Past exam of the mathematics course of the University of Cambridge 2022 ii Paper 3 34D Solution Created 2026-09-23 Updated 2026-09-25
Every primitive basis of the Bravais lattice has the formwhere andThe unimodular determinant condition is exactly what makes the integer change of basis invertible over .
Let be the primitive vectors of the reciprocal lattice, withSolving these four equations givesThe six reciprocal-lattice points nearest the origin areall at distance .
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 th Brillouin zone consists of wavevectors reached from the origin after crossing exactly 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.
Past exam of the mathematics course of the University of Cambridge 2025 ii Paper 3 34B b Solution Created 2026-09-24 Updated 2026-09-25
Translation by one site multiplies a lattice wave by , so the coefficients of a finite periodic tight-binding ring arePeriodicity after three sites requiresChoosing representatives in the Brillouin zonegivesThe corresponding states areThey reproduce respectively the coefficient vectors , and .
The real-space carbon honeycomb has two atoms in a primitive cell of its triangular Bravais lattice. The reciprocal lattice is also triangular, and its Wigner-Seitz cell is a hexagonal first Brillouin zone. With a consistent orientation convention, the reciprocal-zone hexagon is rotated by relative to a real-space carbon-ring hexagon; reciprocal lengths scale inversely with the direct lattice spacing.
Rolling into a tube imposes , leaving parallel allowed lines in graphene's two-dimensional reciprocal space. They run along the tube-axis direction, perpendicular to the circumferential wrapping vector. The nanotube's actual one-dimensional Brillouin zone is an interval of axial wave vectors with period , where is its axial translation. The parent hexagon is therefore a zone-folding construction, not a literal two-dimensional Brillouin zone for the finished one-dimensional tube.
Let the graphene lattice constant be , and choose direct primitive vectors and . Reciprocal primitive vectors satisfying areThe first Brillouin zone is a hexagon. Its corners K and K' are inequivalent valleys; M denotes an edge midpoint. In the nearest-neighbour model, , where and are the three bond vectors. The band separation is : it vanishes at K/K' and is at M, varying continuously along each edge.
Graphene reciprocal primitive vectors and the orientation-dependent band separation on its zone boundary
. Source. Details. This graphene zone-boundary band separation is a local conduction-to-valence separation. Ideal graphene itself remains globally gapless because the bands touch at the corners. Circumferential quantization in a nanotube selects lines through this two-dimensional band structure; whether they pass through a valley determines the ideal metallic versus semiconducting distinction.
A photonic crystal has a spatially periodic dielectric response on the scale of the optical wavelength. For a nonmagnetic, lossless, nondispersive dielectric, the source-free Maxwell equations can be written as the eigenproblemThe Bloch theorem gives with periodic . The allowed frequencies form photonic band structures. Coupling of counterpropagating waves near a Brillouin zone boundary opens a stop band. A one-dimensional dielectric multilayer is a simple example; two-dimensional hole arrays and three-dimensional dielectric networks allow more extensive control.
A complete photonic band gap excludes bulk propagating modes for every direction and polarization in a frequency interval. Periodicity alone does not guarantee such a gap: dimensionality, symmetry and refractive-index contrast matter. A missing or altered element can support a localized defect state in the gap, making a cavity; a line defect can guide light. Finite crystals have leakage and wavelength-dependent transmission rather than an infinitely sharp exclusion. Applications include selective mirrors, optical filters, small resonators and integrated waveguides.[ref]
Mathematical Methods in Nanophotonics, MIT 18.369, Spring 2016 by Steven G. Johnson. Lecture 2: Maxwell eigenproblem; lectures 9–14: Bloch states, one-dimensional gaps and defect localization
A three-phonon normal process conserves crystal wavevector without a reciprocal-lattice shift, . An Umklapp process instead hasThe crystal can absorb the reciprocal-lattice momentum; after reduction to the first Brillouin zone the phonon drift transporting heat is relaxed. A normal process redistributes phonons while preserving their total crystal momentum, so it is not by itself the same resistive mechanism. energy must also be conserved in either process.
