产生算符和湮灭算符

波函数$\ket{n_1n_2\cdots n_m}$表示在态$\lambda$上粒子数为$n_\lambda$，

定义产生算符$a_\lambda^\dagger$，湮灭算符$a_\lambda$，占有数算符$N_\lambda$

$$\ket{n_1n_2\cdots n_m}=\prod_{\lambda=1}^m\frac{1}{\displaystyle\sqrt{n_\lambda!}}(a_\lambda^\dagger)^{n_\lambda}\ket{0}$$

$$\bra{n_1n_2\cdots n_m}=\bra{0}\prod_{\lambda=m}^1\frac{1}{\displaystyle\sqrt{n_\lambda!}}(a_\lambda)^{n_\lambda}$$

$$N_\lambda=a_\lambda^\dagger a_\lambda=n_\lambda$$

玻色子算符对易关系

$$[a_\lambda,a_{\lambda’}]=[a_\lambda^\dagger,a_{\lambda’}^\dagger]=0$$

$$[a_\lambda,a_{\lambda’}^\dagger]=\delta_{\lambda\lambda’}$$

费米子算符对易关系

$$\lbrace a_\lambda,a_{\lambda’}\rbrace=\lbrace a_\lambda^\dagger,a_{\lambda’}^\dagger\rbrace=0$$

$$\lbrace a_\lambda,a_{\lambda’}^\dagger\rbrace=\delta_{\lambda\lambda’}$$

$$$$

算符二次量子化

单体作用

$$\hat{O}_\text{2nd}=\sum_{\mu\nu}\bra{\mu}\hat{O}\ket{\nu}a_\mu^\dagger a_\nu$$

二体作用

$$\hat{V}_\text{2nd}=\frac{1}{2}\sum_{\mu\mu’}\sum_{\nu\nu’}(\mu\nu|\hat{V}|\mu’\nu’)a_\mu^\dagger a_\nu^\dagger a_{\nu’}a_{\mu’}$$

自旋算符

$$\hat{S}_z=\frac{\hbar}{2}(a_+^\dagger a_+^\dagger-a_-^\dagger a_-)$$

$$\hat{S}=\sum_{\alpha\alpha’}\bra{\alpha}\sigma\ket{\alpha’}a_\alpha^\dagger a_{\alpha’}$$

哈密顿量

$$\begin{aligned}\hat{H}&=\int \text{d}^3 \boldsymbol{x}\; \psi(\boldsymbol{x})^\dagger \left[-\frac{\hbar^2}{2m}\nabla^2+U(\boldsymbol{x})\right]\psi(\boldsymbol{x})\\
\quad\\
&+\frac{1}{2}\int \text{d}^3 \boldsymbol{x}\text{d}^3 \boldsymbol{x}’\; V(\boldsymbol{x}-\boldsymbol{x}’)\psi(\boldsymbol{x})^\dagger\psi(\boldsymbol{x’})^\dagger\psi(\boldsymbol{x’})\psi(\boldsymbol{x})
\end{aligned}$$

$\psi(\boldsymbol{x})^\dagger,\psi(\boldsymbol{x})$为产生湮灭算符的连续化，满足与$a_\lambda,a_\lambda’$相似的对易关系，只需将$\delta_{\lambda\lambda’}$替换为三维$\delta$函数$\delta(\boldsymbol{x}-\boldsymbol{x’})$

电子气

电子动量与自旋共同本征态$\ket{\boldsymbol{k}\lambda}$

$$\braket{\boldsymbol{x}\lambda}{\boldsymbol{k}\lambda}=\frac{1}{\sqrt{V}}e^{i\boldsymbol{k}\cdot\boldsymbol{x}}\eta_\lambda$$

电子、正电荷背景与二者相互作用哈密顿量，先将库伦势加上指数衰减因子$\mu$防止积分发散，再令$\mu\to 0$

$$H=H_e+H_b+H_{eb}$$

$$H_b=\frac{e^2}{2}\int \text{d}^3 \boldsymbol{x}\text{d}^3 \boldsymbol{x}’ \frac{n(\boldsymbol{x})n(\boldsymbol{x’})\exp(-\mu|\boldsymbol{x}-\boldsymbol{x’}|)}{|\boldsymbol{x}-\boldsymbol{x’}|}=\frac{2\pi N^2 e^2}{\mu^2 V}$$

$$H_{eb}=-e^2\sum_{i=1}^N\int \text{d}^3 \boldsymbol{x}\frac{n(\boldsymbol{x})\exp(-\mu|\boldsymbol{x}-\boldsymbol{r}_i|)}{|\boldsymbol{x}-\boldsymbol{r}_i|}=\frac{-4\pi N^2 e^2}{\mu^2 V}$$

$$H_e=T+U$$

$$T=\sum_\sigma\int\text{d}^3\boldsymbol{k} \frac{\hbar^2 k^2}{2m}a_{\boldsymbol{k}\sigma}^\dagger a_{\boldsymbol{k}\sigma}$$

$$\begin{aligned}
U&=\frac{e^2}{2}\sum_{ij}\sum_{\sigma\sigma’}\sum_{\lambda\lambda’}\int\text{d}^3\boldsymbol{k}\text{d}^3\boldsymbol{k’}\text{d}^3\boldsymbol{p}\text{d}^3\boldsymbol{p’}(k\sigma,p\lambda|\frac{\exp(-\mu|\boldsymbol{r}_i-\boldsymbol{r}_j|)}{|\boldsymbol{r}_i-\boldsymbol{r}_j|}|k’\sigma’,p’\lambda’)a_{k\sigma}^\dagger a_{p\lambda}^\dagger a_{p’\lambda’} a_{k’\sigma’}\\
\quad\\
&=\frac{2\pi (N^2-N) e^2}{\mu^2 V}+\frac{2\pi e^2}{V}\sum_{\sigma\lambda}\int_{q\neq 0} \frac{1}{q^2} a_{k+q,\sigma}^\dagger a_{k’-q,\lambda}^\dagger a_{k’\lambda}a_{k\sigma}
\end{aligned}$$

$$H=H_0+H_1=\sum_\sigma\int\text{d}^3\boldsymbol{k} \frac{\hbar^2 k^2}{2m}a_{\boldsymbol{k}\sigma}^\dagger a_{\boldsymbol{k}\sigma}+2\pi e^2 \sum_{\sigma\lambda}\int_{q\neq 0} \text{d}^3 q\text{d}^3 k\text{d}^3 k’\; \frac{1}{q^2} a_{k+q,\sigma}^\dagger a_{k’-q,\lambda}^\dagger a_{k’\lambda}a_{k\sigma}$$

$$\langle H\rangle=\frac{3\hbar^2k_F^2N}{10m}$$