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For an ordinary quark-antiquark meson, opposite intrinsic parities and orbital parity give $P=(-1{)}^{L+1}$. For a self-conjugate flavor state, charge conjugation exchanges the two constituents. Reversal of their relative momentum gives $(-1{)}^L$, spin exchange gives $(-1{)}^{S+1}$, and reordering the two fermionic creation operators gives one extra minus sign. Their product is $C=(-1{)}^{L+S}$. Electrical neutrality alone does not imply that a state is an eigenstate of charge conjugation; charged mesons have no individual C eigenvalue.

Missing transverse momentum is the negative vector sum of reconstructed visible transverse momenta, with detector corrections specified. Under the ideal hypotheses of zero total incoming transverse momentum, complete visible reconstruction and exactly one invisible neutrino, momentum conservation givesThe neutrino is not directly seen. Even under these ideal hypotheses, its longitudinal momentum $p_{z\nu }$ is not measured. Beam remnants can carry substantial unobserved longitudinal momentum; the colliding quarks' longitudinal momenta are also unknown event by event. Imposing an assumed parent mass can constrain $p_{z\nu }$, but that imports a hypothesis rather than measuring it. Finite coverage, poorly reconstructed particle jets and other invisible particles can also generate or modify missing transverse momentum.

Yoichiro Nambu's superconductivity work led to the spontaneous symmetry breaking ideas recognized by his share of the 2008 Nobel Prize in Physics.^[ref] His 1960 treatment made the Meissner effect response consistent with gauge invariance by including collective pair contributions alongside the quasiparticles. The resulting Ward identity relates the interaction correction to the current and the self-energy; retaining the quasiparticles alone omits part of the response.^[ref]

The individual Nobel lecture explains the bridge to particle: the superconductivity gap suggested a dynamically generated fermion mass, and the accompanying collective excitation suggested a light pion. It also notes that the long-range Coulomb interaction turns the charged phase response into a plasmon. A neutral global phase mode and a charged material therefore need different low-momentum descriptions.^[ref]

In 1961, Yoichiro Nambu and Giovanni Jona-Lasinio supplied a four-fermion model with a self-consistent mass and a massless pseudoscalar bound state: an idealized pion. Their original constituents were nucleon fields. Their introduction explicitly distinguishes the zero-mass neutral collective excitation from its charged plasma counterpart and allows approximate chiral symmetry as the explanation of a nonzero pion mass. This is a model of dynamical symmetry breaking, rather than a microscopic derivation of Quantum chromodynamics.^[ref]

In the modern two-light-quark account, chiral symmetry breaking leaves isospin intact and produces three Goldstone bosons in the exact non-anomalous chiral limit. Small quark masses explicitly break that chiral symmetry, so the observed pions are pseudo-Goldstone bosons. The chiral protection of a fermion mass identifies the distinct obstruction to inserting an explicit fermion mass into a chirally symmetric equation. The useful lesson is that massive individual excitations and much lighter collective excitations can coexist because they describe different motions of an interacting state.

A concrete experimental motivation for treating the electromagnetic response carefully is the Field-cooled Meissner experiment. MIT's realization cools lead in a solenoid field using helium gas. A nearby thermometer follows the cooling; a test coil feeds an integrator and oscilloscope. Expulsion gives a pulse, warming reverses it, and switching off the field while the sample is normal checks its sign. A change in alternating-field response alone can reflect normal electrical conductivity. This readout tests magnetic flux expulsion, not the spectrum or existence of a neutral Goldstone boson.^[ref]

For a global symmetry in field theory, different orientations can be physically distinct states. A local gauge symmetry instead includes redundant descriptions of the same physical state. A gauge-dependent field expectation alone is consequently insufficient as a physical criterion for spontaneous symmetry breaking. The Higgs mechanism changes the physical mode content when the gauge field is dynamical; the global Goldstone theorem alone does not prove that mechanism. In particular, the massless direction of a gauge-fixed scalar potential need not appear as a separately observable massless particle.

The number of quarks of quark flavor $f$ minus the number of corresponding antiquarks. The strong interaction and electromagnetic interaction conserve each such difference even when they create or annihilate pairs. Weak charged currents can change these differences. For the strange quark, strangeness is $S=-F_s$.

For an isoscalar valence target and massless leading-order scattering, $R_{\nu }=G_L^2+G_R^2/3$, where $G_{L/R}^2=(g_{L/R}^u{)}^2+(g_{L/R}^d{)}^2$. The factor $1/3$ is ${\int }_0^1(1-y{)}^{2}dy$. With Standard Model quark couplings and $w={\sin }^2{\theta }_W$, this becomes $1/2-w+20w^2/27$. Sea quarks and threshold, nuclear and QCD corrections matter beyond this limit.

