Landau Diamagnetism and Landau Levels

There is a theorem that says this article should not exist. The Bohr–van Leeuwen theorem proves that in classical statistical mechanics the orbital magnetic response of any charge distribution in thermal equilibrium is exactly zero: the vector potential shifts away under a change of integration variable, and the partition function forgets the field entirely. And yet a real electron gas in a magnetic field is diamagnetic — it develops a moment opposing the field, purely from orbital motion. Landau showed in 1930, at the age of twenty-two, that quantum mechanics restores the effect: the discreteness of the cyclotron spectrum, invisible to classical phase-space integrals, gives a small, negative orbital susceptibility. We take the spectrum itself from the quantum-mechanics track. The article aq-qm-charged-particle-em-field-minimal-coupling solves a charged particle in a uniform field by minimal coupling $\mathbf p \to \mathbf p - q\mathbf A$ and finds the Landau-level ladder $E_n = \hbar\omega_c(n+\tfrac12)$ with $\omega_c = eB/m_e$ (in SI). That ladder is what we borrow. What this article owns is the thermodynamics: each level carries an enormous degeneracy $eB/h$ per unit area — supplied here — and summing the grand potential over the ladder produces the magnetic response. The weak-field expansion gives $\chi_{\rm L} = -\tfrac13\mu_0\mu_B^2 g(\varepsilon_F)$ , exactly one-third of the Pauli spin susceptibility and opposite in sign — the Landau–Peierls $-\tfrac13$ ratio, taken by name against aq-sm-pauli-paramagnetism. At low temperature and strong field the smooth $B^2$ response is overlaid by oscillations periodic in $1/B$ , the de Haas–van Alphen effect, whose period measures extremal cross-sections of the Fermi surface — one of the sharpest probes of a metal.

Key takeaways

Quantum mechanics quantizes a charged particle's transverse motion into Landau levels $E_n = \hbar\omega_c(n+\tfrac12)$ ; this article takes that spectrum from aq-qm-charged-particle-em-field-minimal-coupling and supplies each level's huge degeneracy $eB/h$ per unit area.
Classically the orbital response is exactly zero (Bohr–van Leeuwen); the effect is intrinsically quantum, born of the $B$ -dependent zero-point energy $\tfrac12\hbar\omega_c$ .
Summing the grand potential over Landau levels (Euler–Maclaurin, to order $B^2$ ) gives a negative orbital susceptibility $\chi_{\rm L} = -\tfrac13\mu_0\mu_B^2\,g(\varepsilon_F)$ .
This is exactly $-\tfrac13$ of the Pauli spin susceptibility: $\chi_{\rm L} = -\tfrac13\chi_P$ , so a free-electron gas is net paramagnetic with $\chi_{\rm tot} = \tfrac23\chi_P$ .
At low $T$ and strong $B$ the grand potential oscillates periodically in $1/B$ — the de Haas–van Alphen effect — with period $\Delta(1/B) = 2\pi e/(\hbar A_{\rm ext})$ set by the extremal Fermi-surface cross-section $A_{\rm ext}$ .
The magnetic length $\ell_B = \sqrt{\hbar/eB} \approx 25.7$ nm at 1 T sets the scale; the areal degeneracy is $eB/h \approx 2.4\times10^{10}\ \mathrm{cm^{-2}}$ per tesla.
Key idea: Landau quantization gives a $B$ -dependent zero-point energy; the Euler–Maclaurin correction to the level sum yields $\chi_{\rm L} = -\tfrac13\chi_P$ , and the discrete crossing of levels through $\varepsilon_F$ produces de Haas–van Alphen oscillations periodic in $1/B$ .

The core insight

Quantization of cyclotron orbits into Landau levels with degeneracy $\propto B$ produces a $B$ -dependent zero-point energy; expanding the grand potential to order $B^2$ gives a negative, diamagnetic susceptibility $\chi_{\rm L} = -\tfrac13\chi_P$ , while the discrete crossing of Landau levels through the Fermi energy yields de Haas–van Alphen oscillations periodic in $1/B$ .

Where this lives today

Landau quantization is the engine of modern fermiology. Measuring de Haas–van Alphen frequencies $F = \hbar A_{\rm ext}/(2\pi e)$ versus field orientation reconstructs the complete Fermi surface through the extremal areas it presents; the temperature damping yields cyclotron effective masses and the Dingle damping yields quasiparticle lifetimes. This is the backbone of band-structure verification, and in current research it detects Fermi-surface reconstruction in cuprate superconductors and maps the Fermi surfaces of topological semimetals, where the oscillation phase carries a Berry-phase signature. The same Landau levels, in a two-dimensional electron gas where longitudinal motion is frozen, become the flat, macroscopically degenerate ladders behind the quantum Hall effect: integer filling gives the integer Hall plateaus, partial filling with strong interactions the fractional effect. In astrophysics, magnetized white dwarfs and neutron stars reach fields where $\hbar\omega_c$ rivals $\varepsilon_F$ , pushing electrons into the ultra-quantum lowest-Landau-level limit that reshapes the equation of state. The same level-counting that gives a milligram of copper its tiny diamagnetism controls the structure of the densest stars.

