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