I Rotational excitation of quantum particle

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I was watching a lecture and there was a connection drawn between classical rotational energy and quantum rotational excitation. The energy of a rotating system is $$E = (L^2) / 2 I $$ with L being the angular momentum and I the moment of Inertia. Then to make it quantum$$ n^2 * ħ^2$$ was substituted for## L^2 ##on top so it just becomes $$E rot = (n^2 * ħ^2) / 2 I $$ It was then stated that for a given mass, because the Moment of inertia grows with the radius squared, then the smaller a particle (such as an electron) then the larger the energy require to rotationally excite it. What I don't understand is why is " I " substituted to quantized energy in the numerator but left in the denominator? Then conceptually, how can a smaller radius of a particle require more energy to spin? This goes against the classical metaphor and confuses me. Thanks!
 
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Because in your equation the angular momentum is constant while the moment of inertia is decreasing. That means it must spin faster the smaller it is and your equation is telling you that to make it do that you must give it more energy.

Your intuition is running counter to that because it is assuming that a smaller object has less angular momentum (which is true classically for constant energy).

So you are holding two mutually contradictory ideas at once probably due to your intuition becoming confused between angular momentum and moment of inertia.

Incidentally, this highlights the problem with thinking of the spin of a point particle (such as an electron is usually thought of) as classical angular momentum. Because such a particle would have infinite energy. Rather to treat spin as classical angular momentum in your equation we must think of it as having a fixed moment of inertia and then the classical expression for moment of inertia breaks down. A more general way of thinking about it is that when discussing the properties of a quantum object at a space-time location, classical mechanics breaks down.
 
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