Charge conjugation of Complex Klein Gordon Lagrangian

[orentago]

Homework Statement

Show that the complex Klein-Gordon Lagrangian density:

L=N\left(\partial_\alpha\phi^{\dagger}(x)\partial^\alpha\phi(x)-\mu^2\phi^{\dagger}(x)\phi(x)\right)

is invariant under charge conjugation:

\phi(x)\rightarrow C\phi(x)C^{-1}=\eta_c \phi^\dagger (x)

Where C is a unitary operator and \eta_c is a phase factor.

Homework Equations

The Attempt at a Solution

The transformation can also be written as follows: \phi^\dagger (x) \rightarrow \eta_c^{-1} \phi(x)

Hence performing the transformations on \phi(x) and \phi^\dagger (x) gives:

N\left(\partial_\alpha(\eta_c^{-1}\phi(x))\partial^\alpha(\eta_c\phi^\dagger (x))-\mu^2(\eta_c^{-1}\phi(x))(\eta_c \phi^\dagger(x))\right)=N\left(\partial_\alpha\phi(x)\partial^\alpha \phi^\dagger (x)-\mu^2\phi(x) \phi^\dagger(x)\right)=N\left(\partial^\alpha\phi(x)\partial_\alpha \phi^\dagger (x)-\mu^2\phi(x) \phi^\dagger(x)\right)

Where the final step can be made fairly easily by raising and lowering indices. I'm a little unsure over my first assumption about how \phi^\dagger (x) transforms, but otherwise I'm fairly confident in the rest of my steps. Is this solution valid?

 

[orentago]

I have another question. It can be shown that C a(\mathbf{k})C^{-1}=\eta_c b(\mathbf{k}) and C b(\mathbf{k})C^{-1}=\eta_c^* a(\mathbf{k}).

I can show the first one fairly easily by just substituting the full expressions of the fields into the transformation condition above. I'm a little uncertain about the second one. I can't see how to derive it using the same method as the first one. My other idea was that it could be derived from the first expression, by multiplying both sides from the left by C^{-1} and from the right by C, to get rid of the operators around the a(\mathbf{k}), but then this wouldn't give the right expression on the RHS.

Any thoughts?

 

[orentago]

Does anyone have any ideas for the second question, or should I supply more information?

 

[orentago]

Actually no worries I've sorted it: applying charge conjugation twice takes one back to the original state. It's so simple I don't know why I didn't think of it!