In Higgs potential, there is, among various terms, the kinematic term : $(D_{\mu}\Phi)^{\dagger}(D^{\mu}\Phi)$.

where $\Phi$ is the $SU(2)$ isospin Higgs doublet : $\Phi=\begin{pmatrix}\Phi^+\\\Phi^0\end{pmatrix}=\begin{pmatrix}\phi_1+i\phi_2\\\phi_3+i\phi_4\end{pmatrix}$

My question is : what is the complex conjugate of the Higgs doublet that enter in this formula :

  • Is it the "easy" basic formula $\Phi^\dagger =\begin{pmatrix}\phi_1-i\phi_2 ; \phi_3-i\phi_4\end{pmatrix}$, as discussed here:

How to demonstrate *mathematically* than charge of components of Higgs complex conjugate $\phi_1-i\phi_2$ is -1?

  • or is it : $\tilde{\Phi}=i\sigma^2\Phi^*=i\left(\begin{array}{cc}0 & -i\\i &0\end{array}\right)\left(\begin{array}{c}\Phi^-\\ \Phi^{0*}\end{array}\right)=\left(\begin{array}{c}\Phi^{0*}\\ -\Phi^-\end{array}\right)$ ?

as they say on page 13, formula 64 : "The conjugate Higgs doublet is given by ... [the formula that I put just above]" https://arxiv.org/abs/1406.1786

Comments welcome.


1 Answer 1

  • It is the "easy" adjoint, in the "conventional doublet" representation of SU(2). (Unless you really wanted to use the tilded conjugate one and its adjoint throughout, perversely, but consistently.) As a rule, you don't mix conventional untilded Higgs doublets with conjugate doublets (tilded) in the same term: You save the conjugate (tilded) ones for Yukawa couplings to fermions, to move the v.e.v. to another rung and give up-type quarks a mass, and not only down-type ones.

The point is that the conventional Higgs SU(2)-isodoublet $$\Phi=\begin{pmatrix}\Phi^+\\\Phi^0\end{pmatrix}=\begin{pmatrix}\phi_1+i\phi_2\\\phi_3+i\phi_4\end{pmatrix},\\ \Phi^\dagger =\begin{pmatrix}\phi_1-i\phi_2 , ~~~~~ \phi_3-i\phi_4\end{pmatrix} $$ transforms identically (show this: you haven't lived if you haven't!) to the differently charged conjugate isodoublet $$\tilde{\Phi}=i\sigma^2\Phi^*=i\left(\begin{array}{cc}0 & -i\\i &0\end{array}\right)\left(\begin{array}{c}\Phi^-\\ \Phi^{0*}\end{array}\right)=\left(\begin{array}{c}\Phi^{0*}\\ -\Phi^-\end{array}\right)= \begin{pmatrix}\phi_3-i\phi_4\\-\phi_1+i\phi_2\end{pmatrix},\\ \tilde{\Phi}^\dagger =\begin{pmatrix}\phi_3+i\phi_4 , ~~~~~ -\phi_1-i\phi_2\end{pmatrix}. $$ Note how the components of these two are related in a synchronized swimming display, but their components are placed in different rungs. In particular, $\langle \phi_3\rangle=v$ dictates that the v.e.v. is in the lower component for the conventional doublet, and in the upper component for the conjugate doublet, so they give masses to different fermions in their Yukawas.

However, when it comes to Higgs bilinears $\partial_{\mu}\Phi^{\dagger}\partial^{\mu}\Phi$, or $ \Phi^{\dagger} \Phi$, we could use either doublet (as long as we don't mix them!), $$ \Phi^{\dagger} \Phi= \phi_1^2+ \phi_2^2+ \phi_3^2+ \phi_4^2= \tilde{\Phi}^\dagger \tilde{\Phi} . $$

  • $\begingroup$ As usual, a big thank you to you Cosmas. You are the "bible" of particle physics. $\endgroup$ Jan 31, 2021 at 18:00

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