University Physics IV · Molecular and Solid-State Physics · 12.6
Bloch's Theorem & Energy Bands
You cannot solve 10²³ coupled Schrödinger equations, and you do not have to. Periodicity gives you a symmetry, the symmetry gives you a quantum number, and one unit cell then carries the whole crystal — with bands, gaps and effective masses arriving as consequences rather than assumptions.
Build the model
Connect the measurement to the mechanism.
A crystal's potential repeats, and that one fact organises everything else. Translation by a lattice period commutes with the Hamiltonian, so the two share eigenstates; the translation eigenvalue must be a pure phase, or the state would blow up across the crystal. Every stationary state therefore obeys ψ(x + a) = e(ika)ψ(x), equivalently ψₖ(x) = uₖ(x)e(ikx) with uₖ(x + a) = uₖ(x) — and that last clause is the entire content, because with u unconstrained any function whatever can be written that way.
What it buys is enormous: an infinite crystal collapses to one unit cell plus a label k; the label is defined only modulo 2π/a, so it folds into a single Brillouin zone; and the cell problem at each k is a Hermitian eigenvalue problem with a discrete ladder of solutions — the bands. The gaps then follow from Bragg reflection rather than decree: at k = ±π/a two counter-propagating waves are degenerate, the Fourier component UG mixes them into standing waves, and their energies separate by exactly 2|UG|. What it costs is three assumptions — independent electrons, a rigid perfect lattice, and a k that is a label rather than a momentum.
The third is where most of the confusion in this unit lives; the second is why a perfect crystal has no resistance at all.
- Simple definition
- Bloch's theorem says that in a potential with the period of the lattice, every stationary electron state is a plane wave multiplied by a function carrying that same period: ψₖ(x) = uₖ(x)e(ikx) with uₖ(x + a) = uₖ(x).
- Example
- On a chain of period a = 0.30 nm, the state labelled k = π/2a = 0.524 Å⁻¹ obeys ψ(x + a) = e(iπ/2)ψ(x) = iψ(x): the probability density repeats exactly in every cell, and only the phase advances, by a quarter turn per cell.
Equivalent form ψₖ(x + a) = e(ika)ψₖ(x): |ψ|² repeats cell by cell while the phase advances by ka.
a is the lattice period, in Å or nm; k in Å⁻¹. The periodicity of uₖ is the whole claim, not the exponential.
Two labels differing by G name one state, so ħk is not a momentum: k = 8.00 Å⁻¹ is the state at k = −0.378 Å⁻¹.
G is a reciprocal-lattice vector in Å⁻¹; for a = 0.30 nm, G = 2.094 Å⁻¹ and the zone edge sits at 1.047 Å⁻¹.
One orbital state per cell per band — the count that decides metal or insulator once the electrons are poured in.
N is the number of primitive cells in a Born–von Kármán ring; the spacing 2π/Na is 6 × 10⁻⁷ Å⁻¹ for a 1 mm crystal.
a = 0.30 nm puts the free edge at 4.18 eV; a 1.10 eV component splits it into 3.08 eV and 5.28 eV.
UG is the Fourier coefficient of U(x) at G, in eV; ħ²/2m = 3.810 eV⋅Ų turns an inverse ångström into an energy.
Bandwidth is overlap, not crystal size: 4f bands stay atomically narrow while 4s bands spread over several eV.
β is the nearest-neighbour hopping integral and α the on-site shift, both in eV; a is the same lattice period.
vg = 0 at the zone boundary, because that state is a standing wave; negative curvature at a band top gives holes.
dE/dk in eV⋅Å gives vg in m s⁻¹ after dividing by ħ = 6.582 × 10⁻¹⁶ eV⋅s; ħ²/m = 7.620 eV⋅Ų.
