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The first Brillouin zone is $|k_x|<K,|k_y|<L$. The second consists of four disconnected lobes. Its right lobe has vertices $(K,-L),(2K,-L+K^2/L),(2K,L-K^2/L),(K,L)$; its upper lobe has vertices $(-K,L),(K,L),(0,L+K^2/L)$. Reflection in the coordinate axes gives the other two. The bisectors to $(2K,2L)$ and $(4K,0)$ truncate the lobes. Their combined area is $4KL$, equal to the first zone's area. Counting the Bragg planes crossed from the origin proves the construction.

Construct Bragg planes by taking the midpoints between the reciprocal origin and the other reciprocal lattice points. In one dimension these are points at $k=m\pi /a$, for nonzero integer $m$.

The first Brillouin zone is the interval bounded by the nearest two such points:The second Brillouin zone consists of the next intervals on both sides:Boundary points may be assigned to one adjacent zone to avoid double counting. The first zone is reached from the origin without crossing a Bragg point; the second is reached after crossing one. Each zone has total length $2\pi /a$. In particular, the two separated pieces of the second zone are not the entire translated first-zone interval centered at $2\pi /a$.

A Wigner-Seitz cell is the region of real space at least as close to a chosen Bravais lattice point as to any other. It is bounded by perpendicular bisectors of vectors to neighboring lattice points and is a primitive unit cell.

A reciprocal lattice vector $\mathbf{G}$ satisfies $e^{i\mathbf{G}\cdot \mathbf{R}}=1$ for every lattice translation $\mathbf{R}$. With primitive vectors ${\mathbf{a}}_i$, reciprocal primitive vectors satisfy ${\mathbf{a}}_i\cdot {\mathbf{b}}_j=2\pi {\delta }_{ij}$, and $\mathbf{G}={\sum }_j{n}_j{\mathbf{b}}_j$ for integers $n_j$.

The first Brillouin zone is the Wigner-Seitz cell of that reciprocal lattice, centered at $\mathbf{k}=0$. Its boundaries are Bragg planes, $\mathbf{k}\cdot \mathbf{G}=G^2/2$. Higher Brillouin zones are the regions separated from the origin by successively more such planes: an interior point of zone $j$ has crossed exactly $j-1$ planes. A second Brillouin zone is generally disconnected in this extended-zone representation; it is not just a translated first-zone rectangle.

For the first two Brillouin zones of a rectangular lattice, write $K=\pi /a$ and $L=\pi /b$, so $0<K<L$ and the reciprocal lattice vectors are $(2nK,2mL)$. The perpendicular bisectors to the four nearest axial reciprocal points encloseThe second Brillouin zone consists of points beyond exactly one Bragg plane. In the first quadrant, it is the union of the following two regions:Reflect these regions in both axes. The first is a trapezoidal side lobe; the second is half of a triangular top lobe. The sloping boundary is the bisector for $\mathbf{G}=(2K,2L)$. On the side lobe, $k_x<2K$ is the bisector condition for $\mathbf{G}=(4K,0)$: crossing that plane too would enter the third or a higher zone. These additional bisectors are why four rectangles pasted onto the first zone would be incorrect.

Equivalently, the full right lobe has verticesand the top lobe has vertices $(-K,L),(K,L),(0,L+K^2/L)$. Add their left and bottom reflections. Their total area isequal to the first-zone area, as required by reciprocal-space state counting. These formulas remain valid for every $a>b$, including very elongated rectangles; boundaries themselves can be shared by convention.

The cubic Miller indices give an interplanar spacing $d_{hkl}=a/\sqrt{h^2+k^2+l^2}$. Equivalently, the reciprocal lattice vector normal to these lattice planes has length $2\pi /d_{hkl}$. HenceThe eight $\{111\}$ reciprocal lattice vectors are the nearest nonzero reciprocal points. Their perpendicular bisectors form the eight hexagonal faces of the first Brillouin zone, a truncated octahedron. For the positive $(111)$ direction, that Bragg plane is $\mathbf{k}\cdot {\mathbf{G}}_{111}=|{\mathbf{G}}_{111}{|}^2/2$, or $k_x+k_y+k_z=3\pi /a$. It is parallel to the direct $(111)$ lattice planes. The six remaining square zone faces correspond to the $\{200\}$ reciprocal points. The zone boundary is halfway to a reciprocal point, not the point itself.

The nearly-free electron model treats a weak periodic potential as a perturbation of plane waves. Near a Bragg plane, states with wavevectors $\mathbf{k}$ and $\mathbf{k}-\mathbf{G}$ are nearly degenerate and must be mixed together. Their two-state Hamiltonian isAt the unperturbed crossing, the band gap is $2|V_{\mathbf{G}}|$. The corresponding standing waves place different charge densities near the ions and between them, giving different potential energies. Far from a crossing, ordinary perturbation theory is adequate; near it, the degenerate two-wave calculation is essential. In three dimensions, gaps at individual zone boundaries do not guarantee a global insulating gap because bands from different directions can overlap.