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Ok for real photons there is the formula when summing over the polarizations:

$$ \sum_{\lambda=\pm}\epsilon^{*\mu}_\lambda\epsilon^\nu_\lambda = -\eta^{\mu\nu}$$

But if I have a matrix element of the form: $$\epsilon^{*\mu}M_\mu\qquad \epsilon^\mu\bar{M}_\mu$$

So when I take the absoulte squared of that I have: $$|M|^2 = \epsilon^{*\mu}M_\mu \epsilon^\nu\bar{M}_\nu= -M_\mu\eta^{\mu\nu}M_\nu = -M \bar{M}$$

But know the absolute squared is negative. So I know I have a heavy mistake somewhere here. Can someone help me understand? Thanks a lot.

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  • $\begingroup$ You are using the $(1,-1,-1,-1)$ signature right? $\endgroup$ – MannyC Apr 19 at 13:08
  • $\begingroup$ Yes. Sorry I didn't mention. $\endgroup$ – higgshunter Apr 19 at 13:16
  • $\begingroup$ Why must $M_\mu M^{*\mu}>0$? $\endgroup$ – WAH Apr 19 at 14:35
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    $\begingroup$ Well I thought that now we have $|M|^2 = -M_\mu M^\mu$ and the total amplitude should be real of course and positive? Or am I missing something? $\endgroup$ – higgshunter Apr 19 at 14:41
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$$-M_\mu\eta^{\mu\nu}M_\nu=-M_0^2+M_1^2+M_2^2+M_3^2$$ Since due to Ward identity $M_0=M_3$ (choosing $x^3$ as the direction of motion), the above result is positive.

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  • $\begingroup$ Seems logical. Since it should be positive and also in my actual calculations if I evaluate the resulting Trace I get something positive. I was just looking for a general explanation to see why this should be the case. $\endgroup$ – higgshunter Apr 20 at 9:02

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