Pauli Paramagnetism

Take the classical law for a gas of magnetic moments — align them with a field and you get a susceptibility $\chi \propto 1/T$ , growing without bound as you cool, and set by the total number of moments. Now apply it to the $\sim 10^{23}$ conduction electrons in a piece of sodium. The prediction is a strong, strongly temperature-dependent paramagnetism. The measurement is almost nothing: a feeble susceptibility of order $10^{-6}$ that barely moves as you change the temperature. That mismatch, by a factor of a hundred, is the puzzle Wolfgang Pauli resolved in 1927 — and its resolution was one of the first hard proofs that electrons in a metal obey Fermi–Dirac statistics. We take one fact from the quantum-mechanics track and build everything else on it. The article aq-qm-spin-half-pauli-matrices establishes that a spin- $\tfrac12$ electron has $\hat S_z = \tfrac{\hbar}{2}\,\mathrm{diag}(1,-1)$ with eigenvalues $\pm\hbar/2$ — a genuine two-outcome degree of freedom, up or down along the field. That two-level structure is all we borrow. What this article owns is the thermodynamics: attach a magnetic moment $\mu_B$ to each spin, drop the electrons into a degenerate Fermi sea, and compute how the gas responds. The exclusion principle throttles that response. Deep in the Fermi sea an electron cannot flip its spin, because the state it would flip into is already occupied. Only the thin shell of electrons within $\sim\mu_B B$ of the Fermi energy can rearrange, so the magnetization is proportional to the density of states at the Fermi level $g(\varepsilon_F)$ — not to the total electron count. The result is the Pauli susceptibility $\chi_P = \mu_0\mu_B^2\,g(\varepsilon_F)$ : small, temperature-independent, and a direct fingerprint of the degenerate Fermi gas. Its orbital partner, Landau diamagnetism, is handed forward by name.

Key takeaways

A magnetic field lowers the energy of field-aligned electron spins by $\mu_B B$ and raises anti-aligned spins by $\mu_B B$ ; the gas re-equilibrates by moving electrons from the anti-aligned band into the aligned band.
Only electrons within $\sim\mu_B B$ of the Fermi energy can move — the rest are locked by exclusion — so the induced moment is set by the density of states at the Fermi level, $g(\varepsilon_F)$ , not by the total electron number $n$ .
The susceptibility is $\chi_P = \mu_0\mu_B^2\,g(\varepsilon_F)$ , or in free-electron form $\chi_P = \tfrac32\,\mu_0 n\mu_B^2/\varepsilon_F$ — small and, at leading order, temperature-independent.
It is quenched below the classical Curie estimate by a factor $\tfrac32\,T/T_F \sim 10^{-2}$ , which is exactly why simple metals are only feebly paramagnetic.
The Sommerfeld expansion supplies a tiny correction $\chi(T) = \chi_P\,[1 - \tfrac{\pi^2}{12}(k_BT/\varepsilon_F)^2 + \cdots]$ ; the susceptibility falls very slightly as $T$ rises.
Pauli paramagnetism ( $+\chi_P$ ) and Landau diamagnetism ( $-\tfrac13\chi_P$ for free electrons) add to a net paramagnetism $\tfrac23\chi_P$ ; exchange interactions (Stoner) can enhance $\chi_P$ toward a ferromagnetic instability.
Key idea: a field splits the spin-up and spin-down density-of-states curves by $2\mu_B B$ ; the gas transfers electrons from the higher band to the lower one, producing a net moment $\propto g(\varepsilon_F)$ , so $\chi_P = \mu_0\mu_B^2\,g(\varepsilon_F)$ — small and $T$ -independent, the mark of degeneracy.

The core insight

A magnetic field splits the spin-up and spin-down density-of-states curves by $2\mu_B B$ ; the degenerate gas re-equilibrates by transferring electrons from the higher (anti-aligned) band to the lower (aligned) band, producing a net moment proportional to the number of states near the Fermi energy, $g(\varepsilon_F)$ . Hence $\chi_P = \mu_0\mu_B^2\,g(\varepsilon_F)$ — small and temperature-independent, a direct signature of the degenerate Fermi sea.

