as far as I understand, the canonical partition function of a single particle can be expressed as follows: $$ z = \sum_i e^{-\beta\cdot\epsilon_i} $$ Where $i$ are the micro states, $\beta$ is the inverse of the product of the Boltzman constant and the temperature, and $\epsilon_i$ is the energy at that state. Which can be expanded for a system of N particles as: $ Z=z^N$. Where Z can be expressed $$ Z=\sum_i e^{-\beta\cdot E_i} = \sum_j w(E_j)\cdot e^{-\beta\cdot E_j}$$ Where j represents the energy level, and $w$ is the degeneracy of that energy level. So at this point I have my first question:

  1. Are $\epsilon$ from the single particle partition function and $E$, from the system's partition function different?

Then the probability of occupancy of a micro state is given by: $$ P_i(E_i)= \frac{e^{-\beta\cdot E_i}}{Z} $$ And the probability of an energy level being occupied is: $$ P(E_j)= \frac{w(E_j)\cdot e^{-\beta\cdot E_j}}{Z} $$

And finally, the average energy can be equated to the internal energy. Which can be calculated as follows: $$ U=\bar{E}=\sum_i P_i(E_i)\cdot E_i $$

And here the second question:

  1. What would be the equivalent of the internal energy in terms of a summation over the energy levels ($j$) instead of the micro states ($i$)?

Thanks for any help in advance.

  • $\begingroup$ But partition function for N particles $Z=z^n$ is correct for only maxwell-boltzmann statistics, right ? For quantum statistics, we cannot do that. $\endgroup$ Mar 15, 2023 at 1:39

1 Answer 1


Maybe expanding the equation $Z=z^N$ explicitly will be helpful $$ \begin{aligned} Z&=z^N\\ &=(\sum_ie^{-\beta \epsilon_i})^N\\ &=(\sum_{i_1}e^{-\beta \epsilon_{i_1}})\cdot(\sum_{i_2}e^{-\beta \epsilon_{i_2}})\cdots(\sum_{i_N}e^{-\beta \epsilon_{i_N}})\\ &=\sum_{i_1,i_2,\cdots,i_N}e^{-\beta(\epsilon_{i_1}+\epsilon_{i_2}+\cdots+\epsilon_{i_N})} \end{aligned} $$ And now we can see $$ E_i = \epsilon_{i_1}+\epsilon_{i_2}+\cdots+\epsilon_{i_N} $$ So for the first question, $\epsilon$ is different from $E$.

For the second question, I think $$ \begin{aligned} U=\bar{E}&=\sum_iP_i(E_i)E_i\\ &=\sum_jP(E_j)E_j\\ &=\sum_j\frac{w(E_j)e^{-\beta E_j}}{Z}E_j \end{aligned} $$

  • $\begingroup$ Makes sense, thanks! $\endgroup$
    – HWIK
    Dec 6, 2022 at 18:48
  • 1
    $\begingroup$ you're welcome. $\endgroup$
    – lbyshare
    Dec 7, 2022 at 2:29
  • $\begingroup$ For $N$ particle system, should not you get $U = N \bar{E} $ ? $\endgroup$ Mar 15, 2023 at 2:06
  • $\begingroup$ The $\bar{E}$ in your equation is the energy of a single particle, but in our equations it means the energy of the total $N$ particles. $\endgroup$
    – lbyshare
    Mar 17, 2023 at 3:23

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