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The coupling of the Higgs boson to the electroweak gauge bosons in the Standard model is given by

$$\mathcal{L}_{\text{H-g}} = - \left( \frac{H}{v} + \frac{H^{2}}{2v^{2}} \right) \left(2M_{W}^{2}W_{\mu}^{+}W^{-\mu} + M_{Z}^{2}Z_{\mu}Z^{\mu} \right).$$

However, in Cliff Burgess' textbook 'The Standard Model: A Primer,' the author suppresses the contribution $H^{2}/2v^{2}$ in equation (4.55) on page 146 when he discusses the decay of the Higgs boson to electroweak gauge bosons.

Is there a reason why the contribution $H^{2}/2v^{2}$ is suppressed?

nightmarish
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    ...is $H/v$ "small" at the scale of the processes considered? There's not really enough information here to say something concrete, imo. – ACuriousMind Feb 18 '17 at 22:34

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The reason is very simple -- term of the form $H A_\mu A^\mu$ corresponds to the vertex with one Higgs and two bosons -- directly the Feynman diagram giving the decay. The term $H^2 A_\mu A^\mu$ is the vertex with two Higgs bosons and two gauge bosons. At tree level this is some scattering process (say, $2H\to 2Z$), but not decay. It contributes to decay only at higher order loop diagrams, thus it is significantly suppressed.

$HZ_\mu Z^\mu$: Higgs decay, $H^2 Z_\mu Z^\mu$: 2H scattering

Fedxa
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