As a homework exercise for Advanced Statistical Mechanics I need to derive the canonical partition functions for the following systems:

  • Single component ideal gas on a square lattice
  • Single component ideal gas in real space
  • Partition function for a lattice gas on a square lattice where each lattice site can only have a single particle, but neighbouring particles do not interact

I do not want any final answers as I would like to derive them myself, but I'd like to know a strategy on how to solve this problem. My guess is that I have to use the general formula for the canonical partition function $$ Z(N,V,T) = \sum_mexp(-\beta \epsilon_m) $$ Where $e_m$ is the energy of state $m$ and $\beta = 1/k_bT$.

Normally when solving these kind of problems I simply determine $e_m$ but I'm not sure how to do this for this problem.


1 Answer 1


The correct formula for the canonical partition function is

$Z = \sum_{\sigma \in \text{conf}} e^{- \beta E(\sigma)} $,

where $\sigma$ is a configuration and $E$ is the energy functional. Now if you want to want to transfer this to a sum over energies, one obtains

$Z = \sum_{E \in \text{Spec}} N(E) e^{-\beta E} $

in the case where Spec, the set of possible energies is discrete and $N(E)$ is the number of states with a given energy; and

$Z = \int_0^\infty \rho(E) e^{-\beta E} d E $

for a continuous set of energies, and w.l.o.g. the lowest energy was set to zero. Here $\rho(E) \geq 0$ is the density of states.

Now to your problems. Recall that "ideal" means that the particles are non-interacting. That means that the partition function $Z_N$ of $N$ particles is given as

$Z_N = (Z_1)^N , $

where $Z_1$ is the partition function of one particle (try proving this if you are not sure why). Then in the first two cases it might be good idea to enumerate the particles by their momentum or their crystal momentum. For the last problem you simply have to restrict the configurations to those with no particles at the same lattice point.

I hope i haven't given away too much and have still helped you!


Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge that you have read and understand our privacy policy and code of conduct.

Not the answer you're looking for? Browse other questions tagged or ask your own question.