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Operators, quantization and rotations. Angular momentum is the generator of spatial rotations. In quantum mechanics, its components are Hermitian operators on their appropriate domains and obeyThese angular momentum commutation relations prevent all three components from having simultaneous sharp values. For example, the Robertson uncertainty principle gives $\Delta J_x\Delta J_y\ge \hslash |⟨J_z⟩|/2$. One conventionally diagonalizes ${\hat{J}}^2$ and a selected component ${\hat{J}}_z$:The allowed labels are $j=0,\frac{1}{2},1,\frac{3}{2},\dots$ and $m=-j,-j+1,\dots ,j$. An angular-momentum multiplet has $2j+1$ states. The magnitude $\hslash \sqrt{j(j+1)}$ differs from the largest possible component $\hslash j$; the remaining components fluctuate even in a state with maximal $m$.

The angular momentum lowering operator and its raising counterpart, ${\hat{J}}_{\pm }={\hat{J}}_x\pm i{\hat{J}}_y$, satisfyThe ladder terminates at $m=\pm j$. A rotation through angle $\theta$ about unit vector $\hat{\mathbf{n}}$ acts through the unitary operator $\exp [-i\theta \hat{\mathbf{n}}\cdot \hat{\mathbf{J}}/\hslash ]$. Thus quantized components and the continuous rotation of a state are compatible.

Orbital angular momentum. The orbital contribution is associated with motion in space:For a spinless spatial wavefunction, the simultaneous angular eigenfunctions are the spherical harmonics $Y_{\ell m}(\theta ,\varphi )$, with $\ell =0,1,2,\dots$ and $m=-\ell ,\dots ,\ell$. Single-valued ordinary scalar spatial wavefunctions give integer orbital labels. In a central potential energy, $[\hat{H},{\hat{L}}^2]=[\hat{H},{\hat{L}}_z]=0$, so stationary states may be labelled by these angular quantum numbers. The angular kinetic contribution appears in the radial equation as $\ell (\ell +1){\hslash }^2/(2m{r}^2)$, the centrifugal potential. Noncentral interactions need not conserve these orbital labels.

Intrinsic spin and addition. Spin angular momentum is an intrinsic internal degree of freedom with the same commutation algebra. It is not the angular momentum of a small classical charged sphere literally rotating about its centre. For an electron, $s=1/2$ and $m_s=\pm 1/2$, so ${\hat{S}}_i=(\hslash /2){\sigma }_i$ in terms of the Pauli matrices. Its squared spin is $3{\hslash }^2/4$. Other particles can have integer or half-integer spin, including spin zero.

For one particle with orbital and spin degrees of freedom, $\hat{\mathbf{J}}=\hat{\mathbf{L}}+\hat{\mathbf{S}}$; the two component sets commute with each other. Addition of angular momentum permits $j=|\ell -s|,|\ell -s|+1,\dots ,\ell +s$. The coupled states $|j,m_j⟩$ are linear combinations of the uncoupled $|\ell ,m_{\ell }⟩|s,m_s⟩$ with $m_j=m_{\ell }+m_s$, determined by Clebsch-Gordan coefficients. A rotationally invariant spin-orbit interaction can conserve $J^2,J_z$ while mixing the separate component labels.

Angular momentum also carries a magnetic moment. For an electron's orbital motion and intrinsic spin,The minus signs reflect negative electric charge. With the Bohr magneton ${\mu }_B=e\hslash /(2m_e)$, a spin projection $m_s$ has ${\mu }_z=-g_s{\mu }_B{m}_s$. A magnetic field can therefore distinguish angular-momentum states through their interaction energy $-\mathbf{\mu }\cdot \mathbf{B}$.

Stern–Gerlach measurement. In the Stern-Gerlach experiment, atoms emerge from a heated source, pass through collimating apertures, traverse a strongly inhomogeneous magnetic field, and strike a position-sensitive screen. The beam is neutral, so its separation is interpreted through the atomic magnetic moment rather than the ordinary charged-particle Lorentz force. Near a chosen quantization direction, the interaction givesA uniform field can exert a torque but would not separate the beam by this translational force; the gradient is essential. Different allowed magnetic-moment components become correlated with different trajectories. A classical ensemble with continuously varying orientations would yield a continuous spread, whereas discrete quantum components yield separated channels.

