# Scattering Greens function exactly at energy of bound state

I have a small bit of confusion about the expansion I am seeing in literature for the Greens function in time independent scattering theory. For example here is an excerpt from Scattering Theory of Waves and Particles by R.G. Newton:

My question is quite simple. Why in equation 7.26 is there no $$\pm i\epsilon$$ in the denominator in the first term, i.e. the discrete sum over bound states. I understand that for most energies the $$\epsilon$$ in this term may be simply ignored, but what if $$E$$ is exactly equal to some $$E_n$$, then we have a pole at this energy right? So does this equation have some implicit assumption that we do not take $$E$$ exactly at the bound state energy? Or is there some other reason?

Two different situations are possible here, depending on whether one deals with eigenstates or a scattering problem. The former seems to be the case here - i.e., the spectrum of a Hamiltonian contains both bound and extended states. The examples of this a ubiquitous - a hydrogen atom has bound states for energies $$E<0$$ and extended for $$E>0$$. An easily solvable toy example is a negative delta-function potential. Since these are eigenstates of a Hamiltonian, there is no mixing between them (i.e. their eigenfunctions are mutually orthogonal). This would also hold, if a bound state lies within a continuum spectrum, although it is hard to come up with a realistic example.