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

Its the first one. This is exactly what the "dagger" does. It transposes the spinor, converting it from a column spinor to a row spinor, and takes every entry to its complex conjugate, i.e:

$$\psi=\begin{pmatrix}\psi_L\\\psi_R\end{pmatrix} \xrightarrow{\dagger} \begin{pmatrix}(\psi^T_L)^* (\psi^T_R)^*\end{pmatrix} = \begin{pmatrix}\psi_L^\dagger \psi_R^\dagger\end{pmatrix}$$

• Note that the spinor is a 4-component one. – Ryan Unger Apr 9 '15 at 15:37
• @0celo7, yeah I missed that. Fixed. – Constandinos Damalas Apr 9 '15 at 16:10
• And in the case that $\psi_{L/R}$ only contains 1 element - is the first option true? – user77345 Apr 9 '15 at 21:05
• @RobinWang, yes since the transpose of a scalar is the scalar itself. – Constandinos Damalas Apr 9 '15 at 21:06