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For a charged spin one-half particle with the Dirac equation reference value $g=2$, the anomalous magnetic moment is conventionally expressed by the dimensionless quantity $a=(g-2)/2$. For the electron, write $a_e=(g_s-2)/2$. A stipulated $g_s=2.0024$ would therefore mean $a_e=0.0012$: a $0.12\%$ excess of the spin-component moment over one Bohr magneton. This is a teaching input, not a quoted measurement.

The University of Iowa's Electron Spin Resonance, University of Iowa laboratory protocol (revised July 2007, Experiment and Resonance Data Analysis, printed pp. 2–6) specifies DPPH on cotton in a tube inside an RF coil, between Helmholtz coils. A bridge monitors RF absorption; an oscilloscope records this output and the voltage across a $1\Omega$ current-monitor resistor. Sweeping the coil current passes through the resonant magnetic field. The protocol describes two dips when the sweep revisits resonance; these are repeated crossings, not two distinct spectral lines.

Its calibration replaces the sample with a Hall effect probe, maps the magnetic field in $2\mathrm{mm}$ steps, and accounts for sample dimensions. At least four RF frequencies test the resonance relation. Sweep speed is adjusted to obtain comparable passages. Only a local peak-plus-background fit is prescribed, because the full shape is not Gaussian. The protocol also flags unexplained trailing spikes and possible electronics contributions. These controls constrain field and sweep errors; an empty-tube/cotton blank would be an additional background control, not a reported measurement in this protocol.^[ref]

For an isolated effective spin one-half with dimensionless positive factor $g$, the basic resonance model isHere ${\mu }_{\mathrm{B}}$ is the Bohr magneton, $h$ is the Planck constant, $\nu$ is the applied frequency, and $B_{\mathrm{res}}$ is the calibrated resonant magnetic field. A change in frequency should shift the resonance field proportionally in this model. The factor $g$ describes the electronic magnetic moment coupling, not the species' net electric charge.

A resonance supports the presence of detectable paramagnetism. It does not, by itself, identify an unknown radical: other species exhibiting paramagnetism, impurities, overlapping responses, and interactions with atomic nuclei can complicate interpretation. Conversely, absence of a detectable resonance need not mean absence of free radicals; a low concentration, a short lifetime, or an unsuitable measurement range can conceal them. The occupancy and chain diagrams are original bookkeeping illustrations; neither is a measured or simulated DPPH spectrum.

The magnetic moment inferred from the weak-field Curie constant of a localized angular-momentum multiplet. Uniform zero-field occupation of its $2J+1$ substates gives $⟨m_J^2⟩=J(J+1)/3$, yielding ${\mu }_{\mathrm{eff}}=|g_J|{\mu }_B\sqrt{J(J+1)}$. Here $g_J$ is the Landé g-factor and ${\mu }_B$ the Bohr magneton. The dimensionless number $m_{\mathrm{eff}}={\mu }_{\mathrm{eff}}/{\mu }_B$ differs from the saturated projection $|g_J|J$.

The electron magnetic moment couples an electron's angular momentum to a magnetic field. Its intrinsic spin contribution isHere ${\mu }_B$ is the Bohr magneton, and $g_s$ is the positive electron spin g-factor under this sign convention. The minus sign makes magnetic dipole moment antiparallel to spin angular momentum. An electron in a chemical atom can also contribute an orbital magnetic dipole moment; the measured response of a whole atomic level combines these contributions.

Measuring a tiny departure of $g_s$ from $2$ tests how well a theory predicts an electron's interaction with its surroundings. The Dirac equation gives $g_s=2$ before radiative corrections; the anomalous magnetic moment makes a small correction experimentally accessible.^[ref] A useful strategy is to compare two atomic energy levels whose orbital and spin contributions have different weights, rather than trying to calibrate an absolute magnetic field more accurately.

The dimensionless coefficient in the spin part of the electron magnetic moment. It is distinct from a gyromagnetic ratio, which has units: in the angular convention, ${\gamma }_e=-g_{s}e/(2m_e)$. It is also distinct from the Landé g-factor $g_J$ of a combined atomic angular momentum and from a nuclear or hyperfine g-factor.

