Past exam of the mathematics course of the University of Cambridge / 2003 / iii / Paper 49 (The Standard Model)

Assume $\lambda >0$ and $v>0$. The minima of the scalar potential have $|\varphi |=v$. Choose the vacuum expectation value ${\varphi }_0=v{e}_3$. Its stabilizer subgroup consists exactly of matrices $\mathrm{diag}(R_2,1)$ with $R_2\in O(2)$, soThe continuous spontaneous symmetry breaking is $SO(3)\to SO(2)$ and has two broken generators. For the full orthogonal group, the cross-product notation requires the gauge connection to transform in the adjoint, as an axial vector: $R[A{]}_{\times }{R}^{-1}=[(\det R)RA{]}_{\times }$. The scalar field transforms as the ordinary vector $R\varphi$. This axial transformation of orthogonal gauge connections makes the stated covariant derivative transform correctly even for $\det R=-1$; assigning an ordinary-vector transformation to the connection would fail.

Near this nonzero vacuum, unitary gauge removes the two angular Goldstone bosons and gives $\varphi =(0,0,v+h)$. PutThese are the physical charged gauge bosons, neutral massless gauge boson, and radial scalar field. Directly substituting into the original gauge field strengths givesThe scalar field kinetic term has no radial-vector cross term, because ${\partial }_{\mu }\varphi$ and ${\mathbf{A}}_{\mu }\times \varphi$ lie in perpendicular internal directions. The complete physical gauge-scalar Lagrangian becomesFor each real vector the mass term is $m^2{A}_{\mu }{A}^{\mu }/2$, while for the complex pair it is $m_W^2{W}_{\mu }^{+}{W}^{-\mu }$. Hence the adjoint triplet Higgs spectrum isThe radial scalar also has $m_h^2=\lambda v^2$ in the question's potential normalization. The two lost scalar modes supply the longitudinal polarizations of the massive vectors through the Higgs mechanism.

For the lepton kinetic terms let $P_L=(1-{\gamma }^5)/2$, $P_R=(1+{\gamma }^5)/2$, and denote the left neutral triplet entry by $n_L=c_{\alpha }{\nu }_L+s_{\alpha }{N}_L$. The missing singlet is the perpendicular combinationIts overall sign is immaterial. The real orthogonal matrix $\left(\begin{array}{cc}{c}_{\alpha }& s_{\alpha }\\ -s_{\alpha }& c_{\alpha }\end{array}\right)$ is orthogonal, so the sum of the kinetic terms of $n_L$ and $L^S$ is ${\bar{\nu }}_{L}i/\partial {\nu }_L+{\bar{N}}_{L}i/\partial N_L$ with no mixed terms. Combining with the right triplet gives ordinary Dirac field kinetic terms for $E^{+},N,e$, and only a left-handed kinetic term for the massless neutrino. This is orthogonal completion of mixed neutral lepton kinetic terms. The heavy charged field is $E^{+}$, as printed in the PDF, not the TeX's $N^{+}$.

To compute the charge commutator, the canonical anticommutation relations implyThe quartic terms cancel on reordering the fermions. Thus well-defined normal-ordered integrated bilinears obey fermionic bilinear charge algebra, $[Q_C,Q_D]=Q_{[C,D]}$. Left/right mixed commutators vanish since $P_L{P}_R=0$.

In either triplet, writeThe displayed current has a factor two relative to the projected bilinears, since $1\mp {\gamma }^5=2P_{L,R}$. Consequently its charges are $T^{+}=2(Q_K^L+Q_K^R)$ and $T^{-}=(T^{+}{)}^{†}$. The neutral combinations have canonical anticommutators because $c_{\alpha }^2+s_{\alpha }^2=1$. Applying the matrix commutator giveswhere $Q_{\ell }=e\int :(E^{+†}{E}^{+}-e^{†}e):$ is the electromagnetic charge in the lepton sector. The neutrino, heavy neutral lepton and singlet contribute zero. This establishes triplet charged-current closure on electromagnetic charge, with the requested proportionality independent of normalization choices for the weak Noether charges. The PDF correctly writes $J^{-}=(J^{+}{)}^{†}$; the TeX repeats $J^{+}$ incorrectly. Equal-time algebra does not require conservation of the separate fermionic weak currents after symmetry breaking. For full gauge-theory charges, the corresponding charged-boson contributions must also be included.

