# Does a Buckyball spin like an electron or like a baseball?

Does a Buckyball spin like an electron or like a baseball?

We are often told that an electron does not really spin like a baseball. Only one (or two, if you count up and down) spin states, for example.

How about a Buckyball? Does it spin more like an electron, or more like a baseball?

Where is the dividing line? How can you measure the difference?

• A buckyball is a fairly large molecule. Why couldn't it spin like a baseball? – Asher Aug 27 '15 at 21:49

The fundamental difference between an electron's spin and that of a baseball is that the electron is (as far as we know) a point particle. It therefore cannot rotate in the usual sense, where individual parts move relative to the center of mass; we say that its angular momentum is intrinsic. The magnitude $\lvert\vec{S}\rvert^2$ of a particle's intrinsic angular momentum $\vec{S}$ is fixed, which is the sense in which an electron has "only one" spin state. (The direction is not fixed, so, as you say, the spin can be up or down.)
A buckyball, like a baseball, has internal structure; the carbon atoms can be set in motion around the center of mass, giving it angular momentum. (The fermions inside the carbon atoms have intrinsic angular momentum, but in the ground state of the molecule these cancel out.) The magnitude of the buckyball's angular momentum $\vec{J}$ is not fixed, so in this sense it is more like a baseball.
But angular momentum is quantized, and while this is utterly irrelevant for a baseball, it has measurable consequences for even large molecules like fullerenes. The total angular momentum obeys $\lvert\vec{J}\rvert^2 = \hbar^2 J(J+1)$, where $J$ is an integer (assuming your buckyball has an even number of fermions, e.g., 60 ${}^{12}$C atoms), while the projection on any axis is restricted to integers from $-J$ to $+J$.
The kinetic energy associated with rotation is $\frac{\lvert\vec{J}\rvert^2}{2I}$, where $I$ is the moment of inertia, so this implies (unequally spaced) steps in the allowed energy. For C$_{60}$, $I$ is small enough that these steps have been measured using Raman spectroscopy.