# Question about Gluon self-energy diagram/symmetry factor

I am a qft beginner and have a question: In Peskin & Schroeder chapter 16.5 there is given an expression for a gluon self energy diagram:

According to the feynman rules we get

$$\frac{1}{2} \int \frac{d^4p}{(2\pi)^4} \frac{-i}{p^2}\frac{-i}{(p+q)^2}g^2 f^{abc}f^{bcd} N^{\mu\nu},$$

where

$$N^{\mu\nu} = [g^{\mu\rho}(q-p)^{\sigma}+g^{\rho\sigma}(2p+q)^{\mu}+g^{\sigma\mu}(-p-2q)^{\rho}]*[\delta_{\rho}^{\nu}(p-q)_{\sigma}+g_{\rho\sigma}(-2p-q)^{\nu}+\delta_{\sigma}^{\nu}(p+2q)_{\rho}]$$

Where does the symmetry factor $$\frac{1}{2}$$ come from? I guess it has to do with some double counting when applying Feynman rules, but I can not figure out why. I also tried to reproduce this expression by doing the contractions explicitly, but I get a far simpler expression, which must be wrong. Something like:

$$N^{\mu\nu} = g^{\mu\nu} (q^2-p^2)-q^{\mu}q^{\nu}+2q^{\mu}p^{\nu}-p^{\mu}p^{\nu}$$

I must be making a big systematic error in my calculation, because it is very different from the correct expression in peskin & schroeder.

Edit: The Problem in my calculation might be due to an error when fouriertransforming. A particular contraction might look like:

$$g^2f^{cij}f^{dlm}g^{\alpha\lambda}g^{\rho\kappa}g^{\tau\beta}g^{\gamma\sigma} \int d^4x \int d^4y (\partial_\alpha^x D_{\mu\rho}^{ac})(x_0-x) D_{\nu\tau}^{bl}(y_0-y) (\partial_{\beta}^y D_{\lambda\sigma}^{id})(x-y) D_{\kappa\gamma}^{jm}(x-y).$$

When we Fouriertransform we apply $$\int d^4x_0 \int d^4y_0 e^{-iqx_0} e^{iq'y_0}$$. The term then becomes after some intermediate steps:

$$g^2f^{cij}f^{dlm}g^{\alpha\lambda}g^{\rho\kappa}g^{\tau\beta}g^{\gamma\sigma} \int \frac{d^4p}{(2\pi)^4} \int {d^4p'} \delta(q+p-p')\int d^4y e^{i(q'+p-p')y} (-iq_{\alpha}) \tilde{D}_{\mu\rho}^{ac}(q) \tilde{D}_{\nu\tau}^{bl}(q') (-ip_{\beta}) \tilde{D}_{\lambda\sigma}^{id}(p) \tilde{D}_{\kappa\gamma}^{jm}(p') = -(2\pi)^4 \delta(q-q') g^2f^{cij}f^{dlm}g^{\alpha\lambda}g^{\rho\kappa}g^{\tau\beta}g^{\gamma\sigma} \int \frac{d^4p}{(2\pi)^4}q_{\alpha}p_{\beta} \tilde{D}_{\mu\rho}^{ac}(q) \tilde{D}_{\nu\tau}^{bl}(q') \tilde{D}_{\lambda\sigma}^{id}(p) \tilde{D}_{\kappa\gamma}^{jm}(p+q)$$

where we transformed $$D_{\lambda\sigma}^{id}(x-y)$$ with $$-p$$ and $$D_{\kappa\gamma}^{jm}(x-y)$$ with $$p'$$ to get the desired momentum convention in the diagram. For me this is amiguous however, since I could have transformed these two propagators the other way around and the result would then be different. Then we obtain $$ip'_{\beta}=i(p+q)_{\beta}$$ instead of $$-ip_{\beta}$$ in the above expression (and we would have instead $$\tilde{D}_{\lambda\sigma}^{id}(p+q) \tilde{D}_{\kappa\gamma}^{jm}(p)$$). Does this make a difference when considering the sum of all contractions? If yes, what is the right way to do it?