In class, we have defined the Charge Conjugation Operator ($C$) such that:

\begin{equation}

C \left(\gamma^\mu\right)^T C^{-1} = – \gamma ^\mu ,

\end{equation}

\begin{equation}

\psi^C \equiv C\,\overline{\psi}^T ,

\end{equation}

\begin{equation}

(\psi^C)^C = \psi,

\end{equation}

with $\overline{\psi} = \psi^\dagger \gamma ^0$.

I would like to demonstrate that $C^T = – C$ in every representation, but I am not sure how to proceed.

** What I tried**: Applying the first equation twice we get:

\begin{eqnarray}

\gamma ^\mu &\; = C \left[ C (\gamma^\mu)^T C^{-1} \right]^T C^{-1} \\

&= C (C^{-1})^T \gamma ^\mu C^T C^{-1}.

\end{eqnarray}

This can only be true if

\begin{equation}

C^T = \pm C.

\end{equation}

Now I have to restrict C to be antisymmetric. I tried using the third equation, but I've got to:

\begin{equation}

(\psi ^C)^C = \psi = C (\gamma ^0 )^T C^* (\gamma ^0)^T \gamma^0 \psi \gamma ^0,

\end{equation}

Now I am not sure how to proceed. Any ideas?

## Best Answer

I'll post the answer n case anybody needs it.

We have that $$\psi ^C = C \;\overline{\psi} ^T = C \left( \psi ^\dagger \gamma ^0\right)^T = C (\gamma ^0 )^T \psi^*,$$

using the first property in the question: $$C (\gamma ^0) ^T C^{-1} = - \gamma ^0 \implies C (\gamma^0)^T = - \gamma ^0 C$$

we have $$\psi = (\psi ^C)^C = -\left( \gamma ^0 C \psi ^* \right)^C = - C \left(\overline{\gamma ^0 C \psi ^*}\right)^T. $$

Since $\overline{\gamma}^0 = \gamma ^0$ and $\overline{C} = \gamma^0 C ^\dagger \gamma ^0$, then $$\psi= - C \left( \psi^T \gamma ^0 \gamma^0 C^\dagger \gamma^0 \gamma ^0 \right)^T = - C C^* \psi.$$

For this to be true, we must have $$- C C ^* = \mathbb{1} \implies - C^\dagger C^T = \mathbb{1}. $$

Since it is true that $C C^\dagger = \mathbb{1}$, we have that $C^T = - C$.