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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 I have made a mistake?

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2 Answers 2

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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.

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  • $\begingroup$ 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, 2015 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, 2015 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, 2015 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, 2019 at 21:13
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    $\begingroup$ @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, 2019 at 21:27
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While the answer by @Clever is correct, it glosses over an important subtlety, namely the fact that when calculating $(\overline \psi \psi)^\dagger$ the exchange of the spinors happens via the complex conjugate and not via the transpose (as one might think)!

Therefore in the calculation of $(\overline \psi \psi)^\dagger$ one does not need to make use of the anticommutation relations, and there will be no minus sign in the result. Why? Because for Grassmann numbers $\theta_i$ the complex conjugate is defined such that it exchanges the order of the mutliplication:

$$ (\theta_1 \theta_2)^* := \theta_2^* \theta_1^*$$

This is standard for Grassmann variables because it ensures the product $\theta_1^* \theta_1$ being both real and not Grassmann. It allows us to see what is going on behind the scenes:

$$(\bar \psi \psi)^\dagger = (\bar \psi \psi)^* = (\psi^\dagger \gamma^0 \psi)^* = (\psi^*_i \gamma^0_{ij} \psi_j)^* = \psi_j^* (\gamma^0_{ij})^* \psi_i = \psi_j^* \gamma^0_{ji}\psi_i = \psi^\dagger \gamma_0 \psi = \bar \psi \psi$$

What happened?

  • First we use the fact that for a number the $\dagger$ operation is the same as the complex conjugate
  • Then we use the definition of $\bar \psi = \psi^\dagger \gamma^0$ and write out everything with spinor/Dirac indices
  • Next we distribute the complex conjugate over the product and use the fact that this interchanges Grassmann numbers
  • Then we use $(\gamma^0)^\dagger = \gamma^0$ in index notation
  • Finally we rewrite everything with $\bar \psi $ and $\psi$

Side note: When computing transposes instead of adjoints of fermion bilinears one has to actually anticommute and generate a minus sign. This is needed e.g. in the context of Majorana spinors.

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