Past exam of the quantum field theory course of the University of Oxford / 2020 / 3

Keep $\lambda$ for the cubic coupling and $j$ for the external source. Expanding the interaction part of the scalar field path integral yieldsA functional derivative of the source exponential inserts $i\varphi (y)$. In particular, ${\delta }_j(y{)}^3{e}^{i\int j\varphi }=-i\varphi (y{)}^3{e}^{i\int j\varphi }$, so $-i\lambda \varphi (y{)}^3$ can be replaced by $\lambda {\delta }_j(y{)}^3$. The functional derivatives commute and can be taken outside the free path integral. Resumming the exponential series givesThis is a formal perturbation series; a real cubic scalar potential is unbounded and the expression does not assert a stable nonperturbative vacuum state.

To evaluate the free Gaussian path integral, use integration by parts to writeThe inverse kernel isCompleting the square with $\varphi =\chi -K^{-1}j$ changes the exponent to $i\int \chi K\chi /2-i\int j{K}^{-1}j/2$. The translation-invariant functional measure makes the remaining Gaussian path integral precisely $Z[0;0]$, henceFor $ϵ>0$, the exponent includes $-ϵ\int {\varphi }^2/2$, which damps the regulated finite-dimensional integrals. The Feynman i-epsilon prescription displaces the positive-energy pole below the real $p^0$ axis and the negative-energy pole above it, selecting the time-ordered vacuum expectation value. Indeed the Feynman propagator is ${\Delta }_F=iD$. Thus $iϵ$ specifies both convergence and the vacuum state boundary conditions, with $ϵ\to 0^{+}$ taken after integration.

For the interacting vacuum state defined perturbatively by the same Feynman i-epsilon prescription, the normalized time-ordered product isEach functional derivative inserts $i\varphi$, accounting for $i^{-m}$. The denominator implements cancellation of vacuum bubbles, while disconnected pieces carrying external insertions still contribute. The lower limit $i=i$ in the printed product is a typographical error; the product here runs from 1 to $m$.

At order $V$ in the cubic scalar field interaction term, the expansion of the generating functional supplies $1/[V!(3!{)}^V]$. The $V!$ counts permutations of indistinguishable integrated interaction vertices, and $3!$ counts permutations of the three field slots at each interaction vertex. Applying the Wick theorem pairs fields into Feynman propagators. Several pairings produce the same Feynman diagram; their number $N_G$ multiplies the expansion coefficient. For fixed labelled external points, the net combinatorial weight isHere $S_G$ is the order of the graph's automorphism group when the external labels are fixed, with permutations of identical parallel lines and reversal of a self-loop included as applicable. This is the Feynman-diagram symmetry factor convention in which a Feynman diagram is divided by $S_G$. Each cubic interaction vertex then contributes $-i\lambda$, each line contributes ${\Delta }_F$, and each internal vertex is integrated over its position. External labelled points are not divided by $m!$.

Hold the external endpoints fixed and distinguish the three Feynman diagrams by their left-to-right order in the source. Write $S$ for the divisor and $w=1/S$ for the multiplying combinatorial weight. Each line below includes external Feynman propagators; amputation leaves the same $w$.

Left: two-point bubble. There are two cubic interaction vertices and two parallel internal lines. Interchanging those lines preserves the labelled external endpoints, giving $S=2$. Direct Wick contraction counting gives two assignments of the external endpoints to the vertices, $3^2$ choices of their field slots, and $2!$ pairings of the remaining slots:With $\Delta ={\Delta }_F$ and external points $x_1,x_2$, its contribution is

Middle: three-point triangle. The three labelled external endpoints fix their three adjacent cubic interaction vertices, so no nonidentity graph automorphism remains. Counting gives $3!$ assignments of endpoints to vertices, $3^3$ choices of external field slots, and $2^3$ assignments of remaining slots to the two specified neighbours:Its contribution isThe apparent rotational symmetry of an unlabelled triangle does not survive fixing three distinct external endpoints.

Right: two-point graph with a chord. There are four cubic interaction vertices: the left and right attachment vertices $A,B$ and the two chord ends $C,D$. Its internal edges are $AC,AD,BC,BD,CD$. Reflection interchanges $C,D$, fixes $A,B$, and preserves the two external labels. This is the only nontrivial graph automorphism, so $S=2$.

For a separate Wick contraction count, choose the ordered external attachment vertices in $4\cdot 3$ ways and their field slots in $3^2$ ways. Assign their two remaining slots to $C,D$ in $2^2$ ways. At each of $C,D$, choose the slot joining $A$ and then the slot joining $B$ in $3\cdot 2$ ways; the last slots join each other. ThusIts contribution isConsequently the three multiplying weights are $\boxed{(1/2,1,1/2)}$, or divisors $\boxed{(2,1,2)}$. The central chord is a single internal line joining two vertices, not a vertex at the centre of the circle.