Past exam of the mathematics course of the University of Cambridge 2014 iii Paper 44 4 c Solution Created 2026-10-03 Updated 2026-10-06
Use the Minkowski metric, path-integral weight and the Abelian gauge theory transformation . The gauge functional varies asThus the Faddeev-Popov operator is . It depends on the gauge field despite the gauge group being Abelian: the ghosts interact because this gauge condition is nonlinear.
Choose the gauge-fixing fermion . The gauge-fixed action isHere the tensor is distinct from the scalar gauge functional . With , integrating over imposes the exact printed constraint. For nonzero , eliminating instead givesThis version displays the additional gauge-dependent cubic and quartic gauge-field vertices as well as the ghost interaction; the strict condition is its limit.
For the Fourier transform convention , the quadratic ghost kernel is . With the ordering ,The term has Fourier coefficient , where is the incoming ghost momentum. Multiplication by in the Feynman rule givesThese signs refer to the displayed action, Fourier convention and ghost ordering. Reversing the ghost/antighost convention changes corresponding signs consistently. A closed ghost loop has the additional minus sign from Grassmann variables.
Past exam of the mathematics course of the University of Cambridge 2014 iii Paper 46 1 Solution Created 2026-10-03 Updated 2026-10-06
Use the mostly-plus Minkowski metric, and write for equality on the constraint surface. Assume that the mechanical constraints are locally independent. They are first-class constraints whenThus their Poisson brackets vanish on the constraint surface, and their Hamiltonian flows preserve that surface. The structure functions of a constraint algebra may depend on the phase space point. The finite real span of the constraints is a Lie algebra if it closes with constant structure coefficients, in a suitable choice of generators. The Jacobi identity then gives the usual conditions on the structure constants of a Lie algebra. With general structure functions the finite real span need not close, even though the Poisson bracket of all smooth functions is itself a Lie bracket.
To see the gauge invariance directly, let generate a canonical gauge transformation:The variation of the phase-space action integrand isThe second term cancels without using the equations of motion. Taking to vanish at the temporal boundaries leaves the action invariant. Arbitrary functions therefore relate different descriptions of the same physical motion. This reasoning also works with structure functions; constant structure coefficients are only needed for the finite-dimensional Lie algebra claim.
For a closed string, choose and periodic fields. A convenient Nambu-Goto phase-space action isHere and are Lagrange multipliers. The Nambu–Goto phase-space constraints are and , with canonical Poisson bracketsLet . Differentiating the periodic Dirac delta function givesThe opposite signs in and cancel these terms, so . Replace the original constraints by the equivalent chiral densitiesTheir mixed Poisson brackets vanish. Choose opposite Fourier orientations for the two sectors:The chiral constraint algebra of a closed string isEach is the Witt algebra: the vector fields on a circle satisfy . Fourier expansion identifies each real algebra, with , with the Lie algebra of vector fields on the circle. The two commuting copies give , not a quantum central extension.
For an open string, allowed boundary conditions must remove the endpoint term in the variation of the action, consistently with the allowed endpoint variations. The spatial boundary term isIt expresses the open-string endpoint momentum flux. In the temporal gauge for a string , take a boundary-adapted parametrization with at the ends. Fixing gives , a Dirichlet boundary condition. At the other end allow arbitrary spatial variations; for nonzero these require , a Neumann boundary condition. Also in this temporal gauge for a string, so . The constraint at this free-end string boundary condition reduces to . Hamilton's equation consequently gives there. Since , the free endpoint has spatial speed one. This is the null motion of a free string endpoint.
A straight rotating string with one fixed endpoint supplies the required solution in at least two spatial dimensions. Set , , andTake . The Hamilton's equations become , which holds because both second derivatives give . The Nambu–Goto phase-space constraints are satisfied byThe endpoint at stays at the origin, while at and the endpoint moves around a circle of radius with angular speed . At each time the whole string lies on a straight radial segment. Its spatial proper length isThe velocity is everywhere perpendicular to the segment, so this also equals the sum of local rest-frame lengths. The induced worldsheet metric becomes degenerate at the null free endpoint, as expected for the limiting free-end solution.
Past exam of the mathematics course of the University of Cambridge 2014 iii Paper 46 2 Solution Created 2026-10-03 Updated 2026-10-06
Use , transverse coordinates , , and the Minkowski metricThe relativistic particle phase-space action becomesIn the light-cone gauge , solve the mass-shell condition for , assuming . The reduced phase-space action is , withThe last equality selects the future-directed momentum sector and makes positivity transparent. With and , the Schrodinger equation isThe inverse acts only on Fourier modes with nonzero . Multiplication by gives . ThereforeThe light-cone Hamiltonian thus gives the same Klein-Gordon equation as covariant quantization.
For the massive two-form field, take . Apply to its field equation. Antisymmetry of makes , soExpanding , the other divergence terms vanish by this condition, leaving . The light-cone decomposition of a massive two-form makes its dependent components explicit. The divergence equation isTaking and , respectively, givesThe equation follows from these expressions: the two terms containing cancel and . Consequently and are independent, each satisfying the Klein-Gordon equation with mass . The number of independent particle polarizations isThis is the exterior square of the vector representation of the massive little group . In the analogous Proca equation, determines from and , leaving components. A massive field has no gauge freedom that would justify setting these longitudinal components to zero. If , instead use the two-form gauge field symmetry : the light-cone gauge for a two-form removes , leaving transverse particle polarizations. The massive and massless counts are different.
