Is the Dirac operator $i \not D$ or $\gamma^0 i \not D$ and nonabelian gauge field $A_\mu = A_\mu^\alpha T^\alpha$ hermitian? In quantum mechanics, it is common to write the momentum operator
$$P = i \partial_x.$$
It turns out that $p$ is hermitian although $i^\dagger = -i$ we also have $\partial_x ^ \dagger=-\partial_x$. It is a subtle issue to me to see $\partial_x ^ \dagger=-\partial_x$; I normally try to write a discrete matrix form of $\partial_x$ to understand $\partial_x ^ \dagger=-\partial_x$. It really requires us to define where are we taking the hermitian conjugate.
Let us consider Dirac operator acts on multiplet of matter fields with a generic nonabelian gauge field $A_\mu^\alpha T^\alpha$

*

*Is the Dirac operator with gauge field hermitian?

$$
i \not D = i \gamma^\mu (\partial_\mu - i g A_\mu)
$$
In the 4 dim Lorentz spacetime, the standard convention like in Peskin shows that $\gamma^{0  \dagger}=\gamma^{0}$ but $\gamma^{j \dagger}=-\gamma^{j}$. My question really requires us to define where are we taking the hermitian conjugate.


*Is the nonabelian gauge field $A_\mu = A_\mu^\alpha T^\alpha$ hermitian? It seems that $A_\mu^\alpha$ is always real (why is that?) and $T^{\alpha \dagger}=T^{\alpha}$. But why is that? Is there a proof on guaranteeing those properties?


*Now I see that as I am writing it, $i \not D$ is not hermitian, but $\gamma^0 i \not D $ is hermitian. Some explanations will be great --- it seems in contradiction to what we normally write Dirac lagrangian as $\bar{\psi} i \not D \psi$ instead of $\psi^\dagger (\gamma^0 i \not D) \psi$.
 A: *

*There seems to be some serious misunderstandings here about Hermiticity. That is a property of operators which act on the Hilbert space. In usual QM, we say $\partial_x^\dagger = - \partial_x$ because we are working in the ${\hat x}$ basis and states are described as wave-functions $\psi(x)$ and $\partial_x$ is a linear operator which acts on this wave-function (which is a state in the Hilbert space).
In this case, the Dirac field $\psi(x)$ is NOT a wave-function. It is an operator. $\partial_x$ in this context is NOT a linear operator on the Hilbert space, it is a operation which acts on operators. Therefore, it is not meaningful to talk about hermiticity of $\partial_x$. What you can do is ask if the operator $\partial_x \psi(x)$ is Hermitian or not and you can work this out using the limit definition of the derivative and it is easy to show that $(\partial_x \psi(x) )^\dagger = \partial_x \psi^\dagger(x)$.
In the same way, it is not meaningful to ask if $\not\!\!\!D$ is Hermitian because it is NOT operator which acts on wave-functions.


*You have to be careful about the gauge field $A_\mu = A_\mu^a T^a$ since it is a "matrix" of operators. So when you talk about $\dagger$ you have to distinguish between the adjoint operation for operators AND the conjugate transpose (conjugation acts on operators via the adjoint) operation which acts on matrices. Just so we can be super clear about everything, let me denote conjugate transpose by $CT$ and distinguish it from the adjoint $\dagger$. Then, we have
$$
(A_\mu)^{CT} = (A_\mu^a T^a)^{CT} = [ (A_\mu^a T^a)^\dagger ]^T = [ (A_\mu^a)^\dagger (T^a)^* ]^T = (A_\mu^a)^\dagger (T^a)^{CT} 
$$
The component gauge field $(A_\mu^a)^\dagger$ is Hermitian and it is often convenient to work in basis on the Lie algebra such that $(T^a)^{CT} = T^a$. In this basis it follows that
$$
(A_\mu)^{CT} = A_\mu^a T^a = A_\mu
$$
Thus, the correct statement is that $A_\mu^a$ is Hermitian whereas the matrix of operators $A_\mu$ is invariant under conjugate transpose. In many texts, you will see authors are a bit sloppy with notation and use $\dagger$ to denote BOTH the adjoint AND the conjugate transpose operation.
For OP's edification, let's work out the Hermitian conjugate of $\not\!\!D\psi$,
\begin{align}
[( \not \!\! D \psi )_i]^\dagger &= [\gamma^\mu_{ij} \partial_\mu \psi_j - i g A_\mu \psi_i ]^\dagger \\
&= [( \gamma^\mu_{ij} )^* \partial_\mu \psi_j^\dagger + i g A_\mu \psi_i^\dagger ] \\
&= [ \partial_\mu \psi_j^\dagger ( (\gamma^\mu)^{CT} )_{ji}  + i g A_\mu \psi_i^\dagger ] \\
&= [ \partial_\mu \psi^\dagger (\gamma^\mu)^{CT}  + i g A_\mu \psi^\dagger ]_i
\end{align}
Now, we use the fact that $(\gamma^\mu)^{CT} = \gamma^0 \gamma^\mu (\gamma^0)^{-1}$. Then,
\begin{align}
( \not \!\! D \psi )^\dagger &= [ \partial_\mu \psi^\dagger \gamma^0 \gamma^\mu  + i g A_\mu \psi^\dagger \gamma^0 ] (\gamma^0)^{-1} \\
&= [ \partial_\mu {\bar \psi} \gamma^\mu  + i g A_\mu {\bar \psi}  ] (\gamma^0)^{-1}
\end{align}
It is conventional notation to write
$$
 \partial_\mu {\bar \psi} \gamma^\mu  + i g A_\mu {\bar \psi} = {\bar \psi} \overleftarrow{\not\!\!\!D}
$$
We finally get
$$
( \not \!\! D \psi )^\dagger  = {\bar \psi} \overleftarrow{\not\!\!\!D} (\gamma^0)^{-1} .
$$
That's all you can say about Hermitian conjugation. As mentioned previously, you CANNOT ask if $\not\!\!D$ itself is a Hermitian operator or not. That question does not make sense.
