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In the explanation for the fermion masses, the term $$\bar{L}\phi R$$ is added to the Lagrangian where $L$ is a left-handed $SU(2)$ doublet, $\phi$ is the Higgs left-handed $SU(2)$ doublet and $R$ is a SU(2) singlet. The $\bar{L}\phi R$ is said to be invariant under $SU(2)_L \times U(1)_Y$ symmetry gauge transformations. I understand that $\bar{L}\phi$ is invariant under $SU(2)_L$ but if we include $R$ as above then under its gauge transformation under $U(1)_Y$, won't we be left with a term of the form $e^{iY\theta(x)}$ that we cannot cancel hence breaking the gauge invariance ?

Kaitou1412
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    Note that in the standard model, the left handed leptons have hypercharge $-1/2$, the right handed leptons have hypercharge $-1$, the left handed quarks have hypercharge $1/6$, the right handed 'up' quarks have hypercharge $2/3$, the right handed 'down' quarks have hypercharge $-1/3$, and the Higgs has hypercharge $1/2$. Does this help? Arguably the trickiest term to write down is that coupling the Higgs to the right handed 'up' quarks to the left handed quarks. – gj255 Oct 12 '17 at 16:19

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The coupling term is a Yukawa term, $$\mathcal{L}_\mathrm{Yuk}\sim-(\bar{L}\phi R+\bar{R}\phi^\dagger L)$$ up to a coupling constant. There is the h.c. part.

Now, to make the term $\bar{L}\phi R+\bar{R}\phi^\dagger L$ symmetric under $SU(2)_L\otimes U(1)_Y$, you need to give proper charges to the Higgs doublet. It turns out that assigning $Q=0$ and $Y_3=1$ assures that the term is invariant under the gauge transformation.

Sayan Mandal
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