# Commutation of squared angular momentum operators

1. Jan 26, 2015

### jorgdv

Hello there. I am trying to proove in a general way that

[Lx2,Lz2]=[Ly2,Lz2]=[Lz2,Lx2]

But I am a little bit stuck. I've tried to apply the commutator algebra but I'm not geting very far, and by any means near of a general proof. Any help would be greatly appreciated.

Thank you.

2. Jan 26, 2015

### kith

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

3. Jan 26, 2015

### jorgdv

Of course! We can show $[L^2,L_i^2]=0$ for $i \in \{x,y,z\}$

so

$[L_x^2,L_i^2]+[L_y^2,L_i^2]+[L_z^2,L_i^2]=0$, and for $i=z$ and $i=x$ we have the equalities.

Thank you very much for the hint, I should have seen that sooner