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?

1 Answer

Hints:

• P&S is considering the free fermion propagator in eq. (16.4), i.e. the cubic $$\bar{\psi} A\psi$$ interaction term does not contribute.

• In the Lagrangian density (16.1) there are implicitly written sums over

1. Dirac spinor indices and

2. fermion species indices, i.e. color and flavor indices.

• Okay thank you, but still, I don't see how we get the $\alpha,\beta$ indices and the $\delta^i_j$ factor. May 17, 2022 at 11:55
• I updated the answer. May 17, 2022 at 12:02
• Thank you! Is there any resource where this is explicitly shown? I'm still a little bit confused on where the indices would be in my derivation(in the end I would have indices in my $\chi$s but then how do I get the form of the book). May 17, 2022 at 12:43
• As @Qmechanic says, you find the propagator from the free field theory. The calculation should be entirely straightforward if you know how to do it for a single Dirac fermion. May 17, 2022 at 15:56
• @Oбжорoв as I have shown above, I did derive the single Dirac fermion propagator. I'm not sure how to generalise it though. May 17, 2022 at 18:09