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. For a free scalar particle with ordinary periodic boundary conditions and no flux, this gives integer angular momentum and equal energies for clockwise and counterclockwise modes.

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.

Helpful background. Time-Independent Schrödinger Equation to formulate an energy eigenvalue problem.

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

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

In the ordinary periodic representation, Hamiltonian-domain wavefunctions are single-valued on the circle (generic $L^2$ states need not have pointwise endpoint values):

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

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:

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

The free ring combines continuous rotational symmetry with time reversal: the rotations give distinct angular-momentum states, and time reversal pairs their equal energies. Time reversal alone does not protect a twofold degeneracy for a spinless particle. A real angle-dependent potential can mix the two modes and split them, as Exercise 3 shows. 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 canonical angular momentum remains $\hbar m$, while mechanical angular momentum is $\hbar(m-\alpha)$ and carries the current. 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.

For a normalized mode, the probability current and circulating charge current are

$$J_{\theta,m}(\Phi)=\frac{\hbar(m-\alpha)}{2\pi I}, \qquad \mathcal I_m=qJ_{\theta,m}=-\frac{\partial E_m}{\partial\Phi}.$$

The derivative is taken on a fixed eigenstate branch. The spectrum as a set is periodic in $\alpha$ with period $1$, since $E_{m+1}(\alpha+1)=E_m(\alpha)$; a fixed branch need not be periodic.

Equivalently, one may remove the vector potential locally and impose a twisted boundary condition. In this convention $\chi(\theta)=e^{-i\alpha\theta}\psi(\theta)$ obeys $\chi(\theta+2\pi)=e^{-2\pi i\alpha}\chi(\theta)$ and the same twist for its derivative. Exercise 1 verifies the sign. 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. Verify the equivalence between the periodic-gauge flux Hamiltonian and the free differential operator acting on twisted functions $\chi=e^{-i\alpha\theta}\psi$. Why can the twist be discarded only at integer $\alpha$?

Solution

Write $\psi=e^{i\alpha\theta}\chi$. Direct differentiation gives

$$(-i\partial_\theta-\alpha)\psi=e^{i\alpha\theta}(-i\partial_\theta)\chi,$$

so $H\psi=e^{i\alpha\theta}[-\hbar^2\chi''/(2I)]$. Periodicity of $\psi$ and $\psi'$ gives the twist $e^{-2\pi i\alpha}$ for both $\chi$ and $\chi'$. The allowed free-operator modes are therefore $e^{i(m-\alpha)\theta}$, with the same flux-shifted energies. Their ordinary derivative current equals the mechanical current in the periodic gauge.

Only when $e^{-2\pi i\alpha}=1$ is this the ordinary periodic domain. Removing the vector potential while silently keeping periodic functions at noninteger $\alpha$ changes the physical problem and loses the flux dependence.

2. For one particle in the lowest energy level, determine the circulating charge current as a function of $\alpha$. What happens at $\alpha=1/2$? Compare an equal mixture of $u_0,u_1$ with their coherent equal superposition at that flux.

Solution

Away from half-integers, the minimizing label $m_*$ is the nearest integer to $\alpha$, giving

$$\mathcal I_{\rm gs}=\frac{q\hbar}{2\pi I}(m_*-\alpha).$$

This is a piecewise linear periodic ground-state current. At $\alpha=1/2$, $u_0$ and $u_1$ are degenerate and carry opposite currents $\mp q\hbar/(4\pi I)$. The energy minimum has a cusp there, so its derivative does not assign a unique current to the degenerate ground space. The preparation matters.

Both the equal incoherent mixture and $\psi=(u_0+u_1)/\sqrt2$ have zero current. Their densities differ:

$$\rho_{\rm mix}=\frac{1}{2\pi}, \qquad \rho_{\rm coherent}=\frac{1+\cos\theta}{2\pi}.$$

For the coherent state, substitution into $J_\theta=(\hbar/I)[\operatorname{Im}(\psi^*\partial_\theta\psi)-\alpha|\psi|^2]$ shows that the current vanishes pointwise at half flux. The nonuniform density is stationary because its two components have the same energy.

3. With no flux, add a weak real potential $V(\theta)=\varepsilon\cos(2\theta)$, where $|\varepsilon|\ll B=\hbar^2/(2I)$. Does time reversal keep the first excited $m=\pm1$ pair degenerate?

Solution

The Fourier components of $V$ change $m$ by $\pm2$. Its matrix in the ordered pair $(u_1,u_{-1})$ is

$$V_{\rm pair}=\frac{\varepsilon}{2}\begin{pmatrix}0&1\\1&0\end{pmatrix}.$$

The degenerate perturbation prescription therefore gives energies $B\pm\varepsilon/2+O(\varepsilon^2/B)$. The leading eigenfunctions can be chosen as $\cos\theta/\sqrt\pi$ and $\sin\theta/\sqrt\pi$, which are real standing waves with zero current.

The real potential preserves spinless time reversal but breaks continuous rotation symmetry. There is no protected doublet: time reversal can leave each real eigenstate invariant rather than map it to an independent equal-energy state.

References

• Y. Aharonov and D. Bohm, “Significance of Electromagnetic Potentials in the Quantum Theory,” Physical Review 115, 485–491 (1959), doi:10.1103/PhysRev.115.485.
• D. J. Griffiths and D. F. Schroeter, Introduction to Quantum Mechanics, 3rd ed., Cambridge University Press, 2018.
• J. J. Sakurai and J. Napolitano, Modern Quantum Mechanics, 3rd ed., Cambridge University Press, 2020.
• R. Shankar, Principles of Quantum Mechanics, 2nd ed., Springer, 1994.