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An asymptotic scalar field is a free field describing stable incoming or outgoing particles of an interacting scalar field theory. With a field that overlaps the stable one-particle state, the scattering asymptotic condition is $\varphi \to \sqrt{Z}{\varphi }_{\mathrm{in}}$ in the remote past and $\varphi \to \sqrt{Z}{\varphi }_{\mathrm{out}}$ in the remote future. These are scattering limits of suitably smeared matrix elements, not unrestricted strong pointwise limits of local fields. With covariant state normalization $⟨p^{\prime }|p⟩=(2\pi {)}^{3}2E_p{\delta }^3(\mathbf{p}-{\mathbf{p}}^{\prime })$, the free-field vacuum-to-one-particle matrix element is one. Translation invariance and the asymptotic condition then give $⟨p|\varphi (0)|0⟩=\sqrt{Z}$ after choosing its phase positive. The wave-function renormalization residue $Z$ measures that overlap.

A simultaneous rescaling of spacetime coordinates and a field with specified weight $\Delta$. Its infinitesimal variation at a fixed argument is $\bar{\delta }\varphi =\alpha (\Delta \varphi +x^{\mu }{\partial }_{\mu }\varphi )$. For a canonical real scalar field in $D$ spacetime dimensions, $\Delta =(D-2)/2$ keeps the kinetic action invariant. Prescribing $\Delta =1$ therefore restricts kinetic invariance to $D=4$. In four dimensions the classical massless quartic scalar field theory has a conserved dilatation current $j^{\mu }=\varphi {\partial }^{\mu }\varphi +x_{\nu }{T}^{\mu \nu }$.

The two-point function Feynman integral with two interaction vertices joined by five parallel scalar propagators has four independent loop momenta:This formula uses Euclidean momenta. In three dimensions the convolution theorem and $G(x)=e^{-m|x|}/(4\pi |x|)$ givewith cutoff regularization required at $\rho =0$. With a lower radial cutoff $a$, the coefficient of $p^2$ has ultraviolet divergence $-\log (1/(ma))/(1536{\pi }^4)$. It therefore generates a divergent wave-function renormalization in three-dimensional sextic scalar field theory. By contrast the three-line sunset diagram has only a momentum-independent logarithmic ultraviolet divergence in three dimensions. This illustrates why a superficial degree of divergence of two allows both a mass counterterm and a kinetic counterterm, while degree zero allows only the former in a two-point function.

Take $M^2>0$ initially to avoid an infrared divergence. A Schwinger parameterization and the Euclidean Gaussian integral give, in the convergent range $0<\mathrm{Re}d<4$,Integration over $t$ supplies the Gamma function. Its analytic continuation defines the answer in dimensional regularization:Use $d=4-ϵ$ near four dimensions and $d=6-ϵ$ near six dimensions. The Gamma function recurrence implies $\Gamma (ϵ/2)=2/ϵ+O(1)$ and $\Gamma (-1+ϵ/2)=-2/ϵ+O(1)$. ThereforeBoth are simple poles. Equivalently their residues as functions of $d$ are $-2/(4\pi {)}^2$ at $d=4$ and $2M^2/(4\pi {)}^3$ at $d=6$. The negative pole near six dimensions is a feature of analytic continuation; it is not a negative value of the original convergent positive integral. A massless zero-momentum integral is scaleless and cannot be used indiscriminately to extract these ultraviolet poles, because dimensional regularization then combines ultraviolet and infrared contributions.

For the following Feynman diagrams, specify the Euclidean action and additive counterterms explicitly so their signs are unambiguous. The four-dimensional interaction is ${\mu }^{ϵ}\lambda {\varphi }^4/4!$, the six-dimensional interaction is ${\mu }^{ϵ/2}g{\varphi }^3/3!$, and the common free inverse scalar propagator is $p^2+m^2$. Here $ϵ=4-d$ or $6-d$, respectively. The factors of $\mu$ give dimensionless renormalized couplings and change no leading pole residues. The cubic Euclidean theory is treated as a formal perturbation expansion about its massive quadratic scalar field theory; its real cubic potential is not bounded below.

