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Invariance of H^daggerH

  1. Jul 19, 2013 #1
    Hi,

    Since H^DaggerH is invariant under SU(2) X U(1), does this mean that H^DaggerH is invariant under rotations and translations?

    Thanks
     
  2. jcsd
  3. Jul 19, 2013 #2

    fzero

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    From the limited amount of context you give, I am guessing that ##H## is supposed to be the Higgs field? If so, then ##H(x)## is a scalar field, so by definition, ##H(x)## is already invariant under rotations and translations. As a consequence, ##H^\dagger H## is also invariant under these.

    Note that ##SU(2)\times U(1)## is a gauge or "internal" symmetry group. It therefore has nothing at all to do with rotations and translations (collectively called the Poincare group). So the transformation properties under ##SU(2)\times U(1)## are completely independent from the transformation properties under the Poincare symmetries. The latter are determined by the type of field we're dealing with: scalar, spin 1/2, vector, etc.
     
  4. Jul 19, 2013 #3
    Thanks, fzero. I'm sorry, yes, you are correct that my question is about the Higgs Field. I think I will use a different notation or state that part in any future posts... I am very happy you answered my poorly written question! This is very interesting information you have provided that I was not aware of!!
     
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