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.

  • 2
    $\begingroup$ Expanding on that comment: $(\gamma^\mu)^\dagger = \gamma^0 \gamma^\mu \gamma^0$. $\endgroup$ – MannyC Mar 5 '19 at 4:50
  • $\begingroup$ @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). $\endgroup$ – Arturo don Juan Mar 5 '19 at 20:17
  • $\begingroup$ @MannyC If you would like to form that comment into an answer, I could accept it. $\endgroup$ – Arturo don Juan Jun 27 '19 at 18:55
  • $\begingroup$ 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. $\endgroup$ – lux Jun 28 '19 at 4:25

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 $(\gamma^0)^4 = \mathbb{1}$.

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  • $\begingroup$ 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. $\endgroup$ – lux Jun 28 '19 at 4:27
  • $\begingroup$ 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. $\endgroup$ – MannyC Jun 28 '19 at 5:20

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