University Quantum Mechanics II · Spin and Charge in Electromagnetic Fields · 3.9
Landau levels in a uniform magnetic field
This is where the section's machinery pays off. Minimal coupling, a well-chosen gauge and the oscillator you already know combine into an exact solution: a charge in a uniform field climbs a ladder spaced by ħωc, and every rung holds one state per flux quantum. Those two numbers organise electrons in traps, metals and semiconductors.
Build the model
Connect the measurement to the mechanism.
A charge in a uniform field is one of the few magnetic problems quantum mechanics solves exactly, and the reason is a choice of gauge. Minimal coupling gives H = (p − qA)²/2m; choosing A = (−By, 0, 0) leaves no x or z in H, so pₓ and pz are conserved and become the numbers ħkₓ and ħkz. What remains in y is the oscillator of Quantum Mechanics I, with frequency ωc = |q|B/m and centre y₀ = −ħkₓ/(qB).
Motion along B stays free; motion across it is quantised into Landau levels (n + ½)ħωc, the quantum cyclotron orbit with its zero-point energy. The striking feature is what the energy leaves out. kₓ only says where the orbit sits, so each level is hugely degenerate, holding |q|B/h states per unit area, one for every flux quantum h/|q| through the sample. The gauge is scaffolding: the commutator [vₓ, vy] = iħqB/m² rebuilds the same ladder without one.
The model assumes a non-relativistic particle, a field uniform over many magnetic lengths ℓ = √(ħ/(|q|B)), and a large clean sample. It fails as ħωc approaches mc², near 4.4 × 10⁹ T for the electron; in a crystal m and g give way to measured effective values; and edges, disorder and interactions, which build the quantum Hall effect, belong to condensed matter.
- Simple definition
- Landau levels are the discrete energies (n + ½)ħωc of a charged particle's motion perpendicular to a uniform magnetic field, each shared by |q|B/h states per unit area.
- Example
- For a free electron at 10 T, ωc = eB/mₑ = 1.76 × 10¹² rad s⁻¹, so the levels are ħωc = 1.16 meV apart, 13.4 K in temperature units, and each level holds 2.42 × 10¹⁵ states per square metre per spin state.
No x or z appears, so pₓ and pz commute with H and ψ = e(ikₓx) e(ikz z) φ(y) reduces the problem to one equation in y.
q in C (−e for the electron), B in T, p in kg m s⁻¹, m in kg; (∇ × A)z = −∂Aₓ/∂y = B
kₓ says where the orbit sits across the sample, not how fast it moves: ⟨vₓ⟩ = 0 in every Landau state, whatever kₓ is.
y₀ in m, kₓ in m⁻¹; for the electron, q = −e, so y₀ = +ℓ²kₓ
Quantised across B, free along it, and blind to kₓ. As B → 0 the spacing closes and the free-particle continuum returns.
n = 0, 1, 2, …; ωc = |q|B/m in rad s⁻¹; ħωc = 0.1158 meV per tesla for a free electron
One state per flux quantum through the sample, the same for every n and independent of the mass.
Φ₀ = h/|q| = 4.136 × 10⁻¹⁵ T m² for the electron, so e/h = 2.418 × 10¹⁴ m⁻² T⁻¹, per spin state
The one length in the problem: it sets the width of every Landau wavefunction, and ℓ²kₓ turns a wavenumber into a position.
ℓ in m: 25.66 nm at 1 T for the electron, falling as B(−1/2); 2πℓ² is the area per state
mvₓ/(qB) and mvy act like a position and its momentum, so H⊥ = ½m(vₓ² + vy²) is an oscillator of frequency ωc in every gauge.
v = (p − qA)/m is the kinetic velocity in m s⁻¹; no gauge has been chosen
Pick the gauge that keeps a symmetry
Minimal coupling gives H = (p − qA)²/2m but does not say which A to use, and that choice decides how hard the algebra is. A uniform field Bẑ is unchanged by translations in x and y and by rotations about z, yet no single vector potential displays all of that. The symmetric gauge A = ½B × r, used for the weak-field terms, keeps the rotation and labels states by Lz. The Landau gauge A = (−By, 0, 0) keeps translation in x instead; check (∇ × A)z = ∂Ay/∂x − ∂Aₓ/∂y = 0 − (−B) = B. Now H = [(pₓ + qBy)² + py² + pz²]/2m contains no x and no z, so [H, pₓ] = [H, pz] = 0 and the eigenfunctions can be taken as ψ = e(ikₓx) e(ikz z) φ(y). Translation in y is still a symmetry of the physics, but in this gauge it must be paired with a gauge transformation, so it no longer looks like one. A gauge that exposes one symmetry hides another.
