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在运动方程中, 经常要计算各种产生湮灭算符之间的对易关系. 而对于 Fermion 来说, 它满足的是反对易关系, 因此在最后一步中, 需要用下述关系, 写成反对易子的形式 $$ \begin{split} [AB, C] =& A{B, C} - {A,C}B\ [A,BC] =&{A, B} C -B{A,C} \end{split} $$

Boson

Boson 满足对易关系 $$ [a_k, a^{\dagger}{k'}] = \delta{kk'} $$

$$ \sum_{k'}[a^{\dagger}{k3}a{k3},a^{\dagger}{k'2}a{k'3}] = -a^{\dagger}{k2}a{k3} $$

$$ \sum_{k'}[a^{\dagger}{k3}a{k3},a^{\dagger}{k'3}a{k'2}] = a^{\dagger}{k3}a{k2} $$

Fermion

Fermion 满足反对易关系 $$ {a_k, a^{\dagger}{k'}} = \delta{kk'} $$

$$ \sum_{k'}[a^{\dagger}{k3}a{k3},a^{\dagger}{k'2}a{k'3}] = -a^{\dagger}{k2}a{k3} $$

$$ \sum_{k'}[a^{\dagger}{k3}a{k3},a^{\dagger}{k'3}a{k'2}] = a^{\dagger}{k3}a{k2} $$