The physical picture

Begin with the spectrum the quantum-mechanics track supplies. In aq-qm-charged-particle-em-field-minimal-coupling, a charged particle in a uniform field $B\hat z$ is solved by minimal coupling $\mathbf p \to \mathbf p - q\mathbf A$ ; the transverse motion becomes a harmonic oscillator and the energy is the Landau-level ladder $E_n = \hbar\omega_c(n+\tfrac12) + p_z^2/2m$ , with the cyclotron frequency $\omega_c = eB/m_e$ in SI. Motion along the field stays free. We take this ladder as given; what we add is everything thermodynamic. The first new fact is the degeneracy. A classical electron gas fills a continuum of transverse states; the field gathers that continuum into a discrete set of highly degenerate Landau levels, each holding $eB/h$ states per unit area. As $B$ grows, these levels — think of them as nested "tubes" in momentum space — fatten and march outward. The lowest level sits at the zero-point energy $\tfrac12\hbar\omega_c$ , which rises with $B$ . That rising floor is the microscopic origin of diamagnetism: increasing the field costs energy, so the gas responds by opposing the field.

Figure

Landau-level fan — the levels E_n = ℏω_c(n+½) rise linearly from the origin with slope ℏω_c = ℏeB/m_e; as B grows they sweep upward through the Fermi energy ε_F, each crossing producing a kink in the free energy

Why exactly $-\tfrac13$ ? The number is a kinematic accident of free electrons. Spin aligns with the field (paramagnetic, Pauli); orbital motion opposes it (diamagnetic); for free electrons the orbital response is exactly one-third the spin response, opposite in sign. The two partly cancel, leaving a net paramagnet, $\chi_{\rm tot} = \tfrac23\chi_P$ . In real metals the band mass $m^*$ shifts the ratio, so the net sign varies — bismuth, with very light carriers, is famously strongly diamagnetic. The whole effect is non-analytic in a way Bohr–van Leeuwen anticipates: no finite classical phase-space integral can produce it. That non-analyticity reappears as the de Haas–van Alphen oscillations. As the field increases, each Landau tube crosses the Fermi energy in turn; every crossing hiccups the free energy, and the magnetization oscillates with a period set by the tube spacing — the geometry of the Fermi surface itself.

Figure

de Haas–van Alphen oscillations — the magnetization oscillates periodically in 1/B with period Δ(1/B) = 2πe/(ℏA_ext); each period marks one Landau tube crossing ε_F, so the period measures the extremal Fermi-surface area

The core equations

The starting point is the single-particle spectrum, taken from the quantum-mechanics owner. Written in SI, a charged particle in a uniform field has quantized transverse motion and free longitudinal motion.

\varepsilon_{n,p_z} = \hbar\omega_c\!\left(n+\tfrac12\right) + \frac{p_z^2}{2m},\qquad \omega_c = \frac{eB}{m_e},\quad n = 0,1,2,\dots \qquad (1)

where: $n$ — Landau-level index; $\omega_c = eB/m_e$ — cyclotron frequency (SI); $p_z$ — momentum along the field, free. This is the ladder from aq-qm-charged-particle-em-field-minimal-coupling, which obtains it by minimal coupling $\mathbf p \to \mathbf p - q\mathbf A$ and the harmonic-oscillator structure of the transverse motion. We do not rederive it; we take it and count states. Each Landau level is enormously degenerate, and counting that degeneracy is the first thing this article owns. The number of allowed orbit-centre positions in a sample sets it.

\ell_B = \sqrt{\frac{\hbar}{eB}},\qquad n_L = \frac{eB}{h} = \frac{1}{2\pi\ell_B^2}\ \text{(states per unit area per level)} \qquad (2)

where: $\ell_B$ — magnetic length (the size of the lowest cyclotron orbit); $n_L$ — areal degeneracy per Landau level; $h = 2\pi\hbar$ . One state occupies an area $2\pi\ell_B^2 = h/eB$ , so raising $B$ crams more states into each level and empties the levels above — this degeneracy, $eB/h$ , is not stated in the quantum-mechanics owner and is supplied here. The thermodynamics lives in the grand potential, summed over the ladder and integrated over the free longitudinal motion.

