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The closed string already contains the particle required for quantum gravity. For bosonic string theory in its critical dimension of string theory, the mass-shell condition and closed-string level matching areAt $N=\tilde{N}=1$, the states ${ϵ}_{mn}{\alpha }_{-1}^m{\tilde{\alpha }}_{-1}^n|0;k⟩$ are massless. After imposing the Virasoro constraints and quotienting null string states, the transverse polarization tensor decomposes into symmetric trace-free, antisymmetric and scalar pieces under $SO(D-2)$. They describe a graviton, the Kalb–Ramond field and the dilaton. In particular, the symmetric trace-free sector is a massless spin-two graviton. Its number of degrees of freedom is $D(D-3)/2$.

The state–operator correspondence associates each physical string state with a string vertex operator. For example, the matter part of a massless closed-string vertex operator isIts conformal weights are $(1,1)$, so its integrated form $\int d^{2}z{V}_{ϵ,k}$ is invariant under changes of string worldsheet coordinates. At fixed insertion positions it is accompanied by $c\bar{c}$ from the worldsheet ghost fields. Physical vertices represent BRST cohomology classes; longitudinal changes of the polarization tensor are BRST-exact operators and decouple from physical scattering amplitudes. For the graviton, this string-state gauge redundancy becomes the linearized target-space diffeomorphism $h_{mn}\mapsto h_{mn}+{\partial }_m{\xi }_n+{\partial }_n{\xi }_m$.

The Polyakov path integral computes a closed-string scattering amplitude by inserting the external string vertex operators, integrating their unfixed positions and the inequivalent worldsheet moduli, and including the worldsheet ghost fields. On a Riemann sphere three positions are fixed by the residual conformal group. For example, contractions of tachyon vertex operators produce the Koba-Nielsen factor, whose position integral gives the Virasoro–Shapiro amplitude.

The string dual resonance property means that one crossing-symmetric scattering amplitude has equivalent expansions in the different channels: its poles exhibit intermediate string states in each channel, rather than separate channel contributions being added again. The scattering-amplitude factorization at a pole identifies the intermediate particle and its couplings. Set $8\pi T=1$, so ${\alpha }^{\prime }=4$ and the closed-string tachyon has $M^2=-1$. For four such external states the Mandelstam variables obey $s+t+u=-4$. The Virasoro–Shapiro amplitude has generic $s$-channel poles at $s=-1,0,1,\dots$, exactly the bosonic string mass spectrum in these units.

The massless pole gives an especially direct check. With overall normalization $C$ and $u=-4-s-t$, the Gamma function recurrence givesThusThe quadratic residue in $t$ contains a spin-two exchange. A scalar exchange alone could not give this angular dependence. Combined with the known massless spectrum and scattering-amplitude factorization, it identifies exchange of the graviton, with possible scalar contributions from the dilaton; the antisymmetric Kalb–Ramond field does not couple to two identical scalar tachyons here. Consequently the graviton participates in interactions, rather than being an isolated free state.

Decoupling longitudinal graviton polarizations forces a universal coupling to the conserved stress-energy tensor. Consistency of this massless spin-two gauge invariance extends the linearized coupling to the nonlinear dynamics of general relativity. At distances large compared with $\sqrt{{\alpha }^{\prime }}$, the gravitational sector of the effective action begins with the Einstein-Hilbert action, alongside dilaton and Kalb–Ramond field terms and higher-derivative string corrections. The infinite tower of string states supplies the short-distance completion of this quantum gravity expansion.

Interactions are organized by string worldsheet topology. A constant dilaton gives the string coupling $g_s=e^{{\Phi }_0}$, and a connected oriented surface of genus $g$ has Euler characteristic $\chi =2-2g$. Its weight is $g_s^{-\chi }$; with $n$ normalized external vertices,The sphere is tree order, the torus is one loop, and each additional handle adds a factor $g_s^2$. Each coefficient is an integral over worldsheet moduli, so this is string perturbation theory in $g_s$, with an independent low-energy expansion in ${\alpha }^{\prime }$.

