Dirac propagator in Non-Abelian Theory
I am trying to derive equation (16.4) from chapter 16.1 page 506 of Peskin&Schroeder. Here is my derivation
My Attempt
We start here by considering the dirac spinor part of the Non-Abelian lagrangian $$ \mathcal{L}D=\bar{\psi}(i\gamma^\mu \partial\mu+g\gamma^\mu A_\mu^at^a-m)\psi.\tag{16.1} $$ Then for the derivation, I will first add some sources, fourier transform the action and through some transformations reach to the desired Dirac Feynamn propagator $S_F(x-y)$.
First let's write down the action including 2 sources since we can consider $\psi,\bar{\psi}$ as seperate fields. $$ \mathcal{S}[\psi(x),\bar{\psi}(x)]=\int{d^4x\left[\bar{\psi}(x)(i\gamma^\mu(\partial_\mu-igA^a_\mu t^a)-m)\psi(x)+\bar{\eta}(x)\psi(x)+\eta(x)\bar{\psi}(x)\right]}\Rightarrow\ S[\psi(k),\bar{\psi}(k)]=\int{\frac{d^4k}{(2\pi)^4}\left[\tilde{\bar{\psi}}(k)(\gamma^\mu k_\mu+g\gamma^\mu A^a_\mu t^a-m)\tilde{\psi}(k)+\tilde{\bar{\eta}}(k)\tilde{\psi}(k)+\tilde{\bar{\psi}}(k)\tilde{\eta}(k) \right]} $$ where the definitions are as follows $$ \psi(x)=\int{\frac{d^4k}{(2\pi)^4}e^{-ikx}\tilde{\psi}(k)}\;\;\;\bar{\psi}(x)=\int{\frac{d^4k}{(2\pi)^4}e^{ikx}\tilde{\bar{\psi}}(k)}\ \eta(x)=\int{\frac{d^4k}{(2\pi)^4}e^{-ikx}\tilde{\eta}(k)}\;\;\;\bar{\psi}(x)=\int{\frac{d^4k}{(2\pi)^4}e^{ikx}\tilde{\bar{\eta}}(k)}\ $$ Then if we also do the following transformation $$ \chi(k)\equiv\tilde{\psi}(k)+\frac{\tilde{\eta}(k)}{\gamma^\mu k_\mu+g\gamma^\mu A^a_\mu t^a-m} $$ we get the following expression(after simplifying) $$ S=\int{d^4x\bar{\chi}(x)(i\gamma^\mu D_\mu-m)\chi(x)-\int{\frac{d^4k}{(2\pi)^4}d^4x d^4y\frac{\bar{\eta}(x)e^{-ik(x-y)}\eta(y)}{\gamma^\mu k_\mu+g\gamma^\mu A_\mu^a t^a-m}}} $$ which would give the propagator $$ S_F(x-y)=\int{\frac{d^4k}{(2\pi)^4}\frac{ie^{-ik(x-y)}}{\gamma^\mu k_\mu+g\gamma^\mu A_\mu^a t^a-m}} $$ Why am I not getting the correct form? How can I include the appropriate indices in my derivation? 818