\Omega = -k_BT\,\frac{eB}{h}\,\frac{L_z}{2\pi\hbar}\,A\sum_{n=0}^{\infty}\int_{-\infty}^{\infty}\!\!\mathrm{d}p_z\,\ln\!\left[1 + z\,e^{-\beta\varepsilon_{n,p_z}}\right] \qquad (3)

where: $\Omega = -k_BT\ln\mathcal Z$ — grand potential; $z = e^{\beta\mu}$ — fugacity; $A, L_z$ — transverse area and length along the field, $V = AL_z$ . The prefactor $eB/h$ is the areal degeneracy from Eq. (2); the entire $B$ -dependence of the thermodynamics is now encoded in this sum over $n$ . Comparing the discrete sum to its continuum integral (next card) extracts a smooth $B^2$ term, and the result is the boxed Landau susceptibility.

\boxed{\;\chi_{\rm L} = -\frac{1}{3}\,\mu_0\mu_B^2\,g(\varepsilon_F)\;},\qquad \mu_B = \frac{e\hbar}{2m_e},\quad g(\varepsilon_F) = \frac{3n}{2\varepsilon_F} \qquad (4)

where: $\chi_{\rm L}$ — orbital (diamagnetic) volume susceptibility; $\mu_B$ — Bohr magneton; $g(\varepsilon_F)$ — total DOS per unit volume at the Fermi energy. The minus sign is the whole point: the orbital response opposes the field, in stark contrast to the Pauli spin response. Setting it beside the Pauli result gives the exact ratio and the net response of a free-electron gas.

\chi_{\rm L} = -\tfrac13\,\chi_P,\qquad \chi_P = \mu_0\mu_B^2\,g(\varepsilon_F),\qquad \chi_{\rm tot} = \chi_P + \chi_{\rm L} = \tfrac23\,\chi_P \qquad (5)

where: $\chi_P$ — the Pauli spin susceptibility owned by aq-sm-pauli-paramagnetism; $\chi_{\rm tot}$ — net free-electron susceptibility, still paramagnetic because the spin term wins three-to-one. The $-\tfrac13$ is the Landau–Peierls ratio, exact for free electrons with $g$ -factor $2$ and bare mass. At low temperature and strong field the smooth $B^2$ law is not the whole story: the level sum develops an oscillatory remainder, periodic in $1/B$ .

\Delta\!\left(\frac{1}{B}\right) = \frac{2\pi e}{\hbar\,A_{\rm ext}},\qquad A_{\rm ext} = \pi k_F^2\ (\text{free sphere}) \qquad (6)

where: $A_{\rm ext}$ — extremal cross-sectional area of the Fermi surface perpendicular to $B$ ; $k_F$ — Fermi wavevector. Each increment $\Delta(1/B)$ marks one Landau tube passing through $\varepsilon_F$ , so measuring the period maps the Fermi surface directly — the Onsager relation behind de Haas–van Alphen fermiology.

Derivation ABohr–van Leeuwen: classical orbital magnetism vanishes

The goal is to prove that in classical statistical mechanics the orbital magnetic moment of a charge distribution in thermal equilibrium is exactly zero — establishing that Landau diamagnetism is an irreducibly quantum effect. The strategy is short: write the classical partition function with the field entering through minimal coupling, then change the momentum integration variable to absorb the vector potential. Because that shift has unit Jacobian and leaves the infinite limits unchanged, the field disappears entirely. The lesson to watch for is that the theorem does not say the effect is small — it says it is identically zero classically, so the quantized, $B$ -dependent spectrum is the entire origin of the phenomenon.

Step 1. Classical Hamiltonian. With minimal coupling, $H = \sum_i \tfrac{1}{2m}(\mathbf p_i - e\mathbf A(\mathbf r_i))^2 + U(\{\mathbf r\})$ . Why: the field acts on charges through $\mathbf A$ , which sits solely inside the kinetic term.

Step 2. Write the partition function. $Z_{\rm cl} = \frac{1}{N!h^{3N}}\int \prod_i \mathrm{d}^3r_i\,\mathrm{d}^3p_i\,e^{-\beta H}$ . Why: any field dependence of the classical thermodynamics must show up in this phase-space integral.

Step 3. Shift the momenta. At each fixed $\mathbf r_i$ , substitute $\mathbf p_i' = \mathbf p_i - e\mathbf A(\mathbf r_i)$ ; the Jacobian is unity and the limits stay $(-\infty,\infty)$ . Why: $\mathbf A(\mathbf r_i)$ is constant in the $\mathbf p_i$ integration, a pure translation that changes neither measure nor domain.

