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Quick noob question: commutative of eigenstates

  1. Aug 19, 2010 #1
    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 =)
     
  2. jcsd
  3. Aug 19, 2010 #2
    It is true that if A and B commute, then they have a common complete set of eigenvectors. So obviously if A and C commute, then they also have a complete set of common eigenvectors, but nobody says these two sets must be the same! This of course means that some of the eigenvalues of A must be degenerate, because otherwise the eigenvectors are uniquely determined. This is exactly the case of L^2, whose eigenvalues j = 0, 1/2, 1, 3/2, 2, ... show (at least) a j(j+1)-fold degeneracy. In fact L_z and L_x don't commute, generally: their commutator is iL_y, which is most of the times different from zero.
     
  4. Aug 19, 2010 #3
    ah thanks for the clarification! i notice [Lx,Lz] !=0 and that's why i posted this. i must've forgot about degenerate eigenvalues.
     
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