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A uniformly charged circular ring hasHere $k$ is the spring constant for extension of its full circumference and the symmetric gauge is used. The canonical angular momentum $J=m{r}^2\dot{\varphi }+qB{r}^2/2$ is conserved. Eliminating the angle gives $m\ddot{r}=J^2/(m{r}^3)-q^2{B}^{2}r/(4m)-4{\pi }^{2}k(r-r_0)$. This model neglects electric self-interaction and noncircular deformation.

Robert Laughlin's 1983 variational method uses pairwise zeros to suppress close electron encounters within the lowest Landau level. For aligned spin, odd $m$ gives fermionic exchange symmetry.^[ref]

Here $\mathcal{N}$ enforces wavefunction normalization, ${\ell }_B$ is the magnetic length, and complex coordinates $z_j=x_j-i{y}_j$ describe electrons in a magnetic field $B\hat{\mathbf{z}}$, $B>0$, with the symmetric gauge. Reversing the orientation conjugates the polynomial without changing probability densities. The Gaussian function confines the wavefunction.

The exchange check is especially simple for two electrons. Writing $w=z_1-z_2$, exchange sends $w\mapsto -w$, so $w^m\mapsto (-1{)}^m{w}^m$. An aligned two-electron spin state is symmetric, requiring the spatial wavefunction to change sign. Hence $m=1,3,5,\dots$ works, whereas $m=2$ would give the wrong sign. At a close encounter, the probability density vanishes as $|w{|}^{2m}$: $m=3$ imposes a sixth-order zero, stronger than the quadratic zero at $m=1$. This is the local geometric mechanism for avoiding a large repulsive Coulomb potential energy.

A useful large-$N$ counting outline follows directly from the polynomial. Each complex coordinate has maximum degree $m(N-1)$. A lowest Landau level orbital proportional to $z^k{e}^{-|z{|}^2/(4{\ell }_B^2)}$ has its areal probability density maximum at $r^2=2k{\ell }_B^2$, by differentiating its logarithm. This suggests droplet area $A\simeq 2\pi mN{\ell }_B^2$, neglecting the boundary correction. Dividing $N/A$ by the degeneracy of a Landau level per area, $1/(2\pi {\ell }_B^2)$, gives filling factor $\nu \simeq 1/m$. This counting is a motivation for the fraction, not an independent proof that the wavefunction minimizes the interacting Hamiltonian.

For a charged particle in a uniform magnetic field, the symmetric gauge $\mathbf{A}=\mathbf{B}\times \mathbf{x}/2$ makes the displayed magnetic pseudomomentum a conserved quantity when no other position-dependent potential breaks translation symmetry. It differs from the kinetic momentum $\mathbf{\pi }=\mathbf{p}-e\mathbf{A}=m\dot{\mathbf{x}}$. With $\mathbf{B}=B{\mathbf{e}}_3$, Hamilton's equations for $H=|\mathbf{p}-e\mathbf{A}{|}^2/(2m)$ give ${\dot{p}}_1=(eB/2){\dot{x}}_2$ and ${\dot{p}}_2=-(eB/2){\dot{x}}_1$, proving conservation of $K_1=p_1-eB{x}_2/2$ and $K_2=p_2+eB{x}_1/2$. The charge factor is necessary unless it has already been absorbed into the field convention. In quantum mechanics the transverse components need not commute, even though each commutes with this Hamiltonian operator; magnetic translation algebra records that distinction.

With no electric force, the Lorentz force equation isIts scalar product with velocity proves conservation of energy for the particle's kinetic energy:If $\mathbf{B}=\mathcal{B}(\mathbf{x},t)\hat{\mathbf{b}}$ with a fixed unit direction $\hat{\mathbf{b}}$, then $m\ddot{\mathbf{x}}\cdot \hat{\mathbf{b}}=0$, so the parallel component of velocity is constant even when the field magnitude varies.

The angular momentum about the origin does not generally remain constant, becauseFor example, at an instant with $\mathbf{x}=(r_0,0,0)$, $\dot{\mathbf{x}}=(v_0,0,0)$ and $\mathbf{B}=(0,0,B)$, its derivative is $(0,0,-eB{r}_0{v}_0)$, nonzero when those factors are nonzero. A force perpendicular to velocity can preserve kinetic energy while exerting a nonzero torque.

