这里分析一下两体相互作用在二次量子化下的形式

两体相互作用

我们要推导如下恒等式：

$\hat{n}_\alpha \hat{n}_\beta - \delta_{\alpha,\beta} \hat{n}_\alpha = \hat{\psi}^\dagger_\alpha \hat{\psi}^\dagger_\beta \hat{\psi}_\beta \hat{\psi}_\alpha,$

这是在 二次量子化（second quantization） 框架下关于 粒子数算符和反对易关系的一个恒等式。

对于费米子，有：

$\hat{n}_\alpha = \hat{\psi}^\dagger_\alpha \hat{\psi}_\alpha, \quad \hat{n}_\beta = \hat{\psi}^\dagger_\beta \hat{\psi}_\beta$

情况一：$\alpha \ne \beta$

$\begin{aligned} \hat{n}_\alpha \hat{n}_\beta &= (\hat{\psi}^\dagger_\alpha \hat{\psi}_\alpha)(\hat{\psi}^\dagger_\beta \hat{\psi}_\beta) \\ &= \hat{\psi}^\dagger_\alpha \hat{\psi}_\alpha \hat{\psi}^\dagger_\beta \hat{\psi}_\beta \\ &= \hat{\psi}^\dagger_\alpha (\delta_{\alpha\beta} - \hat{\psi}^\dagger_\beta \hat{\psi}_\alpha) \hat{\psi}_\beta \quad \text{(反对易)} \\ &= -\hat{\psi}^\dagger_\alpha \hat{\psi}^\dagger_\beta \hat{\psi}_\alpha \hat{\psi}_\beta \\ &= \hat{\psi}^\dagger_\alpha \hat{\psi}^\dagger_\beta \hat{\psi}_\beta \hat{\psi}_\alpha \quad \text{(交换两次符号抵消)} \end{aligned}$

结论：

$\hat{n}_\alpha \hat{n}_\beta = \hat{\psi}^\dagger_\alpha \hat{\psi}^\dagger_\beta \hat{\psi}_\beta \hat{\psi}_\alpha \quad (\alpha \ne \beta)$

情况二：$\alpha = \beta$

$\begin{aligned} \hat{n}_\alpha^2&= (\hat{\psi}^\dagger_\alpha \hat{\psi}_\alpha)^2 = \hat{\psi}^\dagger_\alpha \hat{\psi}_\alpha \hat{\psi}^\dagger_\alpha \hat{\psi}_\alpha= \hat{\psi}^\dagger_\alpha (1 - \hat{\psi}^\dagger_\alpha \hat{\psi}_\alpha) \hat{\psi}_\alpha \\ &= \hat{n}_\alpha - \hat{n}_\alpha^2 \Rightarrow 2\hat{n}_\alpha^2 = \hat{n}_\alpha \Rightarrow \hat{n}_\alpha^2 = \hat{n}_\alpha \end{aligned}$

另一方面：

$\hat{\psi}^\dagger_\alpha \hat{\psi}^\dagger_\alpha = 0 \Rightarrow \hat{\psi}^\dagger_\alpha \hat{\psi}^\dagger_\alpha \hat{\psi}_\alpha \hat{\psi}_\alpha = 0$

所以有：

$\hat{n}_\alpha^2 - \hat{n}_\alpha = -\hat{n}_\alpha = \hat{\psi}^\dagger_\alpha \hat{\psi}^\dagger_\alpha \hat{\psi}_\alpha \hat{\psi}_\alpha$

因此，对所有 $\alpha, \beta$ 都有：

$\hat{n}_\alpha \hat{n}_\beta - \delta_{\alpha\beta} \hat{n}_\alpha = \hat{\psi}^\dagger_\alpha \hat{\psi}^\dagger_\beta \hat{\psi}_\beta \hat{\psi}_\alpha$