During class we arrived to the statement that the $(1/2,0)$ or right-handed representation of SU(2) is realized by: $$ R_R=\exp\left(i\theta^a\frac{\sigma^a}{2}-\eta^a\frac{\sigma^a}{2}\right) $$ whereas for the (0,1/2) or left-handed we have: $$ R_L=\exp\left(i\theta^a\frac{\sigma^a}{2}+\eta^a\frac{\sigma^a}{2}\right) $$ where $\theta^a$ and $\eta^a$ indicate the structure constants and $\sigma^a$ are the Pauli matrices.
We also said that the two component spinor $\xi_R$ is right-handed if it transforms under the action of the Lorentz group according o the $(1/2,0)$ and same for the left-handed spinor $\eta_L$ and the $(0,1/2)$.
However I encountered a different approach in Landau-Lifshitz vol.IV of theoretical physics. Here it's stated that a spinor is a two component quantity $\xi^\alpha$ Lorentz-transforming according to a matrix $B$ with unit determinant: $$ \xi'^\alpha = B^\alpha_\beta \xi^\beta. $$ Then spinors with dotted indices are introduced, which transform according to the conjugate complex of the previous formula: $$ \eta^{\dot{\alpha}} \sim \xi^{\alpha\ast}. $$ The two are claimed to be algebraically independent and the statement is formulated that symmetric four-spinors of higher rank such as: $$ \zeta^{\alpha_1 \alpha_2 \ldots \alpha_n \dot{\beta}_1 \dot{\beta}_2 \ldots \dot{\beta}_m} $$ build up $(n+1)(m+1)$ irreducible representations of the Lorentz group. This means the $\xi$ and the $\eta$ of Landau should be the same as my class' $\xi_R$ and $\eta_L$. However the transformation laws we used are not the conjugate complex of one another!
Furthermore we proved that $R_R^\ast$ is unitarily equivalent to $R_R$ through $\exp(i\sigma^2)$.
So how can the two definitions match?
Thanks
EDIT: I think I figured out something. The thing with dotted indices is that while, under a Lorentz transformation, the spinors with non-dotted or dotted TOP indices transform as: $$ \xi^{{\alpha}}=R^{{\alpha}}_{{\beta}}\xi^\beta $$ $$ \eta'^{\dot{\alpha}}=R^{\ast\dot{\alpha}}_{\ \ \dot{\beta}}\eta^{\dot{\alpha}} $$ so this is indeed the complex conjugate representation, the spinors with non-dotted or dotted BOTTOM indices follow the rule: $$ \xi_\alpha = g_{\alpha\beta}\xi^\beta;\qquad \eta_{\dot{\alpha}}=g_{\dot{\alpha}\dot{\beta}}\eta^{\dot{\beta}} $$ where $$ g_{\alpha\beta}=g_{\dot{\alpha}\dot{\beta}}=\left( \begin{array}{cc} 0 & 1 \\ -1& 0 \\ \end{array} \right). $$ Furthermore the bi-spinor or Dirac spinor, which is made up of Right and Left component is defined to be: $$ \psi = \left( \begin{array}{c} \xi^\alpha \\ \eta_{\dot{\alpha}} \\ \end{array} \right) $$ which means the correct identification should be: $$ \xi_R=\xi^\alpha; \qquad \eta_L=\eta_{\dot{\alpha}}. $$ Am I getting this right?