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The Hamiltonian has the discrete symmetry described in solution, whereas either selected easy-axis ground state has a nonzero order parameter $⟨S^x⟩$ and is not invariant under it. This is spontaneous symmetry breaking: the equations preserve the symmetry while the selected macroscopic quantum state does not.

An infinitesimal longitudinal magnetic field selects one orientation. The defining order of limits isin the ordered zero-temperature phase, with the opposite sign for $h\to 0^{-}$. A symmetric finite-volume quantum state can instead have zero magnetization. Quantum ordered states are generally dressed by fluctuations rather than exact fully polarized product states, because unequal transverse exchange can create pairs of spin waves. Discrete easy-axis symmetry does not imply a gapless Goldstone mode; indeed solution finds a nonzero spin wave gap. Nor does this argument assert positive-temperature order in a one-dimensional short-range chain.^[ref]

Interpret the two displayed exchange components as belonging to the common nearest-neighbour sum with coefficient $-J$. Write $H_f=|h|$ and, for $h\ne 0$, $s=\mathrm{sgn}h$. The source explicitly fixes $J>0$ but does not separately constrain $h$ in this question; keeping $H_f$ covers either field direction.

For classical vectors of length $S$, define ${\mathbf{q}}_j=(S_j^x,S_j^y)$. The Cauchy-Schwarz inequality gives ${\mathbf{q}}_j\cdot {\mathbf{q}}_{j+1}\le |{\mathbf{q}}_j||{\mathbf{q}}_{j+1}|\le S^2$, and $h{S}_j^x\le H_{f}S$. Each term is therefore bounded below, soUniform alignment along the field saturates every bound:Equality requires all planar vectors to be parallel with length $S$, and the nonzero field then fixes their direction. At $h=0$, every uniformly oriented vector in the $xy$ plane is a classical ground state:A nonuniform configuration cannot improve these simultaneously saturated bond bounds, including on the periodic ring.

Choose the local positive $\tilde{z}$ axis along the classical spin. For nonzero field, a proper rotation isIt preserves the Spin commutation relations and turns the field term into $-H_{f}S{\sum }_j{\tilde{S}}_j^z$. At zero field choose the classical direction $+x$, so $s=1$. Expanding about the original $z$ axis without this rotation would put the reference state away from the classical minimum and would not give a controlled small-deviation expansion.

To the required order,The longitudinal exchange gives $-J{S}^2+JS(n_j+n_{j+1})+O(S^0)$ on each bond. The transverse exchange givesFinally the field contributes its classical constant and $H_{f}S{n}_j$. Each site belongs to two bonds, so after removing the order-$S^2$ constant,These are the order-$S$ terms; the discarded products of number operators and the next square-root terms are order $S^0$ for occupations held fixed as $S\to \infty$. A different transverse-axis phase can reverse the sign of $B$ without changing the spectrum, but the sign above follows from the stated rotation and oscillator convention.

Use the unitary discrete Fourier transformThe transformed operators obey $[a_k,a_q^{†}]={\delta }_{kq}$. The hopping terms contribute $2A\cos k$ to the number coefficient. In the pairing sum, the odd sine part cancels between $k$ and $-k$. ThuswhereThe factor one half in the full momentum sum avoids counting an anomalous pair twice. For $H_f>0$, ${\alpha }_k>|{\Delta }_k|$ for every allowed momentum. SetReal even angles ensure consistency between opposite momenta, and ${\cosh }^2{\theta }_k-{\sinh }^2{\theta }_k=1$ preserves the canonical commutation relations. The anomalous coefficient becomes ${\Delta }_k\cosh 2{\theta }_k-{\alpha }_k\sinh 2{\theta }_k=0$. Normal ordering then giveswithThe field-polarized planar ferromagnetic spin wave has energy gap $ϵ(0)=S\sqrt{H_f(H_f+2J)}$. The momenta run over the whole Brillouin zone once, including the self-inverse momenta $0$ and, for even $L$, $\pi$; these use single-mode squeezing rather than introducing an extra independent partner.