Where this lives today

Pauli paramagnetism is a workhorse of experimental condensed-matter physics because it is a direct meter for the density of states at the Fermi level. Since both $\chi_P$ and the electronic heat-capacity coefficient $\gamma$ are proportional to $g(\varepsilon_F)$ , their ratio — the Wilson ratio — is a pure number, equal to one for free electrons; measured deviations quantify the strength of electron–electron interactions and are a standard diagnostic for correlated metals and heavy-fermion compounds. Knight-shift NMR isolates the spin susceptibility alone (it couples only to spin), separating it cleanly from the orbital Landau piece. The same physics is now reproduced in ultracold atomic Fermi gases, where a two-component gas realizes a tunable, impurity-free electron gas. Its "spin susceptibility" — the response of the population imbalance to a chemical-potential difference between hyperfine states — follows the Pauli form and its $(T/T_F)^2$ correction, a clean test of Fermi-liquid theory free of the band-structure complications of real solids. And when exchange pushes $Ug(\varepsilon_F)$ toward $2$ , the enhanced susceptibility diverges and the metal turns ferromagnetic — the Stoner mechanism behind iron, cobalt, and nickel, and the reason nearly-ferromagnetic metals like palladium have anomalously large $\chi$ .

The physical picture

Start with what the quantum-mechanics track hands us. In aq-qm-spin-half-pauli-matrices the electron's spin along the field is a two-level observable: $\hat S_z = \tfrac{\hbar}{2}\,\mathrm{diag}(1,-1)$ , with eigenvalues $\pm\hbar/2$ . Attach to each of those two states a magnetic moment of one Bohr magneton, $\mu_B = e\hbar/2m_e$ , pointing along or against the field. That coupling — the Zeeman energy $\mp\mu_B B$ — is what this article adds; the spin- $\tfrac12$ structure itself is borrowed, not rederived. Now picture two density-of-states parabolas, one for each spin, drawn back to back. Turning on $B$ slides the spin-up (aligned) curve down by $\mu_B B$ and the spin-down (anti-aligned) curve up by $\mu_B B$ . The two bands used to fill to the same Fermi level; now the aligned band's states sit lower, so electrons spill from the anti-aligned band into the aligned band until a single common chemical potential is restored. The number that move is the number of states in an energy slab of width $\mu_B B$ at the top of the sea — that is $g(\varepsilon_F)\,\mu_B B$ — and each carries a moment $\mu_B$ . This is the whole mechanism, and the first figure shows it.

Figure

Spin-split density of states — the field slides the aligned band down by μ_B B and the anti-aligned band up by μ_B B; electrons in the shaded sliver of width μ_B B at ε_F transfer, leaving a net moment

Why is the metal so weakly magnetic? Because the Pauli principle locks up every electron deep in the sea. Flipping a deep spin means sending it into an already-occupied state, which is forbidden. Only the sliver within $\sim\mu_B B$ (and, at finite $T$ , within $\sim k_BT$ ) of $\varepsilon_F$ is free to respond, so the response is choked down from the classical value by roughly $T/T_F \sim 1/100$ . The very feature that makes the naive Curie law fail — a huge reservoir of frozen electrons — is what the exclusion principle enforces. The second figure is the punchline: susceptibility versus temperature. The classical Curie prediction is a curve that diverges as $1/T$ toward low temperature. The measured metallic susceptibility is a nearly flat line at the small Pauli value $\chi_P$ , crossing the Curie curve only near $T \sim T_F$ — a temperature of tens of thousands of kelvin that no metal ever reaches. Everywhere a metal actually lives, on the far left of the plot, degeneracy has flattened the response.

Figure

Susceptibility vs temperature — the classical Curie 1/T curve diverges at low T while the Pauli line stays flat at χ_P; they cross only near T ≈ T_F, far above any real metal's temperature

The temperature independence is the subtle signature. As $T$ rises the Fermi step smears over a width $k_BT$ , but $g(\varepsilon)$ is nearly flat across that width, so the integrated imbalance barely changes. Only the curvature of $g(\varepsilon)$ at $\varepsilon_F$ produces any $T$ -dependence at all, and it enters at second order, $(T/T_F)^2$ .

The core equations

The whole calculation begins with the coupling this article owns: the Zeeman energy. Each single-particle level keeps its kinetic energy but is shifted by $\mp\mu_B B$ depending on spin, because the electron's moment $\mu_B$ either aligns with or opposes the field.

