Physics Stack Exchange is a question and answer site for active researchers, academics and students of physics. Join them; it only takes a minute:

Sign up
Here's how it works:
  1. Anybody can ask a question
  2. Anybody can answer
  3. The best answers are voted up and rise to the top

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

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


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

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.

share|cite|improve this question
up vote 4 down vote accepted

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).

share|cite|improve this answer
Wait a minute! I think I got it! – Wildcat Mar 7 '13 at 9:50

Your Answer


By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.