In the closed-string mode expansion, are center-of-mass canonical variables, while are independent left- and right-moving transverse string oscillators. Their complex conjugates are . The two zero-mode Lagrange multipliers impose the remaining mass-shell condition and closed-string level matching. The string level operators areTheir quantum definitions use normal ordering. The symplectic terms in the phase-space action givewith all brackets between distinct sectors zero. The nonzero-index string oscillators obey and similarly for the right-moving sector. Define the momentum-labelled oscillator vacuum byFor , has . HenceStarting with , a finite product with creation operators of mode has eigenvalue . The Fock space is generated by these products; both level operators have nonnegative integer eigenvalues. This establishes the integer string oscillator level property. Subtracting their physical zero-mode constraints enforces .
There is a distinction between the displayed classical zero modes and their quantum constraints. With the normal-ordering constant of a string , these areAt the massless first closed-string level, the states areTheir transverse polarization tensor splits into a symmetric trace-free part, an antisymmetric part, and its trace. These are the graviton, Kalb–Ramond field, and dilaton, with respective particle polarization counts , , and one. They have the transverse little group representations of massless particles. In a Lorentz-consistent bosonic string theory, the first chiral level is a massless vector, not a massive vector with one missing physical polarization; the closed-string products are therefore massless. This fixes . Equivalently, regularized transverse zero-point energy gives , and Lorentz consistency fixes the critical dimension of the bosonic string .
It follows that the bosonic string mass spectrum isThe ground state has and is a tachyon; level one is massless; for the mass is . The masslessness claim uses the consistent quantum theory, rather than an unshifted reading of the classical .
A massive two-form at closed-string level two is present. To see it without confusing it with the level-one massless Kalb–Ramond field, the level-two states in one chiral sector areThey have components and assemble into the symmetric traceless square of the massive little group vector space . The full closed-string level is . For two symmetric trace-free matrices , the mapis an equivariant map onto antisymmetric matrices. To verify surjectivity, take diagonal with distinct entries in positions and with only its symmetric entry nonzero. Their commutator gives the antisymmetric basis element. Finite-dimensional representations of the compact little group are completely reducible, so this quotient representation is also a subrepresentation. It has exactly particle polarizations and is described by the massive field equation with . At this gives 300 particle polarizations, consisting in light-cone coordinates of 24 components and 276 components .
Past exam of the mathematics course of the University of Cambridge 2014 iii Paper 46 4 Solution Created 2026-10-03 Updated 2026-10-06
Take an oriented closed string in flat, critical bosonic string theory, with and a mostly-plus target Minkowski metric. After continuation of the worldsheet to Euclidean signature, the Polyakov path integral sums over embeddings and worldsheet metrics, divided by worldsheet diffeomorphisms and Weyl transformations. In a flat target its kinetic action isTarget-time continuation or analytic continuation of external momenta defines the Lorentzian scattering amplitude; a naive real Euclidean Gaussian for timelike would not be convergent.
An external tachyon is represented by the tachyon vertex operator , with conformal weights . The physical integrated vertex has , so . Schematically its tree amplitude isThe Faddeev-Popov determinant from conformal gauge is represented by the worldsheet ghost fields. At tree level the worldsheet is a Riemann sphere, whose unpunctured complex structure has no moduli. Its residual conformal automorphisms are the Möbius transformations, . Fix three insertion points, accompanying their unintegrated vertices by the required ghost factors. The remaining complex insertion positions are integrated over the sphere; equivalently one integrates all positions and divides by the residual conformal group.
The embedding fields are free, withTheir zero-mode integral gives momentum conservation, and their nonzero-mode Gaussian integral gives the Koba-Nielsen factorThis explains the sphere tachyon position integral without needing its evaluation. For four tachyons the resulting Virasoro–Shapiro amplitude displays the exchanged string spectrum directly.
Use all-incoming external momenta and introduce alongside the two printed dimensionless Mandelstam variables. Since and , one obtains . The physical center-of-mass energy squared in the channel is . The channel measures the analogous crossed momentum transfer, with sign determined by the mostly-plus convention. Rewriting the Gamma factors in a symmetric form givesThe Gamma function poles imply, at generic fixed values of the other invariant,The denominator Gamma factors can remove residues at special intersecting channel kinematics; the statement concerns a generic single-channel limit. An -channel pole occurs when the intermediate momentum satisfies the mass-shell condition for a closed-string state:The pole at exchanges the ground-state tachyon, the pole at exchanges massless states, and the positive integer poles exchange the infinite massive tower. The -channel interpretation is the crossed version. Factorization means that each residue is a sum of products of couplings to intermediate physical states that couple to the chosen external particles. It need not expose every representation at that mass.
For example, the Gamma function recurrence and Gamma function residue at a nonpositive integer give the dimensionless Virasoro–Shapiro amplitude pole residueThe residue polynomial has degree , consistent with maximum spin in the exchanged level. The amplitude also has the corresponding -channel poles by crossing symmetry.
The massless fields can be treated as target backgrounds rather than separate asymptotic insertions. Write the target metric as , introduce a Kalb–Ramond field , and a dilaton . In conventional Euclidean signs their string nonlinear sigma model action isHere is the antisymmetric tensor density. The three backgrounds correspond to the graviton, antisymmetric tensor and scalar states at closed-string level one. Expanding the vacuum functional in , then Fourier expanding the backgrounds, produces exactly their integrated string vertex operators. Its functional derivatives therefore generate the amplitudes with massless external strings. The connected vacuum functional organizes connected amplitudes; the spacetime effective action organizes the corresponding vertices after treating massless propagation consistently.
