Solving the gauged Dirac equation perturbatively

[Thoros]

Homework Statement

Given the gauge invariant Dirac equation$$(i\hbar \gamma^\mu D_{\mu} - mc)\psi(x, A) = 0$$Show that the following holds: $$\psi(x, A - \frac{\hbar}{e} \partial\alpha) = e^{i\alpha}\psi(x, A)$$

Homework Equations

The covariant derivative is $$D_\mu = \partial_{\mu} + i\frac{e}{\hbar} A$$ And the Dirac equation expanded gives $$(i\hbar \gamma^\mu \partial_{\mu} - mc)\psi(x, A) = e\gamma^\mu A_{\mu}(x)\psi(x, A)$$
The free field Feynman propagator $$S_{F}(x-x')$$ satisfies $$(i\hbar \gamma^\mu \partial_{\mu} - mc)S_{F}(x-x') = i\hbar\delta^{(4)}(x - x')$$

The Attempt at a Solution

So i separate the hamiltonian density of the system into the unperturbed and the interaction terms $$\mathcal{H} = \mathcal{H}_{0} + \mathcal{H}_{interaction}$$
giving $$\mathcal{H}_{0} = c\overline{\psi}(-i\hbar \gamma^\mu \partial_{\mu} + mc)\psi$$
and $$\mathcal{H}_{int} = e\overline{\psi}(\gamma^\mu A_{\mu})\psi$$

All i can quess now is that the interaction term should give a contribution to a perturbative series, but i fail to see and accomplish this. Also, where does the free field Feynman propagator come into play?

 

[Oxvillian]

Are you sure the question involves perturbation theory?

Surely all you have to do is show that the wavefunction
$$e^{i\alpha}\psi$$
satisfies the Dirac equation with the replacement
$$A_{\mu} → A_{\mu} - \frac{\hbar}{e} \partial_{\mu}\alpha.$$
That's the "gauge symmetry" of the equation - a gauge transformation on the vector potential induces a change of phase in ψ.

 

[Thoros]

Are you sure the question involves perturbation theory?

Good point, i think the word "iteratively" was used by my professor. But i discussed it with others and they were also confused about the perturbative/iterative part. As of now, i still have no real progress.

Edit:
I must thank you for brining this up. I ignored that part of the question and showed it by just applying the U(1) symmetry to the equation with the covariant derivative. However, i ended up with an opposite sign. I'm just going to take it as a sign mistake for now.