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Integrating finite Grassmann variables out of a quadratic action gives $\det [B-gC(x)]$. If $B$ is invertible, divide by $\det B$ and expand its logarithm to obtain $V_{\mathrm{eff}}={\sum }_{m\ge 1}{g}^m\mathrm{Tr}[B^{-1}C(x){]}^m/m$. Since the bosonic variables commute, an $m$-leg effective Feynman rule uses the fully symmetrized trace tensor, with weight $-g^m(m-1)!\mathrm{Sym}\mathrm{Tr}[B^{-1}{C}^{(k_1)}\cdots B^{-1}{C}^{(k_m)}]$. The minus sign comes from $e^{-V_{\mathrm{eff}}}$. This organizes closed fermion loops into bosonic vertices. It is a formal perturbative identity; a singular $B$ prevents the normalized expansion, even though the unfactored determinant polynomial still exists.

The gauge-covariant kinetic term of the Higgs doublet gives $M_W^2(1+H/v{)}^2{W}_{\mu }^{+}{W}^{-\mu }+M_Z^2(1+H/v{)}^2{Z}_{\mu }{Z}^{\mu }/2$. Differentiation with respect to the external fields yields Feynman rules $2i{M}_V^2{g}_{\mu \nu }/v$ for one Higgs boson and two matching massive electroweak gauge bosons, and $2i{M}_V^2{g}_{\mu \nu }/v^2$ for two Higgs bosons and the same pair. Identical neutral fields supply combinatorial factors. The neutral radial Higgs boson has no corresponding elementary coupling to photons at tree level.

The neutral vector mass term is $\frac{1}{2}{m}_Z^2{Z}_{\mu }{Z}^{\mu }(1+H/v{)}^2$. Expanding it gives the cubic $HZZ$ and quartic $HHZZ$ interactions in unitary gauge. Differentiating the action with respect to identical fields gives Feynman rules $2i{m}_Z^2{g}_{\mu \nu }/v$ and $2i{m}_Z^2{g}_{\mu \nu }/v^2$. A single $Z$ does not couple to two copies of the real neutral radial Higgs boson through an elementary vertex.

An interaction vertex joins the field insertions from one interaction term in a Feynman diagram. Its number of incident lines is the number of fields in that term, and its momentum-space Feynman rule includes the interaction coefficient and momentum conservation.

Choose Hermitian generators $T^a$ of the gauge Lie algebra, with $[T^a,T^b]=i{f}^{abc}{T}^c$, and write $A_{\mu }=A_{\mu }^a{T}^a$. For the usual compact gauge algebra choose $\mathrm{tr}(T^a{T}^b)={\delta }^{ab}/2$. More generally use a nondegenerate invariant symmetric bilinear form on the algebra in place of this trace contraction; a completely arbitrary component metric would not give gauge invariance.

A matter vector is useful to fix conventions: with ${\psi }^{\prime }=U(x)\psi$ and $D_{\mu }={\partial }_{\mu }-ig{A}_{\mu }$, require $D_{\mu }^{\prime }{\psi }^{\prime }=U{D}_{\mu }\psi$. Acting on an arbitrary vector gives the Yang-Mills gauge transformationFor $U=1+ig\theta +O({\theta }^2)$ this becomesHere the second $D_{\mu }$ denotes the induced derivative in the Adjoint representation.

Define the curvature by $[D_{\mu },D_{\nu }]=-ig{F}_{\mu \nu }$. Expanding the commutator derivesSince commutators of the transformed derivatives conjugate, $F_{\mu \nu }^{\prime }=U{F}_{\mu \nu }{U}^{-1}$. Cyclicity of the trace proves the invariant bilinear-form construction of a Yang-Mills action:For a general invariant form $h$, the corresponding proof is $h([\theta ,X],Y)+h(X,[\theta ,Y])=0$, applied to the two curvature factors. No matter field is required in this pure-gauge Lagrangian.