For the one-loop vacuum contribution, torus modular invariance identifies $\tau$ with $(a\tau +b)/(c\tau +d)$, $ad-bc=1$. Integration is over the standard fundamental domain of the modular group,The potential short-proper-time region ${\tau }_2\to 0$, responsible for a point-particle ultraviolet divergence, is absent. It would count metrics already represented elsewhere in $\mathcal{F}$. Torus modular invariance is therefore the geometric reason for one-loop ultraviolet finiteness. More explicitly, in $D=26$ the vacuum integrand is proportional to $d^2\tau {\tau }_2^{-14}|\eta (\tau ){|}^{-48}$, with $\eta$ the Dedekind eta function; the complete expression is invariant under the modular group.

Ultraviolet finiteness does not make the bosonic one-loop vacuum energy finite. The remaining long-tube region ${\tau }_2\to \infty$ is an infrared divergence from the tachyon: $|\eta (\tau ){|}^{-48}\sim e^{4\pi {\tau }_2}$. It signals instability of the bosonic vacuum. Tachyon-free consistent backgrounds remove this particular obstruction, though other infrared effects must still be treated. The ultraviolet improvement comes from the extended string and the complete spectrum together with the worldsheet gauge identifications; the genus expansion remains a perturbative description of quantum gravity.

Use the canonical Kähler potential and normalize superspace integration by $\int d^2\theta {\theta }^2=1$. The superspace Lagrangian density isThe full superspace integration produces a D-term, while the chiral superfield integral produces an F-term. Work in chiral coordinates for the chiral-superfield component expansion, so the omitted spacetime-derivative terms do not enter the extraction of $F$.

Set $\delta \Phi =\sqrt{2}\theta \psi +{\theta }^{2}F$. A Taylor expansion of the superpotential givesHigher terms vanish because there are only two Grassmann variables ${\theta }^{\alpha }$. With $\theta \psi ={\theta }^{\alpha }{\psi }_{\alpha }$ and $\psi \psi ={\psi }^{\alpha }{\psi }_{\alpha }$, their anticommutation gives $(\theta \psi {)}^2=-\frac{1}{2}{\theta }^2\psi \psi$. ThusThe canonical Kähler potential contributes $F^{\ast }F$. Consequently the part of the Lagrangian density containing the auxiliary field isThe fermion terms coming from the superpotential areThe Weyl spinor kinetic term and the complex scalar field kinetic term are canonical because $K={\Phi }^{†}\Phi$.

Treat $F$ and $F^{\ast }$ as independent variables in their algebraic Euler-Lagrange equations. They giveEquivalently,After eliminating the auxiliary field, the Lagrangian density contains $-V$, so the F-term scalar potential isIt is nonnegative for arbitrary complex $m,g,\phi$, since it is a modulus squared. A supersymmetric vacuum satisfies F-flatness, so its zeros are:If $m=0$ and $g\ne 0$ these coincide; if $g=0$ and $m\ne 0$ only the origin remains; if $m=g=0$ every constant $\phi$ is a supersymmetric vacuum.

To compare masses, expand around a supersymmetric vacuum $v$ satisfying $W^{\prime }(v)=0$, writing $\phi =v+\eta$. With $M=W^{\prime \prime }(v)=m+2gv$,Writing $\eta =(A+iB)/\sqrt{2}$ gives the scalar field mass term $-\frac{1}{2}|M{|}^2(A^2+B^2)$; the Weyl spinor has physical mass $|M|$. Therefore the supersymmetric mass degeneracy isIn particular, at $v=0$ the boson and fermion masses are both $|m|$. At the second supersymmetric vacuum $v=-m/g$ the sign of $M$ reverses, but the physical mass is again $|m|$. Equality here concerns fluctuations about a supersymmetric vacuum, not arbitrary backgrounds with $W^{\prime }(v)\ne 0$.

In the convention $V\supset \lambda |\phi {|}^4$ and ${\mathcal{L}}_{\mathrm{Yukawa}}=-y\phi \psi \psi +h.c.$, the same superpotential coefficient gives $\lambda =|g{|}^2$ and $y=g$. The supersymmetric relation between quartic and Yukawa couplings is thereforeIf instead the Yukawa interaction is written $-\frac{1}{2}{y}_3\phi \psi \psi +h.c.$, then $y_3=2g$ and the same physical relation is $\lambda =|y_3{|}^2/4$. The numerical factor depends on the definition of the Yukawa coupling.

