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The three real components form the Adjoint representation of $SU(2)$. Around a nonzero vacuum, a local gauge transformation rotates the direction of $\varphi (x)$ to the third axis. In this unitary gauge, the two angular fields are removed from the scalar parametrization and become longitudinal polarizations of two gauge bosons, through the Higgs mechanism. Thus $\varphi =(0,0,v+\eta {)}^T$ describes the perturbative physical fluctuations, with $v^2=-m^2/\lambda$.

The subgroup generated by $t^3$ fixes $v{e}_3$, giving $SU(2)\to U(1)$. Define the physical vector fieldsThe gauge covariant derivative on the radial configuration isso the scalar kinetic term becomes $\frac{1}{2}(\partial \eta {)}^2+g^2(v+\eta {)}^2{W}_{\mu }^{+}{W}^{-\mu }$.

Here is an exact form of the Lagrangian in these physical fields, up to its additive vacuum constant. DefineandThe minus sign in the printed non-Abelian gauge field strength gives precisely these signs. ThenIts quadratic terms give $m_A=0$, $m_{W^{+}}=m_{W^{-}}=gv$, and $m_{\eta }=\sqrt{2\lambda }v$. The two massive vectors have six real polarization degrees of freedom in total; the massless vector has two and the scalar one. Their total of nine equals the original six vector plus three scalar degrees of freedom.

Under the surviving $U(1)$, $W^{+}$ and $W^{-}$ have opposite charges and $\eta$ is neutral. The Yang-Mills theory supplies $A{W}^{+}{W}^{-}$ and $AA{W}^{+}{W}^{-}$ couplings and quartic charged-vector self-interactions. The scalar term supplies $2g^{2}v\eta W^{+}{W}^{-}$ and $g^2{\eta }^2{W}^{+}{W}^{-}$, together with the cubic and quartic scalar self-interactions. There is no direct $\eta AA$ term.

With suitable fermions, this spectrum offers a massless photon and massive charged mediators of short-range weak charged currents, so it captures part of the weak/electromagnetic pattern. It differs crucially from the electroweak interaction of the Standard Model: the latter gauges $SU(2{)}_L\times U(1{)}_Y$, uses a complex Higgs doublet, and breaks to $U(1{)}_{\mathrm{em}}$. Three rather than two generators are broken, leaving a massive neutral Z boson as well as $W^{\pm }$ and a photon. Its charge operator $Q=T^3+Y$ and independent hypercharge coupling give a nontrivial weak mixing angle. The present theory has no massive neutral mediator, no such mixing angle, and does not reproduce the observed chiral fermion charge assignments or their usual Yukawa couplings from the standard doublet/singlet assignments.

First assume global four-dimensional N=1 supersymmetry, canonical positive Kähler potential, and tree level. Write $W_i={\partial }_{i}W$ and $m_{ij}=W_{ij}$ at the stationary vacuum. The chiral-superfield fermion mass matrix is the symmetric matrix $m$, and the sum of squared Weyl spinor masses is $\mathrm{tr}(m^{†}m)$. From the F-term scalar potential $V={\sum }_k|W_k{|}^2$, its mixed and holomorphic Hessian matrix blocks areFor canonically normalized real and imaginary parts of the scalar fields, the real scalar mass matrix has trace twice the mixed trace. Its holomorphic blocks can split the two real scalar masses but do not change their sum. HenceA Weyl spinor has two spin states, so this proves the tree-level supertrace mass sum rule,The paper defines its supertrace with $(-1{)}^{2j+1}$, the negative of the conventional boson-minus-fermion supertrace used here. The asserted zero is the same in either convention. The result requires the stated canonical tree-level hypotheses; a noncanonical Kähler metric, supergravity, or radiative corrections can change the sum rule.

If vector multiplets are also present, the gauge contributions to the F-term supertrace cancel rather than being omitted. At $D_a=0$, define $C={\sum }_a{g}_a^2\Vert T_{a}v{\Vert }^2$ for the scalar expectation vector $v$. Differentiating $\frac{1}{2}{\sum }_a{D}_a^2$ adds $2C$ to the real-scalar trace. The symmetric gaugino-matter mass matrix has off-diagonal entries $\sqrt{2}{g}_a{T}_{a}v$, giving an extra $4C$ in its fermion squared-mass trace. The covariant scalar kinetic term gives vector squared-mass trace $2C$. Thus their contribution is $2C-2(4C)+3(2C)=0$. The gauge-theory result uses all spin states, including the vector weight three; it is not obtained by dropping massive gauge partners.

