Past exam of the mathematics course of the University of Cambridge / 2002 / iii / Paper 66 (The Standard Model)

Use $Q=T_3+Y$ and $T_a={\tau }_a/2$. For each generation $r=e,\mu ,\tau$, the leptons areBoth are color singlets. The conjugate antilepton fields transform in the conjugate representations and have opposite hypercharge; they are already included by the conjugate fields in the Lagrangian. The two components of $L_r$ have electric charges $0,-1$, and $R_r$ has charge $-1$. There is no right-handed neutrino representation in this field content. These assignments use the electroweak representation and hypercharge table in the same hypercharge normalization as the scalar doublet.

For a doublet of hypercharge $Y$, defineThe $SU(2)$ term is absent for a singlet. Under the local transformation $\psi \mapsto e^{iY\omega }U\psi$, the gauge fields transform so that $D_{\mu }\psi \mapsto e^{iY\omega }U{D}_{\mu }\psi$. Consequently the required gauge-covariant kinetic terms areThe fields of definite chirality are defined with $P_{L,R}=(1\mp {\gamma }_5)/2$. Since $\bar{L}$ has hypercharge $+1/2$, the Yukawa interactionis invariant: its hypercharges add to zero, and $\bar{L}\varphi$ contracts the two doublet indices. A bare charged-lepton mass violates these assignments. No renormalizable neutrino Yukawa term is available without a right-handed neutrino. A gauge-invariant scalar potential depending on ${\varphi }^{†}\varphi$ can select the nonzero vacuum, without altering these kinetic and Yukawa constructions.

There is an explicit local unitary gauge transformation whenever $r_{\varphi }=(|{\varphi }_1{|}^2+|{\varphi }_2{|}^2{)}^{1/2}\ne 0$:Thus, with the real field $\rho =\sqrt{2}{r}_{\varphi }-v$,This proves the polar coordinates for the electroweak Higgs doublet constructively. The gauge choice is local about the nonzero vacuum; it need not be nonsingular at zeros of the doublet.

The three identical-generation kinetic terms permit unitary flavor changes. A singular value decomposition gives $U_L^{†}{Y}_{\ell }{U}_R=\mathrm{diag}(y_e,y_{\mu },y_{\tau })$ with $y_r\ge 0$. Rotate both components of each left doublet together, and the right singlets independently. In this basis,The left/right cross terms combine into this Dirac mass. Since the neutrinos have no mass matrix, their simultaneous left rotation preserves a diagonal charged current. This is charged-lepton Yukawa matrix diagonalization; the neutrinos remain massless in the stated renormalizable model.

The scalar kinetic term givesIntroduce $s_W=g^{\prime }/\sqrt{g^2+g^{\prime 2}}$ and $c_W=g/\sqrt{g^2+g^{\prime 2}}$. The combinations $W^{+}=(A_1-i{A}_2)/\sqrt{2}$, $W^{-}=(W^{+}{)}^{\ast }$, $Z=c_W{A}_3-s_{W}B$ and $A_{\gamma }=s_W{A}_3+c_{W}B$ therefore diagonalize the quadratic terms:The factor $1/2$ applies to the real neutral field, not to the complex charged field. The massless photon corresponds to the unbroken generator $T_3+Y$, which annihilates the vacuum. This gives the electroweak doublet gauge-boson mass matrix and its diagonal spectrum.

Take metric $g_{\mu \nu }=\mathrm{diag}(1,-1,-1,-1)$ and write $M=M_W$. The outgoing electron and antineutrino momenta are $k$ and $k^{\prime }$. Up to an irrelevant overall phase, the amplitude isA fermion spin sum givesAnticommuting ${\gamma }_5$ through the massless momentum and gamma matrix, and using $(1-{\gamma }_5{)}^2=2(1-{\gamma }_5)$, reduces the trace to twice the four-gamma trace minus twice its ${\gamma }_5$ counterpart. In particular its symmetric part isThe epsilon-tensor part is antisymmetric and contributes neither to the unpolarized polarization sum nor to the real linear polarization used below. There is no extra factor for selecting the neutrino helicity: the chiral vertex has already done so.

The external massless equations imply $p_{\mu }{J}^{\mu }=0$ for $p=k+k^{\prime }$, so $p_{\mu }{T}^{\mu \nu }=0$. With $k\cdot k^{\prime }=M^2/2$, averaging over the three initial spin states givesIndeed $-g_{\mu \nu }{T}_S^{\mu \nu }=16k\cdot k^{\prime }=8M^2$. This is a massive-vector spin average, not an average over the final spins.

In the rest frame, eliminating the momentum delta function and integrating the remaining radial delta function $\delta (M-2|\mathbf{k}|)$ reduces the two-body Lorentz-invariant phase space to $d{\Phi }_2=d\Omega /(32{\pi }^2)$. ThereforeThis derives the massless leptonic W decay width without invoking a precomputed decay formula.

