QFT Example Sheet 4 Solutions - Quantum Field Theory Example Sheet 4 Michelmas Term 2011 Solutions by Johannes Hofmann Laurence Perreault Levasseur Dave

QFT Example Sheet 4 Solutions - Quantum Field Theory...

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Quantum Field Theory Example Sheet 4 Michelmas Term 2011 Solutions by: Johannes Hofmann [email protected] Laurence Perreault Levasseur [email protected] Dave M. Morris [email protected] Marcel Schmittfull [email protected] Exercise 1 In the previous example sheet we considered 2-particle to 2-particle scattering in φ 4 theory, L = 1 2 μ φ∂ μ - 1 2 m 2 φ 2 - λ 4! φ 4 . In this exercise, we shall compute the leading order contribution to the amplitude for the scattering of 3-particles to produce 3-particles. Giving labels to the momenta of these particles, our initial and final states therefore look like | i = 8 E p 1 E p 2 E p 3 a p 1 a p 2 a p 3 | 0 , | f = 8 E p 1 E p 2 E p 3 a p 1 a p 2 a p 3 | 0 . As in Exercise 11 of example sheet 3, we determine the S -matrix element lim t →∞ f | U ( t, - t ) | i = f | T exp - 4! d 4 x φ ( x ) 4 | i . (1) As before there are a number of diagrams which the prescription for calculating this amplitude tells us we must throw away: these are, first, the O (1) term as this contains no interactions. Second, there being 6 external lines but only 4 fields φ appearing in (1) at O ( λ ), not all the external lines in | i and | f can be contracted with fields in the time order product. The diagrams below correspond to such terms and, as they are disconnected, are consequently ignored. The quantity of interest is thus the term 1 2 - 4! 2 d 4 x d 4 y f | T φ 4 ( x ) φ 4 ( y ) | i Via 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 time ordered product, leaving at most four to be contracted with external states, and, consequently, at least one in-going line is matched with an out-going line. Discarding these, we are left with those of the form φ ( k 1 ) φ ( k 2 ) φ ( k 3 ) | : φ ( x ) φ ( x ) φ ( x ) φ ( x ) φ ( y ) φ ( y ) φ ( y ) φ ( y ) : | φ ( k 1 ) φ ( k 2 ) φ ( k 3 ) , (2) giving the diagram 1
k 1 k 2 k 2 k 1 + k 2 + k 3 k 1 k 2 k 2 and three similar contractions in which one in-going leg is instead contracted with a φ ( x ) and one out-going leg with a φ ( y ). These latter contractions have diagrams of the form The contractions (2) may easily be calculated via the methods detailed in the solutions to Exercise 11 of the previous example sheet. After performing the two integrals, this short calculation gives 1 2 - 4! 2 d 4 x d 4 y φ ( k 1 ) φ ( k 2 ) φ ( k 3 ) | : φ ( x ) φ ( x ) φ ( x ) φ ( x ) φ ( y ) φ ( y ) φ ( y ) φ ( y ) : | φ ( k 1 ) φ ( k 2 ) φ ( k 3 ) = 1 2 - 4! 2 i ( p 1 + p 2 + p 3 ) 2 - m 2 × (2 π ) 4 δ (4) ( p 1 + p 2 + p 3 - p 1 - p 2 - p 3 ) , and we claim that an overall combinatorial factor of 2 × (4!) 2 arises 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 variables x and y in (2), since two contractions differing in only this respect give identical expressions when these integrals are evaluated.