How to show that $\sigma^2\psi_L^*$ transforms as a right-handed spinor? (Peskin&Schroeder) In Peskin & Schroeder, it is written that the quantity $\sigma^2\psi_L^*$ transforms as a right-handed spinor. What confuses me is that I only get the correct result when considering the following:
\begin{equation}
\psi_L^*\to(\psi_L')^*=\exp\left((i\boldsymbol{\theta}-\boldsymbol{\beta})\cdot\frac{\boldsymbol{\sigma^*}}{2}\right)\psi_L^*,
\end{equation}
from which the desired result follows by multiplying with $\sigma^2$ and using $\sigma^2\boldsymbol{\sigma}^*=-\boldsymbol{\sigma}\sigma^2$. However, to me, this looks like we considered the transformation of the spinor $\psi_L$ and not the transformation of the quantity $\sigma^2\psi_L^*$, where I understand the transformation of the latter as:
\begin{equation}
\sigma^2\psi_L^*\to(\sigma^2\psi_L^*)'=\exp\left((-i\boldsymbol{\theta}-\boldsymbol{\beta})\cdot\frac{\boldsymbol{\sigma}}{2}\right)\sigma^2\psi_L^*=\sigma^2\exp\left((i\boldsymbol{\theta}+\boldsymbol{\beta})\cdot\frac{\boldsymbol{\sigma}^*}{2}\right)\psi_L^*=\sigma^2(\psi_L')^*,
\end{equation}
which does look odd to me. What I expect when reading "show that $\sigma^2\psi_L^*$ transforms as a right-handed spinor" is something along the line of:
\begin{equation}
\sigma^2\psi_L^*\to(\sigma^2\psi_L^*)'=S[\Lambda]\sigma^2\psi_L^*,
\end{equation}
which is not what I get when doing the calculations, as shown above.
I hope that I made it clear what confuses me.
 A: Left handed spinors $\psi_{\text{L}}$ transform as
$$
\psi_{\text{L}} \mapsto \exp\left((-\mathrm{i}\boldsymbol{\theta} +\boldsymbol{\beta})\cdot\frac{\boldsymbol{\sigma}}{2} \right)\psi_{\text{L}}\tag{1}
$$
and right handed spinors $\psi_{\text{R}}$ transform as
$$
\psi_{\text{R}} \mapsto \exp\left((-\mathrm{i}\boldsymbol{\theta} -\boldsymbol{\beta})\cdot\frac{\boldsymbol{\sigma}}{2} \right)\psi_{\text{R}}. \tag{2}
$$
Using equation (1), the object $\sigma_2\psi_{\text{L}}^*$ transforms as
\begin{align}
\sigma_2\psi_{\text{L}}^* \quad\mapsto & \quad\sigma_2\left[\exp\left((-\mathrm{i}\boldsymbol{\theta} +\boldsymbol{\beta})\cdot\frac{\boldsymbol{\sigma}}{2} \right)\psi_{\text{L}}\right]^* \\
&= \sigma_2\exp\left((\mathrm{i}\boldsymbol{\theta} +\boldsymbol{\beta})\cdot\frac{\boldsymbol{\sigma}^*}{2} \right)\psi_{\text{L}}^* \\
&= \sigma_2\exp\left((\mathrm{i}\boldsymbol{\theta} +\boldsymbol{\beta})\cdot\frac{-\sigma_2\boldsymbol{\sigma}\sigma_2}{2} \right)\psi_{\text{L}}^*\\
&= \underbrace{\sigma_2^2}_{1}\exp\left(-(\mathrm{i}\boldsymbol{\theta} +\boldsymbol{\beta})\cdot\frac{\boldsymbol{\sigma}}{2} \right)\sigma_2\psi_{\text{L}}^*,
\end{align}
that is, in the same way as the right handed spinor in equation (2).