For spin one-half, the largest magnitude of a spin-component magnetic dipole moment is $g_s{\mu }_B/2$, not $g_s{\mu }_B$. Thus the ideal value $g_s=2$ gives one Bohr magneton. This follows directly from the spin angular momentum projections $S_z=\pm \hslash /2$.

Consider a stipulated dilute collection of independent two-level systems, each with longitudinal magnetic moment $s\mu$, $s=\pm 1$, and energy $E_s=-s\mu B_0$. The prescribed vacuum/applied magnetic flux density is $B_0={\mu }_0{H}_{\mathrm{app}}$. Ignore interactions, orbital magnetic response, and the small sample-field correction. This is the magnetic version of the existing independent spin-one-half two-level system, with alignment probabilities rather than an interacting Curie–Weiss model.

For temperature $T>0$, the canonical ensemble gives the one-moment partition function $Z_1=2\cosh x$, with $x=\mu B_0/(k_{B}T)$. Taking the expected value over its two Boltzmann weights givesHere $n$ is the number density, $k_B$ the Boltzmann constant, and $\tanh$ the hyperbolic tangent. For $|x|\ll 1$, its small-argument expansion of the hyperbolic tangent gives $M\simeq {\mu }_{0}n{\mu }^2{H}_{\mathrm{app}}/(k_{B}T)$, a Curie's law response in the applied-$H$ convention. For large $|x|$, $M$ approaches the saturation magnetization $n\mu$ with sign set by the field, and its differential magnetic susceptibility falls towards zero.

Both plots use stipulated $n=10^{26}{\mathrm{m}}^{-3}$, $\mu =9.2740101\times 10^{-24}\mathrm{J}{\mathrm{T}}^{-1}$, $k_B=1.380649\times 10^{-23}\mathrm{J}{\mathrm{K}}^{-1}$, and ${\mu }_0=1.25663706\times 10^{-6}\mathrm{T}\mathrm{m}{\mathrm{A}}^{-1}$. The moment is approximately one Bohr magneton; the chosen vacuum permeability is a rounded model input. At $T=2\mathrm{K}$, $M_s=927.401\mathrm{A}{\mathrm{m}}^{-1}$ and ${\chi }_{\mathrm{model}}(0)=3.91410\times 10^{-4}$. The linear approximation overestimates $M$ by less than $3\%$ for $|x|\le 0.3$, corresponding to $|{\mu }_0{H}_{\mathrm{app}}|\le 0.89324\mathrm{T}$ or $|H_{\mathrm{app}}|\le 0.71082\mathrm{MA}{\mathrm{m}}^{-1}$. These parameters keep the modeled response weak; neglected magnetic feedback is of relative order ${\chi }_{\mathrm{model}}(0)$, rather than an explanation of a material's ordered phase.

At $B_0=0$, cooling increases the zero-field slope as $1/T$. At fixed $B_0\ne 0$, cooling instead gives ${\chi }_{\mathrm{model}}\sim 4{\mu }_{0}n{\mu }^2{e}^{-2\mu |B_0|/(k_{B}T)}/(k_{B}T)$: a saturated moment can hardly change further. The order of the zero-field and zero-temperature limits matters. This independent model has no finite-temperature collective phase transition; the diverging zero-field slope alone does not establish one.

The nuclear magneton is the unit of magnetic momentwhere $m_p$ is the proton mass. For a nuclear spin operator $\mathbf{I}$, the nuclear magnetic moment is $\mathbf{\mu }=g_I{\mu }_N\mathbf{I}/\hslash$. Consequently its gyromagnetic ratio is ${\gamma }_I=g_I{\mu }_N/\hslash$, and its Larmor frequency in a uniform magnetic field of magnitude $B$ is $|{\gamma }_I|B$. The nuclear magneton is smaller than the Bohr magneton by the electron-to-proton mass ratio.

Take the field along $z$ and write $x={\mu }_{B}B$, with Bohr magneton ${\mu }_B$. The printed Hamiltonian uses dimensionless angular-momentum operators, with eigenvalues $L(L+1)$ and $m_L$; physical angular momenta carry an additional $\hslash$. Thus $\beta$ and $x$ are energies. Assume the usual positive spin-orbit interaction coefficient $\beta >0$. Reversing its sign reverses the zero-field ordering but not the state count.