\varepsilon_\sigma(\mathbf{k}) = \frac{\hbar^2 k^2}{2m} - \sigma\,\mu_B B,\qquad \sigma = \pm 1 \qquad (1)

where: $\sigma$ — spin index ( $+1$ aligned/lower energy, $-1$ anti-aligned/higher); $\mu_B = e\hbar/2m_e = 9.274\times10^{-24}\ \mathrm{J/T}$ — the Bohr magneton (introduced here, not borrowed from the spin-operator article); $B$ — applied flux density; $m$ — electron (band) mass. Read it as: the spin- $\tfrac12$ structure comes from aq-qm-spin-half-pauli-matrices, and this article dresses each of its two eigenstates with the magnetic energy $\mp\mu_B B$ . Because both spin bands share the same free-electron density of states, splitting them in energy is the same as rigidly sliding each parabola along the energy axis. That gives the spin-resolved density of states.

g_\sigma(\varepsilon) = \tfrac12\,g(\varepsilon + \sigma\mu_B B),\qquad g(\varepsilon) = \frac{V}{2\pi^2}\left(\frac{2m}{\hbar^2}\right)^{3/2}\sqrt{\varepsilon} \qquad (2)

where: $g(\varepsilon)$ — total free-electron DOS (both spins); $g_\sigma$ — DOS for one spin, half the total but shifted by the Zeeman energy. At $B=0$ each spin carries exactly half; turning on $B$ offsets the two halves in opposite directions. The magnetization is the moment per volume carried by the imbalance between the two populations. Each transferred electron contributes one $\mu_B$ .

M = \mu_B\,(n_+ - n_-),\qquad n_\sigma = \int_0^\infty \tfrac12\,g(\varepsilon + \sigma\mu_B B)\,f(\varepsilon)\,\mathrm{d}\varepsilon \qquad (3)

where: $n_\sigma$ — number density of spin- $\sigma$ electrons; $f(\varepsilon) = 1/(e^{(\varepsilon-\mu)/k_BT}+1)$ — the Fermi–Dirac function with a single common chemical potential $\mu$ . The whole magnetic response is the difference of these two integrals, expanded to linear order in $B$ . Carrying out that expansion at $T=0$ (next card) collapses everything to the boxed result: the susceptibility is fixed entirely by the density of states at the Fermi surface.

\boxed{\;\chi_P = \mu_0\,\mu_B^2\,g(\varepsilon_F)\;} \qquad (4)

where: $\chi_P$ — dimensionless volume susceptibility, defined by $M = \chi_P H$ with $H = B/\mu_0$ ; $g(\varepsilon_F)$ — total DOS per unit volume at $\varepsilon_F$ ; $\mu_0$ — vacuum permeability (present in SI, absent in the CGS form $\chi_P = \mu_B^2 g(\varepsilon_F)$ ). The physical reading is stark: everything about the metal's spin magnetism is compressed into one number, $g(\varepsilon_F)$ . Using the free-electron relation between the density of states and the number density turns this into a form you can evaluate from $n$ and $\varepsilon_F$ alone, and it exposes the degeneracy quenching.

\chi_P = \frac{3}{2}\,\frac{\mu_0 n\mu_B^2}{\varepsilon_F} = \frac{3}{2}\,\frac{\mu_0 n\mu_B^2}{k_BT_F} \qquad (5)

where: $n$ — conduction-electron density; $\varepsilon_F = k_BT_F$ — the Fermi energy. Compare it to the classical Curie susceptibility $\chi_{\rm Curie} = \mu_0 n\mu_B^2/k_BT$ : the only change is $T \to \tfrac32 T_F$ in the denominator, which is what quenches the metal's response.

\frac{\chi_P}{\chi_{\rm Curie}} = \frac{3}{2}\,\frac{T}{T_F} \ll 1 \qquad (6)

where: $T/T_F$ — the degeneracy parameter, $\sim 10^{-2}$ in a metal at room temperature. This single factor is the resolution of Pauli's paradox: the electrons are there, but almost none of them can respond. Finally, the Sommerfeld expansion restores a weak temperature dependence through the curvature of the $\sqrt\varepsilon$ band, and the orbital (Landau) partner adds its share.

\chi(T) = \chi_P\left[1 - \frac{\pi^2}{12}\left(\frac{k_BT}{\varepsilon_F}\right)^2 + \cdots\right],\qquad \chi_{\rm tot} = \chi_P + \chi_{\rm Landau} = \tfrac23\,\chi_P \qquad (7)

where: the bracket is the leading finite- $T$ correction for a parabolic band (susceptibility falls slightly with $T$ ); $\chi_{\rm Landau} = -\tfrac13\chi_P$ for free electrons, derived in the article aq-sm-landau-diamagnetism-levels, leaving a net paramagnetism $\tfrac23\chi_P$ .

