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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.

Where is my confusion about this?

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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.

By extracting the Higgs VEV you obtain the mass term. Since the initial construction is explicitly gauge invariant on the level of lagrangian, such term obtained in described way doesn't violate gauge invariance. However, by choosing concrete VEV you choose the given vacuum state of theory, which is not gauge invariant.

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  • $\begingroup$ 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? $\endgroup$ – Martin Ueding Feb 12 '16 at 16:47
  • $\begingroup$ Sorry, here is the misprint. I've corrected an answer. $\endgroup$ – Name YYY Feb 12 '16 at 21:58

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