# Show that when angular momentum $L_x$ and $L_y$ commute with operator $G$, then $L_z$ also commutes with $G$

I want to prove that if Angular momentum $$L_x$$ and $$L_y$$ commute with an operator $$G$$, angular momentum $$L_z$$ also commutes with $$G$$.

if $$[L_x , G] = [L_y, G] = 0$$

then $$[L_z , G] = 0$$

I know that $$[L_x, L_y] = ih(L_z)$$ and $$L^2 = L_x^2 + L_y^2 +L_z^2$$

• Hint: Jacobi identity. – Qmechanic Jun 11 at 11:07
• @Qmechanic Thanks, I've tried using it but don't know what aproach I should take. Should I write Lx and Ly using momentum p? – Wouter A Jun 11 at 11:53
• @WouterA Replace $L_z$ with $[L_x, L_y]$ and expand the final expression. – gented Jun 11 at 12:17
• @gented Thanks, I found the answer! – Wouter A Jun 11 at 12:40
• @WouterA If you have found the answer, please post it. After some days, you can accept your own answer. – Wrichik Basu Jun 11 at 13:23

We know that, $$[L_x,L_y]=2 \;i \;\hslash \;L_z$$ Consider an operator G, and relations as mentioned $$[G,L_x]=0 =[L_x,G]$$ $$[G,L_y]=0=[L_y,G]$$ Note: G is a member of the Lie algebra as well because we have defined a Lie bracket for it.
Qmechanic gave a really nice hint. If there are three operators A,B and C in a lie algebra, the Jacobi Identity is as follows: $$[A,[B,C]]+[B,[C,A]]+[C,[A,B]] = 0$$ Let, $$A = G$$ and $$B = L_x$$ and $$C = L_y$$, we thus obtain, $$[G,[L_x,L_y]] = 0$$ We can now use the commutation relation and obtain the result that you wanted, i.e $$[G,L_z] = 0$$