Step 4. The field drops out. After the shift, $Z_{\rm cl} = \frac{1}{N!h^{3N}}\int \prod_i \mathrm{d}^3r_i\,\mathrm{d}^3p_i'\,e^{-\beta[\sum_i p_i'^2/2m + U]}$ , with no $\mathbf A$ . Why: the kinetic term is field-free and $U$ never depended on it, so $Z_{\rm cl}$ is independent of $B$ .

Step 5. Conclude. $M = -\partial F/\partial B = k_BT\,\partial\ln Z_{\rm cl}/\partial B = 0$ . Why: a $B$ -independent free energy has zero field derivative; only quantization can restore a finite response. *Verified: the momentum shift is volume-preserving (unit Jacobian) and the integration domain is invariant, so $\partial\ln Z_{\rm cl}/\partial B = 0$ ; checked in verify_landau-diamagnetism-levels.py, matching Pathria & Beale §8.2 and Landau–Lifshitz Vol. 5.*

Go deeper · Derivation B · 2 minWeak-field expansion of the grand potential and the minus-one-third susceptibility

Derivation BWeak-field expansion of the grand potential and the minus-one-third susceptibility

The goal is to derive $\chi_{\rm L} = -\tfrac13\mu_0\mu_B^2 g(\varepsilon_F)$ from the Landau-level grand potential, obtaining both the sign and the exact factor $\tfrac13$ . The strategy is to compare the discrete sum over levels to its continuum integral by the Euler–Maclaurin formula: the leading correction scales as $(\hbar\omega_c)^2 \propto B^2$ , and since $\hbar\omega_c = 2\mu_B B$ this becomes the diamagnetic susceptibility. The subtle point is the half-integer offset in $E_n = \hbar\omega_c(n+\tfrac12)$ : it makes the midpoint form of Euler–Maclaurin apply, cancelling the endpoint term and leaving the clean $\tfrac{1}{24}f'(0)$ correction that carries the whole result.

Step 1. Isolate the sum. Write $\Omega = -\frac{eBV}{(2\pi)^2\hbar^2}\int\mathrm{d}p_z\,k_BT\sum_n \ln[1 + z\,e^{-\beta\varepsilon_{n,p_z}}]$ and define $f(x) = k_BT\ln[1 + z\,e^{-\beta\hbar\omega_c x - \beta p_z^2/2m}]$ , so the level sum is $\sum_{n\ge0} f(n+\tfrac12)$ . Why: the areal degeneracy $eB/h$ is pulled out front, and the entire field dependence of the sum sits in the argument $\hbar\omega_c(n+\tfrac12)$ .

Step 2. Apply midpoint Euler–Maclaurin. $\sum_{n\ge0} f(n+\tfrac12) = \int_0^\infty f(x)\,\mathrm{d}x + \tfrac{1}{24}f'(0) + \cdots$ ; the half-integer offset cancels the $\tfrac12 f(0)$ endpoint term, leaving $\tfrac{1}{24}f'(0)$ as the leading correction. Why: the midpoint form of Euler–Maclaurin has no first-order endpoint term, so the discrete-minus-continuum gap begins at $\tfrac{1}{24}f'(0)$ — this is the term that carries the diamagnetism.

Step 3. Identify the field scaling. The integral term reproduces the $B=0$ grand potential $\Omega_0$ ; the correction scales as $(\hbar\omega_c)^2 = (2\mu_B B)^2$ , since $\mu_B = e\hbar/2m_e$ gives $\hbar\omega_c = 2\mu_B B$ . Collecting, $\Omega(B) = \Omega_0 + \tfrac{1}{6}(\mu_B B)^2\,\dfrac{\partial^2\Omega_0}{\partial\mu^2}$ . Why: $\tfrac{1}{24}(2\mu_B B)^2 = \tfrac{1}{6}(\mu_B B)^2$ , and $f'$ evaluated across the band converts to the second $\mu$ -derivative of the field-free grand potential.

Step 4. Convert to susceptibility. With $-\tfrac1V\partial^2\Omega_0/\partial\mu^2 = \partial n/\partial\mu = g(\mu)$ , the magnetization $M = -\tfrac1V\partial\Omega/\partial B$ gives $\chi = \partial M/\partial B|_{B\to0} = -\tfrac13\mu_0\mu_B^2\,g(\mu)$ . Why: differentiating the $B^2$ term twice pulls down a factor $2$ , and the $\tfrac16$ becomes $\tfrac13$ ; the overall minus sign comes from $-\partial^2\Omega_0/\partial\mu^2 = -Vg(\mu)$ combined with the outer $-\tfrac1V$ , leaving a genuinely negative (diamagnetic) response.