One half went to Yoichiro Nambu for discovering the mechanism of spontaneous symmetry breaking in subatomic physics. One quarter each went to Makoto Kobayashi and Toshihide Maskawa for explaining broken symmetry through a mechanism requiring at least three families of quarks.^[ref]

A weak interaction decay of a baryon into final hadrons without final leptons. A weak charged current changes quark flavour, while spectator quarks and the produced quarks undergo hadronization. A quark-flow Feynman diagram specifies a possible weak topology; it does not calculate the full hadronic decay amplitude.

The residual strong interaction between nucleons. Its long-distance part involves pion exchange and depends on spin and isospin. It is distinct from directly exchanging gluons between color-singlet nucleons.

Why do protons and neutrons bind into an atomic nucleus, even though protons repel electrically? An attractive nuclear force can lower the total energy below that of separated nucleons, producing nuclear binding energy. The deuteron, containing one proton and one neutron, is the simplest such bound state.

In Quantum chromodynamics, nucleons are composite colour singlets built from quarks and gluons. Their residual nuclear force has a longest-range strong contribution from one-pion exchange, with additional interactions at shorter distances.^[ref]

The $+11$ contribution to ${\beta }_0$ is the net Non-Abelian gauge theory contribution; the $-2n_f/3$ term comes from quark loops. Thus quarks screen color charge, opposing asymptotic freedom. It is the net gluon contribution that produces antiscreening: the inferred QCD coupling decreases toward shorter distances. Unlike the electrically neutral photon, a gluon carries color charge and has self-interactions. Describing this as vacuum polarization must not imply that all vacuum contributions have the screening sign of quantum electrodynamics.^[ref]

Individual gauge-dependent Feynman diagrams do not separately define observable screening clouds. A proof of the coefficient sums the gauge fixing, Faddeev-Popov ghost and matter contributions and extracts the charge renormalization; the final one-loop order coefficient is independent of the gauge choice. The beta function is negative near zero for integer $0\le n_f\le 16$. This is a statement about weak coupling, not a determination of the low-energy particle spectrum.^[ref]

Hadronic processes with disconnected quark-flavour lines are often suppressed relative to connected quark rearrangement. A predominantly strange phi meson can decay to kaons through light-pair creation while its original strange lines survive, whereas a purely pionic final state requires strange-pair annihilation. The rule is an approximate dynamical pattern, not an exact conservation law.

A high-energy hadron is treated as constituents scattering incoherently. In electromagnetic deep inelastic scattering, the charged constituents are quarks and antiquarks.

There is one tree-level Feynman diagram: annihilation of the up quark and its antiquark into an $s$-channel gluon, followed by production of the down quark and its antiquark. A gluon interaction vertex preserves quark flavor, so a $t$-channel line joining the incoming up quark to the outgoing down quark is impossible.

Let $i,j,k,l$ label the initial quark, initial antiquark, final quark, final antiquark color charges. Up to an irrelevant common sign, the scattering amplitude isThe Mandelstam variables are $s=(p_1+p_2{)}^2$, $t=(p_1-p_3{)}^2$, $u=(p_1-p_4{)}^2$, with $s+t+u=0$ in the zero-mass limit. The spin sum uses $\sum u\bar{u}=\sum v\bar{v}=/p$, and the initial spin average contributes $1/4$. The resulting two gamma matrix traces contract toThe initial color average contributes $1/N_c^2$. Summing final color charges gives the distinct-flavor quark scattering color factorThereforeEquivalently $C=64{\pi }^2{\alpha }_s^2/9$ for the QCD coupling ${\alpha }_s=g^2/(4\pi )$. There is no initial gluon spin average: the gluon is internal.

