Past exam of the particle physics course of the University of Oxford / 2018 (A11047W1)

Source paper: www.whatdotheyknow.com/request/oxford_physics_past_papers_2/response/2340696/attach/6/Physics%20Papers%202017.pdf?cookie_passthrough=1. The held cover identifies C4 on 14 June 2018; the annual bundle's filename does not determine the year. Natural units $\hslash =c=1$ are used except in the silicon-detector calculations, where SI units and the momentum conversion are stated explicitly. All six questions are solved.

Work in the W boson rest frame, taking the spin-quantization axis along its laboratory motion. Neglect charged-lepton and neutrino masses. The weak charged current produces a negative-helicity ${\ell }^{-}$ and a positive-helicity antineutrino. Their momenta are opposite, so both spin Vectors point opposite to the charged lepton's direction $\hat{\mathbf{n}}$; the final total spin projection along $\hat{\mathbf{n}}$ is $-1$.

Let ${\lambda }_W=+1,0,-1$ denote the parent spin projection along $z$. In the given probability amplitude, use $j=1$ and the rotation element $d_{-1,{\lambda }_W}^1(\theta )$. The supplied spin-one rotation elements giveThe transverse states have spin projections $\pm 1$; the longitudinal state has projection zero, not total spin zero. The original diagram below distinguishes momenta from spins and shows the corresponding normalized angular profiles.

With relativistically normalized one-particle states, Fermi's golden rule becomesIn the parent rest frame, the three-momentum delta function fixes ${\mathbf{p}}_2=-{\mathbf{p}}_1$. The energy delta function fixes the common magnitude $p^{\ast }$; its radial Jacobian converts the remaining phase-space factor intoOnly one momentum magnitude remains because the two final momenta are constrained, not independent. Explicitly,For leptonic W boson decay both final masses are small compared with $m_W$, including the tau mass at the precision requested, so $p^{\ast }\simeq m_W/2$.

For the longitudinal state,Insert this and $p^{\ast }=m_W/2$ into the phase-space formula:Rotational invariance makes the integrated decay probability independent of the orientation of the initial spin. Both transverse polarizations therefore have the same rate. An unpolarized sample averages these three equal rates; it does not add three copies. Hence its width into one specified charged-lepton flavour is also $g^2{m}_W/(48\pi )$.

At a hadron collider, select $W\to \ell \nu$ events and reconstruct the transverse massusing missing transverse momentum for the neutrino. Fit the shape, especially the region above the nominal Jacobian edge, with finite-width resonance templates convolved with detector resolution, recoil modelling and backgrounds. The Breit-Wigner width controls this off-shell tail. Calibrated momentum and missing-momentum resolutions must be separated from the physical width. This is a lineshape measurement; the lifetime is far too short for a direct flight-distance measurement.

Assume only Standard Model decay channels, negligible final-state masses, three lepton families, three quark colours and a unitary CKM matrix. The top-quark channels are kinematically closed. Summing the allowed quark channels gives $3{\sum }_{i=u,c}{\sum }_{j=d,s,b}|V_{ij}{|}^2\simeq 6$ times one leptonic width, while the leptons contribute three. Neglecting QCD and electroweak radiative corrections,The source's rounded CKM entries give almost the same factor; their tiny nonunitarity is not physically significant. ConsequentlyThe propagated uncertainty uses $\delta g/g=\frac{1}{2}\delta {\Gamma }_W/{\Gamma }_W$, taking $m_W$ as fixed. It is the measurement uncertainty within the stated tree-level approximation; omitted radiative corrections are an additional modelling limitation. In particular QCD increases the hadronic width and lowers the inferred coupling when included.

The printed electromagnetic comparison $4\pi /137$ is $e^2$, not the probability amplitude coupling $e$. Compare like quantities:Equivalently ${\alpha }_W=g^2/(4\pi )\simeq 0.0347$ exceeds ${\alpha }_{\mathrm{EM}}\simeq 1/137$. The weak interaction is weak at low energy primarily because of the massive propagator, rather than an exceptionally small fundamental coupling.

