 Solution

Choose the doublet bases so that the column of barred fields $B=({\bar{\psi }}_L^1,{\bar{\psi }}_L^2{)}^T$ and $R=({\psi }_R^1,{\psi }_R^2{)}^T$ both transform by the doublet matrices in d. This choice follows the source's assignment of the barred doublet itself. Since $H$ transforms in $\epsilon$, its Yukawa coupling must multiply a fermion bilinear in $\epsilon$.

There is exactly one such bilinear in $2\otimes 2$: the Clebsch-Gordan coefficients in d give $B_1{R}_2-B_2{R}_1$. Equivalently, writing the upper block of the Yukawa matrix as $A$, invariance requiresThe first equation sets $A_{11}=A_{22}=0$, because ${\alpha }^2\ne 1$; the second sets $A_{21}=-A_{12}$.

A mixed doublet-singlet bilinear remains a doublet after multiplication by $H$, so all four mixed entries vanish. Finally ${\bar{\psi }}_L^3{\psi }_R^3$ transforms in $\epsilon \otimes 1=\epsilon$, and multiplication by $H$ makes it invariant. ThereforeThe allowed Yukawa interaction is $H\{a({\bar{\psi }}_L^1{\psi }_R^2-{\bar{\psi }}_L^2{\psi }_R^1)+b{\bar{\psi }}_L^3{\psi }_R^3\}$, together with its Hermitian conjugate when forming a real Lagrangian density.

The displayed texture depends on basis. If instead the unbarred left doublet is assigned $D$ in the same ordering as the right doublet, its barred doublet has the dual representation $D^{\ast }$ and the equivalent upper block is $a\mathrm{diag}(1,-1)$. Exchanging the two barred components converts between these descriptions. In either convention there are exactly two independent complex Yukawa couplings.