In the modern electronic description of ground-state silver, filled shells cancel their electronic angular momenta and the unpaired $5s$ electron has $L=0$, $S=1/2$, $J=1/2$. The effective electronic measurement therefore has two components, approximately ${\mu }_z=\pm {\mu }_B$, rather than an orbital $\ell =1/2$, which is not allowed. For positive field gradient the electronic $m_J=+1/2$ component has negative ${\mu }_z$ and deflects downward in this coordinate convention. Nuclear spin and hyperfine structure are neglected in this simplified two-channel account. Historically the original splitting preceded the electron-spin interpretation, so the modern explanation should not be confused with the original investigators' interpretation.

More generally a prepared spin state $a|+z⟩+b|-z⟩$ gives the two outcomes with probabilities $|a{|}^2,|b{|}^2$. Selecting one channel prepares a state for further measurements. An ideal second analyser along the same axis then gives a definite result, whereas one along a perpendicular axis gives two equally likely outcomes for a spin-one-half eigenstate. This connects the apparatus directly to incompatible spin components and the Born rule. Angular momentum is quantized through its operator algebra; the Stern–Gerlach gradient turns that discrete internal information into spatially distinguishable outcomes.

The Bohr model combines classical circular motion in the Coulomb potential with explicitly quantum postulates. A one-electron atom of nuclear charge $+Ze$ has electron–nucleus reduced mass $\mu$. With $C=Z{e}^2/(4\pi {\epsilon }_0)$, the relative circular orbit obeysBohr postulates permitted nonradiating stationary orbits with $L=n\hslash$, for $n=1,2,\dots$. These assumptions giveFor hydrogen with the nucleus taken infinitely heavy, the $n=1$ radius is the Bohr radius, $a_0\simeq 5.29\times 10^{-11}\mathrm{m}$. The finite reduced mass gives the small nuclear-motion correction, and increasing $Z$ contracts the one-electron orbit.

The circular-motion equation gives $T=C/(2r)$ while the potential energy is $-C/r$, so the total energy is $-C/(2r)$. Substituting the quantized radius yieldsFor the heavy-nucleus hydrogen approximation, $E_n\simeq -13.6\mathrm{eV}/n^2$. The ionization threshold is zero energy, so removing the ground-state electron requires about $13.6\mathrm{eV}$.

The model further postulates that radiation is emitted or absorbed only in transitions between stationary orbits, with photon energy $h\nu =|E_i-E_f|$. Emission from $n_i>n_f$ givesThus it reproduces the principal hydrogen spectral series, their limiting wavelengths, and the gross hydrogenic energy levels. For example, $n=3\to 2$ in hydrogen has $\Delta E\simeq 1.89\mathrm{eV}$ and $\lambda \simeq 656\mathrm{nm}$, a visible line. Absorption reverses the transition when the photon matches the energy gap.

The achievement is that quantized energies explain discrete spectra, but the model's stationary orbits are postulates rather than consequences of classical electrodynamics: an accelerating classical charge would radiate. It does not provide a general calculation of line intensities or transition selection rules and fails as a quantitative theory of many-electron atoms. Spin, fine structure and the full Zeeman effect require further physics. Modern quantum mechanics replaces definite electron orbits by wavefunctions and quantum probabilities. In particular, the true hydrogen ground state has $\ell =0$, so its orbital angular momentum vanishes; the model's rule $L=n\hslash$ should not be identified with the modern orbital eigenvalue rule $L^2=\ell (\ell +1){\hslash }^2$. Bohr's energy quantization remains a successful hydrogenic approximation, while its literal orbital picture does not.

For a neutral atom, count both the nucleons and the electrons. Protons, neutrons and electrons are all fermions, and exchanging two identical intact atoms produces the sign $(-1{)}^{N_f}$ from exchanging their fermionic constituents. An odd number gives a composite fermion; an even number gives a composite boson, consistently with the Spin-statistics theorem.

