From Grand Canonical Ensemble to Ideal Gas

Ideal Gas System

The energy and particle number of ideal gas are \begin{equation} E = \sum_{p} \epsilon_p n_p \end{equation} \begin{equation} N = \sum_{p} n_p \end{equation} where \(\epsilon_p\) is the energy of a particle with momentum \(p\), \(n_p\) is the number of particles with momentum \(p\).

Grand Partition Function

We can write down the grand partition function of ideal gas: \begin{equation} \Xi (z,V,T) = \sum_{N=0}^{\infty} z^N Z_N (V,T) = \sum_{N=0}^{\infty} \sum_{\lbrace n_p \rbrace} z^N e^{-\beta \sum_{p} \epsilon_p n_p} \end{equation} where \begin{equation} z = e^{\beta \mu} \end{equation} is the fugacity. After calculating the double sum symbols, we obtain \begin{equation} \Xi (z,V,T) = \prod_{p} [\sum_{n_p} (z e^{-\beta \epsilon_p})^{n_p}] \end{equation} For Boson, \(n_p = 0,1,2...\); for Fermion, \(n_p = 0,1\). So the formula can be written as \begin{equation} \Xi (z, V,T) = \prod_{p} \frac{1}{1-z e^{-\beta \epsilon_p}}, \text{for Boson} \end{equation} \begin{equation} \Xi (z, V,T) = \prod_{p} （1+z e^{-\beta \epsilon_p}）, \text{for Fermion} \end{equation}

Parameter Form of EOS

Using the formula \(P = -(\frac{\partial J}{\partial V})_{\beta ,\mu}\), we can derive the pressure of GCE: \begin{equation} \label{f1} \frac{PV}{k_B T} = \ln \Xi (z,V,T) = -\zeta \sum_{p} \ln (1-\zeta ze^{-\beta \epsilon_p}) \end{equation} where \(\zeta = -1\) for Fermion and \(+1\) for Boson.

The total number of particles is \begin{equation} \label{f2} N = z\frac{\partial}{\partial z} \ln \Xi (z,V,T) = \sum_{p} \frac{ze^{-\beta \epsilon_p}}{1-\zeta ze^{-beta\epsilon_p}} \end{equation} The \eqref{f1} and \eqref{f2} are the parameter form of the equation of state (EOS). And the number of particles with momentum \(p\) is \begin{equation} n_p = tr(\hat n_p \frac{1}{\Xi} e^{-\beta (\hat H - \mu \hat N)}) = \frac{1}{e^{\beta (\epsilon_p - \mu)} - \zeta} \end{equation}

Ideal Fermi Gas

Now we let \(V\) become \(\infty\), so the sum w.r.t. \(p\) should be changed into: \begin{equation} \sum_{p} … = \frac{V}{(2\pi)^3 \hbar^3} \int d^3 p… \end{equation} We first consider the ideal Fermi gas, the parameter form equation of state can be written into: \begin{equation} \frac{P}{k_B T} = \frac{1}{\lambda^3} f_{5/2}(z) \end{equation} \begin{equation} \frac{1}{v} = \frac{1}{\lambda^3}f_{3/2}(z) \end{equation} where \(\lambda = \sqrt{2\pi \hbar^2/mk_B T}\), while \(f_{5/2}(z)\) and \(f_{3/2}(z)\) are defined as \begin{equation} f_{5/2}(z) = \sum_{l=1}^{\infty} \frac{(-1)^{l+1} z^l}{l^{5/2}} \end{equation} \begin{equation} f_{3/2}(z) = z\frac{\partial}{\partial z}f_{5/2}(z)=\sum_{l=1}^{\infty} \frac{(-1)^{l+1} z^l}{l^{3/2}} \end{equation}

Ideal Bose Gas

Now we consider about the ideal Bose gas, \begin{equation} \frac{P}{k_B T} = \frac{1}{\lambda^3} g_{5/2}(z)-\frac{1}{V}\ln (1-z) \end{equation} \begin{equation} \label{f3} \frac{1}{v} = \frac{1}{\lambda^3} g_{3/2}(z) + \frac{1}{V}\frac{z}{1-z} \end{equation} where \(g_{5/2}(z)\) and \(g_{3/2}(z)\) are \begin{equation} g_{5/2}(z) = \sum_{l=1}^{\infty} \frac{z^l}{l^{5/2}} \end{equation} \begin{equation} g_{3/2}(z) = z \frac{\partial}{\partial z} g_{5/2}(z) = \sum_{l=1}^{\infty} \frac{z^l}{l^{3/2}} \end{equation} We know that \begin{equation} \langle n_0 \rangle = \frac{z}{1-z} \end{equation} Obviously, if \(\langle n_0\rangle /N\) is not a small quantity, it will influence the formula \eqref{f3}, which means a finite part of particles occupied the state \(p=0\). This is the Bose-Einstein Condensation.

Internal Energy

The internal energy of ideal Bose and Fermi gas is \begin{equation} U(z,V,T) = -\frac{\partial}{\partial \beta} \ln \Xi (z,V,T) \end{equation} As \(\ln \Xi (z,V,T) = PV/k_BT\), we can derive \begin{equation} \frac{U(z,V,T)}{V}=\frac{3k_B T}{2\lambda^3}f_{5/2}(z) \text{, for Fermi} \end{equation} \begin{equation} \frac{U(z,V,T)}{V}=\frac{3k_B T}{2\lambda^3}g_{5/2}(z) \text{, for Bose} \end{equation} We can obtain that \begin{equation} U=\frac{3}{2} PV \end{equation} which also correct for the classical ideal gas.