When imposing local gauge invariance for a simple lagrangian describing a free Dirac fermion field: $$\mathcal{L}=\bar{\psi}\left(i\gamma^{\mu}\partial_{\mu}-m\right)\psi$$
Is there a particular reason for choosing the local gauge transformations to be: $$\psi\rightarrow \psi'=e^{-i\alpha(x)}\psi'$$ $$\bar{\psi}\rightarrow \bar{\psi'}=\bar{\psi'}e^{i\alpha(x)}$$ and not: $$\psi\rightarrow \psi'=e^{i\alpha(x)}\psi'$$ $$\bar{\psi}\rightarrow \bar{\psi'}=\bar{\psi'}e^{-i\alpha(x)}$$
I encountered the first convention more often, and I am wondering whether there is a physical or mathematical reason for that, or just a pure convention. Since it is trivial to see that no matter what convention we use, the symmetry will not be broken. $$e^{-i\alpha(x)}\cdot e^{i\alpha(x)} = e^{i\alpha(x)}\cdot e^{-i\alpha(x)} =1$$
Edit:
- Dirac adjoint transformation fixed
- Sorry for not being clear. I was not suggesting that this lagrangian is gauge invariant. In order to maintain the gauge symmetry, the ordinary derivative has to be replaced by the covariant derivative $D_{\mu}$. $$D_{\mu}:=\partial_{\mu}+iqA_{\mu}$$
This leads to the introduction of the field $A_{\mu}$, and the new lagrangian will have by construction an interaction term between the Dirac field (that was initially considered) and the scalar field $A_{\mu}$.
I was really just thinking only at the form of the gauge transformations (the $\pm$ convention). I could as well picked the global invariance case, where there is no need of introducing the covariant derivative, because the lagrangian already has global gauge symmetry.