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I am trying to calculate the tree level scattering cross section for Moeller scattering using QED. I have got some traces which I cannot evaluate.

e.g.$$Tr(\gamma^\nu \not{p_2}\gamma^\mu \gamma_\nu \not{p_1\gamma_{\mu}}).$$

I am able to evaluate traces which don't contain any slashed momentum in between. I know I have to use the trace formulas of Dirac matrices to manipulate the given trace but I don't know exactly how I can apply the trace formulas for slashed momentum in between. No book does an example explicit calculation. How to proceed?

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  • $\begingroup$ Related: physics.stackexchange.com/q/255243 $\endgroup$
    – Manas Dogra
    Commented Oct 8, 2020 at 14:38
  • $\begingroup$ Recall that $\not{p} = p_{\mu} \gamma^{\mu}$ (with the implicit sum over $\mu$ of course). The factor $p_{\mu}$ can be pulled out of the trace, since it is just a number, and you can use the usual trace formulas (you will also need formulas for the contraction of gamma matrices). $\endgroup$
    – jkb1603
    Commented Oct 8, 2020 at 14:43
  • $\begingroup$ Isn't $p_{\mu}$ a covariant 4-vector because it has one index $\mu$ attached to it? $\endgroup$
    – Manas Dogra
    Commented Oct 8, 2020 at 14:45
  • $\begingroup$ Yes. It is a four vector, but its components $p_0,p_1,p_2,p_3$ are numbers in the internal (four-dimensional) vector space in which the Dirac matrices act. The trace is to be taken with respect to this vector space. $\endgroup$
    – jkb1603
    Commented Oct 8, 2020 at 14:47
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    $\begingroup$ Yes, $\gamma^{\mu} p^{\nu} = p^{\nu} \gamma^{\mu}$ for any $\mu, \nu \in \{0,1,2,3\}$, since $\gamma^{\mu}$ is a matrix and $p^{\nu}$ (I mean the components here, not the vector) is a number (therefore $p^{\nu}$ can be pulled out of the trace). $\endgroup$
    – jkb1603
    Commented Oct 8, 2020 at 15:08

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It can be simplified as Tr$(\gamma^\nu \not{p_2}\gamma^\mu \gamma_\nu \not{p_1\gamma_{\mu}})$=Tr$(\gamma^\nu \gamma^{\rho}p_{2\rho}\gamma^\mu \gamma_\nu \gamma^{\sigma}p_{1\sigma}\gamma_{\mu})$.

Since $p$ doesn't carry spinor indices.

$p_{2\rho}p_{1\sigma}$Tr$(\gamma^\nu \gamma^{\rho}\gamma^\mu \gamma_\nu \gamma^{\sigma}\gamma_{\mu})$ using ${\gamma^\nu \gamma^{\rho}\gamma^\mu \gamma_\nu}=4g^{\rho\mu}$

Tr$(\gamma^\nu \gamma^{\rho}\gamma^\mu \gamma_\nu \gamma^{\sigma}\gamma_{\mu})$= Tr$(4g^{\rho\mu} \gamma^{\sigma}\gamma_{\mu})$=$2$ Tr$( \{ \gamma^{\rho},\gamma^{\mu}\} \gamma^{\sigma}\gamma_{\mu})$

$\implies2$ Tr$( \gamma^{\rho}\gamma^{\mu}\gamma^{\sigma}\gamma_{\mu})$ $+$ $2$ Tr($\gamma^{\mu}\gamma^{\rho} \gamma^{\sigma}\gamma_{\mu})$=$2$Tr$(\gamma^{\rho}\times-2\gamma^{\sigma})$+$2$Tr$(4g^{\rho\sigma})$ where I have used $\gamma^{\mu}\gamma^{\sigma}\gamma_{\mu}=-2\gamma^{\sigma}$

Since $g^{\rho\sigma}$ doesn't have any spinor indicies we can take it out of second trace.

$\implies-4$Tr$(\gamma^\rho\gamma^\sigma)+8g^{\rho\sigma}$Tr$(\delta_{\alpha\beta})=-16g^{\rho\sigma}+32g^{\rho\sigma}$

Orignal expression then becomes $p_{2\rho}p_{1\sigma}\times16g^{\rho\sigma}$ =$16p_1\cdot p_2$

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  • $\begingroup$ In 2nd last step, $Tr(\delta_{\alpha\beta})=4$ right? $\endgroup$
    – Manas Dogra
    Commented Oct 8, 2020 at 15:32
  • $\begingroup$ @ManasDogra yes $\endgroup$
    – aitfel
    Commented Oct 8, 2020 at 15:33

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