Section 22.5 Motion on a Ring
Now we want to solve (22.4.2) for \(\Phi(\phi)\) obtained from the separation of variables procedure in Section 22.4 for a quantum mechanical particle confined to a ring. Since the coefficient of \(\Phi\) on the right-hand-side of (22.4.2) is a constant
\begin{equation}
\sqrt{\frac{2I}{\hbar^2}(E-U(r_0))}=\hbox{constant}\tag{22.5.1}
\end{equation}
this equation is identical to the equation for an undamped harmonic oscillator, with a solution basis that we can choose to be either sines and cosines or imaginary exponentials. For reasons that will become clear later, we will choose exponentials
\begin{equation}
\Phi_m(\phi)\defeq N\,e^{im\phi}\tag{22.5.2}
\end{equation}
where
\begin{equation}
m = \pm\sqrt{\frac{2I}{\hbar^2}(E-U(r_0))}\tag{22.5.3}
\end{equation}
and \(N\) is a normalization constant.
There is no spatial “boundary” on the ring on which we can impose boundary conditions. However, there is one very important property of the wave function that we can invoke: it must be single-valued. The variable \(\phi\) is geometrically an angle, so that \(\phi+2\pi\) is physically the same point as \(\phi\text{.}\) If we go once around the ring and return to our starting point, the value of the wave function must remain the same. Therefore the solutions must satisfy the periodicity condition \(\Phi_m(\phi+2\pi)=\Phi_m(\phi)\text{.}\) This is impossible unless \(m\) is real so that the solutions are oscillatory, i.e. \(E-U(r_0)>0\text{.}\) Furthermore, the solutions must have the correct period, i.e.
\begin{equation}
m\in\{0,\pm1,\pm2,...\} .\tag{22.5.4}
\end{equation}
The quantum number \(m\) is called the azimuthal or magnetic quantum number. Note that the solution permits both positive and negative values of m as well as zero.
Solving (22.5.3) for the possible eigenvalues of energy, we obtain
\begin{equation}
E_m = \frac{\hbar^2}{2I}\, m^2 + U(r_0)\tag{22.5.5}
\end{equation}
For this simplified ring problem, we can choose the potential energy \(U(r_0)\) to be zero, but we will have to remember that we should not make this choice when we are working on the full hydrogen atom problem. There is a degeneracy that arises in this calculation. Note that the wave functions corresponding to \(+|m|\) and \(-|m|\) have the same energy but, (as we will see in Section 22.7), represent different states of the motion.
This is a one-dimensional problem, just like the problem of a particle-in-a-box (now using the independent variable \(\phi\) instead of \(x\)) and the solutions have the same oscillatory form. Everything that you learned about a particle-in-a-box is immediately applicable here. As in that problem, the energy eigenvalues are discrete because of a boundary condition. The only differences that arise come from the different boundary conditions. In the ring case, the appropriate boundary condition is periodicity, since \(\phi\) is a physical angle. Therefore, the eigenstates have to fit an integer number of wavelengths into the ring and the energy eigenvalues are degenerate. For a particle-in-a-box, \(\Psi(x)=0\) at the boundaries, appropriate to an infinite potential. Therefore, the eigenstates can fit a half integer number of wavelengths into the box and the energy eigenvalues are not degenerate.
