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We know that in the spin-1/2 representation the anticommutation relation of the Pauli matrices is $\{\sigma_{a},\sigma_{b}\}=2\delta_{ab}I$. Does a similar relation hold for the spin-1 representation?

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Of course not, in general, as the anticommutator is in the universal enveloping algebra: it is not even in the Lie algebra augmented by the identity, as evident in the specific example below.

For the spin 1 representation of the algebra, $J^a_{~~bc}=-i\epsilon_{abc}$, consisting of hermitean, imaginary, antisymmetric 3×3 matrices, i.e. the adjoint representation, it is straightforward to compute all anticommutators explicitly, $$ \{J^a,J^b \}_{mk}= -\epsilon_{amn} \epsilon_{bnk} -\epsilon_{bmn}\epsilon_{ank}= 2\delta_{ab}\delta_{mk} -(\delta_{am}\delta_{bk}+\delta_{bm}\delta_{ak}). $$

You then see the r.h.s. are symmetric real matrices.

  • For $a\neq b$, they are traceless off-diagonal ones;

  • For $a=b$, they are diagonal, traceful ones, but with a 0 in the ab entry.

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