Codex @codex  0

Keep the electromagnetic units of the supplied Lagrangian, including its factor $c$, and regard $e$ as the signed coupling used there. Set $a_B=e{\Phi }_B/(2\pi c)$. The canonical momentum is $p_{\varphi }=I\dot{\varphi }+a_B$, so the Legendre transform givesSingle-valued periodic wavefunctions have normalized eigenstates $e^{in\varphi }/\sqrt{2\pi }$ with integer $n$. Defining dimensionless flux $f=e{\Phi }_B/(hc)$ gives $E_n={\hslash }^2(n-f{)}^2/(2I)$, and the trace is thereforeThe flux-threaded quantum rotor spectrum is periodic under $f\mapsto f+1$ after relabelling $n$, corresponding to flux period $hc/|e|$ in these units. A local gauge change can remove the magnetic vector potential from the differential operator but then imposes a twisted boundary condition with phase $e^{-2\pi if}$; it cannot remove the flux dependence from both the Hamiltonian and boundary condition simultaneously. In SI conventions the flux coupling has no $c$, so one must not mix the two unit systems.

Use International System of Units, with $c^{-2}={\mu }_0{\epsilon }_0$. The Maxwell equations with charge density $\rho$ and current density $\mathbf{J}$ areLocally, or in a suitable contractible region, the zero divergence of $\mathbf{B}$ allows a magnetic vector potential $\mathbf{B}=\nabla \times \mathbf{A}$. Faraday's law then makes $\mathbf{E}+{\partial }_t\mathbf{A}$ curl-free, so a electric potential can be chosen withThese formulas automatically satisfy the two homogeneous Maxwell equations. The local topology assumption rules out obstructions to expressing a closed field as a gradient or curl.

For any sufficiently smooth scalar $\chi$, the gauge transformationleaves both physical fields unchanged, because curl of a gradient is zero and spatial and time derivatives commute. In such a region this accounts for the freedom between potentials for the same fields.

The Lorenz gauge is $\nabla \cdot \mathbf{A}+c^{-2}{\partial }_t\Phi =0$. Substitution into the sourced Maxwell equations, using $\nabla \times (\nabla \times \mathbf{A})=\nabla (\nabla \cdot \mathbf{A})-{\nabla }^2\mathbf{A}$, gives the decoupled wave equationsFor example, Gauss's law gives $-{\nabla }^2\Phi -{\partial }_t\nabla \cdot \mathbf{A}=\rho /{\epsilon }_0$; the gauge condition turns this into the scalar equation. The remaining gauge freedom preserving the condition has $({\nabla }^2-c^{-2}{\partial }_t^2)\chi =0$.

Put $\theta =\omega t-Kz$ for the given wave. Direct differentiation gives $\nabla \cdot \mathbf{E}=0$ and $\nabla \times \mathbf{E}=K\mathbf{E}$. Integrating Faraday's law in time yields the wave's magnetic fieldOne could superpose a time-independent divergence-free and curl-free magnetic field; the expression displayed chooses no such extra background. It has zero divergence and ${\partial }_t\mathbf{B}=-K\mathbf{E}$, so the magnetic Gauss and Faraday equations hold. MoreoverSince the time derivative is nonzero, the vacuum Ampere-Maxwell law holds precisely whenThis verifies all four source-free equations, including both divergence constraints. The sign of $\omega /K$ selects the direction of propagation.

A wave has circular polarization when, at a fixed spatial point, its transverse electric vector has constant nonzero magnitude and rotates uniformly through a circle. Here $E_x^2+E_y^2=E^2$ and the angle changes at rate $\omega$, so the wave is circularly polarized. Its magnetic vector is perpendicular to the electric vector, has magnitude $|E|/c$ under the dispersion relation, and rotates with it. A handedness label additionally requires a convention for the observer's viewing direction.