An electron has $S=1/2$. addition of angular momentum givesThe coupled angular-momentum state count can be checked directly. The product basis contains $(2L+1)\times 2=4L+2$ states. The coupled basis containsCoupling therefore reorganizes the states rather than creating or losing them. There is no hyperfine multiplicity because $I=0$.

Use $\mathbf{J}=\mathbf{L}+\mathbf{S}$ to obtainThe two spin-orbit interaction energies are $\beta L/2$ and $-\beta (L+1)/2$. In a weak field, project the Zeeman effect onto a fixed $J$ multiplet:This is the first-order energy formula, not a statement that $H_Z$ has no matrix elements between different $J$ values. For a common $m=m_J$ and a conventional choice of relative phase, its complete coupled-basis block isThe inter-multiplet Zeeman coupling terms give second-order weak-field corrections because the two spin-orbit interaction multiplets are already separated.

In the strong-field basis the Zeeman part is diagonal, $E_Z=x(m_L+2m_S)$. WriteIts diagonal matrix elements are $\beta m_L{m}_S$. Its off-diagonal matrix elements connect $(m_L,+1/2)$ to $(m_L+1,-1/2)$ with coefficientand the reverse spin change has $m_L+1$ replaced by $m_L-1$ in the corresponding ladder coefficient. These terms preserve $m_J$ but mix the two product states. To first order in $\beta /x$, the energy correction is simply $\beta m_L{m}_S$.

For the P state, $L=1$. Measured relative to its gross-structure centroid, the six low-field levels areThus the upper quartet starts at $\beta /2$ with slopes $+2,+2/3,-2/3,-2$ in a plot against $x$; the lower doublet starts at $-\beta$ with slopes $\pm 1/3$. Their zero-field separation is $3\beta /2$.

At high field all six product-state energies, retaining the term linear in $\beta$, areThe two zero-slope levels have different $m_J$ and no mutual matrix element, so their first-order degeneracy does not require an additional degenerate perturbation. At finite field they split at order ${\beta }^2/x$.

An exact interpolation makes the requested whole-field sketch unambiguous. The stretched levels areFor $m=\pm 1/2$, use product states $|m-1/2,+1/2⟩$ and $|m+1/2,-1/2⟩$. Their Hamiltonian block issoThese energies reduce to the quartet and doublet at zero field and to the six strong-field expressions above. Within a fixed $m_J$, the nonzero off-diagonal element prevents a crossing of the two branches. Levels in different $m_J$ sectors can cross without hybridizing. For the S state, $L=0$ and the spin-orbit interaction term vanishes, leavingthroughout this Hamiltonian's range of validity.

For electric dipole transitions at low field, $\Delta J=0,\pm 1$ and $\Delta m_J=0,\pm 1$ apply, with the usual exclusion of $J=0\to 0$. Each P state has opposite parity to the S state. The $P_{1/2}\to S_{1/2}$ transition has four allowed sublevel pairs. For $P_{3/2}\to S_{1/2}$, the two stretched upper states each have one destination and the other two each have two, giving six. Hence there are 10 allowed state-to-state channels at weak nonzero field. Relative to the two zero-field transition energies, their Zeeman shifts are

In the decoupled high-field limit, an electric dipole acts on orbital coordinates, so $\Delta m_S=0$. Each of the six P product states connects to the S state with the same spin and $m_L^{\prime }=0$. Thus there are 6 allowed channels in the limiting product-state description. Their photon energies areThis Paschen-Back transition-channel count is not the count of distinct frequencies: to first order in $\beta$ the two $m_L=0$ channels coincide, giving 5 distinct frequencies, with shifts $0,\pm (x+\beta /2),\pm (x-\beta /2)$. If the residual spin-orbit interaction splitting is unresolved there are the three usual orbital Zeeman groups.