Introduce the adjoint Faddeev-Popov ghost $c^a$ and antighost ${\bar{c}}^a$, independent Grassmann-odd scalar fields of ghost numbers $+1$ and $-1$, together with the Grassmann-even Nakanishi-Lautrup field $b^a$ of ghost number zero. The bar labels the independent antighost, not an additional propagating complex-conjugate matter field. Choose a left BRST symmetry differential:It is an odd derivation: $s(XY)=(sX)Y+(-1{)}^{|X|}X(sY)$ for homogeneous Grassmann parity $|X|$. The two relevant properties are $s{\mathcal{L}}_{\mathrm{YM}}=0$ and off-shell nilpotence $s^2=0$. The first is gauge invariance with parameter replaced by the ghost, and the second follows from the Lie algebra's graded Jacobi identities. ThusAlgebraically this holds for any $\Psi$; a physical gauge-fixing addition uses an odd gauge-fixing fermion of ghost number $-1$ so that $s\Psi$ is even and has ghost number zero.

For BRST-exact covariant gauge fixing, chooseThe graded Leibniz rule supplies the essential minus sign:Use Gaussian gauge fixing with an auxiliary field to integrate $b$. Pointwise,Translation of the regulated oscillatory Gaussian leaves only a field-independent normalization. The remaining Lagrangian is thereforeThis is covariant Feynman gauge, with gauge parameter one. Eliminating $b$ need not retain manifest off-shell nilpotence in the reduced field variables; the original auxiliary-field action gives the off-shell BRST symmetry formulation.

Finally, the ungauge-fixed quadratic momentum kernel isGauge directions are null vectors, so there is no inverse defining a vector propagator, and the functional integral also counts an infinite gauge orbit volume. The added gauge-fixing term cancels the longitudinal part, leaving $Q_{\mu \nu }^{ab}=-{\delta }^{ab}{p}^2{\eta }_{\mu \nu }$. With the vacuum prescription, the propagators areThe ghost action represents the Faddeev-Popov determinant and supplies ghost-vector vertices; closed ghost loops have the fermionic minus sign. Expanding the cubic and quartic Yang-Mills terms and this ghost interaction now defines consistent perturbative Feynman rules. Physical mass-shell poles remain, whereas the gauge degeneracy has been removed. Possible global gauge-fixing obstructions and boundary zero modes are separate from the perturbative inverse around the trivial vacuum.

For a finite-dimensional integral, $u_i{e}^{b\cdot u}={\partial }_{b_i}{e}^{b\cdot u}$. Expanding a polynomial interaction $V(u)$ formally and interchanging each finite-order derivative with the regulated integral giveswhere $I_0$ is the free Gaussian integral. This is a perturbation-series identity; it does not assert convergence of the full expansion for an arbitrary potential.

For the Minkowski functional integral, each field insertion is generated byExpand $e^{i{S}_I[\varphi ]}$, replace its field factors by these functional derivatives, and take the derivatives outside the free integral. The result is polynomial interactions as source differential operators:If the field interaction term contains derivatives of the field, the corresponding spacetime derivatives act on the source-insertion operators as well. For the scalar polynomial interactions the operator in the exponent isThe free quadratic source exponential then supplies all Wick contractions, reproducing the Feynman rules. For normalized correlation functions, divide the resulting interacting functional by $Z[0]$; the interaction operator changes this denominator too, so it cannot simply be ignored.

Let a gauge transformation act infinitesimally as $\delta A_{\mu }^a=(D_{\mu }\omega {)}^a$, and define the Faddeev-Popov operator by variation of the chosen gauge functional:Work in a perturbative regular gauge slice, with boundary conditions that remove residual zero modes and with one local representative of each gauge orbit. The Faddeev-Popov gauge-orbit identity is $1={\Delta }_{\mathrm{FP}}[A]\int \mathcal{D}\omega \delta [\mathcal{F}[A^{\omega }]]$. Inserting it and factoring out the formal gauge-group volume givesThe functional delta imposes the gauge condition at every point and for every color. The determinant compensates the gauge-orbit Jacobian. For ordinary unoriented real gauge-orbit integration the local Jacobian is $|\det \mathcal{M}|$; perturbatively its sign is fixed and absorbed into $\mathcal{N}$, leaving the displayed determinant. Multiple global intersections, or a Gribov ambiguity, require more care; the local perturbative formula is not a global uniqueness assertion.

The complex Grassmann Gaussian integral now generalizes to spacetime and color indices. Introduce independent anticommuting Faddeev-Popov ghost fields $c^a(x),{\bar{c}}^a(x)$ and writeThe factors $i$ in the functional determinant differ from a real exponential only by a regulated, field-independent constant. These Grassmann fields are Lorentz scalars in the adjoint color representation; the antighost is independent of the ghost rather than an ordinary complex-conjugate commuting variable. They are unphysical fields, not external observable particles. Representing the functional determinant by a local ghost action when $\mathcal{F}$ is local makes ordinary Feynman rules available. Anticommutation gives the minus sign of a closed ghost loop, enabling cancellation of unphysical gauge contributions and maintaining the gauge identities in perturbation theory.

