Hermitian properties of Dirac operator

Question

I am trying to understand the Hermiticity of the (massless) Dirac operator in both (flat) Minkowski space and Euclidean space.

Let us define the Dirac operator as $D\!\!\!/=\gamma^\mu D_\mu$, where $D_\mu = \partial_\mu-igA_\mu$, where in general $A_\mu$ is a non-Abelian gauge field. For completeness, they us assume the gauge fields are members of SU(2), and we are working in the Weyl representation for $\gamma$'s).

I have read in a number of sources on Lattice QCD that in Euclidean space $D\!\!\!/^\dagger =-D\!\!\!/$ , however I wish to show this.

Generally $D\!\!\!/=\gamma^0(\partial_0-igA_0)+\gamma^i (\partial_i-igA_i)$.

Then noting that $A_\mu^\dagger=A_\mu$, $\gamma_\mu^\dagger=\gamma_\mu$:

$D\!\!\!/^\dagger=(\partial_0^\dagger+igA_0)\gamma^0+ (\partial_i^\dagger+igA_i)\gamma^i $. Obviously for this to be true, $\partial_\mu^\dagger=-\partial_\mu$, but why? My understanding was that $\partial_\mu$ really represents $\mathbf{I}_{2x2} \partial_\mu$ for SU(2).

I am then further interested in understanding if in Minkowski space the Dirac operator is Hermitian, anti-Hermitian, or none of the above.

Similar to above, working in the (+,-,-,-) metric, noting in this case $\gamma_0^\dagger=\gamma_0$ and $\gamma_i^\dagger=-\gamma_i$,

$D\!\!\!/=\gamma^0(\partial_0-igA_0)-\gamma^i(\partial_i-igA_i) $,

so

$D\!\!\!/^\dagger=(\partial_0^\dagger+igA_0)\gamma^0 -(\partial_i^\dagger+igA_i)(-\gamma^i)=(-\partial_0+igA_0)\gamma^0 -(-\partial_i+igA_i)(-\gamma^i)=-\big((\partial_0-igA_0)\gamma^0+(\partial^i-igA_i)\gamma^i \big)\neq -D\!\!\!/ ~~\text{or}~~D\!\!\!/ $

Edit After a helpful comment, I see that $\partial_\mu^\dagger=-\partial_\mu$, however I believe I made a mistake in my original Minkowski space derivation, and I don't think it is non-Hermitian generally. Can anyone clarify this?

Answer 1

Hint: yes, $\partial^\dagger=-\partial$. One way to see this is that $\hat p=-i\partial$, and $\hat p$ is self-adjoint, which means that $(-i\partial)=(-i\partial)^\dagger=+i\partial^\dagger$. Or put it another way, check the definition of $\dagger$: $$ \langle f,\partial g\rangle=\int f\partial g=-\int \partial f\ g=-\langle \partial f,g\rangle $$ where I used integration by parts. We usually say that $i\partial$ is hermitian instead of saying that $\partial$ is anti-hermitian, but these are obvioulsy equivalent. You'll hear the former more often though.

With this, I believe you can easily prove that $iD\!\!\!/\ $ is hermitian. For that you'll need to use the fact that the gamma matrices are self-adjoint (meaning $\bar\gamma^\mu=\gamma^\mu$, not $\gamma^{\mu\dagger}=\gamma^\mu$, which is false). If you need more details say so and I'll elaborate.

EDIT

As $iD=i\partial+gA$ is the sum of two terms, it suffices to prove the hermicity properties of both of them independently. I believe you know how to deal with the second term: $$ g\gamma^\mu A_\mu $$ is self-adjoint because $\bar\gamma^\mu=\gamma^\mu$.

You can prove that $i\not\partial$ is hermitian by proving that so is $i\bar\psi\not\partial\psi$. This is easier because $$ (i\bar\psi \not\partial \psi)^\dagger=\overline{i\bar\psi\not\partial\psi}=-i\bar\psi \bar{ \not\partial}\psi $$ so you've got to prove $\bar{\not\partial}=-\not\partial$ instead of $\not\partial^\dagger=-\not\partial$. Now, $$ \bar{\not\partial}=\overline{\gamma^\mu\partial_\mu}=\bar\gamma^\mu \partial_\mu^\dagger=-\gamma^\mu\partial_\mu=-\not\partial $$ where I used $\bar\gamma=\gamma$ and $\partial^\dagger=-\partial$.

I hope its more clear now. You should be able to fill in the details.