The apparent loss of four channels is a change in eigenstate mixing, not a change in Hilbert-space dimension. At finite large field, four weak spin-forbidden Paschen-Back lines arise from small opposite-spin admixtures, with intensities vanishing as $(\beta /x{)}^2$. A generic finite-field calculation therefore retains ten nonzero channels; six carry finite strength in the complete-decoupling limit. Observation direction and polarization select which of these channels a detector actually receives.

Take the energy of an isolated, field-free gas atom as zero, and let $E_b>0$ be the field-independent binding energy at one adsorption site. The two adsorbed spin energies areHere ${\mu }_{\mathrm{B}}$ is the Bohr magneton. The lower state is stabilized by increasing $|B|$; the assignment of the two signs to names for the spin projections is a convention.

For $n$ unpaired electrons, the spin-only magnetic moment iswhere ${\mu }_{\mathrm{B}}$ is the Bohr magneton and the second expression uses $g\simeq 2$.

For the square, a large separation of the highest $B_{1g}$ level favours pairing in the lower four molecular orbitals: $S=0$, $n=0$, ${\mu }_{\mathrm{eff}}=0$. If the cost of occupying $B_{1g}$ were smaller than the relevant electron pairing energy, a high-spin complex with two unpaired electrons and $S=1$ could instead occur, giving $\sqrt{8}{\mu }_{\mathrm{B}}\simeq 2.83{\mu }_{\mathrm{B}}$.

For ideal tetrahedral $d^8$, the occupation is $e^4{t}_2^4$. Hund's rules distribute four electrons over the three degenerate $t_2$ molecular orbitals as one pair and two single occupations. Thus $S=1$ and ${\mu }_{\mathrm{eff}}\simeq 2.83{\mu }_{\mathrm{B}}$. Merely increasing the $e$–$t_2$ splitting does not create a conventional paired tetrahedral $d^8$ alternative: four electrons must still occupy $t_2$.

Let $g_v(E)$ denote the spin-summed density of states per volume. An electron moment with magnitude the Bohr magneton ${\mu }_B$ has Zeeman energies $\mp {\mu }_{B}B$. At low temperature, the two spin populations areThe common chemical potential changes only at order $B^2$, so it does not affect this linear imbalance. The magnetization is $M={\mu }_B(n_{+}-n_{-})={\mu }_B^2{g}_v(E_F)B$. Define SI volume magnetic susceptibility by $M={\chi }_{p}H$ and take the weak applied field as $B\simeq {\mu }_{0}H$. Then Pauli paramagnetism givesThe last equality uses the three-dimensional parabolic free electron model, $g_v(E_F)=3n/(2E_F)$. This susceptibility is nearly independent of temperature for $T\ll T_F$, unlike the Curie law for localized independent moments. The displayed result excludes orbital diamagnetism and interaction enhancements. If susceptibility instead denotes $\partial M/\partial B$, omit ${\mu }_0$; confusing that convention with $\partial M/\partial H$ changes the units.

In the one-band electron Hall effect model, $R_H=-1/(ne)$, with $e>0$. The measured Hall coefficient therefore givesA monovalent body-centered cubic lattice has two conduction electrons per conventional unit cell. Thus $n{a}^3=2$, andCounting two spin states within the Fermi sphere gives $k_F=(3{\pi }^{2}n{)}^{1/3}=9.0418\times 10^9{\mathrm{m}}^{-1}$. Using the specified effective mass in the dispersion,There is one mole of conduction electrons per mole of sodium. Its molar density of states at $E_F$ is $3N_A/(2E_F)$, givingUsing the volume density of states in c givesThe effective mass changes the dispersion and density of states; the electron Bohr magneton used for the spin coupling retains the bare electron mass. Replacing it by $e\hslash /(2m^{\ast })$ would be an additional, unjustified magnetic-moment assumption. A Hall factor or a multiband metal would also invalidate the simple density inference; the stated calculation uses the requested one-band approximation.