The superficial degree of divergence is the power of a common large-momentum scale obtained by scaling all independent internal momenta together, before cancellations or subtraction of divergent subdiagrams. In a scalar field theory with a standard quadratic kinetic term and non-derivative interactions, each loop momentum integral supplies $d$ powers of momentum and each internal propagator supplies minus two. ThusFor a connected Feynman diagram, counting vertex half-edges and applying the stated identity for the loop order givesIt follows that $D=(d-2)L-2V+2$. Meanwhile, ${\sum }_n(n-4)V_n=2I+E-4V=2L-2V+E-2$. Substitution provesThis power counting in quantum field theory concerns the overall scaling: even a Feynman diagram with $D<0$ can contain divergent subdiagrams, and cancellations may improve convergence when $D\ge 0$.

The canonical mass dimensions are $[\varphi ]=(d-2)/2$ and $[g_n]=d-n(d-2)/2$. A finite polynomial scalar field theory has power-counting renormalizability when its interactions have nonnegative mass dimensions and the allowed local counterterms form a finite set closed under subtraction of divergent subdiagrams. For $d>2$, non-derivative interactions therefore satisfy $n\le 2d/(d-2)$. Equivalently,so adding vertices of such interactions cannot force an indefinitely increasing number of external legs or derivatives in divergent counterterms. In dimensions $d\le 2$ a fixed finite polynomial still has this favorable power counting in quantum field theory; the displayed upper bound on $n$ is intended only for $d>2$.

For six-dimensional cubic scalar field theory, $3V=2I+E$ and $L=I-V+1$ yieldThe coupling $g$ has zero mass dimension. The possible overall divergences have $E\le 3$: vacuum energy, a linear tadpole diagram term, a mass term, a kinetic term, and a cubic interaction. These are finitely many local counterterms. Divergent subdiagrams of higher-point Feynman diagrams are removed by the same counterterms, establishing perturbative renormalizability of cubic scalar theory in six dimensions. A real cubic potential is unbounded below, so this claim is one of perturbative quantum field theory, not a claim that the polynomial defines a stable nonperturbative vacuum.

After Wick rotation, use the Euclidean action with positive kinetic and mass terms and interaction $g{\varphi }^3/3!$. The Euclidean momentum-space Feynman rules are: an internal scalar propagator $1/(k^2+m^2)$, a vertex $-g$, momentum conservation at each vertex, integration $d^{6}k/(2\pi {)}^6$ for each independent loop, and division by the Feynman-diagram symmetry factor. Strip the overall momentum-conserving delta function and external propagators when computing an amputated Green's function.

The one-particle-irreducible two-point vertex at one loop is the bubble:

Interchanging its two internal lines is an automorphism, giving Feynman-diagram symmetry factor two. In the normalization of the printed Minkowski amputated Green's function, the two vertices contribute $(-ig{)}^2$, the two internal propagators contribute $(-i{)}^2$, and Wick rotation supplies the overall $i$ already factored out there. Thus the Euclidean bubble contribution isFor this self-energy calculation we impose a vanishing one-point function, so tadpole diagram insertions are canceled by the linear counterterm. Without that convention a connected but one-particle-reducible Feynman diagram with a tadpole insertion is also possible: both external legs attach to one vertex, a zero-momentum line joins it to a second vertex, and that second vertex has a self-loop. Its amputated contribution is $g^2/(2m^2)\int d^{6}k/[(2\pi {)}^6(k^2+m^2)]$, independent of $p$. It is not the bubble drawn above; keeping it without a tadpole subtraction would add this constant to the full amputated answer.

To find the bubble's local divergence, put $K=k^2+m^2$ and expand at large $|k|$:Odd terms vanish under the angular integral. Rotational symmetry gives $⟨(k\cdot p{)}^2⟩=k^2{p}^2/6$. The terms through order $p^2$ are thereforeand the $p^2$ coefficient approaches $-1/(3k^6)$. Terms of order $p^4$ and higher are integrable at infinity in six dimensions; subtracting the two displayed terms leaves an ultraviolet-convergent integral. Hence the divergent part is a local polynomial $A{p}^2+B$, with momentum-independent coefficients. For example, when $m^2>0$ its logarithmically divergent coefficient isChanging the reference scale in this logarithm only changes a finite local term.

At zero external momentum the spherical ultraviolet cutoff and $\mathrm{Vol}(S^5)={\pi }^3$ giveWith $u=k^2$, integrate $u^2/(u+m^2{)}^2=1-2m^2/(u+m^2)+m^4/(u+m^2{)}^2$. This gives the momentum-cutoff two-point function in six-dimensional cubic scalar theoryIn particular, the divergent part of this constant iswhile the last term in $B_{\Lambda }$ tends to a finite constant. For $m=0$ the zero-momentum integral gives $B_{\Lambda }=g^2{\Lambda }^2/(256{\pi }^3)$ directly. The finite parts of a local subtraction can depend on the regularization prescription; the conclusion about mass and kinetic counterterms does not.

