Double sided arrow notation in Dirac Field Lagrangian

[TAKEDA Hiroki]

In a thesis, I found double sided arrow notation in the lagrangian of a Dirac field (lepton, quark etc) as follows.
\begin{equation}
L=\frac{1}{2}i\overline{\psi}\gamma^{\mu}\overset{\leftrightarrow}{D_{\mu}}\psi
\end{equation}
In the thesis, Double sided arrow is defined as follows.
\begin{equation}
A\overset{\leftrightarrow}{\partial_{\mu}}B:=A(\partial_{\mu} B)-(\partial_{\mu} A)B
\end{equation}
If covariant derivative is normal partial derivative $D_{\mu}=\partial_{\mu}$,both
$$L=\frac{1}{2}i\overline{\psi}\gamma^{\mu}\overset{\leftrightarrow}{\partial_{\mu}}\psi$$ and $$L=\frac{1}{2}i\overline{\psi}\gamma^{\mu}{\partial_{\mu}}\psi$$ can give the same physical results, because the difference between two expressions is a total derivative.
But, in a normal covariant derivative case, I can not understand if these two expression give the same physical results. For example, when a covariant derivative is given by
$$D_{\mu}=\partial_{\mu}+ieA_{\mu}$$
,this action is defined by
\begin{equation}
\overline{\psi}\gamma^{\mu}\overset{\leftrightarrow}{D_{\mu}}\psi=\overline{\psi}\gamma^{\mu}(D_{\mu}\psi)-(D_{\mu}\overline{\psi})\gamma^{\mu}\psi
\end{equation}
?
In that case, how this action $(D_{\mu}\overline{\psi})$ is defined?

 

[vanhees71]

The Lagrangian density is defined up to total 4D divergences, as you correctly wrote above. Thus your Lagrangian should be equivalent to
$$\mathcal{L}=\overline{\psi} (\mathrm{i} \mathrm{D}_{\mu} \gamma^{\mu}+m) \psi.$$
It's a slight advantage to have a symmetric form of the Lagrangian, and that's why sometimes you find the symmetrized version of the Lagrangian you quoted.

I've not done the calculation, but I think as noted it's indeed not clear what the 2nd term in (3) should be. I guess to clarify this you should try to reformulate it in terms of
$$\overline{D_{\mu} \gamma^{\mu} \psi}=(D_{\mu} \gamma^{\mu} \psi)^{\dagger} \gamma^0.$$