Angular Momentum and Position/Momentum Commutation Relations
Formula
The commutator of angular momentum and position is as follows.
$$ \begin{align*} [L_{z}, x] &= \i \hbar y \\ [L_{z}, y] &= -\i \hbar x \\ [L_{z}, z] &= 0 \end{align*} $$
The commutator of angular momentum and momentum is as follows.
$$ \begin{align*} [L_{z}, p_{x}] &= \i \hbar p_{y} \\ [L_{z}, p_{y}] &= -\i \hbar p_{x} \\ [L_{z}, p_{z}] &= 0 \end{align*} $$
The squares of angular momentum and position, and the squares of momentum are commutative. That is, the following equation holds.
$$ \begin{align*} [L_{z}, r^{2}] &= 0 \\ [L_{z}, p^{2}] &= 0 \end{align*} $$
Here, it is $r^{2} = x^{2} + y^{2} + z^{2}$ and $p^{2} = p_{x}^{2} + p_{y}^{2} + p_{z}^{2}$. Additionally, the following holds.
$$ \begin{align*} [L_{z}, x^{2}] &= 2\i\hbar xy \\ [L_{z}, y^{2}] &= -2\i\hbar xy \\ [L_{z}, p_{x}^{2}] &= 2\i\hbar p_{x}p_{y} \\ [L_{z}, p_{y}^{2}] &= -2\i\hbar p_{x}p_{y} \\ \end{align*} $$
Explanation
From the above formula, we can see that angular momentum is commutative with position and momentum of the same coordinate.
Proof
The following formulas are useful.
Commutator of position and momentum
$$ [x, p_{x}] = \i \hbar $$
$$ \begin{align*} [ A + B, C ] &= [ A, C ] + [ B, C ] \\[0.5em] [AB, C] &= A[ B, C ] + [ A, C] B \\[0.5em] [A, BC] &= B[ A, C ] + [ A, B] C \end{align*} $$
$[L_{z}, x]$
First, expanding the commutator $[L_{z}, x]$,
$$ \begin{align*} [L_{z}, x] &= [xp_{y}-yp_{x}, x] \\ &= (xp_{y}x - xxp_{y}) - (yp_{x}x - xyp_{x}) \end{align*} $$
Here, since position and momentum of different coordinates are commutative, the first two terms are $0$.
$$ xp_{y}x-xxp_{y} = xxp_{y}-xxp_{y}=0 $$
Thus, we can organize as follows.
$$ \begin{align*} [L_{z},x] &= -yp_{x}x + xyp_{x} \\ &= -yp_{x}x + yxp_{x} \\ &= y(xp_{x} - p_{x}x) \\ &= y[x, p_{x}] \\ &= y(\i \hbar) \\ &= \i \hbar y \end{align*} $$
Similarly, calculating $[L_{z}, y]$, we get the following.
$$ \begin{align*} [L_{z},y] &= [xp_{y}-yp_{x},y] \\ &= xp_{y}y-yxp_{y}-yp_{x}y+yyp_{x} \\ &= xp_{y}y-yxp_{y} \\ &= xp_{y}y-xyp_{y} \\ &= x(p_{y}y-yp_{y}) \\ &= x[p_{y},y] \\ &= x(-\i\hbar) \\ &= -\i\hbar x \end{align*} $$
Also, since $z$ is commutative with $x$, $y$, $p_{x}$, and $p_{z}$, we obtain the following.
$$ \begin{align*} [L_{z},z] &= [xp_{y}-yp_{x},z] \\ &= 0 \end{align*} $$
$[L_{z}, p_{x}]$
By the properties of commutators, $[L_{z},p_{x}]$ is as follows.
$$ \begin{align*} [L_{z},p_{x}] &= [xp_{y}-yp_{x},p_{x}] \\ &= xp_{y}p_{x}-p_{x}xp_{y}-yp_{x}p_{x}+p_{x}yp_{x} \\ &= xp_{y}p_{x}-p_{x}xp_{y} \\ &= xp_{x}p_{y}-p_{x}xp_{y} \\ &= (xp_{x}-p_{x}x)p_{y} \\ &= [x,p_{x}]p_{y} \\ &= \i\hbar p_{y} \end{align*} $$
The third equality holds because $y$ and $p_{x}$ are commutative. The fourth equality holds because $p_{x}$ and $p_{y}$ are commutative. Similarly, we obtain the following.
$$ \begin{align*} [L_{z},p_{y}] &= [xp_{y}-yp_{x},p_{y}] \\ &= xp_{y}p_{y}-p_{y}xp_{y}-yp_{x}p_{y}+p_{y}yp_{x} \\ &= -yp_{x}p_{y}+p_{y}yp_{x} \\ &= [p_{y},y]p_{x} \\ &= - \i \hbar p_{x} \end{align*} $$
Since $p_{z}$ is commutative with $x$, $y$, $p_{x}$, and $p_{y}$, the following equation holds.
$$ \begin{align*} [L_{z},p_{z}] &= [xp_{y}-yp_{x},p_{z}] \\ &= 0 \end{align*} $$
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$[L_{z}, r^{2}]$, $[L_{z}, p^{2}]$
By the properties of commutators, we obtain the following.
$$ \begin{align*} [L_{z}, x^{2}] &= x[L_{z},x] + [L_{z},x]x \\ &= x\i\hbar y + \i\hbar yx \\ &= 2\i\hbar xy \end{align*} $$
Similarly, we can calculate as follows.
$$ \begin{align*} [L_{z}, y^{2}] &= y[L_{z},y] + [L_{z},y]y \\ &= y(-\i\hbar x) - \i\hbar xy \\ &= -2\i\hbar xy \end{align*} $$
$$ \begin{align*} [L_{z},z^{2}] &= z[L_{z},z] + [L_{z},z]z \\ &= 0 \end{align*} $$
Thus, we obtain the following.
$$ [L_{z}, r^{2}] = [L_{z}, x^{2} + y^{2} + z^{2}] = 2\i\hbar xy - 2\i\hbar xy + 0 = 0 $$
By the same method, we can calculate as follows.
$$ \begin{align*} [L_{z},p_{x}^{2}] &= p_{x}[L_{z},p_{x}]+[L_{z},p_{x}]p_{x} \\ &= 2\i\hbar p_{x}p_{y} \end{align*} $$
$$ \begin{align*} [L_{z},p_{y}^{2}] &= p_{y}[L_{z},p_{y}]+[L_{z},p_{y}]p_{y} \\ &= -2\i\hbar p_{x}p_{y} \end{align*} $$
$$ \begin{align*} [L_{z},p_{z}^{2}] &= p_{z}[L_{z},p_{z}]+[L_{z},p_{z}]p_{z} \\ &= 0 \end{align*} $$
Thus, we obtain the following.
$$ [L_{z},p^{2}] = [L_{z},p_{x}^{2}+p_{y}^{2}+p_{z}^{2}] = 2\i\hbar p_{x}p_{y} - 2\i\hbar p_{x}p_{y} + 0 = 0 $$
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