# Partition function of the QHO

1. Jun 18, 2012

### unchained1978

I've derived Z for the quantum harmonic oscillator and was wondering if anyone could verify I did everything correctly. I don't have any experience working with exponential traces so I want to make sure I'm using them correctly.
Z is defined as $\mathcal{Z}= tr(e^{-\beta H})$.
So the natural thing to do is write the exponential as a power series $e^{-\beta H}=\sum \frac{(-\beta H)^{n}}{n!}$ and using schrodinger's equation $H|\psi\rangle = E |\psi\rangle$ this gives you $e^{-\beta H}|\psi\rangle=\sum \frac{(-\beta E)^{n}}{n!}|\psi\rangle→tr(e^{-\beta H})=\sum_{n} e^{-\beta E_{n}}$
Writing out the energy levels this gives $e^{-\frac{1}{2}\beta \hbar\omega}\sum_{n} e^{-\beta\hbar\omega n}$ Which when summed over gives $\mathcal{Z}=\frac{e^{\frac{1}{2}\beta \hbar\omega}}{e^{\beta\hbar\omega}-1}$
Which is the right result I think. I'm just a bit nervous about the trace argument.

2. Jun 18, 2012

### jfy4

I got
$$\frac{e^{-\frac{1}{2}\beta\hbar \omega}}{1-e^{-\beta\hbar\omega}}$$
also this should be in the Quantum sub-forum, not Classical.

3. Jun 18, 2012

### Dickfore

Both are equal. Multiply the numerator and denominator by $e^{\beta \hbar \omega}$ to get the first result from the second.