Complete the square and the oscillator appears
With pₓ → ħkₓ the y-equation reads −(ħ²/2m)φ″ + (q²B²/2m)(y + ħkₓ/(qB))²φ = (E − ħ²kz²/2m)φ. Match the potential to ½mω²(y − y₀)²: then ω = |q|B/m = ωc, the classical cyclotron frequency, and y₀ = −ħkₓ/(qB). Quantum Mechanics I supplies the rest: φ is the n-th oscillator eigenfunction displaced to y₀, of width ℓ = √(ħ/(mωc)) = √(ħ/(|q|B)), and E = (n + ½)ħωc + ħ²kz²/2m. Checks: ħωc is J s × s⁻¹, an energy; as B → 0 the spacing closes onto the free continuum; and (n + ½)ħωc = ½mωc²r² for r² = (2n + 1)ℓ², a classical orbit enclosing exactly (n + ½) flux quanta. For a free electron ωc/B = e/mₑ = 1.759 × 10¹¹ rad s⁻¹ T⁻¹, so at 10 T the rungs are 1.16 meV apart, about 13 K: the ladder shows only in cold, strong-field experiments.
kₓ is a position label, not a momentum you can feel
The energy ignores kₓ, and that needs digesting. In a field p is the canonical momentum and the velocity is v = (p − qA)/m. In a Landau state m⟨vₓ⟩ = ⟨pₓ + qBy⟩ = qB⟨y − y₀⟩ = 0, because the displaced oscillator is centred on y₀. So a state with large ħkₓ is not drifting along x; it circulates about a guiding centre at y₀. That is the quantum form of a classical fact: a cyclotron orbit of given speed can be drawn anywhere in the plane at the same energy. Each Landau-gauge state is a strip, a plane wave in x times a band about ℓ wide in y, and superposing many kₓ rebuilds a state localised about one guiding centre, the quantum cyclotron orbit. A different gauge would label the same level by a different conserved momentum, a further sign that kₓ carries no energy of its own. For an electron at 2.0 T, ℓ = 18.1 nm, and kₓ = 1.0 × 10⁸ m⁻¹ puts the strip at y₀ = ℓ²kₓ = 32.9 nm.
Count the guiding centres: one state per flux quantum
Put the particles in a rectangle Lₓ by Ly, periodic in x. Then kₓ = 2πj/Lₓ for integer j, and the guiding centres form a comb of spacing Δy₀ = 2πħ/(|q|BLₓ) = h/(|q|BLₓ). Requiring 0 ≤ y₀ ≤ Ly, safe away from the edges when Ly ≫ ℓ, allows N = Ly/Δy₀ = |q|BLₓLy/h states per level. Per unit area that is |q|B/h = B/Φ₀, one state for each flux quantum Φ₀ = h/|q| through the sample, independent of n and of m, since only ħ, q and B build the comb. For the electron e/h = 2.418 × 10¹⁴ m⁻² T⁻¹, so a 1 mm² film at 1 T holds 2.4 × 10⁸ states in every level for each spin state, each claiming 2πℓ² = 4.14 × 10⁻¹⁵ m². The count is measured: in a two-dimensional electron gas the resistance oscillates periodically in 1/B as levels empty one by one, and the density read from that period agrees with the density from the classical Hall voltage.
The same ladder without choosing a gauge
Wavefunctions change with the gauge; the spectrum cannot. Work with the kinetic momenta Π = p − qA = mv, which are gauge-invariant. Using [pⱼ, f(r)] = −iħ ∂f/∂xⱼ, [Πₓ, Πy] = iħq(∂Ay/∂x − ∂Aₓ/∂y) = iħqB, so [vₓ, vy] = iħqB/m². Set X′ = Πₓ/(qB), keeping the sign of q, and P′ = Πy; then [X′, P′] = iħ for either sign of charge, and H⊥ = (Πₓ² + Πy²)/2m = P′²/2m + ½mωc²X′², an oscillator whatever A is. The degeneracy has a gauge-free form too. The guiding-centre coordinates X = x + Πy/(qB) and Y = y − Πₓ/(qB) commute with H but not with each other: [X, Y] = −iħ/(qB). Like x and p they fill their plane at one state per area 2πħ/(|q|B) = 2πℓ², the same |q|B/h. In the Landau gauge Y = −pₓ/(qB), which is exactly y₀.
