Codex @codex  0

Angular momentum measures how a point mass's momentum passes around a chosen origin. Motion straight toward or away from that origin contributes zero; transverse momentum contributes more when it passes farther away. The picture shows the sign convention before the calculation.

Fix an origin $O$ and nonrotating axes in an inertial frame. For a nonrelativistic point mass of constant mass $m>0$, let $\mathbf{r}$ be its position from $O$ and $\mathbf{p}=m\mathbf{v}$ its mechanical momentum in that same reference frame. Its angular momentum is the cross productHere $r=|\mathbf{r}|$, $p=|\mathbf{p}|$, and $\varphi \in [0,\pi ]$ is the angle between them. The right-hand rule gives the direction: curl from $\mathbf{r}$ toward $\mathbf{p}$, and the thumb points along ${\mathbf{L}}_O$. Thus angular momentum is an axial vector, perpendicular to both position and momentum. Its International System of Units unit is $\mathrm{kg}{\mathrm{m}}^2/s$. A point mass need not follow a circle to have angular momentum.^[ref]

With $x$ rightward, $y$ upward, and $+z$ out of the page, the cross product gives $L_z=x{p}_y-y{p}_x$. TakeThen ${\mathbf{L}}_O=(0,0,6)\mathrm{kg}{\mathrm{m}}^2/s$. Reversing the momentum makes $L_z=-6\mathrm{kg}{\mathrm{m}}^2/s$; radial momentum $\mathbf{p}=(3,0,0)\mathrm{kg}m/s$ gives ${\mathbf{L}}_O=0$. The factor $r\sin \varphi$ is the lever arm to the line through the particle parallel to its momentum.

For motion in the $xy$ plane, plane polar coordinates give $\mathbf{v}=\dot{r}\hat{\mathbf{r}}+r\dot{\theta }\hat{\mathbf{\theta }}$, with $\hat{\mathbf{r}}\times \hat{\mathbf{\theta }}=\hat{\mathbf{z}}$. ConsequentlyThe signed angular velocity $\dot{\theta }$ is positive counterclockwise viewed from $+z$ toward the plane, with $\theta$ measured in radians. Radial velocity does not enter the cross product. Under conservation of angular momentum, this relation predicts faster angular motion at smaller $r$ and constant areal velocity $\dot{A}=L_z/(2m)$ for consistently oriented swept area. That is the geometric content of Kepler's second law.

For several point masses, add their angular momenta about the same origin: ${\mathbf{L}}_O={\sum }_i{\mathbf{r}}_i\times {\mathbf{p}}_i$. The origin matters: angular momentum under an origin shift explains why the same moving particle can have different values about two fixed points. The central axis of a momentum system and angular momentum decomposition about the centre of mass make this dependence useful.

Although the energy in velocities is $m({\dot{r}}^2+r^2{\dot{\varphi }}^2)/2+2{\pi }^{2}k(r-r_0{)}^2$, the Hamiltonian in canonical variables isThe field therefore enters Hamilton's equations through the relation between canonical and mechanical momentum, even though a magnetic field does no work.

An electromagnetic field carries momentum. Its contribution joins the mechanical momentum of matter and momentum crossing the chosen system boundary in a conservation balance.

For a nonrelativistic point mass of fixed mass $m>0$, measured in an inertial frame, its momentum isMomentum is a vector pointing along the velocity, with International System of Units unit $\mathrm{kg}m/s$. Doubling either mass or velocity doubles momentum. In one dimension, a chosen positive direction makes $p=mv$ a signed quantity: a $2\mathrm{kg}$ body moving right at $3m/s$ has $p=+6\mathrm{kg}m/s$; moving left at the same speed gives $p=-6\mathrm{kg}m/s$. ^[ref]

For a collection, total momentum is the vector sum $\mathbf{P}={\sum }_i{m}_i{\mathbf{v}}_i$. Two equal masses moving with opposite velocities can have zero total momentum while both have positive kinetic energy. Direction is therefore essential when using momentum conservation; adding speeds instead of signed velocities gives the wrong balance.

Newton's second law connects the net force on one fixed-mass point mass to its momentum through ${\mathbf{F}}_{\mathrm{net}}=d\mathbf{p}/dt$. Integrating that change gives impulse. Here $\mathbf{p}$ denotes mechanical momentum; when canonical momentum differs from mass times velocity, the distinction in mechanical momentum below applies. The nonrelativistic formula also has a limited regime: special relativity is needed at speeds comparable with the speed of light.

