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For an overall logarithmically divergent one-loop Feynman integral, its leading large-momentum term fixes its ultraviolet pole. Giving its denominators a positive auxiliary mass $M$ regulates the artificial small-momentum singularity produced by setting external scales to zero. The difference from the original leading expansion must be ultraviolet convergent. This justifies extracting its ultraviolet pole without mistaking a scaleless integral for an ultraviolet-finite one; multiloop subdivergences require additional treatment.

With insertion convention $i\Pi$, ghost propagator $i/p^2$, and oriented ghost-gluon Feynman rule $g{f}^{abc}{p}^{\mu }$ with $p$ pointing away along the ghost arrow, the ghost loop is $i{\Pi }_{\mathrm{gh}}=-g^2{C}_A{\delta }^{ab}\int d^{D}l(2\pi {)}^{-D}(l+k{)}^{\mu }{l}^{\nu }/[l^2(l+k{)}^2]$. In dimensional regularization with $D=4-ϵ$, a Feynman parameter and a shift give the displayed ultraviolet pole. Its contraction is $g^2{C}_A{\delta }^{ab}{k}^2{k}^{\nu }/(32{\pi }^2ϵ)$; it cancels the nontransverse part of the two-three-gluon-vertex contribution. A different insertion or ghost-vertex convention must be translated before comparing signs.

For the QED proper Feynman vertex with both external four-momenta zero, a massive internal fermion and a massless internal photon in Feynman gauge, the even numerator reduces under Lorentz covariance to ${\gamma }^{\mu }[\alpha k^2+\beta m^2]$. The gamma matrices identity ${\gamma }^{\nu }{\gamma }^{\mu }{\gamma }_{\nu }=(2-D){\gamma }^{\mu }$ and the integrated replacement $k_{\rho }{k}_{\sigma }\mapsto g_{\rho \sigma }{k}^2/D$ prove the coefficients. The term $m^2/[k^2(k^2-m^2+i0{)}^2]$ has neither an infrared divergence nor an ultraviolet divergence for $m>0$; its integral in four dimensions is $-i/(16{\pi }^2)$. Only the remaining squared massive denominator supplies the ultraviolet pole.

Set the external four-momenta to zero after identifying the proper diagrams. In each box or triangle, at least one endpoint of the internal photon is a cubic vertex adjacent to an external scalar of zero momentum. That vertex is proportional to $l_{\rho }$, whereas the supplied numerator obeys $l^{\rho }{P}_{\rho \sigma }(l)=0$. All seven such graphs vanish algebraically. This use of the transverse numerator is stronger than merely saying that one has chosen a generic Lorenz gauge.

The two surviving seagull vertex bubbles have vertex factors $2i{e}_D^2{g}^{\mu \rho }$ and $2i{e}_D^2{g}^{\sigma \nu }$. Combining their two propagators givesThis scalar-QED four-point vertex pole in transverse gauge calculation determines the proportionality constant as well as the required tensor structure. Keep $m>0$ while extracting the ultraviolet pole. Lorentz covariance inside the regulated integral replaces $l^{\mu }{l}^{\nu }$ by $g^{\mu \nu }{l}^2/D$, giving the factor $(D-1)g^{\mu \nu }/D$.

For its scalar integral, Feynman parameter givesThere is no infrared divergence at $l=0$ with nonzero $m$. At pole order the two photon assignments and the factor $3/4$ therefore yieldThe common engineering scale factor is included in $e_D^2$; finite logarithms are immaterial to the pole. Adding $2i{e}_D^2{\delta }_4{g}^{\mu \nu }$ cancels it whenThis is minimal subtraction scheme, up to optional finite terms. It agrees with the transverse-gauge scalar QED wave-function pole and $Z_4=Z_2$. Setting $m=0$ at the outset would instead make the zero-momentum integral scaleless and obscure its separate ultraviolet divergence.

