Past exam of the mathematics course of the University of Cambridge 2006 ii Paper 1 33A Solution Created 2026-09-24 Updated 2026-09-24
Past exam of the mathematics course of the University of Cambridge 2010 ii Paper 2 34B Solution Created 2026-09-24 Updated 2026-09-24
Past exam of the mathematics course of the University of Cambridge 2012 iii Paper 46 4 Solution Created 2026-10-03 Updated 2026-10-07
Take the muon mass to be , and use the quantum electrodynamics convention with . Let be the incoming muon and antimuon momenta and the outgoing photon momenta, so . External particles are on shell: and . The two tree-level Feynman diagrams are the two orders in which the photon legs attach to the muon line. There is no three-photon QED vertex, and therefore no single-photon annihilation diagram into two photons at this order.
The QED Feynman rules needed here are the vertex , the Dirac propagatoran incoming muon spinor , an incoming antimuon adjoint , and an outgoing photon polarization . At a vertex enforce four-momentum conservation. Spinor products are ordered along the fermion line; exchanging the external photons introduces no relative minus sign. Define and .
With the convention that a diagram contributes , the two muon-antimuon annihilation amplitudes areTheir denominators are respectively and , where , . A different consistent overall amplitude-phase convention has no physical effect.
A physical photon polarization vector obeys , and represents an equivalence class . For a real null momentum, adding preserves transversality. One can additionally choose transverse spatial vectors with and ; there are two independent physical polarizations. The equivalence class removes the unphysical longitudinal direction, rather than imposing four independent physical polarization states.
The invariance of the total amplitude is a two-photon fermion Ward identity. Replace by and put , . The external Dirac equations giveAway from propagator poles, ; the identity extends with the common Feynman prescription. ThereforeThe two terms in the bracket then give and , which cancel. Exchanging photon labels proves the second Ward identity. Thus the sum is unchanged under for either photon. Neither diagram is generally gauge-invariant separately. Physically, longitudinal pure-gauge polarization does not couple to the observable scattering amplitude; only the two transverse photon degrees of freedom contribute.
For Compton scattering write the incoming momenta as and outgoing momenta as , with . The outgoing muon contributes , the incoming photon contributes without conjugation, and the outgoing photon contributes . The two tree-level Compton amplitudes areThese are the lower two diagrams. Their intermediate muon momenta are and , with denominators and . The first attaches the incoming photon before the outgoing photon along fermion flow; the second reverses that attachment order. Both contribute at order in the amplitude.
Compton scattering has one incoming and one outgoing muon and one photon on each side; annihilation has an incoming particle-antiparticle pair and two outgoing photons. Accordingly the external spinors/polarizations and physical Mandelstam channels differ, although both arise from the same two-vertex QED matrix element. Crossing symmetry relates them by continuing and one outgoing photon momentum , with the corresponding external wavefunction replacements. The annihilation channel becomes the Compton channel and the other channel becomes its channel. Identical final photons require a factor in an annihilation phase-space integral over an otherwise double-counted full final-state region; there is no analogous final-state identical-particle factor for the muon-photon Compton final state. Gauge cancellation holds for the summed Compton amplitude as well.
Past exam of the mathematics course of the University of Cambridge 2013 iii Paper 41 3 Solution Created 2026-10-03 Updated 2026-10-07
The conjugate complex scalar field has the opposite charge. Thus the derivative on in the kinetic term must mean , while the derivative on is . This is the conjugate gauge covariant derivative. Applying the same plus-charge differential operator to both fields would not give the stated gauge-invariant theory. For example, take , and ; the literal plus-charge kinetic product changes from to . The opposite-charge interpretation removes this term.
For an arbitrary real function , takeThen and the conjugate derivative transforms oppositely. Also because mixed derivatives commute. The kinetic contraction and are unchanged, proving local U(1) gauge symmetry.
