Hi guys I'm currently trying to solve a mock exam for an exam in a few days and am a bit confused by the solutions they gave us for this exercise:


A solid is composed of N atoms which are each localized at points in a lattice. Excursions about the equilibrium position of each results in each atom behaving as a 1-dimensional harmonic oscillator. Write down the partition function for an individual atomic harmonic oscillator, and for the collection, assuming that they have arrived in thermal equilibrium with each other at temperature T.

$ Z_{SP} = \sum\nolimits_{n=1}^\infty e^{(-E_n \beta)}$ where $\beta$ is $1/(k_B T)$ and the Energy levels of the quantum harmonic oscillators are $E_n = \hbar \omega (n + 1/2)$. I now tried to use geometric series to evaluate the sum:

$ Z_{SP} = e^{(-\beta \hbar \omega / 2)}\sum\nolimits_{n=1}^\infty e^{(-\beta \hbar \omega)n} = e^{(-\beta \hbar \omega / 2)} \frac{1}{(1 - e^{(-\beta \hbar \omega)})}$ With the substitution $\theta = \hbar \omega \beta $ it says the solution for $Z_{SP}$ should be $e^{(-\theta /2)} \frac{1}{e^{\theta} - 1)}$ which isn't quite the expression I got.

Can anybody tell me where I went wrong or is the solution just false?

  1. The quantum number $n$ of the harmonical oscillator runs from $0$ to $\infty$. (Your sum starts at 1.)
  2. $\sum_{n=0}^{\infty} e^{-\theta(n+1/2)} = \frac{e^{-\theta/2}}{1-e^{-\theta}} = \frac{e^{\theta/2}}{e^\theta - 1} = \frac{1}{e^{\theta/2}-e^{-\theta/2}}$. I guess there just is an error in your exercise. (TAs make mistakes, too.)
  3. The FAQ says no homework questions. Let's hope they don't tar and feather us. ;-)
  • $\begingroup$ Oops yes the calculation I did was only correct for starting with n=0, I wasn't sure for a second if the ground state is n=1 oder n=0... Now I know again:) $\endgroup$ – user17574 Jan 30 '13 at 10:37
  • $\begingroup$ The FAQ and moreover this meta question allow homework questions, provided they show work on the part of the person asking and they are clear about what the conceptual issue is. $\endgroup$ – user10851 Jan 30 '13 at 18:18

When evaluating the geometric series you have to remember that the sum starts at n = 1. The formula for the geometric series is easy to remember, see it comes from just pulling out a factor of $x$ and isolating the sum (2 sec. derivation):

$$ \Sigma_{n=0}^\infty x^n = 1 + x + x^2 + ... = 1 + x (1 + x + x^2 + ...) = 1 + x\, \Sigma_{n=0}^\infty x^n\\ \Leftrightarrow(1 - x) \, \Sigma_{n=0}^\infty x^n = 1 \\ \Leftrightarrow \Sigma_{n=0}^\infty x^n = \frac{1}{(1 - x)} $$

You have to pull out a single factor of $e^{-\beta \hbar \omega}$ to use this formula,

$$ Z_{SP} = e^{-\beta\hbar\omega/2} \Sigma_{n=1}^\infty (e^{-\beta \hbar \omega})^n\\ = e^{-\beta\hbar\omega/2} e^{-\beta \hbar \omega} \Sigma_{n=0}^\infty (e^{-\beta \hbar \omega})^n\\ = e^{-\beta\hbar\omega/2} e^{-\beta \hbar \omega} \frac{1}{1 - e^{-\beta \hbar \omega}}\\ = \frac{e^{-\beta\hbar\omega/2}}{e^{\beta \hbar \omega} - 1}.\\ $$

with your $\theta = \hbar\omega\beta$ you get what the solution says,

$$ Z_{SP} = \frac{e^{-\theta/2}}{e^{\theta} - 1}. $$

  • $\begingroup$ I just messed up, the sum goes from 0 to $\infty$, so no need to single out a factor. But thanks anyway! $\endgroup$ – user17574 Jan 30 '13 at 10:38
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    $\begingroup$ @Morten, but the sum should start at n = 0, not n = 1. $\endgroup$ – Kenshin Jan 30 '13 at 10:44
  • $\begingroup$ @Chris. I actually thought it was on purpose (for some obscure reason) since it was consistent with the provided solution. But of course you are right :)! $\endgroup$ – Morten Jan 30 '13 at 10:49

Your solution looks correct to me. Note that it can be rewritten as $Z_{SP} = \frac{e^{\theta/2}}{e^\theta -1}$, which suggests that the exam solution simply has a sign error in the argument of the numerator's exponential.

  • $\begingroup$ Agreed. The calculation user17574 did was correct. $\endgroup$ – Kenshin Jan 30 '13 at 10:23
  • $\begingroup$ Okay that was what I was thinking! Thanks a lot :) $\endgroup$ – user17574 Jan 30 '13 at 10:28

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