\[\newcommand{\bra}[1]{\langle#1|} \newcommand{\ket}[1]{|#1\rangle} \newcommand{\braket}[2]{ \langle #1 | #2 \rangle}\]

Factorisation of the partition functions

In the previous lectures, we described the right mathematical space, Fock space, in which to consider a system made of a varying number of quantum particles. We will now focus on the special case of a system made of $\mathcal{N}$ identical and non-interacting quantum particles. The single particle Hilbert space has $N$ accessible states, each having an energy $\epsilon_\lambda$. On Fock space, the density matrix which maximize the entropy under (grand canonical) constraints is the operator:

\[\hat{\rho}= \frac{1}{\Xi}e^{-\beta\left( \hat{H} - \mu \hat{\mathcal{N}}\right)}\]

with

\[\Xi = \text{Tr}(e^{-\beta(\hat{H} - \mu \hat{\mathcal{N}})})\]

If the system is made of a varying $\mathcal{N}$ number of identical particles,

\[\boxed{\Xi = \prod_\lambda \xi_\lambda}\]

With the partition function for a single particle being:

\[\boxed{\xi_\lambda = \sum_{n_\lambda=0}^\infty e^{-\beta n_\lambda (\epsilon_\lambda - \mu)}}\]
Proof: $$ \begin{align} \Xi &= \text{Tr}(e^{-\beta(\hat{H} - \mu \hat{\mathcal{N}})}) \\ &=\sum_{n_\lambda} \bra{n_1,n_2,...,n_N} e^{-\beta(\hat{H} - \mu \hat{\mathcal{N}})}\ket{n_1,n_2,...,n_N} \\ &= \sum_{\lambda=0} e^{-\beta\sum_\lambda n_\lambda(\epsilon_\lambda -\mu)}\\ &= \prod_\lambda\left( \sum_{n_\lambda} e^{-\beta n_\lambda(\epsilon_\lambda -\mu)}\right)\\ &= \prod_\lambda \xi_\lambda \end{align} $$

The average number of particle occuping a quantum state is given by:

\[\boxed{\langle n_{\lambda} \rangle = \frac{\partial \ln(\xi_\lambda)}{\partial \alpha}}\]

with $\alpha := \beta \mu$

Proof:

Partition functions and occupation number

Bosons

Due to spin-statistics theorem, Bosons have no limits on their occupation number, each state can be populate infinitly. They follow the Bose-Einstein (BE) statistics:

\[\xi^{BE}_\lambda = \sum_{n_\lambda=0}^\infty \left(e^{-\beta (\epsilon_\lambda - \mu)}\right)^{n_\lambda} = \frac{1}{1-e^{-\beta(\epsilon_\lambda - \mu)}}\]

Remembering the expression for an infinite geometric sum:

\[\sum_{k=0}^\infty q^k = \frac{1}{1-q}\]
Proof:

We then get:

\[\boxed{\langle n^{BE}_\lambda \rangle = \frac{1}{e^{\beta(\epsilon_\lambda - \mu)}-1}}\]

Fermions

For fermions, Fermi-Dirac (FD) statistics:

\[\xi^{FD}_\lambda = \sum_{n_\lambda=0}^1 \left(e^{-\beta n_\lambda (\epsilon_\lambda - \mu)}\right) = 1 + e^{-\beta (\epsilon_\lambda - \mu)}\] \[\boxed{\langle n^{FD}_\lambda \rangle = \frac{1}{e^{\beta(\epsilon_\lambda - \mu)}+1}}\]

PLOTS

Classical limit