Let's think about a system that has a two-fold degeneracy for some given energy level. That is, two states $ \psi_{a} $ and $ \psi_{b} $, both of which correspond to energy $ E_{0} $. An example would be a spin-1/2 particle with a Hamiltonian that is spin-independent.
Now imagine that when we apply a perturbation, H', to the system, the degeneracy breaks into two distinct energy levels, $ E_{1} $ and $ E_{2}$.
The subtlety is that these two distinct energy levels do not necessarily correspond to the two degenerate states (it is not necessarily the case that $ H'\psi_{a} = E_{1}\psi_{a} $ and $ H'\psi_{b} = E_{2}\psi_{b} $). It is possible that the two distinct perturbed energies correspond to linear combinations of the two degenerate states, i.e. $ H'(\alpha \psi_{a} + \beta \psi_{b}) = E_{1}(\alpha \psi_{a} + \beta \psi_{b}) $, and similarly for some other linear combination (orthogonal to the first).
This subtlety is the reason that in degenerate perturbation theory, the first-order corrections to the energy require calculation of off-the-diagonal elements, i.e. things that look like $ \langle \psi_{a} | H' | \psi_{b} \rangle $. This is annoying, because in first order perturbation theory, we only needed to calculate one inner product. In degenerate perturbation theory, we have to compute a whole matrix of inner products to calculate the correction to the energy.
Since it is tedious to compute inner products, it'd be nice to know a trick to find out if the off-the-diagonal elements of the perturbing Hamiltonian are 0. The trick is given and proven on pg. 259-260 of Griffiths.
In the case of your question, the original Hamiltonian is spherically symmetric (Coulomb potential has no angular dependence). Also, the the perturbing Hamiltonian is spherically symmetric. If you look at the form of the angular momentum operator, you will note that it only involves things with $ \theta $ and $ \phi$ (derivatives and cosines and stuff). The nice thing about that, is that if I have a Hamiltonian that is purely radial (depends only on r), than it definitely commutes with L.
All Griffiths is doing is showing that = the conditions of the theorem on pg. 259 are satisfied, which shows that degenerate perturbation theory collapses to non-degenerate perturbation theory.
I'm not familiar with the idea of states being "connected" by a Hamiltonian. I would speculate that two states are connected by a Hamiltonian if the expected value of one state is nonzero given that it is initially in the other state.
Hope this helps!
I think what is going on here is that "old-fashioned" perturbation theory can make the power counting of higher-order terms manifest. This is similar to the use of effective field theories like NRQED instead of QED to make the power counting manifest.
The appearance of bound states at weak coupling (as in the Coulomb problem) means that the suppression of higher-order diagrams by $\alpha$ must be compensated by an IR enhancement involving inverse powers of $m_e/|\mathbf{q}|$. In order to see this we would like to have an a-prior estimate ("power counting") of diagrams in terms of $m_e/|\mathbf{q}|$.
Some guidance is provided by the fact that we already know that these diagrams are summed by the Schrodinger equation. If we can find a perturbative scheme that corresponds to solving the Schroedinger (Lippmann-Schwinger) equation by iterating diagrams, then the power counting should be manifest. This is what old-fashioned perturbation theory does. By isolating the pieces that correspond to the non-relativistic limit, we can isolate the terms that are summed by the Schroedinger equation, and indeed, we can estimate the magnitude of diagrams without explicitly computing them, and find the $m_e/|\mathbf{q}|$ enhancement.
Of course, fully covariant perturbation theory can be used to compute the same diagrams, but the $m_e/|\mathbf{q}|$ is not manifest (you can't see it without an explicit computation), and there is no straightforward way to sum the $(e^2m_e/|\mathbf{q}|)^n$ terms.
Best Answer
The scattering states must be included in the perturbative calculations if the result is to be highly accurate. In particular, it is not justified to ignore the continuous spectrum at energies close to the dissociation threshold.
The Hilbert space in the position representation is the space of square integrable functions on $R^3\setminus\{0\}$ with respect to the inner product $$\langle\phi|\psi\rangle:=\int \frac{dx}{|x|}\phi(x)^*\psi(x).$$ The bound states alone are not dense in this space.
For thorough treatments of the hydrogen spectrum see the books
G R. Gilmore, Lie Groups, Lie Algebras & Some of Their Applications, Wiley 1974, Dover, 2002. pp. 427-430
and
A.O. Barut and R. Raczka, Theory of group representations and applications, 2nd. ed., Warzawa 1980. Chapter 12.2.