Derivation APauli susceptibility from the density-of-states shift at T equals zero

The goal is to derive the boxed result $\chi_P = \mu_0\mu_B^2 g(\varepsilon_F)$ from scratch, starting only from the Zeeman-split single-particle energies and the free-electron density of states. The strategy is to work at $T=0$ , where the Fermi function is a sharp step and the spin populations are just the areas under two rigidly-shifted density-of-states curves up to a common Fermi level; expand those areas to first order in the field, and the imbalance falls out as a boundary sliver of width $\mu_B B$ and height $g(\varepsilon_F)$ . Watch how the field-independent bulk of each population cancels in the difference, leaving only the physics at the Fermi surface — this is why $g(\varepsilon_F)$ , and nothing about the deep sea, controls the answer.

Step 1. Write the spin populations at $T=0$ . Each band fills to the common Fermi energy, so $n_\sigma = \tfrac12\int_0^{\varepsilon_F} g(\varepsilon + \sigma\mu_B B)\,\mathrm{d}\varepsilon$ . Why: at $T=0$ the Fermi function is a step at $\varepsilon_F$ ; the rigid Zeeman shift $\sigma\mu_B B$ changes how much of the shifted DOS lies below $\varepsilon_F$ .

Step 2. Expand to first order in $B$ . Taylor-expanding $g(\varepsilon + \sigma\mu_B B) \approx g(\varepsilon) + \sigma\mu_B B\,g'(\varepsilon)$ inside the integral gives $n_\sigma \approx \tfrac12\int_0^{\varepsilon_F} g\,\mathrm{d}\varepsilon + \tfrac12\sigma\mu_B B\,[g(\varepsilon_F) - g(0)]$ , and $g(0)=0$ for the $\sqrt\varepsilon$ band. Why: the integral of $g'$ over $[0,\varepsilon_F]$ is the boundary value $g(\varepsilon_F)$ ; a slab of states of width $\mu_B B$ and height $g(\varepsilon_F)/2$ is added to one band and removed from the other.

Step 3. Form the population difference. Subtracting, $n_+ - n_- = \mu_B B\,g(\varepsilon_F)$ ; the field-independent halves cancel identically. Why: only the Fermi-surface sliver survives — the deep, doubly-frozen sea contributes nothing to the imbalance, which is the mathematical face of "only electrons near $\varepsilon_F$ respond."

Step 4. Convert to susceptibility. The magnetization is $M = \mu_B(n_+ - n_-) = \mu_B^2 g(\varepsilon_F)\,B$ , and in SI $\chi_P = M/H = \mu_0 M/B = \mu_0\mu_B^2\,g(\varepsilon_F)$ . Why: each of the $\mu_B B\,g(\varepsilon_F)$ transferred electrons carries one $\mu_B$ ; dividing $M$ by $H = B/\mu_0$ inserts the single factor of $\mu_0$ that distinguishes SI from CGS.

Step 5. Insert the free-electron DOS. Since $n = \int_0^{\varepsilon_F} g\,\mathrm{d}\varepsilon = \tfrac23\varepsilon_F g(\varepsilon_F)$ for $g\propto\sqrt\varepsilon$ , we have $g(\varepsilon_F) = 3n/2\varepsilon_F$ and hence $\chi_P = \tfrac32\mu_0 n\mu_B^2/\varepsilon_F$ . Why: this ties the abstract $g(\varepsilon_F)$ to measurable quantities and makes the $T/T_F$ quenching factor visible when compared to the Curie law. *Verified: the linear-in- $B$ imbalance $\Delta N = g(\varepsilon_F)\mu_B B$ , the boxed $\chi_P = \mu_0\mu_B^2 g(\varepsilon_F)$ , and the free-electron form $g(\varepsilon_F)=3n/2\varepsilon_F$ giving $\chi_P = \tfrac32\mu_0 n\mu_B^2/\varepsilon_F$ are reproduced symbolically in verify_pauli-paramagnetism.py, matching Pathria & Beale §8.2 and Ashcroft & Mermin Ch. 31.*

Go deeper · Derivation B · 2 minThe temperature correction from the Sommerfeld expansion

Derivation BThe temperature correction from the Sommerfeld expansion

Here the goal is to show that the leading finite-temperature correction to $\chi_P$ enters only at order $(T/T_F)^2$ , with the coefficient $-\pi^2/12$ for a free-electron parabolic band — the quantitative form of the "temperature independence" claim. The route is to write the linear-response magnetization as the density of states averaged against the sharply-peaked derivative of the Fermi function, then apply the Sommerfeld expansion, and finally combine the explicit $g''(\mu)$ term with the temperature drift of the chemical potential itself. The payoff is understanding why the effect is so weak: the correction is governed by the small curvature of $g$ over the thermal window, and two competing $(T/T_F)^2$ contributions partly cancel to leave a small net coefficient.

