# How do we know that renormalization doesn't change the form of the ghost action in Yang-Mills theory?

In field theory, we typically construct a Lagrangian by only specifying its (global or gauge) symmetries, then writing down all renormalizable terms that respect those symmetries with arbitrary couplings, whose values must be determined by experiment. The reason that we don't usually consider "fine-tuned" Lagrangians in which symmetry-allowed couplings happen to vanish is that under renormalization group flow, these couplings will typically reappear anyway. So even if a coupling happens to vanish at some "bare" energy scale, it will appear at lower energy scales, so we might as well include it right from the get-go. (Although RG flow cannot change the linear symmetries of a Lagrangian, so fortunately we don't need to worry about needing to keep track of new symmetry-breaking couplings.)

But in Yang-Mills theory, we take a very different approach in constructing the Faddeev-Popov ghost action for the path integral. I don't want to say that it's "fine-tuned" in the technical sense, but let's say it's "very precisely and non-generically cooked up" in order to gauge-fix the path integral and eliminate redundant path integration over gauge-equivalent field configurations. Specfically, it takes the form

$$S_\text{ghost} = -\partial_\mu \bar{c}^a \partial^\mu c^a + g f^{abc} (\partial^\mu \bar{c}^a) A_\mu^b c^c,$$

where $c$ and $\bar{c}$ are Grassmann-valued non-complex scalar fields ("ghosts" and "antighosts") in the adjoint reprentation of the gauge group, and $g$ and $f^{abc}$ are the Yang-Mills coupling and structure constants respectively.

How do we know that this form is preserved under RG flow? Of course the coupling $g$ flows under renormalization, but it's not entirely obvious to me why RG flow doesn't generate completely new, unwanted renormalizable terms like, say, ghost mass terms $-m^2 \bar{c}^a c^a$, as there is no obvious symmetry prohibiting such terms (indeed, the ghost action explicitly breaks the gauge symmetry anyway). On physical grounds, it makes sense that no such terms should appear, because we made no reference to any particular energy scale in deriving the ghost action, so if such unwanted couplings vanish at one scale then they should vanish at all scales. But is there a more rigorous way to see this?

(Note: I'm not talking about ghosts gaining mass through the Higgs mechanism. I'm just talking about RG flow.)

• Isn't that related to BRST ? One could check whether your mass term is consistent with that. – Adam Jul 12 '17 at 9:05
• 1) It is typically claimed that the gauge-fixed YM Lagrangian is the most general renormalisable and BRST-invariant Lagrangian, so it is not "very precisely and non-generically cooked up". I don't know of any reference where this is explicitly proved though. [see e.g., Srednicki's book, chapter 74; the book is for free on his webpage]. 2) Note that $S_\mathrm{ghost}$ has that particular form only for the gauge condition $\delta(\partial_\mu A^\mu-f)$; other gauge conditions give rise to different ghost actions. This doesn't affect physical amplitudes, as you already know. – AccidentalFourierTransform Jul 12 '17 at 12:46
• @AccidentalFourierTransform Yes, I agree that the YM Lagrangian is quite generic, but $S_\text{ghost}$ isn't part of that Lagrangian; it's part of the path integral measure. (E.g. it doesn't come up in canonical quantization). We have \begin{align*}Z &= \int \mathcal (DA)_\text{gauge-fixed}\, e^{i S_{YM}} \\ &= \int DA\, (\text{Jacobian determinant}) \times (\text{gauge-fixing function}) \times e^{i S_{YM}} \\ &= \int DA\ e^{i S_\text{ghost}}\, e^{i S_\text{gauge-fixing}}\, e^{i S_{YM}}. \end{align*} – tparker Jul 12 '17 at 14:41