At low temperature, large-wavevector phonons are rare, suppressing Umklapp scattering. Boundary and defect scattering then limit the mean free path; with a nearly constant boundary-limited path, grows as . As temperature increases, heat capacity rises but Umklapp scattering shortens , often producing a electrical conductivity maximum. At sufficiently high temperature a simple anharmonic model has nearly constant and , giving . These are regime estimates, not a universal law for every material or size.
Past exam of the physics A course of the University of Cambridge 2013 Paper 2 Section C C11 Solution 2026-10-20
One mass per cell. If is the displacement from equilibrium, the two springs give . Substituting yieldsThe first Brillouin zone is ; adding leaves the sampled displacements unchanged. The branch is linear near zero, with sound speed , and reaches at either zone edge.
Two masses per cell. Take a cell of length , with light displacement and heavy displacement . The coupled equations reduce toA nonzero displacement requires determinant zero:Solving this quadratic in givesNow the first zone is . The minus branch is acoustic: as , both sublattices move together and . Its sound speed is . The plus branch is optical: at the atoms move oppositely with , giving finite frequency . This terminology describes relative motion; optical activity itself also requires coupling to the electromagnetic field.
diatomic-chain zone-edge polarizations and dimensionality. At the edge, , so the heavy and light sublattices decouple: the acoustic edge frequency is and the optical edge frequency is . HenceThe acoustic edge mode moves only the heavy sublattice; the optical edge mode only the light one. If , the gap closes and the two branches are the zone-folded monatomic branch.
Original monatomic and diatomic chain dispersions, zone-edge gap and long-wavelength displacement patterns
. Details. A three-dimensional crystal with two atoms per primitive cell has six branches: three acoustic and three optical. For propagation along a suitable symmetry direction these resolve into one longitudinal and two transverse branches of each type. The wavevector and Brillouin zone are three-dimensional; anisotropic force constants determine the dispersions, and polar crystals can have longitudinal–transverse optical splitting. A universal six-branch curve cannot be specified from masses alone.
Past exam of the physics A course of the University of Cambridge 2015 Paper 2 Section A 5 Solution 2026-10-20
Write for the primitive cubic lattice spacing. With wavevector convention , reciprocal-lattice primitive vectors have length . The first Brillouin zone is the cube , henceIt has the same volume as any primitive reciprocal unit cell. A reciprocal coordinate defined in cycles per metre would instead give ; that is a different convention from this -space convention.
Past exam of the physics A course of the University of Cambridge 2015 Paper 2 Section C C11 Solution 2026-10-20
Write the real displacement as , choosing real positive peak amplitude . Nearest-neighbour harmonic springs store energy in their extension , each spring counted once. Therefore the instantaneous physical energy isThe printed complex displacement is a calculation device: squaring the complex phasor itself would not produce physical energy. The chain sketch in the accompanying figure shows masses at equilibrium spacing , with longitudinal displacements along the chain.
The two spring forces on site giveSubstituting the normal-mode phasor and cancelling its nonzero amplitude givesThe long-wavelength limit is acoustic, , and the first Brillouin zone ends at . The maximum angular frequency is .
Time averaging uses . ConsequentlyThe dispersion relation makes these two contributions equal, givingThis is equality of time-averaged kinetic and potential energies in a harmonic normal mode; the instantaneous energy partition need not be equal.
Use periodic boundary conditions on length to count modes: . The density on the signed axis is . Using folded wave-number counting for a monatomic chain combines and at positive magnitude and counts both directions, so the convention used by the requested expression isThis is a large- continuum density, not a second independent branch. It gives modes after integrating over that half-zone; endpoint counting changes only finite-size corrections.