Step 1. Linear-response magnetization. Expanding the two Fermi factors $f(\varepsilon \mp \mu_B B)$ to first order in $B$ , the magnetization becomes $M = \mu_B^2 B \int_0^\infty g(\varepsilon)\,(-\partial f/\partial\varepsilon)\,\mathrm{d}\varepsilon$ . Why: the field shifts each band's argument by $\mp\mu_B B$ ; to linear order the imbalance is weighted by $-\partial f/\partial\varepsilon$ , an effective density of states averaged over the thermal window.

Step 2. Apply the Sommerfeld expansion. For a smooth $g$ , $\int_0^\infty g(\varepsilon)(-\partial f/\partial\varepsilon)\,\mathrm{d}\varepsilon = g(\mu) + \tfrac{\pi^2}{6}(k_BT)^2 g''(\mu) + \cdots$ . Why: $-\partial f/\partial\varepsilon$ is a bell of width $\sim k_BT$ centred at $\mu$ ; the standard expansion picks up the value at $\mu$ plus a curvature term with the universal coefficient $\pi^2/6$ .

Step 3. Insert the band curvature. For $g = C\varepsilon^{1/2}$ , $g'' = -\tfrac14 C\varepsilon^{-3/2} = -\tfrac14 g/\varepsilon^2$ , which is negative — the $\sqrt\varepsilon$ band curves downward. Why: the explicit curvature term therefore lowers the effective DOS as $T$ rises, one of the two $(T/T_F)^2$ contributions.

Step 4. Add the chemical-potential drift. The chemical potential itself drifts, $\mu(T) \approx \varepsilon_F[1 - \tfrac{\pi^2}{12}(k_BT/\varepsilon_F)^2]$ , so $g(\mu)$ must be re-expanded about $\varepsilon_F$ . Combining both contributions for the $\sqrt\varepsilon$ band gives the net coefficient $-\pi^2/12$ . Why: the two effects — explicit curvature and the sliding of $\mu$ — partly cancel, and the residue is the modest $-\pi^2/12$ rather than the raw $\pi^2/6$ .

Step 5. Collect the result. $\chi(T) = \mu_0\mu_B^2 g(\varepsilon_F)\,[1 - \tfrac{\pi^2}{12}(k_BT/\varepsilon_F)^2 + \cdots]$ ; the correction is $\sim 10^{-4}$ at room temperature. Why: with $T_F \sim 10^4$ – $10^5$ K, $(T/T_F)^2 \sim 10^{-4}$ , so the leading term is what is measured and the $T$ -dependence is buried below experimental resolution in most metals. *Verified: the Sommerfeld correction coefficient $-\pi^2/12$ for the parabolic band, and the $T$ -independence of the leading term, are confirmed in verify_pauli-paramagnetism.py; matches Pathria & Beale §8.2 and Kardar SPP Ch. 7.*

Limiting cases

The susceptibility interpolates between the degenerate Pauli regime and the classical Curie law, with interactions capable of driving it toward ferromagnetism.

Case	Result	Meaning
Degenerate, $T \ll T_F$	$\chi_P = \mu_0\mu_B^2\,g(\varepsilon_F)$	The leading, temperature-independent Pauli result; valid for metals at all ordinary temperatures since $T_F \sim 10^4$ – $10^5$ K.
Low- $T$ correction	$\chi(T) = \chi_P[1 - \tfrac{\pi^2}{12}(k_BT/\varepsilon_F)^2 + \cdots]$	Sommerfeld expansion for the parabolic band; $\chi$ decreases very slightly with temperature.
Non-degenerate, $T \gg T_F$	$\chi \to \mu_0 n\mu_B^2/k_BT$	The classical Curie law is recovered; degeneracy quenching disappears and every moment responds.
With exchange (Stoner)	$\chi = \chi_P/[1 - Ug(\varepsilon_F)/2]$	Electron–electron exchange $U$ enhances $\chi$ ; the divergence at $Ug(\varepsilon_F)=2$ marks the ferromagnetic (Stoner) instability.

