Let's have Dirac mass term in lagrangian: $$ L_{M} = \bar{\Psi}\Psi $$ Lagrangian must be real-valued, i.e., its Hermitian conjugation doesn't change it. But due to Grassmann nature of spinor fields, $[\psi_{a}^{*}, \psi_{b}]_{+} = 0$, $$ L_{M}^{\dagger} = -\bar{\Psi}\Psi $$ Where have I make a mistake?


Taking the Hermitian conjugate reverses the order of the $\psi$'s. You have

$$ L_M^\dagger = \left( \bar{\psi}\psi\right)^\dagger = \left( \psi^\dagger \gamma^0\psi\right)^\dagger = \psi^\dagger{\gamma^0}^\dagger \psi = \psi^\dagger\gamma^0\psi = \bar{\psi}\psi = L_M \ , $$

where we use that $\gamma^0$ is Hermitian.

  • $\begingroup$ Let's check hermicidity of $\bar{\psi}\kappa + h.c.$. Hermite conjugation gives $(\bar{\psi}\kappa)^{\dagger} = -\bar{\kappa}\psi + h.c.$, because for $\kappa^{a}, \psi^{b}$ anticommutator $[\kappa_a, \psi_b]_{+}=0$ holds. How to deal with the minus sign? $\endgroup$ – Name YYY Aug 28 '15 at 11:54
  • $\begingroup$ Thus if $\kappa =\psi$, then hermitian terms takes the form $\bar{\psi}\psi - \bar{\psi}\psi=0$, which is true for all bilinear form of identical grassmannian numbers. $\endgroup$ – Name YYY Aug 28 '15 at 12:02
  • $\begingroup$ It doesn't matter what the (anti)commutation behavior of the fields/matrices is, the prescription of taking the Hermitian conjugate (the transpose) is to reverse the order. If you want to bring it back into the initial 'order' (of the fields/matrices), that's where you have to take into account that they are Grassmannian etc. $\endgroup$ – Clever Aug 31 '15 at 6:31
  • $\begingroup$ @Clever But if you look at this post what you find is that extra minus sign is needed: physics.stackexchange.com/questions/458451/… Something is not clear about all this $\endgroup$ – Vicky Apr 30 at 21:13
  • 1
    $\begingroup$ @Vicky It's perfectly consistent. You need a minus sign for the transpose, but not a minus sign for the Hermitian conjugate. $\endgroup$ – knzhou Apr 30 at 21:27

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