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GR0877 Problem 100

PROBLEM STATEMENT: The partition function $Z$ in statistical mechanics can be written as

$\displaystyle Z = \sum_{r} e^{- E_r /kT}$,

where the index $r$ ranges over all possible microstates of a system and $E_r$ is the energy of microstate $r$. For a single quantum mechanical harmonic oscillator with energies

$E_{n} = \left( n + \frac{1}{2} \right)\hbar \omega$, where $n = 0, 1, 2,\dots,$

the partition function $Z$ is given by which of the following?
(A) $\displaystyle Z = e^{- \frac{1}{2}\hbar \omega /kT}$

(B) $\displaystyle Z = e^{\frac{1}{2}\hbar \omega /kT}$

(C) $\displaystyle Z = e^{\frac{1}{2}\hbar \omega /kT} - 1$

(D) $\displaystyle Z = e^{\frac{1}{2}\hbar \omega /kT} + 1$

(E) $\displaystyle Z = \frac{e^{\frac{1}{2}\hbar \omega /kT}}{e^{\hbar \omega /kT}-1}$

SOLUTION: (E) The partition function for a single QM harmonic oscillator is:
$\displaystyle Z = \sum_{n=0}^{\infty} \exp\left( -\frac{E_n}{kT}\right) = \exp\left( -\frac{\hbar \omega}{2kT}\right) \sum_{n=0}^{\infty} \exp\left( -\frac{n \hbar \omega}{kT}\right)$.
$\displaystyle \sum_{n=0}^{\infty} \exp\left( -\frac{n \hbar \omega}{kT}\right) = 1 + \exp\left( -\frac{\hbar \omega}{kT}\right) + \exp\left( -\frac{2 \hbar \omega}{kT}\right) + \dots$ which is an infinite geometric series:
$\displaystyle \sum_{n=0}^{\infty} \exp\left( -\frac{n \hbar \omega}{kT}\right) = \frac{1}{1 - \exp\left( -\frac{\hbar \omega}{kT}\right)}$. Thus, $\displaystyle Z = \frac{\exp\left( -\frac{\hbar \omega}{2kT}\right)}{1 - \exp\left( -\frac{\hbar \omega}{kT}\right)} = \frac{\exp\left( \frac{\hbar \omega}{2kT}\right)}{\exp\left( \frac{\hbar \omega}{kT}\right) -1}$.

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