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I've been trying to learn quantum chemistry at an introductionary level. While reading I've found out that the Born-Oppenheimer approximation seems to be the reason for the basic and crucial model of molecules for chemists; molecular structure in space as nuclei occupying some positions with chemical bonds between them. Thus, I've tried to learn more about this approximation. However, I've gotten stuck at the reasoning for wanting to try to separate the wavefunction of the molecule into a product of two functions, one depending on the positions of the nuclei and one depending on the positions of the electrons as well as depending parametrically on the positions of the nuclei. All resources state that the reasoning is because nuclei move considerably slower than electrons, but I have not found one that could elaborate why this is the case.

Most books and other resources simply states that "Because the nuclei is over thousand times heavier than an electron, in the timescale of the movements of the electron the nuclei can be considered to be standing still.", or "Because of the great difference in mass, the electrons can respond almost instantaneously to the movements of the nuclei", and then quickly moves on. However, in this answer on "Born–Oppenheimer adiabaticity" it was said that "nuclei at average are moving considerably slower than electrons" because of the equipartition theorem.

In the book Introduction to Statistical Physics by S. R. A. Salinas, the theory of equipartition of energy is formulated by:

Consider a classical system of $n$ degrees of freedom given by the Hamiltonian $$H = H_0 + \phi p_j^2$$ where (i) $H_0$ and $\phi$ are functions that do not depend on the particular coordinate $p_j$; (ii) the function $\phi$ is always positive; and (iii) the coordinate $p_j$ is defined from $-\infty$ to $+\infty$.

It then shows how the expectation value of $\phi p_j^2$ equals $\frac{1}{2}k_BT$, where $k_B$ is the Boltzmann constant and $T$ temperature. Thus, we can conclude that for any system (in thermal equilibrium) that has energy contributions that are quadratic in such a variable, the average energy is $\frac{1}{2}k_BT$ for each of those degrees of freedom. Since kinetic energy is always quadratic in the momentum, it will have an average energy of $\frac{1}{2}k_BT$ in thermal equilibrium in each degree of freedom if the rest of the Hamiltonian does not depend on the momentum (which is the case for the Hamiltonian of a molecule). This elaborates the reasoning in the linked answer. However, this seems to "only" apply for macroscopic systems of molecules which also have to be in thermal equilibrium. I am not completely satisfied with this, especially considering when this is only about the average kinetic energy. I would assume that the distribution of kinetic energy for the different degrees of freedom in a general system at room temperature is such that a decent portion are thousands of times higher than others. I could not find any answer that rectified this.

One book I read mentioned something different than others; that the nuclei move considerably slower than electrons because of conservation of momentum. However, it gave no elaboration. One text I remember reading somewhere on a website said something along the lines of: "The electrostatic forces on nuclei and electrons from their pairwise interaction are the same, and so the changes in their momenta from these forces must be the same. Therefore, one might assume that the actual momenta of the nuclei and electrons are of similar magnitude." This is the explanation for the reasoning that makes the most sense to me, but I have not seen it anywhere else. Since it came from a source that wasn't necessarily trustworthy it made me unsure, until I saw the statement of conservation of momentum which felt related.

Does the statement of equipartition make sense as the explanation for the reasoning? If yes, why? Does the statement of changes in momenta make sense as the explanation for the reasoning? Or maybe there is something different that is the sensible explanation? Please do help me understand more clearly, I've been at this for weeks and I cannot seem to get any answer from books and other texts.

Atsjo
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    +1 just for "trying to learn quantum chemistry" :-) Welcome to Stack Exchange! – uhoh Dec 30 '22 at 03:04
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    As kinetic energy is proportional to mass and velocity as $mv^2/2$ with the same energy the heavier particle moves more slowly. – porphyrin Dec 30 '22 at 14:35
  • If you consider the simplest atom, the mass of a proton is aproximately 1836 times the mass of an electron, so if both particles have the same kinetic energy, the ratio of their speeds is $v_e = \sqrt{1836} v_e$. Considering that in a stationary state both particles will have a constant energy and that you can partition it between as E = K + V, the electron responds faster to a change in the fraction of potential energy. – PAEP Dec 30 '22 at 18:47
  • Considering zero total momentum of the atom, $\sum{m_i \vec v_i} = 0$ – Poutnik Dec 30 '22 at 20:25
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    @uhoh Thank you. (: – Atsjo Dec 31 '22 at 13:07
  • @porphyrin I understand that part, but how do you know nuclei have the same kinetic energy as electrons? – Atsjo Dec 31 '22 at 13:09
  • If you hit a ping-pong ball with a paddle, and if you hit a billiard ball, just as hard with that paddle, imparting the same amount of energy, which ball will move faster? – DrMoishe Pippik Jan 01 '23 at 03:38
  • @Atsjo, equipartition of energy – porphyrin Jan 01 '23 at 12:15

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