Route a directed ghost loop so that its internal four-momenta are $l$ and $l+k$. At the external gluon with indices $(\mu ,a)$, let the incoming ghost carry colour $c$ and the outgoing ghost carry colour $d$; the other Feynman vertex has the reverse order. The stated ghost-gluon Feynman rules supply $g_D^2{f}^{acd}{f}^{bdc}(l+k{)}^{\mu }{l}^{\nu }$. Their colour factor isThe two ghost propagators supply $i^2=-1$, while the closed ghost loop supplies another $-1$. Thus, with the insertion convention stated above,This establishes every prefactor before integration. To isolate its ultraviolet pole, take generic off shell $k$, initially $k^2<0$, so the massless integral has no overlapping infrared divergence. Introduce a Feynman parameter, shift $r=l+xk$, and set $\Delta =-x(1-x)k^2>0$. Odd powers of $r$ vanish, givingThe needed dimensional regularization integrals follow from the supplied gamma function formula. In $D=4-ϵ$,Here $\Gamma (ϵ/2)=2/ϵ+O(1)$, rather than $1/ϵ$. Using $1/D=1/4+O(ϵ)$ and ${\int }_0^{1}x(1-x)dx=1/6$, obtainTherefore the ghost-loop pole of the gluon self-energy isThe factor ${\mu }^{ϵ}$ in $g_D^2$ affects only finite terms at this order. The isolated ghost loop is not transverse: contracting this result gives $g^2{C}_A{\delta }^{ab}{k}^2{k}^{\nu }/(32{\pi }^2ϵ)$. Transversality is a condition on the complete gluon self-energy, including the other loop graphs.

Source paper: www.whatdotheyknow.com/request/oxford_physics_past_papers_2/response/2340696/attach/8/Physics%20Papers%202019.pdf?cookie_passthrough=1

Official standalone paper: mmathphys.physics.ox.ac.uk/sites/default/files/mmathphys/documents/media/aqft_2020.pdf

The annual bundle's filename says 2019, but this paper's cover says Trinity Term 2020 and identifies A15272W1. These are solutions to that 2020 paper. Use natural units, the Minkowski metric $g=\mathrm{diag}(1,-1,-1,-1)$ and the Feynman i-epsilon prescription. Write $/p={\gamma }^{\mu }{p}_{\mu }$. In question 1, $D=4-ϵ$, $e_D={\mu }^{ϵ/2}e$, and $\mu$ is the renormalization scale; keeping the dimensionful coupling $e_D$ explicit prevents a factor-of-two ambiguity in the ultraviolet pole. In question 2(a), the factors of $i$ in the stated generating functional fix the normalization of its kernels, as explained in the solution.

The Clifford algebra of the gamma matrices gives ${\gamma }^{\nu }{\gamma }^{\mu }=2g^{\nu \mu }-{\gamma }^{\mu }{\gamma }^{\nu }$ and ${\gamma }^{\nu }{\gamma }_{\nu }=D1$. ThereforeThe contraction $g^{\nu }{}_{\nu }=D$ must be retained until the dimensional regularization calculation is complete; replacing it prematurely by four can change finite terms multiplying a ultraviolet pole.

Only the term proportional to $\alpha$ has an ultraviolet pole. Its $k^2$ cancels the massless denominator, leaving the standard Feynman integralIn the present dimensional regularization convention, $\Gamma (ϵ/2)=2/ϵ+O(1)$ and $\alpha =1+O(ϵ)$. ThereforeThe counterterm contribution is $-i{e}_D{\delta }_1{\gamma }^{\mu }$. Cancellation of the ultraviolet pole fixeswhere the displayed expression uses the minimal subtraction scheme; a different renormalization scheme may add finite $O(e^2)$ terms. In the frequently used convention $D=4-2\epsilon$, the same result is $Z_1=1-e^2/(16{\pi }^2\epsilon )+O(e^4)$. The sign agrees with $Z_1=Z_2$ from the Ward identity for QED renormalization constants in Feynman gauge.

Keep $m>0$ and set $p=p^{\prime }=0$, which is an off shell choice. The numerator isTerms odd in $k$ integrate to zero. Lorentz covariance and dimensional regularization give, inside this integral,Applying the gamma matrices identity from part (ii) once more yieldsConsequentlyThis massive zero-momentum QED vertex integral uses an integrated tensor identity for the replacement of $N^{\mu }$, not an equality for each fixed $k$. The physical photon remains massless. The retained fermion mass supplies a scale and makes this zero-external-four-momentum calculation suitable for extracting the ultraviolet pole without a scaleless integral.

