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Under the electroweak gauge group $SU(2)_LU(1)_Y$ one identifies the 4 gauge fields $W^+, W^-, W^0, B$. After symmetry breaking $W^0$ and $B$ mix to give the observed fields $Z^0$ and $A$. Is there an intuitive argument showing immediatly that $A$ can not be identified with $B$?

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up vote 2 down vote accepted

Yes--- the electric charge is unbroken, from the zero mass of the photon, so B would have to be unbroken. But B commutes with all the generators of the SU(2), so all the electroweak doublets would have the same electric charge. But this is impossible, as the only things in a family with a given electric charge are unique--- they can't make an SU(2) doublet with anything else.

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