The symmetry first, then what the theorem actually claims
Define the translation operator by Tₐ ψ(x) = ψ(x + a). The kinetic term never notices a shift and U(x + a) = U(x) by assumption, so [Tₐ, H] = 0 and the two operators share a complete set of eigenstates. Now ask what Tₐ's eigenvalue λ can be. Wrap the chain into a ring of N cells, so TₐN = 1 and λN = 1: λ is a pure phase, λ = e(ika) with k = 2πm/Na. Any other modulus would make |ψ|² grow or shrink without limit along the crystal. Then define uₖ(x) = ψₖ(x)e(−ikx) and test it: uₖ(x + a) = ψₖ(x + a)e(−ikx)e(−ika) = e(ika)ψₖ(x)e(−ikx)e(−ika) = uₖ(x). That is Bloch's theorem, and the periodicity of uₖ is all of it — the factorisation ψ = u e(ikx) on its own says nothing, since any function at all admits it. The physical payload is that |ψₖ|² repeats with the lattice: an electron in a Bloch state is spread evenly across every cell, not localised on one atom.
k is a label, and it is defined only modulo 2π/a
Replace k by k + G, where G = 2πn/a. The extra factor e(iGx) is itself lattice-periodic, so it can be absorbed into uₖ without disturbing the theorem: the same physical state now carries a different label. Labels are therefore unique only modulo G, and the convention is to fold every k into the first Brillouin zone, −π/a < k ≤ π/a. With a = 0.30 nm, G = 2.094 Å⁻¹, so a state prepared at k = 8.00 Å⁻¹ is the state at 8.00 − 4G = −0.378 Å⁻¹. Two consequences follow. First, ħk is not a momentum: expanding the periodic uₖ in its Fourier series gives ψₖ = ΣG cG e(i(k + G)x), a superposition of many plane waves, so ψₖ is no eigenstate of p̂ at all. Call ħk the crystal momentum; it obeys ħ dk/dt = Fₑₓₜ for external forces and is conserved only modulo ħG. Second, folding stacks solutions at each k, so a second index is needed — the band index n.
Counting the states one band can hold
The ring condition ψ(x + Na) = ψ(x) forces e(ikNa) = 1, so k = 2πm/(Na) with m an integer. Inside one Brillouin zone, of width 2π/a, those values number exactly (2π/a) ÷ (2π/Na) = N. Each band therefore holds N orbital states — one per primitive cell — or 2N counting spin, and that is the arithmetic which later decides whether a solid conducts. The spacing is minute: a 1 mm crystal of period 0.30 nm has N = 3.3 × 10⁶ cells, so adjacent k differ by 6 × 10⁻⁷ Å⁻¹ against a zone half-width of 1.047 Å⁻¹. Bands look continuous for every practical purpose, yet the count stays exact and finite. The band index has a separate origin: at each fixed k, the equation for uₖ is solved on one cell under periodic boundary conditions, and a Hermitian eigenvalue problem on a bounded domain returns a discrete ladder E₁(k) < E₂(k) < … . Bands are that ladder, traced out as k sweeps the zone.
The gap comes from Bragg reflection, not from decree
Switch the potential on weakly. A plane wave e(ikx) is coupled by U only to e(i(k − G)x), and the mixing stays small while the two free energies differ — except where they are equal. That happens when |k| = |k − G|, that is k = ±G/2 = ±π/a, which is the one-dimensional Bragg condition λ = 2a: the wave reflects off the lattice and comes back. There, degenerate perturbation theory replaces the estimate with a 2 × 2 problem, (E₀ − E)² = |UG|², giving E± = E₀ ± |UG| and a gap of exactly 2|UG|. The eigenvectors are the symmetric and antisymmetric mixtures — standing waves cos(πx/a) and sin(πx/a), carrying no current at all. The cosine piles |ψ|² onto the ion cores where U is most negative and takes the lower root; the sine puts nodes there and takes the upper. With a = 0.30 nm the free edge sits at ħ²π²/2ma² = 4.18 eV, so a Fourier component of 1.10 eV opens band edges at 3.08 eV and 5.28 eV, and no state at any k has an energy between them.
