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If I have the hamiltonian of the simple harmonic oscillator

$$H = \frac{p^2}{2m} + \frac{1}{2} m \omega ^2 x^2 $$

Then it's partition function is:

$$Z = \frac{k_b T}{\hbar \omega} $$

You can get the average energy using $$U = \frac{\partial \ln{Z}}{\partial \beta} $$

where $\beta = k_b T $. in 1D , $$ U = kb * T $$

My question is, can you integrate U back up to obtain the original partition function, $Z = \frac{k_b T}{\hbar \omega} $? The closest I get is

$$\int U d\beta = \beta (T) - \beta(0) = k_b T - \hbar \omega $$

where $\hbar \omega $ is the zero point energy. The problem is that I then have to exponentiate to recover $Z$, and I get no where close to the original partition function. Any advice here?

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closed as unclear what you're asking by Aaron Stevens, ZeroTheHero, Gert, user191954, noah Feb 20 at 16:19

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  • $\begingroup$ Wow I'm an idiot. You're right, beta is 1/kbT $\endgroup$ – Drew Lilley Feb 15 at 1:53
  • $\begingroup$ How are you getting that partition function? $\endgroup$ – Aaron Stevens Feb 15 at 2:55
  • $\begingroup$ Just integrated through the p and x coordinates from - infinity to infinity $\endgroup$ – Drew Lilley Feb 15 at 4:02
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As Aaron pointed out, I miswrote $\beta$. I should have $\beta = \frac{1}{k_b T}$ , and then it works out because:

$$- \int U d\beta = -\frac{1}{\beta} d\beta = \log{\frac{1}{\beta}} - \log{0} = \log{\frac{1}{ \beta \hbar \omega}} = \log{\frac{k_b T}{\hbar \omega}}$$

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