Commutation of squared angular momentum operators

  • #1
29
0
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
Hint: [itex]L^2 = L_x^2 + L_y^2 + L_z^2[/itex]
 
  • #3
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
 

Suggested for: Commutation of squared angular momentum operators

Replies
4
Views
630
Replies
2
Views
584
Replies
14
Views
688
Replies
10
Views
713
Replies
1
Views
462
Replies
0
Views
552
Replies
14
Views
695
Back
Top