A lithium atom has three protons and three electrons. Lithium-6 also has three neutrons, giving $3+3+3=9$ fermionic constituents; lithium-7 has four neutrons, giving $3+4+3=10$. ThereforeThe composite statistics of neutral atoms cannot be inferred from the mass number alone: counting only nucleons would classify the nuclei and give the opposite answers here. The atom's total spin includes its electrons as well as its nucleus.

Indistinguishable particles have the same intrinsic properties and cannot be assigned physically observable permanent labels. Exchanging their mathematical coordinates leaves all physical predictions unchanged. In ordinary three-dimensional quantum mechanics the total state, including spin, obeysThe Spin-statistics theorem associates integer spin with bosons and half-integer spin with fermions. Exchange does not mean physically transporting two particles along some prescribed trajectories; it is the symmetry of the many-particle state under permutation of its arguments.

For orthonormal one-particle states $u,v$, normalized two-particle spatial combinations are $[u(1)v(2)\pm v(1)u(2)]/\sqrt{2}$. They illustrate coherent exchange amplitudes. If both fermions have the same complete one-particle state, the antisymmetric combination vanishes. This is the Pauli exclusion principle: a fermionic mode has occupation zero or one. Bosonic modes can have arbitrarily large occupation. A many-fermion product basis is antisymmetrized by a Slater determinant.

It is the full spin-spatial state that must have the appropriate symmetry. For two electrons, an antisymmetric spin singlet state can accompany a symmetric spatial orbital, so opposite-spin electrons can occupy the same spatial orbital. A symmetric spin triplet instead needs an antisymmetric spatial state. Thus exclusion concerns identical complete states, not an unconditional prohibition on sharing a position or spatial orbital.

Counting allowed occupations gives Bose-Einstein statistics and Fermi-Dirac statistics. For an ideal gas in thermal equilibrium,Bosonic accumulation permits Bose-Einstein condensation; exclusion gives a filled Fermi sea and degeneracy pressure even at zero temperature. In the dilute classical limit the occupations are small and both expressions approach the Maxwell-Boltzmann distribution. Exchange effects become important when wavepackets overlap appreciably, often characterized by number density times the cube of the thermal wavelength.

Composite particles follow their total exchange statistics while remaining intact. The composite statistics of neutral atoms depend on electrons as well as nucleons, as in the two lithium isotopes above. Symmetrization can produce bunching or antibunching in joint detection probabilities even in the absence of an interaction potential; it is not an additional classical force.

A Stern-Gerlach experiment uses an oven and slits to produce a collimated atomic beam, conventionally of neutral silver atoms, and sends it through an inhomogeneous magnetic field onto a position-sensitive detector. The unpaired outer electron supplies the effective spin one-half magnetic dipole moment; neutral atoms avoid the ordinary Lorentz deflection of a charged beam. For an analysing direction $z$, the spatial gradient produces a force approximately $F_z={\mu }_z{\partial }_z{B}_z$. The resulting separated beam positions measure the allowed projections of angular momentum. A spin one-half beam gives two branches, rather than the continuous distribution expected from classical dipoles with arbitrary orientations. A uniform magnetic field can rotate a spin but does not itself provide this spatial separation.

To measure the rotation, first use a $z$-directed Stern-Gerlach experiment and an aperture to transmit only the $+z$ branch. Pass this prepared spin eigenstate through a region of approximately uniform field in the $x$ direction, then use a second $z$ analyser to separate and count the two populations. For a field region of length $\ell$ and beam speed $v$, the precession angle is $\theta =\lambda B\ell /v$, and the predicted fractions areVarying $B$ or the transit time gives complementary oscillations, demonstrating coherent Larmor precession. Velocity selection and a sufficiently uniform rotation field keep different particles from accumulating different phases and washing out the contrast. Detection should compare the branch fractions after allowing for losses and analyser efficiency.

The orbital angular momentum of a particle is the operator $\mathbf{L}=\mathbf{r}\times \mathbf{p}$, with $\mathbf{p}=-i\hslash \nabla$ in the position representation. Its angular momentum commutation relations areThus one may specify $L^2$ and one component, conventionally $L_z$, simultaneously. One generally cannot assign sharp values to all three components, because they do not commute. The common angular eigenfunctions are the spherical harmonics:For a scalar orbital wavefunction, single-valuedness in the azimuthal angle gives integer $m$, and regularity on the sphere and termination of the ladder give integer $l\ge |m|$. Half-integer angular momenta describe intrinsic spin rather than scalar orbital motion.

