quantum mechanics - Nature of perturbed state in perturbation theory?

I'm interested in the “nature” of the perturbed state in perturbation theory in quantum physics.

The first order perturbed state is given by

$$ \psi^{(1)}_{n}=\Sigma_{m}a_{m}\psi^{(0)}_{m} ~, $$

where the perturbed state $\psi^{(1)}_{n}$ is expanded as a sum over unperturbed states $\psi^{(0)}_{m}$ (i.e., eigenstates of the unperturbed Hamiltonian).

Now as the Hamiltonian changes upon applying the perturbation, its basis should also change accordingly. My question is, why the basis does not change when we add a perturbation to the system? More precisely, what is the justification for expanding the perturbed state as a sum of unperturbed states?

Answer

Suppose that the Hamiltonian is written as $$ H = H^{(0)} + V $$ It is assumed that both $H^{(0)}$, the unperturbed hamiltonian, and $H$, the perturbed hamiltonian, are self-adjoint. In particular, there is an orthonormal basis $\psi^{(0)}_m$ for the Hilbert space consisting of eigenvectors of $H^{(0)}$, and a an orthonormal basis $\psi_m$ consisting of eigenvectors of $H$.

One can choose to write any state $\psi$ in either basis. In particular, one can choose to write the first order perturbed state in the basis $\psi^{(0)}_m$ of $H^{(0)}$ and then use this to derive lots of useful things.

Addendum. One thing I forgot to mention is why one would want to write the perturbed state in the unperturbed basis. The simple, essential answer is that perturbation theory is so useful precisely because the unperturbed Hamiltonian can be easily diagonalized; namely it's eigenvectors and eigenvalues are known exactly, whereas those of the unperturbed Hamiltonian are not.