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By Stetz A.W.

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We will have to evaluate the time-ordered product T [: ϕ3 (x1 ) :: ϕ(x2 ) :]. To do this by brute force we would split ϕ as before ϕ = ϕ− + ϕ+ , and then rearrange the terms to that all ϕ− ’s stood to the left of all ϕ+ ’s. 3. THE WICK EXPANSION 43 a formidable calculation. Fortunately there is a theorem from G. C. Wick that makes it possible to do this calculation without much work. 108) in rather formal terms that make it easy to generalize to situations in which there are more than two fields.

51) 52CHAPTER 3. THE INTERACTION PICTURE AND THE S-MATRIX The vacuum is stable, but we must allow the possibility that phase of the vacuum state evolves with time. 2. One-particle states are also stable. 53) 3. Proof: β, out|ϕout |α, in = β, in|Sϕout |α, in But β, in|ϕout is itself an in state, so we can write β, in|ϕout |α, in = β, out|ϕin S|α, in Comparing these two equations gives Sϕout = ϕin S 4. S is unitary. 7 The LSZ Reduction Scheme What is the relation between the fully-interacting field ϕ and ϕin ?

27), consequently the only non-vanishing terms in the Wick expansion are those with four normally-ordered fields and one pair of contracted fields. There are nine ways of contracting two groups of three fields. Consequently T [: ϕ3 (x1 ) :: ϕ3 (x2 ) :] = 9 ϕ(x1 )ϕ(x2 ) : ϕ2 (x1 )ϕ2 (x2 ) : + · · · Sβα = p1 p2 |S|k1 k2 = (−i)2 2! λ 3! 42) and do the integrals over d4 x1 and d4 x2 . This will give six terms which collapse into three after a bit of algebra. The result is rather simple, so let’s just look at the integrals 1 If you want a proof see J.

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