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Is it true that the QED Lagrangian

$$\mathcal{L} = \bar{\psi}(i\gamma^\mu D_\mu-m) \psi $$

is invariant under charge conjugation?

$$\begin{align} \psi &\mapsto -i(\gamma^0 \gamma^2 \psi)^T\\ \bar{\psi} &\mapsto -i(\bar{\psi} \gamma^0 \gamma^2)^T\\ A_\mu &\mapsto -A_\mu \end{align}$$

I keep finding that $\mathcal{L}\mapsto-\mathcal{L}^*$. It's clear that the equations of motion are the same, but is it actually a symmetry of the action?

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As to your question, yes, the QED Lagrangian is indeed invariant under charge conjugation.

You may have found differently because your transformations under charge conjugation are faulty. The prefactors are correct, however, under charge conjugation $\psi$ goes to $\bar{\psi}$ and vice versa, i.e.

$$ \hat{C} \, \psi \, \hat{C} = -i(\bar\psi \gamma^0 \gamma^2)^T, \\ \hat{C} \, \bar{\psi} \, \hat{C} = -i(\gamma^0 \gamma^2 \psi)^T, \\ \hat{C} \, A^\mu \, \hat{C} = -A^\mu. $$

Using these relations, you should find

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

where we define $\hat{C} \, A_\mu \, \hat{C} = -A_\mu$. (Reasons for choosing this particular transformational behavior can be found on page 9 here.)

From the first, third and fourth line, it is clear now that $\mathcal{L}_\text{QED}$ is invariant under charge conjugation $\hat{C}$. However, as you rightly pointed out, whether a certain transformation marks a symmetry of our theory is determined by the invariance of the action $S = \int \! d^dx \, \mathcal{L}$, not the Lagrangian. MarkWayne also briefly talks about this in his answer to a similar question.

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    $\begingroup$ Can you add how one determined $C\partial_\mu C$ and $C A_\mu C$for completeness? $\endgroup$ – Mikkel Rev Oct 26 '16 at 22:48

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