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Angular momentum operator

  1. Feb 1, 2015 #1
    One can represent the mean of the angular momentum operator as a vector. But what is the (mathematical) justification to represent the operator by a vector which has a direction that the operator has not. Yet worse, l(l+1) h2 is the proper value of operator L^2 and from such result it is assumed that (l(l+1))^1/2 is the length of operator L . Such trick-representation seems to be a strong convention, but it does not make it mathematically correct.
     
  2. jcsd
  3. Feb 1, 2015 #2
    Correct me if I'm wrong, but aren't l(l+2)h^2 the eigenvalues of the momentum operator? Operators don't have length, but certainly their eigenvalues do.
     
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