# What is the complex conjugate of two fermionic fields coupled? $(\bar{\psi} \chi)^{\ast} =$ ____?

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).

$$(\psi_{a}^{\ast} \gamma^0_{ab} \chi_{b})^{\ast} = \psi_{a} \gamma^{0\ast}_{ab} \chi_{b}^{\ast}$$ 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.
• 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) – QuantumEyedea Feb 24 '20 at 19:57