Charge conjugation on spinors: Am I missing a (-1)?

Question

I'm trying to prove the transformation rules for Dirac Bilinears under charge conjugation as given in "Fundamentals of neutrino physics and astrohysics" by Carlo Giunti et.al. According to them:

$$\psi_b\stackrel{C}{\longrightarrow}\,{\psi_b}^C=\xi_b\,C\,\overline{\psi_b}^{\,T}\,,$$ $$\overline{\psi_a}\stackrel{C}{\longrightarrow}\,\overline{{\psi_a}^C}=-{\xi_a}^*\,{\psi_a}^{\,T}\,C^T\,,$$ $$C^\dagger=C^T=C^{-1}=-C\,,$$

so I tried to use these formulas to compute the transformation rule of the most basic scalar and I got

\begin{align*} S_{ab}\equiv\overline{\psi_a}\psi_b\longrightarrow\overline{{\psi_a}^C}{\psi_b}^C&= -\,{\xi_a}^*\,\xi_b\,{\psi_a}^{\,T}\,\overline{\psi_b}^{\,T}= -\,{\xi_a}^*\,\xi_b\,(\overline{\psi_b}\,{\psi_a})^{\,T}\\&= -\,{\xi_a}^*\,\xi_b\,\overline{\psi_b}\,{\psi_a}= -\,{\xi_a}^*\,\xi_b\,S_{ba}\end{align*}

but, apparently, the minus sign is wrong. Does it has something to do with the components of the spinors being C-valued (or Grassman) numbers? and, if so, which of my steps is wrong?. The book clearly states that it should be + instead of -

Answer 1

Yes, you are missing a minus sign because the component of $\psi$ are Grassmann variables, namely they anticommute.

First notice:

\begin{equation} \left(\bar\psi^{T}\right)_b=\left[\left(\psi^{\dagger}\gamma^{0}\right)^T\right]_b= \left(\gamma^0\right)_{bc}\left(\psi^{\dagger}\right)^T_c=\gamma^{0}_{bc}\psi_c^{\dagger}=\gamma^{0}_{cb}\psi_c^{\dagger} \end{equation} where the last equality holds since $\gamma^0$ is symmetric. Hence:

\begin{equation} S_{ab}=\bar{\psi}_a\psi_b\to-\xi_a^*\xi_b\psi_a\gamma^0_{bc}\psi^{\dagger}_c=+\xi_a^*\xi_b\psi^{\dagger}_c\gamma^0_{cb}\psi_a=+\xi_a^*\xi_b\bar{\psi}_b\psi_a=+\xi_a^*\xi_b S_{ba} \end{equation}