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Hund's rule and angular momentum coupling

  1. May 25, 2010 #1
    Hi.

    In Hund's second rule, it seems that we calculate the value of L simply by summing the [itex]L_z[/itex] components of the individual electrons. But L has to do with the eigenvalue of the L^2 operator, i.e. the eigenvalue is L(L+1). So how can this be correct?
     
  2. jcsd
  3. May 25, 2010 #2

    Meir Achuz

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    L is the maximum eigenvalue of L_z.
    L(L+1) is the eigenvalue of a different operator, L^2.
     
  4. May 25, 2010 #3
    Yeah that is exactly my point. We know that that L_z has some particular value. Now why is L=L_z ?
     
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