For example, the linear covariant choice ${\mathcal{F}}^a={\partial }^{\mu }{A}_{\mu }^a$ gives ${\mathcal{M}}^{ab}={\partial }^{\mu }{D}_{\mu }^{ab}$, so non-abelian ghost-gauge interactions remain. In an axial gauge ${\mathcal{F}}^a=n^{\mu }{A}_{\mu }^a$ with constant $n$, however,on the exact gauge slice. Its determinant is field-independent, so axial-gauge ghosts decouple and contribute only a normalization factor. This statement uses the strict delta-functional constraint; a finite-width Gaussian gauge weight is not the same exact slice.

The disadvantage is that choosing $n$ obscures manifest Lorentz covariance and the axial-gauge propagator contains spurious $1/(n\cdot k)$ poles. An axial-gauge pole prescription and control of residual transformations satisfying $n\cdot \partial \omega =0$ are necessary; formal decoupling alone does not fix those problems.

Abelian ghosts do not always decouple. In a field-independent linear Lorenz gauge, $\delta A_{\mu }={\partial }_{\mu }\omega$ gives $\mathcal{M}=\square$ and they are free. But the quadratic Abelian gauge fixing $\mathcal{F}[A]=\partial \cdot A+\kappa A_{\mu }{A}^{\mu }$ givesIts Faddeev-Popov operator depends on $A$, and the ghost action contains $2\kappa \bar{c}{A}^{\mu }{\partial }_{\mu }c$. This is an explicit abelian ghost-gauge interaction. Abelianity removes the commutator term in the gauge transformation, not the field dependence of an arbitrary nonlinear gauge condition.

With signature $(+---)$, the momentum-space Feynman rules for phi-fourth theory follow by expanding the interaction exponential and using free-field Wick contractions. Every internal scalar propagator supplies $i/(k^2-m_0^2+i0)$, every quartic Feynman vertex supplies $-i{\lambda }_0$, and each vertex conserves four-momentum. Integrate every independent loop momentum with $d^{4}k/(2\pi {)}^4$ and divide by the Feynman-diagram symmetry factor $S$. Equivalently, assign momenta to all lines and attach $(2\pi {)}^4{\delta }^4(\sum k)$ at each vertex before eliminating constrained integrations. A connected amputated graph therefore has the formwith the remaining overall momentum-conservation delta function understood. For an unamputated correlation function, retain the external propagators; scattering amplitudes instead use LSZ reduction. The factor $4!$ in the interaction cancels the number of ways to attach the four identical fields at a vertex, leaving the stated vertex factor.

For a connected Feynman diagram, $L=N_l-N_v+1$. The superficial degree of divergence is the power under simultaneous large rescaling of all independent loop momenta before considering cancellations or divergent subgraphs. Since each integration contributes four powers and each internal propagator removes two,Counting vertex half-edges gives $4N_v=2N_l+N_e$. Substitution yieldsThe connectedness assumption matters: a disconnected interaction graph with $C$ components instead has the corresponding count $D=4C-N_e$. Tree graphs have no loop integral even when this formal power count is nonnegative.

In four dimensions the possible primitively divergent proper vertices have $N_e=2$ or $4$, with $D=2$ or $0$; odd vertices vanish by the $\varphi \mapsto -\varphi$ symmetry. The divergent local structures are a mass term, a kinetic term and a quartic coupling term. If vacuum diagrams are retained, a vacuum-energy term is also required. Thus one can renormalize using a finite family of operators already allowed by the theory's symmetries: four-dimensional phi-fourth theory is perturbatively renormalizable. The superficial count alone is not an all-orders convergence proof. After divergent subgraphs are subtracted with these counterterms, higher-point graphs have no new overall ultraviolet divergence; a negative overall $D$ does not by itself exclude subdivergences.

