 On charge conjugation of Dirac spinor

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Suppose we have Weyl spinor $\psi_{a}$, which transforms under irreducible representation $\left( \frac{1}{2}, 0\right)$ of the Lorentz group,
$$\psi_{a} \to (T(g))_{a}^{\ b}\psi_{b},$$and complex conjugated spinor $\kappa^{\dot{b}}$, which transforms under $\left(0, \frac{1}{2}\right)$ representation,
$$\kappa^{\dot{b}} \to (T(g))^{\dot{b}}_{\ \dot{a}}\kappa^{\dot{a}}$$
Dirac spinor corresponds to the direct sum $\left( \frac{1}{2}, 0\right) \oplus \left( 0, \frac{1}{2}\right)$:
$$\Psi = \begin{pmatrix} \psi_{a} \\ \kappa^{\dot{b}}\end{pmatrix}$$
Charge conjugation of Dirac spinor is determined as
$$\tag 1 \Psi \to \hat{C}\Psi = \begin{pmatrix} \kappa_{a} \\ \psi^{\dot{b}}\end{pmatrix}$$
Sometimes charge conjugation is called "the Dirac version of  of complex conjugation".

The question. Why do we need to introduce complex conjugation in a form $(1)$, not in a form of ordinary complex conjugation, $\Psi \to \Psi^{*}$?

asked Sep 3, 2015

Because complex conjugation is not a spinor-representation independent concept (and so is not the way you introduce charge conjugation here). A representation-independent definition is obtained through considering the effect of charge conjugation on the states and consequently causal field operators.

(1) is the ordinary complex conjugation. It takes this form simply because the complex conjugate representation of $(0,1/2)$ is $(1/2,0)$ (and vice-versa).
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