The ladder operators $L_{\pm }=L_x\pm i{L}_y$ obey $[L_z,L_{\pm }]=\pm \hslash L_{\pm }$. Their normalized action isThe coefficient vanishes at the top or bottom of the multiplet. Its squared magnitude follows from $L_{\mp }{L}_{\pm }=L^2-L_z^2\mp \hslash L_z$, and positivity of the resulting norm constrains the allowed $m$ values.

For a central potential, the Hamiltonian commutes with $L^2$ and every component of $\mathbf{L}$. Stationary states can be separated as $R_{nl}(r)Y_l^m(\theta ,\varphi )$, and rotational symmetry makes the energy independent of $m$, giving a $2l+1$ quantum degeneracy at fixed radial label and $l$. Additional degeneracy between distinct $l$ values requires additional structure, as in the Coulomb potential. The parity of $Y_l^m$ is $(-1{)}^l$. These results organize atomic orbitals and their selection rules. They describe quantized rotation and symmetry, without assigning the electron a classical orbit or a simultaneously definite three-component angular momentum vector.

A Stern-Gerlach experiment sends a collimated beam through an inhomogeneous magnetic field and then measures its transverse position after a drift region. The neutron is electrically neutral but has a nonzero magnetic moment proportional to its spin, $\mathbf{\mu }={\gamma }_n\mathbf{S}$. A predominantly $z$-directed field gives the interaction Hamiltonian operator and transverse forceSince a neutron has spin one-half, its spin component along the local field has only two eigenvalues, $S_z=\pm \hslash /2$. Hence there are two opposite force branches, corresponding to ${\mu }_z=\pm {\gamma }_n\hslash /2$, and a beam containing both spin components gives two displaced detector spots. This is neutron Stern-Gerlach splitting, not an electric-charge Lorentz force. A uniform magnetic field can cause Larmor precession but does not provide the required separating magnetic field gradient.

The neutron's gyromagnetic ratio is negative, so its magnetic moment points opposite to its spin; this reverses which spin label goes to which spot without changing the two-way splitting. Equal populations occur for an initially unpolarized beam. A beam prepared entirely in one field-axis spin eigenstate gives only its corresponding branch, explaining the qualification “in general”.

Let $N=L/a$ be the integer number of primitive unit cells. A Bloch state can be written as ${\psi }_k(x)=e^{ikx}{u}_k(x)$, with $u_k(x+a)=u_k(x)$. The periodic boundary condition ${\psi }_k(x+L)={\psi }_k(x)$ therefore requiresThe spacing of allowed wavevectors is $\Delta k=2\pi /L$. The first Brillouin zone has width $2\pi /a$, soFor an exact endpoint convention take a half-open Brillouin zone, such as $[-\pi /a,\pi /a)$. Its two endpoints differ by a reciprocal lattice vector and must not both be counted as distinct values of crystal momentum. There are exactly $N$ inequivalent integers $j$ modulo $N$, irrespective of whether $N$ is even or odd. This is state counting in a Bloch band: one allowed wavevector per primitive unit cell per energy band. A separate twofold spin multiplicity gives $2N$ electronic states per energy band, without doubling the number of wavevectors.

In a metal, a partially filled energy band or overlapping energy bands supplies electrons with nearby unoccupied states. At thermal equilibrium, oppositely directed group velocities are equally populated and the net electric current vanishes. An applied electric field distorts the occupation distribution, producing a small mean drift velocity. A filled energy band alone has no such low-field intraband response because its velocity contributions cancel over the entire Brillouin zone.

