Rotation energy in a ball symmertical molecule, why only two axis?

DemoniWaari
Messages
15
Reaction score
0
I was just wondering that when we have a molecule, and you introduce heat, the molecule starts to get translation, rotation and vibration energy. I'm considering just a molecule with two atoms and one axis which binds them together.

Now the real question is that why is there only two axis in rotation energy? Can't you have three axis there? I mean that you can rotate in the x, y and z direction. And yes, I do know that the third one is parallel with the axis which binds the atoms together, but can't you still make it rotate?

I do realize that if the molecule is NOT symmetrical in that axis THEN it sure is considered a third axis on which the molecule can rotate. But even when it is symmetrical can't you pump energy in it?

On a macroscopic level I do see that when you make something like that rotate around it's own axis it doens't "change" when you look at it, but it STILL can have energy on that rotational axis, which rose this question in my mind.

I hope my explanation makes any sense, thanks for your answers!

Thanks.
 
Physics news on Phys.org
I think you can rotate the molecules. However, the moment of inertia is really small in that direction, which makes the energy required for the rotation really large.
 
To elaborate on that, quantum mechanics says that to get a rotation, you need at least h-bar worth of angular momentum (action is quantized in h-bar bundles). The connection between angular momentum and energy is that the rotational energy is the square of the angular momentum divided by twice the moment of inertia, so even if you use the minimum angular momentum, the energy you need will scale like the inverse of the momentum of inertia. So as mfb said, this requires more energy than is easily obtainable when the moment of inertia is very small, and the last rotational mode is "frozen out." The same thing happens to the other two rotational modes at very low temperature-- they don't get excited, and don't show up in the specific heat of the molecules at very low T.
 
Consider rotating just a single atom. What does that even mean? You have a cloud of electrons surrounding a nucleus, which may or may not have some angular momentum around the center of mass. The energy levels of the atom are already quantized with respect to angular momentum. In a sense, the rotation degree of freedom is already taken into account at the atomic level.

And as mfb and Ken were getting at, the energy splitting due to angular momentum is large compared with the molecular rotation levels.
 
Last edited:
Oh thank you all for your answers! This cleared up my thoughts :)
 
Not an expert in QM. AFAIK, Schrödinger's equation is quite different from the classical wave equation. The former is an equation for the dynamics of the state of a (quantum?) system, the latter is an equation for the dynamics of a (classical) degree of freedom. As a matter of fact, Schrödinger's equation is first order in time derivatives, while the classical wave equation is second order. But, AFAIK, Schrödinger's equation is a wave equation; only its interpretation makes it non-classical...
Insights auto threads is broken atm, so I'm manually creating these for new Insight articles. Towards the end of the first lecture for the Qiskit Global Summer School 2025, Foundations of Quantum Mechanics, Olivia Lanes (Global Lead, Content and Education IBM) stated... Source: https://www.physicsforums.com/insights/quantum-entanglement-is-a-kinematic-fact-not-a-dynamical-effect/ by @RUTA
Is it possible, and fruitful, to use certain conceptual and technical tools from effective field theory (coarse-graining/integrating-out, power-counting, matching, RG) to think about the relationship between the fundamental (quantum) and the emergent (classical), both to account for the quasi-autonomy of the classical level and to quantify residual quantum corrections? By “emergent,” I mean the following: after integrating out fast/irrelevant quantum degrees of freedom (high-energy modes...
Back
Top