# Charge-conjugation of Weyl spinors

I am having trouble reconciling two facts I am aware of: the fact that the charge conjugate of a spinor tranforms in the same representation as the original spinor, and the fact that (in certain, dimensions, in particular, in $D=4$), the charge conjugate of a left-handed spinor is right-handed, and vice versa.

To be clear, I introduce the relevant notation and terminology. Let $\gamma _\mu$ satisfy the Clifford algebra: $$\{ \gamma _\mu ,\gamma _\nu \} =2\eta _{\mu \nu},$$ let $C$ be the charge conjugation matrix, a unitary operator defined by $$C\gamma _\mu C^{-1}=-(\gamma _\mu )^T.$$ One can show that (see, e.g. West's Introduction to Strings and Branes, Section 5.2) that $C^T=-\epsilon C$ for $$\epsilon =\begin{cases}1 & \text{if }D\equiv 2,4\, (\mathrm{mod}\; 8) \\ -1 & \text{if }D\equiv 0,6\, (\mathrm{mod}\; 8)\end{cases}.$$ Define $B:=-\epsilon \mathrm{i}\, C\gamma _0$. Then, the charge conjugate of a spinor $\psi$ and an operator $M$ on spinor space are defined by $$\psi ^c:=B^{-1}\overline{\psi}\text{ and }M^c:=B^{-1}\overline{M}B,$$ where the bar denotes simply complex conjugation. We define $$\gamma :=\mathrm{i}^{-\left( D(D-1)/2+1\right)}\, \gamma _0\cdots \gamma _{D-1},$$ and $$P_L:=\frac{1}{2}(1+\gamma )\text{ and }P_R:=\frac{1}{2}(1-\gamma ).$$ We then say that $\psi$ is left-handed if $P_L\psi =\psi$ (similarly for right-handed). Finally, the transformation law for a spinor $\psi$ is given by $$\delta \psi =-\frac{1}{4}\lambda ^{\mu \nu}\gamma _{\mu \nu}\psi.\qquad\qquad(1)$$

Now that that's out of the way, I believe I am able to show two things: $$\delta \psi ^c=-\frac{1}{4}\lambda ^{\mu \nu}\gamma _{\mu \nu}\psi ^c \qquad\qquad(2)$$ and $$(P_L\psi )^c=P_R\psi ^c\text{ (for }D\equiv 0,4\, (\mathrm{mod}\; 8)\text{)}.\qquad\qquad(3)$$ The first of these says that $\psi ^c$ transforms in the same way as $\psi$ and the second implies that, if $\psi$ is left-handed, then $\psi ^c$ is right-handed (in these appropriate dimensions).

I'm having trouble reconciling these two facts. I was under the impression that when say say a Fermion is left-handed, we mean that it transforms under the (1/2,0) representation of $SL(2,\mathbb{C})$ (obviously, I am now just restricting to $D=4$). It's charge-conjugate, being right-handed, would then transform under the $(0,1/2)$ representation, contradicting the first fact. The only way I seem to be able to come to terms with this is that the two notions of handedness, while related, are not the same. That is, given a Fermion that transforms under $(1/2,0)$ and satisfies $P_L\psi =\psi$, then $\psi ^c$ will transform as $(1/2,0)$ and satisfy $P_R\psi =\psi$. That is, the handedness determined in the sense of $P_L$ and $P_R$ is independent of the handedness determined by what representation the Weyl Fermion lives in.

Could someone please elucidate this for me?

Your equations (1) (2), saying $\delta \psi =-\frac{1}{4}\lambda ^{\mu \nu}\gamma _{\mu \nu}\psi$ with or without $^c$, just says that both $\psi$ and $\psi^c$ are in the same representation, namely $(1/2,0) + (0,1/2)$.
The third equation (3), saying $(P_L\psi)^c=P_R \psi^c$, just says that the charge conjugation swaps the two irreducible components of the reducible representation that is the Dirac spinor.
It looks correct and free of contradictions to me. In a basis compatible with the decomposition of a spinor into its left and right Weyl components, $\lambda_{\mu\nu}$ can be brought in a block diagonal form, corresponding to the two $SL(2,\mathbb{C})$ factors. One block acts trivially on left spinors, the other trivially on right spinors. Applying charge conjugation exchanges the blocks, but does not change the transformation law.