The grand partition function of non interacting hamiltonians

Question

In the case of non interacting particles I know we can write the Hamiltonian as $$H(\mathbf{q}_1,\dots,\mathbf{p}_1,\dots)=\sum_{i=1}^N h(\mathbf{q}_i,\mathbf{p}_i)$$ but I am having trouble calculating its grand partition function.

$$\begin{align}\mathcal{Z} &= \sum_{\Gamma}e^{\beta\mu N(\Gamma)-\beta\mathcal{H}(\Gamma)} \\ &= \sum_{\Gamma}e^{\beta\mu N(\Gamma)-\beta\sum_{i=1}^N h(\mathbf{q}_i,\mathbf{p}_i)} \\ &= \sum_{\Gamma}e^{\beta\mu }\prod_{i=1}^N e^{-\beta h(\mathbf{q}_i,\mathbf{p}_i)} \end{align}$$

I cannot see how the sum of microstates would be done. Im not sure what the microstates of a general classical system will be. Moreover, I believe it may be possible to set $N(\Gamma)=1$ as each h represents one particle. Please keep as general as possible.

Answer 1

The first thing we can do is to split up $\Gamma$ according to the number of particles in the given states. Let $\gamma_N$ be a state with $N$ particles. The grand canonical partition function is then \begin{align} \mathcal{Z} = & \sum_\Gamma \exp\left(-\beta(\mathcal{H} - \mu N)\right)\\ =& \sum_{N=0}^\infty\exp\left(\beta \mu N \right)\sum_{\gamma_N}\exp\left( -\beta\mathcal{H}\right)\\ =& \sum_{N=0}^\infty\exp\left(\beta \mu N \right) Z_N \end{align} where $Z_N$ is the canonical partition function for an ensemble of $N$ particles.

What to do next depends on the type of statistics your particles obey. The simplest case is for distinguishable, non-interacting, particles, in which case $Z_N = Z_1^N$, so the sum for $\mathcal{Z}$ is a geometric series, so, given some requirements to ensure convergence, you get \begin{equation} \mathcal{Z} = \frac{1}{1-\exp\left(\beta\mu\right)Z_1}. \end{equation}

There is a particularly neat result for the case of indistinguishable, non-interacting, particles obeying Maxwell-Boltzmann (i.e. classical) statistics. In this case $Z_N = \frac{1}{N!}Z_1^N$ where the factor of $N!$ takes care of double counting states that differ by the exchange of identical particles. You then get \begin{align} \mathcal{Z} =& \sum_{N=0}^\infty\exp\left(\beta\mu N\right)\frac{Z_1^N}{N!}\\ =& \;\exp\left(Z_1\exp\left[\beta\mu\right]\right). \end{align}

For quantum particles obeying Fermi-Dirac or Bose-Einstein statistics life is more complicated and I do not know of a simple way of writing the canonical partition function. Generally in these cases other methods are used.