For the usual scalar field theory with a nonnegative potential $U$, shifted to zero in the vacuum, write its static energy in $d$ dimensions asApply Derrick scaling ${\varphi }_{\lambda }(x)=\varphi (\lambda x)$. The gradient acquires a factor $\lambda$ and the volume a factor ${\lambda }^{-d}$, givingA smooth finite-energy static solution is a stationary point of this energy, so its Derrick virial identity isFor $d\ge 3$, both terms are nonnegative, forcing $T=P=0$. The field is a constant vacuum: there is no nontrivial finite-energy static soliton in this scalar-gradient-plus-nonnegative-potential model. The assumptions matter: extra derivative terms, gauge fields, time dependence or a potential not bounded below by its vacuum value can change the argument.

For $d=2$, the relation becomes $P=0$ and says nothing further about $T$. A gradient-only energy is scale invariant, so Derrick theorem alone does not rule out nonconstant solutions; sigma-model lumps are examples. For a scalar field with isolated potential minima, $P=0$ constrains the continuous field to one vacuum and hence makes it constant. If the vacuum values form a nontrivial manifold, the scaling relation alone leaves open a nonconstant map into it. In neither case does the two-dimensional scaling argument itself prove existence.

For the usual canonical scalar field theory, subtract the vacuum energy so that the potential $U$ is nonnegative, and write the static energy in $d$ space dimensions asAssume finite energy and sufficient decay to justify the scaling variation, with a fixed vacuum at infinity. Under Derrick scaling ${\varphi }_{\lambda }(x)=\varphi (\lambda x)$,A static solution must be stationary under this admissible variation. Thus the Derrick virial identity isFor $d\ge 3$ both terms are nonpositive, and the coefficient of $T$ is strictly negative. Therefore $T=W=0$: the field is a constant vacuum. This proves the Derrick theorem under its hypotheses. Negative potentials, additional derivative terms, gauge fields or time dependence can invalidate this particular obstruction; it is not a theorem about arbitrary energy functionals.

A non-topological soliton can instead be supported by a conserved ordinary Noether charge. For a complex scalar field with global $U(1)$ invariance, take $\varphi =e^{i\omega t}f(x)$ and let $I=\int f^2$. In the normalization with kinetic density $|\partial \varphi {|}^2/2$, its charge is $Q=\omega I$. At fixed charge, eliminating $\omega$ givesThe charge-constrained scalar-field scaling now readsThe positive scaling power of the charge term can balance the others even for $d\ge 3$. Such a Q-ball is time-dependent although its energy density is stationary, and the fixed-charge variational problem is different from the static one.

Suitable potentials permit this mechanism in any spatial dimension. For example, suppose $U(f)>0$ for $f\ne 0$ and the ratio below attains a strictly positive minimum:Such a potential can favor a large region of nearly constant nonzero amplitude. At its preferred amplitude, the bulk terms minimize at $I\approx |Q|/{\omega }_0$, with energy about ${\omega }_0|Q|$. A wall has surface energy of order $|Q{|}^{(d-1)/d}$, lower order than the bulk term for every fixed $d\ge 1$. For large charge the energy can therefore be below the free-particle threshold $m|Q|$. This explains how a localized fixed-charge solution may exist and resist dispersal. Existence and stability still require a suitable potential and charge range; not every complex-field theory or every member of a rotating family is a constrained minimum.

For the particular one-dimensional profile, substituting the rotating field into the equation givesMultiplying by $f^{\prime }$ shows thatBoth $f$ and $f^{\prime }$ vanish at infinity, so the integration constant is zero. HenceHere $U(f)=f^2/2-f^4/4$. This potential is not globally bounded below; the existence of the displayed localized solutions alone does not prove their nonlinear stability.

Use metric signature $(+,-)$ and setThe Lorentz boost of a Q-ball givesThe peak follows $x=x_0+vt$. Because the equation is Lorentz invariant and the field is a scalar, this is again a solution.

