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) = 0Show 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.