Past exam of the mathematics course of the University of Cambridge / 2026 / iii / Paper 305 (The Standard Model)

A global symmetry is spontaneously broken when it preserves the action but does not preserve a chosen ground state. If $G$ is broken to the stabilizer $H$, the degenerate vacua form a vacuum manifold $G/H$. The Goldstone theorem states that a relativistic theory has one massless scalar mode for each broken continuous internal generator, so the standard counting gives $\dim G-\dim H$ Goldstone bosons.

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The continuous symmetry isStrictly, the diagonal central $U(1)$ acts trivially, so the faithful group is the quotient by that common center. Since $L$ and $R$ are unitary matrices,Every term is therefore unchanged by cyclicity of the matrix trace:and the same conjugation argument applies to $\mathrm{Tr}(M^{†}M)$ and $\mathrm{Tr}[(M^{†}M{)}^2]$.

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Let $x_a\ge 0$ be the eigenvalues of the positive semidefinite matrix $M^{†}M$. The potential isFor $k<0$ and $\lambda >0$, each summand is minimized atso $M_0^{†}{M}_0=v^{2}1$. A symmetry transformation preserves the representative $M_0=v1$ precisely when $L=R$, givingThe number of broken generators is $2N^2-N^2=N^2$, so the Goldstone theorem predicts $N^2$ modes, each a Goldstone boson.

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Every vacuum can be written as $M=vU$ with $U(x)\in U(N)$. Freezing the massive radial modes and substituting this parameterization into the action gives, up to a constant vacuum energy,Writing $U=\exp (i{\pi }^a{T}^a/v)$ displays the $N^2$ Goldstone fields ${\pi }^a$. The omitted terms contain more derivatives or arise from integrating out radial excitations, so this is the leading nonlinear sigma model on the vacuum manifold $U(N)$.

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For Hermitian $M$, the continuous transformations preserving the field space act by conjugation,with the central $U(1)$ acting trivially; there is also the discrete symmetry $M\mapsto -M$. The vacuum equation is $M^2=v^{2}1$, so every vacuum is unitarily conjugate toBecause the integer $p$ cannot change continuously, the vacuum manifold has disconnected componentsOn the $p$th component the unbroken continuous group is $U(p)\times U(N-p)$, and the Goldstone theorem givesGoldstone bosons. The discrete sign symmetry exchanges the components $p$ and $N-p$ but produces no Goldstone mode.

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Varying the Yang-Mills theory field strength and using $\delta A_{\mu }=D_{\mu }\alpha$ givesThus $F_{\mu \nu }$ transforms covariantly in the adjoint representation.

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With matrix-valued fields, charge conjugation acts asUnder parity, $A_0$ is even and $A_i$ is odd:Consequently $F_{0i}$ is odd and $F_{ij}$ is even. Under time-reversal symmetry, which is antiunitary, $A_0$ is even and $A_i$ is odd after $t\mapsto -t$; hence $F_{0i}$ is even and $F_{ij}$ is odd. Equivalently, the non-Abelian electric and magnetic fields transform as

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Since $\delta F_{\mu \nu }=-ig[F_{\mu \nu },\alpha ]$, both $F_{\mu \nu }{F}^{\mu \nu }$ and $F_{\mu \nu }{}^{\star }{F}^{\mu \nu }$ vary by a commutator. Cyclicity of the matrix trace therefore makes both terms gauge invariant. In electric and magnetic variables,The Yang-Mills kinetic term is even under $C$, $P$, and $T$. The Yang-Mills theta term is even under $C$ because both fields change sign and transpose, but odd under $P$ and under $T$ because exactly one of $\mathbf{E},\mathbf{B}$ changes sign.

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Expanding $F_{\mu \nu }$, using antisymmetry of ${ϵ}^{\mu \nu \rho \sigma }$, and cyclically rearranging the matrix trace giveswhere the Chern-Simons current isThe theta contribution to the action is therefore a boundary term. For variations that vanish at the spacetime boundary, its variation vanishes, so it contributes nothing to the classical Euler-Lagrange field equations. Its integral can still distinguish topologically nontrivial gauge configurations in the quantum theory.

