Particle on a Ring

Introduction

So far our quantum particles have moved along straight lines between hard walls. Many real systems are round instead: the π electrons of benzene circulate around a ring, a molecule rotates about an axis, and an electron in an atom has angular motion around the nucleus. The simplest model of rotation is a particle constrained to move on a circle of fixed radius. It introduces a new kind of boundary condition and the quantisation of angular momentum.

Core explanation

The model. A particle of mass m moves on a circle of radius r in the xy-plane, with zero potential energy everywhere on the circle. Its position is described by a single angle φ. Because r is fixed, the only energy is kinetic, and the Hamiltonian becomes

Ĥ = −(ħ²/2I) d²/dφ²

where I = mr² is the moment of inertia. The Schrödinger equation is −(ħ²/2I) d²ψ/dφ² = Eψ.

Solutions. The equation is satisfied by complex exponentials, ψ = A e^(imₗφ), with E = mₗ²ħ²/(2I). At this stage mₗ could be any number.

The cyclic boundary condition. Unlike a box, a ring has no walls. Instead, after travelling once around (φ increasing by 2π) the particle returns to the same point, so the wavefunction must return to the same value: ψ(φ + 2π) = ψ(φ). This requires e^(imₗ2π) = 1, which is true only when mₗ is an integer:

mₗ = 0, ±1, ±2, ±3, …

Energies. The allowed energies are therefore E = mₗ²ħ²/(2I). Three features stand out. First, mₗ = 0 is allowed and gives E = 0: there is no zero-point energy, because the wavefunction for mₗ = 0 is a constant, which spreads the particle evenly around the ring with no curvature. The uncertainty principle is not violated, since the angle is then completely unknown. Second, every level with mₗ ≠ 0 is doubly degenerate, as +mₗ and −mₗ give the same energy. Third, levels spread apart as mₗ² grows.

Normalisation. Requiring the integral of ψ ² from 0 to 2π to equal 1 gives A = 1/(2π)^½, so ψ = (1/2π)^½ e^(imₗφ). Note that ψ ² = 1/(2π) for every state: the probability density is uniform around the ring even though the wavefunction itself oscillates in its real and imaginary parts.

Angular momentum. The z-component of angular momentum is represented by the operator L̂ z = −iħ d/dφ. Applying it to ψ gives mₗħψ, so each state has a definite angular momentum L z = mₗħ. The sign of mₗ indicates direction: positive for anticlockwise motion viewed from above, negative for clockwise. The energy can be written classically as E = L z²/(2I), which is why opposite directions share the same energy.

Formulae

ψ mₗ(φ) = (1/2π)^½ e^(imₗφ); E = mₗ²ħ²/(2I); I = mr²; L z = mₗħ; mₗ = 0, ±1, ±2, …

Step-by-step reasoning

To find the levels of a particle on a ring:

1. Write the kinetic-energy-only Hamiltonian in terms of the angle φ and the moment of inertia I. 2. Solve with e^(imₗφ). 3. Impose ψ(φ + 2π) = ψ(φ), forcing mₗ to be an integer. 4. Substitute to get E = mₗ²ħ²/(2I) and note the degeneracies. 5. Normalise over one full turn.

Visual explanation

Draw a circle and wrap a wave around it. For mₗ = 1 exactly one wavelength fits the circumference; for mₗ = 2, two wavelengths fit. A wave that does not fit a whole number of times would meet itself out of step and cancel, which is the picture behind the boundary condition. The energy ladder shows a single rung at zero and pairs of rungs at 1, 4 and 9 units.

Real-world analogy

Think of a circular running track with markers painted at equal spacing. A pattern repeating around the track must join up seamlessly at the start line; you can have one, two or three repeats per lap, but never two and a half. Clockwise and anticlockwise versions of the same pattern cost the same effort.

Real-world example

Benzene's six π electrons can be modelled as particles on a ring of radius about 139 pm. They fill mₗ = 0 (two electrons) and mₗ = ±1 (four electrons), giving a closed shell that helps explain aromatic stability. The same model underlies the 4n + 2 rule: closed shells occur for 2, 6, 10, 14 … π electrons.

Why?

Why is a ring's lowest energy zero while a box's is not? In a box the walls force the wavefunction to zero at the edges, so it must curve, and curvature means kinetic energy. On a ring there are no edges, so a perfectly flat, constant wavefunction is allowed, and a flat wavefunction has no kinetic energy.

Common misconception

"A state with mₗ = +1 has its electron on one side of the ring and mₗ = −1 on the other." Both have uniform probability density all the way round. They differ in the direction of circulation, which shows up in the sign of the angular momentum, not in where the particle is found.

Worked example

Question: Treat benzene's π electrons as particles on a ring of radius 139 pm. Estimate the wavelength of the lowest-energy transition, from mₗ = 1 to mₗ = 2.

Reasoning: ΔE = (2² − 1²)ħ²/(2m e r²) = 3ħ²/(2m e r²). ħ² = (1.055 × 10⁻³⁴)² = 1.112 × 10⁻⁶⁸ J² s². m e r² = 9.109 × 10⁻³¹ × (1.39 × 10⁻¹⁰)² = 1.760 × 10⁻⁵⁰ kg m². So ΔE = 3 × 1.112 × 10⁻⁶⁸/(3.520 × 10⁻⁵⁰) = 9.48 × 10⁻¹⁹ J. λ = hc/ΔE = 1.986 × 10⁻²⁵/9.48 × 10⁻¹⁹ = 2.10 × 10⁻⁷ m.

Answer: About 210 nm, in the ultraviolet, in the same region as benzene's observed absorption bands.

Quick check

1. How many states of a particle on a ring have an energy of 9ħ²/(2I), and what are their mₗ values? Answer: Two states, with mₗ = +3 and mₗ = −3, corresponding to opposite directions of circulation.

Exam focus

Be ready to state the cyclic boundary condition and derive from it that mₗ is an integer. Know E = mₗ²ħ²/(2I), the absence of zero-point energy, and the double degeneracy of every level except mₗ = 0. Contrasting ring and box results is a favourite short-answer question.

Advanced insight

Because the energy depends on mₗ², complex wavefunctions with ±mₗ can be combined into real ones, cos(mₗφ) and sin(mₗφ), which have 2 mₗ nodes around the ring. These real combinations are standing waves without definite angular momentum, whereas the complex ones are travelling waves with L z = ± mₗ ħ. The same choice between real and complex forms reappears for p and d orbitals.

Summary

A particle on a ring of radius r has Hamiltonian −(ħ²/2I)d²/dφ² with I = mr². The cyclic condition ψ(φ + 2π) = ψ(φ) forces mₗ = 0, ±1, ±2, …, giving E = mₗ²ħ²/(2I) and angular momentum L z = mₗħ. The ground state has zero energy, all other levels are doubly degenerate, and the probability density is uniform around the ring. The model explains closed π shells in benzene.

Practice questions

1. State the boundary condition for a particle on a ring and explain why it leads to quantisation. Answer: ψ(φ + 2π) = ψ(φ); for e^(imₗφ) this holds only if mₗ is an integer, so only discrete energies are allowed. 2. How does the energy of each level change if the radius of the ring is doubled? Answer: I = mr² becomes four times larger, so every energy falls to one quarter of its value. 3. What angular momentum does an electron in the mₗ = −2 state have about the axis? Answer: L z = −2ħ, meaning a magnitude of 2ħ with clockwise circulation viewed from above. 4. Why does a particle on a ring have no zero-point energy? Answer: The mₗ = 0 wavefunction is a constant, which satisfies the cyclic condition and has no curvature, so its kinetic energy is zero.