Add the spin, then find where the model stops
An electron adds the spin energy gₑμB B mₛ = (gₑ/2)ħωc mₛ, since μB B = ħωc/2. With gₑ = 2, the Dirac value (quoted from relativistic quantum mechanics, not derived here), E = (n + ½ + mₛ)ħωc: (n, ↑) and (n + 1, ↓) coincide, and (0, ↓) sits at exactly zero. QED, also quoted, gives gₑ/2 = 1 + aₑ with aₑ ≈ α/2π = 1.1614 × 10⁻³, first order in α, against a measured 1.15965 × 10⁻³; single-electron traps measure it from that small split between the two ladders. The model itself stops in three places. It needs ħωc ≪ mc², that is B ≪ mₑ²c²/(eħ) = 4.41 × 10⁹ T, reached only near magnetars. Atoms sit α² lower: ħωc equals the Hartree energy α²mₑc² = 27.2 eV only at 2.35 × 10⁵ T, which is why a laboratory field merely perturbs an atom. And in a crystal m and g become measured effective values (0.067 mₑ in GaAs), while edges, disorder and the quantum Hall effect are condensed-matter subjects.
Change one variable at a time
Make the relationship visible.
Drag kₓ and the strip state slides across the sample while the dot on the fan never moves: that is the degeneracy. Then hold kₓ at 360 μm⁻¹ and raise B from 1 T to 10 T: the packet narrows as ℓ ∝ B(−1/2) and y₀ shrinks tenfold, so ten times as many guiding centres fit in the strip.
LEVEL ENERGY (n + ½)ħωc0.347 meV
GUIDING CENTRE y₀ = ℓ²kₓ79.0 nm
MAGNETIC LENGTH ℓ18.1 nm
STATES PER LEVEL eB/h484 μm⁻²
Live interpretationLEVEL ENERGY (n + ½)ħωc: 0.347 meV. GUIDING CENTRE y₀ = ℓ²kₓ: 79.0 nm. MAGNETIC LENGTH ℓ: 18.1 nm. STATES PER LEVEL eB/h: 484 μm⁻²
Catch the common trap
Explain before calculating.
An electron moves in a 2.0 μm × 2.0 μm film with a uniform 3.0 T field perpendicular to it. Ignoring spin and the edges, how many states does the n = 2 Landau level hold?
Choose an answer to test the model.
Practice & worked examples
Reason from the model, then test the result.
EasyAn electron in vacuum sits in a uniform 1.5 T field. Find the Landau-level spacing ħωc in meV, the magnetic length, and the number of states per square micrometre in each level for one spin state. Is the ladder resolved at the 4.2 K of liquid helium?
- ωc = eB/mₑ = (1.759 × 10¹¹ C kg⁻¹)(1.5 T) = 2.638 × 10¹¹ rad s⁻¹, so ħωc = (1.0546 × 10⁻³⁴ J s)(2.638 × 10¹¹ s⁻¹) = 2.782 × 10⁻²³ J = 0.1737 meV. Units: C kg⁻¹ × T = s⁻¹, since 1 T = 1 kg s⁻¹ C⁻¹.
- ℓ = √(ħ/(eB)) = √(6.582 × 10⁻¹⁶ T m² ÷ 1.5 T) = √(4.388 × 10⁻¹⁶ m²) = 2.095 × 10⁻⁸ m = 20.9 nm.
- States per level: eB/h = (2.418 × 10¹⁴ m⁻² T⁻¹)(1.5 T) = 3.627 × 10¹⁴ m⁻² = 363 μm⁻². Check: 1/(2πℓ²) = 1/(2π × 4.388 × 10⁻¹⁶ m²) = 3.627 × 10¹⁴ m⁻², the same.
- Thermal energy at 4.2 K: kB T = (8.617 × 10⁻⁵ eV K⁻¹)(4.2 K) = 0.362 meV, so ħωc/kB T = 0.48 and the n = 1 level is populated e(−0.48) = 0.62 times as densely as n = 0. The rungs are washed out; ħωc = kB T only at 2.0 K.
Answerħωc = 0.174 meV, ℓ = 20.9 nm and 363 states per μm² in each level. At 4.2 K, kB T = 0.36 meV is about twice ħωc, so the ladder is not resolved.
