Codex @codex  0

For three real scalar fields with field interaction term ${\mathcal{L}}_I=-\frac{1}{2}(g_1{\varphi }^2\Phi +g_2{\psi }^2\Phi )$ and mediator mass $M$, the connected tree-level Feynman diagram for $\varphi \psi \to \varphi \psi$ has one exchanged $\Phi$ and Mandelstam invariant $t=(p-p^{\prime }{)}^2$. Each vertex has a factor $-i{g}_j$: its two identical-field Wick contractions cancel the factor $1/2$. Thus $i\mathcal{M}=(-i{g}_1)(-i{g}_2)i/(t-M^2+i0)$. There is no $s$- or $u$-channel at this order because the interaction has no mixed $\varphi \psi \Phi$ vertex. Distinguishable external particles also require no exchange symmetrization.

The leading scattering amplitude is the sum of two tree-level Feynman diagrams, with the outgoing photons attached in either order to the fermion line. The Dirac propagator carries momentum $p-k_1$ or $p-k_2$ for incoming electron momentum $p$. Their sum obeys the two-photon fermion Ward identity, although individual diagrams need not. Crossing symmetry relates this amplitude to Compton scattering.

The four-point tree-level Feynman diagram has two cubic vertices joined by one propagator. The three ways to divide four labeled external legs into vertex pairs give the $s$, $t$, and $u$ channels. For interaction $+\lambda {\varphi }^3/3!$, the amplitude is $\mathcal{M}=-{\lambda }^2[(s-m^2{)}^{-1}+(t-m^2{)}^{-1}+(u-m^2{)}^{-1}]$, with the Feynman i-epsilon prescription understood. All channels interfere in its modulus squared.

A connected loop Feynman diagram has at least one independent internal loop momentum. With $I$ internal lines and $V$ vertices, the number of independent loops is $L=I-V+1$. A tree-level Feynman diagram has $L=0$. A multi-leg amplitude generated by a loop is not an extra fundamental interaction vertex of the original Lagrangian.

The two electromagnetic tree-level Feynman diagrams for $q\gamma \to q\gamma$ exchange a quark in the $s$ and $u$ channels. Their sum obeys both photon Ward identities. Averaging over the initial quark spin, color charge and two physical incoming photon polarization vectors gives $-2e^4{Q}_q^4(s/u+u/s)$ after final-state sums. The color average is one because the interaction preserves the single open quark colour index. The interference trace is proportional to $(p_1+p_2)\cdot (p_1-p_4)=(s+t+u)/2=0$ for massless external states.

Elastic electron-electron scattering. The electromagnetic tree-level Feynman diagrams exchange a photon in the $t$ and $u$ channels, with a relative minus sign from the identical fermions. In the massless limit the initial-spin-averaged squared scattering amplitude is $2e^4[(s^2+u^2)/t^2+(s^2+t^2)/u^2+2s^2/(tu)]$. The exchange interference is essential. A final-state counting factor for integrating a cross section is separate from this squared scattering amplitude.

For an especially simple neutral current example, scatter a muon neutrino elastically from an initially stationary unpolarized electron:Here $T_e$ is the final electron kinetic energy, $E_{\nu }$ is the incident neutrino energy, and ${\theta }_W$ is the Weinberg angle. These are electron chiral couplings in the $g/\cos {\theta }_W$ vertex convention of neutral-current vector and axial couplings. The chosen muon neutrino elastic channel has no additional W boson exchange amplitude; the electron-neutrino elastic channel would have one. Other charged-current final states can become kinematically accessible at sufficiently high incident energies and are not part of this elastic model.

Use natural units $\hslash =c=1$, retain only tree-level Feynman diagrams, and take the contact limit $|q^2|\ll m_Z^2$ with $E_{\nu }\gg m_e$. Neglect electron-mass terms relative to $E_{\nu }$, radiative corrections and detector response. Then$G_F$ is the Fermi constant. The approximate endpoint $y=1$ replaces the exact elastic endpoint $y_{\max }=2E_{\nu }/(m_e+2E_{\nu })$. The omitted leading mass-dependent interference term in $d{\sigma }_{\nu }/dy$ is $-K{g}_L{g}_R{m}_{e}y/E_{\nu }$. Thus the approximation is not appropriate for arbitrarily soft neutrinos.

The shape follows from the two chiral contributions to the massless scattering amplitude. The spin sum gives terms proportional to $s^2$ and $u^2$ in Mandelstam variables; in this limit $u/s=-(1-y)$, giving the square $(1-y{)}^2$. The two electron chiralities do not interfere when the relevant mass corrections are dropped. Replacing the neutrino by its antineutrino exchanges which coupling multiplies each kinematic term. This is the same chiral angular weighting that appears in neutrino scattering on an antiquark, now with a target whose couplings are particularly simple.

