Section 14.4 Angular Momentum of the Particle on a Ring
Classically, a particle moving in a circle has an angular momentum perpendicular to the plane of the circle, which for a ring in the \(x,y\)–plane would be in the \(z\) direction. Since angular momentum is defined by \(\vf{L}=\rr\times\vf{p}\text{.}\) To make the transition to quantum mechanics, we replace \(p_x\) and \(p_y\) by their operator equivalents:
\begin{equation}
L_z = xp_y - y p_x
\Longrightarrow
\Lhat_z = x\frac{\hbar}{i}\frac{\partial}{\partial y}
- y\frac{\hbar}{i}\frac{\partial}{\partial x}\tag{14.4.1}
\end{equation}
Using a straightforward application of the chain rule (see Practice Problems, below) to replace the Cartesian partial derivatives with their polar representations, we obtain
\begin{equation}
\Lhat_z = \frac{\hbar}{i}\frac{\partial}{\partial\phi}\tag{14.4.2}
\end{equation}
The effect of operating on the ring eigenfunctions with this operator is:
\begin{equation}
\frac{\hbar}{i}\frac{\partial}{\partial\phi}
\left( \frac{1}{\sqrt{2\pi}} e^{im\phi} \right)
= m\hbar \left( \frac{1}{\sqrt{2\pi}} e^{im\phi} \right)\tag{14.4.3}
\end{equation}
The energy eigenfunctions \(\Phi_m(\phi)\) are thus also eigenfunctions of \(\Lhat_z\) with eigenvalues \(m\hbar\text{.}\) Because the \(\Phi_m(\phi)\) are eigenfunctions of both energy and angular momentum, we can make simultaneous determinations of the eigenvalues of energy and angular momentum.
Considering the angular momentum helps us understand the degeneracy of the eigenfunctions with respect to energy. The \(\pm m\) degeneracy of the energy eigenstates corresponds to \(L_z=+m\hbar\) and \(L_z=-m\hbar\text{.}\) That is, the two degenerate states represent particles rotating in opposite directions around the ring.
For a classical particle rotating in a circular path in the \(x,y\)-plane, the kinetic energy is \(T=\frac12
I\omega^2=L_z^2/2I\text{,}\) where \(I\) is the rotational inertia (moment of inertia). The rotational inertia of a single particle of mass \(\mu\) moving in a circle of radius \(r_0\) is \(I=\mu r_0^2\text{.}\) The Hamiltonian for the system is thus
\begin{equation}
H = T+U = \frac{L_z^2}{2I}+U
= \frac{\hbar^2}{2\mu r_0^2} \frac{\partial^2}{\partial\phi^2} + U_0\tag{14.4.4}
\end{equation}
It is apparent from this approach that the energy and the angular momentum have simultaneous eigenvalues because they are commuting operators. Clearly \([L_z^2,L_z]=0\text{,}\) so that \(E\) and \(L_z\) have the same eigenfunctions. Therefore, we see that (14.1.2) and (14.4.3) are the position-space representations of the eigenvalue equations
\begin{align}
\Hhat|m\rangle \amp= E_m |m\rangle\tag{14.4.5}\\
\Lhat|m\rangle \amp= \hbar m |m\rangle\tag{14.4.6}
\end{align}
Because the \(\Phi_m\) are simultaneous eigenstates of both \(\Hhat\) and \(\Lhat_z\text{,}\) it is possible to make simultaneous measurements of both the energy and the \(z\)-component of angular momentum.
In setting up the problem of the particle on the ring, we constrained the motion to the \(x,y\)-plane, so that the angular momentum vector is in the \(z\) direction. However, according to quantum mechanics (yet another form of the Heisenberg uncertainty relationships) it is not possible to know the direction of the angular momentum vector. Our knowledge of the angular momentum vector is limited to its length and any one component. If the vector lies along the \(z\)-axis, then we would know all three of its components (the \(x\) and \(y\) components being zero). We’ll see how the three-dimensional problem solves this contradiction.