MediumA GaAs heterostructure holds a two-dimensional electron gas of density 4.0 × 10¹⁵ m⁻², with effective mass m* = 0.067 mₑ and g* = −0.44, both measured values. At B = 5.0 T, treating each orbital Landau level as holding 2eB/h electrons per unit area (both spins), find ħωc, how many levels are occupied and how full the top one is, check that the spin splitting may be neglected, and find the lowest field that fits every electron into n = 0.
- The free-electron spacing is 0.1158 meV T⁻¹, and ħωc scales as 1/m: ħωc = (0.1158 meV T⁻¹)(5.0 T)/0.067 = 8.64 meV, about 100 K in temperature units, so the ladder is sharp at liquid-helium temperature.
- Capacity of one orbital level: 2eB/h = 2(2.418 × 10¹⁴ m⁻² T⁻¹)(5.0 T) = 2.418 × 10¹⁵ m⁻², independent of m*, since the flux-quantum count involves only e, h and B.
- Filling: (4.0 × 10¹⁵ m⁻²)/(2.418 × 10¹⁵ m⁻²) = 1.654. The n = 0 level is full and n = 1 is 65% full.
- Spin check: |g*|μB B = 0.44 × (5.788 × 10⁻² meV T⁻¹)(5.0 T) = 0.127 meV, 1.5% of ħωc, so lumping both spins into one level is fair here.
- All in n = 0 needs 2eB/h ≥ 4.0 × 10¹⁵ m⁻², so B = (4.0 × 10¹⁵ m⁻²)(h/2e) = (4.0 × 10¹⁵ m⁻²)(2.068 × 10⁻¹⁵ T m²) = 8.3 T. Limiting check: as B → 0 the filling grows without bound and the ladder merges into the continuum.
Answerħωc = 8.64 meV; n = 0 is full and n = 1 is 65% full (filling 1.65 levels); the spin splitting is 0.13 meV, 1.5% of ħωc; all electrons fit in n = 0 from B = 8.3 T.
HardA single electron is held in a Penning trap with a uniform 5.36 T field, cooled to 0.100 K; ignore the trap's weak electric field. Find νc = ωc/2π and ħωc/kB, the probability that the electron is not in n = 0, and the rms radius of its n = 0 motion. Then, with gₑ/2 = 1 + aₑ and the measured aₑ = 1.15965 × 10⁻³, find the frequency νₐ separating (n, ↑) from (n + 1, ↓), and compare it with the first-order QED estimate aₑ ≈ α/2π.
- νc = eB/(2πmₑ) = (27.992 GHz T⁻¹)(5.36 T) = 150.04 GHz. Then ħωc = hνc = (6.626 × 10⁻³⁴ J s)(1.5004 × 10¹¹ s⁻¹) = 9.942 × 10⁻²³ J, and ħωc/kB = 7.201 K.
- The rungs are equally spaced and equally degenerate, so thermal occupation follows the oscillator: P(n ≥ 1) = e(−ħωc/kB T) = e(−7.201/0.100) = e(−72.01) = 5.3 × 10⁻³². An exponent this large needs four figures: rounded to 72.0 it would give 5.4 × 10⁻³². The electron is in the cyclotron ground state for all practical purposes.
- The cyclotron radius, r minus the guiding centre (X, Y), is (−Πy, Πₓ)/(qB), whose square is (Πₓ² + Πy²)/(qB)² = 2mH⊥/(qB)² = (2n + 1)ℓ² in level n, so for n = 0 the rms radius is ℓ = √(ħ/(eB)) = 25.66 nm/√5.36 = 11.1 nm, pure zero-point motion.
- Spin energy (gₑ/2)ħωc mₛ gives E(n, ↑) − E(n + 1, ↓) = (gₑ/2 − 1)ħωc = aₑħωc. So νₐ = aₑνc = (1.15965 × 10⁻³)(150.04 GHz) = 174.0 MHz. With gₑ = 2 exactly, νₐ would be zero and the two ladders would coincide.
- First-order QED: α/2π = (7.2974 × 10⁻³)/(2π) = 1.1614 × 10⁻³, giving νₐ = 174.3 MHz, 0.15% above the measured-aₑ value. The gap, 1.8 × 10⁻⁶ in aₑ, is of order (α/π)² = 5.4 × 10⁻⁶, the size of the next QED term, which is quoted, not derived here.
Answerνc = 150.0 GHz and ħωc/kB = 7.20 K; P(n ≥ 1) = 5.3 × 10⁻³²; rms radius 11.1 nm; νₐ = 174.0 MHz, against 174.3 MHz from α/2π alone.