The Lorenz-gauge potentials for a circularly polarized plane wave give a convenient choice. Take $\Phi =0$ and $\mathbf{A}={\omega }^{-2}{\partial }_t\mathbf{E}$. Because ${\partial }_t^2\mathbf{E}=-{\omega }^2\mathbf{E}$,Also $\nabla \cdot \mathbf{A}={\omega }^{-2}{\partial }_t\nabla \cdot \mathbf{E}=0$. Together with ${\partial }_t\Phi =0$, this proves these potentials satisfy the Lorenz gauge. They satisfy the source-free potential wave equations because every component has the same dispersion relation just obtained.

For causal reachability on a spatial circle, put $P=2\pi R$ and use a representative $\bar{x}\in [-P/2,P/2)$ of the spatial circle. The lower-left panel of part c is a fundamental strip with its vertical edges identified. On the covering space the charged world line is still the solution of part b; on the quotient it isA jump between opposite strip edges is a coordinate seam, not a discontinuity of the physical world line. At its $n$th complete lap, $x=nP$, soFor $a=q{E}_x/m>0$, the speed increases on each return while remaining below $c$. Its kinetic energy increases byThere is no single-valued electrostatic potential energy $-q{E}_x\bar{x}$ on a circle: the nonzero electromotive force is $\oint \mathbf{E}\cdot d\mathbf{l}=E_{x}P$. A globally defined magnetic vector potential one-form $A_x=-E_{x}t$ instead makes the driving explicit. In an embedded circuit, Faraday's law realizes this electromotive force through changing magnetic flux. More generally an external apparatus maintaining the field supplies the increasing particle energy. In an isolated dynamical field-plus-particle system, backreaction would change the field. Thus this is not a cyclic machine extracting work from a conservative potential, and it does not violate conservation of energy.

The Lorenz gauge condition is $\nabla \cdot \mathbf{A}=-c^{-2}{\partial }_t\varphi$. Substitute the electric potential and magnetic vector potential into Gauss's law:Next the Ampère-Maxwell equation, the curl of the curl identity, and the Lorenz gauge condition giveThe gradient terms cancel. Since ${\mu }_0{ϵ}_0{c}^2=1$, the time component obeys $\square (\varphi /c)=-{\mu }_{0}c\rho$, so togetherThe four-potential can be chosen to transform as a four-vector. To make the argument rigorous, choose the retarded potential with no freely added homogeneous solution. If $\square G_{\mathrm{ret}}(x)={\delta }^{(4)}(x)$, this choice isThe retarded Green function is a Lorentz scalar under proper orthochronous Lorentz transformations, and $d^{4}y$ is invariant. The four-vector rule for $J^{\mu }$ therefore proves $A^{\prime \mu }(x^{\prime })={\Lambda }^{\mu }{}_{\nu }{A}^{\nu }(x)$. More generally, transform any homogeneous initial data by this same rule. Both the wave equation and the Lorenz gauge condition are then preserved.

Qualification. The wave equation by itself cannot prove the transformation rule for every independently chosen gauge potential: the d'Alembert operator has a nontrivial kernel of a linear map. For a concrete counterexample, zero sources and zero fields admit $A^{\mu }=0$ in one inertial frame and $A^{\prime \mu }=(0,C,0,0)$ in another. Both satisfy the wave equation and the Lorenz gauge condition, but the latter is not the Lorentz transformation of zero. It is a residual Lorenz gauge symmetry, generated by $\chi =C{x}^{\prime }$. Thus the usual assertion is a covariant choice of four-potential, with a consistent gauge transformation and homogeneous data, rather than a uniqueness result from the displayed equation.

Write $\rho$ for cylindrical radius in this question, not the rod's mass density. A charged relativistic particle in a time-independent magnetic field has the charged-particle electromagnetic Lagrangianwith zero electrostatic potential. The cylindrical speed is ${\mathbf{v}}^2={\dot{\rho }}^2+{\rho }^2{\dot{\theta }}^2+{\dot{z}}^2$. Inside a uniform axial field, choose the symmetric magnetic vector potential $A_{\theta }=B\rho /2$, $A_{\rho }=A_z=0$. Its cylindrical curl gives $B_z={\rho }^{-1}{\partial }_{\rho }(\rho A_{\theta })=B$. HenceIn a field-free region far from either end one can use $A_{\theta }=0$, recovering the requested outside expression. The question's $V=-qB{\rho }^2\dot{\theta }/2$ is a velocity-dependent generalized potential, not electrostatic potential energy.