The valence quark distributions describe net quark content: $\int u_v(x)dx=2$ and $\int d_v(x)dx=1$ for a proton. Their momentum densities $x{u}_v,x{d}_v$ are small at very small $x$, peak at moderate momentum fractions, and vanish toward $x=1$, where one parton carrying almost all proton momentum is unlikely. The larger $u$ distribution reflects two valence up quarks and its comparatively harder shape; its pointwise value need not be exactly twice the down distribution.

The sea quark distribution is generated largely by gluon radiation and $g\to q\bar{q}$ splitting. It grows strongly toward small $x$ and falls at large $x$. The plotted sea curve is scaled by $0.05$, so it must be multiplied by twenty before numerical use. The total up-quark PDF isusing the valence-up distribution and the up-sea distribution. The summed sea curve alone does not fix $\bar{u}$ without a flavour assumption. Under the later stated four-flavour symmetry, $\bar{u}=\Sigma /8$.

At leading order, neglect the transverse momentum of the incoming partons and of the produced on-shell boson. In the hadron centre-of-mass frame,Then $m_W^2=(p_{+}+p_{-}{)}^2=x_{+}{x}_{-}s$, andSolving these product and ratio equations givesOver $-2<y<2$, each momentum fraction ranges betweenThey are anticorrelated: increasing $y$ increases $x_{+}$ while decreasing $x_{-}$. Higher-order recoil changes the simple two-to-one kinematics.

The leading subprocess is $u\bar{d}\to W^{+}$. There are two beam assignments, and charge conjugation relates antiproton PDFs to proton PDFs. ThereforeIn a narrow-width leading-order normalization the prefactor is $\pi g^2/(12s)$ when $|V_{ud}|=1$; only the PDF dependence is needed for the fractions below.

Write the common sea PDF as $S=\bar{u}=\bar{d}=\bar{s}=\bar{c}=\Sigma /8$. The Valence and sea decomposition of proton-antiproton W production gives the three parton luminositiesThe factor two in the sea-sea term includes both beam assignments. The antiproton's valence antiquarks make valence-valence Annihilation possible.

At $y=0$, both fractions equal $x=0.0402$. Reading the original plot near this value gives representative momentum densitiesThese are approximate graphical readings, not a precision PDF fit. Undo the sea-curve scale and divide by eight flavours/antiflavours:All products have the same $1/x^2$ factor, which cancels in the fractions. The three relative weights areTheir sum is $0.3128$, givingReading the central curves more finely changes these percentages by about a percentage point, well within what the plot supports. Omitting its factor $0.05$ would substantially underestimate the sea contribution.

Use leptonic decays: an isolated high transverse momentum positive electron or muon, together with substantial missing transverse momentum from the neutrino, provides a clean $W^{+}$ signature. The lepton transverse momentum has a Jacobian scale near $m_W/2$, and the lepton–missing-momentum transverse mass has an edge near $m_W$, broadened by width, recoil and resolution. Lepton identification, isolation and a prompt primary-vertex origin suppress Jets faking leptons and heavy-flavour decays. Missing-momentum quality requirements suppress mismeasured multijet events; a veto on a second oppositely charged lepton suppresses $Z\to {\ell }^{+}{\ell }^{-}$ when both are reconstructed. Hadronic $W$ decays have a much larger ordinary-QCD background.

The flavour states $K^0=d\bar{s}$ and ${\bar{K}}^0=s\bar{d}$ have definite Strangeness and are distinguished by flavour-sensitive production or decay. The propagation states diagonalize the effective Hamiltonian $H_{\mathrm{eff}}=M-i\Gamma /2$; their complex eigenvalues give a pole mass and a decay width. The common flavour combinations specified in the source assume the usual CPT-symmetric two-state description and are normalized with $|p{|}^2+|q{|}^2=1$.

