Past exam of the advanced quantum field theory for particle physics course of the University of Oxford / 2017 / 3

The gauge-invariant functional integral counts every gauge orbit repeatedly. Insert a gauge-orbit identityDividing by the gauge-group volume and averaging $f$ with a Gaussian weight produces the term $-G^2/(2\xi )$. The determinant is represented by a pair of independent anticommuting scalar fields $c,\bar{c}$, giving the prescribed ghost density $-\bar{c}\mathcal{M}c$, where $\mathcal{M}=\delta G/\delta \theta$. The Grassmann integral gives a determinant rather than the inverse determinant of a commuting Gaussian integral; field-independent constants and overall determinant signs are absorbed in normalization.

Although the gauge group is Abelian, the determinant here is field dependent because the gauge condition contains the scalar fluctuation $b$. Thus Abelian ghosts need not decouple in a broken-phase $R_{\xi }$ gauge. The fields implement gauge-orbit counting and are not physical external particles.

For the ghost coupling in a linear Cartesian Abelian Higgs gauge, the common factor in the scalar parametrization cancels from the gauge transformation. Separating real and imaginary parts givesConsequentlyFor the variables and gauge function actually printed, the ghost mass and Higgs-ghost vertex are thereforeFor the usual nonnegative gauge parameter, $m_c=\sqrt{\xi }g|v|$. The momentum-space ghost propagator is $-i/(k^2+m_c^2-i0)$ in this kinetic convention. There is no direct ghost-photon vertex or ghost-$b$ vertex. At $\xi =0$ the ghost becomes massless and decouples.

The scalar prefactor $1/4$ is incompatible with the stated canonical broken-vacuum interpretation. At $h=b=0$ it gives $|\varphi {|}^2=v^2/16$, whereas the potential is minimized at $|\varphi {|}^2=v^2/2$. In fact the printed parametrization givesIts $h,b$ kinetic terms have coefficient $-1/16$, not $-1/2$, and its vector mass parameter is $g^2{v}^2/8$. The given gauge function consequently does not cancel the vector-scalar mixing in those literal variables. The ghost operator computed above nevertheless follows algebraically from that gauge function without changing the printed fields.

With a canonical Abelian Higgs shift $\varphi =(v+h+ib)/\sqrt{2}$, the expansion is about a true minimum, the vector mass is $m_A=g|v|$, the gauge fixing cancels the mixing, and the same ghost formulas acquire the usual interpretation $m_c^2=\xi m_A^2$. For the literal $1/4$ parametrization, a canonical fluctuation is $h_{\mathrm{can}}=h/\sqrt{8}$, so its ghost vertex would instead be $-i\sqrt{8}\xi g^{2}v$; the non-vacuum background problem still remains. These two interpretations must not be silently mixed.

Gauge invariance of the scalar potential implies the infinitesimal identityThe irrelevant common factor $i$ from the real representation's infinitesimal rotation has been divided out. Differentiate with respect to ${\varphi }_k$ and evaluate at the stationary vacuum $v$:The second term vanishes by stationarity, leavingWhenever $T^{a}v\ne 0$, this is a nonzero vector in the kernel of the scalar mass-squared matrix, so it is a zero-eigenvalue direction. It is tangent to the degenerate vacuum orbit. In a global theory it is a Goldstone direction; in the gauged theory the corresponding would-be Goldstone field supplies a longitudinal polarization of a massive vector.

For the specific potential, stationarity at a nonzero vacuum gives $v^2=m^2/(4\lambda )$, assuming $m^2>0$ and $\lambda >0$. Its Hessian isThe antisymmetry of the triplet generators gives $v_i(T^{a}v{)}_i=0$, verifying the claimed kernel directly.

Since the hypercharge is zero, electric neutrality means $T^{3}v=0$. Choose the real-triplet vacuum $v_i=v{\delta }_{i3}$, with $v^2=m^2/(4\lambda )$. With canonical real-scalar kinetic density $-\frac{1}{2}(D_{\mu }\Phi {)}^T{D}^{\mu }\Phi$, insert this vacuum into the covariant derivative. The hypercharge field $B_{\mu }$ has no coupling to this scalar, while only $T^{1}v,T^{2}v$ are nonzero. ThusThe charged combinations $W^{\pm }=(W^1\mp i{W}^2)/\sqrt{2}$ haveBoth $T^3$ and the independent hypercharge generator annihilate the vacuum. The connected unbroken subgroup is thereforeThe electromagnetic generator $Q=T^3+Y/2$ belongs to it, but it is only one linear combination of two surviving neutral generators. This model has two massless neutral gauge fields, rather than the single massless photon and massive neutral vector of the usual scalar-doublet electroweak model. A neutral-field rotation does not create a massive eigenvalue from this zero block.

The two transverse triplet components are eaten, leaving one real, electrically neutral radial scalar $H={\varphi }_3-v$. The radial eigenvalue of the Hessian found in part i isIndeed the radial potential is a constant plus $4\lambda v^2{H}^2+4\lambda v{H}^3+\lambda H^4$, fixing the factor of two between its quadratic coefficient and its mass squared. The physical degree-of-freedom count is preserved: three real scalars plus four massless vectors give $3+4\times 2=11$, while one scalar, two massive vectors and two massless vectors give $1+2\times 3+2\times 2=11$.