The zero-field case has a zero-mode obstruction to bosonic Bogoliubov diagonalization. Its nonzero-momentum modes have $ϵ(k)=2\sqrt{2}JS|\sin (k/2)|$, but ${\alpha }_0={\Delta }_0=JS$, so no finite squeezing angle exists for the uniform mode. With $Q_0=(a_0+a_0^{†})/\sqrt{2}$, its Hamiltonian is $JS{Q}_0^2-JS/2$, not an identically zero oscillator. For example, its matrix element between the vacuum and the two-boson number state is $JS/\sqrt{2}$, whereas $ϵ(0)b_0^{†}{b}_0$ would vanish. Consequently the requested oscillator diagonalization is literal for nonzero field; at zero field treat the collective orientation separately or introduce a field regulator before taking a limit.

For nonzero field, the spin-wave ground state is the vacuum of all $b_k$. The inverse Bogoliubov transformation gives $⟨a_k^{†}{a}_q⟩={\delta }_{kq}{\sinh }^2{\theta }_k$ and vanishing one-point oscillator averages. Translational invariance therefore gives the depletionRotating back to the physical axes yields, to the retained spin-wave order,The zero transverse components also follow from rotation by $\pi$ about the physical $x$ axis in a symmetry-preserving ground state. The nonzero depletion describes zero-point quantum fluctuations: the original boson vacuum is not the ground state because the anomalous terms create pairs. It is nonnegative and must satisfy $\delta S\ll S$ for the expansion to be reliable. At strong field it tends to zero, with $\delta S\sim J^2/(8H_f^2)$ in the infinite-chain limit.

For fixed $H_f>0$, the momentum sum has a finite thermodynamic limit obtained by $L^{-1}{\sum }_k\to (2\pi {)}^{-1}{\int }_{-\pi }^{\pi }dk$. At zero field the uniform mode already has no oscillator vacuum, as explained in part d. Even after removing that mode, the nonzero modes behave asThus assigning a finite ordered moment by simply setting $h=0$ in the infinite momentum sum is invalid. The divergence indicates failure of this expansion about a fixed planar orientation; it is not a negative physical magnetization. At finite size an exact zero-field state chosen to have definite total $S^z$ has zero planar one-point magnetization by the continuous rotation symmetry. Establishing its full correlations requires a treatment beyond the divergent ordered-state approximation.

Choose a coarse-grained magnetization $m(\mathbf{x})$ as the order parameter: it vanishes in the paramagnetism phase and develops a spontaneous nonzero value in the ferromagnetism phase. In zero applied magnetic field, spin reversal sends $m\mapsto -m$, so the local Landau free energy contains only even powers. A minimal spatially inhomogeneous free-energy functional isThe positive gradient coefficient penalizes rapid spatial variations. Near an ordinary transition take $a(T)=a_0(T-T_0)$, $a_0>0$, and positive $b$; retain the sextic term when $b$ can approach or cross zero. The source $h$ is conjugate to the magnetization and explicitly breaks spin reversal.

Determine equilibrium by minimizing this free-energy functional, imposing the appropriate boundary conditions. Its Euler-Lagrange equation isFor a uniform system at $h=0$, compare the values of all stable stationary solutions, rather than accepting every solution of this equation. The selected global minimum gives the equilibrium magnetization; derivatives of its free energy give entropy, heat capacity and magnetic susceptibility. At an ordinary transition the two ordered minima emerge continuously from zero. Spatial dependence also permits domain walls and an estimate of the correlation length.

This is a mean-field approximation: it selects a minimizing profile and neglects statistical fluctuations around it. The symmetry constrains the expansion, but its numerical coefficients are phenomenological. Close enough to a transition, thermal fluctuations near a critical point can invalidate the resulting mean-field critical exponents.^[ref]

Component chemical potentials. Define the per-spin quantum concentration $n_Q={\lambda }_T^{-3}=(m{k}_{B}T/(2\pi {\hslash }^2){)}^{3/2}$. For a classical component with one-particle potential energy $U_{\sigma }(\mathbf{r})$, the local Helmholtz free energy density is $k_{B}T{n}_{\sigma }[\ln (n_{\sigma }/n_Q)-1]+n_{\sigma }{U}_{\sigma }$. Its derivative with respect to number density gives the full local chemical potential, including the magnetic energy. With the up state defined as the lower level,The Bohr magneton ${\mu }_B$ is a magnetic moment; ${\mu }_{B}B$ is an energy. The factor two for the two spin states is not already included in $n_Q$.