At momenta small compared with , massive string propagators can be expanded in powers of momenta over their masses. The analytic part of the amplitudes consequently determines local higher-derivative interactions, ordered by powers of . Massless exchange poles are retained through propagation of the massless fields, rather than expanded into local contact terms. Up to field redefinitions, the leading massless-sector action in the string-frame metric isThe terms denoted contain additional derivatives, including curvature-squared terms in the bosonic theory. The expansion concerns the massless sector around the perturbative bosonic background; the tachyon instability remains and is not cured by omitting its field from this displayed action. Thus this is a formal perturbative effective description, not a claim of a stable bosonic vacuum.
Finally split the dilaton into a constant and its variation, . By the Gauss-Bonnet theorem, its dilaton Euler-characteristic weighting on a connected closed oriented surface follows fromDefine the string coupling by . A genus- path integral is weighted byThe sphere carries , the torus carries , and each extra handle adds . At higher genus one integrates over complex-structure moduli as well as insertion points, with the associated antighost insertions supplying the correct moduli measure. With canonically normalized external vertices an -point genus- amplitude scales as .
Summing connected worldsheets of every genus of a surface yields the string-loop effective action expansionEach coefficient has its own low-energy expansion. The two parameters have different roles: resolves finite string size through higher derivatives, while counts additional string loops.
Past exam of the mathematics course of the University of Cambridge 2014 iii Paper 47 1 Solution Created 2026-10-03 Updated 2026-10-06
Use the Minkowski metric and the angular field . The Euler-Lagrange equation becomes . For the dimensionless light-cone coordinatesthis is . This normalization keeps the coupling in the relation between the physical field and its angle, rather than silently setting .
A Bäcklund transformation is a system of first-order differential relations that maps a solution to another solution. One convention for the Sine-Gordon Bäcklund transformation uses a nonzero parameter and defines byA compatible initial value or integration constant selects a particular transformed solution. Put and . Differentiating gives and . Their sum and difference yieldThus compatibility of the first-order relations contains the field equations for both fields, and the transformed physical field also solves Sine-Gordon theory.
The transformation provides a generating conservation law. On the branch close to the original field write . Its first equation isIt recursively determines a formal small- expansion of in local derivatives of :At every subsequent order the new coefficient occurs linearly, so this recursion continues indefinitely. The second Bäcklund equation and the first imply the exact identityIndeed and . Comparing powers of therefore gives local conserved currents. If , our coordinate convention givesFor localized fields approaching vacua at spatial infinity, the boundary flux vanishes and is conserved. The formal expansion need not converge: each coefficient is a separately exact local conservation law.
For example , , a light-cone combination of energy and momentum. The next coefficient is a derivative improvement, so it contributes no independent charge under the same decay conditions. At order , remove the improvement generated by and multiply by . One obtains the genuinely higher conservation lawIt can also be checked directly using . Continuing the recursion, and using the opposite light-cone construction, produces the local conserved-charge hierarchy of sine-Gordon theory, with infinitely many nontrivial higher-spin charges after derivative improvements are removed. This is the Bäcklund generating current for sine-Gordon conserved charges. The hierarchy is the characteristic field-theory form of classical integrability; an ordinary energy conservation law alone would not supply these constraints.
The same transformation constructs solutions rather than only currents. Starting from the vacuum , its two first-order equations integrate toFor the exponent is . This is a Sine-Gordon kink with velocity , center set by and classical rest mass . Negative parameters can supply the corresponding opposite-orientation seeds.
The allowed Bianchi permutability for sine-Gordon Bäcklund transformations gives the two-step field algebraically. For vacuum seed and ,Branches of the inverse tangent must be continued smoothly; its principal value alone does not specify the vacuum labels of a multi-kink field. All angles here are . In the physical-field version each field difference in the superposition formula carries ; the formula printed without it implicitly uses the angular-field convention.
To display two real scattering solutions, take , set , and put , . Choosing , with and zero phase constants yields the Sine-Gordon kink-antikink scattering solutionChoosing instead yields the Sine-Gordon two-kink solutionThe latter has net angular winding , while the former has zero net winding. At large positive or negative time they separate into localized kinks with velocities . Repeated commuting transformations give general multi-soliton fields. Continuing the kink-antikink velocity to an imaginary value also gives a real Sine-Gordon breather:up to the irrelevant overall field sign and translations.
The scattering interpretation ties the construction to the conserved hierarchy. In an exact multi-soliton sector, the incoming species and rapidities reappear after the collision, up to permutation: the solitons change positions, not their asymptotic shapes or velocities, and the collision emits no radiation. In the two-kink example the large-time centers obey , so a right-moving trajectory acquires a shift . These shifts encode the interaction even though the collision is elastic. The commuting construction makes the net displacement in a many-soliton collision the sum of its pairwise displacements, independent of how the collisions are ordered; this is pairwise additivity of soliton shifts and the classical counterpart of factorized scattering.
