A simple calculation in Peskin's and Schroeder's QFT book on page 608 chapter 18

I am trying to calculate the term: $$(t^a)_{ij} (t^a)_{kl}$$

In the book it's written that it equals to $$A\delta_{il}\delta_{kj}+B\delta_{ij}\delta_{kl}$$ and from using equation (18.40) $$tr[t^a](t^a)_{kl}=0$$ $$(t^a t^a)_{il}=4/3 \delta_{il}$$ we get: $$(t^a)_{ij}(t^a)_{kl}=1/2(\delta_{il}\delta_{kj}-1/3 \delta_{ij}\delta_{kl})$$

Now, I tried to prove the last equality, but I am not sure how to continue, we have: For $$k=l, i=j , i\ne l, k\ne j$$: $$B= (t^a)_{jj}(t^a)_{ll}$$. and for $$i=l, k=j , i\ne j , k\ne l$$ $$A= (t^a)_{lj}(t^a)_{jl}$$.

And as far as I can tell there's no summation convention here, am I right?

So how to get eq. (18.41) in the book?

• Which one is 18.41? – user207455 Jul 7 at 11:52
• both $$tr[t^a](t^a)_{kl}=0$$ $$(t^a t^a)_{il}=4/3 \delta_{il}$$ are eq. (18.40) in the book. – MathematicalPhysicist Jul 7 at 11:59

There is summation convention used in those formulae. In particular they are all summed over all values of $$a$$.
Using summation convention on both $$a$$ and other repeating indices: \begin{align} tr[t^a](t^a)_{kl} & = (t^a)_{ii} (t^a)_{kl} = 0, \\ (t^a)_{ii} (t^a)_{kl} & = A \delta_{il} \delta_{ki} + B \delta_{ii} \delta_{kl} = A \delta_{kl} + 3 B \delta_{kl} = 0. \end{align} Therefore: \begin{align} A = -3 B. \end{align}
For the second constraint: \begin{align} (t^a)_{ij} (t^a)_{jl} = A \delta_{il} \delta_{jj} + B \delta_{ij} \delta_{jl} = 3 A \delta_{il} + B \delta_{il} = \frac{4}{3} \delta_{il}, \end{align} or: \begin{align} 3A + B & = \frac{4}{3}. \end{align}
Solving the system of equations we get $$A = \frac{1}{2}$$, $$B = -\frac{1}{6}$$.