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Differentiating the tight binding dispersion produced by the Hückel method givesHenceUnlike a free particle, the band effective mass depends on wavevector and can change sign. At an inflection point $\cos (ka)=0$, the inverse mass vanishes and the curvature-based mass diverges; a quadratic approximation then ceases to describe the dispersion locally. For the usual $\beta <0$ convention, the bottom at $k=0$ has positive mass and the top at $k=\pi /a$ has negative electron mass. The constant $\alpha$ has no effect on curvature.

Expand a single-electron wavefunction as $|\psi ⟩={\sum }_n{\psi }_n|n⟩$. Project the Schrödinger equation onto $⟨n|$ to obtainTranslation symmetry suggests the Bloch state ${\psi }_n=N^{-1/2}{e}^{ikna}$ on a periodic chain of $N$ sites. Substitution givesTake the conventional positive hopping $t>0$. The first Brillouin zone is $-\pi /a\le k<\pi /a$; the band has minimum ${ϵ}_0-2t$ at $k=0$ and maximum ${ϵ}_0+2t$ at the zone edges. Its width is $4t$. A negative hopping interchanges the extrema but does not change the density of states after replacing $t$ by $|t|$.

There are $N$ allowed wavevectors and two spin states per wavevector. For the total spin-summed density of states $D(E)$, each interior band energy has two roots, $\pm k$. Using the spacing $2\pi /(Na)$ givesThus the density of states per atom isIt is symmetric about ${ϵ}_0$ and has integrable inverse-square-root Van Hove singularities at both band edges. Its integral over the band is two states per atom. The plot labels the dispersion extrema and both singular edges.

Monovalency gives one electron per atom, so the two-spin-state band is half filled. At zero temperature, occupying $-k_F<k<k_F$ gives $2(Na/2\pi )(2k_F)=N$. HenceRelative to the band minimum, the Fermi energy is $2t$.

The density of states obeys $g({ϵ}_0+x)=g({ϵ}_0-x)$. At $\mu ={ϵ}_0$, the Fermi-Dirac distribution satisfiesPairing the energies above and below the band centre therefore leaves exactly half of the two available states per atom occupied, at every positive temperature. Increasing $\mu$ strictly increases the mean particle number, so this solution at fixed half filling is unique. ConsequentlyThis particle-hole symmetry of a half-filled bipartite band argument uses the symmetric tight binding band and fixed half filling; it is not a general temperature-independence property of chemical potential in metals with additional bands or an asymmetric density of states.

Both models respect Bloch's theorem and describe electrons in a periodic lattice, but they start from different useful limits.

In the nearly-free electron model, start with extended plane waves and free-electron energies ${\epsilon }_{\mathbf{k}}^0={\hslash }^2{k}^2/(2m)$. A weak periodic potential has Fourier coefficients $V_{\mathbf{G}}$ at reciprocal lattice vectors and couples $\mathbf{k}$ to $\mathbf{k}-\mathbf{G}$. Away from degeneracies its corrections are small. Near a Bragg scattering, retain the two resonant plane waves and diagonalizeThe resulting eigenvalues areAt the crossing the gap is $\boxed{2|V_{\mathbf{G}}|}$. The two standing-wave combinations place electronic density differently relative to the periodic potential and therefore acquire different energies. Folding the many free parabolas into the first Brillouin zone produces the repeated band labels; the nonzero potential opens avoided crossings between coupled states. A Fourier component forbidden by symmetry need not open its corresponding first-order gap.

