In Higgs-Kibble mechanism, if we consider a $SU(2)_L$ doublet of complex scalar fields, then one of them is charged and the other neutral. Why does the neutral field acquire vev and not the charged one?
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2$\begingroup$ Possible duplicate of why do the electroweak vacuum have to be charge and color neutral?. See also Why cannot fermions have non-zero vacuum expectation value? $\endgroup$– AccidentalFourierTransformCommented Dec 28, 2016 at 19:57
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Charged scalars can't have a non-zero vev, because after spontaneous symmetry breaking the $U(1)_{em}$ subgroup of $SU(2)_L\times U(1)_Y$ survives, which means that the vacuum is invariant under it.
Take field $\phi$ with charge $q$ under $U(1)_{em}$ with a vacuum expectation value $\left<\phi\right>=v$. A $U(1)_{em}$ transformation acts on $\phi$ as $\phi\mapsto e^{iq\alpha}\phi$, and so the vev transforms as $v\mapsto e^{iq\alpha}v$. So the vacuum is invariant only if $q=0$ (neutral scalar) or $v=0$ (no vev).
