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?


2 Answers 2


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, 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
  • 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, 2019 at 21:27

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.


Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge you have read our privacy policy.

Not the answer you're looking for? Browse other questions tagged or ask your own question.