统计力学一捅就寄（4）——Bose统计与Fermi统计理论
Bose统计与Fermi统计理论
平衡分布： \[ a_{i}=\frac{\omega_{i}}{\exp \left(\alpha+\beta \varepsilon_{i}\right) \pm 1}\left\{\begin{array}{l} +\text { Fermi } \\ -\text { Bose } \end{array}\right. \] 非简并条件（Bose,Fermi→Boltzmann）： \[ x=n\lambda_T^3\ll1,\quad\lambda_{T} \equiv\left(\frac{h^{2}}{2 \pi m k_{B} T}\right)^{1 / 2} \] 不满足非简并条件的气体
- \(x<1\)，弱简并，宏观量可对\(x\)展开，量子修正
- \(x>1\)，简并，性质显著不同
宏观量统计表达式
巨配分函数\(\Xi\): \[ \Xi(\alpha, \beta, y)=\prod_{i}\left(1 \pm e^{-\alpha-\beta \varepsilon_{i}}\right)^{\pm \omega_{i}}=\prod_{i}\Xi_i \]
\[ \ln \Xi=\pm \sum_{i} \omega_{i} \ln \left(1 \pm e^{-\alpha-\beta \varepsilon_{i}}\right) \]
（体系粒子数可变）
一些宏观量的表达式： \[ \begin{aligned} &N=\sum_{i} a_{i}=\sum_{i} \frac{\omega_{i}}{\exp \left(\alpha+\beta \varepsilon_{i}\right) \pm 1}=-\frac{\partial \ln \Xi}{\partial \alpha}\\ &U=\sum_ia_i\varepsilon_i=-\frac{\partial\ln\Xi}{\partial\beta}\\ &Y_{k}=\sum_{i} a_{i} \frac{\partial \varepsilon_{i}}{\partial y_{k}}=-\frac{1}{\beta} \frac{\partial \ln \Xi}{\partial y_{k}}, \quad \left(p=\frac{1}{\beta} \frac{\partial \ln \Xi}{\partial V}\right) \\ &S=k\left(\ln\Xi+N\alpha+U\beta\right)=k\left[\ln \Xi-\alpha \frac{\partial \ln \Xi}{\partial \alpha}-\beta \frac{\partial \ln \Xi}{\partial \beta}\right]\\ &S'=0\\ &\mu=-\frac{\alpha}{\beta}\\ \end{aligned} \]
弱简并理想玻色气体和费米气体
对\(x\)展开 \[ \begin{aligned} \ln \Xi&=\pm \int_0^\infty g(\varepsilon) \ln \left(1 \pm e^{-\alpha-\beta \varepsilon}\right)d\varepsilon\\ &=\sum_{n=1}^{\infty}(\mp 1)^{n-1} \frac{1}{n} \int_{0}^{\infty} g(\varepsilon) e^{-n(\alpha+\beta \varepsilon)}d \varepsilon\\ &=C(\beta)Vf(\alpha) \end{aligned} \] 求\(N\)，解出\(x\)、\(\alpha\)的关系，把\(f(\alpha)\)用\(x\)展开，算出别的宏观量
算就完了
Bose-Einstein凝聚
\(x\sim1\)，理想简并玻色气体
\[ N=\sum_{i} a_{i}=\sum_{i} \frac{\omega_{i}}{\exp \left(\beta (\varepsilon_{i}-\mu)\right) - 1}\\ \Rightarrow \varepsilon_i-\mu\le0\\ \Rightarrow \mu\le\varepsilon_0 \]
\(N\)恒定，则\(T\)↓，\(\mu\)↑，当\(T=T_c\)时，\(\mu=\varepsilon_0\)，此时 \[ N=N_{\varepsilon>0}=\int_0^\infty \frac{g(\varepsilon)}{\exp \left(\beta \varepsilon\right) - 1}d\varepsilon\\ \] \(T<T_c\)时，\(\mu=\varepsilon_0\)，\(N>N_{\varepsilon>0}\)，多出来的粒子全在基态凝聚 \[ N_{\varepsilon=0}=N-N_{\varepsilon>0} \] \(\mu=\varepsilon_0\)下所有宏观量都有解析解，非常舒服
奇怪的积分公式： \[ I(n)=\int_0^\infty dx\frac{x^{n-1}}{e^x-1}=\Gamma(n)\xi(n)\\ \xi(4)=\frac{\pi^4}{90} \]
光子气体
理想简并玻色气体，相对论性，\(m=0\)，\(\alpha=0\)，\(g_s=2\)
\[ g(\nu)d\nu=g_s\frac{4\pi V}{c^3}\nu^2d\nu \]
能量频谱 \[ U(\nu)d\nu=\frac{g(\nu)h\nu}{e^{\beta h\nu}-1}d\nu=\frac{8\pi Vh}{c^3}\frac{\nu^3}{e^{\beta h\nu}-1}d\nu \]
\[ \ln\Xi(\beta,V)=\frac{8\pi^5V}{45h^3c^3\beta^3} \]
声子气体
理想简并玻色气体，类相对论性\(\varepsilon=\hbar\omega= vp\)，\(m=0\)，\(\alpha=0\)，\(g_s=\)横1+纵2
存在截止频率\(\omega_D\)（声子能量上限）， \[ g(\omega)=\left\{\begin{array}{l} 9N\omega^2/\omega_D^3\\ 0 \end{array}\right.\quad \begin{array}{l} 0\le\omega\le\omega_D\\ \omega>\omega_D \end{array} \]
\[ 3N=\int_0^{\omega_D}g(\omega)d\omega \]
\(N\)为固体的原子数
有限定积分没有解析解，只能近似
Fermi气体
\[ f_i=\frac{a_{i}}{\omega_{i}}=\frac{1}{\exp \left(\beta (\varepsilon_{i}-\mu)\right)+1} \]
零温情形，完全简并Fermi气
\[ \begin{align} \lim_{\beta\to\infty}f_i=\left\{\begin{array}{l} 1\\ 0 \end{array}\right.\quad \begin{array}{l} \varepsilon_{i}>\mu_0\\ \varepsilon_{i}<\mu_0 \end{array} \end{align} \]
Fermi能：\(\varepsilon_F=\mu_0=\mu(T=0)\) \[ N=\int_0^\infty \frac{g(\varepsilon)}{\exp \left(\beta (\varepsilon-\mu_0)\right)+1}d\varepsilon=\int_0^{\mu_0} g(\varepsilon)d\varepsilon \] 零点宏观量： \[ \begin{align} &U_0=\int_0^{\mu_0} \varepsilon g(\varepsilon)d\varepsilon\\ &P_0=-\frac{\partial U_0}{\partial V}\\ &S_0=k\ln\Omega_F=0 \end{align} \]
低温情形，强简并Fermi气
把\(f'(\varepsilon)\)和\(\varepsilon^n\)在\(\mu\)附近展开，硬算
结论：低温下电子气热容\(C_V\sim\frac{\pi^2}{2}Nk\frac{kT}{\mu_0}\)
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