Past exam of the mathematics course of the University of Cambridge / 2013 / iii / Paper 45 (The Standard Model)

Take real field coordinates ${\varphi }^I$; for complex scalar fields, split them into real and imaginary parts. Assume a smooth scalar potential and positive, canonically normalized kinetic terms in a relativistic theory. At a classical vacuum, $V_I({\varphi }_0)=0$. The Taylor expansion isThus the scalar mass matrix is the Hessian matrix, and its eigenvalues give squared masses of the linearized scalar excitations.

Put $X_a(\varphi )=i{t}^a\varphi$. Infinitesimal invariance of the scalar potential gives the identity $V_I{X}_a^I=0$ at every field value. Differentiate with respect to ${\varphi }^J$ and evaluate at the classical vacuum:Every nonzero infinitesimal symmetry direction therefore lies in the kernel of the scalar mass matrix. This is the classical Goldstone theorem expressed through Goldstone directions in the scalar mass matrix.

Let $H_0$ be the full stabilizer subgroup of ${\varphi }_0$. The linear map from the Lie algebra of $G$ to field space, $T\mapsto iT{\varphi }_0$, has kernel equal to the Lie algebra of $H_0$. The rank-nullity theorem givesThere are consequently at least $\dim G-\dim H_0$ massless scalar directions, namely the tangent directions to the symmetry orbit through the classical vacuum. If the intended $H$ is $H_0$, these are the $\dim G-\dim H$ symmetry-required Goldstone bosons. Exactly that many massless modes occur if the scalar mass matrix is positive definite on a complement of those tangent directions. A nonsingular positive field-space kinetic metric changes normalization, but not the number of zero masses.

Two qualifications are needed for the literal assumptions. A subgroup fixing the classical vacuum need not be the full stabilizer subgroup. For example, take $G=SO(3)$ acting on a real triplet and $V=\lambda (|\varphi {|}^2-v^2{)}^2/8$ with $\lambda ,v>0$. At ${\varphi }_0=v{e}_3$, the scalar mass matrix is $\lambda v^2\mathrm{diag}(0,0,1)$: there are two massless modes. Choosing $H=\{1\}$ satisfies the printed invariance condition but would incorrectly predict three. The full stabilizer subgroup is $SO(2)$.

Even with the full stabilizer subgroup, symmetry does not exclude an accidental massless scalar. Take $G=SO(2)$ rotating $(x,y)$, with an invariant singlet $z$, andAt $(v,0,0)$ the full continuous stabilizer is trivial, but the scalar mass matrix is $\mathrm{diag}(2\lambda v^2,0,0)$. One zero direction is the Goldstone boson; the other is an accidental massless scalar at quadratic order. The proof establishes the symmetry-required count, not unconditional equality with the total number of massless fields. The statement concerns global internal symmetry; gauging it changes the physical interpretation through the Higgs mechanism.

Assume $\lambda >0$ and $v>0$. The vacuum manifold is the sphere $|\varphi |=v$. Choose ${\varphi }_0=v{e}_3$. For the given generators, $t^3{\varphi }_0=0$ whereas $t^1{\varphi }_0$ and $t^2{\varphi }_0$ are nonzero and independent. The Adjoint representation of SU(2) rotates this sphere, and the full connected stabilizer subgroup of $v{e}_3$ is generated by $t^3$. Hence the classical symmetry pattern isThe central element of SU(2) acts trivially on the triplet and belongs to this stabilizer; it does not add a separate broken direction. The gauge-field-free representative has zero gauge field strength and constant scalar magnitude. Locally near this nonzero classical vacuum, unitary gauge aligns the triplet along the third axis.

First make the gauge normalization explicit. Write the gauge kinetic term as $-\mathcal{N}{F}_{\mu \nu }^a{F}^{a\mu \nu }/4$, where $\mathrm{Tr}(t^a{t}^b)=\mathcal{N}{\delta }^{ab}$ if $F=F^a{t}^a$ is the matrix used inside the printed trace. Canonical normalization of a gauge kinetic term givesFor a conventionally normalized component trace, $\mathcal{N}=1$ and $g_c=g$. If the trace is the ordinary matrix trace in the displayed three-dimensional representation, $\mathrm{Tr}(t^a{t}^b)=2{\delta }^{ab}$ and $\mathcal{N}=2$. The PDF does not specify which trace convention is intended; both are covered by this formula.

In unitary gauge, the gauge covariant derivative isExpanding this kinetic term and the scalar potential gives the complete physical-field LagrangianHere a square of a vector means its contraction with the Minkowski metric, in signature $(+---)$. The labels 1 and 2 represent massive vectors; 3 represents the surviving Abelian vector.

