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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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  • In 2, "symmetry left" might be code for "symmetry realized linearly". The SSB generators still correspond to conserved currents, but their symmetries are realized nonlinearly : they are hidden... It's important to say exactly what you mean and spell out the factual statements. Start by reviewing the global sigma model. – Cosmas Zachos Sep 26 '21 at 18:33
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    It is written in terms that I usually hear/read on notes and lectures. So maybe the language choices have make me confused... – Pipe Sep 26 '21 at 18:43
  • Linked. – Cosmas Zachos Sep 26 '21 at 19:11
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    So SSB allows to pass from linearly realized $G_{SM}$ to linearly realized $G_{SM}^{SSB}$, so that $\frac{G_{SM}}{G_{SM}^{SSB}}$ is realized nonlinearly (hidden)? – Pipe Sep 26 '21 at 19:36
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    Yes! And the Goldstone bosons, ultimately eaten, are the projective coordinates for this coset space. – Cosmas Zachos Sep 26 '21 at 20:52

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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    I am very confused by the second answer. Because following https://physics.stackexchange.com/questions/190416/what-role-does-spontaneously-symmetry-breaking-played-in-the-higgs-mechanism and https://physics.stackexchange.com/questions/105375/understanding-elitzurs-theorem-from-polyakovs-simple-argument/105397#105397 it seems that no gauge symmetry is broken. – Pipe Sep 26 '21 at 18:00
  • Ok, I have looked at the two stack exchange links that you posted and I think that I see where the problem is. Basically the question is whether you like to think of spontaneous symmetry breaking in unitary gauge or without any gauge fixing. If you fix unitary gauge, then you arrive at a gauge theory with smaller gauge group. If you choose not to do so, then of course the gauge symmetry of the Lagrangian cannot magically change just because you chose to rewrite the Lagrangian using some new fields. – SPHerical Sep 26 '21 at 19:35
  • Thanks. So, without choosing any gauge, how $G_{SM}^{SSB}$ is interpreted? As the remaining global (linearly realized) symmetry? – Pipe Sep 26 '21 at 19:42
  • I am not certain whether this is what you ask for, but the situation is the following: If you have a theory with gauge group $G$, then this means that all your fields are maps from space-time into some space that $G$ acts on (usually this action will indeed be a linear representation). In theories with spontaneous symmetry breaking you can rewrite the Lagrangian with new fields, but these still take values in spaces that $G$ acts on. So it is still a theory with "local $G$ gauge symmetry". By fixing unitary gauge you reformulate the theory in terms of something with a "smaller" gauge group. – SPHerical Sep 26 '21 at 19:53