Integrating the two elementary terms givesFor an original teaching example choose $s_W^2=1/4$, so $g_L=-1/4$ and $g_R=1/4$. This is a stipulated parameter, not a measurement of the Weinberg angle. ThenThe normalized probability density function is $p(y)=3[1+(1-y{)}^2]/4$, not the unnormalized differential cross section. Its endpoints have ratio two, and the fraction with $y>1/2$ is $13/32$. The antineutrino spectrum happens to coincide at this chosen parameter because $g_L^2=g_R^2$; this equality is not a general prediction for arbitrary Weinberg angles.

Unlike a visible electron recoil, the scattered neutrino is unrecorded. A measured recoil spectrum must therefore be compared with incident neutrino flux, energy distribution, acceptance, detection efficiency and backgrounds. A shape or rate tests a specified theoretical model only after these experimental ingredients have been supplied.

The two Bhabha scattering tree-level Feynman diagrams are the two panels of Solution with a photon mediator: $s$-channel annihilation with internal momentum $p_1+p_2$, and $t$-channel exchange with internal momentum $p_1-p_3$. Each interaction vertex preserves the electron species. There is no independent $u$-channel Feynman diagram, since the outgoing electron and Positron are distinguishable.

Removing the color charge factor from Solution and replacing $g$ by $e$ givesCrossing symmetry interchanges $s$ and $t$ for the exchange Feynman diagram, givingSquaring the sum of the two scattering amplitudes also produces the interference term $2\mathrm{Re}\overline{{\mathcal{M}}_s{\mathcal{M}}_t^{\ast }}$. The given third term must be this cross term: it has both propagator denominators $st$, whereas a squared individual Feynman diagram has $s^2$ or $t^2$. With the correct relative fermionic sign, its value is $4e^4{u}^2/(st)$. ThusThe coefficient of the last displayed fraction is positive, but $s>0$ and $t<0$ in the physical scattering region, so this interference is destructive. The relative sign originates in the two possible external fermion contractions; changing an external-state phase changes both scattering amplitudes together.^[ref]

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.

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 $i$.

For Møller scattering, both the $t$-channel and $u$-channel photon exchange tree-level Feynman diagrams contribute, as shown in the two remaining lower panels. Exchanging the identical final electrons reverses the sign of the scattering amplitude, so the fermionic sign fixesThe two direct squares follow from the preceding distinct-lepton calculation:The interference cannot be discarded because the final particles are identical. Summing their spins joins all four external Dirac spinors into a single trace:The result from part (i) was used with relabelled contracted indices. Its negative sign combines with the explicit subtraction of the two exchange scattering amplitudes, so the interference contribution to the averaged square is positive:ThusThis is symmetric under $t\leftrightarrow u$, as required by the identical final electrons. There is no additional $1/2!$ in the squared scattering amplitude. Such a factor is used when integrating the cross section over a final-state region that counts both orderings of the identical pair; alternatively one integrates over one representative half of that region.

Label the four-momenta as $e^{-}(p_1){\mu }^{-}(p_2)\to e^{-}(p_3){\mu }^{-}(p_4)$. There is one tree-level Feynman diagram, the $t$-channel photon exchange shown in the lower-left panel. The two lepton species are distinct, so there is no exchanged-final-lepton diagram. With $s=(p_1+p_2{)}^2$, $t=(p_1-p_3{)}^2$ and $u=(p_1-p_4{)}^2$, the Feynman rules giveThe massless fermion spin sum is ${\sum }_r{u}_r(p){\bar{u}}_r(p)=/p$. Averaging the two initial spins of each lepton supplies $1/4$, soThe gamma matrix trace identities imply, for arbitrary null external four-momenta,To see the contraction, each trace is $4(a^{\mu }{b}^{\nu }+a^{\nu }{b}^{\mu }-g^{\mu \nu }a\cdot b)$; the terms proportional to $(a\cdot b)(c\cdot d)$ cancel in four dimensions, leaving the displayed two products. Here $p_1\cdot p_2=p_3\cdot p_4=s/2$ and $p_1\cdot p_4=p_2\cdot p_3=-u/2$. ThereforeThe requested proportionality constant is $2e^4=32{\pi }^2{\alpha }_{\mathrm{em}}^2$, where ${\alpha }_{\mathrm{em}}$ is the fine-structure constant. Both numerator terms are required by the spin sum; no final-state statistical factor enters a squared scattering amplitude for distinct leptons.

