Particle on a Ring

A particle on a ring is the canonical quantum system with one periodic angular coordinate. The configuration space is the circle $S^1$, not an interval with two separate ends. That single geometric fact forces integer angular momentum, periodic boundary conditions, and a spectrum symmetric under clockwise and counterclockwise circulation.

This page is the angular-systems treatment. For a more introductory route from one-dimensional boxes to periodic motion, see Particle on a Ring: First Encounter.

Configuration Space

Let a particle of mass $M$ be constrained to a circle of fixed radius $R$. Its position is specified by one angle,

$$0\le\theta\lt2\pi,$$

with the identification

$$\theta\sim\theta+2\pi.$$

The Hilbert space is $L^2(S^1,d\theta)$, with inner product

$$\langle \phi\vert\psi\rangle = \int_0^{2\pi} \phi^*(\theta)\psi(\theta)\,d\theta.$$

A scalar wavefunction must be single-valued on the circle:

$$\psi(\theta+2\pi)=\psi(\theta).$$

For the free second-derivative Hamiltonian, the derivative is periodic as part of the same self-adjoint domain:

$$\psi'(\theta+2\pi)=\psi'(\theta).$$

The probability density $\lvert\psi(\theta)\rvert^2$ is density per unit angle. Density per unit arc length $s=R\theta$ differs by the factor $ds=R\,d\theta$.

Hamiltonian and Angular Momentum

The moment of inertia is

$$I=MR^2.$$

Classically, the kinetic energy is

$$T=\frac{L_z^2}{2I}.$$

In the angle representation,

$$\hat L_z = -i\hbar\frac{d}{d\theta},$$

so the free-ring Hamiltonian is

$$\hat H = \frac{\hat L_z^2}{2I} = - \frac{\hbar^2}{2I} \frac{d^2}{d\theta^2}.$$

Thus the ring is a free particle with a compact coordinate. The differential expression resembles a particle in a box, but the boundary condition is periodic rather than Dirichlet or Neumann.

The ring has one periodic coordinate $\theta$, and the free spectrum is quadratic in the integer angular-momentum label: $E_m=Bm^2$ with $B=\hbar^2/(2I)$. The states $m$ and $-m$ are degenerate when no flux or external field selects an orientation.

Eigenfunctions and Integer Quantum Number

The angular-momentum eigenvalue equation is

$$\hat L_z u(\theta)=\lambda u(\theta).$$

Solving gives

$$u(\theta)=Ce^{i\lambda\theta/\hbar}.$$

Single-valuedness requires

$$u(\theta+2\pi)=u(\theta),$$

so

$$e^{i2\pi\lambda/\hbar}=1.$$

Therefore

$$\lambda=\hbar m, \qquad m\in\mathbb Z.$$

The normalized eigenfunctions are

$$u_m(\theta) = \frac{1}{\sqrt{2\pi}} e^{im\theta}, \qquad m\in\mathbb Z.$$

The integer $m$ can be positive, negative, or zero. Positive and negative values represent opposite orientations of angular momentum around the ring.

Energy Spectrum and Degeneracy

Acting with the Hamiltonian gives

$$\hat H u_m = \frac{\hbar^2m^2}{2I}u_m.$$

Define

$$B=\frac{\hbar^2}{2I}.$$

Then

$$E_m=Bm^2, \qquad m\in\mathbb Z.$$

The $m=0$ level is nondegenerate. Every nonzero level is twofold degenerate:

$$E_m=E_{-m}, \qquad m\ne0.$$

This degeneracy is a time-reversal and orientation symmetry of the free ring. The two states have equal kinetic energy but opposite angular momentum:

$$\hat L_z u_m=\hbar m u_m, \qquad \hat L_z u_{-m}=-\hbar m u_{-m}.$$

The ground state has $E_0=0$ because the constant wavefunction has no angular variation and therefore no kinetic energy. This does not contradict the harmonic oscillator’s zero-point energy; the ring has no confining angular potential with a finite-width minimum.

Probability Current Around the Ring

For the angular Schrödinger equation,

$$i\hbar\frac{\partial\psi}{\partial t} = - \frac{\hbar^2}{2I} \frac{\partial^2\psi}{\partial\theta^2},$$

the angular probability density $\rho=\lvert\psi\rvert^2$ obeys

$$\frac{\partial\rho}{\partial t} + \frac{\partial J_\theta}{\partial\theta} =0.$$

The angular probability current is

$$J_\theta = \frac{\hbar}{I} \operatorname{Im} \left( \psi^* \frac{\partial\psi}{\partial\theta} \right).$$

For the eigenstate $u_m$,

$$J_{\theta,m} = \frac{\hbar m}{2\pi I}.$$

The density of $u_m$ is constant, but the current is not zero unless $m=0$. This is a useful warning: a stationary probability density does not imply the absence of probability flow.

Relation to Translations on a Compact Space

The ring is also the periodic-box basis written geometrically. If $s=R\theta$ is arc length and $L=2\pi R$ is the circumference, then periodicity gives

$$k_m = \frac{2\pi m}{L} = \frac{m}{R}.$$

The tangential momentum is

$$p_m=\hbar k_m=\frac{\hbar m}{R},$$

and the angular momentum is

$$L_z=Rp_m=\hbar m.$$

The energy

$$\frac{p_m^2}{2M} = \frac{\hbar^2m^2}{2MR^2} = \frac{\hbar^2m^2}{2I}$$

is the same spectrum written in linear or angular language. This is the simplest example of how compact configuration spaces quantize generator eigenvalues.