In the Drude model, let $n$ be the mobile electron number density, let $e>0$ be the elementary charge, and let $\tau$ be the momentum relaxation time. The mean motion obeysThe steady state is ${\mathbf{v}}_d=-e\tau \mathbf{E}/m$. The current density is opposite to the electron drift velocity:This gives microscopic Ohm's law, with electrical conductivity $\sigma$ in $\mathrm{S}{\mathrm{m}}^{-1}$ and electrical resistivity ${\rho }_{\mathrm{el}}$ in $\Omega \mathrm{m}$. For a uniform wire of length $\ell$ and cross-sectional area $A$, its electrical resistance is $R={\rho }_{\mathrm{el}}\ell /A$. The dissipated power per volume is $\mathbf{J}\cdot \mathbf{E}=\sigma E^2$, the Joule heating supplied by the external source.

The free electron model with Fermi-Dirac statistics improves the microscopic picture. The Pauli exclusion principle fills states up to the Fermi energy at low temperature. For a spherical, parabolic energy band with two spin states,Only states near the Fermi surface change their occupation in a weak-field response; the large equilibrium Fermi velocity is not the small drift velocity. The same $n{e}^2\tau /m$ formula results for this isotropic parabolic model when $\tau$ is evaluated for transport near the Fermi surface. In a crystal, the energy band group velocity, effective mass and possibly direction-dependent scattering replace the simplest bare-particle picture. The mean free path is approximately ${\ell }_{\mathrm{mfp}}=v_F\tau$.

A perfectly periodic static lattice is already included in Bloch states and does not by itself give a finite electrical resistivity. Momentum relaxation comes from defects, impurities, boundaries, and electron-phonon scattering. Independent scattering mechanisms approximately add their rates, giving Matthiessen's rule,The approximately temperature-independent ${\rho }_0$ is the residual resistivity. In ordinary simple metals, the phonon contribution is approximately proportional to temperature at high temperature; a three-dimensional acoustic-phonon transport model gives a $T^5$ contribution well below the relevant Debye temperature. Other mechanisms can dominate in particular materials or regimes, so neither power law is universal.

The classical Drude model incorrectly treats the equilibrium carrier statistics. Fermi-Dirac statistics explains the small electronic heat capacity, while electrons carrying both charge and heat give the approximate Wiedemann-Franz law ${\kappa }_e/(\sigma T)={\pi }^2{k}_B^2/(3e^2)$ for degenerate carriers with appropriate nearly elastic scattering. Both the simple electrical conductivity formula and this thermal relation depend on the assumptions about energy bands and scattering, rather than being exact for all metals.

Identical spin one-half particles are fermions. By fermionic exchange symmetry, the complete spatial-and-spin wavefunction must change sign when the two particle labels are exchanged. Define the normalized antisymmetric spin-one-half singlet stateand the three symmetric spin-one-half triplet statesThe Hamiltonian is independent of spin, so these spin functions carry no additional energy.

Both particles can occupy $u_0$ if their spin factor is antisymmetric. The unique ground state wavefunction isTwo equal-spin particles could not share this orbital, by the Pauli exclusion principle.

At the first excited energy, $E=ϵ={\hslash }^2/(2m{R}^2)$, one orbital is $u_0$ and the other is $u_s$ with $s=+1$ or $-1$. Define symmetric and antisymmetric normalized spatial functionsFor each $s$, all allowed wavefunctions are spanned byThere are $2(1+3)$ orthonormal states, hence first-excited-state degeneracy $g=8$. In each product, exactly one factor is antisymmetric, ensuring the required total fermionic exchange symmetry.

For a plane wave wavefunction ${\psi }_k(x)\propto e^{ikx}$, impose the cyclic or periodic boundary condition ${\psi }_k(x+L)={\psi }_k(x)$. It requiresAdjacent allowed wavenumbers on the full signed $k$ axis are spaced by $\Delta k=2\pi /L$. Ignoring spin, the density of states per unit $k$ is consequentlyBoth positive and negative $k$ are included in this density. It is not yet the density per unit magnitude $|k|$, which has an extra factor of two away from zero. In $d$ independent dimensions the same periodic boundary conditions give one spatial state per $k$-space cell of measure $(2\pi /L{)}^d$.