Tight binding reaches the same bands from the opposite limit
Start from the other end, with atomic orbitals φ(x − na) that barely overlap. Build the combination ψₖ = N(−1/2) Σₙ e(ikna) φ(x − na); it satisfies Bloch's theorem by construction, since shifting x by a merely relabels the sum and pulls out e(ika). Evaluating ⟨H⟩ with nearest-neighbour overlap only gives E(k) = Eₐₜ − α − 2β cos(ka): a band of width 4|β|, centred on the atomic level and shifted by α. Note what does not appear — N. Bandwidth is set by how strongly one site talks to its neighbour, not by how large the crystal is. Take β = 0.85 eV and the band spans 3.40 eV; core 1s orbitals, whose overlap is negligible, give bands narrower than a µeV, which is why core levels stay atomic and sharp while valence levels smear into bands. Weak potential and strong potential are two limits of one picture: the first explains the gaps, the second the widths, and they describe the same E(k).
Velocity, mass, and why a perfect crystal is transparent
Two derivatives extract the physics from E(k). The slope gives the group velocity, vg = (1/ħ)dE/dk; for the tight-binding band that is (2βa/ħ) sin(ka), which peaks at 7.75 × 10⁵ m s⁻¹ for β = 0.85 eV at k = π/2a, and vanishes at both k = 0 and k = ±π/a — the boundary state is a standing wave and goes nowhere. The curvature gives the effective mass, m* = ħ²/(d²E/dk²): with the same numbers, 2βa² = 15.3 eV⋅Ų against ħ²/m = 7.620 eV⋅Ų makes m* = +0.50 m at the band bottom and −0.50 m at the top, which is exactly why a nearly full band is bookkept with positive holes. The deepest consequence is a negative one. Bloch states are stationary states of the periodic Hamiltonian, so an electron placed in one propagates forever: a rigid, perfect lattice cannot scatter it. Every ohm of resistance comes from something that breaks the periodicity — phonons, impurities, defects, surfaces.
Change one variable at a time
Make the relationship visible.
Raise |UG| from zero: the free parabola splits, but only where it meets the zone boundary at ±π/a, and the forbidden strip widens as 2|UG|. Then walk the marker out to k = π/a and watch the group velocity fall to zero — that state is a standing wave, going nowhere.
GAP 2|UG|2.00 eV
E AT THE MARKER0.93 eV
GROUP VELOCITY573 km/s
LOWER BAND WIDTH3.24 eV
Live interpretationGAP 2|UG|: 2.00 eV. E AT THE MARKER: 0.93 eV. GROUP VELOCITY: 573 km/s. LOWER BAND WIDTH: 3.24 eV
Catch the common trap
Explain before calculating.
A one-dimensional crystal has U(x + a) = U(x). Which statement is the actual content of Bloch's theorem for its stationary states?
Choose an answer to test the model.
Practice & worked examples
Reason from the model, then test the result.
EasyA one-dimensional crystal has period a = 0.30 nm and one significant Fourier component in its potential, UG = −1.10 eV at G = 2π/a. Find G and the zone boundary, the free-electron energy there, and the two band edges. Then give the reduced label of the state prepared at k = 8.00 Å⁻¹.
- The reciprocal-lattice vector is G = 2π/a = 2π/3.00 Å⁻¹ = 2.094 Å⁻¹, and the first Brillouin zone runs from −G/2 to +G/2, so its boundary sits at k = ±π/a = ±1.047 Å⁻¹.
- The free-electron energy there is E₀ = ħ²k²/2m = 3.810 eV⋅Ų × (1.047 Å⁻¹)² = 3.810 × 1.097 = 4.18 eV.
- At that boundary the states at +π/a and −π/a are degenerate and differ by exactly G, so UG couples them. The 2 × 2 secular equation (E₀ − E)² = |UG|² gives E± = E₀ ± |UG| = 4.18 ± 1.10 eV.
- So the band edges are 3.08 eV and 5.28 eV, the gap is 2|UG| = 2.20 eV, and no Bloch state at any k in the crystal has an energy inside it.
- k is defined only modulo G. Since 8.00/2.094 = 3.82, subtract 4G: 8.00 − 4(2.094) = 8.00 − 8.378 = −0.378 Å⁻¹, which lies inside the zone. Same state, correct label.
AnswerG = 2.094 Å⁻¹ and the zone edge is ±1.047 Å⁻¹; E₀ = 4.18 eV, band edges 3.08 eV and 5.28 eV with a 2.20 eV gap; and k = 8.00 Å⁻¹ is the state at k = −0.378 Å⁻¹.