The canonical stress-energy tensor gives the energy and signed physical momentum at fixed time:The minus sign agrees with the scalar-field momentum flux in the chosen signature. For the boosted field,andThe first integral gives $U(f)=\frac{1}{2}[(f^{\prime }{)}^2+{\omega }^2{f}^2]$. Changing variable to $\xi$, so $dx=d\xi /\gamma$, yieldsThus the rest-energy normalization and the relativistic energy-momentum relation are

Let $D_{xy}=⟨0|T\varphi (x)\varphi (y)|0{⟩}_0$ be the free Feynman propagator, and abbreviate $v_3=-i\kappa$, $v_4=-i\lambda$. Expand the normalized Dyson series, so vacuum bubbles cancel between numerator and denominator, and retain the connected correlation function. Wick's theorem yieldsThese are respectively the free line, the quartic tadpole diagram, the two-cubic-vertex bubble diagram, and an tadpole contribution to a cubic-theory two-point function, attached by a bridge to the vertex carrying both external legs. The last graph is connected but is a one-particle-reducible Feynman diagram; removing its bridge disconnects it. It must be included when all connected graphs are requested. It disappears if an additional vanishing-one-point-function renormalization condition and its corresponding tadpole subtraction are imposed; that convention was not specified here.

The factors can be counted directly. The quartic term has $4\cdot 3=12$ attachments for the two labelled external legs, divided by $4!=24$. At second order in the cubic interaction the expansion denominator is $2!(3!{)}^2=72$. There are 36 Wick contractions for the bubble, and 36 for the attached-tadpole graph, giving $1/2$ for each. Thus each loop topology shown has symmetry factor $S=2$. Disconnected products of one-point functions and vacuum bubbles are not part of $G_c$.^[ref]

The one-loop cubic-quartic scalar vertex classification follows without truncating arbitrarily in powers of the couplings. A connected graph with $E$ external legs, $V_3$ cubic vertices, $V_4$ quartic vertices and loop number $L$ satisfiesFor $E=2,L=1$, the only possibilities are $(V_3,V_4)=(2,0)$ or $(0,1)$, exactly those just listed.

Source paper: www.whatdotheyknow.com/request/oxford_physics_past_papers_2/response/2340696/attach/9/Physics%20Papers%202020.pdf?cookie_passthrough=1

The source cover identifies Hilary Term 2021 despite the attachment's filename. Use natural units $\hslash =c=1$, the Minkowski metric $\eta =\mathrm{diag}(1,-1,-1,-1)$, and $/p={\gamma }^{\mu }{p}_{\mu }$. The three-dimensional scalar field theory in 3 has two spatial dimensions and one time dimension. Results are at the stated order of perturbative quantum field theory.

Set $g=g_1$ and $g_2=0$. The superficial degree of divergence is then $\omega =6-2n$, independently of the number of cubic interaction vertices. After subtracting ultraviolet subdivergences, the divergent part of a one-particle irreducible Feynman diagram is a local polynomial in external momenta of degree at most $\omega$. Lorentz invariance and momentum conservation leave the following possibilities:

Thus a finite set of counterterms closes under perturbation theory. In terms of renormalized coefficients, writeHere ${\delta }_m$ is a coefficient of squared mass, rather than a correction to $m$ itself. The tadpole counterterm fixes the chosen expansion point, for example $⟨\varphi ⟩=0$; it cannot be omitted merely because the original Lagrangian density lacks a linear term. The vacuum energy density term cancels out of normalized Green functions, but is required if the unnormalized vacuum functional is also to be renormalized. This establishes perturbative renormalizability; the unbounded scalar potential of a real cubic scalar field theory does not by itself define a stable nonperturbative vacuum state.

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.

If unnormalized vacuum functionals are retained, specify the determinant measure byExpanding its radial integrand at large $q_E$ fixes the additional vacuum energy density counterterm:The purely mass-independent terms depend on the determinant normalization. Normalized Green functions eliminate this entire constant.

For a connected scalar field theory graph without derivative vertices, each independent loop momentum supplies $d$ powers and each internal scalar propagator removes two. Thus $D=dL-2I$. The graph identities $L=I-{\sum }_n{V}_n+1$ and $2I+E={\sum }_{n}n{V}_n$ give the displayed formula. The mass dimension of the field is $(d-2)/2$ and that of $g_n$ is $n-d(n/2-1)$. Negative-dimensional couplings allow the superficial degree to grow with vertex count. This is overall power counting in quantum field theory; subgraphs can diverge even when the whole graph has negative degree.