# Why can we not apply perturbation theory in Born-Oppenheimer approximation

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.

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