# Commutator of $L^{2}$ and $L_{z}$

I'm trying to work through a proof of why $$[L^{2},L_{z}]=0$$, and am getting lost on this step:

We can use the commutation relation $$[\hat{L}_{z},\hat{L}_{x}]=i\hbar\hat{L}_{y}$$ to rewrite the term as:

$$\hat{L}_{x}\hat{L}_{x}\hat{L}_{z} = \hat{L}_{x}\hat{L}_{z}\hat{L}_{x} - i\hbar\hat{L}_{x}\hat{L}_{y}$$

I understand that they are trying to get the first term on the RHS to cancel with the next term in the commutator, and that the Levi-Cevita symbol dictates the negative sign for the second term, I'm just unsure how they come about from the commutation relationship. Any help would be appreciated.

• $[L^{2},L_{z}]=[L_{x}L_{x}+L_{y}L_{y}+L_{z}L_{z},L_{z}]=L_{x}[L_{x},L_{z}]+[L_{x},L_{z}]L_{x}+L_{y}[L_{y},L_{z}]+[L_{y},L_{z}]L_{y}=0$ – Cinaed Simson Nov 15 '19 at 6:34

Given

$$[ L_z, L_x ] = i \hbar L_y$$

just multiply both sides by $$L_x$$:

$$L_x \bigg( L_z L_x - L_x L_z \bigg) = i \hbar L_x L_y$$

which gives

$$L_x L_x L_z = L_x L_z L_x - i \hbar L_x L_y$$

• Oh, that's incredibly obvious. Thanks. – h_bear Nov 15 '19 at 4:43