Discussion
The partition function of a two-dimensional oscillator whose energy \( E_{n_{x}n_{y}} = (n_{x} + n_{y} + 1)\hbar\omega \), \( n_{x} = 0, 1, 2, \ldots \); \( n_{y} = 0, 1, 2, \ldots \), is :
Answer: C
The partition function is \( \frac{e^{\hbar\omega/k_{B}T}}{\left(e^{\hbar\omega/k_{B}T} - 1\right)^{2}} \).
The two quantum numbers are independent, so the sum factorises into two geometric series, with \( x = e^{-\hbar\omega/k_{B}T} \):
\[ \begin{aligned} Z &= x \sum_{n_{x}} x^{n_{x}} \sum_{n_{y}} x^{n_{y}} = \frac{x}{(1 - x)^{2}} \\ &= \frac{e^{\hbar\omega/k_{B}T}}{\left(e^{\hbar\omega/k_{B}T} - 1\right)^{2}} \end{aligned} \]
the last step multiplying top and bottom by \( x^{-2} \). The options with \( 2k_{B}T \) in the exponents halve the level spacing \( \hbar\omega \), and the option with \( +1 \) has the wrong sign in the denominator; none of them is the sum of these geometric series.
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