Hubbard 模型对易关系推导 - $[c_//sigma, H]$ 和 $[n_//sigma, H]$

///'首先，我们来推导 //$[c_//sigma, H]$ 的对易关系。//n//n根据哈密顿量的定义，我们有 $H = //sum//limits_//sigma //varepsilon_0 c_//sigma^//dagger c_//sigma + U n_//uparrow n_//downarrow$。//n//n考虑 //$[c_//sigma, H]$，我们需要将 $c_//sigma$ 移到 $H$ 的左边：//n//n//$[c_//sigma, H] = c_//sigma H - H c_//sigma$//n//n将 $H$ 展开，得到：//n//n//$[c_//sigma, H] = c_//sigma //left(//sum//limits_//sigma //varepsilon_0 c_//sigma^//dagger c_//sigma + U n_//uparrow n_//downarrow//right) - //left(//sum//limits_//sigma //varepsilon_0 c_//sigma^//dagger c_//sigma + U n_//uparrow n_//downarrow//right) c_//sigma$//n//n由于 $c_//sigma$ 是费米子算符，它满足反对易关系 $c_//sigma c_//sigma^//dagger + c_//sigma^//dagger c_//sigma = 1$。因此，我们可以使用反对易关系将 $c_//sigma$ 移到右边：//n//n//$[c_//sigma, H] = c_//sigma //left(//sum//limits_//sigma //varepsilon_0 c_//sigma^//dagger c_//sigma + U n_//uparrow n_//downarrow//right) - //left(//sum//limits_//sigma //varepsilon_0 c_//sigma^//dagger c_//sigma + U n_//uparrow n_//downarrow//right) c_//sigma$//n//n$= //sum//limits_//sigma //varepsilon_0 c_//sigma c_//sigma^//dagger c_//sigma - //sum//limits_//sigma //varepsilon_0 c_//sigma^//dagger c_//sigma c_//sigma + U c_//sigma n_//uparrow n_//downarrow - U n_//uparrow n_//downarrow c_//sigma$//n//n由于费米子算符的反对易关系，我们有 $c_//sigma c_//sigma^//dagger = c_//sigma^//dagger c_//sigma + 1$，因此：//n//n//$[c_//sigma, H] = //sum//limits_//sigma //varepsilon_0 (c_//sigma^//dagger c_//sigma c_//sigma - c_//sigma^//dagger c_//sigma c_//sigma) + U c_//sigma n_//uparrow n_//downarrow - U n_//uparrow n_//downarrow c_//sigma$//n//n$= U c_//sigma n_//uparrow n_//downarrow - U n_//uparrow n_//downarrow c_//sigma$//n//n$= U(c_//sigma n_//uparrow n_//downarrow - n_//uparrow n_//downarrow c_//sigma)$//n//n$= U(c_//sigma n_//uparrow n_//downarrow - n_//downarrow c_//sigma n_//uparrow)$//n//n由于 $n_//uparrow n_//downarrow$ 是对易的，我们可以将 $n_//uparrow n_//downarrow$ 移到左边：//n//n//$[c_//sigma, H] = U(n_//uparrow n_//downarrow c_//sigma - n_//downarrow c_//sigma n_//uparrow)$//n//n接下来，我们来推导 //$[n_//sigma, H]$ 的对易关系。//n//n根据哈密顿量的定义，我们有 $H = //sum//limits_//sigma //varepsilon_0 c_//sigma^//dagger c_//sigma + U n_//uparrow n_//downarrow$。//n//n考虑 //$[n_//sigma, H]$，我们需要将 $n_//sigma$ 移到 $H$ 的左边：//n//n//$[n_//sigma, H] = n_//sigma H - H n_//sigma$//n//n将 $H$ 展开，得到：//n//n//$[n_//sigma, H] = n_//sigma //left(//sum//limits_//sigma //varepsilon_0 c_//sigma^//dagger c_//sigma + U n_//uparrow n_//downarrow//right) - //left(//sum//limits_//sigma //varepsilon_0 c_//sigma^//dagger c_//sigma + U n_//uparrow n_//downarrow//right) n_//sigma$//n//n由于 $n_//sigma$ 是算符，我们可以将其展开为 $n_//sigma = c_//sigma^//dagger c_//sigma$。同时，利用反对易关系 $c_//sigma c_//sigma^//dagger + c_//sigma^//dagger c_//sigma = 1$，我们可以将 $c_//sigma^//dagger$ 和 $c_//sigma$ 的位置交换：//n//n//$[n_//sigma, H] = (c_//sigma^//dagger c_//sigma) //left(//sum//limits_//sigma //varepsilon_0 c_//sigma^//dagger c_//sigma + U n_//uparrow n_//downarrow//right) - //left(//sum//limits_//sigma //varepsilon_0 c_//sigma^//dagger c_//sigma + U n_//uparrow n_//downarrow//right) (c_//sigma^//dagger c_//sigma)$//n//n$= //sum//limits_//sigma //varepsilon_0 (c_//sigma^//dagger c_//sigma) (c_//sigma^//dagger c_//sigma) - //sum//limits_//sigma //varepsilon_0 (c_//sigma^//dagger c_//sigma) (c_//sigma^//dagger c_//sigma) + U (c_//sigma^//dagger c_//sigma) n_//uparrow n_//downarrow - U n_//uparrow n_//downarrow (c_//sigma^//dagger c_//sigma)$//n//n$= U (c_//sigma^//dagger c_//sigma) n_//uparrow n_//downarrow - U n_//uparrow n_//downarrow (c_//sigma^//dagger c_//sigma)$//n//n$= U(c_//sigma^//dagger c_//sigma n_//uparrow n_//downarrow - n_//uparrow n_//downarrow c_//sigma^//dagger c_//sigma)$//n//n$= U(c_//sigma^//dagger c_//sigma n_//uparrow n_//downarrow - n_//downarrow c_//sigma^//dagger c_//sigma n_//uparrow)$//n//n由于 $n_//uparrow n_//downarrow$ 是对易的，我们可以将 $n_//uparrow n_//downarrow$ 移到左边：//n//n//$[n_//sigma, H] = U(n_//uparrow n_//downarrow c_//sigma^//dagger c_//sigma - n_//downarrow c_//sigma^//dagger c_//sigma n_//uparrow)$//n//n综上所述，我们得到了 //$[c_//sigma, H]$ 和 //$[n_//sigma, H]$ 的对易关系：//n//n//$[c_//sigma, H] = U(n_//uparrow n_//downarrow c_//sigma - n_//downarrow c_//sigma n_//uparrow)$//n//n//$[n_//sigma, H] = U(n_//uparrow n_//downarrow c_//sigma^//dagger c_//sigma - n_//downarrow c_//sigma^//dagger c_//sigma n_//uparrow)$///