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Partition function of the QHO

  1. Jun 18, 2012 #1
    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 schrodinger'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.
     
  2. jcsd
  3. Jun 18, 2012 #2
    I got
    [tex]
    \frac{e^{-\frac{1}{2}\beta\hbar \omega}}{1-e^{-\beta\hbar\omega}}
    [/tex]
    also this should be in the Quantum sub-forum, not Classical.
     
  4. Jun 18, 2012 #3
    Both are equal. Multiply the numerator and denominator by [itex]e^{\beta \hbar \omega}[/itex] to get the first result from the second.
     
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