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Consider $d$-dimensional gamma matrix structures. I have an expression like $$ \sum_{h_2=\pm}\text{Tr}(\not{\xi}_2\not{p}_3\bar{\not{\xi}}_2\not{p}_1), $$ where $\not p=p^\mu \eta_{\mu\nu}\gamma^\nu$ are the momenta contracted with the gamma matrices and $\not \xi=\xi^\mu \eta_{\mu\nu}\gamma^\nu$ are the helicity vectors contracted with the gamma matrices. $h_2$ is the helicity corresponding to $\xi_2$. $\bar{\not\xi}_i$ is the same helicity vector as $\not\xi_i$, but with opposite helicity.

We have $p_i^\mu\eta_{\mu\nu}\xi^\nu_i=0$.

I know things like the anti-commutation relation $\{\gamma^\mu,\gamma^\nu\}=-2\eta^{\mu\nu}$ or the completeness relation $$ \sum_{h_i=\pm}\not{\xi}_i\bar{\not{\xi}}_i=2\cdot\text{Id}, $$ but I don't know how to handle the $\xi$'s with another expression between them. The final expression should not be depending on the $\xi$'s anymore. The process I'm looking at is a tensor structure involving fermion and gluon scattering. The helicity vectors belong to gluons.

I would be very happy if anybody could help me with the evaluation of the first expression.

Edit: After I have mentioned it in the comments, the term I have a problem with in particular is $$ \text{Tr}(\not{p}_3\not{\xi}_3\not{\xi}_4\not{p}_4\not{p}_1\not{p}_3\bar{\not{\xi}}_3\bar{\not{p}}_4\not{p}_4\not{p}_1). $$ As far as I know, the solution of this expression depends on the dimension $d$. Hence, I do not think the usual commutation relations lead to a result. Currently, I am spending my time figuering out how FeynCalc, mentioned by Luc J. Bourhis, works. But for the possibility of somebody providing more ideas, I wanted to mention this in the original question as well. What I have been told recently, is that $$ \sum_{h_i=\pm}\text{Tr}(\dots\not\xi_i\not p_j\bar{\not\xi_i}\dots)=f(d)\cdot\sum_{h_i=\pm}\text{Tr}(\dots\not p_j\dots) \text{ for }i\neq j $$ with $f(d)$ being some rational function of the dimension $d$. I have some guesses about $f$, but I will not write them here, because I have many dobts concerning them.

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  • $\begingroup$ I would just use $$\newcommand{\slashed}[1]{#1\!\!\!/}\mathrm{Tr}(\slashed{a}\slashed{b}\slashed{c}\slashed{d})=4(ab)(cd)-4(ac)(bd)+4(ad)(bc)$$ $\endgroup$ – user154997 Jul 7 '17 at 11:41
  • $\begingroup$ Thank you! I must admit that I have forgotten about that identity. Still, what I was hoping for, was an identity for the $\xi$'s. The actual expression I wanted to handle is $$\text{Tr}(\not{p}_3\not{\xi}_3\not{\xi}_4\not{p}_4\not{p}_1\not{p}_3\bar{\not{\xi}}_3\bar{\not{p}}_4\not{p}_4\not{p}_1)$$ which is much more complicated. I was hoping there would be an identity for the polarisation vectors such that this expression simplifies... I am looking at light like momenta with $\sum_{i=1}^4p_i=0$ btw. $\endgroup$ – Fred Jul 7 '17 at 15:07
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For the calculation in your question, you could just use:

$$\newcommand{\slashed}[1]{#1\!\!\!/}\mathrm{Tr}(\slashed{a}\slashed{b}\slashed{c}\slashed{d})=4(ab)(cd)-4(ac)(bd)+4(ad)(bc)$$

For the longer one in the comments, you could do what every does these days: use a program! I am afraid I don't remember a specific result for such expressions with polarisation but I can show you how to proceed by hand. Note that all the formula I will use are valid in dimensional regularisation, with dimension $d=4-2\epsilon$ typically then.

So I will move $\not{\bar{\xi}}_3$ to the left.

$$\begin{gather}\text{Tr}(\not{p}_3\not{\xi}_3\not{\xi}_4\not{p}_4\not{p}_1\not{p}_3\bar{\not{\xi}}_3\not{p}_4\not{p}_4\not{p}_1) = \\ -\text{Tr}(\not{p}_3\not{\xi}_3\not{\xi}_4\not{p}_4\not{p}_1\bar{\not{\xi}}_3\not{p}_3\not{p}_4\not{p}_4\not{p}_1)\\ -2(\xi_3p_3)\text{Tr}(\not{p}_3\not{\xi}_3\not{\xi}_4\not{p}_4\not{p}_1\not{p}_4\not{p}_4\not{p}_1) \end{gather}$$

You can then repeat that. When $\not{\bar{\xi}}_3$ will pass through $\not{\bar{\xi}}_4$, you will get a factor $(\xi_3\xi_4)$, and then you will be left with

$$\sum_{h_3=\pm} \text{Tr}(\cdots \not{\xi}_3\bar{\not{\xi}}_3\cdots)$$

which you can simplify with the completeness relation.

You will still be left with traces of 8 gamma matrices of course. They can be evaluated with the recurrence trick:

$$\begin{align}\text{Tr}(\not{a}_1\cdots\not{a}_{2n})&= (a_1a_2)\text{Tr}(\not{a}_3\cdots\not{a}_{2n})\\ &-(a_1a_3)\text{Tr}(\not{a}_2\not{a}_4\cdots\not{a}_{2n})\\ &+\cdots\\ &+(a_1a_{2n})\text{Tr}(\not{a}_2\cdots\not{a}_{2n-1}) \end{align}$$

I think it would be clever to use one of the leftover $\xi$ for $a_1$, after a circular permutation of the trace.

As I mentioned above, I recommend using a program. Nowadays everybody seems to use the Mathematic package FeynCalc. I toyed a bit with it a few years ago and I found it pretty comprehensive if a bit short on the documentation but there is a significant community to ask questions to. Note that I used to rely on an ancient package called Tracer. I would not recommend it (not even sure it still runs on recent Mathematica). I have just noticed that FeynCalc has even a command Polarization but I haven't tried it!

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  • $\begingroup$ They are valid in $d=4-2\epsilon$ dimensional regularisation. $\endgroup$ – user154997 Jul 7 '17 at 17:02
  • $\begingroup$ Thank you very much for this answer. This is what I've been afraid to do, but I guess I won't really have a choice. So thank you very much! Unfortunately, I haven't yet found a good program to handle it in $d$ dimensions, but anyway, I will probably be able to handle it. I will wait one or two days to see whether somebody else has an even more beneficial answer, but I guess that's it then. $\endgroup$ – Fred Jul 8 '17 at 11:59
  • $\begingroup$ What CAS do you have access to? Mathematica? Maple? None? $\endgroup$ – user154997 Jul 8 '17 at 12:02
  • $\begingroup$ Mathematica, Maple, Matlab (all from university), maybe more. But I only have experience with Mathematica. $\endgroup$ – Fred Jul 8 '17 at 12:04
  • $\begingroup$ Great! I amended my answer with some recommendations then. $\endgroup$ – user154997 Jul 8 '17 at 14:43

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