Past exam of the quantum field theory course of the University of Oxford / 2016 / 3

Let brackets denote mass dimension. In natural units, $[x]=-1$ and $[{\partial }_{\mu }]=1$, so a dimensionless action requires $[\mathcal{L}]=D$. The kinetic term fixesThe cubic scalar field theory interaction then requires $[\lambda ]+3[\varphi ]=D$, givingThe coupling constant is dimensionless at this critical dimension of a monomial scalar interaction. For $D<6$ it has positive mass dimension, while for $D>6$ it has negative mass dimension, the usual power counting distinction between super-renormalizable and non-renormalizable cubic interactions.

Write $S=S_0+S_I$, where $S_0$ contains the quadratic kinetic term and mass term, and $S_I=-\lambda \int d^{D}x\varphi (x{)}^3/3!$. ThenLet $Z_0=\int D\varphi e^{i{S}_0}$ and let $⟨\cdot {⟩}_0$ denote the unit-normalized free Gaussian vacuum expectation value. The unnormalized numerator ${\mathcal{N}}_N$ is consequentlyThe Minkowski path integral is understood with the usual scalar Feynman propagator pole prescription, as a formal perturbation series. A real cubic potential energy is unbounded below; no nonperturbative stable vacuum is being assumed here.

Set $\Delta (x-z)=⟨T\varphi (x)\varphi (z){⟩}_0$, $K_i(x)=\Delta (y_i-x)$, and $t=\Delta (0)$, with a common regularization understood for coincident points. The zero-order one-point term has one field and vanishes. At first order, Wick's theorem gives three choices for the contraction with the external field, leaving a tadpole diagram:The second-order term contains seven fields, so it vanishes by the vanishing odd correlators of a free Gaussian field. HenceThe coefficient is $3/3!=1/2$, or symmetry factor $S=2$ when each diagram is weighted by $1/S$. No spacetime integral has been evaluated. The one-point tadpole diagram is shown in the first panel of the following figure; its other panels organize solution.

Use the notation ${\Delta }_{12}=\Delta (y_1-y_2)$, ${\Delta }_{xz}=\Delta (x-z)$ and ${\int }_{xz}=\int d^{D}x{d}^{D}z$. The order-$\lambda$ term vanishes because it has five fields. At order ${\lambda }^2$, enumerate the eight-field Wick contractions, whose expansion denominator is $2!(3!{)}^2=72$.

If the two external fields contract with each other, the remaining vacuum diagrams are a pair of cubic vertices joined by three lines, with six Wick contractions, and a pair joined by one line with a tadpole diagram at each end, with nine Wick contractions. Define their combined integrated weight byThus $Z/Z_0=1+(-i\lambda {)}^2{V}_2+O({\lambda }^4)$. The numerator isThe three latter terms have respectively 36, 36 and 18 Wick contractions. Together with the six and nine Wick contractions, they exhaust all $7!!=105$ pairings.

Dividing by $Z/Z_0$ cancels the two vacuum diagrams, givingThe first correction is the two-line bubble diagram, with symmetry factor $S=2$. The second is the connected tadpole contribution to a cubic-theory two-point function: both external lines meet one vertex, joined by a single line to a second vertex bearing a tadpole diagram; its symmetry factor is also $S=2$. Interchanging the two integrated vertices is already included in the 36 contractions and gives no extra factor.

The final term is the disconnected Feynman diagram consisting of two separate one-point tadpole diagrams, with coefficient $1/4$. It survives cancellation of vacuum bubbles, since each component touches an external point. Indeed it is precisely the leading $⟨\varphi (y_1)⟩⟨\varphi (y_2)⟩$. It would disappear from the connected two-point correlation function, but the question defines the full correlation function. No normal ordering or tadpole subtraction of the interaction was specified. All terms are depicted in solution, and all spacetime integrals remain unevaluated.

For a connected cubic Feynman diagram with $V$ internal vertices, $I$ internal lines and three external lines, $3V=2I+3$. Hence $V$ is odd and the loop order is $L=I-V+1=(V-1)/2$. Through order ${\lambda }^3$, there is one tree-level Feynman diagram at $V=1$ and seven labelled one-loop Feynman diagrams at $V=3$:

For an explicit check, use the propagator notation of solution and define, with $\{i,j,k\}=\{1,2,3\}$,There is no ordering of $j,k$ in this notation. The complete connected correlation function through this order isFor completeness of the classification, a one-loop connected internal graph either has a three-edge cycle, a pair of parallel edges and a bridge, or a self-loop and two bridges. The cubic degree condition forces exactly the external attachments just listed. Direct Wick contraction counting gives 1296 pairings for the triangle and 648 for each of the six other labelled graphs, divided by $3!(3!{)}^3=1296$.

Connected does not mean one-particle irreducible. After amputation of external propagators, only the triangle is one-particle irreducible at this order; the six other loop graphs belong to the requested connected answer. The tadpole diagrams cannot be discarded without an additional subtraction prescription.