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?


1 Answer 1


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 $\chi_m$

  • $\begingroup$ 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 $R$. $\endgroup$
    – user28157
    Commented Oct 14, 2021 at 16:46
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    $\begingroup$ 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. $\endgroup$ Commented Oct 14, 2021 at 19:45
  • $\begingroup$ @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? $\endgroup$ Commented Jan 12, 2022 at 21:28
  • $\begingroup$ @AntoniosSarikas $\phi_m$ is an ($R$ dependent) expansion coefficient and not a solution to the Schrodinger equation $\endgroup$ Commented Jan 12, 2022 at 23:16
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    $\begingroup$ @AntoniosSarikas You could, if you are interested in doing something with the full set of solutions. If you are just interested in some generic solution or the ground state then I wouldn't bother. I just used the notation in the question $\endgroup$ Commented Jan 14, 2022 at 13:18

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