The perturbative non-renormalization theorem applies to the local holomorphic superpotential in the Wilsonian effective action, using regularization in quantum field theory that preserves supersymmetry. There are two complementary ways to see why it applies here.

First, in supergraph calculations in perturbative quantum field theory, the algebra of supersymmetric covariant derivatives puts any candidate local correction to the superpotential into a full $d^4\theta$ superspace integration. It therefore corrects a D-term, such as the Kähler potential, rather than a local chiral $d^2\theta$ F-term. Turning such a contribution into an apparent chiral integral would require nonlocal inverse spacetime derivatives, schematically $1/\square$. These are excluded from the local Wilsonian effective action with a fixed nonzero momentum cutoff. This gives the diagrammatic non-renormalization theorem.

Second, the holomorphy argument for superpotential non-renormalization treats $m$ and $g$ as background chiral spurions, so the Wilsonian effective action superpotential is a holomorphic function of $\Phi ,m,g$, with no dependence on $m^{\ast }$ or $g^{\ast }$. Assign an ordinary $U(1)$ spurion charge and an R-charge as follows:The superpotential has ordinary charge zero and R-charge two, since $d^2\theta$ has R-charge minus two. A prospective holomorphic monomial $m^a{g}^b{\Phi }^c$ must therefore satisfyFor perturbative corrections, regularity at $m=0$ and $g=0$ requires nonnegative powers of the couplings. Regularity in $m$ is justified by keeping a nonzero Wilsonian momentum cutoff and retaining $\Phi$ as a light field. Hence only $c=2$ and $c=3$ survive: the existing $m{\Phi }^2$ and $g{\Phi }^3$ structures. The quadratic coefficient is fixed by the free theory at $g=0$, where there are no interaction loops. The cubic coefficient is fixed at first order in $g$, where a connected three-field interaction has only its tree-level Feynman diagram; loop corrections would require further powers of the interaction couplings, which the holomorphy argument for superpotential non-renormalization charge constraints forbid. ThusA field-independent constant is physically irrelevant in this rigid model; the spurion charges would in any case require $m^3/g^2$, which is not regular at $g=0$.

The Kähler potential can still receive wave-function renormalization. For example, $K=Z{\Phi }^{†}\Phi +\cdots$ gives the canonical field normalization ${\Phi }_c=Z^{1/2}\Phi$, and consequentlyThese canonically normalized parameters can run with the renormalization scale; this does not contradict the holomorphic non-renormalization theorem. When massless modes are included in the full quantum effective action, nonlocal terms associated with infrared divergences can also imitate an F-term. Distinguishing that action from the local Wilsonian effective action is part of the theorem's convention.

For bosonic Matsubara frequencies ${\omega }_n=2\pi nT$, set $a=vk/(2\pi T)$. Applying the residue theorem to $\pi \coth (\pi z)/(z^2+a^2)$, whose integer poles reproduce the desired summands, givesIt follows thatUsing the area $S_{d-1}=2{\pi }^{d/2}/\Gamma (d/2)$ of the unit sphere, the thermal phase fluctuation becomesAt small $k$, $\coth (vk/(2T))\sim 2T/(vk)$, so the infrared divergence is governed by ${\int }_{0}dk{k}^{d-3}$. It diverges for $d=1,2$ and is finite for $d=3$. Therefore short-range systems cannot have true finite-temperature breaking of this continuous symmetry in one or two dimensions, in agreement with the Mermin-Wagner theorem, whereas it is allowed in three dimensions. In two dimensions a Berezinskii–Kosterlitz–Thouless transition may still produce quasi-long-range order.

A representative Feynman diagram annihilates $u\bar{d}$ into $W^{+\ast }$, producing ${\nu }_{\mu }$ and an off-shell ${\mu }^{+}$; the charged antilepton emits a photon and becomes the outgoing ${\mu }^{+}$. The top panel below shows this contribution. A complete gauge-invariant decay amplitude includes radiation from the charged hadronic side, together with structure-dependent terms. The physical separation into inner bremsstrahlung, structure-dependent radiation, and their interference is explicit in the standard radiative pion decay description.^[ref]

