In Weinberg's Lectures on Quantum Mechanics, he mentions

Unfortunately, we cannot simply use first-order perturbation theory, with $T_{nuc}$ taken as the perturbation and the state vectors $\Phi_{a,X}$ taken as unperturbed energy eigenstates. This is because we are looking for discrete eigenvalues of the full Hamiltonian, for which the eigenvectors $\Psi$ would be normalizable, in the sense that $(\Psi,\Psi)$ is finite, while $(\Phi_{a,X},\Phi_{a,X})$ is infinite. We cannot expand in powers of a perturbation that converts a state vector with continuum normalization into one that is normalizable as a discrete state.

I do not get this. Why does perturbation theory fail here?

About the notations: follow from http://en.wikipedia.org/wiki/Born%E2%80%93Oppenheimer_approximation, $T_{nuc}$ is the kinetic energy of nucleus, i.e. $T_n$ in wiki. $\Psi$ in Weinberg is the same as $\Psi$ in wiki. $\Phi_{a,X}$ is $\chi_k (\mathbf{r}; \mathbf{R})$ in wiki, where $k\leftrightarrow a$ is the label of energy eigenstates.

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    $\begingroup$ Could you explain the notations ? What is $T_{nuc}$, $\Phi_{a,X}$ (and $a$ and $X$), etc. ? $\endgroup$
    – Adam
    Mar 15 '14 at 18:55
  • $\begingroup$ There are explanations of notations now in the question. $\endgroup$
    – Andy
    Mar 16 '14 at 7:36

Weinberg said that if we have a hamiltonian with continuous spectra (free hamiltonian), when we add an interaction, we cannot represent a bound state produced by this interaction with superposition of the free states with perturbative corrections. This happens because the free states have a singular norm (continuous norm), and this singularity is compensate by divergence of the perturbative series (Dyson's series), generating a state with finite norm, the bound state.


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