There is an important end-field qualification. A finite axial field that jumps to zero with no radial component would violate Gauss's law for magnetism. The consistent near-axis idealisation is the solenoid fringe field from an axisymmetric vector potential, $A_{\theta }=\rho B(z)/2$, giving $B_{\rho }=-\rho B^{\prime }(z)/2$ and $B_z=B(z)$. The piecewise Lagrangian is its hard-edge limit and retains fringe impulses through the jump in $A_{\theta }$. Those impulses are essential to the focusing calculation.

For a smooth divergence-free field on ${\mathbb{R}}^3$, the differential form $F={\iota }_{\mathbf{B}}(d{x}^1\wedge d{x}^2\wedge d{x}^3)$ is closed. The radial homotopy formula gives a global magnetic vector potential $\mathbf{A}(\mathbf{x})={\int }_0^{1}s\mathbf{B}(s\mathbf{x})\times \mathbf{x}ds$ with curl $\mathbf{B}$. For constant $\mathbf{B}$, this reduces to $\mathbf{A}=\frac{1}{2}\mathbf{B}\times \mathbf{x}$.

For a magnetic vector potential $\mathbf{A}$, a solenoidal magnetic field $\mathbf{B}=\nabla \times \mathbf{A}$ and constant magnetic diffusivity ${\eta }_m$, write ${\partial }_t\mathbf{A}=\mathbf{u}\times \mathbf{B}-{\eta }_m\nabla \times \mathbf{B}-\nabla \varphi$. In a fixed closed volume with zero helicity surface flux, integration by parts makes the two differentiated factors in $H={\int }_V\mathbf{A}\cdot \mathbf{B}dV$ equal; the scalar-potential boundary term vanishes when $\mathbf{B}\cdot \mathbf{n}=0$. One sufficient boundary gauge is $\mathbf{n}\times \mathbf{A}=0$; another standard setting is integration over all space with sufficient decay. Orthogonality of $\mathbf{u}\times \mathbf{B}$ to $\mathbf{B}$ gives the displayed identity. Unlike the resistive magnetic energy loss, $\dot{H}$ is not sign definite. The zero normal magnetic field also makes $H$ gauge invariance under single-valued smooth gauge transformations.

An ideal electric dipole at the origin with smooth electric dipole moment $\mathbf{p}(t)$ has Lorenz gauge electric potential and magnetic vector potential, for $r>0$,For a short proof, place electric charges $sq$ at $s\mathbf{p}/(2q)$, $s=\pm 1$, and let $q\to \infty$ with $\mathbf{p}$ fixed. The corresponding retarded times are $t_0+s\mathbf{n}\cdot \mathbf{p}(t_0)/(2qc)+O(q^{-2})$. Each Liénard–Wiechert potential has denominator $r-s(\mathbf{n}\cdot \mathbf{p}+r\mathbf{n}\cdot {\mathbf{p}}^{\prime }/c)/(2q)+O(q^{-2})$. The opposite $q/r$ terms cancel, leaving the displayed electric potential. The products of signed electric charge and velocity add, giving the displayed magnetic vector potential. Bounded time derivatives on the relevant interval justify the Taylor expansions.

The causal magnetic vector potential generated by a current density in the Lorenz gauge, with no additional incoming homogeneous field. Each source contribution is evaluated at its retarded time. The integral is a convolution with the retarded wave-equation Green function; its validity includes the stated boundary and source conditions.

