As I understand it, the main (or at least an important) statement of the Born-Oppenheimer Approximation is that the electronic and nuclear motion are separated and that the total wavefunction $\Psi$ is a product of the electronic wavefunction $\psi$ and the nuclear wavefunction $\chi$:
$\Psi = \psi\chi$
Usually, when this is derived in textbooks, one sees the "clamped-nuclei" electronic Schrödinger equation
$\left( T_e(r) + V(r,R) - W_n(R) \right) \psi_n(r,R) = 0$
where $W_n(R)$ is the electronic energy and the nuclear Schrödinger equation
$\left( T_N(R) + W_n(R) - E \right) \chi_n(R) = 0$
where the non-adiabatic coupling $\Lambda$ has been set to zero.
But how one gets from the above two to $\Psi=\psi\chi$ is not explicitly shown, just some explanation along the lines of uncoupling and separability. So either it must be very simple or quite difficult I guess.
I occurred to me that if you rewrite the two equation as follows:
$\left( T_e(r) + V(r,R) \right) \psi_n(r,R) = W_n(R) \psi_n(r,R)$
$T_N(R) \chi_n(R) = \left(E - W_n(R)\right)\chi_n(R)$
Then on the left we have the Hamiltonias, whose sum $T_e + V + T_N$ is the full Hamiltonian, and on the right we'd have the eigenvalues whose sum $W_n + E - W_n = E$ is the total energy. As both are sums, I believe this implies that $\Psi$ is the product of $\psi$ and $\chi$.
I wonder whether this is a valid way to arrive at the initial equation, and if not, whether there is any other way to express the first in terms of the others?