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 1



  • 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.

  • $\begingroup$ Okay thank you, but still, I don't see how we get the $\alpha,\beta$ indices and the $\delta^i_j$ factor. $\endgroup$ May 17, 2022 at 11:55
  • $\begingroup$ I updated the answer. $\endgroup$
    – Qmechanic
    May 17, 2022 at 12:02
  • $\begingroup$ 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). $\endgroup$ May 17, 2022 at 12:43
  • $\begingroup$ 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. $\endgroup$ May 17, 2022 at 15:56
  • $\begingroup$ @Oбжорoв as I have shown above, I did derive the single Dirac fermion propagator. I'm not sure how to generalise it though. $\endgroup$ May 17, 2022 at 18:09

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