Choose the phase convention $CP|K^0⟩=|{\bar{K}}^0⟩$. If mixing conserves CP, $p=q$ can be chosen, and the symmetric and antisymmetric combinations are CP even and odd. More generally the equality must be understood with the corresponding flavour-state phase convention; $|p|=|q|$ alone is not a statement fixing an arbitrary written phase.

The almost CP-even state decays rapidly to two pions, whereas the almost CP-odd state is largely restricted to slower channels such as three pions. Their very different lifetimes are easier to distinguish than their tiny mass splitting, motivating the names $K_S^0$ and $K_L^0$. In the source's first notation the subscript $L$ means light, not long-lived: the light state is identified with the short-lived kaon, and the heavy state with the long-lived kaon. Below, $\ell ,h$ label light/heavy to avoid that collision.

The two indistinguishable paths to the same final state must be added as probability amplitudes:Taking the squared modulus gives the time-dependent decay rate, with the common final-state phase-space normalization absorbed into the probability amplitudes:The cross term is the interference between direct decay and decay after flavour conversion. For reference, diagonal propagation gives $g_{\pm }=(e_{\ell }\pm e_h)/2$, where $e_j=e^{-i{m}_{j}t-{\Gamma }_{j}t/2}$ and $t$ is proper time. The unfactored expression remains usable if $A_f=0$, when the ratio $\lambda$ itself is undefined.

Starting with an antikaon instead givesThusThis factored form assumes $A_f,{\bar{A}}_f\ne 0$; otherwise use the preceding probability amplitude. The three mechanisms are separated below. Comparisons for distinct final states must pair $f$ with its CP conjugate $\bar{f}$.

Direct CP violation means unequal magnitudes of CP-conjugate decay probability amplitudes:It changes decay probabilities even at $t=0$, before any mixing. For a common CP eigenstate final state, this reduces to $|A_f|\ne |{\bar{A}}_f|$. Usually it requires interfering decay contributions with different weak phases and different strong phases; it is not equivalent to a nonzero weak phase in one isolated probability amplitude.

CP violation in mixing occurs whenThe flavour-transition probabilities are proportional to $|q/p{|}^2|g_{-}{|}^2$ for $K^0\to {\bar{K}}^0$, and $|p/q{|}^2|g_{-}{|}^2$ for the reverse. Their inequality remains when decay probability amplitudes themselves conserve CP. The coefficients multiplying the mixed-decay terms in the rate equations encode this asymmetry.

Interference CP violation concerns the relative phase between direct decay and decay after mixing. For a common CP eigenstate final state, it can occur with both $|q/p|=1$ and $|{\bar{A}}_f/A_f|=1$, so neither of the first two mechanisms is needed. In that limit $|\lambda |=1$, butproduces different interference terms for $\lambda$ and ${\lambda }^{-1}={\lambda }^{\ast }$. With $\Delta m=m_h-m_{\ell }$,The term proportional to $\mathrm{Im}\lambda \sin (\Delta mt)$ consequently reverses sign. In the CP-conserving limit, $\lambda$ is the real CP parity of the final state in a consistent phase convention. For non-CP-eigenstate final states, compare the full coefficients in CP-conjugate rates rather than imposing this simplified criterion.

Introduce all four physical probability amplitudes: $A_f,{\bar{A}}_f,A_{\bar{f}},{\bar{A}}_{\bar{f}}$. Under exchange of the flavour and final-state labels, defineThis CP-conjugate decay-amplitude ratio gives the conjugate-channel ratesCP is a unitary discrete transformation, so exchanging state labels does not mean complex-conjugating every strong phase by fiat. Nor should one assume that CP-conjugate probability amplitudes are equal while studying CP violation.

The equality $\bar{\lambda }={\lambda }^{-1}$ holds whenin particular when $\bar{f}$ and $f$ are the same CP eigenstate ray, with a consistent state phase. It is not automatic for two distinct CP-conjugate final states. For example, $p=q$, $A_f={\bar{A}}_{\bar{f}}=2$ and ${\bar{A}}_f=A_{\bar{f}}=1$ conserve the paired decay probability amplitudes but give $\lambda =\bar{\lambda }=1/2$, not inverse values.

