advanced26 min studyLesson 492 in Physics

Reciprocal lattices and Brillouin zones

Construct reciprocal vectors with an explicit 2π convention, fold wavevectors consistently, and define the Brillouin zone as a reciprocal Wigner–Seitz cell.

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What you’ll understand

  • Compute reciprocal vectors from direct-lattice scalar products.
  • Find Brillouin-zone boundaries from perpendicular bisectors.
  • Distinguish reciprocal translations from changes of physical wavelength or cell symmetry.

A crystal selects special phase patterns because translating by a lattice vector can leave a plane wave unchanged. Those patterns form the reciprocal lattice. We will include a factor of two pi in its definition throughout this module, so wavevectors are angular spatial frequencies. This convention connects lattice translations directly to phase factors, diffraction conditions, and phonon wavevectors without silently switching between cycles and radians per unit length.

The reciprocal lattice is a phase condition

Define reciprocal primitive vectors by ai⋅bj=2πδij\mathbf a_i\cdot\mathbf b_j=2\pi\delta_{ij}. An integer combination G of the b vectors then satisfies eiG⋅R=1e^{i\mathbf G\cdot\mathbf R}=1 for every direct-lattice translation R. The dot product is dimensionless, with a in length units and b in inverse-length units. Some crystallographic conventions omit two pi and put it into the Fourier exponential instead. Both are legitimate, but combining a no-two-pi reciprocal vector with our exponential produces incorrect diffraction angles and zone widths.

In three dimensions a reciprocal vector is obtained from a cross product divided by the oriented primitive volume. Its direction is perpendicular to the other two direct vectors, not generally parallel to its matching direct vector. Orthogonal rectangular lattices make the construction look simpler than it is. For a sheared cell the reciprocal grid shears oppositely so the dual scalar products remain fixed. Computing those products explicitly is an effective check on a proposed reciprocal basis, including signs and factors of two pi.

b1=2πa2×a3a1⋅(a2×a3),G=∑jhjbj,hj∈Z,eiG⋅R=1\mathbf b_1=2\pi\frac{\mathbf a_2\times\mathbf a_3}{\mathbf a_1\cdot(\mathbf a_2\times\mathbf a_3)},\qquad \mathbf G=\sum_j h_j\mathbf b_j,\quad h_j\in\mathbb Z,\qquad e^{i\mathbf G\cdot\mathbf R}=1
The other reciprocal primitive vectors follow by cyclic permutation. These angular wavevectors have units rad per length, with radians dimensionless. The denominator is an oriented nonzero volume.↔ Scroll wide equations horizontally.

A Brillouin zone chooses representatives

The first Brillouin zone is the Wigner–Seitz cell around the reciprocal origin. Its boundary points are equally distant from the origin and a neighboring reciprocal point. Squaring that distance condition gives a plane q⋅G=∣G∣2/2\mathbf q\cdot\mathbf G=|\mathbf G|^2/2. Intersecting the relevant half-spaces constructs the zone. For a rectangular lattice the boundaries are simple coordinate limits; for an oblique lattice the shortest reciprocal vectors can produce a polygon whose faces are not obtained by taking half of each basis vector independently.

Wavevectors differing by a reciprocal vector give identical phase factors on the Bravais sites. Choosing one representative in the first zone removes that redundancy for lattice amplitudes. A multi-atom basis can acquire corresponding internal phase conventions, so eigenvectors may need to transform when a wavevector is folded. The physical displacement pattern remains equivalent when these conventions are handled consistently. Folding does not mean that a freely propagating probe in empty space has literally changed its wavevector; it is a representation of periodic-medium states.

∣q∣2=∣q−G∣2⟺q⋅G=∣G∣22,VBZ=(2π)3vp|\mathbf q|^2=|\mathbf q-\mathbf G|^2\quad\Longleftrightarrow\quad \mathbf q\cdot\mathbf G=\frac{|\mathbf G|^2}{2},\qquad V_{\mathrm{BZ}}=\frac{(2\pi)^3}{v_{\mathrm p}}
The first relation defines a candidate reciprocal bisector, while the closest-neighbor half-spaces define the actual first zone. The volume formula is three dimensional; in two dimensions replace volumes by areas and the third power by a square.↔ Scroll wide equations horizontally.

Cell choice preserves total state counting

For a periodic sample with N primitive cells, there are N allowed wavevectors in one Brillouin zone per branch, ignoring the bookkeeping of equivalent boundary points. The allowed spacing follows from the finite sample's periodic boundary condition. Increasing real-space cell volume by an integer supercell factor shrinks the zone by that factor and increases the number of folded branches. The total count of states or vibrational coordinates is unchanged. Counting both folded branches and the original full zone would therefore double-count physical modes.

Reciprocal geometry alone does not specify an energy or frequency dispersion. The zone tells us the domain of independent wavevector labels, while forces and masses determine phonon frequencies. Similarly, the reciprocal lattice gives candidate diffraction locations, but basis interference can remove their intensity. Separating these roles avoids drawing a band gap simply because a zone boundary exists or assuming every reciprocal point produces an observable peak. Geometry constrains the calculation; it does not replace the dynamical or scattering-amplitude information.

