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I have a question about the self-adjointness of the gradient in spinor space.

In the derivation of the Dirac adjoint equation, as in Hermitian adjoint of 4-gradient in Dirac equation , it has been pointed out that the adjoint does not influence the gradient $\partial_\mu$ since the vector space we are considering is $\mathbb{C}^4$, and not ${L^2[\mathbb{C}]}$, so $\partial_\mu$ is hermitian.

Nevertheless, the condition for the Dirac Lagrangian \begin{equation} \mathcal{L}_D = \bar{\Psi}(i \gamma^\mu{\partial_\mu} - m)\Psi \end{equation} to be hermitian seems to imply that $\partial_\mu$ is anti-hermitian.

I am a little confused about this: are we applying two different adjoint transformations, in two different spaces? If so, what is the criterion to choose which one to apply?

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Ultimately, it is Hermitian conjugation in both senses: Both wrt. Dirac indices and as differential operators. In particular we still have to integrate by parts. Also note that the complex conjugate of a product two supernumbers $z,w$ is by definition the opposite order: $(z w)^{\ast}=w^{\ast} z^{\ast}$.

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  • $\begingroup$ If I understood that correctly, this operation does the conjugation only wrt Dirac indices: $$ \mathcal{L}_D^\dagger = \Psi^\dagger(-i {\gamma^\dagger}^\mu \overleftarrow{\partial_\mu} - m)\gamma^0 \Psi = \dots = \bar\Psi (-i\gamma^\mu \overleftarrow{\partial_\mu} - m) \Psi $$ and then, after an integration by part, we find the original $\mathcal{L}_D$ with the right (no pun intended) verse of the derivative? And the reason why we don't integrate by part in the adjoint Dirac equation is that we want the derivative to act on the adjoint spinor, which is on the left? $\endgroup$ – Carlo Cepollaro May 1 at 19:45
  • $\begingroup$ It seems you understand correctly. $\endgroup$ – Qmechanic May 1 at 20:23

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