- #1
Ykjh98Hu
- 10
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let L be angular momentum operator.
[L^2 , Lz] = 0
[L^2 , Lx] = 0 (I haven't prove this, but appearantly it's correct according to lecturer)
does it imply that [Lx , Lz] = 0?
this is just one interesting thoughts that cross my mind because I recalled that if 2 matrix [A,B] =0, A and B will have same eigenvectors (ie same basis that diagonalise them). Does this apply to above case because:
if L^2 and Lz = 0, we can spell ALL eigenstates of them.
then Lx suppose to share ALL those eigenstates since it commutes with L^2 too.
And...it violate uncertainty principle! (impossible.) so someone point out my error please! thanks =)
[L^2 , Lz] = 0
[L^2 , Lx] = 0 (I haven't prove this, but appearantly it's correct according to lecturer)
does it imply that [Lx , Lz] = 0?
this is just one interesting thoughts that cross my mind because I recalled that if 2 matrix [A,B] =0, A and B will have same eigenvectors (ie same basis that diagonalise them). Does this apply to above case because:
if L^2 and Lz = 0, we can spell ALL eigenstates of them.
then Lx suppose to share ALL those eigenstates since it commutes with L^2 too.
And...it violate uncertainty principle! (impossible.) so someone point out my error please! thanks =)