It is known that the standard model is $SU(3)_c \times SU(2)_L \times U(1)_Y$ gauge invariant. But, my question is: This is possible, when we write the Lagrangian in terms of weak eigenstates or in other words, our particles are all massless. The only way to give them mass is to have Higgs mechanism which breaks the symmetry and now the standard model is only $SU(3)_C \times U(1)_{EM}$ symmetric. Then why do we say that standard model is still $SU(3)_c \times SU(2)_L \times U(1)_Y$ gauge invariant as in nature we do have massive particles.
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Perhaps the related 190224 dismisses the question. The Higgs mechanism merely hides (realizes in the Nambu-Goldstone mode) the symmetry. – Cosmas Zachos Jul 05 '17 at 15:39
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The point is that it is only the vacuum expectation value of the Higgs that break the gauge symmetry. However, the lagrangian still has the original symmetry (allthough in the massive basis it looks rather messy), which is why SSB is sometimes called hidden symmetry .
In calculations the broken symmetry takes the form of generalized Ward identities that are especially important for longitudinal W & Z boson scattering.
Lunaron
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