Higgs doublet: a charged and a neutral component?

Question

In the book 'Modern Particle Physics' byM. thomson the Higgs doublet is written as

$$\phi = \left(\begin{matrix} \phi^+ \\ \phi^0 \end{matrix}\right)=\phi = \left(\begin{matrix} \phi_1 +i\phi_2 \\ \phi_3+i\phi_4 \end{matrix}\right) $$

With the comment

Because the Higgs mechanism is required to generate the masses of the electroweak gauge bosons, one of the scalar fields must be neutral, written as $\phi^0$, and the other must be charged such that $\phi^+$ and $(\phi^+)^*$ = $\phi^-$ give the longitudinal degrees of freedom of the $W^+$ and $W^-$.

I don't understand why can we interpret these component is being charged? Ok by assuming that the degrees of freedom are being 'eaten' by the weak-bosons corresponding to $\phi_i$, $i\in \lbrace 1,2,3\rbrace$. Which, from the argument above, the first two should be eaten by the charged $W$'s and the last one by the neutral $Z$. However this seems to be somewhat ad hoc...

I don't see why the fact that their degree of freedom is transferred to the bosons is a longitudinal polarization implies that these components must be charged...

Answer 1

There are various layers here.

First, the Higgs boson is a EW doublet with a certain hypercharge $Y$. Defining as usual the electric charge as $Q=T^3+Y$ you see that $\phi^0$ is indeed neutral for the choice $Y=1/2$. But you see also that for the very same choice, for which a vev of $\phi_3$ doesn't break the charge $Q$, the other component $\phi^+$ must be charged. In fact, must have charge exactly $+1$, as given by $Q=T^3+Y=\mathrm{diag}(1,0)$.

Second, even without knowing what the charges are, as long as the charge $Q$ is not broken, the 3 polarizations for every $W$ must have the same electric charge.

Finally, you can see that $\phi_{1,2,4}$ are indeed ``eaten'' by changing variables and using instead $$\phi=e^{i\pi(x)/v \hat{T}}\left(\begin{array}{cc}0\\ \frac{v+h}{\sqrt{2}}\end{array}\right)$$ where $\hat{T}$ are the broken generators. Expanding in the $\pi(x)$ field the expression above for $\phi(x)$ you see that they are, at leading order, the $\phi_{1,2,4}$ (that is, the $\pi(x)$ generate on the vacuum states with the same quantum numbers as the $\phi_{1,2,4}$ do). By doing a gauge transformation that schematically looks like $$W_\mu\rightarrow e^{i\pi(x)/v \hat{T}}W_\mu e^{-i\pi(x)/v \hat{T}}-e^{i\pi(x)/v \hat{T}}\partial_\mu e^{-i\pi(x)/v \hat{T}}$$ (hoping that I dind't mess up the signs and other factors) you can remove the $\pi(x)$ fields, that is removing the $\phi_{1,2,4}$. As you can see from the gauge transformation, they give rise to non-vanishing longitudinal terms, such as $\partial_\mu\pi(x)/v$, in the expression for the $W$. In other words, a massive $W$ emerges via the so-called Higgs mechanism.

Answer 2

What we mean by particle in quantum field theory is a field belonging to an irreducible representation of the Lorentz group and an irreducible representation of the gauge groups. In the electroweak model, since electromagnetic gauge symmetry remains unbroken, the $W^{\pm}$ ($Z$) bosons belong to the four-vector representation and the charge $\pm 1$ ($0$) representation of the electromagnetic gauge group. Thus to give a longitudinal mode to the $W^\pm$ and $Z$ you need fields with these charges. Otherwise we would not call these new fields modes of existing particles, but they would be new particles in their own right.

You can also see it by directly considering the structure of the gauge group. The gauge group is $SU(2) \times U(1)$, however the second factor is not the electromagnetic $U(1)$. Indeed the element $(A, e^{i\theta})$ where $A \in SU(2)$ acts on the Higgs like $$(A, e^{i\theta})\cdot \begin{pmatrix} \phi^+ \\ \phi^0 \end{pmatrix} = e^{i\theta} A \begin{pmatrix}\phi^+ \\ \phi^0 \end{pmatrix}$$ and so has no fixed elements. However a 1-parameter subgroup $$(A, e^{i\theta}) = (\begin{pmatrix} e^{i\theta/2} & 0 \\ 0 & e^{-i\theta/2} \end{pmatrix}, e^{i\theta/2})$$ fixes the lower component. Since after SSB the condensate is electrically neutral, it must be a one-parameter subgroup like this that remains unbroken. The subgroup is clearly isomorphic to $U(1)$, and we see that the top component transforms according to the charge $+1$ representation.