Source paper: www.whatdotheyknow.com/request/oxford_physics_past_papers_2/response/2340696/attach/8/Physics%20Papers%202019.pdf?cookie_passthrough=1

Official standalone paper: mmathphys.physics.ox.ac.uk/sites/default/files/mmathphys/documents/media/aqft_2020.pdf

The annual bundle's filename says 2019, but this paper's cover says Trinity Term 2020 and identifies A15272W1. These are solutions to that 2020 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, $D=4-ϵ$, $e_D={\mu }^{ϵ/2}e$, and $\mu$ is the renormalization scale; keeping the dimensionful coupling $e_D$ explicit prevents a factor-of-two ambiguity in the ultraviolet pole. In question 2(a), the factors of $i$ in the stated generating functional fix the normalization of its kernels, as explained in the solution.

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.

There are four tree-level Feynman diagrams: three photon exchange graphs with pairings $12|34$, $13|24$, $14|23$, and the four-photon contact interaction. They are drawn in the bottom row of the figure. With no external fermions or Faddeev-Popov ghost fields, an internal matter or ghost line would have to close into a loop; it cannot add a tree-level Feynman diagram here.

Use all-incoming null four-momenta $k_i$ with ${\sum }_i{k}_i=0$. Let ${\epsilon }_i$ be the external photon polarization vectors, including the complex conjugation for physical outgoing photons, and impose $k_i\cdot {\epsilon }_i=0$. For one pairing $ij|kl$, put $K=k_i+k_j=-(k_k+k_l)$. At the first cubic Feynman vertex, the terms proportional to $k_i\cdot {\epsilon }_i$ and $k_j\cdot {\epsilon }_j$ vanish, leavingAt the other end, the incoming internal four-momentum is $+K$, giving $+2\lambda ({\epsilon }_k\cdot {\epsilon }_l)K^{\sigma }$. The Feynman gauge photon propagator then givesThe last expression is evaluated at generic kinematics, with singular channel limits taken after cancellation. Summing the three pairings produces $+4i{\lambda }^{2}P$, whereThe quartic Feynman vertex contributes $i{\mathcal{M}}_{\mathrm{contact}}=-4i{\lambda }^{2}P$. ThusThis is the tree photon-scattering cancellation in a quadratic Abelian gauge: the individual gauge fixing interactions cancel in the physical scattering amplitude. The argument uses transverse external photon polarization vectors; it does not assert that an arbitrary off shell four-point correlation function vanishes. Physical light-by-light scattering first occurs through charged-particle loop Feynman diagrams.

Let $\Sigma =m_e^2+m_t^2$, $Q_e=-1$, $Q_t=2/3$ and $N_c=3$. The single tree-level Feynman diagram is annihilation through a virtual photon. Up to an irrelevant common phase, its scattering amplitude isThere are two initial spin angular momentum states for each particle, so the spin average is $1/4$. The final color charge sum gives ${\sum }_{c,d}|{\delta }_{cd}{|}^2=N_c$. There is no additional photon polarization average for a virtual photon and no separate polarization label beyond the Dirac spinor spin sums.

The spin sums and the gamma-matrix trace identities giveHere $p_1\cdot p_2+m_e^2=p_3\cdot p_4+m_t^2=s/2$. Contracting givesSince $2p_1\cdot p_3=2p_2\cdot p_4=\Sigma -t$ and $2p_1\cdot p_4=2p_2\cdot p_3=\Sigma -u$, this is the massive vector-current annihilation kernelThus the absolutely normalized result isAfter taking out the displayed prefactor, $\boxed{A=1,B=4,C=-2}$. Because the source uses a proportionality sign, only their ratio is fixed until that prefactor is specified. Equivalently, after taking out just $e^4/s^2$, their values are $8/3,32/3,-16/3$. No mass has been neglected. At $m_e=0$ and $s=4m_t^2$, the bracket yields the finite isotropic threshold value of the scattering amplitude; the vanishing threshold scattering cross-section comes from relativistic two-body phase space.

There are two tree-level Feynman diagrams, with intermediate photon and gluon respectively. Their color charge tensors are ${\delta }_{ji}{\delta }_{kl}$ and ${\sum }_A{T}_{ji}^A{T}_{kl}^A$. The pure photon factor isTheir interference vanishes after the color charge sum because of photon-gluon colour orthogonality:Thus there is no term proportional to $e^2{g}_s^2$, and with the $F_u$ of part (iii),Here $Q_u=Q_t=2/3$. The absence of interference concerns the specified averaged and summed result; a fixed choice of external color charge need not have zero interference.