Here is an explicit finite model of phase space alone, with its assumptions exposed. Neglect the neutrino mass, treat the photon as massless, and replace the spin-summed invariant decay amplitudes by constants over the available Lorentz-invariant phase space. Write $M=m_{\pi }$, $m=m_{\mu }$, $r=m^2/M^2$, and use natural units. The three-body phase space with one massive daughter follows by factoring through an intermediate invariant squared mass $s$:The two-body comparison has ${\Phi }_2(M;m,0)=(1-r)/(8\pi )$. Since ${\Gamma }_n=\overline{|{\mathcal{M}}_n{|}^2}{\Phi }_n/(2M)$, assuming $\overline{|{\mathcal{M}}_3{|}^2}/\overline{|{\mathcal{M}}_2{|}^2}=\kappa e^2/M^2$ givesHere $\kappa$ is a dimensionless constant encapsulating the assumed matrix element ratio, and ${\alpha }_{\mathrm{em}}$ is the fine-structure constant. If the two-body mode dominates, this ratio approximates the radiative branching fraction. Near $m=M$, $1-r^2+2r\ln r=(1-r{)}^3/3+O((1-r{)}^4)$, displaying the extra phase space suppression.

This model is not a prediction of a physical uncut radiative branching fraction. Actual inner bremsstrahlung has a soft-photon singularity, so a count of events with a separately observed photon requires a positive photon energy cut. To see both the missing input and the mass dependence explicitly, the leading soft-photon result for $E_{\mathrm{cut}}<E_{\gamma }<E_{\mathrm{soft}}\ll E_{\max }$ isThe coefficient follows from the radiation factor $e[p_{\mu }\cdot ϵ/(p_{\mu }\cdot k)-p_{\pi }\cdot ϵ/(p_{\pi }\cdot k)]$: angular integration gives ${\beta }^{-1}\ln [(1+\beta )/(1-\beta )]-2$, where $\beta =(1-r)/(1+r)$, and the soft energy integral is $\int d{E}_{\gamma }/E_{\gamma }$. Thus a finite physical number cannot be deduced from the two masses and phase space alone. The infrared divergence cancels only when the corresponding unresolved real radiation and virtual radiative corrections are combined in an inclusive observable.

The Holstein–Primakoff transformation converts a spin algebra into bosonic operators while retaining the finite physical spin Hilbert space. With $[a_n,a_m^{†}]={\delta }_{nm}$, writeThe Holstein–Primakoff occupation constraint is $0\le n_n\le 2S$. Acting on $|n⟩$, the raising matrix element is $\sqrt{n(2S-n+1)}$ and the lowering element is $\sqrt{(n+1)(2S-n)}$. Their squared difference is $2(S-n)$, reproducing $[S^{+},S^{-}]=2S^z$. Operator ordering inside the square roots is essential. For low boson occupation relative to $S$, expand them to obtain $S^{+}\simeq \sqrt{2S}a$ and $S^{-}\simeq \sqrt{2S}{a}^{†}$. Keeping quadratic terms gives the linear spin-wave approximation; higher terms describe magnon interactions. Large $S$ helps control the expansion, but a low-dimensional infrared divergence can invalidate it even at large finite $S$.

For the ferromagnet choose $H_F=-J{\sum }_n{\mathbf{S}}_n\cdot {\mathbf{S}}_{n+1}$, $J>0$, with unit lattice spacing and $\hslash =1$. Expand around the all-up state. A longitudinal bond contributes $-J{S}^2+JS(n_n+n_{n+1})$, and the transverse exchange contributes $-JS(a_n^{†}{a}_{n+1}+a_{n+1}^{†}{a}_n)$. After a Fourier transform,These are ferromagnetic magnons. The polarized vacuum is an exact ground state of the full ferromagnetic Hamiltonian: every bond is in its maximum total-spin sector, minimizing that bond energy. It therefore has no quantum zero-point reduction of its magnetization. The one-magnon dispersion is also exact, although interacting many-magnon states need not be free bosons. Rotating the polarization generates the maximum-total-spin ground-state multiplet. Selecting a definite direction in a thermodynamic symmetry-breaking description breaks $SU(2)$ to $U(1)$; the two transverse spin components form a canonical pair, producing a single quadratic type-B Goldstone boson. A small aligning field gaps it by the Zeeman energy.

