- #1
faklif
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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?
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