# How does the Dirac operator change under adjoint transformation?

I'm trying to demonstrate an identity $$\int \overline{\psi}D\phi = \int \overline{D\psi}\phi$$ by substituting in the dirac operator as $$D = i\gamma^{a}\partial_{a}$$ and $$\overline{\psi} = \psi^{\dagger} \gamma^{0}$$ but I have no idea what properties the partial differential operator is supposed to have with respect to transpose, complex conjugation etc or its commutation relations to the other terms. Any ideas on what I need to proceed here or what this identity is called?

• theres clearly some kind of 'flip' operation that happens to put the adjoint spinor on the other side but I don't know how to motivate it – user2992309 Dec 22 '18 at 20:43
• Do you know how to adjoint - flip a pure derivative in integration by parts? A gamma matrix? – Cosmas Zachos Dec 22 '18 at 21:00
• No have no idea what those are. edit: strike that, I know that gamma matrices pick up a minus under transpose and gamma zero doesnt – user2992309 Dec 22 '18 at 21:14
• Indeed, $\left( \gamma^\mu \right)^\dagger = \gamma^0 \gamma^\mu \gamma^0$, and i times the derivative is hermitian... could you go from there? – Cosmas Zachos Dec 22 '18 at 22:31
• A hint: $\gamma^0_{ik}\gamma^a_{kl}\gamma^0_{lj}=\left(\gamma^a_{ji}\right)^*$ – octonion Dec 22 '18 at 22:32

It's but an application of the adjoint of the Dirac operator $$D = i\gamma^{a}\partial_{a}$$.
Given $$\overline{\psi} = \psi^{\dagger} \gamma^{0}$$ and $$\gamma^\mu ~^\dagger = \gamma^0 \gamma^\mu \gamma^0$$, you have, mindful of the transposition in the adjoint of a product of operators, $$\overline{\gamma^a\psi} = (\gamma^a\psi)^\dagger \gamma^0= \psi^\dagger (\gamma^a)^\dagger \gamma^0= \psi^\dagger \gamma^0 \gamma^a = \overline{ \psi} \gamma^a .$$
Consequently, (no summation over repeated indices), $$\overline{ \psi} \gamma^a i \partial_a \phi = \overline{\gamma^a\psi} ~ i \partial_a \phi = i \partial_a (\overline{\gamma^a\psi} \phi) - i ~\overline{\gamma^a\partial_a\psi} \phi \\ = i \partial_a (\overline{\gamma^a\psi} \phi) + \overline{~i\gamma^a\partial_a\psi} \phi~ .$$
Integrating both sides, the first term of the rightmost side turns into an ignorable surface term, being a total divergence, so that $$\int \overline{ \psi} \gamma^a i \partial_a \phi = \int \overline{~i\gamma^a\partial_a\psi} \phi~ .$$ Now summing over repeated Lorentz indices yields $$\int \overline{\psi}D\phi = \int \overline{D\psi}\phi.$$