The specified two-direction constraint permits magnetic moments ${\mu }_z=\pm \mu$, withwhere ${\mu }_B$ is the Bohr magneton. There are four magnetic gadolinium ions per conventional face-centered cubic lattice cell; the closed-shell nitrogen ions add no local spin moment. Hence the magnetic-ion number density isAt high temperature, neglect interactions and use $B_z\simeq {\mu }_0{H}_z$. The one-ion partition function and mean magnetic dipole moment areIn the weak-magnetic field limit, $\tanh x\simeq x$, so $M_z=n⟨{\mu }_z⟩={\mu }_{0}n{\mu }^2{H}_z/(k_{B}T)$. Thus the dimensionless SI magnetic susceptibility is $\chi =C/T$, withThe coefficient has units of kelvin. The unrestricted-spin expression proportional to $S(S+1)/3$ is inappropriate here: the source explicitly retains only the two orientations with $S_z=\pm S$. In this two-state Curie law, the squared axial moment is $(g_{J}S{\mu }_B{)}^2$.

In the hydrogenic donor state approximation, the screened Coulomb potential energy is $-e^2/(4\pi {ϵ}_0{ϵ}_{r}r)$ and the kinetic term uses $m^{\ast }=0.067m_e$. Rescaling the hydrogen atom gives the effective Bohr radius of a hydrogenic donor and the positive binding energyThereforeThe radius here is the exponential length of the ground-state wavefunction, not its root-mean-square radius. With $\psi (r)=(\pi a_b^3{)}^{-1/2}{e}^{-r/a_b}$, radial integration gives $⟨r^2⟩=3a_b^2$ and $⟨x^2+y^2⟩=2a_b^2$.

For dilute independent donors of number density $n$, a bound spin one-half electron has magnetic moment magnitude ${\mu }_s=|g|{\mu }_B/2$, with the Bohr magneton defined using the bare electron mass. Its weak-field Curie law response isFor the orbital response, expand the effective mass Hamiltonian $(\mathbf{p}+e\mathbf{A}{)}^2/(2m^{\ast })$ in a uniform magnetic field, using $\mathbf{A}=\mathbf{B}\times \mathbf{r}/2$. The spherical ground state has zero orbital angular momentum; the quadratic term raises its energy byDifferentiating this energy twice and using $B\simeq {\mu }_{0}H$ givesThe donor-only magnetic susceptibility cancels whenThe negative sign of $g=-0.45$ drops out of the linear magnetic susceptibility. This crossover is far below $E_b/k_B\simeq 63.6\mathrm{K}$, consistent with donor binding. It assumes sufficiently weak magnetic field, negligible exchange between donors and a thermal scale above any inter-donor splitting; sufficiently dilute donors can satisfy those assumptions. The background diamagnetism from part (b) is excluded from this cancellation.

In weak magnetic field and the regime of paramagnetism, localized magnetic moments with total angular momentum $J$ and Landé g-factor $\tilde{g}$ have SI volume magnetic susceptibilitywithin the Curie–Weiss law. Here $n_m$ counts magnetic ions per volume, ${\mu }_0$ is the vacuum permeability, ${\mu }_B$ is the Bohr magneton, and $\Theta$ is the Weiss temperature. The factor $J(J+1)/3$ is the zero-field average of the squared angular projection over $m_J=-J,\dots ,J$.

For one magnetic ion per formula unit, a molar volume $V_m$ contains $N_A$ magnetic ions, where $N_A$ is the Avogadro constant. Consequently $n_m{V}_m=N_A$, and the molar magnetic susceptibility isDefining the dimensionless effective paramagnetic moment $m_{\mathrm{eff}}$ by ${\mu }_{\mathrm{eff}}=m_{\mathrm{eff}}{\mu }_B$, numerical substitution givesThe normalization is per mole of formula units, which here also means per mole of magnetic ions.

Let $g(E)$ denote the total density of states, including both spin directions. The field lowers one spin energy by ${\mu }_{B}B$ and raises the other by the same amount. To first order in $B$, each spin population changes by the one-spin density $g(E_F)/2$ times the displacement of its Fermi edge. The population imbalance is thereforeMultiplying by the Bohr magneton carried by each excess aligned electron givesThus, with this total-density convention,The zero-temperature magnetic susceptibility isThis weak positive response is Pauli paramagnetism. If susceptibility per unit volume is required, both $M$ and $g$ are divided by $V$.

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.

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.

