What are Degenerate Rotational Energy Levels Physically?

I'm in introductory chem, so I don't have much of the physical background to really understand what's going on yet, but we're learning about rotational energy levels, and how some rotational energy levels for molecules are degenerate.

As I understand it, degenerate energy levels correspond to different "ways in which the molecule could rotate" such that it gains the same amount of potential energy.

I have two questions...

1. Why does a molecule "gain potential energy" when it rotates? Does it want to stop rotating for some reson?

2. What do degenerate energy levels correspond to physically in terms of the molecule's rotation?

Thanks!

• When I put you on a carousel, I surely need energy to start rotating you, and your hand gets hot when you put it on the ground to decelerate. Right? OK, that was the stupid part of your question. 2. makes more sense. ;)
– Karl
Jan 16, 2020 at 22:57
• and if your teacher calls rotational energy potentional energy, he has rightfully spotted a likeness, but otherwise is imo an idiot. omg
– Karl
Jan 16, 2020 at 23:01
• That's a rather... strong way of putting it, but I agree that calling it "potential energy" is wrong. I'm fairly sure that the Hamiltonian for rotational motion involves just the kinetic energy term, $-(\hbar^2/2m) \nabla^2$. See, e.g. en.wikipedia.org/wiki/… Unlike, say, vibrational motion, there isn't a potential energy term. If you've never heard of a Hamiltonian, then arguably you shouldn't be taught QM (yet)... but well. Jan 17, 2020 at 0:40

I have two questions...

Why does a molecule "gain potential energy" when it rotates? Does it want to stop rotating for some reason?

What do degenerate energy levels correspond to physically in terms of the molecule's rotation?

Do you remember the Newton's law of motion? A body in motion will always remain in motion until and unless there is a force acting on it. So you can imagine, that once a force caused the molecule to rotate, the molecule will continue to rotate forever. As a crude analogy can you recall the time when the Earth stopped spinning on its axis? It is spinning for billions of years. In the same way you can think that the isolated molecule, once excited, will rotate for a pretty long time (on a spectroscopic time scale).

You can think of degenerate rotation as a a way of physical rotation of a molecule in such a way that if you calculate the rotational energy in each case, it turns out to be the same number.

Third point, as pointed out by orthocresol, rotational energy is kinetic energy not potential energy. So no need to invoke potential energy.

Sometimes, we should not actually try to visualize what is happening physically at very small molecular levels. This is where mathematics enters. Many results originate out of pure mathematical calculations which can be hard to imagine mentally. For instance, Bohr's got correct H2 spectrum results by assuming the electron is rotating around the proton. The question is it actually rotating like a planet around a proton? A short story is worth sharing which will help you in future. Feynman (Physics Nobel) used to think a lot about of photons as a student and discussed them with his father. When he finished his PhD in physics, he realized, he still could not explain what a photon is. Sorry I cannot recall the reference. But mathematics has no problem. Yet on the web we can "see" photons in pictures as balls, which are obviously pure but meaningless imaginations.

• That's a great answer @M. Farooq. Would you consider helping us launch a stack exchange just for spectroscopy and materials modeling? We would very much appreciate your support here: Materials Modeling Feb 8, 2020 at 22:07

As pointed out in the comments the rigid rotating molecule only gains kinetic energy when it rotates not potential energy.

The degeneracy describes the fact that some levels have exactly the same energy and this depends the value of the angular momentum rotational quantum number $$J$$. The number of degenerate levels is given by the multiplicity $$2J+1$$. The $$J$$ quantum number has values $$J=0,1, 2,3,\cdots$$. A second quantum number $$m_j=-J\cdots J$$ describes each of the degenerate levels and 'orientation in space'. Thus if $$J=2$$ then $$m_j$$ has five values $$m_j=-2,-1,0, 1,2$$. Normally the $$m_j$$ levels cannot be distinguished by experiment because they are degenerate in energy, but in a magnetic or electric field these energy levels can be made to split (degeneracy is removed but multiplicity remains the same) and transitions between them measured. These effects are called the Zeeman and Stark effect respectively. (Historically the Zeeman effect was of great importance and showed that electrons have spin angular momentum, unknown hitherto.)

As the molecule is a quantum object some care is needed in interpreting what is meant by 'rotation'. When $$J=0$$ the molecule has no kinetic energy, i.e. is stationary and so has a wavefunction like that of an s orbital. To satisfy the Heisenberg Uncertainty Principle the molecule has an equal probability of pointing in any direction in space, i.e. the wavefunction looks like a sphere. When $$J >0$$ the molecule has some rotational kinetic energy but it now becomes more difficult to imagine but when $$J=1$$ the molecule can 'rotate' in any of three directions, with wavefunctions shaped like p orbitals. When $$J=2$$ the wavefunctions look like those of d orbitals and so on.

(Quantum mechanics often presents problems because we are so familiar in thinking of how every day objects move, e.g. a rotating wheel, and its easy and sometimes useful to imagine molecules do the same thing, but this is misleading as the wavefunction describes what happens and this often does not seem to make physical sense. )