Past exam of the mathematics course of the University of Cambridge / 2010 / iii / Paper 45 (The Standard Model)

Take $T_i={\tau }_i/2$ and the hypercharge generator $Y=I_2/2$ on the scalar. The two commuting factors of the electroweak interaction have independent invariant gauge kinetic terms and therefore independent coupling constants. Non-Abelian gauge invariance requires the same $g$ for all three $SU(2)$ generators, while the $U(1)$ factor has its own $g^{\prime }$; no symmetry of this product group requires $g=g^{\prime }$.

A nonzero Higgs doublet can be rotated by $SU(2)$ toThe fundamental representation acts transitively on normalized complex doublets, so this fixes an internal direction without changing the vacuum norm. A general infinitesimal generator annihilates this vacuum only ifFor real parameters this means ${\alpha }_1={\alpha }_2=0$ and $\beta ={\alpha }_3$. Thus the connected stabilizer subgroup is one-dimensional:This proves the claimed electroweak symmetry breaking and identifies the unbroken electric charge generator.

The gauge covariant derivative isWriting the connection as ${\mathcal{A}}_{\mu }=g{A}_{\mu }^i{T}_i+g^{\prime }{B}_{\mu }Y$, its local transformation is ${\mathcal{A}}_{\mu }^{\prime }=U{\mathcal{A}}_{\mu }{U}^{-1}-i({\partial }_{\mu }U)U^{-1}$. Direct substitution gives $D_{\mu }^{\prime }(U\varphi )=U{D}_{\mu }\varphi$. Unitarity then provesso the scalar kinetic term is gauge-invariant.

At the constant vacuum the gauge covariant derivative becomesConsequentlywhere $\mathcal{V}=(A^1,A^2,A^3,B{)}^T$ andThe factor $1/2$ is the canonical real-vector mass-term normalization; the displayed electroweak doublet gauge-boson mass matrix contains squared masses when the gauge kinetic terms are canonical.

The electric charge generator acts on gauge fields through the Adjoint representation, not through the scalar's two-by-two matrix. With $U=e^{i\alpha Q}$, the brackets $[T_3,T_1]=i{T}_2$ and $[T_3,T_2]=-i{T}_1$ give $\delta A^1=\alpha A^2$, $\delta A^2=-\alpha A^1$, and neutral $A^3,B$. Hence the electromagnetic charge matrix of electroweak gauge fields in the requested basis isChoose positive gauge couplings by the gauge-field sign convention. Set $s_W=g^{\prime }/\sqrt{g^2+g^{\prime 2}}$ and $c_W=g/\sqrt{g^2+g^{\prime 2}}$. The Weinberg angle rotation diagonalizes the neutral block, while $W^{\pm }=(A^1\mp i{A}^2)/\sqrt{2}$ obey $\delta W^{\pm }=\pm i\alpha W^{\pm }$. The physical electroweak gauge bosons areThe charges are in units of $e=g{s}_W=g^{\prime }{c}_W$. Thus the tree-level electroweak gauge-boson masses satisfy $m_W=m_Z{c}_W$, and the unbroken photon is massless.

The electron and neutrino form the left-chiral lepton doublet, while the right-chiral electron is a weak singlet. In four-component notation use chiral projectors $P_{L,R}=(1\mp {\gamma }_5)/2$ and packageHere each entry can be obtained by projecting a Dirac field; only the indicated chirality propagates. The hypercharges follow from $Q=T_3+Y$: $(0,-1)=(1/2,-1/2)+Y_{\Psi }$ gives $Y_{\Psi }=-1/2$, and $T_3=0,Q=-1$ gives $Y_{\psi }=-1$. The minimal theory has no right-handed neutrino.

The gauge-invariant lepton kinetic Lagrangian isIf unprojected Dirac spinors are used as notation instead, insert $P_L$ in the first kinetic term and $P_R$ in the second. The inactive components are not additional physical degrees of freedom. Each term is invariant because the gauge covariant derivative transforms in the same gauge group representation as its field.

No bare Dirac mass term is possible: $\bar{\Psi }\Psi$ and $\bar{\psi }\psi$ vanish by opposite adjoint projectors, while a cross term $\bar{\Psi }\psi$ has an uncontracted weak-doublet index and hypercharge $-1/2$. A neutrino Dirac mass would additionally require the absent right-handed field. A Majorana mass term pairs a chiral field with itself; ${\Psi }^T{C}^{-1}\Psi$ has hypercharge $-1$ and ${\psi }^T{C}^{-1}\psi$ has $-2$, so neither is invariant. These statements concern quadratic masses before electroweak symmetry breaking, not masses involving scalar insertions.