In the tight-binding model, instead start with localized orbitals on well-separated atoms or sites, with discrete on-site energies. Their overlap permits tunneling represented by hopping integrals. Bloch sums of these orbitals diagonalize the lattice translation symmetry. For one orthonormal orbital per site in a one-dimensional nearest-neighbor chain,The many initially degenerate copies of one atomic orbital broaden into a band. Other atomic orbitals or multiple sites per primitive cell produce additional bands; their hybridization creates bonding and antibonding combinations and may open gaps. For two orbitals per cell, one illustrative Hamiltonian iswith eigenvalues $(E_A+E_B)/2\pm \sqrt{(E_A-E_B{)}^2/4+|t_1+t_2{e}^{-ika}{|}^2}$. This shows explicitly how two bands arise from the two cell degrees of freedom, even when the isolated sites have equal on-site energies.

For each orbital degree of freedom per primitive cell there is one band per spin; band filling and the presence of a gap at the Fermi level determine metallic or insulating behavior. The nearly-free limit emphasizes weak Bragg scattering of extended states, while tight binding emphasizes hopping between localized states. They are complementary approximations to electronic band structure, rather than two incompatible explanations of why periodic solids have bands.

The periodic boundary condition requires $e^{ikM}=1$, so there are exactly $M$ independent wavevectors in a Brillouin zone:The inverse discrete Fourier transform, with the sign convention of the paper, isFourier orthogonality gives the tight binding dispersion relation ${ϵ}_k=-2t\cos k$. For the interaction, first use the canonical anticommutation relations at distinct sites to write $n_{\sigma ,j}{n}_{\sigma ,j+1}=c_{\sigma ,j}^{†}{c}_{\sigma ,j+1}^{†}{c}_{\sigma ,j+1}{c}_{\sigma ,j}$. Summing its discrete Fourier transform over $j$ enforces crystal momentum conservation, givingEvery wavevector is reduced modulo $2\pi$. The phase $e^{iq}$ comes from the displacement of the second site by one lattice spacing. Equivalently replace $e^{iq}$ by $\cos q$: exchanging the two incoming and the two outgoing fermions changes $q$ to $-q$ with two cancelling fermionic signs. Thus the usual interaction vertex is $2V\cos q$ accompanied by the factor $1/(2M)$.

For each spin let ${\mathcal{F}}_{\sigma }$ contain the $N_{\sigma }$ lowest tight binding energy levels, and set $n_{\sigma }=N_{\sigma }/M$ and $C_{\sigma }=M^{-1}{\sum }_{k\in {\mathcal{F}}_{\sigma }}{e}^{ik}$. The normalized trial Slater determinant isIts nearest-neighbour one-body density matrix is $⟨c_{\sigma ,j}^{†}{c}_{\sigma ,j+1}⟩=C_{\sigma }$. By Wick's theorem, fermionic exchange symmetry subtracts the squared off-diagonal one-body density matrix from the direct contributionThis formula also specifies the finite-$M$ answer exactly, including partially filled terminal energy levels through the chosen occupied set. For $N_{\sigma }$ consecutive allowed wavevectors, $|C_{\sigma }|=|\sin (\pi N_{\sigma }/M)|/[M\sin (\pi /M)]$; its phase locates the centre of that occupied interval.

Write $\tau =|t|$. In the large-$M$ limit the Fermi sea is centred at $k_0=0$ for $t>0$ and $k_0=\pi$ for $t<0$, with Fermi wavevector $k_{F,\sigma }=\pi n_{\sigma }$ measured from $k_0$. ConsequentlyThe subtraction due to fermionic exchange symmetry is essential: replacing $⟨n_j{n}_{j+1}⟩$ by $n^2$ would discard the suppression of neighbouring equal-spin occupation. A completely filled spin band has zero kinetic energy and interaction energy $MV$, as the formula also shows.

A tight binding chain of one species of fermions with nearest-neighbour hopping and density interaction. At half filling and $V/|t|\to +\infty$, minimizing the interaction requires alternating occupied and empty sites. An even periodic chain has two such configurations at $t=0$; finite hopping admixes defects through virtual hopping. A finite ring can select superpositions rather than one ordered product state. Two uncoupled copies have four limiting ordered choices when each is half filled, without a general claim of exact finite-hopping fourfold state degeneracy.