The MSSM tree-level sfermion mass constraint explains the phenomenological difficulty with direct visible-sector breaking. If F-term breaking is neutral under an unbroken electric and color gauge symmetry, and no D-term shifts are present, the same trace argument applies within each conserved-charge fermion block. Its corresponding scalar partners cannot all have squared masses above the mean of the light fermion squared masses. Thus canonical tree-level visible-sector supersymmetry breaking alone cannot make every squark and other scalar partner heavy while leaving the observed fermions light. The usual effective soft supersymmetry breaking terms arise after communicating breaking from a separate sector; integrating out that sector, noncanonical interactions and radiative effects evade the hypotheses. The global trace identity alone would not identify a particular light squark without the conserved-charge block argument.

For the specified O'Raifeartaigh model, phases may be chosen so that $\lambda ,\mu ,M$ are real and positive. Its superpotential derivatives areand its F-term scalar potential isFor fixed $x,z$, choose $y=-2\lambda xz/M$ to minimize the final square. Since $|z^2-{\mu }^2{|}^2\ge (|z{|}^2-{\mu }^2{)}^2$,Thus $M^2>2{\lambda }^2{\mu }^2$ makes $z=y=0$ a global minimum with arbitrary $x$. The hierarchy $M\gg \mu$ ensures this for fixed perturbative $\lambda$, rather than for arbitrarily large $\lambda$. The origin is one member of this classically flat family, with $V_0={\lambda }^2{\mu }^4>0$ and $F_X=\lambda {\mu }^2$. The complex field $x$ is a pseudomodulus.

At the origin the chiral-superfield fermion mass matrix, in the $({\psi }_X,{\psi }_Y,{\psi }_Z)$ basis, isIts physical fermion masses are $0,M,M$; the two massive Weyl spinors form one massive four-component fermion. The massless ${\psi }_X$ is the goldstino. More generally, stationarity gives $W_{ij}\overline{W_j}=0$, so the nonzero auxiliary field direction is a null vector of $m$, proving the goldstino zero mode from vacuum stationarity.

Writing $z=(z_R+i{z}_I)/\sqrt{2}$ and similarly for $x,y$, the quadratic scalar potential isThe mass spectrum of the quadratic-cubic O'Raifeartaigh model is thereforeThe scalar squared-mass sum is $4M^2$, while twice the fermion squared-mass sum is also $4M^2$, so the supertrace is zero, as required. The two massless real $x$ modes are tree-level pseudomoduli, which can be lifted by a quantum effective potential; their tree-level masslessness is not the exact symmetry protection enjoyed by the goldstino.

Use a local first-derivative Lagrangian density $\mathcal{L}(\varphi ,{\partial }_{\mu }\varphi ,x)$ with sufficient differentiability, and let variations have compact support or vanish at the spacetime boundary. If the phrase “a function of the field” were read literally as forbidding derivatives, the equation below would reduce to $\partial \mathcal{L}/\partial \varphi =0$; a propagating field requires derivative dependence. Varying the action givesIntegration by parts and arbitrary interior variations yield the Euler-Lagrange field equation,The boundary condition is part of this derivation; if boundary variations are allowed, their separate boundary equations must also be imposed.

A variational symmetry of a Lagrangian density is an infinitesimal field change $\delta \varphi =\epsilon X$ for which $\delta \mathcal{L}=\epsilon {\partial }_{\mu }{K}^{\mu }$ off shell. Equality to zero is sufficient but is not necessary: a divergence changes only the boundary contribution to the action. For fixed-coordinate field variations,Equating the two expressions proves the scalar-field form of the Noether theorem:Thus every differentiable one-parameter variational symmetry gives a conserved current, and its Noether charge $Q=\int d^{3}x{j}^0$ is constant when the spatial boundary flux vanishes. This is the Noether current for a first-derivative scalar field; the same identity applies to several real components by summing over them. Identically conserved improvement terms can change its local expression without changing the charge under the same boundary assumptions.

