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 $$ \psi^\dagger =\begin{pmatrix}\psi_L^* \psi_R^*\end{pmatrix}~? $$

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

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

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

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