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A Deriving matrix element from Lagrangian

  1. Mar 15, 2017 #1
    Consider the following tree-level Feynman diagrams for the ##W^{+}W^{-} \to W^{+}W^{-}## scattering process.

    94a15c798a.png

    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.
     
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  3. Mar 17, 2017 #2

    vanhees71

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    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.
     
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