Expanding the covariant kinetic term identifies the interactions:The first term gives the scalar electrodynamics three-point vertex, and the second gives the seagull vertex. For scalar charge flow from an incoming particle of momentum to an outgoing one of momentum , the factors areThe factor two in the second rule comes from the two identical photon fields. Equivalently, with all momenta incoming, the three-point rule is for a leg of momentum and a leg of momentum . An antiparticle line has the opposite charge-flow sign. The vertices are shown with dashed scalar lines and wavy photon lines:
The one-photon scalar vertex and the two-photon seagull vertex in scalar electrodynamics, including their momentum-space factors
. For particle-antiparticle scattering, use incoming and outgoing , with particle momenta. At order , there are exactly two tree-level Feynman diagrams: -channel photon exchange and -channel annihilation. The seagull vertex has only two scalar legs and cannot alone provide four external scalar legs.
The leading t-channel photon exchange and s-channel annihilation diagrams for scalar particle-antiparticle scattering
. Set and . The external currents areAll are transverse to the corresponding exchanged momentum: for instance and . This is the on-shell scalar quantum electrodynamics Ward identity. With the vertex convention above,The relative sign comes from the opposite particle/antiparticle charge in the exchange diagram and the two equal annihilation-vertex signs.
To see how the Coulomb gauge expression becomes covariant, let be either pair, , and initially take . The displayed photon propagator givesCurrent conservation gives and the same identity for . Since , the temporal coefficient simplifies asThereforeThe common Feynman pole prescription is understood in these formulas. This Coulomb-gauge propagator between conserved currents identity proves equality of the physical amplitudes despite the noncovariant individual propagator components. It should be formed before taking limits such as : individual instantaneous and transverse terms can be undefined separately in that limit while their sum has a finite covariant limit away from a physical pole.
Substitution gives the final scattering amplitudeHere are the Mandelstam variables for the equal-mass external scalars. This is scalar particle-antiparticle tree scattering; overall external-state phases do not affect its relative channel sign.
Past exam of the mathematics course of the University of Cambridge 2013 iii Paper 45 3 Solution Created 2026-10-03 Updated 2026-10-07
Let , neglect , and average over the two spin states of each incoming particle. Only the single-photon channel is being considered. The spectral-density symbol is not defined in the question, so specify its normalization before calculating. A useful photon-field convention isHere is the product of final-particle measures , including the appropriate sums and symmetry factors. Lorentz invariance and current conservation give the transverse tensor form. In this convention is the hadronic photon spectral density, with mass dimension , rather than a density of a stationary time series.
The electron annihilation amplitude can be written , up to an overall phase. The leptonic tensor after the initial spin average isIt obeys and . In the center-of-momentum frame, the flux denominator in the hint is , since the relative speed is 2 and each beam energy is . Contracting the tensors gives the inclusive relativistic cross-sectionHere is the fine-structure constant. The in the initial spin average is essential because the hint's particles were spinless.
For the other common convention, define the hadronic electromagnetic current with the coupling omitted, , and write its inclusive tensor asEquivalently, for , one has . At leading order in the electromagnetic interaction, is proportional to . ThusIf the symbol is instead used for this dimensionless hadronic electromagnetic-current spectral density, the last formula applies with renamed . The normalization must not be silently switched between these formulas.
At well above the strong-coupling scale of Quantum chromodynamics, asymptotic freedom makes production over distances of order perturbative. The electromagnetic current initially creates a quark-antiquark pair. Subsequent strong interactions produce hadrons, but an inclusive sum over all hadronic final states is much less sensitive to this rearrangement than an exclusive channel. This is the regime in which quark-hadron duality motivates a leading parton model calculation, with radiative and power-suppressed corrections. It is not a pointwise theorem at individual resonances or near thresholds; the comparison is most reliable for sufficiently inclusive or suitably averaged high-energy observables. This argument remains restricted to photon exchange, even where additional electroweak channels could also contribute.
For one active massless quark flavor with charge , the tree-level scattering amplitude isThe two gamma-matrix traces, the initial spin average, and the final color charge sum yieldwhere and . The massless two-body Lorentz-invariant phase space then givesSumming the distinct final flavors, rather than interfering amplitudes for them, givesAccording to the permitted approximation in the paper, active flavors satisfy and are treated as massless; the others are omitted. This is the stipulated step approximation, not the exact pair-production threshold .
