# Time independent perturbation of continuous spectrum

I'm reading Landau's QM book and finding this paragraph confusing. The first order correction solution to wavefunction reads

$$\psi^{(1)}=\sum_m \frac{V_{mn}}{E_n^{(0)}-E_m^{(0)}}\psi_m^{(0)}$$

which i have no problem to understand. However, Landau then said "the results can be generalized to the case where the operator $H_0$ (the unperturbed Hamiltonian) has a continuous spectrum, but the perturbation is applied to a state of the discrete spectrum as before. Then, for instance, we write

$$\psi^{(1)}=\sum_m \frac{V_{mn}}{E_n^{(0)}-E_m^{(0)}}\psi_m^{(0)}+\int \frac{V_{\nu n}}{E_n^{(0)}-E_\nu}\psi_{\nu}^{(0)}d\nu$$

What does it mean to say the perturbation is applied to a state in the discrete spectrum while $H_0$ has a continuous spectrum?

In my copy, it says "... can be generalized at once to the case where the operator $H_0$ has also a continuous spectrum ... ." (My emphasis.)

So he is considering a system where the spectrum has both discrete and continuous parts (e.g. the Coulomb system), and he is applying perturbation theory to one of the discrete states.

• This should probably be a new question (or I should just google better :P) but can you apply perturbation theory to one of the states in the continuous spectrum? I remember reading in Dirac's book that it is the precondition of the whole framework that each corrected eigenvalue (to be obtained by using perturbation theory) lies closest to a unique eigenvalue of the unperturbed Hamiltonian. This made sense to me technically but I never understood if there is a more fundamental reason as to why perturbation theory doesn't make sense for one of the states in the continuous spectrum.
– user87745
Apr 30, 2020 at 8:22