To display their charge and all interactions more clearly, put $a_{\mu }={\mathcal{A}}_{\mu }^3$, $W_{\mu }^{\pm }=({\mathcal{A}}_{\mu }^1\mp i{\mathcal{A}}_{\mu }^2)/\sqrt{2}$, and defineWith the sign of gauge field strength printed in the paper, ${\mathcal{F}}^3=f+i{g}_{c}C$ and $G^{\pm }=({\mathcal{F}}^1\mp i{\mathcal{F}}^2)/\sqrt{2}$. Thus the same Lagrangian becomesThe physical masses areIn particular, literal adjoint matrix trace gives $m_W=gv/\sqrt{2}$; the standard canonical component convention gives $m_W=gv$. These are descriptions with differently normalized couplings, not different physical spectra.

The complex vector pair carries opposite charges under the unbroken U(1) gauge symmetry. The gauge field strength terms contain $a{W}^{+}{W}^{-}$ cubic interactions, $aa{W}^{+}{W}^{-}$ quartic interactions, and four-vector interactions involving the charged fields. There is no pure Abelian cubic or quartic self-interaction. The Higgs mode $\eta$ has cubic and quartic scalar interactions and couples through $2g_c^{2}v\eta W^{+}{W}^{-}+g_c^2{\eta }^2{W}^{+}{W}^{-}$. It is neutral and has no tree-level $\eta aa$ interaction. This is the physical charged-vector Lagrangian for an adjoint SU2 Higgs model.

If the angular fields ${\pi }_1$ and ${\pi }_2$ were retained, their vanishing potential masses would identify the two Goldstone bosons of the ungauged triplet. In the gauge theory they mix with the broken-direction gauge fields and can be removed by gauge fixing; the Higgs mechanism uses them as the longitudinal polarizations of the two massive vectors. They are not additional physical massless scalars. The physical degrees of freedom are conserved: $3+3\times 2=9$ before rearrangement, and $1+2\times 3+2=9$ afterwards.

This is not the Standard Model electroweak interaction. It has three original gauge bosons, leaving two massive charged vectors and one massless neutral vector, with no massive neutral Z boson. The Standard Model instead has $SU(2{)}_L\times U(1{)}_Y$, a complex Higgs doublet, and three massive vectors plus the photon. Its charge is $Q=T_3+Y$, not just the surviving $T_3$. Adding fermions cannot supply the missing gauge generator or turn the surviving neutral vector into both a photon and a Z boson. In particular, the usual right-handed fermions in the Standard Model are SU(2) singlets; they would have zero charge if only $T_3$ were available, instead of the charges produced by hypercharge.

Use natural units and the Minkowski metric $(+---)$, and work at tree level with on-shell final particles. Write $M=m_H$ and, for a final particle of mass $m$, ${\beta }_m=\sqrt{1-4m^2/M^2}$. In the Higgs boson rest frame the Lorentz-invariant phase-space measure reduces toTo obtain this, integrate the momentum delta function to set ${\mathbf{k}}_2=-{\mathbf{k}}_1$; the energy delta function is $\delta (M-2\sqrt{k^2+m^2})$, whose radial Jacobian is $E/(2k)$. Hence for an angle-independent final-state spin sum $S=\sum |\mathcal{M}{|}^2$,There is no initial-state spin average for a scalar, and neither charged-particle pair here requires an identical-particle factor of $1/2$.

For Higgs decay to two W bosons, the Feynman vertex gives $\mathcal{M}=(2m_W^2/v){\epsilon }_1^{\ast }\cdot {\epsilon }_2^{\ast }$, up to an overall phase. Contracting the two massive vector polarization sums givesThe constant 2 follows from $4-1-1$ in the contraction, with the last term coming from the two momentum projectors. Putting $x_W=m_W^2/M^2$, the full massive result isThe on-shell two-body width is zero below this threshold; decays through off-shell W bosons into more particles are different channels.

For Higgs decay to a fermion pair, put $y_b=m_b/v$. The amplitude is $\mathcal{M}=-y_b\bar{u}(k_1)v(k_2)$, up to an overall phase and a color Kronecker delta. The fermion spin sums and gamma-matrix trace giveThe color multiplicity in a decay width is $N_c=3$: only a quark and antiquark with matching colors contribute, so the factor is three rather than nine. ThereforeAgain the on-shell two-body width is zero below threshold. This is a partonic tree-level answer with every mass retained; hadronization is outside the specified calculation.

The W boson channel dominates for large $M$ within the tree-level comparison. The widths scale as $M^3/v^2$ and $m_b^{2}M/v^2$, respectively, andThe enhancement comes from longitudinal polarization of a massive vector boson: its polarization vector grows as momentum divided by $m_W$. Thus the longitudinal pair survives the apparently small $m_W^2$ factor in the interaction. At masses so large that the scalar sector is strongly coupled, the tree-level extrapolation itself needs corrections.