Total equilibrium chemical potential. If spin conversion is allowed, a particle can be added in either spin state at the same equilibrium free-energy cost. Thusnot their sum or zero. The total particle number, or a reservoir, fixes this common value; the supplied information does not specify a numerical chemical potential. A useful local expression is $\mu =k_{B}T\ln [n/(2n_Q\cosh x)]$, with $x={\mu }_{B}B/(k_{B}T)$, whose density dependence is found below.

Why it is uniform and common. Moving a small number $dN$ of one species from ${\mathbf{r}}_1$ to ${\mathbf{r}}_2$ changes the free energy by $[{\mu }_{\sigma }({\mathbf{r}}_2)-{\mu }_{\sigma }({\mathbf{r}}_1)]dN$. If this difference were nonzero, a transfer could lower the free energy. Equilibrium under diffusion therefore requires each full chemical potential to be spatially constant. A spin flip transferring $dN$ from down to up similarly changes free energy by $({\mu }_{\uparrow }-{\mu }_{\downarrow })dN$, proving equality when spin relaxation is permitted. This is chemical equilibrium of a classical spin gas. Mere thermal contact at fixed separate spin populations would not enforce the latter equality; full spin equilibrium is the necessary assumption. A spatially varying magnetic field is balanced by a spatially varying density, not by a varying full chemical potential.

Magnetization. Invert the component equations at the common $\mu$:The two levels have moments $+{\mu }_B$ and $-{\mu }_B$ along the field. The magnetization, magnetic moment per unit volume, is thereforeIts SI units are $\mathrm{A}{\mathrm{m}}^{-1}$. This expression neglects interactions and magnetic feedback of the gas on the specified external field.

High-temperature response. For $|x|\ll 1$, $\sinh x\simeq x$ and $\cosh x\simeq 1$, yieldingAt fixed number density, this is the Curie law: the susceptibility with respect to $B$ is $n{\mu }_B^2/(k_{B}T)$. The conventional dimensionless susceptibility with respect to $H$ is approximately ${\chi }_v={\mu }_{0}n{\mu }_B^2/(k_{B}T)$ when $B\simeq {\mu }_{0}H$. At fixed $\mu$, however, $n_Q{e}^{\beta \mu }$ changes with temperature, so $M$ itself need not scale simply as $1/T$. The moment per particle has the Curie behaviour; the density constraint must be stated.

Total density. Adding the two populations givesFor a finite container and fixed $N$, normalization gives $e^{\beta \mu }=N/[2n_Q\int \cosh x(\mathbf{r})d^{3}r]$. In a weak field the leading density is uniform, with changes only of order $x^2$. Particles preferentially occupy regions where the magnitude of the field lowers one of the spin energies, even though the common chemical potential remains uniform.

Order parameter and the free-energy landscape. In Landau theory, take a scalar magnetization component at fixed volume and composition. It is an order parameter: its nonzero spontaneous value distinguishes an ordered phase with ferromagnetism from the symmetric high-temperature phase. One first constructs a constrained order-parameter free energy at specified magnetization by summing over microscopic states compatible with it. An external magnetic field is an independent control, so it can also appear in that constrained function. For a total magnetic moment the conjugate term is $-BM$; if $M$ is a magnetization density, the total-energy term is $-VBM$, or $-BM$ in free-energy density. Minimizing over $M$ then yields the equilibrium thermodynamic potential as a function of the imposed field.

Thus treating both $M$ and $B$ as arguments of a Landau function does not treat them as independent natural variables of the already-minimized equilibrium free energy. The first is the fluctuating or constrained order parameter, and the second its conjugate field.

