As the vector boson field $W_\mu$ is, together with $Z^0$, the gauge field for the Standard electroweak model, I know it transforms as a connection under the $SU(2)\times U(1)_Y$ group. But, when this simmetry is broken to $U(1)_{e.m.}$, which is the transformation associated to $W_\mu$?
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asked Jun 10, 2015 at 16:51
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$\begingroup$ What do you mean? Having the gauge symmetry broken means that there are no real gauge transformations left. $\endgroup$– ACuriousMind ♦Commented Jun 10, 2015 at 19:36
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$\begingroup$ Well,the gauge Group is broken, but not completely.there is a residual part,that is the elecromagnetic $U(1)$.then my question is: How does the fields defined in theory transform under this subgroup? Photon and leptons are quite straightforward,but what about $W\mu$? $\endgroup$– Riccardo BuscicchioCommented Jun 11, 2015 at 11:55
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$\begingroup$ The $\mathrm{U}(1)_{e.m.}$ trafo is simply specified by the electric charge, $\mathrm{U}(1)$ representations are boring 1D reps which are phase multiplications, after all. $\endgroup$– ACuriousMind ♦Commented Jun 11, 2015 at 12:01
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$\begingroup$ So the $U(1)_{e.m.}$ mixes $W_\mu^+$ and $W_\mu^-$ fields as any other doublet? I wasn't very sure if the $U(1)e.m.$ as a residual of $SU(2)xU(1)$ break would behave so simple. $\endgroup$– Riccardo BuscicchioCommented Jun 11, 2015 at 17:06
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1 Answer
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From the previous comments, I'm quite sure $W_\mu$ transforms with a phase-factor. $$W_\mu \rightarrow e^{i\theta}W_\mu$$ therefore mixing the charged components of the $W$ field.
answered Jun 12, 2015 at 17:49