# Why less gauge fixing conditions in Faddeev Popov method?

According to Dirac theory of constraints systems, to study the dynamics of gauge invariant observables, we can fix the gauge freedom by fixing it using gauge fixing conditions which are equal to number of first class constraints (even though independent parameters of gauge transformation is equal to number of primary first class constraints). So why is the case that in Yang-Mills theory we have two first class constraints but we use only one gauge fixing condition like axial gauge or Lorenz gauge in Faddeev-Popov method?
According to Rothe and Rothe book, phase space path integral for constraint systems is
$$Z = \int dqdp \Pi_{t, r}\delta(\phi^{1}_{r})\Pi_{t, r}\delta(\xi_{r})\det\{\xi_{\alpha}, \phi^{1}_{\beta}\}\exp\left\{\int dt(qp - H_{0})\right\}$$ (we can also extend this to second-class constraints using Senjanovic path integrals)
Should we not need two gauge fixing conditions with property that Poisson bracket with gauge fixing conditions and first class constraints matrix is non-singular i.e. $$\det\{\xi_{\alpha}, \phi^{1}_{\beta}\}=0$$ so that we can use above path integral.

I am unable to find where my reasoning is going wrong with this.

• Which two first class constraints? Which page in Rothe & Rothe? Feb 2, 2021 at 11:31
• @Qmechanic not for Yang Mills but on page 181 section 11.3 the model has two first class constraints but in 11.6 used only one gauge fixing condition? Feb 2, 2021 at 12:56

• $$SU(N)$$ Yang-Mills theory has $$N^2-1$$ first class constraints (ie, the number of generators of the $$SU(N)$$ algebra), not 2.
With that in mind... fixing Lorentz gauge for Yang-Mills $$\partial_\mu A^\mu_a=0$$ does in fact impose a set of $$N^2-1$$ constraints on the $$N^2-1$$ gauge fields $$A_\mu^a$$. This makes sense, since there is one gauge field per generator, and one first class constraint per generator.