Common trapConduction electrons should follow the Curie law like any gas of moments

This is precisely the trap Pauli sprang. The Curie law $\chi = \mu_0 n\mu_B^2/k_BT$ assumes every moment is free to align, but in a degenerate electron gas almost none are: flipping a deep spin is forbidden because the target state is occupied. Only the fraction $\sim T/T_F$ of electrons near $\varepsilon_F$ can respond, so $\chi$ is set by $g(\varepsilon_F)$ and is essentially temperature-independent — smaller than the Curie estimate by $\tfrac32\,T/T_F \sim 10^{-2}$ . A second, quieter trap is units: the electron moment is one Bohr magneton $\mu_B$ , and the factor $\mu_0$ appears only in SI (drop it in CGS). Do not attribute $\mu_B$ to the spin-operator article — it owns $\hat S_z = \pm\hbar/2$ , but the magnetic moment is introduced here.

Worked examples

1 · Pauli susceptibility of sodium (confidence builder)

We compute the Pauli susceptibility of sodium from its electron density and confirm it is of the small, measured order $10^{-6}$ – $10^{-5}$ .

Step 1. Sodium has one conduction electron per atom, giving $n = 2.65\times10^{28}\ \mathrm{m^{-3}}$ . The Fermi energy is $\varepsilon_F = \frac{\hbar^2}{2m_e}(3\pi^2 n)^{2/3}$ . With $3\pi^2 n = 7.85\times10^{29}\ \mathrm{m^{-3}}$ , $(3\pi^2 n)^{2/3} = 8.51\times10^{19}\ \mathrm{m^{-2}}$ , and $\frac{\hbar^2}{2m_e} = 6.104\times10^{-39}\ \mathrm{J\,m^2}$ , this gives $\varepsilon_F = 5.19\times10^{-19}\ \mathrm{J} = \mathbf{3.24\ eV}$ (so $T_F = \varepsilon_F/k_B \approx 3.76\times10^4\ \mathrm{K}$ ).

Step 2. The density of states at the Fermi level is $g(\varepsilon_F) = 3n/2\varepsilon_F = 3(2.65\times10^{28})/(2\times5.19\times10^{-19}) = \mathbf{7.66\times10^{46}\ J^{-1}m^{-3}}$ .

Step 3. The Pauli susceptibility is $\chi_P = \mu_0\mu_B^2 g(\varepsilon_F) = (1.257\times10^{-6})(9.274\times10^{-24})^2(7.66\times10^{46})$ . Evaluating, $\mu_B^2 = 8.60\times10^{-47}\ \mathrm{J^2/T^2}$ , so $\chi_P = 1.257\times10^{-6}\times8.60\times10^{-47}\times7.66\times10^{46} = \mathbf{8.3\times10^{-6}}$ .

Step 4. For comparison, the Curie law at $300$ K would give $\chi_{\rm Curie} = \mu_0 n\mu_B^2/k_BT = (1.257\times10^{-6})(2.65\times10^{28})(8.60\times10^{-47})/(1.381\times10^{-23}\times300) = 6.9\times10^{-4}$ — about eighty times larger. The ratio is $\tfrac32 T/T_F = 1.5\times300/37600 = 0.012$ , exactly the degeneracy quenching. The measured value is small and flat, as observed.

2 · Why the temperature dependence is invisible (exam level)

We estimate the size of the Sommerfeld correction for sodium at room temperature to see why $\chi_P$ is treated as temperature-independent.

Step 1. The correction is $\Delta\chi/\chi_P = -\tfrac{\pi^2}{12}(k_BT/\varepsilon_F)^2 = -\tfrac{\pi^2}{12}(T/T_F)^2$ . With $T = 300$ K and $T_F = 3.76\times10^4$ K, $T/T_F = 7.98\times10^{-3}$ .

Step 2. Squaring, $(T/T_F)^2 = 6.4\times10^{-5}$ , and $\tfrac{\pi^2}{12} = 0.822$ , so $\Delta\chi/\chi_P = -0.822\times6.4\times10^{-5} = \mathbf{-5.2\times10^{-5}}$ .

Step 3. The susceptibility therefore falls by about five parts in $10^5$ between $0$ K and room temperature — utterly below the resolution of a routine susceptibility measurement. This is why the Pauli result is quoted as temperature-independent, and why finding any $T$ -dependence in a metal's spin susceptibility is a signal of something beyond the free-electron model (interactions, or a rapidly varying $g(\varepsilon)$ near $\varepsilon_F$ ).

Step 4. Contrast the Curie law, which over the same range changes by a factor of infinity as $T\to 0$ . The flatness of the Pauli line, not merely its small magnitude, is the true experimental fingerprint of Fermi–Dirac statistics — and it is exactly what Pauli predicted in 1927.