Fold a rectangular-lattice wavevector

A two-dimensional rectangular lattice has translations (2,0) A˚(2,0)\,\mathrm{\mathring A} and (0,4) A˚(0,4)\,\mathrm{\mathring A}. Find reciprocal vectors, first-zone limits, and the first-zone representative of q=(3π/4,0) A˚−1\mathbf q=(3\pi/4,0)\,\mathrm{\mathring A^{-1}}.

  1. Duality gives b1=(π,0) A˚−1\mathbf b_1=(\pi,0)\,\mathrm{\mathring A^{-1}} and b2=(0,π/2) A˚−1\mathbf b_2=(0,\pi/2)\,\mathrm{\mathring A^{-1}}. Their direct-vector dot products are two pi on matching pairs and zero on cross pairs.

  2. The rectangular first zone has ∣qx∣≤π/2|q_x|\le\pi/2 and ∣qy∣≤π/4|q_y|\le\pi/4 in inverse ångströms, with opposite boundaries identified rather than counted twice.

  3. Subtract one b1 from the proposed vector: 3π/4−π=−π/43\pi/4-\pi=-\pi/4. The representative is (−π/4,0) A˚−1(-\pi/4,0)\,\mathrm{\mathring A^{-1}}, and lattice-site phases agree because the difference is a reciprocal vector.

The result

The reciprocal periods are pi and pi/2 per ångström, and the given wavevector folds to negative pi/4 along x inside the first zone.

Explore the model

Interactive lab

Interactive lab · modular

Choosing representatives after reciprocal-like shifts

Reset and set BOTH Modulus m = 12 and Additive step a = 5; predict an orbit visiting all 12 residues. For a fresh run, Reset and set m = 12 and a = 4; predict only residues 0, 4, and 8. Step to compare finite representative classes.

Choosing representatives after reciprocal-like shiftsAddition modulo m starts at zero: rₙ = na mod m. Playback advances two additions per second. The orbit has period m/gcd(m,a), including period one for a = 0. This is not a cryptographic security test or a primality algorithm. Current readings: Residue r0: 0; Orbit period · m/gcd(m,a): 12; Distinct residues visited: 1 of 12. All parameters are adjustable with the labeled controls below.Repeated addition modulo 12: the current residue is 0, and the orbit period is 12.01234567891011+ 5mod 12An additive orbitn = 0rₙ = 0Period = 12Cycle starts at 0:0 → 5 → 10 → 3 →8 → 1 → 6 → 11 →4 → 9 → 2 → 7 → 0Exact residues · two additions per second
Additive orbitCurrent residue / next edge
Residue r0
0
Orbit period · m/gcd(m,a)
12
Distinct residues visited
1 of 12
t = 0 s

Addition modulo m starts at zero: rₙ = na mod m. Playback advances two additions per second. The orbit has period m/gcd(m,a), including period one for a = 0. This is not a cryptographic security test or a primality algorithm. Parameter changes pause playback and recompute the model at the current elapsed time. Reset restores both parameters and returns time to zero. Playback starts paused. Adjust either parameter, then choose Play or Step.

Look forThis exact additive modular orbit illustrates equivalence classes only. It is not a reciprocal-lattice or Brillouin-zone construction and has no length, wavevector, or band-dispersion readout.

Check your understanding

Which equation fixes the reciprocal convention used throughout this module?

Try it yourself

For a1=(2,0) A˚\mathbf a_1=(2,0)\,\mathrm{\mathring A} and a2=(1,3) A˚\mathbf a_2=(1,3)\,\mathrm{\mathring A}, compute the two reciprocal vectors and verify the reciprocal primitive area.

Take a moment to make a prediction or work it out on paper. Open a hint when you need one.

Give me a hint

Solve each pair of scalar-product conditions directly. The reciprocal area must equal the square of two pi divided by the direct area.

Compare with the solution

The first matching condition gives b1x=πb_{1x}=\pi; the cross condition gives π+3b1y=0\pi+3b_{1y}=0, so b1=(π,−π/3) A˚−1\mathbf b_1=(\pi,-\pi/3)\,\mathrm{\mathring A^{-1}}. For b2, b2x=0b_{2x}=0 and 3b2y=2π3b_{2y}=2\pi, so b2=(0,2π/3) A˚−1\mathbf b_2=(0,2\pi/3)\,\mathrm{\mathring A^{-1}}. Their determinant is 2π2/3 A˚−22\pi^2/3\,\mathrm{\mathring A^{-2}}, equal to (2π)2/6(2\pi)^2/6. The reciprocal first vector is not parallel to the first direct vector because the direct cell is oblique.

Take it with you

  • Reciprocal vectors encode phase periodicity and include a declared two-pi convention.
  • The first Brillouin zone is a reciprocal Wigner–Seitz cell, not an arbitrary basis parallelepiped.
  • Wavevector folding preserves total mode counting when branches and basis phases are handled consistently.

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