The short answer to the question, "What happens to the Lagrangian of the Dirac theory under charge conjugation?" is, "Nothing." It is invariant with respect to charge conjugation.

Before getting to the longer exposition, I'd like to point out a potential misunderstanding about the nature of invariance of the equations of motion under symmetry transformations that arises about your statement regarding the parity ($\mathbf{x}\to-\mathbf{x}$). Your equation
\begin{align}
\mathcal{P}\mathcal{L}(t,x^{i})\mathcal{P}^{-1} = \mathcal{L}(t,-x^{i})
\end{align}
is correct. But this, in and of itself, doesn't "say that a theory has a symmetry." In fact, it places a restriction on the theory -- that the theory must be an even function of the position vector. For example, for a real scalar field, $\phi(t,\mathbf{x})$, the kinetic energy operator in the Lagrangian density, $\partial_\mu \partial^\mu \phi(x)$ is invariant under parity. In fact, it is only the *action* ($S=\int d^4x \mathcal{L}$) that need be invariant under the symmetry transformation. (Often this reduces to the invariance of the Lagrangian density.) So each term of the Lagrangian (density) need not be invariant (though this is often the case).

Returning to the Dirac equation -- the full statement, in words, of invariance of the theory of the interaction of electrons with light (QED) under the discrete transformations ($\mathcal{P},\mathcal{C}$, & $\mathcal{T}$) is that the theory is invariant under each of them separately or in any combination. (QED is less "interesting" than the electroweak theory in this regard since the electroweak theory appears to violate all three of these separately -- but perhaps not all simultaneously.)

We have to remember that the invariance under $\mathcal{C}$ requires the transformation not only of the wave function, $\psi_{C} \equiv \mathcal{C}\psi\mathcal{C}^{-1} = C\gamma_{0}^{T}\psi^{*}$ but also the charge, $q\to -q$. In the case of the free Dirac equation/Lagrangian considered above, the charge does not feature, so it's not directly relevant to the present discussion but it's important to keep in mind.

Now the direct answers to your questions. (I won't do the algebra since, if you can carry out the calculations for the Dirac equation itself, then the transformation of the Lagrangian should be straightforward.)

"Suppose we operate C on the Dirac Lagrangian what should we get?"
The corrected relation is:

\begin{align}\mathcal{C}\mathcal{L}_{Dirac}(x^{\mu})\mathcal{C}^{-1} = \mathcal{L}_{Dirac}(x^{\mu})\end{align}

(Incidentally, your relation turns out to be all right since the Lagrangian (density) must be a Hermitian, scalar operator, so $\mathcal{L}^* = \mathcal{L}^\dagger = \mathcal{L}$. *EDIT: Thanks to Omkar for pointing out that this is wrong. $\mathcal{L}^*\ne\mathcal{L}$.*)

On a same note one can ask what equation should C satisfy? Should it
satisfy the conjugated Dirac equation as \begin{align}
> (i\gamma^{\mu}\partial_{\mu} + m)\psi_{C} = 0? \end{align}

If you're using the "West Coast" metric $(+1,-1,-1,-1)$ then the equation that $\psi_C$ should satisfy is the one above with $m\to -m$. That is, the free Dirac equation is the same for $\psi$ and $\psi_C$. This is because the masses of the particle and anti-particle are identical, as first divined by Dirac. (If you're using the metric of the opposite sign, then you're equation is correct.)

This post imported from StackExchange Physics at 2014-07-13 04:41 (UCT), posted by SE-user MarkWayne