The requested lowest-order graphs and the three channels needed for the coupling correction are shown here:

For the two-point function, the free propagator is order one, while its first interacting correction is the tadpole diagram, with $N_v=N_l=L=1$, $N_e=2$, $S=2$ and $D=2$. Its loop integral behaves as $\int d^{4}k/k^2$ and is quadratically divergent. For the connected four-point function, the lowest graph is the order-${\lambda }_0$ contact vertex, with $N_v=1,N_l=L=0$ and formal $D=0$ but no loop divergence. The first loop correction comprises the $s,t,u$ bubble diagrams. Each has $N_v=2,N_l=2,L=1$, $N_e=4$, $S=2$ and $D=0$, matching its logarithmic integral $\int d^{4}k/(k^2{)}^2$.

If the full four-point correlation function rather than its connected part is considered, there is also an order-one contribution: the sum of the three Wick contractions into two free propagators, with pairings $(12)(34),(13)(24),(14)(23)$. Explicitly, $G_0^{(4)}=G_{12}{G}_{34}+G_{13}{G}_{24}+G_{14}{G}_{23}$, with $G_{ij}$ the free scalar propagator between insertions $i,j$. These disconnected pairings are drawn in the bottom row of the plate. They contain no loop integral and are not contributions to the proper four-point vertex.

A six-point amplitude in phi-fourth theory has a one-loop triangle Feynman diagram with two external legs at each of three vertices. It has $N_v=3,N_l=3,L=1$, $N_e=6$, and $D=-2$. For a fixed external-leg partition its loop integral contains three scalar denominators, for exampleAfter Wick rotation, its ultraviolet radial behavior is ${\int }^{\infty }dk{k}^3/k^6={\int }^{\infty }dk/k^3$, which converges. The triangle has no proper loop subgraph, so no ultraviolet subdivergence is hidden in this example. It generates no new six-field counterterm. Nonzero mass avoids a separate infrared issue; with massless propagators exceptional external momenta can produce infrared singularities without changing this ultraviolet conclusion.

The bare mass is a coefficient in the regulated action, whereas the physical mass is the pole position of the full propagator. Write the amputated two-point insertion as $-i\Sigma (p^2)$. Summing its insertions givesAt order ${\lambda }_0$, only the tadpole contributes. Its factor one half follows directly from Wick pairing: of the four fields at the vertex, $4\cdot 3$ choices attach the two labeled external lines and the remaining pair closes the loop, so $12/4!=1/2$. ThusRotate $k^0=i{k}_E^0$ through the Feynman contour. The $i$ from the measure and the $i$ in the propagator, together with the denominator sign, give a positive Euclidean integral. With spherical cutoff $|k_E|<\Lambda$,This is the Wick-rotated cutoff tadpole mass shift, and it is independent of external momentum. Hence to first order,For a positive stable quartic coupling the first-order shift is positive. Keeping the measured pole fixed while changing the cutoff requires adjusting the bare mass. At $m_0=0$ the bracket is understood by its finite limit ${\Lambda }^2$. There is no momentum-dependent wave-function correction from this one-loop tadpole; such corrections begin at higher order.

Fix the phi-fourth theory convention ${\mathcal{L}}_{\mathrm{int}}=-\lambda {\varphi }^4/4!$, with mostly-minus Minkowski spacetime and $\hslash =c=1$. Define the scattering amplitude by stripping the overall momentum-conserving delta function from $S-1=i\mathcal{M}$. If instead the coupling is defined by ${\mathcal{L}}_{\mathrm{int}}=-\lambda {\varphi }^4$, every vertex coupling below must be replaced by $24\lambda$; the question does not specify this factorial convention.

The momentum-space Feynman rules are a Feynman propagator $i/(k^2-m^2+i0)$ for each internal scalar line, a Feynman vertex $-i\lambda$ with four scalar legs, and four-momentum conservation at each vertex. Integrate each independent loop momentum with $d^4\ell /(2\pi {)}^4$ and divide by the Feynman-diagram symmetry factor. For an amputated scattering amplitude there are no external propagators; external states have the usual relativistic normalization. Sum the connected diagrams and all inequivalent assignments of the labelled external momenta. The overall delta function is $(2\pi {)}^4{\delta }^4(\sum p_{\mathrm{in}}-\sum p_{\mathrm{out}})$.