For the Higgs conjugate doublet, put $\epsilon =i{\tau }_2$. The given identity $\epsilon {\tau }_i^{\ast }=-{\tau }_i\epsilon$ exponentiates to $\epsilon U^{\ast }=U\epsilon$ for $U\in SU(2)$. ThereforeIn particular its weak isospin remains $T=1/2$ while its hypercharge reverses. Since $\Psi \mapsto e^{-i\beta /2}U\Psi$, both ${\varphi }^{\prime †}\Psi$ and $\bar{\Psi }{\varphi }^{\prime }$ transform in the trivial representation of the full gauge group. They remain a Lorentz spinor and its adjoint; gauge singlet does not mean Lorentz scalar.

Finally the Yukawa interactionis gauge invariant: the first term has total hypercharge $+1/2+1/2-1=0$, and its doublet indices are contracted. At the Higgs field vacuum it becomes $-(y_{e}v/\sqrt{2}){\bar{e}}_L{e}_R+h.c.$. Rephase $e_R$ to make the coefficient real and positive and assemble $e=e_L+e_R$. ThenThis explicitly generates the gauge-invariant electron Yukawa mass, without a bare mass in the unbroken theory.

Use the metric $g^{\mu \nu }=\mathrm{diag}(1,-1,-1,-1)$ and $\{{\gamma }^{\mu },{\gamma }^{\nu }\}=2g^{\mu \nu }$. Transposing a product reverses its order but does not conjugate the factor $i$. Thus the charge-conjugation matrix identity giveswhere reversing the four distinct gamma matrices requires six exchanges. Similarly,

To fix the transpose sign, work in the Weyl representation of the gamma matrices:The Pauli identity checks $C_0{\gamma }^{\mu T}{C}_0^{-1}=-{\gamma }^{\mu }$ directly, and ${\epsilon }^T=-\epsilon$ gives $C_0^T=-C_0$, with $C_0^{†}{C}_0=I$. Any other intertwiner differs from $C_0$ by a scalar: its ratio with $C_0$ commutes with all gamma matrices, whose products span the full four-by-four matrix algebra. A matrix commuting with all matrix units is scalar. Under a spinor basis change $S$, the intertwiner becomes $S{C}_0{S}^T$, whose transpose is still its negative. Therefore in every such four-component representationThis proves the antisymmetry of the four-dimensional charge-conjugation matrix, rather than leaving the sign undetermined by the general transpose argument.

Hermitian conjugating the charged Dirac equation, using real $A_{\mu },e,m$ and ${\gamma }^0{\gamma }^{\mu †}{\gamma }^0={\gamma }^{\mu }$, gives the adjoint Dirac equationTranspose it and multiply by $C$. Since $C{\gamma }^{\mu T}=-{\gamma }^{\mu }C$, the field ${\psi }^c=C{\bar{\psi }}^T$ obtained by charge conjugation obeysThus algebraic charge conjugation reverses the electromagnetic coupling.

If ${\gamma }_5\psi =\eta \psi$, $\eta =\pm 1$, Hermiticity of ${\gamma }_5$ and its anticommutation with ${\gamma }^0$ implyUsing the first transpose identity,Hence $\boxed{\bar{\psi }{\gamma }_5=-\eta \bar{\psi },{\gamma }_5{\psi }^c=-\eta {\psi }^c}$: algebraic charge conjugation reverses chirality. This concerns conjugating the already projected spinor, not the distinct operation of keeping a chiral projector outside an operator conjugation.

For the Grassmann variation of a chiral Majorana mass term, variations of odd fields must be moved past odd fields with a minus sign. If $A^T=-A$, componentwise reordering givesBoth $C$ and $C^{-1}$ are antisymmetric. Treating $\psi ,\bar{\psi }$ independently, integrating the kinetic variation by parts, and placing varied Grassmann fields on the left givesThe factors two from variation cancel the mass-term factors $1/2$. The first coefficient gives one equation; multiplying the second by $C$ gives the other:The variations can be restricted to their chiral subspaces: the two terms in each equation have the same opposite output chirality, so no inactive component equation has been imposed.

Applying $i/\partial$ to the first equation and substituting the second, using $(i/\partial {)}^2=-\square$, givesA phase rotation removes the phase of $m$, and combining the chiral field with its conjugate then gives a Majorana fermion of this mass, not two independent chiral species.