Step 5. Take the degenerate limit and compare to Pauli. At $T \ll T_F$ , $\mu \to \varepsilon_F$ and $g(\mu) \to g(\varepsilon_F)$ , so $\chi_{\rm L} = -\tfrac13\mu_0\mu_B^2 g(\varepsilon_F) = -\tfrac13\chi_P$ . Why: the Pauli spin susceptibility from aq-sm-pauli-paramagnetism is $\chi_P = \mu_0\mu_B^2 g(\varepsilon_F)$ , so the orbital term is exactly one-third of it and opposite in sign — the Landau–Peierls ratio. *Verified: the midpoint Euler–Maclaurin coefficient, the identity $\hbar\omega_c = 2\mu_B B$ , the $B^2$ coefficient $\tfrac16$ , and the resulting $\chi_{\rm L} = -\tfrac13\mu_0\mu_B^2 g(\varepsilon_F)$ (negative, diamagnetic) are checked symbolically in verify_landau-diamagnetism-levels.py; matches Pathria & Beale §8.2 and Landau–Lifshitz §60.*

Go deeper · Derivation C · 2 minde Haas–van Alphen oscillations from the level crossings

Derivation Cde Haas–van Alphen oscillations from the level crossings

The goal is to extract the oscillatory part of the grand potential — the piece Euler–Maclaurin misses — and show it is periodic in $1/B$ with a period that measures the extremal cross-sectional area of the Fermi surface. The strategy is to replace the smooth Euler–Maclaurin treatment with the Poisson summation formula, which converts the Landau sum into a sum over harmonics; the $p\ne0$ harmonics oscillate as Landau tubes cross $\varepsilon_F$ , and a stationary-phase evaluation of the $p_z$ integral picks out only the extremal cross-sections. The reward is a magnetization oscillation whose frequency is a direct caliper of the Fermi surface.

Step 1. Poisson summation. $\sum_n F(n+\tfrac12) = \sum_{p=-\infty}^{\infty}(-1)^p\int_0^\infty F(x)\,e^{2\pi i p x}\,\mathrm{d}x$ ; the $p=0$ term is the smooth (Landau) part and $p\ne0$ terms are oscillatory. Why: Poisson summation trades the level sum for a sum over harmonics $p$ , isolating the oscillations Euler–Maclaurin smooths away.

Step 2. Stationary phase. The $p$ -th harmonic accumulates a phase $\approx \tfrac{2\pi p}{\hbar\omega_c}\varepsilon_F$ ; only extremal cross-sections $A_{\rm ext}$ survive the $p_z$ integral by stationary phase. Why: away from an extremum the rapid phase averages the contribution to zero, so the signal comes from cross-sections where $A(p_z)$ is stationary.

Step 3. Read off the period. Each increment of $1/B$ by $\Delta(1/B) = 2\pi e/(\hbar A_{\rm ext})$ advances the fundamental phase by $2\pi$ — one Landau tube has crossed $\varepsilon_F$ . Why: the phase is linear in $1/B$ , so equal phase steps correspond to equal steps in $1/B$ ; the oscillations are periodic in inverse field.

Step 4. Temperature damping. Thermal smearing multiplies each harmonic by $R_T = X/\sinh X$ with $X = 2\pi^2 p\,k_BT/\hbar\omega_c$ ; the decay of $R_T$ with $T$ yields the cyclotron effective mass. Why: a Fermi surface blurred over $k_BT$ washes out the oscillations, most strongly for high harmonics, at a rate encoding $m^*$ .

Step 5. Physical reading. $M = -\partial\Omega/\partial B$ inherits the $1/B$ -periodic oscillations; the fundamental frequency $F = \hbar A_{\rm ext}/(2\pi e)$ maps the Fermi surface (Onsager), reducing to $\hbar k_F^2/(2e)$ for a free sphere. Why: rotating the crystal presents different extremal areas, so the full set of frequencies reconstructs the surface. *Verified: the dHvA period $\Delta(1/B) = 2\pi e/(\hbar A_{\rm ext})$ reducing to $2e/(\hbar k_F^2)$ for a free sphere is checked in verify_landau-diamagnetism-levels.py; matches the Onsager relation, Ashcroft & Mermin Ch. 14 and Landau–Lifshitz §60.*

Limiting cases

The orbital response interpolates across temperature, field strength, and degeneracy.

