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It is in fact a very simple matter if you use a different parametrization of the fields. Since we care about the Goldstone bosons only, just send $\lambda\rightarrow \infty$ so that the Higgslike state decouples. Moving to the following parametrization $$\phi_i(x)=U(x)\langle \phi_i\rangle \,,\qquad U(x)=e^{i \hat{T}^a \pi^a(x)}\,,\qquad ... 5 I understand the statement in the following way: Pions, which are pseudo-goldstone bosons of chiral symmetry breaking, are described by the introduction of a unitary matrix U(x), defined as$$U(x)=\text{exp}\left(2i\pi^a(x)T^af_\pi^{-1}\right),$$where \pi^a is the pion field, f_\pi is the pion decay constant and T^a are the generators of the ... 0 The classical situation with no symmetry breaking is the case of the, so-called, isostructural transitions. The word "isostructural" is misleading, since what is meant is "isosymetric". However, historically the term emerged. There is a number of examples of such transiotions. One is the alpha-alpha' transitions in the hydrogen-metal systems, another is ... 3 The correct electroweak gauge group is SU(2)_L \times U(1)_Y where Y denotes the weak hypercharge. After the Higgs field spontaneously breaks this exact symmetry, third generator of SU(2)_L (weak isospin) and weak hypercharge combine to give the remaining unbroken U(1)_{em}. Gauge bosons and fermions fall under different representations of this ... 3 Actually we have the following Lie algebra isomorphism$$u(1)\oplus su(2)\cong u(2),$$and there exists the following Lie group isomorphism$$[U(1)\times SU(2)]/\mathbb{Z}_2 ~\cong~ U(2). In other words, there is a two-to-one map between $U(1)\times SU(2)$ and $U(2)$. So in that sense the Glashow-Salam-Weinberg $U(1)\times SU(2)$ model already contains ...

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Nice question! The short answer is that the group is not $SU(2)\times U(1)$, it is $SU(2)_L \times U(1)_{em}$. In other words the two groups act on different standard model particles differently. For example the left handed neutrino does interact weakly and so transforms under the $SU(2)_L$, but is electrically neutral so it doesn't transform under the ...

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One very general statement is that if you take one of the external momenta to zero, keeping the other momenta fixed, the amplitude will vanish. This is simply a consequence of the non-linear (shift) symmetry acting on the Goldstone bosons. This symmetry implies that they can enter the effective Lagrangian only with derivative couplings. Since the derivative ...

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1） Gauge theory is a theory where we use more than one label to label the same quantum state. 2） Gauge “symmetry” is not a symmetry and can never be broken. This notion of gauge theory is quite unconventional, but true. When two different quantum states $|a\rangle$ and $|b\rangle$ (i.e. $\langle a|b\rangle=0$) have the same properties, we say that there ...

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Let me answer your first question: Phase transition do not necessarily imply a symmetry breaking. This is clear in the example your are mentioning : The liquid-gas transition is characterized by a first order phase transition but there is no symmetry breaking. Indeed, liquid and gas share the same symmetry (translation and rotation invariance) and may be ...

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The worldsheet Weyl anomaly in bosonic string theory is an example. More generally in any dimension you can have trace and Weyl anomalies that break scale or conformal invariance, even in systems with only bosons.

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