These Feynman rules follow by expanding the Dyson series $T\exp (i\int d^{4}x{\mathcal{L}}_{\mathrm{int}})$ and applying the Wick theorem to the free-field correlation functions. A Wick contraction gives the free Feynman propagator; the $4!$ ways of contracting a vertex cancel its factorial denominator. The expansion's $1/V!$ cancels permutations of identical interaction insertions, while any remaining automorphisms give the Feynman-diagram symmetry factor. Fourier transformation produces momentum conservation, and the LSZ reduction formula amputates external propagators to obtain the scattering amplitude.

For six-point amplitudes in phi-fourth theory, an original pair of diagrams is:

In the tree-level Feynman diagram, let the internal four-momentum be $r=p_1+p_2-q_1=q_2+q_3+q_4$. The labelled diagram has symmetry factor one, so its contribution isThis is one channel. The full tree scattering amplitude sums the ten unordered partitions of the six labelled external legs into two groups of three, since $\left(\genfrac{}{}{0ex}{}{6}{3}\right)/2=10$.

For the one-loop triangle Feynman diagram, set $P=p_1+p_2$, $Q_{12}=q_1+q_2$, $Q_{34}=q_3+q_4$, so $P=Q_{12}+Q_{34}$. A consistent routing has denominatorsWith the external labels held fixed, no nontrivial graph automorphism remains, so again the symmetry factor is one. The contribution isThere are other labelled assignments and other one-loop topologies in the full scattering amplitude; the displayed expression belongs to the particular drawn diagram.

One can also express this scalar triangle Feynman integral using Feynman parameters. With $x+y+z=1$, completing the square in $x{D}_0+y{D}_1+z{D}_2$ givesThe identity $1/(D_0{D}_1{D}_2)=2{\int }_{x,y,z\ge 0}dxdydz\delta (1-x-y-z)/(x{D}_0+y{D}_1+z{D}_2{)}^3$ and the shifted momentum integral yieldThus ${\mathcal{M}}_{△}=-{\lambda }^3/(16{\pi }^2)$ times the displayed parameter integral. It is ultraviolet finite in four dimensions, although physical threshold singularities must retain the Feynman i-epsilon prescription.

For the fixed-target production threshold, the incoming four-momenta are $(E,\mathbf{p})$ and $(m,0)$, with $E=\sqrt{m^2+|\mathbf{p}{|}^2}$. Their invariant mass obeysIn the centre-of-momentum frame, four final particles have total energy at least $4m$, attained when all four are at rest in that frame. Therefore $s\ge 16m^2$, equivalently $E\ge 7m$, andThis is the kinematic threshold for $m>0$. Exactly at equality the final phase space has zero volume; a nonzero production rate requires a strict inequality.

For $m^2>0$ and initially $\Re n>d/2$, Schwinger parameterization and the supplied Gaussian integral giveSubstitute $u=\alpha m^2$ to obtain the Euclidean massive loop integralFor a positive integer $n$, $\Gamma (n)=(n-1)!$. Outside the convergence region this expression defines its meromorphic continuation, used in dimensional regularization.

Combine the two denominators with a Feynman parameter and shift $k$ by $(1-x)p$. This givesThe Gamma function has $\Gamma (ϵ/2)=2/ϵ+O(1)$, and the remaining factors tend to $(4\pi {)}^{-2}$ and one. Hence the massive scalar bubble pole in four dimensions isThe pole is independent of the external momentum, as required for a local quartic counterterm.

Use the mostly-plus Minkowski metric and expand $e^{iS}$. The momentum-space Feynman rules are: an internal scalar propagator $i/(-p^2-m^2+i0)$; a four-scalar interaction vertex $-i{\mu }^{ϵ}\lambda$; a momentum conservation delta function at each vertex; and $\int d^{d}k/(2\pi {)}^d$ per independent loop momentum. Divide by the Feynman-diagram symmetry factor. All momenta can be taken incoming. External propagators are omitted when computing an amputated one-particle-irreducible vertex.

The connected one-loop one-particle-irreducible Feynman diagrams are the two-point tadpole diagram and the three four-point bubble diagrams, in the $s,t,u$ channels. Each has symmetry factor $1/2$. A tadpole attached to an external leg through a single bridge would be reducible, so is excluded here.

The Euclidean massive loop integral at $n=1$ has $\Gamma (-1+ϵ/2)=-2/ϵ+O(1)$, so $I_{d,1}=-m^2/(8{\pi }^2ϵ)+O(1)$. Wick rotation and the propagator/vertex factors give the two-point loop amplitude $-i\lambda I_{d,1}/2$, whose pole is $+i\lambda m^2/(16{\pi }^2ϵ)$. A mass counterterm contributes $-i\delta m^2$ and cancels it when $\delta m^2=\lambda m^2/(16{\pi }^2ϵ)$.