At zero field, reversal symmetry requires $F(M)=F(-M)$, excluding odd powers. Near a continuous transition, assume the homogeneous constrained free energy is analytic for small $M$, that the quadratic coefficient crosses zero linearly, and that the quartic coefficient is positive. ThenThe temperature $T_c$ is where the zero-field quadratic coefficient vanishes and spontaneous ordering begins in this approximation. Higher powers are negligible sufficiently close to it; spatial fluctuations are neglected in this mean-field treatment.

Stationarity gives $2M[b(T-T_c)+2d{M}^2]=0$. Above $T_c$, the only minimum is at zero. Below it, zero becomes a maximum and two minima appear:The two signs describe spontaneous symmetry breaking: they are the average magnetizations in the symmetry-broken pure phases. A strictly finite system at exactly zero field samples the two wells symmetrically and has ensemble average $⟨M⟩=0$. A spontaneous pure-phase value is defined by selecting a phase, for example taking the thermodynamic limit before removing a small field. This qualification distinguishes a minimum's location from the average over a symmetric two-well distribution.

Minimum free energy and the heat-capacity jump. At a nonzero minimum, $M^2=-b(T-T_c)/(2d)$. Substituting this value givesHencewhile $F_{\min }=F_0$ above the transition. Take the temperature derivatives only after minimizing, so the equilibrium order parameter is allowed to change with temperature. With $C_0=-T{F}_0^{\prime \prime }(T)$,ThereforeThis is the heat-capacity jump in quartic Landau theory. The entropy is continuous at $T_c$, so there is no latent heat. If $F$ denotes free-energy density rather than total free energy, these expressions denote heat capacity per unit volume with the same normalization.

Stabilizing a tricritical expansion. At a tricritical point, the quartic coefficient is tuned to zero at the same point where the quadratic coefficient changes sign. Reversal symmetry still excludes odd powers. The next available stabilizing term is thereforeA negative or vanishing leading sextic coefficient would not stabilize this truncation for large $|M|$; a higher positive term would then be needed and the present tricritical form would be insufficient. The ordinary quartic theory has a continuous transition for $d>0$, while a stabilized theory with $d<0$ can have a first-order transition. Their meeting at zero field is what makes the simultaneous tuning tricritical.

The tricritical minimum. At zero field and $T<T_c$, stationarity givesThe nonzero solutions are minima: their curvature is $2b(T-T_c)+30d^{\prime }{M}^4=8b(T_c-T)>0$. Put $t=T_c-T>0$. Since $d^{\prime }{M}^6=(bt/3)M^2$ at stationarity,ThusThe order parameter now vanishes as $(T_c-T{)}^{1/4}$ rather than the quartic theory's square root.

Heat-capacity singularity. Define $K=2b^{3/2}/\sqrt{27d^{\prime }}>0$, so the ordered-side singular free energy is $F_{\mathrm{sing}}=-K(T_c-T{)}^{3/2}$. Its derivatives areConsequentlyThis is the tricritical heat-capacity amplitude in a sextic Landau expansion. It is positive on the ordered side and diverges as $T\to T_c^{-}$, rather than remaining a finite jump. This value of the heat-capacity critical exponent is the mean-field result along the path keeping the quartic term zero; a detuning to $d>0$ ultimately restores ordinary quartic behavior. Fluctuation corrections to the mean-field theory are not included in the stipulated polynomial minimization.

Canonical probabilities. A canonical ensemble describes a system of fixed particle number and fixed external parameters such as volume, weakly coupled to a much larger thermal reservoir. They exchange energy and reach thermal equilibrium, while the reservoir's temperature stays effectively constant. The interaction energy is negligible in the state counting. The combined isolated system is described by equal probabilities for its accessible microstates.

