- #1

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- Summary:
- I want to understand how to show that the gauge-invariant, renormalizable term ##\frac 1 2 G_i^{\mu \nu} \varepsilon_{\mu \nu \rho \sigma} G^{\rho \sigma}_i## breaks the CP symmetry of QCD

In "CP violation" book by Bigi and Sanda (section 8.2.1. QCD), I read that

The QCD Lagrangian is given by

\begin{equation}

\mathscr{L}_{QCD} = \bar \Psi^f [i \gamma^{\mu}D_{\mu} -m_f] \Psi^f - \frac 1 4 G_{i \mu \nu}G_i^{\mu \nu} \tag{1}

\end{equation}

Where ##f## stands for quark flavor and

\begin{equation*}

\Psi^f = \begin{pmatrix}

\psi_r^f \\

\psi_g^f \\

\psi_b^f \\

\end{pmatrix}, \quad G_i^{\mu \nu} := F_i^{\mu \nu} + g_s f_{ijk} A_j^{\mu} A_k^{\nu}, \quad F_i^{\mu \nu} := \partial^{\nu} A^{\mu}_i - \partial^{\mu} A^{\nu}_i

\end{equation*}

Where ##f_{ijk}## is the so-called structure constant, which is totally antisymmetric.

The covariant derivative is given by

\begin{equation*}

D_{\mu} \Psi^f = \left[ \partial_{\mu} + ig_s \lambda_j A_j^{\mu}/2 \right] \Psi^f

\end{equation*}

Where ##\lambda## are the Gell-Mann matrices.

In class we were taught CP using a particular representation of the gamma matrices: the Weyl representation (alternative form). So let us use the same here. It follows that

\begin{equation*}

\psi_{cp} = i \gamma_2 \psi^*, \quad \bar \Psi \chi \to \bar \Psi_{cp} \chi_{cp} = \bar \chi \Psi, \quad \bar \Psi \gamma_{\mu} \chi \to \bar \Psi_{cp} \gamma_{\mu} \chi_{cp} = -\bar \chi \gamma_{\mu} \Psi, \quad A^{\mu}_{cp} = -A^{\mu}

\end{equation*}

Using the above transformations I see that ##(1)## is CP invariant (please let me know if you want me to give the proof).

Then Bigi and Sanda asserted that

\begin{equation}

\frac 1 2 G_i^{\mu \nu} \varepsilon_{\mu \nu \rho \sigma} G^{\rho \sigma}_i \tag{2}

\end{equation}

However, I do not see why the presence of the totally antisymmetric tensor ##\varepsilon_{\mu \nu \rho \sigma}## breaks CP. I get that it is CP invariant as well, just as the ##\frac 1 4 G_{i \mu \nu}G_i^{\mu \nu}## term... What am I missing?

Thank you!

__"the QCD Lagrangian is invariant under CP transformations"__and wanted to prove it.The QCD Lagrangian is given by

\begin{equation}

\mathscr{L}_{QCD} = \bar \Psi^f [i \gamma^{\mu}D_{\mu} -m_f] \Psi^f - \frac 1 4 G_{i \mu \nu}G_i^{\mu \nu} \tag{1}

\end{equation}

Where ##f## stands for quark flavor and

\begin{equation*}

\Psi^f = \begin{pmatrix}

\psi_r^f \\

\psi_g^f \\

\psi_b^f \\

\end{pmatrix}, \quad G_i^{\mu \nu} := F_i^{\mu \nu} + g_s f_{ijk} A_j^{\mu} A_k^{\nu}, \quad F_i^{\mu \nu} := \partial^{\nu} A^{\mu}_i - \partial^{\mu} A^{\nu}_i

\end{equation*}

Where ##f_{ijk}## is the so-called structure constant, which is totally antisymmetric.

The covariant derivative is given by

\begin{equation*}

D_{\mu} \Psi^f = \left[ \partial_{\mu} + ig_s \lambda_j A_j^{\mu}/2 \right] \Psi^f

\end{equation*}

Where ##\lambda## are the Gell-Mann matrices.

In class we were taught CP using a particular representation of the gamma matrices: the Weyl representation (alternative form). So let us use the same here. It follows that

\begin{equation*}

\psi_{cp} = i \gamma_2 \psi^*, \quad \bar \Psi \chi \to \bar \Psi_{cp} \chi_{cp} = \bar \chi \Psi, \quad \bar \Psi \gamma_{\mu} \chi \to \bar \Psi_{cp} \gamma_{\mu} \chi_{cp} = -\bar \chi \gamma_{\mu} \Psi, \quad A^{\mu}_{cp} = -A^{\mu}

\end{equation*}

Using the above transformations I see that ##(1)## is CP invariant (please let me know if you want me to give the proof).

Then Bigi and Sanda asserted that

__QCD is actually not CP invariant,__because there exists a gauge-invariant and renormalizable operator that can be added to ##(1)##\begin{equation}

\frac 1 2 G_i^{\mu \nu} \varepsilon_{\mu \nu \rho \sigma} G^{\rho \sigma}_i \tag{2}

\end{equation}

However, I do not see why the presence of the totally antisymmetric tensor ##\varepsilon_{\mu \nu \rho \sigma}## breaks CP. I get that it is CP invariant as well, just as the ##\frac 1 4 G_{i \mu \nu}G_i^{\mu \nu}## term... What am I missing?

Thank you!