# Invariance of H^daggerH

1. Jul 19, 2013

### lonewolf219

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. Jul 19, 2013

### fzero

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.

3. Jul 19, 2013

### lonewolf219

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!!