Find wave function satisfying Schroedinger Equation $[\frac{1}{2m}(\frac{\hbar}{i}\nabla-\frac{e} {c}A(r))^2+V(r)]\psi^{'}(r) = E\psi^{'}(r)$

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Given an "ordinary" wave function $\psi(r)$ that satisfies the "ordinary" stationary Schrodinger equation $[-\frac{\hbar^2}{2m}\Delta+V(r)]\psi(r)= E\psi(r)$, I want to construct a wave function $\psi^{'}(r)$ that satisfies the "electromagnetic" Schroedinger Equation $[\frac{1}{2m}(\frac{\hbar}{i}\nabla-\frac{e} {c}A(r))^2+V(r)]\psi^{'}(r) = E\psi^{'}(r)$, where $A(r)$ is a vector potential.

Okay so my approach is to choose a phase function $\alpha(r)$ such that $[-i\hbar\nabla- \frac{e}{c}A(r)]e^{i\alpha(r)}\psi(r)=e^{i\alpha(r)}[-i\hbar\nabla\psi(r)]$ But to be honest I don't really know how to handle the elctromagnetic Hamiltonian in Quantum Mechanics... What's the angle here?