# Can you integrate internal energy to get original partition function? [closed]

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?

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

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• Wow I'm an idiot. You're right, beta is 1/kbT – Drew Lilley Feb 15 at 1:53
• How are you getting that partition function? – Aaron Stevens Feb 15 at 2:55
• Just integrated through the p and x coordinates from - infinity to infinity – Drew Lilley Feb 15 at 4:02

As Aaron pointed out, I miswrote $$\beta$$. I should have $$\beta = \frac{1}{k_b T}$$ , and then it works out because:
$$- \int U d\beta = -\frac{1}{\beta} d\beta = \log{\frac{1}{\beta}} - \log{0} = \log{\frac{1}{ \beta \hbar \omega}} = \log{\frac{k_b T}{\hbar \omega}}$$