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For a differentiable curve $R(t)$ in the SO(3) group with $R(0)=I$, differentiating $R(t{)}^{T}R(t)=I$ gives $X^T+X=0$, where $X=R^{\prime }(0)$. Conversely, if $X$ is a real skew-symmetric matrix, $e^{tX}$ is an orthogonal matrix and has determinant $e^{t\mathrm{tr}X}=1$. HenceThe displayed generators of the SO(3) Lie algebra areThey are linearly independent and span the three independent entries of a real skew-symmetric matrix. Equivalently, $T_a\mathbf{v}={\mathbf{e}}_a\times \mathbf{v}$. The vector triple-product identity then givesThus the structure constants of a Lie algebra in this basis are

Use a real, antisymmetric convention throughout the SO(3) vector Higgs model. With gauge coupling $g$, the gauge covariant derivative and gauge field strength areThere is no factor of $i$ in these formulas because $T_a$ are real antisymmetric generators. For a local non-Abelian gauge transformation ${\Phi }^{\prime }=R\Phi$, with $R(x)$ in the SO(3) group, the compatible transformation of the gauge potential isIt gives $D_{\mu }^{\prime }{\Phi }^{\prime }=R{D}_{\mu }\Phi$ and $F_{\mu \nu }^{\prime }=R{F}_{\mu \nu }{R}^{-1}$. These covariance laws fix the relative signs in the gauge covariant derivative and gauge field strength.

An SO(3) invariant quartic scalar potential isSince $(R\Phi {)}^T(R\Phi )={\Phi }^T\Phi$, this scalar potential is gauge-invariant. Its minimum is the sphere ${\Phi }^T\Phi =v^2$, rather than a preferred direction in the internal vector space.

Writing ${\mathbf{A}}_{\mu }=(A_{\mu 1},A_{\mu 2},A_{\mu 3})$, the cross product form of the gauge covariant derivative is $D_{\mu }\Phi ={\partial }_{\mu }\Phi +g{\mathbf{A}}_{\mu }\times \Phi$. In components this isCorrespondingly, the gauge field strength has components

A covariantly constant vector Higgs field has constant norm, becauseThe last equality uses antisymmetry of the gauge potential. On a connected region write this norm as $\varphi \ge 0$. If $\varphi =0$, the vector Higgs field is identically zero. If $\varphi >0$, the SO(3) group acts transitively on unit vectors, so choose a smooth local non-Abelian gauge transformation rotating $\Phi /\varphi$ to ${\mathbf{e}}_3$. In this unitary gauge, $\Phi =(0,0,\varphi {)}^T$ with $\varphi$ constant. This is a local construction, and is global on a trivial contractible region; a nontrivial bundle can require more than one gauge patch.

For nonzero $\varphi$, substituting the constant aligned vector Higgs field into $D_{\mu }\Phi =0$ gives $g\varphi (A_{\mu 2},-A_{\mu 1},0{)}^T=0$. For nonzero gauge coupling, the most general compatible gauge potential and gauge field strength arewhere $a_{\mu }$ is an arbitrary real one-form. Thus the surviving gauge group is the $SO(2)$ stabilizer subgroup of ${\mathbf{e}}_3$. A covariantly constant vector Higgs field does not force the gauge field strength to vanish: it only requires the curvature to annihilate that field.

For the Higgs mechanism, choose a vacuum of the SO(3) invariant quartic scalar potential, so the constant norm is $v$. The canonically normalized kinetic term then containsThe two broken-direction gauge bosons have mass $gv$ (or $|g|v$ if the sign of $g$ is unrestricted), while the $T_3$ gauge boson remains massless. The two angular Goldstone bosons supply the massive vectors' longitudinal gauge-boson polarizations. The radial fluctuation remains a physical scalar field, with $m_h^2=2\lambda v^2$ for this normalization. The condition $D_{\mu }\Phi =0$ selects backgrounds without broken-direction gauge potentials; fluctuations about such a vacuum exhibit the two massive modes. Constancy of the norm alone does not require it to minimize the scalar potential.

Let the shell-restricted scalar propagator beWrite $\Delta S=S_0[\hat{\varphi }]+V[\varphi ,\hat{\varphi }]$ and average with the normalized Gaussian shell measure. The cumulant expansion givesThe subtraction removes disconnected Feynman diagrams; the logarithm retains connected Feynman diagrams made from shell contractions. This is the connected shell-contraction expansion. An external line denotes the background low field, not an additional low-momentum propagator.

At order $g$, panel A is the existing four-field interaction vertex. The new shell contractions are B, a tadpole diagram contributing a two-field vertex, and C, a Vacuum Feynman diagram contributing a constant. The Gaussian functional determinant in panel L is another field-independent term, of order $g^0$; the original quadratic kinetic term and mass term are retained as well.

For completeness, all two-vertex topologies from Wick contractions at order $g^2$ are shown. Let $r$ be the number of shell lines joining the vertices and $t_1,t_2$ their numbers of self-contractions. The external-field counts areInterchanging the two vertices identifies the same topology. Enumerating these conditions gives exactly the following eight possibilities:

The support qualification is important. In E or F, a vertex with one external low field and one bridge also has a tadpole whose two momenta cancel. Conservation forces the bridge momentum to equal that single low momentum, outside the shell. Equivalently, convolution by $C_{>}$ annihilates $\varphi$. This is momentum-support exclusion for Wilsonian bridge diagrams. D is different: its bridge carries the sum of three low momenta, which can lie in the shell. Its kernel is proportional to ${\varphi }^3(x)C_{>}(x-y){\varphi }^3(y)$ and need not vanish. Thus one-particle-reducible Feynman diagrams must not be excluded merely because the object being computed is an effective action.

These kernels need not be local before a derivative expansion. Their momentum dependence generates derivative interactions where such an expansion is valid. If all external momenta are sufficiently far below ${\Lambda }^{\prime }$, D vanishes too; in particular it does not produce a zero-momentum local ${\varphi }^6$ coupling at order $g^2$. A local six-field term is allowed and is generated at higher orders, for example by a three-vertex shell triangle.

To all orders, expect every scalar interaction allowed by the original symmetries: arbitrary even powers of the field and their allowed derivative couplings, together with vacuum terms. The Z2 symmetry $\varphi \mapsto -\varphi$ excludes odd-field vertices. The original quartic form is therefore not closed under exact Wilsonian integration.

After integrating out the shell, let a chosen operator's coefficient be $\kappa +\Delta \kappa$ and write the two-derivative quadratic term as $\frac{1}{2}(1+\Delta Z)(\partial \varphi {)}^2$. Here $\Delta \kappa$ is the shell-induced change before rescaling, and $\Delta Z$ is the shell correction to the wavefunction renormalization. Work in an operator basis in which that quadratic term has this form; the other generated operators remain in the Wilsonian effective action.

Under $x^{\prime }=bx$, the volume element changes by $d^{4}x=b^{-4}{d}^4{x}^{\prime }$ and each derivative by ${\partial }_x=b{\partial }_{x^{\prime }}$. Canonical field normalization is restored by definingIndeed, the volume factor, two derivatives and two field factors cancel in the kinetic term. A term with $n$ fields and $m$ derivatives consequently obtains the factor $b^{-4}{b}^m{b}^n(1+\Delta Z{)}^{-n/2}$. HenceThis is Wilsonian rescaling of a scalar coupling. The exponent is minus the coupling's engineering mass dimension, $[\kappa ]=4-n-m$. The rescaling also returns the low-momentum cutoff to $\Lambda$, because $p^{\prime }=p/b$.

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.