Use the source convention $Q=T_3+Y$, with electric charge measured in units of the positron charge. Expanding the covariant derivative in the fermion kinetic term gives interaction $-\bar{f}{\gamma }_{\mu }(g{T}_3{A}_3^{\mu }+g^{\prime }Y{B}^{\mu })f$. Substituting the given neutral-field rotation, the coefficient of $Z^{\mu }$ isThe photon coefficient is $g{s}_W(T_3+Y)=eQ$, with $e=g{s}_W=g^{\prime }{c}_W$. For a Dirac field, $T_3$ acts only on its left-handed component. ThusThe weak neutral current interaction consequently has the requested form, withThe electroweak representation and hypercharge table gives the following values for one generation:

These neutral-current vector and axial couplings use the source's vertex prefactor $g/(2c_W)$. They are twice the coefficients used with prefactor $g/c_W$; the latter convention must not be mixed into this calculation. For the neutrino, the chiral projector removes the inactive right-handed component automatically.

Set $G=g/(2c_W)$ and suppress the species label. For outgoing momenta $k_1,k_2$, the amplitude is, up to an irrelevant phase,The final-state spin sum is the Dirac trace of $/k_1{\gamma }^{\mu }(c_v-c_a{\gamma }^5)/k_2{\gamma }^{\nu }(c_v-c_a{\gamma }^5)$. Anticommutation of ${\gamma }^5$ with the gamma matrices makes its symmetric partThe vector-axial cross term is proportional to the antisymmetric Levi-Civita symbol, so it contracts to zero with the symmetric massive vector polarization sum $-g_{\mu \nu }+p_{\mu }{p}_{\nu }/M_Z^2$. Since $k_1^2=k_2^2=0$, the symmetric tensor is transverse to $p$ as well. Only $-g_{\mu \nu }$ remains, andMultiplying by $G^2$ proves the massless fermion Z decay spin sum:

For the decay width, average over the three initial polarizations. Rotational invariance makes the integrated width the same for each initial spin state, so the massive-vector spin average is legitimate even for a single prepared polarization. In the rest frame, the delta functions in two-body Lorentz-invariant phase space leaveThis follows by setting ${\mathbf{k}}_2=-{\mathbf{k}}_1$ and integrating $\delta (M_Z-2|{\mathbf{k}}_1|)$; the radial integration supplies a factor $1/2$. HenceFor a quark species multiply by three colors, with strong-interaction radiative corrections when comparing with data. For one effectively massless active neutrino, $c_v=c_a=1/2$, so ${\Gamma }_{\nu }=g^2{M}_Z/(96\pi c_W^2)=G_F{M}_Z^3/(12\sqrt{2}\pi )$.

The total resonance width contains visible charged-lepton and hadron decays plus $N_{\nu }{\Gamma }_{\nu }$. Thus the measured invisible contribution determinesEarly resonance measurements supported three light active neutrino species, with experimental uncertainties; the OPAL analysis of its 1989 data explicitly compares the three- and four-species predictions. Under the question's assumption that every possible family has a massless neutrino with ordinary weak couplings, this constrains the number of such families. The inference does not exclude additional neutrinos heavier than $M_Z/2$ or gauge singlet sterile states with no ordinary $Z$ coupling. This is light-neutrino counting from the Z width.

Let $\epsilon =i{\sigma }_2=\left(\begin{array}{cc}0& 1\\ -1& 0\end{array}\right)$, so ${\epsilon }^{-1}=-\epsilon$. Direct multiplication of the Pauli matrices gives $\epsilon {\sigma }_i^{\ast }\epsilon ={\sigma }_i$, equivalently $\epsilon {\sigma }_i^{\ast }{\epsilon }^{-1}=-{\sigma }_i$. For $U=\exp (i{\theta }_i{\sigma }_i/2)\in SU(2)$ this impliesTherefore ${\varphi }^c=\epsilon {\varphi }^{\ast }$ transforms as the same weak fundamental representation doublet as $\varphi$, while complex conjugation reverses its hypercharge to $-1/2$. This is the Higgs conjugate doublet. The Yukawa coupling product $\bar{L}{\varphi }^c{u}_R$ is an $SU(2)$ singlet, and its hypercharges add toIt is consequently gauge-invariant and can supply an up-quark mass.

