Past exam of the mathematics course of the University of Cambridge / 2001 / iii / Paper 63 (The Standard Model)

The unbroken group is the stabilizer subgroup of the chosen vacuum expectation value:Its Lie algebra consists of generators $T$ with $T{\varphi }_0=0$. The vacuum manifold is the orbit $G{\varphi }_0\simeq G/H$ under the stated transitivity assumption. Its tangent space is the image of $T\mapsto T{\varphi }_0$, so its dimension is $r=\dim G-\dim H$.

Let $K_{ij}={\partial }_i{\partial }_{j}V({\varphi }_0)$ be the scalar mass matrix, with canonical scalar kinetic terms. Invariance of the scalar potential gives $({\partial }_{i}V)(T_a\varphi {)}_i=0$ for every $\varphi$. Differentiate with respect to ${\varphi }_j$ and evaluate at the minimum, where ${\partial }_{i}V=0$:Thus every independent tangent direction is a zero eigenvector of $K$. These are the Goldstone directions in the scalar mass matrix; for a global symmetry they are $r$ physical massless Goldstone bosons. For a generic nondegenerate minimum transverse to the orbit, they are precisely all the zero eigenvalues. Without that nondegeneracy assumption, additional massless scalars are possible.

For a local gauge group, use the orthogonal unitary-gauge slice near a scalar vacuum. Since ${\varphi }_0^T{T}_a{\varphi }_0=0$ by antisymmetry, the proposed condition is equivalent toIt sets the gauge-orbit components of the scalar fluctuation to zero. This is an admissible local unitary gauge: on a basis of independent broken tangent vectors $v_a=T_a{\varphi }_0$, the variation of these conditions under the broken gauge parameters is $v_a^T{v}_b$. This Gram matrix is invertible on that basis, so the implicit function theorem selects the required gauge transformation near the vacuum. Unbroken transformations remain as a residual $H$ gauge symmetry; no global gauge slice through zeros of the scalar field is being asserted.

Write $\varphi ={\varphi }_0+h$. To quadratic order, the covariant kinetic term containsThe middle term vanishes in this unitary gauge. The gauge-boson mass rank from a real scalar vacuum is therefore determined byThese are squared masses: the physical gauge boson masses are square roots of the matrix's nonzero eigenvalues after diagonalization. For real $c_a$,Its kernel consists exactly of the unbroken generators. With $g\ne 0$, the rank is $r$, so exactly $\dim G-\dim H$ gauge-field combinations become massive. Each removed scalar Goldstone mode supplies the extra longitudinal polarization of a massive gauge boson.

The remaining physical scalar fluctuations lie in the orthogonal complement of the orbit tangent space. They have no massless modes if $K$ is positive definite on that complement. The unconditional absence of massless scalars does not follow from the printed assumptions. For a concrete counterexample, take a real doublet, $G=SO(2)$ andIts entire minimum set is one circle orbit, as required, but its Hessian matrix vanishes at the minimum. Gauging the rotation gives one vector of squared mass $g^2{v}^2$ and removes the angular scalar. The radial scalar remains massless at quadratic order, since its potential starts at fourth order in the radial fluctuation. This is an accidental massless radial mode beyond Goldstone modes. Thus the intended no-massless-scalar conclusion needs nonzero transverse curvature.

In the Standard Model, the Higgs doublet contains four real components and breaks $SU(2{)}_L\times U(1{)}_Y$ to $U(1{)}_{\mathrm{em}}$. Three Goldstone bosons become the longitudinal components of $W^{\pm }$ and $Z$, while the photon remains massless. The remaining radial Higgs boson is massive for the ordinary quartic potential with positive curvature. With $⟨H⟩=(0,v_{\mathrm{EW}}/\sqrt{2}{)}^T$,The colour gauge group is unbroken. This is the Higgs mechanism, with the generic curvature condition realized by the standard Higgs potential.

Use the current convention in the question, whose vertex prefactor is $g/(2\cos {\theta }_W)$. The neutral-current vector and axial couplings are thenFor a charged lepton, $T_3=-1/2$, $Q=-1$, givingFor an active neutrino, $T_3=1/2$, $Q=0$, so $c_V=c_A=1/2$. The right-handed weak singlet contribution is purely vectorlike in the charge term; combining it with the left-handed contribution gives precisely this Dirac-current form. These coefficients would be halved if the vertex prefactor were $g/\cos {\theta }_W$ instead.

