1

I don't understand the following statement: Any pair of Dirac spinors verifies $(\bar{\Psi}_1\Psi_2)^\dagger=\bar{\Psi}_2\Psi_1$ and it is valid for both commuting and anti-commuting (Grassmann-valued) components.

In the case of Grassmanian number, shouldn't I add an extra sign when I flip them? Is there anything else to take into account in the case of Majorana spinors?

Qmechanic
  • 201,751
Janstillerion
  • 119
  • 2
    There is no minus sign even if $\Psi_1$ and $\Psi_2$ are Grassmann-valued. See related topic here: https://physics.stackexchange.com/q/529496/ – MadMax Jul 08 '20 at 16:53
  • 1
    And see another related question here: https://physics.stackexchange.com/q/532934/ – MadMax Jul 08 '20 at 17:01
  • 1
    Surely the Dirac mass term is $m(\psi^\dagger_L \psi_R+ \psi_R^\dagger \psi_L)$? No $\gamma_0$'s. – mike stone Jul 08 '20 at 17:01

1 Answers1

1

As opertors on Hilbert space, and with hermitian $\gamma_0$, we have $$ (\bar \Psi_1 \Psi_2)^\dagger= (\Psi^\dagger_{1,\alpha}(\gamma_0)_{\alpha\beta} \Psi_{2,\beta})^\dagger\\ = \Psi_{2\beta}^\dagger (\gamma_0)^*_{\alpha\beta}\Psi_{1,\alpha}\\ = \Psi_{2\beta}^\dagger (\gamma_0)_{\beta\alpha}\Psi_{1,\alpha}\\ =\bar \Psi_2\Psi_1. $$ For the Grassmann "c-numbers" we define the dagger operation to coincide with this one.

mike stone
  • 52,996