产生算符和湮灭算符

波函数$|n_1{n}_2\cdots n_m⟩$表示在态$\lambda$上粒子数为$n_{\lambda }$，

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

$$|n_1{n}_2\cdots n_m⟩=\prod _{\lambda =1}^m\frac{1}{{\displaystyle \sqrt{n_{\lambda }!}}}(a_{\lambda }^{†}{)}^{n_{\lambda }}|0⟩$$

$$⟨n_1{n}_2\cdots n_m|=⟨0|\prod _{\lambda =m}^1\frac{1}{{\displaystyle \sqrt{n_{\lambda }!}}}(a_{\lambda }{)}^{n_{\lambda }}$$

$$N_{\lambda }=a_{\lambda }^{†}{a}_{\lambda }=n_{\lambda }$$

玻色子算符对易关系

$$[a_{\lambda },a_{{\lambda }^{\prime }}]=[a_{\lambda }^{†},a_{{\lambda }^{\prime }}^{†}]=0$$

$$[a_{\lambda },a_{{\lambda }^{\prime }}^{†}]={\delta }_{\lambda {\lambda }^{\prime }}$$

费米子算符对易关系

$$\{a_{\lambda },a_{{\lambda }^{\prime }}\}=\{a_{\lambda }^{†},a_{{\lambda }^{\prime }}^{†}\}=0$$

$$\{a_{\lambda },a_{{\lambda }^{\prime }}^{†}\}={\delta }_{\lambda {\lambda }^{\prime }}$$

$$$$

算符二次量子化

单体作用

$${\hat{O}}_{\text{2nd}}=\sum _{\mu \nu }⟨\mu |\hat{O}|\nu ⟩a_{\mu }^{†}{a}_{\nu }$$

二体作用

$${\hat{V}}_{\text{2nd}}=\frac{1}{2}\sum _{\mu {\mu }^{\prime }}\sum _{\nu {\nu }^{\prime }}(\mu \nu |\hat{V}|{\mu }^{\prime }{\nu }^{\prime })a_{\mu }^{†}{a}_{\nu }^{†}{a}_{{\nu }^{\prime }}{a}_{{\mu }^{\prime }}$$

自旋算符

$${\hat{S}}_z=\frac{\hslash }{2}(a_{+}^{†}{a}_{+}^{†}-a_{-}^{†}{a}_{-})$$

$$\hat{S}=\sum _{\alpha {\alpha }^{\prime }}⟨\alpha |\sigma |{\alpha }^{\prime }⟩a_{\alpha }^{†}{a}_{{\alpha }^{\prime }}$$

哈密顿量

$$\begin{array}{rl}\hat{H}& =\int {\text{d}}^3\mathit{x}\psi (\mathit{x}{)}^{†}[-\frac{{\hslash }^2}{2m}{\mathrm{\nabla }}^2+U(\mathit{x})]\psi (\mathit{x})\\ \\ & +\frac{1}{2}\int {\text{d}}^3\mathit{x}{\text{d}}^3{\mathit{x}}^{\prime }V(\mathit{x}-{\mathit{x}}^{\prime })\psi (\mathit{x}{)}^{†}\psi ({\mathit{x}}^{\mathbf{\prime }}{)}^{†}\psi ({\mathit{x}}^{\mathbf{\prime }})\psi (\mathit{x})\end{array}$$

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

电子气

电子动量与自旋共同本征态$|\mathit{k}\lambda ⟩$

$$⟨\mathit{x}\lambda |\mathit{k}\lambda ⟩=\frac{1}{\sqrt{V}}{e}^{i\mathit{k}\cdot \mathit{x}}{\eta }_{\lambda }$$

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$\mu \to 0$

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

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

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

$$H_e=T+U$$

$$T=\sum _{\sigma }\int {\text{d}}^3\mathit{k}\frac{{\hslash }^2{k}^2}{2m}{a}_{\mathit{k}\sigma }^{†}{a}_{\mathit{k}\sigma }$$

$$\begin{array}{rl}U& =\frac{e^2}{2}\sum _{ij}\sum _{\sigma {\sigma }^{\prime }}\sum _{\lambda {\lambda }^{\prime }}\int {\text{d}}^3\mathit{k}{\text{d}}^3{\mathit{k}}^{\mathbf{\prime }}{\text{d}}^3\mathit{p}{\text{d}}^3{\mathit{p}}^{\mathbf{\prime }}(k\sigma ,p\lambda |\frac{\exp (-\mu |{\mathit{r}}_i-{\mathit{r}}_j|)}{|{\mathit{r}}_i-{\mathit{r}}_j|}|k^{\prime }{\sigma }^{\prime },p^{\prime }{\lambda }^{\prime })a_{k\sigma }^{†}{a}_{p\lambda }^{†}{a}_{p^{\prime }{\lambda }^{\prime }}{a}_{k^{\prime }{\sigma }^{\prime }}\\ \\ & =\frac{2\pi (N^2-N)e^2}{{\mu }^{2}V}+\frac{2\pi e^2}{V}\sum _{\sigma \lambda }{\int }_{q\ne 0}\frac{1}{q^2}{a}_{k+q,\sigma }^{†}{a}_{k^{\prime }-q,\lambda }^{†}{a}_{k^{\prime }\lambda }{a}_{k\sigma }\end{array}$$

$$H=H_0+H_1=\sum _{\sigma }\int {\text{d}}^3\mathit{k}\frac{{\hslash }^2{k}^2}{2m}{a}_{\mathit{k}\sigma }^{†}{a}_{\mathit{k}\sigma }+2\pi e^2\sum _{\sigma \lambda }{\int }_{q\ne 0}{\text{d}}^{3}q{\text{d}}^{3}k{\text{d}}^3{k}^{\prime }\frac{1}{q^2}{a}_{k+q,\sigma }^{†}{a}_{k^{\prime }-q,\lambda }^{†}{a}_{k^{\prime }\lambda }{a}_{k\sigma }$$

$$⟨H⟩=\frac{3{\hslash }^2{k}_F^{2}N}{10m}$$