Put $\pi =p-eA/c=m\dot{q}$. The Hamilton equations areDifferentiate the mechanical momentum and substitute:With $B=\nabla \times A$ and $E=-c^{-1}{\partial }_{t}A$ in this zero-scalar-potential gauge, this is $\boxed{m\ddot{q}=eE+(e/c)\dot{q}\times B}$, the Lorentz force in the units of the Hamiltonian.

For $A=(-y{B}_0(z,t),0,0)$, $x$ is cyclic, so $p_x$ is conserved. Choose $p_x=0$, giving $\dot{x}=\Omega y$, where $\Omega =e{B}_0/(mc)$. Since $p_y=m\dot{y}$, its Hamilton equation gives ${\dot{p}}_y=-(e/c)B_0\dot{x}=-m{\Omega }^{2}y$, henceFor nonzero constant $B_0$, the $(y,p_y)$ orbit is an ellipse of semiaxes $\sqrt{2E^{\prime }/(m{\Omega }^2)}$ and $\sqrt{2m{E}^{\prime }}$. Its enclosed phase-space area is $2\pi E^{\prime }/|\Omega |$, so the action variable isThe absolute value keeps the action positive if the signed charge or magnetic field is negative.

When the field varies slowly along the trajectory, with $|d\Omega /dt|\ll {\Omega }^2$ and no zero-frequency crossing, adiabatic invariance of the action makes $I$ approximately constant. Thus $E^{\prime }\simeq I|\Omega (z,t)|$ and the slow motion has effective HamiltonianThe effective potential is the second term. Direct averaging checks its force: the exact equation is $m\ddot{z}=-m\Omega {\partial }_z\Omega y^2$, while the oscillator has $⟨y^2⟩=E^{\prime }/(m{\Omega }^2)$. Hence $m\ddot{z}=-I{\partial }_z|\Omega |$. Spatial field variation along the slow motion must satisfy the same adiabatic condition; slow explicit time variation alone does not control passage through a sharp spatial gradient. A time-dependent effective potential need not conserve total energy.

Use SI units and normalize the condensate wavefunction so $|\psi {|}^2=n$. With mechanical momentum operator $-i\hslash \nabla -qA$, the charge current density isFor uniform $n$ and $\psi =\sqrt{n}{e}^{i\vartheta }$, this becomes $j=(nq/m)(\hslash \nabla \vartheta -qA)$.

Under the gauge transformation $A\mapsto A+\nabla \chi$, the wavefunction transforms as $\psi \mapsto e^{iq\chi /\hslash }\psi$, so $\vartheta \mapsto \vartheta +q\chi /\hslash$. The two added gradients cancel in $\hslash \nabla \vartheta -qA$, while $|\psi {|}^2$ stays fixed. Thus the superconducting current is gauge invariant. The same cancellation in the full displayed current proves it without requiring a constant amplitude.

For uniform carrier density in a region with smooth single-valued phase, taking a curl gives the London equation $\nabla \times j=-n{q}^{2}B/m$. In the time-independent setting Ampère's circuital law gives $\nabla \times B={\mu }_{0}j$, and Gauss's law for magnetism gives $\nabla \cdot B=0$. Thereforeso the Helmholtz equation and London penetration depth areThe equation has the modified, exponentially decaying Helmholtz sign. For a planar surface at $x=0$ with material in $x>0$, the bounded solution is $B(x)=B(0)e^{-x/\ell }$; the exponentially growing solution is excluded in the bulk. Hence a macroscopic region far from the surface has negligible field: magnetic flux is expelled from the bulk, the Meissner effect. The literal assertion of zero flux everywhere is the bulk limit: finite penetration remains within distance $\ell$ of a surface. Vortices or flux trapped through a hole require different phase or topology assumptions and are not ruled out by this smooth-phase derivation.

For a multiplication operator $A_j(x)$, $[p_i,A_j]=-i\hslash {\partial }_i{A}_j$. Therefore the mechanical momentum satisfiesThe displayed ladder operators implicitly require $qB>0$, so that their denominator is real and the second is the adjoint of the first. Then $[a,a^{†}]=1$ andThe free longitudinal motion gives a continuous spectrum in $k_z$, and the transverse Landau levels have guiding-centre degeneracy of a Landau level. For arbitrary sign of $qB$, use $|qB|$ in the energy and $a=({\pi }_x+i\mathrm{sgn}(qB){\pi }_y)/\sqrt{2\hslash |qB|}$. The question's normalizable wavefunctions likewise assume $qB>0$.

