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Selection rules in electric dipole appoximation

  1. Apr 23, 2015 #1

    blue_leaf77

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    Some literatures say that the selection rule in electric dipole approx. for angular momentum ##\Delta j = 0,-1,1## some other say ##\Delta l = -1,1##. I follow the notation used in my references, despite the difference I think since j and l are both angular momenta which fulfill angular momentum commutation relations, I can regard them to be the same. But why are they different?
     
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  3. Apr 23, 2015 #2

    DrClaude

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    ##\Delta l = \pm 1## is due to conservation of angular momentum, as a photon has spin 1. But since ##\hat{\jmath} = \hat{l} + \hat{s}##, you can find combinations of ##l## and ##s## (or ##m_l## and ##m_s##) such that ##l## changes by 1, but ##m_l## changes also such that ##j## doesn't change. Note that this is not possible when ##j = 0##, such that ##j = 0 \rightarrow j'=0## transitions are forbidden (i.e., only for ##j \geq 1## is ##\Delta j = 0## allowed).
     
  4. Apr 23, 2015 #3

    blue_leaf77

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    Did you mean ##s## in place of ##m_l##?
     
  5. Apr 23, 2015 #4

    DrClaude

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    No. The electromagnetic field doesn't couple to spin, so you have ##\Delta S = 0## for many-electron atoms.
     
  6. Apr 23, 2015 #5

    blue_leaf77

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    If I take an example of transition from ##1s_{1/2}## to ##2p_{1/2}##, would that be a good example?
     
  7. Apr 23, 2015 #6

    DrClaude

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    Starting from ##l=0, m_l = 0, s= 1/2, m_s = 1/2, j= 1/2, m_j = 1/2##, the atom can absorb a ##\sigma^-## photon (##\Delta l = 1##, ##\Delta m_l = -1##), to end up in the state with ##l=1, m_l = -1, s= 1/2, m_s = 1/2, j= 1/2, m_j = -1/2##.

    Note that the same photon can also lead to a transition to ##l=1, m_l = -1, s= 1/2, m_s = 1/2, j= 3/2, m_j = -1/2##.
     
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