# Why does the charge conjugation of the spinor transform as a spinor?

I have come across (in QFT Nutshell, A. Zee) how the charge conjugation of the spinor, $$\psi_c \equiv \gamma^2 \psi^*$$, transform (where $$\gamma^2=\sigma^2\otimes i\tau^2$$ is the component of the gamma matrices).

Under a Lorentz transformation, the spinor transforms as

$$\psi \rightarrow e^{-\frac{i}{4}\omega_{\mu\nu}\sigma^{\mu\nu}}\psi$$

where

$$\sigma^{\mu\nu} = \frac{i}{2}[\gamma^\mu,\gamma^\nu]$$

and $$\omega_{\mu\nu}$$ is antisymmetric

Complex conjugating, we have

$$\psi^* \rightarrow e^{+\frac{i}{4}\omega_{\mu\nu}(\sigma^{\mu\nu})^*}\psi^* = e^{-\frac{i}{4}\omega_{\mu\nu}\sigma^{\mu\nu}}\psi^*$$

Hence,

$$\psi_c\rightarrow\gamma^2 e^{-\frac{i}{4}\omega_{\mu\nu}\sigma^{\mu\nu}}\psi^* = e^{-\frac{i}{4}\omega_{\mu\nu}\sigma^{\mu\nu}}\psi_c$$

It was not obvious to me in the last equation at all. I think that in order to make the last equation correct, we must have $$\gamma^2$$ commute with the exponential. Or equivalently,

$$\omega_{\mu\nu}[\sigma^{\mu\nu},\gamma^\lambda] = 0$$

with $$\lambda=2$$.

But this apparently is not correct to me since

$$[\sigma^{\mu\nu},\gamma^\lambda]=2i(\gamma^\mu\eta^{\nu\lambda}-\gamma^\nu\eta^{\mu\lambda})$$

Using the fact that $$\omega_{\mu\nu}$$ is antisymmetric, we approach the result

$$\omega_{\mu\nu}[\sigma^{\mu\nu},\gamma^\lambda] = 4i \gamma^\mu\omega_\mu^{\;\;\lambda}\neq0$$

• Looks like you assumed $(\sigma^{\mu\nu})^*=-\sigma^{\mu\nu}$ in the third equation. You might want to double-check that assumption. – Chiral Anomaly Aug 6 at 1:01

First, note user Chiral Anomaly's comment that $${\sigma^{\mu\nu}}^* \neq -\sigma^{\mu\nu}$$. $$\gamma^2$$ is purely complex in the Dirac basis, so it's the only one that switches sign under complex conjugation. This means that $$(\gamma^\mu)^* = \gamma^\mu(1 - 2\delta^{\mu2})$$. But we won't actually need this, since we can talk about it more abstractly.
Let $$C$$ be the charge conjugation matrix, which is unitary. We first note the definition that $$C^{-1} \gamma^{\mu} C = -({\gamma^{\mu}})^*$$. This means that $$C \sigma^{\mu \nu} = \frac{i}{2} [(\gamma^\mu)^*,(\gamma^\nu)^*] C = - (\sigma^{\mu\nu})^* C$$, where the minus sign comes from the need to conjugate $$i$$.
This means that $$C(-\frac{i}{4}\omega_{\mu\nu}\sigma^{\mu\nu})^* = (-\frac{i}{4}\omega_{\mu\nu}\sigma^{\mu\nu})C$$. And thus that $$C (e^{-\frac{i}{4}\omega_{\mu\nu}\sigma^{\mu\nu}})^* = e^{-\frac{i}{4}\omega_{\mu\nu}\sigma^{\mu\nu}} C$$, which is the necessary property you want.