advanced32 min studyLesson 521 in Physics

Localized moments and Curie susceptibility

Derive the weak-field susceptibility of localized spins while distinguishing angular momentum, magnetic moment, and saturation.

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What you’ll understand

  • Keep the electron moment sign separate from its angular momentum.
  • Derive the Curie constant from zero-field spin fluctuations.
  • Distinguish effective and saturated moments and identify the weak-field regime.

An isolated magnetic ion can retain a discrete angular-momentum multiplet even when its charge motion is frozen by a crystal environment. A field changes the populations of the multiplet's levels, producing paramagnetism without requiring moments to interact. This provides a useful reference against which exchange effects can be recognized. The derivation also exposes two common errors: treating the electron's moment as parallel to its spin and using the saturation moment in a thermal fluctuation formula.

Fix the operator and field conventions

Use dimensionless spin operators s=Sphys/ℏ\mathbf s=\mathbf S_{\mathrm{phys}}/\hbar, with s2=S(S+1)\mathbf s^2=S(S+1) and projection quantum numbers m from minus S to plus S. The positive number g is a spin-only or effective multiplet factor. For an electron-like localized moment, μ=−gμBs\boldsymbol\mu=-g\mu_B\mathbf s. Thus a field along positive z favors negative angular-momentum projection while the magnetic moment aligns with the field. The Bohr magneton has units of joules per tesla.

The Zeeman Hamiltonian is minus moment dotted into magnetic induction B, so its sign becomes positive when expressed through the electron's spin. A spin-half doublet therefore has energies plus and minus half of gμBBg\mu_B B. A crystal-field doublet may require an anisotropic g tensor instead of this scalar approximation. If orbital motion is not quenched, one must first identify the appropriate total-angular-momentum multiplet; simply inserting the bare electron spin can give the wrong Curie constant.

μ=−gμBs,HZ=gμBBsz,Z(B)=∑m=−SSexp⁡ ⁣(−gμBBmkBT)\boldsymbol\mu=-g\mu_B\mathbf s,\qquad H_Z=g\mu_B B s_z,\qquad Z(B)=\sum_{m=-S}^{S}\exp\!\left(-\frac{g\mu_B Bm}{k_BT}\right)
s is dimensionless, S labels its representation, B is magnetic induction in tesla, T is absolute temperature, and kB is Boltzmann’s constant. These levels assume an otherwise degenerate isolated multiplet and a positive scalar g.↔ Scroll wide equations horizontally.

Linear response comes from an isotropic second moment

At zero field all projections in the multiplet have equal statistical weight, so their mean is zero and their mean square is S(S+1)/3S(S+1)/3. Expanding the Boltzmann weights to first order in B makes the mean spin projection negative. Multiplication by the negative moment factor then gives positive magnetization. Equivalently, the susceptibility measures the zero-field variance of the moment component; the factor one third reflects rotational isotropy rather than a classical approximation to the spin length.

For number density n, magnetization M is moment per volume, measured in amperes per metre. The dimensionless SI volume susceptibility is χ=M/H\chi=M/H, where H is the magnetic field strength. In a weakly magnetized sample with negligible demagnetizing correction, B≃μ0HB\simeq\mu_0H, introducing the factor of vacuum permeability. A response defined instead as M divided by B has different units and lacks that factor. Molar susceptibilities introduce the number of moments per mole and must not be substituted silently.

χ=μ0n(gμB)2S(S+1)3kBT=CT,μeff=gμBS(S+1),μsat=gμBS\chi=\frac{\mu_0n(g\mu_B)^2S(S+1)}{3k_BT}=\frac{C}{T},\qquad \mu_{\mathrm{eff}}=g\mu_B\sqrt{S(S+1)},\qquad \mu_{\mathrm{sat}}=g\mu_BS
n has units m−3 and the Curie constant C has units kelvin for dimensionless SI volume susceptibility. The effective moment describes isotropic fluctuations; the saturated moment is the largest field-directed projection per ion.↔ Scroll wide equations horizontally.

Know what a Curie fit does and does not diagnose

The linear expansion requires gμBBS≪kBTg\mu_BBS\ll k_BT. At stronger fields or lower temperatures, populations become unequal enough that magnetization bends toward saturation rather than diverging as an unrestricted linear law. For spin one half the exact independent-moment result is a hyperbolic tangent. Its small-argument expansion recovers the Curie law, while its large-argument limit gives the saturated moment. This limit is an immediate check on signs and factors of two.

