Commutation of squared angular momentum operators

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SUMMARY

The discussion centers on proving the commutation relations of squared angular momentum operators, specifically that the commutators [Lx², Lz²], [Ly², Lz²], and [Lz², Lx²] are equal to zero. The key insight provided is that L² = Lx² + Ly² + Lz², leading to the conclusion that [L², Li²] = 0 for i ∈ {x, y, z}. This establishes the necessary relationships among the squared operators, confirming their mutual commutativity.

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jorgdv
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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.
 
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Hint: [itex]L^2 = L_x^2 + L_y^2 + L_z^2[/itex]
 
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
 

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