Past exam of the mathematics course of the University of Cambridge / 2024 / iii / Paper 305 (The Standard Model)

Write $m=|m|e^{i\theta }$. Parity reverses the spatial coordinates and must exchange the two chiral Weyl spinors. One convenient choice isThe identities ${\sigma }^0={\bar{\sigma }}^0$ and ${\sigma }^i=-{\bar{\sigma }}^i$ exchange the two kinetic terms after $\mathbf{x}\mapsto -\mathbf{x}$. The mass bilinear $m{\bar{\psi }}_R{\psi }_L$ maps to its complex conjugate $m^{\ast }{\bar{\psi }}_L{\psi }_R$, so the two mass terms are exchanged and the action is invariant. More generally, independent unit phases $a,b$ may multiply the two transformations provided $b^{\ast }a=m^{\ast }/m$.

Charge conjugation reverses the gauge charge, soUsing the antisymmetric spinor metric $ϵ=i{\sigma }^2$, a compatible action on the two Weyl fields isComplex conjugation reverses the sign of $i$ and of the gauge representation, while $A_{\mu }\mapsto -A_{\mu }$ restores the original gauge covariant derivative. The identities ${\sigma }^2({\sigma }^{\mu }{)}^{\ast }{\sigma }^2={\bar{\sigma }}^{\mu }$ and ${\sigma }^2({\bar{\sigma }}^{\mu }{)}^{\ast }{\sigma }^2={\sigma }^{\mu }$ then exchange the two equations of motion. Unit phases can be inserted in the two spinor transformations without changing the conclusion, subject to the mass-term relation.

Take the most general parity action compatible with the kinetic and mass terms as ${\psi }_L\mapsto a{\psi }_R$, ${\psi }_R\mapsto b{\psi }_L$, where $|a|=|b|=1$ and $b^{\ast }a=m^{\ast }/m$, and let $\varphi \mapsto c_P\varphi$ with $|c_P|=1$. The two holomorphic Yukawa interactions are exchanged whenup to the common sign convention for antisymmetric spinor contraction. The freely chosen intrinsic phases satisfy these equations exactly when

For charge conjugation, write ${\psi }_L\mapsto a_{C}i{\sigma }^2{\psi }_R^{\ast }$, ${\psi }_R\mapsto b_{C}i{\sigma }^2{\psi }_L^{\ast }$ and $\varphi \mapsto c_C{\varphi }^{\ast }$. It exchanges each Yukawa term with the Hermitian conjugate of the opposite-chirality term. The coefficient equations have the form ${\lambda }_2^{\ast }={\lambda }_1{c}_C{a}_C^2$ and ${\lambda }_1^{\ast }={\lambda }_2{c}_C{b}_C^2$; the available intrinsic phases again solve them exactly whenFor conventional choices of all intrinsic phases these statements reduce to phase-sensitive relations such as ${\lambda }_2={\lambda }_1$ under parity and ${\lambda }_2={\lambda }_1^{\ast }$ under charge conjugation; those phases are conventions, while equality of magnitudes is invariant.

In four spacetime dimensions a Weyl spinor has mass dimension $3/2$, so each chiral current $J_L^{\mu }={\bar{\psi }}_L{\bar{\sigma }}^{\mu }{\psi }_L$ or $J_R^{\mu }={\bar{\psi }}_R{\sigma }^{\mu }{\psi }_R$ has dimension three. Since the interaction is $(g{J}_L+g^{\prime }{J}_R{)}^2$, dimensional homogeneity of the Lagrangian givesEquivalently, the expanded four-fermion interaction has coefficients $g^2,g{g}^{\prime },g^{\prime 2}$ of mass dimension minus two.

Parity exchanges $J_L$ and $J_R$. The square is invariant precisely when the two same-chirality coefficients agree, $g^2=g^{\prime 2}$, while the mixed coefficient is already symmetric. Therefore

Asymptotic freedom means that the gauge running coupling tends to zero as the renormalization scale tends to infinity. In the stated convention this occurs when $b_0>0$, namelyThe high-energy theory then approaches the Gaussian fixed point and perturbation theory becomes increasingly accurate.

The strong-coupling scale $\Lambda$ is defined as the scale at which the one-loop inverse coupling vanishes. Setting $g^{-2}(\Lambda )=0$ givesEquivalently, in terms of the coupling measured at any scale $\mu$,This dimensional transmutation defines ${\Lambda }_{\mathrm{QCD}}$ for Quantum chromodynamics.

For $SU(3{)}_c$, one generation contains two color triplets in the quark doublet and the two right-handed color triplets, so $N_f=4N_g$ and there is no colored scalar. Thuswhich permits at most

For $SU(2{)}_L$, each generation contains three colored copies of the quark doublet and one lepton doublet, again giving $N_f=4N_g$. The one complex Higgs doublet gives $N_s=1$, soHence the maximum is

Let $K=3(4\pi {)}^2$ and $L=\log (M^2/{\mu }^2)$. Running from $M$ down to $\mu$ givesThe assumed unified boundary condition impliesSubtracting the weak equation from the strong and normalized-hypercharge equations therefore givesEliminating $L$ proves

The block-diagonal subgroupis locally $SU(3)\times SU(2)\times U(1)$, with only a finite central quotient distinguishing the global groups. The hypercharge direction in the fundamental representation iswhich is traceless and commutes with the $SU(3)$ and $SU(2)$ blocks.

