# $CP$-transformation for fermionic bilinears

I am trying to derive the transformation of the fermionic bilinear $$\bar{\psi}\psi$$ under $$CP$$ transformation. I know that $$P$$ acts as: $$\psi(t, \vec{x}) \xrightarrow{P} \gamma^0 \psi(t, -\vec{x})$$ while $$C$$ acts as: $$\psi(t, \vec{x}) \xrightarrow{C} C \psi^*(t, \vec{x})$$ where $$C$$ is such that: $$C^{-1}=C^{\dagger}=C^T=-C$$ $$C^{-1}\gamma^{\mu}C = -(\gamma^{\mu})^T$$ (do these hold in any representation?). I have seen a lot of expressions for $$C$$, which differ by each other, such as $$-i\gamma^0 \gamma^2$$, $$-i\gamma^2$$, and also others. First, how do I decide which expression for $$C$$ should I use in my calculations? I have no clue.

Then, I try to perform a CP transformation: $$\psi(t, \vec{x}) \xrightarrow{CP} C(\gamma^0\psi(t,-\vec{x}))^* = C(\gamma^0)^T\psi^*(t,-\vec{x})=C(-C^{\dagger}\gamma^0C)\psi^* = -\gamma^0C\psi^*(t, -\vec{x})$$ since $$(\gamma^0)^T = (\gamma^0)^*$$ by hermiticity of $$\gamma^0$$ and $$C^{-1}\gamma^{\mu}C=-(\gamma^{\mu})^T$$ was used.

Now transform $$\bar{\psi}=\psi^{\dagger}\gamma^0$$. $$\bar{\psi}=\psi^{\dagger}\gamma^0 \xrightarrow{C} (C\psi^*)^{\dagger}\gamma^0 = \psi^TC^{\dagger}\gamma^0 \xrightarrow{P} (\gamma^0\psi)^TC^{\dagger}\gamma^0=\psi^T(\gamma^0)^TC^{\dagger}\gamma^0=\psi^T(-C^{\dagger}\gamma^0C)C^{\dagger}\gamma^0=-\psi^T(t, -\vec{x})C^{\dagger}$$

Now we can transform the bilinear $$\bar{\psi}_i \psi_j$$ under CP: $$\bar{\psi}_i\psi_j \xrightarrow{CP} \psi_i^TC^{\dagger}\gamma^0C\psi_j^* = -\psi_i^T(\gamma^0)^T\psi_j^* = -(\psi_j^{\dagger}\gamma^0\psi_i)^T = -\psi_j^{\dagger}\gamma^0\psi_i=-\bar{\psi}_j\psi_i(t,-\vec{x})$$ where I used the fact that the transpose of a scalar is equal to itself.

However, I know that the result should be without the minus sign (i.e. $$\bar{\psi}_i\psi_j \xrightarrow{CP} \bar{\psi}_j\psi_i$$, with the sign flipped only in the spacial argument). So I must be wrong somewhere. I also tried plugging $$C=-i\gamma^0\gamma^2$$ and redo the calculation, and the same for $$C=-i\gamma^2$$, but I always end up with that minus sign. What am I missing?

The below part of the derivation is wrong: $$-\psi_i^T(\gamma^0)^T\psi_j^* = -(\psi_j^{\dagger}\gamma^0\psi_i)^T$$
The correct derivation has an extra minus sign: $$-\psi_i^T(\gamma^0)^T\psi_j^* = (\psi_j^{\dagger}\gamma^0\psi_i)^T$$ since transposing two Grassmann-odd spinors results in a minus sign: $$(AB)^T=-B^T A^T$$ if $$A$$ and $$B$$ are Grassmann-valued.