Past exam of the advanced quantum field theory for particle physics course of the University of Oxford / 2016 / 3

Work entirely with left-handed Weyl fermions; a right-handed field is counted through its left-handed charge conjugate. For a simple non-Abelian group $G$, defineThe two orientations of a chiral triangle supply the symmetrized trace. Spectator representation dimensions multiply each trace. The perturbative gauge anomaly conditions for $G\times U(1)$ are thereforeThe first concerns $G^3$, the second $G^{2}U(1)$, and the third $U(1{)}^3$. A $GU(1{)}^2$ triangle has group factor ${\sum }_i{Q}_i^2{\mathrm{tr}}_{R_i}{T}^a=0$, because non-Abelian generators of a semisimple group are traceless. If coupling to gravity is required, also impose the mixed gauge-gravitational condition ${\sum }_i\dim (R_i)Q_i=0$. Cancellation of these local triangles is not a universal substitute for checking global anomalies; the Witten SU(2) anomaly is the relevant additional test here.^[ref]

For the Standard Model, one generation consists of $Q_L:(3,2{)}_{1/6}$, $u^c:(\bar{3},1{)}_{-2/3}$, $d^c:(\bar{3},1{)}_{1/3}$, $L_L:(1,2{)}_{-1/2}$ and $e^c:(1,1{)}_1$. These are precisely the all-left-handed representations; conjugating the singlets reverses their hypercharges relative to right-handed particles. Check the requested triangles separately:

For $3$–$3$–$3$, normalize the cubic coefficient of a fundamental to $+1$. Its conjugate has $-1$, so the weak-doublet multiplicity givesFor $2$–$2$–$2$, the doublet generators obey $\{T^b,T^c\}={\delta }^{bc}I/2$, so $\mathrm{tr}(T^a\{T^b,T^c\})=0$. Thus every doublet separately has zero local cubic weak anomaly. For the global Witten SU(2) anomaly, each generation has three coloured quark doublets and one lepton doublet, an even total of four. All three generations give twelve, so the global obstruction also cancels.

For $3$–$3$–$1$, use $T(3)=T(\bar{3})=1/2$. The two conjugate triplets have the same quadratic index, not its negative:For $2$–$2$–$1$, the three quark colours and one lepton doublet giveFor $1$–$1$–$1$, count every coloured and weak component:The mixed gauge-gravitational coefficient also vanishes:Unlisted gauge combinations contain a single generator of at least one simple factor. For example, $3$–$2$–$1$ has ${\mathrm{tr}}_3{T}^a{\mathrm{tr}}_2{t}^{b}Y$, and $3$–$3$–$2$ has ${\mathrm{tr}}_3(T^a{T}^b){\mathrm{tr}}_2{t}^c$; both vanish. The same reasoning covers $3$–$1$–$1$, $2$–$1$–$1$ and $3$–$2$–$2$, including permutations of the legs. Scalars contribute no chiral triangle anomaly. Therefore

For $\lambda >0$, the scalar potential is minimized at ${\varphi }^{†}\varphi =v^2/2$. With the specified upper-component vacuum expectation value ${\varphi }_0=(v,0{)}^T/\sqrt{2}$ and hypercharge $Y=-I/2$, the unbroken generator isThe other three independent generator directions act nontrivially on ${\varphi }_0$. Thus the connected unbroken gauge subgroup is $U(1{)}_{\mathrm{em}}$, while three gauge directions are broken. The normalization here is $Q=T^3+Y$, not the alternative convention in which hypercharge is doubled.

Apply the covariant derivative directly to the vacuum:In the mostly-plus convention the canonical scalar kinetic term is $-(D_{\mu }\varphi {)}^{†}{D}^{\mu }\varphi$. Its quadratic vector terms areDefine $s_W=\sin {\theta }_W=g_1/\sqrt{g_1^2+g_2^2}$ and $c_W=\cos {\theta }_W=g_2/\sqrt{g_1^2+g_2^2}$. Diagonalization givesThe neutral combination $Z$ lies along the nonzero eigenvector $(g_2,-g_1)$; the orthogonal combination $(g_1,g_2)$ is the massless photon. Comparing the mass terms with $-m_W^2{W}^{+}\cdot W^{-}-m_Z^2{Z}^2/2$ yieldsFinally insert the inverse neutral rotation into the coupling:Both coefficients in the photon term agree, givingThe upper vacuum component is electrically neutral and the lower scalar component has charge $-e$, consistently with the source's negative hypercharge. This doublet is the conjugate convention to the commonly displayed positive-hypercharge doublet; it does not change the vector masses or the electromagnetic coupling. The four real scalar components supply three would-be Goldstone modes and one physical radial Higgs mode, preserving the total physical degrees of freedom.

For $M>0$, integrate the vector action by parts, dropping the boundary term, and write it as a quadratic functional:The explicit mass makes this Proca kernel invertible: the displayed action alone has no gauge redundancy left to fix. It is interpreted as unitary gauge when obtained from a Higgs or Stückelberg completion with its would-be Goldstone field removed.

Introduce a source $J_{\mu }$ and complete the square, $A=A^{\prime }-K^{-1}J$. The normalized Gaussian path integral isThe source-independent functional determinant cancels in $Z[J]/Z[0]$. For Fourier momentum $k$, the mixed-index kernel is ${\tilde{K}}^{\mu }{}_{\nu }=-(k^2+M^2){\delta }^{\mu }{}_{\nu }+k^{\mu }{k}_{\nu }$. For $k^2\ne 0$, use the projectorsTheir orthogonality givesCombining the projectors removes their apparent massless denominator. Adding the Feynman prescription gives the unitary-gauge propagatorMultiplication by $\tilde{K}$ verifies that it inverts the quadratic operator. On the physical pole $k^2=-M^2$, the numerator is transverse to $k$, as required for the three physical massive-vector polarizations.

The drawback is its off-shell ultraviolet behaviour: at large spacelike momentum the transverse term falls as $1/k^2$, but the longitudinal tensor term remains of order $1/M^2$. Naive loop power counting therefore looks worse and renormalizability is not manifest. In a finite $R_{\xi }$ gauge the corresponding expression iswhich falls as $1/k^2$ at fixed finite $\xi$. Its $\xi \to \infty$ limit gives the unitary-gauge longitudinal term, so taking that limit before the ultraviolet limit changes power counting.

This is not a claim that a renormalizable Higgs theory ceases to be physically renormalizable in this gauge. Nor does it mean that the free Proca Gaussian integral is inconsistent. For conserved external sources $k\cdot J=0$, the longitudinal numerator drops out of $J^{\mu }{D}_{\mu \nu }{J}^{\nu }$ and the tree exchange again falls as $1/k^2$. The difficulty concerns off-shell Green functions and manifest control of loop divergences, not the number of physical polarizations.