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I know this question has been asked other times, but I am looking for a confirmation of the following.

  1. When we say that the gauge group of the standard model is $G_{SM} = SU(3)_{c} \times SU(2)_{L} \times U(1)_{Y}$ we mean that the standard model is a $\textbf{gauge theory}$ with affine connections (gauge bosons) related to this group $\textbf{and}$ it has a $\textbf{global symmetry}$ $G_{SM}$?
  2. If so, when we say that we have spontaneous symmetry breaking, we intend that the $\textbf{global}$ symmetry is broken, but the gauge structure remain the same? For example, when it is written that after SSB the group is $G_{SM}^{SSB} = SU(2) \times U(1)_{em}$ we mean that $G_{SM}^{SSB}$ is the only left global symmetry (and the underlying gauge structure remain the same)?
  3. In the end, is the Higgs mechanism the same as the Goldstone mechanism with the difference that, because we have gauge freedom, we can reabsorb the massless Goldstone bosons "into" the gauge bosons, providing them with a mass?
Qmechanic
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Pipe
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1 Answers1

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  1. Whenever a theory has a local gauge symmetry, it automatically also has a global symmetry. After all you can just look at the constant gauge transformations. So the answer to this part is yes, but the last comment (i.e. "and it has global symmetry $G_{SM}$") is redundant.

  2. In theories with spontaneous symmetry breaking, we start with a Lagrangian that describes physics at a high energy scale and has a certain gauge symmetry and then discover that at lower energy scales the resulting effective field theory does no longer posses this symmetry. The low energy field theory really has a smaller gauge group (after fixing unitary gauge) so the answer is No.

  3. What do you mean by "Goldstone mechanism"? There is the Goldstone theorem which basically says that massless Goldstone bosons arise when symmetries are spontaneously broken and the Higgs mechanism in which these Goldstone bosons are "eaten up" to give mass to the gauge bosons.

SPHerical
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