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Accordingly to the result

$$ (\bar{\psi}\psi)^\dagger = (-1)\bar{\psi}\psi \tag1$$

coming from the fact that fields with and without bar anti-commutes, you can deduce that the QCD mass term

$$ -m\bar{\psi}\psi, \quad m \in \mathbb{R} \tag2$$

is not hermitian, so QCD Lagrangian is not. Nevertheless, it has to be hermitian since we have the scattering matrix to be unitary. What is happening here?

Vicky
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  • It remains unsolved in that post that I didn't see when I asked mine, so if someone has any idea, I'll appreaciate it – Vicky Apr 30 '19 at 21:15
  • Hi, if you write out the fields in terms of weyl spinors, you get terms $\epsilon ^{ab} \psi_a \chi_b$. Applying the hermitian conjugate to those terms changes the position of $\chi $ and $\psi$. If you move the terms back to the original position you get one minus sign from the anti-commutation, and one from the epsilon – Lunaron Apr 30 '19 at 22:21
  • Forget about matrix and stuff, if you look at the individual terms in the mass, they look like $iab$, where a and b are real Grassmann numbers. Then $(iab)^\dagger = i^\dagger b^\dagger a^\dagger = -iba = iab$, therefore Hermitian. In math jargon, mass is valued in imaginary ring. – MadMax May 01 '19 at 18:17
  • @MadMax What do you think about my answer to this post https://physics.stackexchange.com/questions/476641/cpt-transformation-for-bilinears ? It's quite related to what we are discussing here and in the answer I think that I get to prove that transpose and adjoint of bilinears just swap order but not introduce signs. Everything looks fine, but a second opinion is always useful – Vicky May 01 '19 at 19:08
  • @Vicky, as you mention there, the definition $(AB)^\dagger = B^\dagger A^\dagger$ does not involve any extra sign, regardless of A/B being real or Grassmann, while the fact that the mass term is valued in imaginary (the $i$) ring (the $ab$) is a bit hard to swallow for beginners. – MadMax May 01 '19 at 19:43
  • @MadMax Yes, but the problem is that I've seen some posts about this issues that assert that transpose not only reverses order but introduces a minus sign while adjoint just reverses. See comments on https://physics.stackexchange.com/questions/203019/why-is-the-dirac-mass-term-hermitian-if-grassmann-valued-fields-anticommute/203033?noredirect=1#comment1071504_203033 – Vicky May 01 '19 at 20:02

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