Conservation of the entire hierarchy is much stronger than conservation of energy and momentum: its independent rapidity-weighted sums constrain the whole asymptotic soliton data and rule out particle production in this sector. Generic initial fields may also contain radiation scattering data, so this statement is about the exact soliton collisions, not a claim that every initial field is a pure soliton. The Bäcklund map unifies the conserved hierarchy, explicit soliton construction and elastic, pairwise scattering picture of classical integrability.
Past exam of the mathematics course of the University of Cambridge 2014 iii Paper 50 4 b i Solution Created 2026-10-03 Updated 2026-10-06
Use and . The linearized Ricci tensor and scalar obeyThe background Ricci tensor and Ricci scalar vanish, so varying the metric in the scalar-curvature term contributes nothing at first order beyond this trace. The Einstein tensor perturbation is consequentlyAll contractions use the Minkowski metric. Direct differentiation gives , the linearized contracted Bianchi identity, which checks the relative signs. No time derivatives or gauge-dependent terms have been discarded.
Past exam of the mathematics course of the University of Cambridge 2014 iii Paper 52 1 b Solution Created 2026-10-03 Updated 2026-10-06
Start with the four-dimensional Minkowski metric , where . Choose an arbitrary length , and use the retarded and advanced null coordinates , . For the Minkowski conformal compactification, setBoth lie between and . Since , we have ; the remaining inequalities are . Moreover,Multiply by the square of the conformal factor . The resulting metric is regular on the appropriate boundary pieces and preserves the null directions. Suppressing the angular two-spheres gives a triangular Penrose diagram with radial null geodesics at degrees.
The line is the ordinary timelike centre . The upper sloping edge is future null infinity, reached with and finite ; the lower sloping edge is past null infinity, reached with and finite . The vertices and are future and past timelike infinity, denoted and . The vertex is spacelike infinity, . These are limiting endpoints in the conformal completion, rather than ordinary physical events. In particular, finite diagram coordinates at null infinity do not imply finite physical affine parameter.
The four-dimensional radial diagram is the triangle , . If one instead draws two-dimensional Minkowski spacetime with a signed Cartesian spatial coordinate, the diagram is the full diamond. The centre is a boundary of the radial quotient, not a boundary of the physical four-dimensional Minkowski spacetime.
Kruskal extension and the radial Minkowski Penrose diagram
. Past exam of the mathematics course of the University of Cambridge 2015 ia Paper 4 12C Solution Created 2026-09-24 Updated 2026-10-06
For a Lorentz transformation to an inertial frame moving at speed in the positive direction, write and . Any four-vector transforms asThis convention applies to and uses the same Minkowski metric as the four-momentum calculation.
The definition of proper time gives for . Hence the four-velocity isSince , differentiating with respect to proper time givesHere simplifies the spatial component of the four-acceleration.
At each event, boost with to the instantaneous rest inertial frame. Then and . Thus the signed proper acceleration for one-dimensional motion isThe last equality follows by differentiating ; its magnitude is the usual scalar proper acceleration.
If this signed proper acceleration is constant and the particle starts from rest, integration gives , soIntegrating once more and imposing yields the hyperbolic motion worldlineFor the continuous limiting solution is . Finally, the Taylor series of the square root givesThis recovers constant-acceleration Newtonian mechanics at low speed while the exact relativistic speed remains less than .
Past exam of the mathematics course of the University of Cambridge 2015 ia Paper 4 4C Solution Created 2026-09-24 Updated 2026-10-06
Use the Minkowski metric convention . A particle's four-momentum is , with for rest mass . For a photon, its energy and wavelength obey , and its four-momentum is null. Here is the Planck constant.
The initially stationary electron has . Choose the scattering plane as the plane. The incoming and outgoing photon four-momenta areBy four-momentum conservation, the final electron has . Its rest mass is unchanged, so the relativistic energy-momentum relation gives . Since , expansion yieldsorMultiplying through gives the Compton scattering shift:It is nonnegative and vanishes for forward scattering, consistent with the recoil energy being supplied by the photon.
Past exam of the mathematics course of the University of Cambridge 2015 iii Paper 43 1 Solution Created 2026-10-03 Updated 2026-10-06
Use natural units and the Minkowski metric . A real scalar field assigns a real variable to each spatial point. Its Lagrangian density can be taken to beThe principle of stationary action gives the Euler-Lagrange equation . With , this is the Klein-Gordon equation. An additional nonlinear part of describes interactions.
The canonical momentum is . The Legendre transform in mechanics gives the canonical Hamiltonian density of a real scalar fieldThe Hamiltonian equations and recover the same field equation. In canonical quantization, the fields become operators satisfying the equal-time canonical commutation relationsA spatial lattice makes the analogy with many coupled quantum-mechanical coordinates precise. Each lattice field value is a coordinate, with its own conjugate momentum. The path integral is another representation of the same quantum evolution.
To see its origin, first consider one coordinate with . Split a time interval into steps of length and insert position and momentum resolutions of the identity. The short-time kernel isMultiplying the kernels and integrating over intermediate positions gives the phase-space path integralThe endpoints of are fixed. The momentum integrals are Gaussian integrals; completing the square produces the configuration-space path integralAt finite slicing its normalization contains . This fixes the composition law and the initial delta-function kernel. One sums over all paths, not merely solutions of the classical equation. Restoring replaces the weight by ; stationary phase explains the emergence of classical trajectories.
