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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 $+1$.

Let's consider the adjoint

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

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

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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.

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  • $\begingroup$ 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 $e^{i\theta}Q$ ? $\endgroup$ Jan 31, 2021 at 8:08
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    $\begingroup$ $\Phi^\dagger e^{-i\theta Q}$ 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 $Q|0\rangle=0$. $\endgroup$ Jan 31, 2021 at 10:53
  • $\begingroup$ Thank you for your great help $\endgroup$ Jan 31, 2021 at 11:19

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