Why is the Dirac mass term Hermitian if Grassmann-valued fields anticommute?

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.
• 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? – Name YYY Aug 28 '15 at 11:54
• 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. – Name YYY Aug 28 '15 at 12:02