For the field, use scalar field configuration eigenstates , satisfying . Insert their completeness relations on every time slice. This givesThe endpoint field configurations are fixed. Integrating the Gaussian momentum variables leaves the scalar field path integral . The functional measure means a regulated product over the field variables. A spacetime lattice or another ultraviolet cutoff makes this product finite before the continuum limit; interacting continuum calculations may require renormalization. The oscillatory Minkowski weight is an amplitude, not a positive probability density.
For vacuum expectation values, the boundaries must select the vacuum rather than arbitrary field configurations. Long imaginary-time evolution suppresses excited states: , so after normalization only the lowest-energy component remains as . This is vacuum projection by imaginary time. The corresponding Feynman i-epsilon prescription in the real-time integral specifies the vacuum boundary conditions and the poles of the propagator. With , the Euclidean path integral has the weight , whereIt is often a useful regulated starting point; analytic continuation returns the vacuum time-ordered quantities.
Introduce a classical source and define the normalized vacuum generating functionalwith the same vacuum prescription in numerator and denominator. A functional derivative brings down . The order of the time slices makes the operator insertion time-ordered. Thus source differentiation inserts time-ordered field operators:The denominator removes vacuum diagrams and gives normalized expectation values. It is essential that these are time-ordered products; differentiating this vacuum functional does not directly give every possible operator ordering.
The free theory illustrates the method. Its quadratic kernel is with the vacuum pole prescription, and completing the square gives the Gaussian evaluation of a free scalar generating functionalTwo source derivatives give . Higher derivatives give all pairings, the content of Wick theorem. For an interaction , one may use path-integral perturbation by source derivatives:Expanding this expression generates Feynman diagrams and their Wick contractions. The connected generating functional retains connected contributions; in particular . These functionals turn the computation of field-operator expectations into source differentiation of an ordinary regulated integral.
Past exam of the mathematics course of the University of Cambridge 2015 ii Paper 4 33A Solution Created 2026-09-24 Updated 2026-10-06
Use the Minkowski metric with signature . The source point is the intersection of the worldline with the observation event's past light cone: and . In components this is with . For a timelike worldline the retarded solution is unique in the usual complete-source setting.
Writing , and gives . Thus the Lienard-Wiechert potentials are and . At leading order in and first order in , set and expand the denominator:
Now use and let . Derivatives of and give near-field terms of order ; the radiation terms come from and . The electric field and magnetic field are thereforeThey are transverse to the observation direction. The Poynting vector isIntegrating over a sphere uses , yielding the nonrelativistic Larmor formula . For simple harmonic motion, , whose period average is . Multiplication by the period givesThis is the leading radiation energy in the nonrelativistic and far-field approximation.
Past exam of the mathematics course of the University of Cambridge 2016 ib Paper 2 18D a Solution Created 2026-09-24 Updated 2026-10-06
Use SI units and the Minkowski metric . Write , and the four-current , with Greek indices from to . The electromagnetic field tensor is the antisymmetric tensor whose components areIndices are raised and lowered with . Define the dual electromagnetic field tensor by , with . Then the covariant Maxwell equations areHere and are charge and current density, is the vacuum magnetic permeability, and . For example the equation gives , and the spatial equations give . With the stated dual convention, and ; the second tensor equation gives the two homogeneous Maxwell equations.
Past exam of the mathematics course of the University of Cambridge 2016 iii Paper 301 1 Solution Created 2026-10-03 Updated 2026-10-06
Use units , the Minkowski metric , and the Fourier transform convention . Here gamma matrices satisfy , and is the Dirac adjoint. A single Dirac field describes both Electrons and Positrons; these are the particle and antiparticle sectors of that field.
The free Maxwell Lagrangian and Dirac action, with a linear covariant gauge condition for the photon, giveBefore gauge fixing, the Maxwell Lagrangian has zero modes in field theory along , so its quadratic operator cannot be inverted on all potentials. In the linear Lorenz gauge, the Faddeev-Popov determinant is . It is independent of and may be absorbed into the normalization; the corresponding Faddeev-Popov ghost fields have no interacting vertices in this Abelian linear gauge. A residual gauge symmetry is removed by the specified boundary conditions.
The free generating functional isThe Dirac field variables and their sources are independent Grassmann fields in this path integral. They anticommute; replacing them by ordinary commuting fields would give the wrong statistics and the wrong functional determinant. The bosonic Gaussian functional integral contributes an inverse square root of a determinant, and the Grassmann Gaussian integral contributes a determinant. If is the photon quadratic operator and , completing the square givesIn the last expression and are the photon propagator and Dirac propagator, with Feynman i-epsilon prescription. The products include the appropriate spacetime integrals and index contractions. Functional derivatives with respect to , and consistently ordered left or right Grassmann derivatives with respect to the fermionic sources, generate the time ordering of the corresponding fields. In Feynman gauge, , the momentum-space two-point functions areThe Feynman i-epsilon prescription specifies vacuum boundary conditions rather than an arbitrary inverse of the differential operator.
The covariant photon propagator uses four potential components. Its operator formalism counterpart is Gupta-Bleuler quantization: impose and take the Gupta-Bleuler null-state quotient. This leaves a positive physical state space with two transverse photon polarization vectors. The temporal and longitudinal oscillator components occur in intermediate covariant expressions; the Ward identity removes their dependence from physical amplitudes. The free Dirac field uses the canonical anticommutation relations, producing the same fermionic signs as its Grassmann fields in the path integral.