For positive , andThe density of states of a monatomic harmonic chain vanishes outside the band, is constant to leading order near zero, and has an integrable Van Hove singularity at the upper band edge where group velocity vanishes. Normalization checks the directional counting:
Past exam of the physics A course of the University of Cambridge 2016 Paper 2 Section A 5 Solution 2026-10-20
The effective mass follows from the curvature of the electronic band structure. Differentiating twice givesAt either first Brillouin zone boundary ,The boundary is a band maximum, with negative curvature and zero group velocity. In the semiclassical single-band description, an applied force changes ; the resulting acceleration is opposite the force near that maximum. This is a band-response coefficient, not a negative microscopic electron rest mass. A nearly filled band is often described more conveniently through positively charged electron holes near its maximum.
Put type-A atoms at and type-B atoms at . Their displacements obey Newton's second law,Choose normal modes and . Substitution givesA nonzero displacement vector requires zero determinant:Solving this quadratic in gives the diatomic-chain dispersion relation with equal springs,The primitive unit cell has length , so the first Brillouin zone is .
For a Bravais lattice with primitive translation vectors , define the reciprocal lattice by vectors satisfying for every direct-lattice translation . A convenient basis iswith cyclic formulas for . Then and for integers . The reciprocal primitive volume is .
These vectors are the spatial Fourier series frequencies of a periodic lattice potential or density. Elastic coherent crystal diffraction occurs when the scattering wavevector transfer equals a reciprocal lattice vector, with intensities additionally controlled by the basis crystal lattice structure factor. This is the reciprocal-space version of Bragg's law. A lattice of points and its basis should not be confused: a basis can make allowed reciprocal-lattice diffraction intensities vanish.
The first Brillouin zone is the Wigner-Seitz cell around the reciprocal origin: it consists of points closer to than to any other reciprocal point. Its faces lie on the perpendicular bisectorsTo see the condition, equate and . These are also free-electron Bragg-degeneracy planes, where a periodic potential can open a band gap. Higher zones classify regions beyond successive Bragg planes; the reduced-zone description folds them into the first zone while retaining a band label.
In one dimension with spacing , reciprocal points occur at , and the first zone is . A square lattice of spacing has reciprocal basis , and a square first zone with edges at .
The reciprocal of a face-centered cubic lattice is a body-centered cubic lattice and vice versa, with the scale fixed by the primitive-vector definition. In a Bloch state, wavevectors differing by describe equivalent crystal momenta modulo the lattice periodicity. A crystal with primitive cells has allowed wavevectors in one Brillouin zone, hence states in each band per spin. This state count links reciprocal geometry to band filling, metallic conduction and band insulating behaviour. Reciprocal space also labels phonons and makes conservation of crystal momentum modulo transparent.
Past exam of the physics A course of the University of Cambridge 2018 Paper 2 Section A A4 Solution 2026-10-18
Take nearest-neighbour separation , a common spring constant , and one scalar displacement per atom. For a monatomic chain, the harmonic approximation near a potential minimum, Newton's second law and a normal mode giveThe monatomic-chain dispersion relation has one acoustic phonon branch, with near the origin. The first Brillouin zone ends at , where and the group velocity vanishes.
For alternating masses , choose a two-atom cell of length . The coupled normal modes give the diatomic-chain dispersion relation with equal springs,The first Brillouin zone now ends at . The minus branch is the acoustic phonon branch; its two atoms move together at long wavelength. The plus branch is the optical phonon branch; at the atoms move oppositely with a stationary centre of mass and . For the edge frequencies are and , with a gap between them.
If , the two branches reduce to and , touching at . This is zone folding in an equal-mass diatomic chain: identical sites actually have a primitive unit cell of length . A doubled cell folds the monatomic branch but opens no gap. Equal masses alone do not specify all possible spring models; the sketches assume equal nearest-neighbour spring constants.
Past exam of the physics A course of the University of Cambridge 2019 Paper 2 Section C C11 a Solution Created 2026-10-18 Updated 2026-10-20
The quantum free electron model treats conduction electrons as independent particles moving in an approximately constant potential, with parabolic dispersion relation . Fermi-Dirac statistics successfully explains important features such as a Fermi energy and a small electronic heat capacity, but a constant potential gives no band gaps and cannot by itself explain why different band fillings produce metals, semiconductors and band insulators. It also misses lattice-dependent effective masses, detailed directional responses and the topology of real Fermi surfaces. Scattering and finite electrical resistivity require additional physics.