An original reduced model makes the cancellation explicit. Assume LS coupling, neglect nuclear spin, and write the electronic magnetic dipole moment as $-{\mu }_B(g_L\mathbf{L}+g_s\mathbf{S})/\hslash$. For $L=1$, $S=1/2$, the projection used to derive the Landé g-factor givesThese coefficients follow by substituting $J=3/2$ or $1/2$ into the angular-momentum projection formula; neither level measures pure spin alone. In a weak uniform magnetic field, adjacent electronic projections have spacing $h{f}_J=g_J{\mu }_{B}B$ when $g_J>0$. ThereforeThe common magnetic field and Bohr magneton cancel. The ratio determines $g_s/g_L$, not $g_s$ without further knowledge of $g_L$. For $g_L=1$, it gives $a_e=(R-2)/(R+1)$. Small departures from $R=2$ are magnified differently from small departures of $g_s$ from $2$; the two fractional changes should not be identified.

With the stipulated inputs $g_L=1$, $g_s=2.0024$ and ${\mu }_{B}B/h=100\mathrm{MHz}$, the two electronic frequencies are $66.5867$ and $133.4133\mathrm{MHz}$. Their ratio is approximately $2.0036043$, and inversion recovers $a_e=0.0012$. These are ideal electronic lines, not historical gallium data. Nuclear spin, mixing between levels and other deviations from LS coupling require a more complete model in an actual atom.

Let the atomic beam travel along $x$, with velocity $v>0$, and let $z$ measure displacement along the dominant magnetic field gradient. For a fixed effective magnetic moment component ${\mu }_z$, the magnetic dipole potential energy is $U\simeq -{\mu }_z{B}_z$. Taking its negative partial derivative givesA spatially uniform magnetic field can produce torque and Larmor precession, but cannot produce this translational separation. In a quantum state following a local internal energy eigenvalue $E_n(B)$, the more general expression is $F_z=-{\partial }_z{E}_n$, with effective magnetic moment $-\partial E_n/\partial B$. Treating that moment as constant is an approximation with a physical regime, not an extra measured fact. Assume the local magnetic field direction stays nearly fixed across the sampled beam, and that internal states follow it without transitions, as in the adiabatic theorem. The scalar gradient model describes the beam's neighborhood; it omits the full magnetic field's transverse components and fringe regions.

Assume a constant $G$ over magnet length $\ell$, followed by a field-free drift $D$. Neglect transverse motion at entry and changes in longitudinal velocity. Newton's second law gives transverse acceleration $a_z={\mu }_{z}G/M$ for whole-atom mass $M$. The magnet transit time is $t_m=\ell /v$, and the drift time is $t_d=D/v$. The magnet supplies both a displacement and an exit transverse velocity:The $\ell D$ contribution is motion with the acquired transverse velocity, not continued acceleration outside the magnet. For the plate-at-exit geometry, set $D=0$. For a nonzero entrance transverse velocity $w$, add $w(\ell +D)/v$. These terms show why initial collimation and flight geometry matter when inferring a magnetic moment from a deposit.

Choose hypothetical parameters $M=1.79\times 10^{-25}\mathrm{kg}$, $G=300\mathrm{T}{\mathrm{m}}^{-1}$, $\ell =0.035\mathrm{m}$, $D=0.100\mathrm{m}$, and $v=500\mathrm{m}{\mathrm{s}}^{-1}$. Approximate the two effective magnetic moments by ${\mu }_z=\pm {\mu }_B$, where the Bohr magneton is ${\mu }_B\simeq 9.274\times 10^{-24}\mathrm{J}{\mathrm{T}}^{-1}$. Then $|a_z|\simeq 1.55\times 10^4\mathrm{m}{\mathrm{s}}^{-2}$, $t_m=70\mu \mathrm{s}$ and $t_d=200\mu \mathrm{s}$. The positive displacement is about $0.0381\mathrm{mm}$ inside the magnet plus $0.2176\mathrm{mm}$ during drift: $d=0.2557\mathrm{mm}$, with full separation $2d=0.5114\mathrm{mm}$. These chosen values illustrate the model and are not reconstructed 1922 measurements.