Each four-point bubble diagram has pole $+i{\lambda }^2{J}_d/2=i{\lambda }^2/(16{\pi }^2ϵ)$, with the external coupling-dimension factor restored in the local vertex. Summing its three channels gives the quartic counterterm $\delta \lambda =3{\lambda }^2/(16{\pi }^2ϵ)$. Therefore the one-loop massive phi-fourth counterterms are exactlyThere is no one-loop wave-function renormalization, since the tadpole is independent of external momentum. Vacuum graphs would additionally require a vacuum-energy subtraction.

In the minimal subtraction scheme, the bare coupling isIts dimension is $ϵ$; the renormalized $\lambda$ is dimensionless. Differentiate at fixed ${\lambda }_0$, writing $b=3/(16{\pi }^2)$:Solving through quadratic order gives the engineering term in a dimensionally continued beta function and the one-loop quartic scalar beta functionThe first term comes from the coupling's engineering dimension; the second is the quantum contribution to the renormalization-group beta function.

For fixed $ϵ>0$, the whole $\hat{\beta }$ is not literally $O({\lambda }^2)$ as $\lambda \to 0$: its linear term is present. Interpret the final instruction as keeping terms through quadratic order. Set ${\lambda }_{\ast }=ϵ/b=16{\pi }^2ϵ/3$. Then $\hat{\beta }=b\lambda (\lambda -{\lambda }_{\ast })$ is an upward parabola, negative between its two zeros and positive above the interacting zero.

For positive initial coupling ${\lambda }_r=\lambda ({\mu }_r)$, the quartic running coupling below four dimensions follows by solving the linear equation for $1/\lambda$:If $0<{\lambda }_r<{\lambda }_{\ast }$, then $\lambda (\mu )\to {\lambda }_{\ast }$ as $\mu \to 0$ and $\lambda (\mu )\to 0$ as $\mu \to \infty$. If ${\lambda }_r={\lambda }_{\ast }$ it stays fixed; the zero solution also stays fixed. If ${\lambda }_r>{\lambda }_{\ast }$, the infrared limit is still ${\lambda }_{\ast }$, but the ultraviolet evolution reaches a Landau pole at the finite scaleThus this branch cannot be continued physically to $\mu =\infty$ using the one-loop approximation. These are statements about the quartic coupling direction. The Wilson-Fisher fixed point of the full scalar theory also has a thermal relevant direction at the Wilson-Fisher fixed point; approaching the critical theory requires mass tuning. For negative initial coupling, the quartic potential is unstable: the same formula tends to zero from below in the ultraviolet and reaches a pole at finite scale toward the infrared.

Use the momentum-space Feynman rules with all momenta conserved at each vertex. The free scalar propagator, free Dirac propagator and pseudoscalar Yukawa interaction vertex are respectivelyThe Dirac propagator follows by multiplying the inverse kinetic matrix $/p-m$ by $/p+m$, using the Clifford algebra to obtain $(/p-m)(/p+m)=(p^2-m^2)I_4$. The vertex follows by expanding $e^{i{S}_{\mathrm{int}}}$ once: there is no factorial because it has one scalar and two distinct spinor field factors. An internal scalar has no spinor indices; an internal fermion line carries the displayed matrix with its momentum directed along fermion flow.

These rules use the coefficient exactly as printed. With the standard Hermitian ${\gamma }^5=i{\gamma }^0{\gamma }^1{\gamma }^2{\gamma }^3$, the Hermiticity of a pseudoscalar Dirac bilinear gives $(\bar{\psi }{\gamma }^5\psi {)}^{†}=-\bar{\psi }{\gamma }^5\psi$. A Hermitian physical interaction therefore requires ${\lambda }^{\ast }=-\lambda$; writing $\lambda =ig$ with real $g$ changes the vertex to $g{\gamma }^5$. This convention issue does not alter the formal rules above, and no unstated real-coupling assumption is needed.