In unitary gauge, ${\varphi }^c=((v+H)/\sqrt{2},0{)}^T$. Writing $y_u$ for the full arbitrary coefficient of the Yukawa coupling term, its physical form isA one-generation field rephasing makes the mass positive, and its Higgs boson coupling is $m_u/v$. The printed Yukawa expression has an additional trailing $m$ after $R^{+}$ which is never defined. If it denotes a constant, absorb it into $y_u=f^{+}m$; in the conventional dimensionless Yukawa notation without that factor, $y_u=f^{+}$. It does not affect the gauge-transformation argument.

For three families, use a singular value decomposition of the two mass matrices: $U_{uL}^{†}{M}_u{U}_{uR}=D_u$ and $U_{dL}^{†}{M}_d{U}_{dR}=D_d$, with positive diagonal entries. The original left-handed weak current then becomes a current between mass eigenstates withA complex $3\times 3$ matrix has eighteen real parameters, and $U^{†}U=I$ imposes nine independent real conditions, leaving nine. A generic unitary matrix can be built from three complex plane rotations and three diagonal phases: each plane rotation supplies one angle and one phase. Thus these are three rotation angles and six phases. Rephase each of the six Dirac quark mass eigenfields, applying the same phase to its two chiralities so the diagonal masses stay real. A common phase changes no entry of $V$, leaving five effective rephasings. Removing five of the six phases leaves three angles and one physical phase. This CKM parameter counting assumes generic nondegenerate masses; degeneracies give additional basis freedom.

Exactly massless neutrinos have an arbitrary unitary flavor rotation that changes no mass term. After diagonalizing the charged-lepton mass matrix, choose the neutrino basis with the same left-handed rotation. The apparent lepton charged-current mixing matrix becomes the identity. Thus mixing of exactly massless neutrinos is removable; it is not an analogue of physical CKM mixing.

For the requested CP symmetry calculation, write $j_{ij}^{\mu }={\bar{u}}_i{\gamma }^{\mu }(1-{\gamma }^5)d_j$. The supplied charge conjugation transformations, with the fermion-reordering sign, giveFor $\Gamma ={\gamma }^{\mu }(1-{\gamma }^5)$, use $C({\gamma }^5{)}^T{C}^{-1}={\gamma }^5$ and the given gamma matrix relation to obtainUnder parity, the current acquires the Lorentz parity matrix and $1+{\gamma }^5$ becomes $1-{\gamma }^5$. Under charge conjugation the charged vector also changes sign and becomes its conjugate. The two minus signs cancel, and the parity matrices cancel in the contracted operator. ConsequentlyWith $k=g/(2\sqrt{2})$, the interaction is $-k\sum (V_{ij}{O}_{ij}+V_{ij}^{\ast }{O}_{ij}^{†})$. Thereforewith the transformed density evaluated at $x_P$. This is the CP transformation of a charged quark current.

Physical noninvariance requires a phase that cannot be removed by quark rephasings. The invariant quartetis the Jarlskog invariant, where $s_{ij}=\sin {\theta }_{ij}$ and $c_{ij}=\cos {\theta }_{ij}$ in the standard three-angle parametrization. If $J\ne 0$, no allowed row/column phases can make the matrix real, and the charged-current interaction violates CP symmetry. For example all three angles $\pi /4$ with $\delta =\pi /2$ give $J=1/(8\sqrt{2})\ne 0$.

The printed noninvariance claim is conditional on a nonzero physical phase. Arbitrary Yukawa matrices do not force it: $V=I$, or any real CKM matrix, gives a CP-invariant charged-current interaction. The correct statement is generic CP violation when the irreducible phase is nontrivial, as explained by quark rephasing and CP conservation.

Finally a right-handed neutrino has gauge representation $(1,1{)}_0$. Its Lorentz-invariant same-chirality bilinear with the field obtained by charge conjugation is therefore a gauge singlet and permits a Majorana mass term, conventionallyThe factor $1/2$ is the usual convention for a field paired with itself; it may be absorbed into the printed coefficient. A Higgs Yukawa interaction can independently give a Dirac mass term. The Majorana term violates lepton number by two units but preserves the gauge symmetries. Corresponding terms for the other charged Standard Model fermions carry nonzero electric charge, and quarks also carry nontrivial color charge. Thus this gauge-invariant right-handed neutrino Majorana mass is allowed while those charged-fermion analogues are forbidden. Merely introducing a right-handed neutrino permits mass; vanishing mass coefficients would still leave it massless.

