1
$\begingroup$

In Peskin&Schroeder 16.1, we calculate the tree diagram level scattering ampitude of $q\overline{q}$-$g\overline{g}$. There are three terms: $$i\mathcal{M}^{\mu\nu}_{1,2}=(ig)^2\overline{v}(p_+)\left\{\gamma^\mu t^a\frac{i}{p\!\!\!/-k\!\!\!/_2-m}\gamma^\nu t^b+\gamma^\nu t^a\frac{i}{k\!\!\!/_2-p\!\!\!/_+-m}\gamma^\mu t^b\right\}u(p)\epsilon_\mu^*(k_1)\epsilon^*_\nu(k_2)$$ and $$i\mathcal{M}^{\mu\nu}_{3}=ig^2\overline{v}(p_+)\gamma_\rho t^c u(p)\frac{-i}{k_3^2}\epsilon_\mu^*(k_1)\epsilon^*_\nu(k_2)f^{abc}\left[g^{\mu\nu}(k_2-k_1)^\rho+g^{\nu\rho}(k_2-k_1)^\mu +g^{\rho\mu}(k_2-k_1)^\nu\right]$$

and we have two unphysical polarization named timelike forward and backward polarization $$\epsilon_+(k)=\frac{1}{\sqrt{2}|\textbf{k}|}(k^0,\textbf{k}),\epsilon_-(k)=\frac{1}{\sqrt{2}|\textbf{k}|}(k^0,-{\bf k})$$ and two physical polarization $\epsilon_T(k)$ satisfy $k^\mu\epsilon_{T,\mu}=0$

we could easily prove that $M_{\mu\nu}\epsilon^\mu_+(k_1)\epsilon_T^\nu(k_2)=0$ following the proof in the textbook. Besides, terms like $M_{\mu\nu}\epsilon^\mu_-(k_1)\epsilon_T^\nu(k_2)$ should also be zero (page 516, The same identity holds if $\epsilon_+$ is replaced by $\epsilon_-$). However, I don't know how to calculate it, since unlike the case in $\epsilon^\mu_+(k) \propto k^\mu$ we could act $k\!\!\!/$ on $\overline{v}(p),u(p)$ and use relation of the spinor to cancel the dominator and simplify the expression.

$\endgroup$
2
  • $\begingroup$ Your gluon indices don't make sense. $\endgroup$
    – Buzz
    May 2, 2021 at 4:40
  • 1
    $\begingroup$ As pointed out by @Buzz, your indices don't make sense: $\mu$ and $\nu$ are contracted on he RHS and not on theLHS. As for a possible answer, check my qft notes on hep notes.com, chapter 14. $\endgroup$ May 2, 2021 at 19:29

0

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy