The Fermi Gas

Fermi Energy

In the low-temperature limit \(T \to 0\), the Fermi distribution is going to become a step function. This means that all states with energy below the Fermi energy \begin{equation} \epsilon_F = \mu(n,0) \end{equation} are occupied, and all those above are empty. We know the state density in the momentum space: \begin{equation} g(p) = \frac{V}{(2\pi\hbar)^3} \cdot 4\pi p^2 \end{equation} and consider the spin degeneracy, we can obtain the following number of particles by integral. \begin{equation} \label{fN} N = nV = \frac{V(2S+1)}{(2\pi)^3}\frac{4\pi}{3}k_F^3 \end{equation} where \(k_F = p_F / \hbar\) is the Fermi wave number. Thus the Fermi surface is described by \begin{equation} k_F = (\frac{6\pi^2n}{2S+1})^{1/3} \end{equation} \begin{equation} \epsilon_F =\frac{\hbar^2}{2m}k_F^2= \frac{\hbar^2}{2m}(\frac{6\pi^2n}{2S+1})^{2/3} \end{equation}

Ground State

At absolute zero, the internal energy of the system in ground state is given by \begin{equation} U_0=\sum_{|k|<k_F} \epsilon_k = \frac{V}{(2\pi)^3}(\frac{\hbar^2}{2m})\frac{4\pi}{5}k_F^5 \end{equation} Using the equation \eqref{fN}, we can obtain \begin{equation} \frac{U_0}{N} = \frac{3}{5}\epsilon_F \end{equation} And the pressure at absolute zero is \begin{equation} P_0 = \frac{2}{5}n\epsilon_F \end{equation}

Fermi Temperature

The Fermi temperature is defined by \begin{equation} \label{fFT} \epsilon_F = k_B T_F \end{equation}

Low-Temperature Properties

From the Fermi-Dirac distribution we derived \begin{equation} \label{fF} n\lambda^3=f_{3/2}(z) \end{equation} In the low-temperature region \(z\) is large, we have the asymptotic formula of \(f_{3/2}(z)\): \begin{equation} f_{3/2}(z)=\frac{4}{3\sqrt{\pi}}[(\ln z)^{3/2}+\frac{\pi^2}{8}\frac{1}{\sqrt{\ln z}}+\cdot\cdot\cdot] \end{equation} To the second order, we can rewrite \eqref{fF}: \begin{equation} (\ln z)^{3/2} = \frac{3\sqrt{\pi}}{4}n\lambda^3-\frac{\pi^2}{8}\frac{1}{\sqrt{\ln z}} \end{equation} Substituting the first approximation \(\ln z =\frac{T_F}{T}\), we then obtain \begin{equation} \ln z = \frac{T_F}{T}[1-\frac{\pi^2}{12}(\frac{T}{T_F})^2] \end{equation} \begin{equation} \mu = \epsilon_F[1-\frac{\pi^2}{12}(\frac{T}{T_F})^2] \end{equation} The internal energy is \begin{equation} U=\sum_{\vec k}\epsilon_k n_k = \frac{V}{(2\pi)^3}\frac{4\pi\hbar^2}{2m}\int_{0}^{\infty} dk k^4 n_k \end{equation} After the expansion for \(k^6\) and \(\mu\), we obtain \begin{equation} U = \frac{3}{5}N\epsilon_F[1+\frac{5\pi^2}{12}(\frac{T}{T_F})^2+\cdot\cdot\cdot] \end{equation} from which we can obtain the EOS using \begin{equation} P=\frac{2}{3}\frac{U}{V} \end{equation} and the heat capacity \begin{equation} \frac{C_V}{Nk} = \frac{\pi^2}{2}\frac{T}{T_F}+\cdot\cdot\cdot \end{equation}