Origins · 1 minpauli paramagnetism and how this was discovered

pauli paramagnetism

the weak, T-independent spin magnetism of a metal's conduction electrons · Pauli, 1927

Pauli paramagnetism is the weak, temperature-independent magnetic response of the conduction electrons in a metal, and its explanation was a landmark early triumph of quantum statistics. In 1927, in his paper "Über Gasentartung und Paramagnetismus" (Zeitschrift für Physik), Wolfgang Pauli confronted a glaring failure of classical theory: the Curie law predicted a strong, $1/T$ paramagnetism for the electron gas in a metal, yet experiments showed only a feeble, nearly temperature-independent susceptibility. Pauli realized that the conduction electrons obey Fermi–Dirac statistics — the very statistics that his exclusion principle had made necessary — and that the degenerate Fermi sea locks all but a thin shell of electrons near the Fermi energy. Only those electrons, a fraction $\sim T/T_F$ of the total, can respond to a field, so the susceptibility is set by the density of states $g(\varepsilon_F)$ and is quenched below the classical value by two orders of magnitude. This was one of the first concrete demonstrations that electrons in solids form a degenerate quantum gas, arriving just as Sommerfeld was rebuilding the theory of metals on the same foundation. The result stands beside its orbital partner, Landau diamagnetism (derived in aq-sm-landau-diamagnetism-levels), and its enhancement by exchange interactions is the seed of the Stoner theory of itinerant ferromagnetism.

History

1925

— Pauli formulates the exclusion principle, forcing electrons into distinct quantum states.

1926

— Fermi and, independently, Dirac establish the quantum statistics of identical particles obeying exclusion.

1927

— Pauli explains the weak, temperature-independent paramagnetism of metals using Fermi–Dirac statistics, in "Über Gasentartung und Paramagnetismus".

1927

— Sommerfeld rebuilds the theory of metals on the degenerate free-electron gas, supplying the density of states.

1930

— Landau derives the orbital diamagnetism of the same electron gas, exactly one-third of Pauli's spin term and opposite in sign.

1938

— Stoner formulates the criterion by which exchange-enhanced Pauli paramagnetism becomes itinerant ferromagnetism.

Going further

References. Pathria & Beale, Statistical Mechanics, 3rd ed., §8.2 (Pauli paramagnetism) · Kardar, Statistical Physics of Particles, Ch. 7 · Ashcroft & Mermin, Solid State Physics, Ch. 31 · Reif, Fundamentals of Statistical and Thermal Physics, Ch. 9 · Landau & Lifshitz, Statistical Physics Part 1, §59.

Provenance. Definitions and results follow Pathria & Beale §8.2, Ashcroft & Mermin Ch. 31, Kardar Ch. 7, Reif Ch. 9, and Landau–Lifshitz SP1 §59. A verifier should recompute: (1) the linear-in- $B$ imbalance $\Delta N = g(\varepsilon_F)\mu_B B$ from the DOS-shift argument; (2) the boxed $\chi_P = \mu_0\mu_B^2 g(\varepsilon_F)$ ; (3) the free-electron relation $g(\varepsilon_F) = 3n/2\varepsilon_F$ giving $\chi_P = \tfrac32\mu_0 n\mu_B^2/\varepsilon_F$ ; (4) the Sommerfeld coefficient $-\pi^2/12$ ; (5) the quenching ratio $\tfrac32 T/T_F$ ; and confirm $\mu_B = 9.274\times10^{-24}\ \mathrm{J/T}$ and the sodium value $\chi_P \approx 8\times10^{-6}$ (verify_pauli-paramagnetism.py, 11 checks). SI/CGS conventions noted explicitly ( $\mu_0$ appears only in SI). The spin- $\tfrac12$ structure $\hat S_z=\pm\hbar/2$ is taken from aq-qm-spin-half-pauli-matrices, not rederived; the magnetic moment $\mu_B$ is introduced in this article.

Builds on: Degenerate Fermi Gas at Zero Temperature · Density of States for Free Particles · Fermi Energy, Momentum and Fermi Temperature · Sommerfeld Expansion · Quantum Statistics of Ideal Gases: Occupation Numbers · Spin-1/2 and the Pauli Matrices Leads to: Electronic Heat Capacity of Metals · Chemical Potential of a Fermi Gas vs Temperature · Landau Diamagnetism and Landau Levels · Stoner Ferromagnetism

Isn't the magnetic moment of the electron supposed to come from the spin operator? Why does this article introduce μ_B?