For the constant axial field, Cartesian equations are $m\ddot{x}=eB\dot{y}$, $m\ddot{y}=-eB\dot{x}$, and $\ddot{z}=0$. Use the signed projected areal rate $\dot{A}=(x\dot{y}-y\dot{x})/2=r^2\dot{\theta }/2$. ThenThusThe second equation is the magnetic axial angular-momentum invariant, replacing ordinary conservation of $L_z=m{r}^2\dot{\theta }$. In the symmetric gauge, the added term is the magnetic contribution to canonical angular momentum; it is the axial component that is conserved, not the full mechanical angular-momentum vector.

Let $v^2={\dot{x}}^2+{\dot{y}}^2$ be the constant squared transverse speed. If the motion has a nonzero parallel speed, $v^2=2K/m-{\dot{z}}^2$, where $K$ is total kinetic energy. Solving the invariant and using $v^2={\dot{r}}^2+r^2{\dot{\theta }}^2$ givesThe TeX transcription omits the angular equation, which is present in the PDF and restored here. The PDF's positive radial root describes the outward branch only. An inward branch needs the minus sign, with sign changes at radial turning points; polar coordinates themselves are singular at $r=0$. These qualifications are needed for a global description of a cyclotron trajectory whose radius about an arbitrary origin can both increase and decrease.

The initial time-dependent-field discussion treats the stipulated magnetic force as a prescribed particle model. If one additionally imposes full Maxwell equations with electric field zero throughout a region, Faraday law requires the magnetic field there to be time independent. The energy calculation is nevertheless valid for the stated force model; it does not assert existence of a Maxwell-consistent time-dependent field with zero electric field everywhere.

With the magnetic minimal coupling convention for charge $e$,Here $\nabla \cdot \mathbf{A}=0$, ${\mathbf{A}}^2=B^2(x^2+y^2)/4$, andTherefore in Cartesian coordinatesIn cylindrical coordinates this isThis choice of $\mathbf{A}$ is the symmetric gauge.

In the symmetric gauge,Together with $[p_i,f(x)]=-i\hslash {\partial }_{i}f$, this giveswhereasThusThese are the commutators of the kinetic momentum and magnetic pseudomomentum.

Gauge freedom. The magnetic vector potential satisfies $\mathbf{B}=\nabla \times \mathbf{A}$. The transformationleaves $\mathbf{B}$ unchanged because the curl of a gradient is zero. For time-dependent fields, transform the electric scalar potential as $\varphi \mapsto \varphi -{\partial }_t\chi$ as well, so that $\mathbf{E}=-\nabla \varphi -{\partial }_t\mathbf{A}$ is unchanged. The scalar $\chi$ must have the regularity and single-valuedness appropriate to the physical region and its boundary conditions. Local values of $\mathbf{A}$ therefore depend on the gauge transformation.

Gauge and nonzero component. In the standard axisymmetric Coulomb gauge representative for the long-solenoid approximation,The divergence vanishes since $A_{\varphi }$ is independent of the azimuth. Inside, this agrees with the usual symmetric gauge representative $\mathbf{A}=\frac{1}{2}\mathbf{B}\times \mathbf{r}$. The Coulomb gauge condition alone does not force the other two components to vanish; the symmetry, regularity and customary boundary choice select this representative from the remaining gauge freedom.

Radial dependence. Write the interior magnetic field as $B_0\hat{\mathbf{z}}$, where $B_0={\mu }_{0}nI$ for a vacuum solenoid of solenoid turn density $n$. Its exterior magnetic field vanishes in the infinitely long idealization. Applying Stokes theorem to a coaxial circle gives $2\pi r{A}_{\varphi }(r)=\Phi (r)$. Therefore the Coulomb-gauge vector potential of a long solenoid isIt grows linearly inside, is continuous at $R$, and decays as $r^{-1}$ outside. Although its exterior curl is zero, its circulation is the enclosed magnetic flux $\Phi =B_0\pi R^2$. No regular single-valued gauge function removes it throughout the exterior while preserving this circulation.