At nonzero temperature the one-dimensional ferromagnet is different: its thermal magnon number contains $\int dk[e^{\beta JS{k}^2}-1{]}^{-1}$, with infrared behaviour $\int dkT/k^2$. The divergent magnetization reduction signals the absence of finite-temperature long-range order in an infinite short-range isotropic chain, consistent with the Mermin-Wagner theorem. It does not disallow the polarized zero-temperature ground states just constructed.^[ref]

For the antiferromagnet use $H_{AF}=+J{\sum }_n{\mathbf{S}}_n\cdot {\mathbf{S}}_{n+1}$, still $J>0$, on an even bipartite chain. The classical reference state alternates up and down. Rotate one sublattice through $\pi$ so that both local quantization axes point along their reference spins. In unrotated physical components, to leading order $S_A^z=S-a^{†}a$, $S_B^z=-S+b^{†}b$, $S_A^{+}\simeq \sqrt{2S}a$ and $S_B^{+}\simeq \sqrt{2S}{b}^{†}$. Choosing Fourier phases using the actual site positions givesThe reduced Brillouin zone has $M=N/2$ momenta and length $\pi$. The pair creation and annihilation terms express the fact that an antiparallel reference product state is not an eigenstate. A bosonic Bogoliubov transformation with $u_k=\cosh {\theta }_k$, $v_k=\sinh {\theta }_k$, $\tanh 2{\theta }_k=\cos k$ diagonalizes each two-mode block. With the sign convention $a_k=u_k{\alpha }_k-v_k{\beta }_{-k}^{†}$ and $b_{-k}=u_k{\beta }_{-k}-v_k{\alpha }_k^{†}$, the result isThus the harmonic antiferromagnetic modes have linear, rather than quadratic, long-wave dispersion. In an unfolded Brillouin-zone description the same result is represented by zeros at both zero and the antiferromagnetic wavevector; these are not extra independent degrees of freedom beyond the two-sublattice accounting.

The Bogoliubov vacuum has no $\alpha$ or $\beta$ excitations but contains original spin bosons, with $v_k^2=\frac{1}{2}(|\sin k{|}^{-1}-1)$. Its staggered magnetization is $m_s=S-M^{-1}{\sum }_{k\in \mathrm{RBZ}}{v}_k^2$. This quantum depletion of Néel order diverges logarithmically at small $k$ in one dimension. Linear spin-wave theory therefore does not establish a stable ordered antiferromagnetic ground state for this chain. Its nominal gapless dispersion is a result of expanding around an invalid ordered reference, not a general proof of the exact excitation spectrum. For the nearest-neighbour isotropic chain, integer-spin systems have a Haldane gap, whereas half-integer-spin systems are gapless; the spin-one-half chain in particular has fractional spin excitations rather than a simple dilute gas of ordered-state magnons. This distinction requires physics beyond the harmonic expansion, including the topological term in the continuum theory.^[ref]

At wavelengths with $q^2/(4m{\rho }_0)\ll g$, neglect the density-gradient term in the quadratic density-phase action. Complete the square in its remaining density terms:The convergent Gaussian integral can be performed by a contour shift or by its Fourier-transform identity. Its determinant is independent of $\phi$ at this order. Combining that normalization with the saddle action into a constant $S_0$ givesThe saddle contribution to $S_0$ is $-\beta V{\mu }^2/(2g)$; the additional measure-dependent constant need not be assigned an invented numerical value. The density Gaussian extends the fluctuation integration beyond its exact positivity domain, which is justified only while fluctuations around ${\rho }_0>0$ are small. The smooth zero-winding assumption is also required as discussed above. A constant global phase zero mode gives a finite compact-group volume, handled separately from nonzero phonon modes.

Fourier transforming the Bose-gas phase-only action gives a kernel ${\omega }_n^2/g+({\rho }_0/m)q^2$. Analytically continuing ${\omega }_n\mapsto -i(\omega +i{0}^{+})$ gives its real-frequency poles,In the present units the excitation energy equals $\omega$. Each real Fourier coordinate is a harmonic oscillator with inertia coefficient $1/g$ and spring coefficient ${\rho }_0{q}^2/m$, hence frequency equal to the square root of their ratio. This is the direct relation to a quantum harmonic chain. The mode is a gapless sound phonon; its finite speed reflects both the compressibility $\partial \rho /\partial \mu =1/g$ and the phase stiffness ${\rho }_0/m$.

