Is H^DaggerH invariant under rotations and translations?

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H^DaggerH is invariant under SU(2) X U(1), but this invariance does not imply it is invariant under rotations and translations, which are part of the Poincare group. The Higgs field, represented as H(x), is a scalar field and is inherently invariant under these transformations. Consequently, H^DaggerH also maintains this invariance. The transformation properties under SU(2) X U(1) are independent of those under Poincare symmetries, which depend on the field type. Understanding these distinctions is crucial for discussing the properties of the Higgs field.
lonewolf219
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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
 
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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.
 
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!
 
Time reversal invariant Hamiltonians must satisfy ##[H,\Theta]=0## where ##\Theta## is time reversal operator. However, in some texts (for example see Many-body Quantum Theory in Condensed Matter Physics an introduction, HENRIK BRUUS and KARSTEN FLENSBERG, Corrected version: 14 January 2016, section 7.1.4) the time reversal invariant condition is introduced as ##H=H^*##. How these two conditions are identical?

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