In three dimensions use the $k$-space density of states $(L/2\pi {)}^3$ per spin species and the spherical shell volume $4\pi k^{2}dk$:Thus the three-dimensional free-electron density of states grows as ${ϵ}^{1/2}$. For physical electrons, including both spin states givesAll three total density of states functions have units ${\mathrm{J}}^{-1}$ when energy is in joules; they count states per energy interval. The paper explicitly suppresses spin in (a) but does not repeat that convention in (b), so giving both conventions avoids a hidden factor-of-two ambiguity. Parts (c)--(f) below use the physical two-spin-state electron gas.

In two dimensions the density of states in $k$ space per spin species is $(L/2\pi {)}^2$. A annulus of radius $k$ and thickness $dk$ has area $2\pi kdk$. ThereforeThe two-dimensional free-electron density of states is constant in energy, $g_2^{(1)}\propto {ϵ}^0$, above the band bottom. Including both spin states gives $g_2=L^{2}m/(\pi {\hslash }^2)$. Dividing by $L^2$ gives the density per unit physical area, rather than the total specimen's density.

To preserve the convention of part (a), first count one spin species. Define $g_d^{(1)}(ϵ)dϵ$ as the number of spatial states in the whole specimen in that energy interval. PutIn one dimension there are intervals near both $+k$ and $-k$, so the one-dimensional free-electron density of states isThus $g_1^{(1)}\propto {ϵ}^{-1/2}$. The factor of two here comes from opposite momenta, not spin. For physical electrons with two spin states, the spin-summed density of states is $g_1=2g_1^{(1)}=L\sqrt{2m}{ϵ}^{-1/2}/(\pi \hslash )$. These expressions are continuum approximations away from individual discrete levels.

Let $d=1,2,3$ and take a specimen of length, area or volume $L^d$. One monovalent atom per unit cell supplieselectrons. At zero temperature, the Pauli exclusion principle fills the lowest energy states, with two spin states per spatial wavevector. The occupied region is an interval, disk or ball of radius $k_F$. Using the $k$-space density of states gives, in one dimension,In two dimensions,In three dimensions,Each Fermi energy follows from the free electron model dispersion relation $E_F={\hslash }^2{k}_F^2/(2m)$. The Fermi wavevectors have units ${\mathrm{m}}^{-1}$ and the Fermi energies have units joules for $a$ in metres and $m$ in kilograms. No numerical lattice spacing is supplied, so numerical energies in electronvolts cannot be uniquely calculated. The spin factor of two is necessary here for actual monovalent crystals; the spinless auxiliary count of part (a) was not an instruction to remove electron spin from these crystals.

The length, area and volume of a first Brillouin zone are $(2\pi /a{)}^d$. The corresponding occupied $k$-space measures areHenceThis is state counting, independent of the interval/disk/ball geometry. One Brillouin zone contains one spatial state per unit cell in a single energy band, and that band can hold two electrons per cell because of spin. One electron per cell therefore occupies half of it. The “volume enclosed” is the occupied Fermi sea, not the lower-dimensional area of the Fermi surface itself. Its half-volume does not mean half the distance to the zone boundary in two or three dimensions.

Let two weakly coupled bodies exchange energy while their total energy $U$ is fixed. If body1 has $U_1$, the number of compatible states is ${\Omega }_1(U_1){\Omega }_2(U-U_1)$. Its most probable equilibrium division maximizes the logarithm, givingThis equal intensive quantity labels thermal equilibrium. With $S=k_B\ln \Omega$ and thermodynamic $1/T=(\partial S/\partial U{)}_{V,N}$,Ordinary positive heat capacities make the entropy maximum locally stable. Particle numbers and volumes are held fixed during the exchange.

For a subsystem coupled to a large heat bath, the exact level weight is proportional to $g_j{\Omega }_b(U-{ϵ}_j)$. Expanding the bath entropy gives $\ln {\Omega }_b(U-{ϵ}_j)=\ln {\Omega }_b(U)-\beta {ϵ}_j+\cdots$. Normalization therefore givesThe Boltzmann form uses the reservoir limit; an arbitrary small isolated pair instead retains its exact bath state count. The degeneracy counts distinct subsystem microstates at the same energy, rather than adding another Boltzmann factor for the whole level.

