1
$\begingroup$

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?

$\endgroup$

1 Answer 1

1
$\begingroup$

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.

$\endgroup$
5
  • $\begingroup$ 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

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge you have read our privacy policy.

Not the answer you're looking for? Browse other questions tagged or ask your own question.