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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 ${\mathit{k}}_{\mathbf{B}}\mathit{T}$.

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

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

$E_n=nh{蝇}_0$

Expression into $P$:

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

Use the following expression to solve further.

${}_{n=0}{}^{鈭�}{x}^n=\frac{1}{1-x}$

Applying denominator of $P$:

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

Use the following expression to solve further.

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

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 $\frac{N}{M+N}{e}^{-\ln \left(1+\frac{N}{M}\right)}$.