In the two-level donor occupancy model, set the single bound-state energy to zero and the conduction representative to $\Delta =0.04\mathrm{eV}$ with degeneracy $g$. Ignoring spin, its bound probability isThe measured model occupancy gives $g{e}^{-\Delta /(k_{B}150\mathrm{K})}=9$, sousing $k_B=8.61733\times 10^{-5}\mathrm{eV}{\mathrm{K}}^{-1}$. The supplied occupancy and energy are rounded, so there is no reason to demand an exact nonrounded integer from that inference.

For $P_b=0.9$, the excited-to-bound weight must be $1/9$. Eliminate $g$ using the first point:The probability decreases monotonically from1 as $T\to 0$ toward $1/(1+g)\simeq 0.005$ as $T\to \infty$. The nonzero high-temperature limit follows from the stipulated finite effective degeneracy, not from a full continuum conduction-band calculation. A realistic band with energy-dependent density of states need not retain this two-level limit.

The microscopic magnetic response becomes a macroscopic field problem through the magnetization $\mathbf{M}$, the magnetic dipole moment per unit volume. Definein magnetostatics. Bound currents are represented by ${\mathbf{J}}_b=\nabla \times \mathbf{M}$ and surface current ${\mathbf{K}}_b=\mathbf{M}\times \hat{\mathbf{n}}$. Normal $B$ is continuous across an interface, while $\hat{\mathbf{n}}\times ({\mathbf{H}}_2-{\mathbf{H}}_1)={\mathbf{K}}_{\mathrm{free}}$. It is generally incorrect to impose continuity of every component of both fields. A linear isotropic material has $\mathbf{M}=\chi \mathbf{H}$ and ${\mu }_r=1+\chi$, but a ferromagnet usually requires a nonlinear, history-dependent constitutive relation.

Internal field differs from externally applied field. For uniform magnetization along a principal axis of an ellipsoid, the demagnetizing field is $-N_{d}M$ and $H_{\mathrm{int}}=H_{\mathrm{applied}}-N_{d}M$. A linear specimen therefore has apparent susceptibility $\chi /(1+N_d\chi )$ relative to the applied field. Shape is especially important for large susceptibility. A long closed magnetic circuit reduces stray fields; a bar magnet still has demagnetizing fields despite the absence of magnetic monopoles. Equivalent magnetic-pole descriptions are bookkeeping for magnetization divergence, not free magnetic charge.^[ref]

In paramagnetism, permanent localized moments are preferentially aligned by the field, against thermal disorder. For independent spin-one-half moments with projections $\pm \mu$, the two Zeeman energies givein the dilute weak-field limit with $B\simeq {\mu }_{0}H$. This is a Curie law; the response saturates at $n\mu$ when alignment dominates. More generally, spin $J$ gives $\chi ={\mu }_{0}n(g{\mu }_B{)}^{2}J(J+1)/(3k_{B}T)$. Interactions can produce a Curie–Weiss response $C/(T-\Theta )$ over a suitable range. Conduction-electron Pauli paramagnetism need not show the localized-moment $1/T$ law.

In diamagnetism, the field induces a magnetic moment opposing it, so the linear susceptibility is negative. The microscopic cause is the change in orbital electronic motion, and a closed-shell insulating material has no permanent atomic moment to orient. This weak response is often only mildly temperature dependent away from other transitions. It also exists as a background in materials with a stronger positive response. Perfect flux expulsion in a superconductor is a distinct collective phenomenon and should not be identified with ordinary weak atomic diamagnetism.

In ferromagnetism, exchange interactions stabilize spontaneous magnetic order below the Curie temperature. A specimen can have zero net moment while its magnetic domains remain locally magnetized in different directions. Domain-wall motion and moment rotation produce a large response; anisotropy and pinning lead to remanence, coercivity and hysteresis. Increasing field eventually approaches saturation. Heating above the ordering transition removes spontaneous order, although an applied field still induces a response. Soft magnetic cores favor easy reversal and small cyclic loss; hard permanent-magnet materials favor retained magnetization and resistance to reversal. These properties concern collective order and barriers, not merely independent moments all following a weak field.^[ref]

