Say we have a four component spinor $\psi$: $$\psi=\begin{pmatrix}\psi_L\\\psi_R\end{pmatrix}$$ Is the Hermitian adjoint of this: $$\psi^\dagger =\begin{pmatrix}\psi_L^\dagger \psi_R^\dagger\end{pmatrix}$$ OR $$\psi^\dagger =\begin{pmatrix}\psi_L^* \psi_R^*\end{pmatrix}~?$$

• Well the second is a 2x2 matrix, so it has to be the first. – Ryan Unger Apr 9 '15 at 13:56
• Sorry, edited - question should make more sense now. – user77345 Apr 9 '15 at 13:58

$$\psi=\begin{pmatrix}\psi_L\\\psi_R\end{pmatrix} \xrightarrow{\dagger} \begin{pmatrix}(\psi^T_L)^* (\psi^T_R)^*\end{pmatrix} = \begin{pmatrix}\psi_L^\dagger \psi_R^\dagger\end{pmatrix}$$
• And in the case that $\psi_{L/R}$ only contains 1 element - is the first option true? – user77345 Apr 9 '15 at 21:05