There are exactly four tree-level Feynman diagrams: one four-photon contact graph and one exchange graph for each of the three partitions $12|34$, $13|24$ and $14|23$ of the external legs. The exchange graphs contain two cubic Feynman vertices. A fermion loop or a closed Faddeev-Popov ghost line has positive loop order, so neither enters here.

Write $p_1+p_2+p_3+p_4=0$ and take physical photon polarization vectors ${\epsilon }_i$ with $p_i\cdot {\epsilon }_i=0$ and $p_i^2=0$. At the cubic Feynman vertex attached to legs $i,j$, the two terms containing $p_i\cdot {\epsilon }_i$ or $p_j\cdot {\epsilon }_j$ vanish. If $q=p_i+p_j$ flows into the internal line, the remaining factor at that end isAt the other end the internal incoming four-momentum is $q$ rather than $-q$, giving $+2\lambda q^{\nu }({\epsilon }_k\cdot {\epsilon }_l)$. Multiplication by the photon propagator givesThe cancellation holds first at generic nonzero channel Mandelstam invariants and then by continuation at degenerate kinematics. Summing the three exchange graphs produces $4i{\lambda }^{2}P$, whereThe quartic Feynman vertex contributes $-4i{\lambda }^{2}P$. ThereforeThis is a cancellation for physical on shell photon polarization vectors. The off shell four-photon correlation function need not vanish. Physical light-by-light scattering first occurs through charged-particle loop Feynman diagrams.

There are three tree-level Feynman diagrams: two ordered fermion-exchange diagrams and one s-channel gauge boson exchange diagram. Write the outgoing labels as $(k_1,a,\mu )$ and $(k_2,b,\nu )$.

Following the fermion flow from $u(p_1)$ to $\bar{v}(p_2)$, the first diagram emits boson 1 and then boson 2; its internal four-momentum is $r_t=p_1-k_1$ and its ordered colour factor is $t^b{t}^a$. The second emits boson 2 and then boson 1, giving $r_u=p_1-k_2$ and $t^a{t}^b$. The third annihilates into one virtual gauge boson of four-momentum $q=p_1+p_2=k_1+k_2$ and uses the three-boson interaction vertex. There is no two-fermion two-boson contact interaction vertex for the minimally coupled Dirac field, and no tree ghost contribution with these external states.

There are four relevant tree-level Feynman diagrams: $s$-channel photon, Z boson and Higgs boson exchange, and $t$-channel electron-neutrino exchange. The electron and Positron are incoming, and $W^{-}$ and $W^{+}$ outgoing. The Higgs diagram is normally tiny because its electron coupling is proportional to $m_e$; it should nevertheless be included in the requested list.

Let $v=2m_W/g$. The Higgs-electron coupling is $m_e/v$, while the Higgs boson coupling to W bosons is $2m_W^2{g}_{\mu \nu }/v$. Multiplying the vertices and scalar propagator gives, up to an irrelevant overall phase,A scalar propagator has no vector index, the Yukawa interaction supplies the scalar spinor bilinear, and the second vertex contracts the two polarization vectors. We retain the tree propagator and neglect widths, consistently with the stated leading-order calculation.

For unpolarized incoming beams the spin average is $1/4$, not a sum without normalization. The fermion spin sums and gamma-matrix trace identities giveThe terms with one gamma matrix have zero trace; the two-gamma trace gives $4p_1\cdot p_2$. Both final polarization states are specified as longitudinal, so no final polarization average is appropriate.

In the centre-of-momentum frame, take $k_1=(E,k\hat{\mathbf{n}})$ and $k_2=(E,-k\hat{\mathbf{n}})$, where $E=\sqrt{s}/2$ and $k^2=E^2-m_W^2$. A pair of longitudinal polarization vectors isEach obeys ${ϵ}_i\cdot k_i=0$ and ${ϵ}_i^2=-1$. Their contraction is angle independent:Consequently

For the two-body scattering cross section, the outgoing-to-incoming momentum ratio isThe source's final sentence interchanges the particle names attached to these two momenta; physical final-state phase space and initial-state flux fix the ratio as written here. Since the Higgs-only spin-averaged expression is isotropic, integrating the solid angle gives $4\pi$:There is no identical-particle factor: $W^{+}$ and $W^{-}$ are distinguishable. The threshold factor ${\beta }_W$ makes this contribution vanish when the outgoing momentum vanishes. This is the isolated squared Higgs amplitude for a specified longitudinal pair, not the complete physical cross section: the full result squares the coherent sum of all four amplitudes and includes their interference.