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

Official standalone paper: mmathphys.physics.ox.ac.uk/sites/default/files/mmathphys/documents/media/aqft_2021.pdf

Use natural units, the Minkowski metric $g=\mathrm{diag}(1,-1,-1,-1)$ and the Feynman i-epsilon prescription. In question 1 use dimensional regularization with $d=4-2ϵ$ and write $C_{ϵ}=1/(16{\pi }^2ϵ)$; in the alternative convention $d=4-ϵ$, replace $C_{ϵ}$ by $1/(8{\pi }^2ϵ)$. Only ultraviolet poles are retained there. Figure 1 is unreadable in the released source: the corresponding specific gaps are marked below. All other parts have complete derivations.

TODO (source gap): Figure 1(a), including its internal connections and four-momentum labels, is absent from the recoverable source. The local PDF renderer reports missing image objects, and the matching official PDF's figure could not be retrieved for visual inspection. Therefore the individual Feynman diagram called (a) cannot be assigned an integrand and its separate independence of $\xi$ cannot be verified from the available diagram. A guessed replacement drawing would not resolve this gap.

An independent determination of the complete ultraviolet pole, given in part (iv), remains possible from the scalar QED Ward identity. Together with the textually supplied integral in part (iv), it fixes the sum of the contributions outside that integral to $3i{e}^3{C}_{ϵ}{k}^{\mu }$ in the stated scalar electrodynamics three-point vertex convention. That sum alone does not identify the contents of panel (a).

The superficial degree of divergence of this scalar electrodynamics three-point vertex is one. Its ultraviolet pole is consequently a local expression at most linear in the external four-momenta. The allowed counterterm Lagrangian, with the charge-conjugation and Hermiticity properties of scalar electrodynamics, contains a single one-derivative scalar electrodynamics three-point vertex:Thus the coefficient ${\delta }_1$ is independent of the chosen external four-momenta. Setting $k^{\prime }=0$ leaves $-ie{\delta }_1{k}^{\mu }$, so a nonzero generic $k$ still determines that coefficient. It does not determine the full finite scattering amplitude at other four-momenta.

Keep $m\ne 0$ and generic off shell $k$ while extracting the ultraviolet pole. These choices avoid spurious infrared divergences; setting all scales to zero would instead produce scaleless integrals that can conceal the ultraviolet pole.

The integral printed in the text can be evaluated without reconstructing Figure 1. Put $A=l^2-m^2+i0$ and $B=(l+k{)}^2-m^2+i0$, and let $I^{\mu }$ denote that integral without its factor $\xi$. The numerator contraction responsible for its covariant gauge dependence isWith the scalar electrodynamics three-point vertex $-ie(p+p^{\prime }{)}^{\mu }$, three such vertices, two scalar propagators and one photon propagator have overall factor $(-ie{)}^3{i}^2(-i)=-e^3$. Thus the supplied integrand gives $V_{\mathrm{supplied}}^{\mu }=-e^3\xi I^{\mu }$.

First isolate the numerator term $l^2$. Its contribution isUnder $l\mapsto -l-k$ the two denominators exchange and the numerator changes sign. A shift-invariant dimensional regularization therefore gives $\boxed{I_0^{\mu }=0}$, including the finite part.

For the remaining numerator $2(l\cdot k)(2l+k{)}^{\mu }$, only its term $4l^{\mu }(l\cdot k)$ contributes to the ultraviolet pole. Corrections from expanding the denominators, and the numerator $2k^{\mu }(l\cdot k)$, are ultraviolet finite by power counting. Use auxiliary-mass extraction of ultraviolet poles with any $M>0$ rather than literally integrating a massless scaleless integral. Lorentz covariance then givesTODO (source gap): The algebra of the supplied expression is complete, but its graphical identification with Figure 1(b), including the routing at each internal vertex, cannot be visually verified from the unreadable panel.