For an active spacetime translation, ${\varphi }^{\prime }(x)=\varphi (x-a)$ and $\delta \varphi =-a^{\nu }{\partial }_{\nu }\varphi$. If the Lagrangian density has no explicit coordinate dependence, $\delta \mathcal{L}=-a^{\nu }{\partial }_{\nu }\mathcal{L}$, so $K^{\mu }=-a^{\mu }\mathcal{L}$. The Noether current is $j^{\mu }=-a^{\nu }{T}^{\mu }{}_{\nu }$, whereThese four translation currents are the canonical stress-energy tensor, also called the energy-momentum tensor. The sign in $j=-aT$ follows from the chosen active translation; the translation charges themselves can be labelled by $P_{\nu }=\int T^0{}_{\nu }{d}^{3}x$.

For the free real scalar field, take the standard kinetic term and mass term. Its Lagrangian density and Klein-Gordon equation areHere ${\Pi }^{\mu }={\partial }^{\mu }\varphi$, so $T^{\mu \nu }={\partial }^{\mu }\varphi {\partial }^{\nu }\varphi -{\eta }^{\mu \nu }\mathcal{L}$. Raising the charge index gives the four-momentum of a free real scalar field,The first is total energy, and the three components of the second are physical spatial momentum. With signature $(+,-,-,-)$, $P_i=-P^i$; this explains the opposite sign if the conserved quantities are instead written with lower spatial indices. For example, a plane wave proportional to $\cos (Et-\mathbf{k}\cdot \mathbf{x})$ has momentum density along $\mathbf{k}$, confirming the sign. All charges require convergence of the integrals and vanishing boundary flux, or periodic boundary conditions in a finite box.

For the complex scalar field, treat $\psi$ and ${\psi }^{\ast }$ as independent variables when varying, equivalently use their two real components. The global phase symmetry of a complex scalar field is $\psi \mapsto e^{-i\alpha }\psi$, ${\psi }^{\ast }\mapsto e^{i\alpha }{\psi }^{\ast }$ for constant $\alpha$. Both the kinetic term and $V(|\psi {|}^2)$ are invariant. This is a global internal symmetry of a classical field theory with circle group $U(1)$; a spacetime-dependent phase would require a gauge connection.

Choosing this orientation for the phase parameter gives the Noether charge of a complex scalar field:To check conservation directly, the Euler-Lagrange field equations are $\square \psi +V^{\prime }(|\psi {|}^2)\psi =0$ and their complex conjugates, assuming a real differentiable potential. Hence ${\partial }_{\mu }{j}^{\mu }=i({\psi }^{\ast }\square \psi -\psi \square {\psi }^{\ast })=0$. Reversing the phase-parameter orientation reverses the current and charge, which is merely a convention. For a scalar carrying electric charge $q$, the physical electric charge is $qQ$ after electromagnetic coupling. In a neutral theory the same global charge can instead label an internal conserved quantum number; the density alone does not identify it automatically with electricity.

Use a dimensionless Hamiltonian in the functional weight $e^{-H}$, assume $\alpha >0$, and let $\varphi (\mathbf{x})={\int }_{|\mathbf{p}|\le \Lambda }{d}^{D}p(2\pi {)}^{-D}{e}^{i\mathbf{p}\cdot \mathbf{x}}\tilde{\varphi }(\mathbf{p})$. Reality means $\tilde{\varphi }(-\mathbf{p})=\tilde{\varphi }(\mathbf{p}{)}^{\ast }$. At zero field, the quadratic form isIn the momentum-shell renormalization group, split into ${\varphi }_{<}+{\varphi }_{>}$ below and above $\Lambda /b$. Disjoint Fourier supports make the quadratic form split into $H_0[{\varphi }_{<}]+H_0[{\varphi }_{>}]$. Integration over the shell gives a source-independent Gaussian functional integral multiplying the partition function, or an additive constant in the effective free energy; it leaves the slow-mode quadratic coefficients unchanged before rescaling. A uniform field has support only at zero momentum, so it does not change this shell integration.