Let $s=q^2>0$, neglect $m_e$, and average over the two spin states of each incoming particle. Only the single-photon channel is being considered. The spectral-density symbol ${\rho }_h$ is not defined in the question, so specify its normalization before calculating. A useful photon-field convention isHere $d{\Pi }_X$ is the product of final-particle measures $d^{3}k/[(2\pi {)}^{3}2E]$, including the appropriate sums and symmetry factors. Lorentz invariance and current conservation give the transverse tensor form. In this convention ${\rho }_h$ is the hadronic photon spectral density, with mass dimension $-2$, rather than a density of a stationary time series.

The electron annihilation amplitude can be written $e\bar{v}(p_{+}){\gamma }_{\mu }u(p_{-})⟨X|A^{\mu }|0⟩$, up to an overall phase. The leptonic tensor after the initial spin average isIt obeys $q^{\mu }{L}_{\mu \nu }=0$ and $g^{\mu \nu }{L}_{\mu \nu }=-s$. In the center-of-momentum frame, the flux denominator in the hint is $2s$, since the relative speed is 2 and each beam energy is $\sqrt{s}/2$. Contracting the tensors gives the inclusive relativistic cross-sectionHere $\alpha$ is the fine-structure constant. The $1/4$ 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, $J_h^{\mu }={\sum }_f{Q}_f{\bar{q}}_f{\gamma }^{\mu }{q}_f$, and write its inclusive tensor asEquivalently, for $i\int d^{4}x{e}^{iqx}⟨0|T{J}_h^{\mu }(x)J_h^{\nu }(0)|0⟩=(q^{\mu }{q}^{\nu }-q^2{g}^{\mu \nu }){\Pi }_J(q^2)$, one has ${\rho }_J=\mathrm{Im}{\Pi }_J/\pi$. At leading order in the electromagnetic interaction, $⟨X|A^{\mu }|0⟩$ is proportional to $e⟨X|J_h^{\mu }|0⟩/s$. ThusIf the symbol ${\rho }_h$ is instead used for this dimensionless hadronic electromagnetic-current spectral density, the last formula applies with ${\rho }_J$ renamed ${\rho }_h$. The normalization must not be silently switched between these formulas.

At $\sqrt{s}$ well above the strong-coupling scale of Quantum chromodynamics, asymptotic freedom makes production over distances of order $1/\sqrt{s}$ 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 $Q_{f}e$, the tree-level scattering amplitude isThe two gamma-matrix traces, the initial spin average, and the final color charge sum yieldwhere $t=-s(1-\cos \theta )/2$ and $u=-s(1+\cos \theta )/2$. 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 $m_f^2<s$ and are treated as massless; the others are omitted. This is the stipulated step approximation, not the exact pair-production threshold $s\ge 4m_f^2$.

The hadronic R ratio is $R=N_c\sum Q_f^2$, relative to the massless muon-pair relativistic cross-section $4\pi {\alpha }^2/(3s)$. If $N_u$ and $N_d$ active flavors have charges $2/3$ and $-1/3$, respectively,For $N_c=3$, the usual sets of three, four, five and six active flavors give $R=2,10/3,11/3,5$. In the two spectral conventions the same leading calculation gives

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 $M$, processes with characteristic energy and momentum transfers $E\ll M$ 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 ${\mathcal{O}}_i^{(d)}$ has mass dimension $d$, the dimensionless $C_i^{(d)}$ are Wilson coefficients, and $\mu$ is a renormalization scale. In a relativistic vacuum with canonical fields, a typical insertion of a dimension-$d$ interaction contributes an additional power $(E/M{)}^{d-4}$, 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 $\phi$ and a heavy real scalar field $S$, withThe coupling $a$ has mass dimension one. The light-field symmetry is $\phi \mapsto -\phi$. For $m^2\ge 0$ and $\lambda >3a^2/M^2$, the displayed scalar potential is bounded below: completing the square in $S$ leaves a positive light quartic. Thus the example can be treated as a stable theory around $S=\phi =0$, with weak enough couplings for the tree approximation.

At tree level the heavy equation of motion is $(M^2+\square )S=-a{\phi }^2/2$. 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 ${\lambda }_{\mathrm{eff}}=\lambda -3a^2/M^2$ in the $-{\lambda }_{\mathrm{eff}}{\phi }^4/4!$ convention. After integration by parts, the next term is $+a^2({\partial }_{\mu }{\phi }^2)({\partial }^{\mu }{\phi }^2)/(8M^4)$; it is a dimension-six local operator. Writing $a=g_{\ast }M$ puts its coefficient in the usual $g_{\ast }^2/M^2$ 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 $\phi \phi \to \phi \phi$. With $s,t,u$ the Mandelstam variables, the full theory has three heavy-exchange channels:For $|s|,|t|,|u|\ll M^2$,The effective field theory reproduces these terms in order: a shifted quartic, then local derivative interactions. Since $s+t+u=4m^2$ on shell, the first derivative correction is a light-mass-dependent constant; it vanishes for $m=0$. 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 $s=M^2$, the full propagator is resonant and the truncated effective field theory fails; retaining $S$ 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.