# Mathematical justification of the Born-Huang expansion in the derivation of the Born-Oppenheimer Approximation

In the book K. Huang and M. Born, Dynamical Theory of Crystal Lattices (1954, Appendix VIII) and also in the Wikipedia article https://en.wikipedia.org/wiki/Born%E2%80%93Oppenheimer_approximation#Derivation, the derivation of the Born-Oppenheimer approximation is given. In these derivations, the following expansion of the exact electron-nuclear wave function is used $$\begin{equation} \Psi\left(r,R\right) = \sum_{m} \phi_{m}\left(R\right) \chi_{m}\left(r,R\right). \label{eq:BornHuangExpansion} \end{equation}$$ Here, $$r$$ and $$R$$ refer to all electronic and nuclear variables, respectively. Further, $$\Psi$$ is the wave function satisfying the exact time-independent Schrödinger equation of the exact Hamiltonian of electrons and nuclei (comprising the kinetic energies and Coulombic interactions) and $$\chi_{m}$$ the wave function satisfying the Schrödinger equation for the exact Hamiltonian minus the nuclear kinetic energy. Please check some details from the Wikipedia article, if necessary.

I have not found any discussions from the literature in which this expansion is discussed in detail from a mathematical point of view. Does anyone see how to prove that this expansion is exact or see that it can't be exact?

Let $$H = H_e + T_n$$ Where $$H$$ is the full Hamiltonian and $$T_n$$ is the nuclear kinetic energy. Notice that the only dependence on $$R$$ in $$H_e$$ is through the position operator, i.e. there is no dependence on $$\frac{\partial}{\partial R}$$. This means we can treat $$H_e(R)$$ as a (Hermitian) operator on the space of functions of $$r$$ for a fixed value of $$R$$. Since $$H_e$$ is Hermitian it has, for any given value of $$R$$, a complete basis of eigenfunctions, $$\chi_m(r,R)$$ which can be used to write any other function of $$r$$. In particular they can be used to write the eigenfunctions of the full Hamiltonian, $$H$$, exactly as, $$\Psi(r,R)= \sum_m \phi_m(R)\chi_m(r,R)$$ Note that since both $$\Psi$$ and $$\chi_m$$ are parameterized by $$R$$, so is the basis coefficient $$\phi_m$$, but the $$r$$ dependence is entirely absorbed into the basis functions $$\chi_m$$
• Many thanks for your answer. I agree that ${\chi_{m}(r, R)}$ form a complete basis and can therefore be used to write any function of $r$, for any given $R$. However, I'm still not completely sure why these functions can be used to write the eigenfunctions of the full Hamiltonian, which are also functions of $R$. Could you please clarify this aspect a bit more? My confusion might be related to the fact that $\chi_{m}(r, R)$ are normalized with respect to $r$ and thus, as you state, $\phi_{m}$ are also functions of $R$. Oct 14, 2021 at 16:46
• The point is that $R$ is just a number (or vector), so you can pick your favorite value of $R$ and solve the problem treating $R$ as a constant. In the same way you can take $\Psi$, pick some value of $R$ and write it's dependence on $r$ in terms of whatever basis you like, say the basis of functions $\chi_m$ for that particular $R$. $R$ is treated simply as a parameter used to describe a family of functions of $\Psi(r)$ and a family of bases used to describe them. Since the function being expanded and the basis function both depend on $R$, the expansion coefficients must do as well. Oct 14, 2021 at 19:45
• @BySymmetry But we know that the solution of to the Schrodinger equation will have in general many solutions. In this form of the expansion if we know $\phi_1, \phi_2, \ldots, \phi_\infty$ and $\psi_1, \psi_2, \ldots, \psi_\infty$ then $\Psi_1$ will be the same as $\Psi_2$. Shouldn't be there some expansion coefficients (numerical constants) in order to obtain different $\Psi_1, \Psi_2$ etc? Jan 12, 2022 at 21:28
• @AntoniosSarikas $\phi_m$ is an ($R$ dependent) expansion coefficient and not a solution to the Schrodinger equation Jan 12, 2022 at 23:16