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SU(2) symmetric/antisymmetric combination using young tableaux

  1. Mar 1, 2010 #1
    I am pretty confused about how to construct states to make symmetric / anti-symmetric combination so I would like to ask some questions.

    For example, for SU(2), states of three spin-half particles can be decomposed as 2 x 2 x 2 = 4 + 2 + 2, 3 irreducible combination with dim 4, 2, 2.

    -if I combine three boxes in a row(in young tableaux), making a total symmetric combination,
    then [[a, b], c] = (ab + ba) c + c (ab + ba) = abc + bac + bac + cba (right?) (here the large bracket represents symmetric index)
    And there should be four in general non-vanishing independent index: 111, 112=211, 221=122, 222(right?)
    But as I see, [[1,1],2] != [[1,2],1] for [[1,1],2] = 112 + 112 + 211 + 211 while [[1,2],1] = 121 + 211 + 112 + 121 and [[2,2],1] != [[1,2],2] for similar reason. (it's because c is symmetric only with [a, b], not with the individual a and b). So there seems to be six non-vanishing index.
    I wonder what is wrong with my calculation.

    -And as I understand, [{a, b}, c] = (ab - ba) c + c(ab - ba) can have two non-vanishing index:
    [{1, 2}, 1], [{1, 2}, 2]
    And for {[a, b], c} = (ab + ba) c - c (ab + ba), there are two independent non-vanishing index:
    {[1,1],2}, {[2,2],1} (or equivalently {[1,2],2})
    Is this correct?

    -And if I calculate {[1,1],2} of |+>|->|+>, it's 0. But as I calculated directly using CG table, |+>|->|+> is a part of the state whose {[1,1],2} only is 1. So I guess it shouldn't be zero. Is something wrong with this?

    Actually there are many other things I get confused but first I would like to know these and then figure out the next.

    Thank you very much.
     
    Last edited: Mar 2, 2010
  2. jcsd
  3. Mar 2, 2010 #2

    DrDu

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    You wrote:
    -if I combine three boxes in a row(in young tableaux), making a total symmetric combination,
    then [[a, b], c] = (ab + ba) c + c (ab + ba) = abc + bac + bac + bca (right?)

    First you should explain your nomenclature. The brackets, are they anti-commutators?
    Why should a totally symmetric representation have the form of the multiple anti-commutators in the form given by you? I also don't understand the right hand side of your equation. I get
    abc+bac+cab+cba, so the c never stands in the middle position. Hence it cannot be a totally symmetric combination.
     
  4. Mar 2, 2010 #3
    You are right. They are my mistakes and I corrected them. But I still don't understand. I'll really appreciate if u can give me some hints.
     
  5. Mar 2, 2010 #4

    DrDu

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    Science Advisor

    Hm, but you still did not explain why you think that a totally symmetric Young tableaux should have to do anything with [[a, b], c].
     
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