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ManDay
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Fourier Transform on the "connected part" of QFT transition prob.
Calculate ⟨0|T[ϕ(x₁)ϕ(x₂)ϕ(x₃)ϕ(x₄)]|0⟩ up to order λ from the generating functional Z[J] of λϕ⁴-theory.
Using the connected part, derive the T-matrixelement for the reaction a(p₁) + a(p₂) → a(p₃) + a(p₄) up to this order in λ (a denotes the quant of the scalar fieldoperator ϕ(x)). Calculate the cross section of this reaction up to order λ².
I've calculated ⟨0|T[ϕ(x₁)ϕ(x₂)ϕ(x₃)ϕ(x₄)]|0⟩ to order λ as
⟨0|T[ϕ(x₁)ϕ(x₂)ϕ(x₃)ϕ(x₄)]|0⟩ = -(Δ(x₁,x₂)Δ(x₃,x₄) + Δ(x₁,x₃)Δ(x₂,x₄) + Δ(x₁,x₄)Δ(x₂,x₃)) + i½(Δ(x₁,x₂)D(x₃,x₄) + Δ(x₁,x₃)D(x₂,x₄) + Δ(x₁,x₄)D(x₂,x₃) + D(x₁,x₂)Δ(x₃,x₄) + D(x₁,x₃)Δ(x₂,x₄) + D(x₁,x₄)Δ(x₂,x₃) + ¼∫d⁴zΔ(z,z))λ
where Δ denotes the Feynman propagator and
D(x₁,x₂) := ∫d⁴zΔ(z,z)Δ(z,x₁)Δ(z,x₂)
I don't understand what is meant by the connected part (or more specifically: what it is. I do have an idea what it refers to) or how to perform the Fourier transform on it. Also, I don't know how to calculate the according cross section.
Thanks for your help.
Homework Statement
Calculate ⟨0|T[ϕ(x₁)ϕ(x₂)ϕ(x₃)ϕ(x₄)]|0⟩ up to order λ from the generating functional Z[J] of λϕ⁴-theory.
Using the connected part, derive the T-matrixelement for the reaction a(p₁) + a(p₂) → a(p₃) + a(p₄) up to this order in λ (a denotes the quant of the scalar fieldoperator ϕ(x)). Calculate the cross section of this reaction up to order λ².
The Attempt at a Solution
I've calculated ⟨0|T[ϕ(x₁)ϕ(x₂)ϕ(x₃)ϕ(x₄)]|0⟩ to order λ as
⟨0|T[ϕ(x₁)ϕ(x₂)ϕ(x₃)ϕ(x₄)]|0⟩ = -(Δ(x₁,x₂)Δ(x₃,x₄) + Δ(x₁,x₃)Δ(x₂,x₄) + Δ(x₁,x₄)Δ(x₂,x₃)) + i½(Δ(x₁,x₂)D(x₃,x₄) + Δ(x₁,x₃)D(x₂,x₄) + Δ(x₁,x₄)D(x₂,x₃) + D(x₁,x₂)Δ(x₃,x₄) + D(x₁,x₃)Δ(x₂,x₄) + D(x₁,x₄)Δ(x₂,x₃) + ¼∫d⁴zΔ(z,z))λ
where Δ denotes the Feynman propagator and
D(x₁,x₂) := ∫d⁴zΔ(z,z)Δ(z,x₁)Δ(z,x₂)
I don't understand what is meant by the connected part (or more specifically: what it is. I do have an idea what it refers to) or how to perform the Fourier transform on it. Also, I don't know how to calculate the according cross section.
Thanks for your help.