Quantum Field TheoryExample Sheet 4Michelmas Term 2011Solutions by:Johannes Hofmann[email protected]Laurence Perreault Levasseur[email protected]Dave M. Morris[email protected]Marcel Schmittfull[email protected]Exercise 1In the previous example sheet we considered 2-particle to 2-particle scattering inφ4theory,L=12∂μφ∂μ-12m2φ2-λ4!φ4.In this exercise, we shall compute the leading order contribution to the amplitude for the scattering of 3-particles toproduce 3-particles. Giving labels to the momenta of these particles, our initial and final states therefore look like|i=8Ep1Ep2Ep3a†p1a†p2a†p3|0,|f=8Ep1Ep2Ep3a†p1a†p2a†p3|0.As in Exercise 11 of example sheet 3, we determine theS-matrix elementlimt→∞f|U(t,-t)|i=f|T exp-iλ4!d4x φ(x)4|i .(1)As before there are a number of diagrams which the prescription for calculating this amplitude tells us we must throwaway: these are, first, theO(1) term as this contains no interactions. Second, there being 6 external lines but only 4fieldsφappearing in (1) atO(λ), not all the external lines in|iand|fcan be contracted with fields in the time orderproduct. The diagrams below correspond to such terms and, as they are disconnected, are consequently ignored.The quantity of interest is thus the term12-iλ4!2d4xd4y f|Tφ4(x)φ4(y)|iVia Wick’s theorem, this contains further contractions some of which again correspond to disconnected diagrams.These are those containing two or more propagators, as such terms require at least four contracted fields in the timeordered product, leaving at most four to be contracted with external states, and, consequently, at least one in-goingline is matched with an out-going line. Discarding these, we are left with those of the formφ(k1)φ(k2)φ(k3)|:φ(x)φ(x)φ(x)φ(x)φ(y)φ(y)φ(y)φ(y) :|φ(k1)φ(k2)φ(k3),(2)giving the diagram1
k1k2k2k1+k2+k3k1k2k2and three similar contractions in which one in-going leg is instead contracted with aφ(x) and one out-going leg withaφ(y). These latter contractions have diagrams of the formThe contractions (2) may easily be calculated via the methods detailed in the solutions to Exercise 11 of the previousexample sheet. After performing the two integrals, this short calculation gives12-iλ4!2d4xd4yφ(k1)φ(k2)φ(k3)|:φ(x)φ(x)φ(x)φ(x)φ(y)φ(y)φ(y)φ(y) :|φ(k1)φ(k2)φ(k3)=12-iλ4!2i(p1+p2+p3)2-m2×(2π)4δ(4)(p1+p2+p3-p1-p2-p3),and we claim that an overall combinatorial factor of 2×(4!)2arises when we sum over every contraction of the form (2).To see this, first note that a factor of 2 arises from interchanging the integration variablesxandyin (2), since twocontractions differing in only this respect give identical expressions when these integrals are evaluated.