The hadronic R ratio is , relative to the massless muon-pair relativistic cross-section . If and active flavors have charges and , respectively,For , the usual sets of three, four, five and six active flavors give . In the two spectral conventions the same leading calculation gives
Past exam of the mathematics course of the University of Cambridge 2013 iii Paper 45 4 Solution Created 2026-10-03 Updated 2026-10-07
An effective field theory is a controlled description of specified low-energy degrees of freedom at a chosen accuracy. It does not require knowing all physics at arbitrarily short distances. If new particles or strong dynamics enter at a scale , processes with characteristic energy and momentum transfers can be described using the light fields and interactions consistent with their symmetries. In natural units, four-dimensional power counting in quantum field theory organizes a local Lagrangian asHere has mass dimension , the dimensionless are Wilson coefficients, and is a renormalization scale. In a relativistic vacuum with canonical fields, a typical insertion of a dimension- interaction contributes an additional power , multiplied by its couplings and any light mass ratios. Additional expansions, such as a loop expansion, must also be specified. Symmetry or on-shell identities can postpone a particular observable's first correction.
The theory is useful when its retained states and expansion parameters adequately describe the experiment. It ceases to be a reliable truncated local description near an omitted particle's production threshold, near a heavy propagator pole, or where the retained dynamics become too strongly coupled for an assumed perturbative expansion. A light particle cannot be removed merely because one wants fewer variables: its propagation can produce nonanalytic momentum dependence that must be represented by retained light fields. Heavy-field decoupling can shift renormalizable masses and couplings as well as generate suppressed interactions; those low-energy parameters must be measured or matched, not assumed unchanged. The ultraviolet cutoff is an organizational scale, while a calculation may use a regulator other than a hard momentum cutoff.
Construction starts by identifying the light particles, the hierarchy of scales, and the exact or approximate symmetries relevant to the problem. Form all allowed local field operators through the desired order in power counting in quantum field theory. Choose an operator basis in effective field theory: remove equivalent terms by integration by parts, algebraic identities and allowed field redefinitions. Operators proportional to lower-order equations of motion can be redundant operators for on-shell amplitudes, provided coefficients are consistently transformed. This reduces bookkeeping without imposing additional physics assumptions.
If an ultraviolet completion is known, determine the Wilson coefficients by matching in effective field theory: calculate low-energy amplitudes or appropriate correlation functions in both descriptions using the same infrared conventions, and adjust coefficients so they agree to the chosen order. Equivalently, integrating out a field performs its path integral while retaining the light fields as backgrounds. Heavy propagators have an analytic expansion below their singularities, producing a derivative expansion; heavy loops also generate local terms and logarithms in their coefficients. Without a specified ultraviolet completion, the coefficients are parameters to be constrained by data.
An effective field theory remains predictive even when it contains nonrenormalizable interactions. At each fixed order in energy and loops there are finitely many required coefficients and counterterms. Renormalization absorbs divergences into that order's allowed operators. The renormalization group evolves the Wilson coefficients between matching and measurement scales, compensating scale dependence in matrix elements and, when appropriate, resumming large logarithms. The truncation error in effective field theory is estimated from the first omitted orders, under a stated coupling-size assumption; it is separate from parameter uncertainty and cannot be inferred merely by writing down infinitely many terms.
A concrete example is heavy scalar exchange in effective field theory. Take a light real scalar field and a heavy real scalar field , withThe coupling has mass dimension one. The light-field symmetry is . For and , the displayed scalar potential is bounded below: completing the square in leaves a positive light quartic. Thus the example can be treated as a stable theory around , with weak enough couplings for the tree approximation.
At tree level the heavy equation of motion is . Substitute its solution back into the action, including both its quadratic and source terms, to obtainThis is a derivative expansion valid for small momentum transfers, not an exact local replacement near the pole. The first term gives in the convention. After integration by parts, the next term is ; it is a dimension-six local operator. Writing puts its coefficient in the usual form. These coefficients are a tree-level matching in effective field theory result.
One can directly check the matching through the on-shell scattering amplitude for . With the Mandelstam variables, the full theory has three heavy-exchange channels:For ,The effective field theory reproduces these terms in order: a shifted quartic, then local derivative interactions. Since on shell, the first derivative correction is a light-mass-dependent constant; it vanishes for . This illustrates why an operator basis in effective field theory can trade some derivative operators for mass-dependent or higher-field interactions using field redefinitions. For massless external particles, the first nonconstant correction in this four-point tree amplitude starts at the following order.