Electric field of the rotating charged surface. For clarity retain absolute permittivity $ϵ$ and permeability $\mu$ in the cylinder formulae; setting them to unity gives the rationalized-unit convention in the question. If the stated unity means relative material constants in SI, use ${ϵ}_0,{\mu }_0$ instead. Assume prescribed charge densities in the laboratory frame and the nonrelativistic, steady-rotation model.

The axisymmetric surface charge is stationary as a distribution even while the material rotates. A Gaussian cylinder of length $L$ has enclosed charge zero for $r<R$ and $2\pi RL\sigma$ for $r>R$. Gauss's law therefore givesThe radial electric field jumps from zero to $\sigma /ϵ$ at the surface and then decays as $r^{-1}$. Its direction reverses for negative $\sigma$. The angular velocity does not enter this electric-field result for a specified surface charge.

Magnetic field of the volume-charged cylinder. A shell of radius $a$ and thickness $da$ has surface charge $\rho da$, azimuthal surface current $d{K}_{\varphi }=\rho \omega ada$, and the field of an infinite solenoid: $d{B}_z=\mu \rho \omega ada$ inside that shell and zero outside it. At radius $r<R$, only shells with $a>r$ contribute. Thus the magnetic field of a rotating charged solid cylinder isWith no applied background magnetic field, $\mathbf{B}=0$ outside. The profile is parabolic, largest on the axis and zero at the boundary. The sign follows $\rho \omega$. As a check, the current density is $J_{\varphi }=\rho \omega r$, and Ampère's circuital law gives $-d{B}_z/dr=\mu J_{\varphi }$, as required.

Electron impulse when the current is switched off. Choose positive $I$ so the initial magnetic flux is ${\Phi }_i={\mu }_{0}nI\pi R^2$ along $+z$. Faraday's law of induction gives, outside the solenoid,For an electron of electric charge $q=-e$, treat its position as fixed while the short pulse acts. The inductive impulse outside a switched solenoid isConsequently the nonrelativistic final velocity isThe induced electric field circulates in the positive azimuthal direction while a positive flux decreases, so the electron accelerates in the negative direction. Once the pulse has passed, there is no exterior field and it travels along the tangent at the impulse point, rather than continuing in a circular orbit. If the resulting momentum is relativistic, use ${\mathbf{v}}_f=c{\mathbf{p}}_f/\sqrt{m_e^2{c}^2+|{\mathbf{p}}_f{|}^2}$ instead. The sudden-switch result assumes negligible displacement during the pulse and the usual quasistatic electromagnetic induction approximation; “instantaneous” switching does not mean instantaneous propagation of fields to arbitrary distance.

One can check the sign using conserved canonical angular momentum: axial symmetry conserves $p_{\varphi }^{\mathrm{can}}=L_z+qr{A}_{\varphi }$. Initially $L_z=0$, and finally $A_{\varphi }=0$, so $L_z=q{\Phi }_i/(2\pi )=-e{\Phi }_i/(2\pi )$.

Interferometric measurement at constant current. Use a coherent electron beam split into two paths passing on opposite sides of a shielded long solenoid, then recombine them to form interference fringes. Keep the current constant during the passage and suppress leakage fields on the paths. In the Aharonov-Bohm effect, the phase difference due to the magnetic vector potential is the closed-loop integralThe overall sign reverses if the path orientation is reversed. The displacement in fringe periods is $\Delta \phi /(2\pi )$, so the measured fringe shift is linear in $I$ and repeats whenA gauge transformation adds a gradient whose closed-loop integral vanishes for a single-valued $\chi$. Endpoint changes in electron wavefunction phase cancel in the interference measurement. The measured Aharonov-Bohm phase is therefore gauge invariant: it determines the flux, or equivalently the circulation of $\mathbf{A}$. In the chosen Coulomb gauge this implies $A_{\varphi }(d)={\mu }_{0}nI{R}^2/(2d)$, but the experiment does not select or measure a unique local vector potential independently of gauge.