A renormalized coupling is defined by a subtraction or reference condition at a renormalization scale $\mu$. Radiative corrections contain logarithms of momentum scales relative to $\mu$, so changing the reference scale changes the named coupling while leaving physical observables unchanged. Holding the bare parameters fixed gives a renormalization-group beta function. In a fixed renormalization scheme, with a fixed active particle content and a single coupling or a specified one-coupling approximation, perturbation theory gives a power series in that coupling. Its coefficients depend on the fields, representations and scheme; beyond the universal leading coefficients they need not be scheme independent. A theory with several couplings generally has coupled beta functions, so the question's scalar equation is schematic rather than literally the form of every quantum field theory. Also $d/d\ln {\mu }^2=\frac{1}{2}d/d\ln \mu$, which fixes factors of two.

For the stated leading equation let $t=\ln ({\mu }^2/{\mu }_0^2)$. Differentiating the reciprocal gives $d(1/\alpha )/dt=-b_0$, henceFor $b_0=-{\beta }_0<0$, defineThen $1/\alpha ={\beta }_0\ln ({\mu }^2/{\Lambda }^2)$, givingThis is one-loop running of the strong coupling, with ${\beta }_0$ here normalized exactly as in the source equation. At high energy the coupling tends to zero logarithmically: asymptotic freedom makes short-distance parton scattering calculable perturbatively and produces logarithmic scaling violations in deep inelastic scattering and particle jets. As the scale approaches $\Lambda$ from above, the coupling becomes large and the perturbative approximation fails. Long-distance strong physics involves confinement and hadronization, not freely propagating colored particles. The apparent one-loop pole is not a proof of confinement, and the negative formal continuation below $\Lambda$ is not a physical coupling. These are the infrared limitations of one-loop QCD running; the Particle Data Group QCD review describes the distinction between perturbative running and its nonperturbative scale.

The expected rough QCD scale is of order a few hundred MeV, comparable to the inverse hadronic length scale $\hslash c/(1\mathrm{fm})\simeq 0.2\mathrm{GeV}$. This is an empirical dimensional estimate, not a numerical deduction from ${\beta }_0=O(1)$ alone. Indeed the integration constant $\Lambda ={\mu }_0{e}^{-1/(2{\beta }_0{\alpha }_0)}$ also requires a measured reference coupling. Its precise value depends on flavor number, perturbative order and renormalization scheme.

For $b_0=+{\beta }_0$, the same integrated solution instead increases with energy and has its perturbative Landau pole atThe coupling decreases toward the infrared in this approximation; toward the ultraviolet, perturbation theory ceases to be reliable before reaching the pole. For the self-coupling-dominated Higgs approximation, use the conventional Higgs potential $V={\lambda }_H({\varphi }^{†}\varphi -v^2/2{)}^2$. ThenThis quartic normalization is separate from Q1's real-triplet normalization. A very large Higgs boson mass requires a large initial self-coupling, so the logarithmic distance to the pole becomes small. Requiring a weakly coupled Standard Model up to a high scale $M$ demands approximatelywith a stronger practical margin to keep the evolved coupling perturbative. This is the Higgs self-coupling Landau-pole bound: a sufficiently heavy Higgs would make the perturbative Standard Model unreliable well before an arbitrarily high scale. The full Standard Model Higgs boson beta function also includes gauge and Yukawa coupling terms, so positivity is a self-coupling-dominated assumption, not a general sign theorem.

For the requested bottom threshold, take the physical color number $N_c=3$. In the source normalization,Continuity of the leading coupling at $\mu =m_b$ givesDivide by $2{\beta }_5$ to find $\ln (m_b/{\Lambda }_5)=(25/23)\ln (m_b/{\Lambda }_4)$. Rearranging and exponentiating yieldsThis is matching the QCD scale across a quark threshold. The named integration constant changes when the beta coefficient changes; the matched coupling itself stays continuous at leading order. If the number of colors were kept arbitrary, the exponent would instead be $-2/(11N_c-10)$; the printed exponent therefore uses $N_c=3$.