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Explanation of quadratic divergence in higgs mechanism

  1. Feb 23, 2013 #1
    Hi can anyone explain what a quadratic divergence is? and if so how it effects the mass of the scalar field i.e why [itex] m^2 = m^2_{0} + \delta m^2[/itex], why are these things squared?

    Also how would this divergence affect the standard model as a natural concept, because from reading books it would seem that this introduces an intolerable amount of fine tumning making the SM as a standalone seem ridiculous.
  2. jcsd
  3. Feb 23, 2013 #2


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    How much do you know about quantum field theory?

    Precision measurements and the newly found particle indicate a Higgs mass ##m \approx 125 GeV##, while ##\delta m^2## is of the order of 1019 GeV, and requires ##m_0## to be fine-tuned to get such a small difference.
  4. Feb 23, 2013 #3
    I no a bit but not much, never covered this before though.

    Since the fine tuning is so small then would it imply that the SM is invalid or does it imply that it is valid providing there is a higher energy e.g. super-symmetric model?
  5. Feb 24, 2013 #4
    fine tuning is provided by weinberg mixing angle.it does not imply any invalidity of SM.however it is not a part of SM to derive the angle.beyond the standard model things like supersymmetry and GUT can yield this theoretically.A quadratic divergence is mainly associated with an integral which diverges quadratically in high energy limit.
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