Bilinear terms in QED lagrangian under charge conjugation

[faklif]

Homework Statement

I want to check that the QED lagrangian \mathcal{L}=-\frac{1}{4}F^{\alpha\beta}F_{\alpha\beta} + \bar\Psi(i\displaystyle{\not} D - m)\Psi where F^{\alpha\beta} = \partial^\alpha A^\beta - \partial^\beta A^\alpha, \ D^\mu = \partial^\mu - ieA^\mu is invariant under charge conjugation which is given as A_\mu \rightarrow A'_\mu = -A_\mu, \ \Psi \rightarrow \Psi' = -i(\bar\Psi \gamma^0\gamma^2)^T.

Homework Equations

See above.

The Attempt at a Solution

I have computed \bar\Psi \rightarrow \bar\Psi' = \Psi'\gamma^0 = -i(\gamma^0\gamma^2\Psi)^T which I have then checked in Peskin and Schroeder.

Next I wanted to compute \bar\Psi \Psi \rightarrow \bar\Psi' \Psi' = -i(\gamma^0\gamma^2\Psi)^T(-i\bar\Psi \gamma^0\gamma^2)^T = -(\bar\Psi \gamma^0\gamma^2)(\gamma^0\gamma^2\Psi) = - \bar\Psi \Psi. Where I transpose the whole expression which I thought should be ok since the Lagrangian is 1x1 and use \gamma^0\gamma^2\gamma^0\gamma^2 = I. Checking this in P&S is not as fun since it's wrong, there should be no minus sign. What am I doing wrong?

I also don't quite understand the computation in P&S which is \bar\Psi \Psi \rightarrow \bar\Psi' \Psi' = -i(\gamma^0\gamma^2\Psi)^T(-i\bar\Psi \gamma^0\gamma^2)^T = -\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 = \bar\Psi \Psi. What I don't understand is the step between the two expressions with indices, why does the sign change?

 

[faklif]

I've kept studying and I think that the minus sign which appears is from \{\Psi_a(x),\Psi_b(y)\} = \delta(x-y)\delta_{ab}. What I don't understand is what happens at x=y with a=b? Don't I have to account for this as well since the sum is over all a and b and not limited in space?