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I was looking into statistical and quantum mechanics and their overlap but I can't seem to solve a basic question as to what a partition function,

$$Z = \frac{\exp\left(-\beta\hbar \omega/2\right)}{1 - \exp(-\beta\hbar \omega)} $$ looks like when the temperature goes to infinity, where $\beta = 1/kT$. When $T\to\infty$, $\beta\to 0$. Obviously plugging this into $Z$ just cancels everything out. So I've tried using a Taylor series expansion

$$\exp(x) = 1+x+\cdots$$ to solve it. However I still can't get rid of the $\beta$. Best case scenario I end up with $Z=-1/2$ which is incorrect since I need to manipulate the $Z$ value later on to play with expectation energies.

If you guys have any idea on how to approach this please let me know. I understand rules are quite harsh on stack exchange and that I'm new, so I apologize for any infractions. Thank you for your help.

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  • $\begingroup$ Note: this is the partition function for a harmonic oscillator with the zero-point energy term $\frac12 \hbar\omega$ included. You can remove that by setting the numerator to 1. The property that, as $\beta\to 0$, you see $Z \to \infty,$ is impossible to remove because the open-form of the expression becomes literally $1 + 1 + 1 + 1+\dots$ and heads out to infinity, and the only way to assign it a value is by those weird zeta-function methods that say it has the value $-1/2.$ $\endgroup$ – CR Drost Feb 2 '17 at 21:20
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Why do you want to get rid of the $\beta$? The partition function behaves as $1/\beta\hbar\omega$ to first order in $\beta$ and goes to infinity when $\beta\rightarrow 0$. Because of this, you cannot compute expectation values directly in the limit $\beta\rightarrow 0$. You first need to compute things to lowest order in $\beta$ and then take the limit. For example, the equipartition theorem for the average energy is recovered to first order in $\beta$: $$\langle E\rangle=-{\partial\ln Z\over\partial\beta}={1\over \beta}=k_BT$$ and therefore diverges in the limit $\beta\rightarrow 0$.

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  • $\begingroup$ I do not understand what you mean. Ive seen online that the answer seems to be E=h-bar omega /2. The method they used to solve used free energy, entropy and trace methods that are beyond my scope of mathematics. $\endgroup$ – FitnessRegiment Feb 2 '17 at 21:03
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    $\begingroup$ $E=\hbar\omega /2$ is the energy of the ground-state of the quantum oscillator. The expectation value of the energy is equal to $\hbar\omega /2$ at zero temperature and not at infinite-temperature. $\endgroup$ – Christophe Feb 3 '17 at 21:13
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Since $Z=\dfrac{1}{2}\operatorname{csch}\dfrac{\beta\hbar\omega}{2}$ and $U=\partial_\beta\ln\dfrac{1}{Z}=\dfrac{\hbar\omega}{2}\coth\dfrac{\beta\hbar\omega}{2}$, at high $T$ (low $\beta$) $U\approx\dfrac{1}{\beta}$, and at low $T$ (high $\beta$) $U\approx\dfrac{\hbar\omega}{2}$.

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