Assume the usual semileptonic flavour selection rule $\Delta S=\Delta Q$, negligible wrong-sign decay probability amplitudes, and equal magnitudes for CP-conjugate allowed probability amplitudes. Then $f_{+}={\pi }^{-}{\ell }^{+}\nu$ is reached directly only from $K^0$, and $f_{-}={\pi }^{+}{\ell }^{-}\bar{\nu }$ only from ${\bar{K}}^0$. Put $|A_{f_{+}}|=|{\bar{A}}_{f_{-}}|=|A|$. The four rates areThere is only one path to each tagged semileptonic channel, so no direct/mixed decay interference survives. The semileptonic wrong-sign mixing asymmetry isIt isolates CP violation in mixing under the stated assumptions. Allowing direct semileptonic CP violation or wrong-sign probability amplitudes would spoil that isolation.^[ref]

Use dimensionless physical helicity $h=\mathbf{S}\cdot \hat{\mathbf{p}}$, with spin eigenvalues $\pm 1/2$ in natural units. In the Dirac representation,The entries $1$ in the supplied transformation denote two-by-two identities. Since its matrix acts only on the upper/lower spinor blocks, it commutes with the repeated Pauli block. Therefore $u^{-1}=u$ andThe eigenvalues remain the physical spin projection along the direction of motion, $+1/2$ or $-1/2$. A convention omitting the factor one half instead labels them $\pm 1$; this changes no measurement probabilities. A change of gamma-matrix representation cannot change the spin observable.

The two Weyl spinors transform in the inequivalent left- and right-chiral representations of the proper Lorentz group. They are selected by $P_{L,R}=(1\mp {\gamma }^5)/2$ and have different boost transformation laws; parity exchanges the two. A massive Dirac equation couples them, whereas in the massless limit they can propagate independently.

For the particular $u$ printed in this paper, transforming its supplied Dirac ${\gamma }^5$ givesThus the upper block is right chiral and the lower block left chiral in this convention. Many references use the opposite block order; neither convention changes physics, but they must not be mixed. Chirality is not generally the same as helicity for a massive electron. The weak charged current selects the left-chiral electron component, reducing to negative helicity at high energy.

Put $r=p/(E+m)$ and assume $p>0$. The common plane wave has unit modulus. The component norm isFor unit spinor probability density choosewhere $E^2=p^2+m^2$ was used. This is spinor normalization, not square normalization of a plane wave over infinite space; a box or wavepacket gives the corresponding spatial normalization. If the convention is instead ${\psi }^{†}\psi =2E$, use $N=\sqrt{(E+m)/10}$.

For momentum along $+z$, the positive-helicity components are the first and third, while the second and fourth have negative helicity. HenceBoth are independent of speed. In the source's chiral block order the exact left-chiral weight is $1/2+3p/(10E)$, tending to $4/5$ at high energy; helicity and chirality probabilities agree only in that limit.

The charged-current process iswith a $t$-channel $W^{-}$ emitted at the electron-to-neutrino vertex and absorbed at the antimuon-to-antineutrino vertex. It is not an $s$-channel Annihilation of the unlike-flavour charged leptons into one $W$.

At high momentum transfer the incident electron's left-chiral fraction tends to $4/5$. In the collision centre-of-mass frame, the incoming antimuon moves along $-z$ and its charged-current allowed right helicity has spin along $-z$. A rest-frame spin at Angle $\theta$ to $+z$ projects onto that spin direction with probability ${\sin }^2(\theta /2)$. Therefore, if ${\sigma }_0$ denotes the cross section for a pure left-helicity electron and fully selected antimuon spin,It vanishes at $\theta =0$ in the ultrarelativistic selection limit and is maximal at $\theta =\pi$. Finite-mass corrections can fill the suppressed orientation. The absolute normalization depends on beam energy and the exchanged-boson propagator; the spin-angle dependence is what is fixed here.

