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?

This post imported from StackExchange Physics at 2014-08-23 04:59 (UCT), posted by SE-user Jonathan Gleason