Case	Result	Meaning
Weak field, degenerate ( $\hbar\omega_c \ll k_BT_F$ , $T \ll T_F$ )	$\chi_{\rm L} = -\tfrac13\mu_0\mu_B^2 g(\varepsilon_F) = -\tfrac13\chi_P$	Smooth Landau diamagnetism; one-third of the Pauli spin term, opposite in sign.
Weak field, nondegenerate (Boltzmann gas)	$\chi = -\mu_0 n\mu_B^2/(3k_BT)$	A Curie-like $1/T$ diamagnetism, one-third of a Curie magnitude; still purely quantum in origin.
Strong field, low $T$ (sharp $\varepsilon_F$ )	$\Delta(1/B) = 2\pi e/(\hbar A_{\rm ext})$	de Haas–van Alphen oscillations periodic in $1/B$ ; the period fixes the extremal Fermi-surface area.
Classical (Bohr–van Leeuwen)	$\chi_{\rm orbital} = 0$	Continuous classical phase space lets the vector potential shift away; orbital magnetism vanishes identically.

Common trapOrbital and spin magnetism are the same effect, and the minus one-third is universal

Two traps in one. First, Landau diamagnetism (orbital) and Pauli paramagnetism (spin) are independent contributions that add — they come from different degrees of freedom and neither contains the other. Second, the famous $-\tfrac13$ ratio holds only for free electrons with bare mass and $g$ -factor $2$ ; in a real metal the band mass $m^*$ enters the orbital term as $(m/m^*)^2$ while the spin term is mass-independent, so a light band mass (as in bismuth) can make the diamagnetism dominate and flip the net sign. And never expect any orbital magnetism classically — by Bohr–van Leeuwen it is exactly zero, so the effect is a pure fingerprint of Landau quantization.

Worked examples

1 · Magnetic length and Landau degeneracy at 1 tesla (confidence builder)

We compute the two numbers that set the scale of Landau quantization — the magnetic length and the areal degeneracy — at a field of one tesla, and check they are enormous.

Step 1. The magnetic length is $\ell_B = \sqrt{\hbar/eB}$ . With $\hbar = 1.055\times10^{-34}\ \mathrm{J\,s}$ , $e = 1.602\times10^{-19}\ \mathrm{C}$ , and $B = 1$ T, $\ell_B = \sqrt{1.055\times10^{-34}/(1.602\times10^{-19})} = \sqrt{6.585\times10^{-16}} = \mathbf{2.566\times10^{-8}\ m} = 25.7\ \mathrm{nm}$ .

Step 2. The areal degeneracy per level is $n_L = eB/h = (1.602\times10^{-19})(1)/(6.626\times10^{-34}) = \mathbf{2.42\times10^{14}\ m^{-2}} = 2.42\times10^{10}\ \mathrm{cm^{-2}}$ .

Step 3. Check the consistency: $n_L = 1/(2\pi\ell_B^2) = 1/(2\pi(2.566\times10^{-8})^2) = 1/(4.14\times10^{-15}) = 2.42\times10^{14}\ \mathrm{m^{-2}}$ — the two forms agree. Each Landau level holds about $2.4\times10^{10}$ states per square centimetre per tesla, an immense degeneracy, which is why even modest laboratory fields dramatically restructure the density of states near $\varepsilon_F$ and make the quantum Hall effect and dHvA oscillations observable.

2 · The cyclotron frequency and the net free-electron susceptibility (exam level)

We compute the cyclotron frequency in SI from the quantum-mechanics owner's form and confirm the Landau–Pauli combination.

Step 1. The cyclotron frequency is $\omega_c = eB/m_e$ (the SI form quoted in aq-qm-charged-particle-em-field-minimal-coupling). At $B = 1$ T, $\omega_c = (1.602\times10^{-19})(1)/(9.109\times10^{-31}) = \mathbf{1.76\times10^{11}\ rad/s}$ , so $\hbar\omega_c = 1.16\times10^{-4}$ eV — tiny compared to a metal's $\varepsilon_F \sim$ eV, confirming that laboratory fields are firmly in the weak-field regime $\hbar\omega_c \ll \varepsilon_F$ .

Step 2. Check the identity $\hbar\omega_c = 2\mu_B B$ : the Bohr magneton is $\mu_B = e\hbar/2m_e$ , so $2\mu_B B = e\hbar B/m_e = \hbar(eB/m_e) = \hbar\omega_c$ . Confirmed — this is the identity that turns the Euler–Maclaurin $B^2$ term into a $\mu_B^2$ susceptibility.

