4

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 $(D_{\mu} \phi)$?

Qmechanic
  • 220,844
J L
  • 2,947

1 Answers1

2

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.