The global symmetries of a Lagrangian remain unchanged before and after spontaneous symmetry breaking, specifically for a doublet of scalar fields. The Lagrangian is invariant under the global U(1) transformation, represented as \(\phi \to e^{i \theta} \phi\), and under the global SU(2) transformation, expressed as \(\phi \to e^{i \theta^{a} T_{a}}\phi\). Therefore, the global symmetry group of the Lagrangian \(\mathcal{L}\) is consistently SU(2) × U(1) throughout the process of symmetry breaking.
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The global symmetries of a Lagrangian DON’T change “after the spontaneous symmetry breaking”. The Lagrangian you wrote is invariant under the global U(1) transformation \phi \to e^{i \theta} \phi. It is also invartiant under the global SU(2) transformation \phi \to e^{i \theta^{a} T_{a}}\phi. So, “before and after the spontaneous symmetry breaking” the global symmetry group of \mathcal{L} is always SU(2) \times U(1).wormwoodsilver101 said: