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The reciprocal lattice of a two-dimensional triangular Bravais lattice is triangular, so its Wigner-Seitz cell is a regular hexagon. Opposite boundary points are identified modulo reciprocal-lattice translations, and the six corners fall into two inequivalent classes of three corners each, conventionally denoted $K$ and $K^{\prime }$.

A three-dimensional Bravais lattice is the set of all integer combinations of three linearly independent primitive vectors. Its reciprocal lattice isand the first Brillouin zone is the Wigner-Seitz cell of the origin in reciprocal space.

For the stated FCC lattice, the primitive-cell volume isThe reciprocal-basis formula givesA reciprocal primitive cell, and hence the Brillouin zone, has volume

In the first Born approximation, with momentum transfer $q=k-k^{\prime }$, the scattering amplitude isFor the spherical shell, align the polar axis with $q$ and use its radial Dirac delta function:Therefore

Translation of each atomic potential multiplies its transform by a phase, so the crystal lattice structure factor givesFor an infinite lattice, the lattice Poisson summation formula turns the sum intoThus the amplitude vanishes away from reciprocal-lattice momentum transfers.

In elastic scattering, $|k|=|k^{\prime }|$ and henceThe two shortest relevant reciprocal-vector lengths of this FCC lattice areBoth are kinematically available when $k>2\pi /a$. Assigning their scattering angles ${\theta }_2$ and ${\theta }_1$, respectively, gives

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.

Every primitive basis of the Bravais lattice has the formwhere $p,q,r,s\in \mathbb{Z}$ andThe unimodular determinant condition is exactly what makes the integer change of basis invertible over $\mathbb{Z}$.

Let $b_i$ be the primitive vectors of the reciprocal lattice, withSolving these four equations givesThe six reciprocal-lattice points nearest the origin areall at distance $4\pi /\sqrt{3}$.

The Wigner-Seitz cell is bounded by the perpendicular bisectors of the segments joining the origin to those six points. It is the regular hexagon with verticesIts area equals the reciprocal primitive-cell area:This is the reciprocal lattice and Wigner-Seitz cell of the unit triangular lattice.

For a periodic potential, Bloch theorem identifies wavevectors differing by a reciprocal-lattice vector. The first Brillouin zone is the Wigner--Seitz cell just found. More generally, the $n$th Brillouin zone consists of wavevectors reached from the origin after crossing exactly $n-1$ reciprocal-lattice Bragg planes; its boundaries are the perpendicular bisectorsFor this triangular reciprocal lattice, the first zone is the central regular hexagon. The second is the sixfold-symmetric collection of regions immediately outside its six sides, bounded next by the bisectors associated with the next reciprocal points. These zones organize free-particle states into bands: Bragg coupling is strongest at their boundaries and opens energy gaps when the periodic potential is introduced.