MediumA tight-binding s band on the same chain has E(k) = Eₐₜ − α − 2β cos(ka), with β = 0.85 eV and a = 0.30 nm. Find the bandwidth, the effective mass at the band bottom, the largest group velocity and the k at which it occurs, and the effective mass at the zone boundary.
- cos(ka) runs from +1 at k = 0 to −1 at k = ±π/a, so E sweeps from Eₐₜ − α − 2β to Eₐₜ − α + 2β. The width is 4β = 4 × 0.85 = 3.40 eV, fixed by the overlap integral alone.
- Near the bottom, cos(ka) ≈ 1 − (ka)²/2, so E ≈ (Eₐₜ − α − 2β) + βa²k². The band is parabolic there, which is what makes an effective mass meaningful at all.
- d²E/dk² = 2βa² cos(ka). At k = 0 that is 2 × 0.85 × (3.00 Å)² = 15.3 eV⋅Ų, so with ħ²/m = 7.620 eV⋅Ų, m* = ħ²/(d²E/dk²) = 7.620/15.3 = 0.50 m.
- vg = (1/ħ)dE/dk = (2βa/ħ) sin(ka), largest where sin(ka) = 1, at k = π/2a = 0.524 Å⁻¹. There vg = 2βa/ħ = 5.10 eV⋅Å ÷ (6.582 × 10⁻¹⁶ eV⋅s) = 7.75 × 10⁵ m s⁻¹.
- At the boundary ka = π and cos(ka) = −1, so the curvature flips: m* = 7.620/(−15.3) = −0.50 m, and vg = (2βa/ħ)sin(π) = 0. Negative mass is why a nearly full band is counted in holes.
AnswerWidth 4β = 3.40 eV; m* = +0.50 m at the bottom and −0.50 m at the top; vg is largest at k = π/2a = 0.524 Å⁻¹, where it is 7.75 × 10⁵ m s⁻¹, and zero at both band edges.
HardThe same chain, a = 0.30 nm, carries U(x) = −2.00 eV × cos(2πx/a). Find the gap at k = π/a by degenerate perturbation theory, say which standing wave takes the lower root, then find the effective masses at the top of the lower band and the bottom of the upper band.
- Write the potential in exponentials: U = −1.00 eV × (e(iGx) + e(−iGx)), so the only non-zero Fourier components are U_(±G) = −1.00 eV and |UG| = 1.00 eV.
- At k = π/a the waves e(iπx/a) and e(−iπx/a) are degenerate at E₀ = 3.810 × 1.097 = 4.18 eV and differ by G, so U mixes them. The secular equation (E₀ − E)² = |UG|² gives E± = 3.18 eV and 5.18 eV: a gap of 2.00 eV.
- The eigenvectors are the symmetric and antisymmetric mixtures, √2 cos(πx/a) and √2 sin(πx/a). The cosine concentrates |ψ|² on the ion sites, where this U is most negative, so it takes the 3.18 eV root; the sine has nodes there and takes 5.18 eV.
- Step off the boundary, k = π/a + q, keeping the same two waves. With λ = ħ²/2m their mean is λ(π²/a² + q²) and their half-difference is 2λπq/a, so E± = λ(π²/a² + q²) ± √[(2λπq/a)² + |UG|²].
- For small q, √(|UG|² + (2λπq/a)²) ≈ |UG| + (2λπ/a)²q²/(2|UG|). Collecting q² gives d²E±/dq² = 2λ(1 ± 2E₀/|UG|), and since m* = ħ²/(d²E/dq²) = 2λ/(d²E/dq²), m*/m = 1/(1 ± 2E₀/|UG|).
- Here 2E₀/|UG| = 8.36, so the lower band gives m* = 1/(1 − 8.36) = −0.136 m and the upper 1/(1 + 8.36) = +0.107 m. A narrow gap makes light carriers: InSb, with Egap = 0.17 eV, reaches m* = 0.014 m for exactly this reason.
Answer|UG| = 1.00 eV, so the edges are 3.18 eV (cosine standing wave) and 5.18 eV (sine), a 2.00 eV gap; m* = −0.136 m at the top of the lower band and +0.107 m at the bottom of the upper.