For an incoming ${\mu }^{-}$, a putative $e^{-}{\mu }^{-}\to {\nu }_e{\nu }_{\mu }$ process would change total electric charge from $-2$ to zero. It is forbidden: both negative charged leptons would have to emit a $W^{-}$, and a single exchanged charged boson cannot connect those two emission vertices. Ordinary elastic $e^{-}{\mu }^{-}\to e^{-}{\mu }^{-}$ scattering remains possible through $t$-channel photon or $Z$ exchange, shown in the figure, but this is not the charged-current reaction of part d. The neutral-current weak contribution is distinct from the electromagnetic one.

A proton contains up quarks on which the charged current can act:through $W^{-}$ exchange, weighted by $V_{ud}$. Spectator quarks complete the hadronic final state. At the exclusive nucleon level the charge assignment is $e^{-}p\to {\nu }_{e}n$; at the high momentum transfers specified, the generally dominant description is inclusive $e^{-}p\to {\nu }_{e}X$ with proton structure and hadronic breakup, not an unsuppressed pointlike proton-to-neutron transition. The quark-level diagram in the figure displays this allowed charge flow. The electron's left-chiral selection is the same as in part d.

A fast charged particle loses energy by ionizing and exciting silicon, creating electron–hole pairs along its path. In a reverse-biased microstrip diode, the electric field separates and drifts the electrons and holes toward opposite electrodes. Their motion induces a current on the strips, which is integrated by the readout electronics to locate the hit.

The bias depletes the bulk of mobile equilibrium carriers, reducing conduction and capacitance, and supplies rapid directed charge collection before recombination or trapping. Without a collecting field, diffusion is slow and charge spreads or recombines, degrading signal and position measurement. The signal is generated by the traversing particle's energy deposition, not by the bias creating the ionization charge.

Treat the $p^{+}$ strips as much more strongly doped than the $n$ bulk, so the depletion layer lies predominantly in the latter. Fully ionized donors leave uniform space charge $\rho =e{N}_D$. In one dimension,At the depletion boundary $x=d$, the minimum-bias solution has $E(d)=0$. Thus $E(x)=e{N}_D(x-d)/{ϵ}_{\mathrm{Si}}$, and the magnitude of the potential drop isFor $N_D=10^{18}{\mathrm{m}}^{-3}$, $d=300\mu \mathrm{m}$, and ${ϵ}_r=11.6$, this isThe applied reverse bias is approximately this value; more precisely the built-in junction potential contributes to the total depletion voltage and is subtracted from the required external bias. It is small on the scale of seventy volts and is not specified. Here the stated free-carrier density is used as the effective donor density of the neutral bulk before depletion.

For binary strip readout, report the centre of the hit strip. With uniform impact position across pitch $w$, the position error is uniformly distributed on $[-w/2,w/2]$, soThis assumes no analogue charge sharing between strips.

The magnetic force is $d\mathbf{p}/dt=q\mathbf{v}\times \mathbf{B}$. Since relativistic momentum is parallel to velocity, $d(p^2)/dt=2\mathbf{p}\cdot q(\mathbf{v}\times \mathbf{B})=0$. A uniform transverse field therefore changes momentum direction but not magnitude. If the trajectory turns at angular speed $v/R$, then $|d\mathbf{p}/dt|=p_{T}v/R$. Equating force magnitudes yieldsThis is the positive radius. The sign of $q$ determines the direction of curvature; the source's signed formula can describe an oriented radius. In practical tracking units,

The source labels $L$ as the full chord from $P_1$ to $P_3$, with $L/2$ on each side of its midpoint. The Right triangle between the Circle centre, chord midpoint and endpoint givesFor fixed geometry and field, the inverse transverse momentum isIf the uncertainty is dominated by the central point, ${\sigma }_s\simeq {\sigma }_x$. Linear propagation of uncertainty givesUsing $|q/e|=1$, $B=2\mathrm{T}$, $L=0.6\mathrm{m}$ and ${\sigma }_x=28.87\mu \mathrm{m}$,The inverse-momentum resolution is approximately momentum independent in this spatial-error-only model; the relative momentum resolution grows as $p_T\sigma (1/p_T)$. Endpoint errors and multiple scattering are excluded by the stated assumptions.

