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On charge conjugation of Dirac spinor

+ 3 like - 0 dislike
131 views

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 in Theoretical Physics by NAME_XXX (1,020 points) [ no revision ]

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 Answer

+ 3 like - 0 dislike

(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).

answered Sep 3, 2015 by 40227 (4,690 points) [ no revision ]

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