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Cubic Term in Higgs Boson

  1. Aug 27, 2012 #1
    Why does the standard model have a higgs boson quadratic and a cuartic term but it does not have a cubic term? is there any problem if it happens to have a cubic term?

  2. jcsd
  3. Aug 27, 2012 #2
    The higgs field is a complex field. How do you propose to include a cubic term and leave the Lagrangian u(1) invariant?
  4. Aug 28, 2012 #3
    Ok, thanks!
  5. Aug 29, 2012 #4
  6. Aug 29, 2012 #5
    But that's after we had shifted the field by its VEV, if I'm not mistaken. That occurs even for:
    \mathcal{L}_{4} = \frac{\lambda}{2} \, \left( \vert \varphi \vert^2 - v^2 \right)^2
    if you make the subst:
    \varphi = (v + \rho) \, e^{i \, \chi}
  7. Aug 29, 2012 #6
    Oh, yeah fair enough.
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