Work entirely with left-handed Weyl fermions; a right-handed field is counted through its left-handed charge conjugate. For a simple non-Abelian group $G$, defineThe two orientations of a chiral triangle supply the symmetrized trace. Spectator representation dimensions multiply each trace. The perturbative gauge anomaly conditions for $G\times U(1)$ are thereforeThe first concerns $G^3$, the second $G^{2}U(1)$, and the third $U(1{)}^3$. A $GU(1{)}^2$ triangle has group factor ${\sum }_i{Q}_i^2{\mathrm{tr}}_{R_i}{T}^a=0$, because non-Abelian generators of a semisimple group are traceless. If coupling to gravity is required, also impose the mixed gauge-gravitational condition ${\sum }_i\dim (R_i)Q_i=0$. Cancellation of these local triangles is not a universal substitute for checking global anomalies; the Witten SU(2) anomaly is the relevant additional test here.^[ref]

For the Standard Model, one generation consists of $Q_L:(3,2{)}_{1/6}$, $u^c:(\bar{3},1{)}_{-2/3}$, $d^c:(\bar{3},1{)}_{1/3}$, $L_L:(1,2{)}_{-1/2}$ and $e^c:(1,1{)}_1$. These are precisely the all-left-handed representations; conjugating the singlets reverses their hypercharges relative to right-handed particles. Check the requested triangles separately:

For $3$–$3$–$3$, normalize the cubic coefficient of a fundamental to $+1$. Its conjugate has $-1$, so the weak-doublet multiplicity givesFor $2$–$2$–$2$, the doublet generators obey $\{T^b,T^c\}={\delta }^{bc}I/2$, so $\mathrm{tr}(T^a\{T^b,T^c\})=0$. Thus every doublet separately has zero local cubic weak anomaly. For the global Witten SU(2) anomaly, each generation has three coloured quark doublets and one lepton doublet, an even total of four. All three generations give twelve, so the global obstruction also cancels.

For $3$–$3$–$1$, use $T(3)=T(\bar{3})=1/2$. The two conjugate triplets have the same quadratic index, not its negative:For $2$–$2$–$1$, the three quark colours and one lepton doublet giveFor $1$–$1$–$1$, count every coloured and weak component:The mixed gauge-gravitational coefficient also vanishes:Unlisted gauge combinations contain a single generator of at least one simple factor. For example, $3$–$2$–$1$ has ${\mathrm{tr}}_3{T}^a{\mathrm{tr}}_2{t}^{b}Y$, and $3$–$3$–$2$ has ${\mathrm{tr}}_3(T^a{T}^b){\mathrm{tr}}_2{t}^c$; both vanish. The same reasoning covers $3$–$1$–$1$, $2$–$1$–$1$ and $3$–$2$–$2$, including permutations of the legs. Scalars contribute no chiral triangle anomaly. Therefore

The tree-level Feynman diagrams are the two ordered quark-exchange graphs and the $s$-channel gluon-exchange graph with a three-gluon vertex. There is no two-gluon-two-quark contact term in the ordinary QCD Lagrangian density. Let $q=k_1+k_2=p_1+p_2$, $r_t=p_1-k_1$, $r_u=p_1-k_2$, and $s=q^2$, $t=r_t^2$, $u=r_u^2$. Define $i\mathcal{M}$ as the product of the stated Feynman rules.

Suppressing external colour indices, the quark graphs areThe massless Dirac equations give $\bar{u}/p_1=0$ and $/p_{2}v=0$. ConsequentlyThe last equality uses $\bar{u}/qv=0$. The quark-exchange sum is therefore $-g^2\bar{u}/k_1[T^a,T^b]v$.

For the gluon graph, use all incoming momenta $(k_1,k_2,-q)$ in the three-gluon tensor:Its double contraction is $k_{1\mu }{k}_{2\nu }{V}^{\mu \nu \rho }=s(k_2-k_1{)}^{\rho }/2$. The vertex and gluon propagator giveCombining the graphs yieldsThus the requested spinor proportionality factor is $-g^2\bar{u}(p_1)/k_{1}v(p_2)$ when the colour matrix is displayed separately; retaining it inside the spinor bilinear makes its external colour indices explicit. For a convention calling the Feynman-rule product itself $\mathcal{M}$, rather than $i\mathcal{M}$, the entire result gains one factor of $i$.

Source paper: www.whatdotheyknow.com/request/oxford_physics_past_papers_2/response/2340696/attach/7/Physics%20Papers%202018.pdf?cookie_passthrough=1

The bundle filename says 2018, but the paper's cover establishes Trinity Term 2019, 24 April 2019, and formal code A15272W1. These are solutions to that 2019 paper. Use natural units, the Minkowski metric $g=\mathrm{diag}(1,-1,-1,-1)$, and the Feynman i-epsilon prescription. Write $/p={\gamma }^{\mu }{p}_{\mu }$. In question 1, $D=4-ϵ$, $g_D={\mu }^{ϵ/2}g$, and the amputated gluon self-energy insertion is $i{\Pi }^{\mu \nu ,ab}$. This fixes the overall sign and factor of $i$ in the results below; defining the insertion as $-i\Pi$ instead reverses all corresponding definitions of $\Pi$. The quadratic Casimir operator in the Adjoint representation is $C_A=N_c$ for $SU(N_c)$, hence $C_A=3$ in QCD.

