# A Trace of a product of Dirac Matrices in a Fermion loop

#### RicardoMP

I'm working out the quark loop diagram and I've drawn it as follows: where the greek letters are the Lorentz and Dirac indices for the gluon and quark respectively and the other letters are color indices.
For this diagram I've written:

$$i\Pi^{\mu\nu}=\int_{}^{}\frac{d^4k}{(2\pi)^4}.[ig(t_a)_{ij}\gamma^\mu].[\frac{i[(\not\! k-\frac{\not{q}}{2})+m_1]}{(k-\frac{q}{2})^2-m_1+i\epsilon}\delta_{il}].[ig(t_b)_{kl}\gamma^\nu].[\frac{i[(\not\! k+\frac{\not{q}}{2})+m_2]}{(k+\frac{q}{2})^2-m_2+i\epsilon}\delta_{kj}]$$

In Peskin & Schroeder's Introduction to QFT it is said: "a closed fermion loop always gives a factor of -1 and the trace of a product of Dirac matrices". In short, I can write the above expression as:

$$i\Pi^{\mu\nu}=g^2T_F\delta{ab}\int_{}^{}\frac{d^4k}{(2\pi)^4}\frac{tr[\gamma^\mu.((\not\! k-\frac{\not{q}}{2})+m_1).\gamma^\nu.((\not\! k+\frac{\not{q}}{2})+m_2)]}{((k-\frac{q}{2})^2-m_1+i\epsilon).((k+\frac{q}{2})^2-m_2+i\epsilon)}$$

where $T_F$ is the Dynkin index in the fundamental representation I get after contracting the generators and deltas.
What I'm having a hard time with is on how to explicitly write the Dirac indices in each gamma matrix and slashed momenta in order to get something of the sort, e.g, $(matrix)_{ii}=tr(matrix)$.

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#### Orodruin

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What is tr(AB) in terms of the components of A and B?

Also, I am unsure why you would want to write this out explicitly. It is far (far!) easier to use gamma matrix and trace relations to compute the trace.

#### vanhees71

Gold Member
There are some useful formulae for such traces in the appendix of Peskin&Schroeder. You can derive them by using the anticommutation relations of the Dirac matrices.

#### RicardoMP

What is tr(AB) in terms of the components of A and B?

Also, I am unsure why you would want to write this out explicitly. It is far (far!) easier to use gamma matrix and trace relations to compute the trace.
I might have not been clear, I'm sorry. I do want to use the trace identities in order to do the calculations. I just wanted to write out the indices explicitly so I show clearly that the numerator is indeed a trace.

#### vanhees71

$$(AB)_{ij}=A_{ik} B_{kj}$$
Then the trace is the sum of the diagonal elements. In the Ricci calculus you just have to set $i=j$ in the above formula (implying summation over $i$ then of course):
$$\mathrm{Tr}(AB)=A_{ik} B_{ki}.$$