e^{\frac{E}{k_{B}T}}}$ ……. (4)

An approximation for$e^z$when $z$is very small will be helpful:

$e^z\approx 1+z$ ……. (5)

$N_{BE}\left(E\right)=\frac{1}{\frac{e^{\frac{E}{k_{B}T}}}{1-e^{-\frac{E}{k_{B}T}}}-1}$

If $k_{B}T>>E$then the$E/k_{B}T$in the exponent of the denominator will be very small, meaning that the approximation of equation (5) can be used with$z$being .$E/k_{B}T$

$\begin{array}{c}{N}_{BE}\left(E\right)\approx \frac{1}{\frac{e^{\frac{E}{k_{B}T}}}{1-(1-\frac{E}{k_{B}T}}-1}\\ \approx \frac{1}{\frac{e^{\frac{E}{k_{B}T}}}{1-1+\frac{E}{k_{B}T}}-1}\\ \approx \frac{1}{\frac{k_{B}T}{E}{e}^{\frac{E}{k_{B}T}}-1}\end{array}$

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Further solving,

$\begin{array}{c}{N}_{BE}\left(E\right)\approx \frac{1}{\frac{k_{B}T}{E}{e}^{\frac{E}{k_{B}T}}-1}\\ \approx \frac{1}{\frac{k_{B}T}{E}{e}^{\frac{E}{k_{B}T}}}\\ \approx \frac{E}{k_{B}T}\frac{1}{e^{\frac{E}{k_{B}T}}}\end{array}$

So, the Bose Einstein distribution approaches $N_{BE}\left(E\right)\approx \frac{E}{k_{B}T}\frac{1}{e^{\frac{E}{k_{B}T}}}$and matches the Boltzmann distribution of equation (4), thus showing their correlation at high temperature.

To show that the Fermi-Dirac distribution of equation (2).

$N_{FD}\left(E\right)=\frac{1}{\frac{e^{\frac{E}{k_{B}T}}}{e^{\frac{E}{k_{B}T}}-1}+1}$

It approaches the Boltzmann distribution when$k_{B}T>>E$, the approximation of equation (5) is used, with $z$being $E/k_{B}T$(since that ratio is very small for the given condition).

$\begin{array}{c}{N}_{FD}\left(E\right)\approx \frac{1}{\frac{e^{\frac{E}{k_{B}T}}}{{(1+\frac{E}{k_{B}T})}^{-1}}+1}\\ \approx \frac{1}{\left(\frac{e^{\frac{E}{k_{B}T}}}{\frac{E}{k_{B}T}}\right)+1}\\ \approx \frac{1}{\frac{k_{B}T}{E}{e}^{\frac{E}{k_{B}T}}+1}\end{array}$

Since$k_{B}T$is much greater than$E$then $k_{B}T/E$will be very large, making$(k_{B}T/E)e^{E/k_{B}T}$very large as well. Given that, adding 1 to it won't significantly affect it; so, it can be ignored.

$\begin{array}{c}{N}_{FD}\left(E\right)\approx \frac{1}{\frac{k_{B}T}{E}{e}^{\frac{E}{k_{B}T}}+1}\\ \approx \frac{1}{\frac{k_{B}T}{E}{e}^{\frac{E}{k_{B}T}}}\\ \approx \frac{E}{k_{B}T}\frac{1}{e^{\frac{E}{k_{B}T}}}\end{array}$

So, the Fermi-Dirac distribution approaches to $N_{FB}\left(E\right)\approx \frac{E}{k_{B}T}\frac{1}{e^{\frac{E}{k_{B}T}}}$and matches the Boltzmann distribution of equation (4) thus showing their correlation at high temperature