Route a directed ghost loop so that its internal four-momenta are $l$ and $l+k$. At the external gluon with indices $(\mu ,a)$, let the incoming ghost carry colour $c$ and the outgoing ghost carry colour $d$; the other Feynman vertex has the reverse order. The stated ghost-gluon Feynman rules supply $g_D^2{f}^{acd}{f}^{bdc}(l+k{)}^{\mu }{l}^{\nu }$. Their colour factor isThe two ghost propagators supply $i^2=-1$, while the closed ghost loop supplies another $-1$. Thus, with the insertion convention stated above,This establishes every prefactor before integration. To isolate its ultraviolet pole, take generic off shell $k$, initially $k^2<0$, so the massless integral has no overlapping infrared divergence. Introduce a Feynman parameter, shift $r=l+xk$, and set $\Delta =-x(1-x)k^2>0$. Odd powers of $r$ vanish, givingThe needed dimensional regularization integrals follow from the supplied gamma function formula. In $D=4-ϵ$,Here $\Gamma (ϵ/2)=2/ϵ+O(1)$, rather than $1/ϵ$. Using $1/D=1/4+O(ϵ)$ and ${\int }_0^{1}x(1-x)dx=1/6$, obtainTherefore the ghost-loop pole of the gluon self-energy isThe factor ${\mu }^{ϵ}$ in $g_D^2$ affects only finite terms at this order. The isolated ghost loop is not transverse: contracting this result gives $g^2{C}_A{\delta }^{ab}{k}^2{k}^{\nu }/(32{\pi }^2ϵ)$. Transversality is a condition on the complete gluon self-energy, including the other loop graphs.

There are four topologies for the one-loop gluon self-energy, shown in the upper row of the figure:

The two internal-gluon graphs have symmetry factor $1/2$. The directed quark and ghost loops have symmetry factor one and a minus sign for a closed anticommuting loop. The four-gluon tadpole diagram must be included in the list even though its massless Feynman integral vanishes in dimensional regularization. Local counterterms are excluded here.

At a stationary symmetry-breaking vacuum of a gauge-invariant scalar potential, the angular field has zero potential curvature. Its free kinetic term is $(\partial b{)}^2/2$, while the displayed gauge-fixing term adds $-\xi g^2{v}^2{b}^2/2$. HenceThis is the would-be Goldstone boson, whose gauge-dependent pole accompanies the longitudinal pole in i; it is not an additional physical scalar particle of that mass. For $M_A\ne 0$, fixed finite $p$ and $\xi \to +\infty$,The angular field becomes infinitely heavy in this gauge and is absent from the unitary gauge propagating description. The qualification about fixed momentum matters: loop integrations probe arbitrarily large $p$, so one cannot automatically interchange the $\xi \to \infty$ limit with ultraviolet integration. If $gv=0$, there is no massive Higgs phase and the asserted heavy-field limit does not follow. A separate radial Higgs boson may remain; its mass depends on the unspecified potential and is not determined by this gauge-fixing term.

The quadratic part of the covariant-derivative term contains $g^2{v}^2{A}_{\mu }{A}^{\mu }/2-gv{A}_{\mu }{\partial }^{\mu }b$. Its mixing term cancels the $+gv{A}_{\mu }{\partial }^{\mu }b$ in the stated gauge fixing term. Thus the vector mass and diagonal quadratic action areFor nonzero $p^2$ introduce the transverse and longitudinal transverse and longitudinal momentum projectors $P_T^{\mu \nu }=g^{\mu \nu }-p^{\mu }{p}^{\nu }/p^2$ and $P_L^{\mu \nu }=p^{\mu }{p}^{\nu }/p^2$. After integration by parts, the inverse quadratic kernel isTheir orthogonality makes the inversion immediate. The gauge-boson propagator isEquivalently,The projector formula explains the physical massive pole and the gauge-dependent longitudinal pole. The apparent $1/p^2$ in the projectors cancels in the equivalent expression; it is not an extra massless physical state. At $\xi =1$ the numerator reduces to $g^{\mu \nu }$. At fixed momentum the $\xi \to \infty$ limit gives the unitary gauge propagator $-i(g^{\mu \nu }-p^{\mu }{p}^{\nu }/M_A^2)/(p^2-M_A^2+i0)$. The relation between this limit and ultraviolet power counting is not uniform in momentum.^[ref]

Using ${\mu }^2=\lambda v^2$, expand the potential around the chosen vacuum:The linear term vanishes at the minimum. With canonical real-field kinetic terms, comparison with $m_h^2{h}^2/2+m_{\pi }^2{\pi }^2/2$ givesThe massive radial scalar changes the magnitude of the vacuum field; the massless Goldstone boson changes its orientation. The symmetry is global here, so no gauge field eats that mode.