The operator-path-integral equivalence can be seen directly with a regulator. Divide time into small intervals and insert complete sets of field-coordinate states for bosons, and resolutions in fermionic coherent states for fermions. The bosonic matrix elements produce the phase-space factor ; integrating out the quadratic canonical momentum produces the bosonic action. The fermionic coherent state overlaps produce the first-order term and the Berezin integral measure. Multiplying the short-time kernels recovers the path integral. Projecting the remote endpoints onto the Fock vacuum with an infinitesimal damping selects the same Feynman propagators as the operator formalism. Field insertions become time-ordered products under this construction. Conversely, their quadratic generating functional obeys the canonical free-field equations and has precisely the oscillator two-point functions, so its higher free correlators agree by the Wick theorem. This establishes the equivalence for the regulated free theory and order by order in the perturbation series.
To couple the matter field electromagnetically, promote its global phase symmetry to the local transformationThe gauge covariant derivative obeys . Replacing by in the Dirac action therefore gives the invariant matter densityThe electromagnetic field tensor is unchanged by the local transformation. Thus the unfixed quantum electrodynamics action is gauge-invariant. The Dirac current is conserved by the matter equations, and the Electron and Positron excitations carry opposite charges. The added gauge fixing density selects a representative and is not itself invariant under arbitrary local transformations; it does not change gauge-invariant observables. At a free fermion vertex,For external on-shell Dirac spinors the corresponding current contraction vanishes. The quantum extension is the Ward identity, which makes physical amplitudes insensitive to adding a multiple of the photon momentum to its polarization vector. A gauge-compatible regularization preserves this vector-current identity.
With , expand in powers of . A term of order contains spacetime integrations and . Applying the Wick theorem pairs the free fields: an - Wick contraction supplies a photon propagator, and a - Wick contraction supplies an oriented Dirac propagator. Each insertion supplies an interaction vertex. The permutations of contractions cancel the expansion factorials except for the Feynman-diagram symmetry factor. Interchanging Grassmann fields produces the fermionic sign, including a minus sign for every closed fermion loop. This is how Feynman diagrams arise from expectation values, rather than an extra dynamical assumption.
The normalized vacuum generating functional removes components with no external insertions. In the operator formalism, for an interacting-vacuum expectation value of an inserted product , the same cancellation appears asVacuum projection and the Feynman i-epsilon prescription are implicit. The linked-cluster theorem exponentiates all connected vacuum bubbles into the same factor in numerator and denominator, so it cancels. Normalize by to remove every vacuum component. This cancellation of vacuum bubbles still leaves products of disconnected diagrams that each contain external insertions. If only connected correlators are wanted, differentiate the connected generating functional .
The resulting momentum-space QED Feynman rules for the bare theory can be stated in Feynman gauge as follows:
- An internal oriented fermion line of four-momentum contributes .
- An internal photon line contributes .
- A one-photon two-fermion interaction vertex contributes , with its spinor and photon indices attached to the incident lines.
- Each interaction vertex conserves four-momentum. With all incident momenta treated as incoming, include ; after using these constraints, integrate every independent loop four-momentum as .
- Contract the gamma matrix and Dirac propagator factors in their order along the fermion line. A closed fermion loop has a trace and a factor . A relative odd permutation of external fermions also contributes .
- Divide each labelled diagram by its Feynman-diagram symmetry factor and sum the allowed diagrams. Omit vacuum components as explained above.
- For an amputated scattering amplitude, an incoming Electron has and an outgoing Electron has ; an incoming Positron has and an outgoing Positron has . An incoming photon has and an outgoing photon has . The external polarization vectors are physical and transverse, and external four-momenta are on shell. This last step follows from the LSZ reduction formula and amputation of external propagators; an unamputated correlator keeps its external quantum field theory propagators.
The illustrated interaction vertex has incoming fermion momentum , incoming photon momentum , and outgoing fermion momentum . Its solid-line arrows indicate fermion flow. The photon wavy line has no fermion-flow arrow. The propagators, the vertex , momentum conservation, and the fermionic signs determine the perturbative amplitudes.
Past exam of the mathematics course of the University of Cambridge 2016 iii Paper 301 2 Solution Created 2026-10-03 Updated 2026-10-06
For this question use the mostly-plus Dirac convention: the Minkowski metric is andThis convention matches the printed plane-wave phase and final identity. It is related to the usual mostly-minus Dirac equation by reversing the metric and taking the negatives of the usual gamma matrices. In particular, the resulting Dirac action and its Dirac adjoint describe the same physical massive field. The Dirac gamma matrices are four complex matrices representing the spacetime Clifford algebra with quadratic form . The irreducible complex representation has dimension four. A convenient explicit choice is the negative of the standard Dirac representation of the gamma matrices:where are the Pauli matrices. Their multiplication law verifies the displayed anticommutator.
In this representation the gamma matrix adjoint and transpose identities areandThe invariant way to express the latter pattern uses the charge-conjugation matrix:Individual transpose signs depend on the basis. More generally, a similarity transformation changes the Hermitizing matrix to and the charge-conjugation matrix to . Then and . Thus the simple formula with itself presumes a compatible Hermitian basis, rather than an arbitrary nonunitary similarity transformation.
Applying to the Dirac equation gives . Its mass shell is , so the frequencies are . The Dirac spinor transforms in the four-component Spinor representation of the Lorentz group. Under spatial rotations, the two upper and the two lower components each transform as a two-component spin- representation: the spin angular momentum matrices are . At rest the positive-energy equation selects the upper two components, giving two independent spin polarizations, and the negative-frequency equation selects the lower two.
