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


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 ?


1 Answer 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.

  • $\begingroup$ 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
  • 1
    $\begingroup$ $\Phi^\dagger e^{-i\theta Q}$ 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

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.