The nearly-free electron model introduces a weak periodic lattice potential. Bloch's theorem replaces individual free plane waves by Bloch states, and Fourier components of the potential mix plane waves whose wavevectors differ by a reciprocal lattice vector. Near degeneracy at a Brillouin zone boundary, this coupling opens a band gap; for a two-wave crossing its size is . Away from the crossing, energies remain close to free-particle values. Band curvature produces band effective masses, and band filling determines whether the Fermi energy lies inside an allowed band or a gap.
The improvement is the inclusion of lattice periodicity, rather than a switch from classical to quantum statistics: the quantum free electron model already uses Fermi-Dirac statistics. The strong delta barriers in the later numerical example are treated exactly by the Kronig-Penney model, rather than by assuming that their strength is perturbatively small.
A Bloch state on the loop has , with . Its circumference is , so . Single-valuedness after one complete circuit requires the periodic boundary conditionThereforeThe allowed Bloch wavevectors are spaced by . Values of differing by label the same crystal momentum modulo a reciprocal lattice vector, so there are inequivalent choices in one Brillouin zone. In this notation is the Bloch wavenumber, not a dimensionless phase; is dimensionless. Spin gives additional states but does not alter this condition.
Past exam of the physics A course of the University of Cambridge 2021 Paper 2 Section A 5 Solution 2026-10-18
Let be the integer number of primitive unit cells. A Bloch state can be written as , with . The periodic boundary condition therefore requiresThe spacing of allowed wavevectors is . The first Brillouin zone has width , soFor an exact endpoint convention take a half-open Brillouin zone, such as . Its two endpoints differ by a reciprocal lattice vector and must not both be counted as distinct values of crystal momentum. There are exactly inequivalent integers modulo , irrespective of whether is even or odd. This is state counting in a Bloch band: one allowed wavevector per primitive unit cell per energy band. A separate twofold spin multiplicity gives electronic states per energy band, without doubling the number of wavevectors.
Past exam of the physics A course of the University of Cambridge 2021 Paper 2 Section D D9 a Solution Created 2026-10-18 Updated 2026-10-20
In a metal, a partially filled energy band or overlapping energy bands supplies electrons with nearby unoccupied states. At thermal equilibrium, oppositely directed group velocities are equally populated and the net electric current vanishes. An applied electric field distorts the occupation distribution, producing a small mean drift velocity. A filled energy band alone has no such low-field intraband response because its velocity contributions cancel over the entire Brillouin zone.
In the Drude model, let be the mobile electron number density, let be the elementary charge, and let be the momentum relaxation time. The mean motion obeysThe steady state is . The current density is opposite to the electron drift velocity:This gives microscopic Ohm's law, with electrical conductivity in and electrical resistivity in . For a uniform wire of length and cross-sectional area , its electrical resistance is . The dissipated power per volume is , the Joule heating supplied by the external source.
The free electron model with Fermi-Dirac statistics improves the microscopic picture. The Pauli exclusion principle fills states up to the Fermi energy at low temperature. For a spherical, parabolic energy band with two spin states,Only states near the Fermi surface change their occupation in a weak-field response; the large equilibrium Fermi velocity is not the small drift velocity. The same formula results for this isotropic parabolic model when is evaluated for transport near the Fermi surface. In a crystal, the energy band group velocity, effective mass and possibly direction-dependent scattering replace the simplest bare-particle picture. The mean free path is approximately .
A perfectly periodic static lattice is already included in Bloch states and does not by itself give a finite electrical resistivity. Momentum relaxation comes from defects, impurities, boundaries, and electron-phonon scattering. Independent scattering mechanisms approximately add their rates, giving Matthiessen's rule,The approximately temperature-independent is the residual resistivity. In ordinary simple metals, the phonon contribution is approximately proportional to temperature at high temperature; a three-dimensional acoustic-phonon transport model gives a contribution well below the relevant Debye temperature. Other mechanisms can dominate in particular materials or regimes, so neither power law is universal.
