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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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