A problem about the charge conjugation of scalar $\bar{\psi}\psi$

Question

When I read the Peskin & Schroeder book, I find the charge conjugation for scalar $\bar{\psi}\psi$ is given by $$ C\bar{\psi}\psi C=-\gamma^0_{ab}\gamma^2_{bc}\psi_c\bar{\psi}_d\gamma^0_{de}\gamma^2_{ea}=+\bar{\psi}_d\gamma^0_{de}\gamma^2_{ea}\gamma^0_{ab}\gamma^2_{bc}\psi_c=-\bar{\psi}\gamma^2\gamma^0\gamma^0\gamma^2\psi \tag{3.147}$$
Here the indices are spinor indices. It confused me that the second and third equals sign can hold. I want to know the detail of this derivation.

Answer 1

The second sign $$ -\gamma^0_{ab}\gamma^2_{bc}\psi_c\bar{\psi}_d\gamma^0_{de}\gamma^2_{ea}=\bar{\psi}_d\gamma^0_{de}\gamma^2_{ea}\gamma^0_{ab}\gamma^2_{bc}\psi_c $$ comes from swapping the Grassmann-odd $\psi_c$ and $\psi_d$.

The third sign $$ \bar{\psi}_d\gamma^0_{de}\gamma^2_{ea}\gamma^0_{ab}\gamma^2_{bc}\psi_c=-\bar{\psi}\gamma^2\gamma^0\gamma^0\gamma^2\psi $$ comes from swapping the anti-commuting $\gamma^0$ and $\gamma^2$.