Show that in the Iimit of large numbers, the exact probability of equation (9-9) becomes the Boltzmann probability of (9-17). Use the fact that $\frac{\mathbf{K}\mathbf{!}}{\left(K-k\right)\mathbf{!}}\mathbf{\equiv }{\mathit{K}}^{\mathbf{k}}$, which holds when $\mathit{k}\mathbf{<}\mathbf{<}\mathit{K}$.

The Boltzmann distribution (also called the Gibbs distribution) is a probability distribution or probability measure that gives the probability that a system will be in a certain state as a function of the energy of that state and the temperature of the system.

The Boltzmann probability in terms of ${\mathit{E}}_{\mathbf{n}}$and ${\mathit{k}}_{\mathbf{B}}\mathit{T}$.

$\mathit{P}\left(E_n\right)\mathbf{=}\frac{{\mathbf{e}}^{\frac{\mathbf{-}{\mathbf{E}}_{\mathbf{n}}}{{\mathbf{k}}_{\mathbf{B}}\mathbf{T}}}}{{}_{\mathbf{n}\mathbf{=}\mathbf{0}}{}^{\mathbf{\infty }}{\mathbf{e}}^{\frac{\mathbf{-}{\mathbf{E}}_{\mathbf{n}}}{{\mathbf{k}}_{\mathbf{B}}\mathbf{T}}}}$

For a system of $N$harmonic oscillators, the allowed energy $E_N$for one oscillator is given by:

$E_n=nh{\omega }_0$

Expression into $P$:

role="math" localid="1660135304939" $\begin{array}{c}P\left(E_n\right)=\frac{e^{\frac{-nh{\omega }_0}{k_{B}T}}}{{}_{n=0}{}^{\infty }{e}^{\frac{-nh{\omega }_0}{k_{B}T}}}\\ =\frac{e^{\frac{-nh{\omega }_0}{k_{B}T}}}{{}_{n=0}{}^{\infty }{\left[e^{\frac{-nh{\omega }_0}{k_{B}T}}\right]}^n}\end{array}$

Use the following expression to solve further.

${}_{n=0}{}^{\infty }{x}^n=\frac{1}{1-x}$

Applying denominator of $P$:

$\begin{array}{c}P\left(E_n\right)=e^{\frac{-nh{\omega }_0}{k_{B}T}}\frac{1}{1-e^{\frac{-nh{\omega }_0}{k_{B}T}}}\\ =\left(1-e^{\frac{-nh{\omega }_0}{k_{B}T}}\right)e^{\frac{-nh{\omega }_0}{k_{B}T}}\end{array}$

Use the following expression to solve further.

$\begin{array}{l}{k}_{B}T=\frac{h{\omega }_0}{\ln \left(1+\frac{N}{M}\right)}\\ \frac{h{\omega }_0}{k_{B}T}=\ln \left(1+\frac{N}{M}\right)\end{array}$

Therefore,

role="math" localid="1660136050103" $\begin{array}{c}P\left(E_n\right)=\left(1-e^{-\ln \left(1+\frac{N}{M}\right)}\right)e^{-n\ln \left(1+\frac{N}{M}\right)}\\ =1-\frac{1}{e^{\ln \left(1+\frac{N}{M}\right)}}{e}^{-\ln \left(1+\frac{N}{M}\right)}\\ =1-\frac{1}{1+\frac{N}{M}}{e}^{-\ln \left(1+\frac{N}{M}\right)}\\ =\frac{1+\frac{N}{M}-1}{1+\frac{N}{M}}{e}^{-\ln \left(1+\frac{N}{M}\right)}\end{array}$

$\begin{array}{c}P\left(E_n\right)=\frac{\frac{N}{M}}{\frac{\left(\frac{N}{M}\right)}{M}}{e}^{-\ln \left(1+\frac{N}{M}\right)}\\ =\frac{N}{M+N}{e}^{-\ln \left(1+\frac{N}{M}\right)}\end{array}$

The expression for $P$is $\frac{N}{M+N}{e}^{-\ln \left(1+\frac{N}{M}\right)}$.