If its total energy is $E_{\mathrm{tot}}$, the probability of a particular system microstate $i$ is proportional to the reservoir multiplicity at $E_{\mathrm{tot}}-E_i$. Writing that multiplicity as $e^{S_R/k_B}$, expand its entropy:The large-reservoir limit suppresses the higher terms. Normalization therefore givesThe sum counts individual microstates; when grouping them by energy, their state degeneracies must be included. This $Z$ is the canonical partition function.

Classical orientational heat capacity. Hold the magnetic field fixed. Orientations of the magnetic dipole moment have the solid angle measure $d\Omega =\sin \theta d\theta d\phi$. With $x=\beta mB$, the angular partition function isChoosing a normalized angular measure would remove $4\pi$, with no effect on the energy derivatives. Differentiating at fixed $B$,The angular average $⟨\cos \theta ⟩=\coth x-1/x$ is the Langevin function. Since $dx/dT=-x/T$,This is the heat capacity of a classical magnetic dipole, in $\mathrm{J}{\mathrm{K}}^{-1}$ for one magnetic moment. The word specific here refers to that single moment; $N$ independent magnetic moments have $N$ times this heat capacity. At $B=0$ the energy is independent of orientation and $C_B=0$, obtained by the continuous $x\to 0$ limit. At fixed nonzero $B$, the classical low-temperature limit instead gives $C_B\to k_B$.

Equipartition and its proof. The equipartition theorem states that an independent quadratic term $a{q}^2/2$ in a classical Hamiltonian, with $a>0$, contributes $k_{B}T/2$ to its mean energy. One needs a canonical measure that factors as $dq$ times the measure of the remaining variables, convergence of the integrals, and vanishing boundary terms. The coordinate must not be restricted by a thermally important hard boundary or a coupled constraint.

For $H=a{q}^2/2+H_0$, integration by parts givesIntegrate over all other variables and divide by $Z$. It follows thatThe same reasoning applies to quadratic momenta among the canonical variables. It is the number of quadratic energy contributions that is counted, not an unrestricted count of coordinate labels.

Two soft orientational directions. When $x\gg 1$, the moment is concentrated near alignment. Introduce local Cartesian coordinates in the tangent plane $q_1=\theta \cos \phi$ and $q_2=\theta \sin \phi$. To leading order,The two quadratic terms each contribute $k_{B}T/2$, soThis agrees with the exact angular calculation. Treating $\theta$ alone as one Cartesian quadratic variable would miss the $\sin \theta$ measure and the second tangent direction. There are no extra rotational kinetic terms in the stipulated Hamiltonian; adding such terms would be a different model.

Quantum spin one. The spin projections are $m_s=-1,0,1$, with $S_z=\hslash m_s$. Put $\Delta =\hslash |\gamma |B>0$, where $\gamma$ is the gyromagnetic ratio. The three energies are $-\Delta ,0,+\Delta$, irrespective of the sign of $\gamma$. Relative to the ground state they are $0,\Delta ,2\Delta$. Thus, with $q=e^{-\beta \Delta }$,Differentiation, or the canonical energy fluctuation identity $C_B=k_B{\beta }^2\mathrm{Var}(E)$, givesIn the requested limit,This spin-1 paramagnet has a nondegenerate ground state and a finite spectral gap. Excitations are exponentially rare, so its heat capacity tends to zero. Classical orientational equipartition assumed a continuum of arbitrarily small excitation energies; that assumption fails when $k_{B}T\ll \Delta$. This is a change in the accessible state spectrum, not a violation of the equipartition theorem within its classical domain.

A phase transition is a qualitative change of thermodynamic phase as control parameters such as temperature, pressure or chemical potential are varied. A genuine equilibrium singularity is a nonanalyticity of the appropriate thermodynamic potential in the thermodynamic limit; finite systems generally show rounded crossovers instead.

An order parameter is a macroscopic quantity that distinguishes the phases. It often vanishes in a symmetric disordered phase and becomes nonzero after spontaneous symmetry breaking, although density-driven transitions need not have that simple symmetry interpretation. It can be a scalar, vector or tensor. At a first-order phase transition an appropriate first derivative of the free energy is discontinuous, and the order parameter commonly jumps; at a continuous phase transition the order parameter can approach zero continuously while a susceptibility or a higher derivative becomes singular.

