Questions: Reciprocal Lattice and Brillouin Zones

Question 1

The reciprocal lattice vectors b_i are defined so that a_i · b_j = 2π δ_{ij}. What is the physical significance of this orthogonality condition?

• A. It ensures that the reciprocal lattice has the same symmetry as the direct lattice
• B. It guarantees that plane waves e^{ik·r} with wavevector k = G (a reciprocal lattice vector) have the periodicity of the direct lattice, so e^{iG·R} = 1 for every lattice vector R
• C. It means the reciprocal lattice vectors are perpendicular to the direct lattice vectors
• D. It ensures that the volume of the reciprocal unit cell equals the volume of the direct unit cell

Answer: (B)

The condition a_i · b_j = 2π δ_{ij} ensures that for any reciprocal lattice vector G = h b_1 + k b_2 + l b_3 and any direct lattice vector R = n_1 a_1 + n_2 a_2 + n_3 a_3, we get G · R = 2π(hn_1 + kn_2 + ln_3), which is always 2π times an integer. Therefore e^{iG·R} = 1 for all R. This means plane waves with wavevector G have the same periodicity as the crystal lattice — the defining property that makes the reciprocal lattice useful.

Question 2

The first Brillouin zone of an FCC direct lattice has the same shape as the Wigner-Seitz cell of a BCC direct lattice.

• T. True
• F. False

Answer: True

The reciprocal lattice of FCC is BCC (and vice versa). Since the first Brillouin zone is defined as the Wigner-Seitz cell of the reciprocal lattice, the first Brillouin zone of FCC has the shape of the BCC Wigner-Seitz cell (a truncated octahedron). Conversely, the first Brillouin zone of BCC has the shape of the FCC Wigner-Seitz cell (a rhombic dodecahedron). This FCC ↔ BCC duality is one of the most elegant results in crystallography.

Question 3

Why is the first Brillouin zone, rather than the entire reciprocal space, sufficient for describing the electronic band structure of a crystal?

Model answer: Because of Bloch's theorem, electronic states in a periodic potential are labeled by a crystal momentum k, and states at k and k + G (where G is any reciprocal lattice vector) are physically equivalent — they describe the same Bloch wave. The first Brillouin zone contains exactly one representative from each equivalence class of k-vectors, so all physically distinct states are captured within it. Any information outside the first zone is redundant, just a copy of what's inside shifted by a reciprocal lattice vector.

This is the k-space analog of how the direct lattice's periodicity means you only need to know what happens in one unit cell. The Brillouin zone is the 'unit cell' of reciprocal space, and the redundancy comes from the discrete translational symmetry of the crystal.

Question 4

What is the geometric relationship between Brillouin zone boundaries and Bragg diffraction?

Model answer: A Brillouin zone boundary bisects the line from the origin to a reciprocal lattice point G, forming a plane where the condition 2k · G = |G|^2 is satisfied. This is exactly the Bragg diffraction condition (also called the von Laue condition): an electron with wavevector k at the zone boundary satisfies the condition for constructive interference from lattice planes. This is why energy gaps open at zone boundaries — the two degenerate plane waves e^{ik·r} and e^{i(k-G)·r} mix and split, creating a standing wave with an energy gap.

The deep connection between Brillouin zone geometry and diffraction is not a coincidence — both arise from the same Fourier relationship between direct and reciprocal space. Zone boundaries are where the perturbation from the periodic potential is strongest.