8.6: Particle on a Ring

Learning Objectives

• To be familiar with a quantum system with angular symmetry.
• To be introduced to quantum angular momentum

Introduction

The case of a quantum particle confined a one-dimensional ring is similar to the particle in a 1D box. Consider a variant of the one-dimensional particle in a box problem in which the x-axis is bent into a ring of radius \(R\). We can write the same Schrödinger equation

\[ \frac{-\hbar^2}{2m} \frac{d^2 \psi(x)}{dx^2} = E \psi(x) \label{1} \]

There are no boundary conditions in this case since the x-axis closes upon itself. A more appropriate independent variable for this problem is the angular position on the ring given by, \( \phi = x {/} R \). The Schrödinger equation would then read

\[ -\frac{\hbar^2}{2mR^2} \frac{d^2 \psi (\phi)} {d (\phi)^2} = E \psi (\phi) \label{2} \]

Solving for the wavefunctions (eigenfunctions)

The kinetic energy of a body rotating in the xy-plane can be expressed as

\[ E = \frac{J_z^2}{2I} \label{3} \]

where \(I = mR^2\) is the moment of inertia and \( J_z\), the z-component of angular momentum. (Since \( J = r \times p\), if r and p lie in the xy-plane, J points in the z-direction. The Schrödinger Equation \(\ref{2}\) can now be rewritten as:

\[ -\frac{\hbar^2}{2I} \frac{d^2 \psi (\phi)} {d (\phi)^2} = \frac{J_z^2}{2I} \psi (\phi) \label{4} \]

or

\[ -\frac{d^2 \psi (\phi)} {d (\phi)^2} = \frac{J_z^2}{\hbar^2} \psi (\phi) \label{5} \]

This is usually simplified further by setting the ratio \(\frac{J_z^2}{\hbar^2} = m_l^2\). Yielding

\[ -\frac{d^2 \psi (\phi)} {d (\phi)^2} = m_l^2 \psi (\phi)\label{6} \]

Do not confuse the variable \(m_l\) with the mass of the particle!

By remembering that the derivative of \(e^{ax}\) is \(ae^{ax}\) and the derivative of that would give us \(a^2 e^{ax}\) we can realize that the wavefunctions (\(\psi\)) that satisfy equation \ref{6} are of the form:

\[ \psi (\phi) = \text{const}\, e^{\pm{i}m\phi} \label{7} \]

For this wavefunction to be physically acceptable, it must be single-valued. Since \(\phi \) increased by any multiple of 2\(\pi \) represents the same point on the ring, we must have

\[ \psi (\phi + 2\pi ) = \psi (\phi) \label{8} \]

and therefore

\[e^{{i}m_l (\phi + 2\pi)} = e^{{i}m_l \phi} \label{9} \]

This requires that

\[e^{2\pi {i}m_l} = 1 \nonumber \]

which is true only if m_l is an integer:

\[ m_l = 0, \pm 1, \pm 2... \label{10} \]

Rearranging the definition of m_l² gives us  \(J_z^2 = \hbar^2 m_l^2\), which can be substituted into equation \ref{3}, giving the quantized energy values

\[E_{m_l} = \frac{\hbar^2}{2I} m_l^2  = \frac{\hbar^2}{2mR^2} m_l^2 \label{E_m_l} \]

In contrast to the particle in a box, the eigenfunctions corresponding to \(+m_l \) and \(-m_l \) (Equation \ref{7}) are linearly independent, so both must be accepted. Therefore all eigenvalues, except \(E_0 \), are two-fold (or doubly) degenerate. The eigenfunctions can all be written in the form const ​\( e^{{i}m_l \phi} \), with \(m_l\) allowed to take either positive and negative values (or 0), as in equation \ref{10}. The normalized eigenfunctions are

\[ {\psi _{m_l}} (\phi) = \frac{1}{\sqrt{2 \pi}} e^{\pm im_l\phi} \label{PonRpsi} \]

and can be verified to satisfy the normalization condition containing the complex conjugate

\[ \int\limits_{0}^{2\pi} {\psi_{m_l}^*} (\phi) {\psi _{m_l}} (\phi) d\phi = 1 \nonumber \]

where we have noted that \( {\psi_{m_l}^*} (\phi) = (2\pi)^{-1/2} e^{-{i}m_l\phi} \). The mutual orthogonality of the functions also follows easily, for

\[ \begin{align} \int\limits_{0}^{2\pi} {\psi_{m^\prime}^*} {\psi _{m_l}} (\phi) d\phi &= \dfrac{1}{2\pi} \int\limits_{0}^{2\pi} e^{{i}(m_l-m_l^\prime) \phi} d\phi \\[4pt] &= \frac{1}{2\pi} \int\limits_{0}^{2\pi} [\cos(m_l-m_l^\prime)\phi + {i} \sin(m_l-m_l^\prime)\phi] d\phi =0 \end{align} \nonumber \]

for \( m_l^\prime \neq m_l \).

