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Suppose $\psi$ and $\chi$ are fermionic fields, and suppose I want to calculate the hermitian conjugate of the operator $\bar{\psi} \chi$ $$ h.c.\ \mathrm{of}\ \bar{\psi} \chi = (\bar{\psi} \chi)^{\ast} $$ I recently asked a question here which states that fermion components anti-commute (since they are Grassman-valued). This leads me to calculate $$ (\bar{\psi} \chi)^{\ast} = (\psi^{\dagger} \gamma^0 \chi)^{\ast} = (\psi_{a}^{\ast} \gamma^0_{ab} \chi_{b})^{\ast} = \psi_{a} \gamma^{0\ast}_{ab} \chi_{b}^{\ast} =! - \chi_{b}^{\ast} \gamma^{0\ast}_{ab} \psi_{a} = - \chi^{\dagger} \gamma^{0\dagger} \psi = - \bar{\chi}\psi $$ where I have used $\gamma^{0\dagger} =\gamma^0$, and also I've used the anti-commutation of fermion components where the "!" is. This seems to suggest that $$ \bar{\psi} \chi + h.c.= \bar{\psi} \chi - \bar{\chi} \psi $$ which doesn't seem correct to me, because swapping $\chi \to e_{R}$ and $\psi \to e_{L}$ I know should yield $\bar{e}_{R}e_{L} + h.c. = \bar{e}_{R}e_{L} + \bar{e}_{L}e_{R}$ (see equation (56) of this document, of this document, for example).

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$$ (\psi_{a}^{\ast} \gamma^0_{ab} \chi_{b})^{\ast} = \psi_{a} \gamma^{0\ast}_{ab} \chi_{b}^{\ast} $$ is not correct. Rather $$ (\psi_{a}^{\ast} \gamma^0_{ab} \chi_{b})^{\ast} =\chi_{b}^{\ast} \gamma^{0\ast}_{ab} \psi_{a}= - \psi_{a} \gamma^{0\ast}_{ab} \chi_{b}^{\ast} $$ Note that the Hermitian is defined as $$ (AB)^\dagger = B^\dagger A^\dagger $$ Note that there is no minus sign even if both $A$ and $B$ are Grassmann-valued.

See related post here.

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  • $\begingroup$ Okay from your link I'm concluding that for Grassman-valued objects we have $(AB)^{\dagger} =B^{\dagger}A^{\dagger}$ and $(AB)^{T} = - B^{T} A^{T}$. Is there a simple explanation why the difference? The link physics.stackexchange.com/q/388079 has an explanation for $(AB)^{T} =-B^T A^T$, but why would this same procedure not end up in $(AB)^\dagger = -B^{\dagger}A^{\dagger}$? (I am having trouble following the logic of your provided link) $\endgroup$ – QuantumEyedea Feb 24 '20 at 19:57
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    $\begingroup$ @Greg.Paul, I am with you. The difference seems ad hoc and arbitrary to me too! But most mainstream physicists seem to be OK with it. $\endgroup$ – MadMax Feb 24 '20 at 20:01

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