Let $p=k+k^{\prime }$ be the Z boson momentum and $k,k^{\prime }$ the outgoing massless lepton momenta. The decay amplitude isUse the fermion spin sums $\sum u\bar{u}=/k$ and $\sum v\bar{v}=/k^{\prime }$. Anticommutation of ${\gamma }_5$ with the Dirac gamma matrices gives the symmetric part of the trace,The vector-axial cross term is antisymmetric in $\mu ,\nu$, and hence drops out of the massive vector polarization sum. The massless equations also give $p_{\mu }{L}^{\mu \nu }=0$: in the symmetric part, the first two contractions sum to $p^{\nu }k\cdot k^{\prime }$, canceling the last; the antisymmetric term vanishes because $p=k+k^{\prime }$. The longitudinal $p_{\mu }{p}_{\nu }/m_Z^2$ term therefore makes no contribution.

Since $2k\cdot k^{\prime }=m_Z^2$, contraction with $-g_{\mu \nu }$ gives $4(c_V^2+c_A^2)m_Z^2$. Thus the massless fermion Z decay spin sum, including all three initial polarizations, isTo use this fully summed expression for the decay width of one unpolarized Z boson, apply the massive-vector spin average, dividing by three. The width of each fixed polarization is the same after integration over all directions, so that procedure also gives the physical single-particle width. Omitting the division would sum the widths of three initial states.

For two massless distinguishable final particles, the two-body Lorentz-invariant phase space isConsequentlyThe tree-level electroweak symmetry breaking relation is $m_W=m_Z\cos {\theta }_W$. Combining it with the Fermi constant normalization givesFor an active neutrino, the vertex itself projects out inactive chiralities, so no additional helicity factor is needed. This is the tree-level massless leptonic Z decay width; lepton masses and radiative corrections have been excluded.

First fix the photon momentum convention. With $Q^2=-q^2>0$, $\nu =P\cdot q>0$ and $x=Q^2/(2\nu )>0$, an absorbed photon gives final momentum $p_X=P+q$. This is the physical momentum support of a deep-inelastic hadronic tensor. The original PDF instead prints ${\delta }^{(4)}(P-p_X-q)$ while keeping that positive-$x$ convention. Taken literally, it would requirein the Bjorken scaling limit, leaving no physical final states. The sign is therefore inconsistent with the requested nonzero deep-inelastic structure functions. Below, use the physically consistent absorption delta function ${\delta }^{(4)}(P+q-p_X)$. Alternatively all photon-momentum and scaling-variable conventions would have to be reversed together.

Work in the leading parton model: the target is unpolarized; scattering is incoherent from massless collinear quarks and antiquarks; intrinsic transverse momentum and target-mass effects are neglected. The parton distribution functions count constituents and include their colour multiplicity. At this order, neglect strong radiative corrections and the resulting logarithmic scale evolution.

Let a quark carry momentum $p=\xi P$, using a lightlike reference $P$ at leading order, and let $r=p+q$. The spin-averaged electromagnetic trace isIntegrating its one-particle final Lorentz-invariant phase space, with the $1/(4\pi )$ tensor normalization, supplies ${\delta }_{+}(r^2)/2$. Thus the spin-averaged electromagnetic parton tensor isOn physical support, $r^2=0$ gives $\xi =x$ and $p\cdot q=Q^2/2$. Define the transverse tensorsThe bracket in $w_f$ becomesThis follows by writing $p=\xi \tilde{P}-q/2$ on shell and expanding; it also verifies transversality to $q$.

The correct parton convolution normalization for a hadronic tensor isThe $1/\xi$ converts number weighting into covariant target-state normalization: a parton response per constituent is normalized by $2p^0$, whereas the hadron tensor uses $2P^0$, and $P^0/p^0=1/\xi$. Antiquark charge changes sign but its square and symmetric spin trace are the same.

Use the parton on-shell delta identity,Matching the two tensor coefficients now givesIn particular $F_2=2x{F}_1$, the Callan-Gross relation for spin-one-half partons. Scaling here means the naive leading approximation; full Quantum chromodynamics produces scale-dependent distributions and corrections.

Each quark contributes $+1$ and each antiquark contributes $-1$ to its flavour-vector charge. Sea pairs therefore cancel in the quark number sum rule:Here $N_f$ is the net flavour number, not the total number including sea pairs. This condition by itself fixes no antiquark distribution.

Under the separately imposed valence-only approximation, ${\bar{q}}_f=0$ and only up/down distributions remain. The proton has $(N_u,N_d)=(2,1)$ and the neutron has $(1,2)$. With $Q_u=2/3$, $Q_d=-1/3$, the valence-only proton and neutron structure-function moments areNo extra factor of three belongs here: colour is already included in the constituent distributions and the flavour counts. These are conditional valence-model moments, not unrestricted asymptotic statements in a theory whose evolution generates a sea.

