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Clebsch-Gordan Theorem?

  1. Mar 29, 2009 #1
    Clebsch-Gordan Theorem??

    symmetric spinor tensors are IRR of SU(2), i.e., [tex]T_{\undergroup{\alpha_1\cdots\alpha_r}}[/tex]

    The Clebsch-Gordan theorem says,

    [tex]{\{j_1\}}\otimes{\{j_2\}}={\{j_1+j_2\}}\oplus{\{j_1+j_2-1\}}\oplus\cdots\oplus{\{|j_1-j_2|\}}[/tex].

    Can I prove this theorem by symmetrizing the tensor product,

    [tex]T_{\alpha_1\cdots\alpha_{2j_1}}\otimes T_{\beta_1\cdots\beta_{2j_2}}[/tex]=(express sum of fully symmetric tensors) ??
     
    Last edited: Mar 29, 2009
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  3. Mar 30, 2009 #2
    Re: Clebsch-Gordan Theorem??

    My suspicion is yes, but it will be hard. Smells like something that you'd do with Young tableaux.

    Edit: I think the CG theorem involves both symmetric and antisymmetric tensors. Do you know how to prove the theorem in the usual way (highest weight procedure)?
     
  4. Mar 30, 2009 #3

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    Re: Clebsch-Gordan Theorem??

    The completely symmetric part of the tensor product will give you the higest spin representation. After that, I would guess you contract indices using the invariant symbol:

    [tex] \epsilon^{\alpha \beta} = \left( \begin{array}{cc} 0 &1 \\ -1 & 0 \end{array} \right) [/tex]

    In other words, to couple the rank n representation T with the rank m represenation S, define:

    [tex] A_{\alpha_1 ... \alpha_{n+m-2}} = \epsilon^{\alpha \beta} T_{\alpha \alpha_1 ... \alpha_{n-1}} S_{\beta \alpha_n ... \alpha_{n+m-2} [/tex]

    [tex] B_{\alpha_1 ... \alpha_{n+m-4}} = \epsilon^{\alpha \beta} \epsilon^{\gamma \delta }T_{\alpha \gamma \alpha_1 ... \alpha_{n-2}} S_{\beta \delta \alpha_{n-1} ... \alpha_{n+m-4} [/tex]

    and so on, where you also symmetrize over the [itex]\alpha_i[/itex].
     
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