There are four topologies for the one-loop gluon self-energy, shown in the upper row of the figure:

The two internal-gluon graphs have symmetry factor $1/2$. The directed quark and ghost loops have symmetry factor one and a minus sign for a closed anticommuting loop. The four-gluon tadpole diagram must be included in the list even though its massless Feynman integral vanishes in dimensional regularization. Local counterterms are excluded here.

The Slavnov-Taylor identity implies transversality of the full perturbative gluon self-energy:The quark loop is separately transverse. This follows without evaluating its Feynman integral: contract its vector Feynman vertex with $k_{\mu }$ and use $/k=(/l+/k-m)-(/l-m)$ between its two Dirac propagators. One obtains a difference of two loop integrals that cancel after a shift of $l$ in dimensional regularization. The colour factor is proportional to $\mathrm{tr}(T^a{T}^b)$ and does not obstruct this argument.

The four-gluon tadpole diagram vanishes after removal of $m_g$. Thus the nontransverse parts of the two-three-gluon-vertex loop and the ghost loop must cancel. Contracting the preceding ghost-loop pole of the gluon self-energy givesFor QCD, set $C_A=3$. No evaluation of the remaining three-gluon Feynman diagram is required. Its uncontracted transverse part is not fixed by this identity.

The proper quark-gluon vertex colour factors come from two one-particle-irreducible Feynman diagrams, shown in the upper-middle and upper-right panels. One has a virtual gluon joining the open quark line on opposite sides of the external gluon insertion. The other has two virtual gluons meeting the external gluon at a three-gluon Feynman vertex, with their other ends joined by an internal quark. A Faddeev-Popov ghost field has no direct quark coupling, so it supplies no additional one-loop Feynman vertex graph here.

Let $T^a$ be the Hermitian generators of the fundamental representation of $SU(N)$, with $\mathrm{tr}(T^a{T}^b)={\delta }^{ab}/2$, $[T^a,T^b]=i{f}^{abc}{T}^c$, ${\sum }_b{T}^b{T}^b=C_{F}1$, and $f^{abc}{f}^{dbc}=C_A{\delta }^{ad}$. Here $C_F=(N^2-1)/(2N)$ and $C_A=N$ are quadratic Casimir operators evaluated in the fundamental representation and the Adjoint representation.

For the first graph, the ordered product is ${\sum }_b{T}^b{T}^a{T}^b$. Expand the double commutator:Thus its colour factor isFor the second graph choose the orientation of the three-gluon Feynman vertex so that its group tensor is $f^{abc}$ and the ordered product along the quark line is $T^b{T}^c$. Antisymmetry removes the anticommutator part:The corresponding real group coefficient is $C_A/2=3/2$ after the conventional factor of $i$ is incorporated into the remaining Feynman rules. If the vertex is written with $i{f}^{abc}$, its raw matrix product is instead $-C_A{T}^a/2$. Reversing the ordered product reverses its sign together with the orientation-dependent three-gluon Feynman rule; these conventions describe the same scattering amplitude. External-leg corrections and local counterterms are separate from these two proper one-loop graphs.

The two tree-level Feynman diagrams are the $s$-channel and $u$-channel quark exchange graphs, drawn in the middle row of the figure. Along the directed quark line the incoming photon is absorbed before the outgoing photon is emitted in the first graph; those two insertions are reversed in the second. Their internal four-momenta are $q_s=p_1+p_2$ and $q_u=p_1-p_4$, respectively. There is no three-photon Feynman vertex in ordinary QED, and no two-photon-two-quark contact interaction in the Dirac field Lagrangian density.

The exchanged particle is now a gluon, and the incoming up quark mass replaces $m_e$. DefineThe Dirac trace and spin average are unchanged. Normalize the fundamental gauge generators by $\mathrm{tr}(T^A{T}^B)={\delta }^{AB}/2$. Averaging over the $N_c^2$ initial color charge choices and summing the final color charge choices givesReplace the factor $N_c{e}^4{Q}_e^2{Q}_t^2$ from part (ii) by $g_s^4{C}_g$. ThereforeKeeping $m_u$ is allowed; if the usual negligible up quark mass approximation is wanted, simply set $m_u=0$. The factor $1/N_c^2$ is essential: initial quark color charge states are averaged, whereas final ones are summed.