For the specified polarization there is no initial spin average. Put $\mathbf{k}=(M/2)\hat{\mathbf{n}}$, ${\mathbf{k}}^{\prime }=-(M/2)\hat{\mathbf{n}}$ and ${\hat{n}}_z=\cos \theta$. The contraction selectsIt follows thatHere $\int {\sin }^2\theta d\Omega =8\pi /3$. The total rate agrees with the spin-averaged result, but its normalized polar distribution is $(1/{\Gamma }_z)d{\Gamma }_z/d\cos \theta =3(1-{\cos }^2\theta )/4$ instead of the isotropic unpolarized distribution. This is polarized massless leptonic W decay.

Let $t^A$ be Hermitian color generators with $\mathrm{Tr}(t^A{t}^B)={\delta }^{AB}/2$, and take $D_{\mu }={\partial }_{\mu }-i{g}_s{G}_{\mu }^A{t}^A$. In explicit spinor, color and flavor indices the massless kinetic term isThe full Quantum chromodynamics Lagrangian also has $-G_{\mu \nu }^A{G}^{A\mu \nu }/4$. A local color gauge transformation acts asThus $D_{\mu }q\mapsto V{D}_{\mu }q$ and the kinetic term is gauge-invariant.

Since the gauge coupling is independent of flavor and chirality, the classical internal global transformations are $q_L\mapsto U_L{q}_L$, $q_R\mapsto U_R{q}_R$, with constant flavor matrices $U_L,U_R\in U(2)$. Equivalently these consist of $SU(2{)}_L\times SU(2{)}_R$, a common vector phase, and a singlet axial phase $q\mapsto e^{i\eta {\gamma }_5}q$, modulo shared discrete factors. The vector phase measures quark number, or baryon number after dividing its generator by three. The nonsinglet axial rotations may be written $q\mapsto e^{-i{\theta }_a{\gamma }_5{\tau }_a/2}q$. At the quantum level the singlet axial phase is anomalous; the exact continuous internal symmetry is $SU(2{)}_L\times SU(2{)}_R\times U(1{)}_V$ up to those discrete identifications. The usual spacetime Poincaré transformations also leave the theory invariant; the massless classical theory has dilation symmetry, which is broken by quantum running. These distinctions are massless two-flavor QCD symmetries.

At equal time the canonical relations, including all indices, areNo constraint removes any of the four spinor components of the Dirac field in these equal-time relations.

To evaluate the axial charges, first prove the elementary bilinear identity. For any constant matrix $K$ on the spinor/flavor space,For example $[q_I^{†}{q}_J,q_K]=q_I^{†}\{q_J,q_K\}-\{q_I^{†},q_K\}{q}_J$; the spatial delta function collapses the integral. This proves the charge action directly from the canonical anticommutation relations.

For $K_a={\gamma }_5{\tau }_a/2$, anticommutation of ${\gamma }^0$ and ${\gamma }_5$ then givesThe charge is an even operator, so its commutator obeys the ordinary product rule. ConsequentlySimilarly, using ${\gamma }_5^2=1$ and the Pauli matrices identity $\{{\tau }_a,{\tau }_b\}=2{\delta }_{ab}I$,Composite fields can be defined with a common symmetry-preserving regulator; these nonsinglet identities have no color axial anomaly. This is the nonsinglet axial charge algebra of quark densities.

If every conserved axial charge left the vacuum invariant, the vacuum expectation of each such commutator would vanish. A nonzero quark chiral condensate instead makes $⟨[Q_{5,a},P_b]⟩=-i{\delta }_{ab}⟨S⟩\ne 0$ for $a=b$. It therefore diagnoses spontaneous breaking of the nonsinglet chiral symmetry. In the massless two-flavor limit, the observed hadronic pattern is described by $SU(2{)}_L\times SU(2{)}_R\to SU(2{)}_V$, with three Goldstone boson modes identified with the pions. The vacuum condensate is a dynamical property of QCD, not something derivable from the canonical algebra alone. The light pion multiplet and absence of parity-degenerate light hadrons support this pattern; the singlet axial anomaly prevents interpreting an additional singlet mode as a fourth Goldstone boson.

For equal nonzero masses, add ${\mathcal{L}}_m=-m\bar{q}q$. It is invariant under local color gauge transformations and common vector flavor rotations, but not under independent left/right rotations. The exact continuous flavor symmetry becomes $SU(2{)}_V\times U(1{)}_V$. The kinetic term and canonical anticommutators are unchanged, and the two proved equal-time commutators still hold. The charges, however, are no longer conserved:The first equation follows from the massive Dirac equations, or the last from $H_m=m\int S$ and $\dot{Q}=i[H,Q]$. This is the flavor-nonsinglet axial Ward identity with equal quark masses. The pions become pseudo-Goldstone bosons with $m_{\pi }^2=O(m)$, rather than exact massless modes. To leading order, in the convention $S=\bar{u}u+\bar{d}d$, $f_{\pi }^2{m}_{\pi }^2=-m⟨S⟩+O(m^2)$. The condensate persists for small masses, but the symmetry is now explicitly broken, so the exact-symmetry Goldstone argument is replaced by an approximate one. The singlet anomaly and gauge symmetry are unaffected.

