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An irreducible decomposition in SU(3)

  1. Jul 1, 2013 #1
    Hi,
    I'v got a problem on Chapter 10 of Georgi's "Lie Algebras in Particle Physics".
    I want to prove

    [itex] u^i v^j_k = \frac{1}{2}A_k^{ij} + \frac{1}{4}B_k^{ij} + \frac{1}{8}C_k^{ij} \ \ \ \ \ \ \ (10.29) \\ [/itex]
    i.e.
    [itex] 3 \bigotimes 8 = 15 \bigoplus \bar{6} \bigoplus 3 \\ [/itex]
    where
    [itex] A_k^{ij} \equiv u^i v^j_k - u^j v^i_k - \frac{1}{4}\delta^i_k u^l v^j_l - \frac{1}{4}\delta^j_k u^l v^i_l \\
    B_k^{ij} \equiv \epsilon^{ijl}
    ( \epsilon_{lmn}u^m v^n_k - \epsilon_{kmn}u^m v^n_l
    ) \\
    C_k^{ij} \equiv 3\delta^i_k u^l v^j_l - \delta^j_k u^l v^i_l \\ [/itex]

    I persist in explaining my procedure in detail to check my calculation.
    First, noting that
    [itex] u^i v^j_k = \frac{1}{2}(u^i v^j_k + u^j v^i_k) + \frac{1}{2}(u^i v^j_k - u^j v^i_k) [/itex]
    In order to have a symmetric(with respect to i and j) and traceless tensor from [itex](u^i v^j_k + u^j v^i_k)[/itex] ,I define [itex]A^{ij}_k[/itex] ,
    [itex] A^{ij}_k \equiv u^i v^j_k + u^j v^i_k -\frac{1}{4}Y^{ij}_k [/itex]
    where
    \begin{equation*}
    \begin{split}
    Y^{ij}_k & \equiv \delta^{i}_k (u^l v^j_l + u^j v^l_l) + (i \leftrightarrow j) \\
    & = \delta^{i}_k u^l v^j_l + \delta^{j}_k u^l v^i_l \\
    ( ∴Y^{ij}_i & = 4 u^l v^j_l ) \\
    \end{split}
    \end{equation*}

    Using [itex]A^{ij}_k[/itex],I'v got

    [itex] u^i v^j_k = \frac{1}{2}A^{ij}_k + \frac{1}{2}
    (u^i v^j_k - u^j v^i_k + \frac{1}{4}\delta^i_k u^l v^j_l + \frac{1}{4}\delta^j_k u^l v^i_l ) \\ [/itex]

    Here,
    \begin{equation*}
    \begin{split}
    \frac{1}{2}(u^i v^j_k - u^j v^i_k) & = \frac{1}{2} \epsilon^{ijl}\epsilon_{lmn}u^m v^n_k \\
    & \equiv \frac{1}{2}\epsilon^{ijl}X_{lk} \\
    \end{split}
    \end{equation*}

    Noting that

    [itex] X_{lk} = \frac{1}{2}(X_{lk}+X_{kl}) - \frac{1}{2}(X_{lk}-X_{kl}) \\ [/itex]
    I'v got
    \begin{equation*}
    \begin{split}
    \frac{1}{2}(u^i v^j_k - u^j v^i_k)
    & = \frac{1}{4}\epsilon^{ijl}(X_{lk}+X_{kl}) + \frac{1}{4}\epsilon^{ijl}(X_{lk}-X_{kl}) \\
    & = \frac{1}{4}\epsilon^{ijl}(\epsilon_{lmn}u^m v^n_k + \epsilon_{kmn}u^m v^n_l) +
    \frac{1}{4}\epsilon^{ijl}(\epsilon_{lmn}u^m v^n_k - \epsilon_{kmn} u^m v^n_l) \\
    & \equiv \frac{1}{4}B^{ij}_k + \frac{1}{4} \cdot \frac{1}{2} \tilde{D}^{ij}_k \\
    \end{split}
    \end{equation*}
    where
    [itex] \tilde{D}^{ij}_k=2\epsilon^{ijl}(\epsilon_{lmn}u^m v^n_k - \epsilon_{kmn} u^m v^n_l) \\ [/itex]

    \begin{equation*}
    \begin{split}
    ∴ u^i v^j_k
    & = \frac{1}{2}A^{ij}_k + \frac{1}{4}B^{ij}_k + \frac{1}{8}(\tilde{D}^{ij}_k + \delta^i_k u^l v^j_l + \delta^j_k u^l v^i_l ) \\
    & \equiv \frac{1}{2}A^{ij}_k + \frac{1}{4}B^{ij}_k + \frac{1}{8}D^{ij}_k \\
    \end{split}
    \end{equation*}
    where
    [itex] D_k^{ij} \equiv 2 \epsilon^{ijl}
    ( \epsilon_{lmn}u^m v^n_k - \epsilon_{kmn}u^m v^n_l
    )
    + \delta^i_k u^l v^j_l + \delta^j_k u^l v^i_l [/itex]

    Then the following equation must hold to get (10.29) in the textbook,
    [itex] C_k^{ij} = D_k^{ij} [/itex]
    i.e.
    [itex] \epsilon^{ijl}\epsilon_{kmn}u^m v^n_l = u^i v^j_k - u^j v^i_k - \delta^i_k u^l v^j_l + \delta^j_k u^l v^i_l \\ [/itex]

    Why is this ?
    Or have I mistaken anywhere ?
    Besides,
    [itex] C^{ij}_j =0 , C^{ij}_i \neq 0 \\ [/itex]
    What implication does exist in the background ?

    Please anybody,help me.
    Thanks in advance.
     
  2. jcsd
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