Step 3. For a free-electron gas, the orbital term is $\chi_{\rm L} = -\tfrac13\chi_P$ and the net response is $\chi_{\rm tot} = \chi_P + \chi_{\rm L} = \chi_P(1 - \tfrac13) = \tfrac23\chi_P$ . Using sodium's $\chi_P \approx 8.3\times10^{-6}$ from aq-sm-pauli-paramagnetism, the net free-electron susceptibility is $\chi_{\rm tot} \approx \tfrac23(8.3\times10^{-6}) = \mathbf{5.5\times10^{-6}}$ — still paramagnetic, because spin beats orbit three-to-one.

Step 4. In bismuth, however, the carrier effective mass is $m^* \approx 0.01\,m_e$ , so the orbital term is enhanced by $(m/m^*)^2 \sim 10^4$ while the spin term is not. The diamagnetism then dominates and bismuth is one of the most strongly diamagnetic elements — showing the $-\tfrac13$ ratio is special to free electrons.

Origins · 1 minlandau levels and how this was discovered

landau levels

the quantized cyclotron energy ladder E_n = ℏω_c(n+½) · Landau, 1930

Landau levels are the discrete energy states of a charged particle whose transverse motion in a uniform magnetic field is quantized into a harmonic-oscillator ladder, $E_n = \hbar\omega_c(n+\tfrac12)$ . Lev Landau introduced them in 1930, at the age of twenty-two, in "Diamagnetismus der Metalle," to solve a genuine paradox: the Bohr–van Leeuwen theorem had shown that classical statistical mechanics gives zero orbital magnetism, yet metals are observably diamagnetic. Landau realized that the discreteness of the cyclotron spectrum — invisible to any classical phase-space integral — and the $B$ -dependent zero-point energy $\tfrac12\hbar\omega_c$ restore a finite, negative susceptibility, exactly one-third of Pauli's spin paramagnetism and opposite in sign. Each level carries a colossal degeneracy proportional to the magnetic flux, and as the field grows these levels sweep through the Fermi energy, producing the de Haas–van Alphen oscillations observed the same year by de Haas and van Alphen. Through Onsager's relation those oscillations became the premier tool for mapping metallic Fermi surfaces, and the same quantization later underpinned the quantum Hall effects. The single-particle spectrum is derived in aq-qm-charged-particle-em-field-minimal-coupling. Landau received the 1962 Nobel Prize, cited for his theory of superfluid helium rather than for this early work.

History

1921

— Bohr and, later, van Leeuwen prove that classical statistical mechanics gives exactly zero orbital magnetism.

1930

— Landau quantizes cyclotron motion into Landau levels and derives the orbital diamagnetism of the electron gas.

1930

— de Haas and van Alphen observe oscillatory magnetization in bismuth, periodic in the inverse field.

1933

— Peierls clarifies the weak-field expansion and the role of the level degeneracy.

1952

— Onsager relates the de Haas–van Alphen period to extremal cross-sections of the Fermi surface.

1962

— Landau receives the Nobel Prize, cited for his theory of superfluid helium.

Going further

References. Pathria & Beale, Statistical Mechanics, 3rd ed., §8.2 (Landau diamagnetism and the de Haas–van Alphen effect) · Kardar, Statistical Physics of Particles, Ch. 7 · Landau & Lifshitz, Statistical Physics Part 1 (Vol. 5), §59–60 · Reif, Fundamentals of Statistical and Thermal Physics, Ch. 11 · Ashcroft & Mermin, Solid State Physics, Ch. 14 · Peierls, Quantum Theory of Solids, Ch. 7.

Provenance. Equations and the $-\tfrac13$ Landau–Pauli ratio follow Pathria & Beale §8.2, Landau–Lifshitz Vol. 5 §59–60, and Kardar Ch. 7. This article works in SI units; the base export and some texts use Gaussian units, where $\omega_c = eB/m_ec$ and $\mu_B = e\hbar/2m_ec$ — the physical results are identical. A verifier should: (1) reproduce the momentum-shift argument for Bohr–van Leeuwen; (2) confirm the midpoint Euler–Maclaurin coefficient and the identity $\hbar\omega_c = 2\mu_B B$ , giving the $\tfrac16(\mu_B B)^2$ grand-potential term and the factor $-\tfrac13$ ; (3) check the dHvA period $\Delta(1/B) = 2\pi e/(\hbar A_{\rm ext})$ reduces to $2e/(\hbar k_F^2)$ for a free sphere; (4) confirm numerically $\ell_B = 25.7$ nm, $eB/h = 2.42\times10^{10}\ \mathrm{cm^{-2}}$ , and $\omega_c = 1.76\times10^{11}$ rad/s at 1 T (verify_landau-diamagnetism-levels.py, 13 checks). The Landau spectrum $E_n = \hbar\omega_c(n+\tfrac12)$ with $\omega_c = eB/m_e$ is taken from aq-qm-charged-particle-em-field-minimal-coupling, not rederived; the level degeneracy $eB/h$ is supplied here.

