Angular momentum commutators

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  • #1
dyn
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Hi.
To show that [ L2 , L+ ] uses the following commutators [ L2 , Lx ] = 0 and [ L2 , Ly ] = 0 . But if [ L2 , Lx ] = 0 this shows that L2 and Lx have simultaneous eigenstates ; but then should L2 and Ly not commute ?
Thanks
 

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  • #2
Nugatory
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Hi.
To show that [ L2 , L+ ] uses the following commutators [ L2 , Lx ] = 0 and [ L2 , Ly ] = 0 . But if [ L2 , Lx ] = 0 this shows that L2 and Lx have simultaneous eigenstates ; but then should L2 and Ly not commute ?
Thanks
No. If A and B commute so have a common eigenbasis and B and C commute so have a different common eigenbasis, it does not follow that A and C commute and have a common eigenbasis.
 
  • #3
dyn
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Thanks. So L2 and Lx have a common eigenbasis as the 2 operators commute and L2 and Ly have a different common eigenbasis. But as Lx and Ly do not commute these 2 sets of eigenbases can never be the same ?
 
  • #4
PeroK
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Thanks. So L2 and Lx have a common eigenbasis as the 2 operators commute and L2 and Ly have a different common eigenbasis. But as Lx and Ly do not commute these 2 sets of eigenbases can never be the same ?

A spin-1/2 system is useful because the operators are simple 2x2 matrices. I suggest you work through this example for these AM operators.

Note that the identity operator, ##I##, commutes with everything (and every vector is an eigenvector of ##I##). So, ##I## has a shared eigenbasis with all operators that have an eigenbasis (such as Hermitian operators).
 

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