In a course that I follow, we use the perturbation method to find the eigenvectors and energies to an Hamiltonian written $ H_0 + V $ where $V$ is a weak perturbation.  It is written that as $V$ is a weak perturbation, we can write the perturbed eigenvectors as a combination of the eigenvectors of the unperturbed hamiltonian.  But I thought that as we know the eigenvectors of the unperturbed Hamiltonian, then we have a basis of the Hilbert space so there is no need of this assumption (we always can write any vector as a linear combination of eigenvectors of the unperturbed hamiltonian).  Can you help me to understand this?
I have been wondering why anyone ever thought that we could find an expansion for eigenstates of some perturbed Hamiltonian in terms fo those for the basis Hamiltonian. My lecturer insisted that this was because the eigenstates of the Hamiltonian are complete', however from my (limited) understanding of the mathematical basis of QM:   When you are given a Hamiltonian, it comes with the Hilbert space of functions of which it acts (subject to some conditions we impose, such as the Lebesgue square-integrability of functions) When you are given another Hamiltonian, there is no guarantee that it will act on the same Hilbert space Therefore, that the zeroth order Hamiltonian has eigenstates forming a complete basis for the space on which ti acts does not guarantee that these same eignestates form a complete basis for the Hilbert space on which the purturbed Hamiltonian acts.   As a simple example. I can imagine adding a small perturbing magnetic field to a Hamiltonian in which there initially is none, such that we lose the degeneracy in the eigenstates accounting for different spins of a particle. One could say that we have the same Hilbert space before and after but we lose degeneracy; or I would have been tempted to say that our initial Hilbert space did not have these extra dimensions corresponding to the spin states. So rather than the eigenstates of the basis Hamiltonian also forming a complete basis for the purtubed Hamiltonian, but having eigenvalue degeneracy which is lifted on the purtubration, I would have said that our Hilbert space has changed. After all, if we take the former view (i.e. a basis accouting for spin but with degeneracy) then it seems like for any problem we would need to know and account for EVERY possible system property that can be discerned. This doesn't seem reasonable to me....  NOTE:  1) My question was asked here previously:   although there is only a discussion and no conclusive answer given. I see that the distinction is important in relativistic QM, and the case of adding a potential onto a free particle Hamiltonian seems to be a special one. Although I am still not sure what the answer is for general purturbations.  2) Similar question here:   but it seems to completely avoid answering the question.
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