5
$\begingroup$

In the discussion of the Born Oppenheimer approximation, we think about the electron eigenstates as a function of fixed nuclear positions $R$. It is then assumed that the nuclear positions vary slowly, so that the electron dynamics can be considered adiabatically following the eigenstates $|\psi(R)\rangle$.

My question is, why is it justified to assume the nuclear coordinates fixed? They are subject to quantum fluctuations. When we consider a diatomic molecule for example, we can approximate the ground state of the nuclei as a rigid rotator. The nuclear coordinates are completely delocalized in the angular dimension, even if the modulus of $R$ has small fluctuations.

$\endgroup$

3 Answers 3

1
$\begingroup$

For any system of $N$ particles, the state with zero angular momentum is invariant with respect to rotations of the reference frame. This means that after you decouple the angular coordinates from the internal ones, your angular part of the wavefunction will be simply constant (which is the reason for smearing of the absolute positions of the nuclei). The remaining part will have the nuclei well-localized near their equilibrium positions—in terms of interparticle distances, not absolute positions.

$\endgroup$
1
$\begingroup$

I think I found a partial answer to this. In fact, not all coordinates of the nuclei are assumed fixed, only the modulus of their relative positions. This is justified by the empirical fact that molecules form stable bonds.

The issue is thus the influence of the absolute orientation of the molecule on the electronic wavefunctions. A similar thing happens for the hydrogen atom: The center of mass is described by a plane wave, hence the wavefunction is completely delocalized. However, when calculating the wavefunction of the relative coordinate we assume the proton to be located at a specific point.

$\endgroup$
0
$\begingroup$

The point is that nuclei are much more massive than electrons. That means that for the electronic problem we can assume (this why it is an approximation) the nuclei to be still with respect to the electrons, so that their positions enter as parameters in the electronic problem. Once you have solved this problem, you can turn to find out how nuclei move.

You can fin here a concise description.

$\endgroup$
2
  • $\begingroup$ The issue is not wheter nuclei are slow, but wheter they are localized. You say the position of the nuclei is a parameter. This only works when they are localized. In your link the author also claims that "Another consequence of the mass difference (...) is that the nuclear components of the wave function are spatially more localized than the electronic (...) wave function". This is just not true in the case of a diatomic molecule. Here the nuclei coordinates are delocalized in the angular dimension. Which value of R do you want to take here as your parameter? $\endgroup$
    – curio
    Mar 28, 2021 at 11:45
  • $\begingroup$ I'm sorry, I forgot to answer at the time and it came to me just yesterday. It's true that in general nuclei aren't fixed, but it seems to me that your question went a bit beyond the standard Born-Oppenheimer approximation. The way I see it, the approximation $assumes$ the nuclei are fixed on the basis of the mass difference and because it neglects pure ionic effects since it is usually used to compute the electronic configuration. $\endgroup$ Apr 21, 2021 at 9:29

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge that you have read and understand our privacy policy and code of conduct.