The canonically normalized $SU(5)$ generator must satisfy $\mathrm{Tr}(T_Y^2)=1/2$. Since $\mathrm{Tr}(Y^2)=5/6$,The unified gauge covariant derivative contains $g_5{T}_Y{B}_{\mu }$, whereas the Standard Model convention contains $g_{Y}Y{B}_{\mu }$. Hence $g_Y=\sqrt{3/5}{g}_5$. The non-Abelian generators already have the canonical normalization, so gauge coupling unification in the SU(5) grand unified theory predicts

A gauge anomaly breaks a redundancy required to remove unphysical polarizations. It violates the Ward identities and makes the quantum gauge theory inconsistent, so all gauge anomalies must cancel. A chiral anomaly, or Adler-Bell-Jackiw anomaly, instead breaks a classically conserved global axial symmetry; it is physically allowed and explains effects such as anomalous pseudoscalar decays and instanton-induced charge violation. A 't Hooft anomaly is an obstruction to gauging a global symmetry. It is invariant under renormalization-group flow, so 't Hooft anomaly matching constrains the infrared theory to reproduce it through massless fields, symmetry breaking, or a topological sector.

Count every right-handed field as a left-handed conjugate, which reverses its $U(1)$ charge and conjugates its non-Abelian representation. The nontrivial local anomaly-cancellation conditions areThe pure $SU(3{)}^3$ anomaly cancels because the two triplets in $Q_L$ balance the conjugates of $u_R$ and $d_R$. There is no perturbative $SU(2{)}^3$ anomaly, and the Witten SU(2) anomaly also vanishes because the three colored quark doublets plus one lepton doublet make four. The mixed gauge-gravitational anomaly is excluded at this stage as requested.

The $[SU(3){]}^{2}U(1)$ condition gives $u+d=2q$, which is even because all hypercharges are integers. The sum $u+d$ and difference $u-d$ have the same parity, so $u-d$ is even. ThereforeSolving the sum and difference equations gives

The $[SU(2){]}^{2}U(1)$ condition gives $l=-3q$. Substituting this and $u=q+y$, $d=q-y$ into the cubic anomaly givesUsing $(q+y{)}^3+(q-y{)}^3=2q^3+6q{y}^2$ and multiplying by minus one yields

If $q\ne 0$, divide the cubic equation by $q^3$ and define the rational numbersThen every integer anomaly-free assignment supplies a rational solution of

Use the stated rational change of variablesDirect substitution turns the anomaly equation intoWriting $v=a/c$ and $w=b/c$ in lowest common denominator gives the integer equation $a^3+b^3=c^3$. The supplied special case of Fermat's Last Theorem says that one of $a,b$ must vanish. Thus $(v,w)=(1,0)$ or $(0,1)$, which givesThe two signs merely exchange the names of the up- and down-type singlets. Choosing $q=1$ and the conventional sign $y=3$ gives the unique assignment up to overall scaling and that exchange:These are six times the conventional Standard Model hypercharges.

The coefficient of the mixed gauge-gravitational anomaly is the sum of all left-handed $U(1)$ charges, with right-handed fields counted as conjugates:For the solution above,More invariantly, $l=-3q$, $u+d=2q$, and $x=-6q$ make the expression vanish identically. Thus the unique cubic-anomaly solution automatically cancels the mixed gauge-gravitational anomaly.

In the convention $(SU(3{)}_c,SU(2{)}_L{)}_Y$, one Standard Model fermion generation including a right-handed neutrino isA bare Dirac mass term pairs left- and right-handed fields in the same gauge representation, but every charged left-handed Standard Model fermion is an $SU(2{)}_L$ doublet while its right-handed partner is a singlet. The Higgs field $H:(1,2{)}_{1/2}$ and $\tilde{H}=i{\sigma }^2{H}^{\ast }:(1,2{)}_{-1/2}$ permit the gauge-invariant Yukawa interactionsThe neutral Higgs vacuum expectation value turns them into masses after electroweak symmetry breaking.

Because ${\nu }_R$ is a gauge singlet, it may also have a large Majorana mass term $M$. Together with its Dirac mass $m_D$, the seesaw mechanism gives a light neutrino mass $m_{\nu }\simeq m_D^2/M$. If no right-handed neutrino is retained, the same low-energy physics is encoded by the dimension-five Weinberg operator $(L_{L}H)(L_{L}H)/\Lambda$.

Now let $\varphi$ be an electroweak scalar triplet with $Y=1$. Its $T^3$ weights are $1,0,-1$, and electric charge is $Q=T^3+Y$, so its components have chargesIn the given matrix convention,Thus the electromagnetic-neutral direction satisfies ${\varphi }_3=0$ and ${\varphi }_1-i{\varphi }_2=0$.

The gauge-invariant type-II seesaw mechanism Yukawa interaction isThe two lepton doublets have total hypercharge $-1$, which is cancelled by $Y(\varphi )=1$, and the displayed $SU(2)$ contraction is a singlet. Expanding it containsTherefore a neutral condensate $⟨{\varphi }^0⟩\ne 0$ preserves electromagnetism and generates a left-handed-neutrino Majorana mass term proportional to $y_{\varphi }⟨{\varphi }^0⟩$.