The way I understand the Born-Oppenheimer approximation is as follows:

In the general solid Hamiltonian for $N$ electrons at $3N$ coordinates $r$ and $M$ ions at $3M$ coordinates $R$,

$$ H = T_e + T_I + V_{ee} + V_{II} + V_{eI} = H_0 + T_I\,, $$

with kinetic energies of electrons $T_e$ and ions $T_I$ and interaction energies of ions with ions $V_{II}$, electrons with electrons $V_{ee}$ and electrons with ions $V_{eI}$, we consider the ions to be fixed at $R$ and pretend to have solved

$$ H_0\phi_{Rn}(r) = \epsilon_{Rn}\phi_{Rn}(r)\,, $$

for all parameter sets $R$, where $\phi_{Rn}(r)$ is the electronic wave function and $n$ numbers different possible electronic eigenstates, i.e. the electronic ground state $n=0$ and electronic excitations $n>0$. Then we assume we can expand the full solid wave function for electrons and ions $\psi(r,R)$ in the functions $\phi_{Rn}(r)$, because the latter form a basis of $r$-space for each fixed $R$,

$$ \psi(r,R) = \sum_n \chi_n(R)\phi_{Rn}(r)\,. $$

Using this Ansatz in the full solid eigenvalue equation

$$ H\psi(r,R) = E\psi(r,R)\,, $$

and performing the actual adiabatic approximation leads to the equation

$$ (T_I + \epsilon_{Rn})\chi_n(R) = E\chi_n(R)\,. $$

This sure looks like an eigenvalue equation for ionic wave functions $\chi_n(R)$ in the effective potential $\epsilon_{Rn}$, where $n$ now also numbers different possible ionic eigenstates, and most authors seem to interpret it just like that.

I am confused by what this means for the full system wavefunction $\psi(r,R)$. If the Ansatz would look like

$$ \psi_n(r,R) \stackrel{?}{=} \chi_n(R)\phi_{Rn}(r)\,, $$

I would happily interpret it as meaning that, as part of our approximation, electrons and ions like to be excited together and the total solid is in an excited state whenever electrons and ions are both excited. However, since we sum over $n$ in the Ansatz, it looks to me like even though electrons and ions can be excited separately, the full solid does not have excited states (which can not possibly be true, can it?).

Is the interpretation of $\chi_n(R)$ as ionic wave function flawed? Am I missing something?

  • $\begingroup$ One-particle wave functions are not really the ground state: even if we treat ions as a uniform charged background, interacting electron gas is still a very difficult problem. $\endgroup$
    – Roger V.
    Jul 12, 2021 at 20:35
  • $\begingroup$ There is no one-particle function here though, just N-electron functions, M-ion functions and (N+M)-particle functions. $\endgroup$
    – user301288
    Jul 13, 2021 at 10:59
  • $\begingroup$ You mean $r$ are coordinates of all the electrons together? $\endgroup$
    – Roger V.
    Jul 13, 2021 at 11:09
  • $\begingroup$ Yes, for N electrons and M ions in 3D space r are 3N coordinates and R are 3M coordinates. $\endgroup$
    – user301288
    Jul 13, 2021 at 15:35

1 Answer 1


You do get a second index for the solid eigenfunctions when you solve the Eigenvalue problem $$ (T_I + \epsilon_{Rn})\chi_n(R) = E\chi_n(R)\, . $$ yielding eigenfunctions and values $E_{nm}$, $\varphi_{nm}(R)$. The expansion of the full wavefunction would then be $$ \psi(r,R) = \sum_{nm}c_{nm}\varphi_{nm}(R)\phi_{Rn}(r) $$

Typically, we assume that a system at room temperature is in its groundstate $$ \psi_{GS}(r,R) = \varphi_{00}(R)\phi_{R0}(r) $$ This state would be in its electronic and nuclear groundstate. Excitation then induce changes in the quantum numbers. Transitions that change the electronic state quantum number $n$ are considered electronic transitions. Transitions that change $n$ and $m$ are typically called vibronic transitions and transitions where $n$ is fixed and only $m$ changs are typically considered vibrational transitions.

I also assume that you mean nuclei when you say ions. An ion has lost an electron and is typically considered to be no longer interacting with the lost electron. So in the ionized case, the electron is no longer be part of the Hamiltonian, but here we consider the interaction of the nuclei with bound electrons. And the function that you call ionic wavefunction is typically called nuclear wavefunction.

  • $\begingroup$ Thanks for your answer! I consider a nucleus to be a special case of an ion. The N interacting electrons are then the electrons "lost" by the original atoms making up the solid (read weakly bound electrons) and the electron-ion-interaction is then the energy of N weakly bound electrons in the potential of M nuclei screened by the rest of the original atoms electrons, which are strongly bound and localized to their respective nucleus. $\endgroup$
    – user301288
    Jul 13, 2021 at 11:17
  • $\begingroup$ @quappas Do you have single molecules in mind or is your question based on crystalls/solids with translation symmetry ? My answer is written with a single molecule in mind. $\endgroup$
    – Hans Wurst
    Jul 15, 2021 at 14:18

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