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The $\mathrm{SU(3)}$ structure constants $f_{abc}$ are defined by $$[\lambda^a,\lambda^b] = 2i f^{abc} \lambda^c,$$ with $\lambda^a$ being the Gell-Mann matrices. In three different books, I find its non-zero values to be $$f^{123} = 1, \ f^{147} = - f^{156} = f^{246} = f^{257} = f^{345} = - f^{367} = \frac{1}{2}, \ f^{458} = f^{678} = \frac{\sqrt{3}}{2}.$$ The 1 and $\frac{1}{2}$ values are fine for me, but I can not work out the $f^{458} = f^{678} = \frac{\sqrt{3}}{2}$. Example for $f^{458}$ where I solve each handside separately:

\begin{align} [\lambda^4,\lambda^5] & = 2i f^{458} \lambda^8 \newline \lambda^4 \lambda^5 - \lambda^5\lambda^4 & = 2i \frac{\sqrt{3}}{2} \lambda^8 \newline \begin{pmatrix}0&0&1\newline0&0&0\newline1&0&0\end{pmatrix}\cdot\begin{pmatrix}0&0&-i\newline0&0&0\newline i&0&0\end{pmatrix}-\begin{pmatrix}0&0&-i\newline0&0&0\newline i&0&0\end{pmatrix}\cdot\begin{pmatrix}0&0&1\newline0&0&0\newline1&0&0\end{pmatrix} &= i\begin{pmatrix}1&0&0\newline0&1&0\newline0&0&-2\end{pmatrix} \newline \begin{pmatrix}i&0&0\newline0&0&0\newline0&0&-i\end{pmatrix}-\begin{pmatrix}-i&0&0\newline0&0&0\newline0&0&i\end{pmatrix} &\neq \begin{pmatrix}i&0&0\newline0&i&0\newline0&0&-2i\end{pmatrix} \end{align}

Same happens for $f^{678}$. I am wondering what am I doing wrong?

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  • $\begingroup$ The commutator you are evaluating has a $\lambda_3$ on the rhs as well, but you skipped it…. $\endgroup$ Commented Jul 6, 2023 at 11:42
  • $\begingroup$ You did not solve for each side correctly: do you understand the sum over c 's involved in your first equation? $\endgroup$ Commented Jul 6, 2023 at 11:51
  • $\begingroup$ Hmm...I saw some mentions of the sum, but am not sure how to apply it. I thought that $c$ is fixed to 8 in this case and I do not see where can I do the actual sum? $\endgroup$
    – gasar8
    Commented Jul 6, 2023 at 12:02
  • $\begingroup$ $f^{453}$ is nonzero. $\endgroup$
    – Ghoster
    Commented Jul 6, 2023 at 20:01

1 Answer 1

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In the commutation relation you are meant to sum over all c s (Einstein repeated indices summation convention), but since your fully antisymmetric structure constants are so sparse, you are left with $$ [\lambda_4,\lambda_5]= i\lambda_3 + i\sqrt{3}\lambda_8 = 2i \operatorname{diag}(1,0,-1) . $$ Can you now see the correct expression? How do you extract each structure constant form the trace-orthogonality of these G-M matrices?

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  • $\begingroup$ Great, that makes sense, I now understand the summation over the last index. Thank you very much! $\endgroup$
    – gasar8
    Commented Jul 7, 2023 at 7:44

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