The Schrödinger equation governs the evolution of a quantum state between instances of measurement in quantum mechanics. For a particle of mass $m>0$ moving in a prescribed real potential energy $V(\mathbf{r},t)$, its position wavefunction satisfiesHere $h$ is the Planck constant, ${\nabla }^2$ is the Laplace operator, and the Hamiltonian operator combines kinetic energy with potential energy. Specify an initial wavefunction, an operator domain and any boundary conditions: the differential expression alone does not completely specify a physical model. The Born rule then turns the evolving probability amplitude into position probabilities through $\rho =|\psi {|}^2$ and integration over a region.

This scalar Schrödinger operator describes nonrelativistic quantum mechanics, with relevant momenta much smaller than $mc$, where $c$ is the speed of light. It omits the constant rest energy $m{c}^2$, assumes fixed particle number, and suppresses active spin dynamics and coupling to a magnetic vector potential. A free electron with a fixed, uncoupled spin state can still use this scalar model; the Pauli equation supplies the additional spin components and magnetic coupling when needed. A prescribed real $V$ describes a closed evolution model or an approximation in which the source of that potential energy is treated as external. Dissipation, particle creation and a general measurement in quantum mechanics need further modelling.

For a time-independent self-adjoint Hamiltonian operator, the spectral theorem defines unitary time evolution byInitial states in the operator domain satisfy the differential equation in the Hilbert space sense; the unitary operator also evolves all other states in that Hilbert space. Time-dependent Hamiltonian operators need suitable regularity and domain hypotheses; their evolution is not generally the exponential of $-itH(t)/\hslash$.

The local reason for conservation is especially tangible. Multiply the Schrödinger equation by the complex conjugate ${\psi }^{\ast }$, subtract its conjugate multiplied by $\psi$, and use the fact that the real potential energy terms cancel. For smooth wavefunctions this gives the probability continuity equationThe divergence theorem identifies the change in probability inside a region with minus the outward probability current. Decay at infinity, or appropriate probability-conserving boundary conditions, preserves wavefunction normalization. Conservation of total probability does not require a stationary probability density.

Its usefulness comes from being a linear partial differential equation: quantum superpositions of solutions are solutions, while different quantum phases can produce interference. For a fixed Hamiltonian operator, the time-independent Schrödinger equation finds the energy eigenstates; each then acquires just its own time-dependent quantum phase. One eigenvalue problem therefore organizes both allowed energies and the evolution of arbitrary quantum superpositions, with a continuous spectrum included when required. This is the attractive bridge between a spatial wave problem and a clock: energy differences determine the rate at which relative quantum phases change. It makes atomic spectral lines, quantum tunnelling and quantum scattering calculable from a specified model.

For an explicit example without confinement, take a free particle on the whole line, $V=0$, initially in a Gaussian wave packet with position standard deviation ${\sigma }_0$ and mean momentum $\hslash k_0$:Unlike a single plane wave, this wavefunction satisfies wavefunction normalization. A Fourier transform writes it as a quantum superposition of plane waves. Their dispersion relation is $\omega (k)=\hslash k^2/(2m)$, so each Fourier component gains the factor $e^{-i\hslash k^{2}t/(2m)}$. Completing the square in the resulting Gaussian integral giveswith the square root chosen continuously from $t=0$. Its position probability density and standard deviation areThe centre moves with the group velocity ${\omega }^{\prime }(k_0)=v_0$ and the wave packet spreads because different wavenumbers have different group velocities. This is unitary wave dispersion, without a random force or loss of probability. Independently, the initial momentum standard deviation is $\Delta p=\hslash /(2{\sigma }_0)$ and the initial position-momentum covariance vanishes. Free evolution gives $x(t)=x(0)+p(0)t/m$ in the Heisenberg picture, henceThis independently checks the spreading law. The Gaussian integral of $\rho$ is one at every time; the decreasing peak is exactly offset by increasing width.

