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Ref. 1 does not seem to mention a symmetry-breaking $U(1)$, which must belong to the part of $SU(5)$ which is not in the standard model. In this answer, we will assume that OP is really asking about the weak hypercharge $U(1)$ gauge factor of the standard model. At the Lie algebra level, recall that the Lie algebra $su(n)$ consists of Hermitian traceless ...

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A flaw in Wilczek's argument has been pointed out in PRL 110, 118901 (2013). In short, the rotating soliton exhibited by Wilczek is not the correct ground state of the system. The true ground state can be shown to be stationary, irrespective of the flux. Impossibility of a quantum system with time-crystal-like ground-state has been further addressed in PRL ...

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A short answer to questions 2 and 3: In Mermin-Wagner's paper the short-range condition is stated as $\sum_{\bf R} {\bf R}^2 |J_{\bf R}|<+\infty$. For interactions with (or more precisely majorized by a) power law decay $|J_{\bf R}| \sim R^{-\alpha}$, this requires $\alpha > D+2$, where $D$ is the space dimensionality (i.e., $\alpha >4$ for $D=2$ ...

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Essentially, Higgs self-interactions are allowed because they don't violate any laws or symmetries (i.e. the $\phi^4$ term is gauge invariant, Lorentz invariant etc). Informally, a Lagrangian can (and possibly should) consist of any/all combinations of fields, derivatives of fields etc that respect the symmetries of the theory. In the Standard Model the ...

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An experimentalist's answer, Our observations tell us that baryon and lepton number are conserved, within the accuracies of our experiments and observations. This means we have chosen as a standard model SU(3)xSU(2)xU(1) because in the group structure of the possible representations of all the quantum numbers assigned to the particles and resonances we ...

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This is more of an extended comment, because I don't know the answer, but hopefully this will be useful. Any field with a non-zero spin must have to have a vacuum expectation value (VEV) of zero, because any other value would break Lorentz invariance. So spin 0 bosons like the Higgs are the only ones that can have non-zero VEVs. As for the form of the ...

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From the form of the Higgs potential (which is quartic, the famous Mexican hat) you can see that for $\Re \phi =0$ as well as for $\Im \phi=0$ (the real and imaginary parts of the Higgs field), it is sitting on the unstable top of the hat. Thus, a small perturbation would lead it away from $0$. Since the potential has a smaller value away from zero so does ...

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Note: In the case of gauge symmetry, the degeneracy of the vacuum under gauge transformations leads to topologically inequivalent vacua characterized by the winding number of the gauge fields, in which case the lagrangian in the path integral has a term which indeed depends on which (theta) vacuum you choose. However here we will consider a vacuum degeneracy ...

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You don't have to write the generating functional in the ground state. More generally $$\langle\mathcal{O}\rangle=\frac{\text{tr}(\rho \mathcal{O})}{\text{tr}\;\rho}$$ where the trace is over all states and $\rho$ is the density matrix. This can be written as a path integral in general for any $\rho$, so I don't see a particular formal difficulty in ...

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The consideration of ground states is an equilibrium assumption. Imagine that instead of developing a theory around a ground state we were doing it around any general state. Actually, for simplicity I'll only consider states with constant $\varphi$, but the same remarks will be valid with the appropriate modifications -- modulo some topological ...

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The reason behind the fact that we build our theories around perturbations of a ground state is simply that solving these equations exactly is not feasible. Hence, we try having perturbative solutions that are approximations based on supposing that the interactions are small enough that they don't deform too much the solutions to the case when there are no ...

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