2
$\begingroup$

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?

$\endgroup$
  • 1
    $\begingroup$ ...is $H/v$ "small" at the scale of the processes considered? There's not really enough information here to say something concrete, imo. $\endgroup$ – ACuriousMind Feb 18 '17 at 22:34
2
$\begingroup$

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

$\endgroup$

Your Answer

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

Not the answer you're looking for? Browse other questions tagged or ask your own question.