Use the electroweak hypercharge normalization $Q=T_3+Y$, rather than $Q=T_3+\hat{Y}/2$. In the latter convention all tabulated hypercharges would be doubled. The electroweak representation and hypercharge table for one minimal Standard Model generation isThe quarks are colour triplets, the leptons are colour singlets, and all listed right-handed fields are weak singlets. A sterile right-handed neutrino, if added, has representation $(1,1{)}_0$ and changes none of the anomaly traces.

On each left weak doublet, $T_3=\mathrm{diag}(1/2,-1/2)$. The electric charges are therefore $1/6\pm 1/2=(2/3,-1/3)$ for $Q_L$ and $-1/2\pm 1/2=(0,-1)$ for $L_L$. On each right singlet $T_3=0$, so its electric charge equals its hypercharge. These are the observed quark and lepton charge assignments.

For Standard Model anomaly cancellation, traces must include both components of each doublet and all three colours. Every right weak singlet has $T_3^2=0$, givingOn the left doublets, $T_3^2=I/4$, while ${\mathrm{tr}}_L{T}_3=0$ separately in each doublet. ConsequentlyThe linear trace isEquivalently, ${\mathrm{tr}}_{L}Q=3(2/3-1/3)+(0-1)=0$. This proves cancellation of the mixed weak-hypercharge gauge anomaly within one generation.

The one-generation hypercharge anomaly traces for the cubic condition areThus both required anomaly differences vanish generation by generation. The right-handed subtraction is important: counting all physical fields with the same chirality sign would not give the cubic cancellation.

For massless up and down quarks, the Quantum chromodynamics kinetic term splits into two independent chirality sectors,The colour covariant derivative acts identically on both flavours and commutes with constant flavour matrices. Independent transformations $q_R\mapsto A{q}_R$, $q_L\mapsto B{q}_L$, with $A,B\in SU(2)$, therefore leave the action invariant. This is chiral symmetry. The non-singlet axial transformations have traceless flavour generators, so they are not broken by the colour axial anomaly; the anomalous axial $U(1)$ is not being included in this symmetry.

For $V_{ij}={\bar{q}}_{Lj}{q}_{Ri}$, transforming each index givesThe nonzero condensate $⟨V⟩=-vI$ is unchanged precisely when $A=B$. Thus chiral symmetry breaking leaves the diagonal vector subgroup $SU(2{)}_V$. Three of the six generators are broken, producing three massless Goldstone bosons in the exact massless limit. They are the pions, represented at low energy by the coset coordinate $U\in SU(2)$.

The uniqueness of the two-derivative two-flavour chiral action follows from the symmetry, rather than merely from guessing an invariant. The chiral group acts transitively on $U$, so every derivative-free invariant is constant. At $U=I$, the stabilizer acts on the three tangent directions by conjugation, the irreducible rotation representation. Its invariant symmetric quadratic form is a single multiple of ${\delta }_{ab}$. Transporting that form around the group gives a unique invariant two-derivative metric. Equivalently, the Maurer-Cartan form $U^{†}{\partial }_{\mu }U$ is traceless, so an independent double-trace term vanishes; the remaining contracted single trace is the same metric. Lorentz invariance supplies the spacetime contraction. Up to total derivatives, a constant term and its overall coefficient, the massless action is thereforeThe positive coefficient is conveniently expressed through the pion decay constant $F$.

Use the Pauli matrices with $\mathrm{tr}({\tau }_a{\tau }_b)=2{\delta }_{ab}$. Expanding the matrix exponential gives ${\partial }_{\mu }U=i{\tau }_a{\partial }_{\mu }{\pi }_a/F+\cdots$. Henceso the three fields have canonical kinetic terms. There is no nonconstant invariant potential in the exact chiral theory, consistent with their masslessness.

For real diagonal $M$, the two terms in the mass trace areTheir sum therefore gives the usual Dirac mass term,These terms explicitly break chiral symmetry. In the low-energy chiral perturbation theory prescription, $V\mapsto -vU$ yieldsThe Pauli matrix algebra implies $(\mathbf{\pi }\cdot \mathbf{\tau }{)}^2=|\mathbf{\pi }{|}^{2}I$. ThusandComparing with the canonically normalized scalar mass term gives the leading pion mass from a two-flavour condensate:Positive quark masses therefore give positive pion squared masses. Even if $m_u\ne m_d$, this leading action depends only on their sum and gives equal masses for all three pions. Strong isospin-breaking effects beyond this order and electromagnetic effects can split them. In the condensate convention used here, $⟨\bar{u}u⟩=⟨\bar{d}d⟩=-2v$, which also explains the factor of two in the mass relation. The massive pions are pseudo-Goldstone bosons, becoming massless as both quark masses tend to zero.