Partition Function – Canonical Partition Function of a Boltzmann Gas for Non-Interacting Particles

partition functionstatistical mechanics

In our lecture, for the canonical partition function of a Boltzmann gas made out of $N$ particles which do not interact it's given:

$$Z=\frac 1 {N!}\Sigma_{\{\vec P_i\}}e^{-\beta\Sigma_{i=1}^N \frac{\vec P_i^2}{2m}}$$

Which is then transformed to:

$$Z=\frac 1 {N!} [\Sigma_{\vec P}e^{-\beta \frac {\vec p^2}{2m}}]^N.$$

I tried to transform the initial equation into the 2nd one but I cannot.

Initially I tried to understand the symbolic: $\Sigma_{\{\vec P_i\}}$, which i think it is a sum over different sets of the monetum values of the particles. What I am saying is:

${{\{\vec P_i\}}}_0$ would be the set that contains all the momentum values ($N$) of all $N$ particles, when the system is in the first microstate, or also known as ground state.

${{\{\vec P_i\}}}_1$ would be the set that contains all the momentum values ($N$) of all $N$ particles, when the system is in the second microstate, or also known as the first excited state.

But I cannot reach the 2nd equation. Can someone help me?

Best Answer

The canonical partition function is the sum of the Boltzmann factor of the Hamiltonian over all the possible microstates at fixed volume and number of particles.

The notation $\sum_{\left\{\vec P_i \right\}}$ is equivalent to $$ \sum_{\vec P_1}\sum_{\vec P_2} \dots \sum_{\vec P_N} $$ where each sum is over the whole set of three-dimensional values of each momentum.

Rewriting the starting formula with this more explicit notation makes it easier to understand how one gets the final result: $$ \sum_{\vec P_1}\sum_{\vec P_2} \dots \sum_{\vec P_N} e^{-\beta\Sigma_{i=1}^N \frac{\vec P_i^2}{2m}} =\sum_{\vec P_1}\sum_{\vec P_2} \dots \sum_{\vec P_N} \prod_i^N e^{-\beta \frac{\vec P_i^2}{2m}}=\prod_i^N \left( \sum_{\vec P_i}e^{-\beta \frac{\vec P_i^2}{2m}} \right) =\left( \sum_{\vec P}e^{-\beta \frac{\vec P^2}{2m}} \right)^N. $$ Notice that to rewrite the multiple sums of a product as the product of sums (the second equality), each sum over the momentum values must be independent of the others (no constraint over the possible values of ${\vec P}_i$).

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