A loop integral in a theory with dimensionless classical couplings can contain a logarithmic ultraviolet divergence. After subtraction, an expression such as $\log (-p^2/{\mu }^2)$ remains; one must specify the momentum scale at which the measured or renormalized coupling is fixed. A logarithm of a dimensionful momentum alone is not meaningful. In dimensional regularization, the same necessity appears through ${\lambda }_{\mathrm{bare}}={\mu }^{d_{\lambda }}{Z}_{\lambda }\lambda (\mu )$ in the continued dimension, with $d_{\lambda }$ tending to zero in the physical dimension.

Thus the renormalization prescription introduces an arbitrary renormalization scale, even when no mass or length occurs in the classical action. Physical quantities cannot depend on its arbitrary choice: explicit logarithms cancel the implicit change of the running coupling, governed by $\mu d\lambda /d\mu =\beta (\lambda )$. A nonzero renormalization-group beta function is a quantum scale anomaly. An integration constant may encode the same data as a dimensionful scale, which is dimensional transmutation. The scale labels a renormalization condition, not a new arbitrary measurable parameter. At a true fixed point the coupling need not run; the use of a subtraction scale is then a convention rather than an intrinsic generated physical scale.

SetThe chain rule givesFor the ultraviolet expansion take ${\beta }_0>0$ and $L=\log ({\mu }^2/{\Lambda }^2)\gg 1$. The leading solution is $y\sim {\beta }_{0}L$. Substituting it in the next term gives $dy/dL={\beta }_0+{\beta }_1/({\beta }_{0}L)+O(\log L/L^2)$, whose integration yieldsChoosing $\Lambda$ removes the additive constant. This proves the two displayed leading terms of the two-loop asymptotic running coupling.

The printed remainder is too small in general. This can be checked even for an exactly two-loop renormalization-group beta function, so it does not depend on unknown higher-loop effects. Put $c={\beta }_1/{\beta }_0$. Separating variables and fixing the integration constant givesExpanding this identity one order further givesThe $\log L/L$ term cannot be absorbed into a fixed change of $\Lambda$. For example, with ${\beta }_0={\beta }_1=1$, the exact implicit relation is $y-\log (y+1)=L$, and the next term is $(\log L+1)/L$, which is not $O(1/L)$. The unspecified $O(y^{-2})$ contribution to the differential equation changes the constant coefficient of $1/L$, while the coefficient ${\beta }_1^2/{\beta }_0^3$ of $\log L/L$ is already fixed. Only special cases such as ${\beta }_1=0$ avoid this particular obstruction.

The asymptotic formula also presupposes an asymptotically free ultraviolet regime. If ${\beta }_0=0$, division by it is invalid and the leading scaling changes. If ${\beta }_0<0$, the weak-coupling asymptotic direction is generally toward the infrared and the logarithmic variable must be chosen accordingly. These qualifications distinguish the valid perturbative solution from an unconditional formula.

Use the canonically normalized Abelian coupling $g_1=-\sqrt{5/3}{g}^{\prime }$, with $g_2=g$ and $g_3=g_s$, and write ${\alpha }_i=g_i^2/(4\pi )$. The sign in $g_1$ is a generator convention; its square is what enters running and matching. Let ${\alpha }_5=g_5^2/(4\pi )$ and $L_X=\log (M_X^2/{\mu }^2)$. To leading order,Use exactly the common-flavor coefficients stipulated for this model:The same $n_f$ cancels from their differences. In particular $b_3-b_2=11/3$ and $b_2-b_1=22/3$, so eliminating the unified coupling givesAt the low reference scale, $e=g\sin {\theta }_W=g^{\prime }\cos {\theta }_W$. With $w={\sin }^2{\theta }_W$ this meansThe last difference equation becomes $3(1-w)/(5\alpha )-w/\alpha =2(w/\alpha -1/{\alpha }_s)$. Solving givesSubstitute this into the first difference equation to obtainAll low-scale couplings on the right must be evaluated at the same specified $\mu$. A dimensionful $M_X$ cannot be fixed by dimensionless coupling values without such a reference scale. The result is independent of the common $n_f$ in these differences, although ${\alpha }_5$ is not. An inferred $M_X>\mu$ requires ${\alpha }^{-1}>8/(3{\alpha }_s)$, and perturbative matching requires a positive small ${\alpha }_5$.

This is the one-loop SU(5) prediction with common flavor coefficients. It is the prediction of the supplied simplified coefficient model, not the full matter-and-Higgs calculation for a realistic SU(5) grand unified theory. Actual representations, intermediate thresholds and higher-loop terms change the running and require the corresponding matching calculation.