Retaining the density-gradient term provides a useful check. Its integration instead gives a temporal coefficient $1/A_q$, soThis is the Bogoliubov spectrum. Its small-$q$ limit is the derived sound branch; at large $q$ within the contact model it crosses towards particle-like behaviour. The crossover in the squared dispersion is $q\sim 2\sqrt{mg{\rho }_0}$, precisely the scale beyond which the discarded density gradient is not small.

The same action checks the dimensional restriction in the saddle analysis. At nonzero temperature its zero Matsubara mode gives phase fluctuations proportional to $\int d^{d}q/q^2$; at zero temperature integrating the frequency gives $\int d^{d}q/|q|$. Their infrared divergences invalidate uniform true long-range order for $d\le 2$ at nonzero temperature and for $d=1$ at zero temperature. The long-wave collective description can remain useful for correlations even where a global condensate saddle is not a literal symmetry-broken state. The contact coupling here is an effective low-energy parameter; this calculation does not determine ultraviolet renormalization or trap-specific surface modes.

The positive denominators in this part are Euclidean. Starting with the Minkowski spacetime Feynman rules, first perform a Wick rotation; a bare Minkowski metric integral without a pole prescription would not equal the displayed positive expression. Work initially with $m^2>0$ and $\Re n>D/2$, where the Euclidean massive loop integral converges.

Use Schwinger parameterization followed by the Gaussian integral:The remaining Gamma function integral givesThe factorial form is for positive integral $n$. Dimensional regularization defines the result elsewhere by analytic continuation.

For the scalar bubble integral, combine the two denominators with a Feynman parameter $x$ and shift the integration variable to $\ell =k-(1-x)p_E$. ThenFor $D=4-ϵ$ this isThe Gamma function recurrence gives $\Gamma (ϵ/2)=2/ϵ+O(1)$, and the parameter integral is $1+O(ϵ)$. ConsequentlyThis ultraviolet divergence is independent of the external momentum. The assumption $m^2>0$ avoids an infrared divergence. In particular the completely massless zero-momentum integral is a scaleless integral and vanishes in dimensional regularization; its ultraviolet and infrared poles must be separated before assigning it the displayed ultraviolet residue.

There are three independent nonfactorized momentum integrals, namely a tadpole integral, the three-line sunset diagram integral, and the five-line banana integral. The six displayed Feynman diagrams involve six complete integral expressions:The distinction between independent building blocks and their products removes an ambiguity in counting “different momentum integrals”. For the strict divergent-wave-function renormalization interpretation of ii, the graphs in c reduce to $T,T^2,B_5$, comprising three full integral expressions and two independent nonfactorized building blocks.

Make a Wick rotation and use $m>0$ to separate ultraviolet divergences from infrared divergences. Write $D_E(k)=1/(k^2+m^2)$, and let $T,B_r(P)$ be the Euclidean versions of the integrals above, with their harmless overall factors of $i$ removed. A spherical ultraviolet cutoff givesThus the one-vertex quartic graph is linearly divergent, and the one-vertex sextic graph is quadratically divergent as $T_{\Lambda }^2$.

For the other two building blocks, the spatial Fourier transform is especially transparent. The three-dimensional Euclidean scalar propagator isThe convolution theorem converts the $r$ parallel lines into $G(x{)}^r$. Angular integration gives, with a short-distance cutoff $a={\Lambda }^{-1}$,This is a different convenient cutoff regularization for the connected block; power-divergent coefficients depend on the cutoff convention, whereas the conclusions about powers and the presence of a kinetic logarithm do not. The sinc function is understood to be $1$ at $P=0$.

For $r=3$, the integrand starts as ${\rho }^{-1}+O(1)$, soIts divergent part is independent of external momentum. In particular,The finite difference and its finite derivative show why the pure three-line sunset diagram does not need a divergent wave-function renormalization in three dimensions.

For $r=5$, the small-distance Taylor expansion isConsequentlyIn particular, the nonzero $P^2\log \Lambda$ term requires a genuine divergent wave-function renormalization, in addition to the mass counterterm. The coefficients here refer to the displayed position-space cutoff and have not included Feynman-diagram symmetry factors or bare couplings.

