Bloch's theorem and energy bands

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 Tₐ^N = 1 and λ^N = 1: λ is a pure phase, λ = e^(ika) with k = 2πj/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.

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 = Σ_G c_G 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.

The ring condition ψ(x + Na) = ψ(x) forces e^(ikNa) = 1, so k = 2πj/(Na) with j 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.

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 = |U_G|², giving E± = E₀ ± |U_G| and a gap of exactly 2|U_G|. 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.

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).

Two derivatives extract the physics from E(k). The slope gives the group velocity, v_g = (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. Within the independent-electron picture, every ohm of resistance comes from something that breaks the periodicity — phonons, impurities, defects, surfaces.