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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?

Name YYY
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2 Answers2

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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.

Clever
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  • 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
  • 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. – Clever Aug 31 '15 at 06:31
  • @Clever But if you look at this post what you find is that extra minus sign is needed: https://physics.stackexchange.com/questions/458451/transpose-of-fermion-bilinears?rq=1 Something is not clear about all this – Vicky Apr 30 '19 at 21:13
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    @Vicky It's perfectly consistent. You need a minus sign for the transpose, but not a minus sign for the Hermitian conjugate. – knzhou Apr 30 '19 at 21:27
  • @knzhou How? Hermitian conjugate is transpose together with complex conjugation. I don't see how the complex conjugation can introduce another minus sign in order to cancel the one that transposition introduces – Vicky Apr 30 '19 at 22:18
  • @Vicky Complex conjugation is defined to swap the order. – knzhou May 01 '19 at 04:11
  • @knzhou What? In a scalar product for vector I know that, but in the case of simple product of 2 operators (as fields are) complex conjugation swap the order? So if I have 2 matrices $A, B$, then $(AB)^* = B^* A^*$? I've never seen that – Vicky May 01 '19 at 11:49
  • @Vicky It's just defined that way. If you don't like it, you can define it without the flip, but then your conventions will be different from everybody else's. Of course, Grassmann numbers aren't the same thing as matrices, so I don't see what your objection is getting at. Different objects can have different definitions. – knzhou May 01 '19 at 11:53
  • @knzhou Where can I see that convention? Could you give me a reference to clear my mind? I thought I knew how to work with these objects... poor me haha – Vicky May 01 '19 at 11:57
  • Nevertheless, is bar fields are row vectors and fields are column ones, complex conjugation of $\bar{\psi}\psi$ is giving you $\psi^* \bar{\psi^*}$, so a matrix? – Vicky May 01 '19 at 12:04
  • @Vicky See Peskin, equation 9.65. Don't think about row and column vectors, they'll just make everything more confusing. Just expand everything explicitly in terms of indices. – knzhou May 01 '19 at 13:32
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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.

Luca
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