Restore the ultraviolet cutoff by ${\mathbf{p}}^{\prime }=b\mathbf{p}$, ${\mathbf{x}}^{\prime }=\mathbf{x}/b$, and chooseThe measure, two gradients and two fields have scale factors $b^D$, $b^{-2}$ and $b^{-(D-2)}$, whose product is one. The mass term has factor $b^D{b}^{-(D-2)}=b^2$, while the uniform source term has factor $b^D{b}^{-(D-2)/2}=b^{(D+2)/2}$. Hence the Gaussian momentum-shell scaling isFor a slowly varying nonuniform source the corresponding formula is $h^{\prime }({\mathbf{x}}^{\prime })=b^{(D+2)/2}h(b{\mathbf{x}}^{\prime })$. Here the nonuniform source on the right is its projection onto the retained Fourier modes; an eliminated source component contributes only to the field-independent Gaussian normalization. The field engineering dimension is $(D-2)/2$ and its Gaussian anomalous dimension is zero. The mass and uniform source are relevant perturbations of the Gaussian fixed point. Positivity of $\alpha$ is necessary for a stable kinetic term near this point. One can use a finite volume and positive mass as infrared regulators and then take the critical limit; at exactly zero mass the integral over the zero Fourier mode alone is not a normalized finite-volume Gaussian measure.

Take ${\tau }^a={\sigma }^a/2$, $Y_{\varphi }=1/2$, metric $g_{\mu \nu }=\mathrm{diag}(1,-1,-1,-1)$, and a Higgs potential with ${\mu }_h^2>0$, $\lambda >0$. The gauge-scalar electroweak interaction isHere $B_{\mu \nu }={\partial }_{\mu }{B}_{\nu }-{\partial }_{\nu }{B}_{\mu }$. With the printed plus-sign convention for $D_{\mu }$, define $[D_{\mu },D_{\nu }]=ig{W}_{\mu \nu }^a{\tau }^a+i{g}^{\prime }{Y}_{\varphi }{B}_{\mu \nu }$; thus $W_{\mu \nu }^a={\partial }_{\mu }{W}_{\nu }^a-{\partial }_{\nu }{W}_{\mu }^a-g{ϵ}^{abc}{W}_{\mu }^b{W}_{\nu }^c$. The minus sign in the nonlinear gauge field strength follows from this convention.

The minima have ${\varphi }^{†}\varphi =v^2/2$, with $v^2={\mu }_h^2/\lambda$. A gauge transformation rotates the vacuum to $(0,v{)}^T/\sqrt{2}$. In unitary gauge, the three angular Goldstone bosons are removed, leaving $\varphi =(0,v+H{)}^T/\sqrt{2}$ and one real Higgs boson. The generator $Q=T^3+Y$ annihilates the vacuum, because its lower component has $T^3=-1/2$ and hypercharge $1/2$. This identifies the unbroken electromagnetic gauge group.

The quadratic terms from the Higgs field kinetic term areDefine the Weinberg angle and physical fields byThe neutral combinations areThe neutral gauge-boson mass matrix is $(v^2/4)\left(\begin{array}{cc}{g}^2& -g{g}^{\prime }\\ -g{g}^{\prime }& g^{\prime 2}\end{array}\right)$. It has eigenvalues zero and $(g^2+g^{\prime 2})v^2/4$, with the zero eigenvector giving $A_{\mu }$. Expanding the Higgs potential about its minimum gives $V=\text{constant}+\lambda v^2{H}^2+\lambda v{H}^3+\lambda H^4/4$. Therefore the tree-level electroweak gauge-boson masses and scalar mass areAlso $m_W=m_Z\cos {\theta }_W$ and $e=g\sin {\theta }_W=g^{\prime }\cos {\theta }_W$. The three removed Goldstone bosons supply the longitudinal polarization vector degrees of freedom of the massive electroweak gauge bosons.

Replacing $v^2$ by $(v+H{)}^2$ in the neutral mass term gives all the Higgs boson couplings to Z bosons in unitary gauge:There are exactly the cubic $HZZ$ and quartic $HHZZ$ elementary tree-level Feynman diagrams. Differentiating with respect to the identical fields gives Feynman rules $2i{m}_Z^2{g}_{\mu \nu }/v$ and $2i{m}_Z^2{g}_{\mu \nu }/v^2$, respectively. There is no elementary $ZHH$ vertex for the real neutral radial field.

