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Bloch theorem states that the energy eigenstates of a Hamiltonian invariant under translations by a Bravais lattice can be chosen as Bloch statesIndeed, the unitary operators $T_r$ commute with one another because $T_r{T}_s=T_{r+s}=T_s{T}_r$, and they commute with $H$ by hypothesis. The simultaneous diagonalization theorem therefore lets us diagonalize all translations within each energy eigenspace. Their eigenvalues form a unitary character $\chi$ of the additive lattice:Writing $\chi (a_j)=e^{-ik\cdot a_j}$ on a primitive basis gives $\chi (r)=e^{-ik\cdot r}$. In position space, with $(T_r\psi )(x)=\psi (x-r)$, this implies ${\psi }_k(x+r)=e^{ik\cdot r}{\psi }_k(x)$. Hence $u_k(x)=e^{-ik\cdot x}{\psi }_k(x)$ is lattice-periodic. Adding a reciprocal lattice vector to $k$ leaves the character unchanged, so the crystal momentum lies in a Brillouin zone.

Forthe equations $a_i\cdot b_j=2\pi {\delta }_{ij}$ give the reciprocal basisThe reciprocal lattice is triangular, so the first Brillouin zone is its Wigner-Seitz cell, a regular hexagon. Its six corners areReciprocal-lattice translations identify these corners in two classes of three. Representatives areFor example, $(b_1-b_2)/3=K^{\prime }-b_2$, while the other equivalences follow by symmetry and reciprocal translations. This gives the requested sketch: a regular hexagon with the vertical edge from $K^{\prime }$ to $K$ at $k_x=2\pi /(\sqrt{3}a)$ and alternating $K,K^{\prime }$ corner classes.

For the tight-binding model, introduce the normalized Bloch sumEach hop by $\pm a_j$ multiplies this state by $e^{\mp ik\cdot a_j}$. Thus the two-direction nearest-neighbour tight-binding dispersion isAlong the boundary edge from $K^{\prime }$ to $K$, $k_x=2\pi /(\sqrt{3}a)$ and $|k_y|\le 2\pi /(3a)$, soFor $\lambda >0$, it rises from $E_0+2\lambda$ at either corner to $E_0+4\lambda$ at the midpoint; for $\lambda <0$ the ordering reverses. Globally the two cosines in the first expression can simultaneously equal $1$ or $-1$, and therefore

Each orbital band contains two one-electron states per lattice site because an electron has two spin states. By band filling, a valency of one leaves this band half-filled and the material conducts, whereas a valency of two fills it. A filled band can be insulating only if it is separated from every empty band by a positive band gap.

The first band's maximum is $E_0+4|\lambda |$. If the second band's minimum is $E_0+\Delta$, a gap exists precisely when $\Delta >4|\lambda |$. Consequently, among the nontrivial fillings described here,For valency one the first band is partially filled. If $\Delta \le 4|\lambda |$, the bands overlap or touch, so even at valency two the overlapping energy bands prevent a band insulator.

For a lattice-periodic potential, Bloch's theorem permits energy eigenfunctions ${\psi }_{\mathbf{k}}(\mathbf{r})=e^{i\mathbf{k}\cdot \mathbf{r}}{u}_{\mathbf{k}}(\mathbf{r})$, where $u_{\mathbf{k}}(\mathbf{r}+\mathbf{R})=u_{\mathbf{k}}(\mathbf{r})$ for every lattice translation $\mathbf{R}$. The crystal momentum is $\hslash \mathbf{k}$, defined modulo a reciprocal lattice vector. A reciprocal vector $\mathbf{G}$ satisfies $\mathbf{G}\cdot \mathbf{R}\in 2\pi \mathbb{Z}$, so $e^{i\mathbf{G}\cdot \mathbf{r}}$ is lattice-periodic.

Expand the Bloch wave as plane waves in one momentum class, $\psi ={\sum }_{\mathbf{q}}{c}_{\mathbf{q}}{e}^{i\mathbf{q}\cdot \mathbf{r}}$ with $\mathbf{q}=\mathbf{k}+{\mathbf{G}}^{\prime }$. Substitution into the Schrödinger equation, using $U={\sum }_{\mathbf{G}}{U}_{\mathbf{G}}{e}^{i\mathbf{G}\cdot \mathbf{r}}$, gives a kinetic coefficient ${\hslash }^2{q}^2{c}_{\mathbf{q}}/(2m)$ and potential coefficient ${\sum }_{\mathbf{G}}{U}_{\mathbf{G}}{c}_{\mathbf{q}-\mathbf{G}}$. Independence of the plane waves therefore givesThis uses $\mathbf{q}$ for the paper's running plane-wave index, to distinguish it from the fixed reduced Bloch wavevector.

For $G=2\pi /a$, the sine potential has coefficients $U_G=-iV$, $U_{-G}=iV$. At the Brillouin zone boundary, the free states $e^{i\pi x/a}$ and $e^{-i\pi x/a}$ are degenerate. Keeping these two resonant waves in the nearly-free electron model givesThe determinant is $(\beta -E{)}^2-V^2$, so the two energies are $E_{\pm }=\beta \pm |V|$, separated by $2|V|$. With $V>0$, the lower combination has coefficients $(1,-i)/\sqrt{2}$ and density proportional to $1-\sin Gx$; the upper has $(1,i)/\sqrt{2}$ and density $1+\sin Gx$. The lower wave places less weight where the potential is positive. Coupling to nonresonant plane waves gives higher-order corrections beyond this two-wave approximation.

The first Brillouin zone of a primitive cubic lattice is the cube $|k_i|\le \pi /a$. Its boundary along $(100)$ is the face center $(\pi /a,0,0)$, while along $(111)$ it is the corner $(\pi /a,\pi /a,\pi /a)$. ThereforeA spin-degenerate band accommodates two electrons per cell. To be a band insulator, the filled first band must be separated from the second at every wavevector; a direct local splitting alone is insufficient. Under the paper's uniform-boundary-gap estimate, take the lower edge as $E^{(0)}-V$ and the upper edge as $E^{(0)}+V$. The highest lower edge is $3\beta -V$ and the lowest upper edge is $\beta +V$. Thus the indirect gap in a uniformly split zone-boundary model isEquality gives touching rather than a positive gap; for smaller $V$, overlapping energy bands prevent a band insulator. This final threshold is the stated uniform-splitting band-edge estimate. A quantitative three-dimensional potential mixes several plane waves at corners and requires its full band structure rather than assuming the two-wave sine calculation remains exact there.