Three examples are:

These examples also illustrate why an order parameter must reflect the physical distinction between phases rather than merely be any quantity that changes with temperature.

A magnetic field favors alignment of the magnetic moments of a paramagnet, reducing the number of substantially occupied magnetic configurations at fixed temperature. Its magnetic entropy therefore decreases as the field magnitude increases, explaining the negative sign of $(\partial S/\partial B{)}_T=-\alpha B/T^2$.

Choose the magnetic Helmholtz free energy convention $dF=-SdT-mdB$. Equality of its mixed partial derivatives gives the magnetic Maxwell relationThe magnetic moment inferred from an entropy derivative, integrated at fixed $B$, isThe given entropy derivative alone cannot determine the temperature-independent function $m_0(B)$. With the usual independent-paramagnet condition $m\to 0$ as $T\to \infty$ at fixed $B$, $m_0(B)=0$, giving Curie's law, $m=\alpha B/T$. The sketch uses that additional condition. The inverse-$T$ expression applies in the weak-field regime; its divergence cannot describe the actual low-temperature saturation of a finite set of magnetic moments.

For reversible quasistatic thermal isolation, $dS=0$. With positive field-dependent heat capacity $C_B=T(\partial S/\partial T{)}_B$,Reducing a positive $B$ therefore lowers $T$: this is adiabatic demagnetization. Equivalently, reducing $B^2$ cools for either field sign. Reduced alignment would raise magnetic entropy at fixed $T$; cooling compensates so total entropy remains constant. Insulation without reversibility would allow additional entropy production and need not follow this adiabat exactly.

Measure total magnetic field along the same south–north line with a calibrated magnetometer, recording position, time and sensor height. A simultaneous base-station magnetometer measures temporal variation so that diurnal changes can be removed from the moving survey. Avoid disturbed periods and magnetic contamination from vehicles, fences, power lines and equipment. Choose station spacing much smaller than the expected anomaly width, for example tens to hundreds of metres for a kilometre-scale target.

Subtract the epoch-appropriate International Geomagnetic Reference Field or another stated regional main-field model to obtain a total-field magnetic anomaly. Level overlapping traverses and remove offsets or drift where appropriate. Accurate terrain/sensor geometry is needed for comparison with a geological model, but there is no universal magnetic analogue of a constant-density Bouguer slab correction: the effect depends on the distribution and direction of magnetization. Measure magnetic susceptibility and remanent magnetization on representative greenstones, granite and host rock to test the assumed magnetic sources.

Assume that the intrusion has a positive magnetic susceptibility contrast, that induced magnetization dominates its remanent magnetization, and that the local present-day geomagnetic field points roughly northward and downward, as in this northern mid-latitude setting. Writewhere $\mathbf{b}$ is the anomalous magnetic field and the last expression assumes $|\mathbf{b}|\ll |{\mathbf{B}}_0|$. A total-field magnetic anomaly measures a field projection, so it need not have the same sign everywhere above a high-susceptibility body.^[ref]

The 100 km east–west extent makes an approximately two-dimensional, east–west-striking body a useful model away from its ends. On a profile drawn north on the left, south on the right, the expected normally induced response has a subsidiary negative lobe on the northern side and a stronger positive lobe over the body, displaced towards the south. The northward horizontal field component makes the anomaly asymmetric; its downward component produces the dominant positive response. The field falls back towards the regional value away from the intrusion. Broad bodies can have edge shoulders or a broad central high, so the sketch represents a qualitative shape rather than one uniquely specified curve.

An exactly vertical inducing field would remove this north–south skew for a symmetric induced body. Real magnetic inclination and magnetic declination should be taken from a suitable geomagnetic field model for the survey date and location; no exact local value is supplied here. Strong oblique or reversed remanent magnetization can shift or reverse the lobes, and heterogeneous magnetic susceptibility, body dip and three-dimensional end effects can alter the profile. Higher magnetic susceptibility alone does not determine a unique amplitude or polarity pattern without these qualifications.

