Past exam of the quantum field theory course of the University of Oxford / 2018 / 2

Use mass dimensions with $[x]=-1$, $[{\partial }_{\mu }]=1$ and dimensionless action. The Lagrangian density in Euclidean space has dimension $D$, and the kinetic term fixes $2+2[{\varphi }_0]=D$. ThusA dimensionless coupling constant is thereforeHere $\mu$ is a positive regularization scale with mass dimension one. The interaction is marginal at its upper critical dimension $D_c=2k/(k-1)$, relevant below it and irrelevant above it by this engineering-dimension test. In particular, $d_2=4-D$, while $d_k=4-2k+2(k-1)ϵ$ near $D=4-2ϵ$. At this stage $g_{2k}$ is the dimensionless version of the displayed bare coupling; after introducing counterterms one may use the same scale factor for the renormalized coupling.

Expand the normalized Euclidean path integral with weight $e^{-S_E}$ and apply Wick's theorem. Use labelled external momenta and a fixed Fourier transform convention throughout. A line carrying momentum $q$ contributes $(q^2+m_0^2{)}^{-1}$. Each $2k$-valent interaction vertex contributes $-{\lambda }_{2k}=-{\mu }^{d_k}{g}_{2k}$: its $(2k)!$ possible contractions cancel the factorial in the Lagrangian density. A vertex also carries $(2\pi {)}^D{\delta }^{(D)}(\sum q)$ with all incident momenta incoming. Integrate each independent loop momentum with $d^{D}q/(2\pi {)}^D$ and divide by the Feynman-diagram symmetry factor for fixed external labels. There are no Minkowski factors of $i$.

For the full $N$-point Green's function, include every graph in which all vertices belong to components attached to external insertions. In particular, disconnected components carrying external insertions must be included, and the zeroth-order result is the sum of free pairings. Pure vacuum bubbles cancel against the denominator. The connected Green's function restricts to connected graphs, while a one-particle-irreducible vertex imposes the stronger restriction that removing one internal line cannot disconnect the graph. These are different objects. Odd $N$ vanishes in the symmetric vacuum because every interaction has even valence. At fixed perturbative order, multiply by the appropriate power of the coupling and sum all allowed graph topologies and external attachments.

If the overall momentum-conservation delta function is removed by convention, report the remaining function as its coefficient. The ordinary unamputated Green's function includes external scalar propagators; dividing them out defines an amputated function. The definition in c performs this amputation but does not, by itself, impose the usual one-particle-irreducibility condition.

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.

Use the wave-function renormalization coefficient and anomalous dimension definition stipulated in this part, keeping their signs and normalization. Put $A=1/[6(32{\pi }^2{)}^2]$. To the required order,At fixed bare ${\lambda }_4$, the leading change of the dimensionless coupling is $\mu d{g}_4/d\mu =-2ϵg_4+O(g_4^2)$, because ${\lambda }_4$ has mass dimension $2ϵ$. Therefore the chain rule, applied before setting $ϵ=0$, givesThusThe vanishing engineering contribution $-2ϵg_4$ multiplies a pole and leaves a finite answer. Dropping it at the start would incorrectly give zero. The numerical answer is conditional on the supplied two-loop coefficient. A convention defining the field anomalous dimension as ${\gamma }_{\varphi }=\frac{1}{2}\mu d\log Z/d\mu$ instead gives ${\gamma }_{\varphi }=-\gamma /2$ for the same $Z$; substituting that convention silently would answer a different question.