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The energy of harmonics oscillator, started of [tex]U=-\frac{\partial}{\partial \beta} \ln Z[/tex] is equal to [tex]\frac{\hbar \omega}{2} + \frac{\hbar \omega}{exp(\beta \hbar \omega)-1}[/tex].

At high temperature, i could say that [tex]exp (\beta \hbar \omega ) \approx 1 + (\beta \hbar \omega )[/tex], and then [tex]U=\frac{\hbar \omega}{2} + kT[/tex], therefore at high temperature [tex]\frac{\hbar \omega}{2}[/tex] is negligible compared to [tex]kT[/tex], and then [tex]U \approx k T [/tex].

I need find arguments about why is incorrect say that [tex]\frac{\hbar \omega}{2} + \frac{\hbar \omega}{exp(\beta \hbar \omega)-1}[/tex] at [tex]\beta \to 0[/tex] (high temperature) is equal to [tex]\infty[/tex]. This motivated by the fact that [tex]k T = k \cdot \infty = \infty[/tex]. I understand that at high temperature the energy has a asyntote equal to kT (http://www.av8n.com/physics/oscillator.htm#sec-e-vs-t ), but still need argumens.

Also ¿why the harmonics oscillators need a specific heat at high temperature?. In this case the specific heat is equal to k. But if the energy us infinity, then the specific heat would be zero.

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# Quantum harmonics oscillator at high temperature

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