The genuinely new scalar cubic interactions are $-\lambda v{h}^3-\lambda vh{\pi }^2$. Including identical-field factorials, their Feynman vertex factors areThe quartic interactions remain $-\lambda h^4/4-\lambda h^2{\pi }^2/2-\lambda {\pi }^4/4$, with respective vertex factors $-6i\lambda ,-2i\lambda ,-6i\lambda$. The shifted Yukawa terms from iv give $h\bar{\psi }\psi$ vertex $-ig$ and $\pi \bar{\psi }\psi$ vertex $g{\gamma }_5$. These Yukawa structures were present before the shift; the new fermion bilinear is the mass term, not an additional independent interaction coupling. There are no derivative interactions from the linear Cartesian field shift.

The scalar potential isSince ${\mu }^2,\lambda >0$, all minima form the circle ${\varphi }_1^2+{\varphi }_2^2=v^2$, with $v=\mu /\sqrt{\lambda }>0$. Choose the representative vacuum $⟨{\varphi }_1⟩=v$, $⟨{\varphi }_2⟩=0$, and write ${\varphi }_1=v+h$, ${\varphi }_2=\pi$. A nontrivial scalar rotation moves this vacuum to another minimum, so the continuous symmetry undergoes spontaneous symmetry breaking. The invariant action and the selected ground state have different stabilizers; the circle of equivalent minima is not evidence that the action lost its symmetry.

The Yukawa terms becomeComparison with the free Dirac equation givesFor the usual choice $g>0$, this is simply $gv$. The absolute value matters if the real coupling is negative: the dispersion depends on $m_{\psi }^2$, and a field redefinition can change the mass coefficient's sign. There is no explicit symmetry-breaking fermion mass parameter in the original theory.

The scalar symmetry is broken completely. Because the spinor transformation uses a half angle, $\alpha =2\pi$ leaves the scalars fixed while sending $\psi \to -\psi$; this discrete fermion parity is unbroken in the full combined transformation group. It does not remove the broken continuous generator or the associated Goldstone boson.

The scalar kinetic and potential terms depend only on the rotation-invariant combinations ${\partial }_{\mu }{\varphi }_1{\partial }^{\mu }{\varphi }_1+{\partial }_{\mu }{\varphi }_2{\partial }^{\mu }{\varphi }_2$ and ${\varphi }_1^2+{\varphi }_2^2$. They are unchanged by the constant SO(2) rotation. The Yukawa term is invariant by ii. Finally, anticommutation gives $U{\gamma }^{\mu }={\gamma }^{\mu }{U}^{-1}$, soNo derivative of $U$ appears because the parameter is global. Thus the entire classical Lagrangian density has the combined chiral symmetry.

An explicit Dirac mass term instead changes asIt is not invariant for arbitrary $\alpha$ unless $m=0$. For the concrete choice $\alpha =\pi$, the original mass term reverses sign. The invariant Yukawa term can nevertheless generate a mass after a vacuum is selected, because the scalar field transforms along with the fermions; this is spontaneous rather than explicit breaking.

Put $U=e^{-i\alpha {\gamma }_5/2}$. The transformed Dirac adjoint is ${\bar{\psi }}^{\prime }={\psi }^{†}{U}^{†}{\gamma }^0=\bar{\psi }U$, since ${\gamma }^0$ anticommutes with ${\gamma }_5$. It is therefore incorrect to transform the two sides of a Yukawa bilinear with mutually inverse matrices.

Write $M(\varphi )={\varphi }_1+i{\gamma }_5{\varphi }_2$. The transformed Yukawa interaction is invariant exactly whenSince $M$ commutes with ${\gamma }_5$, this is $M({\varphi }^{\prime })=e^{i\alpha {\gamma }_5}M(\varphi )$. Use i and equate the coefficients of $1$ and $i{\gamma }_5$:This is the global SO(2) rotation of a chiral scalar-doublet Yukawa model. Equivalently, $\Phi ={\varphi }_1+i{\varphi }_2$ transforms as ${\Phi }^{\prime }=e^{i\alpha }\Phi$, while the two chiral spinor components transform with opposite half-angle phases. The sign of the scalar rotation follows from the negative sign in the specified fermion transformation.

