Fermi and Bose Statistics
Fermi Statistics
For fermions, each single-particle state can accommodate at most one particle, and thus each state is either occupied or empty. To place \(n_j\) fermions into \(g_j\) states, the number of ways: \begin{equation} w_j=\frac{g_j!}{n_j!(g_j-n_j)!} \end{equation} Therefore, under a certain seqence of occupation numbers\(\lbrace n_i\rbrace\), we have the microstate number \begin{equation} \Omega\lbrace n_i\rbrace = \prod_{j}\frac{g_j!}{n_j(g_j-n_j)!} \end{equation} Using Lagrage multipliers, we can obtain \begin{equation} \bar n_j = \frac{g_j}{e^{-\alpha+\beta\epsilon_j}+1} \end{equation} This is the Fermi distribution.
Bose Statistics
For Bosons \begin{equation} w_j=\frac{(g_j+n_j-1)!}{n_j!(g_j-1)!} \end{equation} Therefore, the number of microstates: \begin{equation} \Omega\lbrace n_i\rbrace = \prod_{j}\frac{(n_j+g_j-1)!}{n_j(g_j-1)!} \end{equation} Using Lagrage multipliers, we can obtain \begin{equation} \bar n_j = \frac{g_j}{e^{-\alpha+\beta\epsilon_j}-1} \end{equation} This is the Bose distribution.
Physical Quantities under Quantum Statistics
Pressure
The pressure of the quantum ideal gas can be calculated using the particle flux impinging on a wall arising from a momentum distribution \(f(\vec p)\): \begin{equation} P=m\int d^3 p v_x^2 f(\vec p) \end{equation} Using \(f(\vec p) = h^{-3} n_p\), and \(v_x^2=\frac{1}{3} {\vec v}^2\), we can obtain \begin{equation} \frac{P}{k_B T}=\frac{1}{\lambda^3}\frac{8}{3\sqrt{\pi}}\int_{0}^{\infty} \frac{x^4}{z^{-1}e^{x^2} -\zeta} \end{equation} where \(z = e^\alpha = e^{\beta \mu}\) is called the fugacity.We can get the relation: \begin{equation} PV=\frac{2}{3}U \end{equation} This holds for the ideal Fermi and Bose gas, as well as the classical ideal gas.
Entropy
From the definition of entropy we can obtain \begin{equation} \frac{S}{k_B} = \sum_{k} [\frac{1}{\xi \pm 1} \ln (\xi \pm 1) \pm \frac{\xi}{\xi\pm 1}\ln\frac{\xi \pm 1}{\xi}] \end{equation} where \(\xi = z^{-1} e^{-\beta\epsilon}\). Using the definition of \(U\) and \(N\), we have \begin{equation} \frac{S}{k_B}=-N\ln z + \frac{U}{k_B T} \pm \sum_{k}\ln (1\pm z e^{-\beta \epsilon}) \end{equation} with + for Fermi, and - for Bose.
In the thermodynamic limit the last term becomes an integral, so we can calculate(for Fermi): \begin{equation} \frac{V}{2\pi^2}\int_0^{\infty}k^2 \ln (1 + z e^{-\beta\epsilon})=\frac{2}{3}\frac{U}{k_B T}=\frac{PV}{k_B T} \end{equation} The same final result holds for the Bose case. Now we have the relation \begin{equation} S = \frac{1}{T} (U+PV-N\mu) \end{equation} This verifies the thermodynamics relation \(T^{-1} = \partial S/\partial U\).
Free Energy
We can identify the free energy as \begin{equation} A = N\mu - PV \end{equation} From previouse formula we can obtain \begin{equation} A = k_B T \ln z \mp k_B T \sum_{k}\ln (1 \pm ze^{-\beta\epsilon}) \end{equation} where + for Fermi, and - for Bose.