For a free real scalar field, the Lagrangian density contains its quadratic kinetic term and mass term:The field's conjugate momentum is ${\pi }_1=\partial {\mathcal{L}}_1/\partial {\dot{\varphi }}_1={\dot{\varphi }}_1$. The positive sign of the time-derivative term uses the metric convention fixed in Question 1.

In four spacetime dimensions with $\hslash =c=1$, a relativistic scattering cross-section has mass dimension $-2$, since it is an area. The action is dimensionless, so the Lagrangian density has dimension four. The scalar kinetic term then gives $[\varphi ]=1$, and the cubic term gives $[\lambda ]=1$.

Each term of the four-point tree amplitude in phi cubed theory has dimension $[{\lambda }^2]-[s-m^2]=2-2=0$. Hence $[|\mathcal{M}{|}^2]=0$. In the final integral,while the numerical factors $X,Y$ are dimensionless. ThereforeIncluding the identical-particle factor changes only a dimensionless constant and preserves this check.

For every interacting $n\ge 3$, the one-loop scalar effective action already contains counterexamples. On a constant background, with a positive auxiliary mass $\mu$ to control infrared divergences, its field-dependent part isThe term of order ${\lambda }^V$ has $E=(n-2)V$ external scalar fields and is a one-loop polygon one-particle-irreducible Feynman diagram. Choose $V>n/(n-2)$ and $2V>d$. Then $E>n$, its loop momentum integral is ultraviolet convergent, and its coefficient is nonzero. Removing the ultraviolet cutoff does not remove this finite higher-point scalar vertex. For $d>2$, generic nonexceptional external momenta give the same conclusion without retaining the auxiliary mass. In $d\le 2$, the massless theory still needs an infrared prescription; removing the ultraviolet cutoff does not remove that need or the induced vertices. Hence there are no interacting pairs $(n,d)$ under the literal wording. For $n=2$ the functional determinant is field independent; $n=1$ merely shifts a Gaussian integral when an infrared prescription exists.

There is a different conventional interpretation: absence of new independent divergent counterterms. The superficial degree of divergence of a connected diagram isThe standard perturbative counterterm closure criterion is $[\lambda ]\ge 0$, with the same pairs as in part ii, provided one includes all symmetry-allowed lower-degree potential terms, the kinetic term, and the vacuum energy. A pure monomial family need not itself have counterterm closure: a sextic interaction in three dimensions generates a quartic counterterm, and a quartic interaction generates a mass counterterm. This interpretation concerns the local divergent part or the continuum defining action, not the complete quantum effective action with its finite interaction vertices and derivative expansion.

For the massive four-dimensional quartic scalar field theory, write ${\lambda }_B$ for the bare quartic coupling, ${\lambda }_R$ for its renormalized value, and $m>0$ for the mass held fixed by a renormalization condition. Define the scalar bubble integralThe quartic one-particle-irreducible correlation function, defined as a derivative of the Euclidean quantum effective action, isThere are three bubble diagrams, one for each pairing of external momenta, and each has Feynman-diagram symmetry factor $2$. The minus sign follows equivalently from the quadratic term in the functional determinant $\frac{1}{2}\mathrm{Tr}\log [1+({\lambda }_B/2)(-{\partial }^2+m^2{)}^{-1}{\varphi }^2]$.

Radial integration at zero external momentum givesThe bare quartic vertex is not finite at fixed bare coupling. Its logarithmic ultraviolet divergence must be subtracted. Impose the momentum-subtraction scheme condition ${\Gamma }_R^{(4)}(0,0,0,0)={\lambda }_R$. To one loop this requiresFor the finite difference, a Feynman parameter and a shift of loop momentum giveThe boundary error due to shifting a sharp cutoff vanishes in this limit. Substitution yields the finite renormalized quartic scalar vertexThus finiteness requires holding the renormalized coupling fixed and allowing the bare coupling to depend on ${\Lambda }_0$. The unsubtracted assertion would be false.

