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Consider a general $su(3)$ matrix in usual Gell-Mann basis $\lambda = l_a \lambda_a$, with $a=1,...,8$. Are there any resources studying the eigenvalues /eigenvectors of $$\lambda^2 = \frac{2}{3} l^2 I + d^{abc}l_a l_b \lambda_c \quad ?$$ For one, I have computed the eigenvalues (which should be real, given $\lambda$ is hermitian) with the help of Mathematica and certain $su(3)$ structure constant identities to be $$ \lambda_1 = \frac{2}{3} l^2 - \frac{1}{3}R^{1/3}- \frac{1}{3}l^4R^{-1/3}\\ \lambda_2 = \frac{2}{3} l^2 + \frac{1}{6}(1-i \sqrt{3})R^{1/3}+ \frac{1}{6\sqrt{3}}(3 i + \sqrt{3})l^4R^{-1/3}\\ \lambda_3 = \frac{2}{3} l^2 + \frac{1}{6}(1+i \sqrt{3})R^{1/3}+ \frac{1}{6\sqrt{3}}(-3 i + \sqrt{3})l^4 R^{-1/3}, $$ where $$R=-6(d^{abc}l_al_bl_c)^2 +l^6+2|d^{abc}l_al_bl_c|\sqrt{3(d^{abc}l_al_bl_c)^2-l^6}$$ But I'm not entirely sure I can trust these or if there are simpler, less cryptic ways of writing them. I have learned from eq. 3.14 from doi:10.1007/BF01654302 that the argument of the square root is always negative but Mathematica complains about taking cubed roots when I try to give it numerical values respecting that (which I imagine comes from secret assumptions it made while solving the cubic characteristic equation).

finding eigenvectors seems rather difficult..

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  • $\begingroup$ Interesting. But what is the question? $\endgroup$ – lcv May 23 at 1:24
  • $\begingroup$ Principally, I'm looking for resources on this material. If anyone has diagonalised $\lambda^2$ in the literature... $\endgroup$ – Rudyard May 23 at 8:54
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    $\begingroup$ I would state it more clearly in the text what you're looking for. I think you would get more response. $\lambda$ is the most general traceless $3\times3$ matrix. It's eigevalues, in general, will be roots of a cubic (summing to zero). The same holds for $\lambda^2$, except that it's not traceless. You seek an expression in terms of $\ell$s? You can easily do a numerical check yourself. $\endgroup$ – lcv May 23 at 18:09

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