The example exhibits the central logic: keep the light field, encode virtual heavy exchange in matched local coefficients, and control the error by expanding in momentum divided by the heavy scale. No heavy particle is actually produced in the domain of the approximation. Near , the full propagator is resonant and the truncated effective field theory fails; retaining or adopting a different description is then necessary. Light loops are computed within the effective field theory, while higher-order matching supplies the corresponding heavy corrections.
Past exam of the mathematics course of the University of Cambridge 2014 iii Paper 40 2 iii Solution Created 2026-10-03 Updated 2026-10-06
For an unpolarized initial fermion, average over its two spin states and sum over the unobserved final spin:The initial scalar has only one spin state. One can evaluate this sum directly from normalized Dirac spinors, or use consistent fermion spin sums to express it as a trace. It is the squared sum of both tree scattering amplitudes, not the sum of their separate squares.
The relativistic scattering cross-section is obtained by integrating the Lorentz-invariant phase-space measure and dividing by the invariant flux factor:There is no identical-final-particle factor, because the outgoing scalar and fermion are distinct. All energies are positive, with and .
Equivalently, in the centre-of-momentum frame set . Integrating the energy delta function in the relativistic two-body phase space givesFor this elastic process , so integrate over the full solid angle. This supplies the requested prescription without evaluating the angular integral.
Past exam of the mathematics course of the University of Cambridge 2014 iii Paper 40 2 ii Solution Created 2026-10-03 Updated 2026-10-06
There are exactly two tree Feynman diagrams: the incident scalar can be absorbed before the final scalar is emitted, or after it. The internal fermion four-momenta are respectively and . A scalar exchange diagram would require an absent scalar self-interaction, so there is no additional tree channel.
Define the S-matrix convention . The two tree scattering amplitudes areUsing the Dirac propagator from the preceding Feynman rules,The two terms add with the same relative sign: neither diagram exchanges identical external fermions. Their interference must be retained when squaring the complete scattering amplitude. In the phase convention above, , which gives an equivalent simplified numerator. An overall phase depends on the S-matrix convention and does not change a relativistic scattering cross-section.
Past exam of the mathematics course of the University of Cambridge 2014 iii Paper 40 2 i Solution Created 2026-10-03 Updated 2026-10-06
Keep the mostly-plus Minkowski metric and the Fourier transform . Write , with . To fix the otherwise unspecified phase of and the Dirac adjoint, takewhere are ordinary mostly-minus gamma matrices. Then , , and the fermionic time-derivative term is . In these conventions the interaction with real is Hermitian: in mostly-minus notation it is . If one instead calls the square-one chirality matrix, its coefficient must be to represent the same interaction. These phase choices leave physical relativistic scattering cross-sections unchanged.
Expanding yields the following Feynman rules with relativistically normalized external states:
- A real scalar field line carrying four-momentum contributes .
- An oriented Dirac field line contributes the Dirac propagatorThis follows from .
- Each pseudoscalar Yukawa interaction vertex has one scalar leg, one incoming fermion arrow and one outgoing fermion arrow, and contributes . It also contributes with all vertex momenta counted incoming.
- External incoming particles contribute and outgoing particles . External incoming antiparticles contribute and outgoing antiparticles , with the corresponding fermion arrows. External scalar factors are one. Choose , , and ; all external momenta are on the appropriate mass shell.
- Integrate each independent loop four-momentum with . Keep matrix factors in their order along a fermion line, take a trace around a closed fermion loop, and include a minus sign for each closed fermion loop. Permuting external identical fermions contributes the corresponding fermionic sign; the Wick theorem determines the Feynman-diagram symmetry factors.
Strip the overall four-momentum conservation delta function when defining the scattering amplitude. There are no further bare interaction vertices, no gauge fixing and no Faddeev-Popov ghost fields in this theory. Renormalized higher-order calculations add the required counterterms; these are additional to the rules of the displayed classical Lagrangian density.
