- #1
unchained1978
- 91
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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 [itex]\mathcal{Z}= tr(e^{-\beta H})[/itex].
So the natural thing to do is write the exponential as a power series [itex]e^{-\beta H}=\sum \frac{(-\beta H)^{n}}{n!}[/itex] and using Schrödinger's equation [itex]H|\psi\rangle = E |\psi\rangle[/itex] this gives you [itex]e^{-\beta H}|\psi\rangle=\sum \frac{(-\beta E)^{n}}{n!}|\psi\rangle→tr(e^{-\beta H})=\sum_{n} e^{-\beta E_{n}}[/itex]
Writing out the energy levels this gives [itex]e^{-\frac{1}{2}\beta \hbar\omega}\sum_{n} e^{-\beta\hbar\omega n}[/itex] Which when summed over gives [itex]\mathcal{Z}=\frac{e^{\frac{1}{2}\beta \hbar\omega}}{e^{\beta\hbar\omega}-1}[/itex]
Which is the right result I think. I'm just a bit nervous about the trace argument.
Z is defined as [itex]\mathcal{Z}= tr(e^{-\beta H})[/itex].
So the natural thing to do is write the exponential as a power series [itex]e^{-\beta H}=\sum \frac{(-\beta H)^{n}}{n!}[/itex] and using Schrödinger's equation [itex]H|\psi\rangle = E |\psi\rangle[/itex] this gives you [itex]e^{-\beta H}|\psi\rangle=\sum \frac{(-\beta E)^{n}}{n!}|\psi\rangle→tr(e^{-\beta H})=\sum_{n} e^{-\beta E_{n}}[/itex]
Writing out the energy levels this gives [itex]e^{-\frac{1}{2}\beta \hbar\omega}\sum_{n} e^{-\beta\hbar\omega n}[/itex] Which when summed over gives [itex]\mathcal{Z}=\frac{e^{\frac{1}{2}\beta \hbar\omega}}{e^{\beta\hbar\omega}-1}[/itex]
Which is the right result I think. I'm just a bit nervous about the trace argument.