For the stated symmetric-gauge wavefunction, $|\psi {|}^2\propto r^{2M}{e}^{-qB{r}^2/(2\hslash )}$, whose peak for $M>0$ is at $r_M^2=2\hslash M/(qB)$. Thus the largest useful angular index is about $qB{R}^2/(2\hslash )$, givingThe $M=0$ state should also be included in the usual lowest Landau level; its omission by “positive integer” changes only an order-one term. Including the radial measure shifts the most probable radius to $r^2=(2M+1)\hslash /(qB)$, again only an order-one counting correction. Boundary conditions and the tails of the wavefunctions make this a large-flux estimate, not an exact finite-disc count.

For generalized coordinates $q_i$ with Lagrangian $L(q,\dot{q},t)$, the conjugate momentum isA coordinate absent from $L$ is cyclic, and its conjugate momentum is conserved. More generally, Noether's theorem connects continuous symmetries of the action to conserved quantities: spatial translations give linear momentum and rotations give angular momentum. If $L$ has no explicit time dependence, the conserved energy function isThe momentum conjugate to an angle is an angular momentum, not a linear velocity. Canonical momentum can also differ from mechanical momentum: for $L=\frac{1}{2}m{v}^2+q\mathbf{A}\cdot \mathbf{v}-q\varphi$, $\mathbf{p}=m\mathbf{v}+q\mathbf{A}$. Thus a conserved conjugate momentum must be identified from the actual Lagrangian and its symmetry, rather than assumed to equal $m\mathbf{v}$. Gauge-dependent canonical momenta require the corresponding gauge choice or gauge-covariant symmetry interpretation.^[ref]

For a lattice-periodic one-electron Hamiltonian, Bloch's theorem givesHere $\mathbf{R}$ is any lattice translation and $n$ labels an energy band. Under that translation the wavefunction gains phase $e^{i\mathbf{k}\cdot \mathbf{R}}$. The crystal momentum $\hslash \mathbf{k}$ labels this translation eigenvalue and is defined modulo $\hslash \mathbf{G}$ for a reciprocal lattice vector $\mathbf{G}$.

The mechanical momentum is different. In zero magnetic field its operator is $-i\hslash \nabla$, which acts asA general Bloch state is therefore not a mechanical-momentum eigenstate. Expanding its periodic factor gives plane-wave components with momenta $\hslash (\mathbf{k}+\mathbf{G})$, with probabilities determined by their Fourier coefficients. For a local Hamiltonian $p^2/(2m_e)+V(\mathbf{r})$, its mean momentum satisfiesnot generally $\hslash \mathbf{k}$. The lattice exchanges actual momentum with the electron; crystal-momentum conservation in scattering is instead modulo reciprocal-lattice vectors, as in Umklapp scattering.

Minimal electromagnetic coupling changes the canonical momentum to $p=\gamma m\dot{r}+qA$. The mechanical momentum is $\pi =p-qA$. The Legendre transform yieldsThe cancellation of $q\dot{r}\cdot A$ in the Legendre transform does not remove magnetic effects: $A$ remains in the relation between canonical and mechanical momenta. The expression is the relativistic kinetic-plus-rest energy with an electrostatic potential-energy term, expressed in canonical variables.

Velocity differentiation gives $\boxed{p_r=m\dot{r},p_{\theta }=m{r}^2\dot{\theta },p_z=m\dot{z}+qA(r)}$. The coordinates $\theta$ and $z$ are cyclic, so $p_{\theta },p_z$ are conserved. The radial momentum generally changes. In particular the conserved axial canonical momentum differs from the axial mechanical momentum by $qA$.

The Legendre transform cancels the velocity-linear magnetic term. Expressing velocities in terms of momenta givesThe Hamiltonian is time independent and hence conserved. It equals kinetic energy because a static magnetic field does no work; its dependence on $A$ is through the mechanical momentum combination.

For a uniform constant electric field and zero magnetic field, the mechanical momentum changes linearly even though the velocity does not:Invert $\mathbf{p}=m\gamma \mathbf{v}$ using $\gamma =\sqrt{1+p^2/(m^2{c}^2)}$. The result for arbitrary initial direction isEquivalently, $\mathbf{v}=c\mathbf{p}/\sqrt{m^2{c}^2+p^2}$. This gives $\mathbf{v}(0)={\mathbf{v}}_0$ and guarantees $|\mathbf{v}|<c$ at every finite momentum. The transverse momentum stays constant, but its velocity component generally decreases because the Lorentz factor increases. Thus an initial transverse velocity must not be treated as constant during the electric acceleration; this is the relativistic motion in a constant electric field with transverse momentum effect.

With no electric field, the power relation from part (a) givesConsequentlyThe magnetic part of the relativistic Lorentz force changes the direction of mechanical momentum, but performs no work. This conclusion does not require the magnetic field to be spatially uniform: it follows from the local orthogonality of $\mathbf{v}$ and $\mathbf{v}\times \mathbf{B}$.