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For massless fermions the classical axial rotation has Noether currentPromote $\beta$ to a spacetime-dependent parameter. The classical variation, after integration by parts, is $-\int d^{4}x\beta {\partial }_{\mu }{J}_A^{\mu }$. The chiral anomaly is equivalently encoded by the stated shift $\theta \mapsto \theta +2N_f\beta$, which changes the theta action byThe anomalous Ward identity is thereforeso $\alpha =N_f{g}^2/(8{\pi }^2)$ in the trace convention of the question.

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Since ${\partial }_{\mu }{K}^{\mu }=\mathrm{Tr}(F_{\mu \nu }{}^{\star }{F}^{\mu \nu })$, defineThe anomaly equation gives ${\partial }_{\mu }{\tilde{J}}_A^{\mu }=0$. The price is that the Chern-Simons current $K^{\mu }$ is not gauge invariant, so neither is ${\tilde{J}}_A^{\mu }$.

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A gauge anomaly makes the gauge symmetry inconsistent and must cancel. An Adler-Bell-Jackiw anomaly breaks a classical global chiral symmetry quantum mechanically but does not by itself invalidate the gauge theory. A 't Hooft anomaly is an obstruction to gauging a global symmetry; it remains invariant along renormalization-group flow and constrains the infrared spectrum or symmetry-breaking pattern.

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In the normalization supplied, a left-handed Weyl fermion in the two-index antisymmetric representation has cubic $SU(N)$ gauge-anomaly coefficient $N-4$, while an antifundamental has coefficient $-1$. Cancellation of the gauge anomaly requiresso

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Classically the $p$ antifundamentals admit $U(p)$ flavour rotations and $\lambda$ admits an independent phase rotation, giving $SU(p)\times U(1{)}_{\psi }\times U(1{)}_{\lambda }$ up to finite quotients. A $U(1)$ with charges $q_{\lambda },q_{\psi }$ is free of the mixed $SU(N{)}^{2}U(1)$ anomaly whenUsing $p=N-4$, chooseThe orthogonal axial phase has a chiral anomaly, while this combination survives. The continuous quantum global symmetry is therefore $SU(p)\times U(1)$, again up to possible finite central quotients.

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Use the charges $q_{\lambda }=N-4$ and $q_{\psi }=-(N-2)$ and the index convention $I(\mathbf{p})=1$. The $N$ gauge-colour copies of the flavour fundamental $\psi$ giveSumming the charge cubed and the charge over all Weyl components givesEach of these four coefficients is an ultraviolet 't Hooft anomaly.

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The composite ${\chi }^{ij}={\psi }^{(i}(\lambda {\psi }^{j)})$ is a flavour-symmetric left-handed Weyl fermion. Its $U(1)$ charge isBecause $p=N-4$, the supplied symmetric-representation data giveThusThere are $p(p+1)/2=(N-4)(N-3)/2$ components, sowhich exactly equal the ultraviolet values. The proposed confined spectrum therefore satisfies 't Hooft anomaly matching.

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Use unitary gauge to setThenDefineThe propagating fields are $W_{\mu }^{\pm },Z_{\mu },A_{\mu }$, and the Higgs boson $h$. Expandingand reading the quadratic terms givesThe factors $(v+h{)}^2$ in the kinetic term also give the corresponding Higgs-vector interactions.

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The lower Higgs component has $T^3=-1/2$ and hypercharge $Y=1/2$, so the generator annihilating the vacuum isThis is the unbroken electromagnetic $U(1{)}_Q$. Since $[Q,T^{\pm }]=\pm T^{\pm }$, the two W boson fields have charges $Q(W^{\pm })=\pm 1$. The Z boson, photon, and physical Higgs excitation are neutral.

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Under $SU(3{)}_c\times SU(2{)}_L\times U(1{)}_Y$, one Standard Model fermion generation including a right-handed neutrino is

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A bare Dirac bilinear couples left- and right-handed fields, but these fields transform in different electroweak representations, so ${\bar{f}}_L{f}_R$ is not a gauge singlet. Let $\tilde{H}=i{\sigma }^2{H}^{\ast }$. The four gauge-invariant Yukawa interaction terms areReplacing $H$ by its vacuum expectation value gives Dirac masses $m_f=y_{f}v/\sqrt{2}$. Applying $Q=T^3+Y$ givesfor both chiralities, so the generated mass terms preserve the unbroken electromagnetic gauge symmetry.

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