The comparison with the superficial degree of divergence isThus the actual leading power agrees with $\omega$ in every case. The additional logarithms in products come from ultraviolet subdivergences, which uniform power counting alone does not resolve.

For $T{B}_3$ and $T^2{B}_3$, differentiation with respect to $P^2$ removes the logarithmic divergence of $B_3$, but leaves divergent factors $T$ or $T^2$ multiplying a finite, nonpolynomial function of $P^2$. These are subdivergent contributions, not new nonlocal counterterms: subtracting each tadpole diagram subdivergence and the momentum-independent $B_3$ subdivergence removes them. For example, in a factorized subtraction scheme $T{B}_3\mapsto (T-T_{\mathrm{div}})(B_3-B_{3,\mathrm{div}})$, and likewise for $T^2{B}_3$. The remaining $B_5$ kinetic logarithm explains the strict interpretation in ii. A dimensional regularization scheme can suppress power divergences; it does not invalidate this cutoff-based comparison with $\omega$.

Continue with the cubic scalar field theory of c, and use precisely its counterterm signs. Take $m^2>0$ to separate ultraviolet divergences from infrared divergences. Make a Wick rotation and impose the rotationally invariant cutoff regularization $|q_E|<\Lambda$. PutFor example, spherical coordinates give $I_1=C{\int }_0^{{\Lambda }^2}{t}^2/(t+m^2)dt$. Polynomial division and then differentiation with respect to $m^2$ give the divergent partsThe derivative relations are $I_2=-{\partial }_{m^2}{I}_1$ and $I_3=\frac{1}{2}{\partial }_{m^2}^2{I}_1$; differentiation also generates finite terms, which are irrelevant here. The first supplied integral identity is consistent with these divergent parts after Wick rotation.

The second supplied expansion is incorrect. Write $A=p^2-m^2$. Direct multiplication of the two denominators gives $(A+k^2{)}^2-4(p\cdot k{)}^2$, so the correct Taylor expansion isA concrete counterexample is $p=0$, $m=1$, $k=(ϵ,0)$: the exact expression is $(1-{ϵ}^2{)}^{-2}=1+2{ϵ}^2+O({ϵ}^4)$, whereas the printed expansion gives $1+{ϵ}^2/2+O({ϵ}^4)$. We therefore compute the self-energy without that identity.

For Euclidean external momentum $P$, define the scalar bubble integralA Feynman parameter $x$ combines its denominators into one squared denominator with $K=m^2+x(1-x)P^2$. Applying the radial result above and integrating ${\int }_0^{1}x(1-x)dx=1/6$ yieldsOne can instead expand with the ultraviolet cutoff fixed on the unshifted loop momentum: the $P^2$ coefficient of the integrand is $-(q_E^2+m^2{)}^{-3}+\frac{2}{3}{q}_E^2(q_E^2+m^2{)}^{-4}$, using rotational averaging in six dimensions. Its leading radial term is $-1/(3q_E^6)$, giving the same logarithm. Shifting a sharp integration boundary changes finite terms in this scalar bubble integral, and does not change the displayed divergent coefficients.

To fix all signs and symmetry factors together, expand the one-loop Euclidean effective action around a background scalar field:These terms are the tadpole diagram, the two-point bubble diagram, and the triangle Feynman diagram. The tadpole diagram has factor $1/2$; the two-point bubble diagram also has factor $1/2$ after two differentiations with respect to the background scalar field. The local three-point interaction vertex is $g^3{I}_3$. Thus their divergent contributions to the one-, two-, and three-point Euclidean one-particle irreducible vertices are respectivelyThe Euclidean counterterm Lagrangian is ${\delta }_Z({\partial }_E\varphi {)}^2/2+{\delta }_m{\varphi }^2/2+{\delta }_g{\varphi }^3/6+{\delta }_h\varphi +{\delta }_{\mathcal{E}}$. Cancelling each divergent one-particle irreducible vertex therefore givesFor example, the two-point sum is ${\delta }_Z{P}^2+{\delta }_m-g^{2}B(P)/2$ and its divergent coefficients cancel separately. Finite counterterms depend on the chosen renormalization conditions. The zero-mass limit can use a positive subtraction scale in the logarithms after infrared regularization.