Because $({\gamma }_5{)}^2=1$, its even and odd powers are ${\gamma }_5^{2r}=1$ and ${\gamma }_5^{2r+1}={\gamma }_5$. Separate the matrix exponential into these two power series:No representation of the gamma matrices is needed. Hermiticity of ${\gamma }_5$ also makes this exponential unitary for real $\alpha$, while the projectors $(1\pm {\gamma }_5)/2$ acquire opposite phases.

Contract the numerator of the quoted light cone gauge propagator with $n_{\mu }$:Thus every internal gluon emerging from a ghost-gluon vertex contributes a factor $n_{\mu }{\tilde{\Delta }}^{\mu \nu }=0$. Every ghost-dependent contribution to a physical gluon or matter amplitude vanishes. Equivalently, the strict condition $n\cdot A=0$ removes the Faddeev-Popov ghost interaction in i, leaving the field-independent determinant $\det (n\cdot \partial )$, which cancels from normalized correlators.

This is ghost decoupling in strict axial gauge, not a statement that the free Faddeev-Popov ghost propagator itself is zero. The algebra assumes a consistent axial-gauge pole prescription for $n\cdot p=0$ and treatment of residual gauge zero modes. The overall sign assigned to the displayed vector propagator does not enter the contraction proof. With the mostly-minus convention and the usual $-F^2/4$ action, its conventional prefactor is $-i/(p^2+i0)$; the source prints $+i$. Either prefactor gives the same zero contraction, so no sign correction is needed to obtain the requested ghost-decoupling result.

Choose an infinitesimal gauge transformation $\delta A_{\mu }^a=D_{\mu }^{ab}{\omega }^b$, where $D_{\mu }^{ab}={\delta }^{ab}{\partial }_{\mu }+g{f}^{acb}{A}_{\mu }^c$. Varying the axial gauge function at fixed $\sigma$ givesThis is the Faddeev-Popov operator. The Faddeev-Popov determinant has a Grassmann representation $\det M\propto \int \mathcal{D}\bar{c}\mathcal{D}c{e}^{i\int \bar{c}Mc}$. Choosing this normalization for the antighost givesA simultaneous antighost rephasing changes the overall ghost-action sign and vertex conventions, but not the determinant's field-dependent content. The essential facts are the kinetic operator $n\cdot \partial$ and a ghost-gluon vertex proportional to $n^{\mu }$.

Now average the fixed-$\sigma$ constraint with the stated Gaussian:up to a field-independent normalization. ConsequentlyThe Faddeev-Popov ghost determinant is not generally field independent before imposing the strict axial condition: its interaction term contains $n\cdot A$. In the $\xi \to 0$ gauge this quantity vanishes on the integration slice, giving the decoupling established next. Defining the inverse of $n\cdot \partial$ still requires boundary conditions for residual gauge transformations.

For two distinct commuting real scalar fields ${\chi }_1,{\chi }_2$ in place of the labelled fermion pair, with interaction $g\varphi {\chi }_1{\chi }_2$, there is no anticommutation sign on closing the loop and no spinor trace. The two propagators are simply multiplied. In terms of physical time-ordered scalar contractions $C_1,C_2$, the connected contribution isThe statistics and symmetry factors of a two-propagator loop determine the replacement. Relative to the fermionic formula in ii, the loop-statistics sign reverses. Scalar source kernels normalized like the stated Gaussian give the same overall minus sign after the four contraction factors are converted. There is no spin multiplicity.

If both replacement fields instead mean the same real field $\chi$, the coupling normalization and the identical-line symmetry factor must be specified. With interaction $(g/2)\varphi {\chi }^2$, the cubic vertex is $ig$ and its two-identical-line loop carries $1/2$, giving coefficient $-g^2/2$. With the literal coefficient $g\varphi {\chi }^2$, its vertex is $2ig$ and the coefficient is $-2g^2$. These cases are distinct from two separately labelled real fields; suppressing their combinatorics would hide a factor-of-two ambiguity in the replacement.

First fix the propagator kernels in a positive-sign Gaussian generating functional. From its scalar exponent, $i^{-2}{\delta }^2{Z}_0/\delta J\delta J$ gives the contraction $D(x-y)=-i\Delta (x-y)$. The fermionic insertion rules similarly give $F(x-y)=⟨T\psi (x)\bar{\psi }(y){⟩}_0=-iS(x-y)$, writing $S$ for the kernel labelled $S_F$ in that exponent. Thus the kernels in the question do not already include the factors of $i$ in the actual time-ordered contractions.

