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?


2 Answers 2


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 $C^T = - C$.

  • $\begingroup$ Why did you assume that $C$ is Hermitian $CC^\dagger =1$? $\endgroup$
    – Ramtin
    Feb 13, 2019 at 12:11

From your first equation, we can write, $$C^{-1}\gamma^\mu C=(-\gamma^\mu)^T$$ Take the $\mu=0$ component, $$-\gamma^{0T}=C^{-1}\gamma^0 C$$ Threfore, $$\gamma^0=-C\gamma^{0T}C^T=C\gamma^0C^{-1}$$ Now, $$(C\gamma^0)^T=(i\gamma^2)^T=i\gamma^2=C\gamma^0$$ But also, $$(C\gamma^0)T=\gamma^{0T}C^T=\gamma^0C^T$$ Now you can proceed!

  • $\begingroup$ Not sure how that $C^T$ appeared on line three, it seems to be related to $C^{-1}$, but we have not done such connection. Also, on line four you use $ C \gamma^0 = i \gamma ^2$, but that isn't that valid only in Dirac's Representation (or maybe another one)? Usually it is not true that $(\gamma^o )^T = \gamma ^o $, the Majorana representation is an example. $\endgroup$
    – matrp
    Oct 3, 2017 at 23:14

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