Far field of the rotating sphere. Interpret uniform charge as uniform volume charge of total $Q$, and take $\mathbf{\omega }=\omega \hat{\mathbf{z}}$. The magnetic dipole moment of a rigidly rotating charge distribution isIts axial component is $m_z=(\rho \omega /2)\int r_{\perp }^2{d}^{3}r$. For a solid sphere, $\int r_{\perp }^2{d}^{3}r=8\pi R^5/15$ and $\rho =3Q/(4\pi R^3)$, soThe magnetic field of a rotating charged spherical body is a static magnetic dipole field at large distances:In particular, $|\mathbf{B}|={\mu }_0|\mathbf{m}|\sqrt{1+3{\cos }^2\vartheta }/(4\pi r^3)$. Both the angular mean of the magnitude and its root mean square therefore scale as $\boxed{r^{-3}}$; the latter is $B_{\mathrm{rms}}=\sqrt{2}{\mu }_0|\mathbf{m}|/(4\pi r^3)$. The angular average of the vector is zero, which is different from average strength. Steady rotation leaves the charge and current distributions stationary, so this is magnetostatics, not a radiative $r^{-1}$ field. If uniform charge instead means a thin spherical surface, replace the dipole moment by $Q{R}^2\mathbf{\omega }/3$; the angular pattern and $r^{-3}$ scaling are unchanged.

Use the magnetic-coupling convention of the supplied Hamiltonian; electromagnetic unit factors are absorbed into the stated $e$ and $B$. Let $\alpha =m_1/M$ and $\beta =m_2/M$, soThe kinetic energy separates into $M{\dot{\mathbf{R}}}^2/2+\mu {\dot{\mathbf{r}}}^2/2$. The symmetric gauge coupling ${\sum }_i{e}_i{\dot{\mathbf{r}}}_i\cdot (\mathbf{B}\times {\mathbf{r}}_i)/2$ becomesThe scalar triple product identity givesDiscarding $-e/2$ times that total derivative, which changes endpoint terms and canonical momenta but not the trajectories, leaves

Choose electric scalar potential $\Phi =-Ex$ and the symmetric gauge $\mathbf{A}=B(-y,x,0)/2$, whose curl is $B\hat{\mathbf{z}}$. Minimal coupling gives $L=m{\dot{\mathbf{r}}}^2/2-k{\mathbf{r}}^2/2+q\mathbf{A}\cdot \dot{\mathbf{r}}-q\Phi$. Since $x=\rho \cos \theta$ and $x\dot{y}-y\dot{x}={\rho }^2\dot{\theta }$,The magnetic term is velocity dependent; it is not an ordinary position-only potential.

Every element of the uniformly distributed mass has speed squared ${\dot{r}}^2+r^2{\dot{\varphi }}^2$, so the total kinetic energy is $m({\dot{r}}^2+r^2{\dot{\varphi }}^2)/2$. The extension of the ring's full circumference is $2\pi (r-r_0)$, and the stated spring constant therefore givesThe magnetic coupling for each charge element is $dq\mathbf{A}\cdot \mathbf{v}$. With the symmetric gauge $\mathbf{A}=(Br/2)\hat{\mathbf{\varphi }}$ and tangential speed $r\dot{\varphi }$, integration over the ring gives $qB{r}^2\dot{\varphi }/2$. ThusThis idealization neglects charge self-energy, radiation and deformation away from a circular ring. The elastic constant refers to extension of the whole circumference; replacing it by a radial spring constant without the factor $4{\pi }^2$ changes the problem.

Time-translation symmetry conserves the energy $H$, and axial rotational symmetry makes $\varphi$ a cyclic coordinate, conservingThe radial Euler-Lagrange equation before elimination isInsert $\dot{\varphi }=J/(m{r}^2)-qB/(2m)$. The cross terms linear in $JqB$ cancel, leavingEquivalently use the reduced radial potentialThe last term is constant at fixed $J$ and has no radial derivative, but is part of the full conserved energy. Conservation of $J$ is stated in the rotationally symmetric gauge; the conserved mechanical combination can be written in terms of magnetic flux in a gauge-independent way.