# Quantum harmonics oscillator at high temperature

1. Oct 31, 2013

### alejandrito29

Hello

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

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

I need find arguments about why is incorrect say that $$\frac{\hbar \omega}{2} + \frac{\hbar \omega}{exp(\beta \hbar \omega)-1}$$ at $$\beta \to 0$$ (high temperature) is equal to $$\infty$$. This motivated by the fact that $$k T = k \cdot \infty = \infty$$. 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.

2. Oct 31, 2013

### The_Duck

The precise statement would be that $\frac{\hbar \omega}{2} + \frac{\hbar \omega}{exp(\beta \hbar \omega)-1} \to \infty$ as $\beta \to 0$. We always have to speak of limits when discussing infinity. Nothing ever equals infinity.

The energy is never infinity. For any finite temperature, no matter how large, the energy is finite. So you can always evaluate the specific heat $d U / d T$ and it is always nonzero, for any $T > 0$.

3. Oct 31, 2013

### alejandrito29

¿why The energy is never infinity?

but the literature says that at $$T \to \infty$$ , then $$U \to k T$$

sorry, but still i don´t understand

4. Nov 1, 2013

### f95toli

Because nothign is ever infinite. This is why we have to talk about limits. This is as true here as in in every other case in math and physics.

Which is correct, but does not mean that T ever EQUALS infinity.

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