Take the 2-minute tour ×
Physics Stack Exchange is a question and answer site for active researchers, academics and students of physics. It's 100% free, no registration required.

Adiabatic approximation or the Born-Oppenheimer approximation is used whenever the electronic motion is too fast that the electrons effectively see static nuclei and the nuclei, in turn, see an averaged electronic cloud. My question is:

  1. As the word 'adiabatic' suggests no heat/energy transfer, does this mean that the energies associated with nuclear and electronic motions are non-transferable?

  2. If my understanding is right, the limiting criterion of the approximation is when transfer can no longer be neglected. My other question is how to judge when the transfer is significant?

share|improve this question
add comment

1 Answer

  1. Not exactly, since the total wave function is not factored as $\psi_j(q)\phi(R)$ (where $q$ are the electronic coordinates and $R$ the nuclear coordinates) but as $\psi_j(q;R)\phi(R)$. This means that whenever $R$ changes the electronic wave function instantaneously adapts to remain in the same quantum state. That necessarily implies that the electronic energy must change as $E_j(R)$. Only the total energy and the electronic quantum number $j$ are conserved.

  2. Right: the Born-Oppenheimer approximation fails when, as $R$ changes, the probability of moving from $\psi_j$ to $\psi_k$ ($k\neq j$) is not negligible. Very roughly speaking (if you want I can provide you with the exact expressions) this occurs whenever the nuclear energy is of the order of $|E_k(R) - E_j(R)|$. Or, looking at it from the time-scale of motion, whenever the electronic time-scale associated to the electron transfer between the closest (coupled) electronic states is of the order of the time-scale associated to the nuclear motion.

share|improve this answer
add comment

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.