Builds on: Quantum Statistics of Ideal Gases: Occupation Numbers · Degenerate Fermi Gas at Zero Temperature · Density of States for Free Particles · Sommerfeld Expansion · Fermi Energy, Momentum and Fermi Temperature · Charged Particle in an Electromagnetic Field: Minimal Coupling Leads to: Electronic Heat Capacity of Metals · Fermi Surface and Fermi Sphere · Chemical Potential of a Fermi Gas vs Temperature

If the classical result is exactly zero, how can quantum mechanics give something finite — isn't QM supposed to reduce to classical physics?

It does reduce, but not smoothly. Bohr–van Leeuwen says the classical orbital magnetization is identically zero, not merely small. The quantum susceptibility $\chi_{\rm L} \propto \mu_B^2 g(\varepsilon_F)$ carries $\hbar$ through $\mu_B = e\hbar/2m_e$ , so $\hbar \to 0$ makes it vanish — consistent with the classical zero. The effect is non-analytic in $\hbar$ and unreachable by any classical calculation; that is why Landau's 1930 result was a discovery, not a refinement.

Why does the article derive the level degeneracy eB/h but not the Landau levels themselves?

Because they split along the QM–thermodynamics seam. The single-particle spectrum $E_n = \hbar\omega_c(n+\tfrac12)$ is owned by aq-qm-charged-particle-em-field-minimal-coupling, so we take it as given. But that article does not count the states in each level; the areal degeneracy $eB/h$ (one state per area $2\pi\ell_B^2$ ) is a separate statistical-mechanics ingredient needed for the grand potential, so this article supplies it. Quantum mechanics gives the energy ladder; statistical mechanics fills it.

Why is the susceptibility periodic in 1/B rather than in B?

Because the relevant condition is a Landau level crossing the Fermi energy: levels are spaced by $\hbar\omega_c \propto B$ while $\varepsilon_F$ is fixed. The number of filled levels is $\sim\varepsilon_F/\hbar\omega_c \propto 1/B$ , so equal steps in $1/B$ empty one more level each. Each crossing gives one free-energy oscillation, so they repeat at equal intervals of $1/B$ , period $\Delta(1/B) = 2\pi e/(\hbar A_{\rm ext})$ . Against $B$ the oscillations crowd together as $B$ rises; against $1/B$ they are perfectly periodic — why experimentalists always work in inverse field.

Bismuth is diamagnetic and copper is (barely) paramagnetic. Doesn't the −1/3 ratio predict everything is paramagnetic?

The $-\tfrac13$ ratio, and hence the net $\tfrac23\chi_P$ paramagnetism, holds only for free electrons with bare mass $m_e$ and $g$ -factor $2$ . In a real metal the orbital Landau term scales as $(m/m^*)^2$ with the band mass, while the Pauli spin term does not. Bismuth's carriers are extraordinarily light, $m^* \approx 0.01\,m_e$ , so its Landau diamagnetism is enhanced by $\sim 10^4$ and swamps the spin term. Copper's carriers are near the free-electron mass, so $\tfrac23\chi_P$ roughly holds. The net sign is a competition decided by the band mass, not a fixed prediction.

How is the de Haas–van Alphen effect actually used to find the shape of a Fermi surface?

By measuring oscillation frequencies at many field orientations. Each frequency $F = \hbar A_{\rm ext}/(2\pi e)$ is an extremal cross-sectional area of the Fermi surface perpendicular to $B$ . Rotating the crystal brings different extremal areas into play: a sphere gives one orientation-independent frequency, a multi-sheet surface gives many. Fitting the full angular dependence reconstructs the 3D Fermi surface — how the noble metals, and now many correlated and topological materials, were mapped. The temperature damping simultaneously gives the effective mass on each sheet.

The overview mentions the zero-point energy rises with B. Is that really the origin of diamagnetism?

Yes, that is the physical heart of it. The lowest Landau level sits at $\tfrac12\hbar\omega_c = \mu_B B$ , rising linearly with the field, and the whole ladder rides up with it. Raising $B$ raises the energy of the occupied states, so the free energy grows with $B$ — and a free energy that grows with the field means the system wants to expel it, which is diamagnetism ( $M = -\partial F/\partial B < 0$ ). Euler–Maclaurin makes this quantitative: the discrete sum sits above the continuum integral by a positive $B^2$ amount, and that energy cost is the diamagnetic susceptibility.