Past exam of the mathematics course of the University of Cambridge 2014 iii Paper 45 3 Solution Created 2026-10-03 Updated 2026-10-06
The strong-interaction matrix element between two spin-zero pseudoscalar mesons has only and available. The product of the two intrinsic parities is positive. An axial current would require a pseudovector constructed from these momenta, but an expression involving the Levi-Civita symbol needs three independent four-vectors and therefore vanishes. This is a consequence of parity conservation in the hadronic matrix element, not of parity conservation in the weak interaction. The vector current can have the two independent structures and . Its coefficients are Lorentz scalars; with and fixed, their only varying invariant is . HenceThese are the pseudoscalar-to-pseudoscalar form factors. With relativistically normalized states they are dimensionless. The vanishing axial matrix element and this decomposition explain the two equalities separately.
Write and for the outgoing electron and antineutrino momenta. From the Fermi interaction, an invariant scattering amplitude, up to an irrelevant overall sign or phase, isThe CKM matrix element multiplies the quark current in the convention of this interaction. Let . For massless leptons, the massless Dirac equation and chirality matrix anticommutation giveIn the second term move through the chiral projector before applying . Since , this transverse massless leptonic current givesThe disappearance of uses the massless approximation; for a massive charged lepton its contraction is proportional to the lepton mass.
Use the fermion spin sum and the supplied gamma matrix trace identities. The symmetric part of the leptonic tensor isThe Levi-Civita symbol term is antisymmetric and drops out when contracted with . ThusThere is no initial-spin average because the kaon is spinless. For the integrated massless leptonic tensor, keep every factor of explicit and define the unnormalized two-lepton Lorentz-invariant phase spaceThe leptons are massless, so . Its trace is . ThereforeThe three Lorentz-invariant phase-space measures and their momentum delta function contribute , in addition to in the decay rate. Combining them with the spin sum givesThis massless semileptonic pseudoscalar decay rate uses a two-lepton integral over future-timelike and the pion integral is restricted to the physically allowed region. The null endpoint follows by continuity. The coefficient has mass dimension , so the complete expression has mass dimension one, as a decay rate must in natural units.
In the kaon centre-of-momentum frame, put and . ThenThe dimensionally consistent Källén function isThe pion-only mass term must have fourth power: the second power printed in the PDF is dimensionally inconsistent. This repair also follows directly from squaring . Angular integration and the change of variable givewhere the negative sign reverses the endpoints. Combining this with the bracket yieldsThus . The lower limit is the minimum invariant mass of two massless leptons; at the upper limit the pion is at rest. The coefficient has mass dimension , while has dimension eight. The Källén function also shows why the differential decay rate vanishes at zero pion momentum.
Past exam of the mathematics course of the University of Cambridge 2014 iii Paper 47 3 Solution Created 2026-10-03 Updated 2026-10-06
Use the physical rapidity strip for poles of the two-body S-matrix, and write for scattering rapidity to distinguish it from the theta angle in question 2. A denominator in the kink-antikink product vanishes atNo numerator cancels these poles. For two equal-mass constituents with rapidities , their four-momentum vectors sum toThis gives the relativistic bound-state mass from a rapidity pole. The ordered breather spectrum isIt increases strictly with . The hypothetical state would lie at the two-kink threshold and is not a bound state; it is absent from the pole product. At these couplings and . In particular at weak coupling . This is the Sine-Gordon breather spectrum at reflectionless couplings. At there are no breathers; the subsequent processes involving a physical require .
For two identical neutral particles, exchanging the two outgoing labels does not produce a distinguishable channel. In one spatial dimension the elastic final momenta are the incoming pair, up to interchange. Thus there is one scalar identical-particle amplitude, rather than separately observable transmission and reflection amplitudes.
Put , so the basic amplitude uses . Its poles in the physical strip occur at and . The first is the direct bound-state pole. Choosing constituent rapidities gives real total energy-momentumSince ,Both energy and momentum therefore match an on shell with rapidity . The complementary pole is its crossed-channel partner. At the would-be is a threshold state, so this physical fusion interpretation must not be imposed there.