Let $\chi ={\varphi }^{\prime †}\Psi$ and $\bar{\chi }=\bar{\Psi }{\varphi }^{\prime }$. Question 1 shows they are gauge singlets, so both terms in ${\mathcal{L}}_G=G{\chi }^T{C}^{-1}\chi -G^{\ast }\bar{\chi }C{\bar{\chi }}^T$ preserve electroweak gauge invariance. The original PDF has $C^{-1}$ in the first term; the TeX aid incorrectly replaces it by $G^{-1}$. At the Higgs field vacuum,soComparing with the just-derived chiral mass equations yields the Majorana neutrino mass from the Weinberg operator:For nonzero $G$ this generates a mass without introducing a right-handed neutrino. Under lepton number rotation $\Psi \mapsto e^{i\alpha }\Psi$, the first operator acquires $e^{2i\alpha }$ and its conjugate acquires $e^{-2i\alpha }$. Thus the interaction violates lepton number by $\boxed{\Delta L=2}$. Each composite contains one scalar of mass dimension one and one fermion of dimension $3/2$, so the interaction is dimension five and $G$ has dimension $-1$. This is the Weinberg operator, an effective extension of the minimal renormalizable Standard Model, consistent with the absence of bare quadratic masses in Question 1.

The product of leptonic weak charged currents in the Fermi interaction contains four-fermion cross terms such asThis produces ${\mu }^{-}\to e^{-}+{\bar{\nu }}_e+{\nu }_{\mu }$, and the same interaction permits neutrino-electron scattering and inverse muon decay. Pure electromagnetic interactions have no neutrino coupling and preserve the separate charged-lepton species, so cannot produce these processes by themselves. The factor $1-{\gamma }_5=2P_L$ fixes the chiral structure of the weak charged current.

The pion is a pseudoscalar with negative parity, while the strong-interaction vacuum is parity even. Lorentz covariance restricts a vacuum-to-pion vector-current matrix element to $a{p}^{\alpha }$. In the pion rest frame only its temporal component could be nonzero, but a vector current's temporal component is parity even, whereas the pion is parity odd. The matrix element therefore equals its negative and vanishes. The temporal component of an axial current is parity odd, so its matrix element is allowed and Lorentz covariance fixes it to a constant times $p^{\alpha }$. Defining the pion decay constant in the printed normalization givesThis proves both the vector-current exclusion and the axial-current form; $F_{\pi }$ itself is a strong-interaction parameter, not fixed by symmetry.

The cross term $J_{\mathrm{leptons}}^{\alpha †}{J}_{\alpha }^{\mathrm{hadrons}}$ gives, up to an irrelevant overall phase, the decay amplitudeTake the antineutrino mass to be zero. Since $p=k+q$, ${\bar{u}}_e/k=m_e{\bar{u}}_e$, and $/q{v}_{\nu }=0$, anticommutation with ${\gamma }_5$ givesThe amplitude has helicity suppression by one power of the charged-lepton mass.

Using the external spin sums, with no initial spin average for a spin-zero pion,The middle equality follows from $(1-{\gamma }_5)/q=/q(1+{\gamma }_5)$, $(1+{\gamma }_5{)}^2=2(1+{\gamma }_5)$, and the gamma matrix trace identities. Summing a full massless neutrino spin basis introduces no factor of two: the current projector annihilates the inactive helicity.

For completeness, evaluate the two-body Lorentz-invariant phase space in the pion rest frame. Its three-momentum delta function sets $\mathbf{q}=-\mathbf{k}$, and the energy delta function is $\delta (M_{\pi }-\sqrt{|\mathbf{k}{|}^2+m_e^2}-|\mathbf{k}|)$. The radial root isCombining its derivative with $1/(4E_k|\mathbf{k}|)$ and integrating solid angle givesThe invariant squared amplitude is independent of direction, so the supplied decay formula yields the leptonic pseudoscalar decay widthThe denominator is $4\pi$ because the matrix element contains $\sqrt{2}{F}_{\pi }$. With the alternative definition $f_{\pi }=\sqrt{2}{F}_{\pi }$, the same width is $G_F^2{f}_{\pi }^2{M}_{\pi }{m}_e^2(1-m_e^2/M_{\pi }^2{)}^2/(8\pi )$. In either convention it vanishes as $m_e\to 0$.

Replacing the electron by a muon gives the same hadronic factor and weak coupling. Hence the pion electron-to-muon decay ratio isThis is a tree-level result with massless neutrinos. The cancellation of $F_{\pi }$ and $G_F$ makes the ratio a clean test of the predicted helicity suppression and electron–muon universality in the chiral weak charged current. For example, scalar or pseudoscalar four-fermion interactions could remove the charged-lepton mass factor and substantially enhance the electron mode. Radiative corrections are required for a precision experimental comparison.

With two quark generations, the charged hadronic current containsA $K^{-}$ contains $s\bar{u}$, so its annihilation into the lepton pair selects the term proportional to $\sin {\theta }_C$. Strong strangeness conservation excludes the $d$ term from its vacuum matrix element. Its amplitude is thus multiplied by $\sin {\theta }_C$ and its width by ${\sin }^2{\theta }_C$:Here $F_K$ is the pure strong-interaction decay constant in the same square-root-of-two convention. If the pion current's quark mixing coefficient is also shown explicitly, its width carries ${\cos }^2{\theta }_C$; it is already absorbed if the stated pion matrix element uses the full weak current. Thus the isolated weak-factor ratio is ${\tan }^2{\theta }_C$, while the kaon suppression relative to unit weak coupling is the requested Cabibbo suppression factor ${\sin }^2{\theta }_C$.

