Skip to main content
5 events
when toggle format what by license comment
May 1, 2020 at 20:24 vote accept Carlo
May 1, 2020 at 20:23 comment added Qmechanic It seems you understand correctly.
May 1, 2020 at 20:04 history edited Qmechanic CC BY-SA 4.0
added 3 characters in body
May 1, 2020 at 19:45 comment added Carlo 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?
May 1, 2020 at 18:20 history answered Qmechanic CC BY-SA 4.0