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

Let's consider the $$SU(2)$$ isospin Higgs doublet

$$\Phi=\begin{pmatrix}\Phi^+\\\Phi^0\end{pmatrix}=\begin{pmatrix}\phi_1+i\phi_2\\\phi_3+i\phi_4\end{pmatrix}$$

with hypercharge Y=1, and isospin +1/2 for upper line, and -1/2 for lower line.

From formula $$Q=T_3+Y/2$$, we see that $$\phi_1+i\phi_2$$ is of charge $$+1$$.

$$\Phi^\dagger =\begin{pmatrix}\phi_1-i\phi_2 ; \phi_3-i\phi_4\end{pmatrix}$$

How to demonstrate mathematically that $$\phi_1-i\phi_2$$ has a charge -1 ?

It is essential that you think of the indices 1,2,3,4 as mere tags of real components $$\phi$$ and not group indices (yet). $$Q \Phi = \begin{pmatrix}(1/2+1/2)\Phi^+\\ (-1/2+1/2)\Phi^0\end{pmatrix}= \begin{pmatrix} \Phi^+\\ 0\end{pmatrix},$$ so $$\Phi$$ transforms as $$e^{i\theta Q}\Phi$$, where Q is hermitian, so with real eigenvalues.
The adjoint of this vector is $$\Phi^\dagger e^{-i\theta Q}$$, so that the dot product $$\Phi^\dagger \Phi$$ is invariant (neutral).
That is, infinitesimally, $$-\Phi^\dagger Q = ((-1/2-1/2)(\Phi^+)^*, (1/2-1/2)\Phi^{0~*})= (-(\Phi^+)^*, 0),$$ so $$(\Phi^+)^*$$ has charge -1, as required; call it $$\Phi^-$$. Note even the neutral fields $$\Phi^0 \neq (\Phi^0)^*$$, in general.
• Thanks a lot. May you comment why adjoint is is $\Phi^\dagger e^{-i\theta Q}$ and not $\Phi^\dagger$ as you answered to my comment of physics.stackexchange.com/questions/463026/… .Aso, why do you put minus $\Phi^\dagger Q$ : is it coming from the minus of $e^{i\theta}Q$ ? Jan 31, 2021 at 8:08
• $\Phi^\dagger e^{-i\theta Q}$ is the adjoint of the transformed vector $e^{i\theta Q}\Phi$, not the untransformed one, $\Phi$. And, yes, the increment $[Q,\Phi]$ goes to $[\Phi^\dagger,Q]=-[Q,\Phi^\dagger]$, as $Q|0\rangle=0$. Jan 31, 2021 at 10:53