The title is a bit ambiguous. More specifically, I'm asking:

Are all coupling between massive gauge fields (associated with broken generators) and massless gauge fields of the unbroken group are in the form of covariant derivatives of the massive gauge fields. (This covariant derivative is composed of massless gauge fields and the generators of the unbroken group)

  • $\begingroup$ In the renormalizable SM, the couplings of the photon to the $W^{\pm}, Z^0$ are minimal (via photon covariant derivatives); you are asking if there could be non-minimal coupling terms there? $\endgroup$ Mar 31 at 21:10
  • $\begingroup$ @CosmasZachosC Yes, that's what I'm asking. It's true in the case of SM. I wonder if all such coupling are minimal for generic symmetry breaking model. $\endgroup$
    – Bababeluma
    Apr 1 at 6:08
  • $\begingroup$ Non minimal couplings are not renormalizable. $\endgroup$ Apr 1 at 6:33
  • $\begingroup$ @CosmasZachos thanks! Is there a general proof that all coupling of massive gauge field to the massless gauge field associated with unbroken generator must be in the form of minimal coupling covariant derivatives? It might be proved case by case by brute force, as in the case of SM, but I believe there can be a more abstract prove on general grounds, generalizing to all symmetry breaking models. $\endgroup$
    – Bababeluma
    Apr 1 at 6:38

1 Answer 1


I fail to imagine the contrapositive.

Assuming there are no matter fields to be integrated out in an effective theory (in a rearrangement spoiling renormalizability), in a pure gauge theory of G, all gauge fields, including the future soon-to-be massive ones in G/H are acted upon by covariant derivatives including gauge completions in the unbroken H, and this does not alter after breaking: both unbroken and broken theories are always gauge invariant in H; and, since both these theories are renormalizable, there never was a chance for unrenormalizable H-gauge invariant terms outside minimal-coupling covariant derivatives.

Once you see the point in the standard model, you appreciate the absence of unrenormalizable effective Pauli moment terms there is not an accident, as you cannot arrange the contrapositive in the first place.

To give superfluous formal substance to the point, in the language of Classical to Quantum Fields by L Baulieu, J Iliopoulos, & R Seneor, (Oxford University Press, 2017), Ch 14, the original "unbroken" Y-M theory is $$ {\mathbb D}_\mu= \partial_\mu +ig {\mathbb A}_\mu,\\ {\cal L}_{YM}= \tfrac{1}{2g^2}\operatorname{Tr}([{\mathbb D}_\mu,{\mathbb D}_\nu ]^2 ), $$ where a (hidden) metric saturates the spacetime indices in the square, and ${\mathbb A}_\mu$ is a Lie-Algebra (of G) valued gauge field.

Spliting these gauge fields into those valued in the algebra of ${\mathfrak h}$ of the subgroup H and the (breakable) generators ${\mathfrak k}$ of G/H respectively, $$ {\mathbb D}_\mu= \partial_\mu +ig {\mathbb H}_\mu+ig {\mathbb K}_\mu, $$ we know that this is locally adjoint-gauge-variant w.r.t. H, since it is such w.r.t. G.

So, defining $D_\mu= \partial_\mu +ig{\mathbb H}_\mu$, each term in $$ [{\mathbb D}_\mu,{\mathbb D}_\nu ]= [ D_\mu+ ig{\mathbb K}_\mu , D_\nu + ig{\mathbb K}_\nu]\\ = [D_\mu, D_\mu] +ig[ D_\mu, {\mathbb K}_\nu]-ig[ D_\nu, {\mathbb K}_\mu] -g^2[{\mathbb K}_\mu, {\mathbb K}_\nu], $$ is adjoint-gauge-covariant in H as well, the last term not involving any gauge fields of H at all, of course, and all the rest involving H-covariant derivatives $D$.

So, regardless of what happens to G/H and the Higgs-massification of the vectors ${\mathbb K}_\nu$ in future or not, the trace in the action ensures H-gauge invariance, and there are no non-minimal couplings involved.


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