# Reality of Dirac kinetic term

The Dirac kinetic term is

$$\mathscr{L}_{\text{ferm}}=-i\bar{\psi}\gamma^\mu D_\mu\psi$$

where $$\bar{\psi}\equiv \psi^\dagger \gamma^0$$. Here I've assumed the mostly plus metric, so $$\left(\gamma^0\right)^2=-1$$ (following from the Clifford relation $$\left\{\gamma^\mu, \gamma^\nu\right\}=2\eta^{\mu\nu}$$). I simply want to check that this is real/Hermitian.

\begin{align} \left(i\bar{\psi}\gamma^\mu D_\mu\psi\right)^\dagger&= \left(i\psi^\dagger \gamma^0\gamma^\mu D_\mu\psi\right)^\dagger\\ &=-i\left(D_\mu\psi^\dagger\right)\gamma^\mu\gamma^0\psi\\ &=+i\psi^\dagger\gamma^\mu\gamma^0 D_\mu\psi\\ &=-i\bar{\psi}\gamma^0\gamma^\mu\gamma^0 D_\mu\psi\\ &\overset{?}{=}+i\bar{\psi}\left(\gamma^0D_0-\gamma^i D_i\right)\psi \end{align}

What the heck is going on? I have a feeling that I'm making a stupid, small error.

• Expanding on that comment: $(\gamma^\mu)^\dagger = \gamma^0 \gamma^\mu \gamma^0$. – MannyC Mar 5 at 4:50
• @DanYand I had considered it, but incorrectly. I naively assumed that the gamma matrices were Hermitian, but that's obviously not true. Thanks for reminding me about that (you too @Mane.andrea). – Arturo don Juan Mar 5 at 20:17
• @MannyC If you would like to form that comment into an answer, I could accept it. – Arturo don Juan Jun 27 at 18:55
• One should, in fact, antisymmetrise the derivative so it acts half to the right and half to the left. The difference is a total derivative so is normally ignored. – lux Jun 28 at 4:25

## 1 Answer

I don't understand all the manipulations, but I would do \begin{aligned} \left(i\bar{\psi}\gamma^\mu D_\mu\psi\right)^\dagger&= \left(i\psi^\dagger \gamma^0\gamma^\mu D_\mu\psi\right)^\dagger\\ &=-i\left(D_\mu\psi\right)^\dagger(\gamma^\mu)^\dagger(\gamma^0)^\dagger\psi\\ &=+i\psi^\dagger D_\mu\,(\gamma^\mu)^\dagger(\gamma^0)^\dagger\psi\,. \end{aligned} The first step is trivial. Then I used the fact that $$D_\mu = \partial_\mu + i A_\mu$$. When I take the complex conjugation there's a minus sign from $$i A_\mu$$ and then when I decide to apply it on the other $$\psi$$ there is a minus sign from the fact that I'm turning $$\partial_\mu$$ by parts (so a minus sign overall). Equivalently: derivatives are antihermitian.

Next let's use $$\gamma^0\gamma^\mu\gamma^0 = (\gamma^\mu)^\dagger$$. This is true because $$(\gamma^0)^3 = - \gamma^0$$, and $$\gamma^0$$ is antihermitian, and $$\gamma^0 \gamma^i\gamma^0 = \gamma^i$$, and the spatial $$\gamma^i$$ are hermitian. Also, obviously $$D_\mu$$ passes through the $$\gamma$$ matrices. \begin{aligned} &=i\psi^\dagger\gamma^0\gamma^\mu\gamma^0 (\gamma^0)^3D_\mu\psi\\ &=i \bar{\psi}\,\gamma^\mu D_\mu\psi\,, \end{aligned} since $$(\gamma^0)^4 = \mathbb{1}$$.

• Expanding on my comment above, this is somehow the reverse approach where the total derivative is thrown away to show this term is hermitian up to total derivatives. – lux Jun 28 at 4:27
• Yes, invariance up to total derivatives is usually what one has to prove in this context. Many other continuous and discrete symmetries leave the Lagrangian invariant up to total derivatives. – MannyC Jun 28 at 5:20