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Exchange symmetry when adding angular momentum and in LS coupling?

  1. Oct 24, 2014 #1
    When you add two angular momentum states together, you get states which have exchange symmetry i.e. the highest total angular momentum states (L = l1 + l2) will be symmetric under the interchange of the two particles, (L = l1 + l2 - 1) would be anti-symmetric....and the symmetry under exchange will alternate until we reach the states with (L = l1 - l2).
    If this is all correct, then in order to keep the overall state for two electrons fermionic under LS coupling, we can only combine symmetric total orbital angular momentum states with anti-symmetric total spin angular momentum states (and vice-versa). This page pretty much sums up what I'm trying to say http://quantummechanics.ucsd.edu/ph130a/130_notes/node322.html

    But this would mean we can't have states such as:
    3s3p 3P1 as L = 1 (symmetric) and S = 1 (also symmetric)
    However, my lecture notes use this state as an example to show the effect of residual electrostatic Hamiltonian splitting. So what's wrong?
     
  2. jcsd
  3. Oct 25, 2014 #2

    DrDu

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    This problem only occurs if the electrons occupy orbitals from the same subshell (e.g. both 3p). In your case, you can form both symmetric (3s(1)3p(2)+3s(2)3p(1)) and anti-symmetric functions (3s(1)3p(2)+3s(2)3p(1)) with L=1.
     
  4. Oct 25, 2014 #3
    Why does the problem only occur if we have the same subshell? Also, how do we know that these linear combinations that your wrote are eigenstates of the total orbital angular momentum?

    Thanks!
     
  5. Oct 25, 2014 #4

    DrDu

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    The second point is rather trivial. s corresponds to l1=0, and p to l2=1, adding you get the total L=|l1-l2|...l1+l2=l2.
    For the first point compare instead ##3p^2## with e.g. ##2p3p##. In the first case you have less states available, as there is only one states where m1=m2 while there are two in the other case.
     
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