Does the Charge Conjugation Operator Affect Gamma Matrices Similarly?

[The black vegetable]

TL;DR

What does the charge conjugation operator do mathematically?

I have in my notes the charge conjugation operator converts the spinnor into its complex conjugate ,

$C\begin{pmatrix} \varepsilon \\ \eta \end{pmatrix}=\begin{pmatrix} \varepsilon^{*}{} \\ \eta ^{*} \end{pmatrix}$when applied to gamma matrix from dirac equation does it do the same? Trying to prove that

$C\gamma ^{a}=-\gamma ^{a}C$

Any tips appreciated :)

 

[vanhees71]

It depends a bit on the source from which you study, because there are several conventions around in the literature. The usual definition, as to be found in standard textbooks like Peskin&Schroeder they define charge conjugation on the Dirac spinor field by exchanging particles with anti-particles, i.e., if $\hat{a}(\vec{p},\sigma)$ are the annihilation operators for particles and $\hat{b}(\vec{p},s)$ those of antiparticles you define the unitary charge-conjugation operator via
$$\hat{C} \hat{a}(\vec{p},\sigma) \hat{C}^{\dagger}=\hat{b}(\vec{p},\sigma).$$
Using the properties of the mode functions in the mode decomposition of the Dirac-field operator this gives
$$\hat{C} \hat{\psi}(x) \hat{C}^{\dagger}=-\mathrm{i} \left (\bar{\psi} \gamma^0 \gamma^2 \right)^{\text{T}}$$
and
$$\hat{C} \overline{\psi}(x) \hat{C}^{\dagger}=-\mathrm{i} \left (\gamma^0 \gamma^2 \psi \right)^{\dagger}.$$

 

[The black vegetable]

Okay thanks, was thinking I could do it purely by matrices but will have to look more into it as I'm not familiar with some other terms you used :)