The complete renormalization constant can nevertheless be obtained independently. Here is a short scalar QED wave-function pole calculation, to avoid assuming either missing panel's value. For an external four-momentum $p$, the exchange contribution to the scalar self-energy has numeratorIts denominator is $l^2[(p-l{)}^2-m^2+i0]$. The necessary ultraviolet poles, with the measure $d^{d}l/(2\pi {)}^d$, are respectivelyThe first two follow by a Feynman parameter and a shift of $l$; the third is a tadpole integral; the last follows by the same large-$l$ extraction used above. The seagull vertex contributes only a photon tadpole integral, which is a scaleless integral, and in any case cannot affect the coefficient of $p^2$. Multiplication by the exchange graph's factor $-e^2$ gives the scalar self-energy insertionThe kinetic counterterm $i{\delta }_2{p}^2$ therefore requires ${\delta }_2=e^2(3-\xi )C_{ϵ}$. The scalar QED Ward identity sets ${\delta }_1={\delta }_2$, soIndeed the full unrenormalized scalar electrodynamics three-point vertex has ultraviolet pole $i{e}^3(3-\xi )C_{ϵ}{k}^{\mu }$, which is cancelled by $-ie{\delta }_1{k}^{\mu }$. Subtracting the supplied contribution gives the gauge-independent sum stated in part (ii), without assigning either unreadable panel an integrand.

In the right panel of part a, $p$ enters the fermion line, $p^{\prime }=p+q$ leaves it, and a virtual photon of momentum $k$ bridges the external photon vertex. Write ${\Gamma }^{\mu }={\gamma }^{\mu }+{\Lambda }^{\mu }$ and $F_1=1+F_1^{(1)}+\cdots$. The one-loop amplitude isThis Feynman integral has a logarithmic superficial degree of divergence. Its ultraviolet pole is independent of the external momenta and $m$, because every such numerator insertion or denominator correction lowers the large-$k$ degree. Preserve an auxiliary positive mass $M$ while extracting the ultraviolet pole; setting everything to zero would create a scaleless integral and conceal the cancellation of ultraviolet and infrared divergences.

The leading numerator is ${\gamma }^{\alpha }/k{\gamma }^{\mu }/k{\gamma }_{\alpha }$. Symmetric integration replaces $k_{\rho }{k}_{\sigma }$ by $g_{\rho \sigma }{k}^2/d$. Two contractions of gamma matrices then giveFor the auxiliary-mass extraction of ultraviolet poles,The second term is ultraviolet convergent. ThusThe ultraviolet pole of the QED vertex is therefore ${\Lambda }^{\mu }{|}_{\mathrm{UV}}=e^2{\gamma }^{\mu }/(16{\pi }^2ϵ)$. The Pauli structure has zero trace against ${\gamma }_{\mu }$ and no ultraviolet pole at this order.

Form-factor qualification. The two electron electromagnetic vertex form factors in the source describe matrix elements between external on shell Dirac spinors. A general off shell vertex has more structures, so the unprojected trace is not a projector for the complete finite on-shell $F_1$. It does correctly isolate the divergent local ${\gamma }^{\mu }$ coefficient, which is all that is required here.

Source correction: equation 11 on the original page has $\Gamma (\alpha -1/2)$. The dimensionally regulated integral instead contains $\Gamma (\alpha -d/2)$. This is not an interchangeable convention: for $\alpha =2$ and $d=4-ϵ$, the correct factor $\Gamma (ϵ/2)$ has a logarithmic ultraviolet pole, whereas the printed $\Gamma (3/2)$ is finite.

To obtain the correct expression, shift the loop momentum and perform a Wick rotation. For positive Euclidean ${\mathcal{M}}^2$, introduce a Schwinger parameterization:The derivation first holds in the convergent range and defines the result elsewhere by analytic continuation. For integer $\alpha$, rotating back giveswith the branch inherited from $i0$. This is the Lorentzian continuation of the Euclidean massive loop integral.

Set $p=0$, $\alpha =1$, $d=4-ϵ$. Since $\Gamma (-1+ϵ/2)=-2/ϵ+O(1)$,Consequently the directed tadpole in part d givesThe counterterm is $-i{\delta }_{m^2}$, so cancelling only the pole in the minimal subtraction scheme requiresThis is the quartic complex-scalar mass renormalization result. The scale factors and logarithms contribute finite terms at this order and do not change the displayed pole coefficient. A real-field check gives a tadpole coefficient $\frac{1}{2}(3g/2+g/2)=g$, in agreement with the oriented complex-field computation.

The single-vertex scalar two-point graph at first order in $g_{2k}$ has one $2k$-valent interaction vertex. Attach the two external lines to it and pair its remaining $2k-2$ legs into $k-1$ tadpole diagrams. The graph is connected, has $L=k-1$ independent loop momenta, and is momentum independent after amputation apart from the overall Dirac delta function. In particular, first order in the coupling need not mean one loop.

