# QCD Color Structure relation

+ 2 like - 0 dislike
138 views

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
asked Jul 15, 2015
retagged Jul 16, 2015
There's some inconsistency in the notation going on here - do you sum over repeated indices or not? If the answer is "sometimes", you should write explicit sums where you do so.

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

 Please use answers only to (at least partly) answer questions. To comment, discuss, or ask for clarification, leave a comment instead. To mask links under text, please type your text, highlight it, and click the "link" button. You can then enter your link URL. Please consult the FAQ for as to how to format your post. This is the answer box; if you want to write a comment instead, please use the 'add comment' button. Live preview (may slow down editor)   Preview Your name to display (optional): Email me at this address if my answer is selected or commented on: Privacy: Your email address will only be used for sending these notifications. Anti-spam verification: If you are a human please identify the position of the character covered by the symbol $\varnothing$ in the following word:p$\hbar$ysicsOve$\varnothing$flowThen drag the red bullet below over the corresponding character of our banner. When you drop it there, the bullet changes to green (on slow internet connections after a few seconds). To avoid this verification in future, please log in or register.