Commutation of squared angular momentum operators

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The discussion revolves around proving the commutation relations of squared angular momentum operators, specifically that [Lx², Lz²] = [Ly², Lz²] = [Lz², Lx²]. The user expresses difficulty in applying commutator algebra to achieve a general proof. A helpful hint is provided, indicating that L² = Lx² + Ly² + Lz² can be used to show that [L², Li²] = 0 for i in {x, y, z}. This leads to the conclusion that the individual commutation relations can be derived from this property. The user acknowledges the hint and recognizes its significance in their proof.
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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: L^2 = L_x^2 + L_y^2 + L_z^2
 
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
 
Time reversal invariant Hamiltonians must satisfy ##[H,\Theta]=0## where ##\Theta## is time reversal operator. However, in some texts (for example see Many-body Quantum Theory in Condensed Matter Physics an introduction, HENRIK BRUUS and KARSTEN FLENSBERG, Corrected version: 14 January 2016, section 7.1.4) the time reversal invariant condition is introduced as ##H=H^*##. How these two conditions are identical?

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