The bound-state fusion of factorized S-matrices treats a bound particle as its on shell constituents with analytically continued rapidities. If equal-mass particles fuse to at relative rapidity , use constituent rapidities . To scatter a third particle off , multiply its scattering amplitudes with each constituent and take the bound-state residue or projection in the constituent channel. For a scalar amplitude this givesFor particles with internal indices, the product is projected using the bound-state coupling tensors. The heuristic reason is factorized scattering: conserved higher charges prevent particle production and fix the rapidity data, so the third particle scatters through the constituents by successive two-body processes. Consistency of different orders is the Yang-Baxter equation. Without integrability, an independent three-body interaction would invalidate this simple bootstrap product.
For , the constituent shifts are , givingIt is useful to write this Sine-Gordon breather fusion amplitude in explicitly factorized form:To check the reduction, put . Multiplying the shifted factors gives numerator and denominator . Use and to factor them as . The product tends to one at large positive real rapidity, fixing its overall phase in this bootstrap convention.
For , the nearest pole to the real axis is . It is simple and comes from the first factor. In the crossed, or t-channel, the momentum carried between the external particles is their difference. With the metric its invariant isAt the pole, substitute the breather masses:The trigonometric identity is applied with angles and . Thus the exchanged one-particle state is the lightest breather , on its mass shell. This is crossed-channel lightest-breather exchange. The external momenta at a bound-state pole are analytically continued; on shell here means the invariant mass relation and conservation of the continued energy-momentum, not a pole at real physical rapidity. At the more distant central factor has a double pole, but the nearest pole and its interpretation remain unchanged.
Past exam of the mathematics course of the University of Cambridge 2015 iii Paper 47 3 b Solution Created 2026-10-03 Updated 2026-10-06
Flavour issue in the printed effective interaction. The displayed charged-current factor uses an Electron field even for the muon-neutrino term. Taken literally, this introduces an extra charged-current interaction, contradicting the Standard Model diagrams in part (a). In the Standard Model that factor uses the charged lepton of the same flavour; for Electron scattering the charged-current contribution is therefore present only for . The weak flavour selection in neutrino-electron scattering fixes the physically consistent assignment used below. The literal printed consequence is given afterwards, so the discrepancy is explicit.
At leading order in the Electron mass and in the four-fermion regime, define . The weak neutral current yields the scattering amplitudeAn overall amplitude phase has no observable effect. The Fermi interaction is valid when the exchanged invariants are small compared with the squared weak-boson masses.
Average over the two initial Electron spin states, and sum over final spins. There is no averaging over a sterile right-handed incoming neutrino. Using the fermion spin sum, the averaged square isTo display the trace contraction, defineThe supplied gamma matrix trace identities give the two tensors and . Symmetric-antisymmetric mixed contractions vanish. Set and . The remaining contractions are and , using the stated negative epsilon-contraction sign. Their sum givesThe massless Mandelstam variables satisfy , and the two products are and , respectively. Therefore
In the centre-of-momentum frame, momentum conservation gives . If is the angle cosine between and , then . The massless two-body Lorentz-invariant phase space is , while the flux is . ThusIntegrate and :Consequently the intended coefficients areThis is the massless neutrino-electron contact cross-section.
For completeness, the literal printed muon-neutrino charged-current term would instead shift both couplings by one, exactly as in part (c):For arbitrary real , its linear terms cannot be represented by the requested quadratic form with coupling-independent constants. The Standard Model result and the literal printed interaction are not the same problem.
Past exam of the mathematics course of the University of Cambridge 2015 ii Paper 2 32A Solution Created 2026-09-24 Updated 2026-10-06
Normalize the incident plane wave to unit amplitude. The scattering asymptotics are ; the outgoing coefficient is the scattering amplitude. The general axisymmetric free solution for isusing Spherical Bessel functions, spherical Neumann functions and Legendre polynomials.
For a real short-range central potential, the exterior radial solution in each partial wave is proportional to , whose large- form is . This defines the scattering phase shift. Multiplying this combination by matches the incoming spherical component of the incident plane wave: the factor cancels the shifted incoming phase. The outgoing coefficient instead acquires . Subtracting the plane wave's outgoing component therefore givesFor the barrier's -wave, the regular reduced radial function inside is , with . Outside take . Continuity of and its derivative at equates logarithmic derivatives:Use the branch with as . The matching equation gives , so the -wave scattering amplitude tends toIts negative is the repulsive spherical barrier scattering length, positive for this repulsive barrier.


