Sum over real photon polarizations. The minus sign

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.

• You are using the $(1,-1,-1,-1)$ signature right? – MannyC Apr 19 at 13:08
• Yes. Sorry I didn't mention. – higgshunter Apr 19 at 13:16
• Why must $M_\mu M^{*\mu}>0$? – WAH Apr 19 at 14:35
• 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? – higgshunter Apr 19 at 14:41

$$-M_\mu\eta^{\mu\nu}M_\nu=-M_0^2+M_1^2+M_2^2+M_3^2$$ Since due to Ward identity $$M_0=M_3$$ (choosing $$x^3$$ as the direction of motion), the above result is positive.