I Why Don't L2 and Ly Commute When L2 and Lx Do?

[dyn]

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

 

[Nugatory]

Hi.
To show that [ L² , L₊ ] uses the following commutators [ L² , L_x ] = 0 and [ L² , L_y ] = 0 . But if [ L² , L_x ] = 0 this shows that L² and L_x have simultaneous eigenstates ; but then should L² and L_y 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.

 

[dyn]

Thanks. So L² and L_x have a common eigenbasis as the 2 operators commute and L² and L_y have a different common eigenbasis. But as L_x and L_y do not commute these 2 sets of eigenbases can never be the same ?

 

[PeroK]

Thanks. So L² and L_x have a common eigenbasis as the 2 operators commute and L² and L_y have a different common eigenbasis. But as L_x and L_y 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).