I'm trying to understand the Higgs mechanics. For that matter, I'm exploring the possibility of giving mass to the photon in a gauge-invariant way. So, if we introduce a complex scalar field:
$$ \phi=\frac{1}{\sqrt{2}}(\phi_{1}+i\phi_{2}) $$
with the following Lagrangian density (from now on, just Lagrangian)
$$ \mathcal{L}=(\partial_{\mu} \phi)^{\star}(\partial^{\mu} \phi)-\mu^2(\phi^{\star}\phi)+\lambda(\phi^{\star}\phi)^2$$
and $\mu^{2}<0$.
We note that the potential for the scalar particle has an infinity of vacuums all of them in a circle of radius $v$ around (0,0). We introduce two auxiliary fields $\eta,\xi$ to express the perturbations around the vacuum
$$ \phi_0=\frac{1}{\sqrt{2}}[(v+\eta)+i \xi ]$$
Introducing the covariant derivative and the photon field, I have to compute the following thing
$$(D^{\mu} \phi)^{\dagger}(D_{\mu} \phi) $$
The derivatives included in $(D^{\mu}\phi)^{\dagger}$ are supposed to act upon the $(D_{\mu} \phi)$?
Answer
The answer is no. Just as in the case without a gauge field, it is just a product of two derivatives of the field $\phi$. You might be interested in the chapter "Scalar Electrodynamics" in Srednicki's book.
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