Past exam of the groups and representations course of the University of Oxford / 2022 / 1 / d

Use the basis of solution for the two doublets, treating the barred left-handed fields as the components to which the printed representation assignment applies. Write $x=({\bar{\psi }}_L^1,{\bar{\psi }}_L^2{)}^T$ and $y=({\psi }_R^1,{\psi }_R^2{)}^T$. The doublet contribution to the Yukawa interaction is $H{x}^{T}Ly$, so invariant tensors obeyThe first equation forces $L_{11}=L_{22}=0$, since their factors are ${\alpha }^{\pm 2}\ne 1$. The second sets $L_{12}=L_{21}$. Thus the doublet invariant tensor is $x_1{y}_2+x_2{y}_1$. The product of either doublet with $R_1$ has no trivial representation, whereas $R_1\otimes R_1=R_0$. ThereforeIf instead the unbarred left-handed doublet is chosen to transform by $D_1$, its barred components transform by $D_1^{\ast }$; in that equivalent basis the upper block is $y{I}_2$. Exchanging the two left components relates these conventions. The singular value decomposition gives the same physical masses in either basis.

Now the doublet invariant tensor would satisfy $D_1(Q{)}^{T}L{D}_2(Q)=L$. Its four entries acquire factors ${\alpha }^3$, ${\alpha }^{-1}$, $\alpha$, and ${\alpha }^{-3}$, respectively. None equals $1$, so $L=0$. Equivalently, $R_2\otimes R_3=R_2\oplus R_3$ has no trivial representation. The mixed doublet and third-family terms are again forbidden because their tensor product of group representations contain no trivial representation. Only the product of the two $R_1$ fields survives: