Particle on a Ring: First Encounter

A particle on a ring is the simplest angular quantum system. The particle is constrained to move on a circle of radius $R$, so its configuration is described by one angle $\theta$ rather than by a coordinate on the full line.

This page is a first encounter. It explains periodicity, integer angular momentum, and the free-ring spectrum before the fuller angular-systems treatment in Particle on a Ring and Rigid Rotor.

Configuration Space

The angle coordinate satisfies

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

with $\theta=0$ and $\theta=2\pi$ representing the same physical point. For an ordinary scalar wavefunction, this gives the periodic boundary condition

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

For the free kinetic-energy Hamiltonian, the derivative is matched too:

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

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

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

Normalization means

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

The probability density here is density per unit angle. The corresponding density per unit arc length $s=R\theta$ differs by the Jacobian $ds=R\,d\theta$.

Hamiltonian

Let the particle mass be $m_{\mathrm p}$. The moment of inertia for motion around the ring is

$$I=m_{\mathrm p}R^2.$$

The classical kinetic energy can be written as

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

where $L_z$ is the angular momentum about the center of the ring. Quantization gives

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

and therefore

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

This is the same kinetic-energy structure as a free particle in a periodic box, but written in an angular coordinate.

Angular Momentum Eigenfunctions

The angular-momentum eigenvalue equation is

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

With $\hat L_z=-i\hbar d/d\theta$, the solutions are

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

Periodic single-valuedness requires

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

so

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

Thus

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

The normalized eigenfunctions are

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

This page uses $n$ for the ring quantum number to avoid confusing it with the particle mass. Many angular-momentum texts use $m$ for the eigenvalue label.

Energy Spectrum

Since the Hamiltonian is $\hat L_z^2/(2I)$,

$$\hat H u_n = \frac{\hbar^2 n^2}{2I}u_n.$$

The energy levels are

$$E_n = \frac{\hbar^2n^2}{2I} = \frac{\hbar^2n^2}{2m_{\mathrm p}R^2}, \qquad n\in\mathbb Z.$$

The $n=0$ state is the constant wavefunction and has zero kinetic energy in this ideal free-ring model. For $n\ne0$, the states $n$ and $-n$ have the same energy:

$$E_n=E_{-n}.$$

They carry opposite angular momentum,

$$L_z=\hbar n, \qquad L_z=-\hbar n,$$

and represent opposite circulation directions around the ring.

Relation To A Periodic Box

Let $s=R\theta$ be arc length along the ring. The circumference is

$$L=2\pi R.$$

The periodic-box condition

$$\psi(s+L)=\psi(s)$$

gives

$$k_n=\frac{2\pi n}{L} =\frac{n}{R}.$$

The tangential momentum is

$$p_n=\hbar k_n =\frac{\hbar n}{R}.$$

Multiplying by $R$ gives angular momentum:

$$L_z=Rp_n=\hbar n.$$

The energy relation

$$\frac{p_n^2}{2m_{\mathrm p}} = \frac{\hbar^2n^2}{2m_{\mathrm p}R^2}$$

agrees with the angular formula. The ring is therefore the periodic box with its coordinate interpreted geometrically as an angle.

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

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

with angular probability current

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

For the eigenstate $u_n$,

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

Thus $n\gt0$ and $n\lt0$ states have equal and opposite circulating currents. The $n=0$ ground state has no current.

Time Evolution And Superpositions

Each $u_n$ is a stationary state:

$$\psi_n(\theta,t) = u_n(\theta)e^{-iE_nt/\hbar}.$$

A localized wave packet on the ring requires a superposition:

$$\psi(\theta,0) = \sum_{n\in\mathbb Z} c_nu_n(\theta).$$

Its time evolution is

$$\psi(\theta,t) = \sum_{n\in\mathbb Z} c_nu_n(\theta)e^{-iE_nt/\hbar}.$$

Because the spectrum is quadratic in $n$, a localized packet disperses around the ring and can later show revivals. The detailed revival analysis belongs with finite-spectrum and wave-packet dynamics; the main lesson here is that a ring state is naturally expanded in integer angular-momentum modes.

Magnetic Flux Preview

A charged particle on a ring is sensitive to magnetic flux through the ring even when the magnetic field vanishes on the ring itself. This is the one-dimensional seed of the Aharonov–Bohm effect.

For a particle of charge $q$, a magnetic flux $\Phi$ through the ring can be represented by a tangential vector potential. Minimal coupling gives

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

Acting on the same single-valued basis $u_n$, the energies become

$$E_n(\Phi) = \frac{\hbar^2}{2I} \left( n-\frac{\Phi}{\Phi_0} \right)^2,$$

where

$$\Phi_0=\frac{h}{q}$$

is the flux quantum for charge $q$, up to sign conventions for the charge and orientation. The important point is not the sign convention; it is that the flux shifts the angular-momentum spectrum and can split the $n$ and $-n$ degeneracy.

The general electromagnetic-coupling rule is explained in Minimal Coupling in Wave Mechanics. The wave-mechanics phase discussion is Aharonov–Bohm Effect: First Encounter; the deeper topology belongs with geometry and gauge-bundle material.

Common Mistakes

• Treating $\theta=0$ and $\theta=2\pi$ as two independent endpoints.
• Allowing noninteger $n$ for ordinary single-valued scalar wavefunctions without changing the boundary condition.
• Confusing the particle mass $m_{\mathrm p}$ with the angular-momentum quantum number often called $m$.
• Forgetting that $n$ can be negative.
• Saying that the $n=0$ free-ring ground state contradicts zero-point energy; the free ring has no angular confining potential.
• Confusing the one-coordinate ring with the two-coordinate rigid rotor on a sphere.
• Ignoring gauge conventions when writing flux-shifted spectra.

Where This Is Used

• Periodic Boundary Conditions gives the finite-volume plane-wave basis that becomes the ring basis after setting $L=2\pi R$.
• Particle on a Ring gives the canonical angular-systems version of this model, including currents and magnetic-flux shifts.
• Momentum Eigenstates explains the plane-wave eigenvalue equation behind $e^{in\theta}$.
• Orbital Angular Momentum derives $L_z=-i\hbar\partial/\partial\phi$ for angular wavefunctions.
• Angular Momentum Algebra places the integer ring labels in the broader rotation-generator framework.
• Rigid Rotor generalizes from motion on a circle to motion on a sphere.
• Minimal Coupling in Wave Mechanics supplies the electromagnetic rule used in the flux preview.

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).

Exercises

1. Derive the integer quantization of angular momentum on a ring.

Solution

Solve

$$-i\hbar\frac{du}{d\theta} =\lambda u.$$

The solution is

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

Single-valuedness requires

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

so

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

Therefore $\lambda/\hbar$ is an integer:

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

2. Normalize $u_n(\theta)=C e^{in\theta}$ on $0\le\theta\lt2\pi$.

Solution

The normalization condition is

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

Thus

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

With a simple phase convention,

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

3. Show that the ring spectrum follows from the periodic-box spectrum with $L=2\pi R$.

Solution

For a periodic box,

$$k_n=\frac{2\pi n}{L}.$$

On a ring of radius $R$, the circumference is $L=2\pi R$, so

$$k_n=\frac{n}{R}.$$

The free-particle energy is

$$E_n = \frac{\hbar^2k_n^2}{2m_{\mathrm p}} = \frac{\hbar^2n^2}{2m_{\mathrm p}R^2}.$$

Since $I=m_{\mathrm p}R^2$, this is

$$E_n=\frac{\hbar^2n^2}{2I}.$$

4. Which free-ring levels are degenerate?

Solution

The free-ring energy is

$$E_n=\frac{\hbar^2n^2}{2I}.$$

Thus

$$E_n=E_{-n}.$$

For $n\ne0$, the pair $n$ and $-n$ is twofold degenerate. The $n=0$ level is nondegenerate because it is its own negative.

5. A flux $\Phi$ shifts the energy to $E_n(\Phi)=\hbar^2(n-\Phi/\Phi_0)^2/(2I)$. What happens at $\Phi=\Phi_0/2$?

Solution

At half a flux quantum,

$$E_n\left(\frac{\Phi_0}{2}\right) = \frac{\hbar^2}{2I} \left( n-\frac{1}{2} \right)^2.$$

The states $n=0$ and $n=1$ are degenerate because both have squared offset $1/4$:

$$\left(0-\frac{1}{2}\right)^2 = \left(1-\frac{1}{2}\right)^2 =\frac{1}{4}.$$

More generally, the flux shifts which pairs are degenerate.