This is a instance of a fundamental result in quantum mechanics, that any measured component of angular momentum is restricted to integral multiples of \( \hbar \). The Bohr theory of the hydrogen atom can be derived from this principle alone.

Example \(\PageIndex{1}\): Free Electron Model for Aromatic Molecules

The benzene molecule consists of a ring of six carbon atoms around which six delocalized pi-electrons can circulate. A variant of the FEM for rings predicts the ground-state electron configuration which we can write as \( 1\pi^{2} 2\pi^{4} \), as shown here:

Figure \(\PageIndex{1}\): Free electron model for benzene. Dotted arrow shows the lowest-energy excitation.

The enhanced stability the benzene molecule can be attributed to the complete shells of \(\pi \)-electron orbitals, analogous to the way that noble gas electron configurations achieve their stability. Naphthalene, apart from the central C-C bond, can be modeled as a ring containing 10 electrons in the next closed-shell configuration\( 1\pi^{2} 2\pi^{4} 3\pi^{4} \). These molecules fulfill Hückel's "4N+2 rule" for aromatic stability. The molecules cyclobutadiene \( {(1\pi^{2} 2\pi^{2})} \) and cyclooctatetraene\( {(1\pi^{2} 2\pi^{4} 3\pi^{2})} \) , even though they consist of rings with alternating single and double bonds, do not exhibit aromatic stability since they contain partially-filled orbitals.

The longest wavelength absorption in the benzene spectrum can be estimated according to this model as

\[ \dfrac{hc}{\lambda} = E_2 - E_1 = \dfrac{\hbar^2}{2mR^2} {(2^2 -1^2)} \nonumber \]

The ring radius R can be approximated by the C-C distance in benzene, 1.39 Å. We predict \( \lambda \approx \) 210 nm, whereas the experimental absorption has \( \lambda_{max} \approx \) 268 nm.

Understanding the quantization in terms of wavelengths¹

de Bröglie used essentially the following argument to explain why Bohr's model of electrons orbiting a nucleus yield quantized values. We will use this argument to arrive at the quantization and energies of a particle-on-a-ring without formally solving Schrödinger's equations. We, again, start with the energy of a particle of a mass m rotating at a fixed speed at distance R around the origin. In terms of angular momentum about the z axis (J_z) the energy can be expressed as:

\[E = \dfrac{J^2_z}{2mR^2}  = \frac{J^2_z}{2I}\label{1C} \]

• \(J_z\) is the angular momentum in z-axis and
• \(mR^2\) is the particle's moment of inertia I on the z-axis.

We then use the de Broglie equation (\(\lambda = \frac{h}{ p}\implies p = \frac{h}{\lambda}\)) to quantize the energy of rotation. This is done by expressing the angular momentum in wavelengths. Because the rotation is in the xy-plane, J_z = pR, where p is the linear momentum in the deBroglie equation:

\[J_z = pR = \frac{hR}{\lambda} \label{3C} \]

Because the particle must exist on the line its wavelength must be an integer multiple of the circumference, otherwise it would destructively interfere with itself as the phase of the wave shifts on each circuit around the ring. This is equivalent to the requirement to be single valued expressed in equations \(\ref{8}\) and \(\ref{9}\). Figure \(\PageIndex{2}\) illustrates this where the circumference has been unwrapped to form a line.

 

Figure \(\PageIndex{2}\): Two waves on a ring that has been unwrapped to a line for clarity at the angle \(\phi = 0 =2\pi\). One that has a wavelength equal to the circumference so that it creates a standing wave on the ring (\(\psi_a\)) and one that has a wavelength that is not an integer fraction of the circumference so shifts on each circuit around the ring and destructively interferes with itself (\(\psi_b\)).

Thus the only allowed wavelengths (those that create standing waves) are:

\[\lambda = \frac{2\pi R}{m_l}\nonumber\]

where m_l is an integer. This can be substituted into equation \ref{3C} to yield:

\[J_z = pR = \frac{m_l h}{2 \pi} \nonumber \]

when this is substituted int equation \ref{1C} for J_z, we get:

\[E  = \frac{m_l^2h^2}{8\pi^2I} = \frac{\hbar^2}{2I}m_l^2\label{4C} \]

This is the same result as in equation \ref{E_m_l}.

References

1. Atkins, Peter, and Julio de Paula. Physical Chemistry for the Life Sciences. New York. Oxford University Press. 2006, p 358.