I am trying to understand why the Dirac Lagrangian is invariant under charge conjugation.

The Dirac Lagrangian is:

$$\mathcal{L} = i\bar{\psi}\gamma^\mu \partial_\mu\psi - m \bar{\psi}\psi $$

I know that under charge conjugation the flowing formulas are correct:

$$ \hat{C} \, \bar{\psi}\psi \, \hat{C} = +\bar{\psi}\psi, \\ \hat{C} \, \bar{\psi}\gamma^\mu\psi \,\hat{C} = -\bar{\psi}\gamma^\mu\psi, \\ \hat{C} \, \partial_\mu \,\hat{C} = \partial_\mu \\ $$


$$ \hat{C} \, \mathcal{L} \,\hat{C} = \hat{C} \, i\bar{\psi}\gamma^\mu \partial_\mu\psi \,\hat{C} - \hat{C} \, m \bar{\psi}\psi \,\hat{C} $$ Where $$ \hat{C} \, i\bar{\psi}\gamma^\mu \partial_\mu\psi \,\hat{C} = i\hat{C} \, \bar{\psi}\gamma^\mu \hat{C} \hat{C} \,\partial_\mu\psi \,\hat{C} = i\hat{C} \, \bar{\psi}\gamma^\mu \hat{C} \partial_\mu \hat{C} \,\psi \,\hat{C} $$

And because $\partial_\mu$ commutes with all $\gamma^\mu$ we use the same calculations as in $\hat{C} \, \bar{\psi}\gamma^\mu\psi \,\hat{C} = -\bar{\psi}\gamma^\mu\psi$ and get:

$$\hat{C} \, i\bar{\psi}\gamma^\mu \partial_\mu\psi \,\hat{C} = -i\bar{\psi}\gamma^\mu \partial_\mu\psi$$

Therefore overall:

$$ \hat{C} \, \mathcal{L} \,\hat{C} = -i\bar{\psi}\gamma^\mu \partial_\mu\psi - m \bar{\psi}\psi \neq \mathcal{L} $$

What am I missing?


1 Answer 1


Note that charge conjugation interchanges the anticommuting $\bar \psi$ and $\psi$, so after conjugation by $\hat C$ the $\partial_\mu$ acts on the $\bar\psi$, and needs a sign-changing integration by parts.

In detail: $$ \hat C \psi \hat C^{-1}= {\mathcal C}^{-1}\bar\psi^T\\ \hat C \bar \psi \hat C^{-1} = - \psi^T {\mathcal C} $$ where the ${\mathcal C}$ matrix obeys $$ {\mathcal C}\gamma^\mu {\mathcal C}^{-1}= -(\gamma^\mu)^T. $$

  • $\begingroup$ Thanks, I forgot the integration by parts $\endgroup$
    – Joe
    Commented Mar 5 at 15:56

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