A practical distinction requires a controlled field sweep and a separate magnetic readout. A toroidal core with drive winding $N_1$, path length $l$ and current $I$ has approximately $H=N_{1}I/l$. A pickup winding measures ${\mathcal{E}}_2=-N_{2}AdB/dt$; integrating after calibration gives $B$, and $M=B/{\mu }_0-H$. Reversing the sweep reveals a loop whose positive loss area per volume is $\oint HdB$. An air-core/empty-holder reference, known geometry and turn count, offset correction, temperature control and slower sweeps test background, heating and rate-dependent eddy-current losses. A single force or susceptibility reading cannot establish spontaneous order or hysteresis. The original curves below are illustrative responses, not measured material data.

A rapidly spinning gyroscope approximately has angular momentum $\mathbf{L}=I_3{\omega }_3\hat{\mathbf{e}}$ along its axle. With no torque in an inertial frame, this vector retains its direction. A transverse torque changes its direction according to $d\mathbf{L}/dt=\mathbf{\tau }$; it does not initially push the axle in the direction of the applied force.

For a fixed pivot, centre of mass $l\hat{\mathbf{e}}$ from it and gravity $-Mg\hat{\mathbf{z}}$, the torque is $Mgl\hat{\mathbf{z}}\times \hat{\mathbf{e}}$. In slow steady precession about the vertical, $d\mathbf{L}/dt=\Omega \hat{\mathbf{z}}\times \mathbf{L}$, givingThe factor $\sin \theta$ appears both in torque and in the radius traced by the angular-momentum tip, so it cancels for a nonvertical axle. Reversing spin reverses this precession. The full symmetric-top steady condition is $I_1\cos \theta {\Omega }^2-I_3{\omega }_3\Omega +Mgl=0$; the approximation neglects its quadratic term. Sudden release can also produce nutation, so large spin alone does not guarantee a steady sweep.

A wheel-on-pivot experiment can compare precession periods at different measured spin speeds and reverse the spin. Keep the pivot and centre of mass fixed; bearing friction, spin decay, an initial push and support motion can mimic departures from the predicted inverse-spin relation. A freely gimballed direction reference and a gravity-loaded precessing wheel test different torque conditions.^[ref]

For weak fields in a linear isotropic material, magnetization is $\mathbf{M}={\chi }_m\mathbf{H}$, with magnetic susceptibility ${\chi }_m$. Paramagnetism has ${\chi }_m>0$, while diamagnetism has ${\chi }_m<0$. In either case, $\mathbf{B}={\mu }_0(\mathbf{H}+\mathbf{M})={\mu }_0(1+{\chi }_m)\mathbf{H}$. These weak responses generally disappear when the applied field is removed, unlike hysteretic ferromagnetism.

A paramagnet can contain permanent magnetic moments from unpaired electrons or atomic angular momentum. Alignment lowers the energy $-\mathbf{\mu }\cdot \mathbf{B}$, while thermal disorder opposes it. For independent classical moments of magnitude $\mu$, their weak-field average projection is ${\mu }^{2}B/(3k_{B}T)$, soThis is Curie's law; it assumes weak magnetization and no important interactions between moments. With quantum angular momentum $J$, replace ${\mu }^2$ by $(g{\mu }_B{)}^{2}J(J+1)$. Strong-field magnetization saturates instead of growing linearly without limit. Conduction electrons also exhibit Pauli paramagnetism: only occupations near the Fermi surface readily redistribute between spin states, giving a nearly temperature-independent susceptibility when $T\ll T_F$.

Diamagnetism is an induced orbital response opposing the applied magnetic field. For bound electrons, a weak field changes orbital motion and induces an opposing magnetic moment. For an isotropic collection of atoms, the elementary orbital estimate iswhere the sum runs over the electrons of one atom. It is usually small and weakly temperature dependent. Delocalized electrons also have orbital Landau diamagnetism; for an ideal three-dimensional free electron gas with a quadratic isotropic dispersion and electron spin with $g=2$, ${\chi }_L=-{\chi }_P/3$. Both spin and orbital contributions must be included to determine the total response. Diamagnetism is not a consequence of permanent dipoles choosing to align against the field; that orientation would increase their energy.