actually gives the correct definition of covariant derivative for the gauge transformation , is easily verified: the transformed $D_\mu\varphi(q)$ reads in fact, $$\begin{split}
\partial_\mu (U\varphi(q)U^*) &-ie[UA_\mu(g)U^*,U\varphi(q)U^*] + [U\partial_\mu U^*,U\varphi(q)U^*]\\
&=U D_\mu\varphi(q) U^* + \partial_\mu U \varphi(q) U^*  + U\varphi(q)\partial_\mu U^* + [U\partial_\mu U^*,U\varphi(q)U^*] \\
&=U D_\mu\varphi(q) U^*,
\end{split}$$ where the last equation follows from the fact that $$[U\partial_\mu U^*,U\varphi(q)U^*] = -\partial_\mu U \varphi(q) U^*-U\varphi(q)\partial_\mu U^*,$$ which is easily verified using the commutativity of $U$ and $U^*$ and the identity $U\partial_\mu U^* = -(\partial_\mu U)U^*$, consequence of the Leibniz rule for $\partial_\mu$.

We obtain therefore the following Lagrangian covariant under gauge transformations (which therefore gives rise to an invariant action) $$\cL = \frac12\eta^{\mu\nu}D_\mu \varphi(q)D_\nu \varphi(q) - \frac12m^2 \varphi(q)^2,$$ and then, expanding the covariant derivatives, interaction terms between the neutral scalar field and the electromagnetic potential given by $$\label{eq:interaction}
\cL_I = -\frac{ie}{2}\{[A_\mu(q),\varphi(q)],\partial^\mu\varphi(q)\}-\frac{e^2}{2}[A_\mu(q),\varphi(q)][A^\mu(q),\varphi(q)],$$ with curly brackets denoting the anticommutator. We note that on classical spacetime $\cL_I$ vanishes, as it should, since $A_\mu$ and $\varphi$ commute.[^1]

Therefore, we will understand $\cL_I$ as defined through the noncommutative product in $\cE$. This has the drawback that it will depend explicitly on the center.

If we neglect, as customary, the quadratic term in $A_\mu$ (weak field approximation) 