is the momentum and $m$ is the mass. Using Eq. , we obtain

\[rotor2\] $$\begin{aligned}
   \label{rotor2a}
   \hat{r}\times\mathbf{L}=-r_0\mathbf{p}=-mr_0^2\dot{\hat{r}}.
   \end{aligned}$$ We also have $$\label{rotor2b}
    \dot{\mathbf{L}}=0$$ (conservation of angular momentum).

Comparing Eqs. and , we find that the equation of motion for two spins is equivalent to the equation of motion for a free rotor if we identify $\mathbf{L}\rightarrow\mathbf{M}$, $\hat{r}\rightarrow\hat{n}$ and $J=I^{-1}$, where $I=mr_0^2$ is the moment of inertia of the rotor.

The quantum Hamiltonian of the free rotor is $$H_{rotor}={1\over2 I}\mathbf{L}^2,$$ and its solution is well known. The eigenstates are the spherical harmonics $Y_{lm}(\theta,\phi)$ (where $\theta$ and $\phi$ specify the direction of the unit vector $\hat{r}$) with eigenvalues $$\mathbf{L}^2=l(l+1)\hbar^2,\\\\\\\\\\\  L_z=m\hbar,$$ and corresponding energies $E_l=l(l+1)\hbar^2/2I$, where $l$ and $m$ are integers such that $l\geq0$ and $l\geq|m|$. In particular, $\mathbf{L}(\mathbf{M})=0$ for the ground state of the quantum rotor, but the direction of the vector $\mathbf{r}(\mathbf{N})$ is completely uncertain ($Y_{00}(\theta,\phi)={1\over\sqrt{4\pi}}$) as a result of quantum fluctuations, indicating a breakdown of the classical solution, in which $\mathbf{n}$ is fixed in the ground state. (Alternatively, one can gain this understanding from the Heisenberg uncertainty principle, $\langle\delta\hat{r}\rangle\langle\delta\mathbf{L}\rangle>\hbar$. With $\mathbf{L}=0$ in the ground state, $\delta\mathbf{L}\equiv0$ and $\delta\hat{r}\rightarrow\infty$, the direction of the vector $\hat{r}$ becomes completely uncertain.)

A moment of 