# QCD Color Structure relation

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i want to proof the following relation :

$$t^a t^b \otimes t^a t^b = \frac{2}{N_C} \delta^{ab} \mathbb{1} \otimes \mathbb{1} - \frac{1}{N_C} t^a \otimes t^a$$

Right now I calculated the second term, but still have problems to get to the Casimir operator $C_F$ of the first term.

Knowing $t^a = \frac{\lambda^a}{2}$ and $\lambda^a\lambda^b = \frac{2}{N_C}\delta^{ab} + d^{abc}\lambda^c + i f^{abc}\lambda^c$ yields

$$t^a t^b \otimes t^a t^b = \frac{1}{16}\lambda^a \lambda^b \otimes \lambda^a \lambda^b$$ $$= \frac{1}{16} \left[ \frac{2}{N_C}\delta^{ab} + d^{abc}\lambda^c + i f^{abc}\lambda^c \right] \otimes \left[ \frac{2}{N_C}\delta^{ab} + d^{abc}\lambda^c + i f^{abc}\lambda^c \right]$$ I am hopefully right, that $f^{aab} = d^{aab} = 0$, thus $$= \frac{1}{16} \left[ \frac{4}{N_C^2} \delta^{ab} \mathbb{1} \otimes \mathbb{1} - \frac{4}{N} \lambda^c \otimes \lambda^c + i d^{abc}f^{abc} \lambda^c \otimes \lambda^c + i f^{abc} d^{abc} \lambda^c \otimes \lambda^c \right]$$ where I used $f^{abc}f^{abc} = N$ and $d^{abc}d^{abc} = \left( N - \frac{4}{N} \right)$.

The Casimir operator is defined as $$\frac{N_C^2 -1}{2N_C} \equiv C_F$$

And as I said I do not know how to get to the result of the first equation by plugging in the Casimir operator. I have a strong guess that if I would know how to deal with the terms $i d^{abc}f^{abc} \lambda^c \otimes \lambda^c$ the result will be obvious, but until know I appreciate every help.

This post imported from StackExchange Physics at 2015-07-16 09:30 (UTC), posted by SE-user Knowledge
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