Understanding that QCD is not CP invariant

[JD_PM]

So let us now draw our attention back to the main prove. We first see that $\varepsilon_{\mu\nu\rho\sigma} g_s f_{ilm}F_i^{\mu \nu}A_l^{\rho} A_m^{\sigma} + \varepsilon_{\mu\nu\rho\sigma} g_s f_{ijk} F_i^{\rho \sigma} A_j^{\mu} A_k^{\nu} = 2 \varepsilon_{\mu\nu\rho\sigma}g_s f_{ilm}F_i^{\mu \nu}A_l^{\rho} A_m^{\sigma}$, given that $\varepsilon_{\mu\nu\rho\sigma} = \varepsilon_{\rho\sigma\mu\nu}$.

All is left is to work out the term (let me drop the 2 factor for the explicit computation) $\varepsilon_{\mu\nu\rho\sigma}g_s f_{ilm}F_i^{\mu \nu}A_l^{\rho} A_m^{\sigma}$

\begin{align*}
\varepsilon_{\mu\nu\rho\sigma}g_s f_{ilm}F_i^{\mu \nu}A_l^{\rho} A_m^{\sigma} &= \varepsilon_{\mu\nu\rho\sigma}g_s f_{ilm}(\partial^\mu A^\nu_i - \partial^\nu A^\mu_i)A_l^{\rho} A_m^{\sigma} \\
&= 2 \varepsilon_{\mu\nu\rho\sigma} g_s f_{ilm} \partial^\mu A^\nu_i A_l^{\rho} A_m^{\sigma}\\
&= \frac 2 3 \partial^\mu (\varepsilon_{\mu\nu\rho\sigma} g_s f_{ilm} A^\nu_i A_l^{\rho} A_m^{\sigma})\\
&= \partial^\mu (\varepsilon_{\mu\nu\rho\sigma} A^\nu_i A_l^{\rho} A_m^{\sigma}) -\frac 1 3 \partial^\mu (\varepsilon_{\mu\nu\rho\sigma} A^\nu_i A_l^{\rho} A_m^{\sigma})
\end{align*}

This smells really good! (at least to me!)

We finally have

\begin{align*}
G \cdot \tilde G &= \varepsilon_{\mu \nu \rho \sigma} G_i^{\mu \nu}G_i^{\rho \sigma}\\
&=\varepsilon_{\mu \nu \rho \sigma} (F_i^{\mu \nu} + g_s f_{ijk} A_j^{\mu} A_k^{\nu})(F_i^{\rho \sigma} + g_s f_{ilm} A_l^{\rho} A_m^{\sigma})\\
&=\varepsilon_{\mu \nu \rho \sigma} (F_i^{\mu \nu}F_i^{\rho \sigma} + g_s f_{ilm}F_i^{\mu \nu}A_l^{\rho} A_m^{\sigma} + g_s f_{ijk} A_j^{\mu} A_k^{\nu} F_i^{\rho \sigma} + g^2_sf_{ijk}f_{ilm}A_j^{\mu} A_k^{\nu}A_l^{\rho} A_m^{\sigma})\\
&=\varepsilon_{\mu \nu \rho \sigma} (F_i^{\mu \nu}F_i^{\rho \sigma} + g_s f_{ilm}F_i^{\mu \nu}A_l^{\rho} A_m^{\sigma} + g_s f_{ijk} A_j^{\mu} A_k^{\nu} F_i^{\rho \sigma})\\
&= \varepsilon_{\mu \nu \rho \sigma} (F_i^{\mu \nu}F_i^{\rho \sigma} + 2g_s f_{ilm}F_i^{\mu \nu}A_l^{\rho} A_m^{\sigma})\\
&= 2\partial_{\mu}(\varepsilon^{\mu \nu \rho \sigma} A_{\nu}^i [G_{\rho \sigma}^i - \frac{g_s}{3}f_{ijk} A_{j\rho} A_{k\sigma}])
\end{align*}

Mmm so my result has a 2 extra factor...

This issue is simply solved by redefining $G \cdot \tilde G := \frac 1 2 \varepsilon_{\mu \nu \rho \sigma} G_i^{\mu \nu}G_i^{\rho \sigma}$ but I might have missed some detail in the computation and that is why I get the 2 extra factor

What do you think?

 

[Gaussian97]

Ok, I think the factor 2 is due to the definition of $\tilde{G}$, but anyway I don't think it is really important.
Regarding your proof, there are some technical details missing, but I think that you have the general idea.

 

[JD_PM]

Ok, I think the factor 2 is due to the definition of $\tilde{G}$

Oh indeed! The Hodge dual tensor is actually defined as $\tilde{G}_{i\mu \nu} := \frac 1 2 \varepsilon_{\mu \nu \rho\sigma} G_i^{\rho \sigma}$ (I made the analogy from QED from here) so the 2 factor is taken care by the $\frac 1 2$ included in the dual tensor, so the guess was right!

Vamos! (this is a mother-tongue word I love to use when celebrating something; you could translate it as "let's go!" :) ).

I have recently started to learn more about gauge theories in the context of particle physics and it is a truly fascinating subject. I am currently reading these fantastic notes written by Australian physicist Rod Crewther.

Another discussion in which I have learned A LOT. Thank you very much @Gaussian97, I hope to come across you very very soon.

 

[vanhees71]

A nice older review is by Abers and Lee:

E. Abers and B. Lee, Gauge Theories, Phys. Rept. 9, 1
(1973), https://doi.org/10.1016/0370-1573(73)90027-6