Use metric $g_{\mu \nu }=\mathrm{diag}(1,-1,-1,-1)$ and the hypercharge convention $Q=T^3+Y$. Then the Higgs doublet has $Y=1/2$ and transforms under electroweak gauge invariance asThe covariant derivative sign consistent with the requested expression isA convention that instead calls the doublet hypercharge one puts $Y/2$ in the charge generator and derivative; the physical coupling is the same. Here $B_{\mu }\mapsto B_{\mu }-{\partial }_{\mu }\beta /g^{\prime }$ and the matrix-valued $SU(2)$ connection transforms as $A_{\mu }\mapsto U{A}_{\mu }{U}^{-1}+(i/g)({\partial }_{\mu }U)U^{-1}$, making $D_{\mu }\varphi$ transform like $\varphi$.

In unitary gauge, the upper component of the derivative comes from the off-diagonal Pauli matrices:The lower component isWriting $c_W=\cos {\theta }_W$, $s_W=\sin {\theta }_W$, inversion of the neutral rotation gives $A^3=c_{W}Z+s_{W}A$ and $B=-s_{W}Z+c_{W}A$. Since $g^{\prime }=g\tan {\theta }_W$, the coefficient of the photon vanishes, whereas $-g{c}_W-g^{\prime }{s}_W=-g/c_W$. ThusThe zero photon coupling expresses the neutrality of the surviving radial Higgs boson.

The scalar Lagrangian density is ${\mathcal{L}}_{\varphi }=(D_{\mu }\varphi {)}^{†}{D}^{\mu }\varphi -V$. Adding $-F_{\mu \nu }^a{F}^{a\mu \nu }/4-B_{\mu \nu }{B}^{\mu \nu }/4$ gives the complete bosonic electroweak Lagrangian density. Its scalar and vector mass terms, after electroweak symmetry breaking, are obtained directly asIndeed, the shifted quartic Higgs potential normalization gives ${\varphi }^{†}\varphi -v^2/2=vH+H^2/2$. Comparing with a canonically normalized real scalar mass term, a complex vector mass term and a real vector mass term yieldsThe extra factor one half in this paper's Higgs potential is why $M_H^2$ is $\lambda v^2$, rather than $2\lambda v^2$ in another common convention. The tree-level electroweak gauge-boson masses give $\cos {\theta }_W=M_W/M_Z$: measure the two vector masses and take their ratio. Beyond tree level this extraction requires a specified renormalization convention and radiative corrections.

Expanding the vector terms gives the cubic and quartic Higgs couplings to electroweak gauge bosons:The four elementary Feynman diagrams and their all-incoming Feynman rules are shown below. Differentiating with respect to two identical $Z$ fields supplies a factor two; differentiating with respect to two identical Higgs fields supplies another factor two in the quartic rules.

There are no elementary $H\gamma \gamma$, $HZ\gamma$, $HH\gamma \gamma$ or $HHZ\gamma$ vertices in this tree-level Lagrangian density. Loop-induced photon interactions are a different order. The scalar self-interactions, if included, have rules $HHH:-3i{M}_H^2/v$ and $HHHH:-3i{M}_H^2/v^2$.

The prime LEP2 search used the Higgs boson coupling to Z bosons through Higgsstrahlung, $e^{+}{e}^{-}\to Z^{\ast }\to ZH$, shown in the first panel of the process figure in Question 3. This identifies an associated-production interaction, rather than an on-shell decay of a single $Z$ into a heavier Higgs. The channel's historical role is documented in the LEP2 Higgs-search account.

For the free charged W boson, take $W_{\mu }(x)=\int d^{4}p{e}^{-ipx}{W}_{\mu }(p)/(2\pi {)}^4$, with the conjugate field carrying $e^{ipx}$. In the action, multiplying the two field strengths gives $2[p^2{W}_{\mu }^{†}{W}^{\mu }-(p\cdot W^{†})(p\cdot W)]$. ThereforeTo invert it, separate the transverse and longitudinal projectors $P_T=g-pp/p^2$, $P_L=pp/p^2$. The operator is $(p^2-M_W^2)P_T-M_W^2{P}_L$, soSince the quadratic action kernel is $-O$, its Proca propagator isFor all momentum components small relative to $M_W$, the low-momentum limit is $D_{\mu \nu }\simeq i{g}_{\mu \nu }/M_W^2$, with corrections of relative order $p^2/M_W^2$ or $p_{\mu }{p}_{\nu }/M_W^2$. This local limit leads to the Fermi interaction, with $G_F/\sqrt{2}=g^2/(8M_W^2)$ for the charged-current normalization used here.