In the quantum theory a mode expansion isWith the mostly-plus Dirac convention, is positive frequency. The negative-frequency coefficient obeys . The fermionic annihilation operators and satisfy the canonical anticommutation relations, with their respective fermionic creation operators. The excitations are particles; the excitations are antiparticles with the same positive energy, mass and spin- but opposite charge. After normal ordering, the Hamiltonian operator contains positive multiples of . Reinterpreting the negative-frequency part as antiparticle creation supplies a spectrum bounded below rather than a physical tower of negative-energy particles.
For the printed wave, . Substitution givesThe spin label indexes the two states of a spin- particle, rather than varying the particle's total spin. For real on-shell , Hermitian conjugation and giveThese are right and left null-vector equations for the same on-shell matrix.
To obtain the Gordon identity, take both external Dirac spinors to have the same real mass . Their two equations implyDefine . The Clifford algebra relation yieldsThereforeMultiplying the previous null-vector equation by proves the required formula exactly:For it can be solved for the vector-current matrix element, separating a momentum term from the antisymmetric Dirac spinor term. The identity before division also holds at . The signs depend jointly on the metric, Clifford relation, Dirac mass term and plane-wave phase. In the mostly-minus convention of Questions 1 and 3, the printed phase instead gives ; the corresponding identity uses in the antisymmetric term. Mixing that convention with the formula proved here would produce an apparent sign error.
Past exam of the mathematics course of the University of Cambridge 2016 iii Paper 301 4 Solution Created 2026-10-03 Updated 2026-10-06
Return to the mostly-minus Minkowski metric of Question 3, and write . The field is assumed real and nonzero wherever its inverse occurs. Under the usual Abelian gauge transformation , with inert, the electromagnetic field tensor is invariant but . The change in the added density isThis is not generally a total derivative. The action is not invariant under arbitrary gauge transformations. It retains residual Lorenz gauge symmetry for transformations satisfying , with the same boundary conditions imposed before and after the transformation. The action itself is undefined at ; that value can only be considered as a limiting gauge.
To vary the electromagnetic four-potential, use and . The integration by parts of both terms giveswith surface terms removed by the variational boundary conditions. HenceFor variable , the derivative must act on as well as on :Since enters algebraically, its Euler-Lagrange equation isFor a real field this gives . Thus a dynamical gauge-fixing parameter imposes a constraint equation in field theory; it is not an ordinary propagating scalar. On configurations satisfying both equations, , , and therefore . There is no independent kinetic equation that determines .
For the momentum-space equation at a prescribed constant , use . Then , and the linear equation isEquivalently, . If the separately varied equation is also imposed on a real classical solution, then and its nonzero modes lie on the massless mass shell. For constructing a full-field Green function, however, invert the fixed-background quadratic operator before imposing that on-shell constraint. Holding fixed and integrating over is a Gaussian functional integral; integrating over the original variable as well is a different constrained problem.
For , the Lorentzian versions of the transverse projector of a vector field and longitudinal projector of a vector field areThey satisfy , , and . The mixed-index operator isIts inverse follows by inverting these two scalar eigenvalues. Define by . The positive-sign covariant gauge-fixing inverse isThis is the algebraic inverse of the kinetic operator. A vacuum photon propagator instead has the factor and a Feynman i-epsilon prescription for its simple and double poles. If the name is used for the vacuum two-point function, include that factor consistently in its defining source equation. A retarded Green function would use different boundary conditions at the same poles. The original plus sign corresponds to the usual covariant gauge parameter ; in particular, gives Feynman gauge and .
Contracting gives the longitudinal gauge propagator contractionThus the contraction is not identically zero as a function of four-momentum for any , and it is a nonzero vector at every nonzero non-null four-momentum. A particular component may vanish when . At , the displayed inverse has poles and must be interpreted using the chosen Green function prescription, rather than as a pointwise finite matrix. In the limiting Landau gauge, , the longitudinal part vanishes. A covariant photon Green function may have a longitudinal component even though physical photon polarizations are transverse.
Past exam of the mathematics course of the University of Cambridge 2016 iii Paper 306 2 i Solution Created 2026-10-03 Updated 2026-10-06
Use the mostly-plus Minkowski metric, with the last spatial coordinate singled out:Then , as required. In these light-cone coordinates, , and .
Introduce the Kalb-Ramond field strengthThe field equation is . Under , every second-derivative contribution to cancels by commutation of derivatives. Thus and its equation are gauge-invariant. This is the differential-form identity for the two-form gauge field .
To impose light-cone gauge for a two-form, first choose andThis sets to zero. A residual transformation preserving the gauge obeys . With , this implies , including . Such a parameter changes by zero. Therefore no nontrivial gauge transformation of remains in the sector where is invertible. There is still a redundant description of the gauge parameter itself, ; the excluded zero modes would require separate treatment.
The field equations now say , hence . For this givesFor , follows automatically from antisymmetry of . Therefore the independent components and their equation areThe remaining equations follow from this wave equation and the reconstructed longitudinal components.
At the first massless level of the closed bosonic string, states carry a product of two transverse vector polarizations. Its symmetric traceless, antisymmetric and trace parts are respectively the graviton, the Kalb–Ramond field and the dilaton. The antisymmetric part has exactly the polarization count just found.
