Why does internal rotation of ethane give an energy spectrum asymptotically approaching a rotation based on $\nu=1$ rather than $\nu=0$?

I'm a physicist with very little knowledge of chemistry. Recently a chem colleague left a molecular model in my classroom, so I brought it back to her, and as I was walking down the hall with it, a hydroxy group started spontaneously swinging around in circles. This made me curious about whether such motions could actually be observed. After asking around and doing some web searching, my understanding is that this sort of thing is called an internal rotation, and usually there is a barrier to rotation, which is modeled by a potential $V(\phi)=(1/2)\sum V\_n(1-\cos n\phi)$.


If I'm understanding the history correctly, then originally people thought that the spectrum would be that of a free rotor, but that produced incorrect thermodynamic predictions. Then in 1937 Pitzer published a [famous paper](https://books.google.com/books?id=Mb7sCgAAQBAJ&lpg=PA15&ots=HStcC6b2wU&dq=Pitzer%20%22Thermodynamics%20of%20Gaseous%20Hydrocarbons%22%201937&pg=PA15#v=onepage&q=Pitzer%20%22Thermodynamics%20of%20Gaseous%20Hydrocarbons%22%201937&f=false), "Thermodynamics of Gaseous Hydrocarbons," in which he showed that you could get correct thermodynamic predictions if you put in a barrier to rotation. 


The classic textbook example seems to be ethane. I found [this](https://www.sciencedirect.com/topics/chemistry/internal-rotation) explanation of the physics (di Lauro, "Floppy Molecules," in Rotational Structure in Molecular Infrared Spectra, 2013), which I only partly understand. He gives the following theoretical spectrum of energy states:


[![energy spectrum of ethane](https://i.stack.imgur.com/4D061.jpg)](https://i.stack.imgur.com/4D061.jpg)


So at low energies we have roughly a set of vibrational energy levels with energies $\hbar\omega (\nu+1/2)$, which makes sense to me because I guess we're sitting in a minimum of the potential where the geometry is staggered. He says, "The high levels, on the contrary, correlate and eventually almost coincide with the free internal rotor eigenfunctions, whose energy increases parabolically as $AK\_i^2$, where $K\_i$ is the eigenvalue of $J\_γ$, in ℏ units." This also sort of makes sense to me, because if you put in a lot of energy, you shouldn't see the effect of the little hills and valleys in the potential $V(\phi)$, so it should act more like a free rotor.


Do the units of inverse cm mean $1/\lambda$, or $2\pi/\lambda$? If the former, then $1/\lambda= 200\ \text{cm}^{-1}$ corresponds to have $\hbar\omega\_0/k\approx 300$ kelvin, so it makes sense that the barrier to rotation would have significant thermodynamic effects at temperatures on the order of room temperature or below, since the rotational-vibrational modes are not available. Am I getting this right?


But why is it necessary to place this fictitious rotational band on top of the $\nu=1$ state, rather than $\nu=0$? I would think that the fictitious band should be build from the ground state, it wouldn't match the actual energies at low energies, and then at high energies they would match up.