Magnetic Flux Preview

If the particle has charge $q$, magnetic flux $\Phi$ through the ring shifts the spectrum even when the magnetic field vanishes on the ring itself. This is the ring version of the Aharonov–Bohm effect.

With a tangential vector potential representing flux $\Phi$, minimal coupling gives

$$\hat H(\Phi) = \frac{1}{2I} \left( -i\hbar\frac{d}{d\theta} - \frac{q\Phi}{2\pi} \right)^2.$$

Define the dimensionless flux

$$\alpha = \frac{q\Phi}{h}.$$

The single-valued basis $u_m$ remains convenient, and the energies become

$$E_m(\Phi) = \frac{\hbar^2}{2I} \left( m-\alpha \right)^2.$$

The sign of $\alpha$ depends on the sign of the charge and on the chosen orientation for $\Phi$. The invariant lesson is that flux shifts the parabola in $m$ and can split the $m,-m$ degeneracy.

Equivalently, one may remove the vector potential locally and impose a twisted boundary condition. The two descriptions are related by a gauge transformation. The general rule behind this preview is Minimal Coupling in Wave Mechanics, and the full wave-mechanics phase discussion is Aharonov–Bohm Effect: First Encounter.

Common Mistakes

• Treating $\theta=0$ and $\theta=2\pi$ as independent endpoints.
• Allowing arbitrary real $m$ for an ordinary single-valued scalar wavefunction.
• Forgetting the negative angular-momentum states.
• Confusing the particle mass $M$ with the angular quantum number $m$.
• Thinking the constant $m=0$ ground state contradicts oscillator zero-point energy.
• Inferring zero current from a time-independent density.
• Ignoring charge and orientation conventions when writing the flux-shifted spectrum.

Exercises

1. Normalize $u_m(\theta)=Ce^{im\theta}$ on the ring.

Solution

The normalization condition is

$$1= \int_0^{2\pi} \lvert C\rvert^2\,d\theta = 2\pi\lvert C\rvert^2.$$

Thus

$$\lvert C\rvert=\frac{1}{\sqrt{2\pi}}.$$

Choosing the overall phase to be $1$ gives

$$u_m(\theta) = \frac{1}{\sqrt{2\pi}}e^{im\theta}.$$

2. Show that periodicity forces $m$ to be an integer.

Solution

The eigenfunction has the form

$$u(\theta)=Ce^{i\lambda\theta/\hbar}.$$

Periodic single-valuedness gives

$$e^{i\lambda(\theta+2\pi)/\hbar} = e^{i\lambda\theta/\hbar}.$$

After canceling the common factor,

$$e^{i2\pi\lambda/\hbar}=1.$$

Therefore $\lambda/\hbar$ is an integer. Writing $\lambda=\hbar m$ gives

$$m\in\mathbb Z.$$

3. Compute the angular current in the state $u_m$.

Solution

For

$$u_m(\theta)=\frac{1}{\sqrt{2\pi}}e^{im\theta},$$

we have

$$\frac{\partial u_m}{\partial\theta} = imu_m.$$

Thus

$$u_m^* \frac{\partial u_m}{\partial\theta} = i m\lvert u_m\rvert^2 = \frac{im}{2\pi}.$$

The imaginary part is $m/(2\pi)$, so

$$J_{\theta,m} = \frac{\hbar}{I} \frac{m}{2\pi} = \frac{\hbar m}{2\pi I}.$$

4. Which free-ring levels are degenerate?

Solution

The energy is

$$E_m=Bm^2.$$

Therefore

$$E_m=E_{-m}.$$

For $m\ne0$, the pair $m$ and $-m$ is twofold degenerate. The $m=0$ level is nondegenerate because $0=-0$.

5. At what dimensionless flux $\alpha$ are the $m=0$ and $m=1$ states degenerate?

Solution

The flux-shifted energies are

$$E_m(\Phi) = B(m-\alpha)^2.$$

Set $E_0=E_1$:

$$\alpha^2=(1-\alpha)^2.$$

Expanding gives

$$\alpha^2=1-2\alpha+\alpha^2,$$

so

$$\alpha=\frac{1}{2}.$$

Thus the degeneracy occurs at half a flux quantum in the chosen orientation convention.

Where This Is Used

• Particle on a Ring: First Encounter introduces the same model from the periodic-box viewpoint.
• Propagators and Boundary Conditions derives the ring kernel as both an angular-momentum sum and a winding sum.
• Particle on a Sphere generalizes compact angular motion from $S^1$ to $S^2$.
• Rigid Rotor generalizes the one-angle ring to orientation on the sphere.
• Angular Probability Distributions compares the $S^1$ measure $d\theta$ with the $S^2$ measure $d\Omega$.
• Probability Current supplies the current formula and the warning that stationary density need not mean zero flow.
• Minimal Coupling in Wave Mechanics supplies the flux-shifted Hamiltonian.
• Aharonov–Bohm Rings contrasts this closed spectrum with open-ring conductance harmonics, reservoirs, and dephasing.
• Translations and Momentum explains the generator viewpoint behind $-i\hbar\,d/d\theta$.
• Kicked Rotor Preview adds periodic impulsive driving and compares the resulting classical and quantum maps.

References

• D. J. Griffiths and D. F. Schroeter, Introduction to Quantum Mechanics, 3rd ed., Cambridge University Press, 2018.
• R. Shankar, Principles of Quantum Mechanics, 2nd ed., Springer, 1994.
• J. J. Sakurai and J. Napolitano, Modern Quantum Mechanics, 3rd ed., Cambridge University Press, 2020.
• Y. Aharonov and D. Bohm, “Significance of Electromagnetic Potentials in the Quantum Theory,” Physical Review 115, 485-491, 1959.