A proton-precession magnetometer measures the scalar magnitude $F=|\mathbf{B}|$ of the local magnetic field. Its sensor contains a proton-rich liquid, such as a hydrocarbon or suitable water-based fluid. A strong current through a coil briefly polarizes the nuclear magnetic moments, producing a small net magnetization. The polarizing field is directed sufficiently across the local magnetic field to leave a transverse component when the current is switched off.

The protons then undergo Larmor precession about the ambient magnetic field. Their rotating net magnetization induces a decaying alternating voltage in a receiving coil. Measuring the frequency determines the field:For example, $50\mu \mathrm{T}$ gives about $2.13\mathrm{kHz}$. The known proton gyromagnetic ratio provides an absolute frequency-to-field scale; this instrument does not measure the field's direction. Strong spatial gradients cause different parts of the liquid to precess at different rates and shorten the usable signal, which matters very close to a buried steel object.^[ref]

Near the south magnetic pole, the inducing magnetic field and induced magnetic moment are vertically upward. Directly over the car its dipole field is also upward, again reinforcing the main field. The total-field magnetic anomaly is therefore the same positive central high and weak negative side lobes as at the north magnetic pole under the same induced-dipole assumptions. A fixed signed vertical field component would reverse, but the scalar field strength measured by the proton-precession magnetometer does not. If the car's remanent magnetization dominates, neither this symmetry nor these predicted signs is assured.

At the magnetic equator, the inducing magnetic field and induced magnetic moment are horizontal. Let $x$ point along this field, approximately magnetic north, and $y$ across it. Directly over the car, the dipole field opposes the main field, producing a negative central total-field magnetic anomaly. A north–south traverse through the car givesa symmetric central low with positive shoulders. An east–west traverse through the car instead has $\Delta T(y)\propto -(y^2+h^2{)}^{-3/2}$, a single negative trough. The equatorial anomaly therefore needs a stated traverse orientation; there is no universal single equatorial line profile.

Near the north magnetic pole, take the inducing magnetic field and induced magnetic moment as vertically downward. Directly above the car, its anomalous field reinforces the main field: a positive central total-field magnetic anomaly results. Along a horizontal traverse, the vertical-dipole approximation gives a symmetric central high flanked by weaker negative side lobes:where $h$ is the depth from the sensor to the effective dipole centre. The anomaly width scales with $h$; no absolute amplitude can be calculated without the car's magnetic moment.

The Pauli exclusion principle says that each complete one-particle state can have fermionic occupation number zero or one. “Complete” includes spin and any other internal labels, as well as the spatial wavefunction. Thus two electrons can share a spatial orbital with opposite spin projections: these are two different spin orbitals. Equal energy alone does not mean equal quantum state, because degenerate energy levels contain several independent one-particle states.

This distinction explains how an electron shell can accommodate several electrons while exclusion still allows only one per complete one-particle state. It also makes a finite model of fermionic state counting surprisingly simple: two spatial orbitals with two spin choices each give four complete one-particle states, and exactly six two-electron basis states. The following diagram enumerates them; its boxes represent occupation numbers, not locations in space.

Pauli's Exclusion principle and quantum mechanics by Wolfgang Pauli is his retrospective account, delivered on 13 December 1946 for the 1945 Nobel Prize. On printed pp. 29–30, he describes proposing an extra two-valued electron property in autumn 1924 and formulating exclusion in 1925. The crucial clue linked counts of resolved energy eigenstates in alkali spectra under a magnetic field with the populations of closed electron shells. Resolving every label turned a complicated shell capacity into one electron per complete one-particle state. This historical route preceded the modern spin interpretation and the antisymmetric wavefunction formulation; it was not originally a deduction from a Slater determinant.

Pauli paramagnetism is the weak positive magnetic response produced when a magnetic field shifts the two spin Fermi surfaces of a degenerate electron gas. At zero temperature, if $g(E_F)$ includes both spin states, its total magnetic moment and susceptibility are