# Understanding that QCD is not CP invariant

• A
• JD_PM
The strong CP problem is the unresolved problem of why the coupling constant of the strong force is zero.
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?

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.

arivero and JD_PM
Gaussian97 said:
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 and arivero
JD_PM and arivero