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The Higgs field is \begin{equation} \Phi = \frac{1}{\sqrt{2}} \left( \begin{array}{cc} \phi_{1} + i\phi_{2} \\ \phi_{3} + i\phi_{4} \end{array} \right) \tag{1} \end{equation}

with $\phi_{1}$ and $\phi_{2}$ carrying electric charge $+1$ respectively, while $\phi_{3}$ and $\phi_{4}$ are electrically neutral.

Under the entry "Higgs Boson" in Wikipedia, it states:

It (the Higgs field) consists of four components: two neutral ones and two charged component fields. Both of the charged components and one of the neutral fields are Goldstone bosons, which act as the longitudinal third-polarisation components of the massive $W^+$, $W^−$, and $Z$ bosons.

Before interaction between the Higgs field and the gauge bosons, the total electric charge is the total charge of $\phi_{1}$ and $\phi_{2}$, which is $+2$. After the interaction, however, the total electric charge is the sum of the charges of $W^{+}$ and $W^{-}$, which is $0$; the electric charge is not conserved. What is wrong?

Besides, if two components of the Higgs field carry positive electric charge, the whole space (even the whole universe) is electrically positive since the Higgs field permeates the whole space. This is very doubtful and seems not reasonable to me. Is this case true?

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    $\begingroup$ You're about to learn some interesting technicalities of the Higgs mechanism in the standard model! $\endgroup$
    – Mitchell Porter
    Feb 26 '19 at 8:12
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    $\begingroup$ The so called charged components of the Higgs fields are the Goldstone bosons that couple into the $W^\pm$. This is a doublet $(H^+,~H^-)$ that is absorbed into the charged flavor changing gauge bosons $W^\pm$. These Goldstone bosons do not actually carry a charge. The other doublet $(H^0,~h)$ contains the $H^0$ Goldstone boson that couples into the weak neutral boson $W^0$ or $Z$ while the $h$ is what is left over. This left over piece is what is detected at LHC. $\endgroup$
    – Lawrence B. Crowell
    Feb 26 '19 at 10:33
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    $\begingroup$ @Crowell - According to what you wrote, the Higgs field includes two doublets: $(H^{+}, H^{-})$ and $(H^{0}, h)$. However, I read (1) in some textbooks (including Scrednicki's Quantum Field Theory). There seem to be different expressions of the Higgs field, and the electric charges are different in these different expressions. I am puzzled. $\endgroup$
    – Shen
    Feb 26 '19 at 12:02
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    $\begingroup$ Not quite: $𝜑_1+𝑖𝜑_2$ has charge 1 , like the similar expression of the Ws, and its complex conjugate has charge -1, so the real Lagrangian has charge 0. Here, it is the lower component, neutral, that picks up the v.e.v. You might fill up the vacuum with + and - Higgs pairs, electrons and positrons, etc... but the vacuum is neutral and does not break charge. $\endgroup$
    – Cosmas Zachos
    Feb 26 '19 at 12:12
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Simply read the WP article and heed its consistency. It is in P&S conventions, so please do not look at Srednicki, whose opposite conventions evidently confuse you consistently. Now, $$Q=T_3+Y_w/2, $$ so for the Higgs doublet, Y =1, hence Q = +1 for the upper component and 0 for the lower component, the one that picks up the v.e.v., cf. (1) in the WP article. So both the physical Higgs and the goldston pumping into and out of the vacuum are neutral. SSB does not break charge.

The $\phi_1+i\phi_2$ has then charge + , whereas the individual real and imaginary components in it do not have a well-defined charge.

There are the corresponding terms in the lagrangian that involve the conjugate Higgs doublet, with co-equal goldstons $\phi_1-i\phi_2$ of negative charge, just as there are as many electrons as positrons, so to speak, poetically. Check all therms $\Phi ^* \cdot \Phi$ and their functions are neutral in charge, weak isospin and hypercharge.

It is a charge +1 goldston eaten by $(W_1-iW_2)/\sqrt{2}\equiv W^+$, with similar charge properties, perhaps counterintuitively: $W_1-iW_2$ ate $\partial (\phi_1+i\phi_2)$! I don't have a glib maxim for it, but it follows direct calculation from the covariantly completed Higgs kinetic terms: Remind yourself of how the resulting mass term emerges proportional to $W^+ W^-$, skipping Lorentz indices, so it conserves charge, likewise. A good SM text should help.

As always, check the electron-neutrino-Higgs Yukawa on the last line of (6) in the WP article to ensure you understand how it conserves charge, weak hypercharge, and weak isospin.

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Besides, if two components of the Higgs field carry positive electric charge, the whole space (even the whole universe) is electrically positive since the Higgs field permeates the whole space. This is very doubtful and seems not reasonable to me. Is this case true?

You are confusing the fields of quantum field theory, with the particles generated by creation operators on these fields.

After all the whole space in QFT is covered by the electron field and the quark fields etc . these fields have no charges.

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  • $\begingroup$ Other fields (the electron field, the quark fields, etc.) are irrelevant here. They are electrically neutral and do not contribute to the issue we discuss here. $\endgroup$
    – Shen
    Mar 2 '19 at 6:38
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    $\begingroup$ well, all the fields for QFT are defined by the particle table . That the complex Higgs components have charge does not make the charge of the electron neutral. if one worries about charges of the QFT fields ones should worry equally for all constituents, for mathematical consistency, imo $\endgroup$
    – anna v
    Mar 2 '19 at 6:45

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