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Total spin of triplet and singlet states

  1. Aug 19, 2015 #1
    I'm a litte confused about spin triplet and singlet states. How do we know that for ↑↓+↓↑ the total spin S is 1, and for ↑↓-↓↑ the total spin S is 0?
    Also, how is total ms computed for these two states? (I understand that they are both 0, but not sure where that comes from)

    Thank you very much for the help!
     
  2. jcsd
  3. Aug 20, 2015 #2

    blue_leaf77

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    Apply the (squared) total spin operator ##\mathbf{S}^2 = (\mathbf{S_1}+\mathbf{S_2})^2## to those states.
    Apply the operator ##S_z = S_{1z} + S_{2z}## to those states. Alternatively, upon following the theorem of the addition of angular momenta, you will find that the z component of the resultant spin state is equal to the sum of the z components of the individual states appearing in the resultant state's representation in the individual spin state basis.
     
  4. Aug 20, 2015 #3
    Thank you very much!! I realize I never learned the total spin operator... Is there a recommended reading about this?
     
  5. Aug 20, 2015 #4

    blue_leaf77

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  6. Aug 20, 2015 #5
    Thank you! :)
     
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