At order $g^2$, expand the interaction operator to its second term, or equivalently apply Wick's theorem toFor a connected contribution, the external scalar insertions attach to different interaction points. The two assignments are equal after exchanging $x,y$ and cancel $1/2$. The connected fermionic contraction isThe minus sign is the closed fermion loop sign. The alternative pairing gives two separate fermion tadpoles and belongs to disconnected external components, not the requested connected function. A free external scalar line multiplying a vacuum graph also cancels in the normalized result.

Combining $(ig{)}^2=-g^2$ with the loop minus sign givesThere are four contractions, so replacing $D=-i\Delta$ and $F=-iS$ introduces $(-i{)}^4=1$. ThereforeThis is the fermion-loop diagram in the last panel of i. The trace contracts spinor indices around the loop; it is not an additional degeneracy factor. A convention using physical contractions directly obtains the preceding formula in $D,F$ with the same coefficient. Mixing one convention's Gaussian kernels with another's insertion factors would change its apparent signs.

Use the source action $S_{\mathrm{src}}=\int d^{4}x[J\varphi +{\bar{\eta }}_{\alpha }{\psi }_{\alpha }+{\bar{\psi }}_{\alpha }{\eta }_{\alpha }]$, with independent Grassmann variables used as sources $\eta ,\bar{\eta }$ and left functional derivatives. A scalar insertion is $(1/i)\delta /\delta J$. For the odd fields the ordered rules areThe middle sign is not arbitrary: differentiating the rightmost odd source in $\bar{\psi }\eta$ from the left gives ${\delta }_L(\bar{\psi }\eta )/\delta \eta =-\bar{\psi }$. Derivative ordering must consequently match the displayed order of the fields; the rightmost derivative acts first.

Expand $\exp [ig\int \varphi \bar{\psi }\psi ]$ in its power series. Replace each field in every term by its source insertion, preserving the fermion order. These derivatives can be pulled outside the free path integral. Re-exponentiation gives the unnormalized generating functionalSummation over the spinor index is understood. Normalized correlation functions require division by the same interacting $Z[0,0,0]$, which cancels vacuum factors; that division is separate from this operator identity.

The double-longitudinal Ward identity for gluon fusion uses the defining Lie algebra identity $[T^a,T^b]=i{f}^{abc}{T}^c$ makes the boxed amplitude vanish:In QED, both external gauge bosons are photons, the charge matrices commute, and there is no three-photon vertex. The two fermion-exchange graphs then cancel without a third graph. In QCD, omitting the three-gluon interaction would leave a generally nonzero colour commutator: gauge-boson self-interaction is essential to the cancellation.

This is a tree-level manifestation of a Ward identity and, more generally in a gauge-fixed non-Abelian theory, the Slavnov-Taylor identities. They connect the matter representation, the structure constants and the gauge interactions, ensuring that auxiliary longitudinal polarizations do not enter physical probabilities. With all other external legs physical and on shell, replacing one external massless gauge-boson polarization by its momentum leaves zero. The double-contraction result proved here is one consequence, not by itself a proof that every open-index amplitude tensor is transverse off shell. At loop order the corresponding non-Abelian identities also involve Faddeev-Popov ghosts and a consistent symmetry-preserving renormalization.

The tree-level Feynman diagrams are the two ordered quark-exchange graphs and the $s$-channel gluon-exchange graph with a three-gluon vertex. There is no two-gluon-two-quark contact term in the ordinary QCD Lagrangian density. Let $q=k_1+k_2=p_1+p_2$, $r_t=p_1-k_1$, $r_u=p_1-k_2$, and $s=q^2$, $t=r_t^2$, $u=r_u^2$. Define $i\mathcal{M}$ as the product of the stated Feynman rules.

Suppressing external colour indices, the quark graphs areThe massless Dirac equations give $\bar{u}/p_1=0$ and $/p_{2}v=0$. ConsequentlyThe last equality uses $\bar{u}/qv=0$. The quark-exchange sum is therefore $-g^2\bar{u}/k_1[T^a,T^b]v$.

For the gluon graph, use all incoming momenta $(k_1,k_2,-q)$ in the three-gluon tensor:Its double contraction is $k_{1\mu }{k}_{2\nu }{V}^{\mu \nu \rho }=s(k_2-k_1{)}^{\rho }/2$. The vertex and gluon propagator giveCombining the graphs yieldsThus the requested spinor proportionality factor is $-g^2\bar{u}(p_1)/k_{1}v(p_2)$ when the colour matrix is displayed separately; retaining it inside the spinor bilinear makes its external colour indices explicit. For a convention calling the Feynman-rule product itself $\mathcal{M}$, rather than $i\mathcal{M}$, the entire result gains one factor of $i$.

