推导非0温电子气体费米能的索末菲展开公式

## 一个定积分 考虑 $$ I_0=\int_0^\infty \frac{x}{\mathrm e^x + 1}\mathrm dx $$ 注意到几何级数 $$ \frac{1}{\mathrm e^x + 1}=-\sum_{n=1}^\infty\left(-\mathrm e^{-x}\right)^{n} $$ 因此 $$ I_0=-\sum_{n=1}^\infty \int_0^\infty x\left(-\mathrm e^{-x}\right)^n\mathrm dx=\sum_{n=1}^\infty (-1)^{n+1}\frac{1}{n^2}=\frac{\pi^2}{12} $$ ## 索末菲展开 考虑函数$g=g(E)$和积分 $$ I=\int_{-\infty}^\infty \frac{g(E)}{\mathrm e^{\frac{E-E_F}{kT}}+1}\mathrm dE $$ 设变量$x=\dfrac{E-E_F}{kT}$,上述化简为 $$ \begin{aligned} I&=k_BT\int_{-\infty}^\infty \frac{g(E_F+kTx)}{\mathrm e^x+1}\mathrm dx\\ &=k_BT\left(\boxed{\int_{-\infty}^0\frac{g(E_F+kTx)}{\mathrm e^x+1}\mathrm dx}+\int_{0}^\infty\frac{g(E_F+kTx)}{\mathrm e^x+1}\mathrm dx\right)\\ \boxed{I_1}&=\int_{0}^\infty\frac{g(E_F-kTx)}{\mathrm e^{-x}+1}\mathrm dx\\ &=\int_0^\infty g(E_F-kTx)\mathrm dx-\int_0^\infty\frac{g(E_F-kTx)}{\mathrm e^x+1}\mathrm dx \end{aligned} $$ 进而 $$ I=\int_{-\infty}^{E_F}g(E)\mathrm dE+kT\int_0^\infty\frac{g(E_F+kTx)-g(E_F-kTx)}{\mathrm e^x+1}\mathrm dx $$ 考虑到$kT\ll E_F$, 因此 $$ \delta g\equiv g(E_F+kTx)-g(E_F-kTx)\dot =2kTx\cdot\frac{\mathrm d}{\mathrm dE}g(E_F) $$ 从而 $$ \begin{aligned} I&\dot=\int_{-\infty}^{E_F} g(E)\mathrm{d}E+2(kT)^2\frac{\mathrm d}{\mathrm dE}g(E_F)\int_0^\infty\frac{x\mathrm dx}{\mathrm{e}^x+1}\\ &=\int_{-\infty}^{E_F} g(E)\mathrm{d}E+2(kT)^2\frac{\mathrm d}{\mathrm dE}g(E_F)\cdot\frac{\pi^2}{12}\\ &=\int_{-\infty}^{E_F} g(E)\mathrm{d}E+\frac{\pi^2}{6}(kT)^2\frac{\mathrm d}{\mathrm dE}g(E_F)\\ \end{aligned} $$ 此即**索末菲展开**。 ## 低温电子气体的费米能 对于三维电子气体，其态密度 $$ g(E)=A\sqrt E\,,A=\frac{V}{2\pi^2}\left(\frac{2m}{\hbar^2}\right)^{3/2} $$ 能量分布满足费米-狄拉克统计 $$ f(E) = \frac{1}{e^{\frac{E-E_F}{k_B T}} + 1}\,,E>0 $$ 因此电子数目 $$ \begin{aligned} N&=\int_{-\infty}^{\infty}f(E)g(E)\mathrm dE\\ &=\int_{-\infty}^{E_F} g(E)\mathrm{d}E+\frac{\pi^2}{6}(kT)^2\frac{\mathrm d}{\mathrm dE}g(E_F)\\ &=\int_{0}^{E_F} A\sqrt E\mathrm{d}E+\frac{\pi^2}{6}(kT)^2A\frac{1}{2\sqrt {E_F}}\\ &=\frac{2}{3}AE\sqrt E{\huge{|}}_0^{E_F}+\frac{\pi^2}{12}(kT)^2A\frac{1}{\sqrt {E_F}}\\ &=\frac{2}{3}AE_F\sqrt{E_F}+\frac{\pi^2}{12}(kT)^2A\frac{1}{\sqrt {E_F}} \end{aligned} $$ 带入$T=0K$的情形 $$ N=\frac{2}{3}AE_F^0\sqrt{E_F^0} $$ 得到 $$ \frac{2}{3}AE_F^0\sqrt{E_F^0}=\frac{2}{3}AE_F\sqrt{E_F}+\frac{\pi^2}{12}(kT)^2A\frac{1}{\sqrt {E_F}} $$ 不妨假设泰勒展开 $$ E_F=E_F^0+\varepsilon $$ 则 $$ \frac{\pi^2}{12}(kT)^2A\frac{1}{\sqrt {E_F^0+\varepsilon}}+\frac{2}{3}A\left((E_F^0+\varepsilon)^{3/2}-(E_F^0)^{3/2}\right)=0 $$ 继续展开并扔掉高阶项 $$ \frac{\pi^2}{12}(kT)^2\frac{1}{\sqrt {E_F^0}}+\frac{2}{3}\left(\frac{3}{2}\sqrt{E_F^0}\cdot\varepsilon\right)=0 $$ 最终得到 $$ \varepsilon=-\frac{\pi^2}{12}(kT)^2\frac{1}{E_F^0}=-\frac{\pi^2}{12}(\frac{kT}{E_F^0})^2E_F^0 $$ 即 $$ E_F=E_F^0-\frac{\pi^2}{12}(\frac{kT}{E_F^0})^2E_F^0 $$