Born-Oppenheimer approximation, electronic/nuclear wavefunction and product form

Question

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?

Answer 1

You are correct, although the argument is usually done in the reverse direction, i.e. you would write out the full electronic plus nuclear Hamiltonian first, dropping the coupling terms. This is what you have done in the next-to-last paragraph.

This makes the nuclear and electronic degrees of freedom independent; then assume the solution is separable, i.e. assume $\Psi=\psi\,\chi$, and solve by separation of variables. The result of this last step are the two separate equations you have for the nuclear and electronic degrees of freedom. You then put back the coupling term as a perturbation.

Answer 2

$\Psi = \psi\chi$ is not derived, it's an Ansatz. Consider the Schrödinger equation with the molecular Hamiltonian $H$, $$i \hbar \Psi = H \Psi$$ Often, in Physics literature one makes the Ansatz that the total wavefunction is a product of a nuclear and electronic wavefunctions: $\Psi=\psi \chi$. This is of course physics-motivated and simplifies finding solutions. However, this is not an accurate description of what we are doing (hence you are confused). An accurate description is as follow. We are looking for a solution $u$ that approximates the solution $\Psi$ and $u=\psi \chi$. Mathematically, this correspond to looking for solutions on the manifold: $$\mathcal{M} = \{u \in L^2 (\mathbb{R}^{3N} \times \mathbb{R}^{3L}) : u=\psi \chi \}$$ where $N, L$ are the numbers of atoms and electrons respectively. See e.g. From Quantum to Classical Molecular Dynamics by Christian Lubich. Such an approximate solution is easier to find. An important step is also to characterise the error made by such an Ansatz, i.e. how far $u$ from $\Psi$ is.