For negligible fermion masses, the neutral-current vertex connects a fermion and antifermion of opposite helicity. The two incoming choices are $e_L^{-}{e}_R^{+}$ and $e_R^{-}{e}_L^{+}$; the two outgoing choices are $f_L{\bar{f}}_R$ and $f_R{\bar{f}}_L$. Their combinations give four channels, denoted $LL,LR,RL,RR$, where each letter refers to the fermion rather than antifermion.

Take the incoming electron momentum along $+z$ and Positron momentum along $-z$. The initial $L$ pair has both spins along $-z$, giving total projection $-1$; the $R$ pair has both along $+z$, giving projection $+1$. If the outgoing fermion moves along $\hat{\mathbf{n}}$, an outgoing $L$ pair has both spins along $-\hat{\mathbf{n}}$ and an $R$ pair both along $+\hat{\mathbf{n}}$. The figure shows momentum and spin arrows separately for all four channels.

Let $z=\cos {\theta }^{\star }$. Rotating the spin-one intermediate state between initial and final axes gives probability amplitudes proportional to $d_{\pm 1,\pm 1}^1$ for equal fermion helicities and $d_{\mp 1,\pm 1}^1$ for opposite ones. Thus, apart from a common positive normalization,

The common factor $1/4$ from the rotation elements is absorbed into the normalization. Distinct external helicities are orthogonal, so an unpolarized rate adds their squared probability amplitudes incoherently.

The source interchanges the conventional left and right electron coupling labels. With $g_L=T_3-Q{\sin }^2{\theta }_W$ and $g_R=-Q{\sin }^2{\theta }_W$, the physical electron values areAt ${\sin }^2{\theta }_W=0$, for example, the left electron still carries weak isospin $-1/2$, while the right electron is a weak singlet; the printed assignment gives the opposite result. This fixes how the helicity-channel table should be interpreted.^[ref]

For a muon final state, lepton universality gives the same pair of couplings. Crucially, swapping both labels does not change the summed distribution. Put $L=g_L$, $R=g_R$; thenHere $\mathcal{N}$ includes the resonance propagator, initial spin average and common coupling normalization. Equivalently,The source's simultaneous label interchange reverses ${\mathcal{A}}_{\ell }$ but leaves its square unchanged.

For the Pole forward-backward asymmetry for charged leptons, integrating the normalized distribution over the two hemispheres cancels the even term in the numerator. Since ${\int }_{-1}^1(1+z^2)dz=8/3$ and ${\int }_0^{1}zdz-{\int }_{-1}^{0}zdz=1$,Within this single-$Z$, tree-level model, a nonzero asymmetry requires unequal left/right coupling magnitudes, demonstrating parity violation in the vertices. For $g_L=g_R$ the odd angular term vanishes and $A_{FB}=0$; equal magnitudes with opposite signs also give zero. This inference is process-specific: a forward-backward asymmetry is not universally a parity-odd observable, since reversing both beam and outgoing momenta leaves their relative Angle unchanged. Purely electromagnetic Bhabha scattering is a concrete parity-conserving process with a forward-peaked, asymmetric distribution.

For the supplied ${\sin }^2{\theta }_W=0.23153$,The measurement differs by $0.0188-0.01620=0.00260$, or $0.79$ times its quoted uncertainty of $0.0033$, so it is consistent with the prediction. Statistical fluctuations alone preclude exact equality; photon exchange, initial-state radiation and electroweak radiative corrections also modify the idealized pure-resonance prediction and provide physical reasons for small differences.