The same one-loop scalar effective action explains the other effective interactions. For a constant background its expansion isThe $r=1$ tadpole diagram gives a mass correction $\delta m^2={\lambda }_R{I}_1/2$, withIt requires a mass counterterm. The field-independent vacuum energy also requires subtraction. The $r=2$ term is the quartic logarithm already treated. There is no one-loop wave-function renormalization from the quartic tadpole diagram.

For $r\ge 3$ the polygon diagrams generate finite higher-point scalar vertices, withFor example, the induced sextic term in the effective potential is $+{\lambda }_R^3{\varphi }^6/(768{\pi }^2{m}^2)$, or a six-point interaction vertex $15{\lambda }_R^3/(16{\pi }^2{m}^2)$ when normalized by $6!$. These finite terms survive the removal of the ultraviolet cutoff at fixed $m,{\lambda }_R$.

The external-momentum dependence of the bubble diagrams also generates a derivative expansion of quartic interactions. At small $P$, the subtracted scalar bubble integral is $-P^2/(96{\pi }^2{m}^2)+O(P^4/m^4)$, giving finite derivative couplings. At order ${\lambda }_R^2$ these are the new field-dependent interactions beyond the mass and quartic terms; the sextic and higher interaction vertices require higher powers of ${\lambda }_R$, though they still occur at one loop. Such coefficients are suppressed by powers of the physical mass or external momentum, not by powers of ${\Lambda }_0$. The calculation establishes an order-by-order perturbative limit; it does not establish a nonperturbative interacting continuum limit of a quantum field theory in four dimensions.

The kinetic term fixes the mass dimension of the scalar field, and the interaction then fixes that of its coupling:A relevant coupling has positive mass dimension; a marginal coupling has zero mass dimension. ThereforeFor integer dimensions the complete interacting list isIf the noninteracting $n=1,2$ terms are included, their couplings have mass dimensions $(d+2)/2$ and $2$, respectively, and are relevant for every positive $d$. This is a classification by power counting in quantum field theory; it does not assert the stability of odd monomial potentials.

Take integer $n\ge 3$ and positive integer $d$, so the interaction is genuinely non-Gaussian. For a connected two-point Feynman diagram with $V$ interaction vertices and $I$ internal lines, counting half-edges and using the loop order givesAt one loop $I=V$, hence $(n-2)V=2$. There are only two possibilities: a two-interaction vertex bubble diagram for $n=3$, or a one-interaction vertex tadpole diagram for $n=4$. The latter is independent of external momentum and changes the mass, not the kinetic term.

For cubic scalar field theory, introduce a positive auxiliary mass $\mu$ if needed to separate infrared divergences from ultraviolet divergences. The one-loop self-energy in the quantum effective action is $-{\lambda }^2{B}_d(p)/2$, withThis Feynman parameter representation follows by shifting the loop momentum; it is directly valid for a translation-invariant regulator, or for the convergent subtracted integral. Differentiating at $p^2=0$ givesThus the coefficient of the kinetic term changes by ${\lambda }^2\int d^{d}q/[(2\pi {)}^d(q^2+{\mu }^2{)}^3]/6$.

Only the cubic interaction has a momentum-dependent one-loop two-point correction.There is a terminology qualification: if “wave-function renormalization” means a required ultraviolet counterterm, rather than a finite correction to the field normalization, the large-$q$ radial integral behaves as $\int dq{q}^{d-7}$, and the answer isIt is logarithmically divergent at $d=6$ and power divergent above $6$ with a momentum cutoff. Among interactions with a relevant coupling or marginal coupling, the divergent case is uniquely $(n,d)=(3,6)$. The massless low-dimensional amplitude needs an infrared divergence prescription; one cannot take a local expansion at zero momentum without one. The $n=1,2$ cases, if admitted, are a source and a Gaussian mass term and have no interaction-induced wave-function renormalization.

Let the charge parameter depend on position temporarily. Using the gauge-covariant derivative of a charged scalar field, the variation of the matter action is $\delta S=\int d^{4}x{j}^{\mu }{\partial }_{\mu }\alpha$. The resulting Noether current isHere ${\nabla }_{\mu }\bar{\varphi }={\partial }_{\mu }\bar{\varphi }+i{A}_{\mu }\bar{\varphi }$. The potential does not contribute, and $j^{\mu }=-\delta S_{\mathrm{matter}}/\delta A_{\mu }$. Because $A$ has not been rescaled to a canonical kinetic term, there is no factor $e$ in this Noether current or in the matter interaction vertices.