Set the external four-momenta to zero after identifying the proper diagrams. In each box or triangle, at least one endpoint of the internal photon is a cubic vertex adjacent to an external scalar of zero momentum. That vertex is proportional to $l_{\rho }$, whereas the supplied numerator obeys $l^{\rho }{P}_{\rho \sigma }(l)=0$. All seven such graphs vanish algebraically. This use of the transverse numerator is stronger than merely saying that one has chosen a generic Lorenz gauge.

The two surviving seagull vertex bubbles have vertex factors $2i{e}_D^2{g}^{\mu \rho }$ and $2i{e}_D^2{g}^{\sigma \nu }$. Combining their two propagators givesThis scalar-QED four-point vertex pole in transverse gauge calculation determines the proportionality constant as well as the required tensor structure. Keep $m>0$ while extracting the ultraviolet pole. Lorentz covariance inside the regulated integral replaces $l^{\mu }{l}^{\nu }$ by $g^{\mu \nu }{l}^2/D$, giving the factor $(D-1)g^{\mu \nu }/D$.

For its scalar integral, Feynman parameter givesThere is no infrared divergence at $l=0$ with nonzero $m$. At pole order the two photon assignments and the factor $3/4$ therefore yieldThe common engineering scale factor is included in $e_D^2$; finite logarithms are immaterial to the pole. Adding $2i{e}_D^2{\delta }_4{g}^{\mu \nu }$ cancels it whenThis is minimal subtraction scheme, up to optional finite terms. It agrees with the transverse-gauge scalar QED wave-function pole and $Z_4=Z_2$. Setting $m=0$ at the outset would instead make the zero-momentum integral scaleless and obscure its separate ultraviolet divergence.

For the proper four-point interaction vertex, retain one-particle-irreducible Feynman diagrams with two labelled external photons and the two ends of a directed scalar line. Put $C$ for a cubic scalar–photon vertex and $S$ for a seagull vertex. The one-loop counting identities require $V_C+2V_S=4$. There are three possibilities, giving the nine proper scalar-QED seagull-vertex diagrams:

Each graph has Feynman-diagram symmetry factor one with these external labels: the internal photon and scalar are different species, and the charged scalar endpoints cannot be exchanged. The list is complete because the internal photon must join the ends of the open scalar chain for the graph to remain connected after any one internal line is cut. If it joined two interior points, an outside scalar segment would be a bridge and the graph would be one-particle reducible.

External-leg self-energy insertions contribute to an unamputated four-point correlation function, not this proper vertex. The local four-point counterterm from i is included separately when constructing the renormalized answer; it is not another one-loop topology. Abelian Faddeev-Popov ghosts do not interact in this linear gauge and add no diagram.

Let ${\varphi }_0=\sqrt{Z_2}\varphi$, $A_{0\mu }=\sqrt{Z_3}{A}_{\mu }$, and define the renormalized coupling by $e_0=e_D{Z}_1/(Z_2\sqrt{Z_3})$. Write $Z_2{m}_0^2=m^2+\delta m^2$ and $Z_2^2{\lambda }_0={\mu }^{ϵ}(\lambda +\delta \lambda )$. The complete renormalized Lagrangian density, with its counterterms already included, isSeparating off the classical renormalized terms leaves the counterterm LagrangianThe seagull vertex factor is thus $2i{e}_D^2{Z}_4{g}^{\mu \nu }$. The bare covariant derivative fixes $Z_4=Z_1^2/Z_2$, rather than supplying an unrelated fourth charge parameter. The scalar QED Ward identity, for a symmetry-preserving regularization, gives $Z_1=Z_2$ and hence $Z_4=Z_2$.

For an invertible quadratic gauge field operator, also include ${\mathcal{L}}_{\mathrm{gf}}=-(\partial \cdot A{)}^2/(2\xi )$. One may choose ${\xi }_0=Z_3\xi$, so this term needs no independent counterterm. The strict transverse limit is $\xi \to 0$. Even if the renormalized quartic coupling is set to zero at one scale, its counterterm need not vanish: photon graphs generate a four-scalar interaction. The renormalized Lagrangian density must retain this allowed operator.

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 $1+2\times 3+2\times 2=11$.