The gauge-invariant string coupling to a two-form uses the pullback of the background Kalb–Ramond field to the string worldsheet:Its variation is , by Stokes theorem. It vanishes on a closed worldsheet. For a cylindrical propagation surface it is only an initial/final boundary term, handled by fixed-boundary gauge parameters or the corresponding transformation of external states. Thus the closed string is naturally charged under a two-form gauge field.
Past exam of the mathematics course of the University of Cambridge 2016 iii Paper 308 1 Solution Created 2026-10-03 Updated 2026-10-06
Use the Minkowski metric with signature , and write . The Euler-Lagrange equation isFor the static phi-four kink, and , so the field equation is satisfied. The centre is arbitrary by translation invariance, and the hyperbolic tangent profile increases monotonically from to , crossing zero at . The static kink and its endpoint topological charge are
The phi-four kink rises between the two vacuum values and crosses zero at its centre
. Both endpoint values are isolated classical vacua, since only at . A continuous finite-energy deformation preserving the vacuum boundary conditions cannot change either endpoint to the other isolated classical vacuum. The topological charge is therefore unchanged: this kink cannot deform into a homogeneous classical vacuum, whose charge is zero. There is also a direct Bogomolny bound in this sector. The square completion for a one-dimensional kink givesThe phi-four kink saturates the bound, so its mass is in these units and it minimizes the energy within its topological sector. Its arbitrary position is a collective coordinate, not an instability. A kink and an antikink together have total charge zero and can annihilate without contradicting the protection of an isolated kink.
For the momentum, the canonical stress-energy tensor of this scalar field isConsequently the physical spatial momentum density and the spatial momentum flux areThe sign of makes a right-moving translated kink carry positive momentum. Direct use of the field equation, rather than an assumed static field, yields the scalar-field momentum flux identityThe finite-energy field configuration has by . Integrating the stress-energy conservation law over the left half-line gives the boundary forceUnder the usual vacuum falloff, the stress at the left endpoint is zero. More generally, smooth spatial cutoffs with derivative of order remove the left endpoint using the integrable energy density, so no pointwise limit of every derivative at infinity is needed. The identity expresses the force on the field to the left of : positive force transfers momentum to the right. For well separated solitons, a cut between them measures the interaction force on the left soliton.
Take that cut at . The specified symmetric pair has, at the initial time,The printed field profile does not itself specify the initial velocity. If , the exact initial half-line force isFor the intended initially resting pair, or more generally , put and use . The at-rest force for a symmetric phi-four pair isThe leading force is attractive, towards the antikink:The antikink feels the opposite force by the symmetry of the resting pair. This is an initial, large-separation interaction calculation, not a claim that the superposed profile is an exact static two-soliton solution. Without the initial-velocity condition, the additional momentum flux above prevents a unique force from being inferred from the printed profile alone.
Past exam of the mathematics course of the University of Cambridge 2016 iii Paper 309 4 ii Solution Created 2026-10-03 Updated 2026-10-06
Let . Substitution into the linearized Levi-Civita connection cancels four terms and leavesTo first order in the metric perturbation, the products of Christoffel symbols are quadratic and may be discarded. Therefore the change in the Riemann curvature tensor isThis is the gauge invariance of the linearized Riemann tensor. Lowering the first index with the Minkowski metric does not change the conclusion; the change from using the full metric tensor instead is of second order because the background curvature is zero. HenceThe result is invariance at linear order about flat spacetime, not componentwise invariance of arbitrary nonlinear curvature under finite coordinate changes. Directly in the four-term formula, the eight third derivatives of cancel in pairs by commuting partial derivatives.
Past exam of the mathematics course of the University of Cambridge 2017 ib Paper 2 18C Solution Created 2026-09-24 Updated 2026-10-05
Specify the Minkowski metric convention , with and . For the electromagnetic four-potential, . The electromagnetic field tensor isUsing and , its two component matrices areChanging metric/sign conventions changes the component convention, so stating it is necessary. Under a Lorentz transformation, the tensor law is . For the stated Lorentz boost, put and :Multiplication gives, for example, and . Reading off all entries proves Lorentz transformation of electromagnetic fields:For the wire, the lab line charge is zero and the current is . At points off the wire, write . Gauss's law and Ampère's law giveThe boost leaves unchanged, so the field transformations giveIts physical source is the relativistic charge density of counterstreaming beams. Their boosted number densities are and , by transforming each charge-current four-vector. Thus . A line charge has radial field ; using reproduces the boxed field. Oppositely moving populations contract differently under the boost, so neutrality in one frame does not imply neutrality in the other. The ideal wire's fields are singular on its axis.
Past exam of the mathematics course of the University of Cambridge 2017 iii Paper 301 1 a Solution Created 2026-10-03 Updated 2026-10-05
Use natural units and the Minkowski metric throughout. An Abelian gauge transformation is . For a sufficiently regular , commuting the partial derivatives gives , so the Maxwell Lagrangian is invariant pointwise.
Vary the action with respect to the electromagnetic four-potential, using variations of compact support or vanishing on the boundary. The antisymmetry of the electromagnetic field tensor givesAfter integration by parts, . The principle of stationary action therefore gives the source-free Maxwell equations,In the Lorenz gauge, , this becomes with . The PDF's name “Lorentz gauge” denotes the usual Lorenz gauge, named after Lorenz. Residual gauge transformations preserve it when ; this condition is not itself a complete removal of the gauge freedom.


