# 交換子

（重定向自正則對易關係

## 量子力學

$[\hat{A}, \hat{B}] = \hat{A}\hat{B} - \hat{B}\hat{A}$

$[\hat{A},\hat{B}]=-[\hat{B},\hat{A}]$
$[\hat{A},\hat{B}+\hat{C}]=[\hat{A},\hat{B}]+[\hat{A},\hat{C}],\quad[\hat{A}+\hat{B},\hat{C}]=[\hat{A},\hat{C}]+[\hat{B},\hat{C}]$
$[\hat{A},\hat{B}\hat{C}]=[\hat{A},\hat{B}]\hat{C}+\hat{B}[\hat{A},\hat{C}],\quad[\hat{A}\hat{B},\hat{C}]=[\hat{A},\hat{C}]\hat{B}+\hat{A}[\hat{B},\hat{C}]$
$[\hat{A},\hat{A}^n]=0,\quad n=1,2,3...$
$[k\hat{A},\hat{B}]=[\hat{A},k\hat{B}]=k[\hat{A},\hat{B}]$
$[\hat{A}, [\hat{B}, \hat{C}]]+[\hat{C}, [\hat{A}, \hat{B}]]+[\hat{B}, [\hat{C}, \hat{A}]] = 0$

$[\hat{x}_i, \hat{x}_j] = 0$ $[\hat{x}, \hat{x}] = 0$$[\hat{x}, \hat{y}] = 0$
$[\hat{p}_i, \hat{p}_j] = 0$ $[\hat{p}_x, \hat{p}_x] = 0$$[\hat{p}_x, \hat{p}_y] = 0$
$[\hat{x}_i, \hat{p}_j] = i\hbar \delta_{ij}$ $[\hat{x}, \hat{p}_x] = i\hbar$$[\hat{x}, \hat{p}_y] = 0$$[\hat{y}, \hat{p}_x] = 0$$[\hat{y}, \hat{p}_y] = i\hbar$
$[\hat{L}_i, \hat{L}_j] = i\hbar \epsilon_{ijk}\hat{L}_k$ $[\hat{L}_x, \hat{L}_y] = i\hbar \hat{L}_z$$[\hat{L}_y, \hat{L}_z] = i\hbar \hat{L}_x$$[\hat{L}_z, \hat{L}_x] = i\hbar \hat{L}_y$

### 正則對易關係

$[x,p] = i\hbar$

### 與古典力學的關係

$\{x,p\} = 1 \,\!$

$[\hat f,\hat g]= i\hbar\widehat{\{f,g\}} \,$