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  Charge-conjugation of Weyl spinors

+ 3 like - 0 dislike

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
asked Nov 16, 2013 in Theoretical Physics by Jonathan Gleason (265 points) [ no revision ]

2 Answers

+ 3 like - 0 dislike

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.

This post imported from StackExchange Physics at 2014-08-23 04:59 (UCT), posted by SE-user Yuji
answered Nov 16, 2013 by Yuji (1,395 points) [ no revision ]
+ 0 like - 0 dislike

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.

This post imported from StackExchange Physics at 2014-08-23 04:59 (UCT), posted by SE-user Samuel Monnier
answered Nov 16, 2013 by Samuel Monnier (60 points) [ no revision ]

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