A question about charge in the gauge covariant derivatives of the $U(1)$ Higgs model I'm studying the $U(1)$ Higgs model, with the lagrangian:
\begin{gather}
\mathcal{L}=-\frac{1}{4}F^{\mu\nu}F_{\mu\nu}+\left(D_\mu\phi\right)\left(D^\mu\phi\right)^\dagger-\mu^2\left(\phi\phi^\dagger\right)-\lambda\left(\phi\phi^\dagger\right)^2
\end{gather}
with $D_\mu=\partial_\mu+iqA_\mu$ and $A_\mu$ is the electromagnetic potential and $F^{\mu\nu}$ the electromagnetic tensor. I'm following Mark Thomson book: Modern Particle Physics
The question is: to which particle corresponds the charge $q$? Photon and Higgs boson have an electric charge equal to zero.
 A: When you state "Photon and Higgs boson have an electric charge equal to zero" you are probably thinking about experimental observations, however the model you describe is not aiming to be a phenomenological model but a pedagogical one. The Higgs boson in the Standard Model (which is aiming at describing the real world) is not exactly like the one appearing in this model. This is a toy model often used to illustrate the mechanism of spontaneous symmetry breaking, by means of a "toy" Higgs with a simpler symmetry group, $U(1)$. Without considering symmetry breaking, this is just scalar QED with some extra $\phi^4$ interaction and does not having anything to do with the Higgs boson.
So let us first speak about $D_\mu$, what you write down is the covariant derivative for your "to be" Higgs boson. Here it is modeled by a scalar, more over a charged scalar, if it were not charged, it would not couple to your field $A_\mu$, i.e. if you take $q=0$, there is no term having both $A_\mu$ and $\phi$.
The $U(1)$ symmetry is the same type of symmetry behind electromagnetism (see Quantum electrodynamics (QED). In this sense we could call this charge $q$ an analogue of electric charge. We have by construction built a charged scalar $\phi$ for this toy model.
In real life the Higgs field is an $SU_L(2)$ doublet of which the excitations of just one component are what are called higgs bosons, and is thus charged under $SU_L(2)$ (called weak isospin) and also has a component which is not electrically neutral (can be witnessed after symmetry breaking and diagonalization, check the Higgs sector of the Standard Model for more details). It is usually written in the following way
$$\begin{align}
H = \frac{1}{\sqrt{2}}\begin{pmatrix}
\phi^+\\
\phi^0
\end{pmatrix}
\end{align}$$
where the $+$ denotes the charged component and the $0$ the neutral one under $U_{EM}(1)$, that is with respect to the charge operator $Q$ of electromagnetism.
So to answer cleanly, $q$ belongs to $\phi$, $\phi$ in this model is charged and does not correspond to the real Higgs boson.
P.D. Also if you want symmetry breaking you need $+\mu^2$ in your Lagrangian, instead of $-\mu^2$, for $\mu\in\mathbb{R}$.