A local change of variables in the path integral gives $\delta ⟨\varphi (y)\bar{\varphi }(z)⟩=⟨\delta (\varphi (y)\bar{\varphi }(z))-(\delta S)\varphi (y)\bar{\varphi }(z)⟩=0$. Integration by parts in the term containing $\partial \alpha$ produces the two opposite contact terms of the Ward identity.

For clarity about phases, transform the current correlator with $e^{i(p-p^{\prime })\cdot x-ip\cdot y+i{p}^{\prime }\cdot z}$ and suppress its overall momentum conservation Dirac delta function. Denote the result by $C^{\mu }(p,p^{\prime })$. The coordinate Ward identity becomesAmputation of the two external quantum field theory propagators then fixes the longitudinal part of the current interaction vertex. Since the action couples $A_{\mu }$ to $-j^{\mu }$, the gauge interaction vertex has the opposite sign. The phase convention for the paper's gauge interaction vertices is ${\Gamma }^{\mu }=i{G}^{\mu }$, where $G^{\mu }$ is the corresponding derivative of the Euclidean quantum effective action.

One can derive the exact one-particle-irreducible correlation function directly, without assuming the entire connected current correlator is one-particle irreducible. Local gauge invariance of the quantum effective action impliesA linear covariant gauge fixing adds a breaking term independent of the scalars, which disappears after the scalar differentiations below. The identity requires a regularization in quantum field theory and counterterms preserving the Abelian gauge theory Ward identity; scalar quantum electrodynamics has no gauge anomaly.

Differentiate with respect to $\varphi (y)$ and $\bar{\varphi }(z)$, then set all background fields to zero. The two-point derivative is the inverse exact quantum field theory propagator. Use the Fourier transform convention $\varphi (x)={\int }_p{e}^{ip\cdot x}\varphi (p)$, $\bar{\varphi }(x)={\int }_{p^{\prime }}{e}^{-i{p}^{\prime }\cdot x}\bar{\varphi }(p^{\prime })$, with the photon momentum $p^{\prime }-p$ incoming. This givesAt tree level $G^{\mu }=-(p+p^{\prime }{)}^{\mu }$, so ${\Gamma }^{\mu }=-i(p+p^{\prime }{)}^{\mu }$, and the identity reduces to $-i(p^2-p^{\prime 2})=i(p^{\prime 2}-p^2)$.

To obtain the two-photon scalar Ward identity, differentiate the same functional Ward identity also with respect to $A_{\nu }(w)$. In terms of Euclidean action derivatives it readsThis construction automatically includes the seagull vertex: the Noether current depends on $A$, with $\delta j^{\mu }(x)/\delta A_{\nu }(w)=-2{\delta }^{\mu \nu }{\delta }^4(x-w)|\varphi (x){|}^2$. It cannot be discarded when differentiating a current insertion.

Take the first photon momentum to be incoming $k$, and the second incoming $p^{\prime }-p-k$. On the first scalar leg the contact term shifts $p$ to $p+k$; on the other it shifts $p^{\prime }$ to $p^{\prime }-k$. ConsequentlyWith ${\Gamma }^{\mu \nu }=i^2{G}^{\mu \nu }=-G^{\mu \nu }$, the result isThere is a momentum-routing sign error in the printed second identity. With the scalar momentum convention of the first identity, its first shifted argument must be $p^{\prime }-k$, not $k-p^{\prime }$. This is already forced at tree level: the seagull vertex has $G^{\mu \nu }=2{\delta }^{\mu \nu }$ and hence ${\Gamma }^{\mu \nu }=-2{\delta }^{\mu \nu }$. The corrected right-hand side is $-2k^{\nu }$, matching the left-hand side, whereas the printed right-hand side is $-2p^{\prime \nu }$. Changing to an all-incoming scalar convention would also change the first identity, so it does not fix both printed formulas simultaneously.