# Weak isospin transformation of $\bar\psi \psi \phi$

In an old exam I found the following question regarding the Higgs potential:

Write down the gauge invariant Yukawa interaction term in the Lagrangian that gives rise to the electron mass.

The Higgs doublet is given in the unitary gauge as $$\phi(x) = \begin{pmatrix} 0 \\ v + h(x) / \sqrt 2 \end{pmatrix} \,.$$

The electron doublet is then written as $$\psi = \begin{pmatrix} \nu_e \\ e^- \end{pmatrix} \,.$$

The Yukawa vertex probably is $\bar\psi \psi \phi$. From there I do not directly see that it transforms as a singlet under the weak isospin SU(2) transformation. The $\phi$ probably transforms as a doublet, with the $2$ representation. That means that the $\bar\psi \psi$ must transform with the $\bar 2$ transformation for the whole Lagrangian density to transform as a scalar (be invariant).

I expanded the $\bar\psi\psi$ like so: $$\bar\psi\psi = \begin{pmatrix} \psi_\mathrm L \\ \psi_\mathrm R \end{pmatrix}^\dagger \gamma^0 \begin{pmatrix} \psi_\mathrm L \\ \psi_\mathrm R \end{pmatrix} = \psi^\dagger_\mathrm L \psi_\mathrm R + \psi^\dagger_\mathrm R \psi_\mathrm L$$

The $\psi_\mathrm R$ should transform as a singlet. The $\psi_\mathrm L$ should transform with $2$. Which of the $\psi^\dagger$ transforms with $\bar 2$, then? And $\psi^\dagger_\mathrm R \psi_\mathrm L \phi$ seems to transform with $2 \otimes 2$ which is also not invariant.

Just write interaction term in a form $$L_{Yuk} = g\bar{L}H R,$$ where $H$ is Higgs doublet in arbitrary gauge. Since under $SU_{L}(2)$ transformations $\bar{L}$ is transformed as $\bar{2}$, while $H$ is transformed as $2$, then $\bar{L}H$ is $SU_{L}(2)$ invariant; since $R$ is transformed trivially, then $\bar{L}HR$ is also $SU_{L}(2)$ invariant. But $\bar{L}H$ is not invariant under $U_{Y}(1)$ transformation: summary hypercharge of $\bar{L}H$ is twice $-Y_{L}+Y_{H}\neq 0$. Here we need to recall the hypercharge of $R$: it is equal to $-(Y_{H} + Y_{L})$, so that summary hypercharge of $\bar{L}HR$ is zero. Thus $\bar{L}HR$ is completely gauge invariant.
• Thank you for your answer! I do not get the second last sentence of the first paragraph: “But it is not $\mathrm U_Y(1)$ invariant, making the summary charge as minus double charge of $R$.” Could you perhaps reword it a bit? – Martin Ueding Feb 12 '16 at 16:47