The number of contractions with fixed external labels isDividing by the interaction factorial $(2k)!$ gives the inverse Feynman-diagram symmetry factor $1/[2^{k-1}(k-1)!]$. Use $m>0$ for the mass in the lowest-order propagator and define the tadpole integralThe linear-order correction to the amputated function is consequentlyThis is the correction to the paper's amputated connected correlator, not a definition of an inverse-propagator self-energy with an unspecified sign. At this order using $m_0$ instead of $m$ gives the same lowest-order integral.

For completeness, Schwinger parameterization and a Gaussian integral giveThis first converges for $0<D<2$ and then defines its dimensional regularization continuation. At $D=4-2ϵ$, the Gamma function expansion yieldsThus its highest ultraviolet pole isFor $k=2$ the remainder means finite terms. For $k>2$ the unrenormalized graph has a higher-order pole from its independent ultraviolet subdivergences, rather than only a simple pole. A massless tadpole integral is scaleless and vanishes in dimensional regularization; that does not justify discarding the massive ultraviolet divergence by setting $m=0$ before separating ultraviolet and infrared effects.

The change of field and parameters separates the Lagrangian density in Euclidean space intoThe two-leg counterterm interaction vertex is $-(\delta Z{p}^2+\delta m^2)$, with the overall momentum conservation delta function understood. With $D_0(p)=1/(p^2+m^2)$, the explicit first-order contribution to the propagator isThe minus sign follows either from expanding $e^{-S_{\mathrm{ct}}}$ or from expanding $(p^2+m^2+\delta Z{p}^2+\delta m^2{)}^{-1}$.

For $k=2$, the loop correction is $-{\tilde{\lambda }}_4{T}_D(m)D_0(p{)}^2/2$. Its ultraviolet pole from c isThere is no external $p^2$ dependence in the amputated tadpole diagram, so no one-loop wave-function renormalization is required. The minimal subtraction scheme therefore givesOnly the pole is subtracted: including $1-{\gamma }_E+\log 4\pi$ would impose a different finite subtraction prescription. The defined $\delta m^2$ is the coefficient in $Z{m}_0^2=m^2+\delta m^2$, not automatically the difference $m_0^2-m^2$ beyond this order.

In scalar electrodynamics with photon propagator $-i[g_{\mu \nu }-(1-\xi )l_{\mu }{l}_{\nu }/l^2]/(l^2+i0)$ and dimensional regularization $d=4-2ϵ$, the minimal subtraction scheme gives $Z_{\varphi }=1+e^2(3-\xi )/(16{\pi }^2ϵ)+O(e^4)$. For an external four-momentum $p$, expand the exchange numerator as $4p^2-4\xi p\cdot l+\xi l^2-4(1-\xi )(p\cdot l{)}^2/l^2$. The logarithmic ultraviolet poles of the scalar, vector and tensor integrals give the $p^2$ coefficient $4-2\xi -(1-\xi )=3-\xi$. A massless photon tadpole integral does not change it. The scalar QED Ward identity gives the same vertex renormalization constant.

For interaction ${\lambda }_{2k}{\varphi }^{2k}/(2k)!$, one vertex with two labelled external legs has $k-1$ independently paired tadpole diagrams. Counting contractions gives symmetry factor $S=2^{k-1}(k-1)!$. In a normalized Euclidean path integral, its amputated contribution is $-{\lambda }_{2k}{T}_D(m{)}^{k-1}/S$, where $T_D$ is the tadpole integral. It is linear in the coupling but has $k-1$ loops and a highest four-dimensional ultraviolet pole of order $k-1$ for nonzero mass.

For one unit-charge Dirac field in $d=4-2ϵ$, define the insertion as $i{\Pi }_{\mu \nu }=i(q^2{g}_{\mu \nu }-q_{\mu }{q}_{\nu })\Pi$. The one-loop ultraviolet pole is ${\Pi }_{\mathrm{UV}}=-e^2/(12{\pi }^2ϵ)$. The Ward identity cancels any apparent divergent vector mass term. With photon counterterm insertion $-i(q^2{g}_{\mu \nu }-q_{\mu }{q}_{\nu }){\delta }_3$, this gives ${\delta }_3=-e^2/(12{\pi }^2ϵ)$ in the minimal subtraction scheme.