# A Deriving matrix element from Lagrangian

1. Mar 15, 2017

### spaghetti3451

Consider the following tree-level Feynman diagrams for the $W^{+}W^{-} \to W^{+}W^{-}$ scattering process.

The matrix element for this diagram can be read off from the associated quartic term $\mathcal{L}_{WWWW}$ in the electroweak boson self-interactions, where

$\mathcal{L}_{WWWW} = -\frac{1}{2}g^{2}_{2} \left[ (W_{\mu}^{+}W^{-\mu})^{2} - (W_{\mu}^{+}W^{+\mu})(W_{\nu}^{-}W^{-\nu}) \right]$

$= -\frac{1}{2}g^{2}_{2} \left(\eta_{\mu\nu}\eta_{\rho\sigma}-\eta_{\mu\rho}\eta_{\nu\sigma}\right)W^{+\mu}W^{-\nu}W^{+\rho}W^{-\sigma}.$

Why does this mean that the matrix element is

$\mathcal{M} = -g_{2}^{2}(2\eta_{\mu\sigma}\eta_{\nu\rho}-\eta_{\mu\nu}\eta_{\rho\sigma}-\eta_{\mu\rho}\eta_{\nu\sigma})\epsilon^{\mu}(p_{1})\epsilon^{\nu}(p_{2})\epsilon^{\rho}(k_{1})\epsilon^{\sigma}(k_{2})?$

More specifically, I am not able to derive the factor of $(2\eta_{\mu\sigma}\eta_{\nu\rho}-\eta_{\mu\nu}\eta_{\rho\sigma}-\eta_{\mu\rho}\eta_{\nu\sigma})$ in the matrix element.

2. Mar 17, 2017

### vanhees71

You should symmetrize the coefficient in the Lagrangian. Here it's to be symmetrized in $\mu \nu$ and $\rho \sigma$, i.e., you have
$$\mathcal{L}_{WWWW}= -\frac{1}{4}g^{2}_{2} \left(\eta_{\mu\nu}\eta_{\rho\sigma}+\eta_{\rho \nu} \eta_{\mu \sigma}-2\eta_{\mu\rho}\eta_{\nu\sigma}\right)W^{+\mu}W^{-\nu}W^{+\rho}W^{-\sigma}.$$
Now the coefficient in front of the fields leads directly to the Feynman rule.