In the derivation of results for an ideal gas, it is common to calculate the classical partition function for a single particle in a box. This can be done using the Boltzmann factor and integrating over phase space.

$$ Z_1 =\int e^{-\beta E(\{\vec{p},\vec{q}\} )} \frac{d^3p d^3r}{(2\pi\hbar)^3} $$

$$ Z_1 = V \left( \sqrt{\frac{k_BTm}{2\pi\hbar^2}} \right)^3 $$

Now, it is seen that that $Z_1$ can be written as $$ Z_1 = \frac{V}{\lambda^3} $$ where $\lambda = \sqrt{\frac{2\pi\hbar^2}{k_BTm}}$ is the thermal de Broglie wavelength of the particle.

This implies that we might interpret the single particle partition function as the total number of states of a particle in a box. Is it just a happy coincidence that the partition function corresponds to counting the total number states a free classical particle can take inside the box? What are some ways of motivating/explaining this?

I have seen an argument that goes as strictly the free particle in a box has no potential energy so the Boltzmann factor should be equal to $e^0=1$. Hence, the partition function according to its definition $Z = \sum_i e^{\beta E_i}$ should just count the total number of microstates. However, this doesn't seem convincing to me since why then would it have been valid for us to use the translational kinetic energy in the Boltzmann factor in our phase space derivation?

  • $\begingroup$ I'm not sure what you are confused about. By definition, the partition function is a (weighted) sum over all states. $\endgroup$
    – leapsheep
    Apr 13 at 12:40
  • $\begingroup$ Right, the partition function is a weighted sum. My question is about whether there is a deeper physical intuition as to why the partition function turns out to be the number of thermal de Broglie "cubes" $\lambda^3$ that can fit into a volume $V$ - which seems, in some sense, to be a sum over states, rather than a weighted sum. $\endgroup$
    – user246795
    Apr 13 at 14:35
  • $\begingroup$ I'd have to point out that, in this calculation, the thermal de Broglie wavelength is an entirely free-to-choose parameter. Thus, the equality is due to choice, and nothing inherent to the physics. $\endgroup$ Apr 14 at 9:13
  • $\begingroup$ The weight for an ideal gas is independent of position, so every point in space is weighted the same $\endgroup$
    – leapsheep
    Apr 16 at 12:16

1 Answer 1


The ideal gas is a collection of independent particles, i.e., particles that do not interact with each other. In this case the partition function of the collection is the product of the individual partition functions, corrected for distinguishability: $$ Z_N = \frac{Z_1^N}{N!} $$ The factorization of the partition function arises whenever we have non-interacting particles.


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