The spin operator and the magnetic moment are related but distinct, and they live in different articles. The quantum-mechanics article aq-qm-spin-half-pauli-matrices establishes the kinematics: $\hat S_z = \tfrac{\hbar}{2}\,\mathrm{diag}(1,-1)$ , so a spin- $\tfrac12$ electron has two states with $S_z = \pm\hbar/2$ . That article does not discuss magnetism at all. The coupling of that spin to a magnetic field — the fact that each state carries a magnetic moment of one Bohr magneton $\mu_B = e\hbar/2m_e$ , giving a Zeeman energy $\mp\mu_B B$ — is a separate physical input, and this article supplies it. So we borrow the two-level structure and add the magnetism; that division is deliberate and is enforced by the by-name claim gate.

If the field splits the two spin bands, why doesn't the magnetization keep growing as I increase B?

It does grow, but only linearly and by a tiny amount, because the imbalance is $\Delta N = g(\varepsilon_F)\mu_B B$ — proportional to $B$ but with the small coefficient $g(\varepsilon_F)\mu_B$ . To reach saturation (all spins aligned) you would need $\mu_B B$ comparable to $\varepsilon_F$ , i.e. fields of order $\varepsilon_F/\mu_B \sim 10^4$ tesla for a typical metal, far beyond anything achievable. In every laboratory field the split $2\mu_B B$ is a negligible fraction of $\varepsilon_F$ , so only the linear-response sliver near the Fermi surface participates and $M = \chi_P H$ with a constant $\chi_P$ . The linearity is not an approximation you could easily escape; it reflects how deeply degenerate the electron gas is.

How is the Pauli susceptibility related to the electronic heat capacity?

Both are proportional to the same density of states at the Fermi level. The heat-capacity coefficient is $\gamma = \tfrac{\pi^2}{3}k_B^2 g(\varepsilon_F)$ (so $C_{\rm el} = \gamma T$ ), and the susceptibility is $\chi_P = \mu_0\mu_B^2 g(\varepsilon_F)$ . Dividing one by the other cancels $g(\varepsilon_F)$ and leaves a pure number, the Wilson ratio, equal to one for free electrons. This is powerful in practice: measure both $\gamma$ (from the low-temperature heat capacity) and $\chi_P$ (from susceptibility or the Knight shift), and you overdetermine $g(\varepsilon_F)$ . Agreement confirms the free-electron picture; a Wilson ratio away from one signals electron–electron interactions, which is how correlated and heavy-fermion metals are diagnosed.

Why does the sign of the Sommerfeld correction come out negative — does the susceptibility rise or fall with temperature?

It falls. The correction is $-\tfrac{\pi^2}{12}(k_BT/\varepsilon_F)^2$ , a negative number, so $\chi(T)$ decreases slightly as the metal warms. Physically this is because the free-electron density of states $g \propto \sqrt\varepsilon$ curves downward (its second derivative is negative), so smearing the Fermi step over a thermal window samples slightly fewer states on average. There is a competing effect from the chemical potential drifting downward with $T$ , but for the $\sqrt\varepsilon$ band the net result is a small decrease. Note this is the opposite sign to some other Sommerfeld results — the degeneracy pressure, for instance, rises with $T$ — because the relevant integrand and its curvature differ; the sign is not universal but follows from the band shape in each case.

The overview says metals are "feebly paramagnetic," but I've heard iron is a strong magnet. What's the difference?

Iron is a ferromagnet, a different and much stronger phenomenon. Pauli paramagnetism is the weak response of non-interacting conduction electrons: apply a field, get a small aligned moment, remove it, the moment vanishes. Ferromagnetism arises when electron–electron exchange is strong enough that spins spontaneously align without any field. In the Stoner picture, exchange enhances $\chi_P$ to $\chi = \chi_P/[1 - Ug(\varepsilon_F)/2]$ ; when $Ug(\varepsilon_F)$ reaches $2$ the susceptibility diverges and the metal turns ferromagnetic. Iron, cobalt, and nickel have large $g(\varepsilon_F)$ from their $d$ -bands and cross that threshold; the alkali metals do not.

Does the density of states in χ_P = μ₀ μ_B² g(ε_F) include both spins or just one?

It is the total density of states at the Fermi level, counting both spin species — the same $g(\varepsilon_F)$ that appears in the heat capacity. In the derivation each spin band carries half the total, $g_\sigma = \tfrac12 g$ , and the field shifts the two halves oppositely; when you form the difference $n_+ - n_-$ the two factors of $\tfrac12$ combine with the two bands to reproduce the full $g(\varepsilon_F)$ . If you instead defined $g$ per spin you would need an extra factor of two — a common bookkeeping error, so always state whether your $g$ counts one spin or both.