Vibrational energy in molecules

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Hello! I am not sure how to visualize (ideally in a simple, classical way) the vibrational energy levels of a molecule. For the electronic levels (similar to atoms), I usually think of it as the atom having multiple energy levels (fixed in space, in a probability distribution sense), and when the atom gains energy, the electron goes from one shell to another (of course this is a simplified view, but it is useful to visualize what is going on). In the case of vibrational energy, I imagined that when the molecule gains energy, the 2 atoms (say it is a diatomic molecule) start to vibrate like a spring, and the energy levels are like these of a harmonic oscillator. But from what I read about this topic, I have noticed that everywhere people talk about the electron transition between two vibrational states and the detected photon corresponds to the energy difference between these 2 states. So initially I thought that this vibrational energy goes into the nuclei, making them move around an equilibrium point. But it seems that the energy goes to the electron not to the nuclei. And I am not sure what is vibrating anymore. Of course quantum mechanically we have a many-body system, containing both the nuclei and electrons and one in principle solves the Schrodinger equation for everything at once (or within the Born-Oppenheimer approximation). But I was wondering if there is a simple way of seeing what is going on and what is vibrating, the nuclei or the electron. And if it is the electron what does it vibrate around? Also, are the nuclei moving at all? And if so, where is that energy i.e. why are we detecting only the transition energy of the electron? Can someone clarify this for me please? Thank you!
 

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  • #2
TeethWhitener
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Can you give a reference? At least in the Born-Oppenheimer approximation, it’s assumed that electrons react more or less instantaneously to changes in the nuclear positions. This gives a vibrational potential energy surface that is a function only of the nuclear positions.

There are certainly situations where the BO approximation breaks down and non-adiabatic effects have to be considered. Is that what you’re talking about?
 
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Can you give a reference? At least in the Born-Oppenheimer approximation, it’s assumed that electrons react more or less instantaneously to changes in the nuclear positions. This gives a vibrational potential energy surface that is a function only of the nuclear positions.

There are certainly situations where the BO approximation breaks down and non-adiabatic effects have to be considered. Is that what you’re talking about?
Thank you for your reply! I mentioned BO just for completeness. My questions is actually about the most basic approximation one can make in describing the system (not sure if that is BO), so definitely no adiabatic effect that I am interested about. My main question is, in a vibrational excitation, where is the energy going: to the electron, the nuclei or both? And when we measure the emitted photon, is it coming from the electron changing its state? Or from the nuclei? Are the electron vibrating around the nuclei, or they are fixed relative to the nuclei and they vibrate at the same time with the nuclei (I think that is what you implied in your reply?). Basically, I am not sure what is vibrating and what not in the molecule.
 
  • #4
TeethWhitener
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My questions is actually about the most basic approximation one can make in describing the system
I can’t give you a precise answer without a precise formulation of the question. When a molecule vibrates, everything moves. In the BO approximation, it’s assumed that the electrons “track” the nuclear motion perfectly, so to speak; that is, the electrons adjust to the nuclear positions instantaneously. Regardless of whether the BO approximation holds, the electrons will respond rather quickly to changes in nuclear position.
 
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Have stopped to think about where the energy goes in a purely classical case of vibrational excitation. Into the spring? Into the velocity of the masses.? Both of course. So the answer that the vibrationally excited quantum states also "share" the energy and the state is comprises both electronic and nucleus information. There are of course various approximations.
 
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Have stopped to think about where the energy goes in a purely classical case of vibrational excitation. Into the spring? Into the velocity of the masses.? Both of course. So the answer that the vibrationally excited quantum states also "share" the energy and the state is comprises both electronic and nucleus information. There are of course various approximations.
Thank you for this! So when an excited molecule de-excites from a vibrational state, the energy come from both the nuclei and electrons energy simultaneously? So when we talk about the energy levels of a molecule, we don't talk about electrons energy level anymore, as in the case of atoms, but energy levels of the whole molecule?
 
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TeethWhitener
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So when we talk about the energy levels of a molecule, we don't talk about electrons energy level anymore, as in the case of atoms, but energy levels of the whole molecule?
Not really. You can have electronically excited molecules, just as you can have electronically excited atoms. Also, in a real molecule, the excitation of an electron into a higher energy level will change the vibrational state of a molecule as well (google Franck-Condon factor).
 
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The issue here for you is (I think) hierarchy. Relative to the electronic energies (and "speeds" if you will) the molecular vibrations are low and slow.
This allows the adiabatic approximation where the electrons stay in the ground state

https://en.wikipedia.org/wiki/Adiabatic_theorem#The_Landau–Zener_form

Remember that the electrons are still crucial to the process because they supply the electrical neutrality that creates the attractive interatomic potential. But they are not actively exchanging energy quanta because of the above considerations: they are just the "springs". Of course, as @TeethWhitener says, if they become excited the spring constant changes and we have new interesting phenomena.
 
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TeethWhitener
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Basically, the takeaway is that if you have a system of charged particles, you can’t expect to move a subset of them and not the rest of them.
 
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Basically, the takeaway is that if you have a system of charged particles, you can’t expect to move a subset of them and not the rest of them.
Thank you for this! That is what I would normally expect, but this is not what I understand from most papers I read. For example in this paper (I can search for more, this is the first one that came to my mind) they measure the vibrational levels of a molecule. And they refer to that as "vibronic bands of the four electronic transitions". It might be just a wording issue, but from here I get that the vibronic bands, whose energies they measure, come entirely from the electron transition. I see no mention in the paper about the nuclei, or any sort of energy stored in the nuclear motion. All the energy seems to come from the transitions of the electrons. So this is what confuses me a bit. Why do they talk about electrons only when measuring energy and not about nuclei, too?
 
  • #11
TeethWhitener
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Vibronic transitions are specifically instances where the Born-Oppenheimer approximation breaks down and nuclear and electronic motions cannot be treated separately. The word "vibronic" is a portmanteau of "vibrational" and "electronic."
 
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Vibronic transitions are specifically instances where the Born-Oppenheimer approximation breaks down and nuclear and electronic motions cannot be treated separately. The word "vibronic" is a portmanteau of "vibrational" and "electronic."
Oh so the term vibronic automatically implies motion of both electron and nuclei?
 
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TeethWhitener
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Oh so the term vibronic automatically implies motion of both electron and nuclei?
Like I said, electron and nuclei are both going to move. The Born-Oppenheimer approximation lets you treat them separately. In mathematical terms, the molecular wavefunction is separable:
$$\Psi(\mathbf{r},\mathbf{R}) \approx \psi(\mathbf{r})\psi(\mathbf{R})$$
where ##\mathbf{r}## are the electronic coordinates and ##\mathbf{R}## are the nuclear coordinates. What this means is that the nuclei move very slowly compared to the electrons, so that as the nuclei move, the electrons more or less instantaneously adjust to the new nuclear positions.

Edit: And “vibronic” simply denotes that this approximation cannot be made. The electrons and nuclei are both still moving.
 

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