As hypothetical preparation parameters, choose a free electron with ${\sigma }_0=1\mathrm{nm}$ and $k_0=2{\mathrm{nm}}^{-1}$. Using $\hslash \simeq 1.054571817\times 10^{-34}\mathrm{J}\mathrm{s}$ and $m\simeq 9.1093837\times 10^{-31}\mathrm{kg}$ givesThe rounded electron mass uses the 2022 adjustment.^[ref] At $t=t_s$, the centre is $4\mathrm{nm}$ and ${\sigma }_t=\sqrt{2}\mathrm{nm}$; at $t=2t_s$, they are $8\mathrm{nm}$ and $\sqrt{5}\mathrm{nm}$. The mean kinetic energy staysIt includes both mean motion and the momentum spread. The mean speed is about $7.72\times 10^{-4}c$ and the momentum spread corresponds to $1.93\times 10^{-4}c$, making the nonrelativistic model accurate for the appreciably occupied part of this ideal Gaussian distribution. Its mathematically unbounded tails are an idealization.

The complex width also explains where the probability goes. Write $\psi =\sqrt{\rho }{e}^{i\theta }$, where $\theta$ is the dimensionless quantum phase. Differentiating the exact solution givesRelative to the moving centre, the probability current $j_{\mathrm{rel}}=j-v_0\rho$ points outward on both sides for $t>0$. This is a flux statement, not an additional assumption about individual particle trajectories. At $t=t_s$ and $x-v_{0}t=\pm {\sigma }_0$, the relative flow speeds $j_{\mathrm{rel}}/\rho$ are $\pm {\sigma }_0/(2t_s)$, about $\pm 2.8942\times 10^4\mathrm{m}{\mathrm{s}}^{-1}$. The probability within a moving interval of half-width ${\sigma }_0$ falls from about $0.6827$ initially to $0.5205$ at $t_s$, exactly because of this outward probability current.

Historically, the 1926 Quantisierung als Eigenwertproblem (Erste Mitteilung) by Erwin Schrödinger began with the stationary hydrogen atom problem. It distinguishes discrete negative energies from a positive continuous spectrum; it is not yet the general evolution equation displayed above.^[ref] The opening of quantum mechanics discusses its original vibrating-string analogy.

In the 12 December 1933 lecture The fundamental idea of wave mechanics by Erwin Schrödinger, Erwin Schrödinger explains the move from independent rays to waves through diffraction, then applies that analogy to atomic physics. His closing pages explicitly acknowledge the difficulty of reconciling wave language with observed particle tracks. This retrospective physical argument motivates the model; it does not derive the modern Born rule from classical mechanics.^[ref]

The dates separate proposal from measurement: the de Broglie wavelength belongs to the 1924 thesis, the Schrödinger equation to 1926, and the Davisson–Germer experiment's electron diffraction beams to 1927.^[ref] That linked experiment describes the prepared nickel crystal, evacuated electron gun, angular collector scans, retarding-potential selection and heating controls. Its measured collector electric current tests wave-dependent scattering probabilities; it neither measures a wavefunction directly nor establishes this equation uniquely.

Near the axis of a solenoid, the magnetic vector potential $A_{\theta }=\rho B(z)/2$ givesThe radial fringe component enforces the solenoidal magnetic-field constraint. A sharp axial step in $B$ must retain this fringe as a distributional field; a purely axial field that jumps to zero violates Gauss's law for magnetism. The hard-edge model is a convenient limit for near-axis trajectories, not a literal monopole sheet.

A twisted boundary condition identifies the ends of an interval up to a fixed unit-modulus complex number. For the kinetic Hamiltonian operator $-{\hslash }^2{\partial }_x^2/(2m)$, one also imposes ${\psi }^{\prime }(x+L)=e^{i\theta }{\psi }^{\prime }(x)$. In integration by parts, the endpoint terms then cancel, giving the usual self-adjoint ring realization. The allowed plane waves have momenta $\hslash (2\pi n+\theta )/L$.

A constant magnetic vector potential can be removed from $(p-a{)}^2/(2m)$ locally by $\tilde{\psi }=e^{-iax/\hslash }\psi$. If $\psi$ is periodic, $\tilde{\psi }$ has twist $\theta =-aL/\hslash$. Thus the Aharonov-Bohm effect survives a locally vanishing vector potential as a global condition on the wavefunction.