Exchange, crystal-field splittings, itinerant electrons, and temperature-independent orbital contributions can all modify the measured susceptibility. A nonzero Curie–Weiss intercept can suggest interactions, but a fit alone does not establish an ordered phase or identify a microscopic exchange path. Finite collections of independent moments do not undergo a thermal ordering transition. The observed field response in this lesson is population redistribution within a statistical ensemble, not coherent precession of a single pure spin state.

Extract a Curie constant without mixing moment definitions

A dilute spin-half insulator has g=2 and Q=μ0n(gμB)2/kB=0.48 KQ=\mu_0n(g\mu_B)^2/k_B=0.48\,\mathrm K. In the weak-field independent-moment regime, find its SI volume susceptibility at 30 K, its effective moment, and its saturated moment per ion.

  1. For S=1/2, S(S+1)=3/4S(S+1)=3/4, so C=Q(3/4)/3=0.12 KC=Q(3/4)/3=0.12\,\mathrm K. Q already includes the density, permeability, and squared g factor; inserting them again would double count them.

  2. Divide by the absolute temperature: χ=0.12/30=0.0040\chi=0.12/30=0.0040. It is dimensionless because C and T both carry kelvin. This is M/H, not M/B.

  3. The effective moment is 2μB3/4=3μB2\mu_B\sqrt{3/4}=\sqrt3\mu_B, whereas saturation is 2μB(1/2)=μB2\mu_B(1/2)=\mu_B. The larger effective value is a fluctuation scale, not an attainable z projection.

The result

The susceptibility is 0.0040; the effective moment is approximately 1.732 Bohr magnetons and the saturated moment is one Bohr magneton per ion.

Explore the model

Interactive lab

Interactive lab · probability

Independent projections and their fluctuations

For each comparison, Reset and set BOTH Success probability p=0.50 and Trials n. Compare n=10 and n=20. Predict binomial means 5 and 10 and variances 2.5 and 5. If each success represents moment +μ and each failure −μ, the total moment is μ(2K−n), so its variance is nμ². These exact independent probabilities illustrate spin-half population fluctuations; they contain no exchange, field-to-temperature mapping, or coherent spin dynamics.

Independent projections and their fluctuationsThese are exact binomial probabilities for independent trials with one fixed success probability. Playback tours the distribution; it does not resample outcomes. Current readings: Mean · np: 5; Variance · np(1−p): 2.5; Total probability: 1. All parameters are adjustable with the labeled controls below.03581000.070.150.220.3Successes kP(X = k)k = 0 · P = 9.77e-4
Exact probabilityHighlighted outcomeMean np
Mean · np
5
Variance · np(1−p)
2.5
Total probability
1
t = 0 s

These are exact binomial probabilities for independent trials with one fixed success probability. Playback tours the distribution; it does not resample outcomes. Parameter changes pause playback and recompute the model at the current elapsed time. Reset restores both parameters and returns time to zero. Playback starts paused. Adjust either parameter, then choose Play or Step.

Look forThe variance grows with the number of independent moments even when their mean total moment vanishes.

Check your understanding

For an electron-like spin-half moment with g>0 in B along positive z, which Zeeman state has lower energy?

Try it yourself

For a spin-half ion let gμBB=0.20 meVg\mu_BB=0.20\,\mathrm{meV} and kBT=0.50 meVk_BT=0.50\,\mathrm{meV}. Find the two Zeeman energies and the ratio of the lower-level population to the upper-level population. For g=2, compute the mean moment in units of μB using tanh(0.20)≈0.1974.

Take a moment to make a prediction or work it out on paper. Open a hint when you need one.

Give me a hint

The level splitting is the full quantity gμBB, not half of it. The ratio of Boltzmann weights involves that splitting; the hyperbolic tangent instead uses half the splitting divided by thermal energy.

Compare with the solution

The energies are −0.10 meV for m=−1/2 and +0.10 meV for m=+1/2. Their population ratio is exp⁡(0.20/0.50)=e0.4≃1.492\exp(0.20/0.50)=e^{0.4}\simeq1.492. The difference of their normalized populations is tanh⁡[0.20/(2×0.50)]=tanh⁡(0.20)\tanh[0.20/(2\times0.50)]=\tanh(0.20). Since the two moments are ±μB, their average is +0.1974μB along the field. The linear approximation would give 0.20μB, slightly larger, as expected when saturation begins to curve the response.

Take it with you

  • Dimensionless spin and physical angular momentum differ by ħ.
  • Electron-like moments point opposite to angular momentum and align with an applied field.
  • Curie susceptibility uses S(S+1), while saturation uses S.

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