Use an incoming spacelike momentum transfer $q$, so the physical absorption process has $p_X=P+q$, $Q^2=-q^2>0$, and $P\cdot q>0$. There is a sign inconsistency in the original PDF: its delta function has $P-p_X-q$, while its positive Bjorken x definition and target formulas use incoming $q$. With that literal delta function and $x>0$, the final-state invariant would bewhich becomes negative in the stated high-energy limit and cannot represent a physical hadronic final state. We therefore use ${\delta }^4(P+q-p_X)$, preserving the printed $x=Q^2/(2P\cdot q)$ and structure-function conventions. Equivalently, keeping an outgoing momentum variable requires reversing the signs in those definitions as well.

Assume an unpolarized target, incoherent scattering from effectively free spin-one-half partons, and the massless collinear parton approximation. In the Bjorken scaling limit, take $Q^2,P\cdot q\to \infty$ at fixed $0<x<1$, neglecting target and quark masses and intrinsic transverse momentum relative to $Q$. Write the incoming struck-quark momentum as $k=\xi P$ in the leading lightlike-reference approximation. Let $f_f(\xi )$ and ${\bar{f}}_f(\xi )$ be quark and antiquark parton distribution functions, counting all colours. For the naive scaling result also neglect radiative scaling violations; in full QCD these distributions depend logarithmically on the hard scale.

The outgoing struck quark has momentum $k^{\prime }=k+q$. Its on-shell condition givesAverage its initial spin and sum its final spin. The spin-averaged electromagnetic parton tensor has numeratorThe supplied three-gamma identity gives this directly: contract its middle index with $(k+q{)}_{\lambda }$ and trace with $/k$; the ${\gamma }_5$ trace vanishes. Thus no antisymmetric term survives the unpolarized spin average.

Define transverse tensorsOn the cut, $2k\cdot q=Q^2$, so $\tilde{k}=k+q/2=\xi \tilde{P}$. Expanding the two transverse terms givesThis also explicitly verifies electromagnetic current conservation, $q_{\mu }{S}^{\mu \nu }=q_{\nu }{S}^{\mu \nu }=0$.

The one-particle final-state Lorentz-invariant phase-space measure obeysIncluding the current's charge factor and the given $1/(4\pi )$ normalization, a single quark therefore contributesAn antiquark has the same unpolarized tensor and the same squared charge $Q_f^2$.

The parton convolution normalization for a hadronic tensor isThe $1/\xi$ is important: the distributions count parton numbers, whereas covariantly normalized one-particle responses carry the state normalization $2k^0$. The incoherent sum of normalized responses is $W/(2P^0)={\sum }_f\int d\xi (f_f+{\bar{f}}_f)w_f/(2k^0)$; in the collinear frame $k^0=\xi P^0$, giving the factor above. There is no extra colour multiplicity because it is already included in the distributions.

On the physical cut,Comparing the two transverse coefficients in the hadronic tensor gives, with $n_f=f_f+{\bar{f}}_f$,Therefore the requested deep-inelastic structure functions areThe notation $f_f$ here is the question's $f$ with its flavour index made explicit. The calculation proves the Callan-Gross relation, $F_2=2x{F}_1$, rather than merely citing it. Finite masses and radiative corrections modify this leading approximation; at leading QCD order the same formulas use scale-dependent parton distribution functions.

The quark number sum rule follows because a distribution is a number density. Integrating $f_f$ counts quarks and integrating ${\bar{f}}_f$ counts antiquarks; their difference is the net conserved flavour charge:Every sea quark-antiquark pair contributes zero to this difference. More formally the flavour-vector charge counts quarks minus antiquarks and has a fixed eigenvalue on the hadron state. Thus $N_f$ is the valence or net quark number, not the total number including the sea.

Use both final assumptions printed in the question: only up/down flavours and no antiquarks. The second does not follow just from restricting the flavour list. Then $\int f_u=N_u$, $\int f_d=N_d$, with squared charges $Q_u^2=4/9$ and $Q_d^2=1/9$. The proton has $(N_u,N_d)=(2,1)$ and the neutron $(1,2)$. Hence the valence-only proton and neutron structure-function moments areThe integral limits assume the scaling approximation is valid for the indicated weighted moments, for example by dominated convergence of the weighted integrands. The same net-quark sum rule alone would not fix these individual moments in the presence of antiquarks, since $F_2$ involves the sum of quark and antiquark distributions.