Probably not, but I don't understand what your objection is either.
For a hydrogen atom the way the problem is usually presented is to solve the Schrodinger equation for a Coulomb potential [tex]V(r) \propto -1/r[/tex]. Doing this gives rise to a set of eigenstates and eigenenergies, the lowest of which corresponds to the 1s orbital. In my ignorance, I'd blithely assumed that this was the ground state, and that the reason why this state is stable is due to the fact that there is no lower energy state for it to decay to. Are you saying that this is incorrect?
Of course this is not the whole story since you also have corrections to the Hamiltonian due to things like relativity and spin-spin interactions between the electron and proton. But my understanding was that these will only perturb the eigenstates and eigenvalues slightly, and that you would still have a ground state even if you don't include these corrections.
I don't understand what you mean by the "effective potential of the Schrodinger equation" in this problem. Could you elaborate on this please?