Derivative with respect to the nuclear coordinates in the Born–Oppenheimer approximation

Question

Reading few sources on the Born–Oppenheimer approximation I don't understand one particular thing.

If you look for example here (PDF, 70 KB) and focus attention on equations 14 and 15 than it is clear that

$$ \nabla_{A}^2 \left( \psi_k(\mathbf{r}; \mathbf{R}) \chi_k(\mathbf{R}) \right) = \psi_k(\mathbf{r}; \mathbf{R}) \nabla_{A}^2 \chi_k(\mathbf{r}; \mathbf{R}) + 2 \nabla_{A} \psi_k(\mathbf{r}; \mathbf{R}) \nabla_{A} \chi_k(\mathbf{r}; \mathbf{R}) + \chi_k(\mathbf{R}) \nabla_{A}^2 \psi_k(\mathbf{r}; \mathbf{R}) $$

where

$$ \nabla_A^2 = \frac{\partial^2}{\partial X_A^2} + \frac{\partial^2}{\partial Y_A^2} + \frac{\partial^2}{\partial Z_A^2} $$

and $$ \mathbf{R} = \{ \mathbf{R_i} \}_{i=1}^N = \{ (X_i, Y_i,Z_i) \}_{i=1}^N $$ is a set of all nuclear coordinates.

But honestly I don't undestand why is that so. The fact is that $\psi_k$ depends implicitly only on $\mathbf{r}$ and parametrically on $\mathbf{R}$ (that's why I think they are delimited by $;$ and not just $,$). As far as I know this parametric dependence means that for each set of nuclear coordinates $\mathbf{R}$ there is a complete set of electronic wave functions $\{ \psi_k(\mathbf{r}) \}_{k}$ which are functions of electronic coordinates only. And then of course when you differentiate $\psi_k(\mathbf{r}) \chi_k(\mathbf{R})$ twice with respect to $\mathbf{R_A}$ you get just $\psi_k(\mathbf{r}) \nabla_{A}^2 \chi_k(\mathbf{R})$ because $\psi_k(\mathbf{r})$ is constant with respect to $\mathbf{R}$.

And one more thing - the linked resource (and many others) claimed that the chain rule is used going from 14 to 15. I don't see any usage of chain rule but I see a usage of product rule.

Seems like I don't understand what's going on here but this is critical step because non-adiabatic coupling terms come from this expansion.

Answer 1

For each $\mathbf{R}$ there is a complete set of electronic functions $\{\psi_k(\mathbf{r};\mathbf{R})\}_k$, and what these functions are depends on the value of $\mathbf{R}$. As $\mathbf{R}$ is changed continuously, each element of $\{\psi_k(\mathbf{r};\mathbf{R})\}_k$ varies continuously; thus, it is meaningful to speak of the derivative of $\psi_k(\mathbf{r};\mathbf{R})$ with respect to the components of $\mathbf{R}$, and this derivative is generically not zero.

To make this more concrete, consider an infinite chain of atoms in one dimension, with nearest-neighbor distance $a$. We have $\mathbf{R}_n = R_n = na$ for $n$ integer, and one possible set of electronic basis wavefunctions is $\{\sin kx,\,\cos kx\mid k = 2\pi/ma,\,m\text{ integer}\}$. Now imagine varying the interatomic distance by increasing the value of $a$: clearly, each electronic basis function will change (their wavelengths $\lambda = \frac{2\pi}{k} = ma$ will all increase).