Does the Born-Oppenheimer (adiabatic) approximation Ansatz allow excitations? 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?
 A: 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.