The classical Drude model incorrectly treats the equilibrium carrier statistics. Fermi-Dirac statistics explains the small electronic heat capacity, while electrons carrying both charge and heat give the approximate Wiedemann-Franz law for degenerate carriers with appropriate nearly elastic scattering. Both the simple electrical conductivity formula and this thermal relation depend on the assumptions about energy bands and scattering, rather than being exact for all metals.
Past exam of the physics A course of the University of Cambridge 2021 Paper 2 Section D D9 b Solution Created 2026-10-18 Updated 2026-10-20
The Debye model approximates a three-dimensional crystal as an elastic continuum supporting quantized normal modes, or phonons. At long wavelengths, the acoustic phonon branches have approximately linear dispersion relations : one longitudinal branch and two transverse branches. Replace their complicated actual spectra by these linear forms, then introduce a cutoff chosen to preserve the total vibrational degrees of freedom for atoms. This is an approximation to the spectrum, not a claim that a real crystal has no non-acoustic branches.
For periodic boundary conditions in volume , each allowed wavevector occupies in the space of wavevectors. Counting a spherical shell for each branch gives the density of states per unit angular frequencyAn effective speed of sound can be defined by . In the single-speed Debye model, choose so thatThus the Debye wavevector, Debye frequency and Debye temperature satisfyHere has units of seconds when angular frequency is measured in ; its integral is the dimensionless number of normal modes. The spherical cutoff replaces the real Brillouin zone by an equal-count continuum construction.
Each phonon mode is a quantum harmonic oscillator. Since phonon number is not conserved at thermal equilibrium, the chemical potential is zero and its Bose-Einstein distribution gives mean thermal energy . The zero-point energy is independent of temperature in the harmonic, fixed-volume approximation and does not contribute to the heat capacity at constant volume. Excluding that constant term,Differentiating the first integral at fixed volume avoids losing the temperature-dependent upper limit in the second one. The resulting Debye heat capacity isFor , extend the upper limit to infinity. Expanding and integrating term by term gives . ThereforeThis law follows from the three-dimensional density of states and the availability only of low-frequency modes at low temperature. It is a central success of the Debye model.
For , the Bose-Einstein distribution reduces to the classical equipartition theorem: every mode contributes thermal energy . Consequentlythe Dulong-Petit law, or per mole of atoms. The simple model omits detailed phonon dispersion relation, distinct optical phonon branches, and anharmonic effects such as thermal expansion. In a metal, the low-temperature total heat capacity additionally contains a term approximately proportional to from electrons, so the lattice term should not be mistaken for the whole measurement.
For each primitive linear, square or cubic Bravais lattice of spacing , the reciprocal lattice has spacing . The first Brillouin zone is consequently the interval, square or cube with faces at components . The nearest zone boundary is a distance from the centre. The ratios areAll are below one: the ideal free-electron Fermi surface lies wholly inside the first Brillouin zone. Its nearest approach to the boundary is in three dimensions, where it lies only about short of the nearest face. A lattice-induced band gap at a zone face therefore does not by itself make these half-filled systems insulating in the stated free-electron model. Interactions, structural changes and actual electronic band structure can alter that model prediction.
Past exam of the physics A course of the University of Cambridge 2022 Paper 2 Section C C8 e Solution Created 2026-10-18 Updated 2026-10-20
The length, area and volume of a first Brillouin zone are . The corresponding occupied -space measures areHenceThis is state counting, independent of the interval/disk/ball geometry. One Brillouin zone contains one spatial state per unit cell in a single energy band, and that band can hold two electrons per cell because of spin. One electron per cell therefore occupies half of it. The “volume enclosed” is the occupied Fermi sea, not the lower-dimensional area of the Fermi surface itself. Its half-volume does not mean half the distance to the zone boundary in two or three dimensions.





