由 Hubbard 近似i推导如下对易关系：$c_sigmaH$$n_sigmaH$

根据Hubbard模型的哈密顿量形式，我们有：

$H = -t\sum_{\langle ij \rangle, \sigma} (c_{i\sigma}^\dagger c_{j\sigma} + c_{j\sigma}^\dagger c_{i\sigma}) + U\sum_i n_{i\uparrow} n_{i\downarrow}$

其中，$c_{i\sigma}$是在格点$i$上自旋为$\sigma$的费米子算符，$n_{i\sigma} = c_{i\sigma}^\dagger c_{i\sigma}$是对应的粒子数算符。

现在我们来推导对易关系$[c_\sigma,H]$和$[n_\sigma,H]$。

首先，我们来计算$[c_\sigma,H]$，其中$c_\sigma$是一个费米子算符。根据对易关系$[c_{i\sigma},c_{j\sigma'}^\dagger] = \delta_{ij}\delta_{\sigma\sigma'}$，我们有：

$[c_\sigma,H] = -t\sum_{\langle ij \rangle, \sigma'} ( [c_{i\sigma}, c_{j\sigma'}^\dagger c_{j\sigma'}] + [c_{j\sigma'}^\dagger c_{j\sigma'}, c_{i\sigma}]) + U\sum_i [c_{i\sigma}, n_{i\uparrow} n_{i\downarrow}]$

根据对易关系$[c_{i\sigma}, c_{j\sigma'}^\dagger] = \delta_{ij}\delta_{\sigma\sigma'}$和$[c_{j\sigma'}^\dagger c_{j\sigma'}, c_{i\sigma}] = 0$，上式可以简化为：

$[c_\sigma,H] = -t\sum_{\langle ij \rangle, \sigma'} [c_{i\sigma}, c_{j\sigma'}^\dagger] c_{j\sigma'} + U\sum_i [c_{i\sigma}, n_{i\uparrow} n_{i\downarrow}]$

$= -t\sum_{\langle ij \rangle, \sigma'} \delta_{ij}\delta_{\sigma\sigma'} c_{j\sigma'} + U\sum_i [c_{i\sigma}, n_{i\uparrow} n_{i\downarrow}]$

$= -t\sum_{\langle ij \rangle} c_{j\sigma} + U\sum_i [c_{i\sigma}, n_{i\uparrow} n_{i\downarrow}]$

$= -t\sum_{\langle ij \rangle} c_{j\sigma} + U\sum_i (c_{i\sigma} n_{i\uparrow} n_{i\downarrow} - n_{i\uparrow} n_{i\downarrow} c_{i\sigma})$

$= -t\sum_{\langle ij \rangle} c_{j\sigma} + U\sum_i (c_{i\sigma} n_{i\uparrow} n_{i\downarrow} - n_{i\uparrow} c_{i\sigma} n_{i\downarrow} + n_{i\uparrow} c_{i\sigma} n_{i\downarrow})$

$= -t\sum_{\langle ij \rangle} c_{j\sigma} + U\sum_i (c_{i\sigma} n_{i\uparrow} n_{i\downarrow} - n_{i\uparrow} c_{i\sigma} n_{i\downarrow} + n_{i\uparrow} (1 - n_{i\uparrow}) c_{i\sigma} n_{i\downarrow})$

$= -t\sum_{\langle ij \rangle} c_{j\sigma} + U\sum_i (c_{i\sigma} n_{i\uparrow} n_{i\downarrow} - n_{i\uparrow} c_{i\sigma} n_{i\downarrow} + n_{i\uparrow} c_{i\sigma} (1-n_{i\downarrow}))$

$= -t\sum_{\langle ij \rangle} c_{j\sigma} + U\sum_i (c_{i\sigma} n_{i\uparrow} n_{i\downarrow} - n_{i\uparrow} c_{i\sigma} n_{i\downarrow} + n_{i\uparrow} c_{i\sigma} - n_{i\uparrow} c_{i\sigma} n_{i\downarrow})$

$= -t\sum_{\langle ij \rangle} c_{j\sigma} + U\sum_i (c_{i\sigma} n_{i\uparrow} - n_{i\uparrow} c_{i\sigma})$

$= -t\sum_{\langle ij \rangle} c_{j\sigma} + U\sum_i (c_{i\sigma} - n_{i\uparrow} c_{i\sigma})$

$= -t\sum_{\langle ij \rangle} c_{j\sigma} + U\sum_i c_{i\sigma} - U\sum_i n_{i\uparrow} c_{i\sigma}$

$= -t\sum_{\langle ij \rangle} c_{j\sigma} + U\sum_i c_{i\sigma} - U\sum_i (c_{i\sigma}^\dagger c_{i\sigma} - c_{i\sigma} c_{i\sigma}^\dagger) c_{i\sigma}$

$= -t\sum_{\langle ij \rangle} c_{j\sigma} + U\sum_i c_{i\sigma} - U\sum_i (n_{i\sigma} - c_{i\sigma}^\dagger c_{i\sigma}) c_{i\sigma}$

$= -t\sum_{\langle ij \rangle} c_{j\sigma} + U\sum_i c_{i\sigma} - U\sum_i n_{i\sigma} c_{i\sigma} + U\sum_i (c_{i\sigma}^\dagger c_{i\sigma}) c_{i\sigma}$

$= -t\sum_{\langle ij \rangle} c_{j\sigma} + U\sum_i c_{i\sigma} - U\sum_i n_{i\sigma} c_{i\sigma} + U\sum_i c_{i\sigma}^\dagger (c_{i\sigma} c_{i\sigma})$

$= -t\sum_{\langle ij \rangle} c_{j\sigma} + U\sum_i c_{i\sigma} - U\sum_i n_{i\sigma} c_{i\sigma} + U\sum_i c_{i\sigma}^\dagger n_{i\sigma}$

$= -t\sum_{\langle ij \rangle} c_{j\sigma} + U\sum_i c_{i\sigma} - U\sum_i n_{i\sigma} c_{i\sigma} + U\sum_i n_{i\sigma} c_{i\sigma}$

$= -t\sum_{\langle ij \rangle} c_{j\sigma} + U\sum_i c_{i\sigma}$

类似地，我们可以计算$[n_\sigma,H]$：

$[n_\sigma,H] = -t\sum_{\langle ij \rangle, \sigma'} [n_{i\sigma}, c_{j\sigma'}^\dagger c_{j\sigma'}] + U\sum_i [n_{i\sigma}, n_{i\uparrow} n_{i\downarrow}]$

$= -t\sum_{\langle ij \rangle, \sigma'} \delta_{ij}\delta_{\sigma\sigma'} n_{j\sigma'} + U\sum_i [n_{i\sigma}, n_{i\uparrow} n_{i\downarrow}]$

$= -t\sum_{\langle ij \rangle} n_{j\sigma} + U\sum_i [n_{i\sigma}, n_{i\uparrow} n_{i\downarrow}]$

$= -t\sum_{\langle ij \rangle} n_{j\sigma} + U\sum_i (n_{i\sigma} n_{i\uparrow} n_{i\downarrow} - n_{i\sigma} n_{i\uparrow} n_{i\downarrow})$

$= -t\sum_{\langle ij \rangle} n_{j\sigma}$

综上所述，我们得到了对易关系$[c_\sigma,H] = -t\sum_{\langle ij \rangle} c_{j\sigma} + U\sum_i c_{i\sigma}$和$[n_\sigma,H] = -t\sum_{\langle ij \rangle} n_{j\sigma}$