Past exam of the advanced quantum field theory for particle physics course of the University of Oxford / 2021 / 3

Put $r^2={\Phi }^T\Phi$. The potential has its minimum atChoose the vacuum expectation value ${\Phi }_0=(0,0,v{)}^T$. Since the weak hypercharge is zero, $Q=T^3$ on this real scalar triplet, and $T^3{\Phi }_0=0$. Neither $T^1$ nor $T^2$ annihilates ${\Phi }_0$. The stabilizer subgroup is consequentlyIn particular, spontaneous symmetry breaking has not reduced the two neutral gauge generators to a single one: $Y$ annihilates every component and remains unbroken independently of $T^3$.

The scalar mass matrix isWith canonical kinetic normalization $\frac{1}{2}(\partial h{)}^2$, the radial Higgs boson therefore hasThe two tangential zero eigenvalues are Goldstone modes, absorbed by the two massive gauge bosons through the Higgs mechanism.

Let $g$ and $g^{\prime }$ be the two gauge couplings. In unitary gauge, $\Phi =(0,0,v+h{)}^T$ and the weak hypercharge term in the gauge covariant derivative vanishes. Using $(T^a{)}_{bc}=-i{ϵ}_{abc}$, the kinetic term becomesThus, in the basis $(W^1,W^2,W^3,B)$, the gauge-boson mass matrix and its eigenvalues areHere $W^{\pm }=(W^1\mp i{W}^2)/\sqrt{2}$. The two massless neutral gauge bosons may be rotated to any orthonormal basis; no neutral mass term selects a unique mixing angle.

Expanding the same kinetic term to first order in $h$ givesFunctional differentiation gives the Feynman rulesand, in the real basis, $h{W}_{\mu }^a{W}_{\nu }^b:2i{g}^{2}v{\delta }^{ab}{g}_{\mu \nu }$ for $a,b\in \{1,2\}$. All such three-point Feynman vertices involving $W^3$ or $B$ vanish. The factor two in the real basis comes from the two identical gauge boson fields; in the charged basis it is already explicit in ${\mathcal{L}}_{hVV}$.

This cannot be the sole symmetry-breaking sector of the Standard Model. Its gauge-boson mass matrix has two zero eigenvalues, so besides the electromagnetic photon it predicts an extra massless neutral gauge boson. Every linear combination of $W^3$ and $B$, including the combination normally called $Z$, is massless. The observed massive Z boson is a concrete contradiction. Choosing $g,g^{\prime },m,\lambda$ differently cannot change these two zero eigenvalues.

A real scalar triplet can occur as an additional sector, but a further vacuum expectation value, for example that of a Higgs doublet, is then necessary to obtain the observed neutral gauge boson mass pattern. This does not rescue the triplet-only model posed here.

Normalize the generating functional by its value with all sources zero, and take the connected part of the time-ordered correlation function. Source differentiation of the interaction exponential is equivalent to the Dyson series followed by Wick's theorem. Write $D(x-y)=⟨T\varphi (x)\varphi (y){⟩}_0$ and $F_{ij}(x-y)=⟨T{\psi }_i(x){\bar{\psi }}_j(y){⟩}_0$ temporarily, so that every contraction has an unambiguous meaning.

The second-order Dyson series contributesThere are two connected external scalar field attachments:They become equal after exchanging the integration variables $x,y$, cancelling $2!$. The connected fermion contractions have the closed fermion loop sign:The minus sign can also be checked directly: the pairing of the first with the fourth field and the second with the third has positive pairing parity, but $⟨T{\bar{\psi }}_i(x){\psi }_j(y)⟩=-F_{ji}(y-x)$. Multiplying it by $(ig{)}^2=-g^2$ leaves a positive result:A contraction of the two external fields with each other times a vacuum bubble cancels against the normalization. Separate fermion tadpole integrals at $x$ and $y$ belong to a product of one-point functions, removed by taking the connected part.

To match the source's generating functional convention, distinguish its displayed kernels from contractions. The positive $iJ\Delta J/2$ exponent gives $D=-i\Delta$. With the source insertion order in equation (5), the positive $i\bar{\eta }{S}_F\eta$ exponent similarly gives $F=-i{S}_F$. The two scalar propagators and two Dirac propagators supply a total phase $(-i{)}^4=1$. Thus, writing $S=S_F$ as in the requested formula,The matrix trace closes the spinor indices; no extra factor of two or additional symmetry factor remains for a Dirac field with this Yukawa interaction.

For commuting real scalar fields, remove the closed fermion loop sign and replace the Dirac propagators by scalar propagators. There are no spinor indices, hence no matrix trace. With two independent real scalar fields ${\chi }_1,{\chi }_2$ and interaction $g\varphi {\chi }_1{\chi }_2$, the same normalization givesThe overall sign has reversed relative to part (i) because $(ig{)}^2$ is now unaccompanied by a closed fermion loop sign. The same formula holds in the source's kernel convention, since there are again four propagators.

There is an additional convention to specify if the two symbols are intended to denote the same real scalar field. For the literal interaction $g\varphi {\chi }^2$, there are two identical connected Wick contractions between ${\chi }^2(x)$ and ${\chi }^2(y)$, so the coefficient becomes $-2g^2$. If the conventional interaction is instead $(g/2)\varphi {\chi }^2$, each vertex is $ig$ and the loop has symmetry factor $1/2$, giving coefficient $-g^2/2$. In both interpretations the closed fermion loop sign and spinor matrix trace disappear; the differing numerical factor comes from the interaction normalization and field identity.