角運動量交換関係
先に有名なものをまとめて書く。
$[\hat{L}_{x},\hat{L}_{y}]=i\hbar \hat{L}_{z}$、$[\hat{L}_{y},\hat{L}_{z}]=i\hbar \hat{L}_{x}$、$[\hat{L}_{z},\hat{L}_{x}]=i\hbar \hat{L}_{y}$
$[\hat{L}^2,\hat{L}_{z}]=0$
なお、$\hat{L}=\hbar\hat{l}$として、$\hat{l}$について交換関係を書いていることもある。
計算1
$\hat{L_{x}}=[\hat{\boldsymbol{r}}\times \hat{\boldsymbol{p}}]_{x}=-i\hbar\left(y\displaystyle\frac{\partial}{\partial z}-z\displaystyle\frac{\partial}{\partial y}\right)$
※$\hat{p}=-i\hbar\displaystyle\frac{\partial}{\partial \boldsymbol{r}}$を成分ごとに分けて用い、外積をそのまま計算しただけです。
同様に
$\hat{L_{y}}=-i\hbar\left(z\displaystyle\frac{\partial}{\partial x}-x\displaystyle\frac{\partial}{\partial z}\right)$
なので交換関係は
$[\hat{L}_{x},\hat{L}_{y}]=-\hbar^2\left[\left(y\displaystyle\frac{\partial}{\partial z}-z\displaystyle\frac{\partial}{\partial y}\right)\left(z\displaystyle\frac{\partial}{\partial x}-x\displaystyle\frac{\partial}{\partial z}\right)-\left(z\displaystyle\frac{\partial}{\partial x}-x\displaystyle\frac{\partial}{\partial z}\right)\left(y\displaystyle\frac{\partial}{\partial z}-z\displaystyle\frac{\partial}{\partial y}\right)\right]$
$=-\hbar^2\left[yz\displaystyle\frac{\partial^2}{\partial x\partial z}+y\displaystyle\frac{\partial}{\partial x}-z^2\displaystyle\frac{\partial^2}{\partial x\partial y}-xy\displaystyle\frac{\partial^2}{\partial z^2}+xz\displaystyle\frac{\partial^2}{\partial y\partial z}\right]$
$-\hbar^2\left[-yz\displaystyle\frac{\partial^2}{\partial x\partial z}+xy\displaystyle\frac{\partial^2}{\partial z^2}+z^2\displaystyle\frac{\partial^2}{\partial x\partial y}-x\displaystyle\frac{\partial}{\partial y}-xz\displaystyle\frac{\partial^2}{\partial y\partial z}\right]$
$=\hbar^2\left(x\displaystyle\frac{\partial}{\partial y}-y\displaystyle\frac{\partial}{\partial x}\right)$
$=i\hbar \hat{L}_{z}$
展開したときに、$\displaystyle\frac{\partial}{\partial z}\left(z\displaystyle\frac{\partial}{\partial x}\right)=\displaystyle\frac{\partial}{\partial x}+z\displaystyle\frac{\partial^2}{\partial x\partial z}$
などのようになることに注意する！
また、$[L_{y},L_{z}]=i\hbar L_{x}$、$[L_{z},L_{x}]=i\hbar L_{y}$についても同様に示すことが出来る。
計算2
$[\hat{L}^2,\hat{L}_{z}]=0$を示す。
$[\hat{L}^2,\hat{L}_{z}]=[\hat{L}_{x}^2+\hat{L}_{y}^2+\hat{L}_{z}^2,\hat{L}_{z}]$
$=[\hat{L}_{x}^2 ,\hat{L}_{z}]+[\hat{L}_{y}^2 ,\hat{L}_{z}]$
$=2\hat{L}_{x}[\hat{L}_{x},\hat{L}_{z}]+2\hat{L}_{y}[\hat{L}_{y},\hat{L}_{z}]$
$=2\hat{L}_{x}(-i\hbar\hat{L}_{y})+2\hat{L}_{y}(i\hbar\hat{L}_{x})=0$
