Discussion
A simple harmonic oscillator with an angular frequency \(\omega\) is in thermal equilibrium with a reservoir at absolute temperature \(T\), with \(\omega = \frac{2k_{B}T}{\hbar}\). Which one of the following is the partition function of the system?
Answer: A
The partition function is \(\frac{e}{e^{2} - 1}\).
For a quantum oscillator with levels \(E_{n} = (n + \frac{1}{2})\hbar\omega\), the sum is geometric:
\[ \begin{aligned} Z &= \sum_{n=0}^{\infty} e^{-(n + 1/2)x} = \frac{e^{-x/2}}{1 - e^{-x}} \\ x &= \frac{\hbar\omega}{k_{B}T} = 2 \end{aligned} \]
So \(Z = e^{-1}/(1 - e^{-2})\); multiplying top and bottom by \(e^{2}\) gives \(e/(e^{2} - 1)\).
The \(+1\) in the denominator would come from an alternating series, which the oscillator does not have, and \(e/(e - 1)\) is what \(x = 1\) would give, forgetting the factor 2 in \(\omega = 2k_{B}T/\hbar\).
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