Quantum Field Theory – Does the Slavnov-Taylor Identity Hold for Scalar Yang-Mills?

brstquantum-field-theoryrenormalizationward-identityyang-mills

I want to renormalize the minimally-coupled scalar Yang-Mills theory:
$$\mathcal{L}_{YM\phi}=(D_\mu\phi)^\dagger(D^\mu\phi)-\frac{1}{4}F_{\mu\nu}^a{F^{\mu\nu}}^a-\frac{1}{2\xi}(\partial_\mu {A^\mu}^a)^2-\bar{c}^a\partial^\mu D_\mu^{ac}c^c$$ to find the $\beta(g)$ function, and I would like to do so using the vector-scalar-scalar vertex. In a Yang-Mills theory coupled to fermions, I can safely use the vector-fermion-fermion vertex knowing that choosing other vertices would yield the same solution because of the Slavnov-Taylor identity, obtained with BRST symmetry. Now, are there analogous "BRST" symmetry (and corresponding "Slavnov-Taylor" identity) for the scalar case?

Best Answer

  1. Yes, the Slavnov–Taylor (ST) identities are generalized Ward identities for non-Abelian Yang-Mills theory with or without matter. (If there's matter the ST identities contain more terms.)

  2. More generally, any anomaly-free gauge theory can in principle be given a BRST formulation with corresponding generalized Ward identities.

Related Question