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Fill the lower energy band of the dimerized tight-binding chain with both spin states. Take zero on-site energy and measure ground-state energy per doubled primitive unit cell. State counting in a Bloch band and $q=ka$ giveEquivalently, for $|\delta |\le t$, $\mathcal{E}(\delta )=-(8t/\pi )E(\sqrt{1-{\delta }^2/t^2})$, where $E(k)$ is the complete elliptic integral of the second kind in the modulus convention.

For fixed $t>0$ and $|\delta |/t\to 0$, the change isThe logarithm comes from states near the edge of the Brillouin zone. Put $u=\pi /2-q$. To leading order the square root there is $\sqrt{t^2{u}^2+{\delta }^2}$. For a fixed small cutoff $\eta >0$, direct integration givesThe rest of the band, and the higher-order corrections near its edge, contribute only $O({\delta }^2/t)$; multiplying by $-8/\pi$ gives the displayed asymptotic expansion. Here the remainder uses Big O notation. The negative logarithmic term exceeds any fixed quadratic elastic cost $K{\delta }^2$, $K>0$, for sufficiently small nonzero $\delta$. This explains the zero-temperature instability to a Peierls distortion in the ideal chain model.

Let $N=L/a$ be the integer number of primitive unit cells. A Bloch state can be written as ${\psi }_k(x)=e^{ikx}{u}_k(x)$, with $u_k(x+a)=u_k(x)$. The periodic boundary condition ${\psi }_k(x+L)={\psi }_k(x)$ therefore requiresThe spacing of allowed wavevectors is $\Delta k=2\pi /L$. The first Brillouin zone has width $2\pi /a$, soFor an exact endpoint convention take a half-open Brillouin zone, such as $[-\pi /a,\pi /a)$. Its two endpoints differ by a reciprocal lattice vector and must not both be counted as distinct values of crystal momentum. There are exactly $N$ inequivalent integers $j$ modulo $N$, irrespective of whether $N$ is even or odd. This is state counting in a Bloch band: one allowed wavevector per primitive unit cell per energy band. A separate twofold spin multiplicity gives $2N$ electronic states per energy band, without doubling the number of wavevectors.

For state counting in a Bloch band, take a primitive cubic lattice of spacing $a$ and impose periodic boundary conditions on lengths $L_i=N_{i}a$. The allowed wavevectors are $k_i=2\pi l_i/L_i$. The first Brillouin zone has side $2\pi /a$, so it contains $N_i$ distinct values in each direction and hence $N_x{N}_y{N}_z=N$ wavevectors. Each orbital-derived energy band has one state per wavevector per spin orientation. ThereforeThe phrase “per atom” attached to $2N$ in the PDF is a counting error: $2N$ is the total, not the per-atom count.

By the Pauli exclusion principle, a monovalent crystal supplies $N$ valence Electrons and half-fills a spin-degenerate band; its Fermi level lies within allowed states, giving a metal in the independent-electron picture. Two electrons per atom can fill one band. If a band gap separates it from all empty bands, the crystal is a band insulator. An even valence count only permits insulation: overlapping bands can still produce a metal. Conversely, electron interactions can produce a Mott insulator, and symmetry breaking can permit insulation outside this simple spin-degenerate electronic band structure criterion. The relevant count in a crystal with a basis is per primitive unit cell.

For one conduction electron per atom, let $n_a$ be the atom density. State counting in a Bloch band gives the Fermi wavevector $k_F=(3{\pi }^2{n}_a{)}^{1/3}$. The Debye model preserves the three vibrational modes per atom by a spherical Debye wavevector cutoff $q_D=(6{\pi }^2{n}_a{)}^{1/3}$, soBoth wavevectors are of order the inverse atomic spacing. The numerical factor uses the spherical Debye approximation; a real zone boundary and the largest-frequency mode depend on direction and branch. Because $q_D<2k_F$, short-wavelength acoustic phonons can scatter Fermi surface electrons through substantial angles. For nearly elastic scattering within a spherical Fermi surface,Near room temperature, when a substantial fraction of the phonon spectrum is occupied, electron-phonon scattering is often the main source of electrical resistance. For $T\gg {\Theta }_D$, the phonon population and scattering rate are approximately proportional to $T$, giving $\rho \propto T$, $\sigma \propto T^{-1}$ and roughly constant $K_e$ when the Wiedemann-Franz law holds.

For small-angle electron-phonon transport at low temperature, thermally active acoustic modes instead haveThey predominantly scatter forward. The factor $1-\cos \theta$ makes such collisions inefficient at relaxing an electrical current, while a phonon energy of order $k_{B}T$ can still substantially change an electron's thermal excitation. Heat-current relaxation therefore need not share the charge-current relaxation time. In the clean three-dimensional acoustic-phonon model this gives ${\tau }_Q^{-1}\propto T^3$ but ${\tau }_{\sigma }^{-1}\propto T^5$, so the phonon-limited electrical resistivity scales as $T^5$ and $K_e\propto T^{-2}$. These laws describe the regime before residual elastic scattering takes over; at sufficiently low temperature impurities give nearly constant residual resistivity and $K_e\propto T$, restoring the usual Lorenz number.^[ref]

Phonons also transport heat themselves, with $K_{\mathrm{ph}}\simeq c_{\mathrm{ph}}{c}_s{\ell }_{\mathrm{ph}}/3$. At low temperature their population decreases and their phonon mean free path can become boundary- or defect-limited. At higher temperature Umklapp scattering can relax the phonon heat current. Such lattice conduction is an additional contribution, not the quantity constrained by the electronic Wiedemann-Franz law.