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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 $L^2 = L_x^2 + L_y^2 +L_z^2$

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    $\begingroup$ Hint: Jacobi identity. $\endgroup$ – Qmechanic Jun 11 at 11:07
  • $\begingroup$ @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? $\endgroup$ – Wouter A Jun 11 at 11:53
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    $\begingroup$ @WouterA Replace $L_z$ with $[L_x, L_y]$ and expand the final expression. $\endgroup$ – gented Jun 11 at 12:17
  • $\begingroup$ @gented Thanks, I found the answer! $\endgroup$ – Wouter A Jun 11 at 12:40
  • $\begingroup$ @WouterA If you have found the answer, please post it. After some days, you can accept your own answer. $\endgroup$ – Wrichik Basu Jun 11 at 13:23
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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 $$ [G,L_z] = 0 $$

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