# What happens to the gauge covariant derivative if the theory contains multiple fields in different representations?

I'm studying a graduate level course in QFT, and I have a potentially very stupid question that I can't find adressed anywhere (which I guess implies that I'm missing something fundamental). I'm currently working a problem where there is assumed to exist both a real scalar field $$\Phi$$ belonging to the adjoint representation of $$SU(2)$$ (an $$SU(2)$$ triplet), and a fermion field $$\psi$$ belonging to the fundamental representation (an $$SU(2)$$ doublet). As far as I understand, if we consider only the scalar, the Lagrangian for the theory should contain a kinetic term $$\frac{1}{2}D_\mu\Phi D^\mu\Phi,$$ where the covariant derivative should have the form $$D_\mu\phi_a = \partial_\mu\phi_a + g\varepsilon_{abc}A_\mu^b\phi_c,$$ (here $$\Phi = \phi_a\tau_a$$, with $$\tau_a = \sigma_a/2$$, and $$g$$ is the coupling strength).

However, since the fermion transforms in the fundamental representation, to make the theory gauge invariant, the Lagrangian should contain a term $$i\bar{\psi}\gamma^\mu D_\mu \psi,$$ but this time the covariant derivative should be $$D_\mu\psi = \partial_\mu\psi - igA_\mu^a\tau^a\psi,$$ clearly different from what we had before.

Now my question is, am I interpreting this correctly as there being two different covariant derivatives present, one for each different representation present in the theory, each with its own set of associated gauge fields $$A_\mu^a$$? Or should they be combined in some way? Or is something else going on entirely?

• Yes, they are different. May I ask why you'd want to combine them? If there is no interaction going on, it should be completly fine to view them seperately. – Creo Nov 4 '18 at 18:06
• Okay, thank you! I don't necessarily want to combine them, it's rather that I didn't know what the correct approach was. And this Wikipedia article (en.wikipedia.org/wiki/Gauge_covariant_derivative#Standard_Model) refers to "the" covariant derivative of the standard model as what looks to me as something of a combination of different "gauge field terms" or whatever you want to call them. But perhaps this is because the fields in the SM interact? The problem in question does not specify whether the fermion and scalar fields interact or not. – Rolf Alien Nov 4 '18 at 18:58
• Indeed, this derivative takes all ''gauge fields'' into account (there might even be different gauge fields for the same group -not in the standard model though, these are all ''fundamental''). But these can be viewed as one gauge field too (as a $\mathfrak{su}(3) \oplus \mathfrak{su}(2) \oplus \mathfrak{u}(1)$ valued field). But if you are only starting with QFT/Gauge Theory this should be none of your concern right now (and any further discussion might be better off as a new question(?) - there also is a chatroom the ''h-bar''). – Creo Nov 4 '18 at 19:13
• I should add that the covariant derivative of the standard model does not imply that the gauge fields interact, it's used to take the interaction of the fermionic (matter) fields with said gauge (bosonic) fields into account. – Creo Nov 4 '18 at 19:25

A simple example is the $$U(1)_Y$$ ("hypercharge") gauge field in the Standard Model. Different quarks and leptons have different hypercharges (some of which are zero), so they transform differently under the $$U(1)_Y$$ gauge transformation. There is still only one $$U(1)_Y$$ gauge field.
Now consider the $$SU(2)$$ case. Although we often write the covariant derivative $$D_\mu$$ by itself, it's really not defined all by itself — just like $$\partial_\mu$$ isn't defined all by itself. The covariant derivative is defined in terms of how it acts on something else. The components $$A_\mu^a$$ of the gauge field are the same no matter what $$D_\mu$$ is acting on, but they may be arranged differently depending on what it's acting on ($$\phi$$ versus $$\psi$$). The coefficients $$\epsilon_{abc}$$ and the matrix of coefficients $$\tau^a$$ are just compact